Class TJ \s\ Book .L8 GofpgM . COPYRIGHT DEPOSIT. s MACHINISTS' AND DRAFTSMEN'S HANDBOOK: CONTAINING TABLES, RULES AND FORMULAS, WITH NUMEROUS EXAMPLES EXPLAINING THE PRINCIPLES OF MATHEMATICS AND MECHANICS AS APPLIED TO THE MECHANICAL TRADES INTENDED AS A REFERENCE BOOK FOR ALL INTERESTED IN MECHANICAL WORK. BY PEDER LOBBEN, MECHANICAL ENGINEER. Member of the American Society of Mechanical Engineers and Worcester County Mechanics Association. Honorary Member of the N. A. Stationary Engineers. Non-Resident Member of the Franklin Institute. SECOND AND ENLARGED EDITION. NEW YORK D. VAN NOSTRAND COMPANY 23 Murray and 27 Warren Streets IQIO • < Copyright. 1899. BY PEDER LOBBEN Copyright. 1910, ey PEDER LOBBEN: yd- W* (gdA268763 . / K PREFACE TO FIRST EDITION. It is the author's hope and desire that this book, which is the outcome of years of study, work and observation, may be a help to the class of people to which he himself has the honor to belong, — the working mechanics of the world. This is not intended solely as a reference book, but itmay also be studied advantageously by the ambitious young engineer and machinist; and, therefore, as far as believed practical within the scope of the work, the fundamental principles upon which the rules and formulas rest are given and explained. The use of abstruse theories and complicated formulas is avoided, as it is thought preferable to sacrifice scientific hair- splitting and be satisfied with rules and formulas which will give intelligent approximations within practical limits, rather than to go into intricate and complicated formulas which can hardly be handled except by mathematical and mechanical experts. In practical work everyone knows it is far more important to understand the correct principles and requirements of the job in hand than to be able to make elaborate scientific demonstra- tions of the subject; in short, it is only results which count in the commercial world, and every young mechanic must remem- ber that few employers will pay for science only. What they want is practical science. Should, therefore, scientific men, (for whom the author has the greatest respect, as it is to the scien- tific investigators that the working mechanics are indebted for their progress in utilizing the forces of nature), — find nothing of interest in the book, they will kindly remember that the author does not pretend it to be of scientific interest, and they will therefore, in criticizing both the book and the author, remember that the work was not written with the desire to show the reader how vulgarly or how scientifically he could handle the subject, but with the sole desire to promote and assist the ambitious young working mechanic in the world's march of progress. P. Lobben. New York, October, 1899. PREFACE TO SECOND EDITION. The preface to the first edition explains the purpose of the book, and the only thing I have to add is that I wish to express my thanks to the technical press and to the public in general for the friendly spirit in which my work has been received, which has made it possible for me to have the pleasure of getting out this, the second and much enlarged edition. Peder Lobben. Holyoke, Mass., July, 1910. TABLE OF CONTENTS. NOTES ON MATHEMATICS. Signs 1 Formulas 3 ARITHMETIC. Addition 5 Subtraction 5 Multiplication 6 Division 6 Fractions 7 Decimals 11 Ratio 16 Proportions 16 Compound proportions , 17 Interest 19 Compound interest . . 22 Results of small savings 26 Equation of payments 27 Partnership , 27 Square root 29 Cube root. . 30 Radical quantities expressed without the radical sign.... 32 Reciprocals 32 Table of squares, cubes, square roots, cube roots, and reciprocals of fraction and mixed numbers from 1-64 to 10 33 Table of squares, cubes, square root, cube root and recip- rocals of numbers from 0.1 to 1000 37 NOTES ON ALGEBRA. Sign of quality and size of operation 63 Useful formulas and rules 63 Extracting- roots 64 Powers 64 Equations 65 Quadratic equations 67 Progressions 68 The arithmetical mean 70 The geometrical mean 70 LOGARITHMS. To find the logarithms of numbers of four figures 73 I II CONTENTS. To find the logarithms of numbers having more than four figures 73 To find the number corresponding to a given logarithm.. 74 Addition of logarithms 75 Subtraction of logarithms 76 Multiplication of logarithms 77 Division of logarithms 78 Short rules for figuring by logarithms 79 Simple interest by logarithms 81 Compound interest by logarithms 82 Discount or rebate 83 Sinking funds and savings 84 Paying a debt by instalments 89 Table of logarithms of numbers 90 Hyperbolic logarithms 128 Table of hyperbolic logarithms 126 WEIGHTS AND MEASURES. Long measure 133 Square measure 134 Cubic measure 134 Liquid measure 134 Dry measure 134 Avoirdupois weight 134 Troy weight 135 Apothecaries' weight 135 Weights of produce 135 The metric system of weights and measures 135 Table reducing millimeters to inches 137 Table reducing inches to millimeters 137 Table of reduction for pressure per unit of surface 138 Table of reduction for length and weight 138 Weight of water, measure of water 138 Specific gravity 138 Table of soecific gravity, weights and values of various metals 139 Table of specific gravity of various woods 139 Table of specific gravity of various materials 140 Table of specific gravity of liquids 140 To calculate weights of casting from weight of pattern. . . 141 Weight of a bar of rough iron of any shape of cross section 141 To calculate the weight of sheet iron of any thickness. . . 141 To calculate weight of metals not given in the tables 141 To calculate the weight of zinc, copper, lead, etc., in sheets 142 To calculate the weight of cast iron balls 142 The weight of round steel per linear foot 142 CONTENTS. HI Table of weights of square and round bars of wrought iron 143 Table of weight of flat iron in pounds, per foot 144 Table of sizes and numbers of U. S. standard gage for steel and iron in sheets 145 Table of wire gages 146 Table of weight of iron wire 147 Table of decimal equivalents of the numbers of twist drills and the steel wire gage 148 Table of decimal equivalents of Stubs' steel wire gage.. . 148 GEOMETRY. Angles 149 Polygons 149 Triangles 150 Circles 152 Trigonometry 153 The trigonometrical table and its use 157 Trigonometrical tables 158-175 Solution of right angle triangles 176 Solution of oblique angle triangles 178 Problems in geometrical drawings 184-192 MENSURATION. The difference between one square foot and one foot square 193 Area of triangles 193 Area of Parallelograms 196 Area of Trapezoid 196 Area of a circle 196 To change a circle into a square of the same area 197 To find the side of the largest square which can be inscribed in a circle 197 To find the area of any irregular figure 197 To find the area of a sector of a circle 198 To find the length of an arc of a segment of a circle 199 To find the area of a segment of a circle 199 To find the radius corresponding to the arc when the chord and the height of the segment are given 200 Table of areas of segments of a circle 201 To calculate the number of gallons of oil in a tank 202 Circular lune 202 Circular zone 202 To compute the volume of a segment of a sphere 203 To find the volume of a spherical segment when the height of the segment and the diameter of the sphere are known 203 To find the surface of a cylinder 203 To find the volume of a cvlinder 204 To find the solid contents of a cylinder 204 IV CONTENTS. To find the area of the curved surface of a cone 204 To find the volume of a cone 205 To find the area of the curved surface of a frustum of a cone 205 To find the area of the slanted surface of a pyramid 206 To find the area of the slanted surface of a frustum of a pyramid 207 To find the total area of a frustum of a pyramid 207 To compute the volume of a frustum of a pyramid 207 To find the surface of a sphere 207 To find the volume of a sphere 208 To compute the diameter of a sphere when the volume is known 208 To compute the circumference of an ellipse 208 To compute the area of an ellipse 208 Table giving circumferences and areas of circles 209-212 STRENGTH OF MATERIALS. Tensile strength 213 Modulus of elasticity 213 Strength of wrought iron 214 Strength of cast iron 214 Elongation under tension 215 Table of modulus of elasticity and ultimate tensile strength of various materials 216 Formulas for tensile strength 216 To find the diameter of a bolt to resist a given load 218 To find the thickness of a cylinder to resist a given pres- sure 219 Strength of flat cylinder heads 220 Strength of dished cylinder heads 220 Strength of a hollow sphere exposed to internal pressure 221 Strength of chains 222 Strength of wire rope 222 Strength of manila rope 222 Crushing strength 223 Table of modulus of elasticity and ultimate crushing strength of various materials 223 Table of safe load on Pillars 226 Hollow cast iron pillars 228 Weight of cast iron pillars 228 Table of safe load on round cast iron pillars 229 Wrought iron pillars 231 Table of safe load in tons on square pine or spruce posts 231 Transverse strength 233 Formulas for calculating transverse strength of beams. . . 235 To find the transverse strength of beams when their sec- tion is not uniform throughout the whole length 243 CONTENTS. V Square and rectangular wooden beams 245 To calculate the size of beam to carry a given load 247 To find the size of a beam to carry a given load when also the weight of the beam is to be considered 248 Crushing and shearing load of beams crosswise on the fiber 250 Round wooden beams 251 To calculate the size of round beams to carry a given load when span is given 251 Load concentrated at any point not at the center of the beam 252 Beams loaded at several points 253 To figure sizes of beams when placed in an inclined posi- tion 254 Deflection in beams when loaded transversely 254 To calculate deflection in beams under different modes of support and load 261 To find suitable size of beam for a given limit of deflec- tion 263 To find the constant for deflection 264 Modulus of elasticity calculated from the transverse deflection in a beam 265 Allowable deflection 266 Torsional strength 266 Hollow round shafts 270 Square beams exposed to torsional stress 270 Torsional deflection 271 Formula for torsional deflection in hollow round beams 271 Shearing strength 272 Factor of safety 274 Notes on strength of materials 274 MECHANICS. Newton's laws of motion 276 Gravity 276 Acceleration due to gravity 276 Velocity 277 Height of fall 277 Time 278 Distance a body drops during the last second 278 Table of time velocity and height of fall 279 Upward motion 279 Body projected at an angle 279 Motion down an inclined plane 283 Body projected in a horizontal direction from an elevated place 284 To calculate the speed of a bursted fly-wheel from the distance the fragments are thrown 284 VI CONTENTS. Force, energy and power . 285 Inertia 285 Mass 286 Momentum 286 Impulse 286 Kinetic energy 287 To calculate the force of a blow 288 Formulas for force, acceleration and motion 288 Centers , 292 Moments 292 Levers 292 Radius of gyration 293 Moment of inertia 293 Polar moment of inertia 297 Angular velocity 299 Centrifugal force 302 Friction 303 Coefficient of friction 304 Rolling friction 304 Axle friction 305 Horse-power absorbed by friction in bearings 305 Angle of friction 306 Rules for friction 306 Table of coefficient of friction 306 Friction in machinerv 306 Pulley blocks 307 Friction in pulley blocks 307 Differential pulley blocks 308 Inclined plane 308 Screws 311 The parallelogram of forces 316 Horse-power 317 To calculate the horse-power of a steam engine 317 To calculate the horse-power of a compound or triple expansion engine 317 To judge approximately the horse-power which may be developed by any common single cylinder engine.... 318 Horse-power of waterfalls 318 Animal power 318 Hauling a load 318 Power of man 319 Power required to drive various kinds of machinery.... 319 Speed of machinery 320 Calculating sizes of pulleys 321 To calculate the speed of gearing 322 Efficiency of machinery 322 Crane hooks and chain links 323 Cranes 324 Proportions of a two-ton derrick 324 CONTENTS. VII BELTS. Lacing belts 327 Cementing belts 328 Length of belts 329 Horse-power transmitted by belting 329 To calculate size of belt for given horse-power when diameter of pulleys and number of revolutions are known 330 To calculate width of belt when pulleys are of unequal diameter 331 To find the arc of contact of a belt 332 Pressure on the bearings caused by the belt 333 Special arrangement of belts 335 Crossed belts 336 Quarter-turn belts 336 Angle belts 337 Slipping of belts 337 Tighteners on belts 337 Velocity of belts 337 Oiling of belts 338 ROPE TRANSMISSION. Transmission capacity of wire ropes 340 Deflection in wire ropes 342 Transmission of power by manila ropes 344 Weight of manila rope 346 Preservation of manila rope 347 PULLEYS. Stepped pulleys 350 Stepped pulleys for back geared lathes 354 FLY WHEELS. Weight of rim of a fly-wheel 356 Centrifugal force in fly-wheel and pulleys 357 Rules for calculating safe speed of fly-wheels 360 Rule for calculating safe diameter of fly-wheels 360 SHAFTING. Shaft not loaded at the middle between hangers 361 Transverse deflection in shafts 361 Allowable deflection in shafts 362 Torsional strength of shafting 363 Torsional deflection in shafting 364 Horse-power transmitted by shafting 365 Distance between bearings 367 Proportions of keys 368 Dimensions of couplings 369 VIII CONTENTS. BEARINGS. Area of bearing surface 369 Allowable pressure in bearings 369 GEAR TEETH. Circular pitch 375 To calculate diameter of gear according to circular pitch 375 Table of pitch diameter of gears 376 To calculate diameter of gears when distance between centers and ratio of speed is given 376 Diametral pitch ' 377 To calculate the number of teeth when distance between centers and ratio of speed is given 381 Cycloid teeth 3S2 Involute teeth 337 Approximate construction of involute teeth 388 Internal gears with involute teeth 391 Chordal pitch 391 Rim, arms and hub of spur gears 393 Width of gear wheels 397 Bevel gears 397 Cutting bevel gears 401 Worms and worm gears 403 Calculating the size of worm gears 406 Elliptical gear wheels 408 Strength of gear teeth 409 SCREWS. Pitch, inch pitch and lead of screws and worms 417 Screw cutting by the engine lathe 417 Cutting multiple threaded screws and nuts by the engine lathe 417 U. S. Standard screws 418 Diameter of tap drill 420 Coupling bolts 422 Fillister head screws 422 Dimensions of hexagon and square head cap screws 423 Wood screws and machine screws 423 International standard metric screw threads 433 To gear a lathe to cut metric thread when the lead screw is in inches 436 Rivets, nails, eyebolts 438 Lag screws 439 PIPES. Wrought iron pipes 410 Brass and copper tubes 441 Boiler tubes 442 Cold drawn steel tubing 442 CONTENTS. IX NOTES ON HYDRAULICS. Pressure of fluid in a vessel 443 Velocity of efflux 443 Velocity of water in pipes 445 Quantity of water discharged through pipes 447 Notes on water 448 NOTES ON STEAM. Properties of saturated steam 451 Steam heating 453 Value of high and low pressure steam for heating purposes 454 Hot water heating in dwelling houses 454 Quantity of water required to make any quantity of steam at any pressure 454 Weight of water required to condense one pound of steam 455 Weight of steam required to boil water 456 Expansion of steam in a steam engine 457 Thermometers 460 NOTES ON COPPER WIRE. Weight and resistance of copper wire 464 Sizes of cotton covered copper wire 466 Carrying capacity of copper wire 466 NOTES ON ELECTRICAL TERMS. Volt 468 Ampere 468 Coulomb 469 Ohm 469 Watt 469 Joule 470 Farad 471 Henry 471 SHOP NOTES. Standard sizes of American machinery catalogues 473 Shrink fit 473 Press fit 473 Running fit 474 Babbitt metal 474 Helical springs 474 Emery, emery cloth, sandpaper 474 Weight of a grindstone 475 Lathe centers, lathe mandrels 475 Common sizes of steel for lathes and planer tools 475 Angles and corresponding taper per foot 475 Taper per foot and corresponding angles 476 Morse taper 476 Jarno taper 476 X CONTENTS. Marking solution 477 A cheap lubricant for milling and drilling 477 Soda water for drilling 477 Solder 477 Soldering fluids 478 Spelter 478 Alloy which expands in cooling 478 Shrinkage of castings 478 Case hardening wrought iron and soft steel 479 Case hardening boxes 479 To harden with cyanide of potassium 480 BLUE PRINTING. To prepare blue print paper 480 Blue print frame 480 Blue printing 480 Botes on flDatbematics. A Unit is any quantity represented by a single thing, as a magnitude, or a number regarded as one undivided whole. Numbers are the measure of the relation between quanti- ties of things of the same kind and are expressed by figures. Numbers which are capable of being divided by two without a remainder are called even numbers. 2, 4, 6, 8, etc., are even numbers. Numbers which are not capable of division by two without giving a remainder are called odd numbers. 1, 3, 5, 7, 9, etc.. are odd numbers. A number which can not be divided by any whole number but itself and the number 1 without giving a remainder is called a prime number. 1, 2, 3, 5, 7, 11, 13, 17, 19, etc., are prime numbers. All numbers that are not prime are said to be composite nwnbers, because they are composed of two or more factors ; 4, 6, 8, 9, 10, 12, etc., are composite numbers. Whole numbers are called integers. Whole numbers are also called integral numbers. A mixed number is the sum of a whole number and a fraction. The least common multiple of several given numbers is the smallest number that can be divided by each without a remainder. For instance, the least common multiple of 3, 4, 6, and 5 is 60, because 60 is the smallest number that can be divided by those numbers without a remainder. Signs. + (plus) is the sign of addition. — (minus or less) is the sign of subtraction. The signs 4- and — are also used to indicate positive and negative quantities. X (times or multiply) is the sign of multiplication, but in- stead of this sign, sometimes a single point (.) is used, especially in formulas: in algebraic expressions very frequently factors are written without any signs at all between them. For in- stance, aXb or a.b or ab. All these three expressions indicate that the quantity a is to be multiplied by the quantity b. 2 NOTES ON MATHEMATICS. — (divided by) is the sign of division. = (equal). When this sign is placed between two quanti- ties, it indicates that they are of equal value. For instance : 4 + 5 + 2 = 11 8 — 3 -f6 — 2 = 9 5 X 12 = 96 100 -h- 5 = 20 . (decimal point) signifies that the number written after it has some power of 10 for its denominator. ° ' " means degrees, minutes and seconds of an angle. ' " means feet and inches. a f a" a'" reads a prime, a second, a third. a\ cii a% reads a sub 1, a sub 2, a sub 3, and is always used to designate corresponding values of the same element. n s/ This is the radical sign and signifies that a root is to be extracted of the quantity coming under the sign ; this may be square root, cube root, or any other root, according to what there is signified by the number prefixed in place of the letter n . 3 4 For instance : VVeads square root, V reads cube root, VVeads 5 fourth root, V reads fifth root, V64 = 8, because 8 X 8 ,= 64 3 V64 = 4, because 4 X 4 X 4 = 64 V81 = 3, because 3X3X3X3 = 81 The sign that a quantity is to be raised to a certain power is a small number placed at the upper right hand corner of the quantity ; this number is called the exponent. For in- stance, 7 2 signifies that 7 is to be squared or multiplied by itself, that is : T 2 = 7 X 7 = 49 73 =7X7X7 = 343, etc. braces, [ ] brackets, ( ) parentheses, signify that the quantities which they include are to be considered as one quantity. For instance : 35 — (8 -f- 6) is equal to 35 — 14 = 21. In this case the parenthesis indicates that not only 8, but the sum of 8 + 6 is to be subtracted from 35. (vinculum or bar) is a straight line placed over two or more quantities, indicating that they are to be operated upon as one quantity. For instance, V^25 + 11. The vinculum attached to the radical sign indicates that the square root shall be extracted from the sum of 25+11, which is the same as the square root of 36. 35-1-1 5-4-22 In an expression as — 3 — ±. — the bar indicates that 3X8 NOTES ON MATHEMATICS. 3 the sum of 35+15+22 shall be divided by the product of 3X8 which is the same as 72 divided by 24. Whenever a number or a quantity is placed over a line and a number or a quantity is placed under the same line it always indicates that the number or quantity over the line shall be divided by the number or quantity under the line. Such a quantity is called a fraction. The quantity above the line is called the numerator, and the quantity below the line is called the denominator. A frac- tion may be either proper or improper. The fraction is proper when the numerator is smaller than the denominator ; for instance, f ; but improper if the numerator is larger than the denominator, for instance, -^ = If. A fraction can always be considered simply as a problem in division. Formulas. A formula is an algebraic expression for some general rule, law or principle. Formulas are used in mechanical books, because they are much more convenient than rules. Generally speaking, the knowledge of algebra is not required for the use of formulas, because the numerical values corre- sponding to the conditions of the problem are inserted for each letter in the formula except the letter representing the unknown quantity, which then is obtained by simple arithmet- ical calculations. It is generally most convenient to begin the interpretation of formulas from the right-hand side ; for instance, the formula for the velocity of water in long pipes is : v — 8.02 J ^ d - > /. /. In this formula v represents the velocity of the water in feet per second. h represents the "head"* in feet. d represents the diameter of the pipe in feet. f represents the friction factor determined by experiments. / represents the length of the pipe in feet, and 8.02 is a constant equal to the square root of twice acceleration due to gravity. Assume, for instance, that it is required to find the velocity of the flow of water in a pipe of 3 inches diameter ( %. foot) ; the length of the pipe is 1,440 feet, the "head" is 9 feet, and the friction factor is 0.025. Inserting in the formula these numerical values, and for convenience writing the diameter of the pipe in decimals, we have : * In hydraulics the word " head " means the vertical difference between the level of the water at the receiving end of the pipe and the point of discharge, or its equivalent in pressure. See Hydraulics, page 443. NOTES ON MATHEMATICS. v = 8.02 x J 9 X 0.25 * 0.025 X 1440 Solving the problem step by step we have : v = 8.02 xJ 2.25 * 36 v = 8.02 X \A).0625 7/ = 8.02 X 0.25. v = 2.005 feet per second. In mechanical formulas, if not otherwise specified, it is always safe to assume the letter g to mean acceleration due to gravity, usually taken as 32.2 feet or 9.82 meters. In formulas relating to heat the letter / usually signifies the mechanical equivalent of heat = 778 foot pounds of energy ; but in formulas relating to strength of materials the letter J usually signifies the polar moment of inertia, and the letter / the least rectangular moment of inertia. The letter x always expresses the unknown quantity. The following Greek letters are also used more or less. The letter ~, called pi, is used to signify the ratio of the circumference to the diameter of a circle, and is usually taken as 3.1416. % called sigma, usually signifies the sum of a number of quantities. The letter A, called delta, usually signifies small increments of matter. The letter 6, called theta, or the lefter <£, called phi, usually signifies some particular angle, sometimes also the coefficient of friction. But all these letters may be employed to express anything, although it is usually safe, if not otherwise specified. to expect their meaning to be as stated. It is always customary to express known quantities by the first letters in the alphabet, such as a, &, c, etc., and unknown quantities by such letters as x, y, z, etc. Bdtbmetic. Addition. All quantities to be added must be of the same unit; we can not add 3 feet + 8 inches + 2 meters, without first reducing these three terms either to feet, inches or meters. The same also with numbers. Units must be added to units, tens to tens, hundreds to hundreds, etc. Example. 318 -f 5 + 38 + 10 + 115 = 486 Solution : 318 5 38 10 115 486 = Sum. Subtraction. Two quantities to be subtracted must be of the same unit. In subtraction, the same as in addition, the units are placed under each other, and units are subtracted from units, tens from tens, hundreds from hundreds, etc. Example. 2543 — 1828 = = 715 Solution: 2543 . . Minuend. 1828 . Subtrahend 715 . . Difference. Subtrahend -f Difference = Minuend. (5) ARITHMETIC. ilultiplication. A quantity is multiplied by a number by adding it to itself ■as many times as the number indicates. Example. 314 X 3 = 314 + 314 + 314 = 942 Solution: 314 . . Multiplicand)^. 3 . . Multiplier } Factors. 942 . . Product. Product . , . ,. Multiplicand = Multiplier. Product Multiplier = Multiplicand. Division. The quantity or number to be divided is called the dividend. The number by which we divide is called the divisor. The number that shows how many times the divisor is contained in the dividend is called the quotient. Example. 6852 ^ 3 = 2284 ition: 3) 6852 ( 6 8 6 2284 6852 . . Dividend. 3 . . Divisor. 2284 . . Quotient. Divisor X Quotient = Dividend. 25 24 12 12 , ARITHMETIC. FRACTIONS. Addition. Fractions to be added must have a common denominator ; thus we cannot add l / 2 + % -f- % + U unless they be reduced to a common denominator instead of the denominators two, three and four ; in other words, we must find the least common multiple of the numbers 2, 3 and 4, which is 12. Thus we have: i = ft t = ft 1 3 4 = T2 3 9 4 = T2 H = 2ft = 2J Example 2. Add: T V + I + i + A + 1+ I + I + 1 The common denominator is found in the following manner : Write in a line all the denominators, and divide with the prime number, 2, as many numbers as can be divided with- out a remainder. The numbers that cannot be divided without a remainder remain unchanged, and these together with the quotients of the divided numbers, are written in the next line below. Repeat this operation as long as more than one num- ber can be divided without remainder, then try to divide by the next prime number, and so on. These divisors and all those numbers remaining undivided in the last line are multi- plied together, and the product is the least common denominator. 2) w ? i n $ n ^ 9 2) ? i 2 0374-9 2) £21 337?9 3) 211 £271? 211 11713 The common denominator is thus : 2X2X2X3X2X7X3= 1008 Thus 1008 is the least common multiple of 16, 8, 4, 12, 7 and 9. 8 ARITHMETIC. The principle of this solution can probably be better understood by resolving these numbers into prime numbers, and also resolving 1008 into prime numbers ; we then find that 1008 contains all the prime numbers necessary to make 16, 8, 4, 12, 6, 7 and 9. Prime numbers in n 1008 are 2 2 2 2 " 16 u 2 2 2 2 8 " 2 2 2 4 « 2 2 " 12 « 2 2 3 6 " 2 3 7 " 7 " 9 " 3 3 3 Solution of Example 2 : 1008 T V 63X7 = 441 | 126 X 5 = 630 i 252 X 1 = 252 T 7 2 84 X 7 = 588 | 168 X 5 = 840 | 144 X 5 = 720 | 126 X 3 = 378 | 112 X 4 = 448 4297 = mi A 2 6 5 Subtraction. When fractions are to be subtracted, they must first be reduced to a common denominator, the same as in addition. Example. must be reduced to f — f = f 5 Examples. No. 1. No. 2. No. 3. 60 FRACTIONS. Multiplication. Fractions are multiplied by fractions, by multiplying numer- ator by numerator and denominator by denominator ; thus : I X Ta = 96 — 3'2 The correctness of this rule can easily be understood if we consider these two fractions as two problems in division, t X ft will then be 3 divided by 8 and the quotient multiplied by 7 and the product divided by 12; thus, 3 is to be multiplied by 7 and the product is to be divided by 8 times 12. Therefore : 3X7 IX 8 8 X X? 8X4 iZ A mixed number may first be reduced to an improper frac- tion and then multiplied as a common fraction, numerator by numerator and denominator by denominator. For instance : ^i X | = j X f = x — 2f A fraction may be multiplied by a whole number by multi- plying the numerator and letting the denominator remain un- changed. For instance : Ax2 = it = iA = ii This must be correct, because we may consider 7 as indicat- ing the quantity and 12 as indicating what kind of quantity in exactly the same sense as we may say 7 dollars or 7 cents ; if either of those were multiplied by 2 the product would, of course, be either dollars or cents respectively, and for the same reason 7 twelfths multiplied by 2 must be 14 twelfths. A fraction may also be multiplied by a whole number, by dividing the denominator by the number and letting the numer- ator remain unchanged. For instance : ft X 2 = | — 1|, because ft is equal to |, so must ft X 2 Examples. No. 1. 3i X f = V X | = || No. 2. \\ Xli = |X f = f = 2X No. 3. ft X \ — ft No. 4. 14Xft = 3 i IO .FRACTIONS. Division. A fraction is divided by a fraction by writing the fractions after each other, then inverting the divisor (that is, changing its numerator to denominator and its denominator to numer- ator), proceed as in multiplication. For instance : 5 _i_ 3 — 5 V •* 20 5 8 • 4 T X 3 24 6 The reason for this rule can very easily be understood, when we consider the fractions as problems in division. That is to say, 5 shall be divided by 8 and the quotient is to be divided by one-fourth of 3. But if the quantity f is divided by 3 instead of one-fourth of 3, we must, of course, multiply the quotient by 4 to make the result correct. Therefore : 8-4 o 2 _l_q 20 5. "8" • d 24 6 A fraction may be divided by a whole number by dividing the numerator by the number and letting the denominator re- main unchanged. For instance : 9 _i_ q — 3 T6 • ° T6 A fraction may be divided by a whole number by multiply- ing the denominator by the whole number and letting the numerator remain unchanged. For instance : Mixed numbers are reduced to improper fractions the same as in multiplication ; they are then figured the same as if they were proper fractions. Examples. No 1 - 7 - — i -7- y 2 14 — 7 1NU. 1. 16 . 2 16 A : 16 8 Nn 9 - 5 - — & — _5_ v 4 — 20 — io 1MO. Z. is . 4 18*3 54 27 Vf n q oi _i_ 1 l — U v 3 — 51 No. 4. 2i ~ 6 = | -f- 6 = T V No. 5. 3i -f- 4 — -U -5-4=| In No. 4 it will be understood that f divided by 6 must be f ¥ , because T \ is exactly a sixth of J. In No. 5, also, it will be understood that if V 1 is divided by 4, the quotient must be f, because 4 is one-fourth of 16. DECIMALS. I I To Reduce a Fraction of One Denomination to a Fraction of Another Fixed Denomination, and Approx= imately of the Same Value. In mechanical calculations, on drawings, and on other oc- casions, it is very frequently necessary to reduce fractions of other denominations to eighths, sixteenths, thirty-seconds, or sixty-fourths. This may be done by multiplying the numerator and the denominator of the given fraction by the number which is to be the denominator in the new fraction, then dividing this new numerator and denominator by the denominator of the given fraction. Example. Reduce % to eighths, sixteenths, thirty-seconds, sixty- fourths, or to hundredths. f X t = M — o or s / % approximately. o 3 or \\ approximately. 16 32 42- I X 1 4 — I! I =F c / or If approximately. f X m = in = ^~ or T %V approximately. Thus |, instead of f , is considerably too small, namely, ^, but f | is a great deal nearer, only T ^ too large, and 0.67 is -g-^ too large. DECIMALS. In decimal fractions the denominator is always some power of ten, such as tenths, hundredths, thousandths, etc. The denominator is never written, as it is fixed by the rule that it is 1 with as many ciphers annexed as there are figures on the right-hand side of the decimal point. y 2 = 0.5 = five-tenths = T % % = 0.25 — twenty-five hundredths = T 2 o% }i = 0.125 = one hundred and twenty-five thousandths = T V 2 o 5 o 1 y 2 = 1.5 = one and five-tenths = 1 T % \% = 1.25 = one and twenty -five hundredths = 1 T %%, etc. 1 2 DECIMALS. Figures on the left side of the decimal point are whole num- bers. When there are no whole numbers, sometimes a cipher is written on the left side of the decimal point, but this is not always done, as it is common with many writers not to write anything on the left side of the decimal point when there is no whole number. Thus : Yz may be written .5 X " " " -25 yi " " " .125 It is, however, preferable to fill in a cipher on the left- hand side of the decimal point when there is no whole number, as by so doing the mistake of reading a decimal for a whole number is prevented. To Reduce a Vulgar Fraction to a Decimal Fraction. Annex a sufficient number of ciphers to the numerator, divide the numerator by the denominator, and point off as many decimals in the quotient as there are ciphers annexed to the numerator. Example. Reduce ^ to a decimal fraction. Solution : 8 ) 7.000 ( 0.875 64 60 56 40 40 00 Thus, yk is equal to the decimal fraction 0.875. DECIMALS. 13 1 Practions Reduced to Exact Decimals. 1 6~4 .015625 B~i .265625 33 64 .515625 49 6~4 .765625 A .03125 9 32 .28125 1 7 32 .53125 M .78125 3 64 .046875 1 9 6 4 .296875 3 5 6"4 .546875 H .796875 1 T6 .0625 A .3125 A .5625 1 3 T6~ .8125 A .078125 2 1 64 .328125 Si .578125 5 3 64 .828125 A .09375 1 1 32 .34375 1 9 32 .59375 2 7 3 2 .84375 A .109375 2 3 6 4 .359375 3 9 64 .609375 II .859375 "§" .125 3 8 .375 1 .625 .640625 7 8 .875 9 6~4 .140625 H .390625 5 7 64 .890625 A .15625 H .40625 2 1 32 .65625 W .90625 H .171875 2 7 64 .421875 43 64 .671875 « .921875 A .1875 7 29 6 4 .4375 .453125 1 1 T6 .6875 11 .9375 13 64 .203125 H .703125 6 1 6"4 .953125 7 32 .21875 M .46875 M .71875 31 32 .96875 H .234375 fi .484375 H .734375 6 3 6 4 .984375 J .25 2 .5 3 4 .75 1 1. To Reduce a Decimal Fraction to a Vulgar Fraction. Write the decimal as the numerator of the fraction and set under it for the denominator the figure one, followed by as many ciphers as there are decimal places ; then cancel the frac- tion thus written, to its smallest possible terms. Example. Reduce 0.3125 to a vulgar fraction. Solution : 0.3125 = T Vo¥o> cancelling this by five we have a 6 o¥o — iff = to — rV To Reduce a Decimal Fraction to a Given Vulgar Fraction of Approximately the Same Value. Multiply the decimal by the number which is denomi- nator in the fraction to which the decimal shall be reduced, and the product is the numerator in the fraction. Example. Reduce 0.484375 to sixteenths, thirty-seconds and sixty- fourths. 14 DECIMALS. Solution : 0.484375 X 16 = 7.75, gives !££, or J^., approximately. 0.484375 X 32 = 15.5, gives ^^, or i.|, approximately. 0.484375 X 64 = 31, gives 1 1 exactly. If the result does not need to be very exact, probably T %, which is 6 3 ¥ too small, is near enough, or the result, y iY% may be called ^, which is -^ too large. |f is -^ too small, therefore either ]/ 2 or \\ is only ¥ x ¥ different from the true value. The first is ef too large and the last is ^ too small, and which fraction, if either, should be preferred, will depend entirely upon the purpose for which the problem is solved. \\ is the exact value. Addition of Decimal Fractions. In adding decimal fractions, care should be taken to place the decimal points under each other ; then add as if they were whole numbers. Example. Solution : Add 50.5 + 5.05 + 0.505 + 0.0505 50.5 5.05 0.505 0.0505 56.1055 To prevent mistakes and mixing up of the figures during addition, it is preferable to make all the decimal fractions in the problem of the same denomination by annexing ciphers. Thus : 50.5000 5.0500 0.5050 0.0505 56.1055 Subtraction of Decimal Fractions. The decimal point in the subtrahend must be placed under that in the minuend ; the fractions are both brought to the same denomination by annexing ciphers, then the subtraction is performed just as if they were whole numbers, but close attention must be paid to have the decimal point in the same place in the difference as it is in the minuend and subtrahend. Example. 318.05 — 121.6542 DECIMALS. 1 5 Solution : 318.0500 Minuend. 121.6542 Subtrahend. 196.3958 Difference. Hultiplication of Decimal Fractions. Multiply the factors as if they were whole numbers. After multiplication is performed, count the number of decimals in both multiplier and multiplicand and point off (from the right) the same number of decimals in the product. If there are not enough figures in the product to give as many decimals as required, then prefix ciphers on the left until the required number of decimals is obtained. Example 1. 0.08 X 0.065 — 0.00520 = 0.0052 In this example it is necessary after the multiplication is performed, to prefix two ciphers to the product in order to obtain the necessary number of decimals, because the pro- duct, 520, consists of only three figures, but the two numbers, 0.08 and 0.065, contain five decimals. Example No. 2. 3.1416 X 5 = 15.7080 = 15.708 Example No. 3. 3.1416 X 0.5 = 1.57080 = 1.5708 Division of Decimal Fractions. Divide same as in whole numbers, and point off in the quotient as many decimals as the number of decimals in the dividend exceeds the number of decimals in the divisor. If the divisor contains more decimals than the dividend, then before dividing annex ciphers (on the right-hand side) in the dividend until dividend and divisor are both of the same denomination, then the quotient will be a whole number. Example. 43.62 -f- 0.003 = 14,540 Solution : 0.003 ) 43.620 ( 14,540 3 13 12 16 15 12 12 00 1 6 RATIO AND PROPORTION. In this example the dividend consists of only two decimals, but the divisor has three, therefore we have to annex a cipher to the dividend. This brings divisor and dividend to the same denomination, and the quotient is a whole number. Example 2. 43.62 -h 0.3 = 145.4 In this example the dividend has one decimal more than the divisor, therefore the quotient has one decimal. RATIO. The word ratio causes considerable ambiguity in mechani- cal books, as it is frequently used with different meaning by different writers. The common understanding seems to be that the ratio be- tween two quantities is the quotient when the first quantity is divided by the last quantity ; for instance, the ratio between 3 and 12 is %, but the ratio between 12 and 3 is 4. The ratio be- tween the circumference of a circle and its diameter is tc or 3.1416, but the ratio between the diameter and the circumfer- ence is \ or 0.3183, etc. This is the sense in which the word is used in this book, as this seems to agree with the common cus- tom with most mechanical writers. The term ratio is also sometimes applied to the difference of two quantities as well as to their quotient ; in which case the former is called arithmetical ratio, and the latter geometrical ratio. (See Progressions, page 68.) PROPORTION. In simple proportion there are three known quantities by which we are able to find the fourth unknown quantity ; there- fore proportion is also called "the rule of three", and it is either direct or inverse proportion. It is called direct proportion if the terms are in such ratio to one another that if one is doubled then the other will also have to be doubled, or if one is halved the other must also be halved. For instance, if 50 pounds of steel cost $25, how much will 250 pounds cost? 9^0 V 9^ 50 lbs. cost $25; 250 must cost zou A zo = $125. 50 This is direct proportion, because the more steel we buy, the more money we have to pay. In inverse proportion the terms are in such ratio that if one is doubled the other is halved, or if one is halved the other is doubled. proportion. 1 7 Example. Eight men can finish a certain work in 12 days. Howmany men are required to do the same work in 3 days ? Here we see that the fewer days in which the work is to be done, the more men are required. Therefore, this example is in inverse proportion. In 12 days the work was done by 8 men ; therefore, in order to do the work in 3 days it will require - == 32 men. It requires 4 times as many men because the work is to be done in one quarter of the time. Compound Proportion. A proportion is called compound, if to the three terms there are combined other terms which must be taken into considera- tion in solving the problem. A very easy way to solve a compound proportion is to (same as is shown in the following examples) place the con- ditional proposition under the interrogative sentence, term for term, and write x for the unknown quantity in the inter- rogative sentence ; draw a vertical line ; place x at the top at the left-hand side ; then try term for term and see if they are direct or inverse proportionally relative to x, exactly the same way as if each term in the conditional proposition and the correspond- ing term in the interrogative sentence were terms in a simple rule-of-three problem. Arrange each term in the interrogative sentence either on the right or left of the vertical line, according to whether it is found to be either a multiplier or a divisor, when the problem, independent of the other terms, is considered as a simple rule-of-three problem. After all the terms in the interrogative sentence are thus arranged, place each corresponding term in the conditional proposition on the opposite side of the vertical line. Then clear away all fractions by reducing them to improper frac- tions, and let the numerator remain on the same side of the verti- cal line where it is, but transfer the denominator to the opposite side. Now cancel any term with another on the opposite side of the vertical line ; then multiply all the quantities on the right side of the vertical line with each other. Also multiply all the quantities on the left side of the vertical line with each other. Divide the product on the right side by the product on the left, and the quotient is the answer to the problem. Example 1. A certain work is executed by 15 men in 6 days, by work- ing 8 hours each day. How many days would it take to do the same amount of work if 12 men are working 7j^ hours each dav? i8 PROPORTION. Solution : 15 12 1 Men 6 Days 8 Hours. " x " iy 2 " % W X? l 1 n % 8 Days. Example 2. A steam engine of 25 horse power is using 1500 pounds of coal in 1 day of 9*4 working hours. How many pounds of coal in the same proportion will be required for 2 steam engines each having 30 horse power, working 6 days of 12% hours each day ? Solution : 1 Machine 25 Hp. 1,500 pounds 1 Day 9% hours. 2 30 " x " 6 " 12% " « 30 X 1 im 300 2 6 " 1 P l £0 6 ? 2 1 19 P* x?% ^ 2 l 2 2 28,800 pounds of coal. Example 3. A piece of composition metal which is 12 inches long, 3J£ inches thick and 4>£ inches wide, weighs 45 pounds. How many pounds will another piece of the same alloy weigh, if it meas- ures 8 inches long, 1% inches thick and 6U inches wide? 12" long, sy" thick, ±y 2 " wide, 45 pounds. 1* " 6^ X 45 n ? ? $y 2 m i M PA P i ? i i % i 2 45 22 y 2 pounds. INTEREST. 19 INTEREST. The money paid for the use of borrowed capital is called interest. It is usually figured by the year per 100 of the principal. Simple Interest. Simple interest is computed by multiplying the principal by the percentage, by the time, and dividing by 100. What is the interest of #125, for 3 years, at 4% per year ? Solution : 125 X 4 X 3 100 #15 In Table No. 1, under the given rate per cent., find the interest for the number of years, months, and days ; add these together, and multiply by the principal invested, and the product is the interest. Example. What is the interest of #600, invested at 6%, in 5 years, 3 months, and 6 days ? Solution : #1.00 in 5 years at 6% = 0.30 " " 3 months " " = 0.015 " " 6 days " " = 0.001 0.316 600 = Principal. #189.60 = Interest 20 INTEREST. U a o © "- c 1> o, & 3> oo cc m h 55 1- ti* (M o » o :: « c: t- i< « c a o m h a r- i 1 .^ c x c (M»OariC0005(MOt-OCOO!Xi-HTtt-C(NOOOHmOG^lOt-0 o o c ri h h h n in n c: m x ;: ^ Tf ^ o o o o o o c c t- m- oo ooooooooooooooooooooooooooooo ooooooooooooooooooooooooooooo d © © d d d © © © © © © © © © © © ©do© © ©' © © © © © © ir5C5tO©iO©m©»C©0©iC©0©0©»C©iO©*0©0©u1©0 !M ^ £- © OOOOrHHrHH!M(NN(M(MCOMCOO:^^'1f^^iCiOiOiOCCC ©©©©©©©©©©©©©©©©©©©©©©©©©©©©© ©©©©©©©©©©©©©©©©©©©©©©©©©©©©© © © ©©©©©©©©©©©©©©©©©©©©©©©©©© O©00G0ir-t-©©O^TfCCCC(NC05T-(CO»OCCXO!M^CXO(N-<*C OOOOO^MHHH(M'M'Mjq(NKKMMM^'*Tj-^T)*LOiOLOiC ©©©©©©©©©©©©©©©©©©©©©©©©©©©©© ©©©©©©©©©©©©©<^_ ,.- - _ _ Z ©©©©©©©©©©©©©©©©©©©©©©© © © © © © © © © © UN UV '«♦'* w*; v»ij (,v ©©©©©© . _ _ ., _ . . _ ©©©©©©©©©©©©©©© © © © © © © © © © © © © © © © © © © © © ^ i^co©i^co©i^co©r--co©r-co©r-co©r--co©t-co©t~co©ir-co H CO O CC X O rn CO O C X O rn M O C X O h CO LO C X C h M iO C X ©©©©©rHi-Hl-li-ll-l — ci©©^c»0iO^C0^ct-XC OOOOOOOrtrlr-iHi^rtHlM!N!M(M!NN(NC0C0C0C0C0:0C0't ©©©©©©©©©©©©©©©©©©©©©©©©©©©©© ©©©©©©©©©©©©©©©©©©©©©©©©©©©©© © © © © © © © © ©' © ©" © ©' © © © ©* © © © © © © © © ©' ©' © © HiMCO^O>w T-i(MOOTt0!01r«XOiH(NM-*iCOt-XCi-!'M oohhhhhhhhhimnisimininnnncocok ©©©©©©©©©©©©©©©©©© '-" - " ^5 O © © © © © © © O © © © © © © © © ^— ^ ' — ' < — ' ^— ' < — ' ^— ' ^^ ^^ t_j i — . c__; < — ; c_j ^_^ ^^ c — ; ^_, . — , w . — . . — . ©©©©©©©©©©©©©©©©©©©©© © © © © © ©©©©'©'©©*© © © © © © © © X CO iff CO h O CO O lO CO (N O X C O CO h O X C i.O CO rt O X C C CO rt Ori(NCO^»OiCCr-XC7. OOH^CO^OOOt-XOiCOiHCNCO^ OOOCOOOOOOOHHHHriHrtHHHHHO)fq«01IM(N ©©©©©©©©©©©©©©©©©©©©©©©©©©©©© ©©©©©©©©©©©©©©©©©©©©©©©©©©©©© © © © © © © © © © © © © © © © © © © © © © © © © © © © © © >> N NJI' l/iv© t>00 ©\ O INTEREST. Wi-OMt-OMt-OcOb- Vs$ 1 W©OCCOOM©C CO CO "5N OOOiOCOHOODOiOttH © 1 o h (n « ^t io o 3 i- x a M j ooooooooooo 1 d o" © © © © © © d o" o" 1 Us o ©©©©©©©©©© thCM©©CO© Ot-1(N0!C oddo'dddddo t-CO©i— CO©t-CO©r-CO i-l Of iff CO GO O i-H CO iff CO CO Tt - « *5 T ICVO tNOO ©\ © - - « pc ^r icvo tsoo a o 2 2 COMPOUND INTEREST. Compound Interest Computed Annually. If the interest is not withdrawn, but added to the principal, so that it will also draw interest, it is called compound interest. Example. What is the amount of #300, in 3 years, at b% ? The interest is added to the principal at the end of each year. Solution : Principal and interest at the end of first year, 105 X 300 100 Principal and interest at the end of second year, 105 X 315 = 100 Principal and interest at the end of third year, ' 5 = #347.2875, = #347.29 = Amount. 100 When compound interest for a great number of years is to be calculated, the above method of figuring will take too much time, and the following interest tables, No. 2 and No. 3, are computed in order to facilitate such calculations. In Table No. 2, under the given rate-per cent., and opposite the given number of years, find the amount of one dollar in- vested at that rate for the time taken. Multiply this by the principal invested and the product is the amount. Example. #400 is invested at o% compound interest for 17 years, com- puted annually. What is the amount ? Solution: In Table No. 2, under 5%, and opposite 17 years, we find 2.292011. Multiply this by the principal. Thus : 2.292011 400 916.8044 = #916.80 = Amount. COMPOUND INTEREST. 23 "3 C 3 O a E o c • ©o 4> M o 03 O u <** 3 O B o * J3 CO O^OOmQOQOOiOLOffi^COMOHCCCOOjaOCOJr'Tt* b"*(MHOOOHMOO>OOMOiOiOt-HO^CO^MN OObOOOHdHiIJOCiOCHiOOMiHO^f-NO co^wioooo^cocoHaosowt-oocDHOiot-Cioo WH^CCoDcocOOOa)'N(NOOO(N^iOt-OWQQOH 0OOLOHt-^Ot-iO(NHOCOt-t-QOOO(NiOCO(Nt-(M(X) OHH!N(MCOTtt'*iOCit-t-ODC:OH'N-*iO-Ot-QO(NW (N(N!N!M(NN(N(MCOMM ^ O©Q(MOlMHC0C0O'Nbi0C0O00OHK5N«C0i0O OhiQh?1COhOOOOCCh©(NCOM^(»MNOhO'* •^05-*Q'+OC!MOOO(NOt-OCi5«HOOH(NCO»Ot-0 OOHHNCO»-*^lC®CCt-COClOrH!>JM-t>OOt-COO C0OO^C0OO0iO000ilr- C5f-OCMO00t'r-O(Ml^iOi0Ot-00^l0H(MHl0t-t' OyiLOC.-*OO^M-*LOCCtNJ^Tt*^(MCOOOCDaj(MW MOClNLOO^OO^ODNCHiOOOOXJOOHt-WO OOOHHHJKMMMW^TfiOiCOOM-CCCOCJOOO CM CM lOHCOOJ^O^MOfMOCMiOOt-OHWCBiO^^Oa) (NOHOOCCOOCOCOOrtf-OOlMOKJrtOOt-HCC'* CO 00 CO ■* O CO ^ CO O O CO L~ C~* CO 10 CO CO CO CO lO lO CO r~ O 10 o ■© co h o co co co o « -t< « « co -t h q co co ro h -t co o: (NObOMWCOH'*OOH'th'r<'1(X!NiOOW^WOO>C qqOHHHH(NW(NMMM , *Ti|'*ioiaiqq©f-^oooo 24 COMPOUND INTEREST. Compound Interest Computed Semi= Annually. When compound interest is to be computed semi-annually, use Table No. 3. Under the given rate and opposite the given number of years, find the amount of one dollar invested and interest computed semi-annually for the time taken. Multiply this by the principal invested, and the product is the amount. Example. $350 is put in a savings bank paying 4%, computed semi- annually. What is the amount in 10 years ? Solution : Under 4%, and opposite 10 years, we find the number 1.4860. This we multiply by the principal invested. Thus: 1 .485949 350 520.08215 = $520.08 = Amount. To compute compound interest for longer time than is given in the tables, figure the amount for as long a time as the table gives ; then consider this amount as a new princi- pal invested, and use the table and figure again for the rest of the time. Example. What is the amount of $40, left in a savings bank 18 years, at 4%, and the interest computed semi-annually. The table only gives 12 years, therefore we will look opposite 12 years, under 4%, and find the number 1.608440. This we multiply by the principal invested. Thus: 1.608440 40 64.3376 But now we have to compute for 6 years more, therefore under 4%, and opposite 6 years, we find the number 1.268243. Multiplying this by the principal, which is now considered as being invested 6 years more, we have : 1.268243 X 64.3376 = $81.60 = Amount. Thus, $40, invested at 4% interest, computed semi-annually, will, after 18 years of time, amount to $81.60. COMPOUND INTEREST. 2 5 E 09 ■a ■a c 3 o D. E o (A N o ©E it: ° •5 *- !1 <^3 is II £§ In (M^HCNXiOr-OOO^OuOOOXr- COOCO(NOO(N jq t- lc c jq n a) x o r. o cs c « 05 o ^ t)* t- Tji io rt co LOnxt-i^cjtNC^or. h::x^w*i^(nqoh©co cot-o-*x^r-HOHiO'-iCHt-cooLO(NX>o»ox pOr^i^^^(NCCCC^xt;iqO^OI^l^QqOiOip'rHCNC© ONQ^OCOOlCOCMONCCOiaOH^OKMHiiH o^ot-^t-or-i-ic-toccxoO'TCOOOOsoaoo Ot-iO(NOXt-t-0(NJ>iOOait-X'*iOiH(NHCt- ©:NmC:^C10"*CO'*OXi^x ^ X T ©OMMOOiHTj*aOOiOMCH(MOOOOOHOt': OOOOOXOX£-(MC0O^'tCMOt-i0 1 -HO^C»0Xfli HCCtObC:0!M^OM^HCO»OOXON^OXOiM qqqqcqHHHHHH(N«N!N(NMMMWM^'t OOiO^CCOkOtNXCOTHCOCCXXt-OXOiOOiOC: H(NCO»OOt-OOHM-*©t-XCHMiflOMariM't qqqqqqqHHHHHHH(N(N(N8q(N(N(NWMcc (A S3 >- « — fs n po fo ^ ^ic icovoi>r>ooooaov©o-N — (s 26 INTEREST. Table No. 4 gives time in which money will be doubled if it is invested either on simple or compound interest, compounded annually. TABLE No. 4. SIMPLE INTEREST. COMPOUND INTEREST. % Years. Days. % Years. Days. 2 3 4 5 6 7 8 9 10 11 12 50 40 33 28 25 22 20 16 14 12 11 10 9 8 120 206 80 240 103 180 40 33 120 2 2y 2 3 3K 4 5 6 7 8 9 10 11 12 35 28 23 20 17 15 14 11 10 9 8 7 6 6 1 30 162 54 240 168 75 321 89 2 16 98 231 42 Results of Saving Small Amounts of Honey. The following shows how easy it is to accumulate a fortune, provided proper steps are taken. The table gives the result of daily savings, put in a savings bank paying 4 per cent, per year, computed semi-annually : Savings Savings Amount in Amount in Amount in Amount in Amount in perDay. per Mo. 5 years. 10 years. 15 years. 20 years. 25 years. .05 $ 1.20 $ 78.84 $ 174.96 $ 292.07 $ 434.88 $ 608.94 .10 2.40 157.68 349.92 584.15 869.76 1,217.88 .25 6.00 394.20 874.80 1,460.37 2,174.40 3,044.74 .50 12.00 788.40 1,749.60 2,920.74 4,348.80 6,089.48 .75 18.00 1,182.60 2,624.40 4,381.11 6,523.20 9,133.22 $1.00 24.00 1,576.80 3,499.20 5,841.48 8,697.60 12,178.96 Nearly every person wastes an amount in twenty or thirty- years, which, if saved and carefully invested, would make a family quite independent; but the principle of small savings has been lost sight of in the general desire to become wealthy. EQUATION OF PAYMENTS. 27 EQUATION OF PAYMENTS. When several debts are due at different dates the average time when all the debts are due is calculated by the following rule: Multiply each debt separately by the number of days be- tween its own date of maturity and the date of the debt earliest due. Divide the sum of these products by the sum of the debts ; the quotient will express the number of days subsequent to the leading day when the whole debt should be paid in one sum. Example. A owed to B the following sums : $250 due May 12, $120 due July 19, $410 due August 16, and $60 due September 21, all in the same year. When should the whole sum be paid at once in order that neither shall lose any interest? Solution : May 12 $250 May 12 to July 19 is 68 days ; 120 X 68 = 8160 May 12 to Aug. 16 is 96 days ; 410 X 96 = 39360 May 12 to Sept. 21 is 132 day s ; 60 X 132 =r 7920 $840 ) 55440 = 65.9 66 days after May 12 will be July 17. When several debts are due after different lengths of time, the average time is calculated by this rule: Multiply the debt by the time ; divide the sum of the products by the sum of the debts, and the quotient is the time when all the debts may be considered due. Example. A owed B $600, due in 7 months ; $200 due in one month, and $700 due in 3 months. When should the whole debt be paid in one sum in order that neither shall lose any interest ? Solution : 600 X 7 = 4200 700 X 3 = 2100 200 x 1 = 200 1500 ) 6500 = 4^ months. Note : If the debts contain both dollars and cents the cents may, if such refinement is required, be considered as deci- mal parts of a dollar, but practically in such problems the cents may be omitted in the calculation. PARTNERSHIP, or calculating of proportional parts, is the calculation of the parts of a certain quantity in such a way that the ratio between the separate parts is equal to the ratio of certain given numbers. 2 8 partnership. Example 1. A composition for welding cast steel consists of 9 parts of borax and one part of sal-ammoniac. . How much of each, borax and sal-ammoniac, must be taken for a mixture of 5 lbs.? Solution : jo X 5 = ±y 2 lbs. borax. T V X 5 = Yz lb. sal-ammoniac. Example 2. An alloy shall consist of 160 parts of copper, 15 parts of tin and 5 parts of zinc. How much of each will be used for a cast- ing weighing 360 lbs.? Solution : 160 \%% X 360 = 320 lbs. of copper. 15 r¥o X 360 = 30 lbs. of tin. 5 tItt X 360 = 10 lbs. of zinc. 180 Example 3. Four persons — A, B, C and D, are buying a certain amount, of goods together. A's part is $500, B's, $100, C's, $250, and D's, $150. On the undertaking they are clearing a net profit of $120. How much of this is each to have? Solution : 500 A's Part = T %°o°o X 120 = $60 100 B's " = tVA X 120 = 12 250 C's " = T Vo°o X 120 = 30 150 D's " = tWo X 120 = 18 $1000 Example 4. Two persons — A and B, are putting money into business, A, $2,000 and B, $3,000, but A has his money invested in the busi- ness 2 years and B 'i]/ 2 years ; the net profit of the undertaking is $2,300. How much is each to have of the profit ? Solution: A, 2000 X 2 = 4000 B, 3000 X 2% = 7500 11500 A's Part is T \%%% X 2300 = $ 800 B's " " TT 5 5°o°o X 2300 = 1500 In cases like this it must be taken into consideration that the time is not equal ; B has not only had the largest capital in- vested but he has also had the capital at work in the business SQUARE ROOT. 2 9 the longest time, namely, 2 l / 2 years, while A has only had his capital invested 2 years. The ratio is, therefore, not $2,000 to $3,000 but $4,000 to $7,500, because $2,000 in 2 years is equal to $4,000 in one year, and $3,000 in 2 l / 2 years is equal to $7,500 in one vear. SQUARE ROOT. When the square root is to be extracted the number is di- vided into periods consisting of two figures, commencing from the extreme right if the number has no decimals, or from the decimal point towards the left for the whole numbers and towards the right for the decimals. (If the last period of deci- mals should have but one figure then annex a cipher, so that this period also has two figures, but if the period to the extreme left in the integer should happen to have only one figure it makes no difference; leave it as it is.) Ascertain the highest root of the first period and place it to the right of the number as in long division. Square this root and subtract the product of this from the first period. To the remainder annex the next period of numbers. Take for divisor 20 times the part of the root already found* and the quotient is the next figure in the root, if the product of this figure and the divisor added to the square of the figure does not exceed the dividend. To the difference between this sum and the dividend is annexed the next period of numbers. For divisor take again 20 times the part of the root already found, etc. Continue in this manner until the last period is used. If there is any remainder, and a more exact root is required, ciphers may be annexed in pairs and the operation continued until as many decimals in the root are obtained as are wanted. Example 1. Extract the square root of 271,441. Solution : V'27 | 14i41 5 2 = 25 [ 20 X 5 = 100 ) 214 100 X 2 + 2 2 = 204 20 X 52 = 1040 ) 1041 1040 X 1 + I 2 1041 521 0000 Thus: \/271,441 = 521, because 521 X 521 = 271,441. * If this divisor exceeds the dividend, write a cipher in the root; annex the next period of numbers, calculate a new divisor, corresponding to the increased root, and proceed as explained. 30 cube root. Example 2. Extract the square root of 26.6256. Solution : V26|62:56 5 2 = 25 J 20 X 5 = 100 ) 162 100 X 1 + l 2 = 101 1 20 X 51 = 1020 ) 6156 1020 X 6 + 62 = 6156 = 5.16 0000 CUBE ROOT. Wher the cube root is to be extracted, the number is divided into periods consisting of three figures. Commencing from the extreme right if the number has no decimals, or from the decimal point, toward the left, for the whole number, and toward the right for the decimals. (If the last period of deci- mals should not have three figures, then annex ciphers until this period also has three figures, but if the period to the extreme left in the integer should happen to consist of less than three figures it makes no difference ; leave it as it is.) Ascer- tain highest cube root in the first period and place it to the right of the number, the same as in long division. Cube this root and subtract the product from the first period. To the remainder annex next period of numbers. For the divisor in this number take 300 times the square of the part of the root already found,* and the quotient is the next figure in the root, if the product of this figure multiplied by the divisor and added to 30 times the part of the root already found, multiplied by the square of this quotient and added to the cube of the quotient, does not exceed this dividend. To the difference between this sum and the dividend is annexed the next period of numbers. For divisor take again 300 times the square of the part of the root already found, etc. Continue in this manner until the last period is used. If there is any remainder from last period, and a more exact root is required, ciphers may be annexed three at a time, and the operation continued until as many decimals are obtained in the root as are wanted. * If this divisor exceeds the dividend, write a cipher in the root, annex the next period of numbers, calculating a new divisor corresponding to the increased root, and proceed as explained. CUBE ROOT. 31 Example 1. Extract the cube root of 275,894,451. Solution : \/275 894 = 216 300 X 6 2 = 10800 ) 59894 451 10800 X 5 + 30 X 6 X 5 2 +, 5 3 = 58625 300 X 65 2 = 1267500 ) 1269451 1267500 X 1 + 30 X 65 X l 2 + l 3 = 1269451 = 651 0000000 Thus : 3 _ «/275,894,451 = 651, because 651 X 651 X 651 = 275,894,451. Example 2 : Extract the cube root of 551.368. Solution : 3 \/551j 5 = 512 300 X 8 2 = 19200 ) 39368 19200 X 2 + 30 X 8 X 2 2 + 2 3 = 39368 = 8.2 00000 The square root of a number consisting of two figures will never consist of more than one figure, and the square root of a number consisting of four figures will never consist of more than two figures ; hence, the rule to divide numbers into periods consisting of two figures. The cube root of a number consisting of three figures will never consist of more than one figure, and the cube root of a number consisting of six figures will never consist of more than two figures ; hence, the rule to divide the numbers into periods consisting of three figures. There will always be one decimal in the root for each period of decimals in the number of which the root is extracted. This relates to both cube and square root. The root of a fraction may be found by extracting the separate roots of numerator and denominator, or the fraction may be first reduced to a decimal fraction before the root is extracted. The root of a mixed number may be extracted by first re- ducing the number to an improper fraction and then extracting the separate roots of numerator and denominator, or the number may be first reduced to consist of integer and decimal fractions, and the root extracted as usual. 32 CUBE ROOT. Radical Quantities Expressed without the Radical Sign. The radical sign is not always used in signifying radical quantities. Sometimes a quantity expressing a root is written as a quantity to be raised into a fractional power. For instance : V 16 may be written I62. This is the same value ; thus, */W= 4 and 16^ = 4. 3 V 27 may be written, 27^ = 3. 3 3 8$ = V 8 2 =V 64 =4. The denominator in the exponent always indicates which root is to be extracted. Thus, 8t will be square 8 and extract the cube root from the product. Example. 4 4 16l = V16 3 = V4096 = 8. Thus, cube 16 and extract the fourth root of the product. RECIPROCALS. The reciprocal of any number is the quotient which is ob- tained when 1 is divided by the number. For instance, the reciprocal of 4 is % — 0.25 ; the reciprocal of 16 is j 1 ^ = 0.0625, etc. Frequently it is a saving of time when performing long division to use the reciprocal, as multiplying the dividend by the reciprocal of the divisor gives the quotient. For instance, divide 4 by 758. In Table No. 6 the reciprocal of 758 is given as 0.0013193. Multiplying 0.0013193 by 4 gives 0.0052772, which is correct to six decimals. When reducing vulgar frac- tions to decimals the reciprocal may be used with advantage. For instance, reduce £f to decimals. In Table No. 6 the recip- rocal of 64 is given as 0.015625, and 15 X 0.015625 == 0.234375, which is the decimal of £f. Important. — Whenever the exact reciprocal is not ex- pressible by decimals the result obtained by its use, as explained above, is only approximate. SQUARES, CUBES, ROOTS, AND RECIPROCALS. 33 TABLE No. 5. Giving Squares, Cubes, Square Roots, Cube Roots, and Reciprocals of Fractions and nixed Numbers, from ^ to 10. n ;z 2 n 3 \/n V ' n 1 n ft 0.000244 0.0000038 0.125 0.25 64 32 0.000977 0.0000305 0.17678 0.31496 32 3 ^4 0.002196 0.000103 0.21651 0.36056 21.3333 1 T6 0.003906 0.000244 0.25 0.39685 16 ft 0.006104 0.000477 0.27951 0.42750 12.8 3 32 0.008789 0.000823 0.30619 0.45428 10.6667 7 6~4 0.011963 0.001308 0.33072 0.47823 9.1428 1 8 0.015625 0.001953 0.35355 0.5 8 ft 0.01977 0.00278 0.375 0.52002 7.1111 ft 0.02441 0.00381 0.39528 0.53861 6.4 1 1 6~4 0.02954 0.00508 0.41458 0.55599 5.8182 ft 0.03516 0.00659 0.43301 0.57236 5.3333 1 3 6~4 0.04126 0.00838 0.45069 0.58783 4.9231 7 ~5? 0.04785 0.01047 0.46771 0.60254 4.5714 1 5 6"4 0.05493 0.01287 0.48412 0.61655 4.2666 1 4 0.06250 0.01562 0.5 0.62996 4 1 7 B~4 0.07056 0.01874 0.51539 0.64282 3.7647 3% 0.07910 0.02225 0.53033 0.65519 3.5556 H 0.08813 0.02616 0.54482 0.66709 3.3684 5 T6~ 0.09766 0.03052 0.55902 0.67860 3.2 H 0.10766 0.03533 0.57282 0.68973 3.0476 H 0.11816 0.04062 0.58630 0.70051 2.9091 If 0.12915 0.04641 0.59942 0.71097 2.7826 3 8 0.14062 0.05273 0.61237 0.72112 2.6667 II 0.15258 0.05960 0.625 0.73100 2.56 i* 0.16504 0.06705 0.63738 0.74062 2.4615 2 7 64 0.17798 0.07508 • 0.64952 0.75 2.3703 tV 0.19141 0.08374 0.66144 0.75915 2.2857 2 9 64 0.20522 0.09303 0.67314 0.76808 2.2069 1 5 .32 0.21973 0.10300 0.68465 0.77681 2.1333 64 0.23463 0.11364 0.69597 0.78534 2.0645 1 2 0.25 0.12500 0.70711 0.79370 2 34 SQUARES, CUBES, ROOTS, AND RECIPROCALS. n n 2 n 3 v?T 3 V n 1 n 33 64 0.26587 0.13709 0.71807 0.80188 1.9394 1 7 32 0.28225 0.14993 0.72887 0.80990 1.8823 fi 0.29906 0.16356 0.73951 0.81777 1.8286 ft 0.31641 0.17789 0.75000 0.82548 1.7778 3 7 64 0.33423 0.19315 0.76034 0.83306 1.7297 19 32 0.35254 0.20932 0.77055 0.84049 1.6842 II 0.37134 0.22628 0.78062 0.84781 1.6410 5 0.39062 0.24414 0.79057 0.85499 1.6 li 0.41040 0.26291 0.80039 0.86205 1.5610 fi 0.43066 0.28262 0.81009 0.86901 1.5238 43 64 0.45141 0.30330 0.81968 0.87585 1.4884 1 1 TB" 0.47266 0.32495 0.82916 0.88259 1.4545 II 0.49438 0.34761 0.83853 0.88922 1.4222 II 0.51660 0.37131 0.84779 0.89576 1.3913 47 64 0.53931 0.39605 0.85696 0.90221 1.3617 1 0.56250 0.42187 0.86603 0.90856 1.3333 If 0.58618" 0.44880 0.87500 0.91483 1.3061 §1 0.61035 0.47684 0.88388 0.92101 1.2800 51 6~4 0.63501 0.50602 0.89268 0.92711 1.2549 1 3 TB" 0.66016 0.53638 0.90139 0.93313 1.2308 53 ¥4 0.68579 0.56792 0.91001 0.93907 1.2075 fi 0.71191 0.60068 0.91856 0.94494 1.1852 M 0.73853 0.63467 0.92702 0.95074 1.1636 f 0.76562 0.66992 0.93541 0.95647 1.1428 57 6~4 0.79321 0.70646 0.94373 0.96213 1.1228 29 32 0.82129 0.74429 0.95197 0.96772 1.1034 II 0.84985 0.78346 0.96014 0.97325 1.0847 11 0.87891 0.82397 0.96825 0.97872 1.0667 fi 0.90845 0.86586 0.97628 0.98412 1.0492 fi 0.93848 0.90915 0.98425 0.98947 1.0323 6 3 B~4 0.96899 0.95385 0.99216 0.99476 1.01587 1 1 1 1 1 1 1* 1.12891 1.19943 1.03078 1.02041 0.94118 1.26562 1.42323 1.06066 1.04004 0.88889 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 35 n n 2 n 3 V n v n 1 n H 1 5 TB" 1.41016 1.67456 1.08965 1.05896 0.84211 1.5625 1.953125 1.11803 1.07722 0.8 1.72266 2.26099 1.14564 1.09488 0.76190 1-3- 1.89062 2.59961 1.17260 1.11199 0.72727 2.06641 2.97046 1.19896 1.12859 0.69565 *i 2.25 3.375 1.22474 1.14471 0.66667 1 9 T(T 2.44141 3.81470 1.25 1.16040 0.64 !f 2.640625 4.29102 1.27475 1.17567 0.61539 Hi 2.84766 4.80542 1.29904 1.19055 0.59260 if 3.0625 5.35937 1.32288 1.20507 0.57143 Hf 3.2S516 5.95434 1.34630 1.21925 0.55172 H 3.515625 6.59180 1.36931 1.23311 0.53333 i« 3.75391 7.27319 1.39194 1.24666 0.51613 2 4 8 1.41421 1.25992 0.5 2 tV 4.25390 8.77368 1.43614 1.27291 0.48485 2 i 4.515625 9.59582 1.45774 1.28564 0.47059 2 T 3 e 4.78516 10.46753 1.47902 1.29812 0.45714 2 i 5.0625 11.890625 1.5 1.31037 0.44444 2 i% 5.34766 12.36646 1.52069 1.32239 0.43243 2 t 5.640625 13.39648 1.54110 1.33420 0.42105 2 T 7 6 5.94141 14.48217 1.56125 1.34580 0.41026 2 £ 6.25 15.625 1.58114 1.35721 0.4 2 T 9 e 6.56541 16.82641 1.60078 1.36843 0.39024 H 6.890625 18.08789 1.62018 1.37946 0.38095 2 H 7.22266 19.41090 1.63936 1.39032 0.37209 2| 7.5625 20.79687 1.65831 1.40101 0.36364 2 H 7.91016 22.24731 1.67705 1.41155 0.35555 2 i 8.265625 23.76367 1.69558 1.42193 0.34783 91 5 Z T6 8.62891 25.34724 1.71391 1.43216 0.34042 3 9 27 1.73205 - 1.44225 0.33333 »t 9.765625 30.51758 1.76777 1.46201 0.32 »4 10.5625 34.32812 1.80278 1.48125 0.3077 H 11.390625 88.44336 1.83712 1.5 0.2963 36 SQUARES, CUBES, ROOTS, AND RECIPROCALS. n n 2 n 3 V n s/ n 1 n H 12.25 42.875 1.87083 1.51829 0.28571 H 13.140625 47.63476 1.90394 1.53616 0.27586 3| 14.0625 52.73437 1.93649 1.55362 0.26667 n 15.015625 58.18555 1.96850 1.57069 0.25806 4 16 64 2 1.58740 0.25 H 18.0625 76.76562 2.06155 1.61981 0.23529 H 20.25 91.125 2.12132 1.65096 0.22222 *1 22.5625 107.17187 2.17945 1.68099 0.21053 5 25 125 2.23607 1.70998 0.2 5| 27.5625 144.70312 2.291288 1.73801 0.19048 51 30.25 166.375 2.34521 1.76517 0.18182 5| 33.0625 190.10937 2.39792 1.79152 0.17391 6 36 216 2.44949 1.81712 0.16667 «i 39.0625 244.140625 2.5 1.84202 0.16 6i 42.25 274.625 2.54951 1.86626 0.15385 6J 45.5625 307.54687 2.59808 1.88988 0.14815 7 49 343 2.64575 1.91293 0.14286 7i 52.5625 381.07812 2.69258 1.93544 0.13793 H 56.25 421.875 2.73861 1.95743 0.13333 ?f • 60.0625 465.48437 2.78388 1.97895 0.12903 8 64 512 2.82843 2 0.125 Si- 68.0625 561.5156 2.87228 2.02062 0.12121 Si 72.25 614.125 2.91548 2.04083 0.11765 H 76.5625 669.92187 2.95804 2.06064 0.11428 9 81 729 3 2.08008 0.111111 H 85.5625 791.4531 3.04138 2.09917 0.10811 n 90.25 857.375 3.08221 2.11791 0.10526 9| 95.0625 926.8594 3.1225 2.13633 0.10256 10 100 1000 3.16228 2.15443 0.1 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 37 TABLE No. 6. Giving Squares, Cubes, Square Roots, Cube Roots, and Reciprocals of Numbers from o.i to iooo. n » 2 n 3 V ' n 3 ,— v n 1 n 10 0.1 0.01 0.001 0.31623 0.46416 0.2 0.04 0.008 0.44721 0.58480 5 0.3 0.09 0.027 0.54772 0.66943 3.3333 0.4 0.16 0.064 0.63245 0.73681 2.5 0.5 0.25 0.125 0.70711 0.79370 2 0.6 0.36 0.216 0.774597 0.84343 1.66667 0.7 0.49 0.343 0.83666 0.88790 1.42857 0.8 0.64 0.512 0.89443 0.92832 1.25 0.9 0.81 0.729 0.94868 0.96549 1.11111 1 1 1 1 1 1 1.1 1.21 1.331 1.04881 1.03228 0.909091 1.2 1.44 1.728 1.09545 1.06266 0.833333 1.3 1.69 2.197 1.14018 1.09134 0.769231 1.4 1.96 2.744 1.18322 1.11869 0.714286 1.5 2.25 3.375 1.22475 1.14471 0.666667 1.6 2.56 4.096 1.26491 1.16961 0.625 1.7 2.89 4.913 1.30384 1.19347 0.588235 1.8 3.24 5.832 1.34164 1.21644 0.555556 1.9 3.61 6.859 1.37840 1.23S55 0.526316 2 4 8 1.41421 1.25992 0.5 2.1 4.41 9.261 1.449138 1.28058 0.476190 2.2 4.84 10.648 1.48324 1 30059 0.454545 2.3 5.29 12.167 1.51657 1.32001 0.434783 2.4 5.76 13.824 1.54919 1.33887 0.416667 2.5 6.25 15.625 1.58114 1.35721 0.4 2.6 6.76 17.576 1.61245 1.37508 0.384615 2.7 7.29 19.683 1.64317 1.39248 0.37037 2.8 7.84 21.952 1.67332 1.40946 0.357143 2.9 8.41 24.389 1.70294 1.42604 0.344828 3 9 27 1.73205 1.44225 0.333333 3.2 1Q.24 32.768 1.78885 1.47361 0.3125 3.4 11.56 39.304 1.84391 1.50369 0.294118 3.6 12.96 46.656 1.89737 1.53262 0.277778 3.8 14.44 54.872 1.94936 1.56089 0.263158 4 16 64 2 1.58740 0.25 4.2 17.64 74.088 2.04939 1.61343 0.238095 4.4 19.36 85.184 2.09762 1.63868 0.227273 4.6 21.16 97.336 2.14476 1.66310 0.217391 4.8 23.04 110.592 2.19089 1.68687 0.208333 5 25 125 2.23607 1.70998 0.2 3» SQUARES, CUBES, ROOTS, AND RECIPROCALS. n n 2 n 3 \f n 3 *S n 1 n 6 36 216 2.44949 1.81712 0.1666667 7 49 343 2.64575 1.91293 0.142857 8 64 512 2.82843 2 0.125 9 81 729 3 2.08008 0.111111 10 100 1000 3.16228 2.15443 0.1 11 121 1331 3.31662 2.22398 0.0909091 12 144 1728 3.46410 2.28943 0.0833333 13 169 2197 3.60555 2.35133 0.0769231 14 196 2744 3.74166 2.41014 0.0714286 15 225 3375 3.87298 2.46621 0.0666667 16 256 4096 4 2.51984 0.0625 17 289 4913 4.12311 2.57128 0.0588235 18 324 5832 4.24264 2.62074 0.0555556 19 361 6859 4.35890 2.66840 0.0526316 20 400 8000 4.47214 2.71442 0.05 21 441 9261 4.58258 2.75892 0.0476190 22 484 10648 4.69042 2.80204 0.0454545 23 529 12167 4.79583 2.84387 0.0434783 24 576 13824 4.89898 2.88450 0.0416667 25 625 15625 5 2.92402 0.04 26 676 17576 5.09902 2.96250 0.0384615 27 729 19683 5.19615 3 0.0370370 28 784 21952 5.29150 3.03659 0.0357143 29 841 24389 5.38516 3.07232 0.0344828 30 900 27000 5.47723 3.10723 0.0333333 31 961 29791 5.56776 3.14138 0.0322581 32 1024 32768 5.65685 3,17480 0.03125 33 1089 35937 5.74456 3.20753 0.0303030 34 1156 39304 5.83095 3.23961 0.0294118 35 1225 42875 5.91608 3.27107 0.0285714 36 1296 46656 6 3.30193 0.0277778 37 1369 50653 6.08276 3.33222 0.0270270 38 1444 54872 6.16441 3.36198 0.0263158 39 1521 59319 6.245 3.39121 0.0256410 40 1600 64000 6.32456 3.41995 0.025 41 1681 68921 6.40312 3.44822 0.0243902 42 1764 74088 6.48074 3.47603 0.0238095 43 1849 79507 6.55744 3.50340 0.0232558 44 1936 85184 6.63325 3.53035 0.0227273 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 39 n n 2 n 3 Vn V// 1 n 45 2025 91125 6.70820 3.55689 0.0222222 46 2116 97336 6.78233 3.58305 0.0217391 47 2209 103823 6.85565 3.60883 0.0212766 48 2304 110592 6.92820 3.63424 0.0208333 49 2401 117649 7 3.65931 0.0204082 50 2500 125000 7.07107 3.68403 0.02 51 2601 132651 7.14143 3.70843 0.0196078 52 2704 140608 7.21110 3.73251 0.0192308 53 2809 148877 7.28011 3.75629 0.0188679 54 2916 157464 7.34847 3.77976 0.0185185 55 3025 166375 7.41620 3.80295 0.0181818 56 3136 175616 7.48331 3.82586 0.0178571 57 3249 185193 7.54983 3.84852 0.0175439 58 3364 195112 7.61577 3.87088 0.0172414 59 34S1 205379 7.68115 3.89300 0.0169492 60 3600 216000 7.74597 3.91487 0.0166667 61 3721 226981 7.81025 3.93650 0.0163934 62 3844 238328 7.87401 3.95789 0.0161290 63 3969 250047 7.93725 3.97906 0.0158730 64 4096 262144 8 4 0.0156250 65 4225 274625 8.06226 4.02073 0.0153846 66 4356 287496 8.12404 4.04124 0.0151515 67 4489 300763 8.18535 4.06155 0.0149254 68 4624 314432 8.24621 4.08166 0.0147059 69 4761 328509 8.30662 4.10157 0.0144928 70 4900 343000 8.36660 4.12129 0.0142857 71 5041 357911 8.42615 4.14082 0.0140845 72 5184 373248 8.48528 4.16017 0.0138889 73 5329 389017 8.54400 4.17934 0.0136986 74 5476 405224 8.60233 4.19834 0.0135135 75 5625 421875 8.66025 4.21716 0.0133333 76 5776 438976 8.71780 4.23582 0.0131579 77 5929 456533 8.77496 4.25432 0.0129870 78 6084 474552 8.83176 4.27266 0.0128205 79 6241 493039 8.88819 4.29084 0.0126582 80 6400 512000 8.94427 4.30887 0.0125 81 6561 531441 9 4.32675 0.0123457 82 6724 551368 9.05539 4.34448 0.0121951 83 6889 571787 9.11043 4.36207 0.0120482 84 7056 592704 9.16515 4.37952 0.0119048 4o SQUARES, CUBES, ROOTS, AND RECIPROCALS. n n 2 n 3 V« v n 1 n 85 7225 614125 9.21954 4.39683 0.0117647 86 7396 636056 9.27362 4.414 0.0116279 87 7569 658503 9.32738 4.43105 0.0114943 88 7744 681472 9.38083 4.44797 0.0113636 89 7921 704969 9.43398 4.46475 0.0112360 90 8100 729000 9.48683 4.48140 0.0111111 91 8281 753571 9.53939 4.49794 0.0109890 92 8464 778688 9.59166 4.51436 0.0108696 93 8649 804357 9.64365 4.53065 0.0107527 94 8836 830584 9.69536 4.54684 0.0106383 95 9025 857375 9.74679 4.56290 0.0105263 96 9216 884736 9.79796 4.57886 0.0104167 97 9409 912673 9.84886 4.59470 0.0103093 98 9604 941192 9.89949 4.61044 0.0102041 99 9801 970299 9.94987 4.62607 0.0101010 100 10000 1000000 10 4.64159 0.01 101 10201 1030301 10.04988 4.65701 0.0099010 102 10404 1061208 10.09950 4.67233 0.0098039 103 10609 1092727 10.14889 4.68755 0.0097087 104 10816 1124864 10.19804 4.70267 0.0096154 105 11025 1157625 10.24695 4.71769 0.0095238 106 11236 1191016 10.29563 4.73262 0.0094340 107 11449 1225043 10.34408 4.74746 0.0093458 108 11664 1259712 10.39230 4.76220 0.0092593 109 11881 1295029 10.44031 4.77686 0.0091743 110 12100 1331000 10.48809 4.79142 0.0090909 111 12321 1367631 10.53565 4.80590 0.0090090 112 12544 1404928 10.58301 4.82028 0.0089286 113 12769 1442897 10.63015 4.83459 0.0088496 114 12996 1481544 10.67708 4.84881 0.0087719 115 13225 1520S75 10.72381 4.86294 0.0086957 116 13456 1560896 10.77033 4.877 0.0086207 117 13689 1601613 10.81665 4.89097 0.0085470 118 13924 1643032 10.86278 4.90487 0.00S4746 119 14161 1685159 10.90871 4.91868 0.0084034 120 14400 1728000 10.95445 4.93242 0.0083333 121 14641 1771561 11 4.94609 0.0082645 122 14884 1815848 11.04536 4.95968 0.0081967 123 15129 1860867 11.09054 4.97319 0.0081301 124 15376 1906624 11.13553 4.98663 0.0080645 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 41 n n 2 ;/ 3 V n */ n 1 n 125 15625 1953125 11.18034 5 0.008 126 15876 2000376 11.22497 5.01330 0.0079365 127 16129 2048383 11.26943 5.02653 0.0078740 128 16384 2097152 11.31371 5.03968 0.0078125 129 16641 2146689 11.35782 5.05277 0.0077519 130 16900 2197000 11.40175 5.06580 0.0076923 131 17161 2248091 11.44552 5.07875 0.0076336 132 17424 2299968 11.48913 5.09164 0.0075758 133 17689 2352637 11.53256 5.10447 0.0075188 134 17956 2406104 11.57584 5.11723 0.0074627 135 18225 2460375 11.61895 5.12993 0.0074074 136 18496 2515456 11.66190 5.14256 0.0073529 137 18769 2571353 11.70470 5.15514 0.0072993 138 19044 2628072 11.74734 5.16765 0.0072464 139 19321 2685619 11.78983 5.18010 0.0071942 140 19600 2744000 11.83216 5.19249 0.0071429 141 19881 2803221 11.87434 5.20483 0.0070922 142 20164 2863288 11.91638 5.21710 0.0070423 143 20449 2924207 11.95826 5.22932 0.0069930 144 20736 2985984 12 5.24148 0.0069444 145 21025 3048625 12.04159 5.25359 0.0068966 146 21316 3112136 12.08305 5.26564 0.0068493 147 21609 3176523 12.12436 5.27763 0.0068027 148 21904 3241792 12.16553 5.28957 0.0067568 149 22201 3307949 12.20656 5.30146 0.0067114 150 22500 3375000 12.24745 5.31329 0.0066667 151 22801 3442951 •12.28821 5.32507 0.0066225 152 23104 3511808 12.32883 5.33680 0.0065789 153 23409 3581577 12.36932 5.34848 0.0065359 154 23716 3652264 12.40967 5.36011 0.0064935 155 24025 3723875 12.44990 5.37169 0.0064516 156 24336 3796416 12.49 5.38321 0.0064103 157 24649 3869893 12.52996 5.39469 0.0063694 158 24964 3944312 12.56981 5.40612 0.0063291 159 25281 4019679 12.60952 5.41750 0.0062893 160 25600 4096000 12.64911 5.42884 0.00625 161 25921 4173281 12,68858 5.44012 0.0062112 162 26244 4251528 12.72792 5.45136 0.0061728 163 26569 4330747 12.76715 5.46256 0.0061350 164 26896 4410944 12.80625 5.47370 0.0060976 SQUARES, CUBES, ROOTS, AND RECIPROCALS. n n , n 3 V n 3 V' n 1 n 165 27225 4492125 12.84523 5.48481 0.0060606 166 27556 4574296 12.88410 5.49586 0.0060241 167 27889 4657463 12.92285 5.50688 0.0059880 168 28224 4741632 12.96148 5.51785 0.0059524 169 28561 4826809 13 5.52877 0.0059172 170 28900 4913000 13.03840 5.53966 0.0058824 171 29241 5000211 13.07670 5.55050 0.0058480 172 29584 5088448 13.11488 5.56130 0.0058140 173 29929 5177717 13.15295 5.57205 0.0057803 174 30276 5268024 13.19091 5.58277 0.0057471 175 30625 5359375 13.22876 5.59344 0.0057143 176 30976 5451776 13.26650 5.60408 0.0056818 177 31329 5545233 13.30413 5.61467 0.0056497 178 31684 5639752 13.34165 5.62523 0.0056180 179 32041 5735339 13.37909 5.63574 0.0055866 ISO 32400 5832000 13.41641 5.64622 0.0055556 181 32761 5929741 13.45362 5.65665 0.0055249 182 33124 6028568 13.49074 5.66705 0.0054945 183 33489 6128487 13.52775 5.67741 0.0054645 184 33856 6229504 13.56466 5.68773 0.0054348 185 34225 6331625 13.60147 5.69802 0.0054054 186 34596 6434850 13.63818 5.70827 0.0053763 187 34969 6539203 13.67479 5.71848 0.0053476 188 35344 6644672 13.71131 5.72865 0.0053191 189 35721 6751269 13.74773 5.73879 0.0052910 190 36100 6859000 13.13405 5.74890 0.0052632 191 36481 6967871 13.82028 5.75897 0.0052356 192 36864 7077888 13.85641 5.769 0.0052083 193 37249 7189057 13.89244 5.779 0.0051813 194 37636 7301384 13.92839 5.78896 0.0051546 195 38025 7414875 13.96424 5.79889 0.0051282 196 38416 7529536 14 5.80879 0.0051020 197 38809 7645373 14.03567 5.81865 0.0050761 198 39204 7762392 14.07125 5.82848 0.0050505 199 39601 7880599 14.10674 5.83827 0.0050251 200 40000 8000000 14.14214 5.84804 0.005 201 40401 8120601 14.17745 5.85777 0.0049751 202 40804 8242408 14.21267 5.86747 0.0049505 203 41209 8365427 14.24781 5.87713 0.0049261 204 41616 8489664 14.28286 5.88677 0.0049020 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 43 n 205 W 2 n 3 V n v n 1 n 42025 8615125 14.31782 5.89637 0.0048781 20(3 42436 8741816 14.35270 5.90594 0.0048544 207 42849 8869743 14.38749 5.91548 0.0048309 208 43264 8998912 14.42221 5.92499 0.0048077 209 43681 9129320 14.45683 5.93447 0.0047847 210 44100 9261000 14.49138 5.94392 0.0047619 211 44521 9393931 14.52584 5.95334 0.0047393 212 44944 9528128 14.56022 5.96273 0.0047170 213 45369 9663597 14.59452 5.97209 0.0046948 214 45796 9800344 14.62874 5.98142 0.0046729 215 46225 9938375 14.66288 5.99073 0.0046512 216 46656 10077696 14.69694 6 0.0046296 217 47089 10218313 14.73092 6.00925 0.0046083 218 47524 10360232 14.76482 6.01846 0.0045872 219 47961 10503459 14.79865 6.02765 0.0045662 220 48400 10648000 14.83240 6.03681 0.0045455 221 48841 10793861 14.86607 6.04594 0.0045249 222 49284 10941048 14.89966 6.05505 0.0045045 223 49729 11089567 14.93318 6.06413 0.0044843 224 50176 11239424 14.96663 6.07318 0.0044643 225 50625 11390625 15 6.08220 0.0044444 226 51076 11543176 15.03330 6.09120 0.0044248 227 51529 11697083 15.06652 6.10017 0.0044053 228 51984 11852352 15.09967 6.10911 0.0043860 229 52441 12008989 15.13275 6.11803 0.0043668 230 52900 12167000 15.16575 6.12693 0.0043478 231 53361 12326391 15.19868 6.13579 0.0043290 232 53824 12487168 15.23155 6.14463 0.0043103 233 54289 12649337 15.26434 6.15345 0.0042918 234 54756 12812904 15.29706 6.16224 0.0042735 235 55225 12977875 15.32971 6.17101 0.0042553 236 55696 13144256 15.36229 6.17975 0.0042373 237 56169 13312053 15.39480 6.18846 0.0042194 238 56644 13481272 15.42725 6.19715 0.0042017 239 57121 13651919 15.45962 6.20582 0.0041841 240 57600 13824000 15.49193 6.21447 0.0041667 241 58081 13997521 15.52417 6.22308 0.0041494 242 58564 14172488 15.55635 6.23168 0.0041322 243 59049 14348907 15.58846 6.24025 0.0041152 244 59536 14526784 15.62050 6.24880 0.0040984 44 SQUARES, CUBES, ROOTS, AND RECIPROCALS. n n* n* V n 3 1 n 245 60025 14706125 15.65248 6.25732 0.0040816 246 60516 14886936 15.68439 6.26583 0.0040650 247 61009 15069223 15.71623 6.27431 0.0040486 248 61504 15252992 15.74802 6.28276 0.0040323 249 62001 15438249 15.77973 6.29119 0.0040161 250 62500 15625000 15.81139 6.29961 0.004 251 63001 15813251 15.84298 6.30799 0.0039841 252 63504 16003008 15.87451 6.31636 0.0039683 253 64009 16194277 15.90597 6.32470 0.0039526 254 64516 16387064 15.93738 6.33303 0.0039370 255 65025 16581375 15.96872 6.34133 0.0039216 256 65536 16777216 16 6.34960 0.0039062 257 66049 16974593 16.03122 6.35786 0.0038911 258 66564 17173512 16.06238 6.3663 0.0038760 259 67081 17373979 16.09348 6.37431 0.0038610 260 67600 17576000 16.12452 6.38250 0.0038462 26 L 68121 17779581 16.15549 6.39068 0.0038314 262 68644 17984728 16.18641 6.39883 0.0038168 263 69169 18191447 16.21727 6.40696 0.0038023 264 69696 18399744 16.24808 6.41507 0.0037879 265 70225 18609625 16.27882 6.42316 0.0037736 266 70756 18821096 16.30951 6.43123 0.0037594 267 71289 19034163 16.34013 6.43928 0.0037453 268 71824 19248832 16.37071 6.44731 0.0037313 269 72361 19465109 16.40122 6.45531 0.0037175 270 72900 19683000 16.43168 6.46330 0.0037037 271 73441 19902511 16.46208 6.47127 0.00369 272 73984 20123648 16.49242 .6.47922 0.0036765 273 74529 20346417 16.52271 6.48715 0.0036630 274 75076 20570824 16.55295 6.49507 0.0036496 275 75625 20796875 16.58312 6.50296 0.0036364 276 76176 21024576 16.61325 6.51083 0.0036232 277 76729 21253933 16.64332 6.51868 0.0036101 278 77284 21484952 16.67333 6.52652 0.0035971 279 77841 21717639 16.70329 6.53434 0.0035842 280 78400 21952000 16.73320 6.54213 0.0035714 281 78961 22188041 16.76305 6.54991 0.0035587 282 79524 22425768 16.79286 6.55767 0.0035461 283 80089 22665187 16.82260 6.56541 0.0035336 284 80656 22906304 16.85230 6.57314 0.0035211 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 45 n n 2 n 3 s/ n v n 1 n 285 81225 23149125 16.88194 6.58084 0.0035088 286 81796 23393656 16.91153 6.58853 0.0034965 287 82369 23639903 16.94107 6.59620 0.0034843 288 82944 23887872 16.97056 6.60385 0.0034722 289 83521 24137569 17 6.61149 C. 0034602 290 84100 24389000 17.02939 6.61911 0.0034483 291 84681 24642171 17.05872 6.62671 0.0034364 292 85264 24897088 17.08801 6.63429 0.0034247 293 85849 25153757 17.11724 6.64185 0.0034130 294 86436 25412184 17.14643 6.64940 0.0034014 295 87025 25672375 17.17556 6.65693 0.0033898 296 87616 25934336 17.20465 6.66444 0.0033784 297 88209 26198073 17.23369 6.67194 0.0033670 298 88804 26463592 17.26268 6.67942 0.0033557 299 89401 26730899 17.29162 6.68688 0.0033445 300 90000 27000000 17.32051 6.69433 0.0033333 301 90601 27270901 17.34935 6.70176 0.0033223 302 91204 27543608 17.37815 6.70917 0.0033113 303 91809 27818127 17.40690 6.71657 0.0033003 304 92416 28094464 17.43560 6.72395 0.0032895 305 93025 28372625 17.46425 6.73132 0.0032787 306 93636 28652616 17.49286 6.73866 0.0032680 307 94249 28934443 17.52142 6.746 0.0032573 308 94864 29218112 17.54993 6.75331 0.0032468 309 95481 29503629 17.57840 6.76061 0.0032362 310 96100 29791000 17.60682 6.76790 0.0032258 311 96721 30080231 17.63519 6.77517 0.0032154 312 97344 30371328 17.66352 6.78242 0.0032051 313 97969 30664297 17.69181 6.78966 0.0031949 314 98596 30959144 17.72005 6.79688 0.0031847 315 99225 31255875 17.74824 6.80409 0.0031746 316 99856 31554496 17.77639 6.81128 0.0031646 317 100489 31855013 17.80449 6.81846 0.0031546 318 101124 32157432 17.83255 6.82562 0.0031447 319 101761 32461759 17.86057 6.83277 0.0031348 320 102400 32768000 17.88854 6.83990 0.0031250 321 103041 33076161 17.91647 6.84702 0.0031153 322 103684 33386248 17.94436 6.85412 0.0031056 323 104329 33698267 17.97220 6.86121 0.0030960 324 104976 34012224 18 6.86829 0.0030864 4 6 SQUARES, ' CUBES, ROOTS, AND RECIPROCALS. n n 2 n 3 \/n 3 v n 1 n 325 105625 34328125 18.02776 6.87534 0.0030769 326 106276 34645976 18.05547 6.88239 0.0030675 327 106929 34965783 18.08314 6.88942 0.0030581 328 107584 35287552 18.11077 6.89643 0.0030488 329 108241 35611289 18.13836 6.90344 0.0030395 330 108900 35937000 18.16590 6.91042 0.0030303 331 109561 36264691 18.19341 6.91740 0.0030211 332 110224 36594368 18.22087 6.92436 0.0030120 333 110889 36926037 18.24829 6.93131 0.0030030 334 111556 37259704 18.27567 6.93623 0.0029940 335 112225 37595375 18.30301 6.94515 0.0029851 336 112896 37933056 18.33030 6.95205 0.0029762 337 113569 38272753 18.35756 6.95894 0.0029674 338 114244 38614472 18.38478 6.96582 0.0029586 339 114921 38958219 18.41195 6.97268 0.0029499 340 115600 39304000 18.43909 6.97953 0.0029412 341 116281 39651821 18.46619 6.98637 0.0029326 342 116964 40001688 18.49324 6.99319 0.0029240 343 117649 40353607 18.52026 7 0.0029155 344 118336 40707584 18.54724 7.00680 0.0029070 345 119025 41063625 18.57418 7.01358 0.0028986 346 119716 41421736 18.60108 7.02035 0.0028902 347 120409 41781923 18.62794 7.02711 0.0028818 348 121104 42144192 18.65476 7.03385 0.0028736 349 121801 42508549 18.68154 7.04059 0.0028653 350 122500 42875000 18.70829 7.04730 0.0028571 351 123201 43243551 18.73499 7.054 0.0028490 352 123904 43614208 18.76166 7.06070 0.0028409 353 124609 43986977 18.78829 7.06738 0.0028329 354 125316 44361864 18.81489 7.07404 0.0028249 355 126025 44738875 18.84144 7.08070 0.0028169 356 126736 45118016 18.86796 7.08734 0.0028090 357 127449 45499293 18.89444 7.09397 0.0028011 358 128164 45882712 18.92089 7.10059 0.0027933 359 128881 46268279 18.94730 7.10719 0.0027855 360 129600 46656000 18.97367 7.11379 0.0027778 361 130321 47045881 19 7.12037 0.0027701 362 131044 47437928 19.02630 7.12694 0.0027624 363 131769 47832147 19.05256 7.13349 0.0027548 364 132496 48228544 19.07878 7.14004 0.0027473 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 47 n 365 n 2 n 3 \f~n 3 V n 1 n 133225 48627125 19.10497 7.14657 0.0027397 366 133956 49027896 19.13113 7.15309 0.0027322 367 134689 49430863 19.15724 7.15960 0.0027248 368 135424 4983032 19.18333 7.16610 0.0027174 369 136161 50248 i09 19.20937 7.17258 0.0027100 370 136900 50653000 19.23538 7.17905 0.0027027 371 137641 51064811 19.26136 7.18552 0.0026954 372 138384 51478848 19.28730 7.19197 0.0026882 373 139129 51895117 19.31321 7.19841 0.0026810 374 139876 52313624 19.33908 7.20483 0.0026738 375 140625 52734375 19.36492 7.21125 0.0026667 376 141376 53157376 19.39072 7.21765 0.0026596 377 142129 53582633 19.41649 7.22405 0.0026525 378 142884 54010152 19.44222 7.23043 0.0026455 379 143641 54439939 19.46792 7.23680 0.0026385 380 144400 54872000 19.49359 7.24316 0.0026316 381 145161 55306341 19.51922 7.24950 0.0026247 382 145924 55742968 19.54482 7.25584 0.0026178 383 146689 56181887 19.57039 7.26217 0.0026110 384 147456 56623104 19.59592 7.26848 0.0026042 385 148225 57066625 19.62142 7.27479 0.0025974 386 148996 57512456 19.64688 7.28108 0.0025907 387 149769 57960603 19.67232 7.28736 0.0025840 388 150544 58411072 19.69772 7.29363 0.0025773 389 151321 58863869 19.72308 7.29989 0.0025707 390 152100 59319000 19.74842 7.30614 0.0025641 391 152881 59776471 19.77372 7.31238 0.0025575 392 153664 60236288 19.79899 7.31861 0.0025510 393 154449 60698457 19.82423 7.32483 0.0025445 394 155236 61162984 19.84943 7.33104 0.0025381 395 156025 61629875 19.87461 7.33723 0.0025316 396 156816 62099136 19.89975 7.34342 0.0025253 397 157609 62570773 19.92486 7.34960 0.0025189 398 158404 63044792 19.94994 7.35576 0.0025126 399 159201 63521199 19.97498 7.36192 0.0025063 400 160000 64000000 20 7.36806 0.0025 401 160801 64481201 20.02498 7.37420 0.0024938 402 161604 64964808 20.04994 7.38032 0.0024876 403 162409 65450827 20.07486 7.38644 0.0024814 404 163216 65939264 20.09975 7.39254 0.0024752 48 SQUARES, CUBES. ROOTS, AND RECIPROCALS. n 405 n 2 n 3 V n V n 1 n 164025 66430125 20.12461 7.39864 0.0024691 406 164836 66923416 20.14944 7.40472 0.0024631 407 165649 67419143 20.17424 7.41080 0.0024570 408 166464 67917312 20.19901 7.41686 0.0024510 409 167281 68417929 20.22375 7.42291 0.0024450 410 168100 6S921000 20.24846 7.42896 0.0024390 411 168921 69426531 20.27313 7.43499 0.0024331 412 169744 69934528 20.29778 7.44102 0.0024272 413 170569 70444997 20.32240 7.44703 0.00242 J 3 414 171396 70957944 20.34699 7.45304 0.0024155 415 172225 71473375 20.37155 7.45904 0.0024096 416 173056 71991296 20.39608 7.46502 0.0024038 417 173889 72511713 20.42058 7.471 0.0023981 418 174724 73034632 20.44505 7.47697 0.0023923 419 175561 73560059 20.46949 7.48292 0.0023866 420 176400 74088000 20.49390 7.48887 0.0023810 421 177241 74618461 20.51828 7.49481 0.0023753 422 178084 75151448 20.54264 7.50074 0.0023697 423 178929 75686967 20.56696 7.50666 0.0023641 424 179776 76225024 20.59126 7.51257 0.0023585 425 180625 76765625 20.61553 7.51847 0.0023529 426 181476 77308776 20.63977 7.52437 0.0023474 427 182329 77854483 20.66398 7.53025 0.0023419 428 183184 78402752 20.6S816 7.53612 0.0023364 429 184041 78953589 20.71232 7.54199 0.0023310 430 184900 79507000 20.73644 7.54784 0.0023256 431 185761 80062991 20.76054 7.55369 0.0023202 432 186624 80621568 20.78461 7.55953 0.0023148 433 187489 81182737 20.80865 7.56535 0.0023095 434 188356 81746504 20.83267 7.57117 0.0023041 435 189225 82312875 20.85665 7.57698 0.0022989 436 190096 82881856 20.88061 7.58279 0.0022936 437 190969 83453453 20.90455 7.58858 0.0022883 438 191844 84027672 20.92845 7.59436 0.0022831 439 192721 84604519 20.95233 7.60014 0.0022779 440 193600 85184000 20.97618 7.60590 0.0022727 441 194481 85766121 21 7.61166 0.0022676 442 195364 86350888 21.02380 7.61741 0.0022624 443 196249 86938307 21.04757 7.62315 0.0022573 444 197136 87528384 21.07131 7.62888 0.0022523 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 49 n n* n 3 V n 21.09502 v n 1 n 445 198025 88121125 7.63461 0.0022472 446 198916 88716536 21.11871 7.64032 0.0022422 447 199809 89314623 21.14237 7.64603 0.0022371 448 200704 89915392 21.16601 7.65172 0.0022321 449 201601 90518849 21.18962 7.65741 0.0022272 450 202500 91125000 21.21320 7.66309 0.0022222 ' 451 1 203401 91733851 21.23676 7.66877 0.0022173 452 204304 92345408 21.26029 7.67443 0.0022124 453 205209 92959677 21.28380 7.68009 0.0022075 454 206116 93576664 21.30728 7.68573 0.0022026 455 207025 94196375 21.33073 7.69137 0.0021978 456 207936 94818816 21.35416 7.69700 0.0021930 457 208849 95443993 21.37756 7.70262 0.0021882 458 209764 96071912 21.40093 7.70824 0.0021834 459 | 210681 96702579 21.42429 7.71384 0.0021786 460 2116O0> 97336000 21.44761 7.71944 0.0021739 461 212521 97972181 21.47091 7.72503 0.0021692 462 213444 98611128 21.49419 7.73061 0.0021645 463 214369 99252847 21.51743 7.73619 0.0021598 464 215296 99897344 21.54066 7.74175 0.0021552 465 216225 100544625 21.56386 7.74731 0.0021505 466 217156 101194696 21.58703 7.75286 0.0021459 467 218089 101847563 21.61018 7.75840 0.0021413 468 219024 102503232 21.63331 7.76394 0.0021368 469 219961 103161709 21.65641 7.76946 0.0021322 470 220900 103823000 21.67948 7.77498 0.0021277 471 221841 104487111 21.70253 7.78049 0.0021231 472 222784 105154048 21.72556 7.78599 0.0021186 473 223729 105823817 21.74856 7.79149 0.0021142 474 224676 106496424 21.77154 7.79697 0.0021097 475 225625 107171875 21.79449 7.80245 0.0021053 476 ! 226576 107850176 21.81742 7.80793 0.0021008 477 227529 108531333 21.84033 7.81339 0.0020965 478 228484 109215352 21.86321 7.81885 0.0020921 479 229441 109902239 21.88607 7.82429 0.0020877 480 230400 110592000 21.90890 7.82974 0.0020833 481 231361 111284641 21.93171 7.83517 0.0020790 482 232324 111980168 21.95450 7.84059 0.0020747 483 233289 112678587 21.97726 7.84601 0.0020704 484 234256 113379904 22 7.85142 0.0020661 5° SQUARES, CUBES, ROOTS, AND RECIPROCALS. n n 2 n 3 V n 3 ,— V n 1 n 485 235225 114084125 22.02272 7.85683 0.0020619 486 236196 114791256 22.04541 7.86222 0.0020576 487 237169 115501303 22.06808 7.86761 0.0020534 488 238144 116214272 22.09072 7.87299 0.0020492 489 239121 116930169 22.11334 7.87837 0.0020450 490 240100 117649000 22.13594 7.88374 0.0020408 491 241081 118370771 22.15852 7.88909 0.0020367 492 242064 119095488 22.18107 7.89445 0.0020325 493 243049 119823157 22.20360 7.89979 0.0020284 494 244036 120553784 22.22611 7.90513 0.0020243 495 245025 121287375 22.24860 7.91046 0.0020202 496 246016 122023936 22.27106 7.91578 0.0020161 497 247009 122763473 22.29350 7.92110 0.0020121 498 248004 123505992 22.31591 7.92641 0.0020080 499 249001 124251499 22.33831 7.93171 0.0020040 500 250000 125000000 22.36068 7.93701 0.002 501 251001 125751501 22.38303 7.94229 0.0019960 502 252004 126506008 22.40536 7.94757 0.0019920 503 253009 127263527 22.42766 7.95285 0.0019881 504 254016 128024064 22.44994 7.95811 0.0019841 505 255025 128787626 22.47221 7.96337 0.0019802 506 256036 129554216 22.49444 7.96863 0.0019763 507 257049 130323843 22.51666 7.97387 0.0019724 508 258064 131096512 22.53886 7.97911 0.0019685 509 259081 131872229 22.56103 7.98434 0.0019646 510 260100 132651000 22.58318 7.98957 0.0019608 511 261121 133432831 22.60531 7.99479 0.0019569 512 262144 134217728 22.62742 8 0.0019531 513 263169 135005697 22.64950 8.00520 0.0019493 514 264196 135796744 22.67157 8.01040 0.0019455 515 265225 136590875 22.69361 8.01559 0.0019417 516 266256 137388096 22.71563 8.02078 0.0019380 517 267289 138188413 22.73763 8.02596 0.0019342 518 268324 138991832 22.75961 8.03113 0.0019305 519 269361 139798359 22.78157 8.03629 0.0019268 520 270400 140608000 22.80351 8.04145 0.0019231 521 271441 141420761 22.82542 8.04660 0.0019194 522 272484 142236648 22.84732 8.05175 0.0019157 523 273529 143055667 22.86919 8.05689 0.0019120 524 274576 143877824 22.89105 8.06202 0.0019084 SQUARES, CUBES, ROOTS, AND RECIPROCALS. n n 2 n 3 \/~n v n 1 n 525 275625 144703125 22.91288 8.06714 0.0019048 526 276676 145531576 22.93469 8.07226 0.0019011 527 277729 146363183 22.95648 8.07737 0.0018975 528 278784 147197952 22.97825 8.08248 0.0018939 529 279841 148035889 23 8.08758 0.0018904 530 280900 148877000 23.02173 8.09267 0.0018868 531 281961 149721291 23.04344 8.09776 0.0018832 532 283024 150568768 23.06513 8.10284 0.0018797 533 284089 151419437 23.08679 8.10791 0.0018762 534 285156 152273304 23.10844 8.11298 0.0018727 535 286225 153130375 23.13007 8.11804 0.0018692 536 287296 153990656 23.15167 8.12310 0.0018657 537 288369 154854153 23.17326 8.12814 0.0018622 538 289444 155720872 23.19483 8.13319 0.0018587 539 290521 156590819 23.21637 8.13822 0.0018553 540 291600 157464000 23.23790 8.14325 0.0018519 541 292681 158340421 23.25941 8.14828 0.0018484 542 293764 159220088 23.28089 8.15329 0.0018450 543 294849 160103007 23.30236 8.15831 0.0018416 544 295936 160989184 23.32381 • 8.16331 0.0018382 545 297025 161878625 23.34524 8.16831 0.0018349 546 298116 162771336 23.36664 8.17330 0.0018315 547 299209 163667323 23.38803 8.17829 0.0018282 548 300304 164566592 23.40940 8.18327 0.0018248 549 301401 165469149 23.43075 8.18824 0.0018215 550 302500 166375000 23.45208 8.19321 0.0018182 551 303601 167284151 23.47339 8.19818 0.0018149 552 304704 168196608 23.49468 8.20313 0.0018116 553 305809 169112377 23.51595 8.20808 0.0018083 554 306916 170031464 23.53720 8.21303 0.0018051 555 308025 170953875 23.55844 8.21797 0.0018018 556 309136 171879616 23.57965 8.22290 0.0017986 557 310249 172808693 23.60085 8.22783 0.0017953 558 311364 173741112 23.62202 8.23275 0.0017921 559 312481 174676879 23.64318 8.23766 0.0017889 560 313600 175616000 23.66432 8.24257 0.0017857 561 314721 176558481 23.68544 8.24747 0.0017825 562 315844 177504328 23.70654 8.25237 0.0017794 563 316969 178453547 23.72762 8.25726 0.0017762 564 318096 179406144 23.74868 8.26215 0.0017730 5 2 SQUARES, CUBES, ROOTS, AND RECIPROCALS. n ft 2 n 3 sf n w n 1 n 565 319225 180362125 23.76973 8.26703 0.0017699 566 320356 181321496 23.79075 8.27190 0.0017668 567 321489 182284263 23.81176 8.27677 0.0017637 568 322624 183250432 23.83275 8.28163 0.0017606 569 323761 184220009 23.85372 8.28649 0.0017575 570 324900 185193000 23.87467 8.29134 0.0017544 571 326041 186169411 23.89561 8.29619 0.0017513 572 327184 187149248 23.91652 8.30103 0.0017483 573 328329 188132517 23.93742 8.30587 0.0017452 574 329476 189119224 23.95830 8.31069 0.0017422 575 330625 190109375 23.97916 8.31552 0.0017391 576 331776 191102976 24 8.32034 0.0017361 ■ 577 332929 192100033 24.02082 8.32515 0.0017331 578 334084 193100552 24.04163 8.32995 0.0017301 579 335241 194104539 24.06242 8.33476 0.0017271 580 336400 195112000 24.08319 8.33955 0.0017241 581 i 337561 196122941 24.10394 8.34434 0.0017212 582 338724 197137368 24.12468 8.34913 0.0017182 583 '. 339889 198155287 24.14539 8.35390 0.0017153 584 341056 199176704 24.16609 8.35868 0.0017123 585 342225 200201625 24.18677 8.36345 0.0017094 586 343396 201230056 24.20744 8.36821 0.0017065 587 344569 202262003 24.22808 8.37297 0.0017036 588 345744 203297472 24.24871 8.37772 0.0017007 589 346921 204336469 24.26932 8.38247 0.0016978 590 348100 205379000 24.28992 8.38721 0.0016949 591 349281 206425071 24.31049 8.39194 0.0016920 592 350464 207474688 24.33105 8.39667 0.0016892 593 351649 208527857 24.35159 8.40140 0.0016863 594 352836 209584584 24.37212 8.40612 0.0016835 595 354025 210644875 24.39262 8.41083 0.0016807 596 355216 211708736 24.41311 8.41554 0.0016779 597 356409 212776173 24.43358 8.42025 0.0016750 598 357604 213847192 24.45404 8.42494 0.0016722 599 358801 214921799 24.47448 8.42964 0.0016694 600 360000 216000000 24.49490 8.43433 0.0016667 601 361201 217081801 24.51530 8.43901 0.0016639 602 362404 218167208 24.53569 8.44369 0.0016611 603 363609 219256227 24.55606 8.44836 0.0016584 604 364816 220348864 24.57641 8.45303 0.0016556 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 53 n n 2 n 3 s/n v n 1 n 605 366025 221445125 24.59675 8.45769 0.0016529 606 367236 222545016 24.61707 8.46235 0.0016502 607 368449 223648543 24.63737 8.46700 0.0016474 608 369664 224755712 24.65766 8.47165 0.0016447 609 370881 225866529 24.67793 8.47629 0.0016420 610 372100 226981000 24.69818 8.48093 0.0016393 611 373321 228099131 24.71841 8.48556 0.0016367 612 374544 229220928 24.73863 8.49018 0.0016340 613 375769 230346397 24.75884 8.49481 0.0016313 614 376996 231475544 24.77902 8.49942 0.0016287 615 378225 232608375 24.79919 8.50404 0.0016260 616 379456 233744896 24.81935 8.50864 0.0016234 617 380689 234885113 24.83948 8.51324 0.0016207 618 381924 236029032 24.85961 8.51784 0.0016181 619 383161 237176659 24.87971 8.52243 0.0016155 620 384400 238328000 24.89980 8.52702 0.0016129 621 385641 239483061 24.91987 8.53160 0.0016103 622 386884 240641848 24.93993 8.53618 0.0016077 623 388129 241804367 24.95997 8.54075 0.0016051 624 389376 242970624 24.97999 8.54532 0.0016026 625 390625 244140625 25 8.54988 0.0016000 626 391876 245314376 25.01999 8.55444 0.0015974 627 393129 246491883 25.03997 8.55899 0.0015949 628 394384 247673152 25.05993 8.56354 0.0015924 629 395641 248858189 25.07987 8.56808 0.-0015898 630 396900 250047000 25.09980 8.57262 0.0015873 631 398161 251239591 25.11971 8.57715 0.0015848 632 399424 252435968 25.13961 8.58168 0.0015823 633 400689 253636137 25.15949 8.58622 0.0015798 634 401956 254840104 25.17936 8.59072 0.0015773 635 403225 256047875 25.19921 8.59524 0.0015748 636 404496 257259456 25.21904 8.59975 0.0015723 637 405769 258474853 25.23886 8.60425 0.0015699 638 407044 259694072 25.25866 8.60875 0.0015674 639 408321 260917119 25.27845 8.61325 0.0015649 640 409600 262144000 25.29822 8.61774 0.0015625 641 410881 263374721 25.31798 8.62222 0.0015601 642 412164 264609288 25.33772 8.62671 0.0015576 643 413449 265847707 25.35744 8.63118 0.0015552 644 414736 267089984 25.37716 8.63566 0.0015528 54 SQUARES, CUBES, ROOTS, AND RECIPROCALS. n n 2 n 3 V n v n 1 n 645 416025 268336125 25.39685 8.64012 0.0015504 646 417316 269586136 25.41653 8.64459 0.0015480 647 418609 270840023 25.43619 8.64904 0.0015456 648 419904 272097792 25.45584 8.65350 0.0015432 649 421201 273359449 35.47548 8.65795 0.0015408 650 422500 274625000 25.49510 8.66239 0.0015385 651 423801 275894451 25.51470 8.66683 0.0015361 652 425104 277167808 25.53429 8.67127 0.0015337 653 426409 278445077 25.55386 8.67570 0.0015314 654 427716 279726264 25.57342 8.68012 0.0015291 655 429025 281011375 25.59297 8.68455 0.0015267 656 430336 282300416 25.61250 8.68896 0.0015244 657 431649 283593393 25.63201 8.69338 0.0015221 658 432964 284890312 25.65151 8.69778 0.0015198 659 434281 286191179 25.67100 8.70219 0.0015175 660 435600 287496000 25.69047 8.70659 0015152 661 436921 288804781 25.70992 8.71098 0.0015129 662 438244 290117528 25.72936 8.71537 0.0015106 663 439569 291434247 25.74879 8.71976 0.0015083 664 440896 292754944 25.76820 8.72414 0.0015060 665 442225 294079625 25.78749 8.72852 0.0015038 666 443556 295408296 25.80698 8.73289 0.0015015 667 444889 296740963 25.82634 8.73726 0.0014993 668 446224 298077632 25.84570 8.74162 0.0014970 669 447561 299418309 25.86503 8.74598 0.0014948 670 448900 300763000 25.88436 8.75034 0.0014925 671 450241 302111711 25.90367 8.75469 0.0014903 672 451584 303464448 25.92296 8.75904 0.0014881 673 452929 304821217 25.94224 8.76338 0.0014859 674 454276 306182024 25.96151 8.76772 0.0014837 675 455625 307546875 25.98076 8.77205 0.0014815 676 456976 308915776 26 8.77638 0.0014793 677 458329 310288733 26.01922 8.78071 0.0014771 678 459684 311665752 26.03843 8.78503 0.0014749 679 461041 313046839 26.05763 8.78935 0.0014728 680 462400 314432000 26.07681 8.79366 0.0014706 681 463761 315821241 26.09598 8.79797 0.0014684 682 465124 317214568 26.11513 8.80227 0.0014663 683 466489 318611987 26.13427 8.80657 0.0014641 684 467856 320013504 26.15339 8.81087 0.0014620 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 55 n n 2 n 3 \/ n V n 1 n 685 469225 321419125 26.17250 8.81516 0.0014599 686 470596 322828856 26.19160 8.81945 0.0014577 687 471969 324242703 26.21068 8.82373 0.0014556 688 473344 325660672 26.22975 8.82801 0.0014535 689 474721 327082769 26.24881 8.83229 0.0014514 690 476100 328509000 26.26785 8.83656 0.0014493 691 477481 329939371 26.28688 8.84082 0.0014472 692 478864 331373888 26.30589 8.84509 0.0014451 693 480249 332812557 26.32489 8.84934 0.0014430 694 1 481636 334255384 26.34388 8.85360 0.0014409 695 1 483025 335702375 26.36285 8.85785 0.0014388 696 484416 337153536 26.38181 8.86210 0.0014368 697 485809 338608873 26.40076 8.86634 0.0014347 698 487204 340068392 26.41969 8.87058 0.0014327 699 488601 341532099 26.43861 8.87481 0.0014306 700 490000 343000000 26.45751 8.87904 0.0014286 701 491401 344472101 26.47640 8.88327 0.0014265 702 492804 345948408 26.49528 8.88749 0.0014245 703 494209 347428927 26.51415 8.89171 0.0014225 704 495616 348913664 26.53300 8.89592 0.0014205 705 497025 350402625 26.55184 8.90013 0.0014184 706 498436 351895816 26.57066 8.90434 0.0014164 707 499849 353393243 26.58947 8.90854 0.0014144 708 501264 354894912 26.60817 8.91274 0.0014124 709 502681 356400829 26.62705 8.91693 0.0014104 710 504100 357911000 26.64583 8.92112 0.0014085 711 505521 359425431 26.66458 8.92531 0.0014065 712 506944 360944128 26.68333 8.92949 0.0014045 713 508369 362467097 26.70206 8.93367 0.0014025 714 509796 363994344 26.72078 8.93784 0.0014006 715 511225 365525875 26.73948 8.94201 0.0013986 716 512656 367061696 26.75818 8.94618 0.0013966 717 514089 368601813 26.77686 8.95034 0.0013947 718 515524 370146232 26.79552 8.95450 0.0013928 719 516961 371694959 26.81418 8.95866 0.0013908 720 518400 373248000 26.83282 8.96281 0.0013889 721 519841 374805361 26.85144 8.96696 0.0013870 722 521284 376367048 26.87006 8.97110 0.0013850 723 522729 377933067 26.88866 8.97524 0.0013831 724 524176 379503424 26.90725 8.97938 0.0013812 *6 SQUARES, CUBES, ROOTS, AND RECIPROCALS. n 725 n 2 11* V n 3 V 7/ 1 525625 381078125 26.92582 8.98351 0.0013793 726 527076 382657176 26.94439 8.98764 0.0013774 727 528529 384240583 26.96294 8.99176 0.0013755 728 529984 385828352 26.98148 8.99589 0.0013736 729 531441 387420489 27 9 0.0013717 730 532900 389017000 27.01851 9.00411 0.0013699 731 534361 390617891 27.03701 9.00822 0.0013680 732 535824 392223168 27.05550 9.01233 0.0013661 733 537289 393832837 27.07397 9.01643 0.0013643 734 538756 395446904 27.09243 9.02053 0.0013624 735 540225 397065375 27.11088 9.02462 0.0013605 736 541696 398688256 27.12932 9.02871 0.0013587 737 543169 400315553 27.14771 9.03280 0.0013569 738 544644 401947272 27.16616 9.03689 0.0013550 739 546121 403583419 27.18455 9.04097 0.0013532 740 547600 405224000 27.20291 9.04504 0.0013514 741 549081 406869021 27.22132 9.04911 0.0013495 742 550564 408518488 27.23968 9.05318 0.0013477 743 552049 410172407 27.25803 9.05725 0.0013459 744 553536 411830784 27.27636 9.06131 0.0013441 745 555025 413493625 27.29469 9.06537 0.0013423 746 556516 415160936 27.31300 9.06942 0.0013405 747 558009 416832723 27.33130 9.07347 0.0013387 748 559504 418508992 27.34959 9.07752 0.0013369 749 561001 420189749 27.36786 9.08156 0.0013351 750 562500 421875000 27.38613 9.08560 0.0013333 751 564001 423564751 27.40438 9.08964 0.0013316 752 565504 425259008 27.42262 9.09367 0.0013298 753 567009 426957777 27.44085 9.09770 0.0013280 754 568516 428661064 27.45906 9.10173 0.0013263 755 570025 430368875 27.47726 9.10575 0.0013245 756 571536 432081216 27.49545 9.10977 0.0013228 757 573049 433798093 27.51363 9.11378 0.0013210 758 574564 435519512 27.53180 9.11779 0.0013193 759 576081 437245479 27.54995 9.12180 0.0013175 760 577600 43S976000 27.56810 9.12581 0.0013158 761 579121 440711081 27.58623 9.12981 0.0013141 762 580644 442450728 27.60435 9.13380 0.0013123 763 582169 444194947 27.62245 9.13780 0.0013106 764 583696 445943744 27.64055 9.14179 0.0013089 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 57 n n 2 n 3 V n V n 1 11 765 585225 447697125 27.65863 9.14577 0.0013072 766 586756 449455096 27.67671 9.14976 0.0013055 767 588289 451217663 27.69476 9.15374 0.0013038 768 589824 452984832 27.71281 9.15771 0.0013021 769 591361 454756609 27.73085 9.16169 0.0013004 770 592900 456533000 27.74887 9.16566 0.0012987 771 594441 458314011 27.76689 9.16962 0.0012970 772 595984 460099648 27.78489 9.17359 0.0012953 773 597529 461889917 27.80288 9.17754 0.0012937 774 599076 463684824 27.82086 9.18150 0.0012920 775 600625 465484375 27.83882 9.18545 0.0012903 776 602176 467288576 27.85678 9.18940 0.0012887 777 603729 469097433 27.87472 9.19335 0.0012870 778 605284 470910952 27.89265 9.19729 0.0012853 779 606841 472729139 27.91057 9.20123 0.0012837 780 608400 474552000 27.92848 9.20516 0.0012821 781 609961 476379541 27.94638 9.20910 0.0012804 782 611524 478211768 27.96426 9.21303 0.0012788 783 613089 480048687 27.98214 9.21695 0.0012771 784 614656 481890304 28 9.22087 0.0012755 785 616225 483736625 28.01785 9.22479 0.0012739 786 617796 485587656 28.03569 9.22871 0.0012723 787 619369 487443403 28.05352 9.23262 0.0012706 788 620944 489303872 28.07134 9.22653 0.0012690 789 622521 491169069 28.08914 9.24043 0.0012674 790 624100 493039000 28.10694 9.24434 0.0012658 791 625681 494913671 28.12472 9.24823 0.0012642 792 627264 496793088 28.14249 9.25213 0.0012629 793 628849 498677257 28.16026 9.25602 0.0012610 794 630436 500566184 28.17801 9.25991 0.0012594 795 632025 502459875 28.19574 9.26380 0.0012579 796 633616 504358336 28.21347 9.26768 0.0012563 797 635209 506261573 28.23119 9.27156 0.0012547 798 636804 508169592 28.24889 9.27544 0.0012531 799 638401 510082399 28.26659 9.27931 0.0012516 800 640000 512000000 28.28427 9.28318 0.0012500 801 641601 513922401 28.30194 9.28704 0.0012484 802 643204 515849608 28.31960 9.29091 0.0012469 803 644809 517781627 28.33725 9.29477 0.0012453 804 646416 519718464 28.35489 9.29862 0.0012438 58 SQUARES, CUBES. ROOTS, AND RECIPROCALS. 11 805 n 2 ;/ 3 s/ n 3 v n 1 n 648025 521660125 28.37252 9.30248 0.0012422 806 649636 523606616 28.39014 9.30633 0.0012407 807 651249 525557943 28.40775 9.31018 0.0012392 808 652864 527514112 28.42534 9.31402 0.0012376 809 654481 529475129 28.44293 9.31786 0.0012361 810 656100 531441000 28.46050 9.32170 0.0012346 811 657721 533411731 28.47806 9.32553 0.0012330 812 659344 535387328 28.49561 9.32936 0.0012315 813 660969 537367797 28.51315 9.33319 0.0012300 814 662596 539353144 28.53069 9.33702 0.0012285 815 664225 541343375 28.54820 9.34084 0.0012270 816 665856 543338496 28.56571 9.34466 0.0012255 817 1 667489 545338513 28.58321 9.34847 0.0012240 818 669124 547343432 28.60070 9.35229 0.0012225 819 1 670761 549353259 28.61818 9.35610 0.0012210 820 672400 551368000 28.63564 9.35990 0.0012195 821 674041 553387661 28.65310 9.36370 0.0012180 822 ( 675684 555412248 28.67054 9.36751 0.0012165 823 677329 557441767 28.68798 9.37130 0.0012151 824 ! 678976 559476224 28.70540 9.37510 0.0012136 825 ; 680625 561515625 28.72281 9.37889 0.0012121 826 i 682276 563559976 28.74022 9.38268 0.0012107 827 i 683929 565609283 28.75761 9.38646 0.0012092 828 1 685584 567663552 28.77499 9.39024 0.0012077 829 ' 687241 569722789 28.79236 9.39402 0.0012063 830 688900 571787000 28.80972 9.39780 0.0012048 831 690561 573S56191 28.82707 9.40157 0.0012034 832 692224 575930368 28.84441 9.40534 0.0012019 833 693889 578009537 28.86174 9.40911 0.0012005 834 695556 580093704 28.87906 9.41287 0.0011990 835 697225 582182875 28.89637 9.41663 0.0011976 836 698896 584277056 28.91366 9.42039 0.0011962 837 700569 586376253 28.93095 9.42414 0.0011947 838 702244 588480472 28.94823 9.42789 0.0011933 839 703921 590589719 28.96550 9.43164 0.0011919 840 705600 592704000 28.98275 9.43538 0.0011905 841 707281 594823321 29 9.43913 0.0011891 842 708964 596947688 29.01724 9.44287 0.0011876 843 710649 599077107 29.03446 9.44661 0.0011862 844 712336 6012115S4 29.05168 9.45034 0.0011848 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 59 n n 2 n 3 \/ n y. n 1 n 845 714025 603351125 29.06888 9.45407 0.0011834 846 715716 605495736 29.08608 9.45780 0.0011820 847 717409 607645423 29.10326 9.46152 0.0011806 848 719104 609800192 29.12044 9.46525 0.0011792 849 720801 611960049 29.13760 9.46897 0.0011779 850 722500 614125000 29.15476 9.47268 0.0011765 851 724201 616295051 29.17190 9.47640 0.0011751 852 725904 618470208 29.18904 9.48011 0.0011737 853 727609 620650477 29.20616 9.48381 0.0011723 854 729316 622835864 29.22328 9.48752 0.0011710 855 731025 625026375 29.24038 9.49122 0.0011696 856 732736 627222016 29.25748 9.49492 0.0011682 857 734449 629422793 29.27456 9.49861 0.0011669 858 736164 631628712 29.29164 9.50231 0.0011655 859 737881 633839779 29.30870 9.50600 0.0011641 860 739600 636056000 29.32576 9.50969 0.0011628 861 741321 638277381 29.34280 9.51337 0.0011614 862 743044 640503928 29.35984 9.51705 0.0011601 863 744769 642735647 29.37686 9.52073 0.0011587 864 746496 644972544 29.3938S 9.52441 0.0011574 865 748225 647214625 29.41088 9.52808 0.0011561 866 749956 649461896 29.42788 9.53175 0.0011547 867 751689 651714363 29.44486 9.53542 0.0011534 868 753424 653972032 29.46184 9.53908 0.0011521 869 755161 656234909 29.47881 9.54274 0.0011507 870 756900 658503000 29.49576 9.54640 0.0011494 871 758641 660776311 29.51271 9.55006 0.0011481 872 760384 663054848 29.52965 9.55371 0.0011468 873 762129 665338617 29.54657 9.55736 0.0011455 874 763876 667627624 29.56349 9.56101 0.0011442 875 765625 669921875 29.58040 9.56466 0.0011429 876 767376 672221376 29.59730 9.56830 0.0011416 877 769129 674526133 29.61419 9.57194 0.0011403 878 770884 676836152 29.63106 9.57557 0.0011390 879 772641 679151439 29.64793 9.57921 0.0011377 880 774400 681472000 29.66479 9.58284 0.0011364 881 776161 683797841 29.68164 9.58647 0.0011351 882 777924 686128968 29.69848 9.59009 0.0011338 883 779689 688465387 29.71532 9.59372 0.0011325 884 781456 690807104 29.73214 9.59734 0.0011312 6o SQUARES, CUBES, ROOTS, AND RECIPROCALS. n n 2 n 3 V n V n 1 n 885 783225 693154125 29.74895 9.60095 0.0011299 886 784996 695506456 29.76575 9.60457 0.0011287 • 887 786769 697864103 29.78255 9.60818 0.0011274 888 788544 700227072 29.79933 9.61179 0.0011261 889 790321 702595369 29.81610 9.61540 0.0011249 890 792100 704969000 29.83287 9.619 0.0011236 891 793881 707347971 29.84962 9.62260 0.0011223 892 795664 709732288 29.86637 9.62620 0.0011211 893 797449 712121957 29.88311 9.62980 0.0011198 894 799236 714516984 29.89983 9.63339 0.0011186 895 801025 716917375 29.91655 9.63698 0.0011173 896 802816 719323136 29.93326 9.64057 0.0011161 897 804609 721734273 29.94996 9.64415 0.0011148 898 806404 724150792 29.96665 9.64774 0.0011136 899 808201 726572699 29.98333 9.65132 0.0011123 900 810000 729000000 30 9.65489 0.0011111 901 811801 731432701 30.01666 9.65847 0.0011099 902 813604 733870808 30.03331 9.66204 0.0011086 903 815409 736314327 30.04996 9.66561 0.0011074 904 817216 738763264 30.06659 9.66918 0.0011062 905 819025 741217625 30.08322 9.67274 0.0011050 906 820836 743677416 30.09983 9.67630 0.0011038 907 822649 746142(543 30.11644 9.67986 0.0011025 908 824464 748613312 30.13304 9.68342 0.0011013 909 826281 751089429 30.14963 9.68697 0.0011001 910 828100 753571000 30.16621 9.69052 0.0010989 911 829921 756058031 30.18278 9.69407 0.0010977 912 831744 758550528 30.19934 9.69762 0.0010965 913 833569 761048497 30.21589 9.70116 0.0010953 914 835396 763551944 30.23243 9.70470 0.0010941 915 837225 766060875 30.24897 9.70824 0.0010929 916 839056 768575296 30.26549 9.71177 0.0010917 917 840889 771095213 30.28201 9.71531 0.0010905 918 842724 773620632 30.29851 9.71884 0.0010893 919 844561 776151559 30.31501 9.72236 0.0010881 920 846400 778688000 30.33150 9.72589 0.0010870 921 848241 781229961 30.34798 9.72941 0.0010858 922 850084 783777448 30.36445 9.73293 0.0010846 923 851929 786330467 30.38092 9.73645 0.0010834 924 853776 788889024 30.39737 9.73996 0.0010823 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 61 n n 2 n 3 \/n V n 1 n 925 855025 791453125 30.41381 9.74348 0.0010811 020 857470 794022770 30.43025 9.74099 0.0010799 927 859329 790597983 30.44007 9.75049 0.0010787 92S 801184 799178752 30.40309 9.75400 0.0010770 929 803041 801705089 30.47950 9.75750 0.0010704 930 804900 804357000 30.49590 9.70100 0.0010753 931 800701 800954491 30.51229 9.70450 0.0010741 932 808024 809557508 30.52808 9.76799 0.0010730 933 870489 812100237 30.54505 9.77148 0.0010718 934 872350 814780504 30.50141 9.77497 0.0010707 935 874225 817400375 30.57777 9.77846 0.0010095 930 870090 620025856 30.59412 9.78295 0.0010084 937 877909 822650953 30.01040 9.78543 0.0010072 938 879844 825293072 30.02079 9.78891 0.0010001 939 881721 827930019 30.04311 9.79239 0.0010050 940 883000 830584000 30.05942 9.79586 0010038 941 885481 833237021 30.07572 9.79933 0.0010027 942 887304 835890888 30.09202 9.80280 0.0010010 943 889249 838501807 30.70831 9.80027 0.0010004 944 891130 841232384 30.72458 9.80974 0.0010593 945 893025 843908025 30.74085 9.81320 0.0010582 940 894910 840590530 30.75711 9.81066 0.0010571 947 890809 849278123 30.77337 9.82012 0.0010500 948 898704 851971392 30.78901 9.82357 0.0010549 949 900001 854070349 30.80584 9.82703 0.0010537 950 902500 857375000 30.82207 9.83048 0.0010520 951 904401 800085351 30.83829 9.83392 0.0010515 952 900304 802801408 30.85450 9.83737 0.0010504 953 908209 805523177 30.87070 9.84081 0.0010493 954 910110 808250004 30.88089 9.84425 0.0010482 955 912025 870983875 30.90307 9.84709 0.0010471 950 913930 873722810 30.91925 9.85113 0.0010400 957 915849 870407493 30.93542 9.85450 0.0010449 958 917704 879217912 30.95158 9.85799 0.0010438 959 919081 881974079 30.90773 9.86142 0.0010428 900 921000 887430000 30.98387 9.86485 0.0010417 901 923521 887503081 31 9.86827 0.0010400 902 925444 890277128 31.01012 9.87169 0.0010395 903 927309 893050347 31.03224 9.87511 0.0010384 904 929290 895841344 31.04835 9.87853 0.0010373 62 SQUARES, CUBES, ROOTS, AND RECIPROCALS. n n 2 n 3 \/~n v n 1 n 965 931225 898632125 31.06445 9.88195 0.0010363 966 933156 901428696 31.08054 9.88536 0.0010352 967 935089 904231063 31.09662 9.88877 0.0010341 968 937024 907039232 31.11270 9.89217 0.0010331 969 938961 909853209 31.12876 9.89558 0.0010320 970 940900 912673000 31.14482 9.89898 0.0010309 971 942841 915498611 31.16087 9.90238 0.0010299 972 944788 918330048 31.17691 9.90578 0.0010288 973 946729 921167317 31.19295 9.90918 0.0010277 974 948676 924010424 31.20897 9.91257 0.0010267 975 950625 926859375 31.22499 9.91596 0.0010256 976 952576 929714176 31.24100 9.91935 0.0010246 977 954529 932574833 31.25700 9.92274 0.0010235 978 i 956484 935441352 31.27299 9.92612 0.0010225 979 i 958441 938313739 31.28898 9.92950 0.0010215 980 960400 941192000 31.30495 9.93288 0.0010204 981 962361 944076141 31.32092 9.93626 0.0010194 982 964324 946966168 31.33688 9.93964 0.0010183 983 966289 949862087 31.35283 9.94301 0.0010173 984 i 968256 952763904 31.36877 9.94638 0.0010163 985 970225 955671625 31.38471 9.94975 0.0010152 986 : 972196 958585256 31.40064 9.95311 0.0010142 987 974169 961504803 31.41656 9.95648 0.0010132 988 976144 964430272 31.43247 9.95984 0.0010121 989 j 978121 967361669 31.44837 9.96320 0.0010111 990 980100 970299000 31.46427 9.96655 0.0010101 991 982081 973242271 31.48015 9.96991 0.0010091 992 984064 976191488 31.49603 9.97326 0.0010081 993 986049 979146657 31.51190 9.97661 0.0010070 994 1 988036 982107784 31.52777 9.97996 0.0010060 995 990025 985074875 31.54362 9.98331 0.0010050 996 992016 988047936 31.55947 9.98665 0.0010040 997 994009 991026973 31.57531 9.98999 0.0010030 998 996004 994011992 31.59114 9.99333 0.0010020 999 998001 997002999 31.60696 9.99667 0.0010010 Botes on Bloebra. Algebra is that branch of mathematics in which the quan- tities are denoted by letters and the operations to be performed upon them are indicated by signs. The same signs are used to indicate the same operations as in arithmetic. Signs of Quantity and Signs of Operation. If a quantity is written a 4 ( — b), the sign that precedes the parenthesis is called the sign of operation, and the sign within the parenthesis is called the sign of quantity with respect to b y but expressions of this kind can be reduced to have only one sign. Thus, a 4 ( — b) = a — b and this final sign is called the essential sign. a + (+ b) = a 4 b. a -\- ( — b) = a — b. a — (4 b) = a — b. a — (—b) = a -h b. Thus, when the sign of operation and the sign of quantity are alike the essential sign is +, but if they are unlike the es- sential sign is — . In multiplying any two quantities, like signs in the two factors give -f in the product, but unlike signs in the two factors give — in the product; thus, (+ a) X ( — b) = — ab, and ( — a) X (— b) = ab. In division, like signs in dividend and divisor give + in the quotient, but unlike signs in dividend and divisor give — in the quotient; thus: — a a — a , a — b b Useful Formulas and Rules in Algebra. The following rules are very useful to remember in solv- ing practical problems in algebra. Let a and b represent any two quantities; then a -\- b will represent their sum and a — b their difference; then (a + b) X (a 4 b) = a' 2 4 2 ab 4 b 2 . {a 4 b) X (a 4 b) is also written (a 4 b) 2 . (63) 64 NOTES ON ALGEBRA. This rule reads: The square of the sum of any two quantities is equal to the square of the first quantity plus double the product of both quantities, plus the square of the second quantity. (a—b)X (a —b) = (a — b) 2 = a 2 —2 ab -\- b 2 . This rule reads: The square of the difference of any two quantities is equal to the square of the first quantity minus twice the product of both quantities, plus the square of the second quantity. (a + b) X (a — b) = a 2 — b 2 . This rule reads : The sum of any two quantities multiplied by their differ- ence is equal to the difference of their squares. Extracting Roots. An even root of a positive quantity is either + or — . An even root cannot be extracted of a negative quantity, as \Z^ may be either a or — a; but V — a 2 is impossible, because (— a) X (— a) = a 2 and (+ a) X (+ a) = a 2 . An odd root may be extracted as well of a negative quantity as a positive quantity, and the sign of the root is always the same as the sign of the quantity before the root was extracted. 3 3 Thus : \SaJ~= a, but \/ (— af = (— a). Powers. When a number or a quantity is to be multiplied by itself a given number of times, the operation is indicated by a small number at the right-hand corner of the quantity ; for instance, a 2 = a X a. A quantity of this kind is called a power; the small number is called the exponent, or the index of the power. Two powers of the same kind may be multiplied by adding the exponents ; for instance, a 2 a 3 = d MZ = a h . Two powers of the same kind may be divided by subtract- ing their exponents ; for instance, _fL = a h ~ 2 = a 3 = a X a X a. a 2 a 5 —* = a 5-3 = a 2 = a X a. a°~l. NOTES ON ALGEBRA. 65 Thus, any quantity in power must be 1, because always when dividend and divisor are alike the quotient must be 1. — — = # 5-6 = a -1 , but a b divided by a 6 is equal to 1 ; therefore SL must be J_ ; lL.= is a numerical equation. In solving equations, we may, without destroying the equal- ity of the equation, add an equal quantity to both members, subtract an equal quantity from both members, multiply both members by an equal quantity, divide both members by an equal quantity, extract the same root of both members, raise both members to the same power. Quantities inclosed by parentheses, a bar, or under a radical sign, and quantities connected by the sign of multiplication, must always be considered and operated upon as one quantity, Example 1. ^ = 3X8 + 10-3 + 3X10. x = 24 + 10 — 3 + 30. x = 64— 3. * = 61. . 66 notes on algebra. Example 2. 3 + S + 10 10 10 ■* = 2 T V Example 3. = 12 x ( l-f- + e) = 12X( 7 ^-) x = 9S Example 4. 8 — 3 , 8 + 10 = 12 X — 5— + ~ tt- o 12X "3- + -F .r = 20 + 6 x =26 Example 5. = ( — 2 — X (8 — 3)+ V^O 4- 16) X = (— 2~~ x 5 + V:36" J X 2 ^=(6X5 + 6) X 2 .* = 36 X 2 jc = 72 When an equation consists of more than one unknown quan- tity, as many equations may be arranged as there are unknown quantities, and one equation is solved so that the value of one of its unknown quantities is expressed in terms of the other, and this value is substituted in the other equation. Example. Two shafts are to be connected by two gears of 16 diam- etral pitch ; the distance between centers is 6% inches. The ratio of gearing shall be 1 to 3. How many teeth in each gear? NOTES ON ALGEBRA. 67 Call small gear x and large gear y ; then x + y must be 108, because 6^ X 16 = 108, which is the number of teeth in both gears added together. The ratio is 1 to 3; therefore 3x—y, Thus: x + y = 108, transposed to •r-j-3 jr=108. 4x = 108. .r = 143. 4 jr = 27 teeth for small gear. The large gear = 27 X 3 = 81 teeth. Quadratic Equations. Equations containing one or more unknown quantities in the second power are called quadratic equations. If the un- known quantity only exists in the second power the equation may be brought to the form x 2 = a and x = \/ #7~ This square root may be either plus or minus. If the unknown quantity exists in both the first and the second power the equation may be brought to the form x 2 4- a x = d, or it may be brought to the form x 2 — a x = b. The coefficient a may be any number. After the equation is brought to this form, complete the square of the first member by adding to both members of the equation the square of half the coefficient a ; this will make the left member of the equa- tion a complete square. Example. A coal bin is to hold six tons of coal. Allow 40 cubic feet per ton. (It takes 35 to 40 cubic feet to hold a ton of coal in a bin). Make the width of the bin 6 feet, and the length equal to the width and the depth added together. How deep and how long will the bin be? Depth = x and length = y. x y = 240, because 6 times 40 equals 240. 6 + x = y, because width + depth = length. Thus: 6 x (6 + x) = 240. $x 2 + 36 x = 240. Dividing by 6 we have : x 2 -j- Qx = 40. Completing the square : x 2 -\- " in the tables. The difference in this case is given in the table as 119. Example 2. Find logarithm to 1892.5. Solution : The mantissa of the number 1892 is given in the table as 276921. The difference is given as 229. The index for a num- ber consisting of four integers is 3. Thus: Log. 1892 = 3.276921 0.5 X 0.000229 = 0.0001145 = 115 Log. 1892.5 = 3.277036 Example 3. Find logarithm to 85673. Solution: The mantissa for the number 85670 is given in the table as 932829. The difference is given in the table as 51. The index for a number consisting of five integers is 4. When an increase- of 10 in the number increases the log- arithm 0.000051 an increase of 3 must increase the correspond- ing logarithm 0.3 times 0.000051. Thus: Log. 85670 = 4.932829 0.3 X 0.000051 = 0.0000153 == 15 Log. 85673 = 4.932844 These calculations (or interpolations as they are usually called) are based upon the principle that the difference between the numbers and the difference between their corresponding logarithms are directly proportional to each other.^ This, how- ever, is not strictly true ; but within limits, as it is used here, it is near enough for practical results. To* Find the Number Corresponding to a Given Logarithm. Example 1. Find the number corresponding to the logarithm 2.610979. Solution: Always remember when looking for the number not to consider the index, but find the mantissa 610979 in the table. In the same line as this mantissa, under the heading " N," is 408, and on the top of the table in the same column as this mantissa is 3 ; thus, the number corresponding to this mantissa is 4083 and the index of the logarithm is 2 ; consequently the LOGARITHMS. 75 number is to have three figures on the left-hand side of the decimal point ; thus, the number corresponding to the logarithm 2.610979 will be 408.3. , Example 2. Find the number corresponding to the logarithm 3.883991. Solution : This mantissa is not in the table. The nearest smaller mantissa is 883945, and to this mantissa corresponds the number 7055. The nearest larger mantissa is 884002, and to this corresponds the number 7656. Thus, an increment in the mantissa of 57 increases the number by 1, but the difference between the mantissa 883945 and S83991 is 46, therefore the number must increase ff = 0.807. Number of Log. 3.883945 = 7655 Difference 0.000046 = 0.807 Number of Log. 3.883991 = 7655.807 Addition of Logarithms. (MULTIPLICATION.) Where the logarithms of the factors have positive indexes, add as if they were decimal fractions, and the sum is the log- arithm corresponding to the product. Example Multiply 81 by 65 by means of logarithms. Solution : Log. 81 = 1.908485 Log. 65 = 1.812913 3.721398 and to this mantissa corresponds the number 5265. The index is 3; therefore the number has no decimals, as it consists of only four figures. To Add Two Logarithms when One Has a Positive and the Other a Negative Index. Example Multiply 0.58 by 32.6 by means of logarithms. Solution : Log. 0.58 = 9.763428 — 10 Log. 32. 6 = 1.513218 11.276646 — 10 7 6 LOGARITHMS. This reduces to 1.276646 and to this logarithm corresponds the number 18.908. This mantissa, 276646, cannot be found in the table, but the nearest smaller mantissa is 276462, and the differ- ence between this and the next is found by subtraction to be 230, and the difference between this and the given mantissa is 184. Thus: Given logarithm 1.276646 To the tabulated log. 1.276462 corresponds 18.90 Difference 0.000184 gives 0.008 Thus, logarithm 1.276646 gives number 18.908 To Add Two Logarithms, Both Having a Negative Index. Add both logarithms in the same manner as decimal frac- tions, and afterwards subtract 10 from the index on each side of the mantissa. Example. Multiply 0.82 by 0.082 by means of logarithms. Solution : Log. 0.82 = 9.913814 — 10 Log. 0.082 = 8.913814 — 10 18.827628 — 20 By subtracting 10 on each side of the mantissa this logar- ithm reduces to 8.827628 — 10 and to the mantissa 82762S corre- sponds the number 6724, but the negative index 8 ..... . — 10 indicates that this first figure 6 is not a whole number, but that it is six-hundredths ; therefore a cipher must be placed between this 6 and the decimal point in order to give 6 the right value according to the index; thus, to the logarithm 8.827628 — 10 corresponds the number 0.06724. Subtraction of Logarithms. (DIVISION.) Logarithms are subtracted as common decimal fractions. To Subtract Two Logarithms, Both Having a Positive Index. Example. Divide 490 by 70 by means of logarithms. Solution : Log. 490 = 2.690196 Log. 70 — 1.845098 0.845098 LOGARITHMS. 77 and to the mantissa of this logarithm corresponds the number 7 or 70 or 700 or 7000, etc., in the table of logarithms, but the index of this logarithm is a cipher ; therefore the answer must be a number consisting of one figure, thus it must be 7. To Subtract a Larger Logarithm From a Smaller One. This is the same as to divide a smaller number by a larger one. Before the subtraction is commenced add 10 to the index of the smaller logarithm (that is, to the minuend) and place — 10 after the mantissa, then proceed with the subtraction as if they were decimal fractions. Example. Divide 242 by 367 by means of -logarithms. Solution: Log. 242 = 2.383815 = 12.383815 — 10 Log. 367 = 2.564666 9.819149 — 10 and to the mantissa of this logarithm corresponds, according to the table, the number 6594, but the negative index, 9 — 10, indicates it to be 0.6594. Thus, 242 divided by 367 = 0.6594. Multiplication of Logarithms. (involution.) To multiply a logarithm is the same as to raise its corre- sponding number into the power of the multiplier. Logarithms having a positive index are multiplied the same as decimal fractions. Thus : Square 224 by means of logarithms. Solution: 2 X log. 224 = 2 X 2.350248 = 4.700496 = 50176 Logarithms having a negative index are multiplied the same as decimal fractions, but an equal number is subtracted from both the positive and the negative parts of the logarithm, in order to bring the negative part of the index to — 10. Example 1. Square 0.82 by means of logarithms. Solution : 2 X log. 0.82 = 2 X (9.913814 — 10) = 19.827628 — 10, and subtracting 10 from both the positive and the negative parts of the logarithm, the result is 9.827628 — 10 ; this gives the num- ber 0.6724. 7 S LOGARITHMS. Example 2. Raise 0.9 to the 1.41 power. Solution: 1.41 X log. 0.9 = 1.41 X (9.954243 — 10) = 14.085483 — 14.1 In this example 10 cannot be subtracted from both parts of the logarithm, but 4.1 must be subtracted in order to get — 10, after the subtraction is performed. The logarithm will then read 9.935483 — 10, which corresponds to the number 86195, and the negative index, 9 — 10, makes this 0.86195. Division of Logarithms. (EVOLUTION.) To divide a logarithm is the same as to extract a root of the number corresponding to the logarithm. Logarithms having a positive index are divided the same as common decimal fractions. Example. Extract the cube root of 512 by means of logarithms. Solution : log. 512 _ 2.70927 0.90309 3 3 and the number corresponding to this logarithm is 8, 80, 800, 8,000, etc., but the index of this logarithm is a cipher; there- fore the answer must be a number consisting of one integer, consequently it must be 8. To Divide a Logarithm Having a Negative Index. Select and add such a number to the index as will give 10 without a remainder for the quotient in the negative index on the right-hand side of the mantissa after division is per- formed. Example 1. Extract the square root of 0.64 by means of logarithms. Solution: log. 0.64 _ 9.8061 8—10 _ 19.80618 —20 _ 90309 _ 10 Z z z and to this logarithm corresponds the number 0.8. Example 2. Extract the cube root of 0.125 by means of logarithms. LOGARITHMS. Solution log. 0.125 _ 9.09691—10 _ 29.09691—80 _ Q QQ8Qrj _ 1Q 3 3 3 and to this logarithm corresponds the number 0.5. Example 3. Extract the 1.7 root of 0.78. Solution : log. 0.78 9.892095 — 10 L7 ~" L7 We cannot here, as in previous examples, add a multiple of 10 to the index on each side of the mantissa, but 7 must be added in order that the negative quotient shall be — 10 after the division is performed. Thus : 9.892095 — 10 16.892095 — 17 n IW .,,, -m — = 9.9obo2b — 10 1.7 1.7 and to this logarithm corresponds the number 0.864. Short Rules for Figuring by Logarithms. MULTIPLICATION. Add the logarithms of the factors and the sum is the logar- ithm of the product. DIVISION. Subtract divisor's logarithm from the logarithm of the divi- dend and the difference is the logarithm of the quotient. INVOLUTION. Multiply the logarithm of the root by the exponent of the power and the product is the logarithm of the power. Example. Log. 86 2 = 2 X log. 86 = 2 X 1.934498 = 3.868996 and to this logarithm corresponds the number 7396. EVOLUTION. The logarithm of the number or quantity under the radical sign is divided by the index of the root and the quotient is the logarithm of the root. So LOGARITHMS. Example. * log. 2401 3.380392 Log. -v/2401 = — ^ = 1 = 0.845098 and this logarithm corresponds to the number 7. EXPONENTS. The logarithm of a power divided by the logarithm of the .root is equal to the exponent of the power. Example. 8* = 64 log. 64 x = x = log. 8 1.80618 0.90309 2 The logarithm of a quantity under the radical sign divided by the logarithm of the root is equal to the index of the root. Example. x 8 = V512 x _ log. 512 log. 8 x _ 2.70927 0.90309 x =3 The reason for these last rules may be understood by re- ferring to the rules for Involution and Evolution ; for instance : 86 2 = 7396, and this expressed by logarithms is: 2 X log. 86 = log. 7396. Therefore: ^ * 396 = 2. log. 86 FRACTIONS. The logarithm of a common fraction is found, either by first reducing the fraction to a decimal fraction, or by taking the logarithm of the numerator and the logarithm of the denominator and subtracting the logarithm of the denominator from the log- arithm of the numerator ; the difference is the logarithm of the fraction. LOGARITHMS. 8 1 Example. Log . % = log. 8 — log. 4 Log. 3 = 0.477121 = 10.477121 — 10 Log. 4 = 0.602060 Thus, log. % = 9.875061 — 10 This is also the logarithm of the decimal fraction 0.75. RECIPROCALS. Subtract the logarithm of the-number from log. 1, which is 10.000000 — 10, and the difference is the logarithm of the reci- procal. Example. Find the reciprocal of 315. Solution : Log. 1 = 10.000000 — 10 Log. 315 = 2.498311 Log. reciprocal of 315 = 7.501689 — 10 To this logarithm corresponds the decimal fraction 0.0031746, which is, therefore, the reciprocal of 315. Simple Interest by Logarithms. Add logarithm of principal, logarithm of rate of interest, and logarithm of number of years ; from this sum subtract log- arithm of 100. The difference is the logarithm of the interest. Example. Find the interest of $800 at 4% in 5 years. Solution : Log. 800 = 2.90309 Log. 4, = 0.60206 0.69897 Log. 5 = 4.20412 Log. 100 = 2.00000 Log. interest = 2.20412 = $160 = Interest. 82 LOGARITHMS. Compound Interest by Logarithms. When the interest, at the end of each period of time, is added to the principal the amount will increase at a constant rate ; and this rate will be the amount of one dollar invested for one period of the time. For instance : If the periods of time be one year each, then $30 in 3 years at 5 % compound interest will be : #30 X 1.05 = #31.50 at the end of first year. #31.50 X 1.05 = $33,075 at the end of second year. $33,075 X 1.05 = $34.73 at the end of third year. This calculation may be written : #30 X 1.05 X 1.05 X 1.05 = #34.73 which also may be written #30 X (1.05) 3 = $34.73. Thus, compound interest is a form of geometrical progres- sion, and may be calculated by the following formulas : a -- = p X. r n Log. a - = 71 X log. r + log. p p- Log. p ■■ 71 ' a = t^t = log. a — _ log.a- 7i X log. r - log- P Log. r log. _ log. a - r - log. p 71 p = Principal invested. n r= The number of periods of time. a = The amount due after n periods of time. r — The amount of $1 invested one period of time. Note. — The quantity r is always obtained by the rule : Divide the rate of interest per period of time by 100, and add 1 to the quotient. Example. What is the amount of $816 invested 6 years at A.% com- pound interest? LOGARITHMS. 83 Solution by formula : Log. a = n X log. r 4- log. p Log. a = 6 X log. 1.04 + log. 816 Log. a = 6 X 0.017033+ 2.911690 Log. a = 0.102198 + 2.911690 Log. a = 3.013888 a = $1032.50 = Amount. Example. If $750 is invested at 3% compound interest, how many years will it take before the amount will be $950. Solution by formula : n = lo g, a — 'log, p log. r __ log. 950 — log. 750 ~ log. 1.03 2.977724 — 2.875061 = 8 years (nearly). 0.012837 Example. A principal of $3750 is to be invested so that by compound interest it will amount to $5000 in six years. Find rate of interest. Solution by formula : Log _ r = l og.a-log.p n T 3.698970 — 3.574031 Log. r = 6 Log. r = 0.020823 r = 1.0491 Rate of interest = lOOr— 100 = 100 X 1.0491 — 100 = 4.91% ; or 5 % per year (very nearly). Discount or Rebate. When calculating discount or rebate, which is a deduction upon money paid before it is due, use formula : p= JL Log. p = log. a — n X log. r 84 LOGARITHMS. Example. A bill of $500 is due in 3 years. How much cash is it worth if 3% compound interest should be deducted. Log. p = log. a — n X log. r Log.p = 2.698970 — 3 X 0.012837 Log.p = 2.698970 — 0.038511 Log.p = 2.660459 p = $457.57 = Cash payment. Note. — Such examples may be checked to prevent miscal- culations, by multiplying the result (the cash payment), by the tabular number given for corresponding number of years and percentage of interest in table on page 23 ; if calculations are correct, the product will be equal to the original bill. For in- stance, 457.57 X 1.092727 = 499.99909339 = $500.00. Thus, the calculation in the example is correct. Sinking Funds and Savings. If a sum of money denoted by b, set apart or saved dur- ing each period of time, is put at compound interest at the end of each period," the amount will be : a = b at the end of the first period. a = b -f- br at the end of the second period. a = b + br + br 2 at the end of the third period. At the end of n periods the last term in this geometrical series is br n — 1 and the first term is b, while the ratio is r. The sum of the series is the amount which according to the rules for geometrical progression (see page 69) will be : r(br Q - 1 ) — b a = — - r— 1 b(r" — 1) r— 1 Example. At the end of his first year's business a man sets apart $1200 for a sinking fund, which he invests at 4% per year. At the end of each succeeding year he sets apart $1200 which is invested at the same rate. What is the value of the sinking fund after 7 years of business ? Solution : a = 1200 X (1.04 7 — 1) 1.04 — 1 1200 X 0.31593 a = - a = 0.04 a = $9477.90 LOGARITHMS. 85 Example. A man 20 years old commences to save 25 cents every working day, and places this in a savings bank at 4% interest, computed semi-annually. How much will he have in the bank when he is 36 years old? (Note. — 25c. a day = $1.50 a week = 26 X $1.50 = $39 in six months. 4 % per year = 2 % per period of time , 36 — 20 = 16 = 32 periods of time). Solution by formula: n _b (r« -1) a a 39 X (1.02 32 -— 1) 1.02 — 1 39 X (1.8845 — 1) 0.02 a = 39 X 0.8845 X 50 a = 1724.775 = $1724.77 = Amount. Thus, in 16 years a saving of 25c. a day amounts to $1724.77. If the money is paid in advance of the first period of time the terms will be : a = br at the end of the first period. a = br -f br 2 at the end of the second period. a = br + br 2 -4- br s at the end of the third period. At the end of n years the last term in this geometrical series is br a and the first term is br, while the ratio is r. The sum of the series is the amount, which, according to rules for geometrical progressions (see page 69), will be : _ r (br n ) — br br (r n — 1) a = : - Example. Assume that the man mentioned in previous example, instead of commencing to save money when 20 years old, already had $39 to put in the bank at 4 % the first period of time, and that he always kept up paying $39 in advance semi- annually. How much money would he then save in 16 years ? %6 LOGARITHMS. Solution : „ _ 39 X 1.02 X (102 32 — 1) 1.02 — 1 _ 39 X 1.02 X 0.8845 0.02 a = 1759.27 Thus, by paying the money in advance semi-annually, he will gain (1759.27 — 1724.77) = $34.50. If a principal denoted by p is invested at a given rate of compound interest, and successive smaller or larger equal pay- ments denoted by b are made at the end of each period of time so that they will commence to draw interest at the beginning of the following period at the same rate as the principal, the formula will be : a = p x r« + *>■-*> r — 1 but for logarithmic calculations it is more convenient to denote the rate of interest by y % and the formula will read : a = p X a = p X r n -\ b (r n — 1) + y 100 100 b (r n -1) ay- -py X r R 100 ay r n — 100 + 100 £ py -f 100 b , ay + 1003 cog. P y + 100 b log. log. r a y -f 100 b P y + 100 b Note. — Using these formulas it must be understood that n represents the number of periods of time that the principal is invested, and that this first period is considered to be the period at the end of which the first payment, b, is made. Example. A man has ^50 in a savings bank and he also puts in #25 every month, which goes on interest every 6 months ; the bank pays 4% interest, computed semi-annually. How much money LOGARITHMS. 87 can he save in 5 years in this way ? (Note. — \% per year = 2% per 6 months, or per period of time, and #25 a month = #150 every 6 months, or per period of time. The interest is computed semi-annually ; therefore 5 years = 10 periods of time). Solution by formula : , v _ , 100 6 (r n — 1) a = p X r n + ^ '— 50 X 1.02 10 -f- 100 X 150 X (102 10 — 1) a = 50 X 1.219 100 X 150 X 0.219 a = 60.95 + 1642.50 a = #1703.45 = Amount. The original sum of #50 has increased to #60.95, and the monthly payments amounted to #1500. The last six payments did not draw any interest, as they were deposited in the last six months of the fifth year and would commence to draw inter- est at the beginning of the sixth year if the amount had not been withdrawn. Example. A man has #800 invested at 0%. How much must he save and invest at the same interest every year in order to increase it to #3000 in five years ? Interest is computed annually. Solution by formula : ^ __ a y — p y x > n lOO r n — 100 3000 X 5 — 800 X 5 X 1.05 5 b = 6 = b = 100 X 1.05 5 — 100 15000 — 1.2763 X 4000 100 X 1.2763 — 100 15000 — 5105.2 127.63 9894.8 100 27.63 b = 358.118 = #358.12 to be paid in each year. The total payments will be : 800 + 5 X 358.12 = #800 + #1790.60 = #2590.60. The rest of the amount is accumulated interest. The last pay- ment is made at the end of the fifth year ; therefore this money does not draw interest. 88 LOGARITHMS. Example. A man calculates that if he had $1800 he would start in business. He has only $120, but is earning $15 a week and figures that he can save half of his weekly earnings. He puts his money in a savings bank, where it goes on interest every six months, at the rate of 4% a year. How many years will it take him to save the required amount? (Note. — $7.50 a week = 26 X ny 2 = $195 in six months, and 4% per year = 2% per six months, or per period of time). Solution by formula : 7 a y + 100 b log. n = jjz + 100 b log. log. r 1800X2 + 100X195 120 X 2 + 100 X 195 log. log. 1.02 3600 + 19500 240 + 19500 log. 1.02 _ log. 1.1702 0.0683 8 periods of time (nearly). log. 1.02 0.0086 \ 31 One period = 6 months ; 8 periods = 4 years ; therefore, under these conditions it takes four years to save this amount of money. If a certain sum of money is withdrawn instead of added, at the end of each period of time, the formula on page 86 will change to : 100 £ (r n — 1) a = p X r n y Every letter denotes the same value as it had in the formula on page 86, except that b represents the sum withdrawn instead of the sum added. Example. A man has $5000 invested at b% interest compounded an- nually, but at the end of each year he withdraws $200. How much money has he left after six years ? Solution: a = 1.05 6 X 5000 — 100 X 200 X(1.05 6 — 1) a =1.34 X 5000 a = 6700 — 1360 a = $5340 = Amount. 100 X 200 X 0.34 LOGARITHMS. 89 If the deducted sum, b, exceeds the interest due at the first period of time, the amount a will become smaller than the prin- cipal/, and in time the whole principal will be used up. This will be when : 100 b (r n — 1) fix This transposes to :s to y 100 b 100 b log. - —py 100 3 100 3 — py " - log- r Example. A principal of $5000 is invested at 4% per year, but at the end of each year #600 is withdrawn. How long will it take to use the whole principal ? 100 x 600 lo S- 100 X 600 — 5000 X 4 log. log. 1.04 6O000 900— 20000 log. 1.04 log. 1.50 log. 1.04 0.176091 0.017033 n = 10.3 years. Paying a Debt by Instalments. This same formula applies also in this case; for instance: A man uses $1500 every year toward paying a debt of $10,000, and b% interest per year. 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N M ^ l-O O r- X © ^ ©©©©-© 1 X X X X X X X X X © © © © © cs © © CS © ©©©.©© © cs © © © © © © © © © © © © 126 HYPERBOLIC LOGARITHMS. HYPERBOLIC LOGARITHMS. The hyperbolic or Napierian logarithm of any number may be obtained by multiplying the common logarithm by the con- stant 2.3025S5 ; practically 2.3. Table No. 6 gives the hyperbolic logarithms from 1.01 to 30. Thehyperbolic logarithm of numbers intermediate between those which are given in the table may be obtained by inter- polating proportional differences. TABLE No. 6.— Hyperbolic or Napierian Logarithms of Numbers. N Log. N Log. N Log. ! N Log. 1.01 0.0099 1.26 2311 1.51 0.4121 1.76 0.5653 1.02 0.0198 1.27 0.2390 1.52 0.4187 1.77 0.5710 1.03 0.0296 1.28 0.2469 1.53 0.4253 1.78 0.5766 1.04 0.0392 1.29 0.2546 1.54 0.4318 j 1.79 0.5822 I 1.05 0.0488 1.30 0.2624 1.55 0.4383 1.80 0.5878 1.06 0.0583 1.31 0.2700 1.56 0.4447 181 0.5933 1.07 0.0677 1.32 0.2776 1.57 0.4511 1.82 5988 1.08 0.0770 1.33 0.2852 1.58 0.4574 1.83 0.6043 1.09 00862 1.34 0.2927 1.59 0.4637 1.84 0.6098 1.10 0.0953 1.35 0.3001 1.60 0.4700 1.85 0.6152 1.11 0.1044 1.36 0.3075 1.61 0.4762 1.86 0.6206 1.12 0.1133 1.37 0.3148 1.62 0.4824 1.87 0.6259 1.18 0.1222 1.38 0.3221 1.63 0.4886 1.88 0.6313 1.14 0.1310 1.39 3293 1.64 0.4947 1.89 0.6366 1.15 0.1398 1.40 0.3365 1.65 0.5008 1.90 0.6419 1.16 0.1484 1.41 0.3436 1.66 0.5068 1.91 0.6471 1.17 0.1570 1.42 0.3507 1.67 0.5128 1.92 06523 1.18 0.1655 1.43 0.3577 1.68 0.5188 1.93 0.6575 1.19 0.1740 1.44 0.3646 1.69 0.5247 1.94 0.6627 1.20 0.1823 1.45 0.3716 1.70 0.5306 1.95 0.6678 1.21 0.1906 1.46 0.3784 1.71 0.5365 1.96 0.6729 1.22 0.1988 1.47 0.3853 1.72 0.5423 1.97 0.6780 1.23 0.2070 1.48 0.3920 1.73 0.5481 1.98 0.6831 1.24 0.2151 1.49 0.3988 1.74 0.5539 1.99 0.6881 1.25 0.2231 1.50 0.4055 1.76 0.5596 2.00 0.6931 HYPERBOLIC LOGARITHMS. 127 N Log. N Log. N Log. N Log. 2.01 0.6981 2.41 0.8796 2.81 1.0332 3.21 1.1663 2.02 0.7031 2.42 0.8838 2.82 1.0367 3.22 1.1694 2.03 0.7080 2.43 0.8879 2.83 1.0403 3.23 1.1725 2.04 0.7129 2.44 0.8920 2.84 1.0438 3.24 1.1756 2.05 0.7178 2.45 0.8961 2.85 1.0473 3.25 1.1787 2.06 0.7227 2.46 0.9002 2.86 1.0508 3.26 1.1817 2.07 0.7275 2.47 0.9042 2.87 1.0543 3.27 1.1848 2.08 0.7324 2.48 0.9083 2.88 1.0578 3.28 1.1878 2.09 0.7372 2.49 0.9123 2.89 1.0613 3.29 1.1909 2.10 0.7419 2.50 0.9163 2.90 1.0647 3.30 1.1939 2.11 0.7467 2.51 0.9203 2.91 1.0682 3.31 1.1969 2.12 0.7514 2.52 0.9243 2.92 1.0716 3 32 1.2000 2.13 0.7561 2.53 0.9282 2.93 1.0750 3.33 1.2030 2.14 0.7608 2 54 0.9322 2.94 1.0784 3.34 1.2060 2.15 0.7655 2.55 0.9361 2.95 1.0818 3.35 1.2090 2.16 0.7701 2.56 0.9400 2.96 1.0852 3.36 1.2119 2.17 0.7747 2.57 0.9439 2.97 1.0886 3.37 1.2149 2.18 0.7793 2.58 0.9478 2.98 1.0919 I 3.38 1.2179 2.19 0.7839 2.59 0.9517 2.99 1.0953 3.39 1.2208 2.20 0.7885 2.60 0.9555 3. 1.0986 3.40 1.2238 2.21 0.7930 2-61 0.9594 3.01 1.1019 3.41 1.2267 2.22 0.7975 i 2-62 0.9632 3.02 1.1053 3.42 1.2296 2.23 0.8020 2-63 0.9670 3.03 1.1086 3.43 1.2326 2.24 0.8065 264 0.9708 3.04 1.1119 3.44 1.2355 2.25 0.8109 265 0.9746 3.05 1.1151 3.45 1.2384 2.26 0.8154 2.66 0.9783 3.06 1.1184 3.46 1.2413 2.27 0.8198 2.67 9821 3.07 1.1217 3.47 1.2442 2.28 0.8242 2.68 0.9858 3.08 1.1249 3.48 1.2470 2.29 0.8286 2.69 0.9895 3.09 1.1282 3.49 1.2499 2.30 0.8329 2.70 0.9933 3.10 1.1314 3.50 1.2528 2.31 0.8372 2.71 0.9969 3.11 1.1346 3.51 1.2556 2.32 0.8416 2.72 1.0006 3.12 1.1378 3.52 1.2585 2.33 0.8459 2.73 1.0043 3.13 1.1410 1 3.53 1.2613 2.34 0.8502 2.74 1.0080 3.14 1.1442 3.54 1.2641 2.35 0.8544 2.75 1.0116 3.15 1.1474 ! 3.55 1.2669 2.36 0.8587 ! 2.76 1.0152 3.16 1.1506 3.56 1.2698 2.37 0.8629 2.77 1.0188 3.17 1.1537 3.57 1.2726 2.38 0.8671 2.78 1.0225 3.18 1.1569 3.58 1.2754 2.39 0.8713 2.79 1.0260 3.19 1.1600 3.59 1.2782 2.40 0.8755 2.80 1.0296 3.20 1.1632 3.60 1.2809 128 HYPERBOLIC LOGARITHMS. N Log. N Log. N Log. N Log. 3.61 1.2837 4.01 1.3888 4.41 1.4839 481 1.5707 3.62 12865 4.02 1.3913 4.42 1.4861 4.82 1.5728 3.63 1.2892 4.03 1.3938 4.43 1.4884 4.83 1.6748 3 64 1.2920 4.04 1.3962 4.44 1.4907 4.84 1.6769 3.65 1.2947 4.05 1.3987 4.45 1.4929 4.85 1.5790 3.66 1.2975 4.06 1.4012 4.46 1.4951 4.86 1.5810 3.67 1.3002 4.07 1.4036 4.47 1.4974 4.87 1.5831 3.68 1.3029 4.08 1.4061 4.48 1.4996 488 1.5851 3.69 1.3056 4.09 1.4085 4.49 1.5019 4.89 15872 3.70 1.3083 4.10 1.4110 4 50 1.5041 4.90 1.6892 3.71 1.3110 4.11 1.4134 4.51 1.5063 4.91 1.5913 3.72 1.3137 4.12 1.4159 4.52 1.5085 4.92 1.5933 3.73 1.3164 4.13 1.4183 4.53 1.5107 4.93 1.5953 3.74 1.3191 4.14 1.4207 4.54 1.5129 494 1.5974 3.75 1.3218 4.15 1.4231 4.55 1.5151 4.95 1.5994 3.76 1.3244 4.16 1.4255 4.56 1.5173 4.96 1.6014 3.77 1.3271 4.17 1.4279 4.57 1.5195 4.97 1.6034 3.78 1.3297 4.18 1.4303 4.58 1.5217 4.98 1.6054 3.79 1.3324 4.19 1.4327 4.59 1.5239 4.99 1.6074 3.80 1.3350 4.20 1.4351 4.60 1.5261 5. 1.6094 3.81 1.3376 4.21 1.4375 4.61 1.5282 5.01 1.6114 3.82 1.3403 4.22 1.4398 4.62 1.5304 5.02 1.6134 3.83 1.3429 4.23 1.4422 4.63 1.5326 5.03 1.6154 3.84 1.3455 4.24 1.4446 4.64 1.5347 5.04 1.6174 3.85 1.3481 4.25 1.4469 4.65 1.5369 5.05 1.6194 8.86 1.3507 4.26 1.4493 4.66 1.5390 5.06 1.6214 3.87 1.3533 4.27 1.4516 4.67 1.5412 5.07 1.6233 3 88 1.3558 4.28 1.4540 4.68 1.5433 5.08 1.6253 3.89 1.3584 4.29 1.4563 4.69 1.5454 5.09 1.6273 3.90 1.3610 4.30 1.4586 4.70 1.5476 5.10 1.6292 391 1.3635 4.31 1.4609 4.71 1.5497 5.11 1.6312 3.92 1.3661 4.32 1.4633 4.72 1.5518 6.12 1.6332 3.93 1.3686 4.33 1.4656 4.73 1.5539 5.13 1.6351 3.94 1.3712 4.34 1.4679 4.74 1.5560 5.14 1.6371 3.95 1.3737 4.35 1.4702 4.75 1.5581 5.15 1.6390 3.96 1.3762 4.36 1.4725 4.76 1.5602 5.16 1.6409 3.97 1.3788 4.37 1.4748 4.77 1.5623 5.17 1.6429 3.98 1.3S13 4.38 1.4770 4.78 1.5644 5.18 1.6448 3.99 13838 4.39 1.4793 4.79 1.5665 5.19 1.6467 4. 1.3863 4.40 1.4816 4.80 1.5686 5.20 1.6487 HYPERBOLIC LOGARITHMS. 129 l-l Log. N Log. N Log. N Lop 5.21 16506 5.61 1.7246 6.01 1.7934 6.41 1.8579 5.22 1.6525 5.62 1.7263 6.02 1.7951 6.42 1 .8594 5.23 1.6544 5.63 1.7281 6.03 1.7967 6.43 1.8610 5.24 1.6563 5.64 1.7299 6.04 1.7984 6.44 1.8625 5.25 1.6582 5.65 1.7317 6.05 1.8001 6.45 1.8641 5.26 1.6601 5.66 1.7334 6.06 1.8017 6.46 1.8656 5.27 1.6620 5.67 1.7352 6.07 1.8034 6.47 1.8672 5.28 1.6639 5.68 1.7370 6.08 1.8050 6.48 1.8687 6.29 1.6658 5.69 1.7387 6.09 1.8066 6.49 1.8703 5.30 1.6677 5.70 1.7405 6.10 1.8083 6.50 1.8718 5.31 1.6696 5.71 1.7422 6.11 1.8099 6.51 1.8733 5.32 1.6715 5.72 1.7440 6.12 1.8116 6.52 1.8749 5.33 1.6734 5.73 1.7457 6.13 1.8132 6.53 1.8764 6.34 1.6752 5.74 1.7475 6.14 1 8148 6.54 1.8779 5.35 1.6771 5.75 1.7492 6.15 1.8165 6.55 1.8795 5.36 1.6790 5.76 1.7509 6 16 1.8181 6.56 18810 5.37 1.6808 5.77 1 7527 6.17 1.8197 6.57 1.8825 5.38 1.6827 5.78 1.7544 6.18 1.8213 6.58 18840 5.39 1.6845 5.79 1.7561 6.19 1.8229 6.59 1.8856 6.40 1.6864 5.80 1.7579 6.20 1.8245 6.60 1.8871 5 41 1.6882 5.81 1.7596 6.21 1.8262 6.61 1.8886 6.42 1.6901 5.82 1.7613 6.22 1.8278 6 62 1.8901 5.43 1.6919 5 83 1.7630 6.23 1.8294 6 63 1 8916 5.44 1.6938 5.84 1.7647 6.24 1.8310 6.64 1.8931 5.45 1.6956 5.85 1.7664 6.25 1.8326 6.65 1.8946 5.46 1.6974 5.86 1.7681 6.26 1.8342 6.66 1.8961 5.47 1.6993 5.87 1.7699 6.27 1.8358 6.67 1.8976 5.48 1.7011 5.88 1.7716 6.28 1.8374 6.68 1.8991 5.49 1.7029 5.89 1.7733 6.29 1.8390 6.69 1.9006 6.50 1.7047 5.90 1.7750 6.30 1.8405 6.70 1.9021 5.51 1.7066 6.91 1.7766 6.31 1.8421 6.71 1.9036 5.52 1.7084 5.92 1.7783 6.32 1.8437 6.72 1.9*51 5.53 1.7102 5.93 1.7800 6.33 1.8453 6.73 1.9066 5.54 1.7120 5.94 1.7817 6.34 1.8469 6.74 1.9081 5.55 1.7138 5.95 1.7834 6.35 1.8485 6.75 1.9095 5.56 1.7156 5.96 1.7851 6.36 1.8500 6.76 1.9110 5.57 1.7174 5 97 1.7867 6.37 1.8516 6.77 1.9125 5.58 1.7192 5.98 1.7884 6.38 1 8532 6.78 1.9140 r 5.59 1.7210 5.99 1.7901 6 39 1.8547 6.79 1.9155 j 5.60 1.7228 6. 1.7918 6.40 1.8563 6.80 1.9169 130 HYPERBOLIC LOGARITHMS. N Log. N Log. N Log. |» Log. 6.81 1.9184 7.21 1.9755 7.61 2.0295 | 8.01 2.0807 6.82 1.9199 7.22 1.9769 7.62 2.0308 1 8.02 2.0819 6.83 1.9213 7.23 1.9782 7.63 2.0321 i 8.03 2.0832 6.84 1.9228 7.24 1.9796 7.64 2.0334 j 8.04 2.0844 6.85 1.9242 7.25 1.9810 7.65 2.0347 8.05 2.0857 6.86 1.9257 7.26 1.9824 7.66 2.0360 8.06 2.0869 6.87 1.9272 7.27 1.9838 7.67 2 0373 8.07 2.0882 6.88 1.9286 7.28 1.9851 7.68 2.0^86 8.08 2.0894 6.89 1.9301 7.29 1.9865 7.69 2.0399 8.09 2.0906 6.90 1.9315 7.30 1.9879 7.70 2.0412 8.10 2.0919 6.91 1.9330 7.31 1.9892 7.71 2 0425 811 2.0931 6.92 1.9344 7.32 1.9900 7.72 2.0438 8.12 2.0943 6.93 1.9359 7.33 1.9920 7.73 2.0451 8.13 2.0956 6.94 1.9373 7.34 1.9933 7.74 2.0464 8.14 2.0968 6.95 1.9387 7.35 1.9947 7.75 2.0477 ; 8.15 2.0980 6.96 1.9402 7.36 1.9961 7.76 2.0490 8.16 2.0992 6 97 1.9416 7.37 1.9974 7.77 2.0503 8.17 2.1005 6.98 1 9430 7.38 1.9988 7.78 2.0516 1 8.18 2.1017 6 99 1.9445 7.39 2.0001 7.79 2.0528 ! 8.19 2,1029 7. 1.9459 7.40 2.0015 7.80 2.0541 1 8.20 2.1041 7.01 1.9473 7.41 2.0028 7.81 2.0554 8.21 2 1054 7.02 1.9488 7.42 2.0042 7.82 2.0567 8.22 2.1066 7.03 1.9502 7.43 2.0055 7.83 2.0580 8.23 2.1078 7.04 1.9516 7.44 2.0069 7.84 2.0592 8 24 2.1090 7.05 1.9530 7.45 2.0082 7.85 2.0605 8.25 2.1102 7.06 1.9544 7.46 2.0096 7.86 2.0618 8.26 2.1114 7.07 1.9559 7.47 2.0109 7.87 2.0631 8.27 2.1126 7.08 1.9573 7.48 2.0122 7.88 2.0643 8.28 2.1138 7.09 1.9587 7.49 2.0136 7.89 2.0656 8.29 2.1150 7.10 1.9601 7.50 2.0149 7.90 2.0669 8.30 2.1163 7.11 1.9615 7.51 2.0162 7.91 2.0681 8.31 2 1175 7.12 1.9629 7.52 2.0176 7.92 2.0694 8.32 2.1187 7.13 1.9643 7.53 2.0189 7.93 2.0707 8.33 2.1199 7.14 1.9657 7.54 2.0202 7.94 2.0719 8.34 2.1211 7.15 1.9671 7.55 2.0215 7.95 2.0732 8.35 2.1223 7.16 1.9685 7.56 2.0229 7.96 2.0744 8.36 2.1235 7.17 1.9699 7.57 2.0242 7.97 2.0757 8.37 2.1247 7.18 1.9713 7.58 2.0255 7.98 2.0769 8.38 2.1258 7.19 1.9727 7.59 2.0268 7.99 2.0782 8.39 2.1270 7.20 1.9741 7.60 2.0281 8. 2.0794 8.40 2.1282 HYPERBOLIC LOGARITHMS. 131 N Log. N Log. N Log. N Log. 8.41 2.1294 8.81 2.1759 9.21 2.2203 9.61 2.2628 8.42 2.1306 8.82 2.1770 9.22 2.2214 9.62 2.2638 8.43 2.1318 8.83 2.1782 9.23 2.2225 9.63 2.2649 8.44 2.1330 8.84 2.1793 9.24 2.2235 9.64 2.2659 8.45 2.1342 8.85 2.1804 9.25 2.2246 9.65 2.2670 8.46 2.1353 8.86 2.1815 9.26 2.2257 9.66 2.2680 8.47 2.1365 8.87 2.1827 9.27 2.2268 9.67 2.2690 8.48 2.1377 8.88 2.1838 9.28 2.2279 9.68 2.2701 8.49 2.1389 8.89 2.1849 9.29 2.2289 9.69 2.2711 8.50 2.1401 8.90 2.1861 9.30 2.2300 9.70 2.2721 8.51 2.1412 8.91 2.1872 9.31 2.2311 9.71 2.2732 8.52 2.1424 8.92 2.1883 9.32 2.2322 9.72 2.2742 8.53 2.1436 8.93 2.1894 9.33 2.2332 9.73 2.2752 8.54 2.1448 8.94 2.1905 9.34 2.2343 9.74 9.75 2.2762 8.55 2.1459 8.95 2.1917 9.35 2.2354 2.2773 8.56 2.1471 8.96 2.1928 9.36 2.2364 9.76 2.2783 8.57 2.1483 8.97 2.1939 9.37 2.2375 9.77 2.2793 8.58 2.1494 8.98 2.1950 9.38 2.2386 9.78 2.2803 8.59 2.1506 8.99 2.1961 9.39 2.2396 9.79 2.2814 8.60 2.1518 9. 2.1972 9.40 2.2407 9.80 2.2824 8.61 2.1529 9.01 2.1983 9.41 2.2418 9.81 2.2834 8.62 2.1541 9.02 2.1994 9.42 2.2428 9.82 2.2844 8.63 2.1552 9.03 2.2006 9.43 2.2439 9.83 2.2854 8.64 2.1564 9.04 2.2017 9.44 2.2450 9.84 2.2865 8.65 2.1576 9.05 2.2028 9.45 2.2460 9.85 2.2875 8.66 2.1587 9.06 2.2039 9.46 2.2471 9.86 2.2885 8.67 2.1599 9.07 2.2050 9.47 2.2481 9.87 2.2895 8.68 2 1610 9.08 2.2061 9.48 2.2492 9.88 2.2905 8.69 2.1622 9.09 2.2072 9.49 2.2502 9.89 2.2915 8.70 2.1633 9.10 2.2083 9.50 2.2513 9.90 2.2925 8.71 2.1645 9.11 2.2094 9.51 2.2523 9.91 2.2935 8.72 2.1656 9.12 2.2105 9.52 2.2534 9.92 2.2946 8.73 2.1668 9.13 2.2116 9.53 2.2544 9.93 2.2956 8.74 2.1679 9,14 2.2127 9.54 2.2555 9.94 2.2966 8.75 2.1691 9.15 2.2138 9.55 2.2565 9.95 2.2976 8.76 2.1702 9.16 2.2148 9.56 2.2576 9.96 2 2986 8.77 2.1713 9.17 2.2159 9.57 2.2586 9.97 2.2996 8.78 2.1725 9.18 2.2170 9.58 2.2597 9.98 2.3006 8.79 2.1736 9.19 2.2181 9.59 2.2607 9.99 2.3016 8.80 2.1748 9.20 '2.2192 9.60 2.2618 10. 2.3026 132 HYPERBOLIC LOGARITHMS. N Log. N Log. N Log. N . Log. 10.1 2.3126 14.1 2.6462 18.1 2.8959 22.1 3.0956 10.2 2.3225 14.2 2.6532 18.2 2.9014 22 2 3 1001 10.3 2.3322 14.3 2.6602 18 3 2 9069 22.3 3.1046 10.4 2.3419 14.4 2.6672 18.4 2.9123 22.4 3.1090 10.5 2.3515 14.5 2.6741 18.5 2.9178 22.5 3.1135 10.6 2 3609 14.6 2.6810 18.6 29231 226 3.1179 10.7 2 3703 14.7 2.6878 18.7 2.9285 22.7 3.1224 10.8 2 3796 14.8 26946 18.8 2.9338 22.8 3.1267 10.9 2 3888 14.9 "2.7013 18.9 2.9391 22.9 3.1311 11 2.3979 15. 2.7080 19. 2.9444 23. 3.1355 11.1 2.4070 15.1 2 7147 19.1 2.9497 23.1 3.1398 11.2 2.4160 15.2 2.7213 19.2 2.9549 232 3.1441 11.3 2.4249 15.3 2.6279 19.3 2 9601 23.3 3 1484 11.4 2.4337 15.4 2.7344 19.4 2.9653 23.4 3 1 527 11.5 2.4424 15.5 2.7408 19.5 2.9704 23.5 3.1570 11.6 2.4510 15.6 2.7472 19.6 2.9755 23.6 3.1612 11.7 2 4596 15.7 2.7536 19.7 2.9806 23.7 3.1655 11.8 2.4681 15.8 2.7600 198 2.9856 23.8 3.1697 11.9 2.4765 15.9 2.7663 19.9 2.9907 23.9 3.1739 12 2.4849 16. 2.7726 20. 2.9957 24. 3.1780 12.1 2.4932 16.1 2.7788 20.1 3 0007 24.1 3.1822 12.2 2.5014 16.2 2.7850 20.2 3.0057 24.2 3.1863 12.3 2.5096 16.3 2.7912 20.3 3.0L06 24.3 3.1905 12 4 2.5178 16.4 2.7973 20.4 3.0155 24.4 3 1946 12.5 2.5259 16.5 2.8033 20.5 3.0204 24.5 3.1987 12.6 2.5338 16.6 2.8094 20.6 3.0253 24.6 3.2027 12.7 2 5417 16.7 2.8154 20.7 3.0301 24.7 3.2068 12.8 2.5495 16.8 2.8214 20.8 3.0349 24.8 3.2108 12.9 2.5572 16.9 2 8273 20.9 3.0397 24.9 3.2149 13 2.5649 17. 2.8332 21. 3.0445 25. 3.2189 13.1 2.5726 17.1 2.8391 21.1 3.0493 25.5 3.2387 13.2 2.5802 17.2 2.84 21.2 3.0540 26. 3.2581 13 3 2.5877 17.3 2.8507 21 3 3.0587 26.5 3 2771 13.4 2.5952 17.4 2.8565 21.4 3.0634 27. 3.2958 13.5 2.6027 17.5 2.8622 21.5 3.0680 27.5 3.3142 13.6 2.6101 17.6 2.8679 21.6 3.0727 28. 3.3322 13.7 2.6174 17.7 2.8735 21.7 3.0773 28.5 3.3499 13.8 2.6247 17.8 2.8792 21.8 3.0819 29. 3 3673 13.9 2.6319 17.9 2 8848 21 9 3.0865 29.5 3.3844 14. 2.6391 18. 2.8904 22. 3.0910 30. 3.4012 Weights anb Measures. The yard is the standard unit for length in the United States and Great Britain. To determine the length of the yard, a pendulum vibrating seconds in a vacuum at the level of the sea in,the latitude of London, with the Fahrenheit thermometer at 62°, is supposed to be divided into 391,393 equal parts ; 360,- 000 of these parts is the length of the standard yard. Actually, the standard yard in both the United States and Great Britain is a metallic scale made with great care and kept by the respec- tive governments, and from this standard other measures of length have been produced. The standard unit of weight in the United States and Great Britain is the Troy pound, which is equal in weight to 22.2157 cubic inches of distilled water at 62° Fahrenheit, the barometer being 30 inches. The Troy pound contains 5,760 Troy grains; the Pound Avoirdupois, which is the unit of weight used in commercial transactions and mechanical cal- culations in the United States and Great Britain, is equal to 7,000 Troy grains. In the United States the standard unit of liquid measure is the wine gallon, containing 231 cubic inches or 8.3389 pounds avoirdupois of distilled water at a temperature of its greatest density (39°-40° F). In the United States the standard unit for dry measure is the Winchester Bushel, containing 2150.42 cubic inches. In Great Britain the standard measure for both liquid and dry substances is the Imperial Gallon, which is denned as the volume of 10 pounds avoirdupois of distilled water, when weighed at 62° Fahrenheit with the barometer at 30 inches. The Im- perial Gallon contains 277.463 cubic inches. The Imperial Bushel of 8 gallons contains 2219.704 cubic inches. Long Measure. 12 inches = 1 foot = 0.30479 meters. 3 feet = 1 yard = 0.91437 meters. h]/ 2 yards = 1 rod or pole = 16>^ feet = 198 inches. 40 rods = 1 furlong = 220 yards = 660 feet. 8 furlongs = 1 statute or land mile = 320 rods = 1760 yards. 3 miles = 1 league = 24 furlongs = 960 rods. 5280 feet == 1 statute or land mile = 1.609 kilometer. 1 geographical or nautical mile = 1 minute = ■£$ degree. As adopted by the British admiralty,* a nautical mile is 6080 ft. 1 nautical mile = 1.1515 statute or land miles. 1 statute or land mile = 0.869 nautical miles. * See Machinery, page 23, Sept., 1897. (*33) 134 WEIGHTS AND MEASURES. Square Measure. 1 square yard = 9 square feet = 0.83G square meters. 1 square foot = 144 square inches = 929 square centi- meters. 1 square inch = 6.4514 square centimeters. A section of land is 1 mile square = 640 acres. 1 acre = 43,560 square feet = 0.40467 hectare. 1 square acre is 208.71 feet on each side. Cubic Measure. 1 cubic yard = 27 cubic feet = 0.7645 cubic meters. 1 cubic yard = 201.97 (wine) gallons = 7.645 hectoliter. 1 cubic foot = 1728 cubic inches = 28315.3 cubic centi- meters. 1 cubic foot = 7.4805 (wine) gallons = 28.315 liters. Note. — 1 cubic foot contains 6.2355 imperial (English) gal- lons. ■^ A cord of wood = 128 cubic feet, being 4X4X8 feet. A perch of stone = 24^ cubic feet, being 16>£ X l l A XI foot, but it is generally taken as 25 cubic feet. Liquid Heasure. 1 pint = 28.88 cubic inches. 2 pints = 1 quart = 57.75 cubic inches = 0.9463 liter. 4 quarts = 1 gallon = 231 cubic inches = 3.7852 liters. Note. — 1 imperial (English) gallon is 277.463 cubic inches. Dry Measure. 1 standard U. S. bushel = 2150.42 cubic inches. 1 standard U. S. bushel = 4 pecks. 1 peck = 2 gallons = 8 quarts. 1 gallon = 4 quarts = 268f cubic inches. 1 quart = 2 pints = 67| cubic inches. 100 bushels (approximately) = 124>£ cubic feet. 80 bushels (approximately) = 100 cubic feet. Avoirdupois Weight. (Used in business and mechanical calculations.) 1 pound = 16 ounces =7000 grains = 0.45359 kilogram 1 ton = 2240 pounds. A short ton is 2000 pounds. A fluid ounce is a measure of capacity and means in America one-sixteenth of a pint wine measure, — 1.042 ounces avoirdupois, or 455.6 grains of distilled water = 29.52 cubic centimeters. In England one fluid ounce means one-twentieth part of one imperial pint, equals one ounce, or 437.5 grains of distilled water = 28.35 cubic centimeters. WEIGHTS AND MEASURES. 1 35 Troy Weight* (Used when weighing gold, silver and jewelry ) 1 pound = 12 ounces =5760 grains = 0.37324 kilogram. 1 ounce = 20 pennyweights = 1.0971 ounces avdp. 1 pennyweight =24 grains, A caret used in weighing diamonds = 3.168 grains. = O.205 grams. Apothecaries* Weight. 1 pound = 1 pound troy weight = 12 ounces. 1 ounce = 8 drachms. 1 drachm = 3 scruples. 1 scruple = 20 grains. Note. — The pound and ounce are the same in apothecaries' as in troy weight. One ounce avoirdupois is 437-£ grains, and 1 ounce troy is 480 grains, but 1 grain has the same value in troy, apothecaries' and avoirdupois and is equal to 0.0648 gram in the metric system. Weights of Produce. The following are the weights of various articles of produce: Pounds per aushel. Pounds per b ushel. Pounds per bushel Wheat, 60 Oats, 32 White potatoes, 60 Corn in the ear, 70 Peas, 60 Sweet potatoes, 55 Corn shelled, 56 Ground peas, 24 Onions, 57 Rye, 56 Corn meal, 48 Turnips, 57 Buckwheat, 48 Malt, 38 Clover seed, 60 Barley, 48 White beans, 60 Timothy seed, 45 The netric System of Weights and measures. The unit in the metric system is the meter. The length of the meter was intended to be T oooVooo ( one ten-millionth part) of the length of a quadrant of the earth through Paris, which is the same as toooVooo of any quadrant from either pole to equator. By later calculations it has been ascertained that the meter as first adopted and now used is slightly too short according to this theoretical requirement, but this, of course, makes no dif- ference ; because, practically speaking, the length of a meter is the length of a certain standard meter kejDt at Paris in the care of the French government, and it is from this standard meter (and not from the quadrant of the earth) that all other standard meters kept for reference are derived The gram, which is the unit of weight, is the weight of 1 cubic centimeter of water at its maximum density, which is at 4° C. (39° to 40° F.) The commercial denomination used for weight is the kilo- gram == 1000 grams = 1 cubic decimeter = 1 liter of water at maximum density. 136 WEIGHTS AND MEASURES. Weight. 1 gram == 0.0352739 ounces avoirdupois = 15.432 grains. 10 grams = 1 decagram = 0.352739 ounces avoirdupois. 100 grams = 1 hectogram = 3.52739 " " 1000 grams = 1 kilogram = 2.20462 pounds " Length. 1 Meter = 10 Decimeters = 39.37 inches. 1 Decimeter = 10 Centimeters = 3.937 inches. 1 Centimeter = 10 Millimeters = 0.3937 inches. 1 Millimeter = T ^ Meter = 0.03937 inches. 1 Decameter = 10 Meters = 32 feet 9.7 inches. 1 Hektometer= 100 Meters =328 " 1 inch. 1 Kilometer = 1000 Meters = 0.6214 mile. 1 Myriameter = 10000 Meters = 6.214 miles. Area. 1 square millimeter = 0.00155 square inch. 100 square millimeters = 1 square centimeter = 0.155 sq. inch. 100 " centimeters = 1 " decimeter =15.5 sq. inch. 100 " decimeters =1 " meter = 10.764 sq. feet. 1 Centiare =1 " meter = 1550 sq. inches. 1 Are = 100 " meters = 119.6 sq. yards. 1 Hectare = 10000 sq. meters = 2.471 acres. Solids. 1 cubic millimeter = 0.000061 cubic inch. 1000 cubic millimeters = 1 cubic centimeter = 0.061 cubic inch. 1000 " centimeters = 1 " decimeter = 61.027 " " 1000 " decimeters =1 " meter — 35.3 " feet. Liquid. 1 liter = 10 deciliters = 1 cubic decimeter. 1 deciliter = 10 centiliters = 100 cubic centimeter. 1 centiliter = 10 milliliters = 10 cubic centimeters. 1 milliliter = T ^ 7 liters = 1 cubic centimeter. 1 decaliter = 10 liters = 10 cubic decimeters. 1 hectoliter = 100 liters = 100 cubic decimeters. 1 kiloliter = 1000 liters = 1 cubic meter. 1 liter = 61.027 cubic inches = 1.0567 quarts. WEIGHTS AND MEASURES. 1 37 TABLE No. 7. Reducing Millimeters to Inches. Mm. Inches. Mm. Inches. Mm. Inches. 0.02 0.00079 0.52 0.02047 2 0.07874 0.04 0.00157 0.54 0.02126 3 0.11811 0.06 0.00236 0.56 0.02205 4 0.15748 0.08 0.00315 0.58 0.02283 5 0.19685 0.10 0.00394 0.60 0.02362 6 0.23622 0.12 0.00472 0.62 0.02441 7 0.27559 0.14 0.00551 0.64 0.02520 8 0.31496 0.16 0.00630 0.66 0.02598 9 0.35433 0.18 0.00709 0.68 0.02677 10 0.39370 0.20 0.00787 0.70 0.02756 11 0.43307 0.22 0.00866 0.72 0.02835 12 0.47244 0.24 0.00945 0.74 0.02913 13 0.51181 0.26 0.01024 0.76 0.02992 14 0.55118 0.28 0.01102 0.78 0.03071 15 0.59055 0.30 0.01181 0.80 0.03150 16 0.62992 0.32 0.01260 0.82 0.03228 17 0.66929 0.34 0.01339 0.84 0.03307 18 0.70866 0.36 0.01417 0.86 0.03386 19 0.74803 0.38 0.01496 0.88 0.03465 20 0.78740 0.40 0.01575 0.90 0.03543 21 0.82677 0.42 0.01654 0.92 0.03622 22 0.86614 0.44 0.01732 0.94 0.03701 23 0.90551 0.46 0.01811 0.96 0.03780 24 0.94488 0.48 0.01890 0.98 0.03858 25 0.98425 0.50 0.01969 1.00 0.03937 26 1.02362 TABLE No. 8. Reducing Inches f :o nil li meters. Inches. Mm. Inches. Mm. Inches. Mm. 1 T6 1.59 H 20.64 2X 57.15 yt 6.17 X 22.22 2/ 2 63.50 A 4.76 If 23.81 3 76.20 X 6.35 1 25.40 4 101.6 A 7.94 1H 28.57 5 127 x •9.52 IX 31.75 6 152.4 7 T6^ 11.11 IX 34.92 7 177.8 % 12.70 1/2 38.10 8 203.2 9 T6^ 14.29 IX 41.27 9 228.6 X 15.87 IX 44.45 10 254 11 T6 17.46 w 47.62 11 279.4 H 19.05 • 2 50.80 12 304.8 *3 8 WEIGHTS AND MEASURES. Table of Reduction for Pressure per Unit of Surface. 1 kilogram per sq. centimeter = 14.223 pounds per sq. inch. 1 kilogram per sq. centimeter = 0.968 atmosphere. 1 pound per sq. inch = 0.0703 kilograms per sq. centimeter. 1 pound per sq. inch = 0.06S atmosphere. Table of Reduction for Length and Weight. 1 kilogram per kilometer = 3.548 pounds per mile. 1 kilogram per meter = 0.672 pounds per foot. 1 pound per mile = 0.282 kilograms per kilometer. 1 pound per foot = 1.488 kilograms per meter. Weight of Water U° C.) 1 cubic cm. weighs 1 gram. 1 cubic inch weighs 0.036125 pounds = 16.386 grams. 1 liter weighs 1 kilogram = 2.2046 pounds. 1 quart weighs 2.0862 pounds -— 0.9463 kilograms. 1 cubic meter weighs 1000 kilograms = 2204.6 pounds. 1 cubic foot weighs 62.425 pounds ■■= 28.32 kilograms. Measure of Water. 1 kilogram measures 1 liter =1.057 quarts. 1 kilogram measures 0.353 cubic feet = 61.03 cubic inches. 1 pound measures 0.01602 cubic ft. = 0.454 liter. 1 pound measures 27.68 cubic ins. = 453.59 cubic centimeter- SPECIFIC GRAVITY. The specific gravity of a body is its weight as compared with the weight of an equal volume of another body which is adopted as a standard. For all solid substances, water at its maximum density (4° C.) is the usual standard. For instance, the specific gravity of zinc is 7 ; this simply means that one cubic foot of zinc is 7 times as heavy as one cubic foot of water. One cubic foot of water weighs 62.425 pounds. There- fore, by multiplying the specific gravity of any solid body by 62.425 its weight per cubic foot is obtained. In the metric system of measure and weight, one cubic centimeter of water weighs one gram ; therefore the table of specific gravity will also directly give the weight of the material in grams per cubic centimeter, in kilograms per cubic decimeter, or in 1000 kilo- grams (the so-called metric ton) per cubic meter. WEIGHTS AND MEASURES. 39 TABLE No. o. Specific Gravity, Weights and Values. i Metric. American. Metals. Approximate value per pound Kilog. per cubic dec. or Pounds per Pounds per avoirdupois specific cubic inch. cubic foot. gravity. Water .... 1 0.036125 62.425 Gold (24 k) . . 19.361 0.697 1208 #300.00 Platinum . . . 21.531 0.775 1344 310.00 Silver .... 10.474 0.377 654 9.50 Wrought iron . 7.78 0.28 485 0.015 Cast iron . . . 7.21 0.26 450 0.008 Tool Steel . . 7.85 0.284 490 0.10 Zinc 7 0.252 437 0.10 Antimony . . . 6.72 0.242 419 0.12 Copper .... 8.607 0.31 537 0.15 Mercury .... 13.596 0.489 849 Tin . . . 7.291 0.262 455 0.25 Aluminum . . 2.67 0.096 166 Lead 11.36 0.408 708 0.05 TABLE No. 10. Specific Gravity and Weight of Medium Dry Wood. Metric. American. Variety. Kilog. per cubic Pounds per cubic dec. specific gravity. foot. Birch 0.60 to 0.80 37.5 to 50 Ash 0.50 to 0.80 31 to 50 Beech 0.60 to 0.80 37.5 to 50 Oak 0.60 to 0.90 37.5 to 56 Ebony .... 1.19 74 Lignum-vitce 1.33 83 Spanish mahoga ny . . . 0.85 53 Hickory .... 0.50 32 Spruce .... 0.50 32 Pine .... 0.40 to 0.80 25 to 50 Pitch pine .80 50 140 WEIGHTS AND MEASURES. TABLE No. ii. Specific Gravity and Weight per Cubic Foot of Various materials. (The weight may vary according to the properties of the material). Materials. Asphalt .... Brick Gray granite . . Red granite . . Limestone . . . Sand Portland cement Brickwork . . Slate Glass Emery .... Grindstone . . Coal Porcelain . . . Lime Metric. Kilog. per cubic dec. specific gravity. 1.4 1.6 to 2 2.4 2.5 to 3 2.7 1.5 1.26 1.75 2.8 2.52 4.0 2.4 1.5 2.4 0.96 American. Pounds per cubic foot. 87 100 to 125 150 157 to 187 168 94 78 110 175 157 250 150 ©4 150 60 TABLE No. 12. Specific Gravity and Weight of Liquids. Liquids. Water . . . Sea water . Sulphuric acid Muriatic acid Nitric acid . Alcohol . . . Linseed oil . Turpentine . Petroleum Machine oil . 1 1.03 1.841 1.2 1.217 0.833 0.94 0.87 0.878 0.9 Metric. Kilog. per cubic dec. 1 1.03 1.841 1.2 1.217 0.833 0.94 0.87 0.878 0.9 Kilog. per liter. 1 1.03 1.841 1.2 1.217 0.833 0.94 0.87 0.878 0.9 American. Pounds per cub, inch. 0.036125 0.037 0.067 0.043 0.044 0.03 0.034 0.031 0.032 0.0324 Pounds per gallon. 8.33 8.55 15.48 9.93 10.16 6.93 7.85 7.16 7.39 7.5 WEIGHTS AND MEASURES. 141 To Calculate Weight of Casting from Weight of Pattern. When pattern is made from pine and no nails used, the rule is : Multiply the weight of the pattern by 17 and the prod- uct is the weight of the castings. When nails are used in the pattern, multiply its weight by a little less, probably 15 or 16. When the pattern has core prints, their weight must be calculated and also the weight of what there is to be cored out in the casting, which must all be deducted. This mode of cal- culating the weight of castings is, of course, only approxima- tion, but it is frequently very useful. Weight of an Iron Bar of any Shape of Cross Section. A wrought iron bar of 1 square inch area of cross section and one yard long weighs 10 pounds. Therefore, the weight of wrought iron bars of any shape, as, for instance, railroad rails, X beams, etc., may very conveniently be obtained by first mak- ing a correct, full size drawing of the cross section and measur- ing its area by a planimeter, which gives the area in square inches. Multiply this area by 10 and the product is the weight in pounds per yard; or multiply the area by 3.33 and the product is the weight in pounds per foot. To Calculate Weight of Sheet Iron of any Thickness. One square foot of wrought iron, 1 inch thick, weighs very nearly 40 pounds (40.2 pounds) and one square foot ¥ y, which is T ftb thick, weighs 1 pound. Therefore, a practical rule for quick calculation of the weight of sheet iron is: Divide the thickness of the iron as measured by a micrometer calliper in thousandths of inches by 25, and the quotient is the weight in pounds per square foot. To Calculate the Weight of Metals Not Given in the Tables. Find the weight of wrought iron, and multiply by the fol- lowing constants : Weight of wrought iron X 0.928 = cast iron. X 1.014 = steel. " X 0.918 = zinc. " " " "X 1.144 = copper. " X 1.468 = lead. 142 WEIGHTS AND MEASURES. To Calculate the Weight of Zinc, Copper, Lead, etc, in Sheets. First find the weight by the rule given for sheet iron, and multiply by the constant as given in the above table, and the product is the weight of each metal in pounds per square foot. To Calculate the Weight of Cast Iron Balls. Multiply the cube of the diameter in inches by 0.1377, and the product is the weight of the ball in pounds. Thus : W=D*X 0.1377. D = 1.936 ^/~w D = diameter of ball in inches. W = weight of ball in pounds. In metric measure, multiply the cube of the diameter in centimeters by 0.003775, and the product is the weight of the ball in kilograms. Thus : W= M B X 0.003775. M = 6.422 X */w W = weight in kilograms. M = diameter of ball in centimeters. TABLE No. 13. Weight of Round Steel per Lineal Foot. Steel weighing 489 pounds per Cubic Foot. Diameter in Weight Diameter in Weight Per Diameter in Weight Per inches. Per Foot. Inches. Foot. Inches. Foot. _1_ .0104 1 A 3.011 2 % 12.044 % .042 % 3.375 l A 13.503 3 T"6 .094 _3_ 3.761 H 15.045 X .167 % 4.168 % 16. 67 5 T6 .261 5 1 6 4.595 X 18.379 X .375 n 5.043 % 20.171 .511 T% 5.512 7 A 22.047 X .667 y 2 6.001 3 24.005 9 1 6 .844 9 1 6 6.512 X 26.048 X 1.042 H 7.043 % 28.173 11 "16" 1.261 1 1 T6" 7.596 X 30.382 u 1.5 % 8.169 X 32.674 1 3 1.761 T6 8.702 5 A 35.05 X 2.042 y* 9.377 U 37.508 1% 2.344 1 5 10.013 y 40.05 1 2.667 2 10.669 4 42.675 WEIGHTS AND MEASURES. 143 TABLE No. 1 4. Weights of Square and Round Bars of Wrought Iron in Pounds per Lineal Foot. (Iron weighing 480 pounds per cubic foot). Thickness Weight of Weight of Thickness Weight of Weight of or Diameter Square Bar Round Bar or Diameter Square Bar Round Bar in One Foot One Foot in One Foot One Foot Inches. Long. Long. Inches. Long. Long. Vl6 .013 .010 2 9 /l6 21.89 17.19 % .052 .041 % 22.97 18.04 3 /l6 .117 .092 n /l6 24.08 18.91 X - .208 .164 X 25.21 19.80 5 /l6 .326 .256 13 /l6 26.37 20.71 % .469 .368 X 27.55 21.64 %6 .638 .501 15 /l6 28.76 22.59 X .833 .654 3 30 23.56 9 /l6 1.055 .828 \\Q 31.26 24.55 % 1.302 1.023 X 32.55 25.57 n /l6 1.576 1.237 %6 33.87 26.60 % 1.875 1.473 X 35.21 27.65 13 /l6 2.201 1.728 5 /l6 36.58 28.73 % 2.551 2.004 % 37.97 29.82 15 /l6 2.930 2.301 7 /l6 39.39 30.94 1 3.333 2.618 X 40.83 32.07 Vl6 3.763 2.955 %e 42.30 33.23 % 4.219 3.313 % 43.80 34.40 3 /l6 4.701 3.692 Ww 45.33 35.60 X 5.208 4.091 % 46.88 36.82 5 /l6 5.742 4.510 13 /l6 48.45 38.05 % 6.302 4.950 % 50.05 39.31 %6 6.888 5.410 15 /l6 51.68 40.59 X 7.5 5.890 4 53.33 41.89 9 /l6 8.138 6.392 Me 55.01 43.21 % 8.802 6.913 % 56.72 44.55 Hi* 9.492 7.455 3 /l6 58.45 45.91 % 10.21 8.018 X 60.21 47.29 13 /l6 10.95 8.601 5 /l6 61.99 48.69 % 11.72 9.204 % 63.80 50.11 15 /l6 12.51 9.828 %6 65.64 51.55 2 13.33 10.47 X 67.50 53.01 Vie 14.18 11.14 9 /l6 69.39 54.50 % 15.05 11.82 % 71.30 56 3 /l6 15.95 12.53 n/ie 73.24 57.52 X 16.88 13.25 3/ 7A 75.21 59.07 5 /l6 17.83 14 13 /l6 77.20 60.63 % 18.80 14.77 % 79.22 62.22 T /l6 19.80 15.55 15 /ie 81.26 63.82 X 20.83 16.36 5 83.33 65.45 144 WEIGHTS AND MEASURES. TABLI 3 No. 14 . — (Continued). Thickness Weight of Weight of Thickness Weight of Weight of or Diameter Square Bar Round Bar or Diameter Square Bar Round Bar in One Foot One Foot in One Foot One Foot Inches. Long. Long. Inches. Long. Long. 5 Vie 85.43 67.10 ?X 169.2 132.9 X 87.55 68.76 X 175.2 137.6 %6 89.70 70.45 % 181.3 142.4 X 91.88 72.16 X 187.5 147.3 5 /l6 94.08 73.89 % 193.8 152.2 % 96.30 75.64 % 200.2 157.2 Vl6 98.55 77.40 X 206.7 162.4 X 100.8 79.19 8 213.3 167.6 %6 103.1 81.00 X 226.9 178.2 % 105.5 82.83 X 240.8 189.2 "Ae 107.8 84.69 X 255.2 200.4 % 110.2 86.56 9 270.0 212.1 13 /l6 112.6 88.45 X 285.2 224.0 % 115.1 90.36 X 300.8 236.3 15 /l6 117.5 92.29 % 316.9 248.9 6 120.0 94.25 10 333.3 261.8 % 125.1 98.22 3^ 350.2 275.1 X 130.2 102.3 X 367.5 288.6 % 135.5 106.4 Z A 385.2 302.5 X 140.8 110.6 n 403.3 316.8 % 146.3 114.9 X 421.9 331.3 % 151.9 119.3 X 440.8 346.2 % 157.6 123.7 % 460.2 361.4 7 163.3 128.3 12 480. 377,. TABLE No. 15. Weight of Flat Iron in Pounds per Foot. Inches | Vl6 1 X %6 X 5 /l6 3 / 7s %6 X | % 3/ X 1 X ! 0.11 0.21 0.32.| 0.42 0.53 0.63 0.73 0.84 % 0.13 0.26 0. 40 ! 0.53 0.66 0.79 0.92 1.06 1.31 % 0.16 0.32 0.47| 0.63 0.79 0.95 1.11 1.26 1.58 1.90 % 0.18 0.37 0.55 0.74 0.92 1.11 1.29 1.48 1.85 2.22 2.58 1 0.21 0.42 0.63 0.84 1.05 1.26 1.47 1.68 2.11 2.53 2.95 3.37 IX 0.24 0.47 0.71 0.95 1.18 1.42 1.66 1.90 2.37 2.84 3.32 3.79 l 1 ^ 0.26 0.53 0.79 1.05 1.32 1.58 1.84 2.11 2.63 3.16 3.68 4.21 1% 0.29 0.58! 0.87 1.16 1.45 1.74 2.03 2.32 2.89 3.47 4.05 4.63 IX 0.32 0.63 0.95 1.26 1.58 1.90 2.21 2.53 3.16 3.79 4.42 5.05 IX 0-34 0.68 1.03 1.37 1.71 2.05 2.39 2.74 3.42 4.11 4.79 5.47 1% !0.37 0.74 1.11 1.47 1.84 2.21 2.58 2.95 3.68 4.42 5.16 5.89 1% 10.40 0.79 1.18 1.58 1.97 2.37 2.76 3.16 3.95 4.74 5.53 6.32 2 0.42 0.84 1.26 1.68 2.11 2.53 2.95 3.37 4.21 5.05 5.89 6.74 TABLE No. 16. Sizes of Numbers of the U. S. Standard Gage for Sheet and Plate Iron and Steel. (Brown & Sharpe Mfg. Co.) Approximate Approximate Thickness in Weight Per Weight Per Number Thickness in Square Foot Square Foot of Fractions of Decimal Parts in Ounces in Pounds Gage. an Inch. of an Inch. Avoirdupois. Avoirdupois. 0000000 % .5 320 20.00 000000 H .46875 300 18.75 00000 tV .4375 280 17.50 0000 H .40625 260 16.25 000 H .375 240 15. 00 tt .34375 220 13.75 A .3125 200 12.50 1 3 9 2 .28125 180 11.25 2 H .265625 170 10.625 3 X .25 160 10. 4 15 "5T .234375 150 9.375 5 3 7 2 .21875 140 8.75 6 13 ¥¥ .203124 130 8.125 7 T¥ .1875 120 7.5 8 11 ¥4 .171875 110 6.875 9 5 3 2 .15625 100 6.25 10 A .140625 90 5.625 11 H .125 80 5. 12 6% .109375 70 4.375 13 3 ¥2^ .09375 60 3.75 14 5 ¥¥ .078125 50 3.125 15 9 T2 8 .0703125 45 2.8125 16 tV .0625 40 2.5 17 T¥¥ .05625 36 2.25 18 1 "20" .05 32 2. 19 7 T6¥ .04375 28 1.75 20 3 ¥0" .0375 24 1.50 21 3¥o .034375 22 1.375 22 1 "3 2" .03125 20 1.25 23 9 ¥20" .028125 18 1.125 24 1 40 .025 16 1. 25 ¥¥¥ .021875 14 .875 26 T¥¥ .01875 12 .75 27 11 ¥T¥ .0171875 11 .6875 28 1 ¥4 .015625 10 .625 29 9 640 .0140625 9 .5625 30 JL 8 .0125 8 .5 31 ¥4¥ .0109375 7 .4375 32 T¥80 .01015625 6^ .40625 33 3 ¥¥¥ .009375 6 .375 34 .00859375 5^ .34375 35 12 5 8 ° ¥¥¥ .0078125 5 .3125 36 T¥T¥ .00703125 4/2 .28125 37 ¥3"60 .006640625 4X .265625 38 1 T60 .00625 4 .25 (145) 146 WEIGHTS AND MEASURES. TABLE No. 1 7. Different Standards for Wire Gage in Use in the United States. Dimensions of Sizes in Decimal Parts of an Inch. (Brown & Sharpe Mfg. Co.) O M So e g « c « •n * fr « 2 « S 6 re .»- -b £ 1 °'* shburn & tiMfg. Co. orcester. Mass. it t/3 V T3 £ . 1-1 ^ re 6 £ lj* i: 3 CO » £ ° W trig Sg 000000 •46875 000000 00000 .45 •4375 00000 0000 .46 .454 .3938 .4 •40625 0000 000 .40964 .425 .3625 .36 •375 000 00 .3648 .38 .3310 .33 •34375 00 .32486 .34 .3065 .305 •3125 1 .2893 .3 .2830 .285 .227 .28125 1 2 .25763 .284 .2625 .265 .219 •265625 2 3 .22942 .259 .2437 .245 .212 .25 3 4 .20431 .238 .2253 .225 .207 •234375 4 5 .18194 .22 .2070 .205 .204 •21875 5 6 .16202 .203 .1920 .19 .201 •203125 6 7 .14428 .18 .1770 .175 .199 .1875 7 8 .12849 .165 .1620 .16 .197 .171875 8 9 .11443 .148 .1483 .145 .194 .15625 9 10 .10189 .134 .1350 .13 .191 •140625 10 11 .090742 .12 .1205 .1175 .188 .125 11 12 .080808 .109 .1055 .105 .185 .109375 12 13 .071961 .095 .0915 .0925 .182 .09375 13 14 •064084 .083 .0800 .08 .180 .078125 14 15 .057068 .072 .0720 .07 .178 .0703125 15 16 .05082 .065 .0625 .061 .175 .0625 16 17 •045257 .058 .0540 .0525 .172 .05625 17 18 •040303 • 049 .0475 .045 .168 .05 18 19 .03589 .042 .0410 .04 .164 .04375 19 20 .031961 • 035 .0348 .035 .161 .0375 20 21 .028462 .032 .03175 .031 .157 .034375 21 22 .025347 .028 .0286 .028 .155 .03125 22 23 .022571 •025 .0258 .025 .153 .028125 23 24 .0201 .022 .0230 .0225 .151 .025 24 25 .0179 .02 .0204 .02 .148 .021875 25 26 .01594 .018 .0181 .018 .146 .01875 26 27 .014195 .016 .0173 .017 .143 .0171875 27 28 .012641 .014 .0162 .016 .139 .015625 28 29 .011257 .013 .0150 .015 .134 .0140625 29 30 .010025 .012 .0140 .014 .127 .0125 30 WEIGHTS AND MEASURES. 47 TABLE No. 17. — (Continued). M 6 £ HO Cfi D~ *£ 31 .008928 .01 0132 .013 .120 .0109375 31 32 .00795 .009 .0128 .012 .115 .01015625 32 33 .00708 .008 .0118 .011 .112 .009375 33 34 .006304 .007 .0104 .01 .110 .00859375 34 35 .005614 .005 .0095 .0095 .108 .0078125 35 36 .005 .004 .0090 .009 .106 .00703125 36 37 .004453 .0085 .103 .006640625 37 38 .003965 .008 .101 .00625 38 39 .003531 .0075 .099 39 40 .003144 .007 .097 40 TABLI I No. 18. Weight of Iron Wire in Pounds per 1 00 Feet. No. of American Birmingham No. of American Birmingham Wire or Brown & or Wire or Brown & or Gage. Sharpe. Stubs' Wire. Gage. Sharpe. Stubs' Wire. 0000 56.074 54.620 19 0.341 0.467 000 44.4683 47.865 20 0.270 0.324 00 35.265 38.266 21 0.214 0.271 27.966 30.634 22 0.170 0.207 1 22.178 23.850 23 0.135 0.165 2 17.588 21.373 24 0.107 0.128 3 13.948 17.776 25 0.0849 0.106 4 11.061 15.010 26 0.0673 0.0858 5 8.772 12.826 27 0.0534 0.0678 6 6.956 10.920 28 0.0423 0.0519 7 5.516 8.586 29 0.0335 0.0447 8 4.375 7.214 30 0.0266 0.0381 9 3.469 5.804 31 0.0211 0.0265 10 2.751 4.758 32 0.0167 0.0214 11 2.182 3.816 33 0.0132 0.0169 12 1.730 3.148 34 0.0105 0.0129 13 1.372 2.391 35 0.00836 0.00662 14 1.088 1.825 36 0.00662 0.00424 15 0.863 1.372 37 0.00525 16 0.684 1.119 38 0.00416 17 0.542 .0.891 39 0.00330 18 0.430 0.636 40 0.00262 TABLE No. 19.— Decimal Equivalents of the Numbers of Twist Drill and Steel Wire Gage. (Brown & Sharpe Mfg. Co.) No. Size in Deci- mals. No. Size in Deci- mals. No. Size in Deci- mals. No. Size in Deci- mals. No. Size in Deci- mals. i No., Size in Deci- mals. 1 .2280 15 .1800 29 .1360 42 .0935i 55 .0520 68 .0310 2 .2210 16 .1770 30 .1285 43 .0890 56 .0465 69 .02925 3 .2130 17 .1730 31 .1200 44 .0860 57 .0430 70 .0280 4 .2090 18 .1695 32 .1160 45 .0820 58 .0420 71 .0260 5 .2055 19 .1660 33 .1130 46 .0810 59 .0410 72 .0250 6 .2040 20 .1610 34 .1110 47 .0785 60 .0400 73 .0240 7 .2010 21 .1590 35 .1100 48 .0760 61 .0390 74 .0225 8 .1990 22 .1570 36 .1065 49 .0730 62 .0380 75 .0210 9 .1960 23 .1540 37 .1040 50 .0700 63 .0370 76 .0200 10 .1935 24 .1520 38 .1015 51 .0670 64 .0360 77 .0180 11 .1910 25 .1495 39 .0995 52 .0635 65 .0350 78 .0160 12 .1890 26 .1470 40 .0980 53 .0595 66 .0330 79 .0145 13 .1850 27 .1440 41 .0960 54 .0550 67 .0320 80 .0135 14 .1820 28 .1405 rABLE Vo. 2 0— Deci ( tnal Brow Bquivale n & Shar nts < peM >f Stubs' Steel Wire Gage. g. Co.) 4) Size in Deci- mals. S.S Size in Deci- mals. be °o O t> Size in Deci- mals. fc.S: Size in Deci- mals. 6 M Size in Deci- mals. 4J Size in Deci- mals. z .413 H .266 11 .188 29 .134 47 .077 65 .033 Y .404 G .261 12 .185 30 .127 48 .075 66 .032 X .397 F .257 13 .182 31 .120 49 .072 67 .031 w .386 E .250 14 .180 32 .115 50 .069 68 .030 V .377 D .246 15 .178 33 .112 51 .066 69 .029 u .368 C .242 16 .175 34 .110 52 .063 70 .027 T .358 B .238 17 .172 35 .108 53 .058 71 .026 S .348 A .234 18 .168 36 .106 54 .055 72 .024 R .339 1 .227 19 .164 37 .103 55 .050 73 .023 Q .332 2 .219 20 .161 38 .101 56 .045 74 .022 P .323 3 .212 21 .157 39 .099 57 .042 75 .020 .316 4 .207 22 .155 40 .097 58 .041 76 .018 N .302 5 .204 23 .153 41 .095 59 .040 77 .016 M .295 6 .201 24 .151 42 .092 60 .039 78 .015 L .290 7 .199 25 .148 43 .088 61 .038 79 .014 K .281 8 .197 26 .146 44 .085 62 .037 80 .013 J .277 9 .194 27 .143 45 .081 63 .036 I .272 10 .191 28 .139 46 .079 64 .035 In using the gages known as Stubs' Gages, there should be constantly borne in mind the difference between the Stubs Iron Wire Gage and the Stubs Steel Wire Gage. The Stubs Iron Wire Gage is the one commonly known as the English Standard Wire, or Birmingham Gage, and designates the Stubs soft wire sizes. The Stubs Steel Wire Gage is the one that is used in measuring drawn steel wire or drill rods of Stubs' make, and is also used by many makers of American drill rods. (I 4 8) <$eometr\\ Geometry is the science which teaches the properties of lines, angles, surfaces and solids. A point indicates only position and has neither length, breadth or thickness. A point has no magnitude. A line has length, but no breadth or thickness ; it is either straight, curved or mixed. A straight line is the shortest distance between two points. A curved line is continuously changing its position. A mixed line is composed of straight and curved lines. A. surface has length and breadth, but no thickness ; it may be either plane or curved. A solid has length, breadth, and thickness or depth. An angle is the inclination of two lines which intersect or meet each other. The point of intersection is called the vertex of the angle. An angle is either right, acute or obtuse. A right angle contains 90 degrees. An acute angle contains less than 90 degrees. An obtuse angle contains more than 90 degrees. Fig. 1. Right Angle. Acute Angle. Obtuse Angle. * Polygons. Polygons are plane figures bounded on all sides by straight lines, and are either regular or irregular, according to whether their sides and angles are equal or unequal. The points at which the sides meet are called vertices of the polygon. The distance around any polygon is called the perimeter. A figure bounded by three straight lines, forming three angles, is called a triangle. The sum of the three angles in any triangle, independent of its size or shape, makes 180 degrees. All triangles consist of six parts ; namely, three sides and three angles. If three of these parts are known, one at least being a side, the other parts may be calculated. A triangle is called equilateral when all its three sides have equal length. Then all the three angles are equal, namely, 60 degrees, because 60 X 3.= 180. (See Fig. 4). * Some authorities define as polygons only figures having more than four sides. i5° GEOMETRY. A triangle is called a right-angled triangle when one angle is 90 degrees ; the other two angles will then together consist of 90 degrees, because 90 -f 90 = 180. ( See Fig. 5). An acute-angled triangle has all its angles acute. (See Fig. 6). Fig. 4. Equilateral Triangle. Right-Angled Triangle. Acute Triangle. The longest side in a right-angled triangle is called the hypothenuse and the other two sides are called the base and perpendicular. The square of the length of the hypothenuse is equal to the sum of the squares of the lengths of the other two sides. ( See Fig. 5). a°- + b* = c- 2 From this law the third side of a right-angled triangle can always be found, when the length of the other two sides is known. Thus : ( See Fig. 5). a = */ & b = Vc 2 \Za 2 + b* If, instead of the letters a, b, and c, numbers are used, for instance, a = 3 and b = 4; what then is the length of c ? c = */& + 4* a = Vo 2 - 4 2 £ = V5 2 - 3 2 c = V9 + 10 a = \/ 'lo- 10 b = V25 — 9 c = \^ 1 Ih ci = \/9 b = VlT c = 5 = 3 b = ± A square is a plane figure having four right angles and bounded by four straight lines of equal length. (See Fig. 7). Fig. 7. GEOMETRY. 151 A parallelogram is a plane figure whose opposite sides are parallel and of equal length. ( See Fig. 8). A rectangle is a parallelogram having all its angles right angles. ( See Fig. 9). A trapezoid is a plane figure bounded by four straight lines, of which only two are parallel. (See Fig. 10). Fig. 8. Fig. 9. Parallelogram. Fig. 10. Rectangle. Trapezoid. Fig. 11. A trapezium is a plane figure bounded by four sides, all of which have unequal length. (See Fig. 11). Polygons having four sides, and conse- quently four angles, are usually called quad- rangles. Polygons having more than four sides are named from the number of their sides. Trapezium. Thus : A polygon having five " " six " " seven " " eight ' ; " nine " ten " eleven sides is called a pentagon. " " a hexagon. " " a heptagon. " " an octagon. " " a nonagon. " " a decagon. ' ; " an undecagon. twelve a dodecagon. The sum of the degrees of all the angles of any polygon can always be found by subtracting 2 from the number of sides and multiplying the remainder by 180. For instance : The sum of degrees in any quadrangle is always (4 — 2) X 180 = 360 degrees. The sum of degrees in any pentagon will always be (5 — 2 ) X 180 = 540 degrees. This is a useful fact to remember in making drawings, as it may be used for verifying angles of polygons. I 5 2 GEOMETRY. Circles. F 1 5" 12 J / The Circle is a plane figure bound- ed by a curved line called the circum- ference or perphery, which is at all points the same distance from a fixed point in the plane, and this point is called the center of the circle. ( See point c, Fig. 12). A Dia?neter is a straight line passing through the center of a circle or a sphere, terminating at the circumfer- Circle. ence or surface. (See line e-d, Fig. 12). A Radius is a straight line from the cenler to the circum- ference of circle or sphere. ( See line c-f, Fig. 12). Diameter = 2 X radius. The ratio of the circumference to the diameter of a circle is usually denoted by the Greek letter tt and is expressed approximately by the number 3.1416 or 22 qi ~f T° Thus, if the circumference is required, multiply the diameter by 3.1416. If the diameter is required, divide the circumference by 3.1416. A Chord is a straight line terminating at the circumference of a circle but not passing through the center of the circle. ( See line a-b, Fig. 12). The curved line a-b, or any other part of the circumference of a circle, is called an arc. Any surface bounded by the chord and an arc, like the shaded surface a-b, is called a segment. Any surface bounded by an arc and its two radii, like the shaded surface c-f-d, is called a sector. PROPERTIES OF THE CIRCLE. Circumference = Diameter X 3.1416 Area = (Diameter) 2 X 0.7854 Diameter = Circumference X 0.31831 Diameter 0.7854 Diameter = 1.1283 X V area Circumference = 3.5449 X */ area Length of any arc == Number of degrees X 0.017453 X radius. Length of arc of 1 Degree when radius is 1 is 0.017453. Length of an arc of 1 Minute when radius is 1 is 0.000290888. Length of an arc of 1 Second when radius is 1 is 0.000004848. When the length of the arc is equal to the radius the angle is 57° 17' 45" = 57.2957795 degrees. TRIGONOMETRY. 153 TRIGONOMETRY. Trigonometry is that branch of geometry which treats of the solution of triangles by means of the trigonometrical functions. When the circumference of a circle is divided into 360 equal parts each part is called one degree. One fourth of a circle is 90 degrees = right angle, because 4X90 = 360. ( See Fig. 13.) k h "^Nm i Im A \ 'j \ C d Fig. 14 Circle = 4 right angles. 1 degree = 60 minutes (60'.) Circle = 360 degrees (360°.) 1 minute = 60 seconds (60".) Concerning the angle n (see Fig. 14) the following are the trigonometrical functions : c h cosecant (cosec.) gf tangent ( tan.) k h cotangent (cot.) c g radius = 1. d b sine ( sin.) c d cosine (cos.) c f secant (sec.) The complement of an angle is what remains after sub- tracting the angle from 90°. Thus, the complement of an angle of 30° is 60° because 90 — 30 = 60. The sufipleinent of an angle is what remains after subtract- ing the angle from 180°. Thus, the supplement of an angle of 30° is 150° because 180 — 30 = 150. As all circles, regardless of their size, are divided into 360 degrees, the trigonomical functions must always be alike if the radius and the angle that they denote are alike. It is on this basis that the tables of trigonometrical func- tions are calculated, and as radius is used the figure 1 . In Table No. 21, the natural sine of 30° is given as 0.5 ; this means that if the line c g (see Fig. 14) is 1 foot, meter, or any other unit, and the angle n is 30 degrees, the line d b will be 0.5 of the same unit as the line eg. Sine 45° == 0.70711 ; that is, if the the angle n is 45 degrees and the line c g is 1 of any unit, the line d b is 0.70711 of the same unit. Cos. 30° = 0.86603 ; that is, if the angle «.is 30 degrees and the line c g is 1 of any unit, the line a b or c d is 0.86603 of the same unit. J 54 TRIGONOMETRY. Sec. 30° = 1.1547; that is, if the angle n is 30 degrees and the line c gis 1 of any unit, the line cf is 1.1547 of the same unit. Cosec. 30° = 2 ; that is, if the angle n is 30 degrees and the line eg is 1 of any unit, the line eh is 2 of the same unit. Tang. 30° = 0.57735 ; that is, if the angle n is 30 degrees and the line c g is 1 of any unit, the line g f is 0.57735 of the same unit. Cot. 30° = 1.73205; that is, if the angle n is 30 degrees and the line c g is 1 of any unit, the line k h is 1.73205 of the same unit. Increasing the angle n will increase sine, tangent and secant, but will decrease cosine, cotangent and cosecant. When the angle ;z approaches 90°, the tangents g f increase more and more to infinite length. When n actually reaches 90° of course c b coincides with c k and becomes parallel to gf, so that in an angle of 90° both the secant and the tangent have infinite length, which is denoted by the sign Oo ? and cosine and cotangent have vanished. In the first quadrant (that is when angle n does not exceed 90°) the trigonometrical functions are all considered to be positive and are denoted by -f (plus). When the angle n exceeds 90°, only sine and cosecants remain positive ; all the other functions have become negative and are denoted by — (minus). The following table gives the properties of the trigono- metrical functions in the four different quadrants : Degree. Sine. Cosine. 0° to 90° 90 ° to 180 ° 180 ° to 270 ° 270° to 360° Increase from to radius + Decrease from radius to -j- Increase from to radius — Decrease from radius to — Decrease from radius to -j- Increase from to radius — Decrease from radius to — Increase from to radius -f~ Degree. Secant. Cosecant. 0° to 90° 90° to 180° 180° to 270° 270° to 360° Increase from radius to Co -)- Decrease from cx> to radius — Increase from radius to °° — Decrease from 00 to radius + Decrease from 0° to radius -f- Increase from radius to °° -f~ Decrease from Co to radius — Increase from radius to 0° — Degree. Tangent. Cotangent. 0° to 90° 90 ° to 180 ° 180° to 270° 270° to 360° Increase from to 0° + Decrease from 0° to — Increase from to °° -f- Decrease from 0° to — Decrease from Oo to -f- Increase from to 0° — Decrease from °° to -j- Increase from to °0 — From the rule that the square of the hypothenuse is equal to the sum of the squares of the base and the perpendicular, it also follows that: TRIGONOMETRY. 155 Sin. 2 + cos. 2 = radius 2 . Tang. 2 + radius 2 = secant 2 . Cot. 2 + radius 2 = cosecant 2 . But the trigonometrical tables are calculated with radius = 1, hence, sin. 2 + cos. 2 = 1. tang. 2 + 1 = sec. 2 cotang. 2 -j- 1 = cosecant 2 . tang. = cosin. 1 tang, secant secant — cosin. cotang. = cosin. sin. 1 sin. cosec. — sin. sin. cotang. = 1 tang. cosin. = */\ — sin. 2 cosin. cotang. tang. sin. 1 cosec. cosin. cotang. _ vr= tang. FIG. 15. / + \ \ /^\ 1 Fig. 16. Fig. 17. 6 \ Z& (P. X^H i Sine and Cosine of the Sum of Two Angles. (See Fig. 15). Sine (a + b) = sin. a X cos. b -f- cos. a X sin. b. Cos. (a -j- b) = cos. a X cos. b — sin. a X sin. b. Sine and Cosine of Twice any Angle. (See Fig. 16). Sin. 2« = 2X sin. a X cos. a. Cos. 2 a . = cos. 2 a — sin. 2 a. Sine and Cosine of the Difference of Two Angles. (See Fig. 17). Sin. {a — b) = sin. a X cos. b — cos. a X sin. b, Cosin. 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Table No. 22 gives the logarithms corresponding to sine, cosine, tangent and cotang. for angles from to 90 degrees, with intervals of 10 minutes. For sine and tangent find the degree in the column to the left and the minutes at the top of the table. For instance : Log. sine 19° 30' = 9.523495 — 10. This, of course, is also logarithm to the fraction 0.33381, which is sine of 19° 30'. For cosine and cotang. find the degree in the column to the extreme right in the table, and find the minutes at the bottom of table. For instance : Log. cotang. 37° 10' = 10.120259 — 10 = 0.120259. Note. — In this table the index of the logarithm is increased by 10, therefore — 10 must always be annexed in the logarithm. Logarithms to angles between those in the table may be obtained by interpolations. For instance, find log. sine 25° 45'. Solution: Log. sine 25° 50' — 9.639242 — 10 Log. sine 25° 40' — 9.636623 — 10 Difference 0.002619 This difference in the logarithm corresponds to a difference in this angle of 10 minutes ; therefore a difference of 5 minutes in the angle will make a difference of 0.001309 in the logarithm. Thus : Log. sine 25° 40' = 9.636623 — 10 Difference 5' — 0.001309 Log. sine 25° 45' = 9.637932 — 10 Example 2. Find angle corresponding to logarithmic sine 9.894246 — 10. Solution : In the table of logarithms of sine : 9.894546 — 10 corresponds to 51° 40' 9.893 544 — 10 corresponds t o 51° 30' Difference 0.001002 corresponds to 0° 10~' To logarithm 9.894246 — 10 must, therefore, correspond an angle somewhere between 51° 30' and 51° 40', which is found thus: The given logarithm is 9.894246 — 10 Nearest less logarithm 9.893544 — 10 for 51° 30' Difference 0.000702 Therefore, the correction to be added to the angle already found will be : 0.0007Q2 x io _ 0.001002 ~~ u • Thus, the logarithmic sine 9.894246 — 10 gives 51° 37' trigonometry. t 6 7 Example 3. Find log. to tangent of 50° 45' Solution : Log. tangent 50° 50' = 0.089049 Log. tangent 50° 40' = 0.086471 Difference 0° 10' = 0.002578 in the logarithm. There- fore a difference of 5' in the angle will give 0.001289 in the logarithm. Thus: Log. tangent 50° 40' = 0.086471 Difference 5' = 0.001289 Log. tangent 50° 45' = 0.087760 Example 4. Find the angle corresponding to log. tangent 9.899049 — 10. Solution : Log. tangent 38° 30' = 9.900605 — 10 Log. tangent 38° 20' = 9.898010 — 10 Difference 0° 10' corresponds to 0.002595 The given logarithm = 9.899049 — 10 Nearest less logarithm = 9.898010 — 10 gives 38° 20' Difference = 0.001039 The difference to be added to the angle already found will M01QMXW _ De 0.002595 — U 4 . The tabulated logarithm 9.898010 — 10 gives angle 38° 20' Difference 0.001039 gives angle 4' Logarithm 9.899049 — 10 gives single 3S° 24' To Find Logarithm for Secants and Cosecants. Logarithm for secants is found by subtracting log. cosine from log. 1. For instance, find logarithmic secant 30°. Solution : Log. 1 = 10.000000 — 10 Log. cosine 30° = 9.937531 — 10 Log. secant 30° = 0.062469 Logarithm for cosecants is found by subtracting log. sine from log. 1. For instance, find logarithmic cosecant 35°. Solution : Log. 1 -=10.000000 — 10 Log. sine 35° = 9.758591 — 10 Log. co-secant 35° = 0.241409 Note. — What is said concerning interpolations of trigono- metrical functions in general in the note headed "Important" on page 157, will also apply to their logarithms. :68 TRIGONOMETRICAL TABLES. 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C4 c> © ci p c c c r. c; c c o o c ft b CO b aXNCfflCOliOffiCOiOiOCC^COOOOiNfflOM CO^COCOGO^C30»Ot^05x Tru'5iO'*Wr-OJt-'*'-iQO'*OCO(MXWX'*©'*© ©cqooNNt-t-Noqoo^ooooMMffl©©©©© ©©©©©©©©©©©©©©©©©©©©©© b b iC^MHWffliONHtOOCJCOiOHiMHXWNOOiO CNOXXrH(NOCOC5XOrHH-*HiOl>-l>XiX) W©ffi«(Nl>J>«COOMt>©©t-'*ffiWt-©'-iM «Hr--iO00iD'*nXi0HNC!j03iOOOHa:(NI> nONCRHjqi*S)XOi'-'«itON©"fq-*iONX C©OcDl>l>l>t"I>J>0000XQ000X©©©©© < © ©©C3©©©©©©©©©C3©©©©©©C:©© b b (MMCCNffltlM©'*©NI>l>^^OC:TCMl>ffll> OXI>XCOJ>OCONXHX(MX)HH©HffiCOiOCi5 xce-n-iON^t-t-iOffl^NT-iawxH^ox ^XXXt-iOMHOOONXOHNlNOOCOaiTfOJT)* (NtSXO^-l'CNffi'-fQ^Ot-ffiOlNCOOCOX CO CD C C l> l> l> i> l> NMXX^XOO©©©©©© a! as as as as as' a: os as' as' ci as as' a: as" os as as as as" as ci b CO MTjtiOCfX©OHNM^iOiat-M©OHNeiS^t OS CdO fl — O Q tc c »74 TRIGONOMETRICAL TABLES. ^WNr-iOOSOONCDO^MlN-tOOJOONCiO^COW © CD M-*nN©HOK3c:««iccHCDHa)o^a)M>ooo CO "O^MNO!NMCOOOCO'*HLO T-iw^ot-ao^miot-oooNmoi-ciHMOMS OOOOOOt— 1 r- ii-n«rH(MiN(N(NPlNMMO;«iCO o b m-iMQCM^ooioxooa-.iNOcott-oo-i-^wM^© COXONtDOONOOiNN^WONClt-OOCOC: (N^COOOWOTfOffllMQO^rHOOOCCrtCXCOXOOCi HCT'*ioj>a)ODcoooxO'-'«iONXONTtiox qoqoqot-iT-|--H^rHcq(MNrqfq^c!:cccono: oobbbbobooboboooooooc'oo b i COPI^HOHCO©iOOO HiNTri>OM'Gi©'*'Ti0050iCl>«^0^1>CD — ICO OiOOiOrtCCHi-«QCHXOMOXOiOTl"*iOCO mN^iONXOhco-*cOX©h«i0 3XON^^» OOOOOOi— i— ' i— It— i— i-^H(NMiM!M(M«««MW obooobbdbodbbodododdo'ob b b CO XO^OOC3CiNOCOOHHXiQiOCO!NOOO:aH'» Oj>C5'-HOaMOl^l>XNa©aMMC'Mi.ONCDN J>C00®iNffiON05I>iO«N-irtHN ON« | OOQ005r-!W*Ol>a>'-iC0Oi0HCNX*OCDC005!0'*NOOX^XCi ONMiOCOXOJriNTliOt-fflOOltOXS-KiCN 0000000<— i— lHHrHri^lNCM^M?q«K75CO ddddddddddddddddddddddd b b JHOOXiOHQHQ^^iONffiXMl- C ?C « (M X> lOOX^Mt-rJNiOdTHNOlOOtOJtt^OOXSl !Nt^^XCOX-*Oii-OHj>MOr J< — H ffi N O lO ■* ■* »o OHcoi©o^tiJi>c; — coot- OOOOOOOt-it— i i— lr- I-hp-N(N(N(N?)(M«WMCO b b OCO-*MJ>"C-HO^O:«WCOHOHXC'tXt-l>X o<©t©«xccoxcoi-rHX-i«o^(N?:"NH-* OH«OXHfflrtXl-l>01'lMlMOMMXX«^^ OOOlOOOi-iN^X^rtC-^HXCOtMHr-H^l OOOOOCOnHrHnrtHf)(N!N^NlNWMMM b d d d o b d d b d d d d d d d d d d d d d b b CO si >OcDNXCOHNMTfiiO©l>X050'-iN«-*ioat- TRIGONOMETRICAL TABLES. 75 c HOOi»t'«50'*MNH00500t-OvO'*«(NHO NNHrtHnrtHHHHrt 1 m b CO «^00'fH«j00CiC»OCCHJ>NiOO00ffl'*CDffi NMINNCDO-*INCOlM^a)OOa:iOOO'03M(Ma!ocD©THH(Mcoo:iMHmNoow©o: OOCJOi'-NOJOOOiWfflOSHHOMt-iOOHr.N OHCXWOl>«Jl>OOtOMKO?Cr:Nt-fQHCO bbbbbbbbbbbbbbbbi-<'i-Hi-H^'r-H'cOOOCO OS-iaO^OOOlMMOr-iiMMTft-OOTl'^C'-i'N OWiOi>OCOO©WCOCOQOCO©iC(M(NMOOCCCO TfTH^xfOi0»0O©OC0I>l>00a)QC J ^ (M Ct C£ (M dddddddddddo'ddo'dHHHHH cm' b IN b CO fdNHOOOiONlMSDiCNCOCOHHHmcD^NlM?? 0©OOOI>OSH-*^TtCCC0050l>^OqH-.OW'^ fflCSKJO^NOJOWOOHOOMiOOWI-OOSO-i -*NOO^ooi>t-co-*ri(Ncooomcc-*f:a-. ^o: OlMiOt-O^OCOHiQOiMNINCO^HOniCCO'O b 00 b 0(MOOM35l>COlOHO'H01ffi' , *r-i^^iOO - *i003 OM"*l-©WiOOOHr(*00«CC'-'f'MOOOO;«CCW Tl<^TtlifT)jiO>OiOCDCOCI>l>OOOOOJOOr-iWiO© dddddddddddddddd^H —' —f r-5 i-i b b NCD00-iOHNO©^NHrH iO.^H ^H l- 00 00 ^r Jf- (M N05(NNW05NNMN0000'- l Q0HHOOThI-- 05Hr)(ffl©rH'^J>C' , *t^'-lCOOCD(NOt-J>005W OOOOOOOOOOOOOOOOO'-H'-ir-ir-i^H b b o«'*aTij>©oa)0'*!ocs 0(NCO!NW©0^fqM(N^Xl00 03 00T)»Ci cOCD.t~t-t--r~t~t~Jr~t^t--r-00CO00CO00000000a0QO to a c3 O O cS i 7 6 TRIGONOMETRY. Solutions of Right=Angled Triangles. Fig. 18. Right-angled triangles (see Fig. 18) may be solved by the following for- mulas : Solving for Any Side. A = C X sine a — B X tang, a = B — C X cos. a — A X cot. a = C = A X cosec. a = B X sec. « = C cosec. # C sec. # A sine« ^B_ cot. # A tang. B COS. tf ^ = Vc 2 — ^ 2 Solving for Any Function or for Any Angle. Sin. a = —q- Tang. a — -^~ Sec. # = -g- Cos. a = —q- Cot. # = -^- Cosec.£= —^- Sin. a = cos. b Sin. £ = cos. a Tang. a = cot. <£ Tang, b = cot. # Sec. a = cosec. <£ Sec. £ = cosec. a Sin. £ = — (=r Tang. & = -£- Sec. £ = -j- Cos. £ = —q- Cot. £ = —jg- Cosec. b = -g- Angle a = 90° — £. Angl Solving for Area. A X B eb = 90° — a. Area 2 C 2 X sin. # X cos. a C 2 X cos. b X sin. £ B 2 X tang, a >4 2 X tang. 3 TRIGONOMETRY. 1 77 Fig. 1 9 ^ ^1 f Example. Find angles a and b and the side X in the right-angled triangle. (Fig. 19). *~ 12.5 feet * Tangent corresponding to a = 125 = 0.4 12.5 Tangent corresponding to b = — ^ — = 2.5 By the trigonometrical table the angles are obtained thus : Tangent 0.40000 gives 21° 48' Tangent 2.50000 gives 68° 12' Therefore : Angle a = 21° 48' and angle b = 68° 12'. Angle b may also be found by subtracting angle a from 90 c , thus : Angle b = 90° — 21° 48' = 68° 12' The length of the side X may be found thus : 5 x sin. c 5 ■* — 0.37137 x — 13.464 feet long. By means of logarithms the length of the side x is obtained thus : Log. x = log. 5 — log. sin. 21° 48' Log. x = 0.698970 — (9.569804 — 10) Log. x = 1.129166 x — 13.464 feet long. Note. — In a right-angle triangle (see Fig. 18) the side A is called the perpendicular, B the base and C the hypothenuse. Hence, divide the perpendicular by the hypothenuse and the quotient is the sine of the angle between the base and the hypothenuse. Divide the base by the hypothenuse and the quotient is the cosine of the angle between the base and the hypothenuse. Divide the perpendicular by base and the quo- tient is the tangent of the angle between the base and the hypothenuse. Divide the base by the perpendicular and the quotient is the cotangent of angle between the base and the hypothenuse. 178 TRIGONOMETRY. Solution of Oblique=Angled Triangles. Fig. 20. Oblique-angled triangles A (see Figs. 20-21-22) may be solved by the following formulas : Solving for Any Side. C sin. a B sin. a I n „,_ ^ _ A = —■ =— — T = -\ B 2 + C 1 — 2 B Ccos. a Z? = C = sm c C sin. £ in. ^ A sin. 3 sin. £• A sin. 76 + 285.44 C = Vl024 = 32 feet long. Note. — In this example the cos. of 117° 48' 5" is used, which, in numerical value, is equal to cos. of 62° 11' 55" = 0.4664, but cos. in the second quadrant is negative (see page 154); therefore cos. 117° 48 ; 5" = ( — 0.46645) and the essential sign of the last product after it is multiplied by this negative cos. must change from — to +. ( See Algebra, page 63). 1 80 TRIGONOMETRY. Example 5. Find the length of the side A when C is 32 feet long, angle a is 20° 38' 12" and angle c is 117° 48' 15". Note. — Supplement to c is 62° 11' 55". Solution : Csin. a A = A = sin. c 32 X 0.35242 0.88456 A = 12.75 feet long. In this example, as in the preceding one, we use the sup- plement of the angle in obtaining its function, but here it has no influence on the signs because sin. is positive as well in the second as in the first quadrant. Example 6. Find angle a in Fig. 20, when A is 12.75 feet, B is 24 feet and C is 32 feet. Solution : B 2 1 C 2_ A 2 cos. a 2BC 242 _j_ 32 2 _ 12.75 s cos. a = cos. a 2 X 24 X 32 _ 576+ 1024 — 162.5625 _ 1536 cos. a = 0.93583 Angle a = 20° 38' 12" Example 7. Find angle b, Fig. 20, by the same formula. Solution : A 2 + C 2 — B 2 cos. b = - cos. b = cos. b = 2AC 12.75 2 + 32 2 — 24 2 2 X 12.75 X 32 610.5625 816 cos. b = 0.748238 Angle b = 41° 33' 43" TRIGONOMETRY. l8l Example 8. Find angle c, Fig. 20, by the same formula. Solution : COS. C = A 2 + B 2 — C 2 2AB COS. C — 12.75 2 + 24 2 — 32 2 2 X 12.75 X 4 rrvz. 738.5625 — 1024 612 cos. c = — 0.46640 Supplement to angle c = 62° 11' 55", and angle c = 117° 48' 5" Note. — The negative cosine indicates that it is in the sec- ond quadrant, therefore the angle is over 90°. The angle corresponding to this cosine is the supplement of angle c. To obtain angle c, the angle of its supplement must be subtracted from 180°. Example 9. Find angles a, b and c in Fig. 20, when side A is 12.75 feet, B 32 feet, and C 24 feet. cos.. ^ 2 + C2 -^ 2 2BC cos.* = 322 + 242 - 12 - 75 i 2 X 32 X 24 cos.,^ 1437 - 4375 1536 cos. a = 0.93583 Angle a = 20° 38' 12" Angle b may be found by the formula : sin. b = sin. a A sin. b — sin. 20° 38' 12" — A sin. b = 0.35244 X 32 12.75 sin. b — 0.35244 X 1.8824 sin. & = 0.66343 Angle b — 41° 33' 43" 1 8 2 TRIGONOMETRY. Angle c may be found by the formula: c = 180° — (a + b) c = 180° — (20° 38' 12" + 41° 33' 43") c = 180° — 62° 11' 55" c = 117° 48' 5" Example 10. Find the area of a triangle (see Fig. 20), when it is known that side A is 12.75 feet, side B is 24 feet, and the including angle C= 117° 48' 5". Solution : Sin. to supplement of 117° 48' 5" = sin. 62° 11' 55" = 0.88456. C X A X B Area = 2 Area = 0.88456X12.75X24 = 135 . 34 square feet Example 11. Find angle c and the sides X and y in the triangle, Fig. 23. Solution : c = 180° — (40° + 60°) = 80° ThesideX= 25Xsin - 40Q sin. 60° 25 X 0.64279 X = X 0.86603 16.06975 0.86603 *- X = 18.556 meters long. By the use of logarithms the side X is solved thus : Log. X = log. 25 -f log. sin. 40° — log. sin. 60°. Log.X= 1.39794 + (9.808067—10) — (9.937531—10). Log.X— 1.268476 X = 18.556 meters long. The side y = 25 X sin ' 80 ° J sin. 60° _ 25 X 0.98481 y 0.86603 y — 28.429 meters long. By the use of logarithms the sidejy is solved thus: Log. y — log. 25 -f- log. sin. S0° — log. sin. 60°. Log. y — 1.39794 + (9.993351—10) — (9.937531—10). Log. y = 1.45376 y = 28.429 meters long. TRIGONOMETRY. 183 Example 12. Find angles c and b and the length of the side X \n Fig. 24. Sin. c = Sin. c = 42 X sin. 54° 35 42 X 0.80902 35 Sin. c ~ 0.9T082 r- is meiei Angle c — 76° 7' 26" Angle £ = 180° — (54° 0' 0" 4- 76° T 26") = 49° 52' 34' Side^T = X = 35 X sin. 49° 52' 34" sin. 54° 35 X 0.76465 0.80901 = 33.08 meters long. By means of logarithms the side X is solved thus : Log. X = log. 35 4- log. sin. 49° 52' 34" — log. sin. 54°. Log.X= 1.544068 4- (9.883463—10) — (9.90795S— 10). Log.X= 1.519573 X = 33.08 meters long. Note. — The angle c is obtained by interpolation thus : In the table of trigonometrical functions the sine 0.97100 corre- sponds to the angle 76° 10' and the sine 0.97030 corresponds to the angle 76°. Thus, a difference of 0.00070 in the sine gives a difference of 10' = 600" in the angle. The sine to angle c is 0.97082 The nearest less sine in the table is 0.97030 corresponding to angle 76° 0' 0". Difference, 0.00052 Therefore when an increase in sine of 0.00070 corresponds to an increase of 600" in the angle, an increase of 0.00052 will 600 X 0.00052 = 446' 0° 7' 26' increase the angle 0.00070 thus, the angle corresponding to the sine 0.97082 must be 76° 7' 26". 1 84 GEOMETRY. X> PROBLEMS IN GEOMETRICAL DRAWING. To divide a straight line into a given number of equal parts. (See Fig. 1). Given line a b, which is to be divided into a given number of equal parts. Draw the line b c, of indefinite length, and point off from b the re- quired number of equal parts, as h.g, 1 t y H % \y % t ' e s& , V F.G. 1 f, e, d, c'; join c' and a, and draw the other lines parallel to c' a. To erect a perpendicular at a given point on a straight line. (See Fig. 2). Given line a b, and the point x. The required perpendicular is xy. Solution : With x as center and any With 1 and 2 as Fig. 2 'l * '2 radius, as x 1, cut the line a b at 1 and 2 centers and with a radius somewhat greater than 1 to x, describe arcs intersecting each other at y. Draw x y. This will be the required perpendicular. From a given point without a straight line to draw a per- pendicular to the line. (See Fig. 3). Given line a b and the point c. The required perpendicular is x. Solution: ^ With the point c as center and any radius as c 1, strike the arc 1 to 2. With 1 and 2 as centers and any suitable radius, describe arcs intersecting each other at n, lay the straight edge through points n and c and draw the perpendicular x. To erect a perpendicular at the extremity of a straight line. (See Fig. 4). The required perpendicular is x. Solution : From any point, as c, with radius as a c, draw the circle. From point of intersection, «, through center, c, draw the diameter n fi. From the points, through the point of intersec- tion at ft, draw the perpendicular x. The correctness of this con- struction is founded on the principle that inside a half circle no other V, Given line a b. Fig. a S PROBLEMS IN GEOMETRICAL DRAWING. 185 angle but an angle of 90° can simultaneously touch three points in the circumference when two of these points are in the point of intersection with the diameter and the circumference and the third one anywhere on the circumference of the half circle. The pattern maker is mak^ ing practical use of this geometrical principle, when he by a common car- penter's square is trying the correctness of a semi-circular core box, as shown in Fig. 5. Draw a line parallel to a given line. (See Fig. 6). Given line a b. The required line x y. Solution : Describe with the compass a — from the line a b, the arcs 1 and 2 ; draw line x y, touching these arcs x— =- Fig. 6 To divide a given angle into two equal angles. The given angle, a b c, is divid- ed by the line b d. Solution: With b as center and any radius, as b 1, describe the arc 1 to 2. With 1 and 2 as centers and any suitable radius, describe arcs cutting each other at d. Draw line b d, which will divide the angle into two equal parts. To draw an angle equal to a given angle. Given angle a b c. Construct angle x y z. With b as center and any radius, as b 1, describe the arc 1 to 2. UsingjK as center and without alter- ing the compass, describe the arc /, intersecting y z. Measuring the distance from 2 to 1 on the given angle, transfer this measure to the " 2 arc /, through the point of intersection. Draw the line y x, and this angle will be equal to the first angle. Note. — Angles are usually measured by a tool called a pro- tractor, looking somewhat like Fig. 9 or 10, usually made from metal, and supplied by dealers in draughting instruments. A i86 PROBLEMS IN GEOMETRICAL DRAWING. protractor may also be constructed on paper and used for measuring angles, but it should then always be made on as large a scale as convenient. Fig. 10 Fig. 9 To draw a protractor with a division of 5°. (See Fig. 10). Construct an angle of exactly 90 degrees, divide the arc into nine equal parts, then each part is 10 c ; divide each part into two equal parts and each is 5°. Prove that the sum of the three angles in a triangle consists of 180°. (See Fig. 11). Solution : In the triangle a b c, extend the base line to i. Draw the line o fi, parallel to the side a b, thereby the angle g will be equal to the angle d, and the angle h / must be equal to angle c. The 6 angle/is one angle in the triangle and / + g + h = 1S0°, therefore / + d + c must also be 180°. To draw on a given base line a triangle having angles 90°, 30° and 60°. (See Fig. 12). Given line a b, required triangle is a, c, b. Solution : Extend the line a b to twice its length, to the point e. With e and b as centers strike arcs intersecting each other and erect the perpendicular a c. With b as center and any radius as /, draw the arc / ?n. With / as center and with the same radius, describe arc intersecting at m. From b through point of intersection at w, draw line b intersecting the perpendicular at c. This will complete the triangle. Fig. PROBLEMS IN GEOMETRICAL DRAWING. I8 7 To draw a square inside a given circle. (See Fig. 13). Solution : Draw the line a b through the center of the circle. From points of intersection at a and d, describe with any suitable radius arcs inter- secting at n and m. Draw through the points the line c d. Connect the points of intersection on the circle and the required square is constructed. To draw a square outside a given circle. (See Fig. 14). Solution : Draw lines a b and c d, and from points of intersection at b and c, describe half circles; their points of intersection determine the sides of the square. To draw a hexagon within a given circle. (See Fig. 15). Apply the radius as a chord succes- sively about the circle ; the resulting figure will be a hexagon. Fig. 15- Fig. 16, To inscribe in a circle a regular polygon of any given number of sides. Solution : Divide 360 by the number of sides, and the quotient is the number of degrees, minutes, and seconds contained in the center angle of a triangle, of which one side will make one of the sides in the polygon. For instance, draw a hexagon by this method. (See Fig. 16). 360 6 = 60° i88 PROBLEMS IN GEOMETRICAL DRAWING. Fig. 17 To find the center in a given circle. (See Fig. 17). Solution : Draw anywhere on the circumfer- ence of the circle two chords at ap- proximately right angles to each other, bisect these by the perpendiculars x and y, and their point of intersection is the center of the circle. To draw any number of circles between two inclined lines touching themselves and the lines. (See Fig. 18). Solution : Draw center line ef. Draw first circle on line i g. From point of intersection between this circle and the center line draw the line h, perpendicular to a b. Describe with a radius equal to /i, the arc intersecting at g 1 , draw line^ 1 i\ parallel to g i, and its point of intersection with the center line gives the center for the next circle, etc. Fig. .. -x- The point where these circle. To draw a circle through three given points. (See Fig. 19). The given points are a, b, and c. Solution: From a and b as centers with suitable radius, describe arcs inter- secting at e -e. Draw a line through these points. From b and c as cen- ters, describe arcs intersecting at d d\ draw a line through these points, two lines intersect is the center of the To draw two tangents to a circle from a given point without same circle. (See Fig. 20). Given point #, and the circle with the center n. The required tangents are a d, and a b. Solution : Bisect line ;/ a. With c as center and radius a c, describe PROBLEMS IN GEOMETRICAL DRAWING. 189 the arc b d through the center of the circle. The points of intersection at b and d are the points where the required tan- gents a b and a d will touch the circle. To draw a tangent to a given point in a given circle. (See Fig. 21). Given circle and the point /z, x y is required. Solution : The radius is drawn to the point h and a line constructed perpendicular to it at the point k. This perpendicular, touching the circle at h, is called a tangent. Fig. 22. To draw a circle of a certain size that will touch the perphery of two given cir- cles. (See Fig. 22). Given the diameter of circles a, b, and c. Locate the center for circle c, when centers for a and b are given. Solution : From center of a, describe an arc with a radius equal to the sum of radii of a and c. From b as center, describe another arc using a radius equal to the sum of the radii of b and c. The point of intersection of those two arcs is the center of the circle c. Note. — This construction is useful when locating the center for an intermediate gear. For instance, if a and b are the pitch circles of two gears, c would be the pitch circle located in correct position to connect a and b. To draw an ellipse, the longest and shortest diameter being given. The diameters a b and c d are given. The required ellipse is constructed thus : (See Fig. 23). From c as center with a radius a u, describe an arc f 1 f. The points where this arc intersects a b are foci. The distance fn is divided into any number of parts, as 1, 2, 3, 4, 5. With radius 1 to b, and the focus/ as center, describe arcs 6 and 6 1 radius and with f 1 as center describe arcs 6' radius 1 to a and/ 1 as center, describe arcs intersecting at 6 and C 1 ; with the same radius and with / as center, describe arcs intersecting at 6 2 and.6 3 . Continue this operation for points 2, 3, etc., and when all the points for the circumference are in this with the same and 6 3 . With 190 PROBLEMS IN GEOMETRICAL DRAWING. way marked out, draw the ellipse by using a scroll. It is a property with ellipses that the sum of any two lines drawn from the foci to any point in the circumference is equal to the largest diameter. For instance : /1 e +fe, = ab, or/6 1 -f-/ 1 6\ = a b. Cycloids. Suppose that a round disc, c, rolls on a straight line, a b, and that a lead pencil is fastened at the point r ; it will then describe a curved line, a, /, r, n, b. This line is called a cvcloid. (See Fig. 24). This supposed disk is usual- ly called the generating circle. The line a b is the base line of the cycloid and is equal in length to 7T times m r, or practically 3.1416 times the diameter of the generating circle. The length of the curved line a, I, r, «, b, is four times r ?n, (four times as long as the diameter of the generating circle). A circle rolling on a straight line generates a cycloid. (See Figs. 24 and 25). A circle rolling upon another circle is generating an epicycloid. (See Fig. 26). A circle rolling within another circle generates a hyfio- cycloid. (See Fig. 27). To draw a cycloid, the generating circle being given. Solution : Divide the diameter of the rolling circle in 7 equal parts. Set off 11 of these parts on each side of a on the line d e. This will give a base line practically equal to the circumference. Divide the base line from the point a into any number of equal parts; erect the perpendicu- lars, with center-line as centers and a radius equal to the radius of the generating circle describe the arcs. On the first arc from d or e set off one part of the base line. On the second arc set off two parts of the baseline; on the third arc three parts, etc. This will give the points through which to draw the cycloid. PROBLEMS IN GEOMETRICAL DRAWING. I 9 I To draw an epicycloid (see Fig. 26), the generating circle a and the fundamental circle B being given. Solution : Fig. 26 Concentric with the circle B, describe an arc through the center of the generating circle. Divide the circumference of the generating circle into any number of equal parts and set this off on the circumference of the circle B. Through those points draw radial lines extending until they intersect the arc passing through the center of the generating circle. These points of inter- section give the centers for the different positions of the gener- ating circle, and for the rest, the construction is essentially the same as the cycloids. In Fig. 26, the generating circle is shown in seven different positions', and the point n, in the circumfer- ence of the generating circle, may be followed from the position at the extreme left for one full rotation, to the position where it again touches the circle B. To draw a hypocycloid. (See Fig. 27). The hypocycloid is the line generated by a point in a circle rolling within another larger circle, and is constructed thus: (See Fig. 27). Fig. 27. Divide the circumference of the gener- ating circle into any number of equal parts. Set off these on the circumference of the fundamental circle. From each point of division draw radial lines, 1, 2, 3, 4, 5, 6. From 11 as center describe an arc through the center of the generating circle, as the arc c d. The point of intersection between this arc and the radial lines are centers for the different positions of the gener- ating circle. The distance from 1 to a on the fundamental circle is set off from 1 on the generating circle in its first new position ; the distance 2 to a on the fundamental circle is set off from 2 on the generating circle in its second position, etc. For the rest, the construction is substantially the same as Figs. 25 and 26. Note. — If the diameter of the generating circle is equal to the radius of the fundamental circle, the hypocycloid will be a straight line, which is the diameter of the fundamental circle. 192 PROBLEMS IN GEOMETRICAL DRAWING. Involute. An involute is a curved line which may be assumed to be generated in the following manner : Suppose a string be placed around a cylinder from a to b, in the fig. 28 c, direction 01 the arrow (see Fig. 28), and having a pencil attached at b ; keep the string tight and move the pencil toward c, and the involute, b c, is generated. To draw an involute. Solution : From the point b, (see Fig. 28) set off any number of radial lines at equal distances, as 1, 2, 3," 4, 5. From points of intersection draw the tangents (perpendicular to the radial lines). Set off on the first tangent the length of the arc 1 to b ; on the second tangent the arc 2 to b, etc. This will give the points through which to draw the involute. To draw a spiral from a given point, c. Solution : Draw the line a b through the point c. Set off the centers r and S, one-fourth as far from c as the distance is to be between two lines in the spiral. Using r as center, describe the arc from c to 1, and using S as center, describe the arc from 1 to 2 ; using r as center, de- scribe the arc from 2 to 3, etc. If a cone (see Conical Sections. Fig. 30), is cut by a plane on the line a b, which is parallel to the center line, the section will be a hyperbola. If cut by a plane on the line c d, which is parallel to the side, the section will be a parabola. If cut by a plane on the linej^, which is parallel to the base line, the section will be a circle. If cut by a line, e f, which is neither parallel to the side, the center- line nor the base, the section will be an ellipse. MENSURATION. MENSURATION. If each side in a square ( see Fig. 1 ) is two feet long, the area of the figure will be 4 square feet; that is, it contains four squares, each of- which is one square foot. Thus the area of any square or rectangle is fig. 1 • calculated by multiplying the length by the width. Example 1. What is the area of a piece of land having right angles and measuring 108 feet long and 20 feet wide ? Solution : 1 s i 108 X 20 = 2160 square feet. Example 2. What is the area in square meters of a square house-lot 30 meters long and 30 meters wide. Solution : 30 X 30 = 900 square meters. (Square meter is frequently written m 2 and cubic meter is written ?/2 3 ). A square inscribed in a circle is half in area of a square outside the same circle. Divide the side of a square by 0.8862, and the quotient is the diameter of a circle of the same area as the square. The Difference between One Square Foot and One Foot Square. One foot square means one foot long and one foot wide, but one square foot may be any shape, providing the area is one square foot. For instance, Fig. 1 is two feet square, but it con- tains four square feet. One inch square means one inch long and one inch wide, but one square inch may be any shape, provided the area is one square inch. One mile square means one mile long and one mile wide, but one square mile may have any shape, provided the area is one square mile. Area of Triangles. The area of any triangle may be found by multiplying the base by the perpendicular height and dividing the product by 2. 194 mensuration. Example. Find the area of a triangle 16 inches long and 5 inches per- pendicular height. Solution : Area = - = 40 square inches. The perpendicular height in any triangle is equal to the area multiplied by 2 and the product divided by the base. The area of any triangle is equal to half the base multiplied by the perpendicular height. The perpendicular height of any equilateral triangle is equal to one of its sides multiplied by 0.866. The area of any equilateral triangle may be found by mul- tiplying the square of one of the sides by 0.433. Example. Find the area of an equilateral triangle when the sides are 12 inches long. Solution: Area = 12 X 12 X 0.433 = 62.352 The side of any equilateral triangle multiplied by 0.6582 gives the side of a square of the same area. The side of any equilateral triangle divided by 1.3468 gives the diameter of a circle of the same area. To Figure the Area of Any Triangle when Only the Length of the Three Sides is Given. Rule. From half the sum of the three sides subtract each side separately ; multiply these three remainders with each other and the product by half the sum of the sides, and the square root of this result is the area of the triangle. Example. Find the area of a triangle having sides 12 inches, 9 inches and 15 inches long. Solution : Half the sum of the sides = 18 Area = V(18 — 12) X (18 — 9) X (18 — 15) X IS Area = \A) X 9 X 3 X 18 Area = V2916 Area = 54 square inches. MENSURATION. 1 95 To Find the Height in any Triangle when the Length of the Three Sides is Given. (See Fig. 2). The base line is to the sum of the other fig. 2 two sides as the difference of the sides is to the difference between the two parts of the base line, on each side of the line measuring the perpendicular height. If half this dif- ference is either added to or subtracted from half the base line, there will be obtained two right-angled triangles, in which the base and hypothenuse are known and the perpendicular may be calculated thus : Using Fig. 2 for an example, and adding half the difference to half the base line, this may be written in the formula: Rule. Multiply the sum of the sides by their difference and divide this product by twice the base; to the quotient add half the base ; square this sum (that is, multiply it by itself) ; subtract this from the square of the longest side, and the square root of the difference is the perpendicular height of the triangle. Example. In the triangle, Fig. 2, the sides are: c = 12 inches. a = 9 inches. b = 6 inches ; find the perpendicular height x. x = Jos _/ (9 + 6)X(9-6 ) X V 2 X 12 x= ^81-(lt + 6) 2 x = Jsi — 7.S75 2 12 \ 2 81 — 62.015 x— J 18.985 x = 4.357 inches. 196 MENSURATION. To Find the Area of a Parallelogram. Multiply the length by the width, and the product is the area. Note. — The width must not be measured on the slant side, but perpendicular to its length. To Find the Area of a Trapezoid. Add the two parallel sides and divide by two ; multiply the quotient by the width, and the product is the area. (See Fig. 3). Example. Find the area (Fig. 3). Solution : Area = t_i_? X4 = 32 square feet. 1* » xeei. »• 2 Note. — The correctness of this may be best understood by assuming the triangle b cut off and placed in the position « , and the trapezoid will be changed into a rectangle 8 feet long and 4 feet wide. The area of any polygon may be found by dividing it into triangles and calculating the area of each separately, and the sum of the areas of all the triangles is the area of the polygon. Fig. 3. - 7 feet. of a trapezoid. Fig. 4. The Area of a Circle. The area of a circle is equal to the square of the radius multiplied by 3.1-416, which written in a formula is, Area = 3.1416 r 2 . The area of a circle is also equal to the square of the diameter multiplied by 0.7854, which may be written, Area = 0.7854 d* The area of a circle is also equal to its circumference multiplied by the radius and the product divided by 2, which may be written, c X r 2 Area = The correctness of these formulas may be best understood by assuming the circle to be divided into triangles (see Fig. 4), of which the height h = radius and the sum of the bases, 3, of all the triangles is equal to the circumference of the circle. MENSURATION. 197 Therefore, according to the formulas, the area of a triangle base X perpendicular height the area of a circle must be = circumference X radius 2 and from this follow all the other formulas. To Change a Circle into a Square of the Same Area. Rule. Multiply the diameter of the circle by the constant 0.8862 and the product is the length of one side in a square of the same area. Example. A circular water-tank 5 feet in diameter and 3 feet high is to be replaced by a square tank of the same height and volume. How long will each side in the new tank be? Solution : Side = 5 X 0.8862 = 4.431 feet long. To Find the Side of the Largest Square which can be Inscribed in a Circle. Rule. Multiply the diameter of the circle by the constant 0.7071 ; the product is the length of the side of the square. Example. What is the largest square beam which can be cut from a log 30 inches in diameter. Solution : 30 X 0.7071 = 21.213 inches square. Note. — A round log of any diameter will always cut into a square beam having sides seven-tenths the diameter of the round log. For instance, a 10-inch log will cut 7 inches square, a 15-inch log will cut 10.5 inches square, a 20-inch log will cut 14 inches square, etc. Fig. To Find the Area of Any Irregular Figure (See Fig. 5). Divide the figure into any number of equal parts, as shown by the perpendiculars 1, 2, 3, etc. Measure the width of the figure at the middle of each division ; add these measurements together, « I98 MENSURATION. divide this sum by the number of divisions (in Fig. 5 it is 8), multiply this quotient by the length a &, and the product is the area, approximately. Note. — Sometimes the figure is of such shape that it is more convenient to divide some of it into squares, rectangles, or triangles, and figure the rest as explained above. To Find the Area of a Sector of a Circle. The area of a sector of a circle is to the area of the whole circle as the number of degrees in the arc of the sector is to 360 degrees. Thus 2 X £ 360 2 A = r* X 3.1416 X a = .00S727 X r* X * = r ' l=hA_ r I = 3-1416 XflXr _ 0<01 745329 XaXr 180 r= Jj60A_ = JA_ > 3.1416 a * a a = 180/ — 57.2956 / 3.1416 r r A = Area of sector. r = radius of sector. a = number of degrees in arc. /= length of arc in same units as A and r. Example. The arc of the sector (Fig. 6) is 60° na^ and the radius is 6 feet. Find area. 360 ._. 7T r 2 60 " Area 60r 2 7t Area Area = 360 60 X 6 X 6 X 3.1416 360 Area == 18.849 square feet. If the length of the arc is known instead of the number of degrees, multiply the length of the arc by the length of the radius, divide product by 2, and the quotient is the area of the sector. The correctness of this rule will be understood by the rule for area of circles, explained under Fig. 4. MENSURATION. 1 99 To Find the Length of Arc of a Segment of a Circle. The length of the arc may be calculated by the formula,* /__ 8 c — C 3 /= Length of arc, a fb \ c == Length of chord from a tof v ( See Fig. 7). C = Length of chord from a to b ) Rule. Multiply the length of the chord of half the arc by 8 ; from the product subtract the length of the chord of the arc ; divide the remainder by 3, and the quotient is the length of the arc. When chord and height of segment are known, the chord of half the arc is calculated thus : Chord of half the arc = ,y/ w a _j_ ^2 h = Height of segment (see d f, Fig. 7). ;/ = Half the length of chord ( see a d or b d, Fig. 7). When only the radius and the height of the segment are known, the length of the chord of the whole arc expressed in these terms will be : 2 X \/ 2 r h /i' 2 The chord of half the arc will be : y/2 rh Therefore the length of the arc will be : _ 8 X s/lrh— 2 X V 2 r h — h 2 1 — 3 /= length of arc (afb, Fig. 7). h = height of segment ( df, Fig. 7 ). r = radius of circle {cf, Fig. 7). To Find the Area of a Segment of a Circle. (See Fig. 7). Ascertain the area of the whole sector and from this area subtract the area of the triangle, and the rest is the area of the segment. Example. Find the the area of the segment when the radius is 9 inches and the arc 60°. * This formula is called " Huyghens's approximate formula for circular arcs," but it is so close that it may for any practical purpose be considered absolutely cor- rect for arcs having small center angles; for center angles as large as 120 ° , the result is only one quarter of one per cent, too small, and even for half a circle the result is scarcely more than one per cent, small as compared to results calculated by taking ~ as 3.1416. 200 MENSURATION. Solution : Area of segment — A — ^l^L _ 0.433 r 2 360 ^ = 60X9X9X3.1416 _ 0433X|X9 360 A = 42.4116 — 35.073 A = 7.3386 square inches. In this example the arc was 60°, consequently the triangle is equilateral ; therefore its area is found by the formula 0.433 r 2 . (See area of equilateral triangles, page 194). Note. — When the segment is greater than a semicircle, calculate by preceding rules and formulas the area of the lesser portion of the circle; subtract it from the area of the whole circle. The remainder is the area of the segment. To Find the Radius Corresponding to the Arc, when the Chord and the Height of the Segment Are Given. Rule. Add the square of the height to the square of half the chord ; divide this sum by twice the height, and the quotient is the radius. In a formula this may be written : r _ n* -f A* ) v = radius = c b or cf ) ( See Fig. 7). n = half the chord = db h = height = df The above rule and formula may be proved by rules for right-angled triangles; thus, c b or r equals hypothenuse, and «, or half the chord, equals perpendicular, and c d } which is equal to r — h, is the base. From the rule that the square of the hypothenuse is equal to the sum of the square of the base and the square of the perpendicular, we have : r 2 = ;; 2 + {r — hf r 2 = n 2 + r 2 — 2rh + h 2 r 2 ~r 2 -\-2rh — n* + h 2 2 rh = n 2 + /i 2 Zh The perpendicular height of the triangle is always equal to the radius minus the height of the segment. ( See triangle a b c, and height, df, Fig. 7). MENSURATION. 201 TABLE No. 23.— Areas of Segments of a Circle. The diameter of a circle = 1, and it is divided into 100 equal parts. h Area. ±_ Area. h Area. D D D 0.01 0.001329 0.18 0.096135 0.35 0.244980 0.02 0.003749 0.19 0.103900 0.36 0.254551 0.03 0.006866 0.20 0.111824 0.37 0.264179 0.04 0.010538 0.21 0.119898 0.38 0.273861 0.05 0.014681 0.22 0.128114 0.39 0.283593 0.06 0.019239 0.23 0.136465 0.40 0.293370 0.07 0.024168 0.24 0.144945 0.41 0.303187 0.08 0.029435 0.25 0.153546 0.42 0.313042 0.09 0.035012 0.26 0.162263 0.43 0.322928 0.10 0.040875 0.27 0.171090 0.44 0.332843 0.11 0.047006 0.28 0.180020 0.45 0.342783 0.12 0.053385 0.29 0.189048 0.46 0.352742 0.13 ' 0.059999 0.30 0.198168 0.47 0.362717 0.14 0.066833 0.31 0.207376 0.48 0.372704 0.15 0.073875 0.32 0.216666 0.49 0.382700 0.16 0.081112 0.33 0.226034 0.50 0.392699 0.17 0.088536 0.34 0.235473 Table No. 23 gives the areas of segments from 0.01 to 0.5 in height when the diameter of the circle is 1. The area of any segment is computed by the following rule: Divide the height of the segment by the diameter of its corresponding circle. Find in the table in column marked —p the number which is nearest, and multiply the corresponding area by the square of the diameter of the circle, and the product is the area of the segment. Example. Figure the area of a segment of a circle, the height of the segment being 12 inches and the diameter of the circle 40 inches. Solution : 12 divided by 40 = 0.3 In the column marked —p find 0.3; the corresponding area is 0.198168. The area of the segment is 40 X 40 X 0.198168 = 317.0688 square inches, or 317 square inches. 202 MENSURATION. To Calculate the Number of Gallons of Oil in a Tank. Example. A gasoline tank car is standing on a horizontal track, and by putting a stick through its bung-hole on top it is ascertained that the gasoline stands 15 inches high in the tank. The diameter of the tank is GO inches and the length is 25 feet. How many gallons of gasoline are there in the tank? Solution : 15 divided by 60 is 0.25 In Table No. 23, the area corresponding to 0.25 is 0.153546. Area of cross section of the gasoline is 60 X 60 X 0.153546 = 552.7656 square inches. Twenty-five feet is 300 inches ; the tank contains 300 X 552.7656 — 165829.68 cubic inches. One gallon is 231 cubic inches. The tank contains 165829.6S divided by 231 = 717.88, or 718 gallons. Note. — If the tank is more than half full, figure first the cubical contents of the whole tank if full, then figure the cubical contents of the empty space and subtract the last quantity from the first, and the difference is the cubical contents of the fluid in the tank. Circular Lune. The circular lune is a crescent-shaped figure bounded by two arcs, as a b c and a d c. (Fig. 8). Its area is obtained by first finding the area of the segment a dc (having c% for center of the circle), then the area of the segment a b c (having c\ for center of circle), then by subtracting the area of the last segment from the area of the first : .li j-rr • j.i r A i i Circular Lune. the difference is the area of the lune. A practical example of a circular lune is the area of the opening in a straight-way valve when it is partly shut. Fig. 9. Circular Zone. Circular Zone. The shaded part, a b c d, of the figure is called a circular zone. Its area is ob- tained by first finding the area of the circle and then subtracting the area of the two segments ; the difference is the area of the zone. When the zone is narrow in proportion to the diameter, its area is ob- tained very nearly by following the rule : Add line a b or c dto the diameter of the circle, divide the sum by 2 and multiply V MENSURATION. 203 the quotient by the width of the zone, and the product is the area. To Compute the Volume of a Segment of a Sphere. Rule. fig. 10. Square half the length of its base, and multiply by 3. To this product add square of the height. Multiply the sum by the height and by 0.5236. Example. Find volume of the spherical segment shown in Fig. 10 ; base line is 8" and height is 2". Solution: Segment of a Sphere. Volume = v — (3 X 4 2 + 2 2 ) X 2 X 0.5236 v — (3 X 16 4- 4) X 2 X 0.5236 v = 52 X 2 X 0.5236 v = 54.4544 cubic inches. To Find the Volume of a Spherical Segment, when the Height of the Segment and the Diameter of the Sphere are Known. Rule. Multiply the diameter of sphere by 3, and from this product subtract twice the height of segment. Multiply the remainder by the square of the height and the product by 0.5236. Example. # The segment (Fig. 10) is cut from a sphere 10 inches in diameter and it is 2 inches high. Figure it by this last rule. Solution : Volume = v = (10 X 3 — 2 X 2) X 2 2 X 0.5236 v = (30 — 4) X 4 X 0.5236 v = 26 X 4 X 0.5236 v = 54.4544 square inches. To Find the Surface of a Cylinder. Rule. Multiply the circumference by the length, and to this pro- duct add the area of the two ends. A cylinder has the largest volume with the smallest surface when length and diameter are equal to each other. 204 MENSURATION. To Find the Volume of a Cylinder. Rule. Multiply area of end by length of cylinder, and the product is the volume of the cylinder. Example. What is the volume of a cylinder 4 inches in diameter and 9 inches long? Solution : Area of end = r 2 ir Volume = r 2 W = 2X2X 3.1416 X 9 = 113.0976 cubic inches. To Find the Solid Contents of a Hollow Cylinder. Rule. Find area of end according to outside diameter ; also find area according to inside diameter ; subtract the last area from the first and multiply the difference by the length of the cylinder. Formula : (J? 2 2 )tt/ Solid contents^ R = Outside radius. r = Inside radius. / = Length of cylinder. Example. Find the solid contents of a hollow cylinder of 6 feet outside diameter, 4 feet inside diameter and 5 feet long. Solution : Solid contents — x— (3 2 — 2 2 ) X 3.1416 X 5 x — (9 — 4) X 3.1416 X 5 ^=5X 3.1416 X 5 x = 78.54 cubic feet. Fig. 10. To Find the Area of the Curved Surface of a Cone. (See Fig. 10). Rule. Multiply the circumference of the base by the slant height and divide the product by 2 ; the quotient is the area of the curved surface. If the total surface is wanted, the area of the base is added to the curved area. MENSURATION. 205 If the perpendicular height is known, the length of the slant side or the slant height is found by adding the square of the per- pendicular height to the square of the radius and extracting the square root of the sum. Formula: d tt s/ r 2 4- h % , Curved area = x = 1 = r* a^ r 2 + & 2 r = Radius of base. d = Diameter of base. h = Perpendicular height. To Find the Volume of a Cone. Rule. Multiply the area of the base by the perpendicular height, and divide the product by 3. By formula : Volume = **"* To Find the Area of the Curved Surface of a Frustum of a Cone. (See Fig. 11). Rule. Add circumference of small end to circumference of large end, multi- ply this sum by the slant height and divide the product by 2. Formula : c Curved area = (2 R tc -f- 2 r tt) A which reduces to Curved area = {R -f- r) it S If the perpendicular height instead of the slant height is known, we have : Curved area = (R -f- r) tt \/ (j? R = Large radius. r = Small radius. Ficb tl. / 1 l/l k Frustum of a Coae. rf + h* h = Perpendicular height. s = Slant height. 2 06 MENSURATION. To Find the Volume of a Frustum of a Cone. Rule. Square the largest radius ; square the smallest radius. Multiply largest radius by smallest radius ; add these three pro- ducts and multiply their sum by 3.1416 ; multiply this last product by one-third of the perpendicular height. Formula : Volume = (^+rHi?r)^A Example. Find the volume of a frustum of a cone. The largest diameter is 6 feet, the smallest diameter is 4 feet, and perpen- dicular height is 12 feet. Solution : Volume — x— (3 2 + 2 2 + 3 X 2) X 3.1416 X — o x = (9 + 4 + 6) X 3.1416 X 4 x = 19 X 3.1416 X 4 x = 238.7616 cubic feet. Note. — This rule will also apply for finding the solid con- tents of wood in a log. fig. 12. To Find the Area of the slanted _ ,-- ^ Surface of a Pyramid. /\ \ ( See Fig. 12). // ! v\ & Rule. //'/ ! V-\ l% » Multiply the length of the perim- /// ' \^\ ^ eter °^ tne base, by the slant height of / JJ.—) r A\\\ \ the side (not the slant height of the ,^.1 ! LjA ^-' edge). Divide the product by 2, and >I X- — i' the quotient is the area. Pyramid To Find the Total Area of the Surface of a Pyramid. Rule. Find area of the slanted surface as explained above, and to this add the area of a polygon equal to the base of the pyramid. To Find the Volume of a Pyramid. Rule. Multiply the area of the base by one-third of the perpen- dicular height. MENSURATION. 207 £|Q..1Ai iffiastum ** 1.3254 0.13979 ift 3.3870 0.91291 A 1.3744 0.15033 ift 3.4361 0.93956 29 6¥ 1.4235 0.16126 ift 3.4852 0.96660 if 1.4726 0.17258 i# 3.5343 0.99402 « 1.5217 0.18427 Itt 9 - 6 4 3.5834 1.02182 ^ 1.5708 0.19635 iA 3.6325 1.05001 If 1.6199 0.20S81 Hi 3.6816 1.07858 1 7 32 1.6690 0.22166 i T 3 6 3.7306 1.10753 , f * 1.7181 0.23489 H f 3.7797 1.13687 *,* 1.7671 0.24850 1"3 2" 3.8288 1.16659 ft 1.8162 0.26250 lit 3.8779 1.19670 1 9 32 1.8653 0.27688 i* 3.9270 1.22718 3 9 1.9144 0.29165 117 3.9761 1.25806 X 1.9635 0.30680 IA 4.0252 1.28931 n 2.0126 0.32233 i*f 4.0743 1.32095 I* 2.0617 0.33824 IA 4.1233 1.35297 2IO CIRCUMFERENCES AND AREAS OF CIRCLES. Diameter. Circumfer- ence. Area. Diameter. Circumfer- ence. Area. 1ft -■■3 2 4.1724 1.38538 2 l A 6.6759 3.5466 4.2215 1.41817 2ft 6.8722 3.7584 Iff 4.2706 1.45134 2X 7.0686 3.9761 W 4.3197 1.48489 2 T 5 6 7.2649 4.2 1*4 4.3688 1.51883 2H 7.4613 4.4301 m 4.4179 1.55316 2rV 7.6576 4.6664 in 4.4670 1.58786 2y 2 7.8540 4.9087 1t 7 6 4.5160 1.62295 2 T 9 6 8.0503 5.1573 Iff 4.5651 1.65843 2^ 8.2467 5.4119 m 4.6142 1.69428 m 8.4430 5.6727 ifi 4.6633 1.73052 2U 8.6394 5.9396 i>2 4.7124 1.76715 2|f 8.8357 6.2126 . iff 4.7615 1.80415 27/ 8 9.0321 6.4918 itt .4.8106 1.84154 21! 9.2284 6.7772 iff 4.8597 1.87932 3 9.4248 7.0686 1 T 9 6 4.9087 1.91748 StV 9.6211 7.3662 111 4.9578 1.95602 sy 8 9.8175 7.6699 1H 5.0069 1.99494 3 T 3 s 10.0138 7.9798 ill IX 5.0560 2.03425 3X 10.2102 8.2958 5.1051 2.07394 g_5_ 10.4066 8.6179 ifl 5.1542 2.11402 3H 10.6029 8.9462 lfi 5.2033 2.15448 &16 10.7992 9.2807 iff 5.2524 2.19532 3/ 2 10.9956 9.6211 1H 5.3014 2.23654 3 T 9 6 11.1919 9.9678 145 A 6¥ 5.3505 2.27815 3^ 11.3883 10.3206 123 5.3996 2.32015 3 T I 11.5846 10.6796 111 5.4487 2.36252 m 11.7810 11.0447 IX 5.4978 2.40528 3 r | 11.9773 11.4160 Iff 5.5469 2.44843 37/8 12.1737 11.7933 Iff 5.5960 2.49195 qi5 °T6 12.3701 12.1768 151 1 6¥ 5.6450 2.53586 4 12.5664 12.5664 113 x 16 5.6941 2.58016 ±ft 12.7628 12.9622 153 6f 5.7432 2.62483 4^ 12.9591 13.3641 12? X 3T 5.7923 2.66989 4 T 3 6 13.1554 13.7721 Iff 5.8414 2.71534 4X 13.3518 14.1863 1# 5.8905 2.76117 4 T 5 6 13.5481 14.6066 157 5.9396 2.80738 m 13.7445 15.0330 129 1 32 5.9887 2.85397 4 T V 13.9408 15.4656 Iff 6.0377 2.90095 4^ 14.1372 15.9043 l*t 6.0868 2.94831 4 T 9 6 14.3335 16.3492 Ifi 6.1359 2.99606 4^ 14.5299 16.8002 111 J 3 2 6.1850 3.04418 4H 14.7262 17.2573 if! 6.2341 3.0927 4K 14.9226 17.7206 2 6.2832 3.1416 m 15.1189 18.19 *A 6.4795 3.3410 m 15.3153 18.6655 CIRCUMFERENCES AND AREAS OF CIRCLES. Diameter. Circumfer- ence. Area. Diameter. Circumfer- ence. Area. A\ 5 4 T6 15.5116 19.1472 10^ 32.9868 86.5903 5 15.7080 19.6350 10# 33.3795 88.6643 5^ 16.1007 20.6290 iox 33.7722 90.7625 5% 16.4934 21.6476 107/g 34.1649 92.8858 5^ 16.8861 22.6907 11 34.5576 95.0334 5^ 17.2788 23.7583 n i A 34.9503 97.2055 5^ 17.6715 24.8505 nx 35.343 99.4019 5^ 18.0642 25.9673 liH 35.7357 101.6234 5^ 18.4569 27.1084 ny 2 36.1284 103.8691 6 18.8496 28.2744 11 H 36.5211 106.1394 6>S 19.2423 29.4648 HX 36.9138 108.4338 6X 19.635 30.6797 liH 37.3065 110.7537 6tt 20.0277 31.9191 12 37.6992 113.098 6/ 2 20.4204 33.1831 12% 38.4846 117.859 ; ey* 20.8131 34.4717 12% 39.2700 122.719 6X 21.2058 35.7848 V2% 40.0554 127.677 ■ 6^ 21.5985 37.1224 13 40.8408 132.733 j 7 21.9912 38.4846 13X 41.6262 137.887 ?M 22.3839 39.8713 13^ 42.4116 143.139 ?X 22.7766 41.2826 13* 43.1970 148.490 7^ 23.1693 42.7184 14 43.9824 153.938 *# 23.5620 44.1787 14X 44.7678 159.485 nn 23.9547 45.6636 U}4 45.5532 165.130 iu 24.3474 47.1731 U% 46.3386 170.874 1H 24.7401 48.7071 15 47.1240 176.715 8 25.1328 50.2656 15X 47 9094 182.655 8>g 25.5255 51.8487 15^ 48.6948 188.692 SX 25.9182 53.4561 Vo% 49.4802 194.828 • 8^ 26.3109 55.0884 16 50.2656 201.062 8^ 26.7036 56.7451 16X 51.051 207.395 8^ 27.0963 58.4264 16K 51.8364 213.825 8^ 27.489 60.1319 16* 52.6218 220.354 $tt 27.8817 61.8625 17 53.4072 226.981 9 28.2744 63.6174 vi% 54.1926 233.706 9>^ 28.6671 65.3968 VI % 54.9780 240.529 9X 29.0598 67.2008 nx 55.7634 247.450 9^ 29.4525 69.0293 18 56.5488 254.470 9^ 29.8452 70.8823 18X 57.3342 261.587 9^ 30.2379 72.7599 18K 58.1196 268.803 9^ 30.6306 74.6619 18X 58.905 276.117 9^ 31.0233 76.5888 19 59.6904 283.529 10 31.4160 78.5400 19X 60.4758 291.040 ioy s 31.8087 80.5158 19# 61.2612 298.648 iox 32.2014 82.5158 19X 62.0466 306.355 ioh 32.5941 84.5409 20 62.8320 314.16 CIRCUMFERENCES AND AREAS OF CIRCLES. Diameter. Circumfer- ence. Area. Diameter. Circumfer- ence. Area. 21 65.9736 346.361 m 207.34 3421.19 22 69.1152 380.134 67 210.49 3525.65 23 72.2568 415.477 68 213.63 3631.68 24 75.3984 452.39 69 216.77 3739.28 25 78.540 490.87 70 219.91 3848.45 26 SI. 681 530.93 71 223.05 3959.19 27 84.823 572.56 72 226.19 4071.50 28 87.965 615.75 73 229.34 4185.39 29 91.106 660.52 74 232.48 4300.84 30 94.248 706.86 75 235.62 4417.86 31 97.389 754.77 76 238.76 4536.46 32 100.53 804.25 77 241.90 4656.63 33 103.67 855.30 78 245.04 4778.36 34 106.81 907.92 79 248.19 4901.67 35 109.96 962.11 80 251.33 5026.55 36 113.10 1017.88 81 254.47 5153.00 37 116.24 1075.21 82 257.61 5281.02 38 119.38 1134.11 83 260.75 5410.61 39 122.52 1194.59 84 263.89 5541.77 40 125.66 1256.64 85 267.04 5674.50 41 128.81 1320.25 86 270.18 5808.80 42 131.95 1385.44 87 273.32 5944.68 43 135.09 1452.20 88 276.46 6082.12 44 138.23 1520.53 89 279.60 6221.14 45 141.37 1590.43 90 282.74 6361.73 46 144.51 1661.90 91 285.88 6503.88 47 147.65 1734.94 92 289.03 6647.61 48 150.80 1809.56 93 292.17 6792.91 49 153.94 1885.74 94 295.31 6939.78 50 157.08 1963.50 95 298.45 7088.22 51 160.22 2042.82 96 301.59 7238.23 52 163.36 2123.72 97 304.73 7389.81 53 166.50 2206.18 98 307.88 7542.96 54 169.65 2290.22 99 311.02 7697.69 55 172.79 2375.83 100 314.16 7853.98 56 175.93 2463.01 101 317.30 8011.85 57 179.07 2551.76 102 320.44 8171.28 58* 182.21 2642.08 103 323.58 8332.29 59 185.35 2733.97 104 326.73 8494.87 60 188.50 2827.43 105 329.87 8659.01 61 191.64 2922.47 106 333.01 8824.73 62 194.78 3019.07 107 336.15 8992.02 63 197.92 3117.25 108 339.29 9160.88 64 201.06 3216.99 109 342.43 9331.32 65 204.20 3318.31 110 345.58 1 9503.32 Strength of flfcatertals. The strength of materials may be divided into Tensile, Crushing, Transverse, Torsional, or Shearing, and besides this, the elasticity of the material or its resistance against deflection must also be taken into consideration in figuring for strength. Tensile Strength. From experiments it is known that it it will take from 40,000 to 70,000 pounds to tear off a bar of wrought iron one inch square. Therefore we usually say that the tensile strength of wrought iron is from 40,000 to 70,000 pounds, according to quality. The average is 50,000 to 55,000 pounds. The tensile strength of any body is in proportion to its cross sectional area ; thus, if a bar of iron of one square inch area will pull asunder under a load of 40,000 pounds, it will take 80,000 pounds to pull asunder another bar of the same kind of iron but of two square inches area. The tensile strength is independent of the length of the bar, if it is not so long that its own weight must be taken into consideration. Table No. 25 gives the load which will pull asunder one square inch of the most common materials. No part of any machine should be strained to that limit. A high factor of safety must be used, sometimes from 4 to 30 or even more, which will depend upon the kind of stress the mem- ber is exposed to, as dead load, variable load, shocks, etc. Dif- ferent factors of safety are also used for different kinds of material. ( See page 274). nodulus of Elasticity. The modulus of elasticity for any kind of material is usually defined as the amount of force which would be required to stretch a straight bar of one square inch area to double its length or compress it to nothing, if this were possible. But a more comprehensive definition is to say that the modulus of elasticity is the reciprocal of the fractional part of the length which one unit of force will, within elastic limit, stretch or com- press one unit of area. For instance, if the modulus of elasticity for a certain kind of wrought iron is 25,000,000, it means that it would take 25,000,000 pounds of pulling force to stretch a bar of (213) 214 STRENGTH OF MATERIALS. one square inch area to double its length, if this could possibly be done; but it means also — which is exactly equivalent — that one pound of pulling force will stretch a bar of one square inch area one 25-millionth part of its length, or one pound compressive force will shorten the same bar one 25-millionth part of its length, and that two pounds of force will stretch or compress twice as much, three pounds thrice as much, etc. Strength of Wrought Iron. From experiments it is known that wrought iron can not very well be stretched or compressed more than one-thousandth part of its length without destroying its elasticity ; therefore if a bar of wrought iron has 25,000,000 as its modulus of elas- ticity, one pound will stretch it 25 o oV o o o of its length and it would take 25,000 pounds to stretch it toVo of its length. Thus, 25,000 pounds would then be said to be its strength at the limit of elasticity for that kind of iron ; 80 to 100 per cent, more will usually be the ultimate breaking load. The pull or load which such a bar can sustain with safety will depend a great deal on circumstances, but it must never exceed 25,000 pounds per square inch of area. It must not even approach this limit if the structure is of any importance or if the load is to be sustained for any length of time, or if it is, besides the load, also exposed to shocks or jar. Strength of Cast Iron. Cast iron of good quality has a modulus of elasticity of 15,000,000 pounds, but if strained so it will stretch ysW of its length its elasticity is usually destroyed. For instance, a bar of cast iron of one square inch area is exposed to tensile strain, its modulus of elasticity being 15,000,000 pounds and its elas- ticity being destroyed if it stretches jsV of its length, what then would be its strength at limit of elasticity? One pound will stretch it T __ 7J i__^_ of its length, therefore it must take 10,000 pounds to stretch it y^Vo of its length; thus we would say that 10,000 pounds is its strength at limit of elasticity. It is not always that cast iron is of as good quality as that; very frequently its elasticity is destroyed if it is exposed to a tensile stress of 6,000 pounds per square inch of area ; thus the strength of cast iron at its limit of elasticity is often found to be only 6,000 pounds instead of 10,000 pounds. Besides, it is very often found that a pulling force of 10,000 pounds will stretch a bar of one square inch area one twelve-hundredth part of its length, and this, of course, gives the modulus of elasticity 12,000,000 pounds. Frequently cast iron is of such quality that it cannot be stretched over 2 sV ^ * ts l en g tn before its elas- ticity is destroyed. Cast iron is very variable in quality, and STRENGTH OF MATERIALS. 21 5 especially so with regard to its tensile strength. Generally- speaking, we may say that for cast iron the Modulus of elasticity is 12,000,000 to 15,000,000 pounds. Tensile strength at limit of elasticity, 5,000 to 10,000 pounds. Ultimate tensile strength, 10,000 to 20,000 pounds. Elongation Under Tension. The total stretch or elongation of any specimen when exposed to tensile stress within the elastic limit is directly pro- portional to the length of the specimen, but it is inversely proportional to the modulus of elasticity and the cross sectional area of the specimen. The following formulas may, therefore, be used in such calculations : E- P X L s X A p _ E X s X A . _^ X L EX A T _E X sX A A -PXL s X.E p E = Modulus of elasticity in pounds per square inch. P = Load or force in pounds acting to elongate the specimen. s = Total stretch of specimen in inches in the length L. L = Original length of specimen in inches before force is applied. A = Cross-sectional area of specimen in square inches. Example. From experiments it is known that the modulus of elasticity for a certain kind of wrought iron is 28,000,000 ; what will then be the total stretch or elongation in a round boiler stay, 1% inches in diameter and 6 feet long, when exposed to a stress of 5000 pounds? Solution : 1% inches diameter = 1.227 inches area (see table, page 209) 6 feet long = 72 inches. s= P X L E X A s _ 5000 X 72 28000000 X 1227 s = 0.0105 inches = total stretch in the stay. Note. — As already stated, wrought iron can not be stretched as much as one-thousandth part of its original length without danger of destroying its elasticity; thus, for this stay, which is 72 inches, the limit of elasticity will be at a stretch of 0.072 inches; therefore the stretch produced by a load of 5,000 pounds, which is calculated to be 0.0105 inches, is well within the safe limit. 2l6 STRENGTH OF MATERIALS. TABLE No. 25.— Modulus of Elasticity and Ultimate Tensile Strength of Various Materials. Ultimate Modulus of Ultimate Materials. Modulus of Elasticity in Pounds per Square Inch. Tensile Strength in Pounds per Square Inch. Elasticity in Kilograms per Square Centimeter. Tensile Strength in Kilograms per Square Centimeter. Cast steel . . . i 30,000,000 100,000 2.200,000 7,000 Bessemer steel . . 28.000,000 70,000 1,970.000 4.930 Wrought iron bars 25,000,000 55,000 1,700,000 3,850 Wrought iron wire 28,000,000 75,000 1,970,000 5,250 ( 12,000,000 10,000 80,0000 700 Cast iron * . . . < to to to to I 15,000,000 20,000 1,000.000 1,400 Copper bolts . . . 18,000,000 35,000 1,200,000 2,400 Brass . . . 9,000,000 17,700 630,000 1,200 Oak .... 1,500,000 1,400,000 17,000 105,000 1,200 1,400 Hickory- 20,000 98,000 Maple . . . 1,100,000 15,000 77,000 1.000 Pitch pine . 1,600,000 15,000 112,000 1,000 Pine . . . 1,100,000 10,000 77,000 700 Spruce . . 1,100,000 10,000 77,000 700 The two last columns in above table are calculated by the rule : One pound per sq. inch = 0.07031 kilograms pei sq. centi- meter and the result is reduced to the nearest round number. Formulas for Tensile Strength. The ultimate tensile strength of any specimen is in propor- tion to its cross-sectional area, and is expressed by the following formula : P_ A P_ S A = Side of a square bar = -v/ Diameter of a round bar = S X 0.7854 P = Force in pounds which will pull the specimen asunder. S = Ultimate tensile strength in pounds per square inch. (See Table No. 25). A = Cross-sectional area of the specimen in square inches. * Very strong cast iron may have an ultimate tensile strength as high as 30,000 pounds per square inch. STRENGTH OF MATERIALS. 217 Example. A piece of iron y 2 inch square is tested in a testing machine and breaks at a total stress of 14,210 pounds. What is the ultimate tensile strength per square inch ? Solution : A bar y 2 inch square has a cross-sectional area of y X x / 2 " is X square inch. 14^10 S = = 56,840 pounds per square inch. X Example. What will be the breaking load for a wrought iron bar H" X }i" when exposed to tensile stress, the ultimate tensile strength of the iron being 55,000 pounds per square inch, as given in Table No. 25, page 216 ? Solution : A bar y&" X }i" is g 9 T square inches in area. P = fa X 55,000 = 7734 pounds, which will break the bar. In order to obtain the safe working stress introduce a suit- able factor of safety, from 5 to 10, according to circumstances, and calculate by the following formulas : a y, c Side of a square bar = -\i — _/ / a — P X f Diameter of a round bar == \ *^— A s~ ^0.7854 S P = Load in pounds. / = Factor of safety. Example. A load of 24,000 pounds is suspended on a round wrought iron bar. The ultimate tensile strength of the iron is 55,000 pounds per square inch. What should be the diameter of the bar to sustain the load, with 10 as the factor of safety ? Solution: A s - A = ?4000_X10 = 4.363 square inches. 55000 In Table No. 24, we find the nearest larger diameter to be '2yi inches. The diameter may also be calculated directly by the fol- lowing formula : 2l8 STRENGTH OF MATERIALS. =v p xf 6- X 0.7854 » = t 24000 X 10 55000 X 0.7854 P — Vl^Q D = 2.358, or nearly 2^ inches diameter. To Find the Diameter of a Bolt to Resist a Given Load. Rule. Multiply pull in pounds by the factor of safety. Multiply the ultimate tensile strength of the material by 0.7854 ; divide this first product by the last and extract the square root from the quotient which will then be diameter of bolt at the bottom of the thread. D = {~~ ~' P = f = S X 0.7854 D 2 X S X 0.7854 f B 2 X S X 0.7854 P D = Diameter of bolt or screw in the bottom of the thread. P = Load or pull in pounds. f= Factor of safety. •S" = Ultimate tensile strength per square inch. 0.7854 is constant = — 4 Note. — Bolts are frequently exposed to a considerable amount of initial stress, due to the tightening of nuts, which must always be allowed for when deciding upon the load to be considered when calculating their diameter. Example. Find diameter of a bolt to sustain a load of 4,450 pounds, taking 10 as factor of safety and ultimate tensile strength of the iron to be 50,000 pounds per square inch. Solution : , 4450 X 10 50000 X 0.7854 D = 1.064" in the bottom of thread ; thus, a l^" screw, standard thread, which is lj 1 ^ inches in diameter at the bottom of thread, will be the bolt to use. STRENGTH OF MATERIALS. 219 Example 2. What size of bolt is required to sustain the same load as is mentioned in the previous example, if only 5 is wanted as a factor of safety? Solution : * = ^ 4450 X 5 50000 X 0.7854 D = Vo.567 J) = 0.75 inch diameter in bottom of thread. Thus a ^-inch standard screw is too small, as that is only ff" in bottom of thread, but a 1-inch standard screw is suffi- cient, being f f " in bottom of thread. To Find the Thickness of a Cylinder to Resist a Given Pressure. When the walls of cylinders are thin in proportion to their diameters use the formula : p — S X t R x/ f= P XPXf P = S S X / P x/ p = Pressure per square inch. P = Radius of cylinder in inches. / = Thickness of cylinder wall in inches. f= Factor of safety. S = Ultimate tensile strength of material. When cylinder walls are thick in proportion to the diameter, such as hydraulic cylinders, their thickness is usually figured by the formula: P X P t = Thickness of cylinder wall in inches. P = Pressure in pounds per square inch. P = Radius of cylinder. S = Ultimate tensile strength. f= Factor of safety. Example. Find necessary thickness of a hydraulic cylinder of 10-inch inside diameter, made from cast-iron, to stand a pressure of 1000 2 20 STRENGTH OF MATERIALS. pounds per square inch, with 4 as factor of safety. The ulti- mate tensile strength of the iron is, by experiments, found to be 20,000 pounds per square inch. ( See Table No. 25 ). Solution : 10-inch diameter = 5-inch radius. 1000 X 5 / = 20000 — 4 — — 1000 1000 X 5 5000 — 1000 5000 4000 t = IX inch. Strength of Flat Cylinder Heads. The American Machinist, in Question No 147, March 22, 1894, gives the following formula for flat circular heads firmly fixed to the flange of the cylinder : t _ J 2 X r 2 X P ^ 3 X Si t = Thickness of cylinder head in inches, r = Radius of cylinder head in inches. P = Pressure in pounds per square inch. .Si = Allowable working stress in the material. The allowable working stress may be taken as i to T ^ of the ultimate tensile strength and may, for cast iron, be from 1500 to 2500 and for wrought iron from 4000 to 6000. The above formula was used to calculate the thickness of a cast iron cylinder head of 30 inches diameter, to resist a pressure of 100 pounds per square inch. This formula is in that case con- sidered to give sufficient thickness, so that no ribs or braces are needed. The above formula may also be used for wrought iron, by selecting the proper value for Si. Assuming the tensile strength of wrought iron to be 44,000 pounds, and allowing a factor of safety of 8, the value of Si for wrought iron will be 5500. Strength of Dished Cylinder Heads. The American Machinist, in Question 183, April 12, 1894, gives the following formula for dished circular heads, firmly fixed to the flanges of the cylinder : t = P * (P 2 + d 2 ) 4X SiXd STRENGTH OF MATERIALS. 221 /= Thickness of cylinder head in inches. P = Radius of cylinder head in inches. P = Pressure in pounds per square inch. d = Depth in inches of dishing of the head at its center. Si = Allowable working stress in the material, which may be the same as given above. This formula was used to calculate the thickness of a cast iron head 44 inches in diameter, dished 7 inches, steam pressure 75 pounds per square inch. Note. — In these examples the radius of the bolt circle should be considered as the radius of the head when calculat- ing the thickness. The diameter of the bolts ought to be so large that the strain on the bolts will not exceed 3000 to 5000 pounds per square inch calculating the area of the bolt at the bottom of the thread, and the distance from center to center of the bolts ought not to exceed six times their diameter. The above formula may also be used for dished cylinder heads of wrought iron or steel, by allowing the proper value for Si. For soft steel Si may be 9000 to 12,000 pounds, and for wrought iron 5000 to 8000 pounds. Caution. — Cast iron is not a desirable material to use for large unribbed cylinder heads; either flat or dished wrought iron or steel is far superior. Strength of a Hollow Sphere Exposed to Internal Pressure. d 2 7r The pressure acts on a surface equal to — t— and it is re- D + d sisted by a metal area equal to — ~ — X tt X A D = External diameter. d = Internal diameter. / = Thickness of metal. When the difference between inside and outside diameter is small it need not be considered in practice, and the formula will be : d*TT P X — 4— = dX 7v X/X^i which reduces to „ 4X/X^i P X d J '^—d t = -±sT Si — Allowable tensile stress in the material. Note. — This formula only allows for tensile strength ; if it is used for calculating the thickness of the body of a globe valve or anything similar a liberal amount of metal must be added, in order to obtain good results when casting. 222 STRENGTH OF .MATERIALS. Strength of Chains. The following table gives approximately the weight of wrought iron chains, in pounds per foot and kilograms per meter ; and also their strength, with six as factor of safety. Chains ought to be tested with twice the load given in the table. Never lose sight of the fact that a chain in use will wear and consequently become reduced in strength; also, that a chain is no stronger than its weakest link. Diameter Load Weight in Load Weight in Kilo- in Inches. in Pounds. Per Foot. in Kilograms. grams per Meter. _3_ 280 0.42 125 0.625 X 500 0.91 225 1.35 H 1125 1.5 510 2.22 % 2000 2.5 900 3.72 % 4500 5.8 3050 8.63 l 8000 10 3600 14.88 Strength of Iron Wire Rope. The following table gives approximately the strength of iron wire rope, with six as a factor of safety. Diameter in Inches. Load in Pounds. Load in Kilograms. n 1000 2500 3500 453 1134 1588 Wire ropes should not be bent over pulleys of very small diameter. When used for hoisting, the diameter of pulley ought at least to be 40 times the diameter of rope. For further information on wire rope see manufacturers' catalogues. Strength of Manila Rope. The size of manila rope is measured by the circumference, therefore so-called three-inch rope is about one inch in diameter. New manila rope of three inches circumference will usually break for a load of 7,000 to 9,000 pounds. For common use such ropes may be loaded as given in the following table, and the diameter of the pulley ought to be at least eight times the diameter of the rope. Size of Rope. Safe Load in Pounds. Safe Load in Kilograms. 3 ins. circumference. 4 « 5 " " 500 800 1300 227 363 590 STRENGTH OF MATERIALS. CRUSHING STRENGTH. Short posts having square ends well fitted may be considered to give away under pure crushing stress. Their strength is in proportion to their area; therefore, when the length of a post does not exceed four to five times its diameter or smallest side, its strength or size may be calculated by the following formulas : Side of a square post = ^^^ f s VP X f — 0.7854 S S * 0.7854 S P = Safe load in pounds to be supported by the post. A = Area of post in square inches. S = Ultimate crushing strength of the material in pounds per square inch, given in Table No. 26. f = Factor of safety. TABLE No. 26. — Modulus of Elasticity and Ultimate Crushing Strength of Various Materials. Modulus of Ultimate Modulus of Ultimate Elasticity Crushing Elasticity in Crushing Materials. in Pounds Strength in Kilograms per Square Strength in per Square Pounds per Kilograms per Sq. Centimeter Inch. Square Inch Centimeter. Cast steel .... 30,000,000 150,000 2,200,000 10,500 Bessemer steel . . 28,000,000 50,000 45,000 1,970,000 3,500 3,000 Wrought iron . < ( 25,000,000 12,000,000 to 50,000 1,700,000 800,000 to 3,500 Cast iron . . . < to 15,000,000 90,000 to 1,000,000 6,300 Oak (endwise) 1,500,000 9,000 105,000 630 Pitch pine " 1,600,000 9,000 112,000 630 Pine " 1,100,000 6,000 77,000 420 Spruce " 11,000,000 6,000 77,000 420 Brick ...... -j 800 to 2,000 56 to 140 Brick work laid } in 1 part cement > 600 42 and 3 parts sand ) - Brick work laid ) in lime and sand ) 240 16 to 17 Granite 10,000 700 224 STRENGTH OF MATERIALS. When a post or column is long compared to its diameter, its strength will decrease as the length is increased. Anyone will, from every-day observation, know that a short post will support with perfect safety a load which will break a long one. Short columns break under crushing, but long ones break under comparatively light load by the combined effect due to both crushing and flexure. It is, therefore, evident that the strength of long columns follows laws very different from those which apply to short ones. The form of the ends has also F|G# ^ . great influence on the strength of a column when under crushing and deflective stress. ( See Fig. 1). When both ends are round the column has least strength; if one end is round and one end flat it is stronger, but if both ends are flat and square with the center-line, it is strongest. The proportions are ap- proximately as 1, 2 and 3. Eccentric loading on columns will also have a very destruct- ive effect upon their strength. Theoretical calculations regarding the strength of columns and posts are difficult, and such empirical formulas as the well-known Hodgkingson's or Gordon's formulas are usually re- sorted to. The Hodgkingson formulas for long columns having square ends well fitted are : •<^4&y v&%%&> dm^/ n3-55 P = 99,000 X tL— for solid cast iron columns. P — 99,000 X /)3-55 d 3-55 L for hollow cast iron columns. P— 285,000 X D'- for solid wrought iron columns. P = Breaking load in pounds. D = External diameter in inches. d = Internal diameter in inches. L = Length in feet. When the breaking load as calculated by these formulas exceeds one quarter of the crushing load of a short column of the same metal area, the result must be corrected by the for- mula: STRENGTH OF MATERIALS. 225 PX C ^ 1 ~ P + %CX A Pi = Corrected breaking load of column. C= Crushing strength of material (see Table No. 26). A = Metal area of column in square inches. Important. — Applying the last formula, the result, Pi, must always be smaller than P. Table No. 27 was calculated by the following formulas : ( 36000 ) Column I. Safe Load = 0.1 X < , J 2 \ ( I 1 + (^ X 0.00025) J t 36000 X 0.07031 ^ Column II. Safe Load = 0.1 X < ^ \/ L 2 x o 00025 \ 36000 Column III. Safe Load = 0.1 X J. , Z 2 x 0004 D 2 ' " ) 36000 X 0.07031 Column IV. Safe Load = 0.1X I 1 , L 2 x qq^x 80000 Column V. Safe Load = 0.1 X < \ _f_ r^_ x 0.0025 ) ( ^ D 2 80000 X 0.07031 j Column VI. Safe Load = 0.1 X j i -f ( J^ 2 . x 0.0025 \ ( ( 80000 Column VII. Safe Load = 0.1 X ] ; J% 1 +(^X 0.0035) ( 80000 X 0.07031 Column VIII. Safe Load = 0.1 X \ ' 72 x / 1 + (±- X 0.0035 ) D 2 5000 Column IX. Safe Load = 0.1 X \ -. . , L 2 w A AA , x I 2 + (^ X0 - 004 ) S 1 5000 X 0.07031 Column X. Safe Load = 0.1 X ] -± \ / j^l. x 004 "i ( 226 STRENGTH OF MATERIALS. TABLE No. 27.— Safe Load on Pillars Having Square Ends Well Fitted. (10 is used as Factor of Safety). I « . -o „4> •J ft Xi o t& c u Wrought Iron. Cast Iron. Wood. (Spruce or white Pine) 4) V C .2 6 "OT3 ir. rgJJ 13 "« > s ■5c to G 4) Hollow Pillar Solid Pillar. Hollow Pillar. Solid Pillar. o e u »- rt V 3 ft CT 1 m c/a 13 3 O Ph « IS x\ o 4> 4> 3 ft a- T3 C 3 4) 4> Jjjj in u 6^ gg- '3 J? e 4> Jh « ft o* T3 C 3 O Ph 3 V u S i<3 g « ^| Sen c 4> u « ft& men 13 C 3 £ u 4) g « as 1 73 > Safe Load = 0.1 X 1+ /^ x 0.0025) and the result obtained reduced to long tons (2240 pounds). Ten is thus used as a factor of safety ; both ends of the pillar are supposed to be square and evenly loaded. For other shapes of ends, mode of loading, or other factors of safety, propor- tional allowance must be made. For instance, if 15 is required as factor of safety, allow only two-thirds of the load given in the table. If the pillar has only one square end and one round end, allow only two-thirds as much load. If it has both ends rounded, or, which is the same, if the ends have only a very imperfect bearing, allow only one-third as much load. Weight of Cast=Iron Pillars. The weight of a cast-iron pillar may be calculated by the formula : lV=(D 2 — d 2 ) X L X 2.45 W = Weight of pillar in pounds. D = Outside diameter in inches. d = Inside diameter in inches. L = Length of pillar in feet. The weight given in Table No. 28 was calculated by this formula, and the length taken as one foot. STRENGTH OF MATERIALS. 229 TABLE No. 28.— Safe Load on Round Cast-Iron Pillars. a v re"* c.c 'o.c 8.S re 01 (/) £ V 2 Cfl Weight in Pounds per Foot in Length. Length of Pillars in Feet. 4> 00 t3 fa V 1 fa u fa CO fa 00 V fa V fa 83 4 >2 5.49 17.14 11 8.1 6.1 4 % 7.56 23.90 15.2 11.3 8.5 5 X 7.07 22.06 16.8 13.3 10.4 8.3 5 X 10.01 31.23 24 19 15 12 6 X 8.64 26.95 23 19 15.5 12.6 10.4 6 X 12.37 38.59 33 27 22 18 15 6 n 14.09 43.96 37 31 25 21 17 6 1 15.71 49.01 42 35 28 23 19 6 1/ 17.23 53.76 47 40 32 26 22 7 x 12.52 39.06 36 31 26 22 19. 16 7 x 14.73 45.96 42 36 31 26 22 19 7 H 16.84 52.54 48 41 35 29 25 21 7 18.85 58.90 54 46 39 33 29 24 7 IX 20.76 64.77 60 52 44 37 32 27 8 X 17-08 53.29 51 45 39 34 29 25 22 8 n 19.59 61.12 59 52 45 39 34 29 25 8 21.99 68.64 66 58 51 44 38 33 28 8 1/ 24.30 75.82 73 64 56 48 42 36 31 8 IX 26.51 82.71 79 70 61 52 45 39 34 8 i/s 28.62 89.29 86 76 66 57 48 42 37 9 x 19.44 60.65 60 54 49 43 37 33 29 24 9 n 22.33 69.67 69 63 56 49 43 38 33 29 9 22.13 78.40 78 71 63 55 48 42 37 33 9 iy 8 27.83 86.83 87 78 69 62 53 47 41 36 9 IX 30.43 94.94 95 85 76 67 58 51 45 39 9 ix 32.94 102.77 102 92 82 72 63 55 48 43 9 1/ 35.34 110.26 110 99 88 78 68 59 52 46 9 IX 39.86 124.36 126 113 100 90 78 67 60 51 10 r 8 25.09 78.28 80 73 67 60 53 47 42 37 34 10 28.28 88.23 90 83 75 67 60 53 47 42 38 10 *x 31.37 97.87 100 92 83 74 66 58 52 47 42 10 IX 34.37 107.23 110 101 91 82 73 64 57 51 47 10 1/8 37.26 116.25 119 109 98 88 79 69 62 55 51 10 1# 40.06 124.99 128 117 106 95 85 75 67 59 54 10 IX 45.36 141.52 146 133 122 109 97 85 77 67 60 11 31.42 98.04 102 95 87 79 71 64 58 52 48 43 11 1/8 34.90 108.89 114 105 96 88 79 71 64 58 53 48 11 IX 38.29 119.46 125 116 106 97 87 78 70 63 58 52 11 1/8 41.58 129.73 135 126 115 105 94 85 76 68 62 56 11 1# 44.77 139.68 146 136 124 113 102 92 82 74 68 61 11 IX 50.86 158.68 166 156 142 129 178 106 94 86 79 71 11 2 56.55 176.44 186 176 160 147 134 120 106 98 90 81 230 STRENGTH OF MATERIALS. TABLE No. 28. — (Continued). $ 2 a 53 Length of Pillars in Feet. 73 c 13 4;.- |.a| ■4-> O-fj tils i - ^j « *j "5 V *j u ti "5 12 '£ "SI 1 4> ^il 2 # *r J CO 00 fa fa fa 00 O §3 34.46 107.51 115 107 99 92 83 76 68 62 58 53 12 iH 38.34 119.62 128 119 108 102 92 84 78 69 63 58 12 1% 42.12 131.41 141 131 119 112 101 93 84 76 70 64 12 in 45.80 142.90 153 142 129 121 110 101 91 82 75 69 12 in 49.39 154.10 165 154 139 131 119 109 99 89 82 75 12 in 56.26 175.53 189 178 159 150 137 125 115 103 94 85 12 2 62.74 195.75 213 201 179 170 155 141 131 117 106 96 13 1 37.70 117.53 127 119 111 104 97 87 79 73 67 61 13 in 41.94 130.85 142 134 126 117 109 101 91 82 75 69 13 ix 46.11 143.86 158 149 140 130 121 112 101 91 84 77 13 in 50.19 156.59 174 163 154 144 133 122 111 101 93 84 13 in 54.16 168.98 190 178 168 157 145 133 121 110 101 92 13 in 61.82 192.88 214 201 189 176 164 151 137 124 114 104 13 2 69.09 215.56 227 224 210 195 182 168 152 137 126 116 14 1 40.94 127.60 138 131 123 115 109 101 92 85 78 72 14 in 45.50 141.96 153 145 139 130 121 112 103 94 87 80 14 IX 50.07 156.31 168 160 153 143 133 123 113 104 95 88 14 i/ 8 54.54 170.04 183 174 167 156 145 134 123 113 104 95 14 i# 58.78 1S3.67 198 189 180 168 156 145 133 122 112 103 14 IX 67.31 210.00 226 216 206 192 179 166 152 140 128 118 14 2 75.36 235.12 254 242 232 216 201 186 171 157 143 133 15 1 43.99 137.28 150 143 136 128 120 112 104 96 90 82 15 i# 49.04 153.19 167 159 152 143 136 125 116 108 100 93 15 IX 54.00 168.48 184 175 167 157 148 138 128 119 110 102 15 1# 58.85 183.77 201 191 182 172 161 151 140 130 120 111 15 in 63.62 198.74 217 207 197 186 175 163 151 141 130 120 15 IX 72.85 227.45 284 236 225 212 202 190 175 160 148 137 15 2 81.68 254.81 279 265 253 237 2291216 197 179 166 154 16 1 47.12 146.95 160 156 150 140 133 125 117 110 101 93 16 1^ 52.57 164.11 179 173 167 158 148 139 130 122 116 106 16 IX 57.92 180.65 198 191 185 174 163 153 143 135 128 117 16 i# 63.18 197.18 216 209 202 190 178 167 152 147 139 128 16 ik 68.33 213.10 234 227 220 206 193 181 170 160 151 138 16 IX 78.34 244.29 26S 258 250 237 224 207 199 183 173 158 16 2 87.96 274.56 301 290 281 267 255 233 218 206 194 177 The length of cast-iron pillars, as a rule, ought not to ex- ceed 20 to 25 times their diameter. Cast-iron pillars, when heavily loaded, are apt to be broken if struck by a blow sidewise. STRENGTH OF MATERIALS. 231 Wrought Iron Pillars. In important work, cast-iron pillars are rapidly going out of use. Wrought iron pillars are now made which compare favorably in price and are far more reliable than those of cast iron. For full information regarding weight and strength of wrought iron pillars and Z bar columns, see manufacturers' catalogues. Wooden Posts. Table No. 29 was calculated by the formula : r 5000 X Area Safe load = 0.1 X ) , / Z 2 nnnA I 1 + (^rX 0.004 and the result divided by 2240. L = Length of post in inches. D = Side of post in inches. TABLE No. 29.— Safe Load in Tons on Square Pine or Spruce Posts Having Square Ends Well Fitted. (10 as Factor of Safety). )! Side of Post in Inches. 1 in. 2 in. 3 in. 4 in. 5 in. 6 in. 7 in. 8 in. 9 in. 10 in. 12 in. 1 2 3 4 5 6 7 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 0.1 0.0 0.0 4 6 1 | 1 0.7 0.5 0.3 0.3 0.1 0.1 8 6 9 2 9 4 1.89 1.59 1.27 0.99 0.77 0.60 0.49 0.39 0.27 3.45 3.12 2.70 2.27 1.8S 1.56 1.29 1.08 0.78 0.58 0.44 5.00 5.00 4.62 4.08 3.54 3.05 2.62 2.25 1.69 1.29 1.00 0.90 0.80 8.0 8.C 7.C 6.3 5.7 5.0 4.5 4.0 3.0 2.4 1.9 1.5 1.3 1.0 0.9 8 4 9 3 4 8 9 10.00 10.00 10.00 9.21 8.45 7.6S 6.94 6.24 5.03 4.06 3.31 2.73 2.27 1.92 1.63 1.41 1.22 12.50 12.50 12.50 12.50 11.21 10.80 9.91 9.08 7.53 6.25 5.19 4.34 3.66 3.11 2.66 2.31 2.01 1.78 1.56 16.00 16.00 16.00 16.00 15.35 14.35 13.41 12.44 10.58 8.96 7.57 6.43 5.49 4.72 4.17 3.55 3.11 2.68 2.38 2.18 1.96 20.00 20.00 20.00 20.00 19.51 18.44 17.40 16.32 14.18 12.22 10.50 9.04 7.80 6.76 5.89 5.17 4.56 4.05 3.56 3.23 2.91 2.64 2.39 30.00 30.00 30.00 30.00 29.22 28.10 26.87 25.50 22.95 20.40 18.02 15.88 14.00 12.36 10.95 9.73 8.6S 7.77 6.99 6.30 5.71 5.18 4.74 4.35 232 STRENGTH OF MATERIALS. The preceding Table gives the safe load in long tons corre- sponding to a square post of the dimensions of sides given at top of the columns, and lengths given in the first column. For round posts the load should be 0.75 to 0.6 of the given load de- pending upon the length of post. Example. What size of post is required, with 10 as factor of safety, to support a load of five tons, when the length of the post is 16 feet? Solution : In the column headed "Length of post in feet" find 16, and in line with 16 find the numbers nearest to five tons, which are 4.34 and 6.43. Thus, a post 16 feet long and 8 inches square will support 4.34 tons, and a post 16 feet long and 9 inches square will support 6.43 tons. It is, therefore, best to select a post 9 inches square. To Calculate the Strength of Rectangular Posts from the Table. Find, in the Table, the strength of the post according to its smallest side, and increase the tabular value in proportion to the largest side of the post. Example. What is the strength, with 10 as factor of safety, of a spruce post 10 feet long, 6 inches thick, %]/ 2 inches wide, with square ends well fitted, calculated by Table No. 29. Solution : In the Table we find the strength of a post 10 feet long and 6 inches square to be 3.09 tons. Therefore, when the pillar is 6 inches thick and 8% inches wide its corresponding strength will be 3.09 X ^f = 4.38 tons. It is a waste of material to use a post of rectangular cross- section. For example, this post is 6 X 8% inches = 51 square inches of cross-section and will support 4.38 tons, but a post of the same length and 7X7 inches = 49 square inches of cross- section, will support 5.03 tons. ( See Table No. 29). To Obtain the Weight of Pillars in Kilograms per Meter when the Weight in Pounds per Foot is Known. Multiply the weight in pounds per foot by the constant 1.4882, and the product is the weight in kilograms per meter. STRENGTH OF MATERIALS. 233 TRANSVERSE STRENGTH. A beam placed in a horizontal position, fastened at one end and loaded at the other, is exposed to transverse stress, and will usually bend more or less, as shown (exaggerated) in Fig. 1, before it will break. The line F|G - f * a b is called the neutral line, and all ,_____^ ., ^ fibres above the neutral line are exposed " = :^^5^"llj~ to tensile stress, and all fibers below ~ *"~*^\'x^N v are exposed to crushing stress, but the \*Y neutral fiber is neither stretched nor y M^ . compressed. A line drawn in a hori- / \ zontal direction, at right angles to, and ^ ^ through the neutral line, is called the neutral axis with reference to this particular place of the section of the beam. The neutral axis is considered to pass through the center of gravity of the section, which, for beams of round, square or rectangular section, is always in the geometrical cen- ter. Therefore, all beams of such section will have an equal amount of material on the upper and under side of the neutral axis, but it is not always desirable for all materials or for all kinds of load to have an equal amount of material on both the side exposed to compression and that exposed to tension. For instance, cast-iron beams are usually made in T formed sec- tion and should always be laid so that the largest web is exposed _ to tensile stress, because cast-iron offers much more resistance to compression than it does to tension. Cast- iron beams of such section ought, therefore, to be laid in this position (T), if fastened at one end and loaded at the other, but should be laid in this position ( JL ), if they are supported under both ends and loaded between the supports. If this is taken into consideration in placing a cast-iron beam, its ultimate transverse breaking strength is greatly increased, but under a moderate load the deflection will be practically equal in either position, because as long as the load is small, well within the elastic limit, cast-iron will stretch under tensile stress as much as it will compress under an equal amount of crushing stress; therefore, the modulus of elasticity for tension and compression of cast-iron is considered to be equal, but under increased crushing load the compression becomes less in pro- portion to the load until the point is reached when the cast-iron can not compress more, and the casting will break. The ultimate crushing strength of cast-iron is five to six times as much as its ultimate tensile strength. A beam supported under both ends and loaded in the middle will carry four times as great a load as another beam of the same size and material fixed at one end and loaded at the 234 STRENGTH OF MATERIALS. ■ lfoot. 1 25 lbs. Fig. 2. 1 n ... \m&> r -\ y&m* 100 lbs. LA 100 lbs. Fig. 3. L 50 lbs. other. This may be understood by referring to Fig. 2, as when the beam is one foot long and loaded with 100 pounds in the middle, each half of the beam supports only 50 pounds, and this 50 pounds acts only upon an arm y 2 foot long, consequently it exerts no more force toward break- ing this beam than the 25 pounds would upon the end of the other beam one foot long. A beam twice as wide as another and of the same length, thickness, and material, will carry twice the load, because the wide beam could, of course, be split into two equal beams ; consequently it must, as a whole , beam, have twice the strength of \~ i* 2 {eet — \ another one of the same material but of only half the width. A beam twice as long as another will break under half the load. This is seen by referring to Fig. 3, be- cause 50 pounds on an arm two feet long will balance 100 pounds on an arm one foot long. A beam twice as thick as another, of the same material, length and width, will carry four times the load. ( See Fig. 4). Suppose the weight a is act- ing on the arm b, tending to swing it around the center c, and this action being counteracted by the weights g and k, also by the arrows e and f. If the weights is taking hold twice as far from the center as the weighty it will offer twice the resistance against swinging the beam that g will ; and exactly the same with the arrows/" and e. Consider the line c b as the neutral fiber, the arrows e and f as representing the fibers resisting crushing, and the weigh ts^- and h as representing the fibers resisting tensile stress. It will be understood that if the fibers are twice as far above or below the neutral fiber they are in a position to orTer twice the resistance to the breaking action of the load ; but a beam of twice the thickness has not only its average fiber twice as far from the neutral point, but it has also twice the area or twice as many fibers, consequently the result must be that it can resist four times the load. Fig. 4. o ~o o "Q OA STRENGTH OF MATERIALS. 235 For instance : The beam a in Fig- ure 5 is four times as strong as the beam b, if placed on the edge, as shown in the figure, and loaded on the top ; but a would be only twice as strong as b if it was laid on the side and loaded on top. U.4in.„ Fig. 5. 1 W JS1 h-4in. -h 00 | a b Formulas and Rules for Calculating Transverse Strength of Beams. The fundamental formula for transverse stress in beams is: Bending Moment = Resisting Moment. The bending moment for a beam fixed at one end and loaded at the other (see Fig. 1) is obtained by multiplying the load by the horizontal distance from the neutral axis to the point where the load is applied. The distance is taken in inches and the load in pounds. The resisting moment is obtained by multiplying the mo- ment of inertia by the unit stress, tensile or compressive, upon the fiber most remote from the neutral axis, and dividing the product by the distance from this fiber to the neutral axis. The theoretical formula for the transverse strength of a beam fastened in a horizontal position at one end and loaded at the extremity of the other end, as shown in Fig. 6, is, SX / P = L X a When the beam is fastened at one end and loaded evenly throughout its whole length, as shown in Fig. 7, the formula will be, SX I P = 2 X LX a When a beam is placed in a horizontal position and sup- ported under both ends and loaded in the middle (see Fig. 8) the formula is, / = 4X f X l L X a When a beam is placed in a horizontal position and sup- ported under both ends and loaded throughout its whole length (see Fig. 9), the formula will be, S X I P=S X LX a 236 STRENGTH OF MATERIALS. When a beam is laid in a horizontal position, fixed at both ends and loaded in the middle between fastenings (see Fig. 10), the formula will be, P = 8 X ^ L X a When a beam is laid in a horizontal position, fixed at both ends and the load evenly distributed over its whole length (see Fig. 11), the formula will be, P = 12 X S X l L X a P = Breaking load in pounds. S =. Modulus of rupture, which is 72 times the weight, in pounds, which will break a beam one inch square and one foot long when fixed in a horizontal position, as shown in Fig. 6, and loaded at the extreme end, and which may be taken as follows : Cast-iron, 36,000. Wrought Iron, 50,000. Spruce and Pine, 9,000. Pitch Pine, 10,000. These are the nearest values, in round numbers, of 72 times the average value of the constant given in Table No. 30. For the safe load, .S" may be taken as follows : For timber, 1,000 to 1,200 pounds. For cast-iron, 3,000 to 5,000 pounds. For wrought iron, 10,000 to 12,000 pounds. For steel, 12,000 to 20,000 pounds. L = Length of beam in inches. a = The distance in inches from the neutral surface of the section to the most strained fiber. / = Rectangular moment of inertia. The tables on pages 237 and 238 give the moment of inertia about the neutral axis X V, and the distance a, for a few of the most common sections : (For explanation of moment of inertia and center of gravity see page 293). These formulas have the great advantage of being theoretic- ally correct for beams of any shape of cross-section, made from any material, providing the load is within the elastic limit of the beam, and a correct constant is used for .5* and the correct value obtained for the moment of inertia. STRENGTH OF MATERIALS. 237 / B x {H*—h?) a ~ 6 x H 1 = I H 2 x B \ H x V2 O.US// a — £ ^iwMMb?' \ / a = I . 1 1 H B vj _Bxf/* 36 %#; *l = _ ^ x // 2 VzH I _ 12 _BxH* 24 _±_ = 0.0982 x £>3 a I = D x D* 64 £> 0.09S2 £>! £> 2 7 _ {DiD^—did*)^ 64 # = 2 a ~ 32 £> 238 STRENGTH OF MATERIALS. -4ftHK&f- X T1 6 i J -Y :? I — I* B— B B- 12 l- & - 1 -^B Y X 7 = B B 3 + b h* 12 a\ B B 2 — 2Bbh + bh 2 2 (B H — b h) BB 2 — bh 2 2(B H—bh) I _ _{B B 2 - -bh 2 ) 2 — ±BBbk(B- -h) 2 (B B 2 - 12 (B H — bh) _ b k 2 ) 2 — \BBbh{B — h) 2 I _ AB B 2 - B 2 — 2Bbk + bh 2 ) -bh 2 Y — ±BBbh(B- -h) 2 a 6(B B 2 — b h 2 ) STRENGTH OF MATERIALS. 239 Beams of symmetrical section, as square, round, elliptical, or H section, may be calculated on theoretically correct prin- ciples in a simpler way, obviating the use of the moment of inertia and the modulus of rupture, as explained below. For a beam fixed at one end and loaded at the other, CxH 2 XB Fig. 6. H =v P XL CX B . P X L CXH 2 P X L When beam is round, Diameter = s -« / P X L C X 0.589 When beam is square, Side = S J PXL P = Breaking load in pounds. H — Thickness or height of beam in inches. B = Width of beam in inches. L = Length of beam in feet. C = Constant which is obtained from experiments, and is the weight in pounds which will break a beam 1 foot long and 1 inch square fixed at one end and loaded at the other. Con- stant C is given in Table No. 30. A rectangular beam fixed at one end and loaded evenly throughout its whole length will carry twice the load of a beam fixed at one end and loaded at the other; therefore, p _ 2 X C X H 2 X B For a rectangular beam supported under both ends and loaded at the center, fig. 8. 4X CX H 2 X B A rectangular beam supported under both ends and loaded evenly throughout its whole length will carry twice the load of 240 STRENGTH OF MATERIALS. a beam supported under both ends and loaded at the center; therefore, Fig. 9. 8 X CX H*X B For a beam fixed at both ends and loaded at the center, P = 8 X CX H* X B For a beam fixed at both ends and the load distributed evenly throughout its whole length, P = 12 X C X H* X B Each letter in these formulas has the same meaning as in formula for Fig. 6, page 239, and each formula may be transposed the same as that formula. The most convenient way is, in each case, to multiply the numerical value of Cfrom Table No. 30, by its proper coefficient according to mode of loading, before it is inserted in the formula. Note. — A square beam laid in this ■ position has 40 % more transverse strength than the same beam laid in this ♦ position. TABLE No. 30. — Constant C. Giving the weight in pounds which will break a beam one foot long and one inch square which is fastened at one end, in a horizontal position, and loaded at the other end. Material. Very Good. Medium. Poor. Wrought iron,* Cast-iron, Spruce and Pine, Pitch pine, Granite, 750 650 160 225 600 500 125 150 25 500 400 90 100 * A wrought iron beam or bar will not actually break under these conditions, but, as it will bend so much that it becomes useless, it is considered to be equivalent to the breaking point. STRENGTH OF MATERIALS. 241 The following formulas will apply to the strength of beams of the shape shown in the adjacent sectional cuts. These for- mulas pertain only to the ultimate breaking strength of beams, and have nothing to do with deflection, which follows entirely different laws. SOLID SQUARE BEAMS. P = Cxlf* L CXH* Fig. 12. c= PXL HOLLOW SQUARE BEAMS. p= CX {H* — h*) LXH Fig. 13. jj P — L = SOLID RECTANGULAR BEAMS CX B X If 2 FIG. 14 L m CX Bx H 2 " C X B 7 B = %~L B "I PXL CX H 2 PXL B X H 2 HOLLOW RECTANGULAR BEAMS. P= CX ( B X H * — b X HXL r = CX(BXH*—bXh*) P X If SOLID ROUND BEAMS. p _ 0.589 C X D z fig. 16. Fig. 15. Z = 0.589 CX D* D=J PXL * 0.589 C 242 STRENGTH OF MATERIALS. HOLLOW ROUND BEAMS. p _ 0.589 CX(D* — d*) Z X D SOLID ELLIPTICAL OR OVAL BEAMS. p _ 0.589 C X Dx X D 2 L Fig. 19. FIG. 20. •\^ feet long and 5 inches in diameter, next a beam 3 3 feet long (the distance from P to i?).and 4 inches in diameter. Solving for strength at A : Z> 3 0.6 C P = P = L 5 3 X 0.6 X 600 4^ P = 10,000 pounds. Solving for strength at B : p __ 4 3 X 0.6 X 600 P = 64 X P = 7680 pounds. Thus, the weakest point of the beam is at B, where its cal- culated breaking load is only 7860 pounds, while the calculated breaking load at A is 10,000 pounds. When a beam is not of uniform section throughout its whole length and is supported under both ends and loaded somewhere between the supports, calculate first the reaction on each support ; then consider the beam as if it was fastened by the load, and consider the reaction at each support as a load at the free end of a beam of length and section equal to the length and section between its load and support. Example. The largest diameter of a round cast-iron shaft is 3 inches and the smallest diameter is 2 inches. The length, mode of 244 STRENGTH OF MATERIALS. loading and support is as shown in Fig. 22. Where will it break ? and what is the breaking load ? FIG. 22. \ Solution : The reaction at y will be }i of the load and the reaction at x will be ft of the load. The beam will evidently break either at a, b or n. (Find constant C in Table No. 30 and. multiply by 0.6, because the beam is round). Solving for strength at n : 2 3 X 500 X 0.G H P = P = 3 8 X 300 X 8 3X3 P = 213Sy 3 pounds. Solving for strength at b : 3 3 X 500 X 0.0 P 5 27 X 300 X I 5X3 P = 4320 pounds. Solving for strength at a : 2 3 X 500 X 0.0 P 2 p _ 8 X 300 X 8 5X2 P = 1920 pounds. Thus, the weakest place in the beam is at a, where it will break when loaded at b with 1920 pounds. If the load is moved nearer u, it will at a certain point exert the same breaking stress on both a and n. Regular beams of this kind are seldom dealt with, but shafts or spindles of similar shape and loaded in a similar manner are frequently used, and their strength and stiffness may be calcu- lated and their weakest spot ascertained by this way of reason- ing, which applies as well to hollow as to solid shafts and spindles made from wrought iron, steel or cast-iron. STRENGTH OF MATERIALS. 245 Beams fastened at one end and loaded at the other may be reduced in size toward the loaded end and still have the same strength. Suppose the beam to be fastened in the wall at X (Fig. 23) and loaded at the other end with a given load, this load will then have the greatest breaking effect upon the beam at X; at half way between X and d fig. 23. the load has only half the breaking effect, at c only one-quarter the effect. Therefore, the beam may be tapered off toward b in such proportion that the square of the height a is equal to three-quarters the square of the height at X. The square of the thickness at b is one-half the square of the thickness at X, and the square of the thickness at c is one-quarter the square of the height at X. Example. An iron bracket is four feet long, projecting from a wall (as Fig. 23). It is strong enough when 24 inches high at X. How high will it have to be at a, b and c? Solution : X = 24 2 = 576 Height at X = 24" a = V % X 576 = V432 Height at a — 20.78" b — V y 2 X 576 == V288 Height at b = 16.97" c — V X X 576 = \/l44 Height at c — 12" • The curved boundary line of such a beam is a parabolic curve, because the property of a parabola is that the square of the length of any one of the vertical lines (ordinates) is in pro- portion as their distance from the extreme point d. By this construction one-third of the material may be saved and the same strength be maintained. If the load is distributed along the whole length of the bracket instead of at its extreme end, it should "have the form shown in Fig. 24. Square and Rectangular Wooden Beams. The strength increases directly as the width and as the square of the thickness. The strength decreases in the same proportion as the length of the span increases. Example 1. Find the ultimate breaking load in pounds of a spruce beam 6 inches square and 8 feet long, when supported under both ends and loaded at the center. 246 STRENGTH OF MATERIALS. Solution : p _4_C_XJll / >_ 4X125X6X6X6 8 P = 13,500 pounds. Note. — C is obtained from Table No. 30, and is multiplied by 4 because the beam is supported under both ends and loaded at the center. The beam is square ; therefore the cube of the thickness is equal to the square of the thickness multiplied by the width. Consequently, for a square beam (thickness) 3 or (width) 3 or square of thickness multiplied by width is the same thing. Example 2. Find the load which will break a spruce beam 8 inches thick, 4 l / 2 inches wide, and 8 feet long, supported under both ends and loaded at the center. Solution : „_ 4CX B X H 2 ' ~ L 4X125X4^X8X8 8 P = 18,000 pounds. Example 3. Find the load which will break the beam mentioned in Example 2, if beam is laid flatwise. 4X125X8X4^X4^ b P = 10,125 pounds. In the first example the beam is square, 6" X 6" = 36 square inches, and its calculated breaking load is 13,500 pounds. In the second example the beam is rectangular, 8" X 4*4" = 36 square inches, and laid edgewise its figured breaking load is 18,000 pounds. In the third example the same beam is laid flatwise, and its breaking load is only 10,125 pounds. Thus, by making a beam deep it is possible to secure great strength with only a small quantity of material, but the limit is soon reached where it will not be practical to increase the depth at the ex- pense of the width, because the beam will deflect sidewise and twist and break if it is not prevented by suitable means. The STRENGTH OF MATERIALS. 247 strongest beam which can be cut from a round log is one hav- ing the thickness 1% times the width. The stiff est beam cut from a round log has its thickness 1 T \ times its width. The best beam for most practical purposes which can be cut from a round log has its thickness \y 2 times its width; for instance, 4 X 6, or 6 X 9, or 8 X 12, etc. The largest side in a beam having its thickness \y 2 times its width which can be cut from a round log is found by multiplying the diameter by 0.832. The diameter required in a round log to be large enough for such a beam is found by multiplying the largest side of the beam by 1 .2; for instance, the diameter of a round log to cut 6" X 9" will be 9" X 1.2 = 10.8 inches, or the diameter of a round log re- quired to cut 8" X 12" will be 12" X 1.2 = 14.4 inches, etc. To Calculate the Size of Beam to Carry a Given Load. Most frequently the load and the length of span are known and the required size of beam is to be calculated. For a rect- angular beam there would then be two unknown quantities, the width and the thickness, but if it is decided to use a beam having its thickness 1 y 2 times its width, the thickness may be expressed in terms of the width. H = Thickness. B = Width. H=iy 2 B Use formula for rectangular beams, page 239, and it will read, CX (1%B)*XB L This will reduce to, n _ CX2% X B* L This will transpose to, 3 B = ^ P ~ C X 2X Example. Find width and thickness of a spruce beam 10 feet long, when fastened at one end and required to carry, with 8 as factor of safety, a load of 1800 pounds at the other end, the thickness to be \y 2 times the width. When the beam is to carry 1800 pounds, with S as a factor of safety, its breaking load is 8 X 1800 = 14,400 pounds. 248 STRENGTH OF MATERIALS. Solution : 14400 X 10 ^ 2X X 125 3 ^ = V512 B = 8 inches in width. H— 1% X 8" = 12 inches in thickness. The weight of the beam itself is not considered in this problem. To Find the Size of a Beam to Carry a Given Load When Also the Weight of the Beam is to be Considered. Rule. Calculate first the size of beam required to carry the load* then figure what such a beam will weigh and add half of this weight to the load, if the beam is fastened at one end and loaded at the other, or supported under both ends and loaded at the center, but add the whole weight of the beam to the weight of the load if the load is distributed along the whole length of the beam. Then figure the size of the required beam for this new load. Example. Find width and thickness of a pitch pine beam to carry 2000 pounds, with 8 as factor of safety, and a span of 27 feet. The beam is supported under both ends and loaded at the center ; its own weight is also to be taken into consideration. Solution : Find the constant for pitch pine in Table No. 30 to be 150, and find the weight of pitch pine in Table No. 10 to be 50 pounds per cubic foot. When the beam is supported under both ends and loaded at the center it is four times as strong as if fastened at one end and loaded at the other; therefore, constant 150 is multiplied by 4. The load, 2000 pounds, multiplied by 8 as a factor of safety, gives 16,000 pounds as breaking load of the beam. 3 • _ / 16000 X 27 > 2X X 150 X 4 3 B — \/ 320 B = 6.84" = width, and 1% X 6.84" = 10.26" = thickness. The area is 6.84 X 10.26 == 70 square inches ; the weight per foot is 70 times 50 divided by 144, which equals 24.3 pounds, say 25 pounds. The weight of the beam is 25 X 27 = 675 pounds. This STRENGTH OF MATERIALS. 249 weight is distributed along the whole beam and, therefore, it does not have any more effect than if half of it, or 337^ pounds, was placed at the center, but as the beam is to be calculated with 8 as factor of safety, the weight allowed for the beam must be 337J£ X 8 = 2700 pounds. Thus, adding this weight to 16,000 pounds gives 18,700 pounds ; this new weight is used for calcu- ing the size of the required beam. ;o<; ■ ..:, X 150 X 4 3 , V 374 B = 7.2 inches = width, and 1% X 7.2" == 10.8 inches, thickness. This, of course is also a little too small, as only the weight of a beam 6.84 inches by 10.25 inches is taken into account, but if more exactness should be required the weight of this new beam may be calculated and the whole figured over again, and the result will be closer. This operation may be repeated as many times as is wished, and the result will each time be closer and closer, but never exact; but for all practical purposes one calculation, as shown in this example, is sufficient. Example 2. Find width and thickness of a spruce beam to carry 4200 pounds distributed along its whole length. The span is 24 feet ; use 10 as factor of safety, and also allow for the weight of beam. The thickness of the beam is to be l}£ times its width. Solution : 3 *=V-! 4200 X 24 X 10 2X X 125~X8 3 B = V 448 B = 7.65 inches, and £T= 11.48 inches. „ 7 . u ,, 7.65 X 11.48 X 24 X 32 4QO . Weight of beam = — — = 468 pounds. 144 Adding ten times the weight of the beam to ten times the weight to be supported, gives 46,680 pounds. 3 =4 46680 X 24 2% X 125 X 8 3 B = V 497.9 i? = 7.93 inches, and H ■=. \y 2 B= 11.9 inches, or prac- tically, a beam 8 inches by 12 inches is required. 250 STRENGTH OF MATERIALS. Crushing and Shearing Load of Beams Crosswise on the Fiber. Too much crushing load must not be allowed at the ends of the beams where they rest on their supports, as all kinds of wood has comparatively low crushing strength when the load is acting crosswise on the fiber. Approximately, the average ultimate crushing strength of wood, crosswise of the fiber, is as follows : — White oak, 2000 pounds per square inch. Pitch pine, 1400 pounds per square inch. Chestnut, 900 pounds per square inch. Spruce and pine, 500 to 1000 pounds per square inch. Hemlock, 500 to 800 pounds per square inch. . The safe load may be from one-tenth to one-fifth of the ultimate crushing load. When the wood is green or water- soaked, its crushing strength is less than is given above. Example. How much bearing surface must be allowed under each end of the beam mentioned in Example 2, providing it also has 10 as a factor of safety ? The crushing strength of spruce crosswise on the fiber is 500 pounds, and using 10 as factor of safety, the load allowed per square inch must be only 50 pounds. The beam is 8 inches wide, and half of 4685 pounds is sup- ported at each end ; thus the length of bearing required under 2342 each end will be — ~ x g = 5.85 inches. Thus, the least bearing allowable should be about 6 inches long. When beams are heavily loaded and resting on posts, or have supports of small area, either hardwood slabs or cast-iron plates should be placed under their ends, in order to obtain sufficient bearing surface for the soft wood. The same care must be exercised when a beam is loaded at one point; the bearing surface under the load should at least be as long as the bearing surface of both ends added together. Short beams are liable to break from shearing at the point of support, especially when loaded throughout their whole length to the limit of their transverse strength. The ultimate shearing strength for spruce, crosswise of the fiber, is 3000 pounds per square inch (see page 273). Safe load may be 300 pounds per square inch. In the above example the beam is 8" X 12" = 96 square inches, and its center load is 4685 pounds, or 2342^ pounds at 9342 1/ each end. The shearing stress is 96 2 = 24.6 pounds per square inch. Hence, the factor of safety against shearing is about 100, and there is not the least danger that this beam will give way under shearing; but such is not always the result. STRENGTH OF MATERIALS. 25 1 Round Wooden Beams. A round beam has 0-589 times the strength of a square beam of same length and material, when the diameter is equal to the side of the square beam. The area of a square beam compared to the area of the round beam is as 0.7854 to 1 ; there- fore it might seem as if that also should be the proportion be- tween their strength, which is the case for tensile, crush- ing and shearing strength, but not for transverse strength or for deflection, because the material is not applied to such advantage in the round beam as it is in the square one. AH preceding formulas for transverse strength of square beams may also be used for round beams if only constant C is multiplied by 0.589, or, say, 0.6. Thus, the formula for a round beam fastened at one end and loaded at the other will be : 0.6 C X Z> 3 P = ^ Note. — In a round beam, of course, it will be D s instead of Jf 2 B for a rectangular one. Example. Find the load in pounds which will break a spruce beam 12 feet long and 6 inches in diameter when supported under both ends and loaded at the center. (Find constant C in Table No. 30.) Solution : 4 X 0.6 C X D* P — L „ 4 X 0.6 X 125 X 6 X 6 X 6 P = 12 ■ P = 5400 pounds. To Calculate the Size of Round Beams to Carry a Given Load When Span is Known. Where the load and span are known, the diameter of the beam is calculated, when fastened at one end and loaded at the other, by the formula : D = ^ PXLX factor of safety 0.6 C Rule. Multiply together the load in pounds, factor of safety and length of span in feet, divide this product by six-tenths of the 252 STRENGTH OF MATERIALS. constant in Table No. 30, and the cube root of this quotient is the diameter of the beam. Example. A round spruce beam is fastened into a wall, and is to carry 1200 pounds on the free end projecting 4 feet from the wall, with 8 as a factor of safety, the weight of the beam not to be considered. Find diameter of beam. Solution : D - -<- 1200 X 4 X 8 0.6 X 125 D-- 3 =V- 3 38400 75 £> = \/~512~ D = 8 inches diameter. Load Concentrated at Any Point, Not at the Center of a Beam. If a beam is supported at both ends and loaded anywhere between the supports but not at the center (see Fig. 25), it will carry more load than if it was loaded at the center. With regard to breaking, the carrying capacity is inversely as the square of half the beam to the product of the short and the long ends between the load and the support. For instance, a beam 10 feet long is of such size that when it is supported under both ends and loaded at the center it will carry 1400 pounds. How many pounds will the same beam carry if loaded 3 feet from one end and 7 feet from the other ? Solution : x _ 1400 X 5 2 X = 7X3 1400 X 25 21 X= 1666% pounds. If weight of beam is also in eluded in its center-breaking-load the formula will be : FIG. 25. b F-«H« — F -h a X b P = Breaking load (including weight of beam) if applied at the center in pounds. F= Half the length of the span. STRENGTH OF MATERIALS. 2 53 W ' = Weight of beam. Pi = Breaking load applied at n. Load at n X distance b Load on Pier A = Load on Pier B = span Load at n X distance a span Beams Loaded at Several Places. -4-ftT-J -10ft, *\ 16-ftr 2L -19-ffc ►( 24 ft. When a beam is loaded at several places the equivalent center load and the load on each support may be calculated as shown in the following example : (See Fig. 26). The equivalent center load for a — - X 20 X 1000 = 555.6 lbs. 12 X 12 The equivalent center load for b = 10 X 14 X 800 = 777.8 lbs. 12 X 12 The equivalent center load for c — 16 X 8 X 900 = 800 lbs. 12 X 12 The equivalent center load for d — 19 X 5 X 300 = 197.9 lbs. 12 X 12 The equivalent center load for loads a, b, c and d is 2331.3 pounds. The load on Pier A = (5 X 300) + (8 X 900) + (14 X 800) + (20 X 1000) _ 24 1662^ lbs. The load on Pier B = (4 X 1000) + (10 X 800) + (16 X 900) + (19 X 300) _ 1337 y lbg 24 /2 Note. — The sum of the load on supports A and B is always equal to the sum of all the loads ; therefore, by subtracting the 254 STRENGTH OF MATERIALS. calculated load on B from the total load the load on A is ob- tained. By subtracting the calculated load at A from the total load, the load on B is obtained. To each load as calculated above for each support also add half the weight of the beam. To Figure Sizes of Beams When Placed in an Inclined Position. FIG. 23 Figure all calculations concerning the transverse strength from the dis- tance S, and leave the length L out of consideration. If the distance S cannot be obtained by measurement it may be found by multiplying L by cosine of angle a. DEFLECTION IN BEAMS WHEN LOADED TRANSVERSELY. Experiments and theory both prove that if the span is increased and the width of the beam increased in the same pro- portion the transverse strength of the beam is unchanged ; but such is not the case with its stiffness. If a beam is to have the same stiffness its depth must be increased in the same ratio as the span, providing the width is unchanged. Within the elastic limit of the beam the deflection is directly proportional to the load; that is, half the load produces half the deflection, but doubling the load will double the deflection. Deflection is proportional to the cube of the span ; that is, with twice the length of span the same load will, when the other dimensions of the beam are unchanged, produce eight times as much deflection. Deflection is inversely as the cube of the depth (thickness) of the beam. For instance, if the depth of a beam is doubled but the length of span and the width of beam is unchanged, the same load will produce only one-eighth as much deflection. Deflection is inversely as the width of the beam ; for instance, when a beam is twice as wide as another beam of the same material but all the other dimensions are unchanged, the same load will produce only half as much deflection. The deflection in a beam caused by various modes of load- ing is calculated by the following formulas : — For beams laid in a horizontal position and loaded trans- versely, fastened at one end and loaded at the other : (See Fig. 6). s _ PxL? SXE X / STRENGTH OF MATERIALS. 255 For beams laid in a horizontal position, fastened at one end and loaded thoroughout the whole length : (See Fig. 7.) s= p x £S 8 X E X I For beams laid in a horizontal position, supported under both ends and loaded at the center: (See Fig. 8). s _ P X U 48 X E X I This formula may be transposed and used to calculate modulus of elasticity from the results obtained when specimens are tested for transverse stiffness. Deflection should be care- fully measured but the specimen must not be bent beyond its elastic limit ; the modulus of elasticity is calculated by the trans- posed formula : E - P-XV 48 X S X I For a square specimen I is (side of beam) 4 divided by 12. (See moment of inertia, page 237). (Also see rule for calculating modulus of elasticity, page 265). For beams laid in a horizontal position, supported under both ends and loaded uniformly throughout their whole length : (See Fig. 9). s _ 5 X P X Z 3 384X^X7 For beams laid in a horizontal position, fixed at both ends, and loaded at the center : (See Fig. 10). 6-= P x £S 192XEXI For beams laid in a horizontal position, fixed at both ends and loaded uniformly throughout their whole length : (See Fig. 11). S= P x Z3 384 X E X I In these formulas the definitions of the letters are : S = Deflection in inches. P = Load in pounds. L = Length of span in inches. E = Modulus of elasticity in pounds per square inch. /— Rectangular moment of inertia. (See pages 237-238). These formulas are applicable to any shape of section or material, when the load is within the elastic limit. 256 STRENGTH OF MATERIALS. For beams of symmetrical section it is more convenient to use the following equally correct but more practical formulas, by which the deflection is calculated directly from the size of the beam by simply using a constant obtained by experiment and reduced by calculation to a unit beam one foot long and one inch square, thus avoiding both the use of the modulus of elas- ticity and the moment of inertia. When beams are supported under both ends and loaded at the center, and the weight of the beam itself is not considered, the following formulas may be used for solid rectangular beams laid in a horizontal position : 6" = L 3 X P X c 3 H* X B 4 . L L*X B Xc SX B 3 X P X c = 4 H* X B X S P X c S xH*X B Z 3 X P B = L * X P X c P = //S X B X S SX H* L*Xc S = Deflection in inches. //= Thickness of beam in inches. B — Width of beam in inches. L = Length of beam in feet. P = Load in pounds. c = Constant obtained by experiment, and is the deflec- tion, in fractions of an inch, which a beam one foot long and one inch square will have if supported under both ends and loaded at the center; the average value for this constant is given in Table No. 31. For any other mode of loading, see rules and explanations on page 261. In previous formulas and rules, the weight of the beam itself was not considered. The deflection in a beam caused by its own weight when it is of rectangular shape and uniform size, and laid in a horizontal position, is obtained by the formula, ux yjwx c H*X B When both the weight and the load are to be considered, the deflection in a solid rectangular beam laid in a horizontal position, supported under both ends and loaded at the center, is calculated by the formula, L*(P+y8lV)c H*X B STRENGTH OF MATERIALS. 257 6" = Deflection in inches. L = Length of span in feet. P = Load in pounds. IV= Weight of beam in pounds. c = Constant obtained by experiments, and is the deflec- tion in fractions of an inch, which a beam one foot long and one inch square will have if supported under both ends and loaded at the center, and may be found in Table No. 31. //= Thickness of beam in inches. B = Width of beam in inches. Rule. To the load add five-eighths of the weight of beam, mul- tiply this by the cube of the length of the span in feet, and multiply by a constant from Table No. 31. Divide this product by the product of the cube of the thickness and the width of the beam ; the quotient is the deflection in inches. The deflection in a beam supported under both ends and loaded evenly throughout is five-eighths of that of a beam supported under both ends and loaded at the center. Therefore, in the following formulas, the weight of the beam itself is multi- plied by five-eighths to reduce the effect of the weight of the beam to the equivalent of a load placed at its center. FOR SOLID SQUARE BEAMS. Fig. 28 FOR SOLID RECTANGULAR BEAMS. B "I s = FOR HOLLOW SQUARE BEAMS — H c (P + H W) Z 3 // 4 — h± Hrt Fig. 30 2 5 8 STRENGTH OF MATERIALS. FOR HOLLOW RECTANGULAR BEAMS. K6- B H* — b h? A FIG. 31 FOR I BEAMS. S = CJP±_HJV)J^ B H* — 2 b k* lit ♦^*6^ FiG. 32 FOR SOLID ROUND BEAMS. c _ 1.7 <;(/>+ # fF) Z 3 ^ ^ FIG. 33 FOR HOLLOW ROUND BEAMS. s _ IX C (P + ft W) Z 3 Z> 4 — tfT 4 Fig. 34 FOR SOLID ELLIPTICAL OR OVAL BEAMS. £>i X Z> 3 h-M FIG. 35 FOR HOLLOW ELLIPTICAL OR OVAL BEAMS. r._ 1.7*(/> + # ^)Z 3 Z»i Z> 3 — a?i ^ 3 ? 9 FiG. 36 STRENGTH OF MATERIALS. 259 6*= Deflection in inches. L = Length of span in feet. P = Load in pounds. IV= Weight of beam in pounds. c = Constant obtained from experiments, or may be ob- tained from Table No. 31, For meaning of the other letters, see figure opposite each formula. A round beam equal in diameter to the side of a square beam will deflect 1.698 times as much, and for convenience, when the deflection of a square or a rectangular beam, whether solid or hollow, is known, it may be multiplied by 1.7, and the product is the deflection of a corresponding round, oval, or elliptical beam of the same material and diameter and laid in the same relative position and loaded in the same manner as the calculated beam. It is well to remember that a round or elliptical beam weighs a little less than a square or rectangular one, when the sides and diameters are equal, and the deflection due to its own weight is, therefore, a little less. TABLE No. 3 1.— Constant c, Giving deflection in inches per pound of load when a beam one foot long and one inch square is supported at both ends and loaded at the center. Material Constant c. Material. Constant c. Cast steel, Wrought iron, Machinery steel Cast-iron, 0.0000144* 0.0000156* 0.C000156 0.0000288 Pitch pine, Spruce, Pine, OOO odd 000 000 CO CO K) CO Ol ^ Example. A beam 6" X 9" of pitch pine, 10 feet long, supported under both ends, is to be loaded at the center with one-tenth of its breaking load. Find the load and deflection. Solution : 9 2 X 6 X 4 X 150 291600 __.. , 2916 pounds. p 10 X 10 Deflection will be, 10 3 X 2916 X 0.00024 6- = 9 3 X 6 100 699.84 4374 = 016 inch. * The constant 0.0000144 corresponds to a modulus of elasticity of 30,000,000 and the constant 0000156 corresponds to a modulus of elasticity of 28,000,000 pounds per square inch. 2 60 STRENGTH OF MATERIALS. Therefore, if this beam had been curved 0.10 inch upward, by increasing its thickness on the upper side, it would have been straight after the load was applied.* In this example the weight of the beam itself is not considered either in figuring the strength or the deflection, because the beam is comparatively short in proportion to its width and thickness. The weight of the beam itself will only be about 200 pounds, and this will be of no account in propor- tion to the load that the beam will carry, with 10 as a factor of safety. The weight of the beam will increase its deflection oniy 0.006 inch. In such a beam the danger is probably greater from crushing of the ends at the supports, if it has not enough bear- ing surface. In long beams the weight of the beam must not be neglected, either in calculating safe load or in calculating deflection. Example 2. A round bar of wrought iron is 5 feet long and 3 inches in diameter, and loaded at the center with 800 pounds. How much will it deflect? A round bar of iron 3 inches in diameter and 5 feet long weighs 119 pounds. (See table of weights of iron, page 143.) Solution : _ 5 3 X (800 + ' /s X 11 9) X 1.7 X 0.0000156 ^ S - ""3""* S = 0.0359 inch. Thus, such a shaft loaded with 800 pounds will deflect rffj of an inch in the length of 5 feet, or 60 inches. If the deflec- tion must not exceed t ^q of the span (see page 266), then the greatest allowable deflection for this span would be 0.04 inch, and the calculated deflection is within this limit. Note. — y-Vo °f the span is equal to a deflection of 0.008 inch per foot of length. Example 3. A shaft of machinery steel, 11 inches in diameter and 6 feet between bearings, carries in the center a 12-ton fly wheel. How much deflection will the weight of the fly wheel cause ? Note.— Such shafts are usually considered as a beam sup- ported under both ends. (See formula for deflection in solid round beams, page 258.) Solution: 12 tons = 24.000 pounds. (Weight of shaft is not taken into consideration.) * This is a thing frequently done in practice. s = Mkl-.M;i'H OF MATERIALS. 26 1 UP X 1.7* 6" = 6- = 6 3 X 24000 X 1.7 X 0.0000156 ll 4 216 X 24000 X 0.00002652 14641 S = 0.00939 inches. Thus, the calculated deflection caused by the fly wheel is a little less than T ^ of an inch. The deflection per foot of span will be Mo^919 which equals 0.001565 inch. Example 4. Calculate the deflection of shaft mentioned in the previous example, when both the weight of fly wheel and the weight of shaft are to be considered. Solution : 6 3 X (24000 + H X 1920) X 1.7 X 0.0000156 S S = ll 4 216 X 25200 X 0.00002652 14641 6" =0.00986 inch. Practically, the deflection is likely to be a little less than what is figured in the two previous examples, because if the hub of the fly wheel fits well on the shaft, it will stiffen it some. (It is a good practice to make such shafts a little larger in diameter in the place where the hub of the wheel is keyed on ; this enlargement will then compensate for what the shaft is weakened by cutting the key-way.) The weight of the shaft may be obtained by considering a cubic foot of machinery steel to weigh 485 pounds, and a shaft 11 inches in diameter will then weigh 320.1 pounds per foot in length, and 6 feet will weigh 1920 pounds. Multiplying this by s /s gives 1200 pounds, to be added to the weight of the fly wheel, which gives 25,200 pounds. The weight of the shaft may also be found in the table of weight of round iron, page 144. To Calculate Deflection in Beams Under Different Modes of Support and Load. Constant c in Table No. 31 is the deflection in fractions of an inch per pound of load when a beam one foot long and one inch square is supported under both ends and loaded at the center, and when this constant for any given material is known, the deflection for beams subjected to other modes of fastening and loads may be calculated thus : For beams supported under both ends with the load dis- tributed evenly throughout their whole length, multiply c by % . 262 STRENGTH OF MATERIALS, For beams fixed at both ends and loaded at the center, multiply c by %. For beams fixed at both ends with the load distributed evenly throughout their whole length, multiply c by y%. For beams fixed at one end and loaded at the other, mul- tiply c by 16. For beams fixed at one end with load distributed evenly throughout their whole length, multiply c by 6. Example. A square, hollow beam of cast-iron, 8 inches outside and 6 inches inside diameter, and 9-foot span, supported under both ends, is loaded at the center with 8000 pounds. How much will it deflect ? Solution : Weight of beam = 9 X 12 X (8 2 — 6 2 ) X 0.26 = 786 pounds. c _ 9 3 X (8000 + H X 786) X 0.0000288 8 4 — 6 4 y— 729 X 8492 X 0.0000288 4096 — 1296 s _ 178.291 2800 ^=0.064 inch. Example. How much would this same beam deflect if the load had been distributed evenly throughout its whole span ? Solution : y _ 9 8 X 8786 X H X 0.0000288 " _ 8 4 — 6 4 y _ 115.289 2800 .9 = 0.041 inch. Example. A round cast-iron beam of 7 inches outside and 5 inches inside diameter is 4 feet between supports, with a load of 2000 pounds distributed evenly throughout its span. How much will it deflect, the weight of beam itself not being considered in the calculation ? STRENGTH OF MATERIALS. 263 Solution : 4 3 X 2000 X 0.0000288 X 1.7 S = X s 7 4 — 5* 256 X 2000 X 0.0000288 X 1.7 X % 1776 ^ = 0.0085 inch. In this example, 1.7 is used as a multiplier because the beam is round, and $£ because the load is distributed evenly throughout the length of the span. Example. A fly wheel weighing 800 pounds is carried on the free end of a 3-inch shaft, 1 foot from the bearing. How much will the shaft deflect? This is the same as a round beam loaded at one end and fastened at the other; therefore, constant c is multiplied by 16 X 1.7. Solution : o Z 3 P 1.7 c X 16 S 1 X 800 X 1.7 X 0.0000156 X 16 3 4 .5 = 0.0042 inch. FIG. 37 Previous calculations for breaking load and also for deflection are based upon a dead load slowly applied and not exposed to jar and vibrations. If the load is applied suddenly it will have greater effect toward breaking the beam than if applied slowly. For instance, imagine a load having its whole weight hanging on a rope, like Fig. 37, just touching the beam but not actually resting upon it. If that rope was cut off suddenly this load would produce twice as much effect toward breaking the beam and would cause twice as much deflection as if it was loaded on gradually. A railroad train running over a bridge will, for the same reason, strain the bridge more when running fast than it would if running slow. To Find a Suitable Size of Beam for a Given Limit of Deflection. For a square beam supported under both ends and loaded at the center, use the formula : 264 STRENGTH OF MATERIALS. Side of the beam — ^ L * P c A round beam supported under both ends and loaded at the center may be calculated by the formula : Diameter of beam = a/: Z 3 P 1.7 c S A rectangular beam supported under both ends and loaded at the center, and having its depth iy 2 times its width, may be calculated by the formula : 4 V3 L? P c — — - — L = Length of span in feet. P = Center load in pounds. S= Given deflection in inches. c = Constant given in Table No. 31. Note. — These three formulas are only approximate, as the weight of the beam itself is not considered ; but if necessary, after the size of beam is obtained, its weight may be calculated and five-eighths of it added to the center load, P\ and using the same formula again, another beam may be calculated for this new center-load, and this new calculation will give a beam only a mere trifle too small. Constants in Table No. 31 are for beams supported under both ends and loaded at the center. For any other mode of loading or fasten- ing, constant c must be multiplied according to rules on page 261. To Find the Constant for Deflection. If experiments are made upon rectangular beams, use formula, ' S H* B LHP + n W) Example. Calculate the constant c, or deflection in inches per pound of load, for a beam of 1 foot span and 1 inch square, supported under both ends and loaded at the center, when experiments are made upon a pitch pine beam 40 feet long, 12 " by 8", weighing 1200 pounds and deflecting 1% inches for a center-load of 500 pounds. Solution: 1.5 X 12 3 X 8 40 3 X (500 4- H X 1200) c = 0.000259 inch. STRENGTH OF MATERIALS. 265 Modulus of Elasticity Calculated from the Transverse Deflection in a Beam. When experiments are made upon rectangular beams sup- ported under both ends and loaded at the center, the modulus of elasticity may be calculated by the formula, E = IHP±HJV) E = Modulus of elasticity. L = Length of span in inches (not in feet). P = Load in pounds. W '== Weight of beam in pounds. S= Deflection of beam in inches. T= Thickness of beam in inches. B = Width of beam in inches. Example. Calculate the modulus of elasticity for a pitch pine rec- tangular beam weighing 1200 pounds, 40 feet span, and 12" by 8", deflecting \y 2 inches for a center-load of 500 pounds. (This beam and conditions are the same as mentioned in the previous example for calculating constants.) Solution : E = 48 ° 3 x ( 500 ± #_X_1200) 4 X l l A X 12 3 X 8 E _ 138240000000 82944 % E = 1,666,666 pounds per square inch. This deflection was obtained by actual experiments on a pitch pine beam of the dimensions given, and the calculated modulus of elasticity agrees fairly well with what is usually given by different authorities in tables of modulus of elasticity. When experimenting it is necessary to take the average of several experiments with different loads and to try the beam by turning it upside down, as very frequently it will then deflect a different amount under the same load. Care should be taken that the load is not so great as to strain the beam beyond its elastic limit. As long as the deflection increases regularly in proportion to the load, it is a sign that the elastic limit is not reached. It is very difficult to ascertain exactly when deflection will commence to increase faster than the load, because material is never so homogeneous, but that the deflection will be more or less irregular, although by care and patience fairly good re- sults may be obtained. 266 STRENGTH OF MATERIALS. Allowable Deflection. The greatest amount of deflection which may be allowed in different kinds of construction can only be determined by prac- tical experience and good judgment of the designer. As a rule, in iron work the deflection is seldom allowed to exceed j-^Vo °f the span, which is equal to T ^, or 0.008 inch per foot of span. Line shaftings are sometimes allowed to deflect t1 l ¥ of the distance between hangers which is equal to 0.01 inch per foot of span, but head shafts carrying large pulleys are generally not allowed to deflect more than 0.005 per foot of span. In woodwork, considerable more deflection is allowed than in iron structures. Beams in houses are frequently allowed to deflect - ^0, or even t |q of the span; this is equal to 0.024 to 0.025 inch*, per foot of span. Woodwork to which machinery is to be fastened must never be allowed to deflect so much. Such woodwork must always be so stiff that it supports the machinery, and not vice versa j for instance, in beams or posts by which hangers and shafting are supported, it is not all-sufficient that they are strong enough, but they must also always be stiff enough. In factories it is very important that floor beams as well as beams supporting heavy shafting have sufficient stiff- ness as well as strength. Floors in factories are frequently loaded up to 300 pounds per square foot of surface. For floors in public buildings, which are never loaded with more than the weight of the people who can get room, the load will hardly ex- ceed 150 pounds per square foot of surface. Floors in tene- ment houses are seldom loaded more than 60 pounds per square foot. Slate roofs weigh about 8.5 pounds per square foot of surface. Snow may be reckoned, when newly fallen, to weigh 5 to 15 pounds per cubic foot, and when saturated with water it may weigh 40 to 50 pounds per cubic foot. Usual practice is to allow 15 to 20 pounds per square foot for snow and wind on roofs. TORSIONAL STRENGTH. The fundamental formula for torsional strength is, Pm= S — a Pm = Twisting moment, and is the product of the length of the arm, m, in inches and the force, P, in pounds. .5* = Constant computed from experiments, and is some- times called the modulus of torsion; its value usually agrees closely to the ultimate shearing strength per square inch of the material. J == Polar moment of inertia (see page 297). a = The distance in inches from the axis about which the twisting occurs to the most remote part of the cross section. * 0.025 is one-fortieth inch per foot of span, STRENGTH OF MATERIALS. 267 Example 1. A round cast-iron bar 3 inches in diameter, is exposed to torsional stress; the length of the lever, m, is 18 inches. Find the breaking force, P, in pounds when the modulus of torsion for cast-iron is taken as 25,000 pounds. Polar moment of inertia for a circle of diameter, d, is, 32 The distance a = y z d. _ 25000 X 3* X 3.1416 X A is x \y 2 25000 X 81 X 3.1416 P = 27 X 32 P = 7363 yi pounds. The advantage of the above formula is that it may be used for any form of section, because it takes in the polar moment of inertia of the section; but it is seldom that calculations of torsional strength are required for other than beams of round or square section, either hollow or solid, and the strength of such beams may be more conveniently cal- culated in an equally correct, but easier way, obviating the use of both the polar moment of inertia and the modulus of torsion, by reasoning thus : Consider two shafts, a and b, Fig 38. Shaft a has twice the diameter of shaft b, and consequently four times the area ; therefore it has, so to say, four times as many fibers to resist the stress, and for this reason it must be four times as strong as shaft b ; but, further, the outside fibers in a are twice as far from the center, therefore the fibers in shaft a must also have on an average twice the advantage over the fibers m shaft b to resist the twisting effort of any load exerting a twisting stress, and for this reason shaft a must have twice the strength of b\ and taking these two reasons together, shaft a must, conse- quently, be eight times as strong as shaft b, in resisting tor- sional stress. Thus, the strength of a solid shaft /%%%ffi%^ Fig. 38. increases as the cube of the diameter. Shaft a is twice as large in diameter as shaft b, and is, therefore, eight times as strong as b, because 2 3 = 8. If shaft a had been three times as large as b it would have been 27 times as strong, because 3 3 = 27.; if shafts had been four times as large as b, it would have been 64 times as stronsr, because 4 3 = 64. 268 STRENGTH OF MATERIALS. Therefore, if the constant corresponding to a load, which, applied to an arm one foot long will twist off or destroy a bar one inch in diameter, is found, the breaking load for any round shaft of the same material when under torsional stress may be easily calculated. The torsional strength (but not the torsional deflection in degrees) is independent of the length of the shaft. The strength depends only upon the kind and the amount of material, and the form of cross-section. A square shaft having its sides equal to the diameter of a round shaft will have ap- proximately 20% more strength than the round one, but it will take nearly 28% more material. A square shaft of the same area as a round shaft has approximately 15% less torsional strength than the round one. Thus : Formulas for torsional strength relating to solid round shafts will be P — n* D* p »=i P 7/1 < = ^r P = Breaking load in pounds. D = Diameter of shaft in inches. m = Length in feet of the arm on which load P is acting. c = Constant, and it is the load in pounds which, when applied to an arm one foot long, will twist off or destroy a round bar one inch in diameter. This constant is obtained from ex- periments, and is given in Table No. 32. Rule. — Multiply the cube of the diameter in inches by the constant c, in pounds, divide this product by the length of the lever w, in feet, and the quotient is the breaking load in pounds. TABLE No. 32.— Constant c. The ultimate torsional strength in pounds of a round beam one inch in diameter, when load is acting at the end of a lever one foot long. Material. Very Good. Medium Good. Poor. Cast Steel Machinery Steel* . , . Wrought Iron .... Cast-iron 2,000 1,200 800 525 1,000 1,100 580 450 600 700 500 350 * Machinery steel or wrought iron may not actually break at this load, but it will deflect and yield so it will become useless. STRENGTH OF MATERIALS. 269 Example 1. A wrought iron shaft is eight inches in diameter, and the force acts upon a lever two feet long. How much force must be applied in order to twist off or to destroy the shaft? Solution : 8*^5S0_ = 512 X 560 =148?4S0 p 0unds . 2 2 Example 2. A force of 870 pounds is acting with a leverage of four feet in twisting a wrought iron shaft. What must be the diameter of the shaft in order to resist the twisting stress, with 10 as a factor of safety ? Solution : 3 D = \l P m X 10 ^ 580 3 D — V 00 = 3.914, or, practically, a 4-inch shaft. Note. — Ten is used as a multiplier of the twisting moment, P m, because 10 is the factor of safety. Constant 580 is taken from Table No. 32. Example 3. A round bar of cast-iron four inches in diameter is to be twisted off by a force of 3200 pounds. How long a leverage is necessary ? (c for cast-iron, in Table No. 32, is 450). Solution : ... D* c P 4 3 X 450 _ 64 X 450 — -■ 9 feet long. 3200 3200 Example 4. Experiments are made upon a cast-iron round bar 2 inches in diameter with a leverage of h% feet ; the bar is twisted off at a force of 832 pounds. Calculate constant c, or the force in pounds if acting with a leverage of one foot, which will break a round bar of the same material one inch in diameter. Solution : c __ P m , = 832X5X_ 4368 = ^ ds 2 3 8 P 27O STRENGTH OF MATERIALS. Hollow Round Shafts. In proportion to the amount of material used, around hollow shaft has more torsional strength than a solid shaft of the same diameter. This is because the fibers in any shaft exposed to twisting stress only offer resistance to the load in proportion to their stretch. Therefore, the fibers near the center are always in position to offer less resistance than the fibers more remote from the center. The formula for torsional strength in round hollow shafts will be : ( D* — d* \ \ D X m J P — Ultimate breaking load in pounds applied at a leverage of m feet. D = Outside diameter of shaft in inches. d = Inside diameter of shaft in inches. m = Length of lever in feet. c = Constant (same as for a solid shaft). Square Beams Exposed to Torsional Stress. The theoretical formula for twisting strength (on page 266) will apply to square as well as round beams. The proportional strength between a round and a square beam may, therefore, be compared by using that formula. Let S represent the side of a square beam and the polar moment of inertia is -} S*. The distance from the center of the beam to the most remote fiber in a square beam \s S \f l / 2 , and, dividing the polar moment of inertia by this distance, we have, i ^4 * = 0.23 S 3 Let D represent the diameter of a round beam. The polar moment of inertia is = 0.098 Z> 4 The distance from the center to the most remote fiber in the round beam is ]/ 2 D. Dividing the polar moment of inertia by this distance, we have oms ^ = 0.196 D 3 Suppose, now, that S and D are equal, for instance, one inch ; the proportion in torsional strength between the two beams must be 0.23 divided by 0.196, which equals 1.18. Thus, for square beams, use the formulas given for round beams, but multiply constant c, in Table No. 32, by 1.2, and STRENGTH OF MATERIALS. 27 I take the side instead of the diameter. The formula for tor- sional strength in a square beam will be : p_ (Side) 3 X 1.2 X c Length of leverage. c = Constant (same as for a round beam). P = Load in pounds. Side is measured in inches. Length of leverage is measured in feet. Torsional Deflection. The torsional deflection in degrees will increase directly with the length of the shaft and the twisting load, and inversely as the fourth power of the diameter of the shaft ; therefore, the formula for torsional deflection is : r_cXmXLXP S & S = Deflection in degrees for the length of the shaft. m = Length of lever in feet. L = Length of shaft in feet. P = Load in pounds. D — Diameter of shaft in inches. c = Constant obtained from experiments for different kinds of material, and is the deflection in degrees for a shaft one inch in diameter and one foot long, when loaded with one pound on the end of a lever one foot long. The author of this book has made experiments on tor- sional deflection in wrought iron shafts two inches in diameter. The average deflection was lj£ degrees in 10 feet of length, when a load of 50 pounds was applied on a lever h% feet long. Constant c, as calculated from these experiments, will be 0.00914. Using this constant, the formula for torsional deflection for wrought iron will be : C _Z X m X P X 0.00914 * -^ Machinery steel and wrought iron will deflect about the same. Cast-iron will deflect twice as much as wrought iron. A square bar will deflect 0.589 times as much as a round bar when side and diameter are alike. Formula for Torsional Deflection in Hollow Round Shafts. S _ L XwX PXc D = Outside diameter in inches. d = Inside diameter in inches. All the other letters have the same meaning as explained under formulas for solid shafts. 272 STRENGTH OF MATERIALS. TABLE No. 33 Constant c, The torsional deflection in degrees per foot of length for a shaft of one inch side or diameter when loaded with one pound at the end of a lever one foot long : Material. Machinery Steel Wrought Iron . Cast-iron .... Oak ..... . Ash Pine and Spruce Round Square Section. Section. 0.00914° 0.00538° 0.00914° 0.00538° 0.018° 0.0106° 0.795° 0.468° 0.784° 0.460° 1.35° 0.79° Example. A round bar of wrought iron 16 feet long and 3 inches in diameter is fastened at one end and the other is exposed to a twisting load of 1000 pounds, acting with 5 feet leverage. How many degrees will this load deflect the bar ? Solution S~- Sz 16 X 5 X 1000 X 0.00914 3 4 731.2 81 S- 9 degrees. Note.— From Table No. 33, it is seen that only steel and wrought iron are suitable for shafts exposed to torsional stress. Wrought iron is about twice as good as cast-iron, over 80 times better than oak, and about 150 times as good as pine. SHEARING STRENGTH. Sometimes force may act in such a manner that the material is sheared off. For instance, the rivets in a steam boiler are exposed to shearing stress (see Fig. 39) when the boiler is under steam pressure. Fig. 39. When holes are punched or bars of r f? [ iron are cut off under punching presses, ' ,jj Z ] the action of the punch in cutting off the ~* «* *»» »- material is shearing, and the resistance which the material offers is its ultimate shearing strength. The average ultimate shear- ing strength of wrought iron is 40,000 pounds per square inch. STRENGTH OF MATERIALS. 273 In cast-iron the ultimate shearing strength is usually between 20,000 and 30,000 pounds per square inch. In steel the ultimate shearing strength will vary from 40,000 to 80,000 pounds per square inch. The resistance offered to shearing is in proportion to the sheared area. Thus, it will take twice as much force to punch a hole two inches in diameter through a three-eighths inch plate as it would to punch a hole only one inch in diameter through the same plate, and it will take four times as much force to shear off a one-inch bolt as it would to shear off a one- half inch bolt, because the area of a one-inch bolt is four times as large as the area of a one-half inch bolt. Example 1. How much force is required to shear off a wrought iron rivet of one-inch diameter if the shearing strength of the wrought iron is 40,000 pounds. Solution : One-inch diameter = 0.7854 square inches; therefore the force required will be 0.7854 X 40,000 = 31,416 pounds. Example 2. A wrought iron plate is one-quarter of an inch thick and the ultimate shearing strength of the iron is 40,000 pounds per square inch. How much pressure is required to punch a hole three-quarters of an inch in diameter ? Solution : The circumference of a %-inch circle is 2.356 inches. The plate is X-inch thick; therefore the area of shearing surface, 2.3562 X X = 0.58905 ; thus, the force required will be 40,000 X 0.58905 = 23,562 pounds. TABLE No. 34.— Shearing Strength Per Square Inch. Material. Steel Wrought Iron Rivets Cast-iron . . . . • Oak, crosswise . . . Oak, lengthwise . . Pitch Pine, crosswise Pitch Pine, lengthwise Spruce, crosswise . . Spruce, lengthwise . Pounds Per Square Inch. 45,000 to 75,000 35,000 to 55,000 20,000 to 30.000 4,500 to 5,500 400 to 700 4,000 to 5,000 400 to 600 3,000 to 4,000 300 to 500 274 STRENGTH OF MATERIALS. FACTOR OF SAFETY. The factor of safety- can only be fixed upon by the experi- ence and good judgment of the designer. It may vary from 4 to 40. In a temporary structure, when the greatest possible load to which it will be exposed is known, a factor of safety of four may be safe enough, but frequendy a greater factor is necessary. Different factors of safety are also necessary for different materials ; a different factor of safety may also be necessary in different parts of the same machine. The following Table, No. 35, is only offered as a guide in selecting factor of safety : TABLE No. 35.— Factor of Safety. Dead Load, Variable Load, Machinery such as build- such as bridges Machinery Exposed to Material. ings contain- and slow- in hard usage, as ing little or no running General. Rolling Mills, machinery. machinery. etc. Steel, 5 7 10 15 Wrought Iron, 4 6 10 15 Cast-iron, 6 10 15 25 Brickwork, 15 25 30 40 Wood, 8 10 15 20 If a structure is exposed to stress alternately in one direc- tion and then in another, it is necessary to use a higher factor of safety than if it is only exposed to a steady stress one way. A comparatively small load, when applied a sufficient number of times, may break a structure or a machine, although it does not break it the first time. For instance, commence to hammer on a bar of cast-iron and it will break after several blows, although the last blow need not be any more powerful than the first one. It is the same way with anything else; it may break in time, although it is strong enough to resist the stress at the beginning ; therefore, within practical limits, the larger the factor of safety the longer time the structure may last. NOTES ON STRENGTH OF HATERIAL. In steel, the crushing strength usually exceeds the tensile Kxeiigth, but wrought iron has usually a little more tensile than crushing strength, and its shearing strength is about 80 per cent, of its tensile strength. Both steel and wrought iron are suitable to resist any kind of stress, and compared to other materials they are especially adapted for anything exposed to twisting and shearing stress. STRENGTH OF MATERIALS. 275 Cast-iron is variable ; it has usually five to six and a-half times as much crushing as tensile strength, and when loaded transversely it will deflect under the same load nearly twice as much as wrought iron. It is especially useful for short pillars or anything exposed to crushing stress, where there is little danger of breakage by flexure; it is very much less reliable when exposed to tensile or torsional stress. Wood is not adapted to resist torsion, but is useful to resist tensile, crushing and transverse stress, also to resist flexure. It has nearly twice as much tensile as crushing strength.; there- fore, it would seem specially well adapted, in all kinds of con- struction, to be the member exposed to tensile stress, but where wood and iron enter into construction together, iron is always used as the member to take the tensile stress and wood as the compressive member, because wood has suck low shear- ing strength lengthwise with its fibers that, with any kind of fastening at the ends, it will tear and split at the holes under comparatively little stress; but this difficulty is easily overcome when wood is used as the compressive member. Wood has comparatively low tensile and crushing strength crosswise on the fiber. _ This is well to remember with beams loaded transversely and laid on posts. The beams may be sufficiently strong, but under heavy load, if suitable precautions are not taken (see page 250) the top of the post may press into the beam, especially if the lumber is green. Stone has high crushing strength but low tensile strength, and, in consequence, very low transverse strength. It is very well adapted for foundations when supported and laid in such a way that its crushing strength comes into play, but when laid as a beam to resist transverse stress it is very unreliable, as it will break for a comparatively small load and it may break from a blow or jar. Brickwork is only suitable for crushing stress, and there is great difference in the strength of different kinds of brick. In calculating strength and stiffness in any kind of design- ing, it should be remembered that it is only possible to deter- mine the strength of any material by actual test, and that the tabular and constant numbers here given are only an average approximate. flfcecbanfcs. The science which treats of the action of forces upon bodies and the effect they produce is called Mechanics. Newton's Laws of Motion. The three fundamental principles of the relation between force and motion were first stated by Sir Isaac Newton, and are therefore called Newton's laws of motion. NEWTON'S FIRST LAW. All bodies continue in a state of rest or of uniform motion in a straight line, unless acted upon by some external force that compels change. NEWTON'S SECOND LAW. Every motion or change of motion is proportional to the acting force, and the motion always takes place in the direction of a straight line in which the force acts. newton's third law. To every action there is always an equal and contrary re- action. Gravity. The natural attraction of the earth on everything on its surface which will cause any body left free to move to fall in the direction of the center of the earth is called the force of gravity. Acceleration Due to Gravity. If a body is left free to fall from a height, its velocity will not be constant throughout the whole fall, but it will increase at a uniform rate. It is this uniform increment in velocity which is called acceleration of gravity. It is usually reckoned in feet per second. A body falling free will at the end of one second have acquired a velocity of 32% feet, or, practically, 32.2 feet per second; but it has fallen through a space of 16.1 feet, because it started from rest and the velocity was increasing at a uniform rate until, at the end of the second, it was 32,2 feet per second; therefore, the average velocity during the first. second can only be 16.1 feet. At the end of two seconds the velocity has increased to 64.4 feet per second and the space fallen (276) MECHANICS. 277 through is 64.4 feet, because the average velocity per second must be half of the final velocity; therefore, the average velocity is 32.2 feet per second, and, as the time is two seconds the space will be 64.4 feet. At the end of three seconds the final velocity has increased to 3 X 32.2 = 96.6 feet per second and the space fallen through is $%•$- X 3 = 144. 9 feet, etc. This is supposing the body was falling freely in vacuum, but while the air will offer a resistance and somewhat reduce the actual, motion, the principle is the same. Acceleration due to gravity varies but little at different latitudes of the earth. At the equator it is calculated to be 32.088 and at the pole 32.253 feet. Acceleration due to gravity decreases at higher altitudes.* but all these variations on the earth's surface are so small that they hardly need to be considered in any calculation concerning practical problems in mechanics. Velocity. The velocity of falling bodies increases at a uniform rate of 32.2 feet per second ; therefore, when commencing from rest, the final velocity in feet per second must be, Rule. Multiply the time in seconds by 32.2 and the product is the final velocity in feet per second ; or, multiply the height of the fall in feet by 64.4 and the square root of the product is the velocity in feet per second. Example. What final velocity will a body acquire in a free fall during seven seconds ? Solution : v = 1 X 32.2 = 225.4 feet per second. Height of Fall. The average velocity per second is always half of the final velocity per second. Therefore the space fallen through in a given time is found by multiplying half of the final velocity by the number of seconds which produced that velocity. Thus, the formulas : h = *JL — / 0.5 v = v 0.5 t — — — ^—t- 0.5 g 2 2 ? 2 * Above the surface of the earth the weight of a body is inversely propor. tional to the square of its distance from the center of the earth. Below the surface of the earth the weight of a body is directly proportional to its distance from the center of the earth. 278- mechanics. Example. A fly-wheel has a rim speed of 48 feet per second. From what height must a body drop to acquire the same velocity ? Solution : A = ."L = »L = 230 ± = 85.78 feet. 2 g 64.4 644 Time. Rule. Divide the space by 16.1, and the square root of the quotient is the time ; or, divide given velocity by 32.2, and the quotient is the time. Example. How long a time does it take before a body in a free fall acquires a velocity of 100 feet per second ? Solution : t — ^L.— 152 = 3.1 seconds. g 32.2 Distance a Body Drops During the Last Second. The space through which a body will drop in the last second is equal to the final velocity minus half of acceleration due to gravity. Therefore, this space is found by the formula: x — v — y 2 g = g{t — y 2 ) x = Space in feet which the body drops the last second of the fall. / == Time in seconds. v = Final velocity. g = Acceleration of gravity = 32.2 feet. h = Height of fall in feet. Example. 40 feet per Solution A body has in a free fall obtained a final velocity of second. What space did it drop the last second ? x = v — V 2 g — 40 — ^- 2 = 40 — 16.1 = 23.9 feet. Example. A body was falling four seconds. How many feet did it drop the last second ? Solution : x — g{t — y 2 ) = 32.2 X (4 — y 2 ) = 32.2 X 3.5 = 112.7 feet. MECHANICS. 279 TABLE No. 35— Time, Velocity and Height. jr= 32.161 Feet. -T- - c , 1 Velocity in Feet at Tune in Seconds. , the End £ f the Time Height of Fall in Feet. Distance in Feet that the Body Drops in the Last Second. 1 2 3 4 5 32.161 64.322 96.483 128.644 160.805 16.08 64.32 144.72 257.28 402.00 16.08 48.24 80.40 112.56 144.72 Upward Motion. A body thrown perpendicularly upward with a certain velocity will continue the upward movement until it reaches the same height from which it would have to fall in order to get a final velocity equal to the starting velocity. Therefore, a body projected upward with a given velocity will return again with the same velocity. This is theoretical in a vacuum, but actually the body neither continues to the theoretical height nor returns with a final velocity equal to the starting velocity, because the air will always offer considerable resistance. The greater the weight of a body, in proportion to its volume, the nearer the velocity, when it returns, will be equal to its starting velocity. Example. A body is projected upward with a velocity of 45 feet per second. How high will it go before it stops and commences to drop again, the resistance of the air not being considered ? The solution of this problem is simply to find the theoretical height from which a body must drop to attain a final velocity of 45 feet, which is solved by the formula, 45 2 6474 2 025 64.4 11.286 feet. Body Projected at an Angle. If a body is projected in the direction of the line d e (see Fig. 1), with an initial velocity per second equal to the distance from d to 1, no force acting after the body is started, it will con- tinue to move at constant velocity in a straight line indefinitely ; at the end of the first second it would be at 1, at the end of two seconds it would be at 2, at the end of the third second at 3, at the end of the fourth second at 4, etc.; but, on account of the force of gravity, the motion will be entirely different. The force of gravity acts on this body exactly as if it was falling in 28o MECHANICS. a vertical line. At the end of the first second the force of gravity has caused this moving body to drop 16.1 feet out of its path; therefore, instead of being at 1 at the end of the first second, it is at a point 16.1 feet vertically under 1 ; instead of being at 2 at the end of two seconds, it is at a point 2 X 2 X 16.1 = 64.4 feet vertically below 2; instead of being at 3 at the end of the third second, it is at a point 3 X 3 X 16,1 = 144.0 feet vertically below 3 ; and instead of being at 4 at the end of the fourth second, H is at a point 4 X 4 X 16.1 = 257.6 feet vertically below 4, etc. Fig. 1. When a body is projected in a vertical upward direction with an initial velocity of v feet per second, it proceeds to a height -; therefore, when projected at an angle, a (see Fig. 1), with a velocity of v feet per second, it will proceed to the , . , v 2 sin. 2 a height YJ~~ When a body is projected in a vertical upward dirction with a velocity of v feet per second, the time for ascent is - and the time for descent is equal to the time for ascent ; therefore, the total time will be 2 ; but when the body is projected upward at an angle of a degrees, the total time for ascent and 2 v sin. a descent will be S The horizontal distance, or the range from d\a ;/. will be equal to the velocity' in feet per second multiplied by the total MECHANICS. 25 1 number of seconds consumed in the ascent and descent, and this multiplied by cos. of the angle a\ therefore, __ . , /2 z/ sin. a\ 2 v 1 sin. a cos. a Horizontal range = v y 1 cos. a = — but 2 X sin. a X cos. a is always equal to sin. of an angle of twice as many degrees as the angle a. Therefore, the formula . . v 2 sin. 2 a ^educes to horizontal range = Thus, the following formulas will apply to bodies projected at an angle. ( See Fig. 1). The greatest possible height will be, t v 2 sin. 2 a ~ ^ _ The greatest possible range will be, , v 2 sin. 2 a S The time in seconds will be, 2 v sin. a v sin. a g 0.5 £■ v = Velocity in feet per second. g = Acceleration of gravity = 32.2. to find the height to which a body can ascend. Rule. Multiply the velocity in feet per second by the sine of the angle (to the horizontal line), square this product and divide by 64.4, and the quotient is the height in feet. to find the longest possible range. Rule. Multiply the square of the velocity in feet per second by sine of an angle of twice as many degrees as the angle of the throw (to the horizontal line), and divide by 32.2. The quotient is the longest distance the body can be thrown. to find the time of flight. Rule. Multiply the velocity in feet per second by sine of the angle (to the horizontal line), and divide by 1(3.1. The quotient is the time in seconds. Example. A body is projected at an angle of 55° to the'horizontal line, with an initial velocity of 120 feet per second. How high 2 52 MECHANICS. will it go ? How far will it go in a horizontal direction ? How many seconds will it take to finish the flight? Solution for height : _.9 „:„ 9. _ h = h = '2g 120 2 X sin. 2 55 ° 64.4 h — 12 ° 2 X °- 81915 " 2 64.4 h _ 14400 X 0.673 64.4 h = 150.5 feet. Solving for horizontal range : i v 2 sin. 2 a g Twice the angle of 55° is 110 c and sine of 110° will be sine of 70°, because 180° — 110 c = 70° ; therefore, sine of 110° equals sine of 70° in the second quadrant, and the solution will be : , _ 120 2 X sin. 70° 32.2 14400 X 0.93969 32.2 Solving for time of flight: v sin. a 128.4 feet. 0.5 £• 120 X sin. 55° 16.1 ,_ 120 X 0.81915 1 Y(n " — 6>1 seconds. Example. A nozzle on a hose is placed at an angle of 28° to the horizontal line and the spouting water when leaving the nozzle has a volocity of 36 feet per second. How far will it theoretic- ally reach in a horizontal direction? Solution: ?/ 2 sin. 2 a Range g b _ 36 2 X sin. 56° g , 1296 X 0.82904 O — — 33 37 feet 32.2 — mechanics. 283 Example 3. A nozzle on a hose is placed at an angle of 38° to # the horizontal line and is spouting water a distance of 40 feet in a horizontal direction. What is, theoretically, the velocity of the water when leaving the nozzle ? Solution : b g sin. 2 a v — /40 X 32.2 sin. 76 ft — J 40 x 32 - 2 — 36.4 feet per second. > 0.9703 Note. — In Example 2 we multiply by sine of 56 degrees, because water is leaving the nozzle at an angle of 28 degrees, and twice 28 equals 56. In Example 3 we multiply by sine of 76 degrees, because twice 38 equals 76. See previous explanations. The greatest possible height will be reached if the body is thrown perpendicularly upward. The greatest possible range is obtained if the body is thrown at an angle of 45° and will then be : § At an angle of 45° the horizontal range will be twice the greatest possible height which could have been reached if the body had been thrown perpendicularly upward. At this angle the horizontal range is four times the height. For an equal number of degrees over or under 45 degrees the horizontal range will be equal ; for instance, if a body is thrown out at an angle of 30 or 60 degrees, the horizontal distance is the same, but the height of ascension will be much more at 60 degrees than at 30 degrees. It is frequently useful to notice this in practical work. For instance, water under pressure is thrown the farthest distance in a horizontal direction from a hose when the nozzle is held at an angle of 45 degrees to the horizontal line. It is possible by the same pressure to throw water twice as far in a horizontal distance as in vertical height. Motion Down an Inclined Plane. A ball rolling along an incline, as a c (Fig. 2), will have the same velocity when it gets to c as it would have had if dropping freely from a to b, supposing all friction to be left out of consideration. The average velocity will also be half of the final velocity, and the time used in the fall will be the distance a c (the length of the incline), divided by the average velocity per second. 284 MECHANICS. Body Projected in a Horizontal Direction From an Elevated Place. When a body is projected in a horizontal direction from a place which is higher than the one where it strikes the ground, the range in feet in a horizontal direction will be equal to the product of velocity in feet per second and the time in seconds which it will take for a body in a free fall to drop a distance equal to the difference in vertical height between the two places. Thus : Horizontal range = v A/ — * S v = Initial velocity in feet per second. h = Vertical height in feet. jr= Acceleration of gravity = 32.2 feet. Example. Water spouts from a nozzle in a horizontal direction at a velocity of 30 feet per second and the nozzle is placed 12 feet above the ground. What is the horizontal range of the water ? Solution : Horizontal range = v -\j— = 30 \J- X , 2 = 22.45 feet. To Calculate the Speed of a Bursted Fly=Wheel from the Distance the Fragments are Thrown. The angle of 45 degrees is the one most favorable to the range ; therefore, suppose the fragments to leave the wheel at that angle and use the formula, v 2 ■ , Horizontal distance = b = —— which transposes to v = a^/ fr g Rule. Multiply the horizontal distance by 32.2, and the square root of the product is the slowest possible rim-speed the wheel could have had at the time of the accident. Example. A 30-foot fly-wheel bursts from the stress due to centri- fugal force, and fragments were thrown a distance of 300 feet from the place of accident. What was the slowest possible speed the wheel could have had at the time the accident occurred ? and what was the corresponding number of revolu- tions per minute? MECHANICS. 285 Solution : v = \/300 X 32.2 = V 9660 = 98.3 feet per second. The length of the circumference of a 30-foot wheel is 94.25 feet, therefore the fly-wheel was running at a speed not less than 60 X q * 9 - = 62.6 revolutions per minute. This calculation does not prove that the wheel did not run faster than 62.6 revolutions per minute when it burst ; it may have revolved a great deal faster, as it is not at all sure that any fragments left the wheel at an angle of 45 degrees, but it is certain that the speed of the wheel was not slower. Sometimes it may be pos- sible to settle upon the angle at which a certain fragment left the wheel by noticing traces and marks where it went, and, figuring from the angle and the range, a pretty fair idea of the bursting speed may be obtained. (See formula on page 283). Force, Energy and Power. Force is a pressure expressed in a push or a pull. Energy is the ability to do work. It is divided into poten- tial energy and kinetic energy. Potential energy is the ability of a body to perform work at any time when it is set free to do so. Kinetic energy is the ability of a moving body to do work during the time its motion is arrested. Kinetic energy is very frequently called "stored-up energy." Work is overcoming resistance through space. In the English system of weights and measures the common unit of work is the foot-pound. Power is the rate of doing work. Work is an expression entirely independent of time, but power always takes time into consideration. For instance, to lift one pound one foot is one foot-pound of work, no matter in what time it is done, but it takes 60 times as much power to do it in one second as it would "take to do it in one minute. Inertia. Inertia is the inability of dead bodies to change either their state of rest or motion. In order to bring about any change, either of motion or rest, dead bodies must always be acted upon by some outside force. Resistance due to inertia is the resistance which a dead body free to move presents to any external force acting to change either its state of motion or rest. 2 86 MECHANICS. Mass. The mass of a body is the quantity of matter which it con- tains. By common consent the unit of mass is, in mechanics, considered to be that quantity of matter to which one unit of force can give one unit of accelei'ation in one unit of time; therefore, when the weight of a body is divided by acceleration of gravity, the quotient is the mass of the bodv. Thus : W m = § W — m X g w g = — 111 Momentum. The product of the mass of a moving body and its velocity is called its momentum or, also, its quantity of motion. The unit for momentum is the product when unit of mass is multi- plied by unit of velocity per second. In mechanical calcula- tions, using English weights and measures, the unit of mass is weight divided by 32.2; therefore, unit of momentum will be: Weight of the moving body in pounds multiplied by velocity in feet per second and the product divided by 32.2. Thus : q = m X v m = mass = — : therefore, cr £> w q = v g q = ^- W g q = Momentum, or quantity of motion. W = Weight of moving body in pounds. v = Velocity of moving body in feet per second. g= Acceleration of gravity. g the formula by which the time in a free fall is obtained, and, consequently, the momentum of a falling body can also be expressed by the product of the weight of the body in pounds and the time in seconds during the fall. This product is usually called "time effect." Impulse. The product of the force and the time in which it is acting as a blow against a body is called impulse, and it is always of the same numerical value as the momentum of the moving body. MECHANICS. 287 Kinetic Energy. The kinetic energy stored in any moving body is always expressed in foot-pounds, by the product of the force in pounds acting to overcome the inertia of the body, and the distance in feet through which the force was acting in starting the body> and is always equal to the weight of the body multiplied by the square of the velocity and this product divided by twice the acceleration of gravity. Thus : K = Kinetic energy in foot-pounds. W = Weight of the body in pounds. v = Velocity of the body in feet per second. 2^=G4.4. In a free fall the height, /i, corresponding to a given velocity, is found by the formula, -^— ; therefore, A^= W X h. Thus, multiplying the weight of a moving body by the height which in a free fall corresponds to its velocity, the product will be the kinetic energy stored in the body. W X v 1 The formula K — — ^ transposes to K = }4 ni v 2 . Hence the simple rule : Multiply half the mass of a moving body by the square of its velocity in feet per second, and the product is the kinetic energy in foot-pounds stored in the body. The kinetic energy stored in any moving body always represents a corresponding amount of mechanical work which is required in order to again bring the body to rest. Example. A body weighing 1(510 pounds is moving at a constant velocity of IS feet per second. How many foot-pounds of kinetic energy is stored in the body ? Solution : „ WX v 2 1 610 X 18 X 18 K = — g = g^ = 8,100 foot-pounds. If this moving body was brought to rest and all its stored energy could be utilized to do work it could lift 8,100 pounds one foot, or it could lift 81 pounds 100 feet, or any other combi- nation of distance and resistance which, when multiplied by one another, will give 8,100 foot-pounds. It is very important always to keep in mind a clear dis- tinction between -work 2nd. power, as power is the rate of doing work, and time must, therefore, always be considered in the question of power. For instance, when 33,000 foot-pounds of 2 88 MECHANICS. work is performed in one minute it is said to be one horse-power ; therefore, if this 32,400 foot-pounds of energy was utilized to do work and used up in one minute, it would do work at a rate of fff§# = 5-*- horse-power, but if utilized during a time of two minutes it would only do work at a rate of |i horse-power, or if utilized in a second the rate of work would be $£ X 60 — 58 jf horse-power, etc. To Calculate the Force of a Blow. The force of a blow may be calculated by the change it produces. For instance, a drop-hammer weighing S00 pounds drops three feet, and compresses the hot iron on the anvil % inch. How much is the average force? (% inch — *4sfoot). The kinetic energy stored in the hammer at the moment it commences to compress the iron is S00 X 3 = 2400 foot-pounds. The average force = 2400 = 115,200 lbs. Vl N In the above example, friction is neglected. The shorter the duration of the blow the more intense it will be. Therefore the force of the hammer mentioned above, if, instead of striking against hot iron, compressing it X inch, had been struck against cold iron, compressing it only a few thou- sandths, the blow would have been as many times more intense as the duration of the blow had been shorter. Therefore it is entirely meaningless to say that a drop-hammer or any other similar machine is giving a blow of any certain number of pounds by falling a certain number of feet, because the in- tensity of the blow will depend upon its duration. Formulas for Force, Acceleration and Motion. From the laws of gravitation, it is known that when one pound of force acts upon one pound of matter it produces an acceleration of 32.2 feet per second each successive second as long as the force continues to act. From Newton's laws of motion, it is known that the motion is always in proportion to the force by which it is produced ; therefore, when one pound of force acts for one second upon 32.2 pounds of matter, it will produce an acceleration of one foot per second. Hence the following formulas: m = Mass of the moving body, which is considered to be weight divided by 32.2. F = Constant force in pounds acting on a body free to move. G = Constant acceleration in feet per second due to the acting force, F. MECHANICS. 289 T= Time in seconds in which the force F acts upon a body free to move. 7/ — Final velocity acquired by the moving body in the time of T seconds. F=m G 7^ G = JL m v= TG ' = -? ^^ v __ F T m v m '■ = F T _ F r m FT m — : V F _ vm T vm T F When a moving body is arrested the product of the resist- ance and time is equal to its momentum. Thus: R T— v m R T R T T = R R= Constant resistance in pounds acting against the moving body. The average velocity of the moving body is half of the final velocity, and the space passed over by the moving body when acquiring the given velocity is half of the final velocity in feet per second multiplied by the time in seconds. Thus : S — -i- X T 9. F T 2 = 2 S m 'IS S = Space in feet. 26" FT ^ 2 S S = F FT 2 The work in foot-pounds required to overcome the inertia of a given body when brought from a state of rest to a given velocity is equal to the kinetic energy stored in the moving body. Thus : 2 2 2 K = Kinetic energy stored in the moving body. 29O MECHANICS. The force required to obtain a given velocity in a given time, when both resistance due to inertia and resistance due to friction is considered, is calculated by the formula : Force — ^ Clty X Mass ) + (weight X coefficient of friction). V Time / which may be written : Force = / Velocity x Mass \ _j_ (resistance due to friction). V, Time / Important. — Always calculate the force required to over- come the resistance due to inertia and the force required to overcome the resistance due to friction separately, and add the two forces in order to obtain the total force required. It is sometimes assumed that adding so much to the mass, as 3 : 2 - of the product of weight and coefficient of friction, should give the result in one operation; but such an assumption is erroneous, because the correct value for the required force is : F= v X W 4- WXf which cannot be transposed to F _ v X W + WXf TXg _F= Required force ; v = velocity ; T= time ; IV= weight of moving body in pounds ; g-= acceleration due to gravity, or 32.2; f= coefficient of friction. Example. 1. A railroad train weighing 225,400 pounds is started from rest to a velocity of 50 feet per second ; the road is straight and level ; the resistance due to friction is assumed to remain con- stant and to be 1000 pounds. What average constant pull in pounds must be exerted by the locomotive at the draw-bar in order to bring the train up to this speed in 40 seconds ? Solution : For the inertia, velocity X mass _ 50 x 225400 Force time 4() - = 8750 lbs. For friction the force — 1000 Total force, 9750 lbs. Note. — This constant force of 9750 pounds has been acting under a uniformly increasing velocity from rest or nothing at the start, to 50 feet per second at the end of 40 seconds ; therefore, the average velocity has been half of the final velocity, or 25 feet a second. The average work of the locomotive in starting the ( Solution MECHANICS. 29I train during this 40 seconds was 25 X 9750 = 243,750 foot- pounds per second, and the horse-power exerted by the locomo- tive on the draw-bar in starting this train was -VW— — 443.18 horse-power, but the power required to keep this train in motion at a speed of 50 feet per second on a level road will be only 50 x 100 ° = 90.91 horse-power. From this it may be seen what an immense power there has to be produced in order to start heavy machinery in a short time, in comparison to the power required to keep it going after it is started. Example 2. How far did the train move before it got up to the required speed of 50 feet per second ? - 50X40 = 1000 feet. 2 2 Example 3. Suppose that after the train had acquired this speed of 50 feet a second, the locomotive was detached and that the resist- ance due to friction continued to be 1000 pounds. How many seconds would the train be kept in motion by its momentum on a level road ? Solution : 225400 Time = v m — X ~32~X R — 1000 = 350 seconds. Example 4. How many foot-pounds of kinetic energy is stored in this train, which weighs 225,400 pounds and runs at a constant speed of 50 feet a second ? K = v X m = o0 X 32.2 — 8,750,000 foot-pounds. 2 2 Example 5. How far will this kinetic energy drive the train on a hori- zontal road if we suppose the constant resistance due to friction, as in Example 3, to be 1000 pounds? Solution : Distance = ^tic_energy = 8750000 = sm feet resistance 1000 When a body free to move is acted upon by a constant force the space passed over increases as the square of time. 292 MECHANICS. Example 0. Under the influence of a constant force a body moves five feet the first second. How far will it move in eight seconds, friction not considered? Solution : Distance = S 2 X5 = £20 feet. Centers. Center of gravity is the point in a body about which all its parts can be balanced. If a body is supported at its center of gravity the whole body will remain at rest under the action of gravity. Center of gyration is a point in a rotating body at which the whole mass could be concentrated (theoretically) without altering the resistance, due to the inertia of the body, to angular acceleration or retardation. Center of oscillation is a point at which, if the whole matter of a suspended body was collected, the time of oscillation would be the same as it is in the actual form of the body. Center of percussion is a point in a body moving about a fixed axis at which it may strike an obstacle without communi- cating the shock to the axis. Moments. The measures of tendency to produce motion about a fixed point or axis, is called moment. The pro- duct of the length of a lever and the force acting on the end of it. tending to swing it around its center, is called the moment, of force or the statical moment, and may be expressed in either foot-pounds or inch-pounds. In Fig. 3, the arm is 18 inches long and the force is 40 pounds; the moment is 18 X 40 = 720 inch-pounds, or iy 2 X 40 "= 00 foot-pounds. Levers. When a lever is balanced, the distance a, fig. a multiplied by the weight w, is always equal to '^ a - y — &- the distance b, multiplied by the force F. 1 g c In a bent lever (as Fig. 5) it is not the length [w\ ' of the lever but the distance from the fulcrum at right angles to the line in which the force is acting, that must be multiplied. Thus: a x w — b X F. In Fig. 6, the force is acting at a right angle to the lever, and. therefore, the distance a is equal to the length of the long end of the lever. 1, MECHANICS. 293 The force is applied to more advantage in Fig. G than in Fig. 5. As a rule, the force should always be applied so as to act at right angles to the lever. Radius of Gyration. The radius of gyration of a rotating body is the distance from its center of rotation to its center of gyration. ,, ,. r .. A moment of rotation Radius of gyration = \ s — — * mass of rotating body or, for a plane surface: -p. ,. c .- /moment of inertia Radius of gyration = \ ■— . * area of surface The radius of gyration of a round, solid disc, such as a grind- stone, when rotating on its shaft, is equal to its geometrical radius multiplied by */ Yz or radius multiplied by 0.7071 very nearly. The radius of gyration of a round disc, if indefinitely thin and rotating about one of its diameters, is equal to radius divided by 2. The radius of gyration of a ring, of uniform cross- section, rotating about its center, such as a rim of a fly-wheel when rotating on its shaft, is : V7?2 _i_ r 2 — — — 7? = Outside radius. r = Inside radius. The radius of gyration of a hollow circle when rotating about one of its diameters is : Radius of gyration = -v/ ~*~ r R = Outside radius. r = Inside radius. Moment of Inertia. The moment of inertia is a mathematical expression used in mechanical calculations. It is an expression causing con- siderable ambiguity, as it is not always used to mean the same thing. The least rectangular moment of inertia, as used when calculating transverse strength of beams, columns, etc., is the sum of the products of all the elementary areas of cross-sections when multiplied by the square of their distances from the axis of rotation. The axis of rotation is considered to pass through the center of gravity of the section. 2 9 4 MECHANICS. The least rectangular moment of inertia is always equa. to the area of surface of cross-section, multiplied by the square of the radius of gyration, when the surface is assumed to rotate about the neutral axis of the section. Mathematicians calculate the moment of inertia by means of the higher mathematics, but it may also be calculated approximately by dividing the cross-section of the beam into any convenient number of small strips and multiplying the area of each strip by the square, of its distance from its center-line to the neutral axis, and the sum of these products is the moment of inertia, very nearly. The narrower each strip is taken, the more exact the result will be; but it will always be a trifle too small. Example 1. Find approximately the rectangular moment of inertia for a surface (or section of : a beam) 6" X 2", about its axis x y. (See Fig 7,) Divide the surface into narrow strips, as a, fi, c\ d< e, f, g, A, i,j. k, /, and multiply each strip by the square of its distance from the neutral axis, xy, and the sum of these products is the moment of inertia of the surface. h 2- *| FIG. 7. a = 2 X % X (2%f b = 2 x y 2 X (2X)' 2 r = 2 X % X dX) 2 7.5625 5.0625 3.0625 d= 2 X y 2 X (IX)' 2 = 1.5625 e = 2 X V 2 X ( X)' 2 — 0.5625 g — 2 X % X ( X)' 2 = 0.0625 //. — 2 X y 2 X ( % Y — 0.5625 z = 2 X % X (IX)' 2 = 1.5625 j = 2 X y 2 X (IX)' 2 = 3.0625 k = 2 X y 2 X (2X)' 2 = 5.0625 l—2Xy 2 X (2X) 2 = 7.5625 Moment of inertia = 35.75 (approximately). The correct value for the least rectangular moment of inertia for such a surface is obtained by the formula, (Depth)* X width andfQrFi 12 - 12 7 will be 63 X 2 = 36. Thus, the approximate rule gives results a trifle too small, but if the sur- face had been divided into smaller strips, the result would have been more correct. Radius of gyration for this surface, when rotating about the axis xy, is : (moment of inertia area > 12 = 1.73 inches MECHANICS. 295 Example 2. Find by approximation the rectangular moment of inertia for a surface, as Fig. 8, (the sectional area of an I beam) about the axis x y. When the beam is symmetrical, the neutral axis is at an equal distance from the upper and lower side, and the moment of inertia for the upper and lower half of the beam is equal ; consequently, when calculating moment. of inertia for a surface like Figs. 8 and 7, it is only nec- essary to calculate the moment of inertia for half the beam, and multiply by 2 in order to e X — : get the moment of the whole beam. Solution: a = 3 X )4 X (2K) 2 = 11.34375 b — 3 X Yi. X (2X) 2 = 7.59375 c = 1 x % x (IK)' 2 = 1.53125 y will also be (diameter) 4 X ■ 64 (diameter) 4 X 7r and the rectangular moment about ; thus the polar moment will be 32 2Qi» MECHANICS. Example. Find the polar moment of inertia and radius of gyration of a round shaft of 4" diameter. Solution /= gJ5 = 44X3 - 141g = 25.1328 J 32 32 Radius of gyration = -v/ Polar moment of inertia area of section. Radius of gyration — a/_ 25.1328 2 2 X 3.1416 Radius of gyration = 1.414 inches. Moment of Inertia in Rotating Bodies. The term, moynent of inertia, as used in calculating stored energy in revolving bodies, is frequently and certainly more concisely called moment of rotation, and is a mathematical expression by which the effect of the whole mass (theoretically) is transferred to the unit distance from center of rotation. This term (moment of inertia or moment of rotation) is obtained by multiplying the square of radius of gyration by mass of moving body.* In English measure, mass is taken as -^\ v of the weigh t of the revolving body, and the radius of gyration is always taken in feet. Example. A solid disc of cast-iron, rotating about its geometrical center, is six feet in diameter and of such thickness that it will weigh 4073.3 pounds. What is its moment of rotation or moment of inertia ? Radius of gyration = 3 X */ } and (radius of gyration) 2 = 3 2 X I Mass = +™A= 126.5 32.2 Moment of rotation = 126.5 X 3 2 X %. = 569.25. Note. — In all such problems relating to stored energy in rotating bodies, the radius of gyration is usually taken in feet and not in inches, as in previous examples of moment of inertia, when relating to strength of material. * Instead of multiplying the mass of the body by the square of radius of gyration in feet and calling the product moment of inertia, some writers multiply the weight of the body by the square of the radius of gyration in feet and call this product moment of inertia. This last expression for moment of inertia, of course, will have a numerical value of 32.2 times the first one. It does not make any differ- ence in the result of the calculation whether weight or mass is used, but the same unit must be adhered to throughout the whole calculation. MECHANICS. 299 Angular Velocity. When a body revolves about any axis, the parts furthest from the axis of rotation move the fastest. The linear velocity at a radius of one foot fro?n the center of rotation is called the angular velocity of the body. It is usually reckoned in feet per second. The angular velocity of any revolving body is ex- pressed by the formula, F a = 2 7T n V & = Angular velocity in feet per second. n = Number of revolutions per second. Rule. Multiply the number of revolutions per second by 6.2832 and the product is the angular velocity in feet per second. Example. What is the angular velocity of a fly-wheel making 300 revolutions per minute ? Solution : 300 revolutions per minute = 5 revolutions per second, therefore, angular velocity = 6.2832 X 5 =31.416 feet per second. Angular velocity expresses the linear velocity at unit distance from center of rotation and in English measure this unit is feet. As already stated, the moment of rotation is an expres- sion for the mass of the rotating body (theoretically) transferred to unit distance from center of rotation ; the product of angular velocity and moment of rotation will, therefore, be the momentum of the rotating body. The constant resistance which has to be exerted at unit radius in order to bring the body to rest in T seconds will be : R=^l T The resistance which has to be exerted at any radius of r feet to bring the body to rest in T seconds will be : R = ^l r T R = Resistance in pounds. Va. = Angular velocity in feet per second. / = Moment of rotation (also called moment of inertia). The constant force which has to be exerted at unit radius in order to bring the body from a state of rest to an angular velocity V & in T seconds will be : ' F - v* I 300 MECHANICS. The constant force which has to be exerted at any radius, r, in order to bring the body from a state of rest to an angular velocity V A in T seconds will be : r T R = Constant resistance in pounds. F = Constant force in pounds. V & = Angular velocity in feet per second. / = Moment of rotation (also called moment of inertia). r — Radius in feet at which the force is applied. T = Time in seconds that the force is acting. Example. A fly-wheel making 120 revolutions per minute and weigh- ing 483 pounds, is brought to rest in two seconds by a resistance acting at a six-inch radius. The radius of gyration of the fly- wheel is 1.2 feet. What is the average force exerted against the resistance during these two seconds ? Solution : 120 revolutions per minute = 2 revolutions per second. Angular velocity = 6.2832 X 2 = 12.5664 feet per second. Moment of rotation = 1.2 X 1.2 X 488 — =21.6 32.2 Radius of resistance, 6 inches = 0.5 feet. R = 12.5664 X 21.6 = mM ds 2 X 0.5 If a rotating body is not bronght to rest, but only reduced in speed from an angular velocity of Va to Va x in T seconds, then the average force or resistance acting at unit radius is : F = (Fa ~ Fa - } - T The average force which has to be exerted at any radius at t feet to reduce the angular velocity from Va. to Va x * n ^sec- onds will be : F _ (Fa - Fai)/ T r Example. A fly-wheel on a punching machine weighs 644 pounds, its radius of gyration is 1% feet, and it makes at normal speed 300 revolutions per minute, but when the machine is punching the MECHANICS. 30I speed is in £ of a second reduced to a rate of 280 revolutions per minute. What average force has the fly-wheel communi- cated to the pitch-line of a 6-inch gear on the fly-wheel shaft ? Solution : The mass of the fly-wheel = 644 = 20 32.2 The moment of rotation = {\y 2 f X 20 = 45 800 revolutions per minute = 5 revolutions per second. Angular velocity = 5 X 6.2832 = 31.416 280 revolutions per minute — 4 2 / 3 revolutions per second. Corresponding angular velocity = 4% X 6.2832 = 29.3216 6-inch diameter of gear = 3-inch radius = % foot. F _ (31.416 — 29.3216) X 45 F = 2.0944 X 45 X 5 X 4 = 1884.96 pounds. The kinetic energy in foot-pounds stored in the revolving body may be obtained by the formula : F a 2 X = kinetic energy . Decreasing the angular velocity to F a i, the stored-up energy will also decrease to w x 4 and the work done by the revolving body will be (F a 2 — Fal 2 ) X -L Example 1. The moment of rotation in a fly-wheel is 1040 ; its angular velocity is 5 feet per second. What is the stored-up energy in the wheel ? Solution : Kinetic energy = 5 2 X 1040 = 13,000 foot-pounds. Example 2. At certain intervals, when machinery is started, the angular velocity of this fly-wheel is reduced to 4^ feet per second. How many foot-pounds of energy has the fly-wheel given up in helping to drive the machinery ? 302 MECHANICS. Solution : x = (5 2 — (4^) 2 ) X 520 x — (25 — 20X) X 520 x = \% X 520 = 2470 foot-pounds of energy given out by the fly-wheel during this change of speed. Example 3. How much stored energy is left in the wheel after its angu- lar velocity is reduced to 4>i feet per second ? Solution : K = (±/ 2 ) 2 X 520 = 20X X 520 = 10,530 foot-pounds. The same result may be obtained by subtracting, thus : 13,000 — 2470 =10,530 foot-pounds. Centrifugal Force. The centrifugal force is the force with which a revolving body tends to depart from its center of motion and fly in a direction tangent to the path which it describes. The centrip- etal force is the force by which a revolving body is prevented from departing from the center of motion. When the centri- fugal force exceeds the centripetal force the body will move away from the center of motion, but if the centripetal force ex- ceeds the centrifugal force, the body will move toward the center of motion. The centrifugal force in any revolving body is equal to the mass of the body (see page 286) multiplied by the square of its velocity, and this product divided by the radius of the revolving body. , WXv 2 m X v 2 32.2 X r~ r cf= Centrifugal force in pounds. r = Radius in feet. i) — Velocity in feet per second. W = Weight of moving body in pounds. m = Mass of moving body. Example. The weight a, in Fig. 11, is four pounds, and the length of the string is two feet; the weight is made to swing around the center c, three revolutions per second. What is the stress on the string due to centrifugal force ? MECHANICS. 303 Solution : The distance from c, to the center of the ball is two feet, and making three revolutions per second, the velocity will be 2 X 3 X 3.1416 X 2 = 37.7 feet per second. cf= 4 X m X m = 88.2 pounds. 32.2 X 2 In metric measure, W X v 2 cf=. 9.81 X r cf= Centrifugal force in kilograms. r = Radius in meters. v = Velocity in meters per second. W=- Weight of moving body in kilograms. Example. Suppose that the weight a, in Fig. 11, is five kilograms, swinging around the center, c, one revolution per second ; the distance from a to c isl^ meters. What is the stress on the string due to centrifugal force ? Solution : The velocity will be 1.5 X 3.1416 X2 = 9.4248 meters per second. 5 X 9.4248 X 9.4248 = kilograms . J 9.81 x \y 2 Friction. The resistance which a body meets with from the surface on which it moves is called friction. It is called sliding friction when one body slides on another; for instance, a sleigh is pulled along on ice — the friction between the runners of the sleigh and the ice is sliding friction. It is said to be rolling friction when one body is rolling on another so that new surfaces continually are coming into contact; for instance, when a wagon is pulled along a road, the friction between the wheels and the road is roll- ing friction, but the friction between the wheels and their axles is sliding friction. Sliding friction varies greatly between different materials, as everybody knows from daily observation. For instance, a sleigh with iron runners can be pulled with less effort on ice than on sand, even if the road is ever so smooth. This is because the friction between iron and ice is a great deal less than the friction between iron and sand. 304 MECHANICS. Coefficient of Friction. 300 lbs. The ratio between the force required to overcome the re- sistance due to friction and the weight of a body sliding along a horizontal plane is called coefficient of friction. For instance, in Fig. 12 a f»g. 12 piece of iron weighing 300 lbs. ^ rests on a horizontal plate b. A /^, ^ ^ - string fastened to a, goes over a v§~7 ' pulley, c. At the end of the ^~^ string is applied a weight, d. If L^ this weight is increased until r-^-\ 50Jb the body a just starts to move - * along on b, and the weight is found to be 50 pounds, the co- efficient of friction will be 50 — — — 0.1667 300 — 6 When the weight of a moving body is multiplied by the coefficient of friction, the product is the force required to keep the body in motion. Of course, any pressure applied to the moving body, perpendicular to its line of motion, may be sub- stituted for its weight. For instance, the frictional resistance of the slide in a slide-valve engine is not due to the weight of the valve, but to the unbalanced steam pressure on the valve. In all cases the rule is : Multiply the coefficient of friction by the pressure perpen- dicular to the line of motion, and the product is the force required to overcome the frictional resistance. Example. The coefficient of friction is 0.1, and the weight of the sliding body is 800 pounds. What force is required to slide it along a horizontal surface ? Solution : Force = S00 X 0.1 = 80 pounds. Rolling Friction. If the body, a, (see Fig. 12) was lifted up from the plane, b, high enough so that two rollers could be placed between a and b, it would be found that the body would move with much less force than 50 pounds because, instead of sliding friction, as in the first experiment, it would be rolling friction. Suppose it is found that a commenced to move when the load, d, was four pounds, then the coefficient of friction for this particular case would be 4 - = -L — 0.0133 300 75 In these experiments the whole force at d is not used to move the load a, as a small part of it is used to move the pulley at c, but in order to make the principle plain, this loss has not been considered. MECHANICS. 305 Axle Friction. The friction between bearings and shafts is frequently called axle friction. This, of course, is sliding friction, but owing to the fact that the surfaces in question are usually very smooth and well lubricated, the coefficient of friction is smaller than for ordinary slides. Example 2. A fly-wheel weighs 24,000 pounds, the diameter of the shaft is 10 inches, and the coefficient of friction in the bearings is 0.08. What force must be applied 20 inches from the center in order to keep the wheel turning ? Resistance due to friction — 24000 X 0.08 = 1920 pounds. This resistance is acting at a radius of 5 inches, but the force is acting at a radius of 20 inches ; therefore, the required force necessary to overcome friction will be -.— ° x 5 =480 pounds. 20 ^ How much power is absorbed by this frictional resistance if the wheel is moving 72 revolutions per minute ? Solution : The space moved through by the force is 72 x 20 X J x 3 - 1416 = 753.984 feet, and 753.934 X 480 = 361,912.32 foot-pounds and 361912.32 _ 1Q Q7 horse . power> 33000 * Horse=Power Absorbed by Friction in Bearings. The horse-power absorbed by the friction in the bearings for any shaft may be figured directly by the formula, H-P = W X f X n X 3 - 1416 x d 33000 X 12 This reduces to : HP = IV X / X n X d X 0.000008 H-P = Horse-power absorbed by friction. W — Load on bearings in pounds. d= Diameter of shaft in inches. /■=■ Coefficient of friction. n = Number of revolutions per minute. Calculating the previous example by this formula, we have : H-P = 24000 X 0.08 X 72 X 10 X 0.000008 = 11.06 horse- power, which is practically the same as figured before. 306 MECHANICS. Angle of Friction. Suppose, instead iof using the string and the weight d (see Fig. 12), that one end of the plane is lifted until a commences to slide ; the angle between b and the horizontal line, when a com- mences to move, is called the angle of friction. The coefficient of friction may also be calculated from the angle of friction, thus : If the body commences to slide under an angle of a degrees, the coefficient of friction will be sm " a = tang. a. Thus, the coefficient of friction is always equal to tangent of the angle of friction. Rules for Friction. 1. Friction is in direct proportion to the pressure with which the bodies are bearing against each other. 2. Friction is dependent upon the qualities of the surfaces of contact. 3. The velocity has, within ordinary limits, no influence on the value of the coefficient of friction. 4. Sliding friction is greater than rolling friction. 5. Friction offers greater resistance against starting a body than it does after it is set in motion. 6. The area of surfaces of contact has, within ordinary limits, no influence upon the value of the coefficient of friction, but if they are unproportionally large or small the friction will increase. TABLE No. 36.— Coefficient of Friction. Slides. i Bearings. 1 Materials. Well Lubri- cated. Not well Lubri- cated. Well Lubri- cated. Not well Lubri- cated. Cast-iron on wrought iron .... Cast-iron on cast-iron Wrought iron on brass Wrought iron on wrought iron . . 0.08 0.08 0.08 0.10 0.16 0.16 0.20 0.20 ] 0.05 0.05 0.05 0.05 0.075 0.075 0.075 0.075 Friction in Machinery. When the surfaces are good the frictional resistance for slides may be assumed as 10 per cent., more or less, according to the conditions of the surfaces. It is always well not to take the coefficient of friction too small; it is better to be on the safe side and allow power enough for friction. In bearings for machinery, the frictional resistance ought not to absorb over six per cent. If more is wasted in friction, there is a chance for improvement. MECHANICS. 307 150 lbs- Pulley Blocks. When friction is not considered, the fig. 13. force and the load will be equal in a single fixed pulley (as A, Fig. 13). Thus, a single fixed pulley does not accomplish anything further than to change the direction of motion. In a single movable pulley (as fat B, Fig. 13), the force is equal to only half the load; thus, 75 pounds of force will lift 150 pounds of load, but the force must act through twice the space that the load is moved. The tension in any part of the rope in B is half of the load W\ thus, when the load is 150 pounds the tension in the rope is 75 pounds, when arranged at B, but it is 150 pounds when arranged as at A. Fig. 14 shows a pair of single sheave pulley blocks in position to pull a car ; when the blocks are arranged as at A y and friction is not consid- ered, a force of 100 pounds on the hauling part of the rope exerts a force of 300 pounds on the post, but only 200 pounds on the car ; but, turning the blocks end for end, as shown at B, a force of 100 pounds on the hauling part of the rope exerts a force of 300 pounds on the car and 200 pounds on the post. This is a point well worth remembering when using pulley blocks. Suppose, for instance, that a man exerted a force of 100 pounds on the hauling part, and that it required 250 pounds of force to move the car ; if he used the pulley blocks as shown at A, his work would be useless, as far as moving the car is concerned, as he could not do it, but turning his blocks end for end he could accomplish the desired result. Always re- member whenever it is possible to have the hauling part of the rope coming from the movable block and pull in the same direction as the load is moving. Friction in Pulley Blocks. In practical work, friction will have some influence, and, to a certain extent, change these results, because some of the ten- sion in the rope is lost by friction in each sheave the rope passes over, therefore the tension in each following part of the rope is always less than it was in -the preceding part. This loss must be obtained from experiments. In good pulley blocks, having roller bearings, this loss is probably not more than 0.1, and we 3 o8 MECHANICS. get a useful effect of 0.9 of the force from one part of the rope to the next ; therefore, when friction is considered, the useful effect in the following cases will be : In single sheave blocks having the hauling part from the movable block (pulling with the load as in B, Fig. 14). IV=F(1 +0.9 + 0.9 2 ) W=FX 2.71 In single sheave blocks having the hauling part from the fixed block (pulling against the load as in A, Fig. 14), W = F (0.9 + 0.9 2 ) JV = FX 1.71 In double sheave blocks having the hauling part from the movable block, W — F (1 + 0.9 + 0.9 2 + 0.9 3 + 0.9 4 ) IV=F X 4.1 In double sheave blocks having the hauling part from the fixed block, JV=F (0.9 + 0.9 2 + 0.9 3 4- 0.9 4 ) W — F X 3.1 Differential Pulley Blocks. In a differential pulley block (see Fig. 15), the proportion between the force and the weight, when friction is neglected, is expressed by the formula : F = W X(R — r) 2 X R The actual force required to lift a weight by such a pulley block is about three times the theoretical force, as calculated above. Inclined Plane. When a weight is pulled upward on an inclined plane, as shown in Fig. 16> and the force F\% acting parallel to the plane, the required force for moving the body will be F = W X sin. a plus friction, and the perpen- dicular pressure P, against the plane will be IV X cos. a. — Cos. -a- MECHANICS. 309 Example 1. The weight, W, (Fig. 16) is 100 pounds; the angle a is 30°. What force, F, is required to sustain this weight, friction not considered ? Solution : Sin. 30° = 0.5 Thus : F = W X sin. 30° = 100 X 0.5 = 50 pounds. Example 2. What is the perpendicular pressure under conditions stated in Example 1 ? Solution : P = W X cos. a = 100 X 0.86603 = 86.6 pounds. Therefore, the frictional resistance between the sliding body and the inclined plane will be only what is due to 86.6 pounds pressure ; in other words, the force required to over- come friction will be IV X /* X cos. a. Example 3. What force is required to move the body mentioned in Example 1 when friction is also considered, taking coefficient of friction, F, as 0.15? Solution : F ' = W (sin. a + cos. a X /) ^ = 100X (0.5 + 0.86603X0.15)= 100 X 0.6290 =62.99 pounds. Note. — This is the force required for moving the load. In order to put it in motion more force must be applied, varying according to velocity, but after motion is commenced the speed would be, under these conditions, maintained forever by this force of 62.99 pounds. When a load is moving down an inclined plane the force due to W X sin. a will assist in moving the body, and if the product W X sin. a exceeds the product W X cos. a X f the body will slide by itself. For instance, in the body mentioned in the previous example, the force required to overcome gravity, regardless of friction, is 50 pounds, and the force required to overcome friction is 12.99 pounds; thus, if the body should be let down the plane instead of pulled up, it would have to be held back with a force of 50 — 12.99 = 37.01 pounds. Note. — When the incline is less than 1 in 35, cosine is so nearly equal to 1 that it may be neglected, and the force required to overcome friction may be considered to be the same as on a level plane. For instance, a horse is pulling a load and ascend- ing a gradient of 1 in 35 ; if the tractive force required to pull the load on a level road was 30 pounds and the weight of the load was 1400 pounds, when ascending the hill, the horse will first 310 MECHANICS. have to exert a force of 30 pounds, which is all due to friction, but beside that he must also exert a force of ^ times 1400 = 40 pounds ; thus the total pull exerted by the horse will be 70 pounds. Inclined Plane With the Force Acting Parallel to the Base. When the pressure is continually acting in a line parallel to the base of ^/\ T the incline, as F, (see Fig. 17) which is frequently the case in mechanical movements, as for instance, in screws, some kinds of cam motions, etc., it ^^ ? ^_ will require more force to move the body than it would if the force was acting parallel to the incline. When force acts parallel to the base, as in Fig. 17, the force required to move the body, if friction is not considered, will be : F = ^Xsin.* = WXtzng.a cos. a Example 1. What force is required to move 100 pounds upward an incline of 30°, as in Example 1, excepting that the force is acting parallel to the base instead of parallel to the incline ? Solution : F — W X tang. 30° F=^'X100X 0.57735 — 57.74 pounds. When both the friction and the weight of the body are con- sidered, the force required to move the body will be : F = W X sm - a + (/ x cos - a ) cos.a — (f X sin.rt) Example 2. What force is required to move 100 pounds upward an in- cline of 30° (as in Example 1) if the force is acting parallel to the base line instead of parallel to the incline ; coefficient of friction is supposed to be 0.15 ? Solution : F = 100 X sin - 30 ° + (0 - 15 X cos - 3Q °) cos. 30° — (0.15 X sin. 30°) F _ 100 x 0.5 + (0.15 X 0.86603) 0.86603 — (0.15 X 0.5) F= 10Q 0.5 + 0.1277045 0.86603 — 0.075 F= 100 X 0.7936 = 79.36 pounds. MECHANICS. 3 11 Note. — From these calculations it is seen that it is more advantageous to apply the force parallel to the incline than parallel to the base. When force is applied parallel to the in- cline : The force required to overcome gravity = 50 pounds. The force required to overcome friction == 12.99 pounds. Total force = 62.99 pounds. When the force is acting parallel to the base : The force required to overcome gravity = 57.74 pounds. The force required to overcome friction = 21.62 pounds. Total force = 79.36 pounds. II F R X 2 7T Weight of the ;load lifted, or force exerted, if the Screws. When friction is not considered, the force which may be exerted by a screw (see Fig. 18) will be : F X 7?X2tt r , W X P P W screw acts as a press. F = Acting force. R = Radius in inches at which the force acts. P = Pitch of screw in inches. FIG. 18 Regarding friction in screws, the thread of a screw may be considered as an in- clined plane, of which the cos. is the middle circum- ference of the screw, the sin. is the pitch, and the force is acting parallel to the base. Hence the fol- lowing formula : F= W X P +/ X d - r dn — fX P X R F = Force, acting at a radius of R inches. W = Weight. P = Pitch of screw in inches. / — Coefficient of friction, usually taken as 0.15. R = Radius in inches at which the force is acting. r == Middle radius of screw in inches. d = Middle diameter of screw in inches. 312 MECHANICS. Example. Find the force required to act on a lever 30 inches long (see Fig. 18) in order to lift the load W, which is 8000 pounds ? The screw is ^-inch pitch and 1^-inch middle radius : coefficient of friction, 0.15. Solution : F — 8000 X °- 5 + ° 15 x 3.1416 X 2.5 1.25 2.5 X 3.1416 — 0.15 X 0.5 X 30 F= 8000 X i' 6781 X 0.0416 =. 89.6 pounds. 7.779 F When the screw has V thread, the frictional resistance will be increased as -^ of the angle a (see Fig. 18), or equal to secant of half the angle of the thread. For United States standard screws the angle of thread is 60°, half the angle is 30°, and secant of 30° is 1.1547, and the formula will, for United States standard thread, become : d- — \.\bfP R All the letters having the same meaning as in the formulas for the square-threaded screws. The following table is calculated for square-threaded screws, the pitch of the screw being double that of the United States standard screw of same diameter. The depth of the thread is equal to its width. We see no good reason why the depth of a square-threaded screw should be, as frequently given in tech- nical books, \% of the pitch of the screw ; ||, as given in pre- vious tables, is more convenient, and also gives a little more wearing surface to the thread. The use of this table is so plain that it needs very little explanation. In the fourth column is the area of the outside diameter of the screw. In the fifth column, the sectional area of the screw at the bottom of the thread, which may be used in calculating the tensile and crush- ing strength of the screw. Subtracting the fifth column from the fourth gives the sixth column, which is the projected area of one thread; this may be used in calculating the allow- able pressure on the thread, etc. The fourteenth column gives the tangential force which is required to act with a leverage of one foot in order to lift one pound by the screw if there was no friction. The fifteenth column gives the total tangential force required per pound of load when both load and friction are included. The sixteenth column gives the difference between the fourteenth and the fifteenth columns, and is the tangential force absorbed by friction alone. The coefficient of friction in both columns is assumed as 0.16. The last four columns in the table give the load or axial pressure which may be allowed on the screw corresponding to 200, 400, 600 and 1000 pounds pressure per square inch of projected area of screw thread when the length of the nut is twice the diameter of the screw. qjB3Jq} JO E3JB p9JD9fOjd JO qOUI 3JBnbs jad spunod 0001 °* SurpuodsajJOD ajnssajd JBtxy CNb-OOiOOOrtiOOOOOOO lOiOOiNffiCJt-Ot-OMWt-®^ t-MMONOOCOIMM^INNQO HHH(MjqMW-*^l-0 •pB3jq| JO B3JE pajoafoid jo qoui ajBnbs jad spunod 009 °* Suipuodsajioo ajnssaid jEixy OiOrHQOHt-iOOO^INaiOOiOO CS?0«.t-'rHr-?0005COi-ii-ioar- OOOOOOi-tr-iT-i— i— i ri !N iM !N (M CO CC ooppoppopppoppooop •pajapisuoo 3JB UOpOUjJ pUE pBOi qjbq uaqM 3DJOJ JBIlUaguEJ, <© CO "* C5OHM«MMrHX«'#Or-X<-i^OC0 ^ct-oiHo:oa)QrH(Ncoi-oiN^x>c-» OOOOhhhhh(N^(N!N(NCOW^^ pppppppppppppppppp •paispxsuoo jou uoxjoxi^ 'pBoq 3aojai oi pajxnbai 30JOJ p311U3gUBJ, T* iH »0 o ^* © ot x ■* o: Ci o © o c o « a o o x ?)(MNnco^oioio©©t-i-a)xxO'- OOOOOOOOOOOOOOOOt-t-h ppppppppppppppppop •pEMqx j° m^a t-HOK5Wt-OMMOO®©X«MO© t-aoisi i ®ONnci5xxo:;MO^ PprHrHrH-rHiyjCqC^^OaO^O-ICOCOCOThTt' •qOUI UB JO SIBXUI03Q UI T.OOJ, SmpEajqj, jo sssujpxq^ t-HOi0Wt*OWMOOCC©00MMO?0 f-050(N^©0(N'»LOiOXXOCOMO'* OOHHrtH{MiN(N(N!N!N(NMWM'* , t •l°°L SuipBSjqx i° s^ui^MX « l- O CO •qoiX<£ ipui 5##:**;£###£;s*;j:sp3p#;p# •qo UI jad spEsaqj, jo jsquin^j ri\J\ H\ h\h\h\ »\eo\«5\ h\h\h\h\ ©lOlO^MCOfNfN^Oq^HHHHrHHH •AV3J0g JO 'UIEXQ SB SUO^ SB 30XM1 JttN^ B ux spBSjqj^ jo - o^ ©Ol'-XXffiXaOOHO-'r.r-'NiMK •rnjq; aqj ux spBsaqx jo 3Dbj -jng SuxiBag jo B9jy I B:10 X COs cq k ^ in n x © tH rH (N -tf" CO q T-t ?i ^ t-; q ■* x is i-; q © © is ^_ t- x iq OOOOOi-HTHr-Jc^COCO^iH >S SO l~ 00 CO © •AV3JDg JO UOIlD9g JO E3iy © I- .» O t- h O (M © CI O © © H 13 © )C ■* CiO^XSlSO'ft-OrHCDaKN'^COMt- h 05 ■* r- n t-; ■<* r - c q oi q n © q © « dddoHHNMcc^ioVxoJHfNOJaJ HHH.(N '* n N 1° V§u 3 l \^<\M \N \?1 \W \^1 \^< \W HHrt(M5lMo:^Tt'».'5iO©©t-F-XO(M •pBaaqx jo xnojjog ye A\ajog jo j3J3uiexq ©MOOM<©0'*'1'COXXiii*TJ'OX Tf^K5>0©OiCiOOCiO(Mt-XXCOOO .w^iqt-qrtM^xq^^qxqcofNrt dddddrJHHrHNjjiNffi'jiiMMriilo •A\3JDg JO J913UIEIQ rhlVtf- rt^^M-- ri^^j^ ^^^Neo^ rtHHrtffKN^NMMWCO^O© (313) 3M MECHANICS. The table on page 313 was calculated by the following formulas : When friction is not considered : ■n, . , , , , Pitch in inches Pitch in inches Force to balance load = = — — 12 X 2 7T 75.4 When both one pound of load and friction are considered, Force — ( P* tc h * n i" c h es + middle circum. X/ \ s/ / middle radius \ ^■middle r.irr.nm. — nitch in inches X f ^ middle circum. — pitch in inches X f 12 calculations by table on preceding page. Example 1. A j ack screw, as shown in Fig. 19, is 1)4" diameter, three threads per inch. What tan- gential force is required to act with a leverage of 18 inches in order to lift 5000 pounds ? Co- efficient of friction in the thread is assumed as 0.16. Tangen- tial force absorbed in friction by the collar at a is assumed to be equal to force absorbed by friction in the thread of the screw, and may, therefore, be taken from the thirteenth column in the table. Solution : Tangential force per pound at 1 foot radius = 0.0133 Tangential force absorbed by friction in collar = 0.0089 Total force per pound of load at 1 foot radius = 0.0222 The tangential force is acting with 18 inches leverage = \y 2 feet, and the load is 5000 pounds; therefore, the required force will be, F = 0.0222 X 5000 = 74 pounds. Example 2. A load of 16,000 pounds rests on a slide and is moved back and forth on a horizontal plane by a screw. The coefficient of friction between slide and plane is 0.1, and the screw should not be loaded with more than 400 pounds per square inch of projected area or thread. Find the suitable diameter of screw. If a pulley of 20-inch diameter is attached to the end of the screw, also find the tangential force required to act at the rim of the pulley in order to turn the screw. MECHANICS. 315 Solution : The coefficient of friction for the slide is 0.10, therefore the axial pressure on the screw will be 16,000 X Ho = 1600 pounds. The allowable force on a 1 ^-inch screw will be found in the table to be 1742 pounds ; therefore, select a screw of 1% inches diameter and a length of nut of 2^ inches. Assuming the friction due to the reaction of the screw against its collar and bearing to be equal to the friction in the thread, and using the table, we have : Force per pound at one foot radius = 0.0112 Force absorbed by friction in collar = 0.0074 Total force per pound of load at one foot radius = 0.0186 The leverage of a 20-inch pulley is 10 inches = 1 %2 foot, and the axial force is 1600 pounds ; therefore, the tangential force required at the rim of the pulley will be : F= ^Ol^XJ^ =36 . 7pounds> 10 /l2 36.7 pounds is really the force required to keep the body in motion after it is started. To start the body from rest requires somewhat additional force, depending on the time used in over- coming its inertia. It is not certain that the friction due to the reaction of the screw against the collar is equal to the friction in the screw. It may be more or it may be less ; this will, to a certain extent, depend on the size of the collar, and also on the finish of its surfaces, its means of lubrication, etc. Therefore, instead of assuming this resistance to be equal to the friction of the thread as found in column 16, it may be calculated for each individual case by assuming a proper coefficient of friction and assuming that this friction acts as resistance at a radius equal to the middle radius of the collar. If a screw is acting under the circumstances illustrated in Fig. 18, there is no collar to absorb any of the force by friction ; but whenever the screw acts against a shoulder this friction must never be forgotten in calculation. Ball bearings may be used to very good advan- tage in the thrust collar on a screw. If a screw works a load continuously up and down, and the weight of the load always rests on the screw, it is necessary to be very careful and allow only a limited load on the screw (only a fraction of what is given in the table), because the pressure of the load always acts on the same side of the thread, and this is very disadvantageous for lubrication, as it does not give the oil a good chance to get onto the surfaces which rub against each other ; but when the screw works a slide with an alternate push and pull, the wear comes on both sides of the thread, which gives a good chance for lubrication, and an axial pressure of 400 pounds per square inch of projected area of bearing surface in the thread will be 3 1 6 MECHANICS. safe, although, under certain circumstances, for instance, in a mechanism working continuously, such a load may be too much for the best results with regard to wear. For anything working like a jack-screw, when the diameter of the screw is over one inch, the load given in the last column is perfectly safe. It is impossible to give rules which will suit all cases ; the experience and judgment of the designer are the best guide with regard to the selection of the proper load. It may seem too much to use 0.16 as the coefficient of friction in the thread of the screw, but the author believes, from careful experiments made on common square-thread screws, as used in commercial machinery, — not made for experimental purposes, but for every-day use, — that this coefficient of friction is a safe average. It is well to remember that the surfaces of the thread on screws with cast-iron nuts do not always have the best of finish, and the nut especially is liable to be a little rough when new ; therefore, this coefficient of friction may be a little greater than that found in screws in machinery when well lubricated and with surfaces smoothed down and glazed over from wear. The Parallelogram of Forces. A line may be drawn to such fig. 20. scale that its length represents a ^""T\ given force acting in the direction c^' \ \ of the line. Another line is drawn to the same scale, from the same point of application, and its length represents another force acting in the same direction as this line. If these two lines are connected by two auxiliary lines, a parallelogram is formed and the diagonal of the parallelogram will represent both the magnitude and the direction of the resulting force. Example. Let the lines a and b in Fig. 20 represent two forces acting in the direction of the arrows. Draw the lines to any scale, for instance, T \ inch to a pound ; if the force represented by a is 64 pounds, the line a will be 64 X T V = 4" long. If the force rep- resented by b is 50 pounds, this line will be 50 X T X , T = 3>6" long. Completing the parallelogram by drawing lines c and £ horse-power. Circular saws (for wood), 36" di- ameter (light work), 6 horse-power. Fan blower for cupola, melting four tons of iron per hour, 10 horse-power. Fan blower for five blacksmith fires, 1 horse-power. Drop hammer, 800 pounds, 8 horse-power. In machine shops and similar places, from 40% to 70% of the total power required is consumed in running the line shaft- ing and counter-shafts. An average of from 55% to 60% is probably the most common ratio. In exceptionally well-arranged establishments, under favor- able conditions, in light. manufacturing it may be possible that only 30% of the power is consumed in driving line and counter shafting, and that 70% is used for actual work. 320 SPEED OF MACHINERY. SPEED OF MACHINERY. The peripheral velocity of circular saws ought not to exceed 10,000 feet per minute. Table No. 37 gives the number of revo- lutions per minute for circular saws of different diameters. TA BLH No. 37- Diameter of saw in inches. I 8 I 10 12 14 16 20 24 28 32 Number of revolutions per minute. 1 4500 3600 i 1 3000 2585 2222 1800 1500 1285 1125 Band Saws. Small band saws, such as are usually used in carpenter shops, have a velocity of 3600 feet per minute. The reason why band saws are run so much slower than circular saws is that if the band saw is given too much speed the blade will be pulled to pieces in starting and stopping. Drilling Machines for Iron. For drilling steel, the surface speed of a drill should not exceed 15 feet per minute; cast-iron, 22 feet; brass, 27 feet; malleable iron. 25 to 30 feet per minute. The feed will vary according to the hardness of the stock. In cast-iron a %" drill will drill a hole 1" deep in 125 revolutions. A X" drill will drill a hole 1" deep in 120 revolutions. A 1" drill will drill a hole 1" deep in 100 revolutions. Lathes. Cast-iron may be turned at a speed of 32 feet per minute when Muchet steel is used for tools. Thus, lathes are usually calculated to have a velocity of about 30 to 32 feet on the slow- est speed, supposing that as large a diameter as the lathe will swing is turned. For wood-turning the surface speed may be from 3000 to (>000 feet per minute ; but when the article to be turned is out of balance the speed must be considerably slower. Planers. Cast-iron is planed at a speed of 25 to 27 feet per minute ; wrought iron, 21 feet; steel, 16 feet per minute. A planer ought to return at least three times as fast as it goes forward. nilling Machines. Rotating cutters working on Bessemer steel or other mate- rials of about equal hardness usually have a surface speed of * The speed of metal working machines may be greatly increased by using tools made from high speed steel. SPEED OF MACHINERY. 32] about 40 feet per minute. Oil is used for lubrication. Cast- iron is milled without oil. Grindstones. When grindstones are used to grind steel and iron in manu- facturing, they work at a surface speed of 2000 to 2500 feet per minute, but grindstones for common shop use, to grind tools, chisels, etc., run at much slower speed. Emery Wheels and Emery Straps. Emery wheels and straps do good work at a speed of 5000 to 6000 feet per minute, but all such high-speed machinery, especially grindstones and emery wheels, must be used very carefully and special attention paid to the strength, so that they will not break under the stress of centrifugal force. Calculating Size of Pulleys. TO FIND SIZE OF PULLEY ON MAIN SHAFT. Multiply the diameter of pulley on counter-shaft by its number of revolutions per minute, and divide this product by the number of revolutions of the main shaft, and the quotient is the diameter of the pulley on the main shaft. Example. A main shaft makes 150 revolutions per minute ; the counter- shaft has a pulley 9 inches in diameter and is to make 400 revolu- tions per minute. What size of pulley is required on the main shaft? Solution : Diameter of pulley = 400 X 9 = 24 inches. 150 to find size of pulley on counter-shaft. Rule. Multiply the diameter of pulley on the main shaft by its number of revolutions per minute, and divide this product by the number of revolutions of the counter-shaft; the quotient is the diameter of the pulley on the counter-shaft. Example. The pulley on a main shaft is 36 inches in diameter and it makes 150 revolutions per minute ; the counter-shaft is to make 450 revolutions per minute. What size of pulley is required ? Solution: Diameter of pulley = 36 X 150 = 12 inches. 450 322 SPEED OF MACHINERY. TO FIND THE NUMBER OF REVOLUTIONS OF THE COUNTER SHAFT. Rule. Multiply the diameter of pulley on the main shaft by its number of revolutions per minute and divide this product by the diameter of pulley on the counter-shaft, and the quotient is the number of revolutions of the counter-shaft per minute. Example. The pulley on a main shaft is 24 inches in diameter and makes 150 revolutions per minute, and the pulley on the counter- shaft is 15 inches in diameter. How many revolutions per minute will the counter-shaft make ? 24 X 150 Number of revolutions = = 240 revolutions per minute. 15 F To Calculate the Speed of Gearing. In calculating the speed of gearing, use the same rules as for belting, but take the number of teeth instead of the diameter. Example. The back gearing on a lathe consists of a gear and pinion of 8 pitch, 96 teeth and 32 teeth, and the other gear and pinion are 10 pitch, 120 teeth and 40 teeth. How many revolutions will the cone pulley make while the spindle makes one revolution? Solution : Cone pulley makes = - — _ = 9 revolutions. y 32 X 40 Efficiency of Machinery. Divide the energy given out by a machine by the energy put into the same machine ; multiply the quotient by 100, ancl the result is the per cent, of efficiency of the machine. Example. A dynamo requires 15 horse-power, but the electrical power given out is only 12 horse-power. What is the efficiency? Solution : Efficiency = — X 100 = 80% 15 A steam engine is to develop 60 horse-power net. What will be the gross horse-power if the efficiency is 75% ? Solution : Gross power = — = 80 horse-power. CRANE HOOKS. 323 CRANE HOOKS. Fig. 3. Crane hooks, as shown in Figs. 1, 2 and 3. may be designed by the following formulas : P=B 2 D^s/'F P = Load in tons. D = Diameter of iron in inches. e=±y 2 D f=2B g=l%D i —\% D j=7/zD t=%D i = y A D m = iy 2 D k = iy s L> S = Standard screw of diameter r n at the bottom of the thread. When a rectangular iron plate is substituted for a washer, the bearing surface of the plate against the wood should at least be equal to the area of the washer, calculated by the above formula. Chain Links. (See Figure 4.) D = Diameter of iron. L = 4y 2 to 5 D. B = sy 2 n. (For strength of chains, see page 222). 324 CRANES AND DERRICKS. CRANES. Cranes and derricks are machines used for raising and lowering heavy weights. In its simplest form, a crane con- sists of three principal mem- bers : The upright post, the horizontal jib and the diagonal brace. (See Fig. 5). The weight P will produce tensile stress in the jib, compressive stress in the brace, and both compressive and transverse stress in the post. FIG. 5. Tension in jib P X y P X z Compression in brace == y Stress in the upper bearing PX /i When the post is held at both ends, as in Fig. 5, it may. with regard to transverse strength, be considered as a beam of length z", fastened at one end and loaded at the other with a load equal to the force h X P The compression on the post caused by the load is equal to P. The downward pressure on the lower bearing is equal to the sum of the weight of the crane and the load which it supports. Proportions for a Two=Ton Derrick (Of the construction shown in Fig. 6) . Pulley blocks should be double-sheave (only single are shown in the cut). Circumference of manila rope, 3% inches. Mast. 8X8 inches, 26 feet long. Boom, 7 X 7 inches, 20 feet long. CRANES AND DERRICKS. 325 Large gear, 72 teeth, 1- inch circular pitch, 2-inch face. Small pinion, 12 teeth, 1-inch circular pitch, 2-inch face. Crank shaft. \y z inches in diameter. Bearings, 2^ inches long. Crank, 18 inches long, Drum, 7 inches in diame- ter, 24 inches long. Drum- shaft, 2X inches in di- ameter. The drum and large gear are fitted and keyed to the drum shaft and also bolted together, thereby relieving this shaft from twisting stress. The radius of the drum added to the radius of the rope makes four inches, and the force is multiplied five times by the double-sheave pulley block; therefore, when Gear the friction in thec ra nk mechanism is not consid- ered, the force required on the crank in order to lift 4400 pounds will be : r, _ 4 X 12 X 4400 18 X 72 X 5 = 33 pounds, very nearly. Thus, when two men are working the derrick (one at each crank), each man has to exert a force of 16>^ pounds, but, including friction, each man probably exerts a force of 20 to 25 pounds, when the derrick is loaded to its full capacity. For very rapid work it is necessary to have four men (two on each winch-handle) to work the derrick, if it is kept loaded to its maximum capacity, but for ordinary stone work such a derrick is usually worked by two men. Stones as heavy as two tons are seldom handled, except where larger derricks and steam power are used. When the derrick is to be worked constantly, the limit of the average stress on the crank handle to be allowed for each man is 15 pounds. When working an 18-inch crank, 48 turns per minute, this corresponds to a force of 15 pounds acting through a space of a little over 220 feet = 3300 foot-pounds of work per minute = yV horse-power. When the crank swings in a shorter radius a few more turns per minute may be expected, but experience indicates that an 18" radius is the most practical proportion. 326 BELTS. BELTS. Oak-tanned leather is considered the best for belting. The so-called " short lap " is cut lengthwise from the middle of the back of the hide, where it has the most firmness and strength. Single belting more than three inches in width is about T y thick, and weighs 15 to 16 ounces per square foot; when less than three inches in width it is usually f'j" thick and weighs about 13 ounces to the square foot. Light double belts, as used for dynamos and other ma- chinery having pulleys of comparatively small diameter, are about ¥ V ; thick and weigh about 21 ounces per square foot. Double belting, as used for main belts, is a little heavier and weighs from 25 to 28 ounces per square foot. Belts as heavy as 30 ounces per square foot are frequently used, and are usually termed "heavy double." Large engine belts are sometimes made with three thicknesses of leather. Belts should be soft, pliable and of even thickness. When a belt is of uneven thickness and has very long joints, so that it looks as if it was partly single and partly double, it is very doubtful if it will do good service, for this is a sure sign that the thin and flimsy parts of the hide have been taken into the stock in making the belt. The ultimate tensile strength of leather belting is from 2600 to 4800 pounds per square inch of section. Thus, a leather belt T y thick will break at a stress of 500 to 900 pounds per inch of width. The lacing of belts will reduce their strength from 50 to 60 per cent.; therefore, when practicable, belts ought to be made endless by cementing instead of lacing. A belt will transmit more power, wear better and last longer, if it is run with the grain side next to the pulley. Belts should never be tighter than is necessary in order to transmit the power without undue slipping ; too tight belts cause hot bearings, excessive wear and tear, and loss of power in over- coming friction ; but, on the other hand, it is necessary to have a belt tight enough to prevent it from slipping on the pulley, be- cause if a belt slips there is not only a direct loss in velocity, but the belt will wear out in a short time ; it is, therefore, very im- portant to use belts of such proportions that the power shall be transmitted with ease. Belts always run toward the side of the pulley which is largest in diameter (therefore pulleys are crowned, in order to keep the belt running straight). A belt will always run toward the side where the centers of the shafts are nearest together. Open belts will cause two shafts to run in the same direction. fiELTS. 3 2 7 A crossed belt will cause the shafts to run in opposite direc- tions. If the distance between the shafts is short, crossed belts will not work well. A short belt will wear out faster than a long one. Very long and heavy belts should be supported by idlers as well under the slack as under the working side ; if not, the weight of a long belt will cause too much stress on itself and also cause too much pressure on the bearings, as well on the driver as on the driven shaft. Belts should never, when it can be avoided, be run vertically, as the weight of the belt always tends to keep it away from the lower pulley, thereby reducing its transmitting capacity; the longer the belt the worse this is. Belts are most effective when they are run in a horizontal direc- tion and, whenever possible, the lower part of the belt should be the working part, as the slackness in the upper part, by its weight, will cause the belt to lay around the pulley for a longer distance, and this will, in a measure, increase its transmitting capacity; but if the upper part is the working part, the slackness in the lower part tends to keep the belt away from the pulleys, and thereby reduces its transmitting capacity. Lacing Belts. Figure 1 shows a good way of lacing belts ; a is the side run- ning next to the pulley and b is the outside. Holes should be punched and not made by an awl, as punched holes are less lia- ble to tear. The lacing is commenced by putting each end of the lace through holes 1 and 2 from the side next to pulley, and then continuing toward the edges, both sides simultaneously, Fig. i. fi n f\ y y u j u y w J a making a double stitch at the edges and sewing back again un- til holes 1 and 2 are reached ; and, lastly, by drawing each end of the lace through x and y. Each stitch will be double, except- 328 BELTS. ing the middle one. The holes x and y, where the ends of the lacing are finally drawn through for fastening, are made by the belt awl and should always be made small, and the lacing, if laid out rightly, always enters these holes from the inside of the belt ; after it is pulled through, a small cut is made in the lacing on the outside, which will prevent it from drawing back again, then the ends are cut off about l / 2 " long, as shown in the figure at.r and y. It is a bad practice to leave the lace-ends on the inside of belts, because they will then soon wear off, allowing the joint to rip. A 1-inch belt ought to have three lace-holes in each end. Length of lacing, 12 inches. A 2-inch belt ought to have three lace-holes in each end. Length of lacing, 18 inches. A 3-inch belt ought to have five lace-holes in each end. Length of lacing, 24 inches. A 4-inch belt ought to have five lace-holes in each end. Length of lacing, 32 inches. A 5-inch belt ought to have seven lace-holes in each end. Length of lacing, 40 inches. A 6-inch belt ought to have seven lace-holes in each end. Length of lacing, 48 inches. An 8-inch belt ought to have nine lace-holes in each end. Length of lacing, 60 inches. A 10-inch belt ought to have eleven lace-holes in each end. Length of lacing, 72 inches. A 12-inch belt ought to have thirteen lace-holes in each end. Length of lacing, 84 inches. Always have the row having the most holes nearest the end of the belt. Cementing Belts. When belts are cemented together, a 3-inch belt is lapped four inches and a 4-inch belt 4>£ inches. In larger belts the lap is usually made equal to the width of the belt, but it may be made even shorter when the width of the belt is over 12 inches. The two ends are jointed together, so that the thickness is even with the rest of the belt. The A?nerican Machi7iist, in answer to Question No. 430, Dec. 5, 1895, says : " For leather belts take of common glue and American isinglass equal parts ; place them in a glue pot and add water sufficient to just cover the whole. Let it soak 10 hours, then bring the whole to a boiling heat, and add pure tan- nin until the whole appears like the white of an agg. Apply warm. Buff the grain of the leather where it is to be cemented; rub the joint surfaces solidly together, let it dry for a few hours, and the belt will be ready for use. For rubber belts take 16 parts gutta percha, 4 parts India rubber, 2 parts common caulk- er's pitch, 1 part linseed oil ; melt together and use hot. This cement can also be used for leather." BELTS. 329 Length of Belts. Small belts, such as 4 inches wide or less, will work well when the distance between the shafts is from 12 to 15 feet, larger belts when from 20 to 25 feet, and for large main belts 25 to 30 feet distance is satisfactory. Horse=Power Transmitted by Belting. A single belt weighing about 15 ounces per square foot is capable of transmitting one horse-power per inch of width, when running at a speed of 800 feet per minute over pulleys of proper size, both of equal diameter. As one horse-power is 33,000 foot-pounds of work per minute, this will make the tension , , , , • • • 33000 due to the power the belt is transmitting = 8QQ = 41 >< lbs. per inch of width, but the total tension in the belt is, of course, considerably more per inch of width, because the belt must be tight enough to prevent its slipping on the pulley. For belts lighter than 15 ounces per square foot it is better to allow 1000 running feet per horse-power per inch of width of belt. For light double belts weighing 21 ounces per square foot, 600 running feet per horse-power per inch of width may be allowed. For double belts weighing 25 ounces per square foot, 500 running feet per horse-power per inch of width may be allowed. Hence the following formulas : For light single belts weighing less than 15 ounces per square foot, H — _^- X — a - //X 10(j0 • 1000 b ~ v For single belts weighing 15 to 16 ounces per square foot, rr V X b H X 800 tl = h — 800 ° ~ v For light double belts weighing about 21 ounces per square foot, v X b . H X 600 H - 600 v For double belts weighing about 25 ounces per square foot, _ * X_?_ b - HX 50 ° H ~ 500 v H = Horse-power. b = Width of belt in inches. v = Velocity of belt in feet per minute, which will be di- ameter of pulley in inches multiplied by 3.1416 and by the num- ber of revolutions per minute, and the product divided by 12. 33° BELTS. Example 1. A double belt 10 inches wide, weighing 25 ounces per square foot, runs over 50-inch pulleys, making 240 revolutions per min- ute. How many horse-power will it properly transmit ? Solution: „ , . r , , 50 X 3.1416 X 240 velocity of belt = ^ = 3141.6 ft. per minute. 3141.6 X 10 = 500 = ^ 2 '^ norse -P ower - Example 2. One hundred horse-power is to be transmitted by a double belt weighing 25 ounces per square foot. The pulleys are 66 inches in diameter and make 150 revolutions per minute. What is the necessary width of belt ? Solution : Pulleys of 66 inches diameter, running 150 revolutions per minute, will give a belt speed of 15 ° X 3 ' 1416 X 66 = 2 591.8; say, 2592 feet per minute. 100 X 500 , , , , , , b — ^— = 19.3 inches; thus, a double belt 20 inches wide will do the work. Example 3. A light single belt 4 inches wide, weighing 13 ounces per square foot, runs over pulleys of 36 inches diameter, making 100 revolutions per minute. How many horse-power may be trans- mitted ? Solution : TT , . P , , 36 X 3.1416 X 100 Velocity of belt = ^ — 942.48 ft. per minute. The belt is a light single belt and its transmitting capacity 4 X 942.48 will be, H — vwS — 3.76992, about 3% horse-power. To Calculate Size of Belt for Given Horse=Power when Diameter of Pulley and Number of Revolutions of Shaft Are Known. The following formulas may be used for calculating belt transmission, and will give results approximately consistent with previously given rules, but they are more convenient for use, as the velocity of the belt does not need to be first calculat- ed, but the velocity of the belt must not exceed the practical limit. BELTS. 331 This formula will do for either single or double leather belts with cemented joints (no lacing), of any weight from 12 to 30 ounces per square foot and of any width from one to thirty inches, when the pulleys are of suitable size to correspond with the thickness of the belt, and the diameter of both pulleys is equal or nearly so : _ d X nX b Xw HX 50000 50000 L " — fix b Xw H X 50000 , H X 50000 H X 50000 dX b X w d X nX w dXnXb H = Horse-power transmitted by the belt. d = Diameter of pulley in inches. n == Number of revolutions per minute. b = Width of belt in inches. w = Weight of belt in ounces per square foot. 50,000 is constant. Example. Calculate Example 2 by the above formula. Solution : 100 X 50000 b = qq x 15Q x 25 = 20 -2 inches, which, for all practical purposes, is the same as the result when calculated by the other rule. Wide and thin belts are unsatisfactory. It is far better Avhen transmitting power to use double and narrow rather than single and wide belts. It is a very bad practice to run at too slow belt speed, and also to use pulleys of too small diam- eter. The smallest pulley for a light double belt should never be less than 12" in diameter, for a heavy double belt never less than 20" in diameter, and for a triple belt the pulley should not be less than 30" in diameter. To Calculate Width of Belt when Pulleys are of Unequal Diameter. When the pulleys are of different diameters the belt will lay around the smallest pulley less than 180 degrees, and the trans- mitting capacity of the belt is correspondingly reduced. The pressure on the pulley due to the tension of the belt will vary as the sine of half the angle of contact, and the adhesion of the belt to the pulley will vary as the pressure ; consequently, also, the transmitting capacity of the belt will vary as the sine of half of the angle of contact, but it is usually advisable in practice to allow a little more on the width of the belt than is called for by this rule. A practical rule is : First calculate the width of the belt by the above rules and formulas, as though both pulleys had the same diameter, 33 2 BELTS. then multiply the result by the following constants, according to the arc of contact between the belt and the small pulley. When the arc of contact between the belt and the small pulley is 90° multiply by 1.60. 100° " " 1.45 140° multiply by 1.15 110° " " 1.35 150° " " 1.10 120° " " 1.25 160° '• " 1.06 130° " " 1.20 170 C - '•' 1.04 Example. The pulley on a dynamo is 15" in diameter, and it makes 1200 revolutions per minute. The driving pulley is so large that the belt only lays around the dynamo pulley for a distance of 150 degrees. What is the necessary width of a light double belt, weighing 21 ounces per square foot, when it takes 40 horse-power to run the dynamo ? Solution : If the arc of contact had been 180 degrees the belt would 40 X 50000 n . ■ . , , be b ■=■ v>qo x 15 X 21 = inches wide, but as the arc of contact is not 180 degrees, but only 150 degrees, this width is multiplied by the constant 1.10. as given in the preceding table. Thus, the width of the belt will be 5.3 X 1.1 = 5.83 inches or, practically, a belt six inches wide is required. When belts are running in a horizontal direction, and the driven pulley and the driver are of equal diameter and finish, the belt will always, when overloaded, commence to slip on the driver, and when pulleys are of unequal size it is always more favorable for the belt when the driving pulley is the larger than when vice versa. To Find the Arc of Contact of Belts. Make a scale drawing of the pulleys and the belt, and measure the arc of contact from the drawing by means of a protractor, or the arc of contact in degrees on the small pulley for an open belt may be calculated by the formula : Cosine of half the angle = — R = Radius of large pulley in inches. r = Radius of small pulley in inches. / = Distance in inches between centers of the shafts. Example. The distance between centers of two shafts is 16 feet ; the large pulley is 60 inches and the small pulley is 20 inches in diameter. What is the arc of contact of the belt ? BELTS. 333 Solution : 16 feet = 192 inches. 60 inches diameter = 30 inches radius. 20 inches diameter = 10 inches radius. Cos. of half the angle = ( 30 ~ 10 ) = .104 192 In tables of natural cosine (page 158), the corresponding- angle is found to be 84 degrees, very nearly ; thus, the angle for arc of contact will be 2 X 84 = 168 degrees on the small pulley. On the large pulley the arc of contact will be 360 — 108 = 192 degrees. For a crossed belt the arc of contact is always the same on both pulleys, and it may be calculated by the formula: Cos. of half the angle = — — ~, — R = Radius of large pulley. ;• = Radius of small pulley. / = Distance between centers. Example. What will be the arc of contact for the belt on the pulleys in the previous examples if belt is run crossed instead of open? Solution : Cosine of half the angle = — 30 + 10 = — 0.208 : the 192 corresponding angle will be 180 — 77 = 108 degrees, and the arc of contact will be 103 X2 = 206 degrees. Pressure on the Bearings Caused by the Belt. Approximately, the pressure on the bearings caused by the belt may be considered to be three times the force which the belt is transmitting. Therefore, the pressure may be calculated by the formula : p _ 3 X 33000 X H v P = Pressure on the bearings due to pull of belt. H= Number of horse-power transmitted by the belt. v = Velocity of belt in feet per minute. Example 1. A belt is transmitting 60 horse-power and its velocity is 900 feet per minute. What is the pressure in the bearings due to the belt ? 334 BELTS. Solution : 3 X33000X6 ° = 6600 pounds. 900 Example 2. Suppose the diameters of the pulleys are increased until a belt speed of 3000 feet per minute is obtained. What will then be the pressure in the bearings caused by the belt when trans- mitting 60 horse-power ? Solution : t, 3 X 33000 X 60 nnoA , P = = 1980 pounds. 3000 ^ By the above examples it is conclusively shown what a great advantage there is in using pulleys so large in di- ameter that proper belt speed is obtained. (See velocity of belts, page 337). The approximate pressure may also be very conveniently obtained from the width of the belt, thus: For light single belts, allow 1000 feet of belt speed per horse-power transmitted per inch of width of belt. The effective pull in such a belt will be 33 pounds per inch of width, and the pressure on the bearings due to the belt will accordingly be 33 X 3 = 99 pounds per inch of width of belt. For convenience, say 100 pounds pressure in the bearings per inch of width of such belts. For belts where 800 running feet are allowed per horse-power per inch of width of belt, this reasoning will give a pressure on the bearing equal to 123^" pounds per inch of belt. For convenience, say 125 pounds pressure in the bearings per inch of width of such belts. For belts where 600 running feet are allowed per horse- power per inch of width, the pressure in the bearing is equal to 165 pounds per inch of width of belt, and where the belt is so heavy that only 500 feet of belt speed per horse-power per inch of width is allowed, the pressure in the bearings will be 198 pounds per inch of width. A good, practical rule, which can very easily be remembered, is, (when belts are in good order and have the proper size and the proper tension) : Multiply weight of belt in ounces per square foot by eight times the width of the belt in inches, and the product is approximately the pressure in pounds upon the bearings caused by the belt. Example. A belt is calculated with regard to the horse-power it has to transmit under a given velocity, and found to be 8-inch double belting, weighing 25 ounces per square foot. What pres- sure will it cause on the bearings when working at proper tension ? BELTS. 335 Solution, by the last rule : f = 8X25X8 = 1600 pounds. Solution, by the first rule : At a speed of 3000 feet per minute such a belt will transmit X8 30 formula 00 x 8 -g^j — = 48 horse-power, and calculating the pressure by the p _ 3 X 33000 X H p = 3 X 33000 X 48 = 15g4 ^ 3000 Both rules give nearly the same result, and one is just as correct as the other, as all such figuring is nothing more than approximation at the best. The pressure on the bearings may be a great deal more than calculated above. Sometimes the pulleys are roughly made, belts are poor, and consequently the coefficient of friction between belt and pulley is small, and as the belt has to be a great deal tighter in order to do the work, the pressure on the bearing will be greatly increased. Very frequently, from pure ignorance or carelessness, belts .are made very much tighter than necessary, and enormous sums of money may be wasted in this way in large factories, as the steam engines, at the expense of the coal pile, have to furnish power not only to do the useful work, but also to overcome all the friction produced by such over-strained belts, hot bearings, etc. A belt will transmit more power over a good, smooth pulley than over a rough one. When pulleys are covered with leather a belt will transmit about 25% more power than it will when running over bare iron pulleys, and in transmitting the same power a much slacker belt may be used, thereby reducing the friction in the bearings. Special Arrangement of Belts. By the use of suitable guide pulleys it is possible to connect with belts shafts at almost any angle to each other. But experience is required and care must be exercised to do it suc- cessfully. When guide pulleys are used in order to change the direction of a belt, always remember that when the belt is run- ning the most pressure is thrown on the pulley guiding the working part of the belt. This pulley is, therefore, very liable to heat in its bearings,, if not designed to have bearing surface enough and also to have proper means for oiling. 33^ BELTS. Fig. 2 shows an arrangement by which the direction of motion of two shafts may be reversed" when the distance between the shafts is too short for the use of a crossed belt or when a crossed belt, for any other reason, cannot be used. Suppose pulley A to be the driver and to run in the direction of the arrow. C and D are guide pulleys, and the motion of the driven shaft B is in the op- posite direction to the shafts. In this case the guide pulley C is on the working part of the belt, and is the one to which special attention must be paid in regard to heating. If the direction of shaft A is reversed, guide pulley D will be on the working part of the belt. Crossed Belts. If the distance between A and B (Fig. 2) had been long enough, it would have been preferable to reverse the motion of B by means of a crossed belt, instead of by the arrangement shown in Fig. 2. Crossed belts do not work well when running on pulleys small in diameter as compared to the width of the belt. Too short distance between the shafts must be avoided. Wide crossed belts are very unsatisfactory; therefore, instead of running one wide crossed belt it is preferable to use two belts, each of half the width, and run them on two separate pairs of pulleys. Such belts should be of equal thickness, and the pulleys should be crowned, well finished and of correct size, so that each belt will do its share of the work. Quarter =Turn Belts. Fig. 3 shows a so-called quarter- turn belt, used to connect two shafts when running at an angle and laying in different planes. The principal point to look out for is to place the pulleys (as shown in Fig. 3) so that the belt runs straight from the de- livering to the receiving side of each pulley. The pulleys shown in Fig. 3 are set right for belts running in the direction of the arrows. If the mo- tion is reversed, the belt will run off the pulleys. BELTS. 337 Angle Belts. The belt arrangement shown in Fig. 4 is usually called an angle belt and is used to connect two shafts at an angle. Either one, A or B, may be the driver, and there are two guide pulleys (one for each part of the belt at C), one of which, of course, is on the driving part of the belt. Crossed belts, quarter-turn belts, and angle belts must never be wide and thin ; much better results are obtained by narrow, double belts than by wide, single ones. Angle belts and quarter-turn belts are frequently bothersome contrivances. Their running is sometimes improved by making a twist in the belt when joining its ends : that is, lacing the flesh side of one end and the hair side of the other end on the outside. This will prevent one side of the belt from stretching more than the other. Slipping of Belts. Owing to the elasticity of belts, there must always be more or less slip or " creep " of the belts on the pulleys. Under favorable conditions it may be as low as 2%, but frequently the slip is more. Therefore, if two shafts are connected by belts, and both should have very nearly the same speed, the diameter of the driver should be at least 2% larger than the diameter of the driven pulley. When the driver is comparatively large in diameter and the driven pulley is small, it is advisable to have the driver from 2 to b% over size, in order to get the required speed. Tighteners on Belts. If tighteners are used they should always be placed on the slack part of the belt. Velocity of Belts. Belts are run at almost all velocities from less than 500 to 5000 feet per minute, but good practice indicates that whenever possible main belts having to transmit quantities of power are run most economically at a speed of 3000 to 4000 feet per min- ute. At a higher speed both practice and theory seem to agree that the loss due to the action of the centrifugal force in the belt when passing around the pulley, and that the wear and tear is so great when the speed is much over 4000 feet per minute that there is not much practical gain in increasing the speed. But, as a general rule, whenever possible the higher the belt speed the more economical is the transmission as long as the belt speed does not exceed the neighborhood of 4000 feet per minute. 338 BELTS. Oiling of Belts. Belts should be kept soft and pliable and are, therefore, usually oiled with either neat's-foot oil or castor oil. Too much oiling is hurtful, but the right amount of oiling at proper times is very beneficial to the action of the belt and will prolong its utility to a great extent. Remarks. — All previous rules for calculating belting are founded upon good, legitimate practice, but are only offered as a guide, as no rule can be given which will fit all cases. For instance, a belt may be amply large to transmit a given horse-power when running in a horizontal direction, but it may fail to do the same work if running in a vertical di- rection. A belt may be large enough to do its work when run- ning in a vertical direction over pulleys of unequal size with the large pulley on the lower shaft, but it may fail to do the same work satisfactorily with the large pulley on the upper shaft and the small pulley on the lower one. Leather belts should not be used where it is damp or wet, but rubber belting will usually give good service in such places. For information regarding rubber belts, see manufacturers' catalogues. WIRE ROPE TRANSMISSION. Transmitting power by wire ropes running at a high speed over grooved pulleys, or " telodynamic transmission," as it is also called, is the invention of the brothers Hirn of Switzerland. For long distances this mode of transmission is far cheaper than leather belting or lines of shafting. Fig. 1 shows a section of a pulley as used for this kind of transmission ; a is an elastic fill- ing, usually made from leather cut out and packed in edgewise. The groove is made wide, so that the rope will rest entirely against the packing and not touch the iron. This is different from transmission with hemp rope, which is made to wedge into the groove of the pulley. The diameter of the pulley in the groove, where the wire rope runs, ought to be at least 150 times the diameter of the rope ; the larger the better, so long as the velocity of the rope does not exceed 5000 feet per minute. The pulleys must run true and be in balance and in exact line with each other, and the WIRE ROPE TRANSMISSION. 339 shafts must be parallel. The distance between shafts should never be less than 60 feet and should preferably be from 150 to 400 feet.* For distances longer than 400 feet, either carrying pulleys or intermediate jack shafts are generally used, although spans as long as 600 feet or more have been used, but only when it is possible to give the rope the proper deflection without its touching the ground. Usually the speed is from 3000 to 6000 feet per minute. Higher speed would be dangerous from the stress in cast-iron wheels due to centrifugal force. ■^ - « ^ .3 eu + + + ^2 4" II ii 11 CO l! II U£ •^ ^ s ^0 "*s> d ins. b ins. cins. /ins. ^ ins- / ins. § \ * 9 T6 i* | A tI H 9 T6 1A | 2 1 l 5 8 ii * 5 n H 1 1 H 7 8 16 16 3 4 n U 3 4 *i h 7 8 n if H »l 1 2 1 H H 7 8 3 1 2 FIG. 1 Tightening pulleys should not be used, because if the distance between centers of shaft is too short to give the proper tightness to the rope without a tightening pulley, wire rope transmission is not the form best adapted to the circumstances. Guide pulleys or idlers should be avoided as much as possible, but when necessary they should be as carefully made and put up as the main pulleys, and they ought not to be less than half the diameter of the main pulley if on the slack part, but of the same size if they are on the tight part of the rope. Wire rope for transmission is usually made from the best quality of iron, has seven wires to a strand and consists of six strands laid around a hemp core in the center of the rope. The diameter of the wire rope is from nine to ten times the diameter of each single wire. Never use galvanized rope for power transmission, but pre- serve the rope by painting with heavy coats of linseed oil and lampblack. * When distance between shafts is less than 60 feet, leather belts are prefer- able to wire rope. 34° WIRE ROPE TRANSMISSION. Transmission Capacity of Wire Ropes. A one-inch rope running 5000 feet per minute is capable of transmitting 200 horse-power. The transmitting capacity of the rope is in proportion to the square of its diameter, and the power transmitted by the rope when the velocity is less than 5000 feet per minute is practically in proportion to its velocity* Hence the formula: H= d * X VX 200 which reduces to H = 0.04 X d* X V 5000 H = Horse-power transmitted. <-/= Diameter of rope in inches, JS= Velocity of rope in feet per minute. Example. How many horse-power may be transmitted by a wire rope ]/ 2 inch in diameter running over proper pulleys at a velocity of 2500 feet per minute ? Solution : H '= 0.04 X ]/ 2 X y z X 2500 = 25 horse-power. The pressure on the bearings will not be less than three times the force transmitted, and may be calculated thus : Pressure on bearings = g X horse-power X 38000 V elocity in feet per min. Example. What will be the least pressure in bearings for a wire rope transmitting 150 horse-power at a velocity of 5000 feet per minute ? Pressure on bearings = 3 X 150 X 33000 _ 29TQ d 5000 F If there is one bearing on each side at an equal distance from the pulley, the pressure on each bearing will be 29 2 yo = 1485 pounds. This is the calculated pressure, and represents what the pressure should be, but it is not certain that this is the actual pressure. It may be greatly increased by having the rope too tioht.f * When the velocity of the rope exceeds 6000 feet per minute the stress caused by centrifugal force when the rope is bending around the pulley considerably reduces its transmitting capacity. This loss increases very fast above this speed, because the centrifugal force increases as the square of the velocity. It is very doubtful if there is practically any gain to run wire ropes at a speed exceeding 6000 feet per minute when wear and tear, loss due to centrifugal force, etc., are considered. t Sometimes a pulley is put on the free end of a line of shafting projecting through the wall and drawn by a wire rope outside the shop; this will do only when a comparatively small amount of power is to be transmitted. WIRE ROPE TRANSMISSION. 341 The tension of the rope may be calculated from its de- flection when at rest (see Fig. 2), and for a rope running hor- izontally the usual formula is : p _ WK \/a* + P P = K*l (very nearly) a P = Force in pounds at f. W = Weight of rope in pounds from d to f, which is half the span. b = Half the span in feet. a = Twice the deflection in feet. Note. — (See Fig. 2.) If the length of the line a represents the weight of the part of the rope from dtof, the length of the line x represents the tension in the rope at/"; therefore the ten- sion will be as many times the weight as the length of line a is contained in the line x. Example. The horizontal distance between two pulleys is 200 feet ; when standing still the deflection in a wire rope of %" diameter is 5 feet. What is the tension in the rope ? Solution : In Table No. 38 the weight of %" wire rope is given as 1.12 pounds per foot; therefore, 100 feet of %" rope will weigh 112 pounds. 112 X V 100 2 + 10 2 _ 112 X 100.5 P 10 10 1125.6 pounds. This is the tension in each part of the rope ; therefore the force against the pulley, due to the weight of the rope, is 1125.6 X2 = 2251.2 pounds. If this is supported by a bearing on each side of the pulley, the pressure on each bearing, if both are the same distance from the pulley, will be 1125.6 pounds. 34 2 WIRE ROPE TRANSMISSION. The tension is increased by reducing the deflection. For instance, if the deflection is reduced to 4 feet the tension on the rope will be, p _ 112 X V^lOO 2 + 8 2 8 />= 1122^3 = 1404.2 pounds. Thus, the tension might be increased to any amount within the ultimate breaking strength of the rope. Deflection in Wire Ropes. When the rope is in motion the deflection will increase on the slack side and decrease on the tight side ; therefore, if the span is long the rope may touch the ground when running if the pulleys are not placed on sufficiently high towers. There is really nothing else which, within practical limits, determines the length of the span, which may just as well be 1000 feet, or even more, providing the proper deflection can be given to the rope without touching the ground. When possible the lower part of the rope should be the working side, but in a long span this is impossible, because, when running, the lower part of the rope would be tight and the upper part slack, causing the two parts of the rope to strike together, which must never be allowed. When the length of the span exceeds 35 times the diameter of the pulleys it is safest to have the upper part of the rope the working side and the lower part the slack side. When the lower part of the rope is the slack side, the least space allowable for the slack of the rope at the center of the span will (when the rope is as tight as given in Table No. 38), be obtained by the formula : Distance = 0.00015 X (span) 2 but, to allow for contingencies, it is better to have more room. When the lower side of the rope is the tight side, the rope will be clear from the ground when running if the space is 0.0001 X (span) 2 . The deflection in the rope when standing still which will produce a pressure on the bearings and give tension enough to transmit the horse-power given in Table No. 38, may be cal- culated approximately by the formula : d — 0.00009 X / 2 d = Deflection in feet. / = Distance between pulleys in feet. (See Fig. 2). Example. The distance between the pulleys being 400 feet, find the greatest allowable deflection in the rope, when standing still, in order to transmit the horse-power given in Table No. 38. WIRE ROPE TRANSMISSION. 343 Solution d: 0.00009 X 400 X 400 = 14.4 feet. When the rope is new it is always put on with more tension than is necessary to transmit the power, because new rope will stretch. It is, therefore, very important when de- signing such transmission to calculate the maximum pressure which the rope will exert on the bearings when put on with the least deflection ever wanted, and calculate size of bearings and shafting for pulleys according to this stress, with due considera- tion not only for strength but also for heat and wear. (See page 360 and page 367.) The correct amount to allow for stretch will vary with different kinds of rope and also with the tempera- ture. If a rope is spliced on a warm summer day it must be made slacker than if it was spliced on a cold winter day, as the length of the rope will be changed considerably by the difference in temperature ; the only guide is practical experience and good judgment. As a general rule, it may be safe to allow about half of the deflection as previously calculated when splicing a new rope, provided that the shafts and bearings are constructed so as to allow such tension. The rope is always strong enough. The splicing of the rope should be done by a man ex- perienced in that kind of work. The splice itself is usually made at least 240 times the diameter of the rope. TABLE No. 38.— Giving Suitable-Sized Pulleys for Different Sizes of Wire Rope, Weight of Rope, Horse=Power which Different Sizes of Wire Rope nay Transmit at Different Velocities, the least Stress at which it may be done and the Least Corresponding Pressure on the Bearings ; also, the Ultimate Average Strength of Wire Rope. - -C -j *fe O O f*.S o.S p= xt± a CB -) 5 Efc, Horse-power Transmitted at Different Velocities. «J £ inches. The cone pulley on the counter-shaft is to be of the same size as the cone pulley on the spindle, and an even ratio of speed is to be maintained throughout the whole range of the ten changes of speed. The slowest speed, when back gears are in, is 6 revolutions per. minute. Calculate the speed of the counter-shaft, the speed of the spindle for each change, and the diameter of each step on the cone pulley of the spindle. Solution : The ratio of speed for each step will be : 5 log. 8 = 0.90309 = Q lg062 V 8 5 5 The corresponding number is 1.516. With back gears in, the speed of spindle : On first cone is 6 revolutions per minute. On second cone is 6 X 1.516 = 9 revolutions per minute. On third cone is 9 X 1.516 = 14 revolutions per minute. On fourth cone is 14 X 1.516 = 21 revolutions per minute. On fifth cone is 21 X 1.516 = 32 revolutions per minute. With back gears out, speed of spindle : On first cone will be 6X8= 48 revolutions per minute. On second cone will be 9X8= 72 revolutions per minute. On third cone will be 14 X 8 = 112 revolutions per minute. On fourth cone will be 21 X 8 = 168 revolutions per minute. On fifth cone will be 32X8 = 256 revolutions per minute. PULLEYS. 355 The speed of the counter-shaft will be : JV= s/ 4S X 256 = 112 revolutions per minute. As the speed of main lines in factories usually runs at some multiple of 10, we may, for convenience in getting even-sized pulleys for connections between counter and main shaft, in practical work, decide to run the counter-shaft 110 revolutions per minute. (When a pair of cone pulleys has an uneven number of steps, and are cast from the same pattern, the speed of the counter should be equal to the speed of the machine when the belt is run on the middle step). The diameter of the largest step of the cone pulley on the spindle is 10J^ inches. The corresponding step on the counter 10J£ X 48 will be Yiq — = 4.581"; practically, 4)4" diameter. The largest and smallest step on the counter-shaft will also be 10^ and 4 l / 2 inches in diameter. Any of the intermediate steps on the spindle may be cal- culated by the formula : Bi = (D + d) X N n + N Dl = (10^ +4K) X 110 _ 9 065 . practically) 9 in> D 2 = < 10 * + 4 ^> X 110 = n% inches. 110 + no 7 Dz = (ioy 2 + m x no = 5>932 practicall 6 in> 168 + 110 y Thus, assuming the counter-shaft to run 110 revolutions per minute, the speed of the spindle, with back gears out, on the five different steps will be : 110 X 10# __ 265 revo i ut j ons per m i nu te. 4^ 110 X 9 ~6 no x ny 2 110 X 6 9 = 165 revolutions per minute. = 110 revolutions per minute. = 73 revolutions per minute. ^ = 47 revolutions per minute. 10J* 356 PULLEYS. When the back gears are in action the speed will be : zj! = 33^ revolutions per minute. 8 1 65 Jr_ = 20 ^ revolutions per minute. 8 = 13% revolutions per minute. _L = §yi revolutions per minute. 8 = 5% revolutions per minute. 8 These speeds are all within the practical requirements of the problem, and now the next operation is to modify the diam- eters slightly in order to get proper tension on the belt. (See page 352.) FLY=WHEELS. Fly-wheels are used to regulate the motion in machinery by storing up energy during increasing velocity, and giving out energy during decreasing velocity. Fly-wheels cannot perform either of these functions without a corresponding change in velocity. The rim of the wheel may be very heavy and moving at a high velocity, the change in speed may be small and hardly perceptible if the energy absorbed and given out is small, but there must always be a change in velocity to enable a fly-wheel to act. The common expression of gaining power by a heavy fly-wheel is very misleading, to say the least. There is no power gained by a fly-wheel but, on the contrary, considerable power is absorbed by friction in the bearings when a shaft is loaded with a heavy fly-wheel, (see example in calculating fric- tion, page 305). Nevertheless, a fly-wheel performs a very use- ful function in machinery by storing up energy when the supply exceeds the demand and giving it out at the time it is needed to do the work. ( For momentum of fly-wheels see example, page 300. For kinetic energy, see example, page 301 ). Weight of Rim of a Fly=Wheel. The weight of a rim of a cast-iron fly-wheel will be : W—tr 1 X 0.7854 X D X 3.1416 X 0.26 ; this reduces to, W=D X d 2 X 0.64 D = Middle diameter of rim in inches. d= Diameter of section of rim in inches. W = Weight in pounds. fly-wheels. 357 Example. A round rim of a fly-wheel is 4 inches in diameter and the middle diameter of the wheel is 36 inches. What is the weight of the rim ? Solution: ff=36X4X4X 0.64 = 369 pounds. For a rim of rectangular section the weight will be : W = Width X thickness X D X 3.1416 X 0.26 W = Width X thickness X D X 0.816 Example. The width of the rim is six inches, the thickness is two inches, and the middle diameter of the rim is 48 inches. What is the weight of the rim ? Solution : J^=2X6X48X 0.816 = 470 pounds. Centrifugal Force in Fly=Wheels and Pulleys. Pulleys are not only liable to be broken by the stress due to the action of the driving belt, but in fast-running pulleys and fly-wheels the stress due to centrifugal force is far more dan- gerous. This stress increases as the square of the velocity and directly as the weight, therefore there is a limit to the velocity at which fly-wheels and pulleys can be run with safety. Generally speaking, increasing the thickness of the rim does not increase its strength, because the total tensile strength, the total weight of the rim, and, consequently, also the centrifugal force, increase in the same proportion ; but it has great influence upon the strength of the wheel to have the ma- terial in the rim distributed to the best advantage. At the same time it is very important to construct the rim and arms of such proportions that the initial stress due to uneven cooling in the foundry, is avoided. The common formula is : ~ .., , , Mass X (velocity) 2 Centrifugal force = ^ — radius Weight „ . . »XrX2?r Mass = — — -2; — Velocity = — 32.2 50 Therefore, r X 2 7T V ( A W 60 32.2 X r W X « 2 X r 2 X 0.01096628 cf = Cf = 32.2 X r cf = IV X ft 1 X r X 0.00034 cf = Centrifugal force in pounds. 35 8 FLY-WHEELS. W = Weight of revolving body in pounds. n = Number of revolutions per minute. ?- = Middle radius of pulley rim in feet. Thus, for any body whose center of gravity swings in a circle of one foot radius, at a speed of one revolution per min- ute, the centrifugal force will be 0.00034 times the weight of the body. Example. The rim of a fly-wheel is five feet in middle radius and weighs 8000 pounds. It makes 75 revolutions per minute. What is the stress due to centrifugal force ? Solution : Centrifugal force = 8000 X 75 2 X 5 X 0.00034 = 76500 pounds. FIG 3 This is the total centrifugal force tend- 0ing to burst the rim (see arrows in Fig. 3). The force tending to tear the rim asunder in any one of the two opposite points ..& as a, b, is 8 ^ x £ = 12175 pounds. The next question is : Has the sec- tion of the rim tensile strength enough to resist this stress with safety? If not, either decrease the rim speed or make the rim of material having more tensile strength. The centrifugal force for the same number of revolutions increases as the radius, therefore the average centrifugal force acting in the arms is only about half of the centrifugal force acting in the rim, and as the stretch is in proportion to the stress, the rim tends to stretch more than the arms, and, consequently, it can not yield freely to the action of the centrifugal force, but is to a certain extent held back at the junction with the arms. This action is shown in an exaggerated form at a, Fig. 4. In regard to this action, the part of the rim between the arms may be considered F,G 4 ______ as a beam fastened at both ends and uniformly loaded throughout its whole length equal to that amount of cen- trifugal force in the rim which is resisted by the arms ; therefore, the rims of large pulleys should always be ribbed on the inside. (See cross-section of rim at X, Fig. 4). Another bad feature frequently seen in pulleys is the counter-balance. (See b. Fig. 4). This little piece itself, weigh- ing probably only five pounds, holds the pulley neatly in balance FLY-WHEELS. 359 and is very innocent as long as the pulley is standing still, but imagine what stress it will produce on the rim of a 6-foot pulley running at a rim speed of 80 feet per second. Solution: cf= ^JiA = 331 pounds. 3 X 32.2 Thus, when that pulley is running at a speed of 80 feet per second this counter-balance of five pounds will produce the same stress as if it was loaded with 331 pounds when standing still; therefore, it is evident how important it is to turn fast- running pulleys both inside and outside in order to reduce counter-balancing to the least possible amount. The danger of the rim deflecting or breaking from the stress due to the resistance from the arms (as shown in Fig. 4), can be avoided by running ribs on the inside of the rim, and the danger caused by counter-balancing can be entirely eliminated by mak- ing the pulley balance without adding any balancing pieces. Thus, one danger of breaking is avoided by proper designing and the other by good workmanship. The direct action of the centrifugal force on the rim is cal- culated by the formula, cf= w X if- X r X 0.00034, and the weight of the rim of a cast-iron fly-wheel having one square inch of sectional area and a radius of one foot will be 2 X re X 12 X 0.26 pounds, and a ring of r-foot radius will weigh r X 2 X - X 12 X 0.26 pounds. As already stated, the centrifugal force increases as the square of the velocity ; that is, if the number of revolutions is doubled the centrifugal force is increased four times ; thus is the dangerous limit approached very rapidly under increased speed, and in order to prevent accident, if the speed should happen to increase, it is necessary always to use a high factor of safety in such calculations. Thus, using 15 as a factor of safety* and assuming the tensile strength of cast-iron as 12,000t pounds per square inch, the stress in each cross-section at a and b must not exceed 800 pounds per square inch. The allowable total centrifugal force acting in the rim may therefore be 2 n X 800 pounds, and in- serting those values and solving for n the greatest number of revolutions allowable for a cast-iron fly-wheel will be: 2 - X 12 X 0.26 X r X ;z 2 X r X 0.000340568 = 800 X 2 - n 1 r 2 = 752891 n r = VT52891 ;/ = 868 r = 868 r n * 15 as factor of safety with regard to strength, is only y 1 15 = 3.873, or less than 4, as factor of safety with regard to speed. t This tensile strength for cast-iron may seem very low, but it is dangerous to assume more, because oi initial stress in arms or rim already, due to uneven cool- ing of the casting in the founuiy. 3^0 f FLY-WHEELS. Transposing this to diameter in inches, the constant will be 24 X 868 = 20832. The formula will be : Number of revolutions per minute = 20832 Diameter in inches. Diameter in inches - Revolutions per minute. Rule for Calculating Safe Speed. Divide 20832 by the diameter of the fly-wheel in inches, and the quotient is the allowable number of revolutions per minute at which a well-constructed fly-wheel may be run with safety. Rule for Calculating Safe Diameter. Divide 20832 by the number of revolutions per minute, and the quotient is the safe diameter in inches for a well-constructed fly-wheel. These rules will not apply to fly wheels made in sections and bolted together. Frequently such wheels are weaker than a wheel made in one casting. Their strength to resist centrifugal iorce must be carefully calculated, considering as well the joints in the rim as the joints between the arms and rim. SHAFTING. When calculating strength of shafting, both transverse and torsional stress should be considered. Transverse stress is produced by the weight of the shaft itself, the pulleys and the tension of the belts, the effect of which is very severe if the distance between the hangers is too long. Torsional stress is produced by the power which the shaft transmits. Usually the distance between the hangers is made so short that the torsional stress on a shaft is the greater. For transverse stress the shaft may be considered as a round beam, supported under the ends and loaded somewhere between supports. According to Table No. 30 the transverse stress which will destroy a wrought iron beam one inch square, fastened at one end and loaded at the other, is 600 pounds ; the strength of a round beam of the same diameter is (see page 251) 0.6 that of a square beam. When the beam is supported under both ends and loaded in the middle, its breaking load will increase four times; therefore, the constant, c, will be 600 X 4 X 0.6 = 1440. Using 10 as factor of safety, the formula for transverse strength of a wrought iron shaft will be: D = J~Z7W /- 144 D* 144/73 >HU4~ W L D = Diameter of shaft in inches. L = Distance between hangers in feet. W = Transverse load in pounds, supposed to be at the middle, between the hangers. i44=constant for v/rought iron, with 10 as factor of safety for ultimate transverse strength. SHAFTING. 361 Formula 1, expressed as a rule, will be : Multiply the distance between hangers, measured in feet, by the transverse load in pounds ; divide this product by 144. and the cube root of the quotient will be the diameter of the shaft in inches, calculated with 10 as factor of safety for transverse strength, but besides strength it is also absolutely necessary to consider stiffness and allowable deflection. Shaft not Loaded at the Middle Between the Hangers. When a shaft is not loaded at the middle of the span, but somewhere toward one of the hangers, it will carry a heavier load, with the same degree of safety, than it would if loaded in the middle, and the ratio is in inverse proportion as the square of half the distance between hangers to the product of the short and the long ends of the shaft. For instance, a shaft is six feet between hangers and loaded at the middle. What would be the difference in transverse strength if it was loaded two feet from one hanger and four feet from the other? 3X3 = 9 and 2X4 = 8. Thus, find the transverse load for a shaft when loaded in the middle, multiply by 9 and divide by 8, and the quotient is the load which the same shaft will carry with the same degree of safety against transverse stress, if loaded two feet from one end and four feet from the other. This rule only applies to the transverse strength, and not to the transverse stiffness of the shaft. For different shapes of shafts and different modes of loading, see beams, pages 243-244. When shafts are heavily loaded near one hanger, and the hanger on the other side of the pulley is further off, most of the load is thrown on the bearing nearest to the pulley, and this bearing is, therefore, liable to heat and to cause trouble, even if the shaft is both stiff and strong enough. ( See reaction on the support of beams, page 252). Transverse Deflection in Shafts. The transverse deflection in a shaft may be calculated by the formula : 4 3 ^=V £ T C Z H SB' s * w c w — S D± s _ L*W C S = Deflection in inches. D = Diameter of shaft in inches. L = Length of span in feet. W = Load on middle of shaft in pounds. C= Constant = 1.7 X constant in Table No. 31, and for wrought iron or Bessemer steel may be taken as 0.00002652. 362 SHAFTING. Constant C may be calculated from experiments by the formula, c _ S D* Z 3 W S = Deflection in inches noted in the specimen, when supported under both ends and loaded transversely at the middle between supports. D = Diameter of specimen in inches. L = Distance between supports of specimen in feet. W = Experimental load in pounds. Example. A round specimen placed in a testing machine, supported under both ends and loaded at the middle with 2000 pounds, deflects 0.1 inch. The diameter of the specimen is two inches and the distance between supports is three feet. Calculate constant C for this kind of material. Solution : 0.1 X 2 4 C = 3 3 X 2000 C = ±-u = 0.0000296 inch. 54000 Thus, the deflection for this kind of material is 0.0000296 inch per pound of load, applied at the middle, between supports, for a round bar one inch in diameter and one foot between supports. Allowable Deflection in Shafts. The distance between the hangers must always be deter- mined with due consideration to the allowable transverse deflec- tion in the shafting, especially when the shaft is loaded with large pulleys and heavy belts, remembering that the deflection increases directly with the transverse load and with the cube of the length between the bearings, (see page 254). The allowable transverse deflection in shafting ought not to exceed 0.006 to 0.008 inch per foot of span ( see page 266). A beam of wrought iron one foot long and one inch square, when supported under both ends and loaded at the middle, will deflect 0.0000156 inch per pound of load, (see Table No. 31, page 259), and a round beam deflects 1.7 times as much as a square beam, when the diameter and side are equal. A round shaft, one inch in diameter and one foot long, when loaded at the middle with 144 pounds will, therefore, deflect 144 X 1.7 X 0.0000156 = 0.00382 inch. Thus, this load does not give more than an allowable de- flection. But, suppose the distance between bearings is doubled and the load decreased one-half; the ultimate strength of the shaft will be the same, but the deflection will be 72 X 1.7 X 2 3 X 0.0000156 = 0.01528 == 0.0764 inch per foot. SHAFTING. $6$ This calculation shows plainly how very necessary it is to have bearings near the pulleys where shafts are loaded with heavy pulleys and large belts. There is nothing more liable to destroy a shaft than too much deflection, because the shaft is, when running, continually bent back and forth, and at last it must break. The fact must never be lost sight of that strength and stiffness are two entirely different things and follow entirely different laws; therefore, after calculations are made for strength, the stiffness must also be investigated, as stiffness is a very important property in shafting. The best way to over- come too much transverse deflection is to shorten the distance between the bearings. Of course, increasing the diameter of the shaft will also overcome deflection, but shafting should never be larger in diameter than necessary, because the first cost increases with the weight, which increases as the square of the diameter, and the frictional resistance will also increase with the increased diameter ; consequently, also, the running expenses. Torsional Strength of Shafting. Shafting may be considered as a beam fastened at one end and having a torsional load applied at the other end equal to the pull of the belt on an arm of the same length as the radius of the pulley. In Table No. 32, page 268, constant c is given as 580 pounds for wrought iron. The formula for twisting stress, as explained under beams ( see page 267) is, D*c n-J-p P = " c D i m = the length of the lever or arm in feet, and will here be the radius of the pulley and be denoted by r. The length of the shaft has no influence on its torsional strength, but only on its angle of torsional deflection (see page 268). Using 10 as factor of safety, the formula will be : * 58 D = Diameter of shaft in inches. r ■==■ Radius of pulley in feet. W = Pull of belt in pounds. 58 = Constant, with 10 as factor of safety = Vw X 580, taken from Table No. 32, page 268. Frequently it is more convenient to calculate the torsional strength of shafting according to the number of horse-power the shaft is to transmit ( see page 317 ). In the above formula, assume W to be 58 pounds, r to be one foot, and D will be one inch. That is, a shaft one inch in diameter is strong enough to resist, with 10 as factor of safety, 364 SHAFTING. a torsional load of 58 pounds acting on an arm one foot long. Assuming this 58 pounds to act on the rim of a pulley of one foot radius, two feet in diameter, and making one revolution per minute, it will transmit power at a rate of 58 X 6f = 364f foot- pounds per minute ; but one horse-power is 33,000 foot-pounds per minute, and if the shaft should transmit one horse-power it , 33000 , . . „ must make og, 4 = 90.52 revolutions per minute. Hence the practical formulas for torsional strength of shafting: V- rr D B Xn „ _ H X 90 ri = n =■ — — 90 Z> 3 D = Diameter of shaft in inches. H = Number of horse-power transmitted by the shaft. n = Number of revolutions made by the shaft per minute. 90 = Constant, using 10 as factor of safety, and assuming the torsional strength to be as given in Table No. 32. Torsional Deflection in Shafting. In constructing different kinds of machinery it is frequently necessary to consider the torsional deflection. The formula for torsional deflection for wrought iron ( see page 271) will be : e _ 0.00914 X m L P jyi This will transpose to s _ 48 H L S = Deflection in degrees. H = Number of horse-power transmitted. , , , , 0.00914 X 33000 _ A „ 48 = Constant ; calculated thus, — 9 v s 1416 L = Length of shaft in feet between the force and the resistance. n ■=. Number of revolutions made by the shaft per minute. D = Diameter of shaft in inches. Example. How many degrees is the deflection of a shaft two inches in diameter, 50 feet long, making 300 revolutions per minute and transmitting 15 horse-power, applied at one end and taken off at the other ? Solution : 48 HL = 48 X 15 X 50 _ 6 ~ ~VD^ 300 X 2* ~ i/2 de - rees ' SHAFTING. 3<>5 Classification of Shafting. Shafting may be divided into three different kinds. First. — Shafts where the main belts are transmitting the power, or so-called " Jack Shafts." Such shafts must have their boxes as near the pulleys as possible. For torsional strength their diameter may be calculated by the formula, 3 D — ^ / H X 125 u ~ y (See Table No. 40.) Second. — Common shafting in shops and factories, where the power is taken off at different places for driving machinery. Such shafts ought to be supported by hangers as given in Table No. 43, and their supports must also be reinforced by extra hangers, if necessary, where an extraordinary large pulley or heavy belt is carried. For torsional strength the diameter of such shafts may be calculated by the formula, 3 D = V~ 7 ~- 9 - ( (See Table No. 41.) Third. — Shafting having practically no transverse stress, but used simply to transmit power from one place to another. Such shafts ought to be supported by hangers according to Table No. 43, and the diameter may be calculated by the formula, 3 p = J H X 50 (See Table No. 42.) TABLE No. 40.— Giving Horse=Power of Main Shafting at Various Speeds. c« a Revolutions per Minute. 60 2.6 80 3.4 100 125 150 175 200 225 250 275 300 1% 4.3 5.4 6.4 7.5 8.6 9.7 10.7 11.8 12.9 •) 3.8 5.1 6.4 8 9.6 11.2 12.8 14.4 16 17.6 19 2X 5.4 7.3 9.1 11 13 16 18 21 23 25 27 2^ 7.5 10 12.5 15 18 22 25 28 31 34 37 2K 10 13 16 21 25 29 33 37 42 46 50 3 13 17 21 27 32 38 43 49 54 59 65 3X 16 22 27 34 41 48 55 62 69 76 82 3^ 20 27 34 43 51 60 68 77 86 94 103 S% 25 34 42 53 63 74 84 95 105 116 126 4 30 41 51 64 .77 90 102 115 128 141 154 4^ 43 58 73 91 109 128 146 164 182 201 219 5 60 80 100 125 150 175 200 225 250 275 300 3 66 SHAFTING. TABLE No. 41.— Giving Horse=Power of Line Shafting at Various Speeds. Revolutions per Minute. IOC 6 125 7.4 150 175 200 225 250 275 300 325 350 400 1 3 X 9 10.4 12 13.4 15 16.4 18 19.4 21 23.8 1# 7.3 9.1 10.9 12.7 14.5 16 18 20 22 23.8 25 29 2 8.9 11.1 13 15.5 17.7 20 22 24 27 19 31 35 2 Mi 10.6 13.2 16 18.5 21 24 27 29 32 34 37 42 %% 12.6 15.8 19 22 25 28 32 35 38 41 44 50 W 15 18.6 22 26 30 33 37 41 44 48 52 59 2/ 2 17 22 26 30 35 39 43 48 52 56 i 61 69 2U 23 29 34 40 46 52 58 64 69 75 1 81 92 3 30 37 45 52 60 67 75 82 ; 90 97 |105 120 3^ 38 47 57 67 76 86 95 L05 !114 124 133 152 3^ 48 59 71 83 95 107 119 131 143 155 167 190 33^ 58 73 88 102 117 132 146 L61 176 190 205 234 4 71 89 107 125 142 160 [178 L96 12131231 249J284 TABLE No. 42.— Giving Horse=Power of Shafting Used Only for Transmitting Power. « 2-e 6c/2 c Revolutions per Minute. 100 6.7 125 150 1 i 175 200 225 ' 250 1 275 300 325 1 35C 1400 l 1 ^ 8.4 10 11.8 13.5 15.1 16.S 18.5 | 20.3 22 22 27 1# 8.6 10.7 12.8 15 17.1 19.3 21.4 23.6 I 25.8 28 3C 34 IK 10.7 13.4 16 18.7 21.5 24 26.8 29.4 32.1 35 31 43 1# 13.2 16.5 19.7 23 26.4 1 29.7 33 36.2 39.5 43 46 52 2 16 20 24 28 32 36 40 44 48 52 56 64 2/s 19 24 29 33 38 42 48 53 57 62 67 76 2% 23 28 34 40 45 51 57 63 68 74 80 91 2H 27 33 40 47 54 60 67 74 80 87 94 107 2/ 2 31 39 47 55 62 70 78 86 94 102 109 125 m 41 52 62 73 83 93 104 114 125 132 146 166 3 54 67 81 94 108 121 135 148 162 175 189 216 3X 69 86 103 120 137 154 172 189 206 223 240 275 3^ 86 L07 128 150 171 193 114 236 257 i 279 300 343" SHAFTING. 367 Distance Between the Bearings. Jack shafts should always have bearings as near the pulleys as possible. Ordinary line shafts, as given in Table No. 41, and shafts for simply transmitting power, may have the distance between the hangers as given in the following table : TABLE No. 43. Diameter of Shaft in Inches. iy 2 to ik 2 to 2 l / 2 2^ to 4 Distance between bearings in feet 6/ 2 8 10 Shafts for Idlers. Shafts for idlers (see C, Fig. 1) have very little torsional stress and the distance between the bearings may also be very short, so that even with a great transverse load such a shaft may be of comparatively small diameter as far as requirements for strength is concerned. In such a shaft there is great danger of trouble from hot bearings ; therefore, in designing, it is very important to make its diameter and the length of the bearing of such proportions that excessive pressure per square inch of bearing surface is avoided. Example. Twenty-five horse-power is through idler C. ( See Fig. 1 ). are 36 inches in diameter and make 40 revolutions per minute. What is the necessary diameter of shaft C, which is supported by two bearings one foot apart and carrying a gear 48 inches in diameter placed at the mid- dle between the bearings. Solution : The velocity on pitch line of gear A will be 40 X 36 X 3.1416 to be transmitted from A to B The gears on shafts A and B Fig. 12 = 377 feet per minute. 25 horse-power = 33,000 X 25 = 825,000 foot-pounds. The pressure at the pitch line of A transferred to Cwill be 825000 377 2188 pounds. 3^8 SHAFTING. The reaction at the pitch line between C and B, is also 21S8 pounds; therefore, the total pressure (besides the weight of C, which is omitted in this calculation ) on both bearings will be 2 X 2188 = 4376 pounds and the pressure of each bearing of C wil« be 2188 pounds. Allowing a pressure on the bearings of 100 pounds per square inch, the necessary bearing surface will be 91 go — = 21.S8 square inches for each bearing. Assuming the length of the bearing to be twice its diameter, D X 2 D = 21.88 ^21^8 2 D =Vio.y4 D — o.3 inches. Calculating the size required with regard to transverse strength by the formula on page 360, 3 D = A / 1 X 4376 _ = 3.12 inches. 144 Thus, a shaft 3.3 inches in diameter is Of ample size for strength. The surface velocity of this shaft will be, 3.3X3.1416X40 ^,^ 12 per minute, and at that velocity a pressure of 100 pounds per square inch of bearing surface is very safe from liability of heating if the bearing is well made and amply provided with oil. Proportion of Keys. The breadth of the key is usually made to be one-fourth of the diameter of the shaft, and the thickness to be one-sixth of the diameter of the shaft. Keys and key-ways are usually made straight and should always be a very good fit sidewise. Frequently set-screws are used on top of keys in mill gearing. Sometimes in heavy ma- chinery keys are made tapering in thickness, usually one-eighth inch per foot of length. A corresponding taper is made in the depth of the key-way in the hub. Key-ways in shafts are always made straight. For light and fine machinery taper keys are never used. SHAFTING. 369 TABLE No. 44.— Dimensions of Couplings for Shafts. (All dimensions in inches.) Dimensions of Couplings. Diameter of Coupling Bolts. Number of of Shaft. d D L / Coupling Bolts. 1% 2 2* 3 3/ 2 4 4/2 5 3 ±y 5 6 VA 8X 9 7 8 9 10 11* 13 14* 16* 17 4 4* 4^ 5^ 6 6^ 7* 7^ # H U % % % 1 1* IX H % y* % % y* 1 4 4 4 4 5 5 6 6 6 BEARINGS. A satisfactory rule is to make the length of the bearing for line shafting six times the square root of the diameter of the shaft. Example. What is a suitable length of bearings for a shaft of four inches diameter ? Solution : Length of bearing = 6 X VT^= 12 inches. Some designers make the length of the bearing four times its diameter. Area of Bearing Surface. The projected area of any bearing is always considered as its bearing surface. Thus, the length of the bearing multiplied by the diameter of the shaft gives the area of bearing surface. For instance, the length of the box is twelve inches and the diameter of the shaft is four inches ; the area of bearing surface is 12 X 4 = 48 square inches. Allowable Pressure in Bearings. The allowable pressure per square inch of bearing surface will depend on the surface speed of the shaft and the condition 37© BEARINGS. of the bearing, arrangements for oiling, etc. For common line shafting from two to four inches in diameter, not making over 200 revolutions per minute, a pressure not exceeding forty- pounds per square inch ought to work well. Greater pressure or greater speed may make it difficult to keep the bearings cool. Example. What pressure may be allowed on a bearing twelve inches long and four inches in diameter ? Solution : Pressure = 4 X 12 X 40 = 1920 pounds. In well constructed machinery there should not be any trouble from heating, if the surface velocity and the pressure in the bearings does not exceed the values given in the following table :— Metric Measure. English Measure. Kilograms per Square Centimeter. Surface Velocity in Meters per Minute. Pounds per Square Inch. Surface Velocity in Feet per Minute. 5 12 20 100 50 20 75 180 300 300 150 60 The bearings for machinery in general are constructed in various ways and of different proportions, according to the de- signer's judgment, but it is a well-known fact that high-speed machinery must have longer bearings than slow-speed ma- chinery. The length of the bearing will usually vary from one and one-half to six times the diameter. When the shafts are small (less than two inches in diameter), and the speed is from 100 to 1000 revolutions per minute, the following empirical formula may be used as a guide : = ^( 1 + ^o) L = length of bearing, d = diameter of bearing, n = num- ber of revolutions per minute. BEARINGS. 371 Figure 1 shows a cheap, solid cast-iron box used for com- paratively small and less important shafts. Dimensions, suita- ble for bearings from one to two inches in diameter, are given in the following table : — TABLE No. 45.— Giving Dimensions of Fig. 1. (All Dimensions in Inches.) ^ - ^_ as rt D 1-1 hi) hJO i—i i—i St + CO 5 + + ^ II II II II II II || II ^2 J *X) ^ <« *«Sk *c d / ^ b c k e i m 1 1% 15* iK 334 *y* % % 1 iK *A 1/8 *t\ ±y 2 6^ 9 9 IK W2 25/ 8 *K 25/ 8 &x 7^ 5/8 5/8 i* iU »A 2s/ 8 «* 6 8« H 1 1 i*4 2 3/ 2 3 3 1 / 6« 9^ * 3 /i 2 Figure 2 shows a babbitted split box suitable for shafts from one to four inches in diameter, and running at a com- paratively slow speed. FIG d = Diameter of shaft. a = 2^ X d. b = l% X d. c — Zy.d-\-% inch. £ = 4X^+1^ inches. e = X X ^ + X inch. /= 14: x ^+ X inch. f = l^XW. l=2Xd Thickness of babbitt metal, / = Vie d -f ^ inch. Diameter of bolt, h = Ysd + X inch. Diameter of bolt, * = Vsd + ^ inch. Figure 3 shows the same general design of box as Figure 2, excepting that the bearing is longer and the base wider. This box is more suitable for comparatively high-speed shafts. 372 BEARINGS. d = Diameter of shaft. a—l}id-\-l% inches. b = l%d+ y 8 inch. c — 2 l / 2 d+2 inches. ^ — 3^+3 inches. e = %d 4~ X inch. /= % d+ % inch. g=%l _ /=*> xVd wi= d h — yi d+ % inch. i = % d + y s inch. Figure 4 shows a babbitted box or pedestal suitable for comparatively heavy-loaded shafts, from three to eight inches in diameter, such as outer bearings for steam engine shafts, bearings for jack shafts, etc. d = Diameter of shaft ; a = 2 d -f- 1 %. inch ; b = 1 }4 d -f- #j inch; c = 2% ^+2 inches; k =3% d-\- 3 inches; *?= % d-\- % inch; f=%d+ X inch-, g=.\y 2 d; Z=2d; m = d; D — \%d. Diameter of bolts, /i = Q.2d -\- % inch (approximately). Diameter of bolts, t'~ 0.2 d -\- y% inch (approximately). BEARINGS. 373 Figure 5 shows a bearing fitted into the frame of a ma- chine, suitable for shafts from one to three inches in diameter. The cut shows a part of the head-stock of a speed lathe fitted with this kind of a bear- ing. The bearing itself, which may be of gun metal or cast iron, is carefully fitted into the frame by planing and scrap- ing. This kind of a bearing is sometimes lined with babbitt, but more frequently the spindle is carefully fitted into the bear- ing by scraping. \J=*=e±l d = Diameter of bearing; /-- id; 2d; a = 2% d-\- 1 inch; b = 1 y 2 d -f- % inch ; c =1% d; e = f = 1% d + }i inch ; f=e = l%d+y s inch; g=Ud+ %mch; k = l%d+y% inch ; i =2d. Diameter of screws = s Aq d + 8 /i6 inch. Figure 6 shows the form of a self-lining and self-oiling bearing, very suitable for high-speed machinery, and used to a great extent for dynamos and electric motors. The figure shows a part of a dynamo frame with the box in section, cut through the center line of bearing, and also a partly sectional cut from the top. For dimensions see Table No. 46. The bearing n n may be cast in one piece from gun metal, as shown in the cut, or it may be (preferably for the larger size) made in two parts. The seat for this box is turned in spherical form on the outside, and a fit is obtained between this bearing and the frame of the machine either by machining or by casting in type metal or babbitt as shown at m m. The loose rings, n n, are continually dipping into the oil reservoir, and carrying oil to the shaft. (Chains are fre- quently used instead of rings). The stop-rings should be set so that the spindle has room for a little motion lengthwise in the bearing. This will in a great measure prevent heating and cut- ting, and by their peculiar shape the stop-rings will, by the action of centrifugal force, throw the oil off at a a, to return to the oil reservoir ; h h are plugs in the oil hole ; the screw i pre- vents the box from turning with the shaft, and also forms a convenient projection to take hold of when taking the cap off of the bearing. 374 BEARINGS. FIG. 6. TABLE No . 46. — Giving Dimensions of Fig. 6. d Z? D X * / b / Inches.|l£^+tf" l%d+%" 1X/+1" 4X\/^ V&d + l" Inch. 1 1^ 2 7 4 ^A *A 1# HI 2A 734 *4 ' »« % 1# 2/ 8 2/8 sy 2 5 43/ 8 Vs 13^ 2* 3 T % m *Ya 4^4 U 2 234 334 Q 7 V X^ 5S/8 5^ % 2^ »A ±* 10 6 6 T V % 2*4 33/ 8 45/ 8 103/ 8 6* 65/ 8 % 234 8« 5 T V 10|| 6S/ 8 ■7* % 3 4 5^ 11# 7 734 /8 3^ Mr HI 11/8 7^ ** H 3*4 ^ 63/ 8 124 7/ 2 8/8 Vs 334 Hf 6tf 125/ 8 734 Q 7 J T6 1 4 5^ 7* 13 8 10 1 GEAR TEETH. 375 GEAR TEETH. Circular Pitch. The length of the pitch circle or pitch line from center of one tooth to the center of the next is the circular pitch of a gear, or a rack. Cast gear teeth, constructed on the circular pitch system, may be made of the following proportions : Thickness of tooth on pitch line = T 6 f pitch. Space between teeth on pitch line == T 7 ¥ pitch. Height of tooth outside of the pitch line = T 3 o pitch. Depth of space inside of pitch line = T % pitch. -..^ ... £ circular pitch X number of teeth. Pitch diameter of gear = 3.1416 For cut gears, use the following formulas : Thickness of tooth on pitch line = 0.5 pitch. Space between teeth on pitch line = 0.5 pitch. Height of tooth outside pitch line = 0.3183 pitch. Depth of space inside of pitch line == 0.3683 pitch. To Calculate Diameter of Gear According to Circular Pitch. Rule. Multiply the circular pitch by the number of teeth in the gear and divide the product by 3.1416 ; the quotient is the diameter of the pitch circle ; add j% of the circular pitch to obtain the whole diameter of the gear. Example. Find whole diameter of a gear of 48 teeth and three-inch circular pitch. Solution : O \/ AQ Pitch diameter = . . . . _ = 45.84 inches. 3.141b Double the addendum = 3X0.6= 1.80 The whole diameter is 47.64 inches. Table No. 47 is calculated for one-inch circular pitch ; to find the pitch diameter of a gear of any number of teeth given in the table, multiply the diameter given in the table by the circular pitch in the gear, and the product is the pitch diameter of the gear. In order to find the whole diameter, add twice the height of the tooth outside the pitch line, as calculated by the above formula. 376 GEAR TEETH. TABLE No. 47 Giving Pitch Diameter of Gears of One Inch Circular Pitch. Teeth. Dia. Teeth. . Dia. Teeth. Dia. Teeth. Dia. 12 3.82 ! 36 11.46 60 19.10 84 26.74 13 4.14 37 11.78 61 19.42 85 27.06 14 4.46 38 12.10 62 19.74 86 27.38 15 4.78 39 12.42 63 20.06 87 27.70 16 5.09 40 12.73 64 20.37 88 28.01 17 5.41 41 13.05 65 20.69 89 28.33 18 5.73 42 13.37 66 21.01 90 28.65 19 6.05 43 13.69 67 21.33 91 28.97 20 6.37 44 14 68 21.05 92 29.29 21 6.69 45 14.32 69 21.97 93 29.60 22 7 46 14.64 70 22.28 94 29.92 23 7.32 47 14.96 71 22.60 95 30.24 24 7.64 48 15.28 72 22.92 96 30.56 25 7.96 49 15.60 73 23.24 97 30.88 26 8.28 50 15.92 74 23.56 98 31.20 27 8.60 51 16.24 75 23.88 99 31.52 28 8.91 52 16.55 76 24.19 100 31.83 29 9.23 53 16.87 77 24.51 101 32.15 30 9.55 54 17.19 78 24.83 102 32.47 31 9.87 55 17.51 79 25.15 103 32.78 32 10.19 56 17.83 80 25.47 104 33.10 33 10.50 57 18.14 81 25.79 105 33.42 34 10.82 58 18.46 82 26.10 106 33.74 35 11.14 59 18.78 83 26.42 107 34.06 To Calculate Diameter of Gears when Distance Between Centers and Ratio of Speed is Given. When calculating gears to connect two shafts of given dis- tance between centers and at a given ratio of speed, use the formula, _ 2 X S X n d = n + JV 2X S XN n + N D = Diameter of large gear. d = Diameter of small gear. iS" = Distance between centers in inches. JV= Number of revolutions of large gear per minute. n = Number of revolutions of small gear per minute. Note. — The small gear is always on the shaft having the greater speed. GEAR TEETH. 377 Example. What will be the diameter of the gears to connect two shafts when the distance between centers is 82 inches, and one shaft is to make 135 revolutions and the other 105 revolutions per minute ? Solution : _, 2 X 32 X 135 nn . . D = — -7- — . . ... — = 36 inches diameter. 13o -\~ lOo , 2 X 32 X 105 aa . , a= — iOK 1 1A _ — = 28 inches diameter. 13o -j- lOo After the diameter of each gear is calculated, the pitch is decided upon according to the power the gears have to transmit. Frequently the pitch will have to be altered somewhat, and such gears sometimes have teeth of very odd pitch, in order to obtain the right number of teeth to give the required ratio of speed. The ratio between the number of teeth in the gears may always be seen from the ratio of speed between the two shafts. For instance, in the above example, the ratio of speed between the shafts is 135 Ao5, which, reduced to its lowest terms, is 9 /r, therefore, the number of teeth in the two gears may be any multiple of 9 and 7, respectively. For instance, 8 X 9 = 72 teeth for the large gear, and 8 X 7 = 56 teeth for the small gear ; or, 10 X 9 = 90 teeth for the large gear, and 10 X 7 = 70 teeth for the small gear, etc. The dimensions of teeth may be calculated according to rules given on page 375. Diametral Pitch. The dia?netral pitch of a gear is the number of teeth to each inch of its pitch diameter. In cut gearing it is always customary to calculate the gears according to diametral pitch. When gears are calculated according to circular pitch the corre- sponding circumference of the pitch circle is usually an even number, but the diameter will generally be a number having cumbersome fractions, and therefore the distance between the centers of the gears will be a number having fractions which may be very inconvenient to measure with common scales. This is because the circumference of a circle divided by 3.1416 is equal to its diameter and the diameter multi- plied by 3.1416 is equal to the circumference. When gearing is calculated according to diametral pitch this trouble is entirely avoided, as this directly expresses the number of teeth on the circumference of the gear according to its pitch diameter. For instance, "six diametral pitch" means that there are six teeth on the circumference of the gear for each inch of pitch diameter. Thus, a gear of six diametral pitch and forty-eight teeth will be eight inches pitch diameter. A gear of " eight diametral pitch " means that the gear has eight teeth per 37§ GEAR TEETH. inch of pitch diameter. A gear of " ten diametral pitch " means that the gear has ten teeth per inch of pitch diameter. A gear of " twelve diametral pitch " means that the gear has twelve teeth per inch of pitch diameter, etc. Thus, the pitch diameter and, consequently, the distance between the centers, will be a number which may be conven- iently measured, and the dimensions of tooth parts are also much more easily calculated by this system. Rules for Calculating Dimensions of Gears According to Diametral Pitch. The pitch diameter is obtained by dividing the number of teeth by the diametral pitch. Example. What is the pitch diameter of a gear of 48 teeth, 16 pitch ? Solution : 48 divided by 16 = 3, therefore the pitch diameter is 3 inches. The number of teeth is obtained by multiplying the pitch diameter by the diametral pitch. Example. What is the number of teeth in a gear of 5 inches pitch diameter and 12 pitch ? Solution : 5X12 = 60, therefore the gear has 60 teeth. The whole diameter of a spur gear is obtained by adding 2 to the number of teeth and dividing the sum by the diametral pitch. Example. What is the whole diameter of a gear blank for 68 teeth, 10 pitch ? Solution : Whole diameter = — —— = 7 inches. The number of teeth is obtained by multiplying the whole diameter of the gear by the diametral pitch and subtracting 2 from the product. Example. The whole diameter of a gear blank is 8 inches ; it is to be cut 10 diametral pitch. Find the number of teeth. Solution : Number of teeth = (8 X 10) — 2 = 78. The diametral pitch is obtained by adding 2 to the number of teeth and dividing by the whole diameter. GEAR TEETH. 379 Example. A gear has 64 teeth and the whole diameter is 16>£ inches. What is the diametral pitch ? Solution : Diametral pitch = ^ n . = 4. 16J2 Thus, the gear is 4 diametral pitch. Note. — The term diameter of a gear usually means diame- ter of pitch circle. The distance between the centers of two spur gears is ob- tained by dividing half the sum of their teeth by the diametral pitch. Example. What is the distance between centers of two gears of 48 and 64 teeth and 8 diametral pitch ? Solution : 4g _i_ g4 Distance — = 7 inches. 2X8 The circular pitch is obtained by dividing the constant 3.1416 by the diametral pitch. Example. What is the circular pitch of a gear of eight diametral pitch ? Solution : Circular pitch = — '— — = 0.393 inch. 8 The thickness of the tooth on the pitch line is obtained by dividing the constant 1.5708 by the diametral pitch. Example. What is the thickness of the tooth on the pitch line of a gear of 6 diametral pitch ? Solution : Thickness of tooth = - 1 —, — = 0.262 inch, o The working depth of the tooth is obtained by dividing 2 by the diametral pitch. The clearance at the bottom of the teeth is T ^ of the thickness of the tooth on the pitch line. The whole depth to cut the gear is obtained by dividing the constant 2.157 by the diametral pitch. Example. Find the depth to cut a gear of 8 diametral pitch. Solution : Depth = ^4^ = 0-27 inch. 3 8o GEAR TEETH. The whole depth is nearly equal to 0.6866 times the circular pitch. The use of the following tables will facilitate calcula- tions regarding dimensions of teeth in diametral pitch. TABLE No. 48.— Comparing Circular and Diametral Pitch. Diametral Pitch. Circular Pitch. Circular Pitch. Diametral Pitch. 2 1.571 inch. \y 2 inch. 2.094 2^ 1.257 ' ItV 2.185 3 1.047 ' ltt 2.285 3/ 2 0.898 ' lf 5 6 2.394 4 0.785 ' IX 2.513 5 0.628 ' 1A 2.646 6 0.524 ' IX ' 2.793 7 0.449 ' ItV 2.957 8 0.393 ' 1 3.142 9 0.349 ' il 3.351 10 0.314 ' 7 /s 3.590 11 0.286 ' tI 3.867 12 0.262 ' U 4.189 14 0.224 ' 1 1 1 6 4.570 16 0.196 ' n 5.027 18 0.175 ' 9 1 6 5.585 20 0.157 ' % 6.283 22 0.143 ' tV 7.181 24 0.131 ' X 8.378 26 0.121 ' A 10.053 28 0.112 ' % 12.566 30 0.105 ' A 16.755 32 0.098 ' % 25.133 TABLE No. , jo. — Giving Dimensions of Teeth Calculated According to Diametral Pitch. Diametral Pitch. Depth Cut in to be Gear. Thickness of Tooth on Pitch Line. Diametral Pitch. Depth to be Cut in Gear. Thickness of Tooth on Pitch Line. 2 1.078 in. 0.785 in. 12 0.180 in. 0.131 in. 2/ 2 0.863 0.628 14 0.154 0.112 3 0.719 0.523 16 0.135 0.098 sy 2 0.616 0.448 18 0.120 0.087 4 0.539 0.393 20 0.108 0.079 5 0.431 0.314 22 0.098 0.071 6 0.359 0.262 24 0.090 0.065 7 0.307 1 0.224 26 0.083 0.060 8 0.270 0.196 28 0.077 0.056 9 0.240 0.175 30 0.072 0.052 10 0.216 0.157 32 0.067 0.049 11 0.196 0.143 GEAR TEETH. 38 I To Calculate the Number of Teeth when Distance Be= tween Centers and Ratio of Speed is Given. Select for a trial calculation, the diametral pitch which seems most suitable for the work. » Calculate the sum of the number of teeth in both gears corresponding to this pitch by multiplying twice the distance between their centers by the diametral pitch selected. The number of teeth in each gear is obtained by the follow- ing formula : NX A T — t — n +jV n XA_ N -r n T = Number of teeth in large gear. / = Number of teeth in small gear. JV— Number of revolutions of small gear. n = Number of revolutions of large gear. A = Number of teeth in both gears. Example. The center distance between two shafts is 15 inches. The small gear should make 126 and the large gear, 90 revolutions per minute. Calculate the number of teeth in each gear, if 8 diametral pitch is wanted. Solution: The number of teeth in both gears is 2 X 15 X 8 = 240. - 126X240 -140 teeth. 126 + 90 /= 9 <>X 240 ^ 100 teeth. 126 + 90 Frequently it is impossible to get gears of the desired pitch to fit within the given center distance and to give the exact ratio of speed. Some modifications must then be made ; either the exact ratio of speed must be sacrificed, the pitch must be changed, or the distance between centers must be altered. Note. — The ratio of the number of teeth in the gears can be seen from the ratio of the speed. For instance, in the above ex- ample the ratio of speed is 90 /i26, which, reduced to its lowest terms, is % ; therefore, the number of teeth in the two gears may, with regard to speed ratio, be any multiple of 5 and 7, respectively, but in order to fit the given center distance and also to be 8 pitch, they must be 100 and 140, which is 20 X 5 = 100 and 20 X 7 = 140. 382 GEAR TEETH. FIG. 1. The shape of gear teeth is usually either Involute or Cycloid ( also frequently called Epicycloid ). The shape of a cycloid tooth for a rack is four equal cycloid curves, which may be con- structed, so to speak, by letting the generating circle a ( see Fig. 1 ) roll along on the pitch line of the rack, both above and below. Cycloid gears „, — -^ have the curve out- side the pitch circle formed by an Epi- cycloid (see Fig. 26, page 191) and the curve inside the pitch circle by a Hypocycloid. The curves al- ways meet on the pitch line in both gears and racks. The theoretical requirements for correct form of Epi- cycloid gear teeth are that the face of / ' \ _J^'" i££-inch circular pitch ; the generating circle is 0.98 inch diameter, which is equal to half of the pitch diameter of a gear of 12 teeth and V z inch circular pitch. All gears of the same pitch having 12 teeth or more, con- structed by the same generating circle in the same manner as the rack, will match and be interchangeable with the rack, and will also match and be interchangeable with each other. GEAR TEETH 183 Fig. 1 a shows a drawing of a pattern for a gear and rack half inch circular pitch, and cast teeth of the cycloid form. The gear has 48 teeth and the pitch diameter may be obtained by means of the table on page 376, thus: the pitch diameter of a gear of 48 teeth, one inch circular pitch, is given as 15.28 inches, the pitch diameter of a gear of 48 teeth and one half inch circular pitch will, therefore, be 15.28 divided by 2, which is 7.64 inches. The width of the space m in Fig. 1 a may be seven-thirteenths of the pitch and the thickness of the tooth / may be six-thirteenths of the pitch. Fig. 1 a Note : — Gears with so small a pitch as half-inch are now very seldom made with cast teeth, but such gears are usually cut from solid stock and the teeth are usually constructed accord- ing to the involute system and calculated according to dia- metral pitch. 3»4 GEAR TEETH Fig. i b shows a gear with internal teeth constructed ac- cording to the epicycloid system. This gear has 48 teeth \ inch circular pitch and the form of teeth is constructed by a generating circle of the same diameter as the gear in Fig. 1 a. Fig. 1 b GEAR TEETH.. 385 When internal teeth are constructed according to this sys- tem, the difference between the number of teeth in the internal gear and its external pinion must never be less than 1 2 ; practically it is better to limit the difference to 15 or 20 teeth. As interchangeability is seldom required for internal gear- ing, such gears and their mates are generally constructed together and the designer chooses a generating circle of suitable size to give the shape of tooth he considers best, and he may also vary the size of the driving or the driven gear so as to reduce con- tacts when the teeth are approaching each other, etc., according to his own judgment and experience. The difference in pitch diameter of the internal gear and its pinion should never be less than the sum of the diameters of the generating circles, and the diameter of the generating circle of the flanks for the pinion should never be larger than half the pitch diameter, but it should, preferably, be smaller. As a rule, fillets at the bottom of the teeth are not used in internal gears, but if used they should be very small. In order that gears constructed with cycloid teeth should /un smoothly, it is very important to have the distance between centers correct, so that the pitch lines will exactly meet each other. For this reason, there are many kinds of machinery where cycloid gears should not be used : for instance, for change gears on lathes involute teeth are far more suitable. When making patterns, the shape of one tooth is usually carefully drawn on a thin piece of sheet metal, either brass or iron ; this is then filed out and used as a templet in tracing the other teeth on the pattern. Sometimes a fly-cutter is made according to this constructed tooth, and all the teeth in the pat- tern are cut on an index machine or a gear cutting machine ; but if such a machine is not available, the next best way is to set out the pitch line of the gear on this templet and also the center line of the tooth, radially towards the center, then draw the pitch line on the pattern, space off each tooth carefully with a pair of dividers and draw the center line on each tooth prolonged across the rim radially in the direction of the center of the gear, then lay the templet carefully on each of these spacings, making the pitch line and the center line of tooth on the templet to exactly match the pitch line and center line of the tooth drawn on the pattern, then trace around the templet and get the shape of one tooth ; then move the templet to the next spacing and trace the next tooth, and so on for all the teeth on the gear. For small patterns it is convenient to fasten the templet to a strip of metal long enough to reach from the teeth to the cen- ter of the gear wheel, placing a point in the center of the gear, drilling a hole in the strip and letting it swing around this point, then after all the teeth are spaced off on the pattern the tem- plet is swung from one tooth to the other and all the teeth are traced by the templet. This method has the advantage that 386 GEAR TEETH. it will mark all the teeth exactly alike, because the templet being fastened to this strip can not easily get out of position. The distance from the pitch line of the templet to the cen- ter hole in the strip must be laid off according to the shrinkage rule, and is, of course, in numerical value equal to the pitch radius of the gear, which should always be calculated and given on the drawing. When gear patterns are less than six inches in diame- ter it is preferable not to allow anything for shrinkage, as the moulder will usually rap the pattern about as much as the cast- ing will shrink in cooling. When a pair of cycloid gears are constructed without con- sidering in terchangeability with other gears of the same pitch, it is customary to choose a generating circle having a diameter equal to three-fourths of the radius of the pitch circle of the small gear, providing this gear has 24 or more teeth. A large generating circle probably reduces the friction in a small measure but gives teeth of less strength. The largest gen- erating circle used ought never to exceed the radius of the pitch circle of the small gear. Decreasing the generating circle will probably increase friction somewhat in the gears, but it gives teeth of greater strength. The smallest generating circle used in practice is equal to half the diameter of the pitch circle of a gear having 12 teeth of the same pitch as the gear to be con- structed. Many eminent mechanics consider it preferable never to use a generating circle smaller than half of the pitch diameter of a gear of 15 teeth. Cycloid gears are mostly used in large cast gears of one- inch circular pitch or more. Sometimes the driving gear is made of slightly larger di- ameter, and the teeth spaced at a correspondingly greater pitch than the theoretically calculated size. This is done in order that the teeth shall not rub on each other on the approaching side, but only touch as they are passing the center line and commence to slide away from each other. This will make the gears less noisy, but probably gears made in this manner will wear faster, as there are fewer points of contact, although this may be offset by the fact that the friction between the teeth when they are meeting and pushing onto each other is more injurious than the friction produced when they are sliding away from each other. The same idea is sometimes employed when constructing bevel gears, in order to make them run quietly. This mode of sizing gears is not, as a rule, used in modern gear construction, but it is a point well worth remembering, because if either of two gears is over or under size, the gear of over-size should always be used as the driver, and the gear of under-size should always be the driven ; never vice versa. This will apply as well to involute as to epicycloid gears. GEAR TEETH. 387 Involute Teeth. Suppose a strap is fastened at a and b on the two round discs in Fig. 2. If the disc b is turned in the direction of the arrow, the strap will move in a straight line from c, toward d. This motion will cause the disc a to rotate with exactly the same surface speed as the disc b, but in the opposite direction. Suppose, further, that to the under side of the disc a (see Fig. 3) is fastened a piece of sheet brass ft, or other suitable material of somewhat larger diameter than disc a, and that a scratch awl is fastened in the strap at the point m ; then by turning the disc b in the direction of h to b, and the strap moving with it, being kept tight by the resistance of disc a, the scratch awl will trace on the brass plate the curve from m to k, but if the discs are moving in the opposite direction, the scratch awl will trace the curve from m to K. Take an- other brass plate and do the same thing with the other disc, and a similar curve will be produced. In these two brass plates the stock may be filed away carefully, following the curves as shown in Fig. 4. The discs are laid to match each other and free to — -R fig. fig. 3. swing on their centers ; turning the disc a in the direction of the arrow, it will give motion to b, and both discs will move with the same speed in exactly the same manner as if they were connected by the strap as shown in Fig. 2. 388 GEAR TEETH. The curve on these two discs represents the form of a gear tooth in the involute system.* The line h g, Fig. 4, is called the line of pressure or the line of action. The circles, P and P, are the pitch circles. The line B R shows the direction of motion of the teeth at the moment they are passing the center line, c c. Approximate Construction of Involute Teeth. It will be noticed that the line of pressure, h g, forms an angle with the line B R. This angle is usually taken as 14>£ degrees. This makes the diameter of the base circle, £, (see Fig. 5) equal to 0.968 times the diameter of the pitch circle. The base circle gg, in Fig. 5, corresponds to the disc in Fig. 3, and the line of pressure in Figs. 5 and 6 corresponds to the strap in Fig. 2. The line of pressure, h g, Fig. 5, is 75^ degrees to the center line,/c. A perpen- dicular i s ere c t e d from the line h g, through the ■center, c. Using the point of in- tersection at i as cen- ter, the tooth is drawn sim- ply by a cir- cular arc. This will, in practical work for small gears having more than twenty teeth, correspond nearly enough to the true involute, which was illustrated by means of the strap, disc and scratch awl, as explained in Figs. 2, 3 and 4. When the gear has less than twenty teeth, and is constructed by circular arcs, as shown in Fig. 5, the top of the tooth will be too thin ; but the top of the tooth will be too thick to clear in the rack, if the true involute curve is used. When the teeth are of true involute curve, a smaller gear than twenty-five teeth will not run freely in a rack having straight teeth slanting 14^ degrees. (See Figs. 6 and 7). Therefore, *The way to actually draw this curve on paper by means of drawing instru- ments is explained on page 192. This way explained here, using the disc on the strap, is merely for illustrating and explaining principles, and serves well for that purpose, but would be inconvenient to use in actual construction of gear teeth. In actual v/ork one tooth is carefully constructed, and templets and cutters are made and used, as was explained for Cycloid Gears, pa~e 385. GEAR TEETH. 389 when a gear has less than twenty-five teeth it is necessary to round the teeth somewhat outside the pitch circle. By making either a drawing or a templet, it is very easy to see how much to round / ~~—14^^- « /•-. A&* p --*■; Fig. 6. Involute Teeth (Cast.) the teeth to make them clear in the rack. In interchangeable sets of cut involute gears it is customary to cut the rack with a cutter shaped for a gear of 135 teeth. This will make the teeth in the rack slightly curved instead of straight, as shown in Fig. 6, and this will also make it possible to construct the small gears in an interchangeable set nearer to a true involute, and still have them run freely in the rack. 39° GEAR TEETH. When gear teeth are constructed as shown in Figs. 6 and 7, the line g h is 75 j^ degrees to the line c f, and the line c i is \A.y 2 degrees to the line c f. (See Fig. 5), # Fig. 7. Involute Teeth (Cut). The line h g will always be tangent to the base circle, which is concentric to the pitch circle. The diameter of the base circle is always 0.968 times the diameter of the pitch circle. The circle forming the shape of the tooth must always have its center on the circumference of the base circle, and its diameter will be one-fourth of the pitch diameter of the gear. As shown in Fig. 5, the same circle gives the form of tooth for coarser or finer pitch. When gears are drawn by this method the pitch circle is divided into as many teeth and spaces as there are to be teeth in the gear ; then the form of the tooth is simply struck by the dividers, always using the periphery of the base circle as center, and always taking the distance in the dividers equal to one-fourth of the radius of the pitch circle. The diameter of the base circle is 0.968 times the diameter of the pitch circle, because cosine of 14^ degrees is 0.96815. The diameter of the circle forming the shape of the tooth is 0.25 times the diameter of the pitch circle of the gear, because sine of 14^ degrees is 0.25038. If the line of pressure is laid at any other angle GEAR TEETH. 39 1 than 14 J^ degrees, all these other proportions will also change. Fig. 6 shows a pattern for gears and rack constructed ■ with necessary clearance as used for cast gears. All tooth parts are of the same dimensions as used for cycloid gears as given on page 375. Fig. 7 shows a cut gear and rack constructed in the same manner. The advantages of the involute system of gears are in the strength of teeth, and also that the gears will trans- mit uniform motion and run satisfactorily, even if the distance between centers should be slightly incorrect. Internal Gears with involute teeth. Internal gears with involute teeth are constructed by the same method as external gears. It is shown by Fig. 5 that the same circle will form the teeth for an internal gear as well as for an external gear of the same pitch diameter. The only dif- ference is that the teeth in the internal gear will be concave, because what is space in the external gear will be tooth in the internal gear. When the difference between the number of teeth in the internal gear and its external pinion is small, it is necessary to round the point of the teeth in the internal gear in order to make them run free without interference. Frequently the teeth, both in the internal gear and its pinion, are made shorter than standard teeth in order to avoid interference. Sometimes it may be advisable to not only shorten the teeth but also to increase the pressure angle from 14^ degrees to 20 degrees in order to obtain smooth running internal gears. CHORDAL PITCH. The term chordal pitch is not used very much in gear cal- culations, but it is of practical value sometimes in machine work to be able to determine the length of the chord as well as the length of the arc. Fig. 8 shows two teeth in a gear of 18 teeth, 1^ circular pitch and 8.59 in. pitch diameter. The distance from d to e measured on the curved line is the circular pitch and is i\ inches. The distance from d to e measured on a straight line is the chordal pitch of the gear. The chordal pitch is always less than the circular pitch. 392 GEAR TEETH The chordal pitch may be calculated by the formula : Chordal pitch = 2 X sine a X r. Angle a is obtained by dividing 180 degrees by the number of teeth in the gear. r = pitch radius of gear. RIMS, ARMS AND HUBS OF SPUR GEARS 393 Example: > Find chordal pitch in the gear shown in Figure 8 where the gear has 18 teeth and li inch circular pitch. Solution: Pitch radius is 4.295 inches. Angle a = — - = 10 degrees. Sine of 10 deg. is found in the table on page 158 to be 0.17365. Chordal pitch =2X0.17365X4.295 = 1.492 inches. Thus, we see in this case the chordal pitch is 0.008 inch less than the circular pitch. For such a gear all calculations must be made according to circular pitch, but when spacing off the teeth on the pitch circle the dividers are, of course, set according to the chordal pitch. "When a gear has a large number of teeth, the difference between the circular pitch and the chordal pitch is practically nothing. Rim, Arms and Hub of Spur Gears. Figure 9 shows the shape of rim, arms and hub of a cast iron gear as used in the ordinary mill gearing. The thickness of the rim of the gear at the edge, as at t, Fig- ure 9 may be 0.5 X P. The thickness of the rim at the middle, as at S, may be 2 X /, or which is the same, equal to the pitch of the gear. The width of the rim, as at F, Figure 9, may be from two to three times the pitch of the gear. The number of arms in large cast iron gears may be taken as follows : When gears are less than 8 feet in diameter use six arms; from 8 to 16 feet diameter, use eight arms and from 16 to 24 feet diameter, use ten arms. The dimensions of the arms may be calculated by the fol- lowing formulas: For 1 in. circular pitch or more, the width of the arm produced to the center of the gear, as at h Figure 9, may be 3 - V- D X P X F m The width of the arm at the rim as at k lt Figure 9, may be -,, = 0.85 X J. D X P X F it = width of arm in inches produced to the center of the hub. &, = width of arm in inches produced to the rim. D = diameter of the gear in inches. 394 RIMS, ARMS AND HUBS OF SPUR GEARS P = circular pitch of the gear in inches. F = width of the face of the gear in inches. m = number of arms. L~ -h-H — Fig. 9 Example: Calculate the arms in a gear of 80 inches diameter 6 inches face and 2 inches circular pitch and 6 arms. Solution: *- v 80 X 2 X 6 -J 160 = 5.429 inches RIMS, ARMS AND HUBS OF SPUR GEARS 395 Practically, the width at the arms produced to the hub will be 5| inches. h 1 = 0.85 X 5.429 = 4.615 inches. Practically, the width of the arms at the rim will be 4f inches. The thickness of the arms is half of their width and the section is elliptic as shown at A, Figure 9. The diameter of the hub in cast iron gears is frequently made twice the diameter of the shaft and the length of the hub may be from 1| to 2\ times the diameter of the shaft. Arms in Small Gears. Small gears with cut teeth are usually calculated according to the diametral pitch and the size of the arms are largely a matter of judgment with the designer. The arms in small gears with cut teeth are generally larger, in proportion to the size of the teeth, than for gears with cast teeth, both because when the rim is cast solid it is so much larger, as the depth of the teeth' to be cut, and will, therefore, cool off slower, and if the arms are too small in proportion to the rim, it 396 RIMS, ARMS AND HUBS OF SPUR GEARS will cause initial stress due to the unequal shrinkage when the casting is made and also because the arms must not only be strong enough to stand the strain they are to transmit, but they must also be strong enough to stand the strain of turning and cutting in the machine shop without undue chattering and trouble Following tables (Nos. 50 and 51) are offered only as a guide and will often be modified by the designer. They are intended for cast iron gears with teeth calculated according to diame- tral pitch from 2 to 16 pitch. TABLE No. 50. Shape of Gears Diametral Pitch Plain Web 6 Arms 2 pitch not exceeding 1 8 teeth froml8to36teetl 1 over 36 teeth 3 " 20 ' 20"40 ' 40 ' 4 " 24 " ' .24 4< 44 " ' 44 4 5 ' 4 28 " ' 28 4< 48 " 1 48 ' 6 " 32 " 1 32 (< 50 il ' 50 ' 8 " 40 " ' 40"56 " 1 56 ' 10 " " 45 M 45"65 " 4 65 ' 12 " 52 " 1 52 4 '70 " 4 70 4 14 " 58 4< 58"75 " 4 75 ' 16 " 64 " 64' '80 " 41 80 " TABLE No. 51. Size of Arms in Small Gears. a 5 fcj* !o in s2 «® (0
  • "S. O » s§ 1- • *2 u i- ._" (LI C t-i • a~ X fc a~ cfl bfl S ■" a bo E'S « bo fl- rt bo a - : « bo 1 CO v D rt fi o'fe S.S o"£ rt.S «gS «." o'& s.s o'S V B 1 * s «s £ be ^5 »- £ be 58 B bo 1*5 •C (LI ii b0 J2 4J •3? ^ is ii bo 5S in § rt Ck^ 'Si So *i P-I isi Ph O li Ph >i 2 1 - 20 2f 24 21 36 3i 48 3fi 60 4 3 1 - 16 2* 20 2* 24 21 35 2} 48 34 4 ! 12 1* 16 2 20 24 30 2t 36 21 5 i - 10 If 12 ltt 16 Ml 20 2 24 2i 6 1 \ 9 If 12 ii 16 it 20 If 24 2 8 1 ■ 8 li 10 iA 12 it 16 li 20 If 10 1 h 7 1* 8 iA 10 ii 12 iA 16 If 12 i 6 1 8 iA 10 it 12 ii 16 li 14 3 \ 6 15 1^ 8 1 10 iA 12 it 16 It 16 1 % 6 i 8 1 5 ITT 10 1 12 iA 16 li BEVEL GEARS 397 The gears according to preceding table are supposed to have 6 arms. The arms are tapered in width ^ inch per inch toward the rim. The thickness of the arms is half of their width, and the section of the arms elliptical as shown in Fig. 10. The arms are provided with rounded fillets both at the hub and at the rim. Width of Gear Wheels. Gears with cast teeth are usually made narrower than gears with cut teeth. In spur gears with cast teeth it is customary to make the width of the gear four to five times the thickness of the teeth, or twice the circular pitch. Width of Gears With Cut Teeth. The following rule is recommended by Brown & Sharpe Mfg. Co. in their " Practical Treatise on Gearing": Divide eight by the diametral pitch, and add one-fourth inch to the quotient; the sum will be the width of face for the pitch required. Example. What width of face is required for a gear of four pitch? Solution : Face = f + \ = Z% inches. For change gears on lathes where it is desirable not to have faces very wide, the following rule may be used : Divide four by the diametral pitch and add one-half inch. By the latter rule a four-pitch change gear would have but a 1^-inch face. BEVEL GEARS. Fig. 1 1 is a diagram showing how to size bevel gear blanks. First, lay off the pitch diameters of the two gears, which may be calculated according to diametral pitch or to circular pitch ; second, draw the pitch line of teeth ; third, lay off on the back of the gear the line a b, square to the " pitch line of teeth :" fourth, on the line a b, lay off the dimensions of the teeth exactly in the same manner as if it was for a spur gear. If the gear is calculated according to circular pitch, find dimensions of teeth by formulas on page 375, but if^ the gear is calculated according to diametral pitch, find dimensions of teeth in Table No. 49. 398 BEVEL GEARS. Make the drawing carefully to scale ( full size preferable whenever possible ), and measure the outside diameter as shown in the diagram. FIG. II To Calculate Size of Bevel Gear for a Given Ratio of Speed. Ascertain the ratio of speed in its lowest terms. Multiply each term separately by the same number, and the products give the number of teeth in each gear. Example. Two shafts are to be connected by bevel gears, one shaft to make 80 revolutions and the other 170 revolutions per minute. Find the number of teeth in the gears. BEVEL GEARS. 399 Solution : Ratio = 80 /i7o = 8 /i7. For instance, multiplying by 6, the large gear on the shaft making 80 revolutions will have 17 X 6 = 102 teeth. The small gear on the shaft making 170 revolutions will have 8 X 6 = 48 teeth. Assuming that on account of room it is necessary to use smaller gears, a smaller multiplier may be used, but if it is desir- able to have larger gears, use a larger multiplier. Decide on the pitch of the gears according to the work they are required to do. Make a scale drawing and get the dimensions as explained on page 397. Dimensions of Tooth Parts in Bevel Gears. Fig. 1 2 shows a sectional drawing of a pair of bevel gears of sixteen diametral pitch, 18 teeth in the small gear and 30 teeth in the large gear. The pitch diameter of the small gear is || = lVs inches. The pitch diameter of the large gear is f§ = 1% inches. The addendum of -the teeth on the back at a is T ^ inch, the same as for a spur gear of 16 diametral pitch. The thickness 4-00 BEVEL GEARS. and the total depth to cut the gear at a are 0.098 inch and 0.135 inch, respectively. These dimensions are found in Table No. 49, as if it was a spur gear of 16 diametral pitch. All the dimensions of the tooth decrease gradually toward b, as the whole tooth is sup- posed to vanish in a point in the center at c. The dimensions of the teeth at b may be calculated and are always in the same proportion to the dimensions at a as the distance c b is to the distance c a ; thus, if the length of the tooth from a to b is made one-third of the length of the distance c a, the distance b c is two-thirds of the distance a c, and, consequently, all the dimen- sions of the tooth at b are two-thirds of the dimensions at a. Instead of calculating the size of the teeth at<£, the dimensions may be obtained by careful drawing. The depth of the tooth at the smallest end is "then measured directly at b, but the thickness is measured at /; the distance t h is laid off equal to b d. The length of the tooth from a to b is to a certain extent arbitrary, but a good rule is seven inches divided by the diame- tral pitch, but never longer than one- third of the distance from a to c. Example. What is the proper length for the teeth of a bevel gear of 8 diametral pitch ? Solution : Seven inches divided by 8 = % inch, if the gears are of such diameters that this will not make the length of the teeth more than one-third of the distance from a to c. Form of Tooth in Bevel Gears. Extend the line a (see Fig.12), until it intersects the axial center line of the gear, as at h ; use/z as the center, and the shape of tooth at a for the large gear is constructed as if it was a spur gear having a pitch radius as large as a k. The shape of the tooth at b is constructed in the same way, by extending the line b (which always — the same as line #, — is square to the pitch line of the tooth) until it intersects the axial center line of the gear, as at d. Using d as center, the shape of the tooth is constructed as if it was a spur gear having a pitch radius equal to db. The shape of the teeth of the small gear is obtained in the same way. which is shown by the drawing. The form of tooth is shown to be approximately involute, constructed as explained for spur gears, page 388. Measuring the back cone radius, a /i, of the large gear, it is found to be f-§ inch, and the diameter will be f § inch ; thus, the shape of the tooth at a for the large gear will be the same as the shape of the tooth in a spur gear of 58 teeth, sixteen diametral pitch. BEVEL GEARS. 4OI Measuring the back cone radius of the small gear, it is found to be f| inch, and the diameter will be f £ inch ; conse- quently the shape of tooth at a for the small gear is the same as the shape of tooth in a spur gear of 21 teeth, sixteen diametral pitch. Therefore, if this pair of gears is to be cut by a rotary cutter having a fixed curve, a different cutter is required for each gear. When, in a pair of bevel gears, both gears are of the same size and have the same number of teeth, and their axial center lines are at right angles, they are called miter gears, and one cutter, of course, wilt answer for both gears. One cutter will also answer in practice when the difference of the back cone radius of a pair of gears is so small that it comes within the limit of one cutter as used for spur gears of the same size. Bevel gears may also be made with cycloid form of teeth, but when- ever cut by rotary cutters, as usually employed in producing small bevel gears of diametral pitch, the involute form of tooth should always be used. Cutting Bevel Gears. When bevel gear teeth are correctly formed, the tooth curve will constantly change, from one end of the tooth to the other. Therefore, bevel gears of theoretically correct form cannot be produced by a cutter of fixed curve ; but, practically, very satisfactory results are obtained in cutting bevel gears of small and medium size in this way. When a regular gear-cutting machine is not at hand, the Universal milling machine is a very convenient tool for cutting bevel gears of moderate size, and is used in the following way : First, see that the gear blank is turned to correct size and angle, and adjust the machine to the angle corresponding to the bottom of the teeth in the gear. The correct index is set ac- cording to the number of teeth in the gear. Adjust the cutter to come right to the center of the gear, cut the correct depth as marked on the gear at a (see Fig. 12), according to Table No. 49, and when the machine is adjusted to the correct angle, and the correct depth is cut at a, the correct depth at b will, as a matter of fact, be obtained. Second, when a few teeth are cut in the gear (two or three) bring, by means of the index, the first tooth back to the cutter. By means ©f the index, rotate the gear, moving the tooth toward the cutter; but, by the slide, move the gear sidewise away from the cutter, until the cutter coincides with the space at b ; then cut through from a to b. This operation will widen one side of the tooth space at a. Note the position of the machine, and, by the use of the index and slide, return the cutter to its central position and in- 4 02 BEVEL GEARS. dex into the next space, and rotate the other side of the tooth toward the cutter as much as the first side ; but, by the slide, the gear is moved side wise away from the cutter until the cutter coincides with the space at bj then cut through on this side from a to b. Thus, by repeated cutting on each side alternate- ly, one tooth is backed off equally on both sides and measured by a gage, until the correct thickness on the pitch-line at a, according to Table No. 49, is obtained. Be very careful to have the machine set over the same amount on each side of the tooth, or else the tooth will be askew. Third, when one tooth, thus by trial, is correctly cut, note the position of the machine and cut all the teeth through on one side, then set over to the other side in exactly the same position as was found to be right for the first tooth ; cut through again and the gear is finished. Thus, when the correct position of the machine is obtained, any number of gears of the same size and same pitch may be cut, by simply letting the cutter go through twice. Note. — As already stated, bevel gear cutting in this way is only a compromise at the best, but by careful manipulation and good judgment an experienced man is able to do a very creditable job. A cutter is usually selected of the same curve as is correct for a spur gear corresponding to the back cone radius of the gear. Thus, it may be thought that the shape of the tooth should be the shape of the cutter, but by investi- gation it will be found that, on account of the " backing off," the teeth will be of a little more rounding shape at the large end than corresponds to the cutter ; therefore, when the gear has few teeth, — less than 25, — it is usually preferable to make the shape of the cutter to correspond to a gear a little larger than would be called for by the back cone radius of the bevel gear to be cut; but when the gear has more than 25 teeth, a cutter of shape corresponding to the back cone radius of the gear will give good results. For instance, in the pair of bevel gears shown in Fig. 12 the back cone radius of the large gear calls for a cutter corresponding in shape to a cutter for a spur gear of 59 teeth, 16 diametral pitch ; and this shape of cutter will, after the teeth are backed off, make the teeth a trifle too round at the large end, and a trifle too straight on the small end, but if the teeth are not too long the job will be very satisfactory. The back cone radius of the small gear calls for a cutter corresponding in shape to a cutter for a spur gear of 21 teeth, 16 diametral pitch, but when the teeth are backed off they will be a little too rounding on the large end ; therefore a better result is obtained by selecting a cutter having a shape corre- sponding to a little larger spur gear ; for instance, a gear of 24 teeth. Such a cutter will give the teeth a better shape on BEVEL GEARS. 403 the large end, although it may be necessary to round the teeth a little, outside the pitch line on the small end, by filing. Of course, a spur gear cutter cannot be used for cut- ting bevel gears, because, although it may have the correct curve, it would be too thick. The thickness of a bevel gear cutter must be at least 0.005 inch thinner than the space be- tween the teeth at their small end. Large bevel gears are made on theoretically correct prin- ciples by planing on specially constructed machines. WORMS AND WORM GEARS. 13 shows a worm and worm gear. Fig. 13' /"= Pitch diameter of gear. g-= Smallest outside diameter. h = Largest outside diameter. a = Outside diameter of worm. b = Pitch diameter of worm. c = Diameter of worm at bottom of thread. The ratio between the linear pitch and the diameter of the worm is arbitrary. It may be four times the circular pitch of the worm gear for single thread; five times the circular pitch of the worm gear for double thread; six times the circular pitch of the worm gear for triple thread. 404 WORMS AND WORM GEARS Increasing the diameter of the worm decreases the angle of the teeth in the worm gear. Decreasing the diameter of the worm increases the angle of the teeth in the worm gear. This angle is most conveniently obtained by drawing a dia- gram as shown in Fig. 13. Draw a line / m, equal to 3^ times the length of line b; this line will be equal to the length of the circumference of the pitch diameter of the screw. Erect the perpendicular, m o, equal to the lead of thescrew. Connect the points / and by theline / o y and the angle s is the angle of the teeth on the worm gear. Caution. — When cutting a worm gear, be careful not to lay the angle of the teeth in the wrong direction. The diameter of worm gears is usually calculated according to circular pitch, for convenience in cutting the worm with the same gears as used for ordinary screw cutting in a lathe. When a worm gear has comparatively few teeth, the flank of the tooth will be undercut by the hob; to prevent this in a measure, it is customary tojiave the blank somewhat over size, so that from five-eighths to three-fourths of the depth of the tooth may be outside the pitch line. The form of teeth is usually involute, and the thread on a worm screw is constructed of the same shape as the teeth in a rack. Fig. 14 shows the shape of tooth and the table gives the dimensions of finishing tool for the most common pitches. The surface speed of a worm screw ought not to exceed 300 feet per minute. Table No. 5 2 is calculated by the following formulas. (See Fig. 14.) fig. 14 P- - Circular pitch. N- 1 '' P 3.1416 M — P / = 4=- a — : P X 0.3183 d = : P X 0.3683 D = -a + d S = P X 0.5 3 = P X 0.31 C = P X 0.335 h = ^ + 4 k = P X 0.1 WORMS AND WORM GEARS. 405 TABLE No. 52.— Giving Proportions of Parts for Worms and Worm Gears, Calculated According to Circular Pitch. (See Fig. 14.) CQ •UIJOAV. JO jajamtnp jsao qoq jo <£ J3}3UIEip UI 1U3UI3JDUJ >C O O O CO MOOO'MOt-M © t— O ©©©©©© 00000000000000000 O X •q°H u ! L 3DEds jo qiuap 3[oq^i" oq C * C iffl O M ■* 10 h t- M » iW CO -*©OCNO00"*O©^C0r-C00 1; O (N (N !N rt ri H rt O O ^rH-rH OOOOOOO O'OO* OOOO •doj, uo spBSjqx jo ssaujpiqx O t-OOOrtMinOOMrtOtCi-CDW'*?: oioo'0©©^(Mt-I 00 10 ^j^oiocowNCNHHHHHoqq th © © © © d © d © © d © d © © © © Pi O •aurj qoiij uo P E3J HX J° ss'aujpiqx "* \o *— co »o )C © iC © lO O-l © © © CO lO tH CO O} l?-LO!MO[-HOOO't(NH GCOuO f oqi^q>oMM!N(NnrtrtHrtoqq Hoddddddddddddddd •1(311(1 F-I13UI -Bi(j Bui'puodsajjo-) * GO *< T3 (U V i) *o £ inches. The addendum for the thread on the worm can be obtained from Table No. 52, column a, and is 0.2387. The outside diameter of the worm will be 4.5 + 2 X 0.2387 = 4.977 inches, or, practically, 4|^ inches. The cutter to be used in roughing out the gear should have a curve of involute form corresponding to a spur gear cutter for 68 teeth, and its thickness ought to be at least 0.005 inch less than the width of space as given in column S of Table No. 52. Therefore the thickness on the pitch line of the roughing cutter will be 0.37 inch. The angle of the teeth may be obtained from a drawing as shown and explained in Fig. 13, or it may be calculated thus : rj, r 1 c lead of worm Tangent of angle S = — — - — : - ■ pitch circumference Tang. S = 3 X ° 75 = 2 ' 25 = 0.15915 4.5 x 3-1416 14-1372 In Table No. 21 the corresponding angle is given as 9 degrees and 10 minutes, very nearly. The depth to which the gear should be cut is given in col- umn D as 0.515 inch. The gear is finished with a hob, as described below, which is allowed to cut until it touches the bottom of the spaces in the gear. The outside diameter of the hob should be larger than the outside diameter of the worm, in order that the teeth in the hob may reach the bottom of the spaces in the gear and leave clearance for the worm, and at the same time leave the gear tooth of the proper thickness on the pitch line. This increment is obtained in column k, Table No. 5 2 , and for 34 -inch pitch is 0.075 inch ; thus, the outside diameter of the hob is 0.075 inch larger than the outside diameter of the worm, or 4.977 + 0.075 = 5.052 inches. The angle of the finish- ing threading tool for both worm and hob is 14^ degrees, making the angle of space 29°, as shown in Fig. 14. The clear- ance angle of the threading tool must be a little more than the angle of the thread. 408 WORMS AND WORM GEARS. The width of the threading tool at the point is given in Column b, Table No. 52 , as 0.2325 inch. The depth of the space to be cut in the worm is given in Column Z>, as 0.515 inch. The diameter of the worm at the bottom of the thread will be : 4.977 — 2 X 0.515 = 3.947 inches. The depth of the space to be cut in the hob is given in col- umn h in Table No. 52 as 0.5525 inch. The diameter of the hob at bottom of thread will be : 5.052 — 2 X 0.5525 = 3.947 inches. Thus the only difference in size between the hob and the worm is in the outside diameter and in the depth of the cut. Both may be finished by the same tool, as the diameter at the bottom of the thread and the thickness of the teeth at the pitch line should be the same for both hob and worm. Elliptical Gear Wheels. Elliptical gear wheels are sometimes used in order to change a uniform rotary motion of one shaft to an alternately fast and slow motion of the other. See Fix. 15. The pitch line is constructed and calculated the same as the circumference of an ellipse. (See page 189.) The gear is constructed involute the same as for spur gears. If the differ- ence between the minor and the major diameters is large it may be necessary to construct the teeth of different shapes at differ- ent places on the circumference; in other words, the whole cir- cumference of the gear cannot be cut with the same cutter. A cutter of the same pitch, of course, but corresponding to a larger diameter of gear, must be used where the curve of the pitch line is less sharp. The centers of the shafts are in the foci of the ellipse. If two elliptical gear wheels, made from the same pattern, or cut together at the same time, on the same arbor, are to work together they must have an uneven number of teeth so that a space will be diametrically opposite a tooth, as will be seen from Fig. 15. STRENGTH OF GEAR TEETH ^ g STRENGTH OF GEAR TEETH. The strength of the teeth and the horse-power that may be transmitted by a gear depend upon so many variable and uncer- tain factors that it is more a matter of practical experience and judgment of the designer than a problem of theoretical calcula- tion. Consequently, there are a great number of different for- mulas and rules given by different authorities. In the writer's opinion, there are at the present time no form- ulas or rules so well adapted to practical conditions as the formulas constructed by Mr. Wilfred Lewis of Philadelphia.* See American Machinist of May 4, 1893, page 3, and June 22, 1893, page 6. Mr. Lewis in constructing his formulas assumes that the gears are well made and mounted, so that the load is distributed across the tooth and not concentrated on one corner only. Mr. Lewis also assumes that the whole load is taken by one tooth, and from a series of drawings of teeth of the involute, cycloid and radial flank system, he determines the relative strength of gear teeth of various forms and thereby obtains the value of the factor y as given in the following table No. 54. In his discussions upon the subject, Mr. Wilfred Lewis obtains the general formula : W = s X pX/Xy. W = Load in pounds transmitted by the gear. s = Safe working stress of the material, taken as 8000 for cast iron, when the working speed does not exceed 100 feet per minute. (See table No. 55) p = Circular pitch of the gear. / = Face of gear in inches. y = A factor depending on the form of the teeth. (See table No. 54.) * Mr. Wilfred Lewis kindly allowed the author to make use of his for mulas in this book. 4io STRENGTH OF GEAR TEETH TABLE No. 54 . The value of y. See formula on page 409 Number Involute 20 deg. Involute 15 deg. Radial ■ of Teeth Obliquity and Cycloidal Flanks 12 0.078 0.067 0.052 13 0.083 0.070 0.053 14 0.088 0.072 0.054 15 0.092 0.075 0.055 16 0.094 0.077 0.056 17 0.096 0.080 0.057 18 0.098 0.083 0.058 19 0.1 00 0.087 0.059 20 0.102 0.090 0.060 21 0.104 0.092 0.061 23 0.106 0.094 0.062 25 0.108 0.097 0.063 27 O.III 0.1 00 0.064 30 0.114 0.102 0.065 34 0.118 0.104 0.066 38 0.122 0.107 0.067 43 0.126 O.IIO 0.068 50 0.130 O.II2 0.069 6o 0.134 O.II4 0.070 75 0.138 0.II6 0.071 100 0.142 0.II8 0.072 150 0.146 0.120 0.073 300 0.150 O.I22 0.074 Rack 0.154 O.I24 0.075 TABLE No. 55. Giving the working stress s for differ= ent speeds (cast iron) . See formula on page 409 . Speed in feet per minute 100 feet or less 200 300 600 900 1200 1800 24OO Value of 5 8000 6000 4800 4OOO 3000 24OO 2 000 1700 For steel take the value of 5 two and one-half times that of cast iron. Example. Calculate the horse power which with safety may be transmitted by a cast iron gear having 60 teeth, of 2 inches circular pitch, and 5 inches face, at the speed of 600 feet per minute. The form of trig teeth isthe common 14^ degrees involute. STRENGTH OF GEAR TEETH 4 II Solution: Using the formula, W= sXpX/X y. and inserting the values corresponding to the problem, we have, W = 4000 X 2 X 5 X 0,114 = 456o pounds. A force of 4560 pounds acting at a speed of 600 feet will give 4560 X 600 33000 = 83 horse P° wer - For bevel gears Mr. Lewis gives the following formulas, referring to Fig. 16: 7)3 -73 W= sXpX/X yX ■ 3D 2 X (D—d) W = Safe load in pounds. 5 = Allowable stress in the material which is de- pending upon the speed See table No. 55. D = large diameter of gear. d = small diameter of gear. / = Face of bevel gear in inches. P = Circular pitch at the large diameter. n= Actual number of teeth. iV=Formative number of teeth which can be calculated by multiplying the secant of angle a (See Fig. 16) by the actual number of teeth thus: N = n X secant a. The formative number of teeth is also the same number of teeth as would correspond to the radius R. R = Back cone radius. (See explanation on page 399.) y = A factor, depending upon the formative number N and also upon the shape of the teeth, obtained from Table No. To illustrate the use of the formula, Mr. Lewis gives the Find the working strength of a pair of cast iron miter gears of 50 teeth, 2 inch pitch, 5 inch face at 120 revolutions per minute. Radial flank system. In this case D = 31.8 inches, d = 24.8 inches. The angle a is 45 degrees, therefore secant a is 1.4; the formative number N will be 1.4 X 50 = 70. The corresponding value of y is found in table No. 54 to be 0.071. 412 STRENGTH OF GEAR TEETH The speed of the teeth is iooo feet per minute, for which by interpolation in table No. 55, we find the value of s to be 2800 and by substituting these values we have: w = 2800 x 2 x 5 x 0.071 x 3X3 3 I ; 8 8 a 3 x7 3 2 I 4 8 8 l 34 . 8) . w = 1988 x 0.795. W = 1580 pounds. Above formula involves considerable labor in calculations and Mr. Lewis gives a simpler approximate formula: d W= sX pX/X yX p- This formula gives almost identically the same results as the more complicated formula given above, when d is not less than I D, as is the case in good practice Example. Find the safe working horse-power of a pair of cast iron bevel gears of 2 inches circular pitch and 5 inches face, if one gear has 60 teeth and the other 30 teeth. The speed is 900 feet per minute, the form of the teeth is the common 14-i degree involute. In this case d is more than two-thirds and, therefore, we may use the simple approximate formula: Solution: As the gear having the least number of teeth is (when both are of the same material) the weakest of the two, we make the calculation for the gear having 30 teeth. Examining the drawing and calculating we find 3 X 2 . D - 3^6 = ^ l lnches * d = 14! inches. Measuring the back cone radius on the scale drawing and calculating, we find the formative number of teeth to be between 33 and 34. The value of y, corresponding to 34 teeth, is in table No. 54 given as 0.104. Inserting those values in the formula we have: x 4-5 W = 3000 X 2 X 5 X 0.104 X y^~f = 2 369 pounds. At a speed of 900 feet per minute, this will give: 23 69 X 900 , , = 64 horse power. 33000 * F STRENGTH OF GEAR TEETH 413 The following tables, Nos. 56 and 57, are calculated by means of Mr. Lewis' formulas and will in a great measure facilitate the calculation of the strength of gear teeth, both in circular and in diametral pitch. TABLE No. 56. Strength of Cast Iron Gears. Involute form 14£ degrees or cycloidal figur- ed for 1 inch circular pitch and 1 inch face of tooth, according to Mr. Wilfred Lewis' formula. — Calculated by the author. Speed IOC or per feet less min. 200 300 600 900 1200 1800 2400 H c CO c 3 Ph co O X CO •O 3 3 U 1 V CO O X CO -a c 3 u V * CO »~ X 12 536 1.62 402 2-43 322 2.92 268 4.87 20I 5.48 161 5.85 134 7.31 114 8.28 13 560 1.69 420 2-54 336 3.05 280 5-09 2IO 5-72 168 6.10 I4O 7-63 119 8.65 14 |576 1.74 432 2.62 346 3-14 288 5-23 216 5.89 173 6.28 144 7.85 122 8.90 15 600 1.81 450 2.72 36o 3-27 300 5-45 225 6.13 1 So 6.54 I50 8.18 128 9.27 16 616 1.86 462 2.80 370 3-35 308 5-6o 231 6.30 185 6.72 154 8.40 131 9-52 17 640 1-93 480 2.90 384 3-49 320 5.82 24O 6.54 192 6.98 l6o 8.72 136 9.89 18 664 2.01 498 3-OI 398 3.61 332 6.03 249 6.79 199 7-23 166 9.05JI4I 9.48 148 10.26 19 696 2.10 522 3.16 418 3-8o 348 6.32 26l 7.12! 209 7-59 174 10.76 20 7 20 2.18 54° 3-27 432 3-92 360 6-54 270 7.36 216 7-85 l8o 9.81 153 11. 13 21 736 2.23 552 3-34 442 4.01 368 6.69 276 7-52 221 8.03 I84 10.03 156 n-37 23 752 2.27 564 3-4i 451 4.10 376 6.83 282 7.71 226 8.20 188 10.25 160 11.62 25 776 2.35 582 3-52 466 4-23 388 7.05 291 7-93 ^33 8.47 194 10.58 165 11.90 27 800 2.42|600 3-63 480 4-36 4OO 7.27 300 8.18 240 8.72 200 10.90 170 12.36 30 816 2.47,612 3- 70 490 4-45 408 7.41 306 8-34 245 8.90 204 II. 13 173 12.61 34 832 2.52 624 3-78 499 4-53 416 7.56 312 8.51 250 9.08 208 H-34 177 12.86 38 856 2.59 642 3-89 514 4.67 428 7.78 321 8.75 257 9-34 214 11.67 182 13-23 43 88o 2.66 660 4.00 528 4.80 44O 8.00 330 9.00 264 9.60 220 12.00 187 13.60 SO 896 2.71 672 4.07 538 4.88 448 8.14 336 9.16 269 9.77 224 12.21 190 13-85 60 912 2:76 684 4.14 547 4-97 456 8.29 342 9-33 274 9-95 228 12.43 194 14.04 75 928 2.81 696 4.21 557 5-o6 464 8-43 348 9.49 278 10.12 232 12.63 197 14-34 100 944 2.86 708 4.29 566 5.14 472 8.58 354 9-65 2S3 10.29 236 12.82 201 14.58 150 960 2.90 720 4-36j 576 5-23 480 8.72' 360 9.81 28S 10.47 24O 13.09 204 14.83 300 976 2-95 732 4-44 585 5.32 488 8.87 366 9.98 293 10.65 244 I3-30 207 15-08 Rack 992 3.00 744 4- 50! 595 5-4i| 496 9.02] 372 10.14 298 10.82 248 I3-5I 211 15-33 For cut Steel Gears multiply these values by 2J 4 i4 STRENGTH OF GEAR TEETH TABLE No. 57. Strength of Cast Iron Gears. Involute form 14£ degrees or cycloidal figured for "one diametral pitch" and 1 inch face of tooth, according to Mr. Wilfred Lewis' formula. — Calculated by the author. Speed 100 feet or less per min. 200 300 600 900 1200 1 3oo 2400 o o h o 6 03 -a 3 3 5 & 3 X 12 Power transmitted = 3°-7 X 2 X 8^ _ horse-power. 3X12 416 STRENGTH OF GEAR TEETH If the small pitch diameter of a bevel gear is less than two- thirds of the large pitch diameter the strength should not be cal- culated by these tables. In such cases use formula on page 411. Note : The large pitch diameter of a bevel gear is obtained simply by calculating the same as for a spur gear, but the small pitch diameter may be obtained either by a scale drawing ( see Fig. 11, page 398) or by trigonometrical calculations.* Important : When calculating the strength of gear teeth for machinery where the motion is intermittent or in alternate direction or where the load is variable, and where the gears con- sequently are exposed to variable strain and shocks, as for instance in punching machines, geared pumps, air compressors, rolling mill machinery, etc. , the strength of the gear teeth must always be calculated according to the maximum and never according to the average load or horse-power. For additional safety it may in such cases be advisable not to make use of table No. 56 or 57 but to use formulas (see pages 409-412) and take the value of s smaller than what is given in table No. 55. Be very careful in all such calculations, never jump at conclusions, but always check results by what experience has taught to be service- able in common practice. Inspecting these tables we see that the horse-power a gear can transmit with safety does not increase in the same ratio as the speed ; for instance, doubling the speed of a gear from i2oo feet to 2400 feet a minute will increase its horse-power less than 50 per cent. We also become aware of the fact that if we do not change the diameter of a gear, its strength will not increase in propor- tion to the size of the teeth. For instance: Assuming a gear 5 inches pitch diameter, cut 6 pitch, it will have 30 teeth, and its strength given in the table, at 100 feet linear speed, is 2564 divided by 6, which gives 427 pounds. If this gear had been larger in diameter so that it could have been cut with 30 teeth, 4 pitch, instead of 30 teeth, 6 pitch, its strength would have been increased from: -~- =427 pounds, to — — = 641 pounds, or in the ratio of 4 to 6, which is a gain of 50 per cent; but if we keep the same diameter and increase the pitch from 6 to 4, the gear will have 20 teeth instead of 30 teeth and its strength at the same speed is 2262 divided by 4 which is 566 pounds. Although the pitch is increased 50 per cent the strength is increased only about 32 per cent. * A very good book on the subject is "Formulas in Gearing," by Brown & Sharpe Mfg. Co.. Providence, R.I. Also see "Machinery and ita supplement Engineering Edition, New York, Feb 1910. SCREWS. 417 SCREWS. "Pitch," "Inch Pitch" and "Lead" of Screws and Worms. The term " pitch of a screw," as commonly used, means its number of threads per inch, while the " inch pitch " is the dis- tance from the center of one thread to that of the next. For instance, a one-inch screw of standard thread is usually said to be an " eight pitch" screw, because it has eight threads per inch of length ; but it might more correctly be said to be a screw of >s-inch pitch, because it is >£-inch from the center of one thread to the center of the next The " lead " of a worm or a screw means the advancement of the thread in one complete revolution ; therefore, in a single- threaded screw, the inch pitch and the lead is the same thing, but in a double or triple-threaded screw the inch pitch and the lead are two different things. The " lead " in a double-threaded screw will be a distance equal to twice the distance from the center of one thread to the center of the next, but in a triple-threaded screw the lead is three times the distance from the center of one thread to the center of the next. Screw Cutting by the Engine Lathe. When the stud and the spindle run at the same speed ( which they usually do) the ratio between the gears may always be obtained by simply ascertaining the ratio between the num- ber of threads per inch of the lead-screw and the screw to be cut. Example. The lead-screw on a lathe has four threads per inch and the screw to be cut has 11)4 threads per inch ( one-inch pipe- thread). Find the gears to be used when the smallest change gear has 24 teeth and the gears advance by four teeth up to 96. The ratio of the number of threads per inch of the two screws is as 4 to 11^. As the smallest gear has 24 teeth and the gears all advance by four teeth, this ratio of the screws must be mul- tiplied by a number which is a multiple of 4 and which, at least, gives the smallest gear 24 teeth. For instance, multiply by 8 and the result is 8 X 11 J^ = 92 teeth for the gear on the lead- serew ; 8 X 4 = 32 teeth for the gear on the stud. Cutting riultiple=Threaded Screws or Nuts by the Engine Lathe. Calculate the change gears as if it was a single-threaded screw of the same lead. Cut one thread and move the tool the proper distance and cut the next thread. 41 8 SCREWS. The most practical way to move the tool from one thread to another, when cutting double-threaded screws or nuts, is to select a gear for the stud or spindle of the lathe having a number of teeth which is divisible by two, and when one thread is cut make a chalk mark across a tooth in this gear onto the rim of the intermediate gear ; count half way around the gear on the stud and make a chalk mark across that tooth; drop the swing plate enough to separate the gears, pull the belt by hand until the oppo- site mark on the gear on the stud comes in position to match the chalk mark on the intermediate gear; clamp the swing plate again and the tool is in proper position to cut the second thread. When triple threads are to be cut, select a gear for the spindle or stud whose number of teeth is divisible by three, and in changing the tool from one thread to the next, only turn the lathe enough so that the gear on the stud moves one-third of one revolution. If, for any reason, it should be inconvenient to make this change by the gear on the stud, the change may be made by the lead-screw gear. The intermediate gear is first released from the gear on the lead-screw, which is then moved ahead the proper number of teeth, and again connected with the intermediate gear. The proper number of teeth to move the gear on the lead-screw is obtained by the following rule : Multiply the number of teeth in the gear on the lead-screw by the number of threads per inch of the lead-screw ; divide this product by the number of threads per inch of the screw to be cut, and the quotient is the number of teeth that the gear on the lead-screw must be moved ahead. Example. A square-threaded screw is to have ^-inch lead and triple thread. The lead-screw in the lathe has two threads per inch, and the gear on the lead screw has ninety-six teeth. How many teeth must the gear on the lead-screw be moved, when changing from one thread to the next ? Solution : A screw of >^-inch lead with triple thread has six threads 2 X 96 per inch, therefore the gear must be moved — ^ — - = 32 teeth, in order to change the tool from one thread to the next. _ 60 ^ U. S. Standard Screws. /JUk /JH^ ^£* ■*■ shows the shape of _^^^^k_^^^[ thread on United States stand- ^^^^^^^^^P ard screws. The sides are ^^^^^^^^^ straight and form an angle of ^^w^^^ ^ sixty degrees, and the thread is so— to SCREWS. 419 flat at the top and bottom for a distance equal to one-eighth of the pitch, thus the depth of the thread is only three-fourths of a full, sharp thread. (See Fig. 1.) Fig. 2 shows the shape of the Whitworth (the English) sys- tem of thread. As compared © ^ with the American system, ^""".^^^,-552 the principal difference is in the angle between the sides of the thread, which is fifty- five degrees, and one-sixth of the depth of the full, sharp thread is made rounding at FlG " 2 the top and bottom. There is also a difference in the pitch of a few sizes. The common V-thread screws have the angle of thread of sixty degrees, the same as the United States standard screws, but the thread is sharp at both top and bottom. This style of thread is rapidly, as it should be, going out of use. The prin- cipal disadvantages of this thread are that the screw has less tensile strength, and it is also very difficult to keep a sharp- pointed threading tool in order. Diameter of Screw at Bottom of Thread. The diameter of screws at the bottom of thread is obtained by the following formulas : — United States Standard Screws : n For V-threaded screws : n For Whitworth screws : d— D 1.281 n d= Diameter of screw at bottom of thread. D = Outside diameter of screw. n = Number of threads per inch. 1.299 is constant for United States standard thread. 1.733 is constant for sharp V-thread. 1.281 is constant for Whitworth thread. 420 SCREWS. TABLE No. 58.— Dimensions of Whitworth Screws. Diameter of Number of Diameter of Number of Diameter of Number of Screw in Threads per J Screw in Threads per Screw in Threads per Inches. Inch. Inches. : Inch. Inches. Inch. X 40 IX 1 7 sy 2 3X T 3 6 24 1H 6 3X 3 X 20 1J* 6 4 3 A 18 1H 5 4X 27/ s H 16 IX 5 4K 27/ 8 A 14 i# 4^ 4X w % 12 2 4^ 5 2U % 11 2% 4 5X 25/8 % 10 2# • I 4 5^ 2/8 n 9 2X 3# 5X 2/ 2 i 8 3 s/ 2 6 2/ z i% 7 3X 3X Diameter of Tap Drill. The diameter of the drill with which to drill for a tap is, if we want full thread in the nut, equal to the diameter of the screw at the bottom of the thread, and is, therefore, obtained by the same formulas. However, in practical work it is always ad- visable to use a drill a little larger than the diameter of the screw at the bottom of thread, because in threading wrought iron or steel the thread will swell out more or less, and a few thou- sandths must be allowed in the size when drilling the hole. In drilling holes for tapping cast-iron, a little larger drill is used, because it is unnecessary in a cast-iron nut to have exactly full thread. Table No. 59 gives sizes of drill for both wrought and cast-iron, which give good practical results for United States standard screws. Table No. 59 gives sizes of hexagon bolts and nuts. The size of the hexagon is equal to \/ 2 times the diameter of bolt + >£-inch ; the thickness of head is equal to half the hexagon. The thickness of nut is equal to the diameter of the bolt. When heads and nuts are finished they are T Vinch smaller. The table is calculated by the following formulas : d = D 1.299 C = 1.155 A E=±A f= — X B=1AUA F—D SCREWS. 421 TABLE No. 59.— Dimensions of U. S. Standard Screws. "o _s Finished S T3 4 Rough Bolt-Heads and Bolt- 3 Nuts. Heads PQ =• J5 and Nuts O 3 3' jj « P a "0 1 PQ O 3 rt in V PQ 3 « te 3 CS a "0 "o Pi S -j? -a _2 * 3 £ J5 >< . PQ PQ h O u j= Zb v fa 3 2 3 OJ C u 55 s «-3 Q 3 _rt ! 1* ^3 H 1* (7 T6 3 1 6 D ^ 2X 4 T \ 01 1 ^16 3 2.629 Pi 5.4284 3^ 0.03571 4^ C17 O32 5H ^I 5 6 3 4i 9 6 015 ^16 3X 2.879 6.5099 3X 0.0357! 5 «6l" K2 5 °32 2^ 3X 4H ^T 3 6 3j^3.100!3>| 7.5477 3X 0.03841 5^ ^39 'II 6H 2}i 3X 2p 3 T 7 6 3K 3.317 321 8.6414 3 0.0417! 5X Ol 5 °64 6H 2^ 3X ^T6 3H 4 3.5673|| 9.9930 3 0.0417 6^ 8fi T* ^6 4 «A 3M 4X3.798 313 11.3292 2% 0.0435 ox 9r 3 6 ^ 3X 4X 6 T 7 6 ±T 3 6 \% 4.028J4, 3 , 12.7366 2X 0.04551 $y s 911 8 ^tV 4X 611 1t 7 6 4X 4.255 4A 14.2197 2^ 0.0476 TX iox '8^ 3^ 4X 7 T 3 6 *H 5 4.480 4i^ 15.7633 2^ 0.0500 : 7^ loji 8}f »H 5 7 T 9 6 4.1 5 5X 4.730;4X 17.5717 2X 0.0500! 8 HA 9X 4 5X m X 3 ^16 5^|4.953!4ff 19.2676 23/^ 0.0526! $n Hfl »H *A 5^ 8fV 5 T 7 6 5X|5.203 5i| 21.2617 23/ 8 0.0526! 8X 12^ 10eV 4Xs 5X 8^ Xll J T6 6 i5.423 5f f 23.0978 2X 1 0.0556! 9>£ 12ff ioh A 9 6 9 T V 511 422 SCREWS. Note. — In finished work, the thickness of the head of the bolt and the nut is equal, and is Yi6 of an inch less than the diameter of the bolt. Columns B and C in Table No. 59 are very useful for many- purposes ; for instance, in selecting size of counter-bore when finishing castings, to give bearing for screw heads : in turning blanks which are afterwards to be cut into square or hexagon heads, etc. Table No. 60.— Coupling Bolts and Nuts. (Hexagon). (All Dimensions in Inches) u t r Z " K Y ( n s , i 1 * ^ n - 4 F / \ V /U- * tr V" No. A Flat Head Round Head B C E F B | C E F 2 .0842 .1631 .0454 .030 .0151 •1544 .0672 .030 .0403 3 •0973 .1894 •0530 .032 .0177 .1786 .0746 .032 .0448 4 .1105 .2158 .0605 •034 .0202 .2028 .0820 •034 .0492 5 .1236 .2421 .0681 .036 .0227 .2270 .0894 .036 •0536 6 .1368 .2684 •0757 .039 .0252 .2512 .0968 •039 .0580 7 .1500 .2947 .0832 .041 .0277 •2754 .1042 .041 .0625 8 .1631 .3210 .0908 •043 .0303 .2996 .1116 .043 .0670 9 .1763 •3474 .0984 •045 .0328 •3238 .1190 •045 .0714 10 .1894 •3737 .1059 .048 •0353 .3480 .1264 .048 .0758 12 .2158 .4263 .1210 .052 .0403 .3922 .1412 .052 .0847 H .2421 .4790 .1362 .057 •0454 ■4364 .1560 .057 .0936 16 .2684 •53i6 •1513 .061 .0504 .4806 .1708 .061 .1024 18 .2947 .5842 .1665 .066 •0555 •5248 .1856 .066 .1114 20 .3210 .6368 .1816 .070 .0605 .5690 .2004 .070 .1202 22 •3474 .6895 .1967 •075 .0656 .6106 .2152 .075 .1291 24 •3737 .7421 .2118 .079 .0706 .6522 .2300 .079I.1380 26 .4000 .7421 .1967 .084 .0656 .6938 .2448 .084 1 .1469 28 .4263 7948 .2118 .088 .0706 •7354 .2596 .088I. 1558 30 .4526 •8474 .2270 •093 •0757 .7770 .2744 .093 |. 1 646 The dimensions imum, the necessary thread is V thread given in Tables No working variations .63 and No. 64 are being below them. max- The SCREWS. 425 TABLE No. 64. Dimensions of Machine Screws. [American Screw Company] n J / . D y \ < / 1 * t + I f-> 6r- ->f-€ H*= — Fillister Head. No. 2 A 0.0842 B C D E F G •I350 .0549 ; .0126 .030 .0338 .0675 3 0.0973 .1561 .0634 ! .0146 032 .0390 .0780 4 0.1105 .1772 .0720 .0166 034 •0443 .0886 5 0.1236 .1984 .0806 .0186 036 .0496 .0992 6 0.1368 • 2195 .0892 .0205 039 •0549 .1097 7 0.1500 .2406 j .0978 .0225 041 .0602 .1203 8 0.1631 .2617 .1063 .0245 043 • 0654 .1308 9 0.1763 .2828 .1149 .0265 045 • 0707 .1414 10 0.1894 .3040 .1235 .0285 048 .0760 .1520 12 0.2158 .3462 ! .1407 .0324 052 .0866 .1731 H 0.2421 .3884 .1578 • 0364 o57 .0971 .1942 16 0.2684 •4307 .1750 • 0403 061 .1077 • 2153 18 0.2947 .4729 .1921 • 0443 066 .1182 .2364 20 0.3210 .5152 ; .2093 .0483 070 .1288 • 2576 22 0-3474 • 5574 .2267 . 0520 075 •1384 .2787 24 0.3737 • 5996 • 2436 .0562 079 .1499 -2998 26 0.4000 .6419 .2608 .0601 084 .1605 .3209 28 0.4263 .6841 • 2779 ! -0641 088 .1710 .3420 1 30 0.4526 • 7264 .2951 1 .0681 093 .1816 •3632 The screws given in tables No. 63 and No. 64 are made in following pitches : No. Threads No. Threads No. Threads No. Threads per inch per inch 1 Der inch per inch 2 64,56,48 7 32,30 14 24,20,1*: 24 18,16,14 3 56,48 8 36,32,30 16 20,18,16 26 16,14 4 40,36,32 9 32,30,24 18 20,18,16 28 16,14 5 40,36,32 10 32,30,24 20 18,16 30 16,14 6 36.32,30 12 24,20 22 18,16 426 SCREWS TABLE No. 65 -Sizes of Machine Screws and Drills. Diameter of Tap Drills (will give erood but not full Diameter Diameter of Screws ^5 G u thread) of Body Drills I^ight Stock Heavy Stock i-c V .Ha Size in u u u V .a B a 'cj N 5 Fractions ^3 a s Size a 3 Size a 3 Size fc so {H fe & & 2 o . 0842 &+ O.OO7 64 50 O.070O 49 O.073O 44 O . 0860 3 0.0973 3 3 2 + O.O03 56 46 0.0810 45 O.0820 40 O.O982 4 0.1 105 /5+O.OOI 40 44 O.0860 43 . 0890 34 O. IIIO 5 0.1236 f — 0.001 40 4 1 . 0960 38 O.IOI5 30 O.I285 5 36 4 1 . 0960 39 0.0995 6 0.1368 6 9 ¥ - 0.004 36 36 O.IO65 33 0.II30 28 O.I405 6 32 37 O.IO4O 35 0. I 100 7 0.1500 ¥ % — 0.006 32 32 O.II6O 31 0.1200 23 0.1540 7 30 32 2. I l60 31 0. 1200 8 0.1631 3 5 2 + 0.007 36 29 O.I36O 28 O.I405 19 O.I660 8 32 30 O.I285 29 O.I360 9 0.1763 11+0.004 32 28 O.I405 26 O.I470 16 0.1770 10 0.1894 T %-0.002 32 22 O.I570 20 O.I6IO 11 0.I9I0 10 30 23 0.1540 22 O.I570 12 0.2158 A- 0.003 24 18 0.1695 17 0.1730 2 0.2210 14 0.2421 i— 0.008 24 9 O.I96O 7 0.2010 i O.25OO 14 20 13 O.185O 10 0.1935 16 . 2684 H+ 0.003 20 6 . 2040 3 0.2I30 K 0.28I3 18 0.2947 if— 0.002 20 I 0.2280 15// 64 O.2344 5 // 0.3I25 20 0.3210 li+0.002 18 D O.246 i" O.25OO 21// 6I„ O.328I 22 0.3474 ii+o. 003 18 J O.272 9 // ¥2 0.28I3 2 3'/ 0.3594 24 0.3737 f— 0.001 16 L O.29O 5 // T6 0.3125 1" 0.3750 26 . 4000 If— 0.006 16 0.316 2 1// 64 O.328I 13// 32 O.4063 28 0.4263 11+0.004 14 Q 0.332 1 1// 3"2 0.3438 1 II T6 0-4375 30 0.4526 f| — 0.001 14 T 0.358 f" 0.3750 15// 32 O.4688 The size of drill as given for light stock will be found practical when the thickness of the stock is about equal to the diameter of the tap ; but when the thickness of the stock is more, it is better to use a little larger drill as given for heavy stock, because there will then be less strain on the taps, and also because when the hole is deep, the threads will, in most cases, be strong enough even if they are not quite so full. SCREWS 427 A. S. M. E. Standard of Machine Screws.* The Standard of Machine Screws approved by unanimous vote of "The American Society of Mechanical Engineers" at Indianapolis, May 30, 1907, has 21 numbered Sizes from Num- ber o to Number 30 inclusive, with an increment of 0.013 i ncn for each number. Number o being 0.060 inch and Number 30 0.450 inch outside diameter. From Number o to Number 10 all numbers and above Number 10 only the even numbers are used. The old screw gage had an increment of 0.013 16 inch; Num- ber o being 0.05784 inch and Number 30 being 0.45264 inch outside diameter. This new standard, therefore, eliminates many awkward and unnecessary decimal figures. TABLE 66— A. S. M. E. Standard Machine Screws.** u £> S 3 Z -1 OJ >-* HPLh Maximum Screw Diameters Minimum Screw Diameters WQ u 11 u OJ o-~ P4Q *c3 ^ SI OJ OJ u 8.2 2 3 4 5 6 7 8 9 10 12 14 16 18 20 22 24 26 28 30 80 72 64 56 48 44 40 36 36 3 2 3° 28 24 22 20 20 18 16 16 14 14 .060 .073 .086 .099 .112 .125 .138 .151 .164 .177 .190 .216 .242 .268 .294 .320 .346 .372 .398 .424 •450 .0519 .0640 .0759 .0874 .0985 .1102 .1218 .1330 .1460 .1567 . 1684 .1928 .2149 .2385 .2615 .2875 .3099 .3314 • 3574 • 3776 .4036 .0438 .0550 .0657 .0758 .0849 • 0955 .1055 .1149 .1279 .1364 .1467 .1696 .1879 .2090 .2290 .2550 .2738 .2908 .3168 .3312 •3572 .0572 .0700 .0828 • 0955 .1082 .1210 .1338 .1466 .1596 .1723 .1852 .2111 .2368 .2626 .2884 .3H4 .3402 .3660 .3920 .4178 •4438 .0505 0625 0743 0857 0966 1082 1197 1308 1438 1544 1660 1904 2123 2358 2587 2847 3070 3284 3544 3745 4005 .0410 .0520 .0624 .0721 .0807 .0910 .1007 .1097 .1227 .1307 .1407 .1633 .1808 .2014 .2208 .2468 .2649 .2810 .3070 .3204 •3464 * For very complete information regarding machine screws and useful suggestions for system of standard gages for manufacturing standard sizes, see Volume 29 of the transactions of The American Society of Mechanical Engineers. **The Author wishes to express his thanks to the Corbin Screw Corpor- ation, New Britain, Conn., for their very valuable suggestions, data, furnished him for this and the following abridged tables of the dimen- sions for A. S. M. K. standard machine screws. 426 SCREWS TABLE No. 67— A. S. M. E. Standard Taps for Machine Screws. co -3 Minimum Maximum H ° CO CO "3 a u CO u —^ CJ 12 a a co CO CO a "S CO CO "Sq s 3 8 " co a ■S e * s fc e f , s - a e w V x •- a & 8.2 *■■* a « 8.2 •- rt Z Hdn '■si aa KQ «Q WQ SQ «Q ^ i_ 8o .060 . 0609 .0528 .0447 .0632 .0538 .0466 • 0465 I 72 .073 .0740 .0650 .0560 .0765 .0660 .0580 •0595 2 64 .086 .0871 .0770 .0668 .0898 .0781 .0689 .070 3 56 .099 . 1002 .0886 .0770 .1033 .0897 .0793 •0785 4 48 .112 .U33 .0998 .0862 .1168 .1010 .0887 .O89 5 44 • 125 .1263 . 1116 .0968 .130 1 .1129 .0995 •0995 6 40 .138 .1394 .1232 .1069 • 1435 .1246 .1097 . no 7 36 •151 .1525 • 1345 .1164 .1569 .1359 .1193 . 120 8 36 .164 .1655 • 1475 .1294 .1699 .1489 .1323 .136 9 32 •177 .1786 .1583 .1380 .1835 .1598 .1411 • I 4°5 10 30 . 19O .1916 . 1700 .1483 .1968 .1716 .1515 .152 12 28 .216 .2176 • 1944 .1712 .2232 .1961 • 1745 • 173 H 24 .242 .2438 .2167 .1896 .2500 .2184 • I93 1 • 1935 16 22 .268 .2698 .2403 .2108 .2765 .2421 .2144 .213 18 20 •294 .2959 .2634 .2309 .303 1 .2652 • 234 6 • 234 20 20 .320 .3219 .2894 .2569 .3291 .2912 .2606 .261 22 18 •34 6 • 3479 .3118 .2757 •3559 .3^8 .2796 .281 24 16 ■372 .3740 • 3334 .2928 .3828 • 3354 .2968 .2968 26 16 .398 .4000 •3594 .3188 .4088 .3614 .3228 •323 28 14 .424 .4261 •3797 .3333 •4359 .3818 •3374 •339 30 14 •450 •45 21 •4057 •3593 .4619 .4078 •3634 .368 The pitches for both screws and taps are a function of the diameter, as expressed by the formula, 6-5 threads per inch = D + 0.02 and the results are given approximately and in even numbers in order to avoid the use of fractional or odd number threads. In this system the form of thread for machine screws has an included angle of 60 degrees, a truncation or flat at the top of the thread equal to one-eighth of the pitch and a trunca- tion at the bottom of the thread equal to one-sixteenth of the pitch. The form of thread for taps for machine screws has an in- cluded angle of 60 degrees, a truncation or flat at the top of the thread equal to one-sixteenth of the pitch and a truncation or flat at the bottom of the thread equal to one-eighth of the pitch. 429 This will, then, when the screw- is fitted in its nut, secure a clear- ance at the top and bottom and fit on the slant side of the thread, as shown by Figure I. lg The maximum tap shall have a flat at top of the thread equal to one-sixteenth of the pitch and the difference between maximum and minimum external diameter will allow at the top of thread of tap any width of flat between one-sixteenth and one- eighth of the pitch. The difference between the minimum tap and the maximum TABLE 68— A. S. M. E. Special Sizes of Machine Screws. M nimum Screw m-2 Maximum Screw Diameters Diameters - K — 1 fc e f , 6 *E SS ■S s Z E *& a a 8.s *-2 *j.2 8.2 2 HfE yQ so v n a 2 jT Q *Q 6 4 • 073 0629 •0527 ! .0698 06l3 .0494 2 56 .086 0744 .0628 .0825 .0727 .0591 3 48 .099 0855 .0719 .0952 .0836 .0677 4 40 .112 0958 • 0795 .1078 •0937 .0747 36 0940 • 0759 .1076 .0918 .0707 5 40 • 125 • 1088 .0925 .1208 .1067 .0877 36 1070 .0889 .1206 .IO48 .0837 6 36 .138 1200 .1019 • 1336 .1178 .0967 32 1177 .0974 • 1333 •1154 .0917 7 32 ■I5 1 I3°7 .1104 .1463 .1284 .IO47 30 1294 .1077 .1462 .1270 .1017 8 32 .164 1437 • 1234 • 1593 .1414 .1177 30 1424 .1207 .1592 .1400 .1147 9 30 •177 1553 ■1337 .1722 •1529 .1277 24 1499 .1229 .1718 .1473 .1158 10 3 2 . 190 1697 .1494 .1853 .1674 •1437 24 1629 • 1359 .1848 .1603 .1288 12 24 .216 1889 .1619 .2108 .1863 .1548 M 20 .242 2095 .1770 .2364 .2067 .1688 16 20 .268 2355 .2030 .2624 .2327 .1948 18 18 • 294 2579 .2218 .2882 •2550 .2129 20 18 .320 2839 .2478 • 3142 .2810 .2389 22 16 •346 3054 .2648 .3400 .3024 •2550 24 18 • 372 3359 • 2998 .3662 .3330 .2909 26 14 .398 35i6 •3052 .39i8 •3485 •2944 28 16 .424 3834 •3428 .4180 .3804 •3330 30 16 •45° 4094 .3688 .4440 .4064 •3590 43° SCREWS screw provides an allowance for error in pitch, or lead, and for the wear of tap in service. This form of tap thread is recommended as being stronger and more serviceable than the so-called V thread, but it has been suggested that strict adherence to this form might, in the case of small taps, add to their cost. Taps with V threads and with the correct angle and pitch diameters used in connection with tap drills of correct diameter, are permissible. TABLE No. 69— A. S. M. E. Special Sizes of Taps for flachine Screws. u (B-C Minimum Maximum »-■ s (LiS u "3 % in u V U "c3 v V *r1 <-> — > >4 £1 13 C a — Dimensions of Heads of flachine Screws. A. S. M. E. Standard. A = = diameter of body. A = diam. of body. B = = 1.64A — .009 = diam. of head B = 1.64A — .009 and radius for oval. = diam. of head. C = = 0.66A — .002 = height of side. C = .66A — .002 = height of head. D = = .173 A + .015 = width of slot. D = .173A + .015 E = = iF = depth of slot. = width of slot. F = = .134B + C = height of head. E = -JC = depth of slot. "1 ! D 1 B —4 ■1 i ^-^ ns: — r Usy Tft, - ! D ! * »i nHlh T" O / - E /-- L H, "J r H | - A* _> < — + &MZ9 \ FLAT FILLISTER OVAL FILLISTER HEAD. HEAD. Num- ber A B c D E F B c D E o .060 .0894 .0376 .025 .025 .0496 .0894 .0376 .025 .019 i ■ 073 .1107 .0461 .028 .030 .0609 .1107 .0461 .028 .023 2 .086 .1320 .0548 .030 .036 .0725 .1320 .0548 .030 .027 3 .099 • 1530 .0633 .032 .042 .0838 .1530 .0633 .032 .032 4 .112 .1747 .0719 .034 .048 • 0953 • 1747 .0719 .034 .036 5 .125 .i960 .0805 .037 .053 .1068 .i960 .0805 .037 .040 6 .138 .2170 .0890 .039 .059 .1180 .2170 .0890 .039 .044 7 .151 .2386 .0976 .041 .065 .1296 .2386 .0976 .041 .049 8 .164 • 2599 .1062 .043 .071 .1410 • 2599 .1062 .043 • 053 9 .177 .2813 .1148 .046 .076 .1524 .2813 .1148 .046 ■ 057 IO .190 .3026 .1234 .048 .082 .1639 .3026 .1234 .048 .062 12 .216 • 3452 .1405 .052 • 093 .1868 • 3452 • 1405 .052 .070 14 .242 .3879 .1577 • 057 .105 .2097 .3879 • 1577 • 057 .079 16 .268 •4305 .1748 .061 .116 .2325 •4305 .1748 .061 .087 18 .294 • 473i . 1920 .066 .128 •2554 •473i .1920 .066 .096 20 .320 .5158 .2092 .070 .140 •2783 •5158 .2092 .070 . 104 22 •346 •5584 .2263 .075 .150 .3011 •5584 .2263 • 075 •113 24 •372 .6010 • 2435 .079 . 162 .3240 .6010 • 2435 .079 .122 26 .398 •6437 .2606 .084 •173 •3469 •6437 .2606 .084 .130 28 .424 .6863 .2778 .088 .185 .3698 .6863 .2778 .088 •139 30 .450 .7290 .2950 •093 .196 •3927 .7290 .2950 •093 .147 432 SCREWS TABLE No. 71— Dimensions of Heads of Machine Screws. A. S. H. E. Standard. A = diam. of body. A = < diameter of body. B = 1.85A— .005 = B = 2A— .( x>8= diameter of head. diam. of head. C = .7A = height of C = A— .008 =thickness of head head. D = .173A + .015 = !-739 width of slot. D = 1 73 A + .015 = width of slot. E = iC + .01 = depth of slot. E =- £-C = depth of slot. 1 B j -82 Deg^ 1 i — Id ! i ~j^ !hd-r \ ll • V / i/CwJ 4-1 \ a LLl / ' f \ ; / 1 i , — / V- *■ ' ^ 3«S bo a '"a.Sa 11 PQ h a 2 fc 8 2 * *■> u -In fls-s 2 s DHo D P d T t 6 1 4.6 16 50 4-7 6.11 470 7 1 5-6 24 75 5-7 7.11 5-70 8 1-25 6.25 30 100 6.4 8.14 6.38 9 125 7-25 41 125 7-4 9.14 7.38 10 1-5 7-9 49 150 8.1 10.16 8.05 11 i-5 8.9 62 175 9-i 11. 16 9.05 12 1-75 9-55 7i 200 9.8 12.19 9-73 14 2 11. 2 98 300 11.4 14.22 11.40 16 2 13.2 137 400 134 16.22 13.40 18 2.5 14-5 165 500 14.8 18.27 1475 20 2.5 16.5 214 650 16.8 20.27 16.75 22 2.5 18.5 269 800 18.8 22.27 18.75 24 3 19.8 308 925 20.1 24.32 20.10 27 3 22.8 408 1200 23.1 27.32 23.10 30 3-5 251 495 1500 25-5 30.38 25-45 33 3-5 28.1 620 1900 28.5 33-38 28.45 36 4 30-4 726 2200 30.8 36.43 30.80 39 4 33-4 876 2600 33-8 3943 33-8o 42 4-5 35-7 IOOI 3000 36.3 42.49 36.15 45 4.5 38.7 1176 3500 39-3 45-49 39-15 48 5 4i 1320 3900 41-5 48.54 41.50 52 5 45 1590 47oo 45-5 52.54 45-50 56 5-5 48.3 1832 5500 48.9 56.60 48.86 60 5-5 52.3 2148 6400 52.9 60.60 52.86 64 6 55-6 2428 7300 56.2 64.65 56.20 68 6 59-6 2790 8400 60.2 68.65 60.20 72 6-5 62.9 3107 9300 63.6 72.70 63-56 76 6-5 66.9 3515 10500 67.6 76.70 67-56 80 7 70.2 3859 1 1600 7i 80.76 70.91 screws 435 Table No. 73 is calculated by the following formulas: d = Diameter of screw in millimeters. D = 1.4^ + 4 millimeters approximately. Di = 1.155Z?. Z> 2 = 1.414/}. H = d. Hi = o.jd. The thickness of the nut is equal to the diameter of the screw and the thickness of the head is 0.7 of the diameter of the screw. TABLE No. 73. Dimensions for Heads and Nuts Screws in the International Standard System. (All dimensions in millimeters) for 1* nfl o"S -o i cd v tJ °co ed S W t $ 1) 2 Hi if 6 12 13.86 16.97 4.2 6 33 50 57-75 70.70 23.1 33 7 13 15.02 18.38 4-9 7 36 54 62.37 76.36 25.2 36 8 15 17-33 21.21 5-6 8 39 58 66.99 82.01 27.3 39 9 16 18.48 22.63 6-3 9 42 63 72.76 89.08 29.4 42 10 18 20.79 25-45 7.0 10 45 67 77-39 94.74 3i.5 45 11 19 21.94 26.87 7-7 11 48 7i 82.01 100.39 33-6 48 12 21 24.26 29.69 8.4 12 52 77 88.94 108.88 364 52 14 23 26.57 32.52 9.8 14 56 82 94-71 115.95 39-2 •56 16 26 30-03 36.76 11. 2 16 60 88 101.64 124.43 42.0 60 18 29 33-50 41.01 12.6 18 64 94 108.57 132.92 44-8 64 20 32 36.95 45-25 14.0 20 68 100 II5-50 141.40 47-6 68 22 35 40.43 49-49 15-4 22 72 105 121.28 148.47 50.4 72 24 38 43-89 53-73 16.8 24 76 no 127.05 155-54 53-2 76 27 42 48.51 59-39 18.9 27 80 116 133.98 164.02 56.0 80 30 46 53-13 65.04 21.0 30 If screws are used of intermediate sizes from those given in Table No. 72 it is recommended that the heads and nuts should be made of standard sizes and that the pitch of the thread should be that of the nearest smaller size. Note : This improvement in construction of screw threads as recommended for machine screws and for International Stand- ard Threads, could in practical work,when fitting machinery, be taken advantage of, also when using screws having U. S. Stand- 436 SCREWS ard threads, simply by using a tap drill that is a little over size (which usually is done anyhow and using a sizing tap having its external diameter so much larger that the truncation or flat of the top of its threads would be only one-sixteenth of the pitch in- stead of one-eighth of the pitch as it is on U. S. Standard taps. This would prevent the screws from squeezing either at the top or at the bottom of the thread but it would take a bearing against the slant side of the thread. Such taps could then be made according to the following formula: 0.1083 D 1 = Outside diameter at top. D = Outside diameter at U. S. S. screw. n = Number of threads per inch. For instance: The outside diameter of a tap for a 1 inch screw 0.1083 . 8 threads per inch would be 1 + — g — = I-OI35 ln ch. The flat on the top of the tap would then, of course, be only 0.0078 instead of 0.0156 inches. To Gear a Lathe to Cut Metric Thread when the Lead=Screw is in Inches. Use two intermediate gears, one having 100 teeth and the other 127 teeth, fasten these two gears together on the same hub, and gear the 100-tooth gear into the gear on the lead-screw and the 127-tooth gear into the gear on the stud (see Fig 3). The lathe will then cut, practi- cally, one-half the number of threads per centimeter that it originally cut per inch with a common intermediate gear. For instance, the stud gear has 24 teeth, the screw gear has 48 teeth and the lead-screw has four threads per inch ; the lathe will then, with a common intermediate gear, cut eight fig. 3 . "^^^ XS^V threads per inch, but by using such a double intermediate gear as is shown in Fig. 3 the lathe will cut four threads per centimeter, which is the same as one-fourth times 10 and equals 2^ millimeters pitch, which corresponds to a metric standard screw of 18 millimeters diameter. To Calculate the Change Gear when Cutting Hetric Screws by an English Lead= Screw. Divide 20 by the pitch in millimeters, and the quotient is the corresponding number of threads per inch to which the lathe must be geared. SCREWS. 437 Example To gear a lathe in order to cut a metric standard screw 24 millimeters in diameter and of three millimeters pitch, the lead-screw on the lathe having four threads per inch. Solution : Twenty divided by three gives 6%, therefore gear the lathe as if it was to cut 6% threads per inch with a common inter- mediate gear, and throw in the special intermediate gear as shown in the cut, and the lathe will cut a screw of three milli- meters pitch. The gearing is easily obtained, thus: The ratio between the screw gear and the stud gear is as 6% to 4, which is the same as 20 to 12, or, reduced to its lowest terms, 5 to 3. Hence, the gears may have any number of teeth providing the ratio is 5 to 3 ; for instance, multiplying by 9, 45 and 27 could be used, or, multiplying by 10, 50 and 30 could be used, etc. TABLE No. 74 .—How to Gear a Lathe when Cutting Metric Thread, Using inch-Divided Lead=Screw and Intermediate Gears, as Shown in Fig. 3 Screw to be Cut. (.Pitch in Milli- meters). LEAD-SCREW ON LATHE. 2 Threads per Inch. 3 Threads per Inch. 4 Threads per Inch. 1 5 ; Threads per Inch. 6 Threads per Inch. 10 Threads per Inch. 5 w> CO V V — OX) CO CO 3 M CO CO in !20 CO 80 _ re 3 M 24 CO 80 ~ V 3 W) CO 20 £ a) 40 I 1 . , 24 120 1.5 24 80 30 80 36 80 30 40 2 20* 100 24 80 24 60 30 60 24 40 40 40 2.5 20 80 1 24 64 : 24 48 30 48 30 40 50 40 3 24 80 27 60 j 24 40 30 40 36 40 60 40 3.5 28 80 21 40 28 40 35 40 42 40 70 40 4 32 80 24 40 32 40 40 40 48 40 80 40 4.5 27 60 27 40 36 40 45 40 54 40 5 30 60 ; 30 40 40 40 50 40 60 40 . 5.5 33 60 33 40 44 40 55 40 66 40 6 36 60 j 36 40 48 40 60 40 72 40 6.5 39 60 39 40 52 40 65 40 7 28 40 42 40 56 40 70 40 7.5 30 40 ! 45 40 I 60 40 8 32 40 ! 48 40 ! 64 40 8.5 34 40 51 40 j 9 36 40 54 40 9.5 38 40 57 40 10 40 40 60 40 10.5 42 40 11 44 40 11.5 46 40 12 48 40 438 RIVETS AND EYE BOLTS Iron and Copper Rivets. The diameter of large iron rivets, such as are used for bridge work, boiler work, etc., is measured in fractions of an inch, such as three-eighths, seven-sixteenths, one-half inch, etc., and the length of the rivet is measured in the same way. The diameter of small iron rivets is usually measured by the Birmingham wire gage, but the length is measured in inches. For instance, an iron rivet, No. 6, will be about 0.203 inches in diameter and No. 10 will be 0.134 inches in diameter. I ron burrs, fitting the rivets, are also sold by numbers. The diameter of the hole corresponds to the Birmingham gage. Copper rivets and copper burrs, such as used for riveting belt joints, are also measured by the Birmingham wire gage. Nails. The length of nails is usually expressed as so many penny. For instance: Two penny nails are 1 inch long. Three penny nails are iK inches long. Four penny nails are iK inches long. Five penny nails are i z /i inches long. Six penny nails are 2 inches long. Seven penny nails are 2 % inches long. Eight penny nails are 2X inches long. Nine penny nails are 2^ inches long. Ten penny nails are 3 inches long. Twelve penny nails are 3 % inches long. Sixteen penny nails are 3X inches long. Twenty penny nails are 4 inches long. Thirty penny nails are \% inches long. Forty penny nails are 5 inches long. Fifty penny nails are 5K inches long. Sixty penny nails are 6 inches long. Eye Bolts. It is very customary to weld an eye to a lag screw (see Fig. 4) to use in handling heavy weights in shops. The following table (No. 75) gives the holding power of lag screws or eye bolts when screwed into spruce timber a little over the full length of thread. The suitable size of bit for the thread is also given in the table. Fir A TABLE No. 75. Load at Diameter Diameter Which the Safe of Screw. of Bit. Screw Pulled Out. Load. 1 inch Ya. inch 16,000 lbs. 2,000 lbs Vs " \\ " 9,000 " 1,125 " 34 « Vi. " 7,000 " 875 " ti " X " 6,000 " 750 " % " H " 3,500 " 437 " Vs " A " 1,900 " 237 " % " 3 (( T6- 700 " 87 " LAG SCREWS. 439 TABLE No. 76. — Giving the Average Weight in Pounds per 100 Square Head Gimlet-Pointed Lag Screws. LENGTH in Inches. *" A" 3/6" 7 » TF #" ft" s/s" H" 7/8 " 1" 1^ 23/< H 7 10 2 3^ « «* 12 17 24 27i 2^ *X 6* 9| 14 19 26 31 3 Wa. 7* 11 16 21 28 34 51 3^ *>% 8* 12* 18 24 31 38 55 4 *ti n 14 20 26 34 42 60 85 112 ±% ey 2 10* 151 22 28 37 46 65 91 121 5 7 u* 17 24 32 40 50 70 97 130 5^ 7^ 12* 18* 26 34 43 54 76 103 140 6 8 18* 20 28 36 46 58 81 110 150 6^ 24 30 38 49 62 86 117 160 7 23 32 41 52 65 92 125 170 7% 241 34 44 55 69 97 132 180 8 26 36 47 58 73 103 140 190 sy 2 77 108 148 200 9 81 113 156 210 9^ 85 118 164 220 10 89 123 172 230 Size of ^ J * $ J® H ^ N(30 «|eo Head in X X X X X X x X X X Inches. to 1 "H \N \X co|n win OT|0 1 -H «efco i—i n|Qo T— t < GIMLET-POINTED LAG SCREW. Example. What is the weight of 8 lag screws 6" long and ^-inch in diameter ? Solution: Under the heading ^-inch, in the line with 6 in the column of length, is the number 36. Thus, 100 lag screws of this size will average to weigh 36 pounds, and one such screw will weigh 0.36 pound ; 8 such screws will weigh 0.36 X 8 = 2.88 pounds. 44Q TABLE No . 77. — Standard Dimensions for Welded Steam, Gas and Water Pipes. V. 0> M to U _ n - X ft to •v ca V u X. s. flu V) IB co cj < ^ u Sec 2§ c S 12 ft b> 1- j2 P-I.5 o.5 "3 a 5.2 "2 w "3 ^ 5-H As J-Jcj -G S3 II ^ 0.405 O.270 .O68 O.O72 0.057 25I3. 0.24 27 0.19 X 0.540 O.364 .088 O.I25 0.104 1383. 0.42 18 0.29 X O.675 O.494 .091 O. 166 0.192 75* • 0.56 18 0.30 H O.840 O.623. 109 0.249 0.305 472. 0.84 14 0.39 X I.050 O.824 .113 0.333 0.533 270. 1. 11 14 0.40 1 1. 315 I.048 .134 0.495 0.863 167. 1.67 nX 0.51 iX I.660 1. 380I. 140 O.668 1.496 96.25 2.24 IIX 0.54 iK I.900 1. 6ll .145 0.797 2.038 70.66 2.68 nx 0.55 2 2.375 2. 067!. 154 1.074 3.356 42.91 3.61 nx 0.58 2K 2.875 2.468 .204 1.708 4.784 30.10 5-74 8 0.89 3 3-500 3.067 •217 2.243 7.388 19-50 7-54 8 0.95 3K 4- 3-548 .226 2.679 9.887 14.57 9.00 8 I . 4 4.500 4.O26 .237 ■3.174 12.73 11. 31 10.66 8 1.05 4K 5- 4.508 .246 3-674 15.96 9.02 12.34 8 1 .10 5 5-563 5-045 •259 4-316 19.99 7.02 14-50 8 1. 10 6 6.625 6.065 .280 5-584 28.89 4.98 18.76 8 1.26 7 7.625 7.023 .301 6.926 38.74 3-72 23.27 8 1-36; 8 8.625 7.982 .322 8.386 50.04 2.88 28.18 8 1 .46 9 9.625 8.937 •344 10. 62.73 2.29 33 -70 8 1.57 10 10.750 10.019 .366 11. 9 78.84 1.82 40. 8 1.68 11 12. 11.250 •375 13-7 99-4 1.46 46. 8 1.78 12 12.750 12. .375I4-6 113. 1 1.27 49- 8 1.88 Pipes from one-eighth to i-inch inclusive is butt-welded, and proved at 300 pounds pressure for square inch. Pipes iX-inch and larger is lap- welded and proved at 500 pounds pressure per square inch. The threaded end of a pipe is taper at a ratio of X-inch per foot. Fittings are also tapered at the same ratio. The thread on the pipe forms a triangle at 60 degrees angle to the center line of the pipe. That is to say, the threading tool should be set square to the center line of the lathe and not square to the slant side of the tap when making a pipe tap, and a taper attachment is always used when cutting the thread on a pipe tap. Length in inches PIPES 44I The thread on pipes are slightly rounded at top and bottom so the actual depth is about four-fifths of the pitch. The depth of the thread can be calculated by the formula: Depth of thread = — — n The length of perfect thread on a pipe is calculated by the formula: .8 X D + 4.8 n D = Actual outside diameter of pipe in inches. n = Number of threads per inch. Water pipes are made both from steel and iron. In many cases the iron pipe is preferred because it is under certain con- ditions less liable to corrode. The size of steam, gas and water pipe is designated in the trade by the nominal internal diameter. Ordinarily the sizes will vary more or less from the standard as given in Table No. 77. Besides the standard sizes as given in Table No. 77, there are also heavier pipes known in the trade as "extra strong," and still heavier known as "double extra strong." These different sizes measure the same as the standard sizes on the outside diameter but the inside diameter is smaller in these heavier sizes. Table No. 77 gives only the Americam system of standard pipes. The English system is different. TABLE No. 78.— Whitworth Pipe Thread. (English System.) Nominal internal diam- eter in inches £ i f \ f i * 1 Threads per inch 28 19 19 14 14 14 14 11 Brass and Copper Tubes. Brazed brass and copper tubes are measured by the outside diameter and the thickness by the Brown & Sharpe gage. Seamless brass and copper tubing is measured by the outside diameter and the thickness is usually measured by the Birmingham gage. There are also seamless brass and copper tubing in the market known as "Iron Pipe Sizes". This kind of tube is used for plumbing and steam work. Extra heavy iron pipe sizes of brass, bronze and copper pipes are also furnished by the manufacturers. The ultimate tensile strength of seamless brass tubing will not exceed 40,000 pounds per square inch. 442 BRASS AND COPPER TUBES The ultimate tensile strength of seamless copper tubes will not exceed 30,000 pounds per square inch. The safe pressure allowable may be calculated by the formula: p= s Xt r Xf P = Safe pressure in pounds per square inch. s = Ultimate tensile strength of the material in pounds per square inch. / = Thickness of pipe in inches. r = Radius of pipe in inches. / = Factor of safety. Example. Find the safe pressure for a brass tube 3 inches diameter and three -sixteenths inch thick if the ultimate tensile strength of the material is known to be 30,000 pounds per square inch, and 8 is the factor of safety. Solution: p = 30000 X 0.1875 = 468 pounds per square inch. NOTE — When ordering brass or copper tubes it is always best to specify if the measurement given is external or internal diameter ; if the gage is Brown & Sharpe or Birmingham, and if brazed or seamless tube is wanted. It is also advisable to state the use of the tube as it may be had in different grades of hardness. TABLE No. 79.— Standard Sizes of Boiler Tubes. Made from Steel or Charcoal Iron. u u V M M J4 ^ O V) W a a Ho a •3 « 2* rt b Qco So <« b .«J B 2u cd B in B a « 13 -2 2 co b S3 B* 3S CO B 1.5 la CO fl 1 •095 2i • 095 3* .120 6 • 155 ii .095 2i .109 34 .120 7 .165 ii • 095 21 .109 4 •134 8 .165 if •095 3 .109 44 •134 9 .180 2 .095 3± .120 5 .148 10 .203 Cold Drawn Steel Tubing. Cold drawn steel tubing (bicycle tubing) is measured by the outside diameter, and the thickness is usually measured by the Birmingham gage. NOTES ON HYDRAULICS. 443 NOTES ON HYDRAULICS. Hydraulics is the branch of engineering treating on fluid in motion, especially of water, its action in rivers, canals and pipes, the work of machinery for raising water, the work of water as a prime mover, etc. Pressure of Fluid in a Vessel. When fluid is kept in a vessel the pressure will vary directly as the perpendicular height, independent of the shape of the vessel. For water, the pressure is 0.434 pounds per square inch, when measured one foot under the surface. The pressure in pounds per square inch may, therefore, always be obtained by multiplying the head by 0.434. The head corresponding to a given pressure is obtained by either dividing by 0.434 or multi- plying by 2.304. Example. What head corresponds to a pressure of 80 pounds per square inch ? Solution : 80 X 2.304 = 184 feet. Velocity of Efflux. The velocity of the efflux from a hole in a vessel will vary directly as the square root of the vertical distance between the hole and the surface of the water. For instance, if an opening is made in a vessel four feet, and another 25 feet, below the sur- face of the water, and the vessel is kept full, the theoretical velocity of the efflux will be nearly 16 feet and 40 feet per second respectively, friction not considered; or, in other words, the velocity will be as 2 to 5, because \^4 = 2 and \/25 — 5. The velocity of efflux in feet per second may always be calculated theoretically by the formula : v — 8.02 X s/h~. Constant 8.02 is \^2g— V64.4, and v = velocity of efflux. h = Head in feet. Table No. 8o gives the theoretical velocity of efflux and the static pressure corresponding to different heads, and is calcu- lated by the following formulas : h ■= V -— v — s/ h X 64.4 v = V P X 2.3 X 64,4 o4.4 v = V P X 148 P — h X 0.434 h — P X 2.Z 444 NOTES ON HYDRAULICS. TABLE No. 80. -Head, Pressure, and Velocity of Efflux of Water. Head in Feet. Pressure in Velocity in . Pressure in Velocity in Pounds per Feet per xiG3,Q in Feet. Pounds per Feet per Square Inch. Second. Square Inch. Second. h P V h P V 0.1 0.0434 2.54 19 8.246 35 0.2 0.0868 3.59 20 8.68 35.9 0.25 0. 1085 4.01 25 10.85 40.1 0.3 0.1302 4.39 30 13.02 43 0.4 0.1736 5.07 35 15.19 47.4 0.5 0.217 5.67 40 17.36 50.7 0.6 0.2604 6.22 45 19.53 53.8 0.7 0.3038 6.71 50 21.7 56.7 0.75 0.3255 6.95 55 23.87 59.5 0.8 0.3472 7.18 60 26.04 62.1 0.9 0.3906 7.61 65 28.21 64.7 1 0.434 8.02 70 30.38 67.1 1.25 0.5425 8.95 75 32.55 69.5 1.5 0.651 9.83 80 34.72 71.8 1.75 0.7595 10.6 85 36.89 73.9 2 0.868 11.4 90 39.06 76.1 2.25 0.9735 12 95 41.23 78.2 2.5 1.082 12.6 100 43.4 80.2 2.75 1.1905 13.3 110 47.74 84.2 3 1.302 13.9 120 52.08 87.7 3.25 1.4102 14.4 130 56.42 91.5 3.5 1.519 15 140 60.76 94.7 3.75 1.6375 15.5 150 65.1 98.3 4 1.736 16 160 69.44 101.2 4.25 1.8445 16.5 170 73.78 104.5 4.5 1.953 17 180 78.12 107.2 4.75 2.0615 17.5 190 82.46 110.4 5 2.17 17.9 200 86.8 113.5 6 2.604 19.6 225 97.65 120 7 3.038 21.2 250 108.5 126 8 3.472 22.8 275 119.35 133 9 3.906 24.1 300 130.2 139 10 4.34 25.4 325 141.05 144 11 4.774 26.6 350 151.9 150 12 5.208 27.8 375 162. 75 155 13 5.642 *28.9 400 173.6 160 14 6.076 30 425 184.45 165 15 6.51 31.1 450 195.3 170 16 6.944 32.1 475 206.15 174 17 7.378 33.1 500 217 179 18 7.812 34 550 238.7 188 NOTES ON HYDRAULICS. 445 Velocity of Water in Pipes. The theoretical velocity of water discharged from a pipe is calculated by the same formula as is used in calculating veloci- ties of falling bodies. ( See page 277). v — \S'2 g h v = Theoretical velocity of efflux per second. h = Head. 2 g = 64.4 if v and h are reckoned in feet. 2 g = 19.64 if v and h are reckoned in meters. If the water, besides the pressure due to the head, is also acted upon by some additional pressure, for instance, steam, the theoretical velocity of the discharge is obtained by the formula, •H »*(*+«&) P = Pressure in pounds per square inch. The constant 0.434 is used because a column of water one foot high will exert a pressure of 0.434 pounds per square inch; thus, by dividing by 0.434, we actually convert the pressure into its corresponding head in feet. All other quantities in this formula are, of course, taken in English units. Note. — By head is always meant the vertical height in feet, or its equivalent in pressure expressed in feet. Table No. 80 gives the theoretical velocity of the discharge and the pressure corresponding to different heads. The theoretical velocity is never obtained in practice, be- cause part of the total head is used to overcome the resistance at the entrance of the pipe, and part is used to overcome the frictional resistance to the flow of the water in the pipe. Thus, only a part of the total head is left to give velocity to the water, therefore the velocity of the water at dis- charge will only be what is due to the velocity head, after deduc- tions are made for resistance at the entrance and for friction in the pipes. In short pipes, the resistance at the entrance to the pipe is comparatively the larger loss, but in long pipes the fric- tional resistance is the larger. When both the resistance at the entrance and the friction in the pipe are considered the formula will be : Vt 2gh +/4 v = Velocity of discharge in feet per second. 2 g = 64.4 L = Length of pipe in feet. d = Diameter of pipe in feet. f— Coefficient of friction, which is obtained from experi- ments, and will vary according to conditions, from 0.01 to 0.05. It is usually in approximate calculations taken as 0.025. 44 6 NOTES ON HYDRAULICS. Example. Find the velocity of discharge from a pipe six inches in diam- eter. The head is 16 feet and the length of the pipe is 100 feet, and coefficient of friction 0.025. Solution : (Note. 6 inches = 0.5 foot.) VD4. 64.4 X 16 100 025 X -ox v = ^- 1030.4 6.5 v = 12.6 feet per second. In Table No. 8ithe quantity of water discharged per min- ute by a pipe six inches in diameter, when the velocity is one foot per second, is 88.14 gallons. Thus, the quantity of water deliv- ered when the velocity is 12.6 feet per second, is 12.6 X 88.14 = 1110.6 gallons per minute. When the length of the pipe is more than 4,000 diameters the velocity of the water may be calculated by the formula, and the quantity is obtained by multiplying the velocity by the constants given in Table No. 8i. Example. Find the velocity of efflux from a water pipe of three inches diameter and 1200 feet long, having a head of six feet, assuming coefficient of friction as 0.025. Solution : 64.4 X 6 64.4 X 6 ^ r V = V nno- s^OO = *-™ feet P er SeCOnd Discharge in gallons per minute : q — 22.03 X 1.79 = 39.4 gallons per minute. NOTES ON HYDRAULICS. 447 TABLE No; 81 .—Quantity of Water Discharged Through Pipes in One Hinute, when Velocity of Efflux is One Foot per Second. 1-8 V .2 fa ** r, J'* ft O V a « — 3 « cr gen - 2 Discharge in Cubic Feet per Minute at Velocity of One Foot per Second. Discharge in Gallons per Minute at a Velocity of One Foot per Second. M u S U a is I s ft S j o So t« •■* | - J T3 . K-Hg = S ^" ^ C ^ 1 k O 3 1H* 4) ^— "^ 3 8 2 V & H s H 3 H 3 % IX 0.750 0.966 0,737 0.964 0.726 0.962 7 /lO i% 0.700 0.950 0.685 0.947 0.673 0.944 % IX 0.667 0.937 0.650 0.933 0.638 0.930 % 1% 0.625 0.919 0.607 0.914 0.593 0.911 6 /l0 1% 0.600 , 0.906 0.581 0.900 0.567 0.897 % 2 0.500 0.846 0.479 0.839 0.463 0.833 Mo 2^ 0.400 0.766 0.378 0.756 0.361 0.748 % 2% 0.375 0.743 0.353 0.732 0.336 0.723 % 3 0.333 0.700 0.311 0.688 0.295 0.678 3 /l0 3X 0.300 0.662 0.278 0.648 0.262 0.637 X 4 0.250 0.596 0.229 0.582 0.214 0.571 y 5 5 0.200 0.522 0.181 0.506 0.167 0.495 y 6 6 0.1667 0.465 0.149 0.449 0.137 0.437 Vi 7 0.1428 0.421 0.127 0.405 0.115 0.393 % 8 0.125 0.385 0.110 0.369 0.0992 0.357 % 9 0.1111 0.355 0.0968 0.340 0.0862 0.327 Ho 10 0.1000 0.330 0.0865 0.315 0.0774 0.303 yn 11 0.0909 0.309 0.0782 0.293 0.0696 0.282 %2 12 0.0833 0.290 0.0713 0.275 0.0631 0.264 %3 13 0.0769 0.274 0.0655 0.259 0.0578 0.248 %4 14 0.0714 0.260 0.0605 0.245 0.0533 0.234 y i5 15 0.0667 0.247 0.0563 0.232 0.0494 0.222 %6 16 0.0625 0.236 0.0526 0.221 0.0459 0.211 %7 17 0.0588 0.225 0.0493 0.211 0.0429 0.201 %8 18 0.0556 0.216 0.0463 0.202 0.0403 0.192 %9 19 0.0526 0.208 0.0438 0.193 0.0379 0.184 y 20 20 0.0500 0.20( ) 0.0415 0.185 0.0359 0.177 NOTES ON STEAM. 459 To Find the Mean Effective Pressure by the Preceding Table. Find the constant in the column corresponding to tr e con- ditions of expansion, and to the given cut-off. Multiply this by the absolute initial pressure, and the product is the average pressure. Subtract the back pressure and the remainder is the mean effective pressure. Example. Find the mean effective pressure for isothermal expansion when the engine is cutting off at one-quarter stroke. The initial pressure is 90 pounds absolute. The absolute back pressure is 18 pounds. Solution : M. E. P. — 90 X 0.596 — 18 = 53.64 — 18 = 35.64 pounds. Note. — All such calculations must be made from absolute pressure (not gage pressure), and when determining the cut-off the clearance must be considered. Clearance. The clearance of an engine is usually expressed as a per- centage of the piston displacement. The space between the piston and the cylinder head at the end of the stroke, also the cavities due to the steam ports, must be included in considering clearance. In high-class Corliss engines the clearance does not exceed 2>£ to 5 per cent., but in common slide-valve engines the clear- ance may go as high as 5 to 15 per cent. When clearance is taken into account the actual ratio of expansion is r = Actual ratio of expansion. R = Nominal ratio of expansion. c = Clearance, expressed as a fractional part of the length of the stroke. Example. The nominal ratio of expansion is 4, and the clearance is 5 per cent. What is the actual ratio of expansion ? Note. — 5 per cent, is %oo = M20 = 0.05 of the stroke Solution : r _ 1+0.05 4- + 0.05 1 05 r = ' = 3.5 = Actual ratio of expansion. 0. O 460 THERMOMETERS Thermometers. The well known instrument called thermometer is used to measure the common temperatures, as from 40 degrees below zero to 450 degrees above zero, Fahrenheit. There are in common use three different kinds of ther- mometer scales, namely the Fahrenheit, the Celsius (also called the centigrade), and the Reaumur. The Fahrenheit is used in England and in the United States. The Celsius and the Reaumur are used in Europe, outside of England. The Celsius is used the most and is almost universally used in technical books and all kinds of scientific calculations. The ratio between those thermometers are: Thermometers. Boiling point. Freezing point, Fahrenheit 212 32 Reaumur 8o° o° Celsius ioo° o° Following formulas, transpose from one thermometer scale to another; 4X(,F — 32) 4XC *- 9 ■ R= ~v~ - _ 5 X(F — 32) _ 5 X R C- 9 G- 5 Example. If the temperature is — 20 (20 degrees below zero) Fahren- heit. How much is this in Centigrades ? Solution : 5 X (— 20 — 32) 5 X— 52 OR C = = ■ = — 28f degrees, or 28f degrees below zero on the Celsius thermometer. Example. The temperature of a furnace is measured by a pyrometer* and found to be 1560 degrees Fahrenheit. How much is this in Celsius degrees? Solution : c = 5 x (1560 -32) = §2 % ,A s^ s a rt in <# rt e rt a > 3 C w <# re a 3 C M «■ Q £ ^r o o H- « S| ^r o o H- 9 114 126 8.2 17 45 53 18.6 10 101 109 9.1 18 40 48 20.8 11 90 98 10 19 36 43 23.2 12 81 91 11 20 32 38 26.3 13 72 80 12.5 21 28 34 29.3 14 64 73 13.7 22 25 32 31.2 15 57 64 15.6 23 22 29 34.4 16 51 60 16.6 24 20 27 37 Example. How many turns of No. 15 double cotton-covered wire will go in one layer on a spool 5 inches long ? Solution : In table No. 86 we find that for No. 15 double cotton- cov- ered copper wire there are 15.6 turns per inch of spool; thus, 5 X 15.6 = 78 turns in 5 inches. Enameled Copper Wire. Enameled copper wire is insulated by multiple coats of enamel. This insulation will not absorb moisture. It will stand a higher temperature than cotton insulation, as enameled wire will stand a temperature as high as 100° C, 212° F., without injury. Another great advantage of enameled wire is that the insu- lation is so thin it does not add more than 1 to 2 mils to the diameter of the bare wire. This is especially a valuable feature for the smaller sizes of wire, because when a spool is wound with a small size of double cotton-covered wire the insulation takes up a large percentage of the space. Carrying Capacity of Copper Wire. When electricity is conveyed through a wire, it is always necessary to select the size large enough to avoid undue heating. The number of circular mils required for each ampere of current will depend a great deal upon the facilities by which the heat is conveyed away from the circuit. Well ventilated armatures in small dynamos such as 2 horse power or less may have 300 to 400 circular mils area in the wire for each ampere of current they carry. Armatures in dynamos NOTES ON COPPER WIRE. 467 from 5 to 10 horse power may have 400 to 500 circular mils for each ampere of current that they convey. In larger and more expensive dynamos there may be over 600 circular mils in the armature winding for each ampere carried. In field coils and similar work we often find as much as 1200 circular mils for each ampere conveyed. In lines about 400 to 500 circular mils are used per ampere of current when the wire is small as No. 10, but the larger the wire the more circular mils are required per ampere. (See Underwriters' Rules ) Table No. 87 gives the carrying capacity of round copper wire according to the underwriters' rule as the standard adopted for interior wiring.* TABLE No. 87. — Carrying Capacity of Copper Wire. Size Amperes Amperes. 1 Size Amperes. Amperes. B. & S.Gage. B.& S.Gage. 18 3 5 4 65 92 16 6 8 3 76 110 14 12 16 2 90 131 12 17 23 1 107 156 10 24 32 127 185 8 33 46 00 150 220 6 46 65 000 177 262 5 54 77 0000 210 312 The lower limit is specified for rubber-covered wires to prevent gradual deterioration of the high insulations by the heat of the wires, but not from fear of igniting the insulation. The question of drop is not taken into consideration in this table, but this may be calculated by the formulas given on page 463. The carrying capacity of the No. 16 and No. 18 wire is given in the table, but smaller wire than No. 14 is not allowed to be used in interior wiring. Example. What is the smallest size copper wire allowed for a line supplying current for 36 incandescent lamps of 16-candle power at 110 volts ? Solution : A 16-candle power lamp at 110 volts will use about one-half of one ampere of current ; 36 lamps will, therefore, take about 18 amperes, and if we are to follow the lower limit No. 12 is too small as that is only good for 17 amperes, but No. 10 will be of ample size. * For instructions regarding wiring, see " Rules and Requirements of the National Board ot Five Underwriters for the Installation of Electrical Wir- ing," which may be obtained from any local electrical inspection bureau. NOTES ON ELECTRICAL TERMS. NOTES ON ELECTRICAL TERMS. The units used in electrical calculations are different from the well-known units used in mechanics. The name of each unit in electricity is derived from the name of some great scien- tist who has assisted in the world's progress. When we become familiar with their meaning and value, these units are not more difficult to understand than the common well-known terms, feet, inches, pounds, gallons, etc Volt. Volt is the practical unit of electromotive force, and is such an electromotive force as will drive one ampere of current through a resistance of one ohm. The name volt is after an Italian electrician, Alessandro Volta (1745-1827). In practice, the electromotive force in volts is measured by an instrument called a Voltmeter. In calculation we obtain the volts by multiplying the ohms by the amperes. As example, we will say that each cell in a storage battery has an electromotive force of from 2 to 2j£ volts. A common so-called dry cell will give an electromotive force of about 1 to V/z volt. Common 16-candle power incandescent lamps are usually run on a 110 volt circuit, that is to say, the filament in the lamp has such resistance that it requires an electromotive force of 110 volts to drive current enough through the lamp to give good light. Electric street cars are usually run on about 500 volt cir- cuits, that is to say, the windings of the motors are of such pro- portion that it takes an electromotive force of 500 volts to drive sufficient current through the motor so it will drive the car at its proper speed. Ampere. Ampere is the practical unit of current strength and is that current which would circulate in a circuit having one ohm resist- ance when the electromotive force is one volt. The name ampere is after a French electrician, Andre* Marie Ampere (1775-1836). In practice, the current strength in amperes is measured by an instrument called an Amperemeter. In calculation we ob- tain the amperes by dividing the volts by the ohms. As examples, we may say a common 16-candle power in- candescent lamp on a 110 volt circuit is using about one-half of one ampere. A 32-candle power incandescent lamp is using about one ampere. A motor running on a 110 volt circuit will take about 7 amperes to produce 1 horse power. NOTES ON ELECTRICAL TERMS. 469 Coulomb. Coulomb is the unit of quantity in measuring electricity. A coulomb is the amount of electricity conveyed by one ampere in one second. The name coulomb is after a French philosopher, Charles Augustin de Coulomb (1736-1806). In practical work the coulomb is obtained by multiplying the ampere by the time in seconds in which the current is acting. A coulomb is therefore the same as ampere-second. As this quantity is very small it is customary with electricians to measure in ampere-hours. One ampere-hour is, of course, 3600 coulombs. The name coulomb is not much used by practical men but it is very customary to speak of ampere-hours. For instance, if a storage battery could furnish 5 amperes of current for a time of 40 hours it would be said to have a capacity of 5 X 40 = 200 ampere-hours. It could, of course, also be said to have a capacity of 720,000 ampere-seconds or 720,000 coulombs. Ohm. Ohm is the practical unit of electrical resistance. The standard (international ohm) is the resistance at the temperature of degree centigrade of a column of mercury 106.3 centimeters long and 1 square millimeter area. Such a column of mercury will weigh 14.4521 grams. The name ohm is after a German electrician, Georg Simon Ohm (1787-1854). As examples, we may mention that 1000 feet of Xo. 10 cop- per wire have a resistance of about 1 ohm and, remembering the resistance of 1000 feet of No. 10 wire, we can almost by mental calculations get the approximate resistance of 1000 feet of any other size of wire by the rules given on page 462. The resistance is different in different materials. Materials having high resistance are called insulators, such as, for instance, silk, cotton, paper, fiber, glass, etc. Among the common metals copper is the best conductor of electricity; next comes aluminum, having a resistance about twice that of copper. Iron has from 6 to 7 times the resistance of copper. The resistance in ohms of a circuit is obtained by dividing the volts by the amperes, and the quotient is the ohms. Watt. Watt is the unit of power, and is the product of the volts and the amperes. 1000 watts is called a kilowatt. 746 watts = 550 foot-pounds per second = 1 horse power. 736 watts = 75 kilogram-meter per second = 1 metric horse power. 47° NOTES ON ELECTRICAL TERMS. The name watt is from a Scottish inventor, James Watt (1736-1819). James Watt invented the steam engine indicator, the condenser, and made great improvements in the steam engine. The watts are also measured by an instrument called a Wattmeter. Important. It has been stated above that the product of the volts and the amperes gives the watts ; this is only true for direct current, but will not hold good for alternating current, because in alternating current the voltage will be ahead of the current in a circuit having induction, but the current will be ahead of the voltage in a circuit having capacity. In either case the product of the volts and the amperes, measured separately by a voltmeter and an amperemeter, will only give the apparent watts, which will be greater than the real watts as measured by the wattmeter ; therefore we have the fol- lowing rules for calculating the power in an alternating circuit: The actual wattmeter reading is the real watts. The product of the voltmeter reading and the amperemeter reading is the apparent watts. The real watts divided by the apparent watts is the power factor. The real watts divided by the apparent watts also gives the cosine of angle of lag. t Thus: The cosine of angle of lag is the power factor. Multiplying the apparent watts by the cosine of angle of lag gives the real watts. The cosine of an angle has vanished when the angle is 90 degrees ; that is, cosine 90 degrees = 0. Therefore, when the angle of lag is approaching 90 degrees the real watts will be very small, and when the angle of lag becomes 90 degrees we have a wattless current. The only practical way to get the watts of an alternating current circuit is to measure by a correct wattmeter. Joule. Joule is the unit of heat and is the product of the volt and the coulomb. The name is from an English natural philosopher, James Prescott Joule (1818-1889). Joule was the first one to discover and determine the mechanical equivalent of heat. The heat produced in an electric circuit is in proportion to the resistance and the square of the current. 1 joule per second = 1 watt mechanical work. 1 joule = 0.00024 calorie * of heat. * Calorie is the unit of heat used in the metric system and is the heat required to increase the temperature of 1 kilogram of water 1 degree centi- grade at or near the temperature of 4 degrees centigrade. NOTES ON ELECTRICAL TERMS. 471 1 joule = 0.1019 kilogram-meter mechanical work. 1 joule = 0.000948 B. T. U. of heat * 1 joule == 0.7373 foot-pounds of mechanical work. The heat produced by an electric current will be: B. T. U. = 0.000948 X C 2 X R X t. B. T. U. = British Thermal Units. C = Current in amperes. R = Resistance in ohms. t = Time in seconds. Farad. Farad is the unit of electrical capacity. A conductor or a condenser whose capacity is one farad will hold a quantity of electricity equal to one coulomb, when the electromotive force is one volt. The name farad is after an English electrician, Michael Faraday (1791-1867). Henry. Henry is the unit of inductance and is the inductance of a circuit in which the variation of a current at the rate of one ampere per second induces an electromotive force of one volt. The name henry is from an American physicist, Joseph Henry (1797-1878). Derived Units. 1 megohm = 1 million ohms. 1 micro-ohm = 1 millionth of an ohm. 1 milliampere = 1 thousandth of an ampere. 1 microfarad = 1 millionth of a farad. Of all these units the volt, ampere, ohm, watt and kilowatt are the units most used in practical work. Example. How many ohms of resistance are there in 6 pounds of copper wire No. 12 ? Solution : In table No. 85 the resistance of one pound of copper wire No. 12 is given as 0.07993. Therefore, the resistance in 6 pounds of copper wire will be 6 X 0.07993 = 0.48 ohms. * B. T. U. is British Thermal Unit and is the heat required to increase the temperature of one pound of water one degree Fahrenheit at a temperature between 39 and 40 degrees. 47 2 notes on electrical terms. Example. An electrical motor is running on a 110 volt circuit and using a current of 40 amperes. How many kilowatts ? How many horse power ? Solution : 110 X 40 Power = — — — — =4.4 kilowatts. 1000 110 X 40 Power = — = 6 horse power, very nearly. Example. A spool is wound with 3 pounds of No. 18 wire laid 4 layers, 250 turns to each layer. The voltage between its terminals is 10 volts. How many amperes are flowing through the spool ? How many ampere-turns are there on the spool ? How much power is consumed by the spool ? How many circular mils are there in the wire for each ampere ? Solution : In table No. 85 we find that 1 pound of No. 18 wire has a resistance of 1.293 ohms; 3 pounds will, therefore, have a re- sistance of 3X1.293 = 3.879 ohms and when the spool warms up a little the resistance will practically be 4 ohms. Then the current will be : 10 C = —— = 2yi amperes. 4 The spool has 4 X 250 = 1000 turns of wire. It will, therefore, have 1000 X 2% = 2500 ampere-turns. Power consumed will be : W == 10 X 2 % = 25 watts. In table No. 85 we find that the area of No. 18 wire is 1624 circular mils ; dividing 1624 by 2}£ we get 650 circular mils per ampere. It may be found in practice that a spool like this will heat too much and, if so, the difficulty is overcome by winding more turns of wire on the spool. This will not change the ampere- turns and consequently not change the magnetic power in the spool ; but it will increase the resistance, consequently decrease the current flowing and thereby reduce the heat and also the power consumed in the spool. The temperature of any coil or spool depends greatly on the area of the radiating surface, and may be determined by ex- perience, but very seldom more than 1 watt of energy can be allowed for each square inch of round radiating surface of the spool. Frequently as much as 2 square inches of radiating sur- face is required for each watt of energy lost in the spool. SHOP NOTES 473 SHOP NOTES Standard Sizes of American Machinery Catalogs. 9 x 12 inches, largest size. 6 x 9 inches, regular size. 4X x 6 inches, medium size. 3 x \% inches, pocket size. Shrink Fit. The allowance to be made for a shrink fit will vary more or less according to the nature of the work and the judgment of the designer. When shrinking a collar on a shaft or similar work, an allowance of 0.002 inch to 0.003 i ncn wn *l do for a shaft of one inch diameter, and as the shaft is larger in diameter add 0.0005 inch to the allowance for each inch the diameter is increased. For instance, a shaft of 6 inches diameter may for a shrink fit be made 0.007 i ncri larger than the hole. However, there may be cases where it is better to allow less, because frequently a shrink fit, in want of suitable tools, only is used instead of a press fit. Press Fit. The force required to press a shaft into a hole made for a press fit will depend not only on the allowance made on the fit, but also on the kind of material, the length of the fit, the finish, etc. Press fits are frequently made so that a pressure of 5 to 10 tons per inch diameter is required to force the shaft into its hole. When the length of the fit is from one to two times its diameter, and the finish is good and smooth, an allowance of three-quarters to one and one-quarter of a thousandth of an inch may do well for pressing a one-inch shaft of machinery steel into a hole in cast iron or machinery steel, and as the shaft increases in size the allowance may be increased about half of one-thousandth for each inch the shaft is increased in diameter; but there is no hard and fast rule to go by, judgment and experience are the best guides. A shaft of machinery steel 4 inches diameter, 8 inches long, straight fit, made 0.004 mc ^ larger than the hole in the cast iron gear required 40 tons to press it into place. White lead mixed in machinery oil was used as a lubricant. A shaft of machinery steel 2% inches diameter, 5 inches long fit, pressed into a gear of machinery steel at a force of about 10 tons, when the shaft was by actual measurement made 0.00 1 inch larger than the hole in the gear. White lead mixed in machinery oil was used as a lubricant. When pressing shafts into gears or similar things, the strength of the hub must be considered and also the strength of the shaft, because if the shaft is long it may buckle under the pressure. The shaft is in this case under the same condi- tion regarding strength as a long column, see page 224. 474 SHOP NOTES Running Fit. To make a running fit like a bearing, an allowance may be made of about two one-thousandths of an inch for a shaft one inch diameter, and one-thousandth more for each inch the shaft is increased in diameter. A shaft 6 inches diameter, running in self-aligning and self-oiling bearings will work well when the shaft is from 0.005 inch to 0.007 inch smaller than the bearings. Babbitt Metal. There are a great number of Babbitt metals or bearing metals sold in the market under different trade names, and at different prices. The writer's experience is that five pounds of lead to one pound of antimony will give very good lining for bearings. Helical Springs. Helical springs (usually but wrongly called spiral springs) are used either as compression springs or as tension springs. The diameter of the spring may be about 8 times the diam- eter of the wire. Emery. Emery is sold by the pound. The coarseness of emery is graded by numbers, ao: 6, 8, 10, 12, 14, 16, 18, 20, 24, 30, 36, 40, 46, 54, 60, 70, 80, 90, 100, 120, 150, CF, F and FF. The lower the number, the coarser is the emery. No. 60 is often used for cutting down, when polishing metal work; then No. 120 and F or FF for finishing. Emery Cloth. Emery cloth is sold in sheets 9x11 inches, and the coarse- ness is graded by numbers, thus: 00, o, %, 1, iy 2 , 2, 2%, and 3. The lower the number the finer is the emery cloth. For shop use in general, the grades from No. 2 to No. o are mostly used. No. o, of course, is a great deal finer than No. 2. Emery cloth can also be bought in rolls of 50 yards, and as wide as 2j inches. Sand Paper. Sand paper is sold in sheets, 9x11 inches, and the coarse- ness is graded by numbers, thus: 00, o, j4 t 1, i}4, 2, 2}4, 3 and 3>^. The lower the number the finer is the sand paper. The grades mostly used in pattern shops are from No. 1 to No. 2. Number 1 is a great deal finer than No. 2. Sand paper can also be bought in rolls of 50 yards, and as wide as 48 inches. SHOP NOTES 475 Weight of a Grindstone. Multiply the constant 0.064 by the square of the diameter in inches and this product by the thickness in inches; the result is the weight of the grindstone in pounds. Example. Find the weight of a grindstone 30 inches in diameter and six inches thick. Solution : Weight = 0.064 X 30 X 30 X 6 = 346 pounds. Lathe Centers. In this country lathe centers are universally made 60 de- grees, but in Europe the most common practice is to make lathe centers 90 degrees. Lathe Mandrels. Lathe mandrels made from tool steel should be hardened and ground with lapped centers. They are made slightly taper, about 0.006 inch to 0.010 inch per foot. Mandrels less than 1 inch in diameter ought to be about 0.0005 inch under size at the smallest end, and mandrels from 1 inch to 2 inches 0.001 inch under size at the smallest end. The length of mandrel up to 2 inches in diameter may be about 3 inches more than 5 times the diameter. For instance, a mandrel l /z inch diameter may be $% inches long, and a man- drel 2 inches diameter may be 13 inches long. Larger sizes may be little longer but ordinarily mandrels, except for special work, do not need to be more than 15 to 18 inches long. Common Sizes of Steel Used for Lathe and Planer Tools. For forged tools for lathes and planers, the common sizes of steel are Y% X % inch, Y% X K inch, ^Xi inch, Y% X 1 X inches, X inches, KXiK inches. TABLE No. 88.— Angles and Corresponding Taper. Angle one side with Included the center Angle Taper per Inch Taper per Foot line i° r O.OO873" O.IO5" 1 1 O.OI746 O.209 1 a O.O2619 O.314 I 2 O.O3492 O.419 ii 2* O.04364 O.524 1* 3 O.05238 O.629 2 4 O.06984 O.838 476 SHOP NOTES TABLE No. 89 — Tapers per Foot and Corresponding Angles Taper Included Angle with Taper Inc.uaea Aiig.e wuh Per Ft. Angle. Center Line Prr Ft. Ansrle. Center Line. i" o° 3 6' 0°l8' I" 4° 46' 2° 23' 1 4 I 12 O 36 li 7 09 3 35 T°6 i 30 45 If 8 20 4 10 1 i 47 54 2 9 3i 4 46 16 2 05 1 02 2* 11 54 5 57 i 2 23 1 12 3 14 15 7 08 i 3 35 1 47 3i 16 36 8 18 1 5 Tfi 4 28 2 14 4 18 55 9 28 Morse Taper. The Morse Taper, which is so universally used for the shanks of drills and other tools, is given in TABLE No. 90.— Horse Taper. No. of Taper. Standard P' 1 ^** * Hug Depth. J^ Diameter of Plug at Small End. Taper per Foot. 1 2 3 4 5 6 2y 8 inch. Z T6 8* " 75i " 0.475 inch. 0.7 0.938 " 1.231 " 1.748 " 2.494 " 0.369 inch. 0.572 " 0.778 " 1.02 1.475 " 2.116 " 0.600 inch. 0.602 « 0.602 " 0.623 " 0.630 " 0.626 w For very complete information regarding the Morse Taper, see American Machinist, May 14, 1896. No. I 2 3 4 5 6 Socket holds % inch to \% inch inclusive " h " " it " 3 2 ^ „ „?2 ., ..3 „ 016 4 Jarno Taper. In the " Jarno Taper " the number of the taper gives the length of the standard plug in half-inches, and it gives the diame- ter of the small end in tenths of inches and the diameter of the large end in eighths of inches. For instance, a No. 8 "Jarno Taper " is four inches long, one inch diameter at large end, and 0.8 inch diameter at small end. The taper, of course, is 0.6 inch per foot for all numbers. This is a very convenient sys- tem, and deserves adoption for its merits. The same taper is SHOP NOTES. 477 also very well adapted to the metric system, as 0.6 inch per foot is equal to 0.05 millimeter per millimeter. The following table is given to illustrate the system. The table could be extended to as large size tapers as are required for any work. TABLE No. °1,_ Jarno Taper. Number of Taper. Length of Taper. Diameter of Large End of Taper. Diameter of Small End of Taper. 1 2 1 y 8 = 0.125 yi = 0.250 * - 0.2 3 4 2 3/ 8 = 0.375 y 2 = 0.500 T 3 * = 0-3 T 4 o = 0-4 5 6 2*A 3 5/ 8 = 0.625 f4 = 0.750 % = 0.5 t =0.6 7 8 3^ 4 7/ 8 = 0.875 1 = 1.000 t 7 o = 0.7 1 =0.8 9 10 5 iy 8 = i.i25 1% = 1.250 A = 0.9 1 = 1.0 This system of taper is described by " Jarno " in the Amer- ican Machinist, October 31, 1889. Marking Solution. Dissolve one ounce of sulphate of copper (blue vitriol) in four ounces of water and half a teaspoonful of nitric acid. When this solution is applied on bright steel or iron, the surface immediately turns copper color, and marks made by a sharp scratch-awl will be seen very distinctly. A Cheap Lubricant for Milling and Drilling. Dissolve separately in water 10 pounds of whale-oil soap and 15 pounds of sal-soda. Mix this in 40 gallons of clean water. Add two gallons of best lard oil, stir thoroughly, and the solu- tion is ready for use. Soda Water for Drilling. Dissolve three-fourths to one pound of sal-soda in one pail full of water. Solder. Ordinary solder is an alloy consisting of two parts of tin and one part of lead, and melts at 360°. Solder consisting of two parts of lead and one part of tin melts at 475°. For tin work use resin for a flux. 47^ SHOP NOTES. Soldering Fluids. Add pieces of zinc to muriatic acid until the bubbles cease to rise, and the acid may be used for soldering with soft solder. Mix one pint of grain alcohol with two tablespoonfuls of chloride of zinc. Shake well. This solution does not rust the joint as acids are liable to do. When soldering lead use tallow or resin for a flux, and use a solder consisting of one part of tin and \ x / 2 parts of lead. Spelter. Hard spelter consists of one part of copper and one part of zinc. A softer spelter is made from two parts of copper and three parts of zinc. A spelter which will flow very easily at low heat consists of 46% of Copper, 46% of Zinc, and 8% of Silver. When making any of these different kinds of spelter, melt the copper first in a black lead crucible and then put in the zinc after the copper has cooled enough to furnish just sufficient heat to melt the zinc, but not enough to burn it. Stir with an iron rod and after the metals have compounded and the compound is still molten, pour upon a basin of water. The metal in striking the water will form into small globules or shot and will so cool, leaving a coarse granular spelter ready for use. When pouring the metal let a helper keep stirring the water with an old broom. Alloy Which Expands in Cooling. Melt together nine pounds of lead, two pounds of antimony and one pound of bismuth. This alloy may be used in fastening foundation bolts for machinery into foundation stones. In such cases, collars or heads are left on the bolts and after the hole is drilled in the stone a couple of short, small holes are drilled at an angle to the big hole ; when the metal is poured in, it will flow around the bolts and also into these small holes, and it is almost impossible for the bolt to pull out. Caution. — When drilling holes in stone, water is always used, but this must be carefully dried out by the use of red-hot iron rods before the melted metal is poured in. If this pre- caution is not taken the metal will blow out, making a poor job, and it may also cause accident by burning the hands and face of the man who is pouring it in. Shrinkage of Castings. General rule : yi inch per foot for iron. % 6 inch per foot for brass. In small castings the molder generally raps the pattern more than the casting will shrink, therefore no shrinkage is al- lowed. Frequently castings are of such shape that the pressure SHOP NOTES. 479 of the fluid iron on some part of the mould is liable to make the sand yield a little and thereby cause the casting to be as large as, or even larger than the pattern. All such things a practical pattern maker takes into consideration when allowing for shrinkage in patterns. Case Hardening Wrought Iron and Soft Steel. Bone dust specially prepared for the purpose, or burnt leather scrap, is placed in a cast-iron box, together with the article to be hardened. Cover the top of the box with plenty of the hardening material in order to keep the air out. Heat the whole mass slowly in a furnace to a red heat from two to five hours in order that it may be uniformly and thoroughly heated through. A few iron rods about 5 /\q inch in diameter are put in when packing the box, one end of the rod reaching about to the middle of the box, and the other end projecting out through the hardening material on top. When the box appears to have the right heat, these rods are pulled out one at a time, in order to judge of the heat in the center of the mass. When the box has been exposed to the fire the desired length of time, its contents are quickly dumped into cool water. Sieves of iron netting are laid on the bottom of the tub into which the case hardening material is dumped so that the hard- ened articles may be conveniently taken up from the water by one of the sieves. The case hardening material itself is also taken out by another sieve which is of very fine netting and placed under the first one. The material is dried and used over again, and a little new material is added when repacking the boxes. When articles are well finished before hardening, this pro- cess gives a very fine color to both soft steel and wrought iron. Case hardening may also be effected by packing the articles in soot, but this process does not give a nice color. Horn and hoof is also used for case hardening. Malleable iron may also be case hardened, but it requires careful handling in order to prevent its cracking and twisting out of shape. Case Hardening Boxes are made from cast-iron and are of various sizes. Small boxes may be made nine inches long, five inches wide, and four inches deep, and about one-fourth inch thick. They should be pro- vided with legs at least one inch high so that the heat may get under the bottom as at the top. An ear having a rectangular hole through it should be cast under the bottom at each end of the box. This gives a chance to handle the box with a fork having flat prongs instead of taking it out of the hardening fur- nace with a pair of tongs, which is liable to break the box, as cast-iron is very inferior in strength when hot. 480 SHOP NOTES. To Harden with Cyanide of Potassium. Heat the cyanide of potassium in a wrought iron pot until cherry red, and keep it so by a steady fire, immerse the pieces to be hardened from three to five minutes, according to their size and degree of hardness required, then plunge into cold water. Large pieces require more time than small ones, and the longer the article remains in the cyanide the deeper the hardening becomes. New cyanide gives the best color and cyanide previously used for hardening produces a harder surface. BLUE PRINTING. To Prepare Blue Print Paper. Dissolve two ounces of citrate of iron and ammonium in 8X ounces of soft water. Keep this in a dark bottle. Also dissolve 1% ounces of red prussiate of potash in 8% ounces of water and keep in another dark bottle. When about to use, mix (in a dark place) an equal quantity of each solution in a cup and apply with a sponge or a camel's hair brush as evenly as pos- sible on one side of white rag paper (such as used for envelopes). Let it dry and put it away in a dark drawer. The paper must not be prepared in daylight but when taking prints it may be handled then, providing care is used to expose it as little as possible to the light before it is put into the printing frame. Blue Print Frame. Make a strong frame similar to a picture frame having a strong and thick glass. Make a loose back, from boards about Yz inch thick, which is held into the frame by four suitable catches so arranged that they press this back firmly and evenly against the glass. The surface next to the glass should be covered by three thicknesses of flannel in order to make a cushion so that the prepared paper and the tracing are kept close together when put in the frame. Blue Printing. The drawing must be made on transparent material, for instance, tracing cloth or tracing paper. Place the tracing in the frame with the side on which the drawing is made next to the glass. Place the prepared side of the sensitive paper against the back of the tracing. Put the loose back into the frame with the padded side against the prepared paper, and fasten it up so that both paper and tracing are kept firmly against the glass. Expose to sunlight from three to six minutes, according to the brightness of the sun. Take the sensitive paper out of the frame and quickly put it into a tub of clean cool water and wash it off, and the drawing will appear in white lines on blue ground. Hang the print up by one edge so that the water will run off, and let the print hang until dry. INDEX. Acceleration due to gravity, 276. Addition, 5. of decimals, 14. of vulgar fractions, 7. of logarithms, 75. Algebra, notes on, 63. Allowable deflection, 266. Alloy which expands in cooling, 478. Ampere, 468. Animal power, 318. Angular velocity, 299. Angles and corresponding taper, 475. Area of circles, 196. tables of, 209. Area of segments of circles, 199. of parallelograms, 196. of triangles, 193. of triangles, formulas for, 176-177. of trapezoids, 196. of a circular lune, 202. of a zone, 202. Arms in gears, 393-396. B. Babbitt metal, 474. # Beams, deflection in, 254- 266. transverse strength of, 233-254. to calculate size to carry a given load, 247. round wooden, 251. not loaded at the center of span, 252. loaded at several places, 253. placed in an inclined position, 254. Bearings, 369-374. area of, 369. allowable pressure in, 369. proportions of, 370-374. Belts, 326-338. arc of contact of, 332. angle, 337. cementing, 328. lacing, 327. length of, 329. horse-power transmitted by, 329. crossed, 336. oiling of, 338. quarter turn, 336. slipping of, 337. tighteners on, 337. velocity of, 337. Bevel gears, 397. dimensions of tooth parts in 399. strength of, 411. Body projected at an angle upward, 279. in horizontal direction from an elevated place, 284. Burrs, iron, 438. copper, 438. Brass tubing, 441. Blue print paper, how to prepare, 480. Blue printing, 480. 481 482 INDEX. Case hardening, 479. boxes, 479. by cyanide of potassium, 480. Cap screws, 423. Center of gravity, 292. of gyration, 292. of oscillation, 292. of percussion, 292. Centrifugal force, 302. Chain Links, 323. Chordal pitch, 391. Circles, 152. area and circumference of, 209. Circular pitch, 375. Couplings, 369. Coupling bolts, 422. Compound proportions, 17. Cone, surface of, 204. volume of, 205. Constant for deflection, how to find, 264. Copper wire, notes on, 462- 467. resistance of, 465. weight of, 464. carrying capacity of, 466. Copper tubing, 441. strength of, 442. Cutting bevel gears, 401. Crane Hooks, 323. Cylinder, thickness of, 219. surface of, 203. volume of, 204. Cube root, 30. table of, 33. Crushing strength, 223. Clearance, 459. Coulomb, 469. Debt paying bv instalments, 89. Deflection in beams when loaded transverselv, 254. allowable, 266. Derrick, proportions of a two ton, 324. Difference between one square foot and one foot square, 193. Dimensions of U. S. stand- ard screws. 421. of Whitworth standard screws, 419-420. Diameter of tap-drill, 420. of screws at bottom of thread, 419. Diametral pitch, 377. Discount or rebate, 83. Division, 6. of decimal fractions, 15. of vulgar fractions, 10. of logarithms, 78. Drawing, problems in geo- metrical, 184. E. Emery and emery cloth, 474. Efficiency of machinery, 322. Electrical terms, notes on, 468-472. Ellipse, area of, 208. circumference of, 208. Elongation under tension, 215. Energy, kinetic, 287. Eye bolts, 438. Equations, 65. quadratic, 67. Equation of payments, 27. F. Factor of Safety, 274. Farad, 471. INDEX. 483 Force, energy and power, 285. of a blow how to cal- culate, 288. acceleration and motion formulas for, 288. centrifugal, 302. Formulas, 3. Fractions, addition of, 7, 14. subtraction of, 8, 14. multiplication of, 9, 15. division of, 10, 15. to reduce from one de- nomination to another, 11. to reduce a decimal to a vulgar, 13. to reduce a vulgar to a decimal, 12. Friction, 303. axle, 305. angle of, 306. co-efficient of, 304, 306. rolling, 304. rules for, 306. in machinery, 306. in pulley-blocks, 307. Frustum of a cone, 205. surface of, 205. volume of, 206. Frustum of a pyramid, 207. surface of, 207. volume of, 207. Fly wheels, 356. weight of, 356. centrifugal force in, 357. to calculate speed of bursted, 284. safe speed of, 360. safe diameter of, 360. Gage for sheet iron and sheet steel, 145. for wire, 146. for twist drills and steel wire, 148. Stubs', 148. Geometry, 149-152. Geometrical mean, 70. Gravity, 276. specific, 138. Gear teeth, 375-416. circular pitch of, 375. comparing circular and diametral pitch of, 380. cycloid form of, 382. dimensions of teeth, 380. Diametral pitch of, 377. involute form of, 387. strength of, 409. width of, 397. Gearing bevel, 397. how to calculate speed of, 322. dimensions of tooth parts in, 380. dimensions of tooth parts in bevel, 399. worm, 403. dimensions of tooth parts in worm, 403. elliptical, 408. Grindstone, weight of, 140, 475. speed of, 321. H. Hauling a load, 318. Henry, 471. Hubs in gears, 393. Horse power, 317. of a steam engine, 317. compound or triple en- gine, 317. of waterfalls, 318. Hot water heating, 454. Hydraulics, notes on, 443. Hyperbolic logarithms, 71, 126. I beam, strength of, 242. Impulse, 286. 484 INDEX. Interest, compound, comput- ed annually, 22. semi-annually, 24. simple, 19. tables, 20-25. by logarithms, 81. Involute, 192. gear teeth, 387. approximate construction of, 388. Inertia, 285. moment of, 293. Inclined plane, 308. Internal gear, 384, 391. International standard thread, 434. Jarno taper, 476. table of, 477. Joule, 470. K. Keys, proportions of, 368. Kinetic energy, 287. L. Lag screws, 439. Lathe centers, 475. Lathe mandrels, 475. Lathe and planer tools, sizes of, 475. Laws, Newton's, 276. Logarithms, 71. table of, 90. hyperbolic, 71, 126. table of, 126. Levers, 292. Lune, circular, 202. ^ Lubricant for milling and drilling, 477. M. Machine screws, 423-433. Machinery, efficiency of, 322. power required to drive, 319. Machinery, speed of, 32Q. catalogs, 473. Mathematics, notes on, 1. Manila ropes, transmission of power by, 344. transmission capacity of, 345. preservation of, 347. weight of, 346. strength of, 222. Marking solution, 477. Mass, 286. Multiplication, 6. of decimals, 15. of vulgar fractions, 9. of logarithms, 77. Mechanics, 276. Mensuration, 193. Metric system of weights and measures, 135. thread with inch divid- ed lead-screw, how to cut, 436, 437. Modulus of elasticity, 213- 216. how to calculate, 213 3 255, 265. Moments, 292. Moment of inertia, 293. in rotating bodies, 298. polar, 297. Momentum, 286. Morse taper, 476. Motion, Newton's laws of, 276. down an inclined plane, 283. N. Nails, 438. Napierian logarithms, 71, 126. Newton's laws of motion, 276. Ohm, 469. Oiling of belts, 338. index. 485 P. Parallelogram of forces, 316. Partnership, 27. Payments, equation of, 27. Pillars, safe load on ' 226- 229. hollow cast-iron, 228. Pillars, wrought iron, 231. weight of, 228, 232. Pipes, wrought iron, 440. Polar moment of inertia, 297 Planer tools, sizes of, 475. Polygons, 149. Posts, wooden, 231. Power, animal, 318. of man, 319. required to drive ma- chinery, 319. Pulley-blocks, 307. differential, 308. Pulleys, 348. How to calculate size of, 321. stepped, 350. correct diameter of, 352. for back-geared lathes, 354. Pyramid, frustum of a, 207. slanted area of a, 206. total area of a, 206. volume of a, 206. Pressure on bearings caus- ed by the belt, 333. Press fit, 473. Pressure of fluid in a ves- sel, 443. Produce, weight of, 135. Problems in geometrical drawing, 184. Progressions, 68. Proportion, 16. compound, 17. Quadratic equations, 67. Quantity of water discharg- ed by pipes, 447. of water required to make any quantity of steam, 454. R. Radius of gyration, 293. Radical quantities expressed without the radical sign, 32. Ratio, 16. Reciprocals, 32. table of, 33. Results of small savings, 26. Rivets, iron, 483. copper, 483. Rope, manila, 344, 345, 347, 346, and 222. wire, 338, 340, 342, 343, and 222. Running fit, 474. S. Sand paper, 474. Safety, factor of, 274. Savings, results of small, 26. Sector of a circle, area of, 198. Segment of a circle, 199. length of arc of, 199. area of, 199, 201. of a sphere, volume of, 208. Sinking funds and savings, 84. Soda water for drilling, 477. Solder, 477. Soldering fluid, 478. Subtraction, 5. of decimal fractions, 14. of vulgar fractions, 8. of logarithms, 76. 486 INDEX. Screws, 311, 417. A. S. M. E. standard machine, 427-432. "pitch," "inch pitch," of worms and screws, 417. machine, 423. international standard, 433-434. U. S. standard, 418-421. Whitworth standard, 419. wood, 423. diameter at bottom of thread of, 419. round and fillister head, cap, 422. Screw-cutting by the engine lathe, 417. multiple threaded, 417. Shafting, 360. allowable deflection in, 362. classification of, 365. horse power of, 365, 366. not loaded at the middle between bearings, 361. torsional strength of, 363. torsional deflection of, 364. transverse strength of, 360. transverse deflection of, 361. Shafts for idlers, 367. Shearing strength, 272. Shop notes, 473-480. Shrink fit, 473. Shrinkage of castings, 478. Spelter, 478. Speed of machinery, 320. of handsaws, 320. of drilling machines (iron), 320. of emery-wheels and straps, 321. of grindstones, 321. of lathes (for iron and wood), 320. of milling machines, 320. of planers, 320. Specific gravity, 138. Sphere, surface of a, 207. volume of a, 208. Springs, helical, 474. Steam, notes on, 450-459. properties of saturated, 457. expansion of, 457-458. weight of, required to boil water, 456. heating, 453. Strength of materials, 213- 275. tensile strength, 213. crushing, 223. transverse, 233. torsional, 266. shearing, 272. cast iron, 214. wrought iron, 214. cylinders, 219. flat cylinder heads, 220. of dished cylinder heads, 220. of chains, 222. of bolts, 218. of hollow sphere, 221. gear teeth, 409-417. of wire rope, 222. of manila rope, 222. of beams when section is not uniform, 243. of square and rectangu- lar wooden beams, 245. Square root, 29. table of, 33. by logarithms, 78. INDEX. 487 T. Taper per foot and corre- sponding angle, 476. Taper, Jarno, 477. Morse, 476. Tank, to calculate number of gallons in a, 202. Tensile strength, 216. Thermometers, 460, 461. Torsional deflection, 271. in hollow round shafts, 271. strength, 266. strength in hollow round shafts, 270. Transverse strength, 233. Tubes, bicycle, 442. boiler, 442. brass, 441. copper, 441. U. Upward motion, 279. U. S. Standard screws, 402. Units derived, 471. Value of various metals, 139. of the trigonometrical functions for some of the most common angles, 156. low pressure steam for heating purposes, 454. Velocity, angular, 299. of efflux, 443-444. of water in pipes, 445. Volt, 468. W. Water, notes on, 448. Water, measure of, 138. weight of, 138. required to condense one pound of steam, 455 Watt, 469.' Weights and measures, 133. metric system of, 135. Weight of iron, square or round, 143. flat, 144. of any shape of section, 141. of sheet-iron of any thickness, 141. of steel, 142. of casting from weight of pattern, how to calculate, 141. metals not given in the tables, 141. of cast-iron balls, 142. and value of metals, 139. of various materials, 140. of liquids, 140. Wire gages, 146. Wire rope transmission, 338. capacity of, 340. deflection in, 342. weight of, 343. strength of, 343. Wood screws, 423. Z. Zone, circular, 202. AUG 5 One copy del. to Cat. Div. *ys a" wm