«i i! ii PRACTICAL MECHANICS, THE APPRENTICE R FIRST BOOK FOR MECHANICS, MACHINISTS, AND ENGINEERS. BY OLIVER BYRNE, MATHEMATICIAN, CIYIL, MILITARY, AND MECHANICAL EXGiXEER. Author of " The Practical Model Calculator ;^^ Compiler and Editor of the " Dictionary of Machines, Mechanics, Engine-work, and Engineering/;'''' '^ The Pocket Companion, ' for Machinists, Mechanics, and Engineers ;^^ ''The Practical Cotton- Spinner ;^^ '- The Practical Metal-worker^ s Assistant ;^' author and inventor of the " Calculus offl'hi-m,'" a new Scienc-a, a sidjstitute for the Differential and Integral Calculus; '■'■ Th< Doctrine of Proportion ;''^ " The Elements of Euclid, hy Colours, etc , etc., N E W Y RK: PUBLISHED BY PHILIP J. COZANS, No. 107 NASSAU STREET CORNER OF ANN. 1860. ? S^ ^..^i^- ^//J^^ (i-- A Entered according to Act of Congress, in the year 1859. By PHILIP J. COZANS, In the Clerk's Office of the District Court of the United States, for the Southern District of New York. STEREOri'PED BY VINCENT L. DILL, 128 Fulton Street, (Sun Building) . iH-W PREFACE. There being no Elementary treatise on Practical Me- chanics in the English language, the present work is de- signed to supply the deficiency. It may be asked how can this be, while there exist so many elementary treatises on the theory and application of Mechanics, or why not ft-ans- late, or copy the great French works, as the German and other nations do. To understand the elementary treatises on the theory and application of Mechanics employed as class books in our schools and colleges, requires consider- able mathematical skill, and after all, the elementary know ledge acquired is of little value to the practical man. On the general theory of Mechanics, there are also many popu- lar works clear of mathematical formula and serviceable to the general reader, but of no practical value. French ele- mentary writers on practical Mechanics, take for granted that they are writing for those who understand the theory of Mechanics, and when these works are copied or trans- lated, they are far from being elementary. As practical mechanics is based upon what is termed the principle of work, to be able to distinguish with ease, units of work from other units, would greatly facilitate the acqui- sition of this branch of knowledge ; I have, therefore, in- troduced a new and simple notation to effect this object, which FreixCh writers and their copyists have neglected. The unit of work is equal to the labor required to raise VI PREFACE. a pound weight through*tlie space of one foot, against the direct action of the force of gravity. If a man takes a pound weight in his hand, and raise it one foot in the di- rection of a plumb-line, he will perform a unit of work. 1 lb. raised 24 feet high = 24 units of work 2 lbs. raised 12 feet high = 24 units of work . 3 lbs. raised 8 feet high = 24 units of work 4 lbs. raised 6 feet high = 24 units of work 6 lbs. raised 4 feet high = 24 units of work 8 lbs. raised 3 feet high = 24 units of work 12 lbs. raised 2 feet high = 24 units of work 24 lbs. raised 1 foot high = 24 units of work. To distinguish units of work from other units, I simply write or set down the numbers that represent them, inclin- ing to the left, instead of the ordinary way, thus : ^4: represents 24 units of work ; 8 X 3 = *i4 ; that is the same as saying, 8 lbs. raised 3 feet high is equal to 24 units of work. ^4:' ; represents 24 units of work done in a minute. ^4'^ ; represents 24 units of work done in a second. 33000 =^11^; represents 33,000 units of work done in a minute, equal one horse power. I have been constantly asked the question, by those whose design it was to become machinists and engineers, " What is the best elementary work on Practical Me- chanics ?'' ; instead of answering, *' there is no such work in the English language,^' I can now say, buy '' The Ap- PRENTICE,'' the title given to this work, to intimate that it is designed for persons who wish to acquire a knowledge of Practical Mechanics. OLIVER BYRNE. PEACTICAL MECHANICS CHAPTER I. On the unit of work with, and without reference to tae unit of time. Work is measured by a unit, like length, weight, time, &c. To raise one pound a foot high against the earth^s gravity is a unit of work ; it is clear, then, to raise 5 lbs. a foot high is to perform 5 units of work, and to raise 5 lbs. four feet high equal 20 units of work. Since resistance and pressure of every kind may be expressed in pounds, it follows that the unit here described may be made to mea- sure every kind of work. It will be presently shown that a unit of work is performed whenever one pound pressure is exerted through a space of one foot, no matter in what direction the space may lie. In developing this subject it is necessary to distinguish units of work from other units ; this I propose to do by writing or setting down the numbers that represent them inclining to the left instead of in tlic ordinary way, thus, *i5 represents 25 units of work. ^5' will represent 25 units of work done in a minute. 13" will represent 73 units of work done in a second. 8 PRAjCTICAL MECHANICS. Sometimes the minute is taken for the unit of time, and 33 minutes is marked 33^ ; in other cases, the second is taken as the unit of time, and 45 seconds is marked 45". 45" represents 45 seconds ; but ^5" units of work done in a second of time. a' represents a minutes, but a' represents a units of work done in one minute. EXAMPLES. Question I. Required the units of work expended in rais- ing a weight of 50 lbs. to the height of 31 feet. Units of work in raising 1 lb. 81 feet = SI. .-. 50 X 31=155t>. the required units of work. Ques. 2. The ram of a pile-driving engine weighs half a ton, and has a fall of 17 feet, how many units of work are performed in raising this ram ? Half a ton = 1120 lbs. Units of work in raising 1 lb. IT feet = n. .-. 1120X n = 1904c0, the units of work required to raise 1120 lbs. to a height of It feet. Ques. 3. How many units of work are required to raise 7 cwt. of coal from a pit whose depth = 13 fathoms ? 7 X 112 = 784 lbs. 13 X 6 = 78 feet, in 18 fathoms. Hence the work consists in raising 784 lbs. to the height of 78 feet. ••. 784 X 78 = 6115^. Ques. 4. If the weight of a man be 183 lbs., and if he as- cends a perpendicular height of 20 feet, he does work in raising himself, what are the number of units ? In this operation the man raises the weight of his own body, .•. 183 X 20 = 3«6O. amount of work. PRACTICAL MECHANICS. 9 If this man were to descend in a bucket, it is clear, he would perform the same work upon a counterpoise weight when he has descended 20 feet. Ques. 5. How many units of work will be required to pump 8000 cubic feet of water from a mine whose depth = 500 fathoms ? A cubic foot of water weighs 62.5 lbs. 8000 X 62.5 = 500,000 lbs. 500 fathoms = 3000 feet. Consequently, the work is to raise 500,000 lbs. a perpendicular height of 3000 feet. Work = 500,000 X 3000 = 1500000000. Ques. 6. A horse draws 150 lbs. out of a well, by means of a rope going over a fixed pulley, moving at a rate 2| miles an hour, how many units of work does this horse per- form a minute, the friction being neglected ? 2i X 5280 = 220 feet passed over per minute. 60 •. Work per minute = 150 X 220 = 33000'. Work per second := 500'^ THE UNIT OF WORK REFERRED TO A UNII OF TIME. A unit of work, or 1, represents 1 lb. raised 1 foot. A unit of work in a minute, or 1', represents 1 lb. raised a foot high in a minute. A unit of work in a second, or 1'', represents 1 lb. raised a foot high in a second. It has been assumed that a horse is capable of raising 33000 Ids. a foot high in a minute, or to perform 33000 units of work in a minute, Hence a horse power = 33000' = 1 ^• Whether this power is greater or less than the power of a horse it matters little, w^hile it is a power so well defined. To raise 10,000 lbs. a foot high in a minute, would be a more convenient unit to measure bv. 10 PRACTICAL MECHANICS. EXAMPLES. Question 1. How many horse power would it take to raise 3 cwt. of coal per minute from a pit whose depth = 110 fathoms ? Depth =: 1 1 X 6 = 660 feet. 3 cwt. = 112 X 3 = 336 lbs. 660 X336=:^^1'560^ Since a horse power := 3300©' ^^^ ^^ -- & ^. il!i, the required power. Ques. 2. How many horse power is required to raise 2200 cubic feet of water an hour from a mine whose depth = 63 fathoms ? 13^500 — 2200 X 62.5 = weight of water in lbs. 63 X 6 = 378 feet, depth of mine. 137500 X 378 60 866250 84>6*i50'. = ^6i-^- 33000 The proposed notation, must be borne in mind, 866^50' sig- nifies 866250 lbs., raised one foot high in a minute, and *i6i ^^-y represents 26^ horse power, or 26^ times 33000 lbs. raised 1 foot high in a minute. Ques. 3. A winding engine is moved by 4 horses, what weiglit of coal will be raised per hour from a pit whose depth = 200 feet ? Work of the 4 horses in an hour, = 4 X 33000 X 60 = ^»aoooo. Work in raising 1 lb. of coal 200 feet = ^0«. Since it takes 200 units of work to raise a pound of coals, 7920000 ^^_^ „ ^^^ --=: 39600 lbs. Consequently four horses, will raise 39600 lbs. of coal or of any thing else, a height of 200 feet in an hour. PRACTICAL MECHANICS. 11 Ques. 4. In what time will an engine of 10 horse power, raise 5 tons of material from the depth of 132 feet ? 5 tons =11200 lbs. Work of the engine per minute, =: 330000^ = 33000 X 10. Work of raising 5 tons, = 11200 X 132 = 14:T[8>400, Because the engine performs 330000 units of work a minute, 1478400 330000 4.48 minutes. Ques, 5. How many cubic feet of water will an engine of 36 horse power (36 ^^•) raise in an hour from a mine whose • depth is 40 fathoms ? 40 X 6 = 240 feet. Work in an hour, = 36 X 33000 X 60 = 11^80000. Work to raise a cubic foot of water 240 feet, = 240 X 62.5 = 15000. ^l^8oooo 15000 = 4752 cubic feet. Ques, 6. From what depth will an engine of 22 horse ]:)ower, raise 13 tons of coal in an hour ? Work done by the engine in an hour, 22 X 33000 = ^^6000. 13 tons = 29120 lbs. 29120 r-: 24.9 feet. Ques. 7. An engine is observed to raise 7 tons of mate- rial per hour, from a mine whose depth is 85 fathoms ; re- quired the horse power of the engine, supposing | of its work to be lost in transmission ? N „ , . 2240X^X85X6 ..bo^^^ Work per mm. z^^ ^ "^ 133^80. 12 PRACTICAL MECHANICS. Since f of the work of fhe engine only go to raise the material, .*. I of 33000' = ^^500 , the units of useful work of one horse power a minute. *n5o«' 4^ii8i9. Ques. 8. Required the horse power of an engine that would supply the city of Brooklyn with water, working 12 hours a day, the water to be raised to a height of 50 feet ; the number of inhabitants = 130,000, and each person to use 5 gallons of water a day. The standard gallon of the United States weighs 8^ lbs., nearly. 130000 X 5 X 8i _ 8125 X 25 lbs. 12 X 60 ~ 27 the pounds of water to be pumped in a minute. 8125 X 25 27 X 50 = 3^6151' 33000 The horse power required ? Ques. 9. What is the horse power of an engine that pumps from three different levels, whose depths are 40, 70, and 90 fathoms respectively ; from the first 20 cubic feet of water are raised per minute, from the second, 10 cubic feet, and from the third, 35 feet, allowing i the w^ork of the engine to be destroyed by useless resistance ? T \ ^ 33000^ t of 33000' = *a*iOOO', the effective power of the engine. 1st level, work = 62.5 X 20 X 240 = 300000^ 2nd '' '' = 62.5 X 10 X 420 = i6*i500'. 3rd " " = 62.5 X 35 X 540 = 1181^50^ n43150 a^ooo' = ^9^^6. PRACTICAL MECHANICS. 13 Ques. 10. There were 6000 cubic feet of water in a mine whose depth is 60 fathoms, when an engine of 50 horse power began to work the pump ; the engine continued to work 5 hours before the mine was cleared of the water ; required the number of cubic feet of water which had ran into the mine per hour, supposing ^ of the work of the en- gine to be lost by transmission ? 4) 33000^ ^>4^50' effective power of the engine a minute. It must not be forgotten that ^41150^ written in the contrary direction, signifies 24750 units of work in a minute or 24750 lbs., raised one foot high in a minute ; so that writing down the numbers in this peculiar manner saves much written ex* planation. ^4l150' X 50 X 5 X 60 = 31^^50000, effective work of the engine in 5 hours. 62.5 X 60 X 6 = *i^50, work in raising one cubic foot of water to a height of 60 fathoms = 360 feet. 3T^^OOOO_ ^^500 - ^^^^^-^^^ cubic feet of water pumped in 5 hours. 16544.44 6000. 5) 10544.44 Water run in during 5 hours. 2108.88 Cubic feet run in in one hour. Ques. 11. A forge hammer weigns 300 lbs., makes 100 lifts a minute, the perpendicular height of each lift = 2 feet ; what is the horse power of the engine that gives pow- er to 20 such hammers ? Work of each Uft = 300 X 2 X 20 = l^OOO. Work in 100 lifts, that is, in one minute, l^OOO X 100 == l^OOOOO 33001>' 36 \Mi\, 1 ! PRACTICAL MECHANICS. Ques. 12. All engine* of 10 horse power (lO^), raises 4(}00 lbs. of coal from a pit 1200 feet deep in an hour, and also gives motion to a hammer which makes fifty lifts in a minute, each lift having a perpendicular height of 4 feet, what is the weight of the hammer ? Work done by the engine in one minute, = 33000 X 10 = 330000 . The units of work performed, in raising the coals, per minute = '^'\^ ''''- = 80000'. 60 Work engaged in raising the hammer per minute, = 330000' — 80000' = ^50000 . Work per minute, in raising one lb. of hammer 4 feet high, 50 times a minute, = 1 X 4 X 50 = *iOO^ — ^OO^ =1250 lbs, the weight of the hammer required. CHAPTER II. On the work of living agents. The laboring force of animals, varies very much witli the way in which their muscular strength is exerted ; and also, with the rate at wliich they labor. Tlie following little table shows the greatest amount of eflective work that a laboring man can perform under the diffen^nt modes in which he may exert his muscular power. WORK DONE BY A MAN PER MINUTE, WHEN HE WORKS 8 HOURS A DA.Y. A man in raising his own body, >4*i50 . A man in working at a treadmill 31M>0'. PRACTICAL MECHANICS. 15 A man drawing, or pushing horizontally, Sl'lW. A man pushing or pulling vertically, *AS8t>'. A man turning a handle, *i4JOO'. A man working with arms and legs, as rowing, 4000^ UNITS OF WORK DONE BY A MAN PER MINUTE, WHEN HE WORKS 6 HOURS A DAY. A man in raising material with a pulley 1560'. A man in raising material with the hands, 1 4L^O', A man in raising material on back, returning empty,. 'I'liC. UNITS OF WORK DONE BY A MAN PER MINUTE, LABORING 10 HOURS A DAY. Raising materials with a wheel-barrow on ramps. . . .^^O'. Throwing earth to the height of 5 feet ^^O'. UNITS OF USEFUL WORK DONE BY A MAN IN A MINUTE, RAISING WATER BY DIFFERENT ENGINES, WORKING 8 HOURS A DAY. With a windlass from deep w^ells, ^£560^ With an upright chain pump, nSO'. With a tread-mill, SnG'. With a Chinese wheel, *1161^ With an Archimedian screw, 1 505^ Raising water from a well, with rope and pail,. . . .105>4'. ^VVORK OF BEASTS. The imaginary horse by which the power of steam en- gines are measured, is supposed to do more work than the general run of horses are able to perform. The work of a horse of average strength is about 22000 pounds raised one foot high in a minute, which according to my notation, is expressed thus : — A mule will perform f the work of a horse, = fiAtOftGf', 16 PRACTICAL MECHANICS. 4n ass will perform about } the work of a horse = 4:>400^ In a common pumping engine a horse of average strength will do only 17550 useful units of work in a minute. From ^iOOO' Take n550 4l>450', lost in this case by friction and useless resistances. EXAMPLES. Question 1. How many cubic feet of clay, weighing 100 lbs. per cubic foot, will 20 men throw 5 feet high in a day of 10 hours long ? From the foregoing table the number 4l10' is found. Work in a day, = 4:10' X 60 X 10 X 20 = 5640000'. Work in raising one cubic foot, = 100 X 5 = 500'. Consequently the number of cubic feet raised in one day by 20 men, 5640000 500 = 11280. Ques, 2. How many bricks will a man raise in a day six hours long to a height of 30 feet, supposing the weight of a cubic = 125 lbs., and 17 bricks to form a cubic foot? For a man working in this way, the foregoing tables give ll^O'. .-. 11^6' X 60 X 6 = 405360, work done by a man in 6 hours. 125 X 30 = 3150, work done in raising a cubic foot 30 feet high. 405360 ,^^^^^ ••• 3150 ^ '"'-'''^ cubic feet raised in a day of 6 hours. Number of bricks = 108.096 X H ^ 1838. PRACTICAL MECHANICS. 17 Ques. 3. How many cubic feet of water will a laborer working with a bucket and rope, raise from a Avell whose depth is 16 feet? In the table will be found 1054:', for a man working in this manner. t05>4' X 60 X 8 ::^ 505t)'iO, work done in 8 hours. The work done in raising a cubic foot of water 16 feet high, =.-. 6.25 X 16 -^ lOOO. Hence the cubic feet raised in eight hours by a man, 5059'A0 lOOO ^ 505.92. Ques. 4. What weight is a man, working at a tread-mill, able to raise in a day of 8 hours, from a depth of 110 feet ? Tabular number = nSO'. 1130' X 60 X 8 =1 830^00, work done in a day. The work required to raise one cwt. or 112 lbs. a height of 110 feet will be, 112 X 110 = 1^3^ ; the weight that may be raised by a man working at a tread-mill. Qices, 5. How many tons of coal would a man raise, work- ing with a wheel and axle, from a pit whose depth is 20 feet, taking for granted, according to tlie tabulated state- ment, that a man so circumstanced, can perform 2600 units of work in a minute ? The work done in one day of 8 hours, ^ *S600' X 60 X 8 z= l^^HOOO. The work to raise one ton, = 2240 X 20 -:zz >^4800 ; li>48000 ^^^^ ^ •*• ~^A80«- = ^^-^^^ ^^"' raised in a day of eight hours. 18 PRACTICAL MECHANICS. Ques.{&, The ram of a pile engine weighs 7 cwt., and has a fall of 23 feet, how many strokes a day, will four men give, working at a wheel and axle ? Work done in one day of eight hours, =z ^600' X 60 X 8 X 4 = 4:»1>*£000. The work of one stroke, = 112 X t X 23 = 1803^ ; •• ;ftft^*> =277 strokes nearly. The force with which animal pull decreases with their speed ; the relation between the traction and speed of a horse, is express- ed with considerable accuracy by the following formula, ^ =:= 250 — 41f r ; in which t — the traction in lbs. and r = the rate in miles per hour. Hence if the speed be 3 miles an hour, the traction will be 125 lbs., for 250 — 41f X 3 =r 125 lbs. If the speed be one mile an hour, the traction will be =: 208 J lbs., for 250 — 41f X 1 = 208i lbs. From this formula, it is easily shown that a horse will do the greatest amount of work when he travels at the rate of 3 miles an hour, in fact |250 — 41f r i X r becomes a maximum when r = 3. If a body move slowly along a horizontal plane, the resistance to be overcome is called friction. Experiment has determined that this resistance on a given surface, is a fractional part of the weight of the- body moved, and it has also been found that any change in the rate of the motion of the body does not affect the resistance due to friction ; nor is the amount of friction altered by the extent of tiie rubbing surfaces. When a wagon or a cart is drawn along a good common road, the resistance of friction is about 3V of the whole load, so that a horse in order to draw 3120 lbs. along a road, must pull with the force of 104 lbs., for — 104 lbs. 30 PRACTICAL MECHANICS. 19 A horse with this traction, according to the foregoing formula, moves at the rate of about 3 J miles an hour, for ^ = 250 — 41f r, becomes 104 = 260 — 41f r, .*. r = Og. A carriage on a Rail road, only requires a pressure of about 3 J^ part of the moving weight to give it motion, or from 4 to 8 lbs. a ton. The fractions ^\ for common roads, and ^i^ for rail-roads are called "ihe coefficients of friction, as these coefficients become small- er, ibe rubbing surfaces become smoother. Let W be a weight drawn on a hor- izontal plane H R, by means of a \ -^v- weight P, attached to a cord A, go- — jj """^ — ing over a fixed pulley C ; then the weight P just necessary to draw or move W, along the plane, will be equal to the resistance of friction. In the case of a Rail-road, if W = 150 tons, P will be = 900 lbs., when the "" coefficient of friction = j--,3__ or 6 lbs. a ton. It is very evident that whatever distance P descends, the weight W, will be drawn along the plane H R, the same distance ; hence the units of work done in moving W, will be the weight of P iu pounds multiplied by the distance in feet through which it descends, or the resistance of friction in pounds, multiplied by the space in feet over which W, is moved. The work of every machine is consumed by the work done, or by the useful work, together with the useless work, or the work des- troyed by the friction of the parts of the machine. I will here ex- plain one of the most beautiful laws of motion ; — when the work applied exceeds the work consumed, the redundant work goes to increase the speed of the parts of the machine, and at the same time like the fly-wheel, acts as a reservoir of work. T/iis accelera- tion goes on increasing until the work of the resistances + the useful work = the work applied ; and then the motion of the machine he- comes uniform. For example, in a Rail-road engine and train, at first the work of the engine exceeds the work of the resistances, and hence the speed of the engine goes on increasing ; but, as the speed increases, the work of the resistances also increases, so that ultimately the engine attains a nearly uniform motion, which is called the greatest or maximum speed, and then the work destroyed by the resistan- ces, will be exactly equal to the work apphed by the moving power. 20 PRACTICAL MACHANICS. Ques. 7. Required the effective horse power of a loco- motive engine, which moves at a steady speed of 23 miles an hour upon a level rail, the weight of the train being 100 tons, and the friction 5 lbs. a ton ? Put 2: — the reqmred horse power. The work of the engme per minute, =iz X 33000' ; The resistance of friction, = 5 X 100 = 500 lbs. The distance moved per minute, 23 X 5280 _. _ . , . ~ 60~ ' "^ ' Work to overcome the friction, = 2024 X 500 = lOl^tOOO^ But as the speed of the train is uniform, the work of the resistances will be equal to the effective work of the engine : 33000 30 KT Ques, 8. What is the rate in miles per hour, of a train of 80 tons will be drawn by an engine of ^0 ^- ; the friction 8 lbs. to the ton. Call X the uniform speed in miles per hour ; Work used in moving the train x miles, = 80 X 8 X 5280 x ; this is the work done by the engine in an hour. But the work done by the engine in an hour, will also be expressed by 33000' X 70 X 60 ; .•. 33000' X 70 X 60 = 80 X 8 X 5280 x ; 33000' X 70 X 60 80 X 8 X 5280 = 41.02 miles. Ques. 9. An engine of 4:8 ^-^ moves with a maximum speed of 33 miles an hour, on a level rail ; required the gross load of the train, friction 6 lbs. per ton ? PRACTICAL MECHANICS. 21 Let X be the gross weight of the train in tons, then, the work consumed per hour in moving the train, = cc X 6 X 33 X 5*A80. Work of the engine per hour, ^ 48 X 33000^ X GO When the speed is uniform or at its maximum, :r X 6 X 33 X 5^80 = 4^8 X 33000 X 60 ; 48 X 33000 X 60 6 X 30 X 528Q — 90 I?- tons. Qi^s. 10. In what time will an engine of 66 horse power moving a train of 200 tons, complete a journey of 100 miles, friction 5 lbs. per ton ; rails horizontal ? Work expanded in moving the train 100 miles, = 100 X 5280 X 200 X 5 = 5^8000000. Work of the engine per hour, ^ 33000' X 66 X 60 X 430680000 5^8000000 130080000 = 4.04 hours. Ques. 11. What work per minute will a horse perform when traveling at the rate of 2| miles an hour ? I have before shown that the traction of a horse moving at the rate of 2^ miles an hour, = 250 — 41f X 2i z= 145| lbs. 2^ miles ^^- 13200 feet per minute, 13200 = -6^ = 220 feet. Hence the work of this horse per minute, = 145f X 220 =z 3'i083i^ Ques. 12. What load will a horse draw, traveling at 3 miles an hour upon a plank road, whose friction is yjo of the whole load, the road being horizontal ? 22 PRACTICAL MECHAl^ICS. The traction at this speed is 125 lbs., for 250 _ 41f X 3 ::= 125, The gross load must be 100 times this weight, =: 125 X 100 = 12500 lbs. Qms. 13. Suppose a horse to be able to perform 33000 units of work in a minute on a horizontal road, whose fric- tion is sV of the whole road ; required the load and rate per hour ? Traction =(250 — 41fr) pounds. Space passed over in a minute, 5280 r 60 r feet. Then ( 250 — 41fr)88r = the units of work performed by the horse in a minute = 3300tS^ From this quadratic equa- tion, the value of is found to be 3 miles ; .*. Traction =^ 125 lbs, and the load will be 125 X 20 = 2500 lbs. Ques. 14. At what rate will a horse draw a ton, on a road whose coeflBcient of friction is -^^ ? 2240 Traction = =70 lbs. .♦. 10 = 250 — (41f) r; 250 — 10 ... r=— — --=4.32, the rate is 4.32 miles an hour. Ques, 15. When a horse exerts a traction of 41f lbs., his rate of motion is 5 miles an hour, what gross load will he draw on a level road whose coefficient of friction is ^\, and what work will he perform a minute ? Load = 30 X 41f =r 1250 lbs. Distance moved in feet per minute, 5 X 5280 ^^^. ^ = 440 feet. Work = 41| X 440 = t8333i'. PRACTICAL MECHANICS. 23 Ques, 16. V>^liat must be the horse power of an engine, to cut 6600 square feet of planking in a day of 10 hours long ? It requires according to experiment, about 29000 units of Avork to saw a square foot of green oak planking, there- fore, The work in cutting 6600 square feet, = ^«0«0 X 6600 = l»lilOOOOO. ... i?-«-oo=,.^„,, 310000' „ , , *"•* 33000 = ^* Ques. 17. An engine of 24 effective horse power, cuts 144 square feet of American live-oak in 5 minutes, how- many units of work are consumed in cutting a square foot ? The work of the engine in five minutes, = 33000' X 5 X 24 -^ 31>eOOOO, the work destroyed in cutting 144 square feet. Therefore the units of work expended in cutting one square foot, 3060000 144 ^^50o. CHAPTER III. On the moving of bodies on inclined planes^ and the raising of materials. If a surface be supposed without friction, the units of work performed by moving a body along it, is equal to the product of the weight of the body in pounds, by the verti- cal height in feet through which it is raised. 24 PRACTICAL MACHANICS. To move a body along the path A C E G I, may be imagined the same as to move it along and up an infinite number of steps, resembling- A B, B C, C D, D E, &c. And since fric- tion is neglected, there is no work required to move the body in a ho- rizontal direction, all the work is ex- pended in raising the body in the per- pendicular direction : hence the units of work required to raise a body from L to I, equals the units of work re- quired to move it along ths path A C E G I, the friction of the path being neglected. The principle here exDlained holds true for an inclined plane. EXAMPLES. Question 1. A train of 200 tons ascends an incline which has a rise of 5 feet in 1000, with a uniform speed of 30 miles an hour, what is the effective power of the engine, the friction being 5i lbs. to the ton ? The pressure of a body on an inclined plane is nearly equal to its weight, when the inclination is small, hence the work due to friction, may be found bv the method ex- plained in the last chapter. Speed of train per minute =: 2640 feet. Weight of train in lbs. = 448000. 5 1 . , rise 1000 of the rail in one foot. 1 200 200 X 2640 = 13.2 feet, the rise of the rail in 2640 feet. Consequently the wliole weight of the train is raised 13.2 feet every minute in opposition to gravity. Work done to gravity per minute, = 448000 X 13.2 11^ r>«l»«0«^ PRACTICAL MECHANICS. 25 Work done to friction per minute, := 200 X H X 2640 ^ *i»OiiLOOO. Total work of the engine per minute, A Ques, 2. A train of 330 tons, ascends an incline that has a rise of j in 100, what is the maximum speed with an en- gine of 120 horse power ; the friction of the rail 8 lbs. per ton? Let X be the speed of the train in feet per hour. The rise in each foot of rail, i^ I divided by 100 i:^ sio ; the rise of the rail in X feet, X "^ 500 ' The work due to gravity in an hour, = 330 X 2240 X ~r- = 14l18A x ; Work due to friction in an hour, 330 X 8 X :?^ = ^64:0 x. It is evident that the work due to gravity in an hour, added to the work due to friction in an hour, must be equal to the work done by the engine in an hour, .-. 33000' X 120 X 60 := ^640 x x 14:T[8.>4 x, ^3^600000 >4148A ^ 57692 feet. Ques. 3. A engine of 50 horse power ascends a gradient having a rise of f in 100 ; with a steady speed of 20 miles an hour, what is the w^eight of the train in tons, the fric- tion per ton being 8 lbs. ? Let X be the weight of the train in tons ; the work required to overcome friction. 26 PRACTICAL ]\IECHANICS. = 1760 XS X x = 14l«80 x^ ; since a speed of 20 miles an hour =:=: 1760 feet a minute Rise of rail in a foot, = ^ divided by 100 :=:= ^^j^ ; Rise of the rail in 1760 feet, Work due to gravity, = 2240 XxX 13.2 =^ ^9568 x' Work of the engine per minute, = 33000' X 50 = 165000 , ^ Consequently, 14.080 x^ + ^9568 x' = 1650000' 1650000 4l36>48 = 37.8 tons. Ques. 4. A train of 100 tons descends a gradient having a rise of i of a foot in 100 feet, at a uniform speed of 60 miles an hour, what is the horse power of the engine, fric- tion reckoned at 8 lbs. a ton ? 60 miles an hour ^^ 5280 feet a minute ; Work due to friction, = 100 X 5280 X 8 11=: 4l^^>5lOOO^ Rise in one foot, := i divided by 100 = 4 J©. Rise in 5280 feet, or in one minute, Work due to gravity, — 224000 X 13.2 — 'iUoGSOO'. In this case gravity acts with the engine ; .'. 4^*A>4«00' minus 'iOSftSOO' ::= l^GTiOO' •33«iW »8 ^.\. PRACTICAL MECHANICS. 27 Ques. 5. If a horse exerts a traction of 144 lbs., what weight can he pull on a plank road, up a hill that has a rise of 3 feet in 190 feet, supposing the coefficient of fric tion to be 2V ? Work of the horse in moving over 190 feet, = 190 X 144=1^1360. The work of friction in moving x lbs. over 190 feet, -^ X 190 = 9.5 X, supposing X to be the required load ; The work due to gravity, when tne load is moved over 1 90 feet = 32;; .•• 9 2^ + 3:^ = ^1360; il360 1^.5 2188.8 lbs. Ques. 6. What would be the backward pressure of a horse in going down hill, that has a rise of 15 feet in 369 with a load of 2000 lbs. supposing the coefficient of friction to be 3V ? Work due to gravity in moving 2000 lbs. 369 feet, = 2000 X 15 = 30000 ; Work due to friction in moving the load 369 feet, 2000 = ""30^ ^ ^-^^ '"= ^4l600 ; From 30000 Take ^4:600 54lOO 54cOO 369 = 14.6 lbs. Ques. 7. How many horses would it take to draw a load of 6 tons up a hill having a rise of 2^ in 100, supposino; the resistance of friction to be fV of the whole load, and the traction of eacli horse, 160 lbs. Let X be the number of horses. 28 PRACTICAL MECHANICS. Work due to friction in moving 6 tons over 100 feet, 6 X 2240 12 X 100 = ll^OOO. Work of gravity in moving 6 tons over 100 feet, = 6 X 2240 X 21 ^ 33600. The work of a; horses in passing over 100 feet of this road, = 160 X :?: X 100 = I6OOO2; ; .•. 16000 X = ll^OOO + 33600, 14^5600 z = 16000 = 9.1 horses. 11! — r c r ^ -e 1;^ H The work in raising materials, having a given form, will be its weight multiplied by the height to which the centre of gravity is rais- ed. Suppose the body n 6 to be raised from the horizontal line H, R, . and the point C to be the middle or cen- tre of gravity of the body ; let e and r be points equi-distant from C. Now if equal weights, say for exam- ple 3 lbs. be placed at e and r, the centre of gravity of these weights will be at C, then R Work in raising 3 lbs. to n^ 3 X H r. " " '* to e = 3 X H g. Sum = 3 X (Hr'+Hg). But H r + H e :z= 2 H C ; .-. sum = 6 X H C. That is, 6 lbs. raised from H to C is the same amount of work as 3 lbs. raised to r, and 3 lbs. raised to e. The same may be proved of any two equal weights, one of them rais- ed as far above the centre of gravity C, as the other is be- low it. PRACTICAL MECHANICS. 29 Ques. 8. Required the units of work in raising the mate- rial of a wall 22 feet long, 13 feet high, and 2i feet thick ; supposing the weight of a cubic foot of the material to be 140 lbs. ? Contents of the wall in cubic feet, = 22 X 13 X 21 1=: tl5. Weight of the wall in pounds, = 115 X 140 = 10010. The height of the centre of gravity of the whole wall, = -^---^H feet ; Work = 10010 X 61 = 65065. Ques. 9. The shaft of a pit is to be sunk 120 feet deep, ind 6 feet in diameter ; in how many days would a man \Vorking with a wheel and axle, raise the material, suppos- ..ng a cubic foot to weigh 100 lbs. ? Number of cubic feet in the shaft, = 6^ X .1854 X 120 = 3392.928 ; Weight of the material, = 3392.928 X 100 =339292.8 lbs. The units of work to raise this material ^f* feet, 120 = 339292.8 X ~- ^0351568 ; for it is clear that the centre of gravity of the shaft is 60 feet from the top. A man will perform ^600^ working 8 hours a day, turning a handle. Units of work in 8 hours, = ^600' X 60 X 8 = 1^48000. The number of days required, ^0351568 1'2A8000 16.33. 30 PRACTICAL MECHANICS. Ques. 10. Required the work in raising 3 cwt. of coals from a pit whose depth is 120 feet, the circumference of the rope being two inches, allowing the weight of one foot of the rope of 1 inch in circumference to be .046 lbs. ? Weight of one foot of rope, := 2^ X .046 =^ .184 lbs. Weight of the whole rope, = .184 X 120 := 22.08 lbs. Weight in raising the rope, = 22.08 X ^1^ = 13*i>4.8. Work in raising the coals, . . = 112 X 3 X 120 := >5l03^0 ; Total work, Ques. 11. A cistern 22 feet long, 10 feet broad, and 8 feet deep ; required the work in filling it, when the height of the bottom of the cistern from the water in the well is 36 feet? Water in the cistern, = 22 X 10 X 8 X 62.5 = 110000 lbs. The height to which the centre of gravity of the water has to be raised, -f 36 = 40 feet. 2 Work = 110000 X 40 = 4400000. Ques, 12. The side A B of a cube of granite is 6 feet, and the weight of a cubic foot is 170 lbs. ; it is required to find the work necessary to overturn it on the edge of A ? The distance of the centre of gravity g, from the edge A, , 6' :=^ — -^ 4.24 feet. When the centre is about to fall, the centre of gravity is raised from r to 7i, consequently the work in overturning PRACTICAL MECHANICS. 31 the body is tlie same as raising its whole weight a perpen- dicular height = r n, .-. rn — 4.24 — 3 = 1.24. Work = no X 6^ X 1.24 = 4^553^.8. In a body of any form about to fall, c the centre of gravity §-, will be at n in the vertical line A C, the work in bringing the body to this position, is due to the vertical distance r n, through which the centre of gravity has been raised. The work necessa- ry to overturn any body is the true measure of stability. CHAPTER IV. ON THE LEVER, V/HEEL AND PULLEY, INCLINED PLANE, WEDGE AND SCREW. In this chapter I intend to treat of the transmission of work by simple machines, it may be observed, that the ob- ject of machinery, whether simple or compound, is to regu- late the distribution of work, or to change the direction — and not to increase work. If the parts of a machine were not subject to friction or any other resistances, the work given out would be exactly equal to the work applied. Dead matter, by its gravity, produces pressure, and by the intervention of mechanism that pressure may be increased or decreased ; but v)ork is peculiarly the production of active or living agents. 32 PRACTICAL MECHANICS. THE LEVER. Suppose two uniform bars A B, B C, to be suspended from their centres w and m, by means of chords attached to the points 4 and 3 of the lever D E turn- ing on the fulcrum F, which must evidently be in the middle of D E = A C, and over the middle of A C, in order to se- cure equilibrium ; that is, in order that the two parts may rest horizontally as if they were in one piece, and suspended by the middle point S. It is also evident that the equilibrium will nut be destroyed if the bars be hung by their ends at the points 4 and 3. Let the weight of the bar xi. B = 6 lbs., and the weight of B C = 8 lbs., then A B will contain 6 units of length, and B C 8 units of length. It is evident from the figure, that F q, the distance at which A B acts from the fulcrum contains 4 units, and F p, the distance at which B C acts from the fulcrum will contain 3 units ; then since it appears that a weight of 6 lbs. suspended at §', balances a weight of 8 lbs., suspended at j9 ; therefore the following relation exists when equilibrium takes place ; 6 lbs. X 4 = 8 lbs. x 3, generally 1^ X ^F = W X F^?. Levers are divided into three kinds : — In the first kind of lever, the power and weight are on opposite sides of the fulcrum. First order of Lever. A B is a lever of the first or- der, F the fulcrum or prop, P the power acting at A C, and W the weight attached to the point D. If C F be twice D F, 5 lbs. at C will balance 10 lbs. at D ; generally as many times as.C F is longer than F D, so many times will W be greater than P. PRACTICAL MECHANICS. 33 sc Lever of the second kind. — The weight is between the ful- crum and the power. W is the weight, E the fulcrum, P the power. When the lever is supposed to be without weight, then if the length of A F = a, and B F = _p, the power P balances the weight W when, W X i? = P X ^. Third kind or order of Lev- er. — In this case the power is between the fulcrum and the weight. P, represents the pow- S er, F the fulcrum, and W the weight. Generally if B F = p, and P F = g', the power P balances the weight W, when Second order of Lever. m B T w Third order of Lever. nB ^ EXAMPLES. Questimi 1. In a lever of the first kind, let W = 30 lbs. The arm C F = 11 feet, the arm F D Put 2; r:=: P ia Ibs. Then by the equahty of moments, 11 X:2^ — 4 X 30, 120 4 ; required P ? 11 = ioi; lbs. Ques. 2. A man exerts a pressure of 50 lbs. on a crowbar at a distance of 4 feet from the fulcrum, what weight will he balance at the distance of 3 inches from the fulcrum ? W X 3 = 50 X 48, .•. W-^800 lbs. 34 PRACTlt)AL MECHANICS. Ques. 3. In a lever of the second kind, W = 11 lbs., B F = 16 inches, and A F = 100 inches : reauired P ? P X 100 = 11 X 16, .-. P== 1.76 lbs. Ques. 4. In a lever of the third kind, W = 40 ; B F = 60 inches, and P F = 8 ; required P ? P X 8 = 40 X 60, .•. P z=z 300 lbs. Ques. 5. In a lever of the first kind, 5 and 8 lbs. are placed on one side of the fulcrum, at a distance of 6 and 4 inches respectively from the fulcrum ; required the power P, acting at the distance of 9 inches from the fulcrum, to maintain equilibrium ? PX9=-5X6 + 8X4, ••• ^=^='>- T Q Ques. 6. In a combina- tion of three levers of the first order, represented on ^ 9 -^ — 7 3- — 8 — yt tl^^ accompanying diagram, A the long arms are 9, 7, 8, inches respectively, the short arms 2, 3. 1 inches ; If a pres- sure of 10 lbs. be applied at P, what is the pressure at Q to balance it ? 9 X 10 = 2 X the pressure at A, The pressure at A =::: 45 lbs., 7 X 45 = 3 X the pressure at B, Pressure at B = 105 lbs. 105 X 8 — 1 X the pressure at Q. Pressure at Q = 840 lbs. PRACTICAL MECHANICS. 35 Ques. 7. Let P F and F W be the arms of a false balance, a certain weight Q weighs 16 lbs., when put in the scale at- tached to the long arm, and only 9 lbs. when weighed in the opposite scale ; what is the true weight of Q ? By the equality of moments, the two following equations are ob- tained. QXWF = 9XPF; Q X PF= 16 X WF. Multiplying these equations together, and then striking out the common factors, Q^ = 9 X 16; .-. Q:=:12. THE PRINCIPLE OF WORK APPLIED TO THE LEYER. on w Let B F = 10 feet, F A = 2 feet, and P = 3 lbs. ; required W ? It is evident, if P be raised 5 feet, W will be depressed one, because F B is five times the length of F A ; consequently the work of P = 3 X 5, and the work of W = W X 1 ; .-. W X 1 = 3 X 5 ; .% W— 15 lbs. IJence the equality of moments is readily established by tlie principle of work, and conversely. When the motion is extremely small, the principle of work is termed the prin- cIdIc of virtual velocities. 36 PRACTiCAL MECHANICS. OF THE LEVER WHEN ITS WEIGHT IS TAKEN INTO ACCOUNT. The tendency of a unifona beam or lever A B, to turn about the fulcrum F, is just the same as if the whole of its w^eight were collected in its middle point, or centre of gra- vity C. For if the preponderating side A F, be hung from a cord m r, placed so that A m = n B, then this cord will sustain one half the weight of the beam. Now if the whole weight of the beam be collected at C, the centre, it would produce the same strain upon the cord, hence the beam acts as if its whole weight were collected in its middle point. n ~TpB A Ques. 8. The weidit of the lever of the first kind = 10 lbs., the length A B"= 56 inches ; A F = 40 ; C F = 36 ; D F == 5 inches, and W = 150 ll3S ; required P, in order to maintain the lever in equilibrium. In this case the weight of the lever acting at its centre of gravity m, co-op- erates with the power. f?z F := A F — A m^ = 40 — i of 56 =: 12 Then by the equahty of moments, P X 86 + 10 X 12 1=5 X 150. P^=: n.25 lbs. Ques, 9. The weight of the lever S R, of the first order = 31 lbs.; S R = 85 inches ; S F = 55 ; A F = 43 ; B F = 19 ; and W = 34 lbs. Required P '? PRACTICAL MECHANICS. 37 P X 43 + 31 X 3| = W X 19, P=:^ 34 X 19 — 31 X 3 43 2 12* lbs. Ques. 10. The weight of a lever of the second order = 8 lbs. S R = 80 inches ; S F = 76 ; A F = 70 inches ; B F = 2 inches, and P =100 lbs. ; Required W ? In this case, the weight of the lever acts with W ; 76 — 80 2 = 36 W X 2 -|- 8 X 36 = 100 X 70, ,•. W = 3356 lbs. Ques. 11. A beam R S, whose weight is 4 cwt., is sup- ported by props at A and F ; a weight W, of 20 cwt. is placed at B ; it is required to determine the pressures on the props, when R S = 50 feet ; F R = 2 feet ; B F = 12 feet : A F = 30 feet ? Let C be the centre of the beam, then C R = | of 50 = 25 ; C F = 25 — 2 1= 23 ; suppose the beam to turn on F as a ful- crum, the pressure on A, = P X 30 = 20 X 12 + 4 X 23, 33' _ 2 P = 30 = 11 30 cwt. Because the two props support the whole weight 24 cwt., 24 — ll/o — 12}^ cwt. the pressure on F. HOW TO GRADUATE THE LEVER OF A SAFETY VALVE OF A STEAM ENGINE. A F is a graduated lever of the second kind, turning up- on F as a centre ; V the valve of opening or closing, as the 38 PRACTIQAL MECHANICS. case may oe, tlie comuiimication of steam in the boiler with the atmosphere ; the lever A F rests upon the pin Q V of the valve V, and a sliding weight W is suspended from the lever, enabling the engineer to place any amount of pres- sure on the valve ; this pressure measures the elasticity of the steam when it begins to escape. When steam of very high temperature is employed, the admission of atmospheric air is dangerous, and under such circumstances this valve may be properly termed the unsafety valve. In order to graduate the lever, a weight W must be found, so that when it is placed at A it may balance the greatest pressure of the steam. The next thing to be found, is the position of the weight W, to give any proposed pressure to the valve. EXAMPLES. (czues. 12. The length A F of a lever =14 inches ; the dis- tance F Q ==^ 2 inches ; the weight of the valve and pin = 5 lbs.; the weight of the lever F A = 8 lbs., and the area of circular value in the narrowest place = 6 square inclies ; the it is required to find the load W, so that when it is placed at the extremity of the lever, the steam may have a pres- sure of 30 lbs. on the square inch, or 45 lbs. including the pressure of the atmosphere ? Pressure of steam on valve, == 6 X 30 = 180 lbs. Hence the effective pressure on the lever, = 180 — 5 = 175 lbs., since the weight of the lever will act at the middle point C, W X 14 + 8 X t = 175 X 2. .-. W = 21 lbs. PRACTICAL MECHAN1C^ 39 Ques. 13. At what distance from the fulcrum must the load be, in the last example, so that the steam may have a pressure of 16 lbs. over the pressure of the atmosphere ? 16 X 6 — 5 = 91. Suppose D to be the required position of the load, then FD X 21 + 8 X t = 91 X 12; .•. F D = 6 inches. Ques. 14. In a bent lever the perpendicular A, on the direction of the force P is 4 inches ; and the perpendicular B upon the direction of W is 7 inches ; Required P when W = 200 lbs. ? P X 4 = 200 X T, .-. P = 350. Qfms. 15. A rope A D supports a uniform pole D, rest- ing on the ground at 0, and supporting the weight W, sus- pended from D ; required the tension of the rope, when A D = 175 feet, A = 40 feet, D = 145 feet, W = 50 cwt., and weight of D = 10 cwt? In this example D may be regarded as a lever turning on as a centre. Draw P perpendicular to A D, and C R a vertical line passing through the centre of gravity C of the pole, then the moment of the force stretching the 40 PRACTICAL MACHANICS. rope, will be equal to the sum of the moments of W and the weight of the pole, that is Tension of cord X P = W X N + Wt. of pole X R. To find the perpendiculars, OP, DN, CR, in order to do this in the most simple manner, it will be observed, that since the three sides of the triangle A D are given, the area may be found by the common rule ; the area =* 2100 square feet ; but the area is also expressed by ADXOP ^_^ 2 -^'' jy^, .*. P = 24 feet. But the area is also = ^^ J ^^2100, 2 .•. DN = 105 feet 0N = V 145' — 105'=: : 100, .•. OR = 50. .•. Tension of cord X 24 = 50 X 10-f 10 X 50. Consequently the tension of the cord =z 229i cwt. THE CENTRE OF GRAVITY. Every heavy body is composed of an infinite number of particles, each of which is acted on by the force of gravity in a direction perpendicular to the horizon. The sum of PRACTICAL MACHANICS. 41 all the parallel forces is evidently the weight of the body. Now there must be a point, where a single force, equal to the weight of the body, being applied, will produce the same effect, as the force of gravity acting upon the various parti- cles composing the body ; — this point is called the centre of gravity of the body. From this deflfinition it immedi- ately follows ; that if the centre of gravity be supported, the body will stand, and vice versa. That the centre of gravity of all symetrical, or regular bodies is in their centre of magnitude. That if any body, acted on by the force of gravity tend to turn a lever, we may regard the weight of the whole body to be collected in its centre of gravity. Ques, 16. Let A, B and C, be three bodies in the same right line, it is required to determine the position of their common centre of gravity G with respect to any assumed point F, when A = 6 lbs., B = 4 lbs., C = 10 lbs. ; A F = 10 feet, B F = 25 feet, and C F = 42 feet ? Let it be supposed that an inflexible rod, without weight, passes through the bodies, and that the system turns upon F as a fulcrum ; then as the whole mass may be regarded as acting in the point G, by the equality of moments. F G (6 -f 4 -f 10) == 6 X 10 -f 4 X 25 + 10 X 42, .•. FG X 20:= 580, F G = 29 feet. Let A, B, &c., being any number of bodies lying in the same horizontal plane ; it is required to determine the po- sition of the centre of gravity G, referred to two axes or lines, X, O Y, perpendicular to each other, these lines X and Y are termed co-ordinate axes. Conceive the bodies to be connected with the axis X, by the perpendicular rods A d, B 5, &c., then the sum of the moments of the bodies, tending to turn round the axis of 42 PRACTIQAL MECHANICS. O X, will be the same as the moment of the whole mass, collected in the centre of gravity G. From this equality the distance of G- from X is obtained. In precisely the same way, we find the distance of G from Y ; and hence the point G becomes known. After the same manner the centre of gravity may be determined, when the bodies are in space, by referring them to the co-ordinate planes. Ques. 17. The weights of three bodies A, B, C, are 10, 17, and 9 lbs. respectively ; and their distances from X are 8, 16 and 28 inches, and from Y, 32, 24 and 11 inches respectively ; required the position of the common centre of gravity ? Let X be the distance of the centre of gravity from the X ; and y the distance from Y, then (10 -f 17 + 9) r = 10 X 8 + 17 X 16 + 9 X 28, .•. X = 14. (10 + n + 9) 7/ ::= 10 X 38 + 17 X 24 + 9 X 11, 23 3/ = 24 36 PRACTICAL MECHANICS. 43 WHEEL AND AXLHi. This useful machine is only ano- ther form of the lever, where the power is made to act with intermis- sion ; in its most simple form, it consists of a large wheel A, and a cylinder, or axle, B, both of which turn on the same axis 0. If the wheel A', be turned round by a pow- er P, applied to it, the axle B, will coil up the rope by which the weight W hangs. The lever by which P acts, is evidently A 0, and that by which W acts is B; hence when these weights, or pressures, balance each other. PXAO = AVXOB. If the wheel be displaced by a handle, then the machine is called a windlass. Sometimes the handle is made to turn a series of wheels acting on each other by means of teetn : when a machine has this form, it is called a crane. I will now consider the equilibrium, on the principle of work. When the diameter of a circle = 1, the circumference == 3.1416. Wken the wheel makes 1 revolution, the axle also makes one. In one turn P descends a space, ==2 X A X 3.1416, and W ascends a space = 2 X OB X 3.1416. The work done by P in one revolution =1:2 X AO X 3.1416 X P; and the work of W in one revolution, ==:2 X OB X 3.1416 X W. It is clear that the work done in one revolution is equal to the work applied, friction being neglected, 44 PRACTICAL MECHANICS. .-. 2 X A X 3.1416 XP:^2xOBx 3.1416 X W. .-. PX0A = WX0B, which is the relation already established, conversely assuming the equation of equilibrium, the principal of work is readily established. Ques, 18. The handle of a windlass is 18 inches, the ra- dius of the axle = 3 inches, and the power applied = 60 lbs. ; what weight could be raised were there no friction ? Work of P in one revolution, = 60 X ~^^ X 3.1416. Work of W in one revolution, = W X ^^^^ X 3.1416. ^ V 3 ^ X IS ••• W X ^^-- X 3.1416 = 60 X ^^ X 3.1416 ...W — 360 lbs. Ques. 19. Required W in the last example, when the thickness of the cord is 1 inch, supposing that ^ of the work applied to be lost in friction, and the rigidity of the cord ? The cord increases the radius of the wheel I of an inch, hence Work of W in one revolution, _1 1*1 Effective work of P in one revolution, = ^ X GO X 3 X 3.1416, W X -^ X 3.1>116 = -J- X 60 X 3 X 3.1416. .-. W — 2T0 'hs. = W X i^iT X 3.1416 ; PRACTICAL MECHANICS. 45 OF COGGED OR TOOTHED WHEELS. Let D be a cogged wheel turning upon the same axis as the wheel C ; Q another cogged wheel, acted upon by the former and turning upon the same axis as the axle L. From the wheel C is suspended the weight P, and from the axle L the weight W ; then while P descends, the wheel and the cogged wheel D, will be turned from right to left, but as every tooth in the cog-wheel D is being turned round, a corresponding tooth in the cog-wheel Q will be turned in the contrary direction, and thus the cord L W will be coiled up on the axle L, and the weight W will be raised. Ques. 20. Let P = 120 lbs., the diameter of the wheel C = 3 feet, the number of teeth in D = 11, the number in Q = 14, the diameter of the axle L = | a foot ; required W, in order that equilibrium may take place ? For every turn of the wheel C, 11 teeth of the wheel Q will be turned round, therefore as many times as the num- ber of teeth in D can be taken out of the number in Q, so many turns will the cog-wheel D make while the wheel Q makes one. Let the axle L and wheel Q make one revolution, then the revolu-tions made by the wheel C, 46 PRACTICAL MECHANICS. 11= 14 divided by = ji The space moved over by W, = i X 3.1416; The space moved over by P, 14 11 X 3 X 3.1416; Work due to W = -^ X 3.14:16 X W. Work due to P = 3 X 3.1416 X 14 11 X l^O. ••• Since, the work due to P and W are equal during one re- volution of L, neglecting friction, ~ X 3.1416 X W =: ^ X 3 X 3.1416 X 1^6, .•. i W= If X 3 X 120. W = 916 A lbs. COMPOUND WHEEL AND AXLE. In the common wheel and axle, there is a practical limit to the power of the machine ; for we can only increase the power by increasing the size of the wheel, or by decreasing the radius of the axle. But in the compound wheel and axle, a given power may be made to raise any weight what- ever. This machine consists of two axles, A and C, cut upon PRACTICAL MACHANICS. 47 the same block, round which a cord coils in opposite direc- tions ; this cord passes round the moveable pulley D, which carries tbe w^eight W. Now if the handle be turned, one of the cords is coiled upon the large axle A, while the other cord is uncoiled from the small axle C, so that the rate at which W ascends, depends upon the difference of the cir- cumferences of the two axles ; and, consequently, the pow- er of the machine will also depend upon this difference- But this may be decreased to any extent, without altering the length of the handle. This truly ingenious contrivance is due to the Chinese. Ques, 21. Let the diameter of the axle A, = .8 feet, the diameter of axle 0, = .6 feet ; the length of the handle H A, = 3i feet ; and W, = 6090 lbs. Required P ? When the handle makes one turn, the cord A w411 be drawn up a space equal to the circumference of the axle A, while the cord C, will be let down a space equal to the cir- cumference of the axle C ; therefore the whole cord will be shortened a space equal to the difference of the circum- ferences, and because the cord is doubled, the weight W, will be raised a space only equal to the half of this differ- ence. H X 3 i^ie - 6 X 3 i^ie ^ ^^^ . Work done upon P, in one revolution, = T X 3.14=16 X P ; ... t X 3.1416 XP^^^-^ ^^^^ 7 -^^^-^^^ ^X 6090 2 Dividing each side by 3,1416, ... 7 P = -^ ^ '^ X 6000, P = 84 lbs. THE PULLEY. A pulley is a grooved wheel, turning on an axis, and 48 PRACTICAL MECHANICS. placed in a block or case. A cord passes over the groove of the wheel, in order to transmit the force applied, in anj proposed direction. There is no advantage gained by a single fixed pulley ; but when there are moveable pillleys, the weight raised will always be greater than the power applied ; and then the advantage depends upon the number of cords by which the weight is suspended. Ques. 25. In the annexed system of pulleys, if W lbs., required P, when equilibrium takes place ? 500 As W is suspended by two cords, c and J,' each cord will support 250 lbs. ; but as the cord is supposed to have a free motion over the wheels, the portions a, b, c, will have the same stretch or tension ; hence P = 250 lbs. Application of the Principle of Work. When W ascends 1 foot, the cords c and b will each be shortened 1 foot, and and therefore P must descend 2 feet. Hence the work of P :^P X 2; and the work of W :i= W X 1 ; .-. Px*i = wxt. w PRACTICAL MECHANICS. 49 Ques. 23. Let there be two moveable pulleys, each weigh- ing 4 lbs., the if P = 60 lbs. it is required to determine W, on the principle of work ? If P descend 4 feet, the first moveable pulley will ascend 2 feet, and the second 1 foot. Consequently, the work done in raising W and the pul- leys = Wxlx4:Xlx4LX3 = Wxl*i; The work of P = 60 X ^ ; .-. W X Vi = GO X 4t. W =z 228 lbs. OF THE INCLINED PLANE. To find the pressure necessary to support a body on an inclined plane without friction. Let the weight W, be drawn up the inclined plane A C by means of the weight P, acting by a cord parallel to the plane . then whilst W is moved from A to C, the counterpoise weight will have descended a vertical space equal to A C, according to the principle of work. Work in raising W — W X C U ; Work due to the descent of P = P X AC; •. PXAC:=:WXCB; . p - ^^ w 50 rRACJICAL MECHANICS. Ques, 24. The length of an inclined plane is 30 feet, the perpendicular height 10 feet, and the weight of the body W = 20 cwt., what pressure P, will be required to sustain the body on the plane, friction being neglected ? Work in raising W in opposition to gravity. The work of P :=^ P X SO ; .-. P X 30 = ^^>^iOO P = U6|lbs. Ques, 25. The length of an inclined plane is a mile or 5280 feet, the height 38 feet, the Aveight of the body 1760 lbs., and the friction n\i part of the weight, Avhat pressure will be required to move the body up the plane ? In this case the inclination of the plane is small, and hence the pressure upon it is very nearly equal to the weight of the body. .'. Pressure to overcome friction — Ve^V lbs. Work due to friction, =^ "^?* X 5^80 ^ 35^00 ; Work due to gravity, = neO X 88 = 15>4880 ; Work of the pressure P applied to move the body, = P X 5^80 ; .-. P X 5^80 = 35*iOO + 154880. .•. p z=z af) lbs. OF THE WEDGE, OR MOVEABLE INCLINED PLANE. Let ABC be a wedge, {see the last figure,) sliding on the horizontal plane A B, by the action of a pressure P, appli- ed parallel to A B, and thercl)y elevating a weight W which is only free to move in a vertical direction. When PRACTICAL MECHANICS. . 51 the wedge begins to act, the resistance W, rests upon the horizontal plane A B ; but when the wedge has moved over a space equal to its length, W will have been elevated a height equal to the thickness C B of the wedge ; then if A B = 1.2 feet, B C = .2 feet, and W = 60 lbs., not taking friction into account. Work applied by P — P X 1.^. Work of W = GO X .^ ; .-. P X l.'i ^ GO X .'A ; .-. P =: 10 lbs. This calculation shows that the advantage of the power depends upon the thinness of the back of the wedge. How- ever, it is necessary to observe, that the astonishing power of the wedge, as usually employed, is equally resolvable into the force of impact, which is considered hereafter. OF THE SCREW. In this simple machine the pressure applied moves in a circle whose radius is the length of the lever, or arm of the screw, whilst the direction in which the work is done, is a right right. Ques, 26. The lever of a simple screw is 6 feet, the thick- ness of the thread = .04 feet ; If a pressure of 200 lbs. be applied to the lever, what pressure Avill be produced on the press-board ? Space moved over by P, in one revolution, — 2 X 6 X 3.1416. Space moved over by W, in one revolution i:=r .04 feet. Work of P, in one revolution, = 'i X G X 3.14l1« X *100 ; Work of W, in one revolution, = W X .0>i. .-. w X .o>4 -= 'A X « X ;^.i A l« X *AOO. .-. W ::^ ]h84 96 Ibsv 52 PRAGTICAL MECHANICS. The equation obtained above, shows that the efficacy of the screw is obtained by increasing the length of the lever, or by decreasing the thickness of the threads. Ques. 27. The lever of a screw is 4 feet, and the thickness of the threads i inch ; required the pressure that must be exerted on the lever to produce a pressure of 12 tons upon the press-board ? In one revolution of the lever, the work done, -^ 1^ X ^'14lO X ^ = 560 ; for ^ of an inch 1= o of a foot. In one revolution of the lever the work applied, = P X 8 X 3.14.16 ; .-. P X 8 X S.liilft -^ 56t>, .-. P = 22.28 lbs. Ques, 28. The lever of a simple screw is 2 feet, what must be the thickness of the threads, so that a pressure of 5000 lbs. may be produced on the press-boards, by a pres- sure of 100 li3S. on the handle, or lever ? Let X be the thickness of the thread, then, the work applied in one revolution, == 4l X 3.14L16 X 100 = 1^56.64. ; Work done, ^xX 5000 = 1^56 64l ; .-. X ^ .^513 feet. OF THE COMPOUND SCREW. This mechanical power consists of two screws ; one of which screws within tlic otlier, so that wliilst the large one is d*escending, the small one is relatively rising w^ithin the large one. in consequence of this compound motion, one revolution of the lever, causes the press-1)oard only to de- scend a dirHance equal to t!io differ(?i!ce of the thickness ef PRACTICAL MECHANICS. 53 the threads of the screws. Hence the advantage depends upon the smallness of this difference, and not upon the ab- solute size of the threads ; and hence the machine becomes analogous to the compound wheel and axle. Ques,29 . In a compound screw, the length of the lever is 2.5 feet, the distance between the threads of the large or hollow screw is f of an inch, and of the small one half an inch ; if 60 lbs. pressure be applied to the lever, what is the pressure on the press-board ? In one revolution the large screw descends f of a inch, but at the same time, the small screw, by turning Avithin the large one, ascends | an inch ; therefore the press-board must descend a space = f — ^ = ^ of an inch = ^\ of a foot. Work done in one revolution, - W X ^ ; Work applied in one revolution, 60 X *i X ^.5 X 3.1>5.16 ^ »>4*iAH ; .-. W X -^ = »>4tfi A8. W = 67858.56. Ques, 30. If the length of the lever = 3| feet, the thick- ness of the thread of the larger screw ^== J of a inch, what must be the thickness of the thread of the smaller screw, so that 20 lbs. applied to the lever, may produce a pressure of 20 tons = 44800 lbs ? I of an inch == j\ of a foot. Let X be the thickness of the smaller screw in feet. .-. 'iO X ^ X 3.'14cl« ^ 4L4800 (rV — x.) .\ A — xz:^ .0098214 ; .-. .0723214 ::== z. 54 PRACTICAL MECHANICS. OF THE ENDLESS SCREW. In this machine, the threads of a screw cut upon a cyl- inder, are made to act upon the teeth of a cog-wheel having an axle, round which a cord coils, as in the common wind- lass. This combination gives a very slow motion to tlie weight Ques. 31. In an endless screw, the length of the handle is 3 feet, the number of teeth in the cog-wheel is 24, and the radius of the axle | of a foot : if 36 lbs. pressure be applied to the liandle, what weiglit will be raised, friction being neglected ? Since the screw is fixed upon the axis of the handle, one turn will cause one of the teeth of the wheel to be moved round ; consequently the handle must make 36 turns, whilst the cog-wheel and axle make one turn. Work done in one revolution of the axie, z.^ W X |- X ^ X 3.1>ilO ; for W will be raised 4.7124 feet. The worK applied in one revolution of the axle, = 3« X 6 X 3.1>41« X ^4 ; .-. w X -f X ^ X 3.1>41« = 30 X « X 3.14cl6 X ^4; . ^ ^ 36 X 6 X 24 W = 3456 lbs. THE HYDROSTATIC PRESS. This powerful macnine consists of two cylinders of dif- ferent sizes, each of which has a piston, and pressure is transmitted from the small piston to the large one, by means of water. Tlic pressure of tlie power is applied to the small piston by a lever of the second kind ; and the PRACTICAL MECHANICS. 55 advantage of the machine depends upon the length of tliis lever, and the extent of surface of the large piston, compar- ed with that of tlie small one. Ques. 32. In a hydrostatic press, the surface of the pis- tons are 3 and 350 square inches respectively, the lever is 30 inches long, and the piston rod is attached 2 inches from- the fulcrum ; If a pressure of 100 lbs. be applied to the lev- er, what pressure will be produced upon the large piston ? Let the small piston descend i of an inch, or sV part of a foot, then 1 cubic inch of water will be thrown into the large cylinder, and its piston will be raised 3^0 part of an inch, or 4 2V(T of a foot ; Work on the small piston, too X 30 ± ^ . ^ - X 36 = 4l1 3 Work on the large piston W X >4*aoo w >i^OO 3 W:::= 1 75000 Ibs. CHAPTER y. WORK OF STEAM AND THE STEAM ENGINE. If steam in the cylinder A D exert a constant, or mean, pressure upon the piston A B, say of 25 lbs. per square inch, then if a weight of 25 lbs. be placed upon every square inch of surface in the piston, the steam would just be able to move the piston with its weights or pressures through the length of the stroke in opposition to the weight or pressure ; tliereforc tlie work performed upon 1 square CM D 56 PRACTICAL MECHANICS. inch of the piston in one stroke, will be the pressure of the steam upon 1 incli nuil- tiplied by the number of feet in the stroke, and the work upon the whole piston will be tlie work upon 1 inch multiplied by fhe number of square inches of the whole pis- ton. In the high pressure engine, the pres- sure of the atmosphere, about 15 lbs. to the square inch, is opposed to the pressure of tlie steam. Besides this, a considerable portion of the pressure of the steam is required to overcome the friction of the parts of the engine. As a mean estimate 1 lb. to the square inch is allowed for the friction due to the engine when unloaded ; and an additional friction of | the effec- tive pressure, or useful load, for the resistance necessary to overcome the friction of the loaded engine. Thus, if the pressure of the steam =^ 56 lbs. on the square inch, from which, if 15 lbs. be taken for the pressure of the air, and 1 lb. for the resistance of the friction of the unloaded piston, then the remainder 40 lbs. will be taken up by the useful load, and the friction arising from that load, that is n . load ^ „ load 4- — — -^ 40 lbs 8 load or = 40 lbs. .•. load 1-= 35 lbs. This load is termed the effective pressure of the steam. In the condensing eno'ine, tlie pressure of the vapor in the condenser, must be taken in the place of the atmospheric pressure, estimated at a maximum, the pressure of the va- por in the condenser is about 4 lbs. per square inch ; then ^ , , load , , r ^ ' load -|- *— — X -|- 4 :^ total pressure oi the steam. In a condensing engine, let the total pressure of the steam i::^ 61 lbs. per square inch, 7 .'. load -^ 49 lbs ; that is, the effective prcssijfe of the steaiu, in this case — 49 lbs. PRACTICAL MECHANICS. 57 EXAMPLES. Question 1. The area of the piston of a steam engine = 2200 square inches, the mean effective pressure of the steam 15 lbs. per square inch, the length of the stroke 8 feet, the number of strokes per minute = 20 ; what is the horse power of the engine ? Work done upon 1 square inch of the piston in one stroke, = 15 X 8 = l*iO. Work upon the whole piston in one stroke, Work done in one minute, or in 20 strokes, = ^64.000 X *iO = 5^80000'. But the power of a horse being, 33000. 5^80000^ 33000' 160^ Ques. 2. The area of the piston of a high pressure engine is 625 square inches, the length of the stroke 6 feet, the pressure of the steam 40 lbs. per square inch, the number of strokes per minute 20 ; it is required to find the number of cubic feet of water which the engine will pump from a mine whose depth is 396 feet, making the usual allowance for friction and the modulus of the pump. The modulus of a machine, is the fraction which expres- ses the relation of the work done to the work applied. For example, if a machine should only perform one-half the work that is applied to it, then the modulus in this case, would be \ — -5. However perfectly a machine may be constructed, there must always be a certain amount of work destroyed by friction. The work that is thus destroyed depends upon the extent and nature of the rubbing surfaces. In many machines also, the power of the laboring agent may be exerted in a more or less efficient manner. Hence it is, that the modulus of one machine sometimes differs very much from that of another. The following table of 58 PRACTICAL MECHANICS. machines for the ^raising of water is taken from Mori/is Mechanique Pratique. Modulus. Incline Chain pump 38 Upright Chain pump 53 Bucket Wheel 60 Chinese Wheel 58 Archimedian Screw 70 Pumps for draining mine.- 66 Load + -i- load + 1 + 15 = 40 ; 8 load ~~~ y4 • load= 21 lbs. Useful work of the engine per minute, = ^1 X «i5 X 6 X ^O X .66 = 1039500^ Work to pump one cubic feet of water a height of 396 feet, ^ 6^ 5 X 396 = 1a>4^500, Consequently the number of cubic feet that will be pumped per minute, 1639500' ^^ .. _ ^^ - i^ ^ - ^ ^ — -— 42 cubic feet. Ques, 3. The area of the piston of a high pressure engine = 660 square inches, the length of stroke = 6 feet, the number of strokes per minute = 20 ; what must be the mean effective pressure of the steam, so that the engine may do the work of 25 horses ? Let X be the number of lbs. pressure on each square inch of tlie piston, then, the work per minute, -^xX 660 X 6 X *iO ; .-. X X 660 X 6 X taO = *i5 x 33000 ; PRACTICAL MECHANICS. 59 7) 10.4166 1.4880 11.9046 15.0000 1.0000 27.9046 lbs. total pressure of the Ques, 4. The area of a piston of a high pressure engine = 3000 square inches, the length of stroke = 11 feet, the number of strokes per minute = 16 ; required the mean pressure of the steam, so that the engine may perform the work of 224 horses, making the usual allowance for friction. Let X lbs. be the efifective pressure of the steam. Work per minute with z lbs. effective pressure, :^xX 3000 X 11 X 16 ^ 5*18000 X x ; The effective work per minute will also be expressed by 3SOOO X ^^4 == T39iOOO . .-. 5^8000 X ^ ^ 139^000. .•. .r = 14 lbs, effective pressure. l-|-15«|_14^-»ofl4^==32 lbs. mean pressure of steam. Ques. 5. The length of the stroke of a high pressure en- gine = 8 feet, the area of the piston 1000 inches, and the number of strokes per minute = 20 ; what must be the pressure of the steam so that the engine may pump 66 cubic feet of water per minute from a mine whose depth is 896 feet, making the usual allowances for friction and modulus of the pump ? The work done in one minute, = 6G X 6^.5 X 8»6 X 360«0«0' ; The modulus of the pump ^=: .66 ; Hence the w^ork of the engine per minute X .66 1= 36t>000', therefore the work per minute, .66 60 PRACTICAL MECHANICS. The useful work doue one ioch of the piston in one stroke, 5600000' CTseful load ^80 8 'i80. 35 lbs. consequently the pressure of the steam zzz 35 _j_ -^1- ^ 15 4_ 1 zzz: 56 lbs. In the steam engine, the true source of work is the evap- orating power of the boiler. The magnitude of the work not only depends upon the quantity of water evaporated in a given time, but also up- on the temperature, and consequently the pressure at which the steam is formed. Experimental tables have been fram- ed, giving the relation of the volume and pressure of steam from a cubic foot of water ; from these tables may be found the volume of steam when its pressure and volume of water are given, and vice vei^sa. The following vrill serve as a specimen of such tables. Volume of a cubic foot of water in the form of steam at the corresponding 'pressures, and Temperatures, Total pressure in pounds, per square inch. Corresponding temperature, by FahrenheiVs ther- mometer. Volume of the Steam compared with the volume of the icater that has produced it. Total pressure in pounds per sq. inch. Corresponding temperature, by FahrenheiVs ther- mometer. Volume of Steam ; Volume of water = 1. 1 102.9 20954 11 197.0 »2222 2 126.1 10907 12 201.3 2050 3 141.0 7455 13 205.3 1903 4 152.3 5695 14 209.0 1777 5 161.4 4624 15 213.0 1669 6 169.2 3901 16 216.4 1572 7 176.0 3380 17 219.6 1487 8 182.0 2985 18 222.6 1410 9 187.4 2676 19 225.6 1342 10 192.4 2427 20 228.3 1280 PRACTICAL MECHANICS. 61 Total pressure in pounds, per square inch. Corresfonding temperature, by Fahrenheit's ther- mometer. Volume of the Steam compared with the volume of the water that has produced it. Total pressure in pounds per sq. inch. Corresponding temperature, by Fahrenheit's ther- mometer. Volume of Steam ; Volume of water ^'l. 21 231.0 1224 61 294.9 460 22 233.6 1172 62 295.9 453 23 236.1 1125 63 297.0 447 24 238.4 1082 64 298.1 440 25 240-7 1042 65 299.1 434 26 243.0 1005 66 300.1 428 27 245.1 971 67 301.2 422 28 247.2 939 68 302.2 417 29 249.2 909 69 303.2 411 30 251.2 882 70 304 . 2 406 31 253.1 855 71 305.1 401 32 255.0 831 72 306.1 396 33 256.8 808 73 307.1 391 34 258.6 786 74 308.0 386 35 260.3 765 75 308.9 381 36 262.0 746 76 309.9 377 37 263.7 727 77 310.8 372 38 265.3 710 78 311-7 368 39 266.9 693 79 312.6 364 40 268.4 677 80 313-5 359 41 269.9 662 81 314-3 355 42 271.4 647 82 315-2 351 43 272.9 634 83 316-1 348 44 274.3 620 84 316-9 344 45 275.7 608 85 317-8 340 46 277.1 596 86 318-6 337 47 278.4 584 87 319. 4 333 48 279.7 573 88 320.3 330 49 281.0 562 89 321.1 326 60 282.3 552 90 321.9 323 51 283.6 542 91 322-7 320 52 284.8 532 92 323-5 317 53 286.0 523 93 324-3 313 54 287.2 514 94 325-0 310 55 288.4 506 95 325 - 8 307 56 289.6 498 96 326-6 305 57 290.7 490 97 327-3 302 58 291.9 482 98 328 - 1 299 59 293.0 474 99 3-28 . 8 1 296 60 294 . 1 467 100 329.(5 293 1 62 PRACTICAL MECHANICS. Ques. 6. In a high pressure engine, the area of the piston = 120 inches, the length of the stroke == 2.4 feet, the effec- tive evaporation of the boiler = .5 cubic feet per minute, the pressure of the steam in the cylinder 64 lbs. per square inch ; making the usual allowance for the loss due to fric- tion, required the useful load per square inch of piston, and the useful horse power ? f useful load z:z: 64 — 15 — 1 =^ 48. Useful load — 42 lbs. From the table it will be found that a cubic foot of water raised to steam of 64 lbs. pressure, has a volume of 440 cubic feet. Yolume of steam evaporated per minute — 440 X .5 = 220 cubic feet ; Volume discharged at each stroke 120 X 2.4 o K- f f =- — = 2 cubjc feet ; 144 ' The number of strokes each minute The work in one stroke == 4c^ X I'iO X 'i.4 = 1^096. Work in one minute =3 4^0»6 X llO = 1330560 . 1330560 ^,^, ^^ Horse power := *^*^4>C^i\^ '-=^ ^%3^- S^. Ques. 7. The area of the piston of a high pressure engine is 264 inches, the length of stroke 5| feet, the pressure of the steam in the cylinder = 72 lbs. per square inch, the number of strokes per minute = 36 ; required the useful load, the water evaporated per hour, and the useful horse power of the engine ? Lot X be the useful load, then :r+ y -f 15 + 1 = 72 lbs. 1x-{-x = 392. 2: = 49 lbs. PRACTICAL MECHANICS. 63 The volume of steam discharged per raiimte, in cubic feet *^64 X 5 5 ---' X 36 — 363 cubic feet. 144 The cubic feet of steam discharged in an hour = 363 X 60 = 21780. From the table it will be found that one cubic foot of water yields 396 cubic feet of steam at 72 lbs. pressure. Consequently the cubic feet of water evaporated oer hour 21780 — 00. 396 The useful horse power of the engine __ ^64 X 40 X 5.5 X S« = IT^ 616. It has been found by experiment, that whatever may be the pressure at which steam is formed, the quantity of fuel necessary to evaporate a given volume of water, is nearly the same. Hence it follows, that it is most advantageous to employ steam of a high pressure. Ques. 8. Required the duty of the engine, in the last ex- ample, allowing a bushel of coals 94 lbs., to evaporate 11.5 cubic feet of water ? The useful work per hour = ^64i X ^» X 5i X 36 X 60 == 1536^9680 ; which is the work of 55 cubic feet of water, or of 1 bushel of coal, = 1536-i9680 X ^}i^^ = 3il330*iA, which is termed the duty of the engine. Ques, 9. A train of 200 tons, moves along at a uniform speed of 30 miles an hour upon a level rail, the resistance of friction upon the rail is 5| lbs. per ton, the resistance of the atmosphere 33 lbs. upon the whole train when the speed is 10 miles an liour, the diameter of the driving wheel 7 feet, the area of the piston 120 inches, the length of the stroke = | feet, and in addition to the resistances of fi'ic- 64 PRACTICAL MECHANICS. tion, tlie resistance due to the blast pipe is 1| lbs. per sq. inch of the piston, when the speed of tlie train is 10 miles an hour. Required the pressure of the steam, the evapora- tion of the boiler, and the number of bushels of coal neces- sary for a journey of 500 miles, supposing that 1 bushel, or 94 lbs. will evaporate 11| cubic feet of water? When the diameter of a circle = 7 feet, the circumfer- ence = 22 feet nearly, which is the space passed over in one revolution of the driving wheel. Total resistance to the motion of the carriages ^ 200 X 5.5 -f ( Sy X ^^ = 1^^^ ^^s- The work in one revolution = 13in X ^^ = 301341. Work of 1 lb. of pressure per square inch of the piston in one revolution :=. 1 X l^O X I- X ^ = ^^O, for the engine has two cylinders, and each piston makes two strokes while the driving wheel turns once round. Consequently, the efifective pressure on one square inch of the piston, 30^3ii ^^^^„ = ^^.j^Q = 42.78 lbs. It has been found by experiment, that the resistance of the blast pipe increases with the speed of the engine, ••. Resistance due to the blast pipe Hence the total pressure of the steam on the piston 42.78 zz- 42.78-] \-iJ^]o-^i^ = 69.14 lbs. The number of revolutions of the driving wheel per minute _ 30 X 5280 - 60X22 -^^^' therefore, the two oistons will make 120 X 4 — 480 strokes a minute. PRACTICAL MECHANICS. 65 The cubic feet of steam discharged per minute will be ■X — X 480 = 600. 144 ^ 2 From the table it will be found that a cubic foot of water produces 411 cubic feet of steam of 69.14 lbs. pressure. The Dumber of cubic feet of water evaporated per minute __ 600 • "~ 411 • As one bushel of coal evaporates 11.5 cubic feet of water, the number of bushels of coal used per minute - 600 ^ "" 411 X 11.5 ' A journey of 500 miles is performed in I65 hours, or in 1000 minutes, at the rate of 30 miles an hour. Therefore the number of bushels of coal for 1000 minutes 600 X 1000 411 X 11.0 = 126.95 bushels. Ques. 10. In a locomotive engine, the area of the piston is 90 inches, the length of the stroke 16 inches, the pressure of the steam 50 lbs., the effective evaporation of the boiler .7 cubic feet per minute, the diameter of the driving wheel 5 feet ; what is the speed of the train per hour ? At 50 lbs pressure, 1 foot of water forms 554 cubic feet of steam. The volume of steam generated per minute = 554 X .^ = 38t.8. Cubic feet discharged in one revolution of the driving wheel -, . ,, 90X16 _., -^^ 1728 -^''- Therefore the number of revolutions of the drivkig wheel per minute 6(> PRACTICAL MECHANICS. .'. The space moved over by the carriage per minute, in feet 1:= 5 X 3.1416 X 116.3. Consequently tfie miles moved over per hour 5 X 3.1416 X 116.3 X 60 5280 20.7. Ques. 11. The area of the piston of a locomotive engine = 80 square miles, the length of the stroke = 15 miles, the pressure of the steam 48 11)S. per square inch, and the diam- eter of the driving wlieel 5 feet ; required the effective evaporation of the boiler, so that the train may have a speed of 30 inch, an hour ; required also the effective horse power of the engine, and the weight of the train, taking the resistances to the motion of the piston as in former ex- amples ? Number of revolutions of the wheel per minute ^ 30 X 5280 ^ 60 X 5 X 3.1416 Number of strokes of both pistons per minute = 168 X 4 --=--. 672. -, 1 l^^ad , ,^ , ,30 ^, .^„ Load -f — i" ^^ + ^ + "To" X 1. 75 ::::= 48 lbs. load ^ i^^XA, ^ 23.4 lbs. O Tlie effective work per minute will be = ^S./A X 80 X ^ X GTi 1=. 15^^i^80^ Hence, the effective horse power Now the effective work has to support a speed of 30 miles an hour, 2640 feet per minute, in the train in opposition to the resistances of friction and the atmosphere. Work due to the resistance of the atmosphere per minute (ao \2 ^^ ^ X 33 X ^6^0 ^184080 PRACTICAL MECHANICS. 67 Therefore the work due to friction per minute Work of frictiou when the train weighs x tons = T X ^64tO X X ; 188400' ^„_ ••• " ^ T X 'A040 ^ '^•'^ ^'^"^- Cubic feet of steam discharged each stroke _ 90 X 1.25 ^ ~ 144 ' the cubic feet discharged in a minute — 80X1.25x672 ~~ 144 From the table it will be found, that a cubic foot of water yields 573 cubic feet of steam, therefore, the number of cubic feet of water evaporated per minute 80X1.25X672 144 X 573 -1^ 81 cubic feet. Question 12. A railroad train of 60 tons, ascends an in- cline that has a rise of i in 100 ; required the maximum speed per hour, 40 being the effective horse power of the engine, friction, 8 lbs. per ton ? Let X = - the speed of the train in feet per hour. the rise of the rail in x feet. Work due to gravity ftp X ^*a4 X X ^.^^ Work of frictiou, = GO X 8 X :?: -^ ^80 x ; But the work due to gravity per hour, added to the work due to friction per hour, must be equal to the work done by the engine in the same time ; .-. 8lfir = >^4« X S3000 X 60 izir lO^OOOOO • 68 PRACTICAL MECHANICS. Hl« 970.56 ft. .-=: 18.4 miles. xf the engine in the last example move this train, what must be the effective evaporation of the boiler, and the duty of the engine. 91056 Speed per minute, = • = 1617.6 feet. Number of strokes of the pistons per minute, -_-i^'A_ X4-412. The effective work of the engine per minute, = 33000 X >i« = 13*iOOOO'. Suppose y = the pounds effective pressure on each square inch of the piston, then the work of y lbs. effective pres- sure per minute 3/ X 80 X ^- X 4cl^ ^ 13*iOOOO^ .•. 3/ 1=^32 lbs. Pressure of steam z.:: 32 + I X 32 4- 15 4- 1 4- ^ X 1.75 = 55.7 lbs. One cubic foot of water in the form of steanj, at 55.7 lbs. pres- sure, is 504 cubic feet. Xumber of cubic feet discharged per minute, in the form of steam 80 X 15 •*• Number of cubic feet of water evaporated per minute Xow this water performs 13^0000', .*. 11.5 cubic feet of water, or 94 lbs. of coal the duty of the engine. PRACTICAL MECHANICS. 69 OF THE WORK DEVELOPED BY THE CONDENSATION OF STEAM. When water is raised into steam at the boiling point or temperature, of 212*^ Fahr., its volume is increased 1711 times, or a cubic inch of water will very nearly form a cubic foot of steauL Now, if steam at this temperature be allowed to enter the lower part of the cylinder, (see the last figure), then the pressure beneath the piston will just counterpoise the pressure of the air upon the piston, and a small additional force will cause the piston to rise. If, then, the steam be condensed by a jet of cold water, a vacuum will be formed, and the piston will be pressed downward with the whole weight of the atmosphere rest- ing upon the surface of the piston. But, it has been found, that a perfect vacuum cannot be formed in this way, be- cause water gives off vapor at all temperatures. Now, if 14.7 lbs. be taken as the mean pressure of the atmosphere, upon one square inch of surface, then by the condensation of the steam, upon an average, an effective pressure of U.I — i zizz 10.7 lbs. is established upon every square inch of the piston ; takin^ the temperature of the condenser at 150^, at which the cor responding pressure of the vapor is 4 lbs. per square inch. However, when the temperature of the condenser is given, the pressure of the vapor is easily found. EXAMPLES. Ques. 13. Required the work developed, by the conden- sation of a cubic foot of water, supposing 4 "lbs. to be the elastic^* ty of the vapor after condensation ; required also, the duty of the atmospheric engine using the steam in this manner. Cubic feet of vacuum foruied by condensation. ^ 1710= 1711 — 1. Pressure on one square inch of the piston = 14.7 — 4 -=:: 10e7 lbs. o 70 PRACTICAL MECHANICS. Suppose the area of the piston to be one square foot, the length of the stroke will be 1710 feet. Cousequently, the work of one cubic foot of water ^ 14l>4: X lO.T X niO = ^634l168. And the duty or work of 94 lbs. of coal = 1 bushel, =:: duty of 11| cubic feet of water ■^ *4634^68 X 11.5 = 30*il>083^0. Ques. 14. What must be the effective evaporation of the boiler of an atmospheric engine, so that the horse power may be 30, allowing 3 lbs. for the elasticity of the vapor in the condenser. Work of the engine per minute 30 X 33000 -^ »«0000 . Work of one cubic foot of water = (lii.T — 3) X 144 X nio = -assioos. Therefore, the number of cubic feet of water evaporated per minute »»oooo .34. ^881008 OF THE USE OF STEAM WHEN USED EXPANSITELY. When steam is used expansively, it is allowed to enter the cylinder for only a part of the stroke, and then, for the remaining portion, the pi.^ton is moved by the expansive force of the steam. This is the most economical way of employing steam power ; for all the available work is taken out of the elastic vapor before it is condensed. When the volume of steam is inci^eased, its elasticity or pressure is decreased in the same ratio ; that is, if its volume is increased three times, its pressure will be one- third of what it was at first, and so on. This is Boyle^s law, but M. Regnault has shown, that it will not hold in extreme cases. PRACTICAL MECHANICS. 71 Let the steam be cut off when the^ piston is at C D, and let the remaining part of the stroke be divided into any even number of parts ; then the pressure of the steam upon tlic piston when it arrives at the different lines, forming the divisions, may be ascertained bv the law just explained. ^' asis Let de, e 0, rn, &c., be lines containing as many units as there are units of pressure on the piston at the correspond- ing points of the stroke, then the work done from A B to C D, will be the area of the rectangle a d, because the work is the product of the pressure c d, by the space d b ; the woi^k done from C D to F N, will be represented by the units in the curved space c dfk, because the work done from D to g is very nearly the pressure c d. multiplied by the space d o, and the work done from q to t, is very nearly the pressure o e multiplied by o r, and so on. Now the smaller these intervals are taken, the more nearly will the areas represent the work. In order to find the area of this space, take the following Rule : — To the sum of the extreme ordioates, add four times the sum of the even ordinates, and twice the sum of the odd ordinates ; then this sum multiplied by one-third the common distance between the ordinates will give the area of c kfd. EXAMPLES. Ques, 15. In a condensing engine the length of the stroke is 5 feet, the steam is cut off at 2 feet of the stroke, the pressure of the steam in the cylinder 48 lb?., and the elas- ticity of the vapor in the condenser is 1 Ibe^. : required the 72 PRACTICAL MECHANICS. work performed upon one square inch of the piston in one stroke ? Let the space through which the steam acts expansively, be divided into six equal parts, then each interval will be | a foot, by Boyle's law. cd = iS — 48. lbs. 2 t -- — X 48 38.4 lbs. 2 X 48 3 82. lbs. 2 sm = — X 48 ^2 27.4 lbs. v=^ X 48 _ 4 24. lbs. .*=^ X 48 44 21.3 lbs. o XV — — X 48 _ 19.2 lbs. Area oi c d y x^ or work done expansively on 1 square inch of the piston in one stroke c 3^ + ^^*; i == 8^.9 ; The work done before the steam is cut off, that is, from A B to CD, = 48 X ^ = »6 ; Work done against the piston by the vapor in the condenser = >4 X 5 = *iO. Consequently, the total work on one inch of the piston in one stroke ^ 8T.«.» + «G — *i« ^ l«3.1> PRACTICAL MECHANICS. 73 Ques, 16. The area of a piston of a condensing engine is 1440 square inches, the length of the stroke, including the clearance, is 5 feet, the steam is cut oS at 1 foot of the cylinder, the clearance = i foot, the pressure of the steam = 30 lbs., the elasticity of the vapor in the condenser = 4 lbs., the effective evaporation of the boiler .2 cubic feet per minute, and the resistances as before described ; re- quired the useful load, and the useful horse power of the engine ? By dividing the space through which the steam acts expansively into four equal parts, and calculating the pressures by Boyle^s law, the work done expansively on one inch = Now the space through which the piston moves before the steam is cut off = 1 — i = I foot ; Work done before the steam is cut off = SO X I = ^*i.5 ; Total work of steam on 1 square inch in one stroke = 48.66 + ^^Jt.5 = 11.16. 11.16 Mean pressure of the steam = — —^-r — =14.9 lbs. But the resistances = the mean pressure of the steam, .•• Load + i^ + 1 + 4 = 14.9. .•. Load = 8.66 lbs. One cubic foot of water generates 833 cubic feet of steam at 30 lbs. pressure. .*. The volume of steam discharged per minute = .2 X 883 = n6.6. Volume discharged each stroke 1440 , . , X 1 =^ 10 cubic feet. 14:4 74 PRACTICAL M-ECHANICS. .•. The number of strokes per minute 176.6 10 17.66. The useful work per minute will be the continued product of the load, the area of the piston, the length of the stroke, and the number of strokes per minute. .*. The eflTective horse power 8.66 X 14l4.6*X ^.^5 X IT. 66 33000' = 31^.1. Ques. 17. What must be the evaporation of the boiler of the engine in the last example, when the steam has a pressure of 48 lbs., so that the effective horse power = 40 ? Proceeding, as in the last example, the load is found to be 16.6 lbs. Therefore, the effective wDrk in one stroke = 16.5 X 14.4.0 X ^.15 = 11^860. The effective work in one minute -= 40 X 33000 = ISiOOOO. Therefore, the number of strokes per minute 11^860 == 11.69. Since 10 cubic feet of steam is discharged in one stroke, the volume of steam discharged per minute = 10 X 11.69 =117 cubic feet, nearly. One cubic foot of water forms 575 cubic feet of steam at 48 lbs. pressure. 117 .*. Cubic feet of water evaporated per minute = -^— ^ . o7o Ques. 18. Required the duty of the engine in example 16? Useful work per minute :^3l.T X 33000 Now, this work is done by the evaporation of .2 feet of water. PRACTICAL MECHANICS. 75 .*. Work of 11.5 feet of water, or 1 bushel of coal. =: 31.T X 33000 X —^ = 60 millions, units of work, the duty of the engine. When water, or any body falls from a given height, the work which it is capable of performing, will just be that vfhich would be done upon it in raising it to the height from which it has fallen. It is no matter in what way this work is used ; whether the water falls into the buckets of an overshot wheel, or delivers its work upon the paddles of an undershot wheel, the laboring force of the water will always be equal to the work due to the height of the fall. Ques, 19. The breadth of a stream is 4 feet, depth 3 feet, mean velocity of the water 15 feet per minute, and the height of the fall 20 feet ; required the horse power of the water-wheel which does ro of the work of the water, and also the number of bushels of corn which the wheel will grind in a day of 10 hours. Cubic feet of water going over the fall per minute, =: 4 X 3 X 15 = 180 ; weight = 180 X 62.5 = 11250 lbs. As the water descends a height of 20 feet, the work which it is capable to perform =^ ^O X 11^50 = 2^5000. But the wheel does ^^ of this work = ^^5000 X ^ = 15^500' Horse power = ^^^^^^^ = ^^.^T ; But a horse is able to grind a bushel of corn in an hour, there" fore, the number of bushels ground in 10 hours = 4.77 X 10 = 4:1.1. 76 PRACTICAL MECHANICS. CHAPTER VI. OF ACCUMULATED WORK. When a body moves uniformly, the space described is, obviously, equal to the units of time during which the body moves, multiplied by the space passed over in each unit of time ; that is to say, the space is equal to the product of the time by the velocity. If, therefore, a rectangle be con- structed, having the units in the base equal to the units in the velocity, and the units in the perpendicular equal to the units of time ; then the units of surface in the rect- angle, will give the units of space moved over by the body. In like manner, the space described by the body, whose motion is uniformly accelerated, may be represented by the area of a trapezoid, whose parallel sides contain respectively the units of velocity at the commencement and end of the motion, and perpendicular between these sides, the units in the time. This proposition will readily be un- derstood, by observing, that the mean length of the trape- zoid will contain the same number of units as the velocity, which the body has at the middle of the time ; and this will be true, however small the intervals of time may be taken. The best illustration of accelerated motion is afforded in the case of falling bodies. FORCE OF GRAVITY. When bodies fall freely near the surface of the earth, the force of the earth's attraction being constant, communi- cates to them equal additions of velocity in equal intervals of time. Thus at the end of one second, the velocity of the body is 32.155 feet ; at the end of two seconds, 2 times, 32.155 feet ; at the end of 3 seconds, 3 times, 32.155 feet ; at the end of 4 seconds, 4 times, 32.155 feet, and so^ on ; or generally, the velocity acquired by a falling body, is equal to the product of the time of the body's fall in seconds by 32J feet, putting 32J for 32.155. PRACTICAL MECHANICS. 77 This is expressed by the equation, The space described by a body in one second, will be half of B2f feet = IGtV feet ; because, the velocity of the body in the middle of the time, will be the mean velocity with which it moves during that time. In like manner, the space described by the body in 4 seconds, will be 4 times 2 x 32^ feet ; because, 4 x 32^ feet is the velocity at the end of the time, and, therefore, 2 X 326- will be the mean velocitv, or the velocity in the middle of the time. But 4 times "2 X 32i = 16 X 16tV = 4^ X 16rV, that is, the space described by a falling body in 4 seconds, is equal to the square of the time, multiplied by the space described in one second. In the jame manner, any other case may be established. In general, relation of space and time is therefore expressed by ^ == ^^ X 16rV. Ques. 1. What velocity will a falling body have at the end of 5 seconds ? 32^ X 5 ri:= 1601 feet. Ques, 2. In what time will a body acquire a velocity of 193 feet a second. 193 "32r =: 6 seconds. Ques. 3. Through what space will a body fall in 4^ seconds ? (4|)^ X 16A = 325Ufeet. Ques. 4. A body is thrown downward with a velocity of 10 feet per second, how far will it descend in 6 seconds ? It is evident that the body will retain its motion of projection, although acted upon by gravity. Now the space due to the pro- jection rr: 6 X 10 =^ 60 feet, and this added to the space due to gravity will give 60 X 6' X 16i*2 = 253 feet. 78 pracJtical mechanics. Ques. 5. In what time will a body fall 193 feet ? By the geueral equation 193 12. 16tV ,•. ^ = 3.4 seconds. Ques. 6. From what height must a body fall, to ac quire a velocity of 96^ feet ? First it is necessary to find the time which the body will have to fall, to acquire a velocity of 96^ feet. 96i ■^^ 3 seconds. Height = 3' X 16,-^2 = 144| feet. Ques. 7. The velocity of a body is 84 feet. From what height would it have to fall, in order to attain this motion ? Time 84 ; distance or space K- 32^) ^ 16, "5 84' 2 X 32i ■ g = 32i and a := 84, then a' g . 2 a' Put^ .•. The space fallen tlirough to acquh'e the velocity a, is equal to the square of (a,) divided by (2 g). In most modern works, g is put for the number 32^, and n is put for the circumference of a circle when the diameter =:^ 1. OF THE WORK ACCUMULATED BODY. When a body is in a state of motion, it will continue in that state, unless acted upon by some external force. But, in order to give this motion to the body, there must be PRACTICAL MECHANICS. 79 work done upon it. Thus, we may give the velocity of 32i feet to one pound, by raising it IGyV fe^t, and then al- lowing it to fall by the force of gravity. Iq this case, the units of work accumulated in the body will be Again, when a heavy fly-wheel is in rapid motion, a con- siderable portion of the work of the engine must have gone to produce this motion ; and before the engine can come to a state of rest, all the work accumulated in the fly, as well as in the other parts of the machine, must be de- stroyed. In this way a fly-wheel acts as a reservoir of work. In order to estimate the work in a moving body, it is simply necessary to consider the height from which it must fall to acquire the given velocity, and then the work will be found, by multiplying that height in feet by the weight op the body in lbs. ; because, the work expended in raising the body, to the neces- sary height to communicate the given velocity, must be the same as the work, which gravity will per- form upon the body in its descent Ques. 8. The weight of a ram is 600 lbs., and at the end of the blow has a velocity of 32^^ feet, what work has been done in raising it ? The ram must have fallen 16iV feet, the work done upon it therefore = 600 X 16^ = 9650. Ques. 9. How many units of work are accumulated in a body whose weight is 144 lbs., and velocity 200 feet ? The height from which the body must fall, to acquire the given gravity 900 ' — ^.^"V ^. ==621.76 feet. 2 X 32i 80 PRACTICAL MECHANICS. Consequently, the work which must have been done upon the body Ques, 10. Required the work accumulated in a cannon ball, whose weight is 32^ lbs., and velocity of 1500 feet ? The height from which the ball must fall, to acquire a velocity of 1500 _ 1500^ ~2 X 32i ' The work accumulated in the body = 1500^X3^1 _ 11^5000. ■a X 3*Ai This example clearly shows, that the work accumulated in a moving body is equal to the square of the velocity in feet per second, multiplied by the weight of the body in lbs., and divided by 2 X 32^. This very important proposition is expressed by the equation F»2 X W^ U^ ^S Ques. 11. A ball weighing 20 lbs., is projected with a velocity of 60 feet per second, on a bowling green. What space will the ball move over loefore it comes to rest, allow- ing the friction to be yV the weight of the ball ? The units of work, U^ in the ball = 17 =^^1A^ = 1119.11. It is evident, that the ball will not stop until all this work is destroyed, that is, until the work destroyed by friction is equal to the accumulated work. Let X be the number of feet over which the ball moves before it comes to a state of rest, then the work destroyed by friction in moving the ball over x feet = -^ X :r = 1119.n; PRACTICAL MECHANICS. 1 81 ^O 503.6 feet. Ques, 12. A train weighs 193 tons, and has a velocity of 30 miles an hour when the steam is turned off ; how far will the train move upon a level rail, whose friction is 5J lbs. per ton 1 80 miles an hour = 44 feet per second, for, 30 X 5280 60 X 60 ~" • 193 tons ~ 432320 lbs, .•. U — ^ X 3^- "^ 130099^0. When the train stops, the work of friction will be equal to the work accumulated in the train, hence, if x be the number of feet the train will move after the steam is cut off, the work in moving the train over x feet = 193 X ^i XX =- U. .•. 193 X 5i X X = 130099^0. 130099^0 ,,,,,, ••• ^- 193 X5i -^^^^^fe^^> Ques. 13. A train weighing 60 tons, has a velocity of 40 miles per hour when the steam is turned off, how far will it ascend an incline of 1 in 100, taking friction at 8 lbs. per ton ? 40 miles an hour = 58 f feet a second. 60 tons = 134400 lbs. Work in the train when the steam is cut off 2\2 (58 If X 134l400\ *A X 3^i 82 PRACTICAL MECHANICS. put X = the number of feet the train will move after the steam is turned off ; then the work due to the friction = GO X » X a: = U^Q x , 100:1::. :3^ = the rise of the rail in z feet ; .•• The work due to gravity X X 13^4^00 = 13>44l x; (58|)2 X 134L400 100 .% il80 X + 134c4 x = *X3ii (58 1) ^ X 134^00 .-. X = ^— 2^ . — = 3942i feet. Ques. 14. Two balls each weighing 64^ lbs., are placed at the extremities of a horizontal arm, which gives motion to a screw driving a punch, as in the common stamping machine. The velocity given to the balls is 8 feet per second. What is the mean resistance opposed to the punch, when it is just driven through an iron plate | of an inch thick ? The weight of the two balls = 128f lbs. If the velocity of the balls be 8 feet, just as the punch begins to cut, then the units of work accumulated Now, if the thickness of the plate be 1 foot, the uniform resist- ance would obviously be 128 lbs. ; but the thickness is ^\ of a foot. 1^8 •. Mean resistance, — - — = 4096 lbs. PRACTICAL MECHANICS. SfrJ Ques, 15. A ball weighs 24 lbs., and is fired from the mouth of a cannon 12 feet long, with the velocity of 1000 feet per second ; it is required to find the mean pressure )f the elastic vapor upon the ball? Let X be the pressure upon the ball during the passage through the barrel, then ,•. X ::=: 31088 Ibs. mean pressure. Ques. 16. A carriage of 1 ton moves on a level rail with the speed of 8 feet per second, through what space must the carriage move, to have a velocity of 2 feet, supposing friction to be 6 lbs. per ton ? Work lost = — 64^ 64^—"^ *®^®- But this work has been taken up by friction passing over a ^pace which put = x feet then, X X &^ ^089 ; .-. X — 348 feet. Ques. 17. If the carriage in the last example move over 400 feet before it comes to a state of rest, what is the re- sistance of friction per ton ? Work accumulated in the carriage Work of friction ::= friction X 'iOO ; .'. Friction X 400 — ^^^8 ; Friction =:z 5.51 lbs. 84 PRACTICAL MECHANICS. Ques, 18. Two weights, W and w, weigliing 5 and 3 lbs. respectively, are connected by a cord that goes over a fixed pulley, as in Atwood's machine ; through what space must W descend to acquire a velocity of 10 feet ? Since the velocity with which w ascends, is the same as the velocity with which W descends, then Work in w -- ^^ , = ^.^^a ; Work in *(? = — ^- ^ := 4:. 611 ; Total accumulated work = 1^.383. Let X be the space passed over by each of the weights, then Work of gravity on W = 5 X ^ Work of gravity on i^ = 3 X :t' Vk X X difference. But the work performed upon w has been yielded by the work of W ; therefore, the work remaining in the bodies = ax^=1^.383. .•. x=z 6.19 feet. OF THE CENTRE OF GYRATION, AND THE WORK IN A ROTATING BODY. Ques, 19. Two balls, A and B, are connected by a rod, which is made to revolve upon a centre C ; the weight of A is 3 lbs., and that of B is 4 lbs., the distance of A from the axis is 8 feet, and that of B is 5 feet ; if a point in the PRACTICAL MECHANICS. 85 rod, at 1 foot from the axis C, has a velocity of 10 feet per second ; it is required to determine the work in the balls, and the point in the rod where we may suppose the weight of the two balls collected, so that the work may not be altered ? Yelocity A = 8 X 10 ; Velocity of B = 5 X 10 ; Work in A r:. ^^ ^*f f ^ ^ = ^»8.>4>4 Work in B =^ rr^— = 155.^4: Total work = >5l53.88 Let X be the distance from the axis i= C ti, then the work of the two balls collected at n >453.88 803 X 3 X 502x^ • • • 700 a;' = .•. X = 29200 41.11 6.46. C % = 6.46, the point n is called the centre, of gyration. When the centre of gyration in a rotating body is known, the accumu- lated work is readily found by the methods already given. But to find this centre generally, requires the aid of the Integral calculus, which would be foreign to the objects of this work to introduce. The distance of the centre of gyration from the axis, in a few of the most useful cases is as follows : — hi a circular wheel of uniform thickness, is equal to the radius of the wheel X v^ o ? ^'^ ^ ^'^^ revolving about its extremity, it is equal to the length of the rod fj I, and when it revolves about its centre, it is equal to the length X ^/ T 2 ; ^^^d in a plane ring, like the rim of a fly-wheel, it is equal to the square root of one half of the sum of the squares of the radii forming the ring. Que,s, 20. The weight of a fly-wheel is 10 cwt. This weight is collected in a point at a distance of 5 feet, from 86 PRACTICAL MECHANICS. the axis. The wheel makes 20 revolntions per minute, the diameter of the axis is 2 inches, and the friction upon it 4 of the whole weight. How many revolutions will the wheel make before it stops ? Velocity of 10 cwt. or 1120 lbs. per second 10 X 3.1416 X 20 60 Work in the wheel = 10.47 feet. 1908.43. ^ 64cJ Circumference of the axis = -^2 X 3. 1416 = . 5236 feet. Work destroyed in x revolution = .5^36 Xx X Jr = 1908.43. .•, X = 22.78 revolutions before the wheel comes to rest. Ques. 21. The w^eight of a fly-wheel is 4 cwt., the dis- tance of the centre of gyration from the axis 4 feet, and the number of revolutions per minute 40 ; find the number of strokes that the fly-wheel will give a forge hammer weigh- ing 200 lbs., with a lift of 2 feet, friction neglected ? Velocity of the wheel per second = 16.755 feet. Work in the wheel Let the hammer make x lifts. Work of x lifts of the hammer = ^00 X "i X a; = 1954.9. 1954.9 400 4.8. Ques. 22. The diameter of a grindstone is 4 feet, and its weight 300 lbs. The circumference is made to revolve with PRACTICAL MECHANICS. 87 the velocity of 5 feet per second. The circumference of the axis is 6 inches, and the friction upon it \ of the weight. It is required to find the number of revolutions, which the stone will make when left to itself ? Centre of gyration from the axis = 2^1. To find the velocity of the centre of gyration, say, as The square of the velocity of the weight in feet per second _ 25 — 2 ' Work in the stone = ^ ^ ft^^ ' — ^^'^ ' Let X be the number of revolutions, then the work destroyed in X revolutions — X — 2^ — X 2; =:^ o^ . o : .-. x = 3.109. OF VIS VIVA, OR LIVING FORCE. The vis viva of a moving body, is a force which is ex- pressed by the product of the quantity of matter by the square of the velocity. Let the weights of two bodies, A and B, be 4 and 7 lbs. respectivelv, and their velocities 5 and 9 feet per second ; required the accumulated work of tlie bodies m terms ot tl:/^ vis viva ? WorkinA= ^^^^"3^ 5 Work in B = ^T^^^^y . ' . Accumulated work in A and B <4 ^ ■ 3*i 88 PRACTICAL MECHANICS. The latter part of the expression is called the vis viva of the bodies A and B. • • . Accumulated work n^^: i the vis viva. Ques. 23. What must be the weight of another body W, having the velocity of 8 feet per second, so that its accu- mulated work may be the same as that of the two bodies mentioned above ? In this case, the work in W »2 X w ^ X 3*i- but this will be equal to the work in the bodies mentioned above ; therefore, by reduction, 8^ X W = 5^ X 4 + 9^ X t .-. W = 10.4 lbs. In the above examples, if the bodies A, B, and W, are parts of the same machine, and if the velocity of A be in- creased, the velocity of B and W will obviously be increased in the same ratio ; hence, it follows, from the expression, that W will not be altered by any change in the velocity of the machine. OF THE MAXIMUM VELOCITY OF THE PISTON OF AN ENGINE. The maximum velocity will obviously take place when the pressure of the steam in the cylinder is equal to the total pressure of the resistances upon the piston ; for, so long as the pressure of the steam is greater than the resist- ances, the motion of the piston must be accelerated, then the work remaining in the piston, at this point of the stroke, will be equal to the work accumulated in the ma- chine. From this equality the velocity may be found. PRACTICAL MECHANICS. 89 Ques. 24. The pressure of the steam is 40 lbs., the steam is cut off at 2 feet of the stroke, and the gross load upon the piston 16 lbs. per square inch. At what point in the stroke will the velocity of the piston be the greatest ? Let X = the length of the stroke, then according to Boyle^s law, the pressure at this point 2 X 40 but the velocity of the piston is greatest when the pressure i= 16 lbs. per square inch, 16 =z 2 X 40 X . % : ^z. 5 feet. Ques. 25. The length of the stroke in a condensing en- gine is 10 feet, the pressure of the steam 30 lbs., and the steam is cut off at 2 feet of the stroke. Required the gross load upon each inch, and the point at which the velocity of the piston is greatest ? Dividing the space through which the steam acts expansively into four equal parts, and finding the pressures by Boyle's law, the total work of the steam upon 1 inch of the piston I 30 + 6 + 4. (15 + 1.5) + ^ X lO } + 30 -^ •j;^i> + l> + >5^(15 + ^.5) + -J& XlM^+;-5il X ^ = 15T.333. . • . Mean pressure of the steam, or gross load 157.333 10 = 15.73 lbs. Since the mean pressure of the steam is equal to the total re- sistances which the steam has to overcome throughout the stroke, the maximum velocity will take place when the steam, expanding itself, attains a pressure equal to 15.73 lbs. ; hence, by Boyle's law, 30 X 2 = 15.73 X ^, 15.73 : 30 : : 2 : 2:: 90 PRACTICAL MECHANICS. X being the point in the stroke when the velocity is the great- est ; .-. X = 3.813 feet. Ques. 26. If the area of the piston in the last example = 4000 square inches, and the weight of the mass moved, calculated to have the same motion as the piston, = 50000 lbs. ; what will be the maximum velocity of the piston ? Having determined in example 25, the point where the maxi- mum velocity takes place, namely, 3.813 feet, or 4 feet nearly, from the end of the cylinder, the first thing to be done is to find tlie work of the steam up to this point. Total work upon the whole piston to the point where the maximum velocity takes place m 30 + 15 + 4c c^^ -f n.i>4i; + ta X ^o + 30 X ^"1 X 4000 = >4063^8 ; the steam acting expansively over 2 feet of the stroke, and divid- ing this space into four equal parts. Now, the excess of this work over the work which is expended in moving the resistances, will leave us the work which is accumulated in the piston. But the work done upon the resistances = 15.133 X 4l©00 = ^5n33. Work accumulated in the piston = 4c©6318 — ^5n33 == 154^6>45. Put V = the maximum velocity of the piston in feet per second, then ^•2 X 50000 e^i = 15^1645. 14.1 feet. In the same manner the velocity may be found for any part of tlie stroke. The weight of the mass moved in any engine referred to the piston, may be found as in example 23. The mechanical principles evolved ia the last four examples are worthy of particu- lar attention. PRACTICAL MECHANICS. 91 VELOCITY ACQUIRED BY A BODY DESCENDING AN INCLINED PLANE. When a body freely descends an inclined plane by the force of gravity alone, the work accumulated in the body, will be simply that which is due to the perpendicular ele- vation, without any regard to the angle or curve of the plane. Hence, the velocity of the body will be that which it would acquire by falling freely through the perpendicular height. Ques, 27. What velocity will a body acquire in descend- ing an inclined plane, whose perpendicular height is 579 feet? t'' X 16yV = 519 ^^ = 36 ^ mzi 6. seconds. Consequently, the velocity acquired at the end = 6 X 32| = 193 feet. When the friction is taken into account, it will be more con- venient to employ the usual expression for accumulated work. Qiies, 28. A train of 10 tons descends an incline of 200 feet, having a total rise of 3 feet ; what will be the veloc- ity acquired by the train, supposing the friction to be 8 lbs. per ton. Work in the train = lO X ^^>40 X 3 — lO X » X ^OO ^ 51^00. Let V = the velocity. G4, r— = siioo, 3 V = 12.1 feet. 92 PRAdTTCAL MECHANICS. TO FIND THE FRICTION UPON AN AXIS BY EXPERIMENT. Ques. 29. The fly-wheel in example 20, was observed to make 28 revolutions before it came to a state of rest ; what part of the weight of the wheel is the friction upon the axis ? Let z be put for this proportional part, then the work destroyed by the frictioD -^ Z X ll^O X .5^36 X ^8 ; but this work must be equal to the work accumulated in the wheel, hence, Z X 164^^0. 00« = 1909.4.3. .•. z -= .116. OF THE RESISTANCE OF FLUIDS. The resistance of a fluid to the motion of a body is occa- sioned by the force necessary to displace that fluid. Now the fluid displaced must have the same motion given to it as that of the moving body ; hence the work destroyed by the fluid will be equal to the accumulated work in the fluid. Let, for example, the front of the body, presented to the fluid, contain 2 square feet, tlie weiglit of a cubic foot of the fluid 62| lbs., and the velocity of the body 9 feet per second, — then the weight of the fluid displaced every second = 9 X 2 X 62 1 lbs., but this weight has a velocity of 9 feet given to it ; therefore, the work expended in the displacement __ 02 X » X ^ X e^i Now this work has been destroyed, while the body has moved the space of 9 feet ; if 3/ be put for the resistance in lbs. this work will also be represented by y X » ; PRACTICAL MECHANICS. 93 » X \ « t>2 X« X 'i X «*ii J' ^ ^ ^ o¥T y «3 X *a X 6^i tt>4i Hence, the resistance increases with the square of the velocity, as well as with the extent of surface presented to the fluid. In extreme velocities this law does not hold strictly true. It appears also from certain recent railroad experiments, that the resistance of the atmosphere to the motion of the train, depends chiefly upon the length of the train, and not so much upon the extent of the frontage of the carriages. OP THE EFFLUX OF FLUIDS. If an aperture be made in the side of a vessel, kept full of water, then the accumulated work of the fluid, as it is being discharged, will just be equal to the work which gravity would perform upon it, while descending a space equal to the perpendicular depth of the orifice. Hence it follows, that the velocity of the discharge is equal to the velocity which a body would acquire in falling freely through a space equal to the perpendicular depth of the orifice. Ques. 30. The cross sectional area of a pipe is 36 square inches, how many cubic feet will it discharge per minute from a vessel kept constantly full, the depth of the dis- charge pipe being 16 yV feet 1 Let V be the velocity of the discharge per second, and ic the weight of a fluid particle in lbs., then, the accumulated work in the particle Work of the particle in descending IGy'g f^et — 16 j\ X W ; 94 PRACTICAL MECHANICS. 04ti = 1$5 yV X w ; !; = a/ 16 yV X 64^' :=.-. 32 1 feet, iubic feet 321 X 60=::482i. Discharge per minute, in cubic feet 36 144 Ques. 31. If the section of the cistern, in the last ex- ample be 8 square feet, and a pressure of 10000 lbs, be applied to a piston fitting the cistern and in contact with the water ; what will be the velocity of the discharge ? It is evident, that the effect of the pressure of the piston upon the efflux, will be equivalent to an additional column of fluid pro- ducing the given pressure on the piston. Height of additional column 10000 ^^ ^ , - FxW = ^' ''''- Height of the total column of fluid ^16-,V + 20 = 36J^; then proceeding as in the last example, the velocity r, will be found = V 64 i X 36 J^ = 48 1 nearly. OF THE WORK PERFORMED BY A JET OF STEAM. When steam issues from a nozzle, its elasticity is nearly the same as that of the surrounding atmosphere ; and the volume of steam at the nozzle will therefore be 1711 times that of the water from which it is raised. The motion of the particles of the steam at the nozzle, is due to the work of the expansion of the steam. Ques. 32. Steam is discharged from a nozzle, whose area is 8 square inches, with the velocity of 600 feet per second ; PRACTICAL MECHANICS. 95 how many horse power would a wheel perform, that takes up all the work in the steam*? Cubic feet discharged per second 144 Weight of steam discharged per second 12.5 X 62.5 1711 Accumulated work per second 600' X .4566 = .4566 lbs. 64i = ^S58*A ^58^ X 60 , . . Horse power = 330OO " — ^ '^' Ques, 33. If 9 cubic feet of water be evaporated per hour, when the area of the nozzle is 1 square inch ; how many horse power will the wheel haye ? Cubic feet of steam discharged peir second — ^^11 X 9 ^ "" 60 X 60 ' Velocity of the steam per second 3600 ' Weight of the steam discharged per second 3600 ^ Accumulated work per second ^ res./. X 9;j^x :O n x o ^^.^^ ^^s; From this expression it appears, that the work performed by steam in this manner, increases with the cube of the water evapor- ated ; 96 PRACTICAL MECHANICS. Horse power ^ ^^^^ = !>.. .Q. In these calculations the co-efficient of efflux has been neglected. OF poncelet's water wheel. In the common under-shot water wheel, the paddles are flat, whereas, in Poncelet's wheel they have a curved shape, A B ; so that the direction of 3 the curve at A, where the water meets the paddle, is the same as the direction of the stream. By this ingenious contrivance, the water rolls up the curved ir?- cline A B, without meeting with any sudden obstruction calcu- lated to occasion a loss of work. The channel has a depression at the point where the water falls from the paddles. Let Y be the velocity of the steam, and V that of the wheel, then since the point A of the pad- dle is moving away from the stream, the water will flow upon the paddle with the velocity V — v, and will continue to rim up the curved line until it has lost its motion, it will then descend, acquiring in its descent the same velocity as that which it had in its ascent, but in a contrary direction. If the wheel were not to move during its return, V — v would be exactly the velocity of discharge from right to left, but the paddle is moving the water from left to right with the velocity v, therefore, the absolute velocity of the water upon leaving the paddle will be V — v — v ==Y — Now all the work will have been taken out of the water, when its motion upon leaving the paddle is nothing, that is, when V — 2v = o, or V :== 2 V. In this case the water having lost all its motion, will simply drop from the paddle, and the work done upon the wheel will be equal to the work accumulated in the water PRACTICAL MECHANICS 97 of the stream. Moreover it appears, that the maximum condition is fulfilled when the velocity of the stream is double that of the wlieel. However, the distinguished in- ventor states, that, in practice, the velocity of the water, in order to procure it maximum effect, ought to be 2^ times that of the wheel, and that the modulus of the wheel is about y\. This wheel, other things being the same, will perform about twice the work of the common under-shot wheel. When there is motion in the water after leaving the wheel, the work accumulated in the water must be calcu- lated, and subtracted from the whole work originally in it, in order to obtain the work done upon the wheel. Ques. 34. Suppose 193 cubic feet of water flow upon the paddles of a Poncelet wheel per second, with the velocity of 8 feet., when the wheel moves at the rate of 3 feet per second ; required the horse power of the wheel ? This is obviously a case of maximum work. Weight of the water flowing per secoud = 193 X 62.5 = 12062.5 lbs. Work per second Horse power =^ *^*^Ot^O ~ .818. No account is taken of friction and other resistances. Ques, 35. Required the same as in the preceding ex- ample, when the quantity of water is 386 cubic feet per second, the velocity of the stream 8 feet, and of the wheel 2 feet ? 98 PRACTICAL MECHANICS. Hiere the velocity with which the water leaves the paddles = 8 — 2X2 = 4 feet. The work done upon the wheel per second = 18000. Weight of the water = 386 X 62.5 lbs. •i X r386 X 6^.5J »-i X f8 — *1) X "^a = 18000'^ Horse power = ^.^^^^ = 3^^" "^^ . THE FLY-WHEEL. To find the position of the crank corresponding to its maxi- mum and minimum velocity in a single acting engine. Let P, and D, be the required positions of the crank, and let P be supposed to be the constant pressure of the connecting rod, acting always in a vertical line. Let Q be the constant resist- ance, acting at 1 foot from the axis of the fly-w^heel, equivalent to the work of the engine. The motion will be accelerated from P to D. This acceleration will commence when the moving pres- sure is equal to the resisting pressure, and will cease under the same condition. The former will correspond to the position of minimum, the latter to that of the maximum velocity. Hence, at these two points, the moment of P must be equal to the moment of Q, and the point D will be as much below the horizontal line V, as the point P is above it. PRACTICAL MECHANICS. 99 .-. P X B = Q X 1 foot, -= Q X 1. (y)^ Again, by the equality of work, and putting r = Y ; and n = 3.1416, the circumference of a circle when the diameter is unity. Work of P in one revolution = *i r JP ; Work of Q in one revolution = *i rr ^ ; .-. ^ r Jf» = 'i ^ ^ ; (z). Dividing equation (y) by equation (z)^ OB 1 1 ^,^^ then = — = = .3183 ; r n 3.1416 this is the cosine of the angle P V ; hence from the tables, an^leP V=n°..2t'. TO FIND THE DIMENSIONS OF THE FLY-WHEEL. Let d and p be the maximum and minimum velocities of the wheel at the distance of 1 foot from the axis ; w, the weight of the wheel, and k the distance of the centre of gyration from the axis. Work of P from P to D ^PXP n^'LrP sin. ^1°..^T =:^^rPX .t>48 ; Work of Q from P to D -= 36«^ = ^ r JP X .3968, by sub- stituting the value of Q X 2 tt given in equation (z). Now the difference of these will give the work that goes on in- creasing the speed of the wheel between the points P and D, that is, work going into the wheel between P and D = "iir P y .»>i8 — *i r P X .39«8 :::::z r P X LIO*^ ; 100 PRAtJTICAL MECHANICS. Accumulated work at P = Accumulated work at D :=: ^i,g Difference = ^^ ^ x (d^ — p^), work gained from P to D. But this must be equal to the work before found fc3 t^ *ig. (d^ —p^) = r P X l.lOta^ ; (1). Let V be the mean velocity of the wheel at 1 foot from the axis, and let the velocities d and p, at each of the extremes, differ from the mean by the Tith part ; then, d = V -A and p z= v — — : n n .-. d^-p^ = --; (2). Let 17 be the work of the engine, and JV the number of strokes performed per minute ; then V = 2 X 3.1416 X -^ = -10472 X iV^; (3), .. U^'SirPJr .-. rP=-^; (4). Substituting the values given in equations (2), (3), and (4) in equation (1), and reducing the given quantities, there will be ob- tained finally "^^ 1^^ X80H.^. (5). After the same manner may be derived an equation, ex- pressing the relation of the elements for the double acting engine. PRACTICAL MECHANICS. 101 Ques. 36. In a single acting engine of 30 horse power, the number of strokes performed by the piston = 20 per minute, the distance of the centre of gyration of the fly- wheel from the axis = 10 feet ; it is required to find the weight of the wheel, so that the velocity of each revolu- tion may not vary by more than ^ from the mean ? n = 5y 17 = SO X 33000 ; ^ = 10 ; iV^ = 20. 5 X 990000 W lO'^ X 20^ X 808.2 = 500014 lbs. CHAPTER VIL OF THE EQUILIBRIUM OF FORCES AND PRESSURES. Let the board A W R, be placed upon three balls roll- ing freely upon a horizontal table. Take any three points, B, Q, S, in the surface of the board, and let the cords AB, D Q, R S, be attached to the points. These cords pass over pulleys, and have different weights hanging at their extremities. Under this arrano^ement the board, acted 102 PRACTICAL MECHANICS. upon by three forces, will roll upon the table until it comes to a position where the three forces destroy each otlier. When this position has been attained, the following import- ant relations will be observed, between the direction and magnitude of the forces. 1. The directions of the forces will intersect, or meet, in the same point 0. 2. From a scale of equal parts, take off r, equal to the units of pounds in the weight or force, drawing the cord Q D, and in like manner from the same scale of equal parts take off v, equal to the units of pounds in the weight, or force, drawing the cord A B, then if upon the two lines r, V, a parallelogram r ev he constructed, the diag- onal e will be the direction of the third force acting by the cord R S, and the units of length in the diagonal e, will giye the units of pounds in tlie weight, or force, acting by the cord R S. The forces represented by the sides r, and V, are called the component forces, and the force just equivalent to these two and represented by e, is termed the resultant. 3. Instead of three forces being applied to the board, let there by any number. Let P be any point taken in the board, and from this point let fall the perpendiculars P 7i, P 5, P ^, &c., on the directions of the forces, or if neces- sary, on the directions of the forces produced ; then the units of length in any of these perpendiculars, multiplied by the units of pounds in the corresponding force, will be the moment of that force, tending to turn the board upon the point P ; and then the principle of the equality of mo- ments will be obtained. FnXOB-^FtX r-^^F s X e. EXAMPLES. Ques. 1. Let A B, be a platform with a load C upon it, supported by a chain AD; it is required to determine by construction the tension of the chain, and also the amount and direction of the pressure upon the liinge at B, the wciglit of the platform being neglected ? PRACTICAL MECHANICS. 103 CONSTRUCTION. Through the point C, draw the vertical line C 0, inter- secting the direction of the chain in the point 0, join B and 0^ and from a scale of equal parts set off n equal to the units in the weight placed at C ; draw r n parallel to A D, and r e parallel to C 0, then n r e will be the parallel- ogram of pressures. Because, is the intersection of two of flic forces or pressures ; therefore, the direction of the third force must pass through ; but this third force must also pass through B. Hence it follows, that B is the direction of the pressure upon the hinge, and in like man- ner e, will give the units of pressure tending to break the chain. dues. 2. A pole, D, supported by a cord A D, carries a weight W ; it is required to find the tension of the cord, and the pressure on the pole ? CONSTRUCTION. On the line D N, mark off D n equal to the units of 104 PRACTTOAL MECHANICS. weight ill W, draw 7i7n parallel to DA, intersecting D m m, and from m draw m e, parallel to B n ; then Bnme will be the parallelogram of pressures ; therefore, the units m De will give the tension of the cord, and the units in D m will give the pressure upon the pole. Ques. 3. A beam, F B, is supported by a cord FA; it is required to determine by construction, the direction and tension of the cord B H, so that the beam may not change its position ? COXSTRUCTION. Through the center of gravity, C, draw the vertical line C K, produce A F, until it intersect this vertical line in the point K, join K and B, then K B produced, will ^ive the direction of the cord. Take K L, equal to the units in the weight of the beam, and con- struct the parallelogram of pres- sures, K N L V, then the units in K V, will give the tension of the cord H B. Ques. 4. A gate, A H, is supported by a pin turning in a socket at 0, and prevented from falling in the direction A D by a hook and loop at A ; it is required to determine the amount and direction of the pressure upon the pin 0, and the force tending to draw the hook A from the wall ? ' CONSTRUCTION. Through the middle, or center of gravity of the gate draw the vertical line D C, join the points D and 0, then D will be the direction of the pressure upon the pin. Take D B equal to the units in the weight of the gate, and construct the parallelogram of pressures F D B Q, then the PRACTICAL MECHANICS. 105 units in D Q will be the pressure upon the pin, and the units in D F will be the force tending to draw the hook. TO FIND WHETHER A PILLAR WILL STAND OR FALL WHEN ACTED UPON BY A GIVEN PRESSURE. Ques, 5. Let P be the direction and amount of the pres- sure, tending to turn the pillar upon the edge as a cen- tre ; and A Y a vertical line passing through the centre of gravity of the pillar, intersecting the line P, produced in the point A ; from a scale of equal parts, take A C equal to the units in the pressure P, and from the same scale take A B, equal to the units of weight in tjie pillar ; construct tlie parallelogram of forces AB D C, then A D will be the amount and direction of the single force tending to over- turn the pillar. If A D produced, intersect the base within the edge 0, the pillar will stand ; and on the contrary, if the point of intersection Q,fall without the base, the pillar will fall ; but if it intersect at the edge 0, then the pillar will just be upon the point of overturning ; the point Q is called the point of resistance. A structure will be more or less stable, according as the point of resistance^ Q, is more or less distant from the edge 0. Hence, the modulus of stability, may be defined to be, the ratio that Q bears to S. Thus, if Q were in the middle between and S, the modulus would be = -| ; if Q 106 PRACTICAL MECHANICS. were at S, the modulus would be 1, or the most stable pos- sible, under the given conditions ; and if Q were at 0, the modulus would be zero, that is, the structure would be on the point of overturning. Marshal Vauban considered, that the modulus of a good structure ought to be about I. PRESSURE OF ROOFS. Ques. 6. Let C A, and Q A, be the rafters of a roof rest- ing upon the side walls C and Q. It is required to deter- mine the thrust on the points C and Q, the roof being with- out a tie beam ? CONSTRUCTION. ' Let it be supposed, that the weight of the roof is equally distributed over the surface, and that a foot of length of this roof acts upon a foot of length of the side walls. Find the weight of each side of the roof one foot long, and let a weight W, be suspended from A equal to half tlie sum of these weights, then one half of the whole weight upon the rafter, C A, will act perpendicularly upon the wall at C, the other half acting in the weiglit W ; and in like manner one lialf of the whole weight upon the rafter, Q A, will act perpendicularly upon the wall at Q, the other half acting in the weight W ; now take A, equal to the units of weight in W, and construct the parallelogram of pressures A e, n, then A e will give the thrust upon the rafter A C, and A ii the thrust upon the rafter A Q. VRACTICAL MECHANICS. 107 Ques. 7. To find the tension of the tie beam Q C, or of a cord connecting 'he feet of the rafters ? CONSTRUCTION. Take C a = A e, the thurst upon the rafter AC, draw a P perpendicular to C Q, and construct the parallelogram at C P then, the pressure C a is equivalent to the two pres- sures C t, and C P ; now the latter pressure gives the teu-_ sion of the cord which would be produced by the thurst of the rafter C A ; but the rafter A Q, will P^o^^^ce an equa and opposite tension, therefore, the units in 2 X C F, will give the units of tension of the cord C (.^. Ques. 8. To find the amonnt and direction of the pressure of the roof tending to overturn the side walls ? It has been shown, that besides C A, the thrust upon the rafter, we have a vertical pressure upon the wall, arising from one half the weight borne by the rafter, lake ^, equal to the units in this vertical pressure and construct the parallelogram Caqt, then the diagonal C g, will give the amount and direction of the pressure of the roof tend- ing to overturn the wall C. Ques. 9. Pressure of fluids on Embankments. Let H B, be the side of a vessel filled with water, then the pressure of the fluid upon the point P, which may be anv point in the side, will be due to the perpendicular depth H P. That is, in other language, the pressure is in pro- portion to the depth. The reason of this is, since gravity 108 PRACTJCAL MECHANICS. acts upon all the particles of the fluid, each particle presses on that which is next below it, and tlien, from the peculiar property of a fluid, this pressure is transmitted equally in every direction. In the base B C produced, take B A, equal to B H, the perpendicular dei:)th of the fluid, then the pressure upon the point B will De due to the pressure of a column of fluid whose height is B A. Join A and H, and from any point, P, draw P D perpendicular to B H, and because P D = P H, the pressure upon the point P, will be due to a column of fluid whose height is P D, and so on to any other point in the side of the vessel. Let us now suppose, that the depth of the water, H B, is 8 feet, and the length of the side 6 feet, then the whole pressure upon the side will be equivalent to the pressure, or weight, of a mass of fluid of the form of the wedge, A B H ; area of the triangle, A B H = 8 X Y =-32. Contents of the wedge — 6 X 32 = 192 eubic feet. Pressure = 192 X 62.5 = 12000 lbs. It may be observed, that the pounds pressure upon the side, or proposed surface of a vessel, is equal to the weight of a column of fluid, whose base is the a rea of the surface, and perpendicular height, the depth of the centre of gravity, or middle point of that surface. g — - depth of centre of gravity ; 8X6 = area of surface. o • •• 8 X 6 X -5- X 62.5 i^ 12000 lbs. as before. • Ques. 10. Find the pressure on a flood-gate whose breadtk is 8 feet, and depth 6 feet ? /» Pressure = 6 X 8 X -^- X ^ 2.5 =1:: 9000 lbs. Ques, 11. The depth of water pressing against an em- PRACTICAL MECHANICS. 109 bankment is 9 feet ; required the pressure upon each foot of length ? 1 X 9 X -^ X 62A:^ 253Ulbs. Ques. 12. Compare the pressure upon the sides of a cubi- cal vessel filled with water, with the pressure upon the bot- tom, allowing that the side of the base ^= a foot ? Pressure on the base =z a X a X (i X 62.5 ; Pressure on the sides =:= ^ x «- X ^ X 62.5 X 4 ; Pressure on the sides ] o -n. ^ • xi .r, _ • • Pressure on the base [ = ^' ^^^^ ''' ^^'^ P>'^'^"''^ '^'^ ^^^ sides is twice that on the base. Ques. 13. Required the pressure on the staves of a cylin- drical barrel filled with water, the diameter of the base being 3 feet, and the perpendicular height 4 feet ? Pressure = 3 X 3.1416 X 4 X ^ = 4n2.4 lbs. It will be readily seen, that there must be a certain point in the side of the vessel when filled with water, where a single pressure will exactly counterbalance the pressure of the water against the whole side. This point is called the ceiitre of pressure. It must obviously lie in the line n G, passing through tiie centre of gravity G, of the wedge of pressures A H B. But, B n == J B H, that is, the centre of pressure 7i, lies at one third of B H, from the bottom of the vessel. If the staves of a barrel be kept together by a single hoop, and if the barrel be filled with water, then the hoop must be placed at one third from the bottom. Ques. 14. An embankment R D, sustains the pressure of water, whose centre of pressure is at P ; it is required to determine the conditions of equilibrium, i&c, supposing the embankment to turn over upon 0, as a centre ? Let V 0, be the vertical line passing through the centre 110 PBACXriCAL MECHANICS. of gravity of the embankment ; P the centre of pressure of the water. Draw P F perpendicular to R 0, then the product of the pressure of the water by F, will give the moment of the water, tending to turn the embankment over R A H P<- Q V upon as a centre ; and in like manner, the product of the weight of the embankment by V, will give the mo- ment of the embankment, tending to turn itself in a direc- tion opposed to the pressure of the water. When these moments are equal, the embankment is upon the point of overturning. If the moment of the water be greater than that of the embankment, the structure will fall, and vice versa. In the following examples, the length of the em- bankment is taken at one foot, because, if it stand for one foot, it will stand for any other length. Ques, 15. Let R =^ 9 feet, D = 3 feet, and the weight of a cubic foot of the material 150 lbs. ; will the embank- ment stand or fall when the water is up to the top ? Surface upon which the water presses :=: 9 X 1 . 9 Pressure of the water = 9 X 1 X "5- X 62.5 -- 253U lbs. Distance of the centre of pressure from the bottom = P = 'I 1= 3 feet ; o Moment of the water = 253U X 3 -^:= 1593.75. PEACTICAL MECHANICS. HI Wei<-ht of the embankment = 3 X 1 X 9 X 150 = 4050 lbs. V — — =1.5 feet. ^ 2 Moment of the embankment = 4050 X -^ == 6015. So that the moment of the water is greater than that of the embankment ; hence the structure will fall. It will be found when the height is 12 feet, the thickness 5 teet, and the weight of a cubic foot of the material 120 lbs., the em- bankment will be just upon the point of overturning Ques. 16. What must be the height of the water in the last example, so that the embankment may be upon the point of overturning ? Let X be the height, then the moment of the water = :.X1X^X62.5X^= ^ X 62.5 ; ... 4!- X 62.5=6075; .\ X zz:^ 8.3 feet. Ques, 17. Required the base, or thickness of a rectangu- lar embankment, when the height is 15 feet, the Aveight of a cubic foot of the material 140 lbs., and the water stands at the brim, so that the structure may be upon the point of overturning ? Pressure of the water rr^ 1 X 15 X ^ X 62.5 :::== t031i Ibs. Moment = ^ X 103U = 35156.25. Let X be the reqmred thickness, then the moment of the em- bankment ^ 15 X r^ X 1 X 140 X 4- ^" ^' X ^^^^ 5 112 PRACTICAL MECHANICS. .*. x' X 1050 = 35156.25 .•. X -^ 5.79 feet. Ques, 18. Let the embankment liave the form of a tra- pezoid, A H R C, where A B = 3 feet ; B C = 2 feet ; R C = 9 feet, and the weight of a cubic foot of the material = 100 lbs. ; will the embankment stand or fall, when the water is at the brim ? Let the embankment be divided into two parts, namely, the rect- angular R B, and the triangular part A B H. Now, the vertical line passing through the centre of gravity of the triangle, will cut the base at n, 2 feet from the centre of motion A ; and a line through the centre of gravity of the parallelogram H C, will cut A C at the distance of 4 feet from A . Weight of A B H = ^ ^ ^ x 1 X 100 := 1350 lbs. Moment of A B H = 1350 X 2 :zz= 2700 ; Weight ofBHRC=2X9XlX 100= 1800 lbs. Moment of B H R C = 1800 X 4 = 7200 ; Consequently, the moment of the whole embankment = 2700 -f- 7200 = 9900. The moment of the water -^ X 62i X -^ henc&it follows that the embankment will stand. = 9 X 1 X = 15931 PRACTICAL MECHANICS. 113 Ques. 19. Required the modulus of stability of the struc- ture, R H D, (fio-. example U,) when D = 5 feet ; D H = 9 feet ; the weight of a cubic foot of the material in O R H D = 150 lbs. ; the water is up to the top ? The weight of the structure — 5 X 1 X 9 X 150 = 6750 lbs. The pressure of the water -= 253 1^ lbs. Draw the rectangle R D, take D P = 3, draw P F par- allel to D, intersecting P P in* the point C ; from any diaa-onal scale mark off C n = 6750 and C t = 2531 ; con- struct the parallelogram nt, — then the diagonal Ce pro- duced, will intersect the base in Q, the ratio ot Q to U V will be found to be about 13 to 25. Ques 20. The breadth of a flood-gate is 10 feet, and the depth 6 feet, the hinges are placed at 1 foot from the respec- tive extremities of the gate. It is required to find the pressure upon the lower hinge ? In this example, the pressure of the water on each half of the gate MC — 6 X 5 X X 62.5 = 5625 lbs. P ^ ^tnueiit.]y tlie wall will stand Ques. 24. What is the thickness of the wall, in the last example, so that it may be upon the point of overturning ? Let X = the required thickness, then the moment of the wall = 40 X 1 X :?: X 120 X "4- — 2400 x'' ; when the wall is upon the point of overturning, the moment of the wall, must be equal to the moment of the pressure of earth. .-. 2400 x' ^ 183040 ,•, X =-= 8.7 feet. Ques, 25. Let R = 9 feet, D = 3 feet, the weight of a cal)ic foot of the material in R D = 150 lbs., the water is to the top H ; besides, the emljankment is supported by 118 PRACTICAL MECHANICS. R II earth L 0, of mean quality, L == 3 feet, and level at the top ; will the embankment stand, when the weight of a cubic foot of the earth = 120 lbs. ? It has also been found by experiment, that when a wall is supported by the pressure of earth, the weight of the equivalent fluid is about 6 times the weight cf the earth. Consquently, the earth L 0, may be considered a fluid, the weight of a cubic foot of which = 6 X 120 = t20 lbs. .•. The moment of this earth = 3XlX-^X720X^= 3240 ; Moment of the wall OH = 3xlX9xl50X-|-== 60t5 ; Moment of the water zzz: 9 X 1 X -4- X 62.5 X -|- = ^^93.75 ; .-. 3240 + 6075 is greater than 7593.75, and hence tbe wall R H D will stand. In these examples, the earth is supposed to be level at the top, but when the earth has its natural slope, a similar method of calculation will apply. When the wall resists the pressure of the earth, take \ of the weight of the earth, for the weight of the equivalent fluid ; and on the contrary, PRACTICAL MECHANICS. 119 Avheii the earth resists the pressure of the wall, take i the weight of the earth for the weight of the equivalent fluid. Ques. 26. Required the same as in example 23, when the earth has its natural slope Q R, and the thickness of the wall = 8 feet = B C ? tl Q B C On this supposition, the weight of a cubic foot of the equivalent fluid == 100 X i = 12i lbs. The moment of the earth C Q R 40 40 = 40 X 1 X -y- X 12i X -^ == 133333 ; The moment of the wall = 40 X 8 X 1 X 120 X Consequently, the wall will stand. 153600 ; Ques. 27. A revetment wall, F Q B C, is 30 feet high = B F ; B C =6 feet. On one side, G Q, earth of mean quality is sustained level witli the top, and on the other side, F B, the earth has a natural slope A D, and rises to the height B D =^ 5 feet ; will the wall stand or fall, supposing that the weight of a cubic foot of the earth to be 120 lbs., and that of the wall 130 lbs. ? ^-•^ PRACTICAL MECHANICS. levd Of 'tll'rw^l?' "^•^™^"* ?! f't e^rth, Which rises to the le\el ol the wall, is opposed bv the sum of the moments of the^wall, and the earth on the other side having rnatural Weight of the wall = 30 X 1 X 6 X 130 = 23400 lbs. Moment of the wall = 23400 X -|- == 70200 ; theTloptgtar'tir'" '^^^ '' ''' ^^"'^^'^"^^ ^^''' ^^^ -^Pect to = 120 X i = 60 lbs. Pressure of this earth = 5 X 1 X 60 X "4- = 750 lbs. Moment = 750 x — = 1250 ; Total moments sustaining the wall r= 70200 + 1250 = 71450 Moment of the level earth = 30XlX^Xl20x.ni6X^ = 92664; Since the latter moment is greater than the sum of two the former, it may be concluded that the structure will faU. OP FLOATING BODIES, AND SPECIFIC GRAVITY. .nrJof "Ilf/fl*^-^! ^^^*',' ^* i^«»stained by the upward pres- « re, t^l 'V ^""^ ^' *'''^'*' ^^ ^" equilibrium of pres- uies, the upward pressure must be equal to the • c- O H a o ^ S ^ aj « Is S C3 !=^ S S". a ti t) o £ c^ O B^ ^^ 8399 7700 7066 11332 934 657 Ques. 28. A barge, (supposed for the sake of simplicity, to be of a rectangular shape), is 10 feet long, 5 feet broad, and 4 feet deep, outside measure, the thickness of the plank- ing is 2 inches, and the weight of a cubic foot of the tim- ber is 50 lbs., to what depth will the barge sink when loaded with 4 tons 1 Content of the exterior s«lid = 10 X 5 X 4 = 200 cubic feet. Content of the interior solid = 91 X 4f X 3f = 172.92. 200 172.92 Timber in the barge, 27.08 cubic feet. Weight of timber in the barge = 27.08 X 50 = 1353 lbs. 122 PRAfTICAL MECHANICS. Let X be the depth of the water displaced, then the weight of the displaced water = a: X 10 X 5 X 62.5 = 3125 x; .\ 3125:?:= 1353 -f 4 X 2240 ,•, 2; ,= 3.3 feet. Ques, 29. How many tons will just sink the barge in tlie last example ? Let 3/ = the number of tons ; 200 X 62.5 = 12500 ; .•. 12500 = 1353 + y X 2240 .•. y i:^ 4.9t6 tons. Ques. 30. What weight will sink the barge in example 28, to the depth of 3 feet ? 10 X 5 X 3 X 62.5 = 9375. Let the required load = z^ then 9375 ^ 1353 + z X 2240 ,*. 2r := 3.58 tons. Ques. 31. How far will the barge sink when empty ? Put V = the required depth, then 10 X 5 X ^ X 62.5 = 3125 v; .'. 3125^=1353 V = .43296 feet. Ques. 32. A body weighs 49 grains in air, and 42 grains in water ; what is the weight of a cubic foot of the 'sub- stance ? The loss of weight = 49 — 42 = 7. This loss is the weight of water having the same bulk as the body. PRACTICAL MECHANICS. 123 Consequently, the number of -times the body is heavier than water 49 = — -— ^^ 7 times ; • But the weight of a cubic foot of water = 1000 ounces, there- fore, the weight of one cubic foot of the body = 1 times 1000 ounces = 7000 ounces, and this number is called the specific grav- ity of the body. Ques, 33. A cubic foot of timber sinks 9 inches in water ; required the specific gravity ? In this case, the weight of the displaced fluid will be the weight of a cubic foot of the timber. Therefore, the specific gravity = 1 X 1 X ^ X 1000 = 750. Ques. 34. A Solid, whose weight is 60 grains, weighs 40 grains in water, and 30 grains in sulphuric acid ; required the specific gravity of the acid ? Weight lost in water = 60 — 40 = 20 grains • Weight lost in acid = 60 — 80 = 80 grains. Therefore, the number of times the acid is heavier than water — 20 — '^ ' Hence, the specific gravity of the acid = li X 1000 = 1500. Ques. 35. A piece of metal weighing 36 lbs. in air, and 32 ^ lbs. in water, is attached to a piece of wood whose weight is 30 lbs., and then the compound mass is found to weigh 12 lbs. in water ; required the specific gravity of the M'ood ? Weight of water, equal in bulk to the metal = 36 — 82 _:. 4 lbs. 124 PRACTICAL MECHANICS. Weight of water, equal in bulk to the compound = 36 -f 30 — 12 = 54 lbs. Weight of water equal in bulk to the wood = 54 — 4 = 50 lbs. But the weight of the wood is 30 lbs., that is bulk for bulk, the wood will be three-fifths of the weight of the water. Conse- quently, the specific gravity of the wood 3 = -^ X 1000 = 600. OF THE LIMITING ANGLE OF RESISTANCE. Let W be a material particle acted upon by a pressure P, in the direction of and equal in magnitude to P W. Complete the rectangle A C, then will C W be the pres- sure upon the plane, and A. W that part of the force P, which tends to give motion to the particle along the plane. Put a for the angle, P W C, then. Pressure on the plane = C W — P cos. a ; and the resistance of friction =: f P COS. a ; The efi*ective pressure tending to move the particle in opposition to friction — A W = P sin. a ; Motion will or will not take place, according as their pressure is greater or less than the resistance of friction ; PRACTICAL MECHANICS. 125 and when motion is upon the point of taking place, the one must be equal to the other, and then the angle a, is called the limiting angle of resistance ; Therefore in this case / P COS. a ^=:^ P sin. a sin. a .•. / =■ = tan. a, COS. a This equation shows that the co-eflficient of friction is equal to the tangent of the limiting angle of resistance. If the pressure be applied within the angle, then no motion can take place, however great that pressure may be ; and on the contrary, if the pressure be applied without this angle, then motion will take place, however small that pres- sure may be. Foreign writers make this property the basis of their theory of machines. Ques. 36. The co-efficient of friction, / = .683, what is the limiting angle of resistance ? From a table of natural tangents it will be found, that the angle 34°. .20' has a natural tangent nearly equal .683 ; hence, the limit- ing angle of resistance := 34°..20^ Ques, 37. Equilibrium of the lever, and the wheel and axle, taking the friction upon the axis into account ? Let P Q be a lever, having the circular axis C T, turn- ins: within a circular socket. In order that motion should take place in the direction of the pressure P, the resultant of the two pressures P and Q must fall on tlie left side of the centre of the axis, and by the last problem, this re- 126 PRACTICAL MECHANICS. sultant C T, must be inclined to T at the limiting angle of resistance. Taking T, therefore, as the centre of motion, by the equal- ity of moments for the state bordering on motion, PXPF = QXQF. Ques. 38. Let P = 14 inches ; Q =- 12 inches, P = 27 lbs., and angle T F = 30", T = 3 inches ? The natural sine of 30° = -.500000.* OF =sin. a 30° X 3 = 1.5 ; PP = 14 _ 1.5 = 12.5 ; Q F = 12 + 1.5 = 13.5 ; .-. 2t X 12.5 = Q X 13.5 .•. Q = 25 lbs. • Ques. 39. Required the same as in the last example, when the angle T F = 40°, and P = 100 lbs. ? Natural sine of 40° ^ .642^876 ; ,643 will be near enough for practical purposes. P =:= sin. 40 X 3 = .643 X 3 = 1.929 ; PP:^ 14_ 1.929 = 12.0n ; QF = 12+ 1.929= 13.929; .•. 100 X 12.071 = Q X 13.929 ; .•. Q ^ 86.6 lbs. Ques. 40. Let P be the radius of a wheel, and Q the radius of an axle ; put W for the weight of the wheel and axle, then as this weight acts through the centre 0, then, PXPF = QXQF + WX0F; Suppose the radius of the = 20 inches, the axle 4 inclies, and the axis 2 inches ; required Q, when the power applied to the wheel = 100 lbs., the weight of the wheel and axle =: 80 lbs., and the limiting angle of resistance = 21°. * See the " Practical Model Calculator J"* by Oliver Byrne, the author of the present work. PRACTICAL MECHANICS. 127 The natural sine of 2P :=:= .3583679 .358 is near enough in practice. OF — sin. 21° X 2 = .716 ; p F z= 20 — .716 = 19.284 ; F = 4.716 ; P = 100 ; W= 80 ; .•. 100 X 19.284 = Q X 4.716 + 80 X .716 .•. Q = 396.7 lbs. Ques, 41. Let the pressures P and Q be inclined to tlie lever, then produce them till they intersect at C, and join the points C and ; now the resultant of the forces P and Q must be drawn, so that the angle T C = the limiting angle of resistance ; for this purpose, describe a circle upon the chord C 0, so as to contain the limiting angle of resist- ance ? Let this circle intersect the circumference of the axis at the point T, join T and C ; then on the line C P take off a space n, equal to the units in the pressure P, and con- struct the parallelogram of pressures C n m t, then the units in C ^ will give the pressure of Q, when the motion is upon the point of taking place. 128 PRACTICAL MECHANICS. Values of/, or tan. a, according to the experiments of M. Morin. Iron on oak 62 Cast iron on oak 49 Oak on oak, fibres parallel , , , , .43 Oak on oak, greased 10 Cast iron on cast iron ....,, ,15 Wrought iron on wrought iron 15 Brass on iron 16 Brass on brass 20 Wrought iron on cast iron 19 Cast iron on elm 19 Soft limestone on the same , .64 Hard limestone on the same. 38 Leather belts on wooden pulleys 4^ Cast iron on cast iron, greased 10 Brass on iron, greased 08 Pivots, or Axis of wrought iron, or cast iron^ or brass, or cast, iron pillow blocks ; Constantly supplied with oil , , .05 Greased from time to time 08 Without any application, or dry 15 TO DETERMINE GENERALLY THE FORCE OP TRACTION. Let W be a heavy body drawn alone the horizontal plane A W, by the force of the traction P, acting in the direc- tion and with the magnitude W P. This force into W A, and W C, the former force just overcomes the resistance PRACTICAL MECHANICS. 129 of friction, the latter tends to reduce the pressure of the body on the plane ; hence, |putting b = angle P W A. Pressure of W on the plane = W — P sin. b. .•. the resistance of friction = f(W — F sin. b.) But when motion is about to take place, this resistance is equal to the force W A, or = P COS. b ; .•• P COS. b = f W — sin. b) ; . r ^(^' • (z) COS. b + J sin. b ^ In the same manner, the traction may be found when the body is moved upon an inclined plane. Ques. 42. A stone weighing 2 cwt., is drawn along a horizontal plane, by a cord inclined at an angle of 35° ; required the force necessary to move the stone, supposing the co-efficient of friction to be = .70 ? Natural sine of 35° = .809 } '^^^'•^y- Natural cosine of 35° = .T X 2. — ^ = 1.15 cwts. .809+ .1 X .574 TO FIND THE LEAST TRACTION. In equation (Z), substitute the value of/ before given sin. a COS. a tlien by an easy reduction, it will be found that, W sin. a COS. (b — a) It is evident that P will be the least possible, when the denominator of this expression is the greatest possible, which will obviously happen when 130 PRACTICAL MECHANICS. h — a ^= Oy ov b = a ; that is, when the angle of traction is equal to the limiting angle of resistance. Therefore, in this case, P = W si7i. a; (J). Ques. 43. A stone weighs 10 cwt., and is drawn along a horizontal plane ; required the least traction, taking a = 23°; By equation (J) ; Nat. sin. 23° — .3907 P r= 10 X .3907 = 3.907 cwts. Ques. 44. Required the least traction, when the weight Df the mass is 9 cwt., and the rubbing surfaces are soft calcareous stone upon the same, and the limiting angle of resistance = 36°..30'. JSTatural sine of 36°..30' Least traction 1= 5.3532 cwt. .5948 9 Ques. 45. If a heavy body be moved upon a rubbing sur- lace, lying between two given points, by a pressure acting parallel to the surface, the work is always the same, what- ever may be the form, of the surface. Let W be a body moved on the plane A C, then by the resolution of pressures, the pressure of W on the plane = W COS. A ; W qmn, being the parallelogram of pressures, and since W n is perpendicular to A C, and mn parallel PRACTICAL MECHANICS. 131 to it, the angle m W n = the angle A ; therefore, the re- sistance of friction — f W COS. A ; Consequently, the total work on the plane A C z:= / Wcos. AxAC-{-WxCB -_^fWxAB-\- W X C B. This final expression is independent of the inclination of the plane, it being, in fact, the work expended in moving the body over the horizontal distance A B, added to the work due to gravity in elevating the body to the vertical space B C. i . ^ As a curved surface may be regarded as being made up of an infinite number of straight planes, therefore, the work upon the whole curve will be equal to the work done upon the horizontal projection of the curve, added to tlie work done -in opposition to gravity. When bodies descend a f=urface, the work of gravity becomes minus. This total work is obviously, iudependent of the nature of the curve. In general, it may be shown, that whatever may be the zigzag course pursued by the body, the work will always be equal to the work on the projection of this curve, added to the work due to gravity. When the in- clination of the plane is small, the horizontal distance may be taken equal to the length of the plane. See Chapters 1. and II. Ques. 46. What work would a horse be able to perform in drawing a load of one ton to the horizontal distance of two mites, up a curved incline, whose total elevation =-= 500 feet, the co-efificient of friction being ^- 3V ? Work on the horizontal line ^ _:^M®_ ^^2.x 5*180 = -iS«*AOO ; Woik due lo gravity x= '1^4« X 5i>« = IViOOOO, Total work = =53*J*£00 + 11*£000« = 1859*AOO, 132 PRACTICAL MECHANICS. If the body W, be upon the point of descending the plane by the force of gravity alone, then, the work necessary to . move the body is nothing ; hence, fWxAB— WxCB • •# / = tan, A ; a result before obtained. O THE PARALLELOGRAM OF PRESSURES PROVED ON THE PRINCIPLE OF WORK. Let B D, and A C, be two inclined planes intersecting each other at right angles in the point W. Let a body W, be placed between the two planes ; then the force with wnich the body tends to descend the plane D B, will be equal to the pressure which it exerts upon the plane A ; and the tendency to descend the plane C A will be equal to the pressure exerted on the plane D B. Let the vertical line W H be drawn through the body, and take W H = the units of weight in W ; from H let fall H C, perpendicular to the plane A C, and H D perpen- dicular to the plane B D, then putting A and B for the in- clinations of the planes A C and B D, by trigonometry W G -^ w X sin, A = the tendency down the plane A C. PRACTICAL MECHANICS. 133 Similarly, W J) = w sin. B = the tendency down the plane B D. But these forces W C and W D, are generated by the gravity of the body, or a force represented by the diagonal W H. Having thus established the proposition for the case of the rectangle, it may be very readily extended to the general form of the parallelogram. THE EQUILIBRIUM OF THE ARCH. When the centres of an arc are taken away, the crown almost invariably sinks ; this occasions the joint at the crown to open at its lower edge, and at the same time a certain portion, D V P N, of the arc to turn upon D as a centre, thereby producing a rupture, or opening, in the ex- terior edge at this point. The same effect will take place in the other half of the arch. These two equal portions which thus tend to break away from the general mass, exert a horizontal pressure along the line P C, thereby tending to cause the walls of the structure to turn on their outer edges. The arc will under- go a rupture at that point where the portion D A^ P N, so breaking away, will produce the greatest horizontal thrust ; 134 PRACTICAL MECHANICS. for this point must, obviously, be the yielding point of the arc. To find the point of Rupture. Let B Gr, be a* vertical line, passing through the centre of gravity of the mass D V P N, and intersecting P C in the point B ; join the points B and D, take B equal to the units of weight in the mass D Y P N, and construct the rectangle F C B, then the units in B C will be the hori- zontal thrust, if it be greater than any other so determined, and the point D will be the point of rupture. MM. Clapeyron and Lam^e, found that the resultant D B forms a tangent to the intrados of the arch, when the line of rup- ture is assumed to follow the vertical line D V. This prop- erty gives a very easy method for finding the point of rupture ; for if upon constructing the figure, as already described, it is found that B D touches the curve, then I) will be the point of rupture. The most troublesome part of this calculation, consists in finding the centre of gravity of the mass D V P N. However, the following method will be found sujfficiently accurate for practical purposes : If D Q be divided into n equal parts, ai*d lines be drawn from the points of division perpendicular to D Q, cutting the intrados and extrados ; let x and y be put for D Y and N P, the extreme ordinates respectively, and A, B, C, &c., for the first, second, third, &c., intermediate ordinates, or vertical distances between the intrados and extrados, then putting s = the common distance between the ordinates six + (3 7?.— l)Y + 6(A-f2B -f3C + &c.) j ■pv n __-i 3|X+Y+2(A + B+C + &c.) I If that part of the divisor which is within the brackets be multipled by | it will give the area of D Y P N. Having ascertained the horizontal thrust, B C, the centre of gravity of the whole mass, including the semi-arch X P N H, with its load, if any, and the pier H X ; then suppos- ing the pier to turn upon its outer edge, if the moment of this mass exceeds the moment of the horizontal thrust, the structure will stand, and vice versa. PRACTICAL MECHANICS. i^i Ques. 47. The radius of a semicircular arc is 11 feet, the thickness of the crown 18 inches ; the mason work is built level with the crown, and the weight of a cubic foot of the material = 120 lbs. ; required the point of rupture, and the horizontal thrust, taking a depth of 1 foot of the masonry ? Construct a figure on a scale of an inch to the foot, and dividing D Q into six equal parts, it may be found by calcu- lation or measurement, that X = 7.8, A = 5.4, B rr= 3.8, C = 2.8, D = 2, E = 1.6, Y = 1.5, and n = 6, and D Q = 10, and hence s = V^« By the formula it will be found, that D G r=rr 3.53 feet, and the weight D V P N = 4000 lbs. Take B == 4000, and construct the parallelogram 0, then B C = 1900 lbs. nearly ; and as B D forms a tangent to the curve, the point D will be the point of rupture. The point D is 65"^ from N, the crown. Ques. 48. If the piers in the last example be 4 feet thick, and their height 28.5 feet measured to the level of the crown ; will they stand or fall ? 136 PRACTICAL MECHANICS. In this case, tne weight of the whole semi-arch ^^ 5100 lbs., taking as before the depth, one foot. The distance of the centre of gravity from the outer edge of the pier =7.6 feet. Weight of the whole pier :=i: 4 X 1 X 28.5 X 120 = 13680 lbs. Hence the moment of the pier = 13680 X 2 = 21360 ; The moment of the semi-arch = 5100 X 16-^38160 ; and the moment of the horizontal thrust ^ 1900 X 28.5 = 54150 ; The sum of the two first moments being greater than fche moment of the thrust, it follows that the pier will stand. But it will be found in the same manner, if the pier be only three feet thick the structure will fall. When the pier is 3.3 feet thick it will be on the point of over- turning. CHAPTER Vni. WOODEN AND IRON BRIDGES. The equilibrium of wooden and iron bridges will be considered after I give some investigations, illustrating a higher method of investigating mechanical subjects. Variable motion is caused by a force, which, acting con- tinually upon a moving body, goes on at every instant in- creasing or diminishing its velocity. Such force is measured at each instant, by the ratio of that very small increment or decrement of velocity, w^hich it causes in the moving- body, to that very small time in which the increment or decrement is caused. When this ratio is constant, or so that in equal times equal degrees of velocity are added to, or taken from the moving body, the accelerating or retarding PRACTICAL MECHANICS. 137 force is constant, but when this ratio is variable, the force is also variable. It is more convenient to use the symbol [~ instead of the usual symbol d, employed in the calculus ; [ "~ must not be taken for ^, which is the well-known sign of the square root. Instead of dy, dx, &c., I write I y, I X, &c. In the case of a variable motion, Put P for the acceler- ating force, Fthe velocity, S the space, and T the time, the relations between these qualities may be found by the equa- tions V^ ^- (I) • Other general forms may be deduced from these, for ex- ample, from (I), re" [T = 4^, and from (II), — •, and V F .-. F[S =^V (F'. Again eliminating V from (I) and (II), Ffr m. On the supposition that T is the independent variable, the last equation becomes 188 PRACTICAL MECHANICS. 2 The above equations relate to accelerated motions, by making |~F negative, they are adopted to retarded motions. In an equally accelerated motion F, is constant, from (TI) there- fore, f F\f --^ f\v, integrated becomes F T = V; (III), The velocities, therefore increase as the time when the force is constant. It is not necessary to correct (HI), when the body moves from a state of rest, for then when T = o, V =^ 0, and F T == o. But if the body has a velocity a, already impressed, then (III) requires correction and be- comes Y = a + FZ (IV) ; for V, becomes a^ in (IV), as it should do, when T^= o. When the force is retarding, V becomes negative and (IV) becomes Y ^a — F T It must not be forgotten that \~ stands for the symbol d. Again integrating F\s ^=^- T^ | F, S = -T-rr + constant. (T). 2i' ' ^ If tlie body moves from a state of rest (V) requires no correction, or addition of a constant, for in this case S must be equal to 0, when F= 0. But if the body should have a velocity «, before the force F commences to act, then the space passed over in the time T will be 2i^ '^F 7 > PRACTICAL MECHANICS. 139 _ V — a' because, when T = 0, the body had a velocity a, and hence must have passed over a space ft2 ~ 2F , before, ^- 2F commences to be described. (VI), Equation (VI), is readily obtained by integrating and correcting the general equation JP [^ = V [ F , the result in this case is the equation S^ a'' — V 2 F Another sent of general expression may be thus formed, — From (II), p [ T = [1^, integrating and F T — V = j=, according to (I), .-. {~S ^ F T\T integrating, and the result is S = -?^^ + constant ; (VII. ^ If the body move from a state of rest, (VII) requires no correction, for S and T vanish simultaneously, in this case also, by transposing ^ S and consequently the measure of the constant accelerating force F is obtained, when the space, vrhich the moving body has described in a given time T, is knovrn. Near the sur- face of the earth gravity may be considered a constant 140 PRACTICAL MECHANICS. force, and since a heavy body falling from rest describes I&T2 feet in one second F = 32i feet, which is generally represented by the letter g, by most writers and mechanics. It must be remembered that [""" stands for d of the Cal- culus, and must not be taken for ^% the sign of the square root. Correcting (VII) for accelerated and retarded motion, with a constant velocity a, before the force F commences to act, F T" S = a T -\ ~ — , (accelerated.) F T" S^a T — , (retarded.) The reasons for these last expressions are obvious, for supposing the force F to cease to act during the time T then (VII) becomes S =:= + constant ^^ a T the uniform velocity a multiplied by the time T, THE VERTICAL MOTIONS OF HEAVY BODIES THROUGH RESISTING MEDIUMS. For want of sufiBcient experiments, Mathematicians take for granted, that the air and other fluids offer resistances proportional to the squre of the velocity of the body mov- ing in them. Let V be the velocity, then this retarding force may be expressed by v '^ multiplied by a constant co-efficient. A, so that the retarding force is A v ', the value of A will depend on the figure of the body, and upon the ratio of its specific gravity to that of the fluid. Consequently, if a body de- scend from a state of rest through a resisting medium, the PRACTICAL MECHANICS. 141 accelerating force is g — Kv\ whicli put for F in (I) and (II) gives, putting t for the time, and s the space, (g — Av'') \~ ~— \T, and (g — Av') [S ^ [V. For the convemence of calculation put m = the square root of g, and n = the square root of a ; then k.=- gn\ then the two last equations become There will result the two following equations by inte- gration and correcting the results on the supposition, that when t = Oj s = 0, and v = o ; V = ;^^ — . — :r^ and s z:^ ■— —- log -:, ^ — r J a third equation may be found by eliminating v, 1 1 1 ( g'^t —gnt \ CURVILINEAR MOTION. Suppose the position of the moveable point to be refer- red to three coordinate axes, and let the acting forces be reduced to three, P, Q, R, parallel to these coordinates. Then [T being the small arc which the point will de- scribe in the time [T, its velocity at this point will be [^ [t because [T is the diagonal of a parallelepiped, which has for its sides [^, [y, [T, the moving point cannot, in the instant [7, describe the small arc [T, without at the same 142 PRACTICAL MECHANICS. time describing, in the directions of the ordinates of x, y, z, the small spaces ["^, [y, [T; for instead of the velocity 1^ the point may be imagined to move with the veloci- [T ' ties [T ' [T ' [T ' parallel to the ordinates of x, y, z^ hence the equation derived from (I) and (II), gives three following equations, Integrating these equations, the partial velocities \ X \ y \ z are obtained, from the composition of which results the ab- solute velocity of the point. Integrating again, and the ordinates x^ 3/, z, are found in terms of t, and hence the place of the point at any instant. When t is eliminated, there will remain two equations between x, y, z, the equa- tions of the curve described by the moving point. The student must keep in mind that [~" stands for the differential symbol d. When the forces acting on the moving point are all in the same plane, the trajectory will lie wholly in that plane, and its equation will be found by eliminating ^, from the two equations. which may be written -= i ^ nnri O — X ^-. andQ = -i.^, (VIII.) t being considered the independent variable, and f"^ is put PRACTICAL MECHANICS. 143 instead of d'' and, [" ^ instead of {dxY, which is written d X'. Supposing the ordinates to be rectangular, then Multiply (VIII) by \^, [7, &c., and add together, 2 _2_ _ j p [T + Q [7 + i^ F = ^^ ^^ + ^^^-[^ + ^ -^^^ 2 • .-. P [T + Q [7 + J? [T 2 .'. P{-7 + Q\J + R\T=-v\l>, (IX.) because, _ V = 4=^ and r — [ « r * I ^ On ^Ag motion of Cannon Shot. If a body be projected obliquely, with a velocity due to the altitude A, (the resistance of the air being neglected), aud if the abscissa of x be taken in the vertical line drawn through the point of projection, and the ordinates of y parallel to the direction of projection, the equation of the curve described by the projectile will be .y' = 4 hx. Taking the general equations (VIII,) 144 PRACTICAL MECHANICS. P [T = f -[^ and Q [T =- I" ^^ [T - ' ' I [T ' putting P — g and Q,-^ o^ and integrating, then --^ ^^ g t + constant ; -^~r- == constant. To find the constant quantities, it must be observed that when t = Oj the velocity in the direction of x = o, . • . -L^ + constant = o, hence constant o ; \ i _ requires no correction. Again, when t = o, the velocity in the direction of y is that due to the height A, which is constant = (2 gh) ^, the required constant. y [T ••. l^ = (2gh)^, (XL) Integrate (X) and (XI), and consider t, as before, the in- dependent variable, then x = igt'siiidy=^ {^g^)^t, tliese results require no correction, for when t = o, x = o and y = 0,^ condition which the resulting equations satisfy without the aid of a constant quantity. Eliminate t, then y^^ihx, (XII.) an equation to the common parabola, having for its diame- ter the vertical drawn through the point of projection ; and the parameter is 4 times the height due to the initial velocity of the projection. Let the coordinates be changed, and instead of taking A P for the abscissa of x, and P M for the ordinate of y, assume the horizontal line A Q - a:, and the vertical Q M PRACTICAL MECHANICS. = 3/1 ; let the angle of elevation TAB R parallel to A B. 145 0^ and draw P M R zLz: P M sin. ^; P R = P M cos, 6 ; «r a^j + 2; = y sin. ; y^ = y cos. 0, . • , Xj :::^ t/, tan. 6 X But (XII), X : y Vi x^ = y^ tan. 6 — 4 h cos. ^ 3/i 4 k cos. ^ 6 ' (XIII.) THE PATH OF A PROJECTILE IN THE AIR. Let g r express the resistance of the medium. This force being in direction of [T ; now, if the curve be referred to an horizontal axis as abscissas of x, and the ordinates y vertical, the resistance g r may be resolved into two forces \ ^ -, \ y \ s \ s one parallel to the axis of x and the other to that of y. Consequently, the body will be acted upon in the direction 146 PRACTICAL MECHANICS. of X by the force — 9; r —=:- and at the same time in the direction of y by the force — g — gr -~~~ . Then the I « general equations (VIII,) are easily applied, they become -gr-\^\T= [4^, (XIV.) - -^rT-^riI-[T= T-j^, (XV.) From these equations t is to be eliminated. On the sup- position that ["^ is constant __ 2. I » h ... gr [T' = [T [T, (XVI.) _ 2 In works on the calculus \ t^ is written df", and |"T is written d '' t, the symbol d is always represented by [~, and is not to be taken for y, the sign of the square root. 2 gr \ t jC\. 1 ' ±j 11 • [^ > r \J _ 2 [7 [7 [ 2 [T = 2 [7 " [7 r7 gr [7 2 — gr [T PRACTICAL MECHANICS. 147 Consequently, (XV) becomes - g \T' = [7, (xvii.) From (XVII) another equation is readily found, [7 = 2 ^ [T [T, (XVIIL) 2 [T, [T, are easily eliminated from- equations (XVI), (XVII), (XVIII), hence 2r f7^ + \T \J = 0, (XIX); which is the equation of the trajectory ; when the laws of resistance are determined, the curve described by the pro- jectile becomes known from (XIX.) Suppose the resistance to be proportional to the square of the velocity, then ipnt gr = gn' v^ — gn'' — ^ . But from (XVII) 2 —2 — \y g . FT' .«. r — — gn^ — 2- [7 Hence the equation of the trajectory (XIX), becomes 3 Sf.^2gn [T, (XX.) ■4^ divided by -^ 2_ ' • \ X \ X -^~ , hence, Integrating (XX), r7 It is easily observed that f -4^ divided by -L^, gives Loo-. JJi- ::::r. 2 /r ?r S + C 148 PRACTICAL MECHANICS. 2 or,-g, = CE ^Sn's^ ,^^j_^ I ^ C being the constant required after integrating, and log 2 c =^ C. To find c, for [y put its value — g [T\ then Now, if be the angle of elevation, and h the height due to the velocity of projection, at the beginning of the motion, or when 5 = 0, the horizontal velocity ^"^ -^ {2gh)icos.O. .*. c = 2 h CQS.' • • • . (XXI) becomes 2 A ^2gn's In order to abridge, put 2 gn^ ^^ m, and -^-i^ = p, then (XXII) becomes T x Since IT' + [7= = [T' •• • (1 + p')-' R = R, and multiplying (XXIII) by this last equation, r7(i + P-)' --fiJ^T, 2 /I > integration will give the following equation. ^ms PRACTICAL MECHANICS. 149 Putting P for the left hand member of this equation, and introducing the value of -p=- from (XXIII), the last equa- tion becomes _ m I X .-. r^ = ^ ^-^ , and P m — Cj m f s .,- f PH Pm— C,m'^ •^ Vm—Q,m' The constant Ci is found by considering that when s = Oj p =^ tan. 0, If these formulas could be integrated, x and y would be found in terms of p ; then eliminating _p, the equation of the required curve would be found. But since the formulas cannot be integrated in finite terms, the fol- lowing method of approximation may be employed, when the student cannot apply the Theorems of Laplace and Lagrange. INTEGRATION BY APPROXIMATION. Let 2/ = / X ["^ ; if when a: = a, then y = b ; and if when z =^ a, a -{- d, a -{- 2 d, a -\- ^ d .... a + n d ; then X = A, Ai, A 2, A3, ... An. and if the difference d be very small, the value of jf X ["x" from X = a to X ^^ a -j- 71 df will be, y — h + d (iA-|-Ai-}-A2 A3 ....iAw;. 150 PRACTICAL MECHANICS. • The truth of this is readily established, for in tlie small interval between x =^ a and x = a -f d, it may be assumed that the function X remains constant. And because at the beginning of this interval X := A, and at the end X = Aj it may be assumed that this constant value is the arithmet- ical mean, or X = J A + i Aj. .-. For this interval X f ^ i= ^ (i A + J A,). For the next interval, comprised between 2; = ^ -f- ^ ^"d x^^ a -\- '^d^ it may in the same way be shown that X = iA, +iA, ; r. X[T = ^(i A, +iA,). - Adding this value of X [T to the preceding, it appears that the value of y from x = a, to a; = a + 2 d, is 7/==5 + ^(iA + A, + I A,), and so on to the last. To apply this metliod to the determination of x and y in this case, it is known when p = tan. ^, then x = ; first take Now since the tangent p of the angle, which the curve makes with the direction of x, becomes gradually less and less, until, at the vertex of the curve it is nothing, and afterwards becomes negative, and goes on increasing in the descending branch ; make successfully p = tan, ^, tan, {0 — d), tan. {0—2 d), tan. (^ — 3 d), &c., until p = ; and then p = — d, — 2d. — 3d, &c., ad infinitum ; taking for d any number at pleasure : and the smaller it is taken, the more exact will be the discription of the curve. Cal- culate the values of A, Aj, A^, A,, &c., of the quantity P m — C, m when J) becomes successively tan. 0^ tan. {0 — d), tan.'{0 — 2 rf), &c. Then putting successively n = 1, 2, 3, &c., ad infinitum, the values of x will be found that correspond to the angles where j) — tan. {fi — d), tan. {0 — 2 d), &c,, &c. PRACTICAL MECHANICS. 151 Also, by the other equation _. f Pry y J Pm— 0,m ' the values may be found, which V P m — Cm acquires, when p becomes tan, ^, tan. {0 — d), tan. (0 — 2 d), cfec. ; proceeding in like manner, the values of y will be *<:nown, which correspond to the same angles, where p -^^ tan. (0 — d), tan. (0 — 2 d), &c. Hence the abscissas x, and the corresponding ordinates y of all the points of the curve, in which p becomes suc- cessively tan. {0 — d), tan. (0 — 2 d), &c NEW PRACTICAL SOLUTIONS OF IMPORTANT USEFUL PROB- LEMS, FROM COMBINING CONTINUED FRACTIONS AND DIOPHANTINE ANALYSIS. Question 1. It can be shown, that it is impossible to re- present the perpendicular on one of the sides from the opposite angle of an equilateral triangle in whole numbers, when the sides are expressed in whole numbers ; with this fact in view, let it be required to find a triangle, whose sides and a perpendicular may be expressed in whole num- bers, a.nd to have the three sides nearly equal. It is re- quired to have the difference between one of the sides, and either of the other two, less than any fractional part of either of the sides. Take any two numbers x and y, and according to Dio- phantine Analysis, 2xy will be the base, x"" — y"" the per- pendicular, and o:^ + y'' the hypothenuse of a rational right angled triangle. In an equilateral triangle, if the side = 1, the perpendicular = \ ^Y. Let a;'' + 3/^ m^ 1, and 2:^3/ :^i (3)^. 152 PRACTICAL MECHANICS. .•. x + y^ \l + i {3)hU = 1.3660254 X — y = I I — i (S)i U = 0.3660254 .•. X = .8660254 y zzi: .5000000. Now by continued fractions, the nearest whole numbers that will supply the required proportions may be found 8660254 /, . . 5000000 ^^-J"''^ 5000000 ., , 3660254 ^^- ''''''^ 3660254 2679492 (2. third 1339746 ... ,, 980762 ^^- >^^^^ 980762 ,^ .r.r 717968 (^--^/^^ 358984 .. . , 262794 ^^' ^^^^^ St (2- --'^ 96190 /, . ,^, 70414 ^ • - ^ 70414 xo • .7. 51552 (^- ^^^^^^ 25776 ,, , ,, 18862 (^- ^'""^^ 18862 .o 7 ,^ 13828 ^ 6^6^"^^^>^ 5034 {I > twelfth o^nr^ (2. thirteenth 1880 ,, . , ,, 1274- ^ * fourteenth 1212 ^^' fift^'^th 606 PRACTICAL MECHANICS. 153 The continued fraction will be ^ + I + 2 J_ i •^ 1 -4- * I 1 _4_ 1- » 2 J_ i •^ ^ 4- 4- I , + ^ + i &c. summing five of these terms the fraction \\ will be found, then 192 _|_ 112 _ 432 19' — IP = 240 2 X 19 X 11 ^ 418 So that 241, 209, and 120, will form a rational right angled triangle, and two of these triangles placed together will form one nearly equilateral, the three sides will be respec- tively 240, 241, and 241, and the perpendicular will be 209. If eleven terms of the continued fraction be summed, f f f- is obtained, .•. 989' — 571' — 652080 989 + 5n' =^ 1304162 2 X 989 X 5n ^ 1129438. These numbers may be divided by two, and then the three sides of the required triangle will be 652080, 652081, 652081 nearly equilateral, and the length of the perpen- dicular on 652080, will be exactly = 564719, a result easily verified. Ques, 2. It is not possible by plane geometry to divide an angle into three equal angles, yet, the tangent, secant, and radius of one third of any plane angle, may be found, in whole numbers, as near as we please, in the following way. Let the angle to be trisected ^ 41°.. 44' .. 33". 72. 3) 41°.. 44'.. 33". 72 13 .. 54 .. 51 .24. The natural tangent of tliis angle is found from the tables to be .2477386. 154 PRACTICAL MECHANICS. Let 2;^ — 3/^ =-z: .247^386, and '^xy = 1.0000000 .-. X = .7993650, y = 6254964. What follows is by continued fractions ; the alternate quotients are marked, ^rst secondj &c. 1993650 first 6254964 (1. 1738686 third ' 1038906 (1. 6254964 second 5216058 (3. 1038906 fourth 699180 (1. 699180 fifth 618252 (2. 21528 seventh 16206 (1. 339126 21528 . ,. sixth 123846 '^^^• 101640 5322 16206 eighth 15966 (3. 240 377 295 + * + I + + »'^ + -f &c. 377' 4- 295' 377' — 295' 2 X 295 X 377 229154 55104 222430 The three whole numbers 229154, 222430, 55104, forms a right angled triangle, one of whose angles will not differ from the third part of 41°.. 44'.. 33".72 by the third part of a second. Ques, 3. Find a body as near the form of a cube as we please, whose edges, diagonal of the base, and of two oppo- site corners, may be expressed in whole numbers ? Letx' + 7/' ^ (3)^ 2 X y -- (2) ^ and PRACTICAL MECHANICS. 155 ... ^ + 3,= |(3)i + {'2)i \i -Ut37n2 x—y= -j (3) 4 — (2) i !■ i — .5637^05 .♦. X — 1.1687108 y = .6050003. To find the continued fraction ; 1168^08,, . ^ 6050003 ^^- ^"^"^ 6050003 .J ^ 5631105 ^^- '"^'''^ 5631105 412298 1514125 1236894 412298 ., f .^ 211831 (.^•/''^'^'^ (13. third- 268931 (2- ^/'* 134467 8897 45491 44485 (15. sixth ml (^- ---^'^ 801 1 + + t'^ + 1 + + T^ ^lllih eighth 211 + &c., 118.19 6 118' ^''■' &c. this continued fraction summed gives for eight terms 11819'' + 6119'^ = 111118685, (A) 11819'^ — 6119^ = 102258831, (B) 2 X 11819 X 6119 -^ 144611284, (C) (B) is the length of each of the edges, (A) the exact length of the diagonal between two of the opposite corners. Further, if a four sided figure be constructed, each of its sides = (B), and (C), one of its diagonals, any one of the 156 PRACTICAL MECHANICS. four angles of this figure will not dilBFer from a right angle two seconds. All the lines are in whole numbers without decimals. If only five terms of the continued fraction be summed, the result = If , then 85^ + 44^ = 9161 85^ — 44^ zrr 5289 2 X 44 X 85 — 1480 There are four exactly equal square faces, the length of each side = 5289, the distance between two of the most remote corners o€ this approximate cube = 9161 ; the four sided ends are equal, the side of each end = 5289, and one of its diagonals = 7480. None of the angles of the ends differ i of a minute from a right angle. The Mc Galium Inflexible Arched Truss Bridge, ap- proaches nearer the standard of perfection than any other wooden bridge that have fallen under my notice, this fact is easily established by comparison and demonstration. Fig. 1 Fig. 1, is what is known as the Burr Bridge, It is com- posed of lower and upper chords, and posts and braces. The posts are framed into the chords, and the braces are framed into the posts. Arches are placed on each side of the truss, securely fastened thereto, and extending below the lower chords, abut against the masonry. This form of truss was extensively used throughout the United States previous to the introduction of railroads. Many spans were of great length, and in cases where the arches were large, and the masonry sufiBciently permanent, PRACTICAL MECHANICS. 157 this bridge was comparatively successful. Much difficulty was, however, experienced, by reason of the absence of counter braces. A moving load produced a vibratory and undulating motion, tending to loosen the connection of the timbers, which generally resulted in failure. Many of the first railroad bridges in this country were built upon this plan, but much greater difficulty was found in adapting it to the use of railroads, than had been pre- viously experienced in its use upon common roads. This difficulty arose from, 1st, the practical impossibility of per- fectly combining the action of the arch and the truss {Qd^oh system, of itself, being insufficient to carry the whole load) ; and 2d, the absence of counter braces. These de- fects, clearly apparent in their use on common roads, were greatly aggravated under the increased and concentrated nature of the weight, and the rapid transit of trains on railroads. It is true, they were obviated in part by adding largely to the amount of material in the structures ; but, as the difficulty was inherent in the plan, violent contor- tions in shape could not be prevented, and these in time caused failures. These remarks are intended to apply to spans of consid- erable length, as experience has proved that plans of even an inferior grade may be measurably successful in spans of ordinary length ; whereas, nothing short of the most judi- cious distribution of material will insure permanency, in cases where long spans are indispensable, and any arrange- ment which can be made permanent in the latter case, must certainly prove so in the former. It is worthy of remark here, that this particular combi- nation of the arch with the truss, is even now with some, a favorite idea, but it is believed that its warmest advocates will be geneially found among those whose opportunities for practical investigation have been limited, and that it is only necessary that the question be properly presented to them, to produce a change of views in respect to it. This partiality for the combination of the arch and the truss is attributable partly to the fact, that the simple truss nas in many instances failed, and as a last resort, the arch has been added, of such dimensions and strength, as to be competent to carry the truss and load, the truss serving only as a stiffener to the arch, while the latter, thrusting 158 PRACTICAL MECHANICS. upon the masonry, has sustained the whole weight. Be- sides, to the casual observer, who has never studied bridge construction, this combination presents at least an appear- ance of great strength and solidity, which do not in fact exist. That the simple truss without the arch has failed in some instances, is unquestionably true ; but while many of these failures have been caused from inattention to, or ignorance of, the laws regulating the composition and resolution of forces, by far the greater nnmber have arisen from the in- ferior quality, or lack of the requisite amount of material, or from inferior workmanship. The acknowledged failure of the Burr Truss, as applied to railroad purposes, led to the invention of several other plans, all of which were based upon the abandonment of the arch, and were aimed at perfecting a truss, which, of itself, would be sufficient to meet the emergencies of the case. This was in pursuance of what was considered a very resonable hypothesis, namely, that one system properly proportioned, must prove much superior to any method or arrangement in which the attempt was made to combine two distinct principles, in their nature heterogeneous. Among the most prominent plans presented to remedy existing defects, was one invented by Col. Stephen H. Long. This plan of bridge was composed of lower and upper chords, posts and braces, similar in outline and gen- eral arrangement to the Burr Truss, but differing from it in detail. An efficient system of counter braces was intro- duced ; these were made adjustable by wooden wedges, as were also the sustaining braces, by means of which any desirable elevation or deflection might be given to the truss. This plan of truss was rigid to a degree not pre- viously attained : and to such an extent was this true, that when properly adjusted, no perceptible deflection was pro- duced by the passage of the load. It was, however, found difficult to keep it in adjustment, in consequence of the great shrinkage of the wedges and other timbers of the truss. The invention of what is known as the Howe Bridge, fol- lowed. In this, as in Col. Long^s bridge, the idea of com- bining the arch with the truss was originally abandoned, for reasons heretofore given, and it was believed that this PRACTICAL MECHANICS. 159 simple form of truss would prove equal to any reasonable requirement. In the Howe Bridge^ the posts used in the Burr and Long bridges are dispensed with, and iron rods substituted, by- means of which any desirable camber may be given to the truss, thus overcoming the practical difficulty previously experienced in the adjustment of Col. Long's bridge, by the use of wooden wedges. This method of producing camber is certainly an im- provement upon the means adopted in the Long bridge, for that purpose, but is much inferior to the latter in its method of counter bracing, in that they are not adjustable, and perform a negative rather than a positive duty. The Howe Bridge is composed of lower and upper chords, braces and counter braces, vertical rods, and cast iron bearing blocks. The braces abut the bearing blocks, which pass through the chords in such a manner as to permit the rods to bear directly upon them. Spans of considerable length were built upon this plan, but experience proved that even this truss — like all others — had its limit, beyond which it could not be safely ex- tended. In the progress of railroad enterprises, in order to save large expenditures of money for masonry, longer spans than had been previously used became desirable, and in certain locations absolutely indispensable ; besides this, locomotives were largely increased in weight, to meet the demands of traffic, and furnish a more economical mode of working, and thus arose the necessity for the adoption of some other expedient to meet the increased requirements of bridges. As all had been done by way of improving this truss that mechanical skill could devise, and which an extensive practice had amply afforded, it became evident, that some radical change must be made in its arrangment, to enable it to meet the exigencies of the case. In this emergency the arch, heretofore condemned in the Burr TrusSf was again resorted to, for it had been proved from the experience which its use in that truss had afforded, that an arch of sufficient size abutting against permanent masonry, would place the truss in a position of secondary importance. It will be observed that the arch of the Burr Bridge, 160 PRACTICAL MECHANICS. b£> 1 PRACTICAL MECHANICS. 161 Fig. 1, abuts upon the masonry in precisely the same man- ner as the arch of what is denominated the Improved Howe Truss, Fig. 2, and the difference between the two consists simply in the mode of connection with the truss, and not in any change of principle, or method of action. It will be seen that the Burr arch is securely fastened to the posts and braces of the truss, forming a solid adjust- able mass. In Fig. 2, the arches are not fastened to the braces or rods, but have an independent connection with the lower chord of the truss, by means of rods radiating from the former to the latter. By this method it was sup posed that any desirable adjustment could be effected, and that the strain could be put upon either system, or equally upon each. This new arrangement, although plausible in theory, is found impossible in practice, for the following reasons : 1st. The rods from the arch to the lower chord are of various lengths, consequently their contractions and expan- sions must vary proportionately. 2d. Not a single rod in the arch is of the same length as those in the truss, hence the expansion and contraction of the rods in the truss will vary from that in each and all the rods connecting the arch with the lower chord. 3d. This combination is exceedingly liable to maltreat- ment, from the careless or ignorant. 4th. And even if it were everything in practice that is claimed for it in theory, (which is not the fact), it involves a constant expenditure for adjustment, which must continue during the existence of the bridge itself. The Burr Truss, Fig. 1, with all its defects, can be made superior by far to the Improved Howe Truss, Fig. 2. For in the former, there may sometimes be a yielding and com- pression between the parts of the truss and those of the arch, producing a certain degree of united action ; while in the Howe Truss, everything depends upon the length of the rods, which must always change with the temperature, and thus render an approach even to perfect adjustment, a matter of extreme delicacy. But in either Fig. 1 or Fig. 2, it is clearly evident that, in order to have a structure absolutely safe, the arch and the truss — each of itself, independently of the other — should be of sufficient strength to sustain the whole load, that tho strain may be borne alternately by each separate system. 162 PRACTICAL MECHANICS. In order to simplify and make clear the real points of difference existing in the combinations of the various plans of trusses, of the same general outline, it may be stated that the material composing any bridge trussj whether of wood or iron, or of both, is subjected either to tension or thru^t^ and it is upon the proper application of these ele- ments, together with a judicious distribution of the mate- rial, rather than upon any difference in detail, that the perfection of any bridge structure depends : this may be illustrated by reference to Figures 3, 4, 5. Figure 3 is the truss of the Burr Bridge ; in this the upper chord and braces are acted upon by thrust, and the lower chord and posts by tension. Figure 4 is the Howe Truss, without the counter braces ; PRACTICAL MECHANICS. 163 in this also, the upper chord and braces are subjected to thrust, and the lower chord and vertical rods are acted upon by tension. Fig. 4. Figure 5 is a plan of truss sometimes used, the counter rods being omitted ; in this the upper chord and vertical struts are subjected to thrusts, and the lower chords and diagonal rods are acted upon by tension. Fig. 5. Upon a comparison of these plans it will be discovered tliat the variations between the Burr Truss, Fig. 3, and the Howe Truss, Fig. 4, consists in the use of vertical rods and Dcaring blocks in the latter, instead of vertical posts in the former, both having precisely the same duty to perform. 164 PRACTICAL MECHANICS. It will also be seen that Fig. 5 varies from Fig. 4, in that the rods are placed diagonally instead of vertically, changing the element of thrust from the diagonal braces in the latter, to the vertical struts in the former, and trans- ferring the element of tension from the vertical to the diag- onal line. Much importance is sometimes attached to just such modifications in detail as exist in Figures 3, 4, 5, while the nature and intensity of the destroying forces are the same and equal in each. This has been proved by actual experiment, by the cele- brated engineer and bridge builder, D. 0. McCallum, as follows : Models were built, one on each plan, of equal length and height of trusses, containing the same sectional area and kind of material in chords and braces, and of equal perfection in details and workmanship, when it was found that the real difference in strength was unappreci- able, and it may be well to add, that any given amount placed upon each, in progress of the experiments, presented precisely the same characteristics and contortions in shape, until final failure took place. All bridges having their chords parallel, irrespective of the particular method adopted in combining them, and re- gardless of the amount of material used in their construc- tion, when loaded to nearly the point of fracture, present somewhat the sanae appearance, the greatest deflection being PRACTICAL MECHANICS. 165 invariably at points near the abutments. This will be un- derstood by the statement, that the vertical strain is in- creased, as the distance from the centre, to the ends of the trass ; at the centre the vertical strain is nothing, and at each end of the truss it is equal to one half the weight of the structure and its load. In point of strength, the arrangements Figures 3, 4, 5, are not superior to the simple combination Fig. 6. All bridges haying their chords parallel, exhibit tho same uniformity of action, and may be illustrated by refer- ence to Fig. 6 cr, in which A A, is the upper chord ; B B, tlie lower chord ; C, C, tension rods ; D, D, braces. 166 PRACTICAL MECHANICS. When a sufficient weight is applied co any truss of this outline, to cause deflection below the straight line, the up- per ends of the braces D, D, are made to approach each other, and the distance between the ends is diminished, and as the deflection increases, the upper ends of the braces D, D, will describe arcs a 6, of a circle downwards^ the radius of which being the length of the braces D, D. But when the upper chord is arched, as in Fig. 6 J, a sufficient weight will cause tlie braces D, D, to describe an arc upwards, re- presented by c d. Pig. 6 b. When tlic cliord c e becomes PRACTICAL MECHANICS. 167 straight, the arc will then be described downwards, as shown in Fig. & a. As an illustration of the McCallum Lijlexibk Arcfied Truss, see Fig. 7, in which A, A, is lower chord ; B, B, upper chord ; 0, C, tension rods ; D, D, braces ; E, E, struts, and W, weight. Upon an inspection of tliis figure, it will be seen that any deflection produced upon the centre of the arch, by means of the weight W, will cause the points B, B, to sep- arate, by thrusting outward, and in the direction of the ends of the truss, producing an upward movement of the upper chord, at the ends of the braces D, D, the latter describing arcs of a circle upward, and from thence will 1)0 couimuni- 168 PRACTICAL MECHANICS. cated by means of the tension rods C, C, to the centre of the lower chord, raising the latter at the point where the rods 0, C, meet. By removing the weight W. and inserting a Tcrtical strut at F, the upward movement of the chords will be arrested by the weight W. This peculiar action may be described as follows : Any deflection produced in the centre of the arch will cause an outward, and consequently, an upward force, at the upper ends of the braces, which, by means of the tension rods and strut, is transferred directly back to the under side of the arch, producing an upward force at the latter point, equal to the original downward force api^lied on top of the same. PRACTICAL MECHANICS. 169 This combination of forces is in agreement with a well- known law, namely, when two forces of equal powers of resistance are opposed to each other, a state of rest is pro- duced. For a further illustration of the action of this truss, see Fig. 8, in which A, A, are pieces of the lower chord, the centre being removed, B, B, upper chord, deflected by the weight W. C, C, are braces which pass through the lower chord, and rest upon the masonry. D, D, are tension rods. It will be seen that the ends of the pieces of lower chord at E, E, are raised considerably above a horizontal line. This upward tendency will continue until the upper chord be- tween B, B, is deflected below a straight line, when the action will be reversed. 170 PRA^CTICAL MECHANICS. Figure 9 exhibits the forces at a state of rest, in which A., xl, are portions of the lower chord ; B, B, upper chord ; 0, C, arch braces, which pass through the lower chord, and rest in the masonry. D, D, tension rods ; E, E, braces ; W, weight. It will be seen that the strain produced by the weight W, is transferred to the lower chord by means of thrust upon the braces E, E, to the points P, F, and by means of tension on the rods D, D, to the points B, B, and from thence it is brought upon the arch braces C, C, which rest upon the masonry. In this manner, a perfect equilibrium of forces is effected, as it is evident that the point G, cannot change position, unless the points B, B, are thrust outward towards the ends of the truss, which must raise these points, tliis being pre- vented by the strain upon the points F, F, communicated by the weight W, through the braces E, E. For a full plan of McCallum^s Inflexible Arched Truss, the reader is referred to Fig. 10. Upon inspection, it will be observed that the sustaining principle is very much in- creased toward the ends of the truss, not only by the addition to the amount of material at these points, but it will be seen also that the pannels become shorter as the vertical strain increases. The posts are placed upon lines radiating with the arch ; the braces form equal angles with the posts ; and in this way the latter are made to approach more nearly together toward the ends of the truss. The student has already had sufficient evidence of the great strength of this form of truss, and it has also been shown, that the tensile strain upon the lower chord is much less than in any other known plan. In fact, the latter may be entirely severed^ and the structure will still be competent to sustain a heavy load. In this, it differs from all other combinations. Upon referring to Figure 10, which represents a clear span of 180 feet, it will be seen that the arch braces which rest upon the abutments are extended to points on the arch about forty-seven feet from the abutments. From the top of each set of arch braces, running diagonally on each side of the truss, are placed heavy suspension rods, which are connected with the lower chords 12 feet further from th.c masonry. Thus the bridge seat is substantially transferred 17] 172 PRACTICAL MECHANICS. to a point 47 feet towards the centre of the bridge, reduc- ing- a span of 180 to 86 feet, so far as the tensile strain upon tlie lower chord is concerned. For this intermediate space of 86 feet, the arch beam is of sufficient strength to sustain tlie whole load, if required. PRACTICAL MECHANICS. 173 Strength, however is not all that is required, for a Rail- road bi'idge especially, subject as it is to a moving load, there must also be rigidity, stiffness, freedom from vibration. A bridge may be strong yet flexible, rigid yet weak ; in fact, flexibility is incompatible with durability, the struc- ture should be prepared at all times to receive its load, and should not be permitted to change shape in the slighest degree, by its passage over it. To produce this result, an effective system of counter braces, is indispensible. The proper office of counter braces is frequently misun- derstood, as is evident from the manner of their application in many cases in which they are used as check braces only, having a negative rather than a positive action ; this may be readily shown. When the load is applied, the truss is deflected in consequence o\ the yielding of the braces ; this has the effect of shortening the diagonals in the direc- tion of their length, while the diagonals in the direction of the counter braces are correspondingly lengthened ; this, will leave a space between the ends of the latter, and the bearing block in the lower chord. When the truss is in this condition, if wedges are insert- ed between the ends of the counter braces and the lower chord, in such a manner as to fill up the whole space, it is evident that the weight may be removed without at all affecting the shape of the truss, the deflection originally produced by the weight, being maintained by the counter braces, the strain upon the sustaining braces and other portions of the truss remaining precisely the same as when the weight was suspended. Now suppose tlie original weight to have been 200 tons, it is evident that as soon as it is removed, each counter brace will be subjected to an upward thrust, easily found from its position ; the sum of all the thrusts making 200 tons. Now let there be a smaller load applied, tliis load will not produce any additional strain upon any portion of the truss, nor will the deflection be increased in the slightest degree ; the only effect produced by suspending the latter weight, will be the relief of the counter braces, equal to the diSerence between the first and second weights. The inventor has found it very difficult to explain this 174 PRACTICAL MECHANICS. clearly in the course of conversation with some individuals, from the fact that weight and strain were confounded. Now it is true, when the original weight was applied of 200 tons, the abutments were loaded with just 200 tons more . than previously, and the truss was also loaded with ten tons more ; but when the wedges were driven, and the weight removed, wliile the abutments were relieved of 200 tons pressure, the truss still retained the original strain produced, the weight being required to produce the strain, the latter remaining after the former has been removed. In order to make a practical application of the above, the following method of adjusting the Inflexible Arched Truss, is submitted. When these bridges are raised, it is usual to load them with a train of locomotive engines, at- tached closely to each other, and that greater weight may be obtained, the tenders are sometimes detached, and the bridge covered with engines only ; with this load, the lat- ter is strained down to a perfect bearing in all its parts ; by this means the whole structure is more or less deflected, while the counter braces are hanging loosely in their places ; if, therefore, when the bridge is in this condition with its load, the counter braces could be lengthened with consid- erable force, it would not recover its original shape upon removal of the load, but would be held down by the action of the counter braces to very nearly the same position as when loaded. In this plan of bridge, the lower ends of the counter braces rest in iron stirrups, which are attached to the vertical ties or posts at a point near the lower chord by means of castings and nuts, by which they may be lengthened several inches, in, this manner they are made to perform a positive duty. When the bridge is adjusted as above, it is clear that a less load than that originally ap- plied cannot produce any deflection whatever, the only effect of the passage of a train over it will be to relieve the counter braces, and will not add a pound pressure upon any timber of the trusses. In the arrangement of any bridge truss, the attainment of the following requisites is desira,ble : Fii^st, Such equilibrium of forces as will produce uni- formity of action. Second, Such quantity and distribution of material, as will insure a large surplus of sustaining principle thereby guarding the structure against accident. PRACTICAL MECHANICS. 175 Thirds Perfect rigidity, that the combination in all its parts may have permanency equal to the durability of the material composing the same. Fourth, The arrangement of the parts should be such as to be free if possible, from the iiecessity of adjustment. The McCallum Inflexible Arched Truss meets all these re- quirements. PROBLEM. Let it be required to find the equal weights w, w, Fig. 11. Kept in a state of rest by a single weight W, which has caused the arc q C a to assume the chord q Db, the rigidity of the arc being neglected. Let R = the radius of the arc q G a, and put 2 c = its length ; then the chord q a and versine D are readily calculated. Let g a = 2 ri and C D = 2 e. Let f-=^Aq = Ap = Ba=^'Bb and g zzzi B n =^ B w = A 5 i^ A ?*. Also, let A B = 2 A 2 k — 2d , ^ > ^ ^h-d, h — d COS. 6 J putting 6 for the angle a B n = ^ A 5 h — c. / 2 A — 2 c 2h — 2c= AB—qb. Putting (f for the angle bB n ^:= q As, h — c ~ zzi: cos, (p. .•.

>. 193 Put e ^=z the length of the loadline of the vessel, in feet. / =z:z the great immersed section, in square feet. g z=z area of resistance in square feet. h =: displacement of the vessel in tons. i =z horse power required to propel the vessel j miles an hour, (statute miles.) k = co-efficient of the vessel. I == resistance of the vessel in pounds. m ==: tabular index corresponding to k. n = slip of either propeller or paddle-wheels. = the pitch of the propeller. f = the number of revolutions a minute. ^ EXAMPLES. Ques. 1. Required the area of resistance of a vessel when e = 260 feet, = the length of the loadline ; the great- est immersed section 500 square ieei = f ; the displace- ment of the vessel 2500 tons = h, » 35/^ _ 35 X 2500 _ ^, = "77" - 500 X 260 ^ '^ Opposite .67 in the following table, k, will be found = 1.76, the co-eflficient of the vessel. f ^ •'500') ^ Then^ = V j^j^ == s/ 500 XI .76 + (260 7 ^ ''^ square feet, the area of resistance required. 19J: PRACTICAL MECHANICS. TABLE. i in k 711 k .50 1.31 .71 1.52 .51 1.37 .72 1.43 .52 1.45 .73 1.36 .53 1.51 .74 1.29 .54 1.56 .75 1.21 .55 1 62 .76 1.13 .56 1.66 .77 1.05 .57 1.71 .78 .978 .58 1.76 .79 .903 .59 1.83 .80 .835 .60 1.89 .81 .763 .61 1.93 82 .691 .62 1.98 83 .625 .63 2.01 ! 84 .559 .64 1.97 85 .496 .65 1.91 86 .433 .66 1.84 87 .379 .67 1.76 88 .326 .68 1.72 89 .277 .69 1.64 90 .229 .70 1.59 95 .025 ! Qxies, 2. Required the resistailce of the vessel in pounds ^— /, when she is running at the rate of 10 miles an hour --=^ j ; the area of resistance := 40 feet = g, found as in the last example. / ^ 4.22 X gj' = 4.22 X 40 X 100 — 16880 lb- , tlif resistance of the vessel in pounds. The resistance in pounds may be found by the following formula nearer the truth Log. I — .B38468 -f- iog. g -j- 1.15 log. j. Log. J log. 10 --= 1.000000, PRACTICAL MECHANICS. 195 1.75 log, j — 1.750000 log.g = 1.602060 Constant = .633468 Log. 9672.3 '-^ 3.985528 The first rule gives the resistance =: 16880 lbs Ques. 5. Required the horse power i, when the speed is e(][ual 11 knots an hour = j ; and the area of resistance = 33 square feet = g, . . j' X g 11^ X 33 ^^^ . ^ =-^ — gg — — ^ = 49^ horse power. The horse power may be found with greater accuracy by the succeeding logarithmic expression Log. i == 2.75 log. J -f log. g — 1.944483 ' Log. J ^ 1.0413937 ; log.g = 1.518514 2.75 X 1.041393 = 2.863830 log.g = 1.518514 4.382344 Constant — 1.944483 Log. 274.07 -^ 2.437861 The horse power according to this latter method = 274.07 ; there is a discrepancy in these results also ; the cube of the speed is employed in one case, and 2f power ill the other - — — — ^ y put into a logarithmic form oo gives the latter rule. Experiment has not yet settled this point in steam navigation, simple as it is, some employ the 196 PRACTICAL MECHANICS. square of the velocity, others the cube, and a third party the 2f power. Government experiments, both in this coiub trv and in England, are generally government jobs. Ln- o'ineers are selected to superintend them upon the principal that determined who should be the village schoolmaster ; one of those worthies, was chosen to fill the important office of village teacher, merely because he had a large family and a wooden leg. PROBLEM 4. To find the slip of Propeller or Paddle-wheels, when the acting area, and area of resistance are given. If the area of resistance = 49 feet, = g ; and the act- ing area = 225 feet = d ; required the slip = n. 773- _ 343 ^ ^43_ ^ ^ -J^^f-Zf-jrjT ~ 343 + 33T[5 ^3718 * PROBLEM 5. To calculate the power and find the properties of Pad- dle-wheels and Screw Propellers. The pitch of a propeller = 33 feet ; and makes 42 revo- lutions a minute, the slip = .35. What is the speed of the vessel ? .^^ ^ ^^ (1 — .35) -^ \\A knots. 88 ^ ^ Put =-. the pitch ; p = the revolutions a minute ; and n = the slip ; then generally ^-— g— (1--); — 88^ __ •'• ^ ■" "^X (1-^) ' 88 j n = ^ — p PRACTICAL MECHANICS. t97 Let the speed be 11.7 knots the hour, = j ; the slip = .35 = n; the pitch 33 feet = o; required the number of revolutions = p. 88 i ^ 88 X 11.7 _ ^ X {l—n) 33 X (1 — .35 "~ the nuniDer of revolutions a minute. Required the slip, when the speed is 11.7 knots per hour, revolutions 48 per minute, and the pitch = 33 feet. . =. 1 - -^^- = 1 ^ '^ X ^^-^ - 35 X Jp 33 X 48 PROBLEM 6. Any four of the five following quantities being given to find the fifth, namely, the radius of a paddle-wheel from tlie centre of motion to the centre of pressure of the floats, = c ; the .slip, = n ; the immersed angle of the paddle- wheels at the centre of pressure = b ; the statute miles per hour = j ; and the revolutions a minute = p. From the general expression c p COS. i b , any one of the quantities is easily deduced, for i>=-— ^i^ ^.n ^i- (1 — n) c COS. I ' pc COS. i b Suppose the immersed angle of the paddle-wheels at the centre of pressure = 54°.. 33' = 6. The slip = .38 = ^ ; the revolutions per minute = 1& == p ; and the radius of the paddle-wheel from the centre of motion to the centre of pressure of the floats = 21 feet ; required the speed = i ? One third of 54"^ .. 33' = 18° .. 11', the natural cosine of winch IS = .9500629, for practical purposes .95 will be sufficiently near tlie truth. 198 PRACTICAL MECHANICS. ^ _cpcos_^J_ 21 X 16_X^ _ 3 ^ 14.136 miles an hour. Again, let the immersed angle = 54° *. 33' ; the revolu- tion per minute = 16 ; the speed 14.136 miles an hour ; the radius of the paddle-wheel from the centre of motion to the centre of pressure = 21 feet ; required the slip ? 16 X 21 X .95 the required slip. CALCULATIONS RESPECTING THE ANTI-SLIP SCREW PROPELLER. Put ^ ^=: :^ the pitch at the periphery = D E. r = :J the pitch at the hub = R S. s := the assumed slip in a fraction = H R, of 5^. ••• q -^ rJ^ '^• Put t :^ extreme radius =: Q D. u = diameter of the screw = 2t = FI. V =^ pitch of the propeller at the periphery = 4 ^ . w = angle of the blades at the periphery 1::= K Q T. w , i-= angle of the blades at the hub. J M = X zz^L ^ pitch of the propeller at the centre of effort of the blades. B J ^^ y = .725 t, the radius at the centre of effort G, —- OGt ::^ QM. . •. The actual pitch at the centre of effort ::^- 4 x. z 1=^ horse power required to drive the propeller and oc revolutions a minute. PRACTICAL MECHANICS. 199 •^ C 200 PRACTICAL MECHANICS. B 1^ number of blades. y ^zr length of blade parallel with the centre line — U V d z:=: the breadth of the propeller blades over the edge between the corners X z s z=z the circular arc in the angle ^ z^r W Y. 6 znz the projected angle of the blades A =:= the true inclined surface of the blades A = the projected area of the blades. A 2 ::::= tne actlug area of tne propeller. Let 7T = the circumference of a circle diameter ^=: 1 . Cot. w = :=r — ^ — = the cotangent of the n u n 2 t 71 1 angle of the blades at the periphery. V — 71 u cot. Wj the pitch of the propeller at the periphery 360 X r; 6 . 360 7t uy ^ nuy ^ _ 129600 y ' g ' / VQ O ' d is the bypothenuse, and y is the base of a right angled plan triangle. (5 = K C measured over W Y, it is a projection the same as U V ^-^ /. S^J^^IF^-y^ ^ 129600 ' _ .7854 w^ OB ' ~~~ 360 . __ 4 ^ B ((5 + y) PRACTICAL MECHANICS. 201 PROBLEM 1. The pitch of the propeller = 33 feet ; and makes .48 revolutions a minute, the slip ^rzr .35. What is the speed of the vessel ? ^^X^Nl -.35)^11.7 knot. 88 By the introduction of the anti-slip screw, the slip would be reduced from 35 to 15 per cent., the speed will be 33 X 48 ,, _ ,. . , . 88 ^^- ^-^--- Put =:= the pitch ; P :^ the revolutions a minute n = the shp ; and J = the knots the hour ; then generally i -"s^s^C •)^ V 88 j — X (1— «) ' ^^ "~ OX;?' 88 ~^ _p ( 1 — n) 202 PRACTICAL MECHANICS. A ship fitted up with Byrne and Elliott's Screw made 13 knots an hour ; the revolutions were 60 a minute, and the pitch 30 feet, what was the slip ? , 88 i _ 88 X ^^ = .12, or 12 per cent. « = ^ — Vxy "" 30 X 60 PROBLEM 2. Given the diameter of a propeller = 12 feet, = w ; and the angle at the periphery = 51° ..33' = «;; to find the pitch = v1 Cot.bl° ..33' = w; .7940121 v^nu cot. w ^ 3.1416 X 12 X .194 = 29.93 feet. \ PROBLEM 3. Given the diameter = 21 feet = u ; the length of the blade parallel with the centre line = 6 feet = r ; the slip 40 per cent., or s = .40 ; the angle of the blades at the periphery = 63° .. 42' = -w ; to find the horse power neces- sary to drive this screw 68 revolutions a minute == ex. Cosine 63°.. 42' r^ .4430112 11" X 68- (6 X .4 X .443 + i) = 7124. 480000 ^ ' Log. 21 -- log. 68 = -- 1.32221 r 1.832509 3.154128 3 log. 480000 = 9.464184 ^ 5.681241 log.oi 1.1124 = 3.182943 = .069119 Log. of 7 1 24 = == 3.852122 PRACTICAL MECHANICS. 203 As a second example suppose u z= 12 feet ; y = 5 feet ; w = bl"" .. 30' ; 5 = 38 per cent., or 5 = 38 ; what power is required to drive this propeller 40 revolutions a minute ? ^^ ' ^ ^^ ' (5 X .38 X .5373 + *) = 509.4 ~ 480000 ''^ '^ Log, 15 = = 1.176091 log. 40 = - I €02060 2.178151 3 8.334453 log. 1.13198 = .053839 8.388292 log. 480000 = 5.681241 log. 509.4 = 2.707051 PROBLEM 4. Given the diameter of a propeller = 21 feet = u ; the angle of the blades at the periphery = 63^ .. 42' = w ; the length y = 6 feet ; the slip s = .40 or 40 per cent. ; four engines drive this screw of 7124 horse power = 2;; how many revolutions, oc, will this propeller make in a minute ? 1 I *ouuuO z ) ^ ^^ - . oc = — ■! -7-. — , ^^ V •" = 68 revolutions. 480000 . {y s cos. w y s cos. 'M?-f-i = 6x.4x .443 + i = 1.1743. 204 PRACTICAL MECHANICS. Log. 480000 = 5.681241 log. 7124 -^ 8.852t24 9.533965 log. 1.1U3^ = .069179 8)9.464186 3.154729 log, 21 = 1.322219 log. 68. = 1.832510 PROBLEM. 5. Given, the diameter of the propeller = 13 feet =u; the angle of the blades at the periphery = 59° ,A& = w ; the angle of the blades at the hub ^= IP .. 48' = i^, ; the diameter of the hub = 1.25 feet, =■■ twice Q S ; (see figure 16,) to find the pitch at the centre of effort. Cotangent of 59^ .. 46' -^ .582793 Cotangent of 11° .. 48' = 4.786730 The pitch of the periphery = 13 X .583 X 3.1416 = 23.8 feet — v The pitch at the hub 1= 1.25 X 4.787 X 3.1416 = 18.8 feet, v^ Put v^ for the pitch at the centre of effort, u^ the diameter at the centre of effort, and u^ the diameter at the hub. {y — Vg) : (v, — %) : : (^'' — ^^2) • (.7257^ — ^^2) . „ -„ I (^ — ^2) (.725 2^ — -z/^J • • V, '^2 » ' '^ u — U^ PRACTICAL MECHANICS. RECAPITULATION. 205 V = 28 . 8 22 . 3 18 . 8 Vj = 22 . 3 u :^:^ 13 feet u^ — ,125 u = 9.425 feet ^o 1.25 feet. PROBLEM 6. Giyen, the diameter of a propeller =: 13.25 feet == u; and the angle w = 65°, to find the acting area of the pro- peller ? 5 to' but V' ^ \/ 1'^ -j- n" w n"^ w^ cot.^ w, hence 5u' 2 -,, 2 > A, 2 7t 7t cosec, w log, 13.5 =^ 1.122216 log, n z^ .491151 log. cosec. w =: .042724 1.662091 log, 2.5 = .397940 3 log. 13.5 = 3.366648 3.764588 1.662091 log. 126.6 — 2.102497 .•• Acthig area of propeller =:i^ 126.6 square feet = A, 206 PRACTICAL MECHANICS. The Anti-slip Screw Propeller, Patented by the Au" thor of this work, Mr. Oliver Byrne, in conjunction with J. G. Elliott, Esq. 5th October, 1858. This propeller may be constructed of any of the known forms, and operated by any of the known methods of ^eer- ing, but the form found to be the best, is that Avhich has the' centre of effort in the centre of the acting area of each blade ; and when the water is discharged through the backs of the blades, the openings should be opposite, or symmetrically arranged to these centres. It has been found, that the power necessary to turn a solid metal screw and shaft, doing a given work, at a given velocity, needs not to be increased when a hollow shaft and screw through which water flows, are employed to do the same work, un- der the same circumstances. Openings of different forms may be made in the blades, (see A, A, A, A, Fig. 16,) provided the action of the water passing through them, has no tendency to give a rotary motion of the screw. A valve may be placed in any convenient place in the hollow shaft, or pipe, that conducts the water aft the pro- peller, to prevent its return in cases of back action. Fig. 17, is an end view of a four blade screw propeller. A, A, A, A, are openings at the rear of the blades, oppo- site the centres of effort in the acting areas. A, A, A, A, communicate throngli passages with the hollow shaft, A Q, (Fig, 16.) Fig. 16, is a longitudinal view of Fig. 17. The pro- peller is represented as permanently keyed on the shaft, but may be retained by a clutch. PRACTICAL MECHANICS. 207 As the propeller is moved, the advanced half of the blade 6 X W, cuts into the water undisturbed by the action of the screw, and as the water passes e, Fig. 17. the aft part of the blade ceases to be as efEcient as the fore part e W, on account of the thickness of the metal of which tlie propeller is formed ; a supply of water through A, ren- ders the aft part of the blade as effective in propelling the vessel as the far part. When water is conducted aft the propeller through a hollow shaft, or through openings in the blades, or tlirough both shaft and blades at the same time, the slip of the screw is much diminished by the action of the blades against the water so discharged. Several contrivances may be employed to conduct the useless or superfluous water to the rear of the screw propeller, in im- mediate contact with the blades aft, not for the purpose of giving a rotary motion to the propeller, but for the pur- pose of diminishing what is termed slip. THE END. Nov, 3 1860. \\ J 709 li' J7,|BRARY OF CONGRESS 019 713 770 8 i ii: i