Cornell University Library The original of tliis bool< is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031289006 THE ESSENTIAL ELEMENTS OF PRACTICAL MECHANICS. THE ESSENTIAL ELEMENTS OP PRACTICAL MECHANICS, §a;geb on % principle of Morfe; DESIGNED FOB ENGINEEROG STUDENTS. OLIVEE BYRNE, FOBIiEBLT FBOI'ESSOB OF UATHEMATIGS; COLLEGE FOB CIVIL ENOIHEEBS. Author qf *^ Dual Arithmetic, a New Art," 4-c. LONDON: E, & F. N. SPON, 48, CHAEING CROSS. 1867. lOSDON : PBIKTED EI 0. TrHIIIHe, BEAUFORT HOUSE, SIBADD. PREFACE. This work is designed to make good a deficiency much felt by practical engineers atnd engineering Students en- gaged in civil, military, and mechanical pursuits. A stu- dent, to understand the treatises and systematic expositions, developing the theory and application of mechanics, em- ployed as class books in our schools and colleges, requires considerable mathematical skill, and after all, the elemen- tary knowledge acquired by studying such works is. of little value to the practical engineer. On the generail theory of mechanics there are many popular works clear of mathe- matical formulae and serviceable to the general reader, but the practitioner cannot render them available. Again, French investigators and writers on practical mechanics take for granted that they are addressing those who under- stand the theory of mechanics, and when their works are copied or translated they are far from being elementary. The essential elements of practical mechanics is based upon what is termed the principle of work, and as VI PEEFAOE, work is measured by a unit, the student should be able to distinguish with ease units of work from other units. I have, therefore, introduced a simple notation and a new system to effect this object, which French writers and their copyists have neglected: — discarding the denomina- tions foot-pounds and kilogramrmUres, which are evidently iniproper. The unit of work is the exact labour re- quired to raise a pound weight tlirough the space of one foot against the direct action of gravity. If a man takes a pound weight in his hand, and raises it one foot in the direction of the 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 higb=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, the num- bers that represent units of work are printed in broad- faced type instead of the ordinary way, thus : 24 represents 24 units of work ; 8x3=24; that is the same- as saying, 8 lbs. raised 3 feet high is equal to 24 units of work. 24' ; represents 24 units of work done in a minute. 24" ; represents 24 units of work done in a second. 33000=1-; PEErACE. Vli represents 33000 units of work done in a minute, equal one horse power. Formerly, the science of mechanics was one of the last branches of applied mathematics to which a student was induced to turn his particular attention, but now the case is different. Since the wealth and power of nations de- pended so much on the application of the sciences to the mechanical and industrial arts, a knowledge of practical mechanics becomes of paramount importance, not only to engineering students, but also to capitalists, statesmen, and merchants, many of whom are but little skilled in the higher departments of mathematics. While this work is designed to meet the exigencies of these changes and circumstances, and dispense with the higher geometry and an extensive use of analytical symbols, it will be found to extend the boundaries as well as simplify the practical application of this branch of philosophy, and to contain solutions of many important problems which have hitherto defied the skill of mathematicians. This treatise, when divided as the nature of the several subjects seemed to demand, is composed of ten chapters. > The first chapter is designed to exemplify how worh is measured by a unit, both with and without reference to a unit of time. The questions selected for the purpose of illustration in this chapter, as well as those given throughout the work, are designed more to teach the student how to think, than to insist upon what he should think. However, each question may be so varied as to become a model for a variety of cases. The second chapter treats of the work of living agents, explains the influence of friction, and introduces one of the most beautiful laws of motion. In the third chapter the principles expounded in Vlll PREFACE. the first and second are applied to the motion of bodies on inclined planes and to the raising of materials. The fourth chapter treats of the transmission of work by simple machines, establishes the properties of the lever, wheel and axle, pulley, inclined plane, wedge, and screw, on the principle of work. In this chapter the student's attention is directed to the graduation of the safety-valve lever, the centres of gravity of bodies, the hydrostatic press, and the anti-friction cam-press. In the fifth chapter I have given some of the most useful propositions and rules relative to the work of steam and the steam-engine; but steam being a subject in which the most elaborate theory can do but little independent of experiments, I therefore deemed it p^dent, rather than delude the student with the usual display of mathematical accuracy where it is far from being attained, to give but little theoretical matter, and introduce some valuable tabulated results founded on experiments. The constant quantities are often taken in round numbers, but they are easily altered to such particular cases, without affecting the principle; of work upon which the calculations are based. It may be necessary to point particular attention to the problems of this chapter, which, for the. first time, are independently and accurately solved by Dual arithmetic, without the use of tables or other extraneous aids. ThcsiiKi/t chapter is on accumulated work. The subjects stated and discussed in this chapter are: the force of grar vity near the surface of the earth; units of work in a rotating body; the centre of gyration; living forces, > ws viva; the measure of motive forces and inertia; the maxi- mum velocity of the piston of a steam-engine ; a question is solved under this head which defied the skill of mathe- maticians before the introduction of Dual arithmetic. In PEEFACE. IX the sixth chapter I have alsQ treated of the velocity ac- quired by bodies in descending an inclined plane or a curve ; the crank and fly-wheel ; impact ; the ballistic pendulum ; shock of elastic bodies ; resistance of fluids ; centre of oscillation, and other kindred matters. Cliapter seven is appropriated to the examination of the equilibrium of forces, pressures, tensions, and thrusts. The principle of the equality of moments, and the construction of diagrams of forces, are next discussed. The remainder of this chapter is occupied with practical examples of wooden and iron bridges, a correct theory of the strength of girders resting on supports, and with investigations respecting chain and suspension bridges, in which Dual arithmetic is again employed to obtain results not attainable by any pre- viously known. method. The comparisons drawn in this chapter with respect to different methods of construction should be calmly and closely considered. One class of mechanical investigators first establish theories on abstract reasoning, and afterwards correct them by experiments. I have taken Morin and Poncelet as representatives of this class. Another class, without any particular hypothesis, interpolate a series of experiments by empirical rules to suit particular cases within the range of such experiments : General Anstruther, alluded to in the eighth chapter, is a practical engineer of this class. A third class of investi- gators assume constructions which they assert possess parti- cular properties that may or may not exist, and support such claims by high-sounding pretensions and a little mis- applied mathematics to cover the delusion. The expounder of Koch's ingenious system of skeleton structures, which may or may not apply in practice, for all that this investi- gator has shown to the contrary, is one of this class. He PBEFACE. assumes it possible that the structure may be composed of a series of right-angled triangles, perhaps because the calcu- ■ lations respecting such triangles are easily made. But m sustaining a load in motion, although these triangles are but little altered, yet it is in this little, which he disregards or explains away, lies the object of research. Such perform- ances leave the subjects to which they are applied in doubt and abeyance. This class of investigators is very numerous, and have only succeeded in breaking down confidence in this important branch of applied mathematics. The eighth chapter treats of the pressure of water and other fluids ; the position of the centre of gravity in many important forms ; the pressiire of earth against revetment walls ; and the measurement of heights by observations of the barometer and thermometer. In treating of this last subject, 1 have shown a simple method of solving an equation that defied the skill of mathematicians. In this chapter will be found some very important observations on the force of gravity and the constant g, which is the letter usually taken to represent it. The essential elements of practical mechanics terminates with the ninth chapter. How sound theory may be made to approximate to the results of practice ; the re- sistance of water to the motion of paddle-wheels and screw- propellers; description of the indicator; performance of paddle-wheel steamers ; use of Dual logarithms ; and the performance of a screw-propeUer with expanding pitch, are subjects discussed in this chapter. The student, before consulting larger works on the re- sistance of water to the motion of vessels, would do well to read the lectures of Captain E. G. Fishboume on naval architecture. Here it may also be necessary to remind the PEEPACE. xi reader that many productions, valuable to the engineer, are incidentally produced by liberally educated professional men who are not professed engineers ; for example, the lectures of the naval officer just named on naval ordnance and iron ships saved this country from squandering large sums of the public money on guns and ships worse than useless. The tenth chapter is added for the use of those who have neglected the study of common arithmetic ; it contains useful arithmetical processes ; operations that may be for- gotten ; the application of principles often misunderstood ; and an introduction to Dual arithmetic. I have omitted to mention many minor topics connected with the general subjects enimierated that could not be conveniently spe- cified in an ordinary preface; however, those topics are particularised in the Index. OLIVER BYENE. PEACTICAL MECHANICS. CHAPTER I. ox THE UNIT OP WOEK WITH, AND WITHOUT KEFEKENCE TO THE UNIT OP TIME. Work is measured by a unit, like length, weight, time, &c. To raise one pound a foot high against tlie earth's gravity in the direction of a plumb-line 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 measure 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 in black broad-faced letters and figm-es instead of in the ordinary way, thus, 25 represents 25 units of work. 25' will represent 25 units of work done in a minute. 73" will represent 73 units of work done in a second. 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". B ^ PRACTICAL MECHAmCS. 45" represents 45 seconds ; but 45" 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 1. Required the units of work expended in raising a weight of 50 lb. to the height of 31 ft. Units of work in raising 1 lb. 31 feet =31- .-. 50x31 = 1550, 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 atott=11201bs. Units of work in raising lib. 17 feet =17- .. 1120x17=19040, the units of work required to raise 1120 lbs. to a height of 17 feet. Ques. 3. How many units of work are required to raise 7 cwt. of coal from a pit whose depth=13 fathoms ? 7x112=784 lbs. 6 feet being the length of a fathom, 13 x 6=78 feet, in 13 fathoms. Hence the work consists in raising 784 lbs. to the height of 78 feet. .-. 784x78=61152. Ques. 4. If the weight of a man be 183 lbs., and if he ascends 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 .-. 183x20=3660, amount of work. 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. FEACTICAL MECHANICS. 3 Qv£s. 5. How many units of work will be required to pump 8000 cubic feet of water from a mine whose deptli= 500 fathoms? A cubic foot of water weighs 62-5 lbs. 8000x62-5 =500000 lbs. 500 fathoms =3000 feet. Consequently the work is to raise 500000 lbs. a perpendicular height of 3000 feet. Work= 500000 X 3000= 1500000000- Ques. 6. A horse draws 150 lbs. out of a well, by means (rf a rope going over a fixed pulley, moving at a rate of 2^ miles an hour, how many units of work does this horse per- form a minute, the friction being neglected? A mile=5280 feet, then, 2^x5280 „„„ „ , , -.^ =220 leet passed over per nunute. .-. Work per minute=150 X 220=33000'- Work per second=550"- THE UNIT OF WOEK BEFEEEED TO A UNIT OP TIME. A unit of work, or 1, represents 1 lb. raiised 1 foot. A unit of work in a minute, or 1', represents 1 lb. riaised 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 lbs. a foot high in a minute, or to perform 33000 umts of work in a minute. Hence a horse power=33000'=l', or 1 HP. 28 horse power decimal 37 is written 28" -37, or 28 37 HP. Whether this power is greater or less than the power of a horse it matters Httle, while it is a power so well defined. To raise 10000 lbs. a foot high in a minute, would be a more convenient unit to measure by. b2 * PEAOTICAL MECHANICS. EXAMPLES. Question 1. How many horse power would it take to raise 3 cwt. of coal a minute from a pit whose depth = 110 fathoms? Depth=110x 6=660 feet. 3 cwt. = 112x 3=336 lbs. 660x336=221760'. Since a horse power= 33000' SSl'T'fiO' 3300d '^^^^ ^^■' *^^ '^®1"^''^<^ P"^®""- 6'72 HP. stands for 6 horse power decimal 72 ; the student should obserre that the full point which separates whole numbers from decimals is placed above (•72), but when othermse em- ployed it is put below (HP.) •72=the vulgar fraction tM, a cipher being put under each decimal figure and 1 under the full point. Ques. 2. How many horse power is required to raise 2200 cubic feet of water an hour from a mine whose depth = 63 fathoms? 137500=2200 x 62-5=weight of water m lbs, 63 X 6=378 feet, depth of mine. i^^^^i^=866250'. ■ 866250 _ .. -330^5^-261 HP. The proposed notation must be borne in mind, 866250' signifies 866250 lbs., raised one foot high in a minute, and 264'', represents 26^ horse power, or 26i times 33000 lbs. raised 1 foot high in a minute.. Ques,. 3, A winding engine is moved by 4 horses, wiat- weight of coal will be raised an hour from a pit whose depth=200feet? Work of the i horses in an hour, =4 X 33000 X 60=7920000. Work in raising 1 lb. of coal 200 feet=200. Since it takes 200 units of work to raise a pound of coals, 7920000 200 -=39600 lbs. PEAOTIOAL MECHANICS. 5 Consequently, four horses will raise 39600 lbs. of coal, or of anything else, a height of 200 feet in an hour. , , Ques. 4. In what time will an engine of 10-horse power raise 5 tons of material from the depth of 13? feet ? 5 tons = 11200 lbs. Work of the engine per minute, = 330000' = 33000 x 10. Work of raising 5 tons, = 11200 X 132 = 1478400. Because the engine performs 330000 units of work a minute, . 1478400 ..a-4. — =4 48 mmutes. 330000 o"^""""- Qices. 5. How many cubic feet of water will an engine of 36-horse piower (36 ^ ) raise in an hour from a mine whose depth is 40 fathoms ? 40x6=240 feet. Work in an hour, = 36 X 33000 X 60 = 7 1 280000. Work to raise a cubic foot of water 240 feet, =240x62-5^15000. yigSpOOO = 4752 cubic feet. 15000 Ques. 6. From what depth will an engine of 22-horse power raise 13 tons of coal in an hour ? Work done by the engine in an hour, 22x33000=726000. 13 tons = 29120 lbs. 726000 = 24-9 feet. 29120 Ques. 7, An engine is observed to raise 7 tons of mate- rial an hour, from a mine whose depth is 85 fathoms ; re- quired the horse power of the engine, supposbg ^ of its work to be lost in transmission 1 ,„ , 2240x7x85x6 ^ oooon- Work a min.= '^^^-^ — i^ = 133280 . 6 PEACTIOAL MECHANICB. Since | of the work of the engine only go to raise the material, .-. I of 33000=27500', the units of nsefal work of one horse power a minute. Ques. 8. Eequired 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=130000, and each person to use 5 gallons of water a day ? The standard, gallon of the United States weighs 8|lbs., nearly. 130000 x5=8|. _ 8125x25 lbs . 12 X 60 27 the pounds of water to be pumped in a minute. 8125x25 ,^ „^ ^, .-. ^7 — X 50=376157' . 376157'_ , 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 a minute ; from the second, 10 cubic feet ; and from the third, 35 feet, allowing ^ the work of the engine to be destroyed by useless resistance ? 1" =33000'. f of 33000=22000', the effective power of the engine. 1st level, work=62-5x 20x240= 300000'. 2nd „ „ =62-5x10x420= 262500'. 3rd „ „ =62-5 X 35 X 540= 1181250'. 1743750'. 1743750 22000' =79" 26. PBACTICAL MEOHANIOS. 7 Ques. 10. There are 6000 cubic feet of water in a mine whose depth is 60 fathoms, when an engine of 50 horSe power began to w^ork 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 run into -the mine per hour, supposing ^ of the work of the engine to be lost by transmission ? 4)33000' 8250' 24750' effeetiye power of the engine a minute. It must not be forgotten that 24750' written in broad - faced figures, signifies 24750 units of work in a minute or 24750 lbs., raised one foot high' in a minute ; so that setting down the numbers in this pecuUar manner saves much written explanation. 24750' X 50 X 5 X 60=371250000, effective work of the engine in 5 hours. 62-5x60x6=2250, work in raising one cubic foot of water to a height of 60 fathoms =360 feet. 371250000^16500 22500 cubic feet of water pumped in 5 hours. 16500- 6000 - 5) 10500- "Water run in during 5 hours. 21000- Cubic feet run in one hour. Ques. 11. A forge hammer weighs 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 power to 20 such hammers ? Work of each lift=300 X 2 x 20=12000. Work in 100 lifts, that is, in one minute, 12000 X 100=1200000'. 1200000 _3g..3g 33000' 8 PEACTIOAL MECHANICS. Ques. 12. An engine of 10 horse power (10"), raises 4000 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, a minute= ^ 1200x4000 _g^o»^, 60 Work engaged in raising the hammer a minute, =330000—80000=250000'. Work a minute, in raising one lb. of hammer 4 feet high, 50 times a minute, = 1x4x50=200'. 250000'_i 200' the weight of the hammer required, : 1250 lbs., CHAPTER n. ON THE WORK OF LIVING AGENTS. The labouring force of animals varies very much with the way in which their muscular strength is exerted ; and also, with the rate at which they labour. The following little table shows the greatest amount of effective work that a labouring man can perform under the different modes in which he may exert his muscular power. WORK DONE BY A MAN PEE MINUTE, WHEN HE WORKS 8 HOURS A DAT. A man in raising his own body 4250'. A man in working at a treadmill 3900'. PEACTIOAL MECHANICS. 9 A man drawing or pushing horizontally . . . . 3120'. A man pushing or pulling vertically 2380'. A man turning a handle 2600'- A man working with arms and legs, as rowing . 4000'. UNITS OF WOEK DONE BY A MAN PEE MINUTE, WHEN HE WOEKS 6 HOUES A DAY. A man in raising material with a pulley . . . 1560'- A man in raising materials with the hands . . . 1470'- A man in raising material on back, returning empty 1126'- UNITS OF WOEK DONE BY A MAN PER MINUTE, LABOUEING 10 HOURS A DAY. Raising materials with a wheelbarrow on ramps . 720'- Throwing earth to the height of 5 feet .... 470'- UNITS OF USEFUL WOEK DONE BY A MAN IN A MINUTE, EAISING WATEE BY DIFFERENT ENGINES, WORKING 8 HOURS A DAY. With a windlass from deep wells 2560'- With an upright chain pump 1730'- With a treadmill 3176'- With a Chinese wheel 2167'- With an Archimedian screw 1505'- Raising water from a well, with rope and pail . . 1054'- WORK OF BEASTS. The imaginary horse by which the power of steam- engines is 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 mintue, which according to the author's notation, is expressed thus : — 22000'. A mule will perform f the work of a horse=14666§'. 10 PBACTICAL MBeHANICS. An ass will perform about ^ the work of a liorse=4400 • In a common pumping engine a horse of average strength will do only 17550 useful units of work in a minute. From 22000' Take 17550' 4450', lost in this case by friction and useless resistances. EXAMPLES. Question 1. How many cubic feet of clay weighing 100 lbs. the cubic foot, will 20 men throw 5 feet high in a day of 10 hours long? Fram the foregoing table the number 470' is found. Work in a day, =470' X 60 X 10 X 20=5640000'. Work in raising one cubic foot, = 100x5=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 foot=1251bs., and 17 bricks to form a cubic foot? For a man working in this way, the foregoing tables give 1126'. .-. 1126x60x6=405360, work done by a man in 6 hours. 125x30=3750, work done in raising a cubic foot 30 feet high. .-. 4O|||O=l08-096, 3750 cubic feet raised in a day of 6 hours. Number of bricks= 108-096 X 17=1838. PRACTICAL MECHANICS. 14 Qms. 3. How many cubic feet of water will a labourer, .working with a bucket and rope, raise from a well whose depth is 16 feet ? In the table will be found 1054', for a man working in this manner. 1054' X 60 X 8=505920, work done in 8 hours. The work done in raising a cubie foot of water 16 feet high, =6-25x16=1000. Hence the cubic feet raised in 8 hours by a man, =505920^505 92. 1000 Ques. 4. What weight, is a man, working at a treadmill, able to raise in a day of 8 hours, from a depth of 110 feet? Tabular number =1730'. 1730' X 60 x 8=830400, , work done in a day. The work required to raise one cwt. or 112 lbs. a height of 110 feet will be, 112x110=12320; .. 830400=67 4 cwt, 12320 the weight that may be raised by a man working at a treadmill. Ques, 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 the 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, =2600' X 60 X 8= 1248000. The work to raise one ton, =2240x20=44800; 1248000_„7fi 44800 raised in a day of 8 hour's. 274 tons 12 PRACTICAL MECHANICS. Ques. 6. 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 a wheel and axle ? Work done in one day of 8 hours, =2600' X 60 X 8 X 4=4992000- The work of one stroke, = 112x7x23=18032; 4992000_ 277 strokes nearly. 10832 The force with which animals pull decreases with their speed ; the relation between the traction and speed of a horse is ex- pressed with considerable accuracy by the following formula : <=250— 41|. »■ ; in which <=the traction in lbs. arid j-=the rate in miles per hour. Hence, if the speed be 3 miles an hour, the traction will he 125 lbs., for 250— 41|x 3=125 lbs. If the speed be one mile an hour, the traction will be=208^ lbs. for 250— 41|xl=208|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 — 41|r|xr 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 fsactional 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 the rubbing surfaces. When a waggon or a cart is drawn along a good common road, the resistance of friction is about -^^ of ithe 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 PEAOTICAL MECHANICS. 13 A horse with this traction, according to the foregoing formula mores at the rate of about 3^ miles an hour, for ' «=250 — 41f r, becomes 104=250— 41f r, .-. r=3|. A carriage on a railroad only requires a pressure of about j^ part of the moving weight to give it motion, or from 4 to 8 lbs. a ton. The fractions ^ for common roads, aud nfc for railroads, are called the coefficients of friction ; as these coefficients become smaller, the rubbing surfaces become smoother. Let W be a weight drawn on a horizontal plane, H E, by means of a weight, P, attached to a icord. A, going 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 railroad, if W=150tons,Pwill be=900 lbs., when the coefficient of friction = Trro, 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 dis- tance ; hence the units of work done in moving W will be the weight of P in pounds multiplied by the distance in feet through which it descends, or the resistance of friction in pounds, multi- plied by the space in feet over which W is moved. The wprk of every machine is consumed by the work done, or by the useful work, together with the useless work, or the work destroyed by the friction, of the parts of the machine. I will here explain one of the most beautiful laws of motion : When the work applied exceeds the work consumed, the redundant work goes to increase tWe speed of the pai-ts of the machine, and at the same time, like the fly-wheel, acts as a reservoir of work. 77ms acceleration goes on increasing until the work of the resistances -\-the useful work = the work applied ; and then the rhotion of me machine becomes uniform. For example, in a railroad 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 resistances will be exactly equal to the work applied by the moving power. 14 PRACTICAL MECHANICS. QiKs. 7. Required the effective horse power of a locomo- tive 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 a;=the required horse power. The work of the engine per minute, =2; X 33000'; The resistance of friction, =5x100=500 lbs. The distance moved per minute, 23x^280^2024 feet; 60 Work to overcome the friction, =2024x500=1012000'. But as the speed of the train is uniform, the work of the resistances will be equal to the effective work of the engine ; .. XX 33000'= 1012000'. 1012000 33000' =30" 7. Ques. 8. What is the rate in miles per hour of a train of 80 tons, drawn by an engine of 70" ; the friction 8 lbs. to the ton ? Call a the uniform speed in miles per hour ; Work used in moving the train x miles, =80x8x5280x0;; this is the work done by the engine in an hour. Bjit the work done by the engine in an hour will also be expressed by 33000' X 70x60; .-. 33000x70 X 60=80x8x5280x0;; 33000' X 70x60 80 X 8 X 5*280 =41-02 miles. Ques, 9. An engine of 48"" moves with a maximum speed of 33 miles an hour, on a level rail; required the' gross load of the train, friction 6 lbs. a ton ? PBACTIOAL MECHANICS. 15 Let X be the gross weight of the train in tons, then the work consumed an hour in moving the train. =;>; X 6 X 33 X 5280- Work of the engine an hour, =48 X 33000' X 60- When the speed is uniform or at its maximum, a; X 6 X 33 X 5280=48 x 33000 x 60 ; 48x33000x60 omn^ ■• ''= 6x30x5280 =''"^°'^^- Ques. 10. In what time will an engine of 66 horse power, moving a train of 200 tons, complete a j6urney of 100 miles, friction 5 lbs. per ton ; rails horizontal 1 Work expanded in moving the train 100 miles, = 100 X 5280 X.200 x 5 = 528000000. Work of the engine per hour, =33000' X 66 X 60=130680000 528000000. 130680000 =-4-04 hours. Ques. 11. "What work per minute will a horse perform when travelling 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, =250-41|x2|=145flbs. 2imiles=13200; Feet per minute = — — — =220 feet. Hence the work of this horse per minute, =145fx 220=32083^' Ques. 12. What load will a horse draw travelling at 3 miles an hour upon a plank road, whose friction is ^ J^ of the whole load, the road being horizontal ? 16 PEACTIOAL MECHANICS. The traction at this speed is 125 lbs., for 250-41fx3=125. The gross load must be 100 times this weight, = 125x100=12500 lbs. Ques. 13. Suppose a horse to be able to perform 33000 units of work in a minute on a horizontal road, whose fric- tion is -^ of the whole road ; required the load and rate per hour ? Traction=(250-41§ r) pounds. Space passed over in a minute, 60 Then (250 -llf?-) 88r=the units of work performed by the horse in a minute=33000'. From this quadratic equation, the value of r is found to be 3 miles ; Traction = 1 2 5 lbs . , and the load will be 125x20=2500 lbs. Ques. 14. At what rate will a horse draw a ton, on a road whose coefficient of friction is ^^^ ? Tractiott=?||?=701bs. .-. 70=250-(41|)r; ,^ 250-70^^.3 41S the rate is 4-32 miles an hour. Ques. 15. When a horse exerts a traction of 41|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 ^^^j, and what work will he perform a minute ? Load=30 X 41|=1250 lbs. Distance moved in feet per minute, -^21^0 = 440 feet. 60 Work=41fx 440= 18333V- PBAOTIOAL MECHANICS. 17 ■s Ques. 16. What must be the horse power of an engine, to cot #600 square feet of pknkixig in a day of 10 hmxi loi^t U requires, according to experiment, about 29000 units of ■work to saw a square foot of green oak planking, tbere- foMi, !Eh€ Twjrk in cutting 6600 square feet, =29000 X 6600=191400000. 191400000_ 319000', 10x60 M«i^^=9fHP. 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? llie work of the engine in five minutes, =33000' X 5 X 24=3960000, the. work destroyed in cutting 144 square feet. Therefore the units of work e?;pended in cuttmg one square foot, =3960^=27500. 144 CHAPTER m. ON THE MOVING OF BODrBS ON INCLINED PLAJTES,' AND THE RAISING OF MATERIALS. If a surface be sappased without friction, the units of work performed by imoving a body along it is equal to the product of the weight of the body in pounds, by the ver- tical height in feet through which it is raised. 18 PEACTIOAL MECHANICS. Fig. 2. To move a body along the path ACE G.l, may be imagined the same as to move it along and up an infinite number of steps, resembling A B, B 0, CD, D E, &c. And since friction is neglected, there is no work required to move the body in a horizontal direction, all the work, is expended in raising the body in the perpendicular direc- tion ; hence the units of work required to raise a body from L to I equals the units of work required to move it along the path A E G I, the friction of the path being neglected. The principle here explained holds true for inclined planes. 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 5^ 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 by the method ex- plained in the last chapter. Speed of train a minute = 2640 feet. Weight of train ia lbs. = 448000. 1000 200' "^^ of the rail in one foot. PEACTIOAL MECHANICS. 19 ^ X 2640 = 13;2 feet, , the rise of the rail in 2640 feet, Consequently the whole ^ weight of the train is raised 13-2 feet eyery minute in opposition to gravity. Work due to gravity a minute, > =448000x13-2=5913600'. Work due to friction a minute, =200 X 5A X 2640=2904000'. Total work of the engine a minute, =5913600+2904000=8817600'. 8817600 , - 33000 -^^^ ^■ Ques. 2. A train of 330 tons ascends an incline that has a rise of ^ in 100, what is the maximum speed with an engine of 120 horse power ; the friction of the rail 8 lbs. per ton ? Let a; be the speed of the train in feet per hour. The rise in each foot of rail, = 1 divided by 100 =-5^ ; the rise of the rail in x feet, — * 500"' The work due to gravity in an hour, =330 X 2240 X -g^= 1478-4 X a; ; Work due to friction in an hour, 330x8xa;=2640x^. It is evident that the work due to gravity in an hour, added to the work due to friction in annour, must be equal to the work done by the engine in an hour, .-. 33000' X 120 X 60=2640 xa+1478-4x a-, 237600000 .,.„. . .•. iv= — ... ..g.^ =57692 feet. Ques. 3. An epgine of 50 horse power ascends a gradient C2 20 PBACTIOAL MECHANICS. having a rise of | in 100 ; with a steady speed of 20 miles an hour, what is the weight of the train in tons, the friction per ton being 8 lbs. ? Let (c be the weight of the train in tons ; the work required to overcome friction, = 1760 X 8 X a; = 14080 X a; . Since a speed of 20 miles an hour=1760 feet a minute rise of rail in a foot, = J divided by 100=:t§^ ; rise of the rail in 1760 feet, =100 =1^-2 f^«*; work due to gravity, =2240 X a? X 13-2=29568 X ipf work of the engine a minute, = 33000' X 50 = 165000 , Consequently, 14080 ^+29568 a:'= 1650006' 1650000 43648 : 37-8 tons. Qwes. 4. A train of 100 tons descends a gradient having a rise of a J 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 =4224000'- Else in one foot, = i divided by 100 = t4b. Else in 5280 feet, or in one minute, 5280 ,„„„ = TPwr=13-2 feet. 400 Work due to gravity, = 224000 X 13-2 = 2956800'. PBACTICAL MECHANICS. 21 In this case gravity acts with the engine ; .-. 4224000' minus 2956800= 1267200' 1 67200' 33000' =38" 4. 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 coeflScient of friction to bfi -L? "Work of the horse in moving over 190 feet, = 190x144=27360. The work of friction in moving x lbs. over 190 feet, ^x 190 = 9-5 a;, supposing a; to be the required load ; He work due to gravity, when the load is moved over 190 feet=:3x; .-. 9 5 ^+3 a; = 27360; ... a; = ?|i^ = 2188'81bs. 125 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 frictioiQ tobeJ^? Work due to gravity in moving 2000 lbs. 369 feet, = 2000x15 = 30000; "Work due to friction in moving the, load 369 feet, = ?|^x 3-69 = 24600; From 30000 Take 24600 5400 5400 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, supposing the 22 PRACTICAL MECHANICS. resistance of friction to be j^ of the whole load, and the trac- tion of each horse 160 lbs. Let X be the number of horses. Work due to friction in moving 6 tons over 100 feet, = ^ii^x 100 = 112000. Work of gravity in moving 6 tons over 100 feet, = 6x2240x2^.-33600. The work of x horses in'passing over 100 feet of this road, = 160xa;xlOO = 16000ir; ■-■ 16000 X =112000+83600, ■•■^=^j'^g99 = 9-l horses. 16000 Fig. 3. The work in raising materials, having a given form, will be their weight .multiplied by the per- pendicular height to which the centre of gravity is raised. Suppose the body, A B, to be raised from ■ the horizontal line, Q R, and the point, 0, to be the middle or centre of gravity of the body ; let e and r be points equi- distant from 0, and the centres of gravity of two equal solices A D, E B. Now if equal weights, say for example 5 lbs., be placed at e and r, the centre of gravity of these weights will be at C, then "Work in raising 5 lbs. to r = 5 x H r. ;> „ „ to e = 5xH e. Sum =5x(Hr+He) . But HH-He = 2 HC; .-. sum=10xHC. , PEAOTICAL MECHANICS. 23 That is, 10 lbs. raised from H to is the same amount of work as 5 lbs. raised to r, and 5 lbs. raised to e. The same may be proved of any two equal weights, one of them raised as far above the centre of gravity C, as the other is below it. Ques. 8. Required the units of work in raising the mate- rial of a wall 22 feet long, 13 feet high, and 2^ feet thick; supposing the weight of a cubic foot of the material to be 140 lbs.? Contents of the wall in cubic feet, = 22xl3x2|- = 715. Weight of the wall in pounds, = 715x240 = 10010. Tlie height of the centre of gravity of the whole wall, = i£.= 6ifeet; 2 ^ Work = 10010 X 6i = 65065. Ques. 9. The shaft of a pit is to be sunk 120 feet, deep, and 6 feet in diameter ; in how many days would a mail working with a wheel and axle, raise the material, sup- posing a ciibic foot to weigh 100 lbs. ? The area of a circle is found by squaring the diameter rfnd multiplying by '7854: 6 squared 6^ = 36. Number of cubic feet in the shaft, = 62 X -7854 X 120 = 3392-928 ; Weight of the material, = 3392-928x100 = 339292-8 lbs. • 120 f f The units of work to raise this material —^ ^®^"'' = 339292-8 X 1^=20357568; for it is clear that the centre of gravity of the shaft is 60 feet from the top. A man will perform 2600' working 8 hours a day, turning a handle. 24 PBACTIOAL MECHANICS. Units of work in 8 hours, ^ 2600' X 60 X 8 = 1248000- The number of days required, _ 20357568 ^igss 1248000 Ques. 10. Kequired the work in raising 3 cwt. of coals from a pit whose depth is 120 feet, the circumferenGe af the rope being 2 inches, allowing the weight of L foot of the rope of 1 inch in circumference to be "046 lbs. ? The weights of ropes of the same material, and of the same length, are as the squares of their circumferences. Weight of one foot of rope, = 22x-046 = -1841bs. "Weight of the whole rope, = •184x120 = 22-08 lbs. Units of work expended in raising the rope, = 22-08 x 8^=1324-8. Work in raising the coals, = 112x3x120 = 40320; Total work, = 40320 + 1324 8 = 41644 8. 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 weflisSBfeet? Taking 1000 oz. or 62J lbs. as the weight of a cubic foot of water, the 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, = _i- +36 = 40 feet. 2 Work= 110000 X 40 = 4400000- PBAOTIOAIi MECHANICS. 25 Ques. 12. The side A B of a cube of granite is 6 feet, and the wd^t of a cabic foot is 170 lbs.. ; it is required to find the work uecessary to overturn it on tiie edge of A? The distaoc^ of the centre of grayity, g, from the edge, Af = \/— =4'M feet. 2 When the centre is about to fall, the centre of ^avity is raised frcnia r to n, consequ^tly the work in «weiturmng the body i& the same as raimig ite whole wei^t a perpeur dicular h^^tsirn,, .-. rn = 4-24- 3 = 1-24. Woi*:= 170 X 6» X 1-24 = 45532-a In a body of any form about to fall, the centre of gravity, y, will be at n in tibe vertical line, A 0, the work in bringing the body to this position is due to the ver- tical distance, r n, through which the centre of gravity has been raised. The work necessary to overturn any body is the true measure of stability. 26 PBACTICAL MECHANICS. Fig. 6. ■ Ques. 13. Let a round shot, strike a heavy penduluiji block, AB, Fig. 5, and move ;with it to the position ab; let O be the point of siispension, G the centre of gravity of the block, shot, and pendulum taken together; O G = 17 feet 5^ inches ; O P = 12 inches, and the point p to b^ 2| inches lower than P, or s P = 2 J inches. Find the units of "work done by this round shot ; the weight of the com- pound body put in motion being 2 tons 4 cwt. 25 lbs. ; friction and the resistance of th6 air being neglected, O P : Ps :: G : G C, the distance through which the centre of gravity, G, is raised — that is, 12 : 2 J :: 17 feet 5^ inches : 4 feet nearly. 3 tons 4 cwt. 25 lbs. = 7193 lbs. 7193 X 4 = 28772, the units of work required. PEAOTIOAL MECHANICS. 27 CHAPTEE IV. ON THE LEVEE, WHEEL 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 object of machinery, whether simple or compound, is to regulate 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 eixactly 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 work is peculiarly the production of active or living agents. THE LEVER. Suppose two uniform bars, AB, BC, to be suspended from their centres, n and m, by means of cords at- tached to the points 4 and 3 of the lever D E turning on the fulcrum F, which must evi- dently be in the niiddle of D"E=A C, and over the middle of AC, in order to secure equili- brium ; thatis, in order that the two parts may rest hori- ,zontally as if they were in one piece, and suspended by the nilddle point, S. It is also evident that the equilibrium will not be destroyed if the bars be hung by their ends at the points 4 and 3. 2« PRAOTIGAL MECHANICS. Let the weight of the bar, AB=61bs., and the weight of B = 8 lbs., then A B will contain 6 units of length, and B 0, 8 units of length. It is evident from the figure that F q, the distance at which A B acts from the fulcrum, con- tains 4 units, and Yp, 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 q, balances a weight of 8 lbs., suspended at p ; therefore the following relation exists when equilibrium takes place : 6 lbs. X 4 = 8 lbs. = 3, generally wx2'F = WxFp. 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 LEVEE. Kg. 7. A B is a lever of the first order, F the fulcrum or prop, P the power acting at G, and W the weight at- tached to the point D. If CF 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. Lever of the second kind. — The weight is between the fulcnmi ^and the power. W is the weight, F the ful- crum, P the power. when the lever is supposed to be without weight, then if the length of A F=«, and B F=p, the power P balances the weight W, when Wxp-Pxa. SECOND ORDER OF LEVEE. PEACTICAL MECHANICS. 29 THIRD ORDER OF LEVER. Third kind or order Fig. 9. of lever. — ^In this case the power is between the fulcrum and the weight. P represents the power, F the ful- crum, and Wthe weight. Generally if B Fzzp, and FF=:q, the power P balances the weight W, when ■Wx^ = Pxg'. EXAMPLES. QuesUon 1. In a lever of the first kind. Fig. 7, let W= 30 lbs. The arm CF = 11 feet, the anli FD=4; re- quired P ? Puta? = Pinlbs. Then by the equality of moments, llxa:=4x30, 120 11 .= 10 1^ 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 ? •Wx3 = 50x48, ■ .'. 'W = 800 lbs. Ques. 3. In a lever of the second kind, Fig. 8, W=r 1 1 lbs., B F = 1 6 inches, and A F = 100 inches ; required P ? Px 100 = 11x16, .-. P= 1-76 lbs. Ques. 4. In a lever of the third Mnd, Fig, 9, W=:4©; B F=60 inches, and P F=8 ; required P? Px 8=40x60,. .-, P= 300 lbs. 30 PEACTICAL MECHANICS. Ques. 5i 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, 9 9 ^'S- 10. Ques. 6. In a combi- nation of three levers of the first order, repre- sented on the accom- panying diagram. Fig, 10, the long arms are 9, , 7, 8, inches respectively, the; short arms 2, 3, 1 inches ; if a pressure 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. Fig. 11. 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 attached to the long arm, and only , 9 lbs. when weighed in the oppo- site scale ;what is the true weight of Q? PRACTICAL MECHANICS. 31 By the equality of moments, the two following equations are obtained : QxWF=9xPF; QxPF=16xWr. Multiplying these equations together, and then sti-jking out the common factors, QxQ=9xl6; .-. Q=12. THE PRINCIPLE OE WORK APPLIED TO THE LEVEE. . Fie. 12. Let B F = 10 feet, F A=2 feet,andP = 31bs.; requiredW? It is evident, i£ 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='Wx 1 ; .-. "Wxl=3x5; .-. W=151bs. Hence the equality of moments is readily established by the principle of work, and conversely. When the motion is extremely small, the principle of work is termed the prin- ciple of virtual velocities. OF THE LEVEE WHEN ITS WEIGHT IS TAKEN INTO ACCOUNT. The tendency of a uni- ^'S- ^^■ form beam or lever, AB, to turn about the fulcruiij, F, is just the same as if the whole of its weight were collected in its middle point, or centre of gravity, C. For 32 PRACTICAL MECHANICS. if the preponderatiBg side, A F, be hung from a cord, m r, placed so that A >w = wB, then this cord will sustain one- half the weight of the beam, and the fulcrum, F, the other half. 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 col- lected in its middle point. Ques. 8. The weight .of the lever of the first klnd= 10 lbs.. Fig." 7, the length A B=56 inches; AF=40; F=36 ; D F=5 inches ; and W=150 lbs. ; reqiiired 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-operates with the power, m F=A F— A m, =40-1 of 56=12 l%en by the equality of moments, Px36+10x 12=5x150. P= 17-25 lbs. Fig. U. Qws. 9. The weight ofthe lever SK, Fig. 14, of the first order=31. lbs.; S K=85 inches; S F=55; A F=43; BF=19; aiidW=34 lbs. EequiredP? S P=55. S C=5^=42i CP=lli; . Px4S+31xlli = Wxl9, p_ 34xl9-31xlH _g„, 43 6 14 lbs. PBAOTICAL MECHANICS. 33 Fig.,l& Qms, 10. The weight of a lever, Fig. 15, of the second order=81bs. SR= 80 inches; SF=76; AF =70 inches ; B F = 2 inches, and P=1001bs.requiredW? In this, case, the wdght of the lever acts with "Wf; 80 .V 76-^ = 36 = F. 2 Wx 2+8x36 = 100x70, .- ■W= 3356 lbs. Ques. 11. A beam, R S, Fig. 15, whose-wefelit is 4cwt., is supported 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 RS=50 feet; F R=2 feet; B F=12 feet; AF=30feetl Let C be the centre of tbe beam, then CR=^.of 50=25; G F=25 — 2=23 ; suppose the beam to turn on F as a fulorum,, the pressure on A, =Px30=20xl2+4X23, -Because the two- props support the whole weight' 24 cwt., 24—11^=12^ cwt. the pressure on F. now TO GRADUATE THE LEVER OF A< SAFBTT VALVE OF A STEAM ENGINE. A F isi at graduated lever of the second kiiic^ turning on F as a centre; V the- valve of opening or closing the com- munication 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 weighty Wj is suspended from the lever, enabling the engineer to place different amounts of pressure on the valve; this pressure measures the elasticity of the steam D 34 PEACTIOAL MECHANICS. when it begins to escape. When steam of very hi^h tem- perature is employed, the admission of atmospheric air is daiigeroiis, and 'unc^r such circmnstances this valve may be properly termed the imsifety valve. In order to era^ duate the lever, a weight, W, must be found, so that when it is placed, at A.it may balance the greatest pressure of the stfeam.. The n^xt thing to be found, is the position of the weight W, t0 g|^/.any proposed pressure to the valve. Fig. 16. EXAMPLES. Qms. 12. The length A F of a lever =14 inches; the distance F Q=2 inches ; the weight of the valve and pid- =5 lbs.; the weight of the lever F A=8 lbs., and the area of circular valve m the narrowest place =6 square inches ; 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- snre,of 30 lbs., on the square inch, or 45 lbs. including the pressure of the atmosphere ? Pressure of steam on valve, = 6x30 = 1801bs. Hence the effective pressure on the lever, =180 — 54 n") lbs., Kince the weight of the lever will act at the middle point, C, Wxl4+8x7 = I7&^2 .-. W=21 lbs. Ques. 13. At what distance from the fulcmm must the, load be, in the last ^sample, so that the steam may have a. pressure pjf 16 lbs. over the pressure of the atmosphere ? PEACTIOAL MECHANICS. 35 16x6-5 = 91. Suppose D , Kg. 16, to be the required 'position oif the load, then t FDx21+8x7 = 91x2; .-. PD = 6 inches. Ques. li. la. a, bent lever, Fig. 17, the peipeadicular, OA, on tteji^ecilion of the force, P, is 4 indies; and the perpetidi6ular, OB, upon the direction of W, is 7 inches ; required F whien W=2001bs. ? Px4 = 200x7, .-. P=350. Ques. 15. A rope, AD, Fig. 18, supports a uniform pole, OD, resting on the ground at O, and supporting the weight, W, suspended from D ; required the tension of the rope, when A D = 175 feet, O A = 40 feet, O D = 145 feet, W= 50 cwt., and weight of O D = 10 cwt. ? In this example O D may be regarded as a lever turn- ing on O as a centre. Draw O P perpendicular to A D, and G B, a vertical Une, passing through the centre of d2 > ,^6 PRACTICAL MECHANICS. Fig. 18. gravity, G, of the pole, then the moment of the force stretcninethe rope will be equal to the sum of the mo- ments of^V^ and the weight of the pole ; that is. Tension of cord x P = W x O N+ Wt. of pole x R. To find the perpendicniars, 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 O D, are given, the area may be found by the common rule ; the area = 2100 square feet; but the area is also ex- pressed by ADxOP „,„„ 1 -2100, .-. O P=24 feet. But the area is also = AO|M=2ioo, .'. D]Sr= 105 fee t. ON=x/Ub'— 105' r- 100, .-. OR = 50. .-. i;ension of cord x 24 = 50 x 19+ 100 x 50. GonBec[aentIy the tension of the cord -229icwt. rEACTICAl MECHANICS. 3T THE CENTRE OP GRAVITY. Without involving error, we maysuppose a body acted upon by the earth's gravity to be composed of an infinite number or particles, each of which is acted on in a direction perpen- dicular to the horizon. The sum of all the pairallel forces is evidently the weight of the body. There is a point, where a single force, equal to the weight of the, body, being appliec^ wUl produce the same effect as the force of gravity acting upon the various par- ticles coippiosing the body ; this point is called the centre of gravity bi the body. From this definition, it immediately follows that if the centre of gravity be supported, the body will stand, and vice versa. But it must not be for- gotten that a heavy body may be moved with ease on an axis passing through and supporting its centre of gravity. That the centre of gravity of all symmetrical 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.l6. Let A, B, and 0, 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=61bs., B = 4 lbs., (J= 10 lbs.; A:F=10 feet, BF=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 the fulcrum ; then as the whole mass maybe regarded as-acting in the point G, by the equality of moments. 38 PBACTIOAL MEOHANIOS. FGx (6+4+10) = 6x10+4x25+10x42, .-. FGx 20 = 580, FG = 29feet. Let A, B, &c., being any number of bodies lying in the same horizontal plane ; it is required to determine the posi- tion of the centre of gravity, G, referred to two axes or Unes, OX, O Y, perpendicular to each other, these Uiies, O X and O Y, axe termed co-ordinate axes. Conceive the bodies to be connected with the axis, O X, by the perpendicular rods, At?, B6, &c., then the sum of the moments of the bodies, tending to turn round the axis, G 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 Gt from O A is obtained. In precisely the same way, we find the distance of G from O Y; and hence Kg. 20. BH jthe point G becomes known. After the same manner the position of the centre of gravity may be found, when the PEAGTigAL MECHANICS. 39 bodies are in space, by referring them to co-ordinate planes. Q^es. 17. The weights of three bodies, C, D, E, are 10, 17, and 9 lbs. respectively (Fig. 20) ; and their distances from O X are 8, 16, and 28 mches, and from 6 Y, 38, 24, and 11 inches respectively; required the position of the common centre of graivily ? Let x be the distance of the centre of gravity from the axis, O X ; and y the distance from O Y, men (10+17+81) «=10 X 8+17 X 16+9 x 28, (10+17+9) y=10x 38+17 X 24+9 Xlly 23 .• ^=24-3g-. WHEEL AND AXLE. This simple machine is only another form of the lever where the power is made to act cpntmuously; it consists of a .large wheel, A C D, and a Cyunder, or axle, B F, both of which turn on the same axis, O. If. the wheel, A Q D, be turned round by a poWer, P, ap- plied' to it, thfe axle, B'F, will coil up the rope by which the wdght W l^gs. The lever by which P acts is evidently A O, and that by which W acts is O B; Jience when these weightis, or preS-' sures, balance each other, Fig. 21i 1 I 1 1 1 PxAO=WxOB. 40 PBACTIOAL MEOHAITICS. If the wheel he displaced by a handle, then the machine is termed a windlass. Sometimes the handle is made to turn a series of wheels acting on each other by- means of tieeth ; a machine so formed 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. When the wheel makes 1 revolution, the axle also makes one. In one turn P descends a space, =2x0 Ax 31416, and W .ascends a space =2xOBx3-1416. The work done by P in one reTolation =2xA0x31416xF; and the work of W in one reTolution =2 X B X 31416 X W- It is clear that the work done in one revolution is equal to the work applied,, friction being neglected, :. 2xA0x31416xF=2x0Rx31416xW- .-. PxOA=WxOB, which is the relation already established. Conversely, if we assume the equation of equilibrium, the priuciple of' work may be readily established. It must be remembered, in applying the principle of work, that O A, O B,, are taken in feet or decimal parts of a foot, and that W and P are measured in lbs. 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, friction being neglected 1 Work of P in one revolution, = 60 x^^Mx 31416. Work of W in one revolution, =Wx^^x31416. **• Wx^^>« 3 1416 = 60x^^1^x3 1416. ,.•. W = 360 lbs. PBACTIOAL MECHANIOg". 4l Ques. 19. Required W in the last example, when the thickness- of the cord is 1 inch, supposing that i of the work applied to be lost in friction, and the rigidity of the cord? The cord increases the radius of the wheel J of an inch, hence Work of W in one revolution, Effective work of P in. one revolution, =^x 60x3x31416, " ^'*A^31416=|x60x3x31416. .-. ■W= 270 lbs. OP COGGED OR TOOTHED WHEELS. Suppose the cogged wheel D E to turn; upon the same juds, N,(-ias the wheel. ; Q R another, cogged wheel, acted i^ P-BAGTIOAL MECHANICS. •upon by the former and turning upon the same axis as the axle L. Frojm the wheel O Is suspended the weight P, and fromL the weight W; then whUe P descends, the wheel and the cogged wheel D E will be turned from right to Mt, but as each tooth in the cog-wheel D E is bang turned round, a corresponding tooth in the cog-wheel QiR 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 20. Let P=120 lbs., the diameter of the wheel 0.=3 feet, the number of teeth in DE^ll, the number in QR=14, the diameter of the axle L = ^ a foot; re- quired W, in order that equilibrium may take place, friction being neglected ? For every turn of the wheel 0, 11 teeth of the wheel Q will be turned round, therefore as many times as the number of teeth in D can be taken out of the number in Q, so many turns will the cog-wheel D E make while the wheel Q B makes one. : -Let. the axle L and wheel QR make one rerolutioii^ then the revolutions made by the wheel 0, = 14 divided by 11=14. The space moved' over by W, =|X 31416; "Die space moved over by P, i|x3x31416; Work due to W=lx31446x W. Work due to P=3 X 31416 x l^x 120. .'. ^ SiQce the work dtie to P and W are eq^&l during one re- volution of L, neglecting friction, I X 3 1416 X W=1J x 3 X 3 1416 x 120, .-. iW=ifx3xl20. W=916 ^\ Ihs, PBAOTIOAIi MEOHANICS. u Ques. 21. Let the handle of a crane, Fig. 22, describe the circle AC, whose radlus=19 inches, and suppose 12 the number of teeth in D E, and 60 in ,Q B ; the mameter of the axle L=4| inches; it has been found that 200 lbs. pressure applied to the handle will only raise 5000 lbs. ; what is lost by friction? Feet passed over by W in one revolution of Q B, _ 4|x 3-1416 ^,^ 12 because x$=5, the space passed over by P for one revolution of Q E= 38x31416x5 12 feet. .•. Units of work due to P= 38x31416x5^300^41x31416^^^ the friction being neglected, in this case 'W= 8000 lbs. Hence 3000 lbs. is due to tibe friction of the machine, f of the power being lost. COMPOUND WHEEL AOT) AXLE. In the ordinary wheel and axle, there is a practical hmit to the power of the machine; for we can only increase the Kg. 23. H^9 1 PBAOTIOAL MEGHANICS. power by increasing the size of the wheel, or by decreasine the raMas ai titie axle. But in the compound wheel and axle, a given ^twer may be made to raise very great The machine consists of two axles, A and C, cut upon the same block, round which a cord coils in opposite' direc- tions; liiis cord passes round the movable pulley . D, which carries the weight 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 Itoit the rate at which W ascends depends upon the diffe- rence of the circumferences of the two axles ; and, conse-, quently, the power of the machine will also depend upon this difference, which may be decreased to aaj extent, without altering the length of the handle. This mgenious contrivance is due to the Chinese. Ques. 22. Let the diameter of the axle A = *8 feet, the diameter of axle 0='6 feet, the length of the handle' Hm=3i feet, and "W=60901bs. Required P ? When the handle Hw, Fig. 23, is turned once round, the cord A will be drawn up a space equal to the circum- ference of the axle A, while the cord will be let down a space equal to the circumference of the axle ; therefore the whole cord will be shortened a space equal to the dif- ference of the circumferences, and, because the cord is doubled, the weight W will be r Jiised a space only equal to the half of this cufference. Units of work expended on W in one revolution of H« = ' 8x3'1416--6x31416 _--_ 2 ^ 6090 ; Work done upon P, in one revolution, =7x31416xP; .-. 7x3-1416xP=:?:^i52H6=l><321L6x6090 Dividing each sid& by 3*1416, .:, 7P=*=^x6090, P=87 Ihs. •PRACTICAL MECHANICS. 45 THE PULLEY. A pulley is a grooved wheel, turning on an axis, and placed in a block or case. A cord passes over the groove of the Wheel, in order to transmit the force applied m any proposed direbtion. There is no advantage gained by a single fixed pulley, except in changing the direction of the force applied, wmch is diminished by the friction of the pulley; but, when there; are movable pulleys, 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. 23. In the annexed system of pulleys, if W=500 lbs., required P, when equilibrium takes place t Fig. 24 As W is suspended by two cords, c and 6, each cord will support 250 lbs. ; but as the cord is supposed to have a 46 PBAOTIOAL MEOHANIfia. free motion over the wheels, the portions a, b, e, will have the same stretch or tension ; hence P=250 lbs. AppUeaUon of the Pidndph of Work, Let P and W be expressed in pounds, then if W, Fig. 24, ascends 1 foot, the cords c and 6 will each be shortened 1 foot, and therefore P must descend 2 feet. Hence the work of P =Px2; and the work of W='W x 1 » .'. Px2=Wxl. Ques. 24. Let there be two movable pulleys, each weigh- ing 4 lbs., then if P=601bs., required the weight raised independent of the weight of the pulleys, on the principle of work ? If P descend 4 f eietj the first movable puUey will ascend 2 feet, and the second 1 foot. Consequently, the' work done in raising W and the pulleys =Wx l-|:4x l+4x2=^W^l!2 1 The work of F=60 X 4 ; .-. W+12=60x4. W-228 lbs. ■ .-. the weight=2201fes. ' PBAOTIOAI, MEOHANIOS. - Ques. 25. Let the cord I K L M N O P Q E S T, Fig. 25, made fast to the hook I, pass over two fixed pulleys, «, and the two movable pulleys, U, and a pressure of 2001bs. applied at T was found by emeri- ment to raise only a weight, W, of 565 lbs.; what part of the power is lost by friction, the nudity of the cord, and the weight of the mov- ablepulleysl .- .; "Wnen tho weight W is raised 1 foot, each of the cords I K, M, O N, and FB, are shortened 1 foot, and the cord S T lengthened 4 feet. Hence^when friction, &c., are neg- lected, the work done on W m raising it a foot == W x !> which must be equal to the work of T= T X 4, as T moves 4 feet, while W moves over l,yy V Wxl=Tx4. .-. T=^=141Jlbs. in the present case. .'. 200-141^=581, and -_l= -29375 the part of the power lost. 47 Fig. 26. OF THE mCIilKED TIjASE. To find the pressure necessary to support a body on an in- clined plane without friction. Let the weight W, Fig. 26, be drawn up the inclined plane, A C, by means of the wei^t P, acting by a cord parallel to the plane; then whilst W is moved from A to, O, the weight F will have descended 48 pbaotical mechanics. a vertical space equal to A 0. Then, accortog.to the prin- ciple of work, Work in raising ■W= W X C B ; Work due to the descent of P =PxAC; .-. Px A 0=^^x03; CB. P = AC W. Fig. 26. Ques. 26. 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 reqtnred to sustain the body on the plane, friction being neglected ? Work in raising W in^ opposition to gravity, =20 x 112 X 10=22400; The work of P=P x 30 ; .V Px 30=22400. P=746flbB. Ques. 27, The length of an inclined plane is a mile,' or "5280 feet, the height 88 feet, the weight of the body 1760 lbs., and the friction -^-^ part of the weight, what pressure will be required to move the body up the plane 1 PRACTICAL MECHANICS. 49 In this case the inclination of the plane being slight, the pressure upon it is veiy nearly equal to the weight of the body. .'. Pressure to overcome friotion= '^^ lbs. Work due to friction when the body is moved over one mile,== 4^ X 5280=35200; 264 Work due to gravity, = 1760x88=154880; , Work of the pressure, P, applied to move the body, ^p ^ 5280 * .'. Fx 5280=35200+ 154880. .-. P=361bs, OF THB TITEiDGE, OB MOVABLE mCLHilED PLANE. Let A BO be a wedge, Fig. 27, sliding, on the horizontal plane A B, by the action of the pressure, F, applied parallel to A B, and thereby elevating the centre of gravity of a weight, W, in a vertical direction. When 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, the pointy will have been elevated a height equal to uie thickness, G B, of the wedge, and the centre of favity, G, some vertical height m n ; then if A B=l-2 feet, G='2 feet, and the pressure at 9'=601bs., not taking friction into account, Work applied by P=Px4*2. Workofy=60x-2; X 1-2=60 X -2; .-. P=101bs. This calculation shows that the advantage of the power depends upon the thinness of the back of the wedge. But it is necessary to observe, that the astonishing power of the wedge, as usually employed, is equally resolvable into the force of impact. 50. PEACTICAl MECHANICS. F3g.-27. Ques. 28. With a velocity of 38"6 feet a second twenty blows of a sledge-hammer of 30 lbs. weight are delivered on the head of the wedge ABO, Fig. 27, and drives it 3 inches in the direction A B. W =300 tons, its centre of gravity, Otf moves from G to n, and is raised a vertical height m « = 1 inch when q is raised 2 inches ; find what part of the power, P, IS lost to useful effect by. friction when A B=8 times G B. When the wedge is moved 3 inches the point q is raised f of an inch, and m n -f^ of an inch='^7 feet. It will be shown hereafter that the hammer of the given weight and velocity in twenty blows may develop 13896 units of work to drive the wedge 3 inches, or in raising the centre of gravity -^ part of a foot. .-. Wxi =13896. .-. W 889346 lbs.=397J^ tons. Hence the friction in this case amounts to a pressure of 97^ tons. (See how to estimate the units of work in a rotating body.) PBAO^IOAL MEOHAinCS. 51 OF THE SCREW. In this simple machine, Fig. 28, the pressure applied moves in a circle whose radius is :the,lenjgth of the lever CD, or lomof the screw, whilst the dkectiou; in • which the work is done is a right line. ng. 28. ■ 1 1 1 H es. 29. The lever D pf a simple screw is 6. feet, the thickness < of the thread I="04 feet; if a pressure, P, of 200 lbs. be applied to the lever, what pressure, W, will be produced on the press board 6? Space moved over by P, in one revolution, =2x6x31416. Space moved over by W, in one revolution = -04 feet. Work of P, in one revolution, =2x6x31416x200; Work of W, in one revolution, =117 X 04. .'. Wx -04=2 X 6x3-1416 X 200- .-. W= 188496 lbs. The last question shows that the efficacy of the screw is obtained by mcreasing the length of the lever, or by de- creasing the thickness of the threads. Quea. 30. The lever, D, of a screw is 4 feet, and the thickness of the threads ^ inch; required the pressure, P, that must be exerted on the lever to produce a pressure of 12 tons upon the press-board B, Fig. 28 ? In one revolution of the lever, the work done, =12x2240xL=560; for i of an inch= Jf of a foot. f2 52 PBACTIOAL MECHANICS. In one revolution of the lever the work applied, =Px 8x31416; .-. Fx8x 31416=560, .-. P=22-28 lbs. Ques. 31. 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 pressure of 100 lbs. on the handle, or lever ? Let X be the thickness of the thread, then, the work applied in one revolution, =4x31416x100=1256-64; Work done, =xx 5000=1256 64; .-. a;=-2513 feet. OF THE COMPOUND SCREW. This mechanical contrivance consists of two screws; one of which screws within the other, so that whilst the large one is descending, the small one is relatively rising within the large one. In consequence of this compound motion, one revolution of the lever causes the press-board only to descend a distance equal to the difference of the thickness of the threads of the screws. Ques. 32. 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 an inch, but, at the same time, the small screw, by turning within the large one, ascends ^ an inch; therefore the press-board must descend a space = f— J= J of an inch=^ of a foot. EBACTIOAL MEOHANIOS. 53 Work done in one revolution,f Work applied in one revolntion, =60 x2x2-5x31416=942-48; .-. Wxi =942-48. W= 67858-56 lbs. Ques. 33. If the length of the lever = 3^ feet, the thick- ness of the tliread of the larger screw = | of an 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.? ^ of an inch=-^g- of a foot. Let X be the thickness of the smaller screw in feet. 20 X 7 X 3 1416=44800 (tV-*) .-. 3ij^a;=-0098175; ••, -0526825 =;c. THE ENDLESS SCREW. In this machine, Fig. 29, the threads of a screw cut upon a cylinder, B 0, are made to act upon the teeth of a Fipr. 29. - Bi; 1 yiSip 54 PRACTICAL MECHANICS. Cog-wheel, F D, having an axle, H E, round which a cord coils, as in the common windlass. . This comhination gives a very slow motion to the weight,, W. Ques. 34. In an endless screw, the length of the handle, A B, is 3 feet, the number of teeth in the cog-wheel, F D, is 24, and the radius of the axle, EH, | of a foot : if 36 lbs. pressure be applied to the handle, what weight, W, 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 24 turns, whilst the cog-wheel and axle make one turn. Work done in one reTolutiou of the axle, =-W'x.|x2x3-1416i for W will be raised 4-7124 feet. The work applied in one revolution of the axle, =36x6x314:16x24; .*. Wx|x 2 x3-1416=36x 6x31416x24; 36x6x24 •• ^- 2xi W=3456 lbs. THE HYDKOSTATIO PRESS. If there be any number of pistons of different magni- tudes, anyhow applied to apertures in a cylindrical vessel filled with ah incompressible and non-elastic fluid, the forces acting on the pistons to maintain an equilibrium will be to one another as the areas of the respective aper- tures, or the squares of the diameters of the pistons. Let AB D, Fie. 30, represent a section passing along the axis of a cylindrical vessel filled with an incompressible PBAOTIOAL MECHANICS. Pig. 30. 55 and non-elastic fluid, and let £ F be two pistons of different magnitudes, connected with the cylinder and closely fitted to their respective apertures or orifices ; the piston F being applied to the aperture in the side of the vessel, and the piston E occupying an entire section of the cylinder or vessel, by wliich the fluid is contained. Then, because by the nature of fluidity, the pressures on every part of the pistons E and F are mutually trans- mitted to each other through the medium of the intervening fluid, it follows that these pressures will be in a state of equilibrium when they are equal among themselves. The pressure of the power is applied to the small piston ijy a lever of the second kind, and the advantage of the machine depends upon the length of this lever, and the extent of surface of the large piston, compared with that of the small one. Ques, 35. In a hydrostatic press, the surface of the pistons F, E, Fig. 30, are 3 and 350 square inches respec- 56 PKACTIOAL MBOHANIOS. tively, the lever is 30 inches Itog, and the piston-rod is attached 2 inches from the fulcrum: if a pressure of 100 lbs. be applied tothe lever, what pressure wiU be pro- duced upon the large piston? Let the small piston descend t of an inch, or ^ part of a foot, then 1 cubic inch of water will be thrown mto the large cylinder, and its piston will be raised s^tr P^ °^ ^ inch, or j^g'o of a foot ; Work on the small piston, -100x30 ,_.-. Work on the large piston, =Wx^200' • 4200"** » W= 175000 lbs. DICK'S ANTI-FRICTION CAM-PEESS. This powerful machine, invented by David Dick, being almost free from friction and capable of so many useful modifications, may be ranked among the most ingenious contrivances of modern times. It obtained one of the highest prizes at the Paris International Exhibition. In the anti-friction cam-press, Fig. 31, let the power P turn the roller R, and also the cams A Qm, B S a ; suppose the circumference of the roller R=the length of the arc w A= B a ; the points S R Q retained in the same straight line ; the point S stationary, but R and Q movable in the di- rection of the arrow Q. Then it is evident, when P makes one revolution, the press-board, C D, will move through a space equal to the difference of a J and m n, Qto = QA, Sb =SB. PRACTICAL MECHANICS. Fig. 81. 57 But by tiirning one of the cams, as A Q m », so that the side Qm;t take the position of the side QA, then CD will move through a space = ab-\-mn. Q^es. 36. Let a pressure of 210 lbs. = P, perpendicular to KP, describe a circular path =24 feet; QA=Qm= 6S=SB; ab = 2 inches, mn = 2*15 inched, ahd the cir- cumference of R = 9 inches = the length of the arc aB = ji A; required the pressure Q on the press-board D I The units of work developed by. P in one revolution equal 24x210=5040= the units of work expended on the surface of E, or on the in- clined plane whose lengths a b = nA=the circumference of R=: 9 inches =|- feet. The height of this inclined plane = the differ- ence of a b, and mn=-lb inches = -0125 feet ; .-. ax 0125=5040. 5040 58 PEAOTICAL MECHANICS. CHAPTER V. WORK OF STEAM AND THE STEAM ENGINE. Kg. 32. Let A B be a steam cy- linder, Fig. 32, and suppose the steam to exert a mean or con- stant pressure upon the piston CD; if this constant pressure be 35 lbs. to the square inch, then if a weight or pressure of 85 lbs. be placed upon every square inch of the -^eton, the steam would just be capable to move the length of the stroke. Therefore the units of work performed upon each square inch of the piston in one stroke will be found by multiplying the pressure of the steam upon 1 inch by the length of the stroke in feet, and the units of work upon the whole piston will be the work upon 1 inch multiplied by the number of square inches of the whole piston. In high-pressure engines the pressure of the atmosphere is opposed to the pressure of the steam. Besides the pressure of the atmosphere, which is about 14*7 lbs. to . 15 lbs. to the square inch, a pressure of about 1 lb. to the square inch is taken for tiie friction due to the engine when unloaded, with an additional friction of about -^ the useful load or effective pressure to allow for the resistance necessary to overcome the friction of the loaded engine. For example, if the pressure of the steam = 56 lbs. as it comes from the boiler, from which if 15 lbs. be taken for' the pressure of the atmosphere, and 1 lb. for the friction of the unloaded PBACTlOAL MECHANICS. 59) engin^ then the remaining 40 lbs. to the square inch will be taken np by the vseful had and its friction. , , load , „ „ .-. load H — =- =40 lbs. 8 load .... or — = — =40 lbs. .*. in this case, the useful load, or effective pressure of ika steam = 85 lbs. When a condensing engine is employed, the pressure in the condenser is deducted instead of that of the atmosphere ; it is seldom more than 3^ or 4 lbs. to the square inch. .'. For a condensing engine 'f of load-|-l-f-4=total or mean pressure of the sfeam. In a condensing engine let the total pressure of the steam = 61 lbs. to the square inch. .-. ^ load=61-5=56 ; /. load=4i. that is, the effective pressure of the steam in this case— 49 lbs. The allowances given above in round numbers may be modified to suit particular cases' without effecting the principle of work upon which our calculations are based. EXAMPLES. Questim, 1. The area of the piston of a steam engine= 2200 square inches, the mean effective pressure of the steam 15 lb. the square inch, the length of the stroke 8 feet, the number of strokes a ininute=23 ; what is the horse power of the engine? Work done upon 1 square inch of the piston in one stroke, =15x8=120. Work upon the whole piston in one stroke, =120 X 2200=264000 Work done in; one minute, or in 20 strokes, =264000x20=5280000'. GO PBAOTIOAL MECHANICS. But the power of a horse being, =33000'. ■ S280000' _ignb 33000' Qm«s. 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. the square inch, the number of strokes a minute =20; it is required to find the num- ber 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 machioe is the fraction which expresses 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 J = "5. How;ever perfectly a machine may be constructed, there must always be a certain amount of work destroyed by friction. The following table of machines for the raising of water Is taken from MorirCs Mechanique Pratique^ Modulus Incline Chain pump . . . , . •, '38 Upright Chain pump . . . ... . '53 Bucket Wheel . -60 Chinese Wheel - . . -58 Archimedian Screw ;..... -70 Pumps for draining mines . . , . . . . *66 With respect to the present question we take the mo- dulus '66. Then, Load+^ load+l+15=40; 8 load „. -^=24;,..,. load=211bs. Useful work of the engine a minute, =21 X 625 X 6 X 20 X 66=ia39500'. Work to pump one cubic foot of water a heiglit of 396 feet, =62 5 X 396=247500, PBACTICAL MECHANICS. 61 Consequently the number of cubic feet that will be pumped a .1039500' 24750 -=42 cubic feet. Ques. 3. Tte area of the piston of a high-pressure engine = 660 square inches, the length of stroke = 6 feet, the number of strokes a 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 the piston, then, the work a minute, =xx660x6x20; .-. xx660x6x 20=25x33000; 7) 10-4166 1-4880 11-9046 15-0000 - 1-0000 27-9046 lbs, total pressure therefore the work a minnte =3696000' :=5600000'. •66 The useful work done one inch of the piston in one stroke, _ 5600000' _oofv 1000x20 Useful load=??P=35 lbs. ; 8 consequently the pressure of the steam =3b+~ +15+1=56 lbs. In the steam engine, the source of work is the evaporating power of the boiler. The magnitude of the work depends upon the quantity of water evaporated in a given time, and also upon the temperature, and consequently the pressure at which the steam is formed. Experimental tables have been framed, 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 versa. The following will serve as a specimen of such tables : PBACTIOAL MECHANICS. 63 Volume of a cdbiq foot of Water in the pomi of Steam at THE CORRESPONDINa PRESSURES AITD TbBPERATURES. Total Correspond- Volume of the Tntal Correspond- pressure in pounds the square inch> ii^ tempe. Steam com- lUKai ing tempe- Volume of ratnre, by pared with the pressure in pounds the square inch. rature, by Steam; Fahrenheit's thermo- volume of the water thnt has Fahrenheit's thermo- volume of waterrrl. meter. produced it. meter. 1 109-2 2Q954 37 263-7 727 2 126-1 10907 38 265-3 710 3 141-0 7455 39 266-9 693 4 152-3 S695 40 268.4 677 5 161-4 4624 41 269-9 662 6 169-2 3901 42 271-4 647 7 176-0 3380 43 272-9 634 8 182-0 2985 44 274-3 620 9 187-4 2676 45 275-7 608 10 192-4 2427 46 277-1 596 11 197-0 2222 47 278.4 584 12 201-3 2050 48 279-7 573 13 205-3 1903 49 281-0 562 14 209-0 1777 ■ 50 282-3 552 15 . 213-0 1669 51 283-6 542 16 216-4 1572 52 284-8 532 17 219-6 14S7 53 286-0 523 18 222-6 1410 54 287-2 514 19 225.6 1342 55 288-4 506 20 228-3 1280 66 289-6 498 21 231-0 1224 57 290-7 490 22 233-6 1172 58 291-9 482 23 236-1 1125 59 293-0 474 24 238-4 1082 60 294-1 467 26 240-7 1042 61 294-9 460 26 243-0 1005 62 295.9 453 27 245-1 971 63 297-0 447 28 247-2 939 64 298-1 440 29 249-2 909 65 299-1 434 30 251-2 882 66 300-1 428 31 253-1 855 67 301-2 422 32 255-0 831 68 302-2 417 33 256-8 808 69 303-2 .411 34 258-6 786 70 304-2 406 35 200-3 765 71 305-1 401 36 2()20 746 72 306-1 396 64 PEACTICAL MECHANICS. Total Correspond- Volume of the Total Correspond- ing tempe- Steam com- VM>aaani"A ing tempe- Volume of pressaie in pounds rature, by Fahrenheit's pared with the volume of the pressure in pounds rature, by Fahrenheit's Steam ; volume of {he square inch. thermo- water that has the square inch. thermo- water = 1. meter. produced it. meter. 73 307-1 391 87 319-4 333 74 308-0 386 88 320-3 330 75 308-9 381 89 321-1 326 76 309-9 377 90 321-9 323 11 310-8 372 91 322-7 320 78 311-7 368 92 323-5 317 79 312-6 364 93 324-3 313 80 313-5 359 94 325-0 310 81 314-3 355 95 325-8 307 82 315-2 351 96 326-6 305 83 316-1 348 97 327-3 302 84 316-9 344 98 328-1 299 85 317-8 340 99 328-8 296 86 318-6 337 100 329-6 293 Qms. 6. In a high-pressure engine, the area of the piston =120 square inches, the length of the stroke=2'4 feet, the effective evaporation of the boiler ='5 cubic feet a minute, the pressure of the steam in the cylinder 64 lbs. on a square inch ; making the usual allowance for the loss due to friction, required the useful load on each square inch of piston, and the useful horse power ? •f. usefiil Ioad=64-15-l=48. Usefulload=421bs. 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. Volame of steam evaporated a minute =440 X -5=220 cubic feet; Volume discharged at each stroke 120x2-4 o ,. „ . = — =-p: — =2 cubic feet; 144 PEAOTICAL MECHANICS. 65 The number of strokes each minute 220 ,_ = -^=110. The work in one stroke =42 X 120 X 2-4=12096. Work in one minute =12096 X 110=1330560'. Horse pow=l||g||0'=40- 32. Ques. 7. The area of the piston of a high-pressure engine is 264 square inches, the length of stroke 5^ feet, the pressure of the steam in the cylinder =72 lbs. the square inch, the number of strpkes a minute=36 ; required the useful load, the water evaporated each hour, and the useful horse power of the engine ? Let X be the useful load, then x+ £. -[-15+l=721bs. 7a;+a;=392. a;=49 lbs. The volume of steam discharged a minute, in cubic feet _2642<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 each hour 396 The useful horse power of the engine 264x 49x5 5x36 _„„i. .--^ 33000' ^^ ^^^ F 66 PEACTIOAL MECHANICS. 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. Eequired the duty of the engine, that is, the units of work developed by a bushel of coal each hour ; in the last example, allowing a bushel of cx)als 94 lbs. to evapo- rate 11"5 cubic feet of water? The useful work per hoxvc =264x49x5ix36x 60=153679680 i which is the work of 55 cubic feet of water. Therefore the work of a bushel of coals, or 11"5 cubic feet of water, = 153679680 X ^4^=32133024, 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, ihe resistance of friction upon the rail is 5-g- lbs. a tori, the resistance of the atmosphere 33 lbs. upon the whole train when the speed is 10 miles an hour, the diameter of the driving-wheel 7 feet, the area of the piston 120 square inches^ the length of the stroke=|- feet, and in addition to the resistances of friction, the resistance due to the blast pipe is 1| lbs. on the square inch of the piston,, when the speed of the train is 10 miles an hour. Required the pressure of the steam, the evaporation of the boiler^ and the number of bushels of coal necessary 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 circum- ference =22 feet nearly, which is the space passed over in one revolution of the (Mviiig-wheel. Total resistance to the motion of the carriages =200 X 5-5-)- C^y X 33=1397 lbs. PBACTICAt, MECHAlJlCS. QJ The work iJi ohc reTOliiiioii =1397x22=3d1^34. diLJ^er™^^°''*^°'' ^^^^ ' ^ ^ *^^ cimmference when 7 = the one Ztlitil^ ^^' ^''""""^ ''^ """ '^"*'" '""^ *'^ *'»^ Pi^t^Ji in =lxl20x| x4^720, for the engine has two cylinders, and each piston makes two strokes while the dnvuig-wheel turns once round. Consequently, the effective pressure on one square inch of the piston, -^^=42 68%. It has been found fey experiment, that the resistance o^ 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 =42-68+ i^ X l+15+4i=69-03 lbs. The number of rerolutions of the driving wheel a minute _30x5280_j2Q. 60 X 22 therefore, the two pistons will make 130x4=480 strokes a minute. The cubic feet of steam disdiarged a minute will be 120 3 ,„„ „^„ = fjjx^x 480=600. From the tatile it will be found that a cubic foot of water produces 411 cubic feet of steam of 69*03 lbs. pi-esaSitire. The number of cubic feet of water evaporated a minute _600 411" f 2 68 PRACTICAL MECHANICS. As one bushel of coal evaporates 11-5 cubic feet vi water, the number of bushels of coal a minute 600 ■"411x11-5' A journey of 500 miles is performed in 16| hours, of in 1000 minutes, at the rate of 30 miles an hour. Therefore the number of bushels of coal for 1000 minutes 600x1000 ,„„„.. , , = TT^ — TT-K =126-95 bushels. 411x11-5 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 a minute, the diameter of the driving-wheel 5 feet ; what is the speed of the train an hour ? At 50 lbs. pressure, 1 foot of water forms 554 cubic feet of steam. The volume of steam generated a minute =554 X -7=387-8. Cubic feet discharged in one revolution of the driving-wheel 4x90x16 ~ 1728 ~'^^- Therefoi'e the number of revolutions of the driving-wheel a minute _387-8_ • The space moved over by the carriage a minute in feet =5x3-1416x116-3. Consequently, the miles moved over an hour 5x3-1416x116-3x60 5280 -=20-75955. Ques. 11. The area of the piston of a locomotive engine =80 square inches, the length of the stroke =15 -inches, the pressure of the steam 48 lbs. the square inch, and the dia- meter of the driving-wheel 5 feet ; required the effective evaporation of the boiler, so that the train may have a speed of 30 miles an hour ; the effective horse power of PKACTICAL MECHANICS. 69 the engine and the weight of the train are also required, taking the resistances to the motion of the piston as in question 91 Number of revolutions of the wheel a minute _ 30 X 5280 ~60x5x3-1416~^^^- Number of strokes of both pistons a minute = 168x4==672. Load+-^+15 + l+-^x l-75=481bs. ^ , 26-75x7 „.:,,, • Load= 5 =23-4 lbs. The effective work a minute will be =23 4x80xfx 672=1572480', Hence, the effective horse power i' '=47" 65. 1572480'_,^,. 33000 Now the effective work has to support a speed of 30 miles an hour, 2640 feet a minute, opposed by the re- sistances of friction and the atmosphere. Work due to the resistance of the atmosphere a minute '' X 33 X 2640=784080'. _i^30\2 -U0>^ Therefore the work due to friction a minute =1572480'-784080'=788400'. Work of friction a minute when the train weighs x tons = 7x2640xa!; 788400' ,„„, ••• ^= 7x2640 =^^'^ *°'^^- Cubic feet of steam discharged each stroke 80 X 1-25 . - 144 ' the cubic feet discharged in a minute 80x1-25x672 - 144 70 PKACTIOAIj MECHAlflCS. From tjie talile itwilj b© found, tfeat a cuWc foot of water yields 573 cubic feet of steam ; therefore, the nurab^ of cubic fetet of water evaporated a minute 80x1-25x672 „, ... , . = ■ -iAA' c-^b ' -='81 cubic feet. 144 X 573 Ques. 12. A railroad train of 6Q tons ascends an in- cline that has a rise of \ in lOQ ; required the maxiipum speed an hour, 40 being tie effective horse power of the engine, friction 8 lbs. a toil % Let a! = the speed of the train in feet an hour. 100 : i : : . : ^^= the rise of the rail in x feet. Work due to gravity in an hour, 60 X 2240 X a; = 400 =336 a,. "Work of friction, •= 60 X 8 x x=£480 « ; But the work due to gravity each hour, added to the work due to friction each hour, must he equal to the ^oi'k done hy the engine in the same time ; ■ .-. 816 fc=40 X 33000 x 60=79200000. 79200000 „„rtK„„, ,Q, ., .•. x= STft =97059 ft.= 18-4 miles. If the engine in the last example move this train, what must be the effective evaporation of the boiler, and the duty of the en^nef Speed a minute,=:— ^r— =sl617'6 feet. Number of strokes of &e j^gtons a minute, 1617-6' -5 X 3-1416 ^^'""*' The effective work of the engine a minute, =33000 X40=1320000'. Suppose t/= the pound* effective preggHre on each square inch of the piston, then the work of y lbs. effective pressure a minute PKAOTIOAL MECHANICS. '71 yx 80x^1x412 = 1320000'. .-. ^=32 lbs. Pressure of steam =32+,J. X 32+15+1+1^ X 1-75 = 55-7 lbs. ■ One cubie foot of water in the form of steam, at 55-7 lbs. pressure, is 504 cubic feet. Number of cubic feet discharged a mimite, in the form of steam, Number of cubic feet of water eyaporated a minute 287 =50j = -o7. Now this water performs 1320000', .". 11"5 cubic feet of water, or 94 lbs. of coal, =^^ X 1320000'=26631579, the duty of the engine. Under the authority of an act of the American Congress, aj^proVed September 11th, 1841, an extensive series of ex- periments was conducted by Professor Johnson upon the evaporative power of several l^inds of coal. The number of saniples tried was 41, including 9 anthracite frpm Pennsylvania ; 6 foreign bituminous coals, namely : 1 frpm Sydney, Nova Scotia ; 2 of Picton coal j 1 Scotch ; 1 of Newcastle ; 1 6f Liverpool. From 1 to 6 trials were made on each sample ; the average quantity used pe? trial being 9'?8 lbs. The experiments occupied 144 days, during e^ch of which continuous observations were made during 12 to 14 hours. The coals were burnt under a steam boiler with apparatus for pomplete regulation, the supply of water and cqals being determined both by weight aud measure. The standard adopted to measure the heating power of each kind of coal was the weight of watpr which a given weight oi each evaprt*ated from the temperature 212° Fahr. 'The following Table gives the results of the comparisons of six qualities, in each of which, that coal which ranks the highest is stated as 1000, and the others in decimal parts oAhe integer. 72 PRACTICAL MECHANICS. •B)qS;3^ aApBpa "3 nam -usdxs ^q ;ooj aiqno V 0} epnnod m jqSia^ ■HopsnqcaoD JO ss3ua?3[duioo aAijBpg qa«3 I3:|}B e:)BjS aq:; uo aqoa ^njnquit jo epunoj •uopinSi JO S3I?;p;dBI 9AI}BI3a ■sinoq m uoi; -DB jCpBsis 0} jsijoq 3q? Sauq 0^ psimbsj sniix mojj mop33ij eAijBiajj pnB J3J[Ul[3 HI 91SBA4. Jli}0} JO 3S«:)n33-J3J ■JBOS JO BJiinq iBUbs loj aajiod 3Ai)ejodeA3 aApvpg ' 'qora JO ?ooj Diqno 3no iCq p33npoid 'ogjg inojj mB3)S }o Bpnnoj •[BOO JO E)qSi3u [Bnb3 joj isi^od 3Ai]BJ0dBA3 aAi:)Ei3;{ •pnj JO P3SIBI inB3:)8 JO Bpnnoj O M rH (N 1— « v:> t* o o iC C- i> co (N iH <© CO CO W3 «s »o o CO o (M o cn tH I ■ a : H o S w ® S-S o g- ■< « c o o. g PS n » iss o CO O 1-4 ■^^ CO &] es 9» i oi^ «s-^ o C» CO op OC 03 eq lo CO m iQ 00 (N CO CO O O OS l> W 1-1 lO-^ ■* ^ o •«i< O CO l> -* CO O CO lO «o tH t^ t- iH tH t* O CO tH CO lO iH 1-1 no 00 iH C3 -H O 00 X O IM UO US >0 O U3 W -C O) b- 00 o ■* oa i> 00 '■^ O C5 -^ tM T-l CO tH CO . o t- CO CM o 1-1 o eo , lO tH lO uo tH CO 00 eo CO » CO CD i-H (N t" t> t> ^ CO CO w CO as Tji CO op «> lb 00 «5 CO ■«J1 '^ CO CO CO Oi op b* l> w cp -^ CO 00 "^ -tH >0 OS CD -<*< 00 CO C» CO do 00 ^- b-eb ^ : : : d • a : Tj 'O • a o •2 -in 't ewoas cton.. iverpo uinelt :otoh.. rypin iZi f^ i^ui^ fi ! PRACTICAL MECHANICS. 73 UNITS OP WORK DEVELOPED BY CONDENSING STEAM. Experiments show that when water is raised to the temperature of 216*3° Fahr., or 4*3° ahove the boiling point, the volume is increased 1573 times, and the pressure will be 16 lbs. on the square inch. Suppose steam of this temperature to enter the lower part of the cylinder A C, Fig. 33, the piston C wiU rise, as 14*7 lbs. may be taken as the mean pressure of the atmosphere. Then let the steam be condensed by cold water; a vacuum will be formed, and the piston pressed down by the whole weight of the atmosphere on each square inch of the piston, if the vacuum be complete ; but it has been found that a perfect vacuum cannot be formed in this way, as water gives off vapour at all temperatures. If the temperature of the vapour in condenser be 103° Fahr., the pressure will be 1 lb. to the square inch : For 130° pressure 2-129 lbs. 145° )} 3-10 150° 3J 3-64 165° 3J 5-23 170° JJ 5-94 &c. &c. EXAMPLES. Ques. 13. Eequired the work developed bjr expanding and condensing a cubic foot of water, supposing 4 lbs. to be the pressure of the vapour after condensation, and 15 lbs. on the square inch to be the pressure of the steam; find, also, the duty of the atmospheric engine using the steam in this manner. Cubic feet of vacuum formed by condensation = 1710=1711-1. Pressure on one square inch of the piston = 14.7_4= 10-7 lbs. Suppose the area of the piston to be one square foot, the length of the stroke will be 1710 feet. Consequently, the work of one cubic foot of water = 144x 10 7 X 1710=2634768. 74 PKAOTICALMEOHANIOS. And the duty or worfe of 94 lbs. of coal=l bushpl, =duty of 11^ cubic feet of water =2634768x115=30299832. Ques. X4. What must be the effective evappration of the boiler of an atmospheric engine, so that the horse power maybe 30, allowins 3 lbs. for the elasticity of the vapour in the condenser ; the volume of steam bejnff 1700 times the volume of the water. from which it is produced ? Work of the engine a minute 30x33000=990000'. Work of one cubic foot of water =(15-3) X 144 X 1700=2937600. Therefore, tiie number of cubic feet of w^ter evaporated a minute _ 990000' _ q^ 2937600 It has been fo\ind by experiment that 1 lb. of good New- castle coal will evaporate 8^ lbs. of water in an hour; therefore, a bushel of this coal will evaporate 814-|lbs. of water. 'Then, if a cubic foptof waterr:62^ lbs., a bushel of coal will evaporate 13 cubic feet of water an hour. Hence the duty of an engine varies with the nature of the water and coal employed. OP THE USE OF STEAM EMPLOYED EXPANSIVELY. If steam enter the cylipder for only a part of the stroke, and then, for the remaining portion, the piston is moved by allowing the steam to expand, it is said to be used expan- sively. This is the most economical way of enaployiiig steam ; for all the available work may be taken out of the elastic vapour befpre it is condensed. When the volume of steam is increased, its elasticity or pressure is decreased in the same ratio; that is, if its volume is increased three times, its pressure will be oue- third of what it was at first, and so on. This is Bpyle's law : but M. B^gnault has shown that it will npt hofcj in extreme cases. PBACTIOAt MEOHANIOS. »fe. 33, 75 TO FIND THE POWER AND ELASTIC rOKCJp; OF STEAM IN UNITS OF WOEK. When the student does not understand the simplp process of calculating dual logarithms, Thpmas Simpson's rule may be applied to obtain approximate resullts. ftULE. Divide that part of the stroke through which the expan- sion takes place into any even number of equal parts, and calculate the pressure on the square inch upon the piston at each division of the stroke ; take the sum of the extreme pres- sure in pounds on the square inch, four times the sum of the even pressures, and twice the sum of the odd pressures ; mul- tiply the sum of all these by .one-third of ta& common dis- tance between the position? of the piston, jind the result will be the work done upon each square inch of the piston after 76 PEACTIOAL MECHANICS. expansion begins. The work done before expansion begins is evidently equal to the pressure on the square inch, multiphed by the number of feet described before ejjjansion. The whole work done during the stroke is equal to the sum of the works done before and after expansion. EXAMPLE. Qms. 15. The pressure of steam upon the piston, Fig. 33, is 21 lbs. on the square inch ; the length of the stroke^ 12 feet ; the steam cut off at 4 feet : fiiid the number of units of work done upon each square irich of the piston; sup- posing the resistance arising from imperfect condensation 2^ lbs, on the square inch* BY BOTLE'S law. Pressure on the piston from A to B, or for 4 feet of the stroke=21 lbs. The steam being cut off at 4 feet, divide the remaining part of the stroke, namely, 8 feet, into 8 equal parts. To find the pressure at the end of 5 feet, 5 : 4 : : 21 : 16-8 lbs. To find the pressure at the end of 6 feet, 6 : 4: : 21 : 14 lbs. To find the pressure at the end of 7 feet, - 7:4:: 21 : 12 lbs. The remaining pressures are found in the same manner, and are marked on the diagram (see Fig. 33). 16-8 120 14-0 9-3 10-5 21 7-6 8-4 7 45-7 32-9 28 4 2 — 182-8 65-8 PRACTICAL MECHANICS. 77 Fom times the even ordinates = 182-8 T?rice the odd ordinates . . = 65-8 Sum of extreme ordinates . = 280 3) 276-6 Area otab c d = 9{2'2 Areaofae/6 = 21x4 = 84-0 Area of /eadc = 176-2 Besistance of uncondensed vapour =2^x12=30. 176'2— 30=146'2 ^^^^^ of ^o^k done on each square inch of the piston in one stroke. Fig. 34. A B represents a steam cylinder, the steam cut off ^t CD; a X y dbis a, plane figure, a c=2 ; a o=2^ ; a n— 3 ; a s=3^ ; &c., taken from any scale of equal parts, c 7 m 110 49-89 324-40 5-708 839-3 13B-5 170-7 m 120 54-43 244-82 6-222 345-8 139-5 174-3 m 130 58-97 265-28 6-740 353-1 142-8 177-8 m 140 68-50 285-61 7-269 357-9 144-8 181-1 318 150 68-04 306-03 7-778 863-4 147-3 184-1 m 160 72-57 326-42 8-296 368-7 149-6 187-1 m 170 77-11 ■346-80 8-814 378-6 151-8 189-8 183 180 81-65 367-25 9-333 378-4 153-9 193-4 174 190 86-18 387-61 9-851 383-9 156-0 194-9 166 200 90-72 408-04 10-370 387-3 157-9 197-4 158 92 PEACTICAL MECHANICS. CHAPTER VI. Accumulated "Work. If a body moves uniformly, the space described is equal to the product of the time by the velocity. A body moving 3 ft. a second will pass over 15 ft. in 5 seconds. Fig. 35. Therefore, if a rectangle like AB D be constructed, CD=3 the units in the velocity, the. perpendicular B D = 5 the units of time ; then the small squares in the rectangle A D answer as counters to designate 15 feet in length. In a similar way, the space described by a body whose motion is uniformly accelerated may be told by the area of a trapezoid, whose parallel sides contain respectively the units of velocity at the commencement and errd of the motion, the perpendicular between those sides being the units in the intervening time. This last proposition will be easily understood, since the mean breadth of the trapezoid contains the same number of units as the velocity which the body has at the middle of the time. This method of making the area of figure represent a distance is useful, and should be caref uUy examined by a young student. HSh| PEAOTIOAL MECHANICS. 93 Tig. 36- Let E F = velocity of 2 feet a second, then suppose the body uni- formly accelerated for 5 seconds ; at the end suppose its velocity to be 4 feet a second. If L Q = 4 feet, F Q = 5 units from the same or any other scale of equal parts. The area of E F Q L in small squares or parallelograms may be used as counters to designate the space passed over 2+4 ^=3=GH. 5 X 3=15 ft. space passed over. In this way the length of a straight line is represented by a plane area. The best illustration of accelerated motion is afforded in the case of falling bodies near the surface of the earth. FORCE OF GRAVITY NEAR THE SURFACE OF THE EARTH. Bodies being allowed to fall freely near the surface of the earth, thp force of the attraction being constant, will communicate 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 32i feet, putting 32i for 32-155. This is expressed by the equation, v=tx32i. The space described by a body in one second will be half of 32^ feet = IQ^^ieet ; because the velocity of the 94 PBACTIOAL MECHANICS. 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, wiU be 4 times 2 x 32 J feet; because, 4 x 32 J feet is the velocity at the end of the time, and, therefore, 2 x 32|^ will be the mean velocity, or the velocity in the middle of the time. But 4 times~2x 32J=^ 16 X 16,^ = 4^ x Iff^Vtliat is, the space described by a falling body in 4 seconds is equal to the square of the time, midtiplied by the space described in one second! In the same manner, any other case may be established. ' In genferal, the relation of the space S in feet, and' the time t in seconds, is expressed by the equa- tion ' Qwes. 1. What velocity will a falling body have at the end of 5 seconds ? 32ix5=160|feet. Qws. 2. In what time will a body acquire a velocity of 193 feet a second ? Ksr=6 seconds. Ques. 3. Through what space will a body fall in 4 J seconds I (4i)^xl6Jj=325iifeet. Qiies. 4. A body is thrown downward with a velocity of 10 feet per second, how far wiU it descend in 6 seconds 1 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 = 6 X 10 = 60 feet, and this added to the space due to gravity will give 60+6= xl6Jj=639 feet. Ques. 5. In what time will a body fall 193 feet ? PEAOTIOAL MECHANICS. 95 By the general equation '12 .-. <=y'l2=3-464 seconds. Ques. 6. From what height must a body fall to acquire 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 96i feet. Qn\ = 3 seconds. Height=32 x 16Jj=144f 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 I 84 Time = g— ; distance or space 321 (84 \2 84;'' 32l) xl6A=^3^ = 109-7feet. Put g=B2i and a =84, then a" y 9 -_«" :. The space fallen through to acquire the velocity a is equal to the square of (a,) divided by (2 g). In most modem works, g is put for the number 32i, and tt is put for the circumference of a circle when the diameter = 1, or jr = arc of a semicircle, radius = 1. OF THE WOEK ACCUMULATED BODY. AVTien a body is in 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. work done upon it. A velocity of 32^ feet may be given to one pound, by raising it 16^^ feet, and then allowing it to fall by the force of gravity. 96 PEAOTIOAL MECHANIQS. In this case, the units of work accumulated in the body will be = 1^ 16^. As another familial- example, take a fly-wheel in rapid motion ; a 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 of the body in pounds ; because the work expended in raising the body to the necessary height to communicate the given velocity must be the same as the work, which gravity will perform 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 vdocity of 32^ feet ; what work has been done in raising it ? The ram must have fallen 16-^ieet; the>work done upon it therefore =600x16^=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 giyen velocity, 2002 =23r32l=621-76 feet. Consequently, the work which must have been done upon the body =621 76 X 144=89533 68. PBACTIOAL MECHANICS. 9? Ques. 10. Eeoviired 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 Teloeity of 1500, ^ _ 1500" 2x32a' The work accumulated in the body This example clearly shows, tl^&t Oii. '/work accumulated in a moving body is equal to the square of tjie velocity in feet per second, multiplied by the weight of the body in lbs., and divided by 2x32i. This very important proposition is ©pressed, by the eijuatioii ■ ■> tt • • Qms. 11. A ball weighing 20 lbs. Jis projected with a velocity of 60 feet per second, on a Bowling-green. What space will the ball move over before it comes to rest, allow- ing the friction to be -^ the weight of the ball I ,' The units of work, XJ, in the ball 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 baU moves before it comes to a state of rest, then the work destroyed by friction in moving the ball over x feet =-|0-xa;alll917; ,. »="19^J'y>< 18=1007-8 feet. ■ H 98 PEAOTIOAL MEOHAliriCS. Ques. 12. A train weighs 193 tons, and has a velocity of 30 miles an hour when the steam is turned off; ^^^ *^'' will the train move upon a level fail, whose friction is lot lbs. per ton ? 30 miles an hour=44 feet a second, for 30 X 5280 ^^^ 60 X 60 193 tons =432320 lbs. Putting XJ for the units of work, . _ _ 44'x 432320 _ i3009920. ■ " 2x32i ^Tien 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 =193x5ix«=U. .-. 193 x5ixa!= 13009920. 13009920_ 193x5* = 12256 feet. Ques. 13. A train weighing 60 tons has a veltocity of 40 miles an hour when the steam is turned off, how tar will it ascend an incline of 1 in 100, taking friction at 8 lbs. a ton. 40 miles an hour = 58§- feet a second. 60 tons = 134400 lbs. Work in the train when the steam is cut off (58f)'x 134400 ~ 2x32| ' put a;=the number of feet the train will move after the steam is turned off; then the work due to the friction =60x8xa;=480«; 100 : 1 : : « : JL.- 100 the rise of the rail in x feet ; PRACTICAL MECHANICS. (^)+io'^=io« P :. a;a,a,+10B)-10«^=*J,H P Patting M for 10' — , and 'N for J, a+10«; the last ecL»a- tjoD may become X N— M— xc J, (a?). Then a natural iwimber n may be found of which N is the dual logariihro ; and a natural number, m, of wtich M is the dual logapthjn, n* :. ^=3f. m 1_£_/'«Y Taking the — root of botii sides of the last equation, and n substituting ^ for — , then, n ^ m-' ^ n y 10* Il^ = 4«0000000,=t, (w). ••• l,(~ ^='400000000 ; L (7-5) = ?01490302. \m J .: J, (7-5)+10«=i, («) = 30^49030?, J,.^i) = '301490302 .-. - =-049050593 n rEACTICAL SffiOHANlCS. 125 Then 1. ( - V =z \, (z)= '400000000 (-049050593) = '19622037. The general solution of eqnations of the form «!,(«) = '19620237 is given in Dual Arithmetic a New Art, Part II., page 98. By this new art z is readily found to be = the dual number ^l | '34,5, '2'4'6'6'0, which is equal to the natural number -076221291. z=- =(-049050598) a; = -076221291. .-. a;= 1-53932 feet. VELOCITY ACQUIRED BY BODIES IN DESCENBINa AN INCLINED PLANE OR CURVE. When a body descends an inclined pkne by the fofce of fravity, the work accumttlated in the body is that which is ae to the perpendicular elevation, T^ithoOt tegaXA to the angle or curvature of the plane. Therefore, the velocity of the body will be that which it would acquire by falling freely along the pet^adicular height of the curve or plane. Ques. 34. What velocity will a body acquu-e in descend- ing an inclined plane^ whose perpendicnlar height is 579 ft.? Putting t for the time of descent in seconds, /2x 16^3^ = 579. .-. <2 = 36j or < = 6 seconds. CtonseqtfMtily, the velocity acquired at the end will be= 6x32A=196 ft. a second. Ques. 35. A train of 50 tons descends an incline of 1200 feet^ having a total rise of 170 feet; find the velocity acquired by the train supposing the friction to be 6|lbs. a ton? 126 PBAOTIOAL MECHANICS. Work done by gravity = 50 X 2240 X 170=19040000. Work due to friction = 6| X 50 X 1200=390000. Then the units of work in the train at the bottom of the incline will be 19040000-390000=18650000- Putting V for the required velocity, then 50 tons =112000 lbs., and !^1^ = 18650000. i)= 103-5 feet a second. Ques. 36. A train of 20 tons descends an incline, AB, Fig. 57, of 360 feet, having a total rise, A E, of 6 feet, and ascends another incline, B 0, having a rise of 1 foot in 20 ; how far will the train ascend the plane B before returning to B, and what will be the velocity of the train on its return to B, supposing the friction to be 8 lbs. a ton on AB, and Bibs, on B ? . .Fig. ^7, 20 tons = 44800 lbs. Units of work accumulated in descend- ing from A to B = 44800x6-20x8x360 = 211200- The^i, putting a for the length of the plane, B C, to which the train will ascend, then gj, =H C, the height; therefore, 44800x^+20x6xa;=211200- 29 , « 28 .;. ^=89^^and-^ =i ^^. _ On returning to B, the units of work accumulated in the train will be 44800x4 If- 20 x 6 x 89 1| =189422 ^. PEAOTIOAL MECHANICS. 127 Putting V for the velocity at B, then .^x 44800 ,18942^- 64i -*«™« .•. «=16"49 ft. a second. Ques. 37. The weight of a fly-wheel is 2000 lbs., the centre of gyration describes a circle 40 feet in circumfer- ence, the circumference of the axis upon which it turns is 10 inches, the wheel when stopped was making 30 revolu- tions a minute, but came to a state of rest in 48 revolu- tions ; what part of the weight of the wheel is the friction updn the axis f Units of work in the wheel = 2-5i||p=12435 23. Since the velocity of the centre of gyration=20 feet= — g^ — • The circumference of the axis=if =f feet. The length of the path In contact during 48 revolutions=|x 48=40 feet. Putting / for the coefScient, / X 2000x40= 12435. •■• /=A nearly. Since a cubic foot of shaken gunpowder weighs about 960, and a cubic foot of cast iron 7200 ounces, it is readily shown that if x=: diameter of calibre of a cannon in inches, a charge of gunpowder i the weight of the round shot will fill — inches o length of bore. For, let ^ = length of bore filled with the gun- powder, then ff being put =3-14159265, ffa" 960 -J- X3/X ji^ = ounces of gunpowder, , was* 7200 ■ -n t J t, i and — X — r^ = ounces weight 01 round snot ; 6 1728 :• 3X , xyx , 960 1728 6 X 7200 1728 .-. y= bx ' 8 inches = ^f' feet, generally y= -nx, w= =A- 128 PRACTICAL MECHANICS. Ques. 38. Suppose the length of the bore of a cast-iron cannon = 12 a; inches {mx), diameter of bore = a: inches, length occupied by the powder = -77- inches (nai); the {)owder being exploded expands, according to Boyle's law, rom the space first occupied to filj the whole bore, the force thus, developed gives an initial velocity of v feet (1200) a second to a round-shot x inches diatneter. Re- quired the pressure (p) of the gas on the square inch before expansion takes place ? Tlie pressure of the atmosphere on the shot as it passes along the bore is in a great measure counteracted by the gas which escapes around and before the shot. 12 00 divided by — = ;r- = — > •'a 5 m Of* The duallogarithm of - =197408103, D 197408103 ,„„,„„,„„ = 1-97408103.. 108 n Put = the dual.logarithiq of — divided by 10'. ■^ pd, = nxp6= units of work on each square inch of cir- cular section bf, calibre by the expansion of the powder. There- fore the units of work done on the round-shot in passing the length of the hore =: n a p f x '^—=-r'!rx^p^. But the units Of wort in the shot is also feqiial to its weight in pounds multiplied by _ ^ 2g g being .put for 32i. ,, 25 25 ■ Weight of" shot = — - ff a;' ounces =— ; — ^rz "s a? lbs. = c •? a;', 09 ^ lo X t}D_ if c be put for the specific gravity of the-shot divided by 1728 Xl6x6. PEAOTIOAL MECHAJSriOS. 129 c * a* X ;;- is also = the units of work in the shot. 2^ — irx^pik = or ay' — - 4 ^^ 2g '• o-P v — ^' — , an expression independent of both * and x. jp=^-^=14173-31641bs. PROBLEM. ?o find the position of the crank corresponding to its maximum and minimwm. velocity in a single-acting' engine. Let p B and O D be the required positions on the crank, and let P represent the constant pressure of the con- Kg. 58. 1 1 ^^^^^^^^^^Hl ■ fl 130 PBACTIOAL MECHANICS. necting-rod, supposed to act in a vertical liiie. Put Q = the constant resistance, acting at one foot from the axis of the fly-wheel, equivalent to the work of the engine. The motion will be accelerated from B to I). This acceleration will commence when the moving pressure is equal to the pegisting . pressure, and it will cease under the same condition. The former will correspond to the posi- tion of minimum, the. letter, to that of 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 A O as. the point B is above it. .-. PxCO = Qxl. (I). Again, by the equality of work, putting . Units Of work' b^ P in 1 revolution = Zt' p. „ „ Q „ =Qx2x31416 •v 2':Prfftx2x31416 (?). Dividing equation (1) by equation (2), '' ' ~ CO 1 '■"• -^ =3-rTr« =-31831 /. From tie table of natural sines ■■;''■-' •31831=cos'ineof 71° 27';' '"■' ^''iC'O' -! -':■"' "'I ' ■"■■ ■••'■i for is the cosine of the angle B C. r !'.■; '-'I «':if:!' f.;- : ' PEOBLEM. Tojind the, dimensions of the fly-wheel, such that its angular vehcity inay at no point differ from the mean velocity be- yond' h certain Emit. Let d and^feie the mkiimum and mlnihlum velocities of the wheel' at the distance of one foot from thesis ; W tlie weight of the wheel, and k the distance'o^-^ centre of gyration 'from the axis. ' ■ Work of P from B to D = p x BD = ■'Px2»'Xsin 71^' 27'= 2>'Px'948. PBAOTIOAL MECHANICS. 131 Work of the constant pressure Q from D to B = Qx2x31416xl42°54' _ 360° Srpx-soes, by putting for Q X 2 X 3-1416 its value 2 »" P, found in the last problem. Now the difference of these will give the worh that goes to increase the speed of the wheel between the points B and D; that is, worh going into the wheel between B and D = 2»P X 948-2 J- P X -3968 = »-Px 11022. Accumulated work at B=J- —. Accumulated work at D = — _ " . 2g Hence the accumulated work gained from B to D= -= — (cP— p"), but this must be equal to the work before found ; 2» (c?-p^) = »-Px 1-1022 , (3). Let V be the mean velocity of the wheel at one foot from the axis, and let the extreme velocities d and p differ from this mean velocity by the mth part ; then d=V+Iandjp = V-I; n n .•. a'—p'= (4). . n Let N be the number of double strokes performed a minute, then V.?ii^:^^^ = -10472xN, (6). Let XT be the work of the engine,, then U-2J-PN .*. ''P^^ (^)- K'2 132 PRACTICAL MECHANICS. Substituting the values given in the equations (4), (5), (6), in equation (3), and reducing, 2g—S2i 'P N' 4 X (-10472)2 r> TT .-. W=|^x 808-2 (7), which is the expression for the weight of the fly-wheel in pounds. If H be put for the horse power of the engine, then U = 33000 H; substituting this in equation (7), and reducing to tons, W=|^,X 11907 (8), which is the expression in units of tons. Let E = the mean radius of the fly-wheel, e = depth, of the rim, then from a well- known property of the centre of gyration ' substituting this in equation (8), then W=-^ ^iL: X 11907. (B?+\)W 6 Neglecting— as being comparatiTely small, then W=^3X 11907 (9). It may be observed that the weight of the wheel varies inversely as the cube of the number of strokes performed by the engine per minute. If a=the area of the section of the rim in square feet, and 450 lbs. be taken as the weight of a cubic foot of the metal, then W=2?r E a tons nearly. Substituting in equation (9), and solving the resulting equation for E, tT.o„ p-/"^ H ^ 11907x2240 Ni then E=(j^ X 2 X 3-1416 x450 ) PEAOTIOAL MECHANICS. 133 =R, . 21 ^JiH. ■• N a ' which is an expression for the mean radius of a cast-iron fly- wheel of a single-acting engine, when there are given the number of strokes of the piston, the horse power, the area of the mean section of the rim, and the proportional jvariation from a mean velocity. Proceeding in the same way, for the double-acting engine, cosine B O A=2 X -31831, and B, _12 r«JI\| ~N V a -^ Fig. 59. ^^^^^^^^^^^^■J In the fly-wheel B AW, Fig. 59, A=E; B=r, the outer and inner radius of the ring. W=: weight of the ling in pounds ; w the weight of the arms, breadth of each, DE=b, If y be the centre of gyration of the ring and arms, then, putting y—Gy, 12 (W-l-w) In practice, the centre of gyration, including the ring and arms, may be assumed &ty—r the length of the inner 134 PBACTIOAL MECHANICS. radius from the centre, 0. Putting a for the angular velocity or number of revolutions a minute, at the end of the time t in which the fly-wheel would concentrate the same power as the steam engine, t may be taken=128 seconds ; but when the work is irregular, t may be taken as high as 170 seconds. Talcing these average quantities, the weight of a fly-wheel for a given horse power H will be ^ (umiga28H_ a' r Ques. 39. Kequired the weight of a fly-wheel when the engine is of 56 horse power; the inner radius of the ring=: 10 ft. making 42 revolutions a minute? Weight=15?l^^^iM^-5409 lbs. nearly. ^ 4:2' X 10'' IMPACT. It has been before shown that the 7nass (m) of a body is a constant quantity at all heights and in all latitudes, while the weight (w) and the value of g are variable; but m= - under all circumstances. There is much uncertainty and error involved in the methods employed hj philosophers to find the value of ^ in different places. In this work, for the want of knowing better, g is put=32^ft. That is, a body falling from a state of rest is supposed to be moving at the end of the first second with a velocity of 32-^ ft. a second. When we say absolutely and without other explanation that the quantity g, which expresses the acceleration produced by gravity, is the measure of this force, we give to students an mcorrect idea, since g\is in reality only the velocity im- ported to or taken from a body by gravity during each second of its action, and the velocity which is expressed in feet cannot measure a force which should be compared with pounds. PEACTIOAL MEOHANICS, 135 MOMENTUM OK QUANTITY OP MOTION. The products m v, m Y, equal to - u or - V, have 9 9 received the name momentum ; it is a conventional phrase^ to which we attach no other signification than that oiF the product of the mas^ into the velocity imparted to or taken from it. If w = 193 lbs. where g = 32i ft., then the mass will 193 he=-— ;- = 6. In Paris ff is said to bfe = 32-1817 ft., in which plape io would be = 193"0902 lbs. ; but the mass ,^ J „ 192-0902 « , remams unaltered, lor -,„ ■ =d also. It may be further observed that the product m v, mY, is equal to / i or / T of the force, and time during which it has acted. If we consider two forces as" acting for different times upon two bodies of unequal masses, we shall have ft=mv, and F T=M V ; and .: ft : FT :: mv : MV, whence it follows that the quantities of motion m w, M V, imparted to or taken from different bodies in unequal times, are as the product of the forces to which they are due into the time during which these forces have acted. It is only when the times are equal that the quantities of motion impressed or destroyed are proportional to the forces, and can serve for their measure. From these re- marks it follows, as we shall presently explain, that in shocks there is no loss of quantity, of motion, which, is expressed in sayirig that there is ' a preservation of the quantities of motion. But we shall show hereafter that snocks occasion a loss of , work. If the fortes are fequal and act during the same time, the quantities of motion imparted or destroyed, in the two bodies with masses »i and'M, are equal. This occurs in the reaction of two bodies •which press, push, or impinge upon each other. The effects of coiilpression and resistance 136 .PRACTICAL MECHANICS^ being equal, opposed and developed during the same time, it follows that the quantity of motion imparted, in the re- action, to one of the bodies is equal to that which is lost by the other. This is a fact which is a necessary conse- quence of the theory of the shocks of bodies. Thus, for example, when a body with a mass m impressed with a velocity v, impinges on a body with a mass M anima:ted with a velocity V, in the same line, whether in the same or opposite directions, it develops at the point of contact equal and opposite efforts of compression, in an element of time t, taking from the impin^ng body a small degree of velocity Vi and consequently a quantity of motion m Vi, and imparting to the body shocked, if it moves in the same direction as the first, an increase of velocity Vi, and a quantity of motion M Vi. These quantities being equal, we have then, at each instant of the mutual shock or com- pression of bodies, m wi=M Vi. In this case one of the bodies loses a quantity of motion equal to that gained by the other, and the sum of their two quantities of motion remains the same. The same thing transpiring at each instant of the shock, it follows then that the total quantity of motion lost by a body is equal to that gained by the other during the compression, and that at each end of this period, the sum of the quantities of motion 5s the same after the shock as before. This consequence constitutes the principle of the CONSEKVATION OF THE QUANTITIES OV MOTION ; otherwise termed, THE PRINCIPLE OF THE CONSERVATION OF MOTION OF THE CENTRE OF GRAVITY. When operating with soft bodies, whose elasticity is com- pletely impaired by the shock, and which after compression unite and travel together with a common velocity H, the quantity of motion after the shock is (m+M) u, and from what proceeds we should have w i>-|-M V = (to-|-M) u. We derive for the common velocity after the shock m «+M V .y \ PRACTICAL MECHANICS. 137 Qms. 40. Let a body, A, weiehmg 772 lbs. move with a velocity of 7 ft. a second, and impinge on a body, B, weighing 965 lbs., moving the same way in a line joiiiing their centres of gravity with a velocity of 4 ft. a second, and after compression suppose these bodies unite and travel together, what is the common velocity after the shock ? 772 Mass of A= ggj- = 24 (m) ; Mass of B = 11^ =30 (M). Then according to (I.) the common velocity 24x7+30x4 . ., "= 24+30 =^*^*- Now had we employed the weights instead of the masses the result would have been the same, and much unnecessary work avoided ; 772x7+965x4 ... "=— 772+965— =^*^*- In this way we avoid employing the quantity ff. The same result would be obtained if the lb. weight happened to be too light or too heavy. Suppose in the present case the standard pound to be j^ part too light, so that A's 772 correct weight would be 772 + 103-= "^76 lbs,, and B's 965 X -^ =970 lbs. Putting these weights for the 193 masses in (I.), 776x7+970x4 ,, . , »= 776+970 ='-''"'■ If the body shocked was at rest, we should have V=o, and (I.) is reduced to "=^+M- ^"-^ or, substituting the weights, w, W, for the masses (II.) becomes 138 PKAOTIOAL MECHANICS. wv If two bodies strike against each other in opposite direc- tions, a similarity of action exists ; but then, at the end of the compression, either the bodies are both brought to a state of rest, and we have m« = M V, and m = o, in (I.), or one of the two goes backward, and they proceed with a common velocity, u. If it is the body M which gotis backwards, the quantity of motion lost by the body mis m (v — u), and the quan- tity of motion developed during the period of compression by the forces of reaction upon the body M is composed of that which has been destroyed, or M V, plus that imparted in an opposite direction, M u, and because the quantities of motion developed on both sides upon each or the bodies should be equal ; therefore, m (v—u) = M (V-f m) ; . mv-MY .^^^. •• "= m+M (™-) If the weights of the bodies, w, W, be substituted for the masses in (til.), the value of u will not be altered, and becomes Divide the numerator and denominator of (IV.) by w and it becomes, W v-~Y a formula in which we see that the velocity of the imping- ing body will be so much the less changed as its Wi^^ht, W, is greater in its ratio, with that of the b.p.dy shocked. Hence in machines working by shocks we must increase the weight of the impinging pieces in their ratio to the pieces sBpoked^ in a ratio so much greater as it is, desired to maintain a PEACTIOAL MEOHANIOS.. 139 greater regularity of motion. If the body shocked is stt rest, such as a pile driven by a ram, we have V= o ; the common velocity of the descent of the pile and the ram after the shock is 1 u = W ' 1+- w which shows that this velocity will differ so much the less from that of the arrival of the ram upon the head of the pile as the weight w of the ram is greater in its ratio with that of the pile. It is best in this case, then, to increase the weight of the ram rather than its velocity ; for the work employed to raise it increases only with its weight, while its work will be increased proportionally to the height of elevation, or to the square velocity of its descent. Ques. 41, Let a body, A, weighing 1351 lbs. (w) move with a velocity of 10 feet a second and impinge on a body, B, weighing 579 lbs. (W) moving with a velocity of 14 feet a second in an opposite direction in a straight line joining the centres of gravity of A and B ; after compression, sup- pose these bodies to unite and travel together, what is the common velocity after the shock? Mass of A= ggj- =42 ; (m), MassofB=g-g=18 (M). From (III.) we find the common velocity m, 42x10-18x14 „„. ^ ^= 18+42— ='-' ''''■ Had we employed the weights of the bodies instead of the masses of A and B, the result, as in the former case, would not be altered. Thus, 1351x10-579x14 1351+579 =2-8 feet a second. In this way we avoid employing the quantity g= 32^, or any other value. The same result would be obtained if the 140 PEAOTICAL MECHANICS. pound weight that measured the weight of the bodies hap- pened to be too heavy or too light. Suppose, in the pre- sent case, the standard pound to be xhs P^^^t too heavy ; 1351 then the correct weight of A would be 1351 TqiT" 1344 lbs., and the correct weight of B would be 579 — 579 iag= 576 lbs. 1344x10-576x14 „„„ , •■• «=— l34i+576— =2'«^^^*- However, in measuring units of work, the pound weight must be correct. The preceding considerations were proved by direct ex- periments by the justly celebrated French engineer, General Arthur Morin. He employed the following apparatus : A wooden box, in which was placed successively clay, more or less soft, sand, pieces of wood, &c., the whole suspended to a dynamometer having a style and turning plate. The plate was impressed with a uniform motion, which was trans- mitted by a weight and regulated by a fan fly-wheel. When the box was immovable, the resistance of the dynamometer was in equilibrium with the weight of the box and its contents, and the curve of flexure traced by the style upon the plate was a circle. The impinging body was a cannon-ball held by thongs, opening at pleasure, and when it struck the materials in the box it caused compression, immediately after which the two bodies fell together with a common velocity. It is easy to compare in each case the results of these experiments with those of theory. Some of these comparisons are presented here : PKAOTIOAL MECHANICS. 141 Experiments made with clay, whose resistance to penetration of projec- tiles with small velocity was 6080 lbs. the square inch. Experiments made with clay, whose resistance to penetration was 346 lbs. to a square foot. Approxi- mate dura- tion of transmis- sion of piotion. o (N O OS i-H -"J* O S iH (M r-l (M (M n ■* © » t^ CO o ©© OO o 1 -3- 1. ty expe- riment. «5 CO 00 1-1, iH i-H ^ op OS O CO ITS «o O O t^ iH iH Al ta Tfi CO lO CO iH b- H t» OS o (N (f» (N Velocity due to height of fall. OS 00 © CO OS o J iH -^ Weight of fall of sphere. >« . rH -J* lO 00 IH lO *; ep CO CO OS CO CO fH A( O O rH O CO CO OO >0 00 CO CO OS CO OO© Jf + lbs. 146-08 147-97 159-28 177-57 lO ■* O lO CO OS T-l I-l 74-85 76-85 93-14 ^ lbs. J 12-23 26-64 44-73 CO CO 2^ 26-64 26-64 44-73 1 the box and its load. ^ lbs. 132-84 134-74 132-84 132-84 op Op l> 1?- -*^ I-l 1-1 48-45 48-45 48-45 142 •PBAOTICAL MECHANICS. Common Geometrical Problem, in which none of the principles of Mechanics are involved. Given S P= 13*52 ft. the radius of the index P of a bal- listic pendulum which shows the range of the recoil by marking soft compressible matter placed in a box. A P the chord of the arc of recoil= 10"4 ft., S G= 10*14 ft., the dis- tance of the centre of gravity, G, of the pendulum, P Q R S, and ball w. It is required to find the height, D G, that the centre of ^avity is raised when P moves from P to A ? Draw AB and C D perpendicular to S P ; then A P^ divided by twice S P=BP, Fig. 60. 10-42 BP=. 27-04- =4 ft. AndSP: PB::SG: DG, that is 13'52 : 4:: 10-14: 3ft.=DG. ' Fig. 60. BBACTIOAL MEOHANIOS. 143 Mechanical Problem. Let y be the centre of gyration of the pendulum PQKS, Fig. 60, without the ball, 83/= 10 ft.; if W= the weight of the pendulum, and W7=the weight of the ball, b, the point where the ball impinges, 6S = 12ft., what weight must be placed at b, so that the pendulum would receive the same motion as when its weight is placed at y, its centre of gyration ; the position of the centre of gyration of the pendulum and ball together is also required ? W X S u* be We have before shown that if a weight = — o „ placed at b, the pendulum would receive the same mo- tion from a blow as when the weight, W, is placed at y. Now if the weight W of the pendulum = 9 tons = 20160 lbs., WxS/ 20160x102 '86" ~ 12* -=14000 lbs. Supposes the weight of the ball =20 lbs., then if ^=: the distance of the centre of gyration of both the pendu- lum and ball together from the point of suspension S, from what has been shown, page 104, ,, •WxS/+wxS&' « - W+.W - 20160x102+20x122 ,„„„,„„ 20160+20 =100-0436. .'. 2;=10'0022, so that the position of the centre of gyration is not much changed by the weight of the ball. Ques. 42. A ballistic pendulum, P Q E S, Fig. 60, is struck by a cast-iron round-shot b, ;y\feighing M;lbs.=201bs. with a ^velocity V ; .%\Q pendulum being at rest weighing W lbs. = 20160 lbs. ; Sy, the distance of the centre of gyration of the pendulum from the point of Suspension = 10 ft. ; S b, the distance from the point of suspension to where the ball ladges=12ft. Kequired the velocity of the point where 144 PBAOTICAL MECHANICS. the ball strikes and of the centre of g3rration of both pen- dulum and ball combined ? By the last problem, if 14,000 lbs. be placed at h, the pendulum would receive the same motion as when the weight of the pendulum ■W= 20160 lbs. is placed at y, its centre of gyration. We may, therefore, consider the body w to impinge directly with a velocity v upon the body weighing 14000 lbs. at rest, and the two bodies will move on together after impact in the same manner as if they were inelastic. Let x be the common velocity after the stroke, then, since the momentum or quantity of motion is the same before and after impact, we have x(w+UOOO)=tBv; _ wv 20 » ■■ '*'"«)+ 14000" 14020' By the last problem the distance of the centre of gyra- tion of ball and pendulum from the point of suspension = 10-0022 ft. . 12 . 10-0022 •• ^^ . ?5^:5Mi! . . 12 . 10 0022 .. j4()2Q . jgg240 ' the velocity of the centre of gyration of the bodies combined. Ques. 43. A ball weighing 20 lbs. strikes a ballistic pen- dulum, P Q E S, Fig. 60, weighing 9 tons, at a distance of 12 ft. from the point of suspension, making the length of the chord of recoil = A P= 10*4 ft.' ; S P, the radius of the index P= 13-52 ft. The distance S Gt= 8-45 ft. , G being the centre of gravity of the ball and pendulum combined. S j/=10 ft., y beingthe centre of gyration of the pendulum without the ball. Keqiiired v, the velocity with which the ball strikes the pendulum ? By the preceding geometrical problem, PB=4ft., then 13-52: 4:: 8-45 :2-fft.=GD. T? ^1- 1 X . 200-044« , ±rom the last question igaoAQ =t"e velocity of the centre of gyration of the ball and pendulum combined. Therefore, the units of work developed by the ball = / 200044f Y w+W V 168240 / ^"64r* PBAOTICAL MECHANICS. 145 These units of work are wholly expended in raising the centre of ^avity from G to D= 2^^ feet. But the unfts of work required to raise the centre of gravity from G to D isalso= •■ I 168240V ^4r ~ " ^'^+^)' • ,,2- 5x192 / 168240 Y 6 V200-044/ ■■■• "=^¥^S='«^«™'"'- This problem is distorted to show how the principle of work will apply without considering the oscillations of the pendulum or involving the centre of oscillation. SHOCK OF TWO ELASTIC BODIES. If we suppose that the two bodies under consideration are perfectly elastic, moving in the same direction, the effects of compression wiU be at first the same as in one of the preceding cases, pp. 134 to 140, and at the end of this time the body m will have lost a velocity, v—u, and a quan- tity of motion m-(v—u), and the body M will have gained a quantity of motion M (m— V), and these quantities being equal, we have, as before explained, the common velocity «, at the end of the compression, m «>-|-M V ^- m+M • But, after the instant of greatest compression, the elastic bodies regain their primitive form, and in the return to it develop, if the elasticity is powerful, efforts equal to their resistance or compression, and consequently destroy or impart quantities of motion equal to those which they have previously destroyed or imparted. It follows from tms that in the unbending of the molecular springs the body m will 146 PBAOTIOAL MEOHANIOS. further lose a yelocity=u— m, and that its final velocity ■win be v—2 (v—u)=2 u—v, and the body M will receive a new increase of velocity equal to m— V, and will then have a final velocity equal to V+2(m-V)=2m-V. If the body M were at rest at the beginning, supposing it perfectly elastic, it will thenfreceive a velocity =2 m, as Y=o. That is to say, twice that imparted to a soft body Tinder the same circumstances, or 2 mv These reasonings relative to soft and elastic bodies pre- suppose the existence of bodies deprived of aU elasticity,., and others endowed with perfect elasticity. Neither of these hypotheses is exact ; and, according to circumstances in which they are placed, a body may act as if deprived of all elasticity, or as if possessed of only a partial elasticity. So also a body which, in certain conditions, acts as if it were perfectly elastic, will only appear as if partially so in other cases. We will here cite some examples, the results of experiments analogous to the preceding theory, made by Morin, and which were effected by placing at the bottom (rt a movable box a plate of cast iron, upon which fell a spherical body. Morin's Expekiments upon the Teansmission of Motion by THE Shook of a Spheeical Projectile falling upon a oast- lEON Plate i f a body, and a; the perpendicular distance between this axis and another parallel to it, then y' + ai'=the square of the radius of gyration through the second axis, Qws. 55. A cylinder, B C, Fig. 65, moves round a fixed horizontal axis, Q K, parallel to the axis of the cylinder, A X ; required the pbsition of the centre of gyration, S, and the centre of oscillation, O, with respect to the axis Q R ; the distance from the point of suspension, P, to the centre of ^avity G=45 inches (a), the radius of the cylinder, D B= 6 inches (r) ? PBACTIOAL MECHANICS. Fig. 65. 16X When the cylinder devolves round the axis A X, the square cf the radius of gyration D 1/=^ r<' ; but the axis Q R is parallel to A X ; therefore, if S be the centre of gyration for this axis, S P= Va^-t-ir" = v'45=' + l2=45-2, nearly. The distance from P to O, the centre of oscillation, _ (FS)' _ 2025+ 18 ~ PG ~ 45 ■ = 45 -4, exactly. A circular disc whose plane is perpendicular to the axis Q E) with G as centre, radius = r, and G P=a, will have its centre of oscillation at the saipe point O, as the cylinder; that isj PO: a>+^r^ a + 2a M 162 PRACTICAL MECHANICS. CHAPTER VII. EQUILIBBIUM OF PKESSUEES, THRUSTS, AND TENSIONS, rORCES, «S;c. Anything which cannot be presented to the senses must be represented conventionally, or no idea of it can be enter- tained by the mind. We have no objection to represent the magnitudes and directions of forces, velocities, pressures, thrusts, and tensions by straight lines of different lengths and in different positions; but we object to the careless manner in which writers on mechanics represent forces, pressures, velocities, &c., by straight lines, triangles, and parallelograms, out of all proportion, and often, too, without the least regard to the action, direction, or nature of the force considered. In the sixth chapter we gave a conven- tional signification to the areas of plane figures, but in the present chapter the sides and angles of triangles, parallelo- grams, and other plane figures are employed to give an idea of mechanical operations and combinations. The split arrow (^w^ >• ) is put to designate the direc- tions of. tensions, the arrow without feathers ( >) points out the directions of thrusts, while the complete arrow ^) applies to forces and velocities in general. If three forces, P, Q, E, Fig. 66, be represented by the abstract numbers, 6, 4, 3, respectively, then if A B G be a triangle whose side A0 = 6, AB = 4, B0=3, equal parts taken from any scale of equal parts; let P O be a tension, equal and parallel to A 0, Q O a tension equal and parallel PEACTIOAL MECHANICS. Fig. 66. 163 to A B, K, O a tension equal and parallel to B C ; these three tensions will keep the point O at rest. Three thrusts represented by three lines P O, O Q, O E, whose lengths are as 6, 4, and 3, respectively, and parallel to the sides of the triangle A B 0, will also keep the point O at rest. Take A T equal and parallel to B 0, then the two thrusts represented in magnitude and direction by the lines T A, B A, and the tension represented in magnitude and direc- tion by the line A C, will keep the point A at rest. Again, if B S be equal and parallel to A 0, tensions represented by A B, B S, and a thrust, C B, will keep the point B at rest. Let P A, Q E, Fig. 67, be a parallelogram, if PA, Q A, represent, the magnitudes and directions of two forces, two pressures, two thrusts, two tensions, or two velocities acting on the material point A ; then the diagonal E A will repre- m2 164 PEACTICAL MECHANICS. sent the magnitude and direction of a single force, pres- sure, thrust, &c., equivalent to the operations represented by the lengths and directions of PA and Q A. Such ima- ginary parallelograms form a great portion of the stock-in- Kg. 67. trade of writers on mechanics. We have introduced Figs» 66 to 70 to illustrate the manner in which the different forms of arrows are apphed. The properties pointed out in Fig. 66 depend upon the truth asserted of the parallelo- gram P A Q R, Fig. 67, which we will prove presently. Tlie principle of sufficient reason ; first employed hy Archi- medes in demonstrating the fundamental proposiMom of mechanics. Let P and Q, Fig. 68, be two equal forces acting in the same plane on the rigid perpendicular prop A E, the point R fixed ; further, let these forces P, Q mjike equal angles^ SAD, SAB, with the prop A E, which may turn to the right or left round the fixed point R. But as the forces P and Q and the angles B AS and DAS are also equal, there is no reason why the post A R wiU turn to the right or left ; hence it will stand, and the two tensions will be neutralised by the perpendicular thrust of the post A R. This we call a mechanical demonstration, and the principle applied "the PEAOTICAL MECHANICS, Fig. 68. 165 principle of sufficient reason." For there is no reason why the post under consideration should fall to one side more than another, hence we conclude it will fall to neither side. Let A B C D be a parallelogram whose sides are all equal ; if the length of the line A B be taken to represent the mag- nitude of the force Q, then A D will truly represent the force P, and two other equal forces, V and tJ, applied to the £oint 0, represented in magnitude and direction by the nes D, B, will counteract the forces P and Q the same as the post A E, the diagonal A C bein^ rigid. The prin- ciple of sufficient reason applies here as in the former case. It is evident that any number of equal forces may be ap- plied in the directions of the arrows, P=Q=V=tJ, with- out disturbing the form of the equal-sided parallelogram ABCD, or the position of the rigid diagonal A, . The 166 PEACTICAL MECHANICS. abstract truth of the parallelogram of forces, thrusts, ten- sions, pressures, or velocities may be established by reason- ing as follows: Let ABO O be a parallelogram; when its sides can be expressed by numbers, it may be divided into small parallelograms, the sides of which are all equal to one another. We have selected a very simple case, and taken O = 4 equal parts and A = 3 of the same parts. Let the force P be to the force Q as m to w ; in the present example, P : Q : : 4 : 3. The forces P and Q applied to the material point O in the direction I O 0, O A, may evidently be represented by the lines 0, O A ; if a new set of forces, represented by the small parallelograms, be introduced as represented by the arrows in the Iigure, the combined action of the forces P and Q will not be interfered with, nor will the equili- brium of the Figure be disturbed. Take, for example, the Fijj. 69. PBACTICAL MECHANICS. 167 small equal-sided parallelogram D E T L, and for a moment suppose the diaeonal D T rigid ; then the principle of suffi- cietit reason applies to the forces introduced, and no dis- turbance takes place. Let us now view these forces under another aspect. The forces along the sides of the small parallelograms on the line C O are equal to the force P, but in a contrary direction, and hence destroys it. The forced represented on the line 9, 10, also destroy one another. This may be said of the forces on the line 7, 8 ; but the f oi'ces on the line A B are not neutrsdised, and are exactly equal to the force P, but applied to the point B in the diagonal OB. Again, the forces along the sides of the small parallelograms on the line A O are equal to the force Q, buL in a contrary direction, and hence may be said to destroy it. The forces pointed out by the arrows on the line 1, 2, also neutralise one another; the same maybe said of the forces supposed to be introduced on the line 3, 4, and so on until we arrive at the forces pointed out on the last line B 0, which, together, are exactly equal to the force Q, but applied to the point B in the diagonal O B. According as the equal parts be long or short, the point B will change its position; but it will always be in the dia- gonal O Z,and the forces o+6 + c+d=P and g+f+e=Q, will in all cases have their resultant position in the diagonal O Z. Had we supposed the force P to be divided into 1000 equal forces, O C would be represented by 1000 equal partSj and O A=Q would then be composed of 750 such parts, and our reasoning on the 750000 equilateral small parallelograms of forces would result in the same conclu- sion. The equilibrium of forces and pressures may be illustrated in a practical way by a simple mechanical contri- vance, represented in Fig. 70, easily constructed. Suppose P Q E to be a slate or board rolling freely upon three equal spherical balls placed on a horizontal table, ABC. If any three points, P, Q, E, be taken in the surface of the plane movable on the balls, and cords, A P, B Q, C E, be attached, passing over the pulleys, with a different amount of weight attached to the other ends ; the ends with the weights attached are not represented in the cut* The 168 .PBACTICAL MECHANICS. Fig. 70. board P Q E will roll upon the balls, and ultimately come to rest upon the table ABO, where the three forces destroy each other. When this experiment is carefully made, it will be found that the directions of the forces A P, B Q, K, meet in the same point, O. Again, from any scale of equal parts take O r=the pounds in the force drawing the cord <4 B, -and from the same scale of equal parts lay off Oa=the pounds drawing the cord K C ; then, if the par rallelogram O rsahe constructed, the diagonal Os will be in the direction of the third force acting over the pulley A. It will be further found that the units of lengm in 0«, the diagonal, will = the weight or force acting by the cord A P, If Oa=22 equal parts, O r=28, and Os=26, all mea- sured on the same scale of equal p^ts, then, if 7 lbs. be attached to the cord Q B, and 5^ lbs. to the cord K C, a weight of 65 lbs. must be attached to cord A B to maintain this equilibrium. When forces are thus represented by a parallelogram, as Or so, the fprces represented by the sides PKACTICAL MECHANICS. 169 are called the component forces, while the force repre- sented by the diagonal is denominated the resultant. Another experiment may be instituted to determine the in- tensity of forces by this simple apparatus. Instead of three forces being applied to the board, let there be any number ; and suppose H to be any point taken in the movable plane P QE, and from H let fall perpendiculars, Hm, H jw,,H/, «SiC., in the directions, or directions produced, of the forces ; then the units of length, taken from any scale of equal parts, multiplied by the pounds in the corresponding foi^ce, will be what is termed the moment of that force tending to turn the board or movable plane upon the point H. This being done, the principle denominated the principle of the equality of moments may be verified by experiment. Here but three forces are applied, but any given iiumber may be introduced ; in the present case, H/ X Bibs. + H m X C lbs. = H ra Albs. The student must not confound the principle of the equality of moments, just defined, with the intensity of forces usually considered by examining the motions they produce, or the spaces passed over when the motions are uni- formly accelerated. It has been before shown that, generally, (Force) =- ] (I.); t being an extremely small portion of time, denominated an element, during which the velocity v is generated. The above formula shows that if, by observing the laws of motion, we know for each instant the value of the ratio—, we shall then have that of the corresponding effort desig- nated (Force) in the above formula. Suppose we know, from experiment, that the motion is uniformly accelerated, we have, for the space S, S=iV,xT2. V V In which Vj = ^= - as in (I.) ; ,„ , W28 ,„- (Force) =-^ (II.) 170 PEACTICAL MECHANICS. Ques. 56. Suppose a locomotive weighinc 10 tons (22400 lbs.) runs, with a uniform accelerated motion, a distance of 154*4 feet in 4 seconds ; required the force in lbs. capable of imparting this accelerated motion, the friction of the rail not being considered ? From (II.) (Force) = '-^ ^ii§^)= 13440 lbs. (See pp. 93, 94, 116.) Definition. — A bar is called a strut, or a tie, according as a thrust or a pull is exerted along its line of resistance. Ques. 57. Let H B A be a skeleton diagram representing centres and lines of resistances, H B a platform with a load W=47 cwt. upon it, C B a chain supporting both load and platform ; find, by construction, the tension of the chain and the amount and direction of the pressure upon H, the point about which the -platform turns, the weight of which being neglected ? Draw W A perpendicular to H B, intersecting the chain in A ; join A and H, and, from a scale of equal parts, lay off Ap=47, a pai-t for each lb. in W which acts in the direc- tion AW. Draw pr parallel to OB, and rn parallel to A W ; the parallelogram A n rp gives the required tension Fig. 71. PEACTICAL MEOHANIOS. 171 and thrust. In the diagram A r= 37 equal parts, answering to 37 cwt., the pressure on the hinge H; A «=26 equdf parts, answering to 26 cwt., the pressure tending to break the chain. Tjike H «= A r, draw s m parallel to W H, and s t parallel to H ; then H t gives the thrust of W against the wall which is represented by H C, and H m shows the pressure in the direction of the stationary wall H 0. Fig. 72. 172 PRACTICAL MECHANICS. Qties. 58, Given a triangular frame, ABC, Fig. 72, loaded and supported both m the direction of the vertical line D E ; find the relative portions of the forces ? Put A, B, C for the three joints, and a, b, c for the three bars. These things being premised, from any point, P, draw P E, P Q, P S parallel to the lines of resistance a,b,c respectively ; across these lines the vertical line Q K S, then the following proportions apply to the forces : Load on A : supporting force at B : supporting force at C : stress along c : stress along a : stress along b ; so stand in order and proportion the lines QS:8K:QR:PS:PE:PQ. Out of the six proportions thus succinctly stated, to illus- trate, we shall single out one : Load on A : supporting force at £ : ; Q S : S B. Ques. 59. Let a pole, A B, Fig. 73, be supported by a cord, B, and carry a weight, W ; required the tension of the cord, and the pressure on the pole ? From B to » lay off equal parts, equal to the imits of weight in W ; draw m n parallel to B, intersecting B A in m I draw mp parallel to B w ; therefore the units m B^ Fig .73. ^^^^^ 1 nb^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^H^H PRACTICAL MECHANICS. 173 will give the tension of the cord C B, and the units, mea- sured on the same scale of equal parts, in the diagonal B m. •will give the thrust on the pole A.B. Struts may be distinguished from ties thus: In Fig. 74 I* Q, QE, represent two bars of a frame meeting at the joint Q, produce the lines of resistance beyond Q to S and T. Then a force in the direction of the arrow A, between Fig. 74. the angle PQE, both bars are struts; if the force be in the direction of the arrow B, in the angle TQP, then P Q is a strut, and QE a tie. Again, when the force is applied in the direction of the arrow C, both bars are ties ; and, lastly, when the force is in the direction of the arrow D, in the angle S Q E, then P Q is a tie, and Q E is a strut. Ques. 60. A beam, AB, resting on a wall, 0, and sup- Sorted by a cord, Q A; it is required to determine the irection and tension of the cord B P, so that the beam may not change its position when the wall C is removed^ the point B being also given? Let G be the centre of gravity of the beam, through which draw the "vertical line Gs in the direction of the 174 PKACTICAL MECHANICS. Fig. 75. Slumb line, to meet Q A produced in s, join s B, and pro- Tice it; B P in the direction of sB shows the direction a cord fastened at B mustjtake so that the beam will not change its position when the wall is removed. To find whether a pier, or other support, S T, Fig. 76, will overturn by the action of a force P, or the resultant of several forces operating in a given direction, P A. Pro- duce the direction of the force P, and let fall the perpen- dicular O A ; then, in order that the structure may not turn on the point O, we must have, G being the centre of gravity, , Weight of the structure S T x O B greater than P x O A. When_ (weight S T) x O B=P x O A, the pier will be on the point of turning on the edge, O. By Construction. Let B C be the vertical line passing through the centre of gravity of the pier or pillar, intersecting the diree- PRACTICAL MECHANICS. 175 tion of the force P in the point C; from a scale of equal parts take C F=the units in the pressure P, and from the same scale take E=the units in the weight of the structure, S T ; complete the parallelogram of forces, D E F, then D will give the amount and direction of the single force tending to overturn the structure or sup- port, S T. When C D pro- duced intersect the hase with- in the edge O, the structure will stand; but should the point of intersection, R, fall without the base, the struc- ture must fall. Ques.61. AT, Fig. 77, is a pier, or buttress, weigh- ing 860000 lbs., P a pres- sure of 320000 lbs.; AB = OT=6-6ft.;OA=BT = 15 ft. ; the cross section A B T O is a rectangular parallelogram, of which Gt is the centre of gi-avity. The direction of the force P tending to overturn the structure cuts A B 4 ft. from A, and A O at 0, 7 ft. from A. It is required to find whether the struc- ture will stand or fall, and what will be the amount of the pressure, P, just suffi- cient to overturn the solid mass of which A B T O is a cross section ? Fig. 76. Fig. 77. 176 PEACTICAL MECHANICS. , CP= VA CHAP''= V7H4'=8-0622577, CP : PA :: CO : OQ, or, 80622577 : 4 :: 8 : 3-969=0 Q. Since A T is a rectangular parallelogram, a vertical line, S G »w, passing through the centre of gravity, G, divides the base O T into two equal parts. S=ST=3-3 ft. ; then we can compare Moment of the wall=860000 x 8=2838000. Moment of the pressure= 320000 X Q'= 1270080, 2838000 being greater than 1270080. The structure will stand. Putting X for the pressure that will just overturn it, acting ill the direction of P, then, a;xOQ=2838000. 2838000 , , „ ••• ^=T9069- = 7150221bs. If P=715022 lbs., the point E will coincide with the point O on the outer edge of the 'structure. Geometrical Proposidon. Fiff. 78. If the sides, or the sides produced, of one triangle, a be, Fig. 78, be respectively perpendicular to the sides, or sides produced, of another, ABC, these triangles are similar. a Q, 6 P, 6 R, represent the three perpendiculars, Angle C+C S P = a right angle, Angle c+c S Q=a right angle ; But angle C S P = angle c S Q, .". angle c= angle C. In the same way the angle h may be shown to be equal to the angle B ; and consequently the angle a— A. JIJAOTIOAL MECHANIOS. .177 Funicular Poylgon.. Funicular JPe^^on is the term employed to designate a po- lygon formed of i^s, chains, or cords, whose angular points are solicited by anjworces whatever. The equilibrium of such a system of thruks, tensions, «S;c., are subject to rules easily determined. First, let us suppose that the polygon 'Iig;79. has not all its sides in the same plane, and is represented by the contour, A B D E F. Let P, Q, E, S, T, . . . . be the forces acting upon each of the summits A,B,G,D,E; the respective tensions of the sides A B, B 0, C D, DE, EF, represented by T, Ti, Tj, Tg . , respectively. If the entire polygon is in equilibrium, it must be so for all the summits separately, and since Ti must be equal, and oppo- site to the resultant of T and P, it follows that any two sides uniting at the same summit,' and the direction of tJie force acting upon this summit, must be in the same plane ; and the tension Ti must be equal and opposite-to the re- sultant of the force Q and tension Tj, ancf consequently in substituting at the simimit G, for the tension X, its .two components P and T, the forces P and Q and the tensions T and Tj, supposed to be transferred to the point 0, parallel to themselves, must be in equilibrium. We see also that the forces P, Q, K, and the tensions T and T3, supposed to be transmitted to tiie summit D in parallel N 178 PEACTICAL MBCHANIOS. Fig. 80. directioii^ must then produce an equilibrium. By continu- ing this''|iWSC6SSj we arrive at the conclusion, for the equi- librium of the funicular polygon, that when all the external forces and the tensions of the extreme sides are regarded as transferred parallel to themselves to any summit, they must necessarily produce an equilibrium there. These remarks are independent of the direction of the forces and the nature of the sides, and are applicable when the sides are subjected to efforts of compression instead of tension, with the reservation solely that the sides be sufficiently nsd to resist the compressions, without a change of form. Where the forces soliciting the funicular polygon are weights, in which case the forces P, Q, E, S . . are all vertical and parallel, and the polygon are necessarily, in one plane ; for, the direction of each force, and thoSe of the two sides meet- ing at the summit, are in the same vertical plane; and, as through one side we can draw but one vertical plane, it follows that the two vertical plaiies, containing the same side of the polygon^ are coincidentj and so for the other sides. Agam, since all the external forces P, Q> K, . . and the tensions-Ty T4 of the- first and last sides are in equili- PBACTIOAL MEOHAKIOS, 179 briom, T and T4 must also* be in equilibriiun with the single resultajit of all the parajlel forces. Determination- of the Tensions by a Graphical Construction. If, in accordance with the preceding views, we describe upon the side A B, produced, from any convenient scale of equal parts, a length equal, to the tension T, and construct the parallelogram £ a b d, Fig. 80, the side B a, measured on the same scale of equal parts, will represent their tension, T^ of the. side B G, and the sijie Bd, the external force P. Then,, if we draw through the point B, for example, a line B Gr' perpendicidar to the directions of the forces P^Q, R, . . . upon T^diich lay off B 0^ 0' D^ D' E», E^ F\ F^ G\ proportional to the forces P, Q, K, S,IJ, and fpm the same point erect the line B O perpendicular to A B with -a length proportioned to the tension T, or to B 6,. the triangle B O C^ having two sides, B O, B 0^, respec- tively perpendicular and proportional to the sides B b and B d of the triangle B bd, must be similar to it by the geometrical proposition preceding. Then the third side, Fiff. 81. ■ n2 1'80 PBAOTICAL MECHANICS. O CV of thfe first will be proportional to the third side, h d or B a, or to the tension Ti, and will be perpendicular to the side B C of the polygon ; the following triangle will be in the same condition/ in relation to the side CD, to the force Q, and the tension Tj, which will be proportional to the side O D', and so on. Then, if the weight soliciting the different summits of the contour A, B, C, . . . are in equilibrium, and we lay off upon a horizontal line lengths re- spectively proportional to these weights, and then from points of this line corresponding to each summit draw straight lines perpendicular to the directions of the sides of the poly- gon, all these right lines will intersect at the same point, arid their lengths will be proportional to the tensions of the sides of the polygon, which will thus be determined. If a line V Z be drawn parallelto B G', the triangle V O Z answers the same purpose as the triangle B O G^ The triangle B O G', Fig. 81, expresses the same relations as the triangle B O G', Fig. 80, B G* being taken equal to the sum and in the direction of the forces P, Q, E, . . . B O is drawn parallel to the direction of T, and B X parallel to the direction of Ts ; G^ X gives the tensioii of T, and Gi O the tension of Ts, both measured on the same scale of equal parts as B G*. O Bis parallel to,T or AB ; O CHs drawn parallel to BO; OD»toCD;OE'toDE; OFitoEF.. . We have been very particular to pass nothing over that might make the proper relations and actions of forces, thus circumstanced, clear to the mind of the student, since the proper arrangement of thrusts and tensions, the action of braces, striits, and ties, in bridge-building, roofing, and in all sorts of lattice-work, designed to combine strength with lightness, depend npori the extension of the simple principles here presented. If the same Fig. 81 be inverted, then such frame, Fig. 82, consists chiefly of struts, and is, therefore, unsta,ble unless their ends are made fast by suitable stays. In a polygonal frame loaded and suspended vertically, represented by the skeleton diagram. Fig. 81, the bars which are struts in Fig. 82 become ties, and the ' frame ' is stable and yet flexible. The diagram of forces for Fig. 82 may be constructed PBACTIOAL MEOHANIQS. 181 as follows:— Suppose the polygonal frame loaded vertically and supported vertically, let A, B, 0,D, . . . be the bars ; a,biC,d... the joints, of which b, c, d, e,/ 'are loaded, a and g are supported. Take any convenient point, as O, draw O A^ parallel to A ; O B^ parallel to B ; 0^ parallei to C; O D^ parallel to D; O E' parallel to E; QF^ parallel to F ; and O G' parallel to G. Then draw the Kg. 82. vertical line A^ F^ crossing the lines O A', O B*, O C'. . . Then if the whole load on the frame be represented by A^ F^, the parts into which A^ F' is cut by=the lines O A', O B*, O C', . . . will represent the fractional parts of the load that must rest on each of the joints to secufe equilibrium. Ai Bi represents the part of the load to be applied at the joint & ; Bi Oi the part to be applied at c ; Oj Di the part to be applied at d; and so on. The lengths of the lines O Ai, O Bi, O Oi, . . . represent the resistances along the lines A, B, 0, ... to which they are respectively parallel. The two parts Fi Gi, Gj Ai, into which Fi Ai is divided by the line O Gi parallel to G represent the supporting forces at a and g, that is, Fi Gi represents the supporting force at 182 PRAOTIOAL HEOHANICB. ^, and Gi Ai the supporting force at a. The horizontal stress of the frame is represented by the length of the per- pendicular O Ga let fall from O on Fi Ai. If the angles of the polygonal figure A B D . . .be given, the angles of the diagram of forces O A, Fi are also given ; hence, when the length of any one of the lines in O Ai Fj is also given, the lengths of the other lines are readily found by plane trigonometry, to any required demee of accuracy. If the skeleton diagram AB CD . . . , Fig. 82, represent an open frame, the bar G is bmitted ; in this case the ptress along the outer bars, represented in the diagram of f orqes by the Bnes O Ai, O Fi, must be met by oblique forces as^but- ments; for vertical supports would not be sufficient. jThis frame, being, as before observed, composed of struts, iis un- stable, and must be connected with smtable stays. Ques. 62. Let A B and B be a roof resting odt the side-walls A and ; a uniform section of this roof weighs 6000 lbs., the angle B A 0=34°, B A=44° ; it is req|iiired to find the thrust on the points A, 0, the roof being with- out a tie-beam ; what would be the tension of a tie-beam, A p, or of a rod connecting the feet of the rafters i and also find the direction and amount of the presiSHue of the roof to overturn the side walls ? To construct the diaOTam of forces, draw O a parallel to A B ; O c parallel-to € B ; and b parallel to A C, Fig. 83. Draw the vertical line a b c^6000 equal parts to repre- sent 6000 lbs. L^t W=6000 lbs. applied, suppose, at the centre of grar vity^, G, of the section. .-. angle a 06=34°; J c=44°. O a 6=90-34 = 56°; O c 6=90-44=46°. Sin. 78° : 6000 : : sin. 56° : O c=5085-35. Natural sine of 78°= -9781476 ; natural sine of 56°= -8290376. Sin. 78° : 6000 :: sin. 46° : Oa=4411-44. Natural sine of 46°=-7193398. PEAGTIOAL MECHANICS. 183 .•. The thrast on the point C=5085 lbs., and the thrust on the point A=44111bs. As sine of 90° : a : : sin. aOb: a i=2466-8. .-. 6c=6000-2466-8=3533-2. 184 PHACTICAL MECHANICS. .-. The side wall at C supports 3533 lbs. of the weight W, and the side wall at A, 2467 lbs. nearly. As sill. 90' : Oa : : cos. aOb:0 i=3657-3. .-. The horizontal thrust along the side A C = 3657 lbs. Fig. 84. Draw A j? perpendicular to A C, and = a h, the vertical weight at A, Fig. 83 ; make A w = O a, tlie thrust along the rafter AB ; complete the parallelogram Apmti, and draw the diagonal m A, = the amount and direction of the pressure of the roof tending to overturn the wall A. Or, to avoid com- plicating the figure, we reproduce the diagram of forces, Fig. 84 ; taking the line.s O a, a b, complete the parallelo- gram O rba, then the diagonal a r gives the amount. and direction of the pressure as well as .the parallelogram Anmp, Fig. 83. On the lines Oe,'cb, construct the parallelogram Osbc, draw S c=the amount and direction of the pressure of the roof applied to overtiu'n the wall C, when tne tie AC is omitted. PRACTICAL MECHANICS. 185 Given the skeleton diagram A B D E, Fig. 85, of a frame loaded and supported vertically ; A E being hori- zontal* O Bi is parallel to A B ; O Ci parallel to B G ; O D, parallel to C D ; O E, parallel to D E ; and O F, is drawn parallel to A E ; Bi Ei represents the whole load, and, being vertical, is perpendicular to O Fj. Now suppose Fig. 85. 186 PEACTIOAL MECHANICS. the angle Fi00i=12°, the natural tangent of which is •21265=FiOi, when OFi=l; the secant of this angle is 1-0223=0 Gi . FiOBi=37°; F^ ODi=25°; F, O Ei= 50° ; then the comparative lengths of the lines in the dia- gram of forces will stand thus : Tangent. Secant. Fi = 1 , FiGi = -21255 Ci= 1-0223 PiBi = -75355 OBi=l-252l FiDi=-46630 ODi=l-1033 FiEi=l-1917 O El = 1-5557 This frame is vertically loaded with 5000 lbs. ; how is it distiibuted, and what are the particulars of its action when the forces Wance each other ? Fi Bi+Fi Ei=l-94525 = 5000 lbs. 1-94625 : 5000 : : 1-0223 : 2628 lbs. stress along the bar h or EC. ; 1-94525 : 5000 : : V2521 : 3218 lbs. stress along A B or bar a. i-94525 : 5000 :: 1-1033 : 2836 lbs. stress along bar c. 1-94525 : 500O : 1-5557 : 3999 lbs, stress along d. Fi Bi-Fi£!i=Ci B,=-75355--21255p=-541 1-94525 : 5i000 : : -541 : 1391 lbs. load on the joint B, Fj Ci+Fi Di=Ci D,=-67885 1-94525 : 5000 : : -67886 : 1745 lbs. loa,d on the joint C. FiEi-F,Di=-7254 1-94525 : 5000 : : -7254 : 1864 lbs. load on jthe jomt D. 1-94525 : 5000 : : Fj Ej : 3063 lbs., the vertical weight falling on the support at E, the remainder of the weight, or 5000 — 3063 = 1 937 lbs., must fall on the support at A. The horizontal stress being represented i^ the diagram of forces by O Fi=l, we have ' 1-94525 : 5000 : : 1 : 2570 lbs.,,' the horizontal stress of the frame. When the Joad is properly distributed, the necessary stays, C E, AC, require but little strength ; without stays such a frame, however well balanced, would be unstable. Here, as in other places, we have been care- PRACTIOAli MECHANICS'. 187 fill to avoid encumbering the mind of the student with theoretical developments of a complicated nature. However, when these siioaple constructions and prdinary calculations are clearly under- stood, the system is easily generalised, and in practice readily applied to a variety of complicated structures. PRACTICAL EXAMPLES OF WOODEN AND IKON BRIDGES IN WHICH THE ACTIONS OF FORCES EXPLAINED IN THIS CHAPTER HAVE BEEN APPLIED WITH MORE OR LESS SUCCESS. Anali/sis of the Wooden Bridges of the United States and Canada. The McCallum Inflexible Arched Truss Bridge ap- proaches nearer the standard of perfection than any other wooden bridges that have fallen under the author's notice ; this fact is easily established by comparison and demon- stration. Fig. 86. Fig. 86 is what is known as the JSwrr Bridge. It is com- 5086(1 of lower and u^per chords, and posts and braces, 'he posts . are framed into the chords, and the braces are framed into the posts. Arches are placed on each side of 188 PEACTICAL MECHANICS. the truss, securely fastened thereto, and, extending helow 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 sufficiently permanent, 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 connexion of the timbers, which generally resulted in failure. Many of the first railroad bridges, both in Europe and America, were built ujpon this plan ; but much greater dif- ficulty was found in adapting it to the use of railroads, than had been previously experienced in its use upon common roads. This difficulty arose from, 1st, the practical impos- sibility of perfectly combining the action of the arch and the truss (each system, of itseli, being insufficient to carry the whole load) ; and, 2nd, the absence of counter-braces. These defects, clearly apparent in their use on common roads, were greatly aggravated under the increased and con- centrated 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 struc- tures ; but as the difficulty was inherent in the plan, violent contortions in shape could not be prevented, and these in time caused failures. These remarks are intended to apply to spans of consi'- derable 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 ensure 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 coni- bination of the arch with the truss is even now, with some, a favourite idea, but it is believed that its warmest advocates will be generally found among those whose op- PRACTICAL MECHANICS. 189 portdnlties for practical investigation have been limited, and that it is oftly 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 has 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 upon the masciflry, has sustained the whole weight. Be- sides, to the casual observer who has never studied bridge construction, this combination presents at least an ap- pearance of great strength and solidity, which do not m 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 reso- lution of forces, by far the greater number have arisen from the inferior quality, or lack of the requisite amount of mate- rial, or from inferior workmanship. . The acknowledged failure of the Bmr Truss, as applied to railroad purposes, led to the invention, of several other plans, all oi wliich were based upon the abandonment of the arch,, and were aimed at perfecting a truss, which, of itself f would be sufficient to meet the emergencies of the case. This was in pursuance of what was considered a very reasonable hypothesis, namely, that jof the cast iron, 65,137 lbs. ; weight of wronght iron, 33,5:^,? lbs. ; making the total weight of cast and wronght iron, 98,664 lbs. Fig. 99 is an elevation of part of the side, showing one pier and part of the cast-iron stretcher. The cap is re- moved from the pier to show how the rods are secured. Fiff. 99. 1 Fi tr. 100 is an elevation of a pier and fovir panels out of eight of which the bridge is composed. The system of 206 PRACTICAL MECHANICS. arranging the braces and connecting rods is exhibited in this figure.' Fig. 100. ■ ■ Fig. 101 is cross section, showing the floor bracing and the position of the rails. Fig. 101 also shows, in section, the roof and posts. Jig. .101. Fig. 102 shows a plan of the flooring of the bridge, the positions of the rails aind floor bracing. PRACTICAL. MECHANICS. Fiff. 102. 207 Fig. 103 shows two posts, part of the stretcher, and the diagonal i ods in one of the panels. Kg. 103. 208 PBACTICAL MECHANICS. The wrought iron requires little workmanship, the rods from the centre to abutments having but an eye at one and a screw at the other end; with a weld or two between, according to length. The long counter rods have two knuckles and one swivel for adjustment of strain, and con- venience in welding, as well as in raising the whole. The cast-iron stretcher is octagonal without, circular within, and averages one inch of metal. It is cast in lengths! according to the length of panel, and jointed in the simplest manner ; — at one end of each length is a tenon, at the other a socket. The latter is bored out, and the tenon and its shoulder turned off in a lathe to fit the socket; thus, when thoroughly joined, to form one continuous pipe between abutments. The ends of the sections of cylin- ders, inserted to those contiguous, are slightly rounded, to allow a small angular movement without risk of joint fracture. A cast-iron plate or washer sets on a bracket cast with each abutment end of stretcher, and at right angles to the centre acting rods. The tension bars are passed through this washer to receive a screw nut for the erection and ad- justment of the system. The stretcher or stijainihg beam, the vertical posts, and suspension bars compose the essential features of the bridge ; each post being hung by two bars from both ends of the stretcher independently of all the others; and each post and pair of tension bars forming with the stretcher a sepa- rate truss. This system, perfect in itself, is additionally connected by diagonal rods in eachpanel ; also by light hollow cast- ings, acting as struts. The diagonal side rods might be safely dispensed with ; for the peculiar merit of the truss is its perfect independence of such provision. They are there- fore used as a safeguard only in case of the fracture of any of the principal suspension rods. By this combination of cast and wrought iron, the former is in a state of compression, the latter in that of tension ; the proper condition of the two metals. It unites the principles of the Suspension and of the Truss Bridges. PEACTIOAL MECHANICS. 209 Each bar performs its own part in supporting the load in proportion to its distance from the abutment ; so that the entire series of suspending rods transmits the same tension to the points of support as would be equally transmitted from thence to the centre of the bridge. This bridge, it will be seen, is composed of seven inde- pendent trusses, which transfer the weight concentrated on each floor-beam directly to the abutments, without aid from any other connexion ; and not from panel to panel as in general use. The strain on cast and wrought iron, is wholly in direct line ; and the result, the least quantity of metal is required to carry a given weight. The weight of bridge and load has a vertical pressure on the piers, towers, «6;c., the only horizontal thrust being from the expansion of iron, which is accommodated by rollers, sliding on abutment bracket over its pedestal, or by other means : the necessary dimen- sions of masonry may therefore be most moderate. It is evident, from an inspection of the cuts, that no chord is requisite at the bottom of the truss to resist ten- sion ; the only advantage of that employed is to regulate the movement produced by expansion, in the performance of which agency the resistance is one to compression. .Although the abutment bracket casting and its pedestal were so constructed as to admit of accommodation to ex- pansion, by rollers, yet such contrivance was omitted with the view of fully testing the effect of greatest expansion throughout the system. This bridge was inspected by the writer, ten nionths after it was erected at Harper's Ferry ; during which time it had been exposed to extremes of cold and heat, and to an average run of twenty trains daily. From the closest . inspection, we find that the extreme expansion measures, as near as possible, five-sixteenths of an inch on each tower, or five-eighths in the entire length, 128 feet of stretcher; and without the slightest percep- tible derangement of masonry ; the dimensions of which are 4 feet square of base, 12 feet high, and 2 feet 9 inches at top. 210 PEACTIOAL MECHANICS. While on the subject of expansion, it may bie well to notice the effect from difference in expansion of the rods. At the first point of suspension, or where the longest and shortest rods meet, the counter rod is about four and a half times longer than the acting rod ; and the expansion of the counter is four and a half times that of the acting rod. But there is also a proportionate difference in the lengths of stretcher from the point directly over the centre of con- nexion to the extremities of these rods. This has been practically proved in this bridge. The suspender bolt, when the expansion is extreme or five-eighths of an inch in length of stretcher, exhibits a motive difference of three-sixteenths toward the short or acting rod ; which difference is provided for, as seen bV slot-dotted in Elevation, where the vertical suspender bolt moves to accommodate any such difference, and to give that proportion of weight to each rod according to the angle. It affords easy access for repairs ; for instance, should a new floor beam be required, it is but needed to slacken the horizontal rod and the keys in longitudinal strut, remove the washer under point of suspension, and let down the beam to be replaced : which can be done without trestling up any part of the bridge. In case of fire, the floor may be entirely consumed with- out any injury to the side truss. The permanent principle in bridge building, sustained throi:^hout this mode of structure, and in which there is such gain in competition with every other, namely, the direct transfer of weight to the abutments, renders the calcula,tion simple, the expense certain, and facilitates the erection of secure, eeonomicalj and durable structures. Trial made on the 1st day of June, 1852, to prove the capability of this Bridge. Three first-class tonnage engines, with three tenders, were first carefully weighed, and liien run upon the bridge, at the same time nearly covering its whole length, and weighing in the aggregate 273,550 Jbs.,- or 136^^^ tons nett, being over a ton for each foot in length of the bridge. PRACTICAL MECHANICS. 211 This burden was tried at about eight miles per hour, and the deflections, according to gauges properly set and reli-r able in their action, were at centre post, If", and at the first post from abutment nine-sixteenths of an inch. From this test it is found that the load did not cover the entire length of bridge by about 13 feet, yet the excess of weight in the middle, and at a speed of about eight miles per hour, produced no greater. deflection than If of an incn at the centre post, aiS nine^slxteenths of an inch at the first point from abutment. Before proceeding further, it is necessary to point out some serious mistakes made by experimenters and writers on the strength of materials. When discussing the strength of girderiEf resting on supports, the author of the present work, in his new theory of the strength of materials, first pointed put the fallacies involved through introducing an ima^fuy line, termed the neutral axis, and merely inves- tigfjiting upright laminae of the material. We do not pro- pose to discuss this subject thoroughly here, but to show the student how errors may be involved when the strength of girders is considered with respect to forces supposed to act only in parallel upright planes. If a beam, Q E, Fig. 104, rests loosely on two supports, A and B, and is loaded in the middle with a weight, W, which deflects it ; before the weight is placed on the beam ab=pq=cd; and ef=rs=nm; but when the beam is Fig. 104. 212 ' PRACTICAL MECHANICS. deflected hjW, pq is greater than ab or ed, and rs is less ihan either ef or m n. Before the beam is loaded it is supposed to be rectangular ; in most cases this change of form may be detected by experiment. Although the nature of the material and amount of pressure may render this change of form imperceptible, yet these forces acting across the girder, in the directions of » q, r s, are in opera- tion, loosening bolts, buckling and puckering upright sheets, and so on. This action should be carefully at- tended to by engineers in constructing girders, whether solid, hoUow, or composed of skeleton frames. The material at r s is wire-drawn and compressed, while a,t pq the material becomes upset, extended, and loosened, according to the elastic limit and nature of the girder. The current erroneous theory of the strength of materials supposes, when the beam is bent by a weight, W, the fibres are compressed a.t pq and extended at rs, without alteration of breadth ; that is pq remains = a b or cd, and also= r s or ef. When Robert Stephenson was erecting the Menai tubular bridge in Wales, the author of the pre- sent work, Oliver Byrne, published an outline of his new theory of the strength of materials in the Civil Engineer and Architects' Journal, and explained his views to Mr, Stephenson, who directed the author to observe his (Ste- phenson's) final experiment on a model of the bridge, one- sixth the full size. ' Fairbaim and other engineers were present. The direction of the fractiire depends on the molecular action of the material of which the girder is formed; A portion of the body will often be forced out near the line p q ; but when the substance supporting the weight is tough, the separation may take place irregularly and diagonally, with a sliding cutting motion, and not directly through the plane pqrsyvn. the middle. tBACTICAL MECHANICS. 213 METHOD OF CONSTKUCTING SKELETON STRUCTURES, PRO- POSED AND PATENTED BY EDWARD KOCH. Half the suspended beam, Fig. 105, gives a clear repre- sentation of most of the particulars of this mode of con- struction, the practical application of which depends on the connecting bolts filling their respective holes without play. After these bolts are posited with mathematical accuracy, it is also required that neither bolts nor holes FiT. 105. afterwards wear much. The inventor proposes plans to surmount these and other engineering diificulties, in a book by his friend Olaus Henrici, entitled " Skeleton Structures, especially in their Application to the Building of Steel and Iron Bridges." In the diagram of forces, Fig. 105, O Hj is parallel and Hj Di perpendicular to H H ; the diagonal braces, D, D, I>, meet in a point Q ; Tj, Tj, Tg . . . are parallel to T,T,T, . . . respectively; andDy DjjDj, . . . are drawn parallel to D, D, D, . • . respectively. The construction of this diagram will show how the forces act in magnitude and direction, and may be measured on a scale of equal parts, 214 pkaCtioal mechanics. or found by calculation, when the inclinations of the bars are given. But, as the triangles standing on Hj Hj axe right-angled triangles, Mr. Henrici employs the following method of estimating the comparative forces : Fig. 106. In the structure, Fig. 106, A and B C represent bars which are connected at A and B by bolts to a wall piece, and at Q by bolt to each other. A weight P acts on the bolt 0, and we want to ascertain the strain in the two bars. Firstly, we have to apply two rectangular axes through 0, and because A C is horizontal, and C P vertical, we take these directions as the axes. The strain in the bar B C we call S, and the one in A C, T. If S is resolved according to the axesj into its components, we find AC BO' AB in the direction of C X : S CY: S BC The two other forces act in the direction of the axes; hence we get in the direction of C X : BC+ ' CY: °BC ' where P is subtracted, because it acts contrary to the strain S. Both components must disappear ; we have therefore : ;S AB BC" -P=0. From these equations we see that PBACTIGAL MECHANICS. 215 « BC „ ■^- AB" ^' which shows, that in the bar B there is tension and in A C compression, because T is negative. In this example we had only one connexion-point; we shall take, therefore, a more complicated example by join- ing more bars to those, and thus we get a structure like the one presented in Fig. 107, of which we shall proceed Fig. 107. to calculate the strains- We will call here the connexion- points 0, 1, 2, 3, &c., and the bars will be represented by two figures (01), (12), &c. ; the lengths of the bars we call r with the respective figures, Vxi, &c., and the strains likewise, S13, «&c. We take now 3 '•14= J a, and find further : (/•l3)^=a'+«'=2«^ 216 PBAOTIOAL ME0HANI03. = 1-6 «'• M^=a» + (L«).^a»(l+-g) 16 therefore : yi7 5 /•34=« ^. a. By this all bars are determined in length and position, and we can proceed to resolve the forces. For axes we always take one horizontal and one vertical line, and be- cause we have three free connexion-points, 1, 2, and 4, we can form six equations, by which we are able to ascertam all the six strams at once. For point 2 we have the same equations as in the previous example, by introducmg S24 instead of S ; and S12 instead of T ; as likewise r^, r^, and rs4 instead of A B, A 0, and B ; and we get s — ''24 p '•14 s — _ ''12 p • '•14 or by introducing the valijes of r : Sia = — - P2. Looking at point 4, we get : S24 resolved into S24 . — and S24 • — > PRACTICAL MECHANICS. 217 Sg, resolved into B^ . I»l and 834 . *'o»~^^* '"84 Vu "U II »» „ Sj4. By adding these, we get 2 equations : Sm ^ - S34 ^ = 0, The first equation will give or because '"84 ♦'12 H S,4 = ^P, and !^ . lii = ^^, 3 roi r24 5 »34 — -g- Jrj. From the other equation follows Su = - I P*. We have still to ascertain the strains Soi a^^d Su, which we are able to do by the equations of point (1), after having altered them in the same way : Soi + ■• Si3 — S12 = 0, *'os »"18 Si4 + ^ ■ Sx3 = P,. By which, in connexion with the above equations, we find; + J3 = VTPi - I-V2" P,. Sji = — Pi — v^P*' 7_- 12 218 PBACTICAL MECHANICS. It vpill be clearly understood how these calculations are carried on. Taking a structure like that in Fig. 108, the Fie. 108. calculations are done in the same way; one strain is calcu- lated after the other without difficulty. I will mention here at once, that the same calculation can be used for a trussed structure, as in Fig. 109 ; the Fig. 109. structure is exactly the reverse of Fig. 108, and it is not necessary to alter the formulas ; for as all the Ps act here in the contrary direction, all the signs will be reversed. All the strains, therefore, will be the reverse of those in Fig. 108 ; where there was tension in a bar there is now compression ; and where there was compression, tension. PEAOTIOAL MECHANXCg. 219 CHAIN ANB SUSPEiSrSION BRIDGES. Suppose a perfectly flexible chain or cord of uniform density and thickness to be suspended from two fixed points, A and B, Fig. 110, and when in equilibrium to form the curve A O B ; this curve is termed the Catenary. The equations of this curve, of so much importance in mechanics, being of a mixed exponential kind, mathema- ticians were obliged to resort to guessing and all sorts of dodges, to obtain results which were often very far from the truth. The results here referred to can now be ob- tained with the greatest ease by direct processes of the dual calculus, in an endless variety of ways, without the use of tables. Haying thus pointedly arrested the atten- tion of students in immistakable terms, at the same time it may be necessary to inform them, that here, as else- where, we accommodate those inexperienced in such in- quiries with solutions under simple forms easily intelligible. Fig. 110. 1 220 PBACTIOAL MECHANICS. PROPOSITION. To find the Equations of the Catenary Curve. Let O be the lowest point of the chain or cord A O B, Fig. 110, O M=x; M»-=y; and the length of the arc O r=«. Again, let v be the length of a portion of the chain, the weight of which is ^ual to the tension at O. If we suppose the part O r rigid, after it has assumed the form of equilibrium, it will evidently be supported in the same manner, and the tensions at O and p q r will be the same as when it swung loosely suspended from the points A, B Opqr \s therefore kept at rest by three forces, namely, the tension at O acting in the direction of the tangent O Y, the tension at the point p q r, acting in the direction of j> y T, the tangent to the curve at the point pqr, and the weight of the piece of cord or chain O r, acting in a vertical direction. Because the three forces just described are respectively in the directions to the three sides of the triangle T j» M, the forces being in equilibrium will be proportional to these sides. .Hence, (Weight of O r) : (Tension at O) : : T M : Mj». p qr represent a very small right-angled triangle, similar to the larger right-angled triangle T M jo ; in the language of the differential calculus qris represented by d x, rp by d.y, and » qhy d s. Whence, v being put for a piece of cord or chain, the weight of which is equal the tension at O in the direction of the tangent O Y ; therefore, s : V : : dx : dy d y V dx s' In every plane curve (dsy=(dxy+(dyy, and, consequently, dx - V'^dxO • In the catenary ^ = V^^" + ^ ' , or dx = "^' PRACTICAL MECHANICS. 221 Taking the integral of this last equation, and observing that s=zo when x = o, we obtain a:+»=\/o'+s', or 3^=31^ + 2 vx; (I.) When V is determined, (I.) is the equation of the catenarif expressed by a; and s as variables. Again, because -j— = - = / := , which, being in- tegrated, gives - = 'og- — ■ ■^ . e' = ^^ + v/^Ulll?; (II.) £ being put for the base of the hyperbolic system of logarithms, (II.) is an equation to the curve between the variable co-ordinates X and y ; v being unknown, but constant. The dual logarithm of e = 10», written \ ,(e)= 100000000, 6 = 2-718281828.... Transposing , and squaring both sides of (11.), we obtidn ^•C-T^)+^ 1 ... rr + .= |(e'f+eT); (III) Further, because, (I.) **=(a! + »)' — c* 1 a; + «- " (2 a; -t- «\- «■ 222 EEAOTICAL MECHANICS. (IV.) is an equation to the catenary between the variables s and y; it is to be particularly observed that v is unknown in (I.), (II.), (III.), (IV.), but not variable. Suppose t to be the length of a portion of the cord or chain which is equal to the tension of Q, then OE = a;, EQ=^, 0Q = «, and t:s :: QG: GE : : ds: dx s ds .= tdx. Differentiating (I.), or «* = a^ + 2vx, gives sds =:xdx +vdx ^•. tdxz=xdx + vdx; dividing hj dx gives t =x + V. Now, suppose the tension at Q to be balanced by means of Q R, a portion of the chain passing over a pulley at Q and hanging freely ; then QB, = x -\- v = O ^ -{- v; consequently, PR = TD = OC = «, evidently a constant quantity, although unknown. Hence, "if the tension be supposed to be balanced by means of portions of the chain or flexible body hanging over .pulleys at the points pqr, Q, A, the lower ends will be in the same horizontal line, B, D, C. Ques. 63. A heavy flexible chain, each foot in length weighing 20 lbs., is suspended at its extremities to two fixed points, in the same noriizontal line, 106'2 feet asunder, the greatest depth of the curve being 15"8 feet. It is required to find the length of the curve, the tension . at the lowest part, and the tensions at the points of suspension, by direct calculation, without the use of tables, or methods of approximation ? In dual arithmetic we have three corresponding numbers, namely, natural number, dual number, and dual logarithm ; any one of these corresponding numbers being given, the other two may be foundiby a few simple additions and sub- tractions. — See the " Young Dual Arithmetician." We have first to put equation (III.) in form to be operated upon by dual arithmetic. x + v = li ^i + iL; (III.) '{^''M'' PRACTICAL MKCHANICS. 223 When y is put =1, x may be represented c*, x and 1/ being both given to find the value of v. Hence (III.) becomes .-. c \/2\,-^ = 62«- — ; or, 6^" Putting —- =z, the last becomes 2c-/J=e^-ei, (V.) Question the 63rd being the first of the kind solved by an independent and direct process, without employing tables or approximate methods of trial and error, it is there- fore necessary that we should be particular with respect to details. The dual logarithm of 2 is the whole number 69314718, written, I , (2) = 69314718, . This logarithm is of the ascending branch, marked by a comma on the right of the logarithm below, and imme- diately after the arrow. The dual logarithm of the reci- procal of 2' or of ^ is represented by the ^ame whole number '69314718, and written, |,(i)'= ■'69314718.' 224 PBACTICAL MECHANICS, This logarithm is of the descending branch, marked by a comma to left of the logarithm above. When 69314718, is considered positive, 69314718 is considered negative, and vice versa. Natural Numbers. Dual Numbers. Dual Logarithms. 2-00000000 ■50000000 i 7,2,6,0,7,8,2,6, 1 '6'6'0'6'8'2'0'2 69314718, '69314718 1 , (10) = 230258509, and | , (l^ ) ='230258509. The same notation and arrangement is applied to other natural numbers and their reciprocals. See the " Young Dual Arithmetician," and "Dual Arithmetic: a New Art," Parts I. and II. In the question before us, c*= ^^-r =-29755178907 ; .-. I , (2 c)= I , (1-09096616)= \ ,0,8,7,4,e,4,5,5,=8706369, and (V.) becomes ^-e^ = l-090966I6v'2 I ^(e)=io»= 100000000, Then, putting u^, for the first dual digit of « 1 (g) and '«i for the first dual digit of the logarithm of- z I (g) the reciprocal to « J (fi^ I 'ttx is nearly = Ml, but less than %. To place this matter in a clear light, it is only necessary to observe that. PBACTICAt MECHANICS. 225 If x= \ l,Q,O.0,0,0,0,0,then^ = \ '1 1,0,1,0,0,0,1, ; (&631018). » x^= I 2,0,0,0,0,0,0,0, „ ^,= I '22,0,2,0,0,0,2, ; (19062036). "**= I 3.0,0,0,0,0,0,0, „ ^U I -3 3,0,3,0,0,0,3, ; (28593054). I . (aj)= I , (1-1)= \ , 1,0,0,0,0,0,0,0, =9631018, ; I .( -)= I ^(j;j)= I , '1 1,0,1,0,0,0,1,='9631018 ; &c. These things being premised, we have 1 ^ [m,-1 I [-„'j ^ 2c ^^5= 2c V«,X (-09531018); but since «, is nearly equal to «'„ in order to find a convenient value for m, the last equation may be put under the form, +2 ( -) =2 c Vm,x (-09531018) .■• ■j^=c^, (-09531018) ; or «,= 100c» (-09531018) ; which gives m,=(-29755 . . .)' (100) (-0953 . . .)=2-835 Because 2-835 ... is less than a convenient value for u„but greater than the corresponding value of u\ ) consequently | 3, is a convenient value for u^. By pursuing a similar process of rea- soning, z as far as the fourth dual digit will be found to be equal \ 3,0,3,4, Mj, ; the next step gives the four succeeding digits, and at the same time develops the general plan of operating on such equations ; besides, any mistake that may be made in the preceding operations will be detected. I , (1-33553107) = 1 , 3,0,3,4,0,0,0,0, = 28932904, Reciprocal, '28932904= j '2'7'8'2'5'1'6'5 = | , (-74876579). Q 226 PBAOTICAL MECHANICS. Putting Xy for the dual logarithm of \ 3,0,3,4, Mj, . . . divided by I0», or «i=-28932904+-00001000m6 ; now equation (V.) becomes e^ = 2c A^ -28932904 + -00001000 %= 2 c V ^28932904 V i + -00003803 % = ( I 0,8,7,4,6,4,6,5,) V -28932904 (1+-00001902 %)= •58682320 (1 + -00001 902 Mj) •58682 32ft (because 1+ -00001902 may) ^^' ^be taken for the square C ^-^^ (root of 1 + -00003803. ) •ooooine for -0001902 . 2 c v/-28932904+-00001000m5 = -58682320 + -00001116ms 6*^ _-!- = 1-33553107 (1 + [«s)--74876579 (UK); 1-33553107 (1 1 [%) may be put= 1-33553107(1 + -00001000 %), and--74876579 (1 \ \u\) may be put = - -74876579(1 + -00001000 ««) ; . «i _ J_ = -58675628 + 2-08429686 (-00001000 %) 2-084 2|9686 12084 .-. -58675628 + -000O2084»(j=-5O682320+-000011T6«,i PEAOTIOAL MECHANICS-. 227 .5792 •'• "5— q„o —5*983 l(fie= \ , 3,0,3,4,5,9,8,3,=28938887, .-. «='^ =-28938887 .-. «= 1-72777897 true to the "last figure, when ^=1; .-. when 3^=53-1, »=91-745063 ft. From (I), s''=x (ir+2»)=J5-8 (15-8 + 183-490126) .'. 5=56-11395 .-. the length of the whole curve, Q O H, Kg. 110, is 112-2279 ft. a:+«= 107-545 ft. Tension at the lowest part = 91-745 x 20=1834-9 lbs. Tensions at points of suspension= 107-545 X 20=21509 lbs. Ques. 65. The length of a heavy flexible chain A O B, Fig. 110, is just double the horizontal line A B, joining the points of suspension A and B ; required the pressure on these points and the tension at the lowest point O ; the distance of O from the horizontal line A B is also re- quired ? If A X= 1, the length of half the curve, or A Q 0,=; 2, and equation (IV.), becomes 2=1 (ev— -J ) 5 ^^"^ ratting z= i. When « = 2- the dual logarithm of £ is nearly=the dual log. of 8- ; 4 « = 8- also. It maybe remarked here that, in operating with the dual calculus, we may take dual digits much greater or much less than any particular one pointed out, and yet obtain, without tentative artifices, a result as near the truth as we please. Since this method gives the same result as near the truth as we please, by several direct processes, it presents a series of direct operations, and not a succession of approximate 'trials. q2 228 PBACTICAL MECHANICS, I (7-38905604)=200000000,and | ,C13533528)='200000000; consequently we may assume 7-38905604 (1 1 [mi)--13533528(1 \ ^u\)= 4(2-00OO00OO+-09531018 Uj) ; . (A) 200000000, is increased by 9531018, for every unit in Uy ; and 2-00000000 by -09531018. To find a convenient value for Ui, (A) may become 7-25372076+7-52439132^=8-00000000+-38124072mi .-. -37119841 «i=-74627924 .'. «i=+2or|2, In practice, one tenth of the figures here employed \rouId not be required to find that i 2, is a convenient value for %. This may be too great, but that is of no consequence, as the next digit may be negative to compensate for the excess in taking %= 1 2, 200000000,+ 19062036,= | , (8-94076744) and '219062036 = I, (-11184734). To find a convenient value of Wj, the equation now takes the form, 8-96075744 (1 1 [«;,)- -11184734(1 \ [mj) =4(2-19062136 + -00995033 Uj); therefore we may put 8-82891010 + 905260478^^=8-76248144 + -03980132 Wj .*. Mj,= - 1- or = ([ 0, '1. For every unit in Mj, we employ 995033, and -00995033 ; but PBACTICAL MEOHANIOS. 229 for every unit in \ we employ '1005034 and —-01005034. See " The Young Dual Arithemetician," pp. 82, 99. 219062036, '1005034 I 218057002,= \ , (8-85134984) ; '218057002= I , (-112977175) .-. 8-85134984 (1 \ [mj)- -11297718(1 1 [Ma)= 4(2-18057002 + -00099950 Wg) This gives 1*3= —3 or | '0 '0 '3 This process being continued gives e=2 \ 2, '1 '3 '3 5, 0, 5, 2, the log of e* is 217731902, for \ ,2,'1'3'3 5,0,5,2,=17731902. Putting B for the base 1-00000001, then £""'"<"= e*-""^''!'. The reciprocal of 2-17731902=-45928042=», the length of chain = tension at 0. But as 11^+2 vx=s^: a!=l-592779=XO. a!+»=2052059=the length of the chain that amounts to the tension at A and B. Ques. 66. Given the length of a heavy flexible chain R Q O H N=1130 ft, (2a), Fig. lia; tins chain hangs fre^ over two pulleys Q, H in the same horizontal line Q H,=226 ft. (26) ; required the position in v^hich it will rest, the length of Q B, and the length of chain that is equal to the tension at O ? Let Q E : O Q R : : 1 : M or S : a : : 1 : ». It is evident that when the chain is in a state of rest, what in the foregoing questions was the pressure on the points of suspension Q, H, will be equal to the weight of either HN or of QR=«+«, the parts hanging vertically. Putting «=the length of QO, when QE=i=i/; OQR = h; and QF=EO=iB, then a;+u+«=n; i>=FR=O0. 230 PEAOTICAL MEOHANIOS. From (III), x+v^j f ^^ + 1 } From (IV.). ' s=-|-(e'^ - 1^ ) Whence, from adding these equations together, we obtain 1 .-. v\,n=v\ , v+ \,e or _ = v' or Taking the — root of both sides of the last equation n we obtain an equation easily solved by the Dual method. 1_ D (1)" =(f,)~=^, putting zfoT^ Hence we hare the final equation Before the introduction of Dual arithemetic, an inde- Sendent and direct solution of equations of this form efied the skill of mathematicians (see " Dual Arithmetic," Part II., pp. 91, 100.) PBACTIOAIi MEOHANIGS. 231 l.G)= ='100000000; audi =1 =1 ; n 5 a .-. z\, «='20000000. .-, «=f2 I 0.'6 2,5,6,8,3,3,=-078658307 ; .•.«=-393291535, found as at page 124. The value of «,=— , not attainable by any previously known method, may be almost instantly obtained under a variety of forms by the Dual calculus to any required degree of accuracy. In the present case, ^=i I '0-6 2,5,6,8,3,3,1,7, ; ^=i 1 4,7,2,1,9,8,2,2,= •078658307, &c. Under the last form all the dual digits are positive. I ^ (- ) ='254264202 .'. j ^ (|-]| =254264202, .-. - =12-7132147; andhenoe i =2-54264294. But we have before shown that n=vgi •*• I i — J— I (g^) = 254264202, and I T J_^ )='254264202 ; Whence s=— ^ g"S — _^ ^ becomes known. €~' « = -| (12-7132147 --078658307) = (-393291535 x 6-3172782)=2-48448448 ; Because, x+v+s=n=5. .: a:=2-12222398. In these calculations, Q E =y is put=l ; but in the ques- tion, QE= 113 ft. 232 PRACTICAL MECHANICS. .'. X = 239-8113097 ft, = E = QF ; s = 280-74675 ft. = arc QO; and »=:44-441966 ft.=F R=0 C=tlie length of chain that is equal to the tension at 0. i)+«=284-253254 ft.= Q B=the length corresponding to the tension at Q or H. Ques. ()7. In .the Chelsea Suspension Bridge, Thomas Page, Engineer, the central spaii between the piers is 34:8 ft. with a cleflection OX, Fig. Ill, of 29 ft. ; supposing the entire weight of the chain A OB, when the weight of the platform, roadway, and full load is transmitted to it, to be 1'5 tons for each foot of its length, required the strains at the highest points. A and B, and the tension at the lowest point 0, as well as the length of the catenary curve passing through the points A, 0,B ? 348 _^. 29_1_ , .-. 2 c= -81649658. Whence, equation (V.), becomes -81649658 VJ =:^-2-.=2Wl; (a).'. X being put for ^.— ; v representing the length of the chain whose weight is equal to the tension at the lowest point 0. We propose to find the value of v under the form \u1U2U3... To find a convenient value for Uj, assume «= \ u^, then (a) gives (1 f [ui)-(l I ['%)=2 c »/7=2eVui X -09531018' 9531018, being the dual of a logarithm of \ 1, or of each unit in «!. Since we arrive at the exact valve of e whether we assume Mj too great or too small (l + ["«i)— (1— [■«i=)+2-Jg-may be put=2cVMi X -09531018 •■• •^=^ ("1 ^ -09531018) =4(-Mi X -09531018\ PEACTICAL MECHANICS. Fig. 111. 233 234 PBAOTICAL MECHANICS. 9-531018 - ,„ .'. «!= g =1-59 .'. The reciprocal of u^, or 'ui may be put= \ '1'6'9 • • • • • or «i,= 1 1,7, I , (1-17934889)= | ,1,7,0,0,0,0,0,0,=16496249, Reciprocal='16496249= \ ,'r5'9'3'4'5'7'7= | ,(-84792554). It is easily found that o, is the most convenient value for %, hence we may put in (a) z= \ 1,7,0,^4, =2 cv/ -16496249 (1 +-00030310^4) =•33162478 (I + -OOO303IOM4); g"_ — • may also be put under the form 1-17934889 (1 1 [«<)- -84792554(1 f [X) =(1-17934889) (1 + -00010000 M4) - (-84792554)(1 - -00010000 M4)" =•33142335 + -00020373 M4 /. -33142335 + -00020373^4 = -331 62478 (1 + •00030310m4) = •33162478 + -00010052 M4 .-. ,,^ -00020143 ^2_014_3^,.3,^ -00010052 19052 .•. z may be again put= \ l,7,0,l,9,Me, when a higher degree of accuracy is required. I ,(1-17957297)= | ,1,7,0,1,9,0,0,0,=16515249, Eeciprocal, '16515249= \ ,'1'5'9'5'3'6'2'1= | , (-84776406) 1-17957297 1-17957297 -84776406 -84776406 -33180891 2-02733 |703 'OOOOollOSSs PEACTICAL MECHANICS. 235 2c'\/-165l5249 + •00000100m6= 2c\/-16515249 (1 + -OOOOOSOSms) = •33181570 (H--00000303Me) •33180891 +-00000203«e=-33181570 + -00000101w6 •00000679 679 „ ce M.=: = = b*66 •00000102 102 .-. I, «=|,1,7,0,1,9,6,6,6,=16515915, .-16515915 ""^^ - ^.16615915 =(^81649658V-16515915. To find the value of z in such equations as (a), to any re- quired degree of accuracy, and by such direct, independent, and simple means, is, without doubt, a great mathematical achievement. ^= .-. «=3-02738298=three times the length of OC. 2» a!+t)=3^19404964; and the length of the catenaiy curve passing through the points A, 0,=y!c^ -{-ivx, (I.); .'. the length of the curve A 0=1-01828555. Bnt when AX=174 ft., then w=526-7646 ft. ; a;+i;=555-7646 ft.; and the length of the catenary curve passing through the points A, 0, B,=354-36337 ft. (526^7646) X (1-5) = 790^1468 tons, the tension at the lowest point in the direction of the tangent O F, and represented by the lengths of the three lines O C. (555^7646x(l-5)=833^6469tons, the tension at the highest point A, but in the direction of the tangent AT ; this tension is represented by the lengths of the three lines, F E, and the line A F. The sectional area at the centre of the bridge =214 sq. inches, and the sectional area at A and B=230sq. inches ; hence a weight of only 5 tons to the square inch gives a power at of 1070 tons, and at A and B a power to sustain a tension of 1150 tons with ease. 236 PBACTICAL MECHANICS. CHAPTER VIII. PEESSUEE or WATBE AND OTHEE FLUIDS. Let L F BD, Fig. 112, be a vessel of any form what- ever, filled with water, ab any portion of the surface F K B in contact with the water, G the centre of eravity of ab, GB. the ipex-pendicular depth of G below me surface of the water ; then if E G= 10 ft., and the area of the surface a J=3 square feet, the lbs. pressure on this surface will be = 3 x 10x62-5= 1875 lbs. 1875 lbs. is the weight of a column of water whose base is the area a b, and perpendicular height the depth ' of the centre of gravity of a 6. If the area of c e on the bottom of the vesser= 5 square feet, and QGi==14ft. the perpendicular depth of the centre of gravity Gi fcelpw the surface of the water ; then the lbs. pressure on the area c e= 5x14x62-5=4375 lbs. Again, if the area of the surface hf, on the slanting face AL=4 square feet, PG2=11 ft., Ga being the centre of gravity of the area hf; then the pressure on .this slanting surface will be 4xllx62-5=27501bs. PEACTICAL MECHANICS. 237 As in former cases, the weight of a cubic of water is taken = 62^ lbs. It may be further observed that the po- sitions or inclinations of the surfaces ah, ce, hf, are not taken into account, but merely their areas and the perpen- dicular distances of the centres of gravity from the hori- zontal surface of the fluid. On this simple principle rests that department of mechanics termed hydrostatiea ; it is easily demonstrated, as gravity acts on all the particles of the fluid, and each particle presses on that next below it, and, further, because, from the peculiar property of the fluid, this pressure is transmitted m aU directions equally. Ques. 68. Find the lbs. pressure on a floodgate whose breadth is 9 ft., and depth 7 ft. 7 X 9=63 square feet area, 7 depth of centre of gravity=— =3'4 ft. .-. 63 X -I x62i= 13781-25 lbs. Qms, 69. What is the pressure upon 10 ft. length of an embankment, the depth of the water pressing against it being 11 ft. ? 238 PBAOTICAL MECHANICS. 10 X 11 X 5^ X 62-5=37812-5 lbs, Ques. 70. Eequired the relation of the pressure upon the four sides of a cubical vessel filled with water, and the pressure on the bottom which is horizontal 1 Put M=the length of the side of the cube in feet,' »' X 62-5 = pressure on the base; nxnx— X 62-5 = pressure on one of the sides, .'. 2 n» X 62*5 = pressure on the four sides. Then m' x 625 : 2 m» x 62-5 : : 1:2, that is, the pressure on the sides is = to twice the pressure on the base. Li these calculations the fluid is supposed to be at rest, and acted on only by gravity. Let A B C D be a vessel fiUed with water, the pressure Q on any point n, in the side A D, Fig. 113, is due to the perpendicular depth An. If in the oase D produced we take D F=D A, the perpendicular depth of the water, then the pressure upon the point D will be due to the pres- sure of a column of the fluid, whose height is = D F. Draw F A, and from any point n draw m n perpendicular to A D ; hence m n= A n, and the pressure Q on the point Tig. 113. PBAOTIOAL MEOHANIOS. 239 n will be due to a column whose height is >n n ; the same reasoning applies to any other point in the side of the vessel. Let tis take an example and compare this method of viewing the subject with the one previously enunciated and illustrated. Qties. 71. What is the pressure on an embankment whose length is 21 ft. and depth of the water AD= 12 ft. ? The whole pressure upon the side of the embankment is equivalent to the pressure or weight of a mass of fluid of the form of a wedge, AF D, Fig. 113. 12 Area of the triangle ADF=12x r=72 square feet, .'. 72 X 21=1512 cubic feet content of the wedge. .-. 1512 X 62-5 = 94500 lbs. pressure. Before it was stated that the pressure in lbs. bn the side is equal to a column of water whose base is the area of the surface, and perpendicular height the depth of the centre of gravity. It is evident, since the side of the embankment is a parallelogram, the depth of its centre of gravity area of the surfaoe=12x 21=252 square feet. .•. 252 X 6 X 62-5 = 94500 lbs., the pressure before found. It may be easily perceived that there is a certain point in the side A D of an embariknient or vessel filled with water, where a single pressure will counterbalance the pressure of the water against the whole side. This point is called 240 PEACTIOAL MECHANICS. THE CENTBE OF PEESSUEE. The centre of pressure must evidently lie in the line P ^ passing through the centre of cavity, G, of the wedge of pressures^ of which the plane AF D is a cross section. Bisect P D in E, and D A in H, draw A B and F H ; these lines cut one another in the centre of gravity G. D P=-|- D A, that is, the centre of pressure, P, in this case lies at -i- of D A, from the bottom, see page 246. Ques, 72. Required the pressure on the staves of a cylindrical vessel filled with water, the diameter of the base being 10 ft. and the perpendicular height 8 ft. ? 3-1416 X 10=31-416 ft. the circumference of the cylinder, a »-. 31-416 X K- X 8 X 62-5 = 62832 lbs. pressure. If the staves of this barrel are to be kept together by a single hoop, that hoop should be 'J = 3^ ft. from the bottom. Ques. 73. An embankment H D, Fig. 114, resists a pres- sure of water, whose centre of pressure is at P ; it is re- quired to determine by construction the conditions of equi- PEACTICAL MECHANICS. 241 librium, supposing when the pressure is sufficient to over- turn the embankment It will turn upon A, as a centre ? Let F O C be the vertical line drawn through G, the centre of ^avity of the embankment. Draw PL perpen- dicular to F C, Intersecting F in O. Make O n=the lbs. pressure in the embankment, and O m=the pressure of the watef, complete the parallelogram O mp n, then if the diagonal O ;> or O » produced falls as at B Inside the base, the embankment will stand, but if the diagonal cuts out- side of A, embankment will fall by turning over upon O. Otherwise, since the pressure of the water P, in pounds multiplied by the length of A L in feet, gives the moment of the water tending to turn the embankment, H D, on A as a centre ; and the product of the weight of the embank- ment, H D, In pounds by the length of A In feet, elves the momentum of the emba,nkment that acts against the pres- sure of the water ; consequently, when these moments are equal the embankment, A H E D, Is upon the point of turning over the point A ; If the moment of the water be the greater of the two, the structure will fall, but if It be the lesser of the two, it will stand. Ques, 74.. Suppose 10 ft. to be the length of an em- bankment whose height, D E, from the surface of the water at E, is 28 ft., A D = 6 ft., will the embankment stand or fall when a cubic foot of the material of which It is composed = 1 60 lbs. ? Surface upon which the water presses=28x 10=280 square feet. qo Pressure of the water=280 X -^ x 62-5 = 245000 lbs, M QQ _= D P=A L=tlie distance of the centre of pressure P, from 3 the bottom A D. ■ 2R 2 .-. 245000 x^ = 2286666 —the moment of the water, o o Weight of the embankment=28xl0,x 6 x 160=268800 lbs. B 242 PEACTICAL MECHANICS. Moment of the embankment=268800x ■h' =806400. This structure must fall, since the moment of the water is greater than the moment of the embankment. Ques. 75. What must be the height of the water in the last question, so that the embankment may be upon the point of overturning? Putting X for the required height, then the moment of the water will be a;xlOx-^x62-5x^=-|'x625. 2 3d .-. %■ X 625=806400. 6 .-. a!=20-59967 ft., height required. Ques. 76. Bequired the thickness of a rectangular em- bankment that supports a pressure of water rising its full height of 16 ft. when the structure is upon the point of turn- ing over ; the weight of a cubic foot of the material, of which the embankment is composed, 128 lbs, ? We may take the length of the embankment = 1 foot, for if it stands for one foot of length, it will stand for any other length. 1 fi 1 X 16 X -5- X 62-5 = 8000 lbs., the pressure of the water. 8000 X — =the moment of the water. o If X be put for the thickness of the embankment, its moment will be 16xa!xlxl28x ^=a;'x8xl28. £1 1 fi Batting 8 x 128 X a^=8000 x 3-' Gives a^= -5^ ; .-. a; = 6-45497 ft. PBACTICAL MECHANICS. 243 Qmst 77. Let the cross section of the embankment, ABOD, Fig. 115, have the form of a trapezoid, where ■AE=6ft., EB=5ft., BG=15ft., and the weight of a cubic foot of the material=120lbs. ; as in former cases, the cubic foot of water is supposed to weigh 62^ lbs. | Let us consider the circumstances with respect to 8 ft. length of embankment, and suppose the cross section, Ftg. .115. ABOD, to be divided into two jparts, namely, the rectan- fular part, BODE, and the triangular part, A E D. It as been before shown that a vertical line, grL, passing through the centre of gravity,-^, of the triangular part, cuts the base, A L, supposed to be horizontal; so that AL=| of A E= 4 ft. 'llie vertical line, G F, passing through the centre of gravity, G, of the parallelogram, E D B, cuts the base at F ; so that AF=AB+iEB=8-5ft, It is supposed that the pressure, P, of the water tends to turn the embankment over a horizontal line passing through A, perpendicular to the plane of the paper. ^2iL X 8 X 120=43200 lbs. weight of the pait of which A D E 5s a cross section ; the moment of this part will be=4320ff x 4= 172800. b2 244 PBACTIOAL MECHANICS. Weight of B C D E=15 X 5 X 8 X 120=72000 lbs. Moment of this part=72000 x 8-5 = 612000. .*. 612000 + 1 J2800 = 784800, the moment of 8 ft. length of embankment. Since the moment of the water will be 15 X 8 x^x 62-5x^^=281250, it follows that the embankment will stand. Ques. 78. The breadth of a flood-gate is 12 ft. ; the depth, A B= 8 ft. ; the centre of the hinge, Q, is 18 in. from the bottom. A; and the hinge, E, is 18 in. from the surface, B; the pressure on Q, Fig, 116, is required? Tier, llfi Since one-half the pressure of the water on the gate only acts on the hinges Q and R, that pressure in lbs. will be= PEACTICAL MECHANICS. 245; 8x6x-5 X 62-5^12000 lbs. Let P be the centre of pressure of the water, theh AP=| ; QK=8-3=5ft.; o 8 , Because the pressure of the water at P is supported by the PR=PB-BE=-^ X2-I| = 3|ft. a i pressure of the ; hinges at Q and R, then, on the principle of the lever (p. 33), , supposing E to be the fdlcram, Putting X for the pressure on Q, a;xQR = PxPR, that is a:x5 = 12000x3|; .-. a; = 9200 lbs, Ques. 79. If one side of an equilateral triangle, im- mersed In a fluid, be perpendicular to the surface of the fluid, find the relation of the pressures on the three sides? Let the side, A B, be F'^- "7- perpendicular to the sur- face of the fluid L N, Fig. 117. From F , and G, the points of bisec- tion, and therefore the centres of gravity of A 0, C B, draw E F, D C, H Gr, perpendicular to AB. It is evident that the perpendicular depths, M F =AE,AD,Ma=AH; of the centres of gravity of the sides, A (J, A 13, B C, are as 1:2:3. Hence the pressure on the side B C is equal to the sum of the pressures on A B, A 0. 246 PRACTICAL MECHANICS. Geometrical Proposition. *1s.)i8. If from any Of the an- gles of a triangle, ABO, a line, A m, be drawn to m, the middle of opposite side, OB, the point G is the centre of gravity of the triangle if m G ^ | of m A . Draw B n, bisecting A 0, join m, n, then it is evi- dent that all lines, as p q, Parallel to B are bisected y Km; hence the centre or gravity of the triangle must lie in A m. In the same manner it may be shown that B n bisects all lines, as r s, parallel to A ; therefore the centre of gravity is also in B n ; consequently the point G, where A m and B n intersect, is the centre of gravity of the triangle A B 0. . But mn=\ A B, and is also parallel to A B ; and be- cause the triangles m w G and GAB are similar, m G= i G A, whence m G=-|^ m A. As the knowledge of the position of the centre of gravity of a body is of much importance in almost every depart- ment of mechanics, and to save the trouble of distinct in- vestigations in cases that often occur, we have thought it proper to add the succeeding results : II. The centre of gravity, G, of a trapezium, A B D, Fig. 119. Let L be the centre of gravity of the triangle ADO, H of ABO, E of ABD, F of BDO; joinHLand EF; these lines cut in G, the centre of gravity of A B D, PBAOTICAL MECHANICS. Fig. 119. 247. III. To find the centre of gravity, Gr, of a quadrilateral, AB CD, when two sides, AD, BC, are ffanpllel, Fig. 120. a=AL=LD, andBK=KC = S; KL = 3c. b+2a KG=c b + a Fig. 120. IV. To find the centre of gravity, Gr, of any triangular pyramid, AB C D, Fig, 121.' Put A B = a, AC=J, AD=c, andBC=- 248 PEACTIOAL MECHANICS. mg. 121. Or, bisect B C in F, draw F D, F A ; make E F=| of F D, and H F=i of A F, and draw H D, A E. The triangles H G E^ GAD are similar. .-. HG=iGD=iHD; EG=^GA=iEA. CASE It. WhenBC=CD = DB, Then AG*a=^(a«+6'+c'-fl!'). CASE III. When BC=C D=D B, and also A B= A C = A D. Then AG»=-^ (3a»-«?> CASE IV. If all the edges are equal, ABCD becomes a regular tetrsedron. Then AG = ia\/6'. V. To find the centre of gravity of a j»jramid whose lose is any polygon. The centre of grayity will be on the line drawn from the vertex to the centre of gravity of the base, and at the dis- tance of I of its length from the vertex. PRACTICAL MECHANICS. 249 VI. The centres of gravity of the surface of a cylinder, of a cone, and of a conic frustrum, are respectively the centres of eravity of the parallelogram, triangle, and trapezoid, -which are vertical sections of the respective VII. The centre of gravity of the surf ace of a spherical segment is at the middle oif its versed sine, or height. VIII. In the cone, as well as in the pyramid, the dis- tance of the centre of gravity from the vertex is | of the axis. IX. In a conical frustrum, the distance of the centre of gravity, measured on the axis from the centre of the less end _ h 3R'+2B r+r" 4 R2^Rr+r«' A=the height ; R, r, the radii of the greater and lesser ends. X. The last theorem will apply for the frustrum of any regular pyramid, taking R and r for the sides of the two ends. XL In a paraboloid, the distance of the centre of gravity from the vertex =| of the axis. XII. In the frustrum of a paraboloid, the distance of the centre of gravity from the centre of the lesser end, along the axis "~3 R'+r' * h denotes the height, E and r the radii of the greater and lesser ends. Xmr. To find the centre of gravity, G, of a circular arc MAm, Fig. 122. From the middle point of the arc A, draw A O to O, the centre of the circle; put ar=AP, y=MP=Pm, radius A 0=r ; the length of the half-arc, or M A=z:=A m. 250 PEACTIOAL MECHANICS, Pig. 122. r V ThenOG =— • z XIV. When the arc is a semicircle, then y=l; and £=i7r= 1-57079, V 1 or -= ., ,---„ =-63662; and then z 1-57079 ' P G=-63662 r. XV. To find the centre of gravity, G, of a circular seg- ment, M A wi, Fig. 123. Fig. 123. 0G= MP' 3 area of A M P PBAOTICAL MECHANICS. 251 XVI. When the circular seameiit hecomesa semicircle. 3 qiiadraHiSi radius r 2-356194 r* 2-356194"" -42441 r. XVII. In a cireakr sector, O B A C, Fig. 124, the dis- tance from the centre of the circle to the centre of gra- vity, or O G,= -5^; in which »'=OA; c=BO; a=the length of the arc BAG. Fiff. 124. XVIII. To find the centre of gravity of a common pa- rabola, Fig. 124. AG=|AC. To find the centre of gravity, g, of the siemiparabola, CAD, Fig.. 125, take XIX. To find the centre of gravity, G, of tlie sector of a sphere, O BAB, Fig. 12&. 252 PEACTICAL MECHANICS. Fipi 125. Let A C=x, O D=r, then AG=i(2r+3x). "^ When x = r, the sector becomes a hemisphere, then AG=|r. XX. To find the centre of gravity, G, of the segment of a spheroid. Fig. 126. Let A be the Tertex of the fixed axis a, putting c for the length of the revolving axis, then AG= 4a — Sx 6a— 4a; PEACTIOAL MECHANICS. 253 a When the segment becomes a hemispheroid, then x=-- and g _. a;=|a;, for the distance of the centre of gravity from the vertex, /. f a; is its distance from the centre of the base. If c=a, the spheroid becomes a sphere, and, as the theorem is independent of c, it is alike applicable to both solids and their corresponding segments. XXI. To find tlie centre of gravity of a Jiyperbotoid. Putting g=the distance of the centre of gravity G, from the a vertex A, and takings* =-5 (cfx+so^)i a' ia + Sx ^~ 6a+4a; X. XXII. The position of the centre of gravity G, of any irregular body, ABC, may be determined when balanced in the manner represented in Fig., 127, and applyipg the proportion, Fig. 127. 254 PBACTICAL MECHANICS. W : w :: a : z=U Q. z= aw w' E S is put=a, and is horizontal. K E, Q G, 8 T are perpendicular to S E. Ques. 80. Let the shores, a c, support the wall, A B, that sustains a pressure of water up to the top A ; the stay delivers its thrust opposite the centre of pressure, P, of the water ; the thrust on the stays is required, when the em- bankment is upon the point of turning over. Suppose AB=15ft. ; 0D=3ft.; the weight of a cubic foot of the material of which the wall is composed 120 lbs. i ad=FB=5hJ. We will estimate for 8 ft. length of wall, but any other length may be selected. Fig. 128. PRACTICAL MECHANICS, 255 "Weight of wall 8 X 3 x 15 x 120=43200 lbs. Moment of the wall = 43200 x |=:64800. Let 6 d be perpendicular to a c ; 125 S(Z=/vJ =|V'2=3-5355. Put x for the thrust on the stays, ac, that prop 8 ft. length of wall ; the moment of this thrust will be a;x 3-5355. Moment of the pressure of the water will be 15 15 15 X 8 X y X 62-5 X -g =281250. A as X 3-5355 + 64800= 281250 .-. 05=61222 lbs. Ques. 81. Eequired the modulus of stability of the stone structure, A D H O, Fig.129. AD=3ft.;0H = 8 ft. ; B E, drawn from the middle of A D to the middle of O H = 18 ft,; the height of the water, H.D=17-4ft.; weight of a cubic foot of the mate- rial of which the wall is composed =200 lbs.? It is only necessary to in- vestigate the action of the forces on a length of one foot. A D H O is a cross sector of the wall. Fiff. 129. 84 ^®-^^^OR+AB-^i+lI=ll' 6 being the centre of gravity of the wall. 256 PRACTICAL MECHANICS. RB : ET :: EG i RS=1^. ,-. S 0=5^. B T*=B E^ — ET", that is, B T is equal to ike square root of 182 -(1)2=1 VT27i= 17-82554. Pressure of the wall, acting in the line G S, through the centre of gravity G= ii^x 1 X 17-82554x200= 19608-094 lbs. The centre of pressure of the water is at P, and H P= lZi=5-8ft. = CS. 3 17-4 Pressure of the water=17-4 x 1 X -o^X 65-2=9211-25 lbs. .-. 19608-094 : 5-8 :: 9211-25 : 2-724652=0 OT=Sn. In the parallelogram CShwi, if the line OS=5'8j represents the pressure of the action in a vertical line passing through its centre of gravity; 0»i=2'724652 re- presents the whole pressure of the .water acting on P, its centre of pressure. The ratio of S O to S m is termed the modulus of sta- bility, which, in a good structure, should not he much less than ^. In the present case, S» 2-724652 ._. .... . ^, , ^-:r= — - ^ =-5384, which is greater than -5^ Hence the structure is secure. A square, AB D, Fig. 130j is immersed vertically in a, fluid, the side A B coinciding with the surface; if the diagonal, BD, be drawn, compare the" pressure on the triangles A BD,P I) C. Bisect AB, DO,_in E and F; join DE,.:BF; take E G=^ E D, and Fg=l B F ; G and 51 are the centres of gravity of the triangles A B D, B D O. PEAOTIOAL MECHANICS. 257 D E is equal and parallel to B F, B^=2srF=2EGj .". the perpendicular m g^=twice the length of the perpen- dicular GrM. .'. The pressure on the .triangle B C D is double the pressure on the triangle A B D. The same is true in the case of a rec- tangle, and the proportions remain the same whatever be the in- clinations of the immersed planes, proAdded only that A B coin- cides with the surface of the fluid ; for the perpendicular depths of the centres of gravity will be altered in the same ratio. Fig. 130. Given a rectangular parallelogram immersed vertically in water, with one side A B, Fig. 131, coincident with the Fig; 131. 258 PBACJTIOAL MEOHAITICS. surface ; it is required to draw from one of the angles, B, to the base a straidit line, B E, so that the pressures on the parts A D E B, E B 0, may be in the given ratio of 772 to n It is evident that the pressure on the whole parallelogram is to the pressure on the triangle, so is »i +m to n. PutAB=a; AD=B C=J, and E C=a;; 6 1 , 25 .-. ax6x-g = -a;X0X — : m+n : n ab' b' , •'■ "a • '^ 1 ■ * "*"*"" ■ " 3 n 2 m+n To compare the jpressnfes on the rectangles A 0, F, Fig. 132, immersed vertically in water^ A B coinciding with the surface of the water* fig. 132. The pressure on A B C D : thatonABEsF :: ADx|AD : AFxiAE :: AD' : AF; /. The pressure A B C D : the pressure on t) C E F : : AD" : AF-AD' Put A F=& and D A=a;, then D l^=b-x. When the pressure on A C is equal to the pressure on C F, then h "■-'■! Or 1 : v/2 : : x : h. PBAOTICAL MECHANICS. S59- If the pressure on A B D is to be to the pressure on DGEF,5to7, then a!» : 6'-a» : : 5 : 7 ... .=si:^6. Fig. 133. Let A D F, Fig. 133, be a rectangular parallelogram im- mersed in water, the side A coinciding with the surface, AB=a; AF=b. FED is the inscribed parabola; find the ratio of the pressure on the parallelogram and the pressure on the parabola. If G be the centre of gravity of A C D F, then. B G=-r; and the pressnres=a 6 x - x 62^=^ x 62^. The area of the pacafeolassf of the parallelogram ; B g=^\ b, if g be the centre of gravity of the parabola ; / :f£l_X62i=the pressure on the pa- 3 5 rabola -5-x62i 2a6'' 5-x62i =.: /, 5 : 4, is the ratio of the pressure on the parallelogram to the pressure on the parabola. Show that if a hollow sphere be filled with a fluid, whose specific gravity is s, tiiat the whole pressure against the S A 260 PBACTICAL MECHANICS. internal surface is three times the weight of the contained fluid. Let r=:the radius ; then 4* r* X internal surface, ir as usual=3-14159265. .*. the pressure=4:r i^xry. s=4* r^s ; The solid content of the sphere = f * t^, and its weight [4'!r»^s : ^-rt^s :: 1 : -1^ or 3 : 1. Let G be the centre of gravity of a trapezoid A B C O, Fig. 134, A B=a, and parallel to O 0=S ; JB 0=c, and per- pendicular to both A B and O C. The straight line Y IS drawn perpendicular to OCX, it is required to find general expressions for the perpendicular co-ordinates «= TnG=On, and Y=n(jr=Om. Fig. 134. 1 Let p be the centre of gfravity of the rectangle A B C D, p Q = - and Q 0=6-- =^y~- If t be the centre of gravity of the triangle A D, then the co-ordinates of t will be, t R=| A D=-|^ B 0= |-. PBACTIOAL MECHANICS. 261 Area of A B C x G m= Area of A B C D x^j Q+area of A D X « R ; that is a+b e (b—a)e c c__y=acx-+— ^X-g, ■■• ^=iO+^y See pp. 37, 38, area O A B CxG»«=Area of A BX! DxB O; that is h+a 2h—a e(J,—a) 2Ji— a) 39. Agam the area O A B CxG»«=Area of A BX! DxOQ+ Area of A O DxR O; that is -x=:acx ■ -+ ^+«^-.- 2h-a (bxay 2 2 3 _ 26 _ g" 3 3 (a+b)' See pp. 37, 38. Fig. 135. Q^es. 82. Given in ABOO,Fis.l35, which represents the, cross sec- tion of an embankment made of brickwork, a cubic foot of which' weighs 112 lbs.; AB=1 ft., parallel to O = 2 ft. ; find the height, B 0, which is perpendicular. to both A B and O 0, when the embankment is upon the point of overturning upon the edge at O by the pressure of the water which stands at the brim, B ? Put «=B C, the required height, then the pressure of the water on 1 ft. length of embankment= cxlx-l x62-5=^x62-5. 262 PEACTICAL MECHANICS. Moment of the water = — X 62-5 X - = — x 62-5; 2 Ob " If G be the centre of gravity of the trapezoid AB C O, then by the last proposition, OD=5'X2- o 11 3(1 + 2)" 9" Pressure of a length of one foot of the embankment = 1+2 — ——V xll2=168t>lbs.; moment of cmbankinei3it= 168vx-g-. «' 11 = |_x 62-5=168 t>x^; .-. V X \/l9-712=4439821. Weight of embanknient=168 b=745-89 lbs. Ques. 83. Let the embankment A B C O, Fig. 136, be the same as in the previous question ; now if the embank- ^ig. 136. PEACTIOAL MECHANICS. 263 inent be raised until H L=3 ft., what must be the perpen- dicular depth B L, so that when the part A B O be upon the point of overturning at O, at the same time the whole embankment, A B L H^ will be upon the point of turning over the edge through H ? BL=DC=2-— =i 9 9 .-. HE=3-I=!? 9 9 .-. the moment of A B C round the point H, will be 20 745-89 X -g-. If flf be the centre of gravity of L H, then o A no H ^=-o ^ ^ — S(2+S^ ~TK ' P"**™S * for B L, the moment of the part C L H round H as a fulcrum will be -^(.-r)(112)x-. The moment of the water acting in opposition will be ^xlx |-x62-5x-|=-:: x62-5 Z o b :. ~ (62-5)=ii^ (^-4-43982) (112) ?i + 745-89 xH2. o J 15 9 .-. /-46;592«= -47-73689344; (/S). .-. z=Q \ 0^,0,'5'1'9'9'5=6-24037812. By dual arithmetic in any cubic equation the value of the un- knoMm quantity, as z, may be found in a few minutes under an endless variety'of forms, but all reducible to one or other of the roots ; in the present' case z may be found' under the form z= 6 \ 0,3,9,4,3,4,8,5, which is also= 6-240378 12 z may be found under the form 5 \ 2,3,1,1,3,2,0,1,= 6-24037812, When we assume z of the form n f Mj Mj Wg . . . . a convenient 264 PRACTICAL MECHANICS. value for n is readily found by supposing A B H L a trapezoid, then B L must be greater than 6, hence n was put = 6; but we might assume re = 4, 5, 6, or 7, and yet determine « = 6-24037812. The height and dimensions of any number of trapezoids forming an embankment like A L H, of equable strength, can be found in a similar way. A B, O 0, H L, &c., may be given any common difference. The labour and difficulty experienced, before dual arithnietic was invented, in finding the roots of equations higher than a quadratic prevented writers on practical mathematics from introducing pro- blems involving cubic or higher equations. The student should pay particular attention to these remarks. In Bolving equation (j8) we put it under the form .^,^4773^8934_4^4g.592_ z Because, in the dual calculus, r* and — become known at one and the same time. This arrangement being made, it is soon found that z=%\ 0,4,0,'5 'u^'ueW- REVETMENT WALLS. If dry or wet clay, mixed earth, shingle, gravel, peat, &c., be excavated, and placed unprotected by a wall, or other support, the face, O A, in a perpendicular position, OP, Fig;' 137, horizontal, during the wet and winter seasouj a portion of the embankment, AOQ, will shde away, and ultimately assume a permanent slopej as O Q;. then O Q is termed the natural slope. The weight and natural slopes of different kinds of earths are given in most engineering pocket-books ; but the natural slope of earth in engineering works should, if possible, be found in each particular undertaking. The thrust of earth upon a revet- 265 PEACTIOAL MEOHANIOS. ment wall is given by a portion, E O C, of the wei _ R O Q ; the line O is termed the line of rupture. Cou- lomb found that the angle, E O 0, which the line of rupture makes with the vertical, is one-half the angle E O Q, which the line of natural slope, O Q, makes with the vertical. OA. Fig. 137. i' A; % / ■■ -T-r-_- ■ p -- :- To find the pressure of earth when it is level with the top of the revetment wall. Rule. Multiply the weight of a cubic foot of the earth by the square- of the tangent of the angle E O (equal half E O Q), the product is to be taken as the weight of a cubic foot of a fluid pressing like water. It may aid the memory to observe, that the product above mentioned takes the place of the 62:5 lbs., the weight assumed for a cubic foot of water. Ques. 84. A revetment wall, 36 ft. high and 8 ft. thick ; any iinit of length may be taken, for convenience, we select a foot ; this wall has to support the pressure of a mixed earth, the natural slope of which was found to be 34 degrees, and the weight of a cubic foot= 112 lbs. ; the wall is built of sandstone, a cubic foot of which weighs 144 lbs. ; will it stand or fall ? 266 PEACTICAL MECHANICS. i of 34°= 17° ; tangent of 17°= -3057307, taken from a table of natural tangents. -3057 will be near enough for our present purpose. •3057 squared= -09345249, of which we take -09345, ^Isonear enough for our present purpose. ■09345 X 112=10-4664 lbs., the weight of a cubic foot of the supposed fluid, acting at the back of the wall like water. 36 X 1 X — X 10-4664=lbs. pressiure of the earth. 36 Moment of the earth=36 x 18 x 10-4664 x - =81386-7264, o a Moment of the reTetment=36 X 8 X 1 X 144 x 5=82944. Hence, the wall will stand, but it is not very secure, for 82944 is not much greater than 81386-7264. Qiies. 85. When the wall is of brickwork, 16 ft, high, supporting damp clay, whose natural slope is 45° ; find the thicKness of the wall when it is no more than sufficiently strong to resist the pressure of the clay? Natural tangent of 22° 30'=-4142136. •4142 squared, -1716; a cubic foot of brickwork=112lbs.; damp clay weighs 120 lbs. ; -1716 x 120=20-592. /. 16 X 1 X 5- X 20-592 x.j= 14057-472, the moment of the clay. Putting x for the thickness of the wall, we have the mo- ment of the wall equal 16xlxa!Xll2x 5=896 a', which must be equal to 14057-472, when the wall is upon the point of overturning, ... .=^(11^=3-96095. The required thickness of the wall =4 ft., nearly. PBAOTIOAL MEOHAmCS. 267 Ques.SG. Let OA=9-6ft., 0=4 ft.; the weight of a cubic foot of the brickwork in A 0=112 lbs., the water is up to H, the embankment is supported by a bank of earth, D, the natural slope of which is 45°, D=5 ft., ED is level; will the embankment stand or fall, EDO being composed of gravel and shingle, a cubic foot of which weighed 96 lbs.? Fig. 138. In making trial calculations, when a wall is supported by- earth, as in Fig. 138, the weight of the equivalent fluid may be taken =6 times the weight of the earth. On this supposition the earth, EDO, may be considered a fluid, a cubic foot of which= 96 x 6= 576 lbs. 4 Then, 9-6 X 1 X 4 x 112 x 2=8601-6, moment of the wall ; As usual, onie foot is taken for the length. 5 5 5 X 1 X 2 X 576 X g=12000, moment of the earth. 9-6 9'6 9-6 X 1 X -H" X 62'5 ^ F =9216, moment of the water. 2 o. 268 PEACTICAL MECHANICS. Since 8601-6 + 12000 is greater than 9216, the waU, C A B, will stand. _ Ques. 87. A revetment wall, A C, Fig. 138, is 30 ft. high=B ; 00=6 ft. On one side, B, earth of mean quality, 45° natural slope, is sustained level with the top, and on the other side, A O, the earth has a natural slope, F D, and rises to the height O D= 5 ft. ; 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.? When a wall resists the pressure of earth with a natural slope at the top, represented by the line B G, Fig. 138, the weight of the equivalent fluid is roughly taken =f the weight of the earth ; but when the earth with a natural slope, DF, resists the pressure of the revetment, AC, and water or earth pressing on B 0, then the weight of the equivalent fluid is roughly estimated at ^ the weight of the earth. Ft) O. In this case, the moment of the earth, which rises to the level of the wall, is opposed by the sum of the moments of the wall, and the earth on the othier side having a natural slope. Weight of the wall =30 X 1 X 6 X 130=23400 lbs. Moment of the wall =23400 X 1=70200; Weight of a cubic foot of the equivalent fluid, with respect to the sloping earth =120x^=60 lbs. Pressure of this earth =5x1x60x1=750 lbs. 2 Moment=750 x |=1250 ; PBAOTICAL MECHANICS. 269 Total moments Bustaimng the wall =70200+1250=71450. Moment of the level earth = 30xlx^xl20x-1716x -=92664; 2 o Since the latter moment is greater than the sum of two the former, it may be concluded that the structure will fall. Referring to Fig. 137, page 265, and -putting 6 for the angle EO Q, then because tan*" "o— TT i> ^"^d may be employed when a table of natural tangents is not conve- nient. The amount of pressure of a bank of earth, supposed to act in plane parallel layers against a vertical plane a foot in breadth = — I — Y h being the height. 2 VH-cos«-/' ^ ^ When the bank is composed of layers each sloping pa^ rallel to the natural slope, then the amount of pressure of the bank= • h^ sin S 2 But if the earth is made up of layers sloping from the vertical, OR, at an angle, IIOB=0, greater than the natural slope, R O Q=^, the pressure of the earth = h^ . sin 6— Vsin« *— sin* 6 — sm A — i ^ — ■ 2 sm—siD^i MEASUKEMENT OF HEIGHTS BY OBSERVATIONS Or THE BAKOMETEB AND THEBMOMETEB. Much of what has been said respecting the pressure of water applies to all fluids, both liquids and gases ; air and 270 PRACTICAL mechanics; other elastic fluids have some properties peculiar to them- selves ; however, the nature and properties of both atmo- spheric air and mercury have been carefully ascertained. So far, the determination of the heights of mountains by the barometer presents no difficulty ; but to solve the ulti- mate equation defied the skill of mathematicians, and the formulse presented by writers on mechanics to effect this object only gave approximate results. This ultimate equa- tion, which may be presented under the form (I.), can be solved with the greateist ease by dual arithmetic. „'_(i±iL)(. + l)„..[f(l + l)']; (L) In this equation all the quantities except the required height' z are known. The logs, are hyperbolic, and (I.) may be put under the form ■r ., . , . .IH/ 11 \ -^ hCl + ar) In this last equation put x=\— 1 1 + - ) i B=— ^^ ; ^ ^ \ r / rg and put M— = A ; and suppose A= hyperbolic log. of a, that is, log. a = A. These substitutions being made, (I.) becomes « = A + Ba; log. ar' ; (II.) 6=2-71828183, and |'e=10». (II.) may be vmtten g* = ax^' . taking the «th root of both sides, € = a' x^" . Putm*" =6, then 1^ = «'; PSACTICAL MECHANICS. 271 X This last equation, by putting y for T-V, becomes A general solution of exponential equations of this form is given in " Dual Arithmetic a New Art," Part IE., pp. 91 to 101. In measuring heights by the barometer it is necessary to know, to the greatest nicety, the ratio of the density of mercury to that of air. The accurate and indefatigable M. Eegnault foimd, at Palis, that a litre of air at 0° cen- tigrade, under a pressure of 760 millimetres, weighed 1'293187 grammes ; and at the level of the sea, in latitude 45°, it weighed 1*292697 grammes. He also found that a litre of mercury, at the temperature of 0° cent., weighed 13595*93 grammes. A litre of water, at its maximum density, weighs 1000 grammes ; therefore the ratio of the density of mercury to that of air, in latitude 45°, will be= ^^^^^'^^ = 10617-49. 1-292697 At Paris, a Utre of atmospheric air weighs 1*293187 grammes, but this number is only correct for me locality in which the experiments were made — ^that is, the latitude of 48° 50" 14", and a height of 60 mfetres above the level of the sea. Taking 1*292697 grammes for the weight of a litre of air under the parallel of 45° latitude, and at the same distance from the centre of the earth as that at which the experiments were tried, to be 1*292697 ^ammes, then, fmtting w for the weight of the litre of air, in any other atitu£, any other distance from the centre of the earth. 272 PEACTICAL MECHANICS. ^^,_ l-292697 (1-00001885) (1--002837) cos \ .jy , 1 +?2 « 71= 6366198 metres, the mean radius of the earth, q the height of the place of observation above the mean radius, and X the latitude of the place. In applying formula (IV.) to a particular example, the author found that at Philadel- phia, U.S., lat. 39° 56' 51"-5 N., the weight of a litre of air was 1*2914392 grammes ; the ratio of the density of mercury to that of air at the level of the sea was 10527*735 ; and n= 6367653 metres at Philadelphia. Eegnault also foimd, by experiments, that 1'36706 repre- sents the volume of air at 100° centigrade thermom., the volume at 0° being supposed = 1. Before the time of Begnaiilt, these aiia many other constants were greatly in error. Experiments show that air, under the same pressure, expands uniformly for equal increments of temperature ; that the expansion due to the same increase of temperature is not the same for all gases, as many scientific men have supposed. However, in air the expansion, for a unit of bulk is '36706, according to Eegnault, from the freezing to the boiling points ; and therefore the expansion for each degree of Fahrenheit is \^ of -36706. Let a=:the expansion of air for each degree of the thermo- meter, i=the ratio of the elasticity of air to its density at the temperature of melting ice ; then the bulk at the temperature cc will be increased, and therefore the density diminished in the ratio of 1 + a a; to 1 ; consequently k {l-\-a. a;)=the ratio of the elasticity to the density at the temperature x. p =the elasticity, ) at the siirface of g =the force of gravity, i the earth. For the air let< P=the elasticity -j ^^ ^j^^ ^j^^^^^ ^ 1D=the density at temp, a;, >• , ,, _, „ ,, » % ..^ ' I above the surtace I G=the torce of gravity, ) p Then = =;i;(l-)-aa;), and .-. P=DA(l + aa;). Then, for what is conventionally termed the differential of P, in the notation of the differential calculus, put d P ; PBAOTIOAL MECHANIOSi 273^ Then, P— dP=:the pressure at the altitude (z+dz); P=the pressure at the altitude g. D o T^ di z .'. - d P=the difference of preBSures=D (idz= •'. r being put for the radius of the earth where the observation is made. d^_ gi* ^ ( dz A •'• ~ P h{l-\-a.as)\{r+zY :. Intergration gives log. P = TO^^(j^^ + C, .-. log.p = - . ,ff — -xr-^ + C; Consequently, log. |=^^1^— ^(i =-i_) h {l + axy^r+zJ~k {l-\-ax)\r-\-zJ ' '<^ '' Prom experiment it also appears that mercury contracts uni- formly as its temperature decreases. ^ 1 J ^"^''^^* of barometer, 1 ^ ^^^ ^^ For mercury let-J M=density of mercury, V ^f ^^^ ^^^. XjS. = temperature of mercury j ' and Hj, M^, T^ the same quantities respectively at the altitude z ; and - = t^e condensation of mercury for one degree of the thermometer; But ;>=^HM; and P=G H, M,=^-^ (H,) M (1+^) • P =i:t±£Y 7 T-T V • putting A for Q--\ — i-iJH, and equating the hyperbolic loga- 274 PRACTICAL MECHANICS P uxx\i xaou value KJl we haye rithm of the last value of ^ with the value before found (V.), k (1 + aa;) \JTz)=^°^-'l1 V^r J \ The temperature x has been supposed to remain the same throughout the whole column z, whereas it always decreases as we ascend from the surface of the earth ; but, being ignorant of the law of this change, a mean value (r) between the values at the two stations is taken and consi- dered constant ; the mean r being substituted for x in the last equation, (I.) is established. In a discussion on General Anstruther's paper respecting the flight of projectiles, reported in the " Journal of the Koyal United Service Institution," vol. ix., the author of the present work made some observations upon the value of the force of gravity and the constant g, the resistance of fluids to bodies moving in them, and on that due to friction when the moving body described an irregular path ; those remarks and observations are introduced in this place for the student's most careful consideration. On the occasion referred to the author observed : When great minds con- spire to perpetuate a fallacy, it has always been a difficult matter to clear that fallacy away. I know of no subject capable of being submitted to mathematical investigation that has received a greater amount of fallacious treatment, and that, too, by great minds, than the motion of projectiles. Initial velocities nave been little more than guessed at, the resistance of the air overrated, and the force of gravity mis- stated. The system introduced by General Anstruther, which is practical and easily applied, must give correct results within the rangfe of his experiments, without offer- ing any special theory about initial velocities, the resistance of the air, oj the force of gravity; indeed, i|i his system^ are collected all those elements. ^ Let us examine the PEACTICAL MECHANICS. 275 theories put forward with respect to the resistance of the atmosphere to the motion of a ball passing through it. The resistance of a fluid to the motion of a body is said to vaiy as the square of the velocity; the hypothesis upon which this law rests may be thus considered : the fluid dis- placed must have the same motion given to it as that of the moving body ; hence, on this supposition, the units of work destroyed by the fluid wiU be equal to the accumu- lated work in the fluid. Fig. 139. Vdocity=7 feet a second. For example, let the cross section of the body perpendicular to the di- rection in which the body moves = 3 square feet, the weight of a cubic foot of the fluid = 62 lbs., and the velocity of the body 7 feet a second. The weight of the fluid moved at a velocity of 7 feet a second=7x3x621bs. The units of work expended in the displacement will be 7^x7x3x62 in which ^gr is put = 32^, 32 J, 32, &c., and is said to vary- inversely as the square of the distance from the centre of the earth. Then the body in moving through 7 feet destroys 7x 7^ x3x6 2 64 , tmits of work. If T be the resistance of the fluid in pounds, the units of work destroyed will also be represented by T x 7 ; m » .. 7''x3x62 .-. Tx7=7x or T = 64 7''x3x62 64 Hence it is concluded that the resistance increases with the square of the velocity, as well as with the area of the cross t2 276 PRACTICAL MECHANICS. section of the body preserited to the fluid. It is easily per- ceived, especially with high velocities, that this law is not strictly true, and very far from the truth when the motion takes place in air. It has heen found, from experiments on railways, that the resistance of the atmosphere to the motion of the train depends chiefly on the length of the train, and not upon the frontage of the carriages. The re^ sistance resembles more that of friction than the moving of a long parallelopiped of the fluid in -which the body moves. TsSing this, latter view of the motion of a ball through the atmosphere from one given point to another, the units of work will always be the same, whatever may be the f ornl of the curve described. Let W, Fig. 140, be a body moved through the air, then, by the resolution of pressures, the pressure of W, perpen- dicular to the tangent= W cos a ; .". The resistance of friction=/W cos a, putting / for the coeflicient , of the friction of the air on the ball. Hence the total work on the plane a^zs zfWcos a,+Wy; hut scosa=3;; .: The work on each plane =fWx+Wy, this expression being independent of the inclination of the plane. It points out the work employed to move the body over the horizbntar distance ay, added to the work due to the gravity in elevating the body the. vertical space ^y. Since a curve may be supposed made up of an infinite number of straight lines, therefore the work upon the whole curve will be equal io the work done upoii the horizontal projection AB, added to the work done in opposition to gravity in raising the body from B to 0, no matter what form be given to the curve A W W W 0. This demonstra- tion shows that the way in which General Anstruther has treated the resistance of the air to be the most rational yet proposed. In the next place, let us examine the manner in which the value of g, representing the force of gravity, has been estimated. PEACTIOAL MECHANICS. Fig. 140. 277 Let r=CM= radius, Fig. 141; v= the uniform velocity of M on the curve ; <=the time employed in moving from M to m, or in falling from M. to p. If / be considered constant during a short space of time, Tig. 141. /•dt=dv ' ■ ds • ^*=^=dt . s='^=Nm=Mp. m M = V t=uniform velocity X time. 278 PEACTICAIi MECHA1T108. Again, suppose the time so short that the arc is equal to the chord, ib'r arc m M= chord m M. Then MpxMA=(ohord Mm)'=(arc Mmy=(vt)'j but, Mp='^— and MA = 2r, .-, Mp'xMA=-^x2r=v^t= ••• 7=7; (A). If T be the time of an entire revolution, then Tv=2ot .-. r'=-^ and/= -^; (B). Let another body, E, Fig. 141, describe the circumference of a circle, E JP Q, in the time Ti ; and let F represent the accelerating force of gravity at E. Put R=EO, then, from (B) Ti» .'• / : F : : ^ . -^j (C). Now the third lavr established by the experiments of John Kepler shows that the squares of the times are as the cubes of the distances from the centre, C ; hence if we put r» for T", and E' for Tj", then becomes /=F::^,:1 (D) ; and thus it is shown, by the experiments of Kepler, that the force of gravity varies inversely as the squares of the distances. Sir Isaac Newton had little to discover in this matter when Kepler's laws are admitted. The experiments at General Anstruther's command are as accurate as those upon which the law indicated by (D) is founded. The in- . PEACTIOAL MECHANICS. 279 fluence of gravity on the motion of the moon round the earth, or of the earth round the sun, is then compared with the falling of a heavy body near the earth's surface, and near the earth's surface /is represented by g, which is said to be found by experiments. Astronomers apply a variety of corrections to u pho ld (D), and the cooked-up value they have given to g. When g is trimmed up to suit the motions of the heavenly bodies, it is out of gear to apply it to the falling of heavy bodies in air near the eaorth's surface. In changing the value of g, General Anstruther is right, for his system corrects the constant quantities, and secures correct results within the range of his limiting experi- ments. To find whether the resistance of the air or any other fluid medium is proportional to the square of the velocity (V) or not, and also to find whether the value usually given to {g), the force of gravity near the surface of the earth is under or over estimated, Generally ^ is put=31i, 32|, 32, &c. Let «' multiplied by some constant coefiicient express the retarding force ; and to simplify the investigation put this coeiEcient under the form w^g. Then the retarding force-will be expressed by g—gnV ; (I.). In this investigation s is put for the space passed over in the time t. In general terms, if F be any accelerating force, U the velocity, S the space, and T the time, the relations between these quantities are found from the two equations „ dS , _, dU .__. U= — and F= — : (II.\ dT dT' '^ '' When T is eliminated from these equations, then UdU=FdS; (III.). Again eliminate U, then we have FdT = ^; (IV.). 280 PBACtlOAL MECHANICS. Then frdiri (I.) substitute g(l -»»"?)«) for P in (II.), thefl, gdt=_iH_; (V.). ^ 1-nV From (III.), gds=j^; (VL)- When the equations (V.) and (VI.) are integrated and corrected, on tiie supposition that when t=o; v=o and .s=o; the resulting equations will be 1 -28"'_ 1 , 1 , 1 Eliminate v and equation (VII.) is obtained The reduction may be effected as follows : . n^ (?=*+!)" (*""'+ 1)^ Then j_j^2p2- 4^2.g< -V2/ V e"^' ^ _ rl f8°e'+l -.2 (Vn.) may be put under the form 2.'«"' = €S"+-. (VIII.). Now from (VHI.) the values of n and g may be found by dropping cannon-shot from a balloon, at different known heights, the time occupied in f alUng being measured. Two PBACTICAL MECHANICS. 281 diflerent values of s and two corresponding values of t being thus determined, (VIII.) presents two independent equa- tions, from which n and g may be eliminated by. the dual method of solving equations. I have but little doubt that the liberty taken by the General to change the reputed force of gravity near the surface of the earth, to meet the practical necessity, will be warranted when the value of ff and the resistance of the atmosphere are independently and accurately determined. The force of gravity is pulled in all directions by astronomers, to make it fit the motions of the heavenly bodies ; yet, after all the corrections and allow- ances are made, the moon and asteroids will not obey. The practical results given by the system of General 'Anstnither also show that it is right to change the initial velocities previously conjectured, for it is eaisily demon- strated that the initial velocities given, both by the ballistic ?endulum and the pendulum ofNavez, are erroneous, 'hese instruments iriay be employed to indicate any con- jectured initial velocity, for the time incorporated in the calculation is made to depend on a body very much re- strained, falling in air a lew inches. To say the least of the process, it is very ridiculous ; the results may be cor- rected until the time anticipated is obtained. The value of g in or out of air is unknown, and yet it is made to tell all about the pendulum, and in return the pendulum is made to tell all about g. The General points out the errors of other systems in bold relief, and arrives at practical results within the range of his guiding experiments, without assuming initial velocity, a particular value for g, or in- cluding any wild theory about the wonderful resistance of the atmosphere to cover short-comings. With respect to the force of gravity, and G, g, g, see pp. 37, 38, 93, 97, 134. The weight of a body is in a •ratio compounded of its mass and the intensity of gravity in the place where it is situated, see p. 134. Hence, if we- denote this weight by W, the mass by M, and the measure of gravity by g, we have W=gTili.. This quantity g, which is independent of the particular 282 PBACTICAL MECHANICS. nature of the T)ody, is in this position, the weight of what has been arbitrarily assumed to represent the unit of mass. We may also write, W=wY, w denoting the weight of the body, which represents the unit of Tolmne, and V its volume. The weight, w, is what is termed the specific gravity of the body that is considered, a denomination which is evidently improper, for gravity is common to all bodies of different species, and therefore there ought to be substituted for it the denomination specific weight. Finally, if D represents the mass, under the unit of vo- lume of the body which is considered, D will be what is termed the density of the body, and we shall have M = DV, and since 'W=g'M., :, 'W=gJ)Y. The preceding are the equations which obtain between the five quantities W, g, M, D, V, each of which should be expressed numerically, by referring it to a unit of its species. The student should bear in mind that the exact size and form of the earth has not been accurately deter- mined. Morin and other mechanical investigators first establish theories on abstract reasoning, and afterwards correct them by experiments ; while others, like Anstruther, without any particular hypothesis, interpolate a range of experiments by proportion or empirical rules. A third class, which is very numerous, assumes constructions, and asserts that those constructions possess particular properties, that may or may not exist, and supports such claims by high-sounding pre- tensions, and a little mathematics to cover the delusion ; a fair representative of this class is the expounder of Koch's . ingenious system of skeleton structures. PBACTICAL MECHANICS. 283 CHAPTER IX. RESISTANCE OF FLUIDS, RESISTANCE OF WATER TO THE MOTION OF PADDLE-WHEELS, SCREW PROPELLERS, &C. &C. THEORETICAL OBSERVATIONS AND CONSIDERATIONS. When a body moves in a fluid, it necessarily displaces the particles of the fluid, impressing on them velocities in a certain ratio to its own, and it is readily conceived that the inertia of the fluid particles, thus set in motion, develop? a resistance which increases with the velocity of the body. Similar effects are produced when a body is at rest or in motion, and acted upon by a fluid. The manner in which the fluid particles are mvided when coming in contact with the body depends much upon the form of the body, and we perceive, without experiment, that the resistance must vary, considerably with mese circumstances. When a body. Fig. 142, of any form, P Q M N, moves in a stream in a direction X Y, if we suppose each particle of the body transferred to a plane perpendicular to the direction of the motion, it is readily observed that in moving an elementary space s=M m, the body will displace a volume of liquid that may be represented hyAs, found by multiplying the projection of the body upon the plane per- 284 PRACTICAL MECHANICS. pendicular to the direction X Y, by the length of the path = «. Fig. 142. In the two successive positions the body occupies the same volume, and there is a space common to the two positions, which corresponds to M Op r, so that the anterior volume, onm qrM^ is evidently equal to the volume ONP Q rp. It is further evident that each of these volumes is also equal to the volume, N w g' Q, made by the greatest cross section of the body, or by the area of its projection, which we have represented by A. Thus, when the body describeis in rektion to the fluid, or when the fluid describes in relation to the body, an elementary space s, the volume of the deviating fluid which passes from the front to the rear of the body is expressed by A s which we shall put=g' ; the mass will therefore be= d jAs 9 9 putting d for the density or weight of a cubic foot of the fluid. This deviating mass effects its relative displacement with a velocity depending essentially upon that of the body in relation to the fluid, in the case when it is the body that moves, and which it is reasonable to suppose is proportional to the velocity of the body. It will be the same for the units of work, or for the vis viva imparted to the deviating PRAOTIOAl MECHANICS. 285 fluid ; so that in the case of a fluid at rest, in which motes a body impressed with a velocity V, the vis viva imparted to the displaced fluid for- an elementary motion of the body will be proportional to U A. S -r^2 and if we put A=the unknown ratio of the vis viva F, really impressed upon the fluid, to the above expression, we shall ha>ve 9 in which k has to be found by experiment, and ■p ■£=1116 units of work. 2 On the other hand, if we put E=the' total resistance which the inertia of the fluid particles opposes to their displacement, the work of this resistance for the elementary space, s, will be and must, according to the general principle of vis viva (see pp. 116, 117, 118), be equal to one-half the vis viva imparted to the displaced fluid. ... Rs=1a'^VS 2 s P__ MA Y2 ~ % In the case where the fluid displaced by the body, is moving with a velocity of its own, if the body moves in an opposite direction to the motion of the fluid, the relative velocity with which the fluid particles are met and dis- placed is Y+v; and when the two velocities V and v are in the same direction, is Y—v:, similar reasoning to, that already employed wiH give us, then, for the case in which the body moves in an opposite direction to the fluid 286 pbactioal meohaitics. and for that in wHch they move in the same direction T-v ^=kdA(^J^- THEOEETICAIi EXPRESSION FOE THE UNITS OP WORK DE- VELOPED EACH SECOND BY THE EESISTANCE OF WATER, AIR, OE ANT OTHEE FLUID MEDIUM. When all the circumstances of motion remain the same, and the phenomena occur continuously in the same manner, the work developed in each second by the resistance of the medium opposed to the motion of the body is, in the case of a fluid at rest, and the work each second in the case of the fluid being in motion is which shows that, in the first case, the work of resistance increases as the cube of the velocity. BQUlVAIiENT EXPEBSSIONS FOR THE EESISTANCE OF FLUIDS TO MOVING BODIES. In the preceding expression of the resistance applied to a liquid whose density, d, is constant, and in places where JEv7 the value of 2g = Q4:-56B4t, and putting K for it then 9 takes the form E=KAV^ or R = KA.(V+t>)^ PRACTICAL MECHANICS. 287 in which farms the expression for the resistance is fre- quently used hy writers on mechanics. Some authors, and in particular Dubuat, call H the height due to the relative velocity V or V+w, and conse- quently Y2 (J+vy H= ^, or H= — a ' and putting K' =K 180 49' 19' 48" Cylinders of the same proportion terminated by spheres , 2-724 0-41969 1-7715 1-3944 1-0276 0-90868 0-84302 0-77449 0-77487 Observatiom wpon these results. — ^The value of the coeffi- cient K found in the experiments for thin plates is con- siderable, and nearly double that found by Dubuat in causing a vertical plane to move in a horizontal direction, thus producing a displacement of water entirely different from that in our experiments, and occasioning the difference of results. It is remarkable that of all the bodies used, the spheres offered the least resistance, and that cylinders, terminated by half spheres, have experienced less than those with acute cones. This result shows, that in regard to the resistance of a medium, the spherical form for projectiles, and the semi- circular for piers of bridges, are the most favourable. Influence of the acwteness of the angles of cones upon the resistance. — la comparing the values of the half angles at the summit of the cones, expressed in fractions of the semi-circumference, with the values of the resistance, we see that the coefficient K of this resistance increases proportionally with these angles; starting from a certain value answering to the angle zero. It may be given by the formula J^=O-59005+2-2998«, a being the half of the angle at the summit in terms of a fraction of the semi-circumference. The comparison of 290 PEAOTIOAL MECHANICS. the values of K given by this formula, with those deduced directly from the experiment, is established in the following table : Comparison between the values of the coefficient K as de- duced by formulas and by experiments. Half angles at the summit ID fractions of the semi.circimiferenGe. Values of the coefficient K deduced From the formula. 1 From experiment. ,0-500 '0-562 0-262 0-145 0-105 0-080 ' 1-739 1-422 1-192 0-923 0-831 0-774 1-771 1-394 1-027 0-908 0-843 0-774 We see that, with the exception of the case relative to the cone whose half -angle at the summit was measured by an arc equal to 0*262 of the semi-circumference, the results, including even that pertaining to the plane base of the cylinder, are quite correctly represented by the formula, and that we may use it for intermediate cases, which have not been experimented upon. Eapefriments vpon the resistance of water to the motion of projectiles. — Without entering into details, for which this is not the proper place, I must say something about the remark- able results of experiments made by me in common with MM. Piobert and Didion, at Metz, in 1836, upon the pene- tration of projectiles in water. These e^eriments were made at the basin which had served for the beautiful hydraulic researches of MM. Pon- celot and Lesbros, in firing horizontally beneath the surface of the water projectiles which penetrate the water after having traversed an orifice formed by spruce scantling. A horizontal flooring, placed at the bottom of the basin, and marked with strips, received the projectiles, which always reached it with a very, small velocity. PRACTICAL MECHANICS. 291 We found, with this arrangement, the resistance offered to solid balls with diameters of 0-354 ft., 0-328 ft., 0-530 ft., and 0'72 ft., and to shells of the same diameters, having different thicknesses and weights, the initial velocities of the projectiles varying from 229 ft. to 1640 ft. in 1". From the general view of all the experiments made, the results of which are published in No. VII. of the " Memorial de rArtUlerie," we conclude that the resistance of water to the motion of these projectiles may be represented by the formula R=0-453 AVlbs., while the experiments above cited (No. 305^ gave us E=0-4197 AVlbs. On the other hand, the ancient experiments due to New- ton, and made by observing the time pf the fall of spheres in water, lead to the value R=0-46498 AV^ lbs., and those which Dubuat made in causing spheres placed at the end of the arm of a horse-gin to pass in a circle through the water, furnish the formula E=0-4197 AV^ lbs. An inspection of all these researches, made by processes so different, and within limits so extended, enables us to conclude that, in liquids, the law of the proportionality of the resistance to the square of the velocity, is appli- cable to spheres, even when moving with the highest velo- cities. The preceding theoretical formulae apply to boats which navigate the sea, rivers, and canals; but their results are influenced by different circumstances, of which it is im- portant to take an account ; some are permanent, others accidental. u2 292 PEACTICAL MECHANICS. ANALYSIS OP MOEIN's EXPEEIMENTS ON THE EESISTANCE OF WATEK TO THE MOTION OF PADDLE-WHEELS. In the first set of experiments two models were employed, the first wheel was 3*31 ft. diameter, and received the paddles, which were varied in number up to twenty at most. The dimensions of the paddles employed on this wheel were as follows : In length parallel to the axis, -328 ft. ; -656 ft. ; -984 ft. ; 1-968 ft. Breadth, in the direction of the radius •328 ft. ; -659 ft. ; 1-U8ft.; -659 ft. The shaft of the wheel formed a windlass around which rolled a cord, which passed, to the summit of a crane 55'77 ft. in height, which supported a box in which was placed the motive weight. The wheel was established upon a fixed frame, and the depths of immersion were varied at will, in raising or lowering the level of the water. The velocity of rotation of the wheel varied from the lowest in which it was possible to observe a regular motion up to 19'68ft. a second. When the motions became uniform, the velocities were observed by a timepiece indicating tenths of seconds. The second wheel employed was 8*567 ft. in exterior dia- meter, with paddles 2'29 ft. long in the direction of the axis, by a breadth of 1*659 ft. in the direction of the radius in which they were placed. The depth of immersion of these paddles was successively 1*659 ft., 1*325 ft,, and' *937 ft. For each number of paddles, and each depth of immersion, the motive weights, and consequently the velo- cities, were gradually changed, so as to have a series of ex- periments in which one element only was variable. Having thus, for each case, the values of the resistance corresponding with the different velocities, a graphic repre- sentation, Fig. 143, was made of all the results in taking for abscissae, along -AX, the motive weights, and for ordinates, in the direction of AY, the squares of the velocities of the middle point of the immersed section. In JRAOTIOAi MECHANICS. 293 all the series so represeinted it was observed, that up to a certain velocity, all the points we're on a straight line OB, which cut the line of abscissae AX in front of the origin Pig. 143. at a point O, variable for each curve, which shows •that the abscissae or the resistance was in each case represented as for boats moved by horses, by an expression of the form R=KaA+KiAV«, V=the velocity of the middle submerged section of the paddle ; Ki and K^ constant coefficients. The immersed surface A of the paddles was determined from the number of paddles simultaneously submerged, in whole or in part, by calculating the sum of the immersed portions of the floats for many successive positions of the wheel, and taking the mean of the sums of the areas thus found, A thus represents the mean value of the total re- sistance of the paddles acting upon the water. The method of representation. Fig.. 143, gives the value of the constant Kb ; for the abscissae OA of the point O of the straight line expressmg the law of resistance, was that of the term KjA. The following values were found in this manner : 294 PEACTICAL MECHANICS. Dimension of paddles. 20 in number. Total submerged area. Constant resistance. Derived from the trace K^A. A square foot Feet. -65 by -65 •98 by 1-15 ■97 by -65; Square feet. . 1-4693 4-677 : 4-4241 lbs. •2867 ' •88219 , •86013 i •19513 . ^18862 •19442 Mean . . . •19272 The construction, Fig. 143.' also enables us to supply the value of the coefficient Ku of ith^' term assumed to be pro- portional to the square of the^elobity, since *he inclination of the straight line expressingi the law of- resistance is given by the expression k;a=' v R — KaA being the value of the abscissae of this straight line OB, diminished by OA ; and V^ being the values of the ordinates. Dividing in each case the values KiA, given by experiment, by the known surface A, we thus find the values of the coefficient Ki. (' Causes which alter the law of resistance assumed to be ex- pressed under the form KjjA+KiAV^. The student must not forget that Morin assumed the above form. Before giving the values of the coefficient Ki, furnished by the summary of the experiments, it is necessary that we shotdd point out a circumstance, which, in altering the con- ditions of the phenomena, exerts a considerable influence upon the results. In order that with different velocities and depths of immersion, the wheel and its floats may be in comparable conditions, it is necessary, as we have hitherto implicitly admitted, that the void formed by the paddles, which have driven before them the water upon PRAOTIGAL MECHANICS. 895 wMch they have acted, should be constantly replaced, so that the next paddle submerged may meet the same resist- ance. Now, in observing the motion of the return of the water into the void, we readily understand that the refilling must be accomplished by the flowing of the surface, as it were, over a dam at the sides, and that a certain time is re- quired for its operation. If, then, the wheel turns so rapidly that the void has no time to fill, the paddles no longer finding the same quantity of water to drive as in Fig. 144. 10 20 30 40 SO 60 70 80 90 100 110 120 Squares of Velocities. less velocities, the circumstances of the phenomena are changed, and accordingly the law of resistance must be modified. This change increasing , more and more with the velocity, it happens that the paddles meet a less amount of water, which may, at last appear as if the wheel turned fn air instead of water. All these effects are represented by the trace giving the results of the experiments, Fig. 144j relating to a series of experiments made with 20 paddles submerged •344 ft. 296 PBACTICAL MECHANICS. The curved line OT, Fig. 144, representing the law of resistance at first near O, is almost straight ; this straight- ness is prolonged up to a certain velocity, depending upon the depth of the immersion and the distance apart of the paddles ; but beyond this velocity it departs more and more, showing that the resistance no longer maintains its propor- tion with the square of the velocity. All these facts are highly important to steam navigation, for they show that it is necessary to establish between the depths of immersion of the paddles, their distance apart, and the velocity with which they are impelled, such relations that the water may always have time to fill up the voids, and that for each wheel, screw, or paddle, there is a limit of speed adopted to the best effect. To find the value of^^ of the second term of the expression for the law of resistance^ Having regard to the circumstances pointed out, and consequently restricting the law of the resistance within the limits of his ability to verify them, Morin makes known the results relating to experiments in which the velocity of the wheel and the spaces of the paddles allowed a complete return of the water. The values of the coefficient Ki, furnished by the experi- ments, were as follows: Values of the coefficient Kj of the formula E=A(K2+KiV). Number and dimenBlons of the paddles. ' • Values of K.. 10 paddles 0-33 ft. by 0-33 ft 1-990 2-0748 Wheel 3-31 ft. 10 ", 0-66 ft. 1,' 0-66 ft 1-9319 in J. 2-0794 diameter. 5 „ 0*98 ft. „ 1'15 ft 2-2365 10 „ 1-97 ft. „ 0*66 ft 1-9928 2-2460 Great wheel ■" 8-567 ft. ;• in diameter. . < (1-659 8 paddles of 2-29 ft. submerged ... ■{ 1-325 (0-937 General mean 2-226» 2-1698 2-4078 2-1355 . ..-:- ■ i PEACTIOAL MECHANICS. 297 The general mean does not differ over '-^ from the partial results, and we see that when the spaces of the paddles were within the indicated limits, that the effect exerted by wheels with plane floats upon the axle may be represented by the formula R = A [•192724-2-13559V], A being the mean of the surfaces of paddles simulta^ neously submerged at rest, V the absolute velocity of the wheel. Case where the wheel tumedin running water. — ^The wheel which had served for the above experiments having been S laced in a small wooden canal, 3'7 ft. wide by 2 ft. eep, the same course was taken to ascertain the law of resistance. Without going into imnecessary details, Morin states that the results of these new experiments were also represented, with sufficient exactness for practice by the same formula, by adding or subtracting tne velocity v of the current to or from that of V, so that the general ex- pression of effort exerted upon still or running water by the paddles of wheels with plane floats will be R = A [0-1927 + 2-0785 (V±«y], in taking for the coefficient of the second term a number which conforms best to all the experiments. Injhtence of the presence of a boat near the wheels, — ^The experiments in question were made upon isolated wheels, and it was proper to ascertain whether the presence of a boat near the wheel would exert any influence upon the in- tensity and law of the resistance. For this purpose, the experimenter placed near the wheel, at a distance of 0*13ft., parallel to the exterior vertical plane of the floats of the wheel, 8*567 ft. in diameter, a boat submerged an equal depth with the floats, and made two sets of experiments, with depths of immersion of 1*325 ft. and 0"84 ft., to compare the results with those of the series niade in the case when the wheel was isolated, and its paddles immersed 1-325 ft. and 0*937 ft. The restJts of these esroeriments seem to show that by reason of the obstacle which the presence of the boat 298 PRACTICAL MECHANICS. opposed to the return of water into the void formed by the float, the resistance diminished somewhat, but so small a quantity that it may come within the limits of the errors of observation. In fact, we found At the depth of immersion of 1-325 ft. without boat K;i=2-1&98 „ „ „ „ with boat Ki=2-1413 At the depth of immersion of 0-937 ft. without boat Ki=2-4078 „ „ „ 0-84 ft. with boat Ki=2-1580 We see, then, that the preceding formula deriypd from a summary of the experiments may be still applied to tlie case where the wheel is placed at the side of a steamboat. Application to the wheels of steamboats. — The formula of the resistance experienced, and the effect transmitted by the paddles of a wheel turning in water being R=KiAU2 = 2•13559AU^ when the axis of this wheel has no motion of translation, it is clear that if this axis is borne upon a steamboat going with a velocity V, the paddles will only impinge upon the water with a velocity U— V, and that in this case the formula expressing the resistance experienced by the floats will be E=KiA(U-V)» in still water, and, finally, that if the boat bearing the wheel ascends or descends a stream running with a velo- city V, the expression of resistance will be R=KiA(U— V— y)" on the ascent, E=KiA(U— V+«)' on the descent. If we examine particularly the case of navigation in still water, the work or this resistance, or that of the machine moving the wheel in 1", will be RU = Ki A (U - V)' U units of work, and if we express in horse powers of 550 units of work a second, the effective force of the motor wiU be R-XT „ KiA(U-V)' 550 ~^~ 550 PRACTICAL MECHANICS. 299 An observation of existing constructions ■will allow us to judge whether the value of the coefficient Kj derived from the experiments above reported, agrees with the observed facts of navigation. Indsed, we have for each boat the dimensions of the wheels ^md floats, and the number of the latter, from which We can deduce the submerged surface of the paddles. Observation gives us the velocity, U, of the paddles, which, by reason of their small height compared to the radius of the wheels, may be regarded as the point of appli- cation of the resistance, as well as the velocity, V, of the boat ; and if we introduce the value of Ki=2'13559, de- rived from our experiments, the above formula shpuld mve the effective force of the machine, such as observation nas furnished. Direct experiments, made by hauling upon a fixed point, in giving the effort exerted and transmitted by the paddles to set the boat in motion, enable us to verify directly the formula E=2-13559 AU^ by introducing the particular data of each case. In making this comparison upon the steamers the Sphinx, the Mentor of 160 nominal horse power, the Mid&e, and the Viloce of 220 horse power, for which the dimensions and different velocities are given by M. Campaignac, in his work upon steam navigation, we have the following data and results : Names of Horse power of each of f paddles ged for heel. A. y of the of the u. 0} steamers. the two engines. o 1 ie ■f^s •s . N. Area each > a g. II nominal effective sq. ft. ft. • ft. Sphinx ' Mentor 80 80-851 41-327 19-993 15-190 ,2-3074 80 80-854 37-136 20-862 15-528 1-9897 M^d^e 110 110-380 56-371 20-823 16-201 2-4318 Veloce 110 111-350 43-928 20-948 15-948 2-2879 2-3543 300 PEAOTIOAL MECHANICS. It should be observed that the value of the whole simul- taneously submerged area of the paddles was determined by tracings, and on the supposition of the vertical floats being entirely submerged a little below the surface, but probably less than it was in reality ; so that the values of Ki are undoubtedly greater than they should be. It isiiot, then, surprising that the mean of these values surpasses those derived from the experiments superintended by General Morin. PBEFOEMANCE OF PADDLE-WHEEL, STEAMEES. Xiet a = area of all the floats immersed, in square feet ; J=the immersed angle of the paddle-wheels, at the centre of pressure ; c= radius from the centre of motion to the centre of pressure of the floats ; d=: acting area in square feet. d={a) multiplied by the natural cosine of ^* JEx. — Suppose the area of all the immersed paddles, a= 312 square feet, the angle 6=48°; what is the acting area? 48° -g-=16°; Natural cosine=-9612617, 312 X •9612617=299-9136504 square feet, 300 square feet acting area, very nearly. Put e= the length of the load line of the vessel, in feet ; /= the greatest immersed section, in square feet ; =the number of revolutions a minute. PliACTIOAL MECHANIOS- 301 EXAMPLES. Ques. 88. Required the area of resistance of a vessel when e=2Q0 ft.=the length of the load line; the greatest immersed section 500 square feet=/; the displacement of the vessel 2500 t.ons=7). ? the vessel 2500 tons=/i 35A 35 X 2500 = ■67. fe - 500x260" Opposite '67 in the following Table, k will be found= 1*75, the coefficient of the vessel. y /"* / f500V T»'««.y=V/ipi7.=V 500 + 1-76(260)'= the area of resistance requireij. =32 square feet, m. *. m. k. •50 rsi 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 76 113 ■56 1-66 77 1-05 ■57 1-71 78 ■978 ■58 1-76 79 •903 '■59 1-83 SO ■835 ■60 1-89 81 ■763 ■01 1-93 82 ■691 ■62 1-98 83 ■625 •63 201 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 Ques. 89. Required the resistance of the vessel in pounds =i, when she is running at the rate of 10 miles an 302 PBAOTIOAL MECHANICS. ho\ir=j; the area of resistance=40 ieet=g, found as in the last example. The succeeding formula is often employed : Z=4-22><5'/=4-22x 40x100=16880 lbs., the resistance of the vessel iu pounds. The resistance in pounds may be found by the following formula nearer the truth : Z=4-3^j**^^ ; which, when put in dual logarithmic form, be- comes \,j= 1,10 = 230258509, 3 4) 690775527 172693882, (4-3x40=172.) 54232429 ,= | , (1-72) 226926311, 230258509 , , '3332198'= |, (-96722708) .'. Z=9672-3, the &st i-ule gave the resistaiice=16880 lbs. Ques. 90. Enquired, the resistance of a vessel in lbs. (Z), when she is running at the rate of 11-2 miles an hour (j) ; e= 200 feet; /=400 square feet; the displacement of the vessel=2000tons(A)? _35 A 35x2000 "*~ fe -400 X 200- '^^^^ From the table for »ra=-875 the value of k is found to be= '353, the coefficient of the vessel. J /* I 400' ^-M7fK?=M400+:35r(20o)^=66-6 square feet; Then, according to the second rule, the resistance in lbs.= Z=4-3y/-". 4-3^=4-3 x66-6=286-38, dual log.=565731962, PEAOTIOAL MEOHANIOS. 303 Dual logarithm of^'= | , (11-2) = 241591377, 241591377, x (1-75) = 422784910, I , (286-38) = 565731962 , 988516872, Dual log. of 10* 921034037 , Dual log. of 1-96369603 = 67482845, .-. i= 19637 lbs. nearly. The dual logs, may be instantly calculated or found by merely inspectmg the author's tables of dual logarithms, recently published. Ques. 91. Required the horse power i, when the speed =11 knots an hour=^'; and the area of resistance=33 square feet=^. i=J-^= — gg — = 499^ horse power. The horse power may be found with greater accuracy by the succeeding logarithmic expression : a ?^-" i=^^^5— =the horse power ; or oo J,0 = 2-75l,O')+l,O/)-i,(88). L^=?=i=-375. 88 8 88 l,(j') = 239789527, Mult. ^'^■^^ 2^ by I 659421199, I, (-375)= '98082925 561338274, I 460517018 ,= 1.(10") 100821256, 100821256=the dual logarithm of 2-7406977. .-. The horse power=274-07. (t). 304 PBACTICAL MECHANICS. 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 in the other. 88 ' put into a logarithmic form, gives the latter rule. Experi- ment has not yet settled this point in steam navigation, simple as it is ; some employ the square of the velocity, others the cube, and a third party the 2f power. Govern- ment experiments, both in this country and elsewhere, are generally Government jobs. Engineers are selected to superintend them upon the principle 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 I. To find the slip of screw propellers or paddle-wheels, when the acting area and area of resistance are given. If the area of resistance=49ft.=^; and the acting area =225it.=d; required the slip =w. y/F' 343 _ 343 PBOBLEM II. To calculate the power and find the properties of paddle- wheels and screw-propellers. The pitch of a propeller =33 ft., and makes 48 revolu- tions a minute, the slip ="35. What is the speed of the vessel ? ^^11^(1— 35) = 11-7 knots. Put o=the pitch; ^=the revolutions a minute ; and //= the slip ; then generally? PBACTICAL MEOHAKIOS. 305 • „-- i5i_. •' ^-ox(l-«)' .-. W=l ^-^ op Let the speed be 11"7 knots the hour =_;' ; the slip='35= n; the pitch 33 ft.=o; reqiured the number of revolu- tions=p. p= — 7^ — N = 5B — n — Sc\ = ^8, the number of revolutions ^ ox(l—n) 33x(l— -35) ' a nunute. Required the slip, when the speed is 11'7 knots per hour, the revolutions 48 per minute, and the pitch=33 ft. , 88j , 88 X 11-7 „, "=^"^=^"33x48 -■^^- PROBLEM III. Any four of the five following quantities being given, to find the fifth, namely, the radius of a paddle-wheel from the centre of motion to the centre of pressure of the floats =c; the slip=M; the immersed angle of the paddle-wheel at the centre of pressure=6 ; the statute miles per hour=_/; and the revolutions a minute=p. From the general expression . cp COS. ib „ . any one of the quantities is easily deduced, for ^ (1— m)ccos.^' " pccos.\h' Suppose the immersed angle of the paddle-wheel at the centre of pressure = 54° 33'=6; the slip=-38=n; the re- 306 PBACTIOAL MECHANICS. volutions per minute = 16 =^ ; and the radius of the paddle- wheel from the centre of motion to the centre of pressure of the floats=21 ft. ; required the speed=j. One-third of 54° 33' =18° 11', the natural cosine of which is = -9500629; for practical purposes '95 will be sufficiently near the truth, ^=££^*(l_„)=?l><^i±^l-38) = J4-136 imlesan hour. Again, let the immersed angle =54° 33'; the revolutions 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 ft. ; required the slip. , 14x14-136 ■ „- „- ,, . , ,. » = 1 - t:; — K^ — TTe = 1 — -62 = -38, the required shp. 16 X 21 X -95 THE nroioATOE. The Indicator is a simple and beautiful little instrument, by which the power of steam can be ascertained with great accuracy, when a properly constructed instrument is used ; but we should know who employs it, and for what purpose it is employed. It can be made to indicate what is false as well as what is true. With very little cookery it may be made to show that a good' engine is a bad one, and the con- trary. There is a class of engineers, which shall be name- less, who make great use of this little instrument in cooking up their reports. Eichards's steam-engine indicator, al- though not faultless, is the best in use. The nature of this instrument is easily imderstood by referring to Fig. 145. A is a small cylinder, B B open at top, E F fitted with a piston, and communicating with the lower part of the main steam cyUtader by a pipe T, and stop-cock. PBAOTIOAL MECHANICS. Fig. 146. 307 The piston, H, is pressed down into the cylinder by a nicely adjusted spiral spring, S S S, and a pencil, C D, is fixed to the piston-rod, HO; Q isa roller round which a piece of paper is wound ; on the axis of this roller is fitted a pulley, P, connected by a string, P E, with some of the moving parts of the engine,. The roller, Q, is also fitted with a ; spring, like the mainspring of a watch, in such a manner that, after being pulled round in one direction by the motion of the engine communicated through the string, K P, it IS riiade to recoil by the spring. If we suppose the ■ x2 308 PEAOTICAL MECHANICS. stop-cock in the pipe, T, closed, the piston, being pressed on Dy the spring S S S and the atmosphere, will remain at the bottom of the cylinder, B B ; and the pencD, being sta- tionary at its lowest point, B, while the roller is made to rotate backwards and forwards, will describe a line on the paper which would appear straight on its being unfolded from the roller, D Q. But if, while the roller continues its motion, the stop-cock be opened, then the piston will be subjected to the pressure of steam in the main steam cylinder, and will be forced upwards in opposition to the pressure of the spring S S S and of the atmosphere ; and the pencil will trace a Ime on the paper varying in height as the piston H of the indicator rises and falls. But, further, if the spiral spring be so adjusted that we know exactly how many pounds will compress it an inch, and if we know the area of the piston H, we can measure the amount of pressure on it by the height to which the pencil is raised above the neutral line, B, where it remains when subjected to no upward pressure of the steam in the direction of the arrow ; and thus the posi- tion of the pencil on the paper, or the mark left by it at any point, furnishes the measure of the pressure on the main piston of the engine at the corresponding point of its stroke. On unfolding the paper from the roller Q, we should find a figure described on it by the pencil resembling y, 4, k, p, Fig. 146, which, when properly analysed, ^ves the means of Fig. 146. PKACTICAL MECHANICS. 309 reckoning the varpng pressures on the piston, and often points out defects in some of the adjustments, and suggests modes of remedying them. If we suppose that the area of the indicator-piston is one square inch, and that the spring is adjusted so that it requires a force of 10 lbs. to compress it 1 inch in length, or 1 lb. to compress it ^^^th of an inch, we can form a scale of tenths of inches, and apply it to the indicator diagram at a number of points, a, b, e, &c. (Fig. 147), equally distant, and measure off the lengths of ordinates, al, J 2, &c., drawn through these points, and thus estimate the pressures acting on the piston of the indi- cator-cylinder at equidistant points of the stroke through which the paper is made to travel. These pressures corre- spond exactly with those to which the main piston of the engine has been subjected during its stroke, because the small cylinder, E B, of the indicator communicates freely with the cylinder of the engine. If we suppose the indi- cator to be fixed- at the top of the steam cylinder, the upper part of the curve, 1, 2, 3, 4, 5, &c., is that traced during the descent of the piston when the steam is pressing on it. The lower part of me curve, x, t, v, y, is that traced during the ascent of the piston when the steam is escaping from the cylinder. Were the indicator fixed to the oottom of Fig. 147. 310 PBAOTICAL MECHANICS. the cylinder, we should get corresponding curves for the steam pressures there. G-enerally, when the slide gear is properly adjusted, these figures are very nearly alike, and, if so, the upper part of the curve, y, 1, 2, 3, 4, 5, 6, 7, x, may be taken as that traced by the active pressure either above or below the piston, while the lower part of the curve, X, t, V, q, p, 0, n, m, y, may be taken as representing the corresponomg resistance of steam during its egress from the cylinder. Now, as the total height, c 3, of any ordinate measures the total pressure on one side of the piston when it is at the point of its stroke corresponding to c, and as the part c of the same ordinate represents the resisting pressure on the piston at the same point of its stroke, the difference o 3, or the part of the ordinate intercepted be- tween the upper and lower boundaries of the curve, mear- sures the effective pressure on the piston clear oi all resist- ances. The same applies to all the ordinates, and, as we may suppose the whme curved space made up of numerous equal, narrow, vertical strips, each measured in height by an ordinate, we reckon the area of the figure contained within the curve as an expression of the power developed by the piston during its stroke ; or, having taken a consi- derable number of pressure ordinates, and found the average, we consider this the mean effective pressure on the piston. For example, the average of those marked on Fig. 147, found by adding them into one sum and dividing by their number, is ^^=20-7 lbs. the mean pressure on every square inch of the piston. In taking the average it is better to take the first and last ordi- nates, m 1, and i 7, at a distance from the respective ends of the stroke, xy=hali oi ab,lc, c d, &c. If we suppose the engine from which Fig. 147 was taken to have a cy- linder 20 in. diameter, 2 ft. 6 in. stroke, making 40 revolu- tions a minute, the power of the engine is readily'calcu- lated. PEAOTICAL MECHANICS. 311 Area of piston=20» x •7854=314-16 square inches ; 314-16 X 20-7=6503'112 mean pressure on the piston. 6503-112 5 ft. double stroke 32515-560 40 r evolutiona a minute 1500622-400 1500622-24 ,^ , , QQQQQ — =45-5 horse power, nearly. CALCULATION RESPECTING THE SCEEW PBOPELLEB KEPEESENTED IN PIG. 148. This propeller has a particular expanding pitch. Put ?= J the pitch at the periphery=DE. r=^ the pitch at the hub = RS. £=HB, the assumed slip in a fraction of q. :. q=r-\-s. Put <= extreme radius=OI. M= diameter of the sorew=2i=FI. 17= pitch of the propeller at the periphery =4^. w=angle of the blades at the periphery=KQT. «;i= angle of the blades at the hub. JM=a;=;J pitch of the propeller at the centre of effort of the blades. BJ=y=-725f, the radius at the centre of effort, G=OG= QM. The actual pitch at the centre of effort = ix. «=horse power required to drive the propeller and a revolutions a minute. ^= number of blades. y= length of blade parallel with the centre line=UV. d=the breadth of the propeller blades over the edge between the corners XZ; c=the circular arc in the angle e = WOY. G=the projected angle of the blades. A=the true inclined surface of the blades. Ai=the projected area of the blades. A2=the acting area of the propeller. Let cr=the circumference of a circle diameter =1. 312 PBAOTIOAL MECHANICS. Fig. 148. PBAOTIOAL MECHANICS. 313 cot. w;= JL = _X = ?J., the cotangent of the angle of the ■tu ir2t vtt blades at the periphery. v=iru cot. w, the pitch of the propeller at the periphery. 360 V Bsrllr- 0= "^^y . (p^ 12960V 8 is the hypothennse, and y is the base of a right-angled plane triangle. 8=KC measured over WY, it is a projection the same as UV ^ _ -7854M'tf;3 ' 360 4«^(a-|-y) -^ 9 e= (y scos. w+ -^ 480000 ^' FBOBLEM IV. The pitch of a propeller= 33 ft., and makes -48 revolu- tions a minute, the slip= *35. What is the speed of the vessel ? ^^'<*?(l_.35)=ll-7 knots. 88 By the introduction of other screws the slip might be reduced from 35 to 15 per cent., the speed,. when the slip is 15 per cent., will be 314 PEAOTICAL MECHANIOS. 33x48 „, , __ V^i J.Uy XU IF ilUU"! Pato= V= n- 3= :tlie pitch ; =the revolutions a minute ; :the sUp; and the knots the hour ; then generally, _ 88^ ^ ox(I-m)' , 88i . "-l-ox^,' 0- 88i , p (1 —n) A ship fitted up with a screw propeller made 13 knots an hour ; the revolutions were 60 a minute, and the pitch 30 ft. What was the slip I „_i_i?i = 1 _88xl8-.i2, or 12 per cent. "-^ oy.p 30x60 ' ^ PROBLEM V. Given the diameter of a propeller =12 ft.=w; and the angle at the periphery= 51° .. 33'=«;; to find the pitch =v. cot. 51° . . 33'=-7940121 e=* u cot. w=3-1416 x 12 x •794=29-93 ft. FBOBIiEM VI. Given the diameter=21ft.=M; the length of the blade parallel with the centre line=6 ft.=y ; the sup 40 per cent., or S=40 ; the angle of the blades at the periphei7=63° . . 43/ =w; to find me horse power necessary to drive this screw 68 revolutions a minute=a. PEACTIOAL MEOHANIOS. 315 Cosine 63". . 42' = -4430712 ^^21«x68» ^^ •443+l) = 7124. 480000 ^ 9 '^ 6X-4X-443+ i=M7431111 y 21x68 = 1428. 1,(1-428)^ 35627490, 3^ I 106883470, 1,(1-17431111)= 16068179. , 122951649, \ , (-48) '73396917 subtract 196348566, , '230258509 j,(^) I, (-71241162)= '33909943 /. «:=7124-1162, as before. 1-428 mult, by 1000. =1428. -480000 mult, by 1000000=480000. (1000)»=10» and -^ x 10=10*. Hence the decimal point had to be moved four places to the right, which brought -71241162 to 7124-1162. As a second example, suppose m= 12 feet; y=5 feet; w=51° 30' ; «=38 per cent., or s=-38 ; what power is re- quired to drive this propeller 40 revolutions a minute ? z=l^^^ (5 X -38 X -5373 x i) = 509-4. 480000 ^ '^ Calculation by Dual Logarithm. 15x40 = 600. 5 X -38 X -5373+^=1-13198111. 316 PBAOTIOAL MECHANICS. {,(•6)= '51082562 3 '153247686 '153247686 \ , (1-13198111)= 12396933, 187603067 ar. co. subtract | , ('48) '73396917 126603083 I , (-5093915) '67453836 .'. z= 509*4, found by adding three dual numbers together. FBOBLEM VII. Given the diameter of a propeller =21 feet=M; the angle of the blades at the periphery =63° . . 42'=w; the length 7=6 feet; the slip »= '40 or 40 per cent.; four engines drive this screw of 7124 horse power =«; how many revo- lutions, a, will this propeller make in a minute 1 u= — i— -1^=68 revolutions. M (. (^'SCOS. w + i) ) yscos. w+|.=6x-4x-443+i=l-1743. By Dual Logarithms. ' • , (-480000) = '73396917 T , , - , (.7124) = '33909943 ) *'^"- .,(1-17431) = 16068179, subtract. '123375039 , .„.. '460517019 * ' ^^^^■■> 3)'583892058 , ,„.. '194630686 ♦'A^^^ 74193735. '268824421 230258509, \ , (10.) I , (-68000231)= '38565912 .-, a=68. FBOBLEM VIII. Given, the diameter of the propeller=13feet=M; the angle of the blades at the periphery = 59° 46' = w ; the angle FBAOTICAI. MECHANICS. 317 of the blades at the hub = 11° 48'= mi ; the diameter of the hub = 1-25 feet=twice QS; (see Figure 148) to find the pitch at the centre of effort. Cotangent of 59° 46'= -582793 Cotangent of 11° 48'=4-786730 The pitch at the periphery = 13x-583x3-1416 = 23-8 {t. = v The pitch at the hab= 1-25 X 4-787 X 3-1416 = 18-8 fl.=». Put vi for the pitch at the centre of effort, ui the diameter at the centre of effort, and Uz the diameter at the hub. (v—Vj) : (vi—Vi) : : («-«,) : (•725«-Mj) , (v—Vs) (•725m— Mj) U—Uj „,=78-8+^ii^^=22-3 Becapitulation. t>=23-8ft. M=13fli. »i=22-3ft. Mi=-725tt=9-426fli. »j=18-8 ft. Mj=l-25 ft. FBOBLEM IX. Given, the diameter of a propeller = 13*5 feet=a; and the angle io=65°, to find the actmg area of the propeller. 5m» . *~ 2 V«^H^^ ' bat t>'='!r'M'cot.^ w, hence . 5«» 2 but more concisely written, -00 -999; -99; -9; 0; -9; -99; -999; 1, decreasing in magnitude from right to left. This scale of bases approaches 1, but cannot be ^eater than 1. ' PBACTIGAL MEOHANIOS. 343 An example will make clear anything that may seem too abstract in the preceding definitions. The diameter of the earth through the poles is said to be 4 1706091*152 ft., which is a conventional form of e^ressing 4 times (10000000-) + once (1000000-)+ 7 times (10000-) + 6 times (1000-)+nine times (10-)+l-+ ^ -i- ^ + ,^l_, by the notation of common arithmetic, which, according to the ordinary method agreed upon to express powers, be- comes 4(10)T+(10)6+7(10)* + 6(10)»+9(10)i + l + (10)-i + 5(10)-'' + 2(10)-». In this notation of common arithmetic 4, 1, 7, &c., are termed digits, and do not exceed 9. In dual arithmetic the powers of the dual bases are only registered. Thus 41706091-152 is equal to (-99999)X-999999)'C9999999)'(-99999999)'(1 + 1)"(1-01)* (l-OOl)S when multiplied by 10^. The bases being omitted, this dual number is written '0'0'0'0'1'3'3'6 f 1072=" 1 0,4,2,0,0,0,0,0,^ ; (A). The natural number 41706091'152 is also equal to the dual number (10)^(2)=' 1 0,4,1,9,8,6,9,6,; it is likewise equal to the dual number '8'3'r4'6'8'l'0f (10)«; and to an immense number of other dual numbers. Referring to the form (A), '1'3'3'6; 0,4,2, &c.,are called dual digits, and express the powers of the bases involved, and, unlike the digits of ordinary arithmetic, may be greater or less than 9. For exainple, the common number 41706091-152 is equal to the dual number '0'0'1'5'7'0'8'4 f 107 \ 15,0,0,0,0,0,0,0, The zero between | and 4, in (A), shows that no power of I'l is employed; the same remark appUes to zero and other bases. A dual number of positive dual digits has always an exact value in common numbers when no contractions are employed in the reduction. 344 PBACTIOAL MECHANIoe; When eight positions to the right and eight to the left of the signs f | , counting from left to right in both cases^ are occupied by ciphers or other digits^ the sign | being placed before the eight ascending digits and f after the eight descending, yet, with respect to range, the dual nuni- ber is said to be one of eight digits, although sixteen posi- tions, and other positions between the signs f and \ , may be occupied. If one of the signs f or j is omitted, the positions at-- tached to the other are supposed to be occupied!^ by ciphers. ft" When the last dual digit and all that follow are re- jected, and. when the last digit is 5, 6, 7, 8, or 9, the digit preceding may be counted one more, as in decimal arithmetic. For most practical purposes common arithmetical results are required true, to not more than seven places of figures. To obtain this degree of accuracy, eight consecutive dual digits must be employed. In making calculations the al- lowances specified must be attended to. A dual number is easily transformed into another, all of whose digits being reduced to ciphers, except the last. The transformation of a dual number of eight digits into another, whose first seven digits are ciphers, is termed reducing a dual number to the eight position. A dual number reduced to the eight position is called a dual logarithm. 2-= I 7,2,6,0,7,8,2,6,= | 0,0,0,0,0,0,0,69314718, = \ 869314718. In practice the 8 is omitted, and the expression is written 2-= \ 69314718, which represents (1-00000001)'"'""^ Then 69314718 is termed the dual logarithm of 2-, written I, (2-) = 69314718. Then the dual logarithm of 41706091-152 is equal to the whole number 1615789463, or \ , (41706091-152) = 1615789463, . The dual logarithms of common numbers are. easily found, as well as the common numbers corresponding to dual loga-r rithms, without the use of tables. PBAOTICAL MECHANICS'. 345 Dotation. Ascending Bbanch. The notation, although new, is easily remembered, from its symmetry, compactness, and uniformity. 1 is represented by | . One Decimal. First Position, !•! is represented by 1 1, (l-iy n » '2, (1-1)' ,, » +3, &c. &c. Two Deouials. Second Position. (1-01) is represented by ^ 0, 1, or ^ ^1, (1-01)^ (i-oi)» &c. Thbee Decimals. (1-001) is represented by (1-001)'' (1-G02)» &c. (l-l)'(l-Ol) is represented by I 5, 1, (1-1)7(1-01)" „ „ '7,2, (1-1)8(1-01)*(1-001)* „ „ |3, 4, 5, (1-001)«(1-00001)''(1-00000001)* is expressed by 10,0,6,0,2,0,0,8, \ 0,0, in the first and second positions, indicates that no power of (1-1) or of (I'Ol) is involved. The cipher in the fourth position indicates that no power of 1-0001 is involved; the same may be said of the other positions. Notation. Descending Branch. In this branch the arrow points up, and the comma is to the left of the di^t and above, while in the ascending branch the arrow points down, and the comma is to the right of the digit and below, 1 is represented by One decimal (-9) „ „ '1 f in the first position. X-9)» .. " '2 0, 2, orw% 0, 3, or \ "S, &c. Third Position. w 0, 0, 1, or J, n, . , 0, 0, 2, or \ '2, . 0, 0, 3, or I % &c. (-9)' Ac. '3> &c.' » 346 PBACTIOAL MEOHANIOS. Two Decimals. Sbookd Position. (■99) is represented by 'O'l f or 'l^ ' • (•99)2 „ „ '0'2-or'22- (•99)» ' „ „ '0'3 .or'Sj- , (•99)* „ „ '0'4jor'4a- &c. &c. Theeb Decimals. Third Position. (-999) is represented by 'O'O'l | or 'Ig (•999)^ „ „ '0'0'2for'23 (•999)'. „ „ '0'0'3Jor'3g &c. &o. ^" In both branches, if there be n decimals in any base, its powers or dual digits are placed in the nth position. ^ (9)»(^99)'' is written '3'2 \ (9)7099)'' „ '7'5 1 (•999)'(^999999)2(-99999999)6 is written '0'0'3'0'0'2'0'6 | A cipher being in the first and also in the second position shows that no power of "9 or '99 is employed ; the same may be said of other positions occupied by ciphers. '0'0'3'0'0'2'0'6' f may be written '3'0'0'2'0'63 f = •3gt'26t'68| In the author's works, " Dual Arithmetic : a New Art," Parts I. and II., and in the " Young Dual Arithmetician," without the use of tables, in a variety of ways, and under different circumstances, it is shown, by easy, independent, and direct processes, how any two of the three corresponding numbers i (Natueal wumbee) ; (Dual number) ; (Dual logarithm) ; miieht be found, the remaining one being given. Any one of these convertible numbers being ^ven, the other two may be also found by employingTables I, and II. (see "Young Dual Arithmetician"). Table I. is of the ascending branch, in which the natural numbers range from 1- to 2^99161136. Table II. is of the descending branch, of w;liich the na- tural numbers range from PBAOTIOAL MEOHANlOSi 347 •29916114 to 1. Hence, when these tables are emplpjned, the powers of the base 1 + 1 or 2 are dispensed with, and only the powers of 10 retained. When operations are performed with dual numbers in their lowest terms, and tables used, it is not necessary that such tables should range beyond the natural numbers from 1-41421356 to 1, and from 1 to -70710678. (See "Tables of Dual Logarithms and of Angular Magni- tudes and Trigonometrical Lines.") CONSTETJOTION OF TABLES OF DUAL LOGARITHMS. In constructing Tables of dual logarithms, no arithme- tical operations beyond those of addition and subtraction are necessary. , To find the natural numbers corresponding to any dual number of one digit. Adding, Ascending branch. ji,=i-i 11 Jr2, = l-21 121 I 3,=1-331 1331 i4,=l-4641 14641 15, = 1-61051 161051 I, 6, = 1-771561 1771561 J, 7, = 1-9487171 19487171 J, 8, = 2-14358881 ^ 214358881 j 9, =2-357947691 &c. '2 j Subtracting, , Descending branch. '1{ =-9 _9^ •81 81 '3'\ =-729 729 '4j =-6561 6461 '5f =-59049 59049 '6 I =-531441 631441 '7 j =-4782969 4782969 '8 \ =-43046721 &c. 348 tEACTICAL MECHANICS* The above tables are formed by simple addition and sub- traction, each line being obtained from the preceding one by commencing a figure further to the right. To find the natural numbers corresponding to dual num- bers of two digits. Ascending, adding. j 0,1 = 1-011 |lO|l 1 0,2=1-0 I 0,3=1-0 {-0,4=1-0 1 20|1 30 10,5=1-0 510100 5 &C. = <&0. Descending, subtracting. 'O'l I =-99 99 01 98|01 '0'2 I =-98 '0'3 I =-97 99 '0'4f =-96 05 96 01 96 05 96 01 '0'5 } =-95 09 90 04 99 Here commencing with \ 0, 1, = 1-01 and 'O'l \ ="99, we pursue the same process as before, only placing the figures to be added to or subtracted from the preceding lines two figures further to the right, instead of one, neglecting sSl figures after the eight decimal place. _ When the first dual digit is greater than one, we start with the value of that digit. Thus : , Ascending branch. ^6,=1 I 6,1,=1 I 6,2,=1 I 6,3,=1 I 6,4=1 i 6,5,=1 1 5611 1 7 8 61 92 84 1 &c.=&c. Descending branch. '6 I =-53|14|41 53|l4 '6'1 1 =-52 41 59 27 '6'2 I =-52 '6'3f =-51 32 65 67 57 '6'4f =-5105 00 10 51 05 00 '6'5f =-50 53 9510 &c.=&c. PRACTICAL MECHANICS. 349 In tne same way the common numbers corresponding to all dual numbers of two digits may be computed and tabu- lated. Ascending, Again, \ 0,0,1 =1-00|1 jlOOIl \ 0,0,2=1-00 200 100 1 200 1 0,0,3=1-00 {0,0,4=1-00 300 100 300 300 400 100 {0,0,5=1-00 501 100 600 400 000 501 \ 0,0,6=1-00 601 501 &o. = &c. Descending. '0'0'lf=-999 '0'0'2 j =-998 999 001 998 00 'O'O'S f =-997 003 997 00 00 '0'0'4 \ =-996 '0'0'5 \ =-995 096 996 009 995 00 01 99 01 '0'0'6 \ =-994 014 98 &c.=&c. Operating the same as before, only setting the line to be added or subtracted three figures back. Ascending. I 6,5,0=1-86 {6,5,1=1-86 { 6,5,2=1-86 192 186 379 186 565 186 841 193 034 379 413 565 {6,5,3=1-86 751978 &o.=&c. '6'5'0=|-505 395 505 '6'5'l=>(.-504 889 |504 10 40 70 '6'5'2=f -504 384 81 504 38 '6'5'3= j -503 880 43 &o.=&e. In this manner the common numbers of all dual num- bers of three digits can be found and tabulated. Ascending. {0,0,0,1=1-000 { 0,0,0,2=1-000 1 100011 1 2 2000 1000 {0,0,0,3=1-000 3000 3 &o.=&e. Descending. 'O'O'O'l^ = -999910000 |9999 '0'0'0'2 f =-9998 0001 9998 '0'0'0'3|=-9&97 0003 <&c. = &c. 350 PEAOTIOAIi MEOHANICS. Or the natural numbers corresponding to dual numbers of four digits are found as those of three digits, settiag the line to be added or subtracted four figures back. 16,5,2,1=1-865 Ascending. I 6,5,2,0=1-865|6541|3 1 186517 8407 1865 8 0272 8 1866 I 6,5,2,2s;l-866 I 6,5,2,3=1-866 213818 1866|2 16,5,2,4=1-866 4005 1866 I 6,5,2,5= 1-866 58714 Ac.=&c. Descending. '6'5'2'0 f =-5043 '6'5'2'1 \ =-5043 8481 5044 3437 5043 '6'5'2'2 { =-5042 '6'5'2'3 1 =-5042 '6'5'2'4 ^ =-5041 8394 5043 3351 5042 8309 5042 '6'5'2'5f =-5041 3267 &c.=«&c. In this way, by common addition and subtraction alone, the natural numbers corresponding to the dual numbers can be obtained throughout tables of both ascending and descending dual numbers. n NEXT TO COMPUTE THE CORRESPONDING DUAL LOGARITHMS. Ascending. The dual logarithm of 1 1, = 9531018, „ > 0,1, ' = 995033, „ ^0,0,1, = 99950, „ „ 1 0,0,0,1,= 10000, Descending. '10536052=dual log of '1 1 '1005034= „ „ 'O'lf '100050= „ „ 'O'O'll '10001= „ „ 'O'O'O'l^ For the calculation of these numbers, see "Dual Arith- metic," Part n., and the " Young Dual Arithmetician." From the above numbers all the dual logarithms of both branches can be found by simple addition. PBACTIOAL MECHANICS. 351 Ascending. log{l,= 9531018 9531018 log|2,=19062036 9531018 log I 3,=28593054 9531018 log I 4, =38124072 9531018 log I 5, =47655090 9531018 log I 6,=57186108 &c.=&o. Descending. '10536052=log of '1 f 10536052 '2f '21072104= 10536052 '31608156= 10536052 '42144208= 10536052 '31 '4 1 '52680260= „ '5 f 10536052 '63216312= „ '6 \ &o.=&c. Or the dual logarithms of all dual numbers of one digit in the first position may be found by the successive addi- tion of the number 9531018, for the ascending branch, and '10536052 for the descending. log 1 0,1,= 995033 995033 log I 0,2,= 1990066 995033 log I 0,3,= 2985099 995033 log I 0,4,= 3980132 &o. = &o. and log I 6, =57186108 995033 log I 6,1,=58181141 995033 log j 6,2,=69176174 995033 log {fe,3,= 60171207 995033 log|i5,4j=61166240 &c.=&o, Descending. '1005034 = log of O'lf 1005034 'ft'O '0'2| '2010068 = 1005034 '3015102= „ '0'3I 1005034 '4020136= „ '0'4^ &c.=&c. '63216312=logof'6'0* 1005034 '64221346= „ '6'1 1 1005034 '65226380= 1005034 '66231414= „ '«' 1005034 '67236448= „ 6'2\ 6'3| '6'4f Thus the dual logarithms of dual numbers of two digits are derived from those of one digit by the successive addi- 352 PBACTICAL MECHANICS'. tion of the number 995033, for the ascending, and '1005034 for the descending branch. Ascending, log I 0,0,1,= 99950, 99950, log 10,0,2,= 199900, 99950, log i 0,0,3,= 299850, 99950, log \ 0,0,4,= 399800, &c. = &c. but log I 6,4,0,=61166240, 99950, log i 6,4,1,=61266190, 99950, log \ 6,4,2,=61366140, 99950, log I 6,4,3, =61466090, &c. = &c. '0'0'3 \ Descending. '100050=log of 'O'O'l I 100050 '200100= „ '0'0'2 f 1 00050 '300150= „ 100050 '400200= „ '0'0'4 1 &c. = &c. '67236448 = log of '6'4'0 | 100050 '67336498= „ '6'4'1 j 100050 '67436548= „ •6'4'2 f 100050 '67536598= „ '6'4'3 'J' &c. = &c. Thus the dual logarithms of all dual numbers of three digits are derived from those of two digits by the successive addition of the numbers 99950, and '100050. Ascending. log 1 0,0,0,1,= log {0,0,0,2,= log {0,0,0,3,= log 1 0,0,0,4,= &c. = &c. 10000, 10000, 20000, 10000, 30000, 10000, 40000, but log I 6,4,3,0, = 61466090, 10000, log I 6,4,3,1, = 61476090, 10000, log I 6,4,3,2, = 61486090, 10000, log { 6,4,3,-3', = 61496090, &c.=&c. '10001 =log of '0'0'0'lf 10001 '20002= „ '0'0'0'2 \ 10001 '30003= „ 'O'O'O'S + 10001 '4004= „ '0'0'0'4 \ &c.=&c. '67536598= 10001 '67546599= 10001 '6'4'3'0 I '6'4'3'1 f '6'4'3'2 \ '67556600= 10001 '67566601= „ '6'4'3'3f &c.=&c. PBAOTICAL MECHANICS. 353 Or the dual logarithms of all dual numbers of four di^^ts may be derived from those pf three digits by the successive addition of the numbers 10000, and '10001. With the exception, therefore, of the calculation of the dual logarithms of j 1, ; 1 0,1 ; \ 0,0,1, ; \ 0,0,0,1, ; and of 1 f ; 'O'l \ ; 'O'O'l f ; 'O'O'O'l f , and of the dual loga- rithms of 10 and 2 — all of which can be verified by methods shown in the author's works on the art and science of dual arithmetic in ten minutes — every result in the tables of dual numbers, their corresponding natural num- bers, and dual logarithms, can be calculated, as here shown, by simple addition and subtraction, without the aid of any other arithmetical operation. Having said enough to induce a student to examine dual arithmetic ; in conclusion we propose to arouse the attention of accomplished mathematicians, in solving by a direct process the following question. In order that this solution may not be passed over lightly, we have selected a question that could not be solved by all the living mathematicians, and all who ever lived, with the help of their numerous theorems, without the aid of dual arithmetic. Question. Find a number which, when added to the circumference of a circle whose diameter = 10000, the sum and its common logarithm, each, is composed of the same conse- cutive digits? ff=3-14159265 Put a= 10000, and ??i=10*, and let a; be the required number. Then 10 lO' =a v+x .". (10™) = aw+x. ' 1 air+x ;= (lO") a iT-^x 2 a 354 PEAOTICAL MECHANICS. taking a *+a; root of both sides of this equation, we have 1 _J_ , — j— _ / 1 \ aV+x. 10" ~ \a *-f k/ Putting h for — and « for , the last equation be- 10™ comes ^°=5, the general solution of Tvhich is given in " Dual Arithmetic," Part II., pp. 98-100. We find «=15276'54179287965. INDEX; ESSENTIAL ELEMENTS OE PBACTICAL MECHANICS. Accelerated motion, when it be- comes unifonn, 13 Accumulated iroil:, 92 Achievement, great mathematical, 235 Agents, work of living, 8 Air, density of, 271 Air, motion of a tody in, 276 Air, to find the resistance of, 279 Analysis of Morin's experiments on the resistance of water to the mo- tion of paddle-wheels, 292 Animals, force of, 12 Archimedes' principle of sufficient reason, 165 Area of a figure made to represent a distance, 92 Area of floats, acting, 300 Atmosphere, resistance to a train of carriages, 66 Atwood's machine, 101 Axis, friction upon, 127 Axle and wheel, 27 Ball fired from a cannon, units of work in, 99 Ballistic pendulum, geometrical pro- blem, 142 Ballistic pendulum, mechanical pro- blem, 143 Beasts, work of, 9 Bent lever, 35 Black broad-faced figures and letter, what they signify, 1 Blast pipe, resistance of, 66 Bodies descending inclined planes, 126 Body, work accumulated in the falling of, 95 Boiler, evaporation of, 81 Boyle's law, 76 Bridge, Bellman's iron bridge, 205 Bridge, Menai tubular, 212 Bridge of S. H. Long, 189 Bridge, the McCallnm, 1 95 Bridge, the Burr truss, 187 Bridge, the Howe bridge, 190 Bridge, the Howe truss improved, 192 Bridge, the inflexible arched truss, 197 Bridges, requisites desirable in, 203 Bridges, rigidity, stifihess of, 200 Bridges, practical ex:amples of, 1 87 Cannon, pressure of exploded gun- powder, 128 Carriages, friction of, 100 Carriages descending incline(^ planes, 126 Catenary curve, 219 ' Centre of gravity defined, 37 Centre of gravity found geometrically, 246 Centre of gravity found by referring bodies to co-ordinate planes, 39 Centre of gravity in different bodies, 247 Centre of gravity of a number of bodies, 38 Centre of gravity, conservation of the motion of, 136 356 INDEX. Centre of gravity, units of work, re- moTing, 24 Centre of gyration of diflferent bodies, 108 to 115 Centre of pressure, 240 Centres of gravity, gyration, and oscillation, relation between, 161 Chain and suspension bridges, 219 - Chapter I. On the unit of work with or without reference to the unit of time, 1 Chapter II. On the work of living agents, 8 Chapter III. On the motion of bodies on inclined planes, and the raising of materials, 17 Chapter IV. The lever, wheel and axle, pulley, inclined plane, wedge, and screw, 27 Chapter V. Work of steam and the steam engine, 58 Chapter VI. Accumulated work, 92 Chapter VII. Equilibrium of pres- sures, thrusts and tensions, forces, &c., 162 Chapter VIII. Pressure of water and other fluids, 236 Chapter IX. Eesistance of fluids, mo- tion of paddle-wheels, screw-pro- pellers, &c., 283 Chapter X. Useful arithmetical pro- cesses, operations that may be for- gotten, and the application of prin- ciples often misunderstood by those who have neglected the study of arithmetic, 319 Chinese ingenious contrivance, 44 Circular plate, centre of gyration, 108 Circumjprence, centre of gyration, 108 Clearance of engine, 80 Coals, evaporative power of several kinds, 71 Cogged or toothed wheels, 41 Compound wheel and axle, Chinese, 44 Condenser, temperature of the vapour in, 73 Condensing steam, units of work de- veloped by, 73 Conservation of thequantitvof motion, 136 Convenient unit to measure by, 3 Crane, 40 Crane, power of, 43 Crank, position of, for maximum ve- locity, 129 Cube, pressure of water on the sides of, 238 Curved patlis, work in moving a body along, 18 Decimal fractions, 328 Decimal point defined, 4 Decimals, addition, subtraction, mul- tiplication, and division, 333 Decimals, important observation, 334 Decimals, important rule, 330 Diagonal rods, stretchers, 207 Diagram of forces, 181 Diagram, skeleton, 171 Diagram offerees, comparative lengths of the lines in, 186 Dick's anti-ftiotion cam-press, 66 Dual arithmetic applied to find the force of steam, 82 Dual arithmetic employed to calcu- late the power of steam, 120 Dual arithmetic with great ease, 221 Dual arithmetic, introduction to, 342 Dual logarithms applied to calculations of steam navigation, 303 Dual logarithms, calculations simpli- fied by the use of, 87 Dual logarithms, how to construct tables of, 347 ^ Duty of the steam engine, 66 Earth, pressure of, 265 Earth, work required to stop its re- volving on its axis, J.07 Elastic bodies, shock of, 145 Elastic force of steam found exactly by dual arithmetic, 82 Elastic force of steam found in units of work, 75 Embankment, 243 Embankment of equable strength, 264 Embankment, pressure of water on 239 Endless screw, 53 Engines, high pressure, 61 Engines, locomotive, 68 Equations, cubic and higher, 264 Equations of the catenary curve. Pro- INDEX. 357 blems that defied the skill of mathe- maticians, solved by, 220 Equilibrium of pressures, thrusts, and tensions, forces, &c., 162 Krror in the hypothesis respecting in- stantaneous transmission of motion, 143 Falling bodies near the surface of the earth, 95 False balance, 30 Fire-en^e, 158 Flood-gate, 244 Fluid, resistance of, 275 Fluids, efflux of, 155 Fluids, resistance of, 153 Fly-wheel, 96 Fly-wheel, dimensions of, 130 Fly-wheel, weight of, 132 Fly-wheel, friction of axis, 105 Force of animals, 12 Force, accelerating, 279 Forces, diagram of, 180 Forge hammer, work of, 7 Fractions, addition of vulgar, 321 Fractions, division of vulgar, 321 Fractions, multiplication of vulgar, 321 Friction of common roads, 13 Fractions, subtraction of vulgar, 322 Fractions, vulgar, 320 Frames, loaded, 185 Friction of rails, work taken up by, 20 Funicular polygon, 177 Geometrical proportion (useful), VIS Graphical construction, Tensions de- termined by, 179 Gravity, force of, near the surface of the earth, 93 Gravity, work due to, 19 Gravitation, the law of, 278 Grindstone, question respecting, 106 Gunpowder, force of, 148 Gyration, centre of, 105 Hammer, force in lbs. delivered by, upon the head of a nail, 1 50 Hammers, work of fly-wheel in lifting, 106 High-pressure engines, 61 High-pressure engines, horse power, 65 Horse, backward pressure of, 21 Horse, maximum force of, 12 Horse power designated by h, or HP, 3 Horse power estimated in a very simple way, 3 Horse power, examples of, 4 Hydrostatic press, 54 Hydrostatics, principle upon which it rests, 237 Impact, shock of bodies, 134 Impinging bodies, velocity of, 138 Important expression, 97 Important principle, 13 Inclined plane, 47 Inclined plane, taking friction into ac- count, 48 Indicator, figure given by, 83 Indicator, use of, 306 Inertia, measure of, 118 Injection, water, 85 Iron and wooden bridges, 187 Iron bridge, in which the forces are wdl combined to meet the demands of railway traffic, 205 Kepler, the third law of, 278 Lever, 27 Lever, vent, 35 Lever, combination of, 30 Lever, equilibrium of, 32 Lever, of first order, 28 Lever, of second order, 28 Lever, of third order, 29 Lever, when its weight is taken into account, 31 Living agent of work, 8 Living forces, vis viva, 116 Locomotive engine, maximum speed . of, 14 Locomotive engines, 67 Locomotive engines, power of, 14 Locomotives on inclined planes, 18 Machines, modulus of, 60 Machinery, important fact respect- ing, 119 Man, work done by, 9 Marine steam-engines, power of, 304 Mass, a constant quantity, 117 Materials, work in raising, of a given form, 22 358 INDEX. Mathematicians, problem solved that defied the skill of, 270 Mathematicians, interesting tQ, 3S3 Maximum velocity of the piston of a steam-engiiie, 119 McCaUum's inflexible arched truss, 199 Measurement of heights by barome- trical observations, 269 Mechanical powers, Chapter IV., 30 toS7 Mercury, density of, 271 Mixing steam, 87 Mixing of steam of different tempe- ratures, 84 Modulus of a machine, 60 Modulus of stability of a structure, 256 Moment (not mommtmi), 166 Moment of pressures, 241 Momentum, caution respecting the use of the term, 151 Momentum, or quantity of motion, 135 Motion of different bodies in water, 289 Motive forces and inertia, 117 Mountains, measurement of the heights of, 273 Observation, important, 152 Oscillation, centre of, 160 Paddle-wheels, 283 Paddle-wheels and screw-propellers, to calculate the power and find the properties of, 304 Parallelogram of forces, new demon- stration of, 166 Pendulum block struck with round- shot, 26 Pendulum, Xaver's, 281 Piston, maximum velocity of, 119 Piston, velocity found at any point of the stroke, 122 Plane, inclined, 47 Power of marine steam-engines, 302 Power of steam employed in raising coal, water, &c., 5 Pressure, centre of, 240 Pressure of earth composed of sloping layers, 269 Pressure of earth on revetment walls, 265 Principle of work, applied to the lever, 31 Principle of work, applied to pulleys, 46 Press, anti-friction, 57 Press, hydrostatic, 55 Pressure of steam, examples of Boyle's law, 77 Pressure of water and other fluids, 236 Principle of the equality of moments, 169 Principle upon which hydrostatics rests, 237 Principle, important, 96 Principle of moments applied, 1?4 Principle of sufiicient reason, 164 Principle of work, ballistic pendulum, 144 Propellers, 283 Projectiles, the flight of, 276 Proportions in which surcharged and saturated steam combine, 83 Pulley, application of the principle of work, 46 Pulley, d^ned, 45 Pulley, work destroyed by friction, 102 Pumping engines, 6 Punch, work of driving, 99 Quantity of heat in steam and in water, 85 Quantity of moment, 169 Quantity of motion, conservation of, 136 Quantity of motion imparted by a constant force, 147 Quantity of motion, or momentum, 135 Questions solved with ease which hitherto defied the combined skill of mathematicians, 123 Eailroads, friction of, 13 Kailroad trains in motion, work accu- mulated in, 98 Bailroad trains, motion of, 70 Ram, pile-driving, 159 Kam, piles driven by, 139 Ham of pile-driving engine, 2 Beason, principle of sufficient reason, 164 Begnault, M., errors corrected by, 272 INBEX. 359 Resistance of fluids, 154 Resistance of fluids to moving bodies, equivalent expressions, 286 Resistance of water to the motion of projectiles, 290 Revetment walls, 264 Roofs, 183 Rotating body, how to estimate the units of, in, 104 Rotation, vis viva of, 118 Rule of three, 323 Safety valve of steam-engine, how to graduate safety valve of, 33 Safety valve weight, to find the posi- tion of, 34 Screw, 27 Screw, action defined, 51 Screw, the compound, 52 Screw, work developed by, 51 Screw, endless, 53 Screw-propeller with expanding pitch, calculations respecting the work of, 311 Shock of bodies, impact, 137 Shock, common velocity after impact, 137 Shock of elastic bodies, 145 Shores supporting a wall, 264 Simpson, Thomas, rule, 75 Skeleton diagram, 170 Skeleton diagram, relative forces ex- pressed by, 172 Skeleton structures by E. Koch, 213 Sledge-hammer, work developed in lifting, 107 Solution, important, 124 Stability, true measure of, 25 Stays, 182 Steam employed expansively, 74 Steam, units of heat in, 85 Steam, work of, 58 Steamers, paddle-wheel, 300 Strength of materials, new theory of, by the author, 211 Strut and tie, 170 Struts distinguished from tics, 173 Superheated and surcharged steam, 86 Surcharged and saturated steam, 83 Suspension bridges, 219 Table of the elastic forcd of steam, and the corresponding temperature, Eahr., Beau., Cent.; lbs., kilog., inches, and metres, of mercury, 89 Table of the evaporative power of difierent kinds of coal, 72 Table, Morin's experiments upon the transmission of motion by the shock of a body falling oil a cast-iron plate, 146 Table, transmission of motion by shock, Morin's experiments, 141 Table, volume of a cubic foot of water in the form of steam, 63 Temperature, pressure, and volume of steam, 64 Temperature of steam being given, to find the corresponding pressure in lbs. by dual arithmetic, 88 Tensions and thrusts determined by graphical construction, 179 Time, unit of, 1 Thermometer, measurement of heights by, 270 Thrusts and tensions, 163 Tie and stmt, 170 Ties distinguished from stmts, 173 Time, space, velocity of falling bodies 94 Toggle joints, force of, 103 Toothed, or cogged-wheels, 41 Traction, 13 Traction of horses, 16 Treadmill, 11 Trussed skeleton structure, 218 Unit of work, 1 Units of heat in steam, 85 Units of work done expansively and estimated by Thomas Simpson's approximate rule, 79 Units of work done by men, examples, 10 Units of work referred to a unit of time, 3 Units of work done in raising weights, 2 Useful load, 69 Useful load, 80 Useful work defined, 6 Value of g, g, in different latitudes, 116 360 INDEX. Velocity acquired by a body descend- ing an inclined plane, 152 Velocity acquired by bodies in de- scending an inclined plane or curve, 125 Velocity of piston, 120 Vessel, area of resistance, 301 Vessel, coefficient of, 301 Vis viva, 116 Volume of steam of a given pressure, 87 Walls, revetment, 264 Water, formula for tlie resistance of, 298 Water, pressure of, 237 Water, pressure on different surfaces compared, 258 Water, quantity evaporated, 62 Water, resistance of, 297 Water, resistance o^ Morin's experi- ments, 287 Water-wlieel, Poncelet's, 156 AVedge, 27 Wedge, or movable inclined plane, 49 Wedge, power lost to useful effect by friction, 50 Weight, mass, density, volume, 282 Wheels, cogged or toothed, 41 Wheel and axle, compound, 43 Wheel and axle, defined, 39 Wheel and axle, 27 Wheel and axle, work done with, 23 Windlass, 40 Wooden and iron bridges, 187 Work, accumulated, 92 Work, accumulated in a railroad train in motion, 98 Work of beasts, 9 Work due to the resistance of the at- mosphere, 69 Work of every machine, how con- sumed, 13 Work in a moving body, 97 Work, principle of, applied to the lever, 31 Work in raising materials of a given form, 22 Work and time, units combined, 1 Work developed by the resistance of water, 286 THE END. LONDON : PllINTED BY C. WHITIKG, EEADFOBT DOrSE, STBAND. Comall University Library arV17372 The essential elements of practical mech iiiililllil 3 1924 031 289 006 oiin.anx ^ii^ ^"-'i^-^M^C