5 s I 1 UBRARYOc * I i OF-CAll Jfc-AMVJHIlMW 8 ' V/?.AHV}|8n-1 3 ff ISI >> .5i\rUNIVER% ^lOS-AW;nfj> I^ov-ft s p ^ 1^1 sa lira lira 1591 1 CALTFQR^ jclOS-ANCHfj^ 1 ^i 9 V 1 ? ^ g s I 3 A NEW TREATISE ON ELEMENTS OF MECHANICS ESTABLISHING STRICT PRECISION IN THE MEANING OF DYNAMICAL TERMS ACCOMPANIED WITH AN APPENDIX ON DUODENAL ARITHMETIC AND METROLOGY BY JOHN W. (NYSTROM, C. E. NEW YORK G. P. PUTNAM'S SONS 182 FIFTH AVENUE 1877. Entered according to Act of Congress, in the year 1875, by JOHN W. NYSTROM, In the Office of the Librarian of Congress, at Washington. filC ZEIRIR^T^. PAGE Lines Top from Bottm FOR READ 20 7 . . . c d . . . c a 77 4 . .. . 160 . '. . 164 83 15 sin. x tan. x h f h 83 I I sin. x = ~w tan. x = b 95 I 55 -i* Hp= 5280 FM _ FM 3600+550 375 98 4 B 2 ' 4 HP fi _ 777600 HP n d q> n d q> HP ^ nd rrp B n d q> 98 14 24 777600 112 112 7 ... R 2 ... R 117 I I 13 .... 9 . . . . ii 118 II 5 1-5544 38.830 50 | 1.5544 1 3 8 - 86 1 20 16 . . 5 seconds . . 2.23 seconds 140 9 . . . I insert . . . . . . Insert i-ji i . . . db' . , - f'V T *7 T b' f b' c' 1 / i 179 i 6 . 8.0040 . 1 . . . 0.0040 191 5 - - 0.5775 , L . . Q-5775L 206 3 . . . A -b c B . . . . . Aa bB 207 8 . AD . . . ... AB 222 . Fig. i So . Fie 18-? 223 20 9 . . Fig. 181 . S. Ig. J.UJ . . . Fig. 184 233 6 . any . . . and 241 9 . Generatrix . . . Acceleratrix 248 5 . . . patn . . . path 35 4 . . Fig. 209 . . . . . Fig. 212 252 8 . . . Fig. 210 . . . . . Fig. 214 2QI 4 . . . Ion. i__ 292 J 3 4/7 VT* 2 9 2 5 . . 1/4332.6* ' V 43 32.6' 1OI 2 276181 TO WILLIAM SELLERS, ESQ., THIS WORK is JR.ESPECTFULLY DEDICATED, AS A MARK OF APPRECIATION OF HIS DISTINGUISHED ACCOMPLISHMENTS IN THE MECHANIC ARTS, BY THE AUTHOR. PREFACE. THE principal objects in introducing this new treatise on me- chanics are the establishment of strict precision in the meaning of dynamical terms, and the classification of physical quantities into elements and functions. A revision of the principles of mechanics is a necessity long felt, and frequently acknowledged in the discussions of learned men, who have heretofore disagreed as to the true meaning of technical terms and the constitution of dynamical quantities. The prevalent discordance on these topics has caused the delay of this publication for over ten years, in which interim various discussions thereupon have been published, both in Europe and in America, indicating the confused condition of the subject. In scientific periodicals we rarely find a sound article on dy- namics, but the action and combination of physical elements are treated as if governed by individual judgment, instead of the fixed and immutable ordinances of nature. A pamphlet entitled Principles of Dynamics, exposing the con- fusion in the science of dynamics, has been published and pre- sented to the principal libraries and institutions of learning. In addition to the objects above mentioned, this treatise con- tains sufficient new and original matter to warrant its publica- tion. The Appendix is added for the purpose of supplying the student with materials for the coming revision and final establishment of an international system of metrology, which must be attended to sooner or later. i 5 ALPHABETICAL INDEX. A. PAGE Abrasion, limit of . . 88 Aerostatics and dynamics . 17 Aggregate forms, three . 99 Air, compression of . . 266 Air, resistance of . . .167 Air, temperature of com- pressed . . . .267 Animal strength or capabil- ity 78 Appendix .... 307 Arithmetics, octonal and deci- mal .... 308 Arithmetics, duodenal . 310 Arithmetics, senidenal . 311 Ascending bodies against gravity .... 121 Asteroids, planetoids . . 287 Astronomy .... 281 Atoms . . . .99 Attraction between any two. masses . . . .113 Attraction, maximum . . 295 Attraction in and above the earth . 104 Attraction on the earth's sur- face 112 Attraction of a mountain . 114 B. Ballistic pendulum . . 253 Burden, man and animal . 78 C. Camel, capability of .78 Calculus applied to dynamics 154 Catenary . . . 21-23 Centre of gravity . . 50 Centre of gyration . . 184 Centring revolving bodies 189,203 Centrifugal and centripetal forces .... 222 Centrifugal force of a liquid 236 Centrifugal revolution indi- cator . . . .235 Changing direction of a mov- ing body . . . .220 Circular or rotary motion . 76 Circus ring, centrifugal force 233 Circular saw . . .79 ALPHABETICAL INDEX. Clock dial, duodenal . . 325 Collision or impact of bodies 141 Civilization in other worlds 288 Comets . . . .282 Compass, duodenal . . 327 Conical pendulum . . 246 Connecting rods . . . 218 Counting beats of seconds . 280 Crank motion . . 214, 217 Cranes . . . 48, 49 Creation of worlds . . 281 D. Decimal arithmetic and me- trology . . . .308 Decimal system, in perfection of 308 Density and volume of planets 302 Diagrams of heavy ordnance 257 Differential pulley . . 34 Dredging machines . . 80 Duodenal arithmetic and me- trology . . . .311 Dynamical terms, rejected . 15 Dynamics, definition of .17, 57 Dynamic principles . . 57 Dynamics of matter . . 149 Dynamics by the calculus . 154 Dynamics of heavy ordnance 254 Dynamics of sound . . 277 Dvnamometers . 269-276 Earth, the planet . Earthwork . 286 42 PAGE Effect, definition of .64 Elastic and hard bodies 141-148 Elements and functions . 57 Elephant, capability of . 78 Equilibrium arid stability . 36 Experiments with heavy ord- . 256 P. Falling bodies . . .117 Falling weight by rope and pulley . . . .127 Firing a ball through a door . 221 Foot-pounds, different kinds . 69 Foot-tons . . . .68 Flour mills . . . .79 Fly-wheels . . . 198-202 Force, definition of . 17, 58, 59 Force, examples . . .71 Force of falling bodies . . 133 Force of inertia . . .185 Force on a body free to move 130 Force of wind . . .45 Forces, variable . . . 157 Friction, work of . . .176 Friction, rubbing . . .81 Friction, dynamometer . . 269 Friction, gearing . . .91 Friction, screw-propellers . 178 Functions and elements . 57 Governors Gravitation 229-233 . 99 ALPHABETICAL INDEX. Gravity, centre of. . .50 Gunpowder pile-driver,Shaw's 263 Gyration . . . .182 Gyration, centre of . . 184 Hammer, steam . . .135 Hammer, forge . . . 213 Heat generated by light . 285 Heavy ordnance . . . 254 Horse-mill . . .78 Horse-power . . 63, 77-80 Horse-power of machines . 79 Horse-power of steam-engines 77 Hydrodynamics, definition of 17 Hydrostatics, definition of . 17 I, J. Impact of bodies . . . 141 Inhabitation of worlds . . 285 Indicated horse-power . .77 Inclined plane, motion on . 125 Index of contents . . 7 Index of illustrations . . 330 Inertia, definition of . 115-185 Inertia, moment of . . 184 Indicator, revolution . . 235 Jupiter, planet. . . 287 Law of celestial mechanics . 289 Levers, different kinds of 26, 27 Lever of a force . . .24 Lifting a weight vertically . 172 Light generating heat Llama of Peru PAGE 285 78 Man-power . Mars, the planet . Mass, definition of Mass of the earth 63,64 . 286 . 106 . 112 Mass of the earth and moon . 299 Mass of the sun, to find the . 290 Mass of any planet, to find the 300 Mass and weight, conversion of .... 108-111 Matter, definition of .99 Matt, explanation of the new term . . . .106 Mean force, to find the . . 168 Mechanics, definition of .17 Mercury, the planet . . 286 Meteors . . . .282 Metre, folded French . . 310 Metrical system, imperfect . 309 Metrology, octonal . . 308 Metrology, decimal . . 310 Metrology, duodenal . . 321 Metrology, senidenal . . 312 Molecule . . . .99 Momentum, static . . 24 Momentum, dynamic . . 106 Momentum, stability . . 37 Momentum curves . . 219 Moment of inertia . .184 Motion, different kinds of . 59 Motion, straight to rotary . 205 10 ALPHABETICAL INDEX. Moon, the Satellite Music . N. Nebulas Neptune, the planet Newton's laws Noise, dynamics of PAGE . 286 . 300 281-283 . 287 . 289 . 278 Nomenclature of numbers . 314 O. Octonal arithmetics and me- trology . . . .308 Orbits of the earth and moon 298 Ordnance dynamics . 251-254 Oxford pendulum . 247-249 P. Parabolic motion of projec- tiles .... 123, 168 Parallelogram of forces . 18 Particle of matter . . 99 Pendulum .. . . 237-250 Pendulum, Oxford diagrams . 248 Pendulum, ballistic . . 253 Percussion, centre of . . 250 Pile-driver, ordinary . . 135 Pile-driver,gunpowder,Shaw's 263 Planetary system . . 283, 284 Planets in our -system . . 286 Platform scale . . .32 Power, definition of . 58, 62 Power in effects or foot- pounds . . .64, 73 Power of a horse or of man 63, 78 Projectile thrown up verti- cally . . . .121 Prong's dynamometer . . 269 Pulleys . . . 34,35 R. Radius of the earth Radius of gyration Rails, adhesion on Recoil of ordnance . 226 . 184 . 94 . 252 Resistance of air to projec- tiles 164 Resultant of forces . 18-20 Revolution indicator, Brown's 235 Roads . . . . 78, 93 Rolling friction . . .88 Rolling-mills . . 79, 91 S. Safety-valves, graduation of . 31 Saturn, the planet . . 287 Saw-mills . . . .79 Scales, platform and weighing 32 Screw-propellers, friction of . 178 Seconds, counting bea,ts of . 280 Senidenal arithmetics . .311 Slip-railway for ships . . 90 Slope, natural repose . . 42 Sound, dynamics of . . 277 Sound, velocity of . . 279 Space in dynamics . 5865 Spring, elastic, body moving against . . . 138, 174 Stamp-mill . . . .210 Stability . . . 36-47 ALPHABETICAL INDEX. 11 PAGE Statics . . 17, 18 Static momentum . 24, 33 Steam-hammers . . . 135 Steamship performance . 81 Systems of arithmetics . . 312 T. Threshing machine . . 79 Time, definition of . 58, 62 Time, examples of . .72 Traction on roads . . 90 U. Uranus, the planet . . 287 V. Velocity, definition of . 68, 60 Velocity, examples . . 72 Venus, the planet . . 286 Vis-viva, explanation of .170 Vis-viva in a body at rest . 171 Volume and density of planets 302 W. Walls, retaining . . 38-44 Weight, explanation of . 105 Waterworks. ... 78 Wind, force of . . .45 Work, explanation of . 58, 67, 75 Woric of a spring . .174 Work of friction . . . 176 Work, diagrams . . .216 Work of mass in circular mo- tion 182 Workmanday . . 68 Tables. PAGE Addition and multiplication . 319 Adhesion on rails . . 94 Catenary curve . . .22 Distance traveled by sound . 279 Duodenal and decimal num- bers 316 Elements of the planetary system .... 284 Elements for centre of gravity 52 Experiments with heavy ord- nance .... 256 Falling bodies, velocity of .118 Fly-wheels, elements of . 212 Friction, sliding . . .86 Friction, shaft journals . . 87 Friction, governors . . 231 Governors . . . .231 Horse-power, foreign . . 63 Limit of abrasion . . 88 Load of burden . . .78 Manual and animal work . 78 Mass converted to weight . 110 Natural slope of granular sub- stances . . . .42 Nomenclature of numbers . 314 Pendulum oscillations . 241, 245 Pile-driver performance, Shaw's . . . .268 Steamship performance . 81 Traction on roads . . 93 Velocity and force of wind . 45 Velocity of the earth and moon 299 Weight converted into mass . 108 INTRODUCTION. THIS treatise is written for students of Mechanics, and the technical terms herein adopted are those used in the machine-shop, rejecting the ideal vocabulary heretofore used in textbooks and colleges. In order to establish a standard language in Mechanics, it is hoped that institutions of learning will approve and confirm the terms and distinctions of dynamical quantities as adopted and defined in these pages, 57 to 69, inclusive. The distinction between the terms force, power and work has here- tofore not been clearly defined, but either of these terms has been promiscuously applied to either or all those quantities, according to individual caprices. Work has thus been distinguished by a variety of terms indicating different characters of that function. It has generally been main- tained that work is independent of time, and that power is dependent on time, both of which propositions are incorrect. REJECTED TERMS. Energy. This term is used to denote work, but the sen.~e of it conveys an idea of a different virtue namely, that of activity or vigor, which is power. We say that a man has a great deal of energy when he can accomplish much Avork in a short time, which is a virtue of power ; but if he accomplishes the same quantity of work in a much longer time, we do not give him credit for much energy. The term energy, if employed at all, ought to be applied to power alone ; but as we have the expressive term power for that function, it is better to dis- pense with the term energy in dynamics. The term work is the proper name for the function whi:ii has been called energy. 14 ELEMENTS OF MECHANICS. Quantity of Motion is a term also used to denote work, which latter is a different function from that of motion. The sense of this term is inseparably associated with an idea of more or less space, which is a function of velocity and time without regard to force ; and as force is an element of work, the term quantity of motion should be rejected as improper to denote that function. The words Actual, Total, Quantity, Mode, Potential, In- trinsic, Kinetic Effort, etc. are often appended to terms without affecting the nature of the quantity so denoted ; the objection to which is that one and the same quantity is differently defined according to the combination of these appended words. As an illustration of the effect of these appendages to terms in dynamics, we may apply them to geometrical quantities ; for instance, volume in geometry corresponds to work in dynamics, and may be expressed thus : Volume of a cube. Cubical volume of a sphere. Total intrinsic volume of a cone. Actual potential volume of a cylinder. Total intrinsic quantity of volume of a pyramid. The actual total quantity of voluminous cubic inches in an intrinsic cubic foot is 1728. (A cubic foot is 1728 cubic inches.) All these expressions mean simply volume; as the different com- binations of terms denoting work mean simply work, or the product of the three simple elements force, velocity and time. The rejection of the superfluous terms will render the subject of dynamics much easier to teach, learn and remember. There is also an expression generally used in the English language namely, " Consumption of coal per horse-power per hour," which is not correct, or rather it is nonsense. The intended idea should be expressed, " Consumption of coal per hour per horse-power." It is the fuel which is divided by time, and not the power. The consumption of fuel is work, which divided by time is power. There exists no such quantity in dynamics as power per time, but power multiplied by time is work, and work per time is power. The following list shows which terms are herein rejected and adopted : INTROD UCTION. 15 DYNAMICAL TERMS. Rejected Terms. Effort of force. Efficiency of force. Acting force. Force of motion. Working force. Quantity of moving force. Quantity of motion. Mode of motion. Mode of force. Moment of activity. Mechanical power. Mechanical effect. Quantity of action. Efficiency. Rate of work. Dynamic effect. Quantity of work. Actual total quantity of work. Total amount of work. Actuated work. Vis- viva. Living force. Energy. Actual energy. Potential energy. Kinetic energy. Energy of motion. Energy of force. Heat a form of energy. .Heat a mode of motion. Mechanical potential energy. Quantity of energy. Stored energy. Intrinsic energy. Total actual energy. Work of energy. Equation of energy. Equality of energy. Reason for Rejection. force. All forces act. Means motive force. Has no definite meaning. Means simply power. Used for power or work. Means simply work. Formula for work. Primitive and realized work. 1020 ELEMENTS OF MECHANICS. MECHANICS is that branch of natural philosophy which treats of the three simple physical elements, force, motion and time ; with their combinations constituting power, space and work. Mechanics is divided into two distinct parts namely, STATICS AND DYNAMICS. STATICS is the science of forces in equilibrium or at rest : it is subdivided into three branches, treating respectively of solids, liquids and Statics, strictly speaking, refers to forces in regard to solids. Hydrostatics treats of the pressure and equilibrium of liquids. Aerostatics treats of the pressure and equilibrium of air or gases. DYNAMICS is the science of force in motion, producing power and work ; and is also subdivided into three branches, embracing re- spectively, solids, liquids and Dynamics, strictly speaking, refers to power and work of solids. Hydrodynamics treats of the power and work of liquids. Aerodynamics treats of the power and work of air or gases. FORCE. The term Force means any action which can be expressed simply by weight, and which can be realized only by an equal amount of reaction. Force is derived from a great variety of sources, but whenever it is simply force it can invariably be expressed by weight, without regard to motion, time, power or work. A detailed explanation of force is given in Dynamics. 2* B 17 18 ELEMENTS OF MECHANICS. STATICS. Statics is the science of forces in equilibrium ; it embraces the strength of materials, of bridges and of girders; the stability of walls, steeples and towers ; the static momentum of levers, with their combination into weighing scales, windlasses, pulleys, funicular ma- chines, inclined planes, screws, catenaria, and all kinds of gearing. The magnitude and direction of a force can be represented by a straight line ; but no force can be realized without an equal amount of resistance in the opposite direction, which likewise can be repre- sented by a straight line. PROBLEM 1. Let the line F represent the magnitude and direction of a force acting on a point c at rest or in motion ; and let R represent an equal amount of resistance in the opposite direc- tion ; then the force F and resistance R are said to be in equilibrium. When two or more forces act in one or the same direction, as in the case of several men pulling on one rope, the several forces are added together, and the sum is considered as only one force. PROBLEM 2. Let the lines F and /represent -magnitudes and direc- tions of two forces acting at a point c. Complete the parallelogram, and the diagonal R will represent the magnitude and direction of the consolidated action of the two forces jPand/. The diagonal R is called the resultant of F and /. Make R equal and opposite to R; then R represents the magnitude and direction of the resistance at the point c. The lengths of the lines F, f and R in any unit of measure represent the corresponding forces expressed in any unit of weight. When the angle between the forces F and / is known, the magni- tude and direction of the resultant R , or resistance R, can be de- termined by the aid of trigonometry, as follows : STATICS. 19 Let u denote the angle in degrees between the two forces F and/, and v = the angle between .Fand the resultant R , then we have tan.v = f sin.u Ffcos.u Resultant 12 = sec.v(F / cos.u). . 2. Use the sign + when u is less than 90, and - when u>90. Example 1, The force jP=68 and /=42 pounds, acting at the point c, Fig. 2, at an Angle = 65 30'. Required the force and direc- tion of the resultant R or resistance R. Formula 1, ta#i.v 42 68 + 42 x cos. 65 Fig. 3. which is the tangent for 24 6', the direction of the resultant R to the force F. Resultant ^' = ses.65 30') -93.574 pounds, the force of resistance required, " Nystrom's Pocket-Book " contains complete tables of the trigono- metrical functions, both natural and logarithmic. Natural. Logarithm. swt.65 30' = 0.90996 = 9.95902 sec.24 & =L0955 =10.03961 The insertion and use of trigonometrical functions in algebraical formulas can be learned without being well versed in the science of trigonometry. PROBLEM 3. When three or more forces are acting in different directions at a point c, Fig. 3, find first the resultant between either two of them, say F&ndf, which gives the resultant r as a single force. Complete the par- allelogram r, f, and the diagonal R is the resultant of the three forces F,f,f. The Formulas 1 and 2 are applied to each two forces or resultants, as in Example 1. PROBLEM 4. Forces may be applied so that they nearly counter- balance one another, as in Fig. 4. The resultant R is smaller than either of the three forces ^,/and/'. 20 ELEMENTS OF MECHANICS. PROBLEM 5. ' To " dissolve a force into two component parts, Fig. 5. Suspend a rope from a to b, and hang a weight F on it at c; then the force F is dissolved into two parts, acting one on the line c a, and the other on c b. Draw the line c d to represent the magnitude and direction of the force F. Complete the parallelogram, and the side / represents the force acting on c d, and f that on c b. The trigonometrical expression for Fig. 5 will be : T-T ri , ^ n Fsin.u F :f =sin..(u+v) : stn.u. f = - . Fsin.v :.sin.v. f-- f sin.u Fig. 6. PROBLEM . Let a rope be suspended from two points A and -3, and two weigh-ts W and w hung upon it. The rope is- considered to have no weight, but only stretched by W and w. Let j,.!/ F denote the tension of the rope at A, and /=that at B', then F: W= sin.u : sm.(u+x) F- W sin.u PROBLEM 7. Let a rope be suspended from the points A and B, and any number of weights P, Q, R, S, Thung upon it. Let .F denote the force of ten- sion at A, and /=that at B. Draw the lines A C and B D, so that they tangent the rope at A and B. From the point of intersection draw the vertical line W to represent the sum of all the weights CATENARY. 21 P, Q, B, S, T= W. Complete the par- allelogram, of which the side F 1 rep- resents the force of tension at A, and * the side /that at B. F:W= f:w = sin.v : sin.(x + v). /= PROBLEM 8. Now let us suppose an infinite number of weights to be suspended on the rope, which is the same as to consider the weight of the rope itself, or that of a chain. Draw the lines and complete the parallelogram as described in the preceding problem. The vertical diagonal TF represents the weight of the chain, and the forces J^and/ are as in the pre- ceding formulas. PROBLEM 9. THE CATENARY. The curve formed by a flexible rope or chain suspended from two ; points is called a catenary or chain-line. Let v denote the angle of the curve with the vertical in any point P whose abscissa is x and ordinate y ; / = length of the curve P. \ The formulas for the catenary will then be lsin.*v 1 sinh- , , . . hyp - h , ., (cosec.v 1). sin.2v x kyp.log.cot.^v " (cosec.v -I) jj (cosec.v -1) hyp. log. cot. %v sin.v(l - sin.v) 22 ELEMENTS OF MECHANICS. The formulas for the catenary are very difficult to manage, on ac- count of the angle v must be given ; but by the aid of the following fcable the solution becomes very simple : Table for the Catenary Curve. Angle Abscissa Ordinate Curve 7 t. / V. X. y- I. X y' 30 1.00000 1.31690 1.73210 1.3169 1.3153 40 0.55573 1.01068 1.19175 1.8186 1.1792 45 0.41421 0.88137 1.00000 2.1278 1.1346 50 0.30540 0.76291 0.83910 2.4981 1.1000 54 0.22078 0.65284 0.70021 2.9570 1.0725 60 0.15470 0.54930 0.5773o 3^507 1.0511 62 0.13257 0.50940 0.53171 3.8425 1.0438 64 0.11260 0.47021 0.48773 4.1759 1.0372 66 0.09484 0.43169 0.44523 4.5518 1.0314 68 0.07853 0.39376 0.40403 5.0141 1.0261 70 0.06418 0.35637 0.36397 5.5527 1.0213 71 0.05762 0.33786 0.34433 5.8636 1.0192 72 0.05146 0.31946 0.32492 6.2079 1.0171 73 0.04569 0.30116 0.30573 6.5914 1.0152 74 0.04030 0.28296 0.28675 7.0213 1.0134 75 0.03528 0.26484 0.26795 7.5068 1.0117 76 0.03061 0.24681 0.24933 8.0631 1.0102 77 0.02630 0.22887 0.23087 8.7023 1.0088 78 0.02234 0.21099 0.21256 9.4445 1.0073 79 0.01872 0.19318 0.19438 10.820 1.0062 80 0.01543 0.17542 0.17633 11.372 1.0052 81 0.01247 0.15773 OJ5838 12.654 1.0041 82 0.00983 0.14008 0.14054 14.254 1.0033 83 0.00751 0.12248 0.12278 16.309 1.0025 84 0.00551 0.1 0491 0.10510 19.046 1.0018 85 0.00382 9.08738 0.08749 22.874 L0013 86 0.00244 0.06987 0.06993 28.613 1.0008 87 0.00137 0.05238 0.05241 38.171 1.0005 88 0.00061 0.03491 0.03492 57.279 1.0002 89 0.00015 0.01745 0.01745 114.586 1.0000 APPLICATION OF THE CATENARY TABLE. The chain for a suspension bridge of 300 feet span is to hang 60 .feet below its supports on the piers. The chain is to support a weight of 52,000 pounds, uniformly distributed in its length. Kequired the length of the chain and the angle v and strain at the supports ? Half the span or y = 150 feet, for which x = 60 feet. 60 which corresponds nearly with an angle of v = 50 in the table, and the length of half the chain will be 1 = 150 x 1.1 = 165 feet. CATENARY. 23 The strain at the supports will be ,, TFm.i7 52000 x siw.50 . A ,. n F= - < -- - = 40449 pounds. sin.Zv w.lOO The ordinates x and y and the length I for any angle v in the table are found as follows : When v = 50 at the support, find x and y where v = 70 ? 0.30540 : 0.06418 = 60 : x. x = - 06418x60 _ 12 .609 feet. 0.30540 0.76291 : 0.35637 = 150 : y. y = - = 70.068 feet. Length 1 = 70.068 x 1.0213 = 71.56 feet. The ordinates can thus be calculated for a sufficient number of points in the catenary to define the course of the curve. The strain at the lowest point, or centre of the catenary, will be w tan.v = 26000 x tan. 50 = 30984 pounds ; when v = angle at the piers, and w = half the weight on the whole chain. The catenary is not a line of the conic sections ; its figure has the appearance of a parabola, but is a little fuller at the vertex. All the curves of the conic sections are of the second order, or of the exponent n = 2; whilst the exponent of the catenary is nearly The formula for any parabola is y = y'p x, when that of the catenary will approach y = y'p x. Length of the curve OP or I = - PROBLEM 1O. Fig ' l - To find the resultant of two parallel forces F and W, acting at the ends of an inflexible line d, e. Prolong the line W until d d' is equal to F, prolong F until e e' is equal to TF, then join d' and e', which will cut the inflex- ible line at c. Draw from c the resultant R, equal and parallel to W+ F; then R repre- sents the magnitude and direction of the re- sistance which balances the two forces F and W. The distance a = d c is the lever for ELEMENTS OF MECHANICS. the force W, and b = c e is the lever for F, which quantities bear the following relation : F : W= a : b, static momentums W a = F b. Fig. 11. STATIC MOMENTUM. The product of a force multiplied by its lever, is called static 'mo- mentum. The resistance at c is called the fulcrum. It is supposed in this case that the forces W and F act at right angles to the levers a and b. The lever of any force is equal to the right angular distance from the fulcrum to the direction of that force. When the static momentums Wa and Fb are equal, then the forces TFand F&re in equilibrium. PROBLEM 11. To find the resultant and static momentums of two forces, F and W, acting obliquely at the ends of an inflexible line, d e. Extend the lines of the forces until they meet at d'. Move the forces to d', and complete the par- allelogram as shown by Fig. 8 ; then the diag- onal I? is the resultant i ^ ^ e ^ wo f rces i the con- tinuation of which will cut the inflexible line at the fulcrum c ; make R = jR', the resistance balancing the two forces. The levers a and b are drawn from the fulcrum c at right angle to the direction of the re- spective forces. The analogy of the forces and levers is the same as that in Fig. 10 namely, F : T7= a : b, static momentums Wa = Fb. The resistance It is less than the sum of the forces F + W. Either one of the points of pressure, d, c or e, may be considered aa a fulcrum, as represented in Figs. 12 and 13.- STATICS. 25 PROBLEM 12. Two forces, F and W, acting Fig. 12. against one another on the line c d, of which c is the fulcrum ; to find the magnitude and direction of re- sistance, in the fulcrum c. Prolong the forces J^and W until they meet at d', from which set off the respective forces F' and W'\ complete the parallelogram and pro- long d' c' until it cut the line d e in c, which is the fulcrum for the forces, and d' c' = JR is the resistance. The levers a and b are drawn from the fulcrum c at right angles to the directions of the respective forces. F : W= a : b, static momentums Wa = Fb. PROBLEM 13. Two forces, jPand W, acting against one another on the line de, of which c is the fulcrum, to find the magnitude and direction of re- sistance, and the fulcrum c. Prolong F and W until they Fi &- 13 - meet at d'. Set off W and F from d', respectively; complete the parallelogram and continue d' d until it cut the line e d at c, which will be the fulcrum, and d'c' = R is the resistant required. The levers a and b are drawn from the fulcrum c at right angles to the direction of the respective forces. F : W= a : b, static momentums Wa = Fb. ELEMENTS OF MECHANICS. LEVERS. Levers are classified into three different kinds in reference to the relative position of the force F, weight TF, and fulcrum c. When the fulcrum c is between the force F and weight TF, the lever is called the first kind. Fig. 14. When the weight TFis between the force jPand the fulcrum c, the lever is of the second kind. Fig. 15. When the force F is between the weight TFand the fulcrum c, the lever is of the third kind. Fig. 16. The two forces F and TFwill be distinguished by considering F the applied force, acting on its lever b, to lift the weight TF, acting on its lever a. PROBLEM 14. ON THE ELEMENTS OF A LEVER OF THE FIRST KIND. Rg. 14. T The distance between TF and F is denoted by /. F:W=a:b, or Fb=Wa. b a ' F ' W Example 1. Suppose the weight TF=360 pounds acting on a lever a = 2.5 feet, what force Fis required on a lever b = & feet? TFa 360x2.5 f = = = 150 pounds, the answer. o 6 That is to say, the force F= 150 pounds acting on a lever 5 = 6 feet will just balance the weight TF= 360 pounds on the lever a = 2.5 feet, without regard to motion. Example 2. The force .F=450 and the weight TF=990 pounds acting on a lever a = 3.6 feet, required the lever b, for the force F? , TFa 990x3.6 , o = = '- = 7.92 feet, the answer. When the force ^and weight TF, and the distance I between them, are given, to find the fulcrum c and the levers a and b. LEVERS. 27 The lever a = l-b, and the lever b = l-a. Fb=Wa, or F(l-a)=Wa. Fl-Fa=Wa, -Fa=Wa-Fl, or Fa-Fl-Wa. Fa + Wa = Fl, a(F+W) = Fl, and a= -^. F + W Example 3. To prove the correctness of the formula, let us assume the same value of the quantities, as in Example 2. I = 3.6 + 7.92 = 11.52 feet. Find the levers a and b. F I _ 450x11.52 _ 3 _ 6 and b = n _ 52 _ 3 _ 6 F+ W 4504-990 PROBLEM 15. ON THE ELEMENTS OF A LEVER OF THE SECOND KIND. F:W=a:b, Fb F^^, TF=^, b ' a ' Wa Fb Example 4. The weight TF=3696 pounds acting on a lever a = 0.5 feet ; what force Fis required on the lever b = 15 feet? Wa 3696x0.5 . F= - = -- = 123.2 pounds, the answer. b 15 To find the levers and fulcrum when the weight W, force F and distance I are given. Fb = Wa, F(a + T) = Wa. Fa+Fl=Wa, Fa-Wa=-Fl, Wa-Fa = Fl. = Fl, and a= --. Example 5. TF=6396, JF= 150 pound, and the distance =9 feet; required the levers a and b. a= Fl = 15 x9 =0.2115655 feet = 2.53877 inches. W-F 6396-15 The lever b = a+J = 9 + 0.2115655 = 9.2115655 feet. 28 ELEMENTS OF MECHANICS. Example 6. A weight TF=50 tons is to be lifted by a force F= 0.25 tons, acting on a lever b = 45 inches ; at what distance shall the weight be applied from the fulcrum c? Fb 0.25x45 _ OOK . , ,, a = = = 0.225 inches, the answer. F and W can be expressed in any unit of weight, as well as the levers in any unit of length. PROBLEM 16. ON THE ELEMENTS OF A LEVER OF THE THIRD KIND. Fig. 16. F:W=a:b, or Fb = Wa. ?F r ^ Wa W= F_ a a Example 7. A weight TF=50 kilograms, acting on a lever a = 4.5 metres, is to be lifted by a force F= 88 kilograms ; required the lever b ? , Wa 50x4.5 _ Kh _ b = = = 2.557 metres. F 88 Given the force F, weight W, and the distance I, to find the fulcrum and levers a and b. a = b + l, b*=a-l. Fb = Wa, or F(a-l) = W a. Fa-Fl=Wa, Fa-Wa = Fl Fl a(F- W)= Fl, and a = F- W Example 8. The distance Z = 2.3 yards, W=5 hundredweight, and F= 5.5 hundredweight ; required the lever a ? Fl 5.5x2.3 The lever b will be 25.3 -2.3 = 23 yards. LEVERS. 29 Fig. 17. PROBLEM 17. We have now considered the levers a and b to be in a straight line, on which the forces .Fand Ware acting at right angles ; but in prac- tice these elements occur in a great variety of relative positions, of which one is illustrated by Fig. 17. The apparent lever a of the weight inclines under the fulcrum, whilst that of the force F rises above. The real or actual levers for these forces are the right- angular distances a and b ; but even these are not in a straight line. Whatever may be the form of the apparent levers, or whatever may be the direction of the forces, the actual levers must be considered from the fulcrum, at right angles to the directions of the respective forces. The static momentums of equilibrium and the formulas for the elements are precisely the same as that explained for the three kinds of simple levers. Fig. 18. PROBLEM 18. We have heretofore considered the levers to be inflexible lines with- out weight, which will answer when the centre of gravity of the mate- rial levers is in the fulcrum, like that of a weighing bal- ance or that of a wheel ; but this centre of gravity is of- ten located at a considerable distance from the fulcrum, as may be illustrated by Fig. 18, which is a lever of the first kind. The levers a and b are in the form of a beam, resting on the fulcrum c, and its centre of gravity at p. Let the weights of the beam be represented by P, acting on the lever x ; there will be two static momentums, F b and P x, on one side of the ful- crum, against one, W a, on the other side. 30 ELEMENTS OF MECHANICS. Wa = Fb + Px, w Fb + Px Fb = Wa-Px, Wa-Px Px^ Wa-Fb. D Wa-Fb a Fb + Pa b ' x Wa-Px X Wa-Fb a " w ' F ' P ' Fig. 19. The centre of gravity p may be found experimentally by balancing the beam over a sharp edge, when the distance x can be measured from the fulcrum c. It is here supposed that the levers a and b are in a straight and horizontal line. The form of the beam affects the location of the centre of gravity p, but when this centre is known the shape of the beam does not affect the static momentums. The pressure on the fulcrum c is equal to W+P-F. PROBLEM 19. This is a lever of the second kind, in which all the static momentums are on one side of the fulcrum. Px = Fb- Wa. Fb - Wa Fb = Wa+Px, Wa + Px b * Wa + Px a F^-Px F ' Fig. 20. CL r _ n W ' PROBLEM 20. ., f V - Pi ' p= X Fb- Wa This is a lever of the third kind, representing a safety- valve for a steam-boiler. Fb= Wa + Px, Wa+Px p _ Fb - Wa. Fb- Wa LEVERS. 31 , Wa+Px Fb-Px Fb-Wa 0=- - a = - x - F W P The formulas are the same as those for the lever of the second kind, Fig. 19. The centre of gravity of the lever is found by balancing it over a sharp edge, and the distance x is measured from the fulcrum. The weight of the lever is found by weighing it, and thus the momentum Px is obtained simply by multiplying the weight P by the distance x, which is a constant quantity in the graduation of the lever for dif- ferent pressures of steam. Example 9. Suppose the safety-valve to be three inches in diam- eter, and the lever to be graduated between pressures of 20 and 50 pounds to the square inch. Area of the valve is 7 square inches, and 7 x 50 = 350 pounds, which is the force F. The lever is found to weigh P=10 pounds and # = 18 inches, which momentum Px = \Q x 18 = 180. The lever b = 4 inches and a =48 inches where the weight W is expected to indicate a steam pressure of 50 pounds to the square inch. Then the weight will be . Fb-Px 350x4-10x18 ( W= = -- = 27.08 pounds. a 48 Now find the lever a, or at what distance from the fulcrum shall the weight be placed for balancing a steam pressure of 20 pounds to the square inch. F= 20 x 7 = 140 pounds, and the lever Fb-Px 140x4-10x18 Measure off this distance from the fulcrum and mark it with the number 20. Divide the distance between the first and second posi- tions of the weight into 50 - 20 = 30 equal parts, and number them from 20 to 50. The lever is thus graduated for the required pres- sures of steam. For greater accuracy it is necessary to weigh the valve with the lever for the weight P, and it is also necessary to place the valve on the lever at its proper distance from the fulcrum, when the lever is balanced over a sharp edge, for finding its centre of gravity. 32 ELEMENTS OF MECHANICS. PROBLEM 21. ON THE COMBINATION OF LEVERS, AS REPRESENTED BY Fig. 21. Fig. 21. Levers can be combined in a great variety of ways, but are generally arranged so that a short lever acts on a long one. W: w = W= : a a' a", w b b' b" , and or Waa'a" - Waa'a" wbb'b" a a' a" ' b b' b" That is to say, the big weight TPis to the small weight w as the prod- uct of all the long levers b is to the product of all the short levers a. The big weight multiplied by the product of the short levers is equal to the small weight multiplied by the product of the long levers. This rule will hold good for all combinations of levers, but without considering the weight and centre of gravity of the levers. The static momentum of the weights of the levers is determined by hanging small weights at TFuntil it balances the levers without any weight at w. PROBLEM 22. PLATFORM SCALE. This figure represents a platform scale. A B is the platform, rest- Pig 22. ing on levers underneath. The lower figure represents the plan of the levers, with the platform removed. The platform rests on four points, from which the levers extend diag- onally to the centre, and there are also four fulcrums, all of which only serve to give parallel motion to the platform, upon which the weight W can be placed at one corner, or be irregularly distributed, without ef- fecting any difference in the weight w. Although the platform and weight W rest upon four diagonal levers, the real levers of the static momentums must be considered parallel to the LEVEES. 33 sides of the platform, as marked on the figures, and only one of them enters into the formulas and calculations. W:w = bb'b": a a' a", or Wa a' a" = w b Vb". . W wbb' b" c *- w = Wa a' a" The platform and levers ought to be so arranged and proportioned that they balance one another, or that they are in equilibrium with- out the weight W and iv. DIFFERENTIAL BALANCE. Fig. 24 represents a convenient balance-scale for weighing heavy weights. It is much used in iron-foundries, where the balance with the weight is hoisted in a crane for weighing. L W:w = L:l, W- W ^ "~T' wi w The lever l = (a-b}. Wl L w L ~W' The object of the links and the short lever is to bring the weight close to the fulcrum, or, more correctly, to obtain a short lever I. There is not room enough on the main-balance to bring the direc- tion of the action of the weight W sufficiently close k) the fulcrum for weighing heavy weights. The levers a and b ought to be equal to a' and b' , respectively. a + b = a' + b' , and a - b = a' - b' . The difference between a and b is generally not made so great as shown in the illustration. For a lever L = 8 feet, the lever I is only a fraction of an inch, and can be made as small as desired by making (a-b] small. The scale should be well balanced by the ball B, without the weights TFand w. ELEMENTS OF MECHANICS. PULLEYS. Fig. 25. PROBLEM 23. FIXED PULLEYS. Let a force F be applied on a rope Fig. 25, extending over a fixed pulley R R to lift a weight W\ then F: W=R: R, or F = W. The radii of the pulley act as levers for the forces, and as the radii are alike on both sides, the force F will be equal to the weight W. Should the forces F and W act on different radii, Fig. 26, then, F:W=r:R, FR^Wr. Wr It ' Wr F ' W- FR FR W ' Fig. 27. PROBLEM 24. DIFFERENTIAL PULLEY-BLOCKS. The endless rope or chain,// passes over a pulley with two grooves of different radii R and r, as represented by Fig. 27. The weight W is equally divided on the ropes or chains // which tension will consequently be 2/ = W, The momentums on each side of the centre will be R-r Example. The weight TF= 1500 pounds, the radius of the large sheave is R = 6 inches, and the small sheave r = 5 inches. Required the force F? 1500(6-5) ^ = 7T~^ = 125 P oun ds- PULLEYS. 35 Fig. 28. PROBLEM 25. MOVABLE PULLEYS. Single movable pulley, Fig. 28, F: W=l :2. If the force is applied at a and acts upward, the result will be the Fig. 31. Double movable pulleys, Figs. 29 and 30, F:W=l:4:, F=\W, W=4F. The force F pulls on only one rope, whilst the weight W hangs on four ropes ; for which the weight is four times the force. For three, four, etc. movable pulleys there will be six, eight, etc. ropes, and the weight will be so much greater than the force. PROBLEM 26. COMPOUND PULLEYS. Let n denote the number of movable pulleys, then W Example. What weight W can be lifted by n = I or four movable pulleys, the force F= 300 pounds ? W= F2" = 300 x2 3 = 300x8 = 2400 pounds, the an- swer. Fig. 31 represents three, or n = 3 movable pulleys. MOTION OF THE FORCE AND WEIGHT. Let M denote the motion of the force F, and ra = motion of the weight W', then M:m = W:F, Wm FM = Wm. FM F ' W ' This proportion will hold good in all kinds of levers and pulleys. That is to say, the motion of the force is to the motion of the weight as the weight is to the force. 36 ELEMENTS OF MECHANICS. STABILITY AND EQUILIBRIUM. Stability of a body is that state of rest which cannot be disturbed by an infinitely small force. Equilibrium of bodies is that state of rest which can be disturbed by an infinitely small force. Stability and equilibrium are generally referred to the force of gravity acting vertically or at right angles to the surface of the earth. The force with which a body rests on a foundation is equal to the weight of the body acting in the direction of the vertical line passing through its centre of gravity. Should the body rest on only one point in the aforesaid line, it will be in equilibrium, and an infinitely small force can upset it ; the body has therefore no stability. Let the body rest on two points in a vertical plane passing through its centre of gravity, and the two points of support are one on each side of the vertical from the same centre of gravity ; then the body will be in equilibrium in the direction at right angles to the plane, and stable in the- direction of the plane ; the body must therefore rest on more than two points to be absolutely stable. PROBLEM 27. Let a body W be supported on a tripod or three points a, b, d in Fig. 32. a horizontal plane. Draw from the centre of gravity of the system the vertical line c, c' . which must fall inside of the triangle a, 5, d to make the weight and tripod stable, and an appreciable force is required to upset the system. The term system means the tripod and the weight, or any number of bodies rigid together and supported at the points a, b, c. The system can be upset in any direction, but it is generally considered to be upset in the direction over either one of the sides in the base- triangle, which is called the line of fulcrum. The static lever upon which the system acts is the rectangular distance from c' to the line of fulcrum. The static momentum of the force must be equal to the momentum of stability to bring the force and weight in equilibrium, and any additional force, however small, will uppet the system. STABILITY AND EQUILIBRIUM. 37 MOMENTUM OF STABILITY. The weight of the system, multiplied by its static lever, is called the ?nomentum of stability. The height of the centre of gravity c does not affect the momentum of stability. The force required to upset the system depends upon its lever of action. The lever of the force is the rectangular or shortest distance between the direction of the force and the line of fulcrum. PROBLEM 28. A body W, with its centre of gravity at c, is rest- ing on the plane A B. A force F is applied to up- set the body over the fulcrum at A. Continue the direction of the force, and draw the line b at right angles to Ff from the fulcrum, then b is the lever for the force, and a is the lever for the weight of the body. F:W=a: b, Fb = W a. , Wa . Fb Fig. 33. b ' Wa F ' Fb Example. The weight of the body is TF=3600 pounds, and its lever a = 1.8 feet. The direction of the force F is such that its lever b = 5 feet. Re- quired the force F? Wa 3600x1.8 100fi . F = = 1296 pounds. b 5 Let the same body be upset over the fulcrum at B, the lever a = 4 feet; the force F being ap- plied so that the lever b = 7 feet. Required the force F? 3600x4 om , n , F= - = 2059 pounds. Fig. 34. 276181 38 ELEMENTS OF MECHANICS. PROBLEM 29. RETAINING WALLS FOR WATER. Fig. 35 represents a wall of hydraulic masonry erected for re- taining water, which presses on the inside with a tendency to upset the wall over the fulcrum at A . j-gi. 35. For simplicity in the formulas and calcula- tions we will assume a definite length of the wall, say one foot. The pressure of the water Hf E! against the wall increases with the depth d, 7 ^^^^EHj^rfrr so that if the line n o represents the water- pressure per unit of surface at the bottom of the water, then any other horizontal line in the triangle m, n, o represents the correspond- ing pressure at that depth, and the area of the triangle represents the total pressure against the wall. The centre of pressure must therefore be at the same height as that of the centre of gravity of the triangle, which is at ^ d from the bottom, where the resultant of all the forces acts on the lever b to upset the wall. The weight of a cubic foot of water at 60 Fahr. is 62.33 pounds, and the total pressure, or the force F, acting on the side of the wall, will be F= \ x 62.33 rf= 31.16 rf. The lever of this force- is 5 = d. The static momentum of the force will be but F= 31.16 d, or the static momentum = |x 31.16 d*= 10.3866 d' 1 . The weight per cubic foot of hydraulic masonry may be assumed as follows : Brick, 90 pounds, Granite, 140 pounds. Let A denote the height of the Avail, t = its thickness in feet, and w = weight per cubic foot of the materials in the wall. Then, the weight of the wall will be when the momentum of stability will be, without water-pressure, Wa = htwa. STABILITY AND EQUILIBRIUM. 39 From this momentum of stability must be subtracted the static momentum of the water-pressure, and the remainder will be the real momentum of stability of the wall with water-pressure. Let W denote the weight of the real momentum of stability, or that part of the weight of the wall which keeps it in proper position against the water-pressure ; then Example. Suppose the wall to be of brick masonry, for which U' = 90, the height 7i = 12 feet, thickness Z = 2 feet, and the depth of water d=W feet. Required the real momentum of stability? The lever a = 1 foot. 10.3866 x 10 2 = 1038.66 foot-pounds, the static momentum of the water. 12 x 2 x 90 x 1 = 2160 foot-pounds. Real momentum of stability = 2160 - 1038 = 1122 foot-pounds. That is to say, the real static momentum of the wall is slightly more than the static momentum of the water-pressure, or that the momentum of stability of the wall without water-pressure is more than double the static momentum of the water. This rule will hold good for any length of wall when the dimen- sions are in feet and the pressures in pounds. In practice, the mo- mentum of stability of the wall alone should be at least four times the static momentum of the water-pressure. FROBLEM 3O. Fig. 36. Let the materials in the wall in the preceding figure and example be formed into a wall of this figure. That is to say, the materials in the triangle p, q, r are placed at the base s, A, r, and formed into a solid wall A, B, q. The static momentum of the water-pressure, and also the weight of the wall W will be the same as in the preceding example, but the base A, B will here be doubled, or t = 4 feet, instead of two. The lever of the weight TFwill be a = \ x4 = 2.66 feet, instead of one foot, and the momentum of stability will be 2160 x 2.66 = 5745.6 foot-pounds. 40 ELEMENTS OF MECHANICS. The real momentum of stab. = 5745.6 - 1038.66 = 4707 foot-pounds, or over four times that in the preceding example, where the wall of the same amount of materials was vertical on the outside. Now let the same wall, with the same materials and dimensions, be turned so that the water-pressure falls on the inclined side, and the fulcrum of the wall at B. The lever of the weight TTwill now be only x4 = 1.33 feet, and the momentum of stability without water-pressure will be 2160 x 1.33 = 2872.8 foot-pounds. Fig. 37. PROBLEM 31. The water-pressures act at right angles on the inside of the wall A, C, the centre of which will be at of the depth d, or at of A, C from A. Continue the force F, and draw b at right angles to it from the fulcrum B. With the assumed dimen- sions of the wall, 12 feet high and 4 feet base, and depth of water 10 feet, the lever b will be 1.2648, a = 1.333 and the side A, C= 10.54 feet. The force F= 10.54 x 62.33 x 0.5 = 3284.79 pounds. Static moment. = Fb = 3284.79 x 1.2648 = 4154.6 foot-pounds. Real moment, stab. = 4154.6 - 2872.8 = 1281.8 foot-pounds. This proves that a wall' of the assumed dimensions has greater sta- bility when the water presses on the vertical side than when on the inclined side. The retaining wall and its inclined side may be so proportioned that the static momentum of the water-pressure increases the stability of the wall, as in Fig. 38. PROBLEM: 32. ELEMENTS OF THE RETAINING WALL. Fig. 38. h tan.v = . t tan.v t tan.v. lit! Moment, stab. = Wa = \ hi wx% t = hfw. STABILITY AND EQUILIBRIUM. ELEMENTS OF THE WATER-PRESSURE. The inclined side A, C=d cosec.v. The force F= \ d cosec.v. The distance S from A to the centre of pressure will be 8 = d cosec.v. b : S = =p t =t d sec.v : d sec.v. S ( =f= t =t 8 sec.v) =f , = O = ^= t COS.V -i- 5 sec.v sec.-u j 6 = =p t cos.v Subtract the smallest term from the largest, and the remainder is the lever b. When t cos.v > 3 sin.v the fulcrum B, as in Fig. 37. , the direction of the force of When t cos .v < d -, the direction of the force F passes inside of 3 sin.v fulcrum B, as in Fig. 38, and the static momentum F b increases the stability of the wall. The trigonometrical functions can be expressed by the dimensions of the wall, as follows : E = length of the hypotenuse of the wall. h t E cos.v = , sec.v = - , t dE E cosec.v = . h dR, : 3A' PROBLEM 33. Whatever may be the shape and position of a retaining wall, its momentum of stability is W a, and the static pressure of the water acts at right angles to the surface of the wall. The force of the water-pressure in pounds is equal to the area of the pressed surface in square feet, multiplied by half the depth in feet x 62.33. The centre of pressure is at one-third of the depth from the bottom. 42 ELEMENTS OF MECHANICS. PROBLEM 34. RETAINING WALLS FOR EARTHWORK. - 40 - The action of earth or other granular sub- stances, like sand, gravel, grain, etc., on re- taining walls is the same as that described for water, and the momentums are calculated by the same formulas, with the only excep- tion that the natural slope of the granular materials diminishes the force F as the co- sine for that slope. The natural slope of a granular substance is the greatest angle with the horizon at which it will repose in a heap. Let s denote the angle of natural slope and W= weight per cubic foot of the material retained by the wall ; which values for some substances are contained in the following table : Natural Slope and 'Weight of Granular Substances. Granular Substances Loosely Heaped. S S. ope. cos.s. Weight, W. 45 070711 Saw-dust, wheat-flour 44 0.71934 Broken stone or coal 43 0.73135 Malt-flour 40 0.76604 Sand (moist) 39 0.77715 95 Sand (dry) 38 0.78801 94 Malt-corn 37 0.79863 47 Wheat, rye and corn Peas 36 35 0.80902 0.81915 45 48 Gravel and earth 35 to 40 0.8 to 0.75 80 to 100 The force F per foot of length of the wall, Fig. 40, will be F= \w d cos.s. For safe calculation we may assume w = 100 pounds per cubic foot of earth or gravel pressing against the wall, and s = 32 51' the safety angle of natural slope, which cosine is 0.84 ; then the force F will be fx 100 x 0.84 rf-42 STABILITY AND EQUILIBRIUM. 43 The static momentum of the force F will be as described for Fig. 35 namely, Fb = \Fd, but F=42 d. The momentum of stability of the wall alone will be of which the weight w = htw, and the lever a = %t, when the sides of the wall are vertical. w = weight per cubic foot of the materials in the wall, which is 90 for brick, 120 for rubble concrete and 140 pounds for granite. Subtract the static momentum of the earth-pressure from the mo- mentum of stability of the wall alone, and the remainder will be the real stability of the wall. For safety in practice the momentum of stability of the wall alone ought to be at least four times the static momentum of the earth-pres- sure, or or hwf 112 d* 10.583 d h = -- and = -. "When the height h of the retaining wall is equal to the depth d of the earth-pressure, we have w V = 112 d, of which t = Example. The depth of earth-pressure is d = 2Q feet, to be retained by a vertical granite wall of w = 140 pounds to the cubic foot. Re- quired the thickness t of the wall. This example is for a wall of cut granite block and of first-class workmanship. The foundation of the wall should be much thicker, depending on the softness of the ground upon which it stands. A brick wall for retaining earth, of 120 feet high, should be 5 feet thick. 44 ELEMENTS OF MECHANICS. PROBLEM 35. Fig. 41. Walls are often built for re- taining earth of greater height than that of the wall, as illus- trated by the accompanying fig- ure, in which case the centre of pressure of the earth will not be at one-third from the bottom, as in the former cases. Let e denote the height of centre of pressure above the bot- tom of the wall; then 2d 3 " to the when the earth above the wall rises with its natural height d. Draw the force F at right angles to B 2 through the centre of pres- sure, and continue it past the fulcrum at A, as shown by the dotted line, and the static momentum of the force of earth-pressure can thus be found graphically on the drawing. The mean height of the column of earth pressing against the wall is d- \ A, and the area of the base of that column is the line .32x1 foot, and the force F= (B^w(d-\ K). The centre of gravity of an irregular four-sided section of the wall is found as follows : Divide the base A, B and the top 1, 2 each into two equal parts, and join the middle points 7 and 8 with the opposite corners, as shown by the figure 41 ; divide each of these four lines into three equal parts, join 3, 4 and 5, 6, and the intersection at W will be the centre of gravity of the four-sided section of the wall, and the lever a is thus found. The area of the four-sided section is equal to that of the two tri- angles A, B, 7 and 1, 2, 8, which, multiplied by the weight w per cubic foot of the material in the wall, will be the weight W. In practice the momentum of stability of the wall alone should be at least four times the static momentum of earth-pressure, or Fb. STABILITY AND EQUILIBRIUM. 45 PROBLEM 36. ON THE STABILITY OF TOWERS TO THE FORCE OF WIND. The momentum of stability of a tower or any other structure ex- posed to the wind should be greater than the static momentum of the greatest storm to which the object may be exposed. For safety in practice, the stability should be at least four times the momentum of the wind. The static momentum is equal to the force of the wind acting on the object, multiplied by the height of the centre of gravity of the surface acted upon. The following table shows the force of wind in pounds per square foot at different velocities : Table of Velocity and Force of "Wind. Miles per hour. Feet per second. Force per sq. ft.-pound. Common Appellation of the Force of Wind. 1 1.47 0.005 Hardly perceptible. 2 3 2.93 4.4 0.020 0.044 I Just perceptible. 4 5.87 0.079 1 5 6 7.33 8.8 0.123 0.177 j- Gentle pleasant wind. 7 10.25 0.241 J 8 11.75 0.315 9 13.2 0.400 10 14.67 0.492 12 17.6 0.708 Pleasant brisk gale. 14 20.5 0.964 15 22.00 1.107 16 23.45 1.25 j 18 26.4 1.55 1 20 29.34 1.968 [ Very brisk. 25 36.67 3.075 J 30 44.01 4.429 ] 35 51.34 6.027 [ High wind. 40 58.68 7.873 j 45 66.01 9.963 } 50 55 73.35 80.7 12.30 14.9 [ Very high. 60 65 88.02 95.4 17.71 20.85 I Storm or tempest. 70 102.5 24.1 Great storm. 75 80 110. 117.36 27.7 31.49 > Hurricane. 100 146.66 50. Tornado, tearing up trees, etc. The highest known wind is termed " tornado," which moves with a velocity of 100 miles per hour, or 146 feet per second, and exerts a 46 ELEMENTS OF MECHANICS. pressure of 50 pounds per square foot on a surface stationed at right angles to the direction of the wind. Let A denote the area in square feet of the structure exposed to the wind, the greatest force of which will be 50 A ; b = height of the centre of gravity of the exposed surface above the ground. The greatest static momentum of the wind will then be 50 A b, and the minimum stability of the structure should be limited to Fig. 42. -pig. 42 represents a square tower with parallel sides exposed to a tornado. The momentums will then be Stability, Wa = Fb static momentum. Practically, the stability Wa ought to be four times the static momentum Fb. Wa = 4Fb, or Wa = 20Q Ab, 200 A b and W= a in which W= the whole weight of the tower in pounds, A = area of one side of the tower facing the wind. Example. Suppose the tower to be four feet square and 20 feet high, from which a = 2 and b = 10 feet. The area of the side will be A = 4 x 20 = 80 square feet. What weight of the tower is required to maintain it stable to a tornado ? _ 200^6 200x80x10 . = = 80000 pounds. It is supposed that the wind acts at right angles on one side of the tower, but if acting in the direction of the diagonal of the square sec- tion, a greater surface will be exposed, but at such angle to the wind that the acting force will be the same as when blowing directly on only one side. On a round tower of diameter equal to the side of the square the force of the wind is only one-half of that on the square tower. PROBLEM 37. Fig. 43 represents a tower or chimney of the form of a conic frustum of diameters d at the top and D at the base, h = height of the tower; all in feet. The height of the centre of gravity b is calculated from the fol- lowing formula : STABILITY. 47 , A hi D-d \ Example. The height of the tower A = 260 feet, diameter at the top d = 10 feet, and D = 25 feet at the base. Required the height of the centre of gravity of the surface of the tower ? 9fiO 9fiO /9 r \ 1 0\ Fi - 43 - , ^DU ^jDU/^O J-V\ .. .. ^ . rt ., The projecting area to the wind is - (Z>+ d) = (25 + 10) = 4550 square feet. 2 2 Of this area only one-half is effective to the force of the wind, or A = 2275 square feet. The static momentum of a tor- nado will then be ^6 = 50 46 = 50 x 2275 x 111.43 = 12675162.5 foot-pounds. The practical momentum of stability of the tower should be at least four times this static momentum. If the materials in the tower and that in the base it stands on were hard enough to stand the crushing force at the fulcrum in upsetting the tower, the lever of the momentum of stability would be half the diameter D of the base ; but as such is not the case, in practice a de- duction of that lever must be made in accordance with the softness of the materials acted upon at the fulcrum. When the base of the tower is square, and the force of the wind acts at right angles to one of its sides, the fulcrum will be a line ; whilst on a circular base the fulcrum will be a point in which the whole weight of the tower acts to crush the materials. In ordinary good brickwork the lever of the momentum of stability may be taken at 0.7 of the radius of the circular base. The weight of the tower in the preceding example should then be ... Fb 4x12675162.5 , W= = = 5794368 pounds. a 0.7x12.5 A small deduction ought also to be made of the lever in a square base, where it may be taken at 0.9 of half the side of the square. The cohesive force of the materials increases the stability of the structure when the masonry is perfectly solid at the base. 48 ELEMENTS OF MECHANICS. CRANES. The ordinary crane for hoisting purposes is represented by Fig. 44. The crane is held in position or supported by the shoe a and cap b. The dotted lines represent the chain or rope by which the weight W is raised. The chain may run either direct from the barrel d to the blocks c and/, or from d via e, c to/. The body of the crane is composed of the post P, gib G and stay 8. The weight W acts on the lever I, and the reaction of force F in the supports a and b, acts on the lever h. Static momentums F h = W I. A r F' w These are the principal elements of the crane. When the stay 8 is in the direction a, c, the force of compression 8 of the stay will be, when L = length of the stay, WL h ' Let m denote the distance from the post P to where the stay 8 bears the gib G, and n = distance from the gib G to where the stay S is supported on the post P. Then the force of compression of the stay will be /S = . m n Wl The force of tension t on the gib will be t = . n The forces F, 8 and t must be expressed by the same units as that of the weight W. The weight of the materials in the crane is not included in the formulas. The tension of the chain is found by the formulas for pulleys. The strain of the stays which hold the cap b is equal to the force F, or CEANES. 49 The vertical pressure in the shoe a is equal to the sum of the weight TTand weight of the crane. For foundry cranes the block c is moved in or out to suit the loca- tion of the weight to be filled, and the gib Q and stay 8 are both made double, so that the chain can pass between the parts. The lateral strain z on the gib where it bears on the stay will be Wl WHARF CRANES. Wharf cranes for loading and unloading Fig. 45. boats are often constructed like Fig. 45, and for which the static elements are the same as those for the foundry crane. When the di- rection of the chain on which the weight W hangs is parallel with post a, b, the strain F at b is equal to the horizontal pressure at a. Static momentums F h = W I. The lateral strength of the curved part of the post must compen- sate the stay in the foundry crane. SHOP CRANE. This form of cranes is used with differential pulleys. The tension-rod or tie T serves the same purpose as the stay in the foundry crane. L = length of the tension-rod. T= force of tension. F= force or pressure in the post journals. /= force of compression of the gib. = distance between the centres of the block and the post. Static momentums F h W I. Fig. 46. /. m n n ' 50 ELEMENTS OF MECHANICS. CENTRE OF GRAVITY. The centre of gravity of a body, or of a rigid system of bodies, is a point in which, if there suspended, the body will be in equilibrium in any position it may be placed, like that of a wheel or circleplane suspended in the centre. Fig. 47. A body suspended freely from any point a will hang with its centre of gravity in the vertical line a b. Now suspend the body from another point c, and the centre of gravity will be on the line c d ; then when the centre of gravity is on both the lines a b and c d, it must evidently be at z, where the two lines cross one another. The lines a b and c d, or the centre of gravity z, can also be found by balancing the body on a sharp edge. The centre of gravity of any figure or body is thus found by suspending or balancing the same in two dif- ferent positions. CENTRE OF GRAVITY OF A PARALLELOGRAM. The centre of gravity of a square, rectangle or paral- lelogram is the point where the two diagonals cross one another. TO FIND THE CENTRE OF GRAVITY OF A TRIANGLE. Fig- 49 - Bisect either two of the three sides of the triangle, and draw the dotted lines to the opposite angles, as shown by Fig. 49. The crossing of these lines is the centre of gravity z, which is one-third of the dotted line from the side. CENTRE OF GRAVITY OF A SYSTEM OF TWO BODIES. r 'R- - 50 - Let two bodies A and B be placed at a dis- tance c between their centres of gravity, and z = centre of gravity of the system; then A a and B b are moments of gravity, which are alike when the system is supported in its centre of gravity. CENTRE OF GRAVITY. 51 Then, c a = cb andi = c a. = B(c-a) = B c- B a. c, or , , and o = Ac A+B A+ Example. A =6 pounds, B = 10 pounds, and the distance c = 18 inches. Required the distance a? Be 10 x 18 . a = = = 11.25 inches. A + B 6 + 10 CENTRE OF GRAVITY of a system of any number of bodies placed in a straight line o p. Assume any point o from which to refer the different moments of gravity, Z= A + B+C+D+E, or the sum of the weights of the dif- ferent bodies ; z = distance from o to the centre of gravity of the system. Flg- 51- Then the moments of gravity are . A, B, C, etc. are weight, and a, 5, , etc. are the levers of the respective momentums of gravity. CENTRE OF GRAVITY of a system of bodies irregularly placed, with the centres of gravity in one plane. Draw from o the lines op and o q, to form rectangular co-ordinate axes outside of the system. The right-angular moments of gravity are referred to the respective axes, as will be understood by the illus- tration . From o set off the distance z toward p and draw the ordinate /. Z ' / which gives the centre of gravity of the system. 52 ELEMENTS OF MECHANICS. By this method the centre of gravity of a great variety of systems of bodies can be ascertained ; for instance, that of a vessel or steam- boat. The axis op is drawn above the deck, and o q at the stern of the vessel. A, B, C, etc. may represent weights of the propeller, engine, boiler, coal and cargo. The centre of gravity of each part is ascertained separately. When a steamer is constructed, the weight and moments of gravity of each part ought to be calculated and summed up to correspond with the displacement, for which the following form of table is set up. ELEMENTS OF THE STEAMER SHOOTING STAR. Horizontal axis, 20 feet above load water-line. Vertical axis, 10 feet aft the centre of rudder. Details. Weights of the parts. Ho Lerer. Moments c izontal. Moment. f gTfty. Vert Lever. ical. Moment. Rudder Tons. 2.1 12.8 50.1 48 560 500 800 3 Feet. 9 47 75 110 120 50 I5fr 27.5 Foot-tons. 1&* 601.6 3757.5 5280.0' 76200,0 25000J) 12000fcfr 8215 Feet. 26.5 27.5 21 23 25 24 se> IS Foot-tons. 55.6 3-52.6 1052.1 1104.0 14000.0 12000.0 24000.0 239.0 Propeller shaft Boilers with water Coal. Car<*o aft Cargo forward Propeller 2102 248940.5 52803.3 Horizontal centre of gravity, centre of rudder. 2102 = 1 15.1 - 10 = 105.1 feet from Vertical centre of gravity, 52803.3 2102 25.1 - 20 = 5.1 feet below load water-line. In practice the calculation is carried out with more details. The horizontal position of the centre of gravity is required for knowing how the vessel will float in regard to the keel and load water-line. The vertical position of the centre of gravity is required for de- termining the stability of the vessel. CENTRE OF GRAVITY. 53 TO FIND THE CENTRE OF GRAVITY OF ANY IRREGULAR BODY RESTING ON TWO SUPPORTS. Having given the weights p and Fi s- 53 - P bearing on the supports, the sum of which is equal to the weight W of the body, to find the horizontal distance z of the centre of gravity from the support p. W:P:d:z and Pd W CENTRE OF GRAVITY BY THE CALCULUS. By the aid of the calculus the centre of gravity of regular figures can be determined with precision. Any figure or body can be considered as a system of bodies com- posed of its parts. The sum of the moments of gravity of all the parts is equal to the moment of gravity of the body or system. Fig. 54 represents a triangle with the base B and height H. Any element b of the triangle multiplied by the height h is the moment of gravity of that ele- ment. Let A denote the area of the triangle, and z = height of centre of gravity from B. Then we have Fig. 54. b: B = (H-K): H and b = zA ~ h H B(H-k) H zA = B H* B H' 3 B H 1 (*-*) That is to say, the height of the centre of gravity from the base is equal to one-third the height of the triangle. 5* 54 ELEMENTS OF MECHANICS. CENTRE OF GRAVITY OF A QUADRANGLE. The illustration represents a quadrangle, a, c, d, e, of which the top b is parallel with the base B. Prolong the sides d a and e c until they meet at/, which will form the triangle d,f, e of height H. h = height of the quadrangle. A = area of the quadrangle. = area of the triangle a,f, c. Q = area of the whole triangle d t f, e. Height of centre gravity of the whole triangle is - Height of centre gravity of triangle a, /, c is -x)+h. z = height of centre gravity of the quadrangle. Moment of gravity, QQff) = Az+ 0[^(JI- h) + A}. . . 1. K ........ 2. Height^", 3. Area of quadrangle, A= (+b). . . . . .4. Area of triangle, a,/ c, = -(H-K) = A t i. i j f Area ot triangle, a, j, e, 2 2(B-b) Moment of gravity, Az = \Q H- \0 H- f A. ... 7. Insert the values 3, 4, 5 and 6 for the corresponding quantities H,A,0 and Q in Formula 7. = Bh Bh JB-b) Bh B-b Bh This formula reduces itself to & ^ -, which gives the height of centre of gravity of the quadrangle above the base B. CENTRE OF GRAVITY. 55 CENTRE OF GRAVITY OF A CONE. D = diameter of the base of the cone. h = height, x and y are co-ordinates. Volume of cone, C= 12 1G Fig. 56. 12 16 CENTRE OF GRAVITY OF A PARABOLA. The illustration represents a parabolic plane of abscissa x and ordi- nate y. z = height of centre of gravity above the base B. d = depth of centre of gravity from the ver- j tex o, or h = d+ z and z = h d. \ The formula for a parabola is y = 2\/p x. -4 I? The parameter of a parabola is /> = -. The area of a parabola is A The moment of gravity from the vertex is = y x 8x p x -aVfv*- yh - f ,/iJ B A" BJt 2.5 and z = That is to say, the height of the centre of gravity of a parabolic plane is two-fifths of the height of the parabola. 56 ELEMENTS OF MECHANICS. CENTRE OF GRAVITY OF A PARABOLOID. Meaning of letters is the same as in the preceding example. Volume of a paraboloid (7= . 8 Moment of gravity for solidity of the paraboloid is . . 4h 4h 12 h A TT^A* = ~' * ' Z = i The centre of gravity of a paraboloid is one-third of the height from the base. EQUILIBRIUM OF STATIC MOMENTUMS. The illustration represents a body fixed Figj>8. on an axis 0, and several forces A, B, C and D, JE, J^are acting on their respective levers a, b, c and d, e,fto rotate the body in opposite directions. Then the opposing static momentums will ^^ ^"""^^SC be in equilibrium, when ^^^ A When the sum of any number of static momentums acting to rotate a body in one direction, is equal to the sum of the momentums acting in opposite directions, then the opposing forces are in equilibrium, and the body will remain at rest. When the sums of the opposing static momentums are not alike, the body will move with a momentum equal to the difference between the two sums. D YNAMICS. 57 DYNAMICS. 1. Dynamics is that branch of mechanics which treats of forces in motion, producing power and work. It comprehends the action of all kinds of machinery, manual and animal labor in the transforma- tion of physical work. Quantity is any principle or magnitude which can be increased or diminished by augmentation or abatement of homogeneous parts, and which can be expressed by a number. Element is an essential principle which cannot be resolved into two or more different principles. Function is any compound result or product of two or more different elements. A function is resolved by dividing it with one or more of its ele- ments. Force, Velocity and Time are simple physical elements. Power, Space and Work are functions of those elements. The combinations of the elements in the functions are as follows : FUNCTIONS. Power P=F V. Space 8 = V T. ELEMENTS. Force = F. Velocity = V. Time = T. Mass = M. F\M = V: T. MOMENTUM. M V= F T. These are the fundamental principles in dynamics. 58 ELEMENTS OF MECHANICS. DYNAMICS COMPARED WITH GEOMETRY. 2. In geometry we have three fundamental elements, expressed by the terms Length, Breadth and Thickness, which serve to repre- sent to the mind the nature of the several properties of geometrical space. We have in like manner in dynamics three fundamental elements, expressed by the terms Force, Velocity and Time, which repre- sent the nature of physical Power, Space and Work. Force is any action which can be expressed simply by weight without regard to motion or time ; it is an essential principle which cannot be resolved into two or more principles, and is therefore a simple element. Force is the first element in dynamics, and corre- sponds to length in geometry. Velocity is speed or rate of motion, an essential principle which cannot be resolved into two or more principles, and is therefore a simple element. Velocity is the second element in dynamics, and corresponds to breadth in geometry. Time is duration, or that measured by a clock ; it is an essential principle which cannot be resolved into two or more principles, and is therefore a simple element. Time is the third element in dynamics, and corresponds to thickness in geometry. Power is the product of the first and second elements, force and velocity, and is therefore a function. Power is the first function in dynamics, and corresponds to the product of length and breadth, which is surface in geometry. Space is the product of the second and third elements, velocity and time, and is therefore a function. Space is the second function in dynamics, and corresponds to the product of breadth and thickness, which is the area of a cross-section of a solid in geometry. Work is the product of the three simple elements force, velocity and time, and is therefore a function. Work is the third 'function in dynamics, and corresponds to the product of length, breadth and thickness.which is volume in geometry. Work is also the product of the element force and function space, because the function space contains the elements velocity and time : like volume in geometry, it is the product of length and area of cross- section. Work is also the product of the function power and element time, because the function power contains the elements force and velocity : like volume in geometry, it is the product of area and thickness. MOTION. 59 DETAILED EXPLANATIONS OF THE ELEMENTS AND FUNCTIONS. FORCE. 3. Force is any action which can be expressed simply by weight, and is distinguished by a great variety of terms, such as attraction, repulsion, gravity, pressure, tension, compression, cohesion, adhesion, re- sistance, inertia, strain, stress, strength, thrust, burden, load, squeeze, pull, push, pinch, punch, etc., all of which can be measured or ex- pressed by weight without regard to motion, time, power, or work. All bodies in nature possess the incessant virtues of attracting and repelling one another, which action is recognized as force. The physical constitution of force is not yet known, but it appears that the force of repulsion is derived from heat, and that of attraction from cold or absence of heat. It is well known that bodies generally expand when heated, which proves the repulsive force of heat; and that bodies contract by de- crease of temperature shows the superior action of the attractive force in the absence of heat. The two forces act opposite to one another, and when the force of attraction is superior to that of repulsion, the body will maintain the form of solid. When the opposing forces are equal, the body will maintain the form of liquid, and when the force of repulsion is su- perior to that of attraction, the body will have the form of a gas. The force of universal attraction will be explained in its proper place. The unit for measuring force is any assumed weight, as pound or ton. MOTION. 4. Motion is a continuous change of position in regard to as- sumed fixed objects. Motion or rest are only relative ; that is to say, when two bodies change their relative positions, either one of them can be considered at rest and the other in motion. There is no abso- lute rest known in the universe. Our earth revolves around its axis and also in its orbit around the sun, but the sun also revolves both around his axis and in a small irregular orbit in regard to his planet- ary system, whilst the whole planetary system moves bodily in space. We are therefore not able to establish any absolute rest, but must consider motion and rest only as relative. Motion is expressed by the following terms : Move, going, walking, passing, transit, involution and evolution, run, locomotion, flux, rolling, flow, sweep, wander, shift, flight, current, etc. 60 ELEMENTS OF MECHANICS. Motion of Translation is when a body moves with a uniform velocity in a straight line without revolving, or when each particle of the body moves in parallel lines. Motion of Gyration is the same as rotation, or when all the particles in a body describe concentric circles around one common axis. Helicoidal Motion is when a body revolves around an axis, and at the same time moves in the direction of that axis, like a screw or rifle-ball, which is the result of the two motions of translation and gyration. Lateral Motion is that motion of translation which a body gen- erates in a direction out of its greatest extension. Lateral motion may also be referred to a stationary line or plane from or to which a body is moving. Small rivers run into larger ones from lateral directions. Rolling Motion is the combination of rotary motion of a body and lateral motion of the axis of rotation, like a wagon-wheel rolling on the ground. There are also different kinds of motions, designated by the name of the path, of the curve, described by the motion, such as parabolic motion, elliptic motion, cycloidal motion, etc. Space Motion. We say that a body has more or less motion in regard to greater or less space moved through, in which case motion means space. VELOCITY. 5. Velocity is rate of motion. It is obtained by dissolving the function space and eliminating the element time. 8 V T Velocity = = = F", the element. Velocity is independent of space and time, but in order to obtain its value or expression as a quantity, we compare space with time. Thus, when the value of velocity of a moving body is required, we measure a space which the body passes through, and divide that space with the time of passage, and the quotient is the velocity. Velocity is therefore expressed by space per time, as feet per second or miles per hour. A definite velocity can be expressed by any units of space and time, because velocity is an essential principle which cannot be resolved into two or more principles. A velocity of 22 feet per second is equal to a velocity of 15 miles per hour. For a definite velocity we can reduce the space and time to infinitely small, or say VELOCITY. 61 absolutely nothing, without affecting the velocity in the least ; which proves that velocity is an independent virtue or an element. Absolute Velocity is that measured or observed in the real path of motion. Apparent Velocity is that observed from an assumed fixed point, and which can be measured only as angular velocity. The terms velocity and motion are often used for one another, and neither one of them can exist without the other, because it is velocity which generates motion by the aid of time. Suppose a body to move from JP to P' , an ob- server at a would see less velocity or motion than would one at b ; but both motions are ap- parent velocities, which can be measured only by the angles P a P ' and PbP', whilst the abso- lute velocity must be measured in the path PP. Velocity or rate of motion is expressed by a variety of terms, as follows : QUICK MOTION. I, swiftness, rapidity, fleet- ness, speediness, quickness, haste, hurry, race, forced march, gallop, trot, run, rush, scud, dash, spring, etc. SLOW MOTION. Slowness, tardiness, dilatoriness, slackness, drawl, retardation, hob- bling, creeping, lounging, linger, sluggish, crawl, drawl, loiter, glide, langttid, drowsy, etc. Velocity is variously expressed by different units of length and time ; for example, in machinery it is generally estimated in feet per second or minute ; in steamboat and railway traveling, by miles per hour ; the velocity of light and electricity, by miles per second ; whilst the retirement of the Niagara Falls toward Buffalo, and the sinking and rising of land or sea may fee expressed 7 in feet or inches per century. The velocity of light in planetacy space is about 200,000 miles per second, or the same as that of electricity through good conductors. The velocity of the earth in its orlit around the sun in reference to a fixed star is about 19 miles per second; The velocity of a point on the earth's equator in reference to the sun is about 1037 miles per hour, or 1520 feet per second. Angular velocity is an apparent motion referred to a fixed centre, like that of a revolving wheel or an oscillating pendulum ; it is measured by periodical revolutions or oscillations, generally denoted bv the letter n. 62 ELEMENTS OF MECHANICS. TIME. 6. Time implies a continuous perception, recognized as duration. Chronology is the science of time. Instant and moment are points of time. Epoch is the beginning of any time marked with some remarkable events and recorded by historians or chronologists. Era is nearly the same as epoch, except that it is generally fixed by nations or denomi- nations, as the Christian era. Time is expressed by a great variety of units namely, millennium, a thousand years ; century, one hundred years ; score, twenty years ; year, season, month, fortnight, week, day, hour, minute and second. Cycle is a period of time in which similar phenomena of the heavenly bodies perpetually occur, such as the cycle of the sun, a period of twenty-eight years, at which the days of the week return to the same dates of the month. The most ordinary unit for measuring time is derived from the period of one revolution of the earth around its axis, in reference to the sun, which is called a solar day, and divided into 24 hours, each hour 60 minutes and each minute 60 seconds. Another unit of time is the period occupied by the earth in making one revolution around the sun, in reference to an assumed fixed star, which unit is called a sidereal year, and contains 365 days, 6 hours, 9 minutes and 9.6 seconds, mean solar time. We have no positively fixed standard for measuring time, for the period of one revolution of the earth around its axis, as well as that around the sun, are both liable to changes by meteors or heavenly bodies falling on the earth which accelerate or retard its motion. Such changes have, within the time of our astronomical records, been so small as to amount to probably not more than a few seconds in thou- sands of years ; but it may happen at any time that a heavenly body large enough might strike the earth and cause a change of time which would at once be perceived by us all without the aid of instruments for that purpose. (See Astronomy.} POWER. 7. Power is the product of force and velocity ; that is to say, a force multiplied by the velocity with which it is acting, is the power in operation. The English unit for measuring power is a force of one pound acting with a velocity of one foot per second, and called one foot-pound of power ; or one effect. POWER. 63 Man-power is a unit of power established by MORIN to be equivalent to 50 foot-pounds of power, or 50 effects ; that is to say, a man turning a crank with a force of 50 pounds and with a velocity of one foot per second is a standard man-power. An ordinary workingman can exert this power eight hours per day without overstraining himself. Horse-Power is a unit of power established by James Watt, to be equivalent to a force of 33,000 pounds acting with a velocity of one foot per minute, which is the same as a force of 550 pounds acting with a velocity of one foot per second. That is to say, one horse-power is 550 foot-pounds of power or effects, or 11 man-power of 50 effects each. The product of any force in pounds and its velocity in feet per second, divided by 550, gives the horse-power in operation. In Watt's rule for horse-power is given a velocity of only one foot per minute, which is equal to 0.2 or J-th of an inch per second about the velocity of a snail. The force corresponding to this velocity is 33,000 pounds, or about 15 tons, which is too large for a clear con- ception of its magnitude, and a horse can never pull with such a force. A horse can pull 550 pounds with a velocity of one foot per second, which is the most natural expression for horse-power. This expres- sion is used on the continent of Europe. Foreign Terms and Units for Horse-Power. Countries. Terms. Eng. translation Units. Eng. equivalent. English Horse-power. Horse-power. 650 foot-pounds. 550 foot-lbs. French Force de chevaL Force-horse. 75 kilogr. metres. 542.47 foot-lbs. German Pferde-krafte. Horse-force. 513 Fuss-funde. 582.25 foot-lbs. Swedish Hast-kraft. Horse-force. | 600 skalpund-fot. 542.06 foot-lbs. Russian Syl-lochad. Force-horse, j 650 Fyt-funt. 550 foot-lbs. The quantities Force and Power are clearly distinguished by different terms only in the English language. On the continent of Europe horse-power is called horse-force or force-horse, which does not distinguish force from power. Horse-force can be considered to be the force with which a horse can pull. The word force is needed in the Swedish and German languages. Puissance, in the French language, means power, but the term is not generally used in that sense in dynamics. In most of the continental languages there are words which cor- 64 ELEMENTS OF MECHANICS. respond with power, but they are not used in that sense in dynamics, where the term force is used for power. The Swedish and German word Jcraft ought to be used for power only, and not for force, as it is also used. The words expressing work are clear and definite in all languages. EFFECT. The term effect has been used to denote both foot-pounds of power and foot-pounds of work. These two kinds of foot-pounds have here- tofore not been clearly distinguished from one another, for which reason the term effect will hereafter be used only to denote foot- pounds of power. P = simple power in foot-pounds or effects. F= force in pounds. V = Velocity in feet per second. IP = Horse-power. Simple power P=F V. Effects. p Horse-power H? = 550 F V Horse-power H? = . 550 Man-power = . OU F V Man-power = . 50 One horse-power = 11 man-power. Any action of force producing motion is power, which is inde- pendent of time. In lifting a weight vertically the force F is equal to the weight lifted, but in drawing a load on a road the force of traction may be considerably less than the weight of the load. A weight of 1000 pounds lifted vertically with a velocity V= 2 feet per second requires a power of 2000 effects, which is equal to 2000 : 50 = 40 man-power, or 2000 : 550 = 3.63 horse-power. A load of 1000 pounds drawn on a horizontal road with a velocity F= 2 feet per second may require a tractive force of only F= 100 pounds, and the power will be 200 effects, or 4 man-power, which is only one-tenth of the power required in lifting the same weight ver- tically. SPACE. 65 Power is the differential of work, or any action which produces work, whether mental or physical. Power multiplied by the time of action is work work divided by time is power. Writers on dynamics have heretofore assumed that "power is the work done in a unit of time," which is an error. The number which expresses the work done in a unit of time, is equal to the number which expresses the power in operation ; but that does not prove the two quantities to be alike. When we say " in a certain time," which is equivalent to the ex- pression " per unit of time," we divide by the time. Work is the product of the three elements Force, Velocity and Time, and when we say "work per unit of time," we eliminate the time from the work, and the remainder is power, which is the product of force and velocity. Power may be expressed by the following terms : Traction, propulsion, impulsion, capability, puissance, labor, haul, drag, draw, heave, occupation, activity, vigor, energy, etc., or any action which implies force and motion without regard to time. SPACE. 8. Space in dynamics means linear space, which is a function of the second and third elements, velocity V and time T, and may be likened to the cross-section of a solid, which is a function of breadth and thickness. Space is herein denoted by 8- V T, which means that the space 8, expressed in linear feet, is the product obtained by multiplying together the velocity V and time T. Space cannot be generated or conceived without the two elements motion and time. In viewing a short linear space our mind flies so rapidly over it that we miss the conception of motion and time. The length of a piece of wood has been generated by the motion and time required for its growth, and when our mind surveys that length, motion and time are required for passing from one end of it to the other. In like manner, when we at a glance survey a bridge we are unable to appreciate its length without following it in contemplation from pier to pier, with some expenditure of time and velocity from one end to the other. So also when we imagine objects or cities far distant apart, our con- ceptions of the magnitude of the distances between them are exces- 6* E 66 ELEMENTS OF MECHANICS. sively vague and inexact without an imaginary transit over them from point to point, which .also requires both time and velocity. The distance run by a locomotive or steamboat is the product of the velocity and time of the trip, and no distance can be accomplished without either of these elements. Geometrical spaces are magnitudes of three different kinds namely, linear, superficial and voluminous. Linear space is that generated by the product of time and motion of a point. Superficial space is that generated by the product of velocity and time of lateral motion of a line. Voluminous space is that generated by the product of velocity and time of lateral motion of a plane. In determining velocity it appears as if motion were dependent on space and time, because we measure a space and divide it by the time, in order to form a conception of velocity or the rate of motion. Space is the product of time and velocity, and when we divide that product by time, the quotient will be the simple element velocity or rate of motion. In other words, when we divide the space with time we resolve the function space into its constituent elements and eliminate the time, and the quotient is the simple element velocity. If we divide the space with velocity, the latter is eliminated from the former, and the quotient is the simple element time. Space in dynamics means the generation of that space by velocity and time. A line of any kind cannot be drawn without velocity and time. A locomotive running with a uniform velocity of 30 miles per hour will make 2640 feet per minute or 44 feet per second; and if we diminish the space and time to infinitely small, or, say, absolutely nothing, the velocity of 30 miles per hour is still constant when passing that time and space reduced to a point. In geometry length is an element without regard to velocity or time, but in dynamics linear space means a physical function gene- rated by velocity and time. Length is a geometrical element. Space is a physical function. In regard to space being composed of velocity and time, the follow- ing question has been asked : The distance to the moon is space ; what has that to do with velocity and time ? The moon has never been on the earth, and consequently not moved from it with velocity and time ? The answer is as follows : In order to find the distance to the WORK. 67 moon, a space is measured on the earth's surface, and is obtained by velocity and time, which is converted into a base-line namely, the diameter of the earth. The moon's horizontal parallax is next measured, through which the velocity and time in the base-line is multiplied until it reaches the moon. Thus, the space to the moon is composed of velocity and time. WORK. 9. "Work is the product obtained by multiplying together the three elements, force F, velocity Fand time T, or work 7T= F V T. "Work may also be expressed by K= F S, or the product obtained by multiplying together the force F and space S, in which it appears as if work was independent of time ; but the time is included in the space S = V T. A given amount of work may be performed in any desired length of time, but the work is nevertheless dependent on whatever time consumed in its execution. A definite quantity of work is not confined to any definite ratio or relation to either of its constituent elements, for either one or two of them may vary ad libitum, but only at the expense of the remaining two or one. A definite quantity at work, only requires a definite product of the combined actions of the three elements. Work is thus dependent on time as well as on force and velocity, for without either one of these three elements it ceases to be work. If work was independent of time, then any amount of work could be accomplished in no time. The greatest amount of work known to have been accomplished in the shortest time, is that in the explosion of nitro-glycerine, which is instantaneous to our perception ; but it required time notwith- standing. Work may also be expressed by K= P T, or the product of power and time. The work of a steam-engine operating with a constant power, will be directly as the time of operation, and so with all labor, whether it be mechanical or manual. The longer we toil the more work will be done, but if we have no time to do the work it will remain undone. Much of the confusion in dynamics has arisen from misconception of the difference between specific quantities and abstract numbers. When a quantity is multiplied by an abstract number, the product will be the same as the sum of so many concrete quantities added together as indicated by the number, and the operation will change the magnitude, but not the nature, of that quantity. But when a 68 ELEMENTS OF MECHANICS. quantity is multiplied by another quantity, the product will be a third quantity of different nature from that of its constituent quan- tities. A force of 2 pounds working with a velocity of 3 feet per second is a power of 6 foot-pounds, which, multiplied by the abstract number 4, will be a power of 24 foot-pounds. But the same power, 6 foot-pounds, multiplied by the quantity 4 seconds, will be 24 foot-pounds of work. Three square feet multiplied by the abstract number 2 will be 6 square feet, but when three square feet are multiplied by a thickness of 2 feet the product will be 6 cubic feet. The erroneous expression that " power is the work done in a unit of time " implies that power is a portion of work. Power multiplied by a unit of time is work, and work divided by a unit of time is power. In both these cases the unit of time does not change the nu- merical value of the quantities, but it changes their nature from one to the other. A pool of water, say 1000 square feet of surface, is frozen over with ice one foot thick, and there will consequently be 1000 cubic feet of ice in the pool, which though identical in number with that of the surface, does not prove that a square foot is a cubic foot. The surface of the pool represents power; the thickness of the ice represents time, and the volume of the ice represents the work con- sumed in freezing it. UNITS OF WORK. FOOT-POUND. The English unit of work is assumed to be that accomplished by a force of one pound raising an equal weight one foot high, which unit is called a foot-pound. Then a force of 6 pounds working through a space of 4 feet is equivalent to 24 foot-pounds of work. This unit is very convenient for small amounts of work, but it is too small for many purposes in practice. FOOT-TON. English ordnance officers have adopted a larger unit for work, namely, foot-ton, which is used for expressing work of heavy ord- nance. It means the work of lifting one ton one foot high. WORKMANDAY. A laborer working eight hours per day can exert a power of 50 foot-pounds. A day's work will then be 50x8x60x60 = 1,440,000 foot-pounds of work, which may be termed a workmanday. FOOT-POUNDS. 69 All kinds of heavy work can be estimated in workmandays, such as the building of a house, a bridge, a steamboat, canal and railroad excavations and embankments, loading or unloading a ship, powder and steam-boiler explosions, and the capability of heavy ordnance, etc. The magnitude of the unit workmanday is easily conceived, be- cause it is that amount of work which a laborer can accomplish in one day. Work expressed in foot-pounds, divided by 1,440,000, gives the work in workmandays. A work of 20 workmandays can be accomplished by 20 men in one day, by one man in 20 days, by 4 men in 5 days, or by 10 men in two <.. 10. DIFFERENT KINDS OF FOOT-POUNDS. There are four different kinds of foot-pounds in mechanics namely, 1st. Foot-pounds of static momentum, which are force in pounds multiplied by its lever of action in feet. 2d. Foot-pounds of dynamic momentum are mass expressed in pounds, multiplied by velocity in feet per second. 3d. Foot-pounds of power (effects) are force in pounds, multiplied by velocity in feet per second. 4th. Foot-pounds of work are force in pounds, multiplied by space in feet. It will be observed that foot-pounds of static momentum and foot- pounds of work are both the product of force and linear space, from which it would appear that these two functions are substantially alike, but they are of entirely different nature. Static momentum is force multiplied by the geometrical element length, without regard to velocity and time ; in which case the force has nothing to do with the generation of that length. Work is force multiplied by the physical function space, which is generated by the two elements velocity and time. "Work done is expressed by the following terms : Hauled, dragged, raised, heaved, cultivated, tilted, broken, crushed, thrown, wrought, fermented, labored, embroidered, etc., or any expres- sion which implies the three simple elements, force, velocity and time. Power is the differential of work. Work is the integral of power. The following formulas show the different combinations of the dy- namic elements and functions. Should either or both the force and the velocity be variable or irregular, the mean action in the time T must be inserted, and the formulas will answer for any kind of operation. 70 ELEMENTS OF MECHANICS. 11- DYNAMICAL FORMULAS. Force or Pressure in Pounds. p w F= . . 1 JP I = .. . 3 v' # 550 IP ^ V FT' : Velocity in Feet per Second. 550 H> 7 F= " ^ ' ' ' I v p . 6 F' -" . 8 v ... FT Time of Action in Seconds. rr- O . .9 F 8 . 11 T v 550 IP' T=. . . 10 T _ % . 12 P F V' Power in Effects. P=FV. . . 13 P=550IP. . . 15 p _F8 . 14 P- K . 16 T ' T' ' Space Passed Through in the Time T. fa v T 17 a 550 yiP 1Q O V JL. . pr .. Xf 18 . -L (7 20 J* . . lo s p .. i-iW Horse -Power. H> P 21 FS 550' 550 T' ' . 23 rp -^ ^ 00 IP " 550' - 550 7* ' Work in Foot-Pounds. K-FVT. - 25 K=F8. . 27 K= P T. . . 26 JT=550IP 7! . 28 EXAMPLES. 71 It will be observed in the preceding formulas that an element is never divided by an element, but a function is divided by an element only when that function contains the element divided with. Power divided by velocity gives force, because power contains the elements force and velocity ; but power cannot be divided -by time, because time is not a constituent element of power. Work can be divided by either one or two of its three constituent elements. When work is divided by either two of its elements, the product will be the third element. Different elements or functions cannot be added to or subtracted from one another. Power or space cannot be added to or subtracted from work. Force, velocity or time cannot be added to or sub- tracted from space. When a formula contains several terms, all the terms must be of the same kind ; for instance : Work K= T( The terms within the parentheses are all power, which multiplied by time gives work. Mistakes in dynamical formulas are easily detected by the above rules. No element can be converted into an element of a different kind. g 12. EXAMPLES CORRESPONDING WITH THE FORMULAS. Force or Pressure in Pounds. Example 1. A power .P=6400 effects is operating with a velocity of V--= 12 feet per second. Required the force F? Example 2. The piston of a steam-engine of IP = 24 horses is mov- ing at the rate of V = 8 feet per second. Required the force F? 550 IP 550x24 __ F= = - = 1650 pounds. V 8 Example S. A work of K= 3266 foot-pounds is accomplished in a space #=16 feet. Required the force F? = = 204 pounds. 72 ELEMENTS OF MECHANICS. Example 4- A work of _fiT= 183600 foot-pounds was accomplished with a velocity "F"=18 feet per second in a time of 3 minutes, or T= 3 x 60 = 180 seconds. Required the force F? K 183600 Velocity in Feet per Second. Example 5. A body moves through a space of 8= 160 feet in a time of T=> 40 seconds. Required the velocity V? ^ S 160 V = = - = 4 ieet per second. T 40 Example 6. A power of P = 42G6 effects is operating with a force F= 760 pounds. Required the velocity F? P 4266 V= = -- - = 5.6 feet per second. F 760 Example 7. The cylinder of a steam-engine of IP = 160 horse- power is 24 inches in diameter, and the effective steam-pressure is 30 pounds to the square inch. Required the velocity of the steam- piston ? The area of the piston is 452.39 square inches, which multiplied by 30 pounds to the square inch will be a force of F= 13570.8 pounds. 550 IP 550x160 -^- = l35m8- = 6 - 5feetperSeC nd - Example 8. A work of K= 864360 foot-pounds is accomplished with a force of F= 68 pounds in a time of 5 minutes. Required the velocity V? The time T= 5 x 60 = 300 seconds. T . K 864360 = = " = 42 - 4fe Time of Action in Seconds. Example 9. A space of #=2896 feet is generated with a velocity of F= 25 feet per second. Required the time T? 115.84 seconds. 25 EXAMPLES. 73 Example 10. A force of .F= 4596 pounds is working through a space 8= 960 feet. What time is required for the force to generate a power of P = 840680 effects ? , FS 4596x960 . T= - = - 5.25 seconds. P 840680 Example 11 a. The stroke of a steam-piston is four feet, and the ef- fective pressure of steam is F= 46360 pounds. The power of the engine is EP = 500 horses. What time is required of the engine to make 64 double strokes ? The space #= 4 x 2 x 64 = 512 feet. F8 46360x512 r "550g" 550x500 Example 11 b. What time is required to raise a weight of 200 tons to a height of $=50 feet with an engine of IP = 8 horse-power? F= 200 x 2240 = 448000 pounds. _, FS 448000x50 or 8 minutes and 29 seconds. Example 12. What time is required to accomplish a work of JT= 96236000 foot-pounds, with a force .P=88 pounds, moving with a velocity of y=1.5 feet per second? - 96236000 88x1.5 or 202 hours 31 minutes and 6 seconds. Assuming a workmanday to be 1,440,000 foot-pounds, it would re- quire about 67 such units to accomplish the work ; that is to say, one man could do the work in 67 days, or 67 men could accomplish it in one day. Power in Effects or Foot-Pounds. Example 13. A weight of five tons is raised vertically at the rate of 1-j inches per second. Required the power P? The force F= 5 x 2240 =- 11200 pounds. Velocity V= 0.125 feet per second. P= 11200 x 0.125 = 1400 effects. 74 ELEMENTS OF MECHANICS. One man-power is 50 effects, and it would require 1400 : 50 = 28 men to raise five tons with a velocity of 1-J- inches per second at con- tinued work. One horse-power is 550 effects, and it would require 1400 : 550 = 2.55 horse-power for the same work. Example 14- What power is required to lift a weight of three tons a space of 8= 5 feet in a time of 10 minutes ? . FS 3x2240x5 , F= -f--^W- 56effeots ' Example 15. How many effects are there in IP = 30 horse-power? P = 550 x 30 = 16500 effects. Exampk 16. What power is required to do a work of K= 186000 foot-pounds in one minute ? T= 60. p= 186000 = 31000effects Space Passed Through in the Time T, Example 17. A body moving with a velocity of V= 960 feet per second for a time of T = 5 seconds. Required the space passed through ? #= T7T=4800 feet. Example IS. A power of P= 6500 effects is operating for a time of T= 12 seconds with a force F= 240 pounds. Required the space passed through ? Example 19. To what height can a steam-engine of IP = 6 horse- power lift a weight of 25 tons in a time of 5 minutes ? F= 25 x 2240 - 56000 pounds. T= 5x60 = 300 seconds. , 550xIPr 550x8x300 _ . , . The hexght 5- jH- = 56 ooo = 23 ' 6 feet Example 20. A work of K= 7280 foot-pounds is to be accom- plished by a force of F= 24 pounds. In what space can the force do the work ? EXAMPLES. 75 Horse-Power. Example 21. How many horse-power are there in P= 56680 effects? Example 22. A weight of three tons is to be raised with a velocity of F= 6 feet per second. Required the horse-power ? Example 23. A steam-crane is to be constructed to lift 30 tons 12 feet high in 5 minutes. Require 1 the horse-power ? Force F=> 30 x 2240 = 67200 pounds. Time T= 5 x 60 - 300 seconds. F8 67200x12 55 55300 Example 24. What horse-power is required to accomplish a work of K= 346000 foot-pounds in T=5 seconds? K 346000 WORK IN FOOT-POUNDS. Example 25. How much work is accomplished with a force of F= 280 pounds, moving with a velocity of V = 9 feet per second for a time of T= 1200 seconds, or 20 minutes? K= F V T= 280 x 9 x 1200 = 3024000 foot-pounds, or 2.1 workmandays. Example 26. How much work can be accomplished by a power of P=36 effects during T=4 seconds? K= p T= 36 x 4 = 144 foot-pounds. Example 27. A weight of 25 tons is lifted #-18 feet. Required the work 9 K= FS=25x 2240 x 18 = 1008000 foot-pounds. Example 28. How much work is accomplished per minute by an engine of IP = 48 horse-power? K= 550 IP T= 550 x 48 x 60 - 1584000 foot-pounds. 76 ELEMENTS OF MECHANICS. I 13. CIRCULAR OR ROTARY MOTION. In this case it is supposed that the force F is applied in the direc- tion of a tangent to the circle of radius r in feet, like that of a belt or rope over a pulley, or in all kinds of gearing. n = revolutions of the circle per minute. JV= total revolutions in the time T, or for generating a definite cir- cular space S, and also for the accomplishment of a definite work K. DYNAMICAL FORMULAS FOR CIRCULAR MOTION. Velocity in the Periphery of the Circle in Feet per Second. V= 60 = 0. 10472 r n. . 29 . 30 Revolutions of the Circle per Minute. 60 F SI Total Revolutions JV. 8 2 r r K . 33 . 34 Force in Pounds, acting in the Periphery. 5250 IP K . 35 . 36 Horse-Power, acting in the Periphery. fr * . 37 550x60 5250 F2nrN_ Fr N 550 T = 87.5 T' . 38 Work in Foot-Pounds, ac- complished in y Revolu- tions, or in the Time T. . 39 K- 60 T . 40 EXAMPLES FOR CIRCULAR MOTION CORRESPONDING WITH THE FORMULAS. Example 29. The radius of a wheel or a crank-pin is r = 2.5 feet, and makes n = 56 revolutions per minute. Required the velocity in the circumference? 60 - 0.1472 rn = 0.1472 x 2.5 x 56 = 20.6 feet per second. STEAM-ENGINES. Example 80. Required the velocity in the periphery of a fly-wheel of radius r = 8 feet, and making N= 125 revolutions in T- 164 seconds? ,. 2*rN 6.28x8x125 F= - = - = 39.25 feet per second. 14. DYNAMICS OF STEAM-ENGINES. The following formulas are for a double-acting steam-engine, of which the stroke of piston = s in feet. F= force or pressure of steam on the piston. If the steam is expanded in the cylinder, F means the mean pressure on the whole piston throughout the stroke s. n = double-strokes per minute. IP = horse-power of the engine. Velocity of the piston in feet per second, V= = . .41 60 30 __. 2 Fs n Fs n Horse-power of the engme, &-,. . .42 ^ ZFsnT FsnT Work done in the time T, K= . . .43 60 30 Work done in ^double-strokes, K=2Fs& , - ^ .44 Let A denote the area of the steam-piston in square inches, and p = mean steam-pressure in pounds per square inch. Then the force on the piston will be F= A p 45 And the horse-power, EP = . 46 16500 This is the gross horse-power of the engine, including that ex- pended in friction and working the pumps, and is generally called Indicated Horse-Power. The indicated horse-power is calculated from the indicator diagram or card taken for that purpose. 78 ELEMENTS OF MECHANICS. 15. Load of Burden that can be Carried by Man and Animals. Carriers. Eoad, kind of. Force. Burden, in pounds. Velocity. Feet per second. Time. Hours per day. Space, in miles. Man Good level.... 100 3 7 14.3 Man Ordinary . . 95 2.5 7 12 Man 50 35 10 238 Llama of Peru Donkey Mountainous. Good level.... 100 300 3.5 3.5 10 10 23.8 23.8 Donkey Mountainous. 200 3.5 10 23.8 Mule Good level 500 5.0 10 34 Mule.:. Horse Mountainous. Good level 400 300 4.5 6 10 g 40.6 327 Horse 300 45 g 245 Camel Deserts 1000 3 to 4 12 30 to 40 Elephant Ordinary 1800 3 to 4 10 35 Man or Animal "Working a Machine. Working- man or animal. Machine which is worked. Force, in pounds. Elements. Velocity. Feet per second. Time. Hours per day. Space, in feet. Functlo Power, in eflect. ns. Work, in foot- pounds. F V T s P K Man Rope and pulley 50 0.8 6 17280 40 864800 Man Crank 20 2.5 8 72000 50 1440000 Man Tread-wheel* 144 0.5 8 14400 72 2073600 Man Tread- wheel t 30 2.5 8 72000 75 2160000 Man Draws or pushes. 30 2 8 57600 60 1728000 Horse.... Horse-mill 106 3 8 64800 318 6768800 Horse.... Horse-mill 72 9 5 162000 648 11664000 Horse.... 4-wheel carriage. 154 3 10 108000 462 16632000 Horse.. / Revolving . 100 3 8 86400 300 8640000 Mule J mill 66 3 8 64800 198 4276800 Ass I platform. * 33 3 8 64800 99 2138400 * Axis horizontal. t Axis 24 from vertical. HORSE-PO WER. 79 I 16. HORSE-POWER REQUIRED TO DRIVE DIFFERENT MACHINES. WATER-WORKS. For every 100 gallons of water pumped per minute to a vertical IP height of 100 feet, requires 5 For every million (1,000,000) gallons pumped per 24 hours to a height of 100 feet, requires 35 ROLLING-MILLS. IP For every square foot of heated iron plate passing through the rollers, requires ......... 5 Bar-iron mills. Two pairs of rough and two pairs of fin- ishing rollers, six puddle furnaces, two welding furnaces, making 10 tons of bar iron per 24 hours, rollers making 70 revolu- tions per minute, require , . . .80 SAW-MILLALTERNATIVE. For every 100 square feet sawed per hour in dry oak, requires 5 Dry pine, per 100 square feet, requires 3 CIRCULAR-SAWS. A saw 3 feet in diameter, making 300 revolutions per minute, will saw 50 square feet per hour in dry oak, and requires . 2 Dry spruce, 90 square feet per hour ... . . .2 THRESHING-MACHINES. Velocity of feed-rollers at the circumference, 0.55 feet per second. Diameter of threshing-cylinder, 3.5 feet, and 4.5 feet long, making 300 revolutions per hour, can thresh 30 to 40 bushels of oats, and from 25 to 35 bushels of wheat per hour . 4 FLOUR-MILLS. One pair of mill-stones 4 feet in diameter, making 130 revo- lutions per minute, can grind 6 bushels of wheat to fine flour per hour ........... 5 Can grind 6 bushels of rye to coarse flour per hour . . 3 For every 100 pounds of fine flour ground per hour, requires . 1 80 ELEMENTS OF MECHANICS. Fig. 61. 17. DREDGING-MACHINES. The accompanying illustrations of dipper and grapple dredges are furnished by the American Dredging Company of Philadelphia. Dipper-Dredge. Fig. 61 represents the or- dinary dipper-dredge, consisting of one scoop, worked with a triple cnain wound on a 15-inch drum, and driven by a pair of engines 10 inches in diameter by 15 inches stroke of cylinders. Under ordinary work the scoop makes 30 to 40 dips per hour, and takes up about two cubic yards, or three tons, of materials each dip. The dipper-dredge is used in harbors and docks, and also in rail- road excavations. Fig. 62. Grapple-Dredge. Fig. 62 represents the grapple-dredge, consisting of a double scoop opening in the bottom like a mouth, and takes up about five tons of materials each grapple. It is worked by a single chain wound on a drum three feet in diameter, with a pair of engines MTEKT iMPHovEDTBRAppiE-oREDaE.- -^ ^c^eg diameter by 20 inches stroke of cylin- ders. Under ordinary work it makes 50 to 60 grapples per hour. Ladder-Dredge. The ladder-dredge consists of an endless chain upon which a number of buckets are fixed and work continually like a Noria. This appears to be the best form of dredge for deepening harbors, but is not so well suited for docks, where the dipper and grapple dredges are the best. FORMULAS FOR THE LADDER-DREDGE. 700 IP F- 550 Tk IP = horse-power required for excavating and raising the materials. T =tons of materials excavated and raised per hour. h = height in feet to which the excavated materials are raised. F= force in pounds required to feed the dredge ahead. V = velocity of the buckets in feet per second. k = 0.1 for hard clay with gravel. | k = 0.05 for common clay and sand. k = 0.07 for hard pure clay. j k = 0.04 for soft clay and loose sand. STEAMSHIP PERFORMANCE. 81 \ 18. POWER REQUIRED TO PROPEL STEAMBOATS AT DIFFERENT SPEEDS. M= nautical miles or knots per hour. T = displacement in tons, which must be well proportioned for IP = horse-power required to propel the vessel M miles per hour. w ~^f- Tons. 2 4 NAU 6 TICAL ] 8 MCILES C 1O >R KN0 12 rs PER 14 HOUB. 16 18 2O T IP IP IP IP H> IP H> IP IP IP 1 0.035 0.280 0.794 2.240 4.386 7.792 12.03 17.92 25.60 35.09 2 0.055 0.444 1.260 3.555 6.960 10.08 19.09 28.44 40.63 54.68 3 0.075 0.598 1.651 4.787 9.123 13.21 25.00 38.30 53.25 72.98 4 0.084! 0.673 1.900 2.389 11.05 15.20 30.31 43.11 64.51 88.41 5 0.102 0.818 2.207 6.550 12.22 17.65 35.17 52.40 74.85 97.76 6 0.115 0.924 2.620 7.392 14.47 20.96 39.70 59.13 84.48 115.8 7 0.128 1.025 2.906 8.198 16.05 23.25 44.03 65.50 93.70 128.4 8 0.130 1.120 3.176 8.96 17.54 31.17 48.12 68.70 102.4 140.3 9 0.151 11.211 3.430 9.690 19.00 27.44 52.10 77.52 110.1 152.0 10 0.162 1.300 3.684 10.40 20.35 29.47 55.82 83.20 118.8 162.8 15 0.213 1.702 4.827 13.62 26.66 38.62 73.50 108.9 156.0 213.2 20 0.258 2.064 5.845 1651 32.25 46.76 88.88 132.1 188.5 258.0 30 0.338 2.704 7.675 21.63 42.42 60.40 116.5 173.0 248.0 339.4 40 0.409 3.272 9.280 26.17 51.25 74.24 140.5 209.3 300.0 410.0 50 0.474 3.792 10.79 30.33 59.53 86.32 163.5 242.6 346.2 476.2 60 0.538 4.304 12.20 34.43 67.35 97.60 185.0 285.4 393.5 538.8 70 0.597 4.676 13.55 38.21 74.85 108.4 205.5 305.7 437.0 598.8 80 0.650| 5.200 14.8 41.60 81.60 118.4 224.0 332.8 476.0 652.8 90 0.705 5.640 16.00 45.22 88.40 128.0 242.5 361.7 516.0 707.2 100 0.755 6.040 19.40 48.4 94.5 163 259 387 551 756 150 0.990 7.720 22.51 61.76 124.0 180.0 341.5 494.1 724.0 992.0 200 1.200 9.600 32.5 76.9 150 260 412 615 875 1201 i 300 1.575 10.60 42.4 100 196 340 540 806 1146 15731 400 1.910 15.28 51.4 122 238 412 654 976 1402 1907 500 2.213 17.70 59.6 141 276 478 759 1131 1611 2213 600 2.50 20.00 67.2 160 313 540 856 1280 1820 2500 ! 700 2.780 22.24 74.6 177 377 599 938 1417 2016 2770 800 3.025 24.20 81.5 194 388 654 1038 1548 2206 3026 1000 3.500 28.00 94.6 225 439 759 1206 1798 2560 3514 1500 4.6111 36.88 124 295 575 995 1580 2355 3352 4605 2000 5.600 44.80 150| 356 696 1205 1913 2854 4060 5570 3000 7.350 58.80 197 467 913 1582 2508 3740 5318 7300 4000 8.87 70.96 238 567 1105 1912 3038 4530 6444 8847 6000 11.441 91.52 303| 742 1448 2507 3981 5935 8446 11586 82 ELEMENTS OF MECHANICS. 19. FRICTION. Sliding- Friction is the force required to rub or slide one sur- face upon another. For the same kind of surfaces the force of fric- tion is proportionate to the pressure of contact, and independent of the velocity with which the rubbing or sliding body moves. Within certain limits the friction is also independent of the extent of surface in contact. Fig- 63 - Let a b represent a horizontal surface _ on which is placed a body W in close con- _ */y^ tact with a b. The body W is attached to ~\^/l a weight / by a rope over a pulley. Ad- fps just the weight / so that it will barely K ' move the body W; then / is the force of friction of the surfaces in contact. If the body W be started with a certain velocity V, the weight / will con- tinue that velocity uniformly ; but if it is greater than the friction, the velocity will be accelerated. TP= weight of the body in pounds, which is the force or pressure of contact of the friction surfaces. f = force in pounds of the friction. # = ratio of W and/, or the coefficient of friction. V= velocity in feet per second of the motion. S= space of motion in feet. = = n=. W & Friction power, P = f V. Friction horse-power, IP = - . 550 Work of friction, K=f 8. The work expended on friction is generally converted into heat, which is not utilized, but lost. It is therefore of great importance in the working of machinery to reduce the work of friction to the lowest possible amount, for which reason lubricating substances are intro- duced between the friction surfaces, such as powdered graphite or soapstone, and all kinds of fatty substances, such as oil, tallow, lard, soap, etc. ; all of which reduce the friction coefficient, but to a dif- ferent degree, depending on how and on what kind of surfaces the lubrication is used. FRICTION. 83 The force of friction can be ascertained only by experiments which have been made by Coulomb, Vince, G. Rennie, N. Wood ; and the most complete and reliable experiments on friction were made by Arthur Morin in the years 1831, '32, and '33, at the expense of the French government. Friction surfaces must be either plane, cylindrical, spherical, con- ical or any figure concentric with an axis of rotation. 20. PLANE FRICTION SURFACE. The friction coefficient for plane surfaces is best determined by a body sliding on an inclined plane. h = height of the inclined plane. Fig M _ I = length, and b = base. W= weight of the sliding body. f= force of friction, which is equal to the force of gravity acting to draw the body down on the inclined plane. The inclined plane is so elevated or adjusted that the body barely moves by its own weight. Let x denote the angle of the inclined plane with the horizon, then the friction coefficient will be From the law of statics we have W:f=l:h, and Wk=jl. Friction coefficient

f cos.x it requires a force of F=(sin.x-fcos.x)W to hold the load or to prevent it from moving down the incline by its own weight. 3 34. ADHESION ON HORIZONTAL RAILS. Adhesion on rails is the force of sliding friction of the locomotive wheels. This force of adhesion is equal to the weight of the locomo- tive multiplied by the friction coefficient in Table I. MO = weight of the locomotive in pounds. A = force of adhesion, A = w $. This force must be greater than the tractive force required to move the train, in order to enable the locomotive to go ahead without sliding the wheels. W= weight of the train in pounds. /= coefficient of traction, Table V. There must be w $> Wfio produce motion. The locomotive wheels will slide on the rails when Wf> w $. TABLE VI. Coefficient of Adhesion on Rails. Condition of Rails. 0301 Very dry 0224 Under ordinary circumstances 0.20 In wet weather 0141 With snow or frost 0100 ADHESION ON INCLINED RAILS. The inclination of the track diminishes the force of adhesion as the cosine for the angle, x = angle of inclination of the track. Adhesion, A = w $ cos.x. The force of traction of the locomotive is limited to the force of RAILROADS. 95 adhesion, and when a train is running up or down an inclined track the tractive force must be F = Wfcos.x i W sin.x = or < w $ cos.x. + when the traction is upward. when the traction is downward. The traction and gravity of the train is in equilibrium when Wf CQS.X = Wsin.x, or when f cos.x = &in.x. The train will run down the track by its own force of gravity, when sin.x>f cos.x. The resistance of wind to the train is not included in the formulas for force of traction. Experiments have shown that the force of traction increases slightly with the velocity ; which is mostly due to resistance of the air. TRACTION-POWER. The tractive power is equal to the force of traction F multiplied by the velocity in feet per second ; and if desired in horse-power divide the product by 550. When the velocity is expressed in miles M per hour, the horse- power will be 96 ELEMENTS OF MECHANICS. 35. BELT AND PULLEYS. The best and simplest mode of transmitting motion from one shaft to another is by a belt and pulleys, which is very extensively used and it gives the smoothest motion. The motion is transmitted by the frictional adhesion between the surfaces in contact of the belt and pulleys, for which reason that friction must be greater than the ten- sion of the belt, otherwise the belt will slip and fail to transmit all the motion due from the driving pulley. There is always some slip in belt and pulleys, for which reason that mode of transmission is not positive or exact, and cannot be used where precise motions are required. Fig. 81 represents a belt transmitting motion between two parallel shafts a and b. If the motion is transmitted from a to b, the pulley Fi D is called the driving pulley, and d the driven pulley. The diameters of the pulleys can be of any desired proportions to suit the work of the machine. ' D = diameter and R = radius in inches of the driving pulley. d= diameter and r = radius of the driven pulley. L = length and 5= breadth of the belt in inches. F= force of tension in pounds of the pulling side of the belt. f= force of tension on the slack side. V= velocity of the belt in feet per second. 8= distance in inches between the centres of the two pulleys. JVand n = numbers of revolutions per minute of the respective pul- leys D and d.

126500 IP DN(F-f) 126500 dn_DN 230 230 ' 126500 IP ~ d(F-f)' Breadth of Belt from Expe- riments. , 2^/IP n d 2AF d ' 432 F d 4320 IP = ~n d 7.8 FV nd Bnd. " 4320 _B_nd a " 24 Bd " 2.4' ' 10 11 12 13 14 15 16 37. VULCANIZED RUBBER BELTS. The vulcanized rubber belting made by the New York Belting and Packing Company is composed of heavy cotton duck, woven expressly for that purpose and vulcanized between layers of a* metallic alloy, by which process the stretch is entirely taken out and the surface made perfectly smooth. This belting is said to be superior to, and is furnished for about half the price of, that of leather. It will stand a heat of 300 Fahr., and the severest cold will not affect its good quality, even if run in wet places or exposed to damp weather. The friction of the vulcanized rubber belting is about double that of leather, which makes it less liable to slip on the pulley. ATTRACTION. 99 38. MATTER. Matter is that of which bodies are composed, and occupies space. Matter is recognized as substance in contradistinction from geo- metrical quantities and physical phenomena, such as color, shadow, light, heat, electricity and magnetism. We have no knowledge of the origin or source of matter, but only know its existence and obedience to forces. Chemistry has, thus far, dissolved matter into some sixty-five distinct elements, but in the philosophy of mechanics we treat matter only as one simple element in relation to the three physical elements -force, 'motion and time. These four elements force F, motion V, time T and mass M are what constitute nature, and their different combinations cause the phenomena which we study and observe. The three first elements F, V and T are what constitute life, which physical combination with matter constitutes organic bodies. Physics divides matter into atoms, molecules, particles and bodies. Atom is the ultimate portion into which matter can -be divided. Molecule is a group of atoms. Particle is a group of molecules. Body is a group of particles, consisting of molecules and atoms of matter. 39 ATTRACTION OR GRAVITATION. It is a well-known and established fact that all bodies in nature have a mutual tendency to attract each other, the action of which is called universal attraction or gravitation. It is a constant action be- tween all kinds of matter, which cannot be disturbed by any other cause. Attraction, gravitation and gravity mean the san^e physical action, but there exist different kinds of tendencies between different kinds of matter to attract each other, which are independent of the general law of gravitation ; such as cohesion, capillar, molecular, chemical affinity, electric and magnetic attractions and repulsions, which are not called gravitation, and which actions are much under the control of human skill. The force of cohesion can be destroyed by a superior force crushing or tearing the body to pieces, or by the application of heat. The atoms of matter are acted upon by two opposite forces namely, molecular attraction and repulsion which cause bodies to exist in three aggregate forms namely, solid, liquid and gaseous, according to the relation between the two forces. 100 ELEMENTS OF MECHANICS. When the force of attractiou is superior to that of repulsion, the body will have the consistency of a solid ; but when the two forces are in equilibrium, the body will have a liquid form ; and when the force of repulsion is superior to that of attraction, the body will con- sist in the form of a gas. It has been anticipated that matter may be converted into a fourth aggregate form namely, that of an imponderable substance but the suggestion is not realized as a fact. All bodies in nature may be, or are capable of being, converted alter- nately into the three aggregate forms namely, solid, liquid and gase- ous although we have not yet so succeeded with some bodies. Ice, water and steam are the three aggregate forms of one body. It appears that the temperature of heat is the force of repulsion, and that the absence of heat (cold) allows the force of attraction to draw the atoms of matter in closer contact. In the case of liquids, it is said that the. two forces are in equilibrium, but we know that liquids differ widely in temperature, like that of water and molten iron. If the force of repulsion increases in some ratio with the tempera- ture, the force of attraction of the atoms of different elements must differ in accordance with the force of repulsion, for, otherwise, liquids could not vary so much in temperature. The physical constitution of the force of attraction is a mystery to us, and we have yet no hope of its ever being revealed. If there were eye-bolts in each atom of matter, and all were thus connected by elastic strings whose elasticity diminished as the square of their length, some conception could be formed of the nature of attraction ; but as there are no such eye-bolts, how does the force of attraction take hold of the atom ? It is self-evident, however, that something must exist between matter to form the connection of attraction ; and whether this " some- thing" be elastic strings or not, it cannot be cut off or interfered with in the least by any intervening means. Universal attraction, means the general force of attraction be- tween the heavenly bodies. Gravitation, means the same universal force of attraction, but implies that action on or near the surface of the earth or of any other heavenly body. ATTRACTION. 101 |40. LAW OF ATTRACTION. The law of universal attraction was anticipated by Copernicus, Tycho Brahe, Kepler, Fermat, Roberval and Hook, and finally estab- lished by Sir Isaac Newton. It is expressed as follows : The force of attraction is directly as the mass, and inversely as the sqttare of the distance. This expression will hold good when the mass of either one of the attracting bodies is taken as a unit, but, more correctly, the law ought to be expressed thus : The force of attraction between any two bodies is equal to the mass of the one body, 'multiplied by that of the ot/ier, and the product divided by the square of their distance apart. This is the universal law of attraction or gravitation upon which our existence wholly depends. {41.. ILLUSTRATION OF THE LAW OF ATTRACTION. Let a and b, Fig. 84, represent two particles of matter, supposed to constitute one body, and c, d, e and/ four particles, constituting an- other body. Each of the six particles is supposed to contain one unit of matter, and the distance D between the two bodies to be one unit of length. Draw straight lines between the particles, as shown in the illus- tration. The particle a will attract the particle e, as well as d, e and/, each with one unit of force ; the attraction, therefore, between the particle a and the body cdef will be four units of force ; and the particle b will also at- ? . tract the body cdef with four units of force ; so that the attraction between the bodies a b and cdef will be eight units of force, as represented by the eight lines drawn between them. The mass of the body a b is 2, that of c d efis 4, and the product of 2 and 4 is 8, the force of attraction, according to Newton's law. It is assumed in the illustration that the distance between the bodies is one unit of length, but if the distance be two units of length, the force of attraction will be only 2, and if the distance D is only half a unit of length, the force of attraction will be 32. 102 ELEMENTS OF MECHANICS. Let F denote the force of attraction between any two bodies of masses M and TO, D = distance between the bodies, then , Mm All force, power and work are derived from this law. The attraction of the sun draws heavenly bodies into it, and the heat generated by the collision is returned into space. It is the heat and light from the sun which decompose carbonic acid in our atmosphere, and promote the growth of vegetation on the surface of the earth, by which we are supplied with food and fuel for motive-power. The burning of the fuel reproduces carbonic acid, which rises into the air, where it is again decomposed by the heat of the sun. Thus, the work of heat is absorbed and reproduced alternately for ever and ever. These ideas of the sources of work agree with those of George Stephenson, Sir William Thomson, Waterston and Rankin. 42. Suppose three bodies A, B and C, to be fixed in a straight line, Fig. 85. and their distances apart as repre- ., >* . 1, sented by the letters a, b and c. (j\ /^ Let the mass or the real quan- VJ^ "j tity of matter in each body be represented by its letter, say A = 16, B = 8, and C= 4, their distances apart being a = I, b = 2, and Then the forces of attraction between the bodies will be as follows : Between A and B, force of attraction = = = 128. a 1 B C 8x4 Between B and C, force of attraction = = = 8. b* 4 A C 16x4 Between A and C, force of attraction = = = 7.1. c* 9 The attraction between the two bodies A and C is not interfered with by the body B ; nor will other bodies stationed or moving be- tween or about either one or both the masses A and influence the force of attraction between A and C. Therefore, the force of attrac- tion between any two particles of matter, whether embodied in a solid or porous mass or isolated in empty space, is equal to the product of the masses of the particles divided by the square of their distance apart. 43. Suppose a particle of matter, A, Fig. 86, to be enclosed with- in a hollow material sphere, abed, the particle being a unit of ATTRACTION. 103 matter acted upon in all directions by the attraction of the spherical surface. Draw the straight lines a d and b c through the particle A to re- present the sides of the two cones with a common vertex at A. Let 8 represent the height of the large cone, and r that of the small one. (a 6) = di- ameter and (a Z>) 2 the area of the base of the small cone ; (c d) = diameter and (c c?)* the area of the base of the large cone. The particle A is attracted in opposite direc- tions by the matter in the bases of the cones which constitute parts of the material spherical surface. The attraction in the direction of the arrow r is (at? In the direction of the arrow $ the attraction is The angles of the sides of the two cones are alike, or the angles aAb = cAd. . . (a b) : r = (c d) : 8. r S ' r* S* That is to say, the opposing forces of attraction are alike or in equilibrium at the particle A, or that the particle is equally attracted from all sides by the spherical surface. Now let the spherical shell be filled up with matter from the outer to the inner dotted concentric circle A ; the equilibrium of attraction on the particle A will not be disturbed by that matter. 44. If the particle A is located outside of the spherical shell, as represented by Fig. 87, the formulas of attractions will be the same as those for Fig. 86, but both the at- tractions will in this case act in one and the same T- direction on the particle, and the force of attraction will be Fig. 87. Let e denote the distance of the particle from the centre of the sphere, and R = radius of the sphere, then e = r + R, and r = e R. z bf (c dY (a bY (c dY J . \ ) _ V / . V ) (e-R)* e+R 104 ELEMENTS OF MECHANICS. That is to say, if the matter of the bases of the cones (a 6) 2 and (c c?) 2 were located in the centre of the sphere, the force of attraction of that matter on the particle A would be the same as when at the surface of the sphere. Therefore, if the hollow sphere be filled up solid with matter, its force of attraction on the particle A would be equal to the mass of that matter divided by the square of the distance from the centre of the sphere to the particle. 45. Fig. 88 represents a section of a solid sphere of homogeneous matter. A particle at A is attracted toward the centre only by the Fj gg concentric sphere enclosed by the dotted circle. All attractions of the hollow sphere outside of the dotted circle are in equilibrium on the particle A, as proved by 43. Let R denote the radius of the whole sphere, and ?- = that of the inner sphere. The mass or quantity of matter in a sphere can be represented by jR* or r*. Then the force of attraction on the i* particle A will be = ?-. That is to say, the force of attraction on any particle of the matter in a solid sphere is proportionate to the distance r of that particle from the centre. Suppose a hole to be made from the surface to the centre of the earth, and a body let down into it by a rope attached to a balance- scale ; then the weight of the body will decrease with the depth until it reaches the centre of the earth, where it will indicate no weight on the balance-scale placed at the surface, omitting the weight of the rope. r = radius of the earth. d = any depth to which the body is sunk in the hole. W= weight of the body at the surface of the earth, and w = weight of the body at the depth d. The radius r and depth d may be expressed in any unit of length, as well as W and w in any unit of weight. The mean radius of the earth is about r = 3956 miles, or r - 20887680 feet. When r and d are expressed in feet, the weight w will be W (20887680 -d) 20887680 WEIGHT. 105 WEIGHT. 46. The weight of a body is the force of attraction between the earth and that body. The weight of a body is greatest at the surface of the earth, and decreases above or below that surface. Above the surface the weight decreases as the square of its distance from the centre of the earth, and below the surface the weight decreases simply as its distance from the centre. The weight of a body A, Fig. 87, weighing 100 pounds at the sur- face a, b of the earth, would weigh only 25 pounds at a height equal to the radius of the earth above a b. A body A, Fig. 88, weighing 100 pounds at the surface of the earth, would weigh only 50 pounds at a depth of half the radius be- low the surface. Therefore, the weight of a body is not a constant quantity, whilst the quantity of matter in the body or the mass is constant wherever the body is weighed. A wholly isolated body has no weight, but is an inert mass, in- capable within itself of changing its own motion or rest. Any change in motion or rest of a body is derived from external force. The weight of a body is measured by the pressure it produces on its sup- port. Two bodies in equilibrium on a balance-scale at the surface of the earth will also be in equilibrium above or below that surface, be- cause the force of gravity acts equally on both bodies ; but the force supporting the balance-scale varies in accordance with the law of gravity. A spring-balance will indicate the true weight of a body hung upon it wherever it is weighed. The force of attraction between the earth and one gallon of distilled water at the level of the sea, in latitude of London, 51 31' N., is 10 pounds avoirdupois. The standard English gallon contains 277.274 cubic inches. The temperature of the water and of the air in which it is weighed should be 62 Fahr., and the barometer 30 inches. This is, however, not the true force of attraction between one gal- lon of water and the earth, for we must add the weight of 277.274 cubic inches of air which are displaced by the water. The weight of a cubic foot of dry air of temperature 62 Fahr. is 530 grains, which will be 0.01215 pounds for the capacity of one gallon. Therefore, when a gallon of water weighs 10 pounds in air the force of attrac- tion between that water and the earth will be 10.01215 pounds. 106 ELEMENTS OF MECHANICS. MASS. 47. Mass is the real quantity of matter in a body, and is pro- portioned to weight when compared in one or the same locality. Mass is a constant quantity, whilst weight varies with the force of gravity which produces it. The force of gravity accelerates or increases the velocity of a fall- ing body at the rate of 32 feet per second at the surface of the earth. This velocity is called the acceleratrix of gravity, and is generally de- noted by the letter g. When a force acts constantly on a body free to move, and the di- rection of the force passes through the centre of gravity of the body in the direction of motion, the velocity of the body will increase con- stantly as long as the force acts constantly. Let M denote the mass of a body. F= the constant acting force. T=time of action. F= velocity of the body at the end of the time T. These quantities bear the following relation to each other : M : F= T: F, and MV=FT. These functions are termed dynamic momentums, and distinguished as follows : Momentum of motion MV= F T, momentum of time. Neither of these momentums should be termed force. When F is expressed in pounds, T in seconds and F in feet per second, then the unit of mass will be 32.17 pounds, which is equal in number to the acceleratrix g for a falling body at the surface of the earth. MATT. 48. No specific name has yet been given to any unit of mass, the want of which makes this subject somewhat obscure. Although we are told that mass is equal to the weight divided by the acceleratrix <7 = 32.17, it does not make the same impression as if we had a specific name for the unit of mass, for which reason it is proposed to assign a name to it namely, matt.., from the word matter ; that is to say, one matt. = 32.17 pounds, or the mass expressed in matts. multiplied by 32.17 would give the weight of the mass in pounds. There are 69.63 matts. in a ton weight of 2240 pounds of matter. If W denotes the weight of a body in pounds, then its mass ex- pressed in matts. will be M- W - W J/ ~7~3217' MATT. 107 One matt. = the mass of 891 cubic inches of distilled water of temperature 40 Fahr. The adoption of this term matt, will distinguish mass from force. Although the weight of a mass is force of gravity, all forces are not weights of matter. The force which sets in motion a railroad train is independent of the force of gravity, but may be as well expressed by weight, as the mass of the train ; but we cannot solve the dynami- cal action without converting the weight or mass of the train into matte. This is the fundamental principle of dynamics of matter, which should be distinctly understood and remembered, and is of so great importance that it is well worthy of repeating namely, M:F=T:V, and HV=FT. We have heretofore been taught, in text-books and in colleges, that momentum MV is force, which is a great error. Force is only one element of momentum. Dynamic momentum divided by time is force ; that is to say, if a mass M= 4 matts. moves with a velocity of V= 6 feet per second, its momentum is 24. If a force is applied to stop the mass, and can do so in T= 3 seconds, then the force of that momentum is 24 : 3 = 8 pounds. Mass is inert, or incapable of changing its own motion or rest, and can, therefore, not be considered as force whether in motion or at rest. Force is required in bringing a mass from rest to motion, or from motion to rest ; but in either case that force must be applied from external causes independent of the mass. When a body in motion is suddenly stopped, like that of a falling body striking the ground, it is stopped by the force of resistance it meets with, and not by any force within itself. The inertia of a body free to move presents a resistance equal to any force applied on it, whether in motion or at rest. 1 49. TABLE FOR THE CONVERSION OF WEIGHT AND MASS. The following table is for converting weight into mass, or mass into weight, which will be very useful in examples of dynamics of matter. Example. Weight 957 pounds - 29.748 matts. See Table. Example. Convert the mass of M = 3466 matts. into pounds. Matts. = pounds. 8460 = 111310 6= 193 Matts. 3466 = 111503 pounds. 108 CONVERSION OF WEIGHT INTO MASS. O 1 2 TO 3 TS OP 4 POUNI 5 s. 6 7 8 9 Lbs. matts. matts. matts. matts. matts. matts. matts. matts. matts. matts. .03108 .06217 .09325 .12434 .15542 .18651 .21759J.24868 .27976 10 .31085 .34193 .37302 .40410 .43519 .46627 .49736 .52844!. 55953 .59061 20 .62170 .65278 .68387 .71495 .74604 .77712 .80821 .83929 '.87038 .90146 30 .93254 .96363 .99472 1.0258 1.0569 1.0880 1.1191 1.15011.1812 1.2123 40 1.2434 1.2745 1.3056 1.3467 1.3678 1.3989 1.4300 1.4610 1.4921 1.5232 50 1.5542 1.5853 1.6164 1.6575 1.6786 1.7097 1.7408 1.7718 1.8029 1.8340 60 1.8651 1.8962 1.9273 1.9684 1.9895 2.0206 2.0517 2.0827 2.1138 2.1449 70 2.1759 2.2070 2.2381 2.2792 2.3003 2.3314 2.3625 2.3935 2.4246 2.4557 80 2.4868 2.5179 2.5490 2.5801 2.6112 2.6423 2.6734 2.7044 2.7355 2.7666 90 2.7976 2.8287 2.8598 2.8909 2.9220 2.9531 2.9842 3.0152 3.0463 3.0774 100 3.1085 3.1396 3.1707 3.2018 3.2329 3.2640 3.2951 3.3261 3.3572 3.3883 110 3.4193 3.4504 3.4815 3.5126 3.5437 3.5748 3.6059 3.6369 3.6680 3.6991 120 3.7202 3.7612 3.7924 3.8235 3.8546 3.8856 3.9168 3.9478 3.9789 4.0100 130 4.0310 4.0720 4.1032 4.1343 4.1654 4.1964 4.2276 4.2586 4.2897 4.3208 140 4.3419 4.3829 4.4141 4.4452 4.4763 4.5073 4.5385 4.5695 4.6006 4.6317 150 4.6527 4.6937 4.7249 4.7560 4.7871 4.8181 4.8493 4.8803 4.9114 4.9425 160 4.9636 5.0046 5.0358 5.0669 5.0980 5.1290 5.1602 5.1912 5.2223 5.2533 170 5.2744 5.3154 5.3466 5.3777 5.4088 5.4398 5.4710 5.5020 5.5331 5.5641 180 5.5853 5.6263 5.6575 5.6886 5.7197 5.7507 5.7819 5.8129 5.8440 5.8750 190 5.8961 5.9371 5.9683 5.9994 6.0305 6.0615 6.0927 6.1237 6.1548 6.1858 200 6.2170 6.2480 6.2792 6.3103 6.3414 6.3724 6.4036 6.4346 6.4657! 6.4967 210 6.5278 6.5588 6.5900 6.6211 6.6522 6.6832 6.7144 6.7454 6.7765 6.8075 220 6.8387 6.8697 6.9009 6.9320 6.9631 6.9941 7.02537.0563 7.0874 7.1184 230 7.1495 7.1805 7.2117 7.2428 7.2739 7.3049 7.3361 7.3671 7.3982 7.4292 240 7.4604 7.4914 7.5226 7.5537 7.5848 7.6158 7.6470 7.6780 7.7091 7.7401 250 7.8612 7.8022 7.8334 7.8645 7.8956 7.9266 7.9578 7.9888 8.0199 8.0509 260 8.1721J8.1131 8.1443 8.1754 8.2065 8.2375 8.2687 8.2997 8.3308 8.3618 270 8.3930' 8.4239 8.4552 8.4863 8.5174 8.5484 8.5796 8.6106 8.6417 8.6727 280 8.7038 8.7347 8.7660 8.7971 8.8282 8.8592 8.8904 8.9214 8.9525 8.9835 300 9.0146 9.3255 9.0455 9.3564 9.0768 9.3877 9.1079 9.4188 9.1390 9.4499 9.1700 9.4809 9.2012 9.5121 9.2322 9.5431 9.2633 9.5742 9.2943 9.6052 310 320 9.6363 9.9472 9.6672 9.9781 9.6985 10.009 9.7296 10.040 9.7607 10.072 9.7917 10.102 9.8229 10.133 9.8539 10.165 9.8850 10.196 9.9160 10.227 330 110.258 10.289 10.320 10.351 10.383 10.413 10.444 10.476 10.507 10.538 340 10.569 10.600 10.631 10.662 10.694 10.724 10.755 10.787 10.818 10.849 350 10.880 10.911 10.942 10.973 11.005 11.035 11.066 11.098 11.129 11.160 360 11.191 11.222 11.253 11.284 11.316 11.346 11.377 11.409 11.440 11.471 3 7 Oil 1 1.501 11.532 11.563 11.594 11.626 11.656 11.687 11.719 11.750 11.781 380111.812 11.843 11.874 11.905 11.937 11.967 11.998 12.030 12.061 12.092 390 12.123 12.154 12.185 12.216 12.248 12.278 12.309 12.341 12.372 12.403 400 12.434 12.465 12.496 12.527 12.559 12.589 12.620 12.652 12.683 12.714 410 12.745 12.776 12.807 12.838 12.870 12.900 12.931 12.963 12.994 13.025 420 13.056 13.087 13.118 13.149 13.181 13.211 113.242 13.274! 13.305 13.336 430 13.367 13.398 13.429 13.460 13.492 13.522 113.553 13.585 13.616 13.647 440 13.678 13.709 13.740 13.771 13.803 13.833 13.864 13.896 13.927 13.958 450 13.989- 14.020 14.051 14.082 14.113 14.144 14.175 14.207 14.238 14.269 460 14.300 14.331 14.362 14.393 14.424 14.455 14.486 14.518 14.549 14.580 470 14.610 14.642 14.673 14.704 14.735 14.766 14.797 14.829 14.860 14.891 480 490 500 14.922 1 14.953 14.984 15.015 1 15.046 15.232 15.264 15.295 15.396, 15.357 15.5421 15.574 15.605! 15.6361 15.667 15.077! 15 108 15.388 15.419 15.698 15.729 15.140 15.451 15.760 15.171 15.482 15.791 15.202 15.513 15.823 CONVERSION OF WEIGHT INTO MASS. 109 O 1 2 TJ 3 1ITS OF 4 potmr 5 >s. 6 7 8 9 Lbs. matts. matts. matts. matts. matts. matts. matts. matts. matts. matts. 500 15.542 15.574 15.605 15.636 15.667 15.698 15.729 15.760 15.791 15.823 510 15.852 15.8841 15.915 15.946 15.977 16.008' 16.039 16.070 16.101 16.133 520 15.162 16.194 16.225 16.256 16.287 16.318! 16.349 16.380 16.411 16.443 530 540 16.473 16.784 16.505 16.536 16.816 16.847 16.566 16.877 16.598 16.909 16.629 1 16.660 16.940 16.971 16.691 16.722 16.754 17.002 '17.033 17.065 550 17.095 17.127 17.158 17.188 17.220 17.251 17.282 17.313 17.344 17.376 560 17.406 17.438 17.469 17.499 17.531 17.562 17.593 17.624 17.655 17.687 570 580 17.717 18.028 17.748 18.059 17.779 18.090 17.810 17.842 18.121 18.153 17.872 18.183 17.903 18.214 17.934 18.245 17.966 18.277 17.997 18.308 590 18.339 18.370 18.401 18.4321 18.464 18.494 18.525 18.556 18.588 18.619 600 18.651 18.682 18.712 18.744 18.775 18.806 18.837 18.868 18.899 18.930 610 18.962 18.993 19.023 19.055 19.086 19.117 19.148 19.179 19.210 19.241 620 19.273 19.304 19.334 19.366 119.397 19.428 19.459 19.490 19.521 19.552 630 19.584 19.615 19.646 19.677 19.708 19.739 19.770 19.801 19.832 19.863 640 19.895 19.926 19.957 19.988 20.019 20.040 20.081 20.112 20.143 20.174 650 20.206 20.237 2.0.268 20.299 20.330 20.351 20.392 20.423 20.454 20.485 660 20.517 20.548 20.579 20.610 20.641 20.662 20.703 20.734 20.765 20.796 670 20.828 20.859 20.890 20.921 20.952 20.973 21.014 21.045 21.076 21.107 680 21.139 21.170 21.201 21.232 21.263 21.384 21.32521.356 21.387 21.418 690 21.449 21.480 21.511 21.542 21.573 21.694 21.635121.666 21.697 21.728 700 21. 759 121.790 21.821 21.852 21.883 21.914 21.945 21.976 22.007 22.038 710 22.070 22.101 22.132 22.163 22.194 22.225 22.256 22.287 22.318 22.349 720 22.381! 22.41 2 22.443 730, 22.692 22.723 22.754 22.474 22.505 32.785 22.816 22.536 22.847 22.567 22.598 22.878 22.909 22.629 22.940 22.660 22.971 740 23.003; 23.034 23.065 23.096 23.127 23.158 23.18923.220 23.251 23.282 750 23.31423.34523.376 23.407 23.438 23.469 23.50023.531 23.562 23.593 760 23.625 23.656 23.687 23.718 23.749 23.780 23.811 23.842 23.873 23.904 770 23.93623.967 23.998 24.029 24.060 24.091 24.122 '24.153 24.184! 24.215 780 24.246 24.277 24.318 24.339 24.370 24.401 24.432 24.463 24.494 i 24.525 790 24.557 24.588 24.629 24.650 24.681 24.712 24.743 24.774 24.805 24.836 800 24.868 1 24.899 24.930 24.961 24.992 25.023 25.054 25.085 25.115 25.147 810 25.179 1 25.210 25.241 25.272 25.303 25.334 25.365 25.396 25.427 25.458 820 25.49025.521 25.5')2 25.583 25.614 25.645 25.676 25.707 25.738:25.769 830i!25.801 25.832 25.863 840 126.112 26.143 26.174 850 126.423 26.454 26.485 25.894 26.205 26.516 25.925 26.236 26.547 25.956 26.267 26.578 25.987 26.298 26.609 26.018 26.049 126.080 26.32926.36026.391 26.640 26.671 ! 26.7 02 860, 26.73426.765 26.796 26.827 26.858 26.889 26.920 26.951 26.982 127.013 870 27.045 27.076 '27.107 27.138 27.169 27.200 27.231 27.26227.29327.324 880: 27.356 27.387 27.418 27.449 27.480 27.511 27.542 27.573 27.604 27.635 890 27.666 27.697 27.728 27.759 27.790 27.821 27.852 27.883 28.914 27.945 900 27.976.28.008 28.039 28.070 28.101 28.132 28.163 28.194 28.225 28.256 910 28.287 28.318 28.349 28.380 28.411 28.442 28.473 28.504 28.535 28.566 920 28.59828.62928.660 28.691 28.722 28.753 28.784 28.815 28.846 28.877 930 28.90928.93028.971 29.002 29.033 29.06429.095 29.126 29.157 29.188 940 29.220 29.241 29.282 29.313 29.344 29.375-29.406 29.437 ! 29.468 '29.499 950 29.53129.55229.593 29.624 29.655 29.686.29.717 29.748 j 29.779 29.810 960 29.842^29.863 29.904 29.935 29.966 29.997 ! 30.028 30.059 ' 30.090 j 30.121 970 30.153 ' 30.174 30.215 30.246 30.277 30 308 '30.339! 30.370 30.401 30.432 980 30.454 30.485 30.526 30.557 30.588 30.619 30.640 30.681:30.711 30.743 990 30.765 30.796 30.837 30.868 j 30.899 1000 31.085 31.107 31.148 31.179J31.210 30.930'30.961 ! 30.992 31.023 31.054 31.241:31.272 31.303 31.334l31.365 110 CONVERSION OF MASS INTO WEIGHT. o 1 2 UNITS OF 3 4 MATTS. 56 7 8 9 Malts ibs. Ibs. Ibs. Ibs. Ibs. Ibs. | Ibs. Ibs. Ibs. Ibs. 32.17 64.34 96.51 128.68 160.85 193.02 225.19 257.36 289.53 lOi 321.7 353.9 386.0 418.2 450.4 4-32.5 514.7 ! 546.9' 578.0 611.2 20 643.4 675.6 707.7 739.9 772.1 804.2 836.4 868.6 899.7 932.9 30 965.1 997.3 1029.4 1061.6 1093.8 1125.9 1158.1 1190.3 1221.4 1254.6 40 1286.8 1319.0; 1351.1 ; 1383.3 1415.5 1447.6 1479.8 1512.0 1543.1 1576.3 50; 1608.5 1640.7 1672.81 1705.0 1737.2 1769.3 1801.5; 1833.7; 1864.8 1898.0 60 ! ] 1930.2 1962.4 1994.5 2026.7 2058.9 2091.02123.22155.42186.52219.7 70 2251.9 2284.1 ; 2316.2 2348.4 2380.6 2412.7 2444.9 2477.1 25082 2541.4 8012573.6 2605 9 2637.9 2670.1 2702.3 2734.4 2766.6 2798.9 2830.0;2863.1 90IJ2895.3 2927.6 2959.6 1 299 1.8 3024.0 3056.1 3088.3 3120.6 3152.6 3184.8 100 3217 3249.3 3281.3 3313.5 3345.7 3377.83410.0,3442.33474.3,3506.5 110 ! 3538.7 3571.0 3603.0 3635.2 3667. 4 3699.5 3731.7 3764.0 3796.0 3828.2 120 3860.4 3892.7;3924.7 3956.9 3989.1 4021.2 4053.4 4085.7 4117.7 4149.9 130=4182.1 4214.414246.4 4278.6 4310.8 4342.9 4375.1 4407.4 4439.4 I 4471.6 140 4503.8 4536.0 4568.1 4600.3 4632.5 4664.6 4696.8 4729.1 4761.1 4793.3 150; 4825.5 48-37.7 4889.8 4922.0 4954.2 4986.3 5018 5 5050.8 5082.8 5115.0 160 5147.2 5179.4 5211.5 5243.7 J5275.9 5308.0 5340.2 5372.5 5404.5 5436.7 170 5468.9 5501.1 5533.2 5565.4 5597.6 5629.7 5661.9 5694.2 5726.2 5758.4 ISO 5790.6 190' 61 12.3 5822.8 5854.9 5887.1 5919.3 6144.5 6176.6 6203.8 6241.0 5951.4 5983.6 6015.9 6047.9 6080.1 6273.1 6305.3 6337.61 6369.6 1 6401. 8 200 ; 6434.0 6466.2 6498.3 6530.5 6562.7 6594.S 6627.0 6659.3 6691.3 6723.5 210 6755.7 6787.9 6820 6852.2 6884.4 6916.5 6948.7 6981.0 7013.0 7045.2 220| 7077.4 7109.617141.7 J7173.9 7206.1 7238.2 7270.4 7302.7 7334.7 7366.9 230 7399.1 7531.3 ! 7463.4 7495.6 7527.8 7559.9 7492.1 7624.4 7656.4 7688.6 240 7720.8 7753.0 7785.1 7817.37849.5 7881.6 7913.8 7946.1 7978.1 8010.3 250 8042.5 8074.7 8106.8 8139.08171.2 8203.3 8235.5 8267.8 8299.8 8332.0 260' 18364.2 8396.4 8428.5 8460.7 8492.9 8525.0 8557.2 ! 8589.5 '8621. 5 8653.7 270! 8685.9 8718.1 8750.2 8782.4 8814.6 8846.7 8878.9 8911.1 8943.2 8975.4 280 1 9007.6 9039.819071.9 9104.119136.3 9168.4 9200.6 9232.8 9264.9 9297.1 290! 9329.3 9361.5 9393.6 300 9651.0 9683.2 9715.3 9425.8*9458.0 9747.5 9779.7 9490.1 !9522.3 9554.5 9586.6 9618.8 9811.8,9844.0 9876.2 9908.3 9940.5 310 9972.7 10005 10037 10069 10101 10133. 10166 10198 10230 10262 320 10294 10327 10359 10381 10423 104551 10488 10520 10552 10584 330 10616 10649 10681 10703! 10745 10777 10810 10842 10874 10906 340 10938 10970 11002 11034! 11066 11098! 11131 11163 11195 11227 350 11259 11292 11324 11356 11388 11420 11453 11485! 11517 11549 360 11581 11614 11646 11678' 11710 11742 11775 11807 11839 11871 370 11903 11936 11968 12000| 12032 12064 12097 12129 121611 12193 380 12225 12257 12289 12321) 12353 12385 12418 12450 12482 12514 390 12547 12-37<) 15611 12643 12675 12707 12740 12772 12804 12836 400 12868 12901 12933 12965! 12997 13029 13062 13094 13126 13158 410 13189 13222 13254 13286 13318 13350 13383: 13415 13447 13479 420 13510 13543 13575 13607 13639 13671 13704; 13736 13768 13800 430 13832 13865 13897 13929 13961 13993 14026 14058 14090 14122 440 14154 14187 14219 14251 14283 15315 14348 14380 11412 14441 450 14476 14508 14540 15572 14604 14636 14669 14701 14733 14765 460 14798 14830 14862 14894 1 14926 14958 14991 15023 15055 15087 470 15120 15152 15184 15216 15248 15280 15313 15345 15377 15409 480 '15441 15473 15505 15537 15570 15602 15634 15666 15698 15731 490 15763 15795 15827 15*59 15892 15924 15956 15988 16020 16053 500 16085 16117 16149 16181 16214 16246 16278 16310 16342 16375 CONVERSION OF MASS INTO WEIGHT. Ill O 1 2 XJ 3 NITS 01 4 - MATT 5 3. 6 7 8 9 Malts Ibs. Ibs. Ibs. Ibs. Ibs. Ibs. Ibs. Ibs. Ibs. Ibs. 500 16085 16117 16149 16181 16214 16246 16278 16310 16342 16375 510 16406 16438 16470 16502 16535 16567 16599 16631 16663 16696 520 16728 16760 16792 16824 16857 16889 16921 16953 16985 17018 530J 17050 17082 17114 17146 17179 17211 17243 17275 17307 17340 540 17372 17404 17436 17468 17501 175331 17565 17597 17629 17662 550 17693 17725 17757 17789 17822 17854 17886 17918 17950 17983 560118015 18047 18079 18111 18144 18176 18208 18240 18272 18305 570i 18337 18369 18401 18433 18466 18498 18530 18562 18594 18627 580 18059 18691 18723 18755 18798 18820 18852 18884 18916 18949 590 18980 19012 19044 19076 19109 19141 19173 19205 19237 19270 600; j 19302 19334 19366 19398 19431 19463 19495 19527 19559 19592 610 19624 19656 19688 19720 19753 19785 19817 19849 19881 19914 620 19946 19978 20010 20042 20075 20107 20139 20171 20203 20236 630 20-267 20299 20331 20363 20396 20428 20460 20492 20524 20557 640 | 20589 20621 20653 20685 20718 20750 20782 20814 20846 20879 650 20911 20943 20975 21007 21040 21072 21104 21136 21168 21201 660 21233 21265 21297 21329 21362 21394 21426 21458 21490 21523 670 21554 21586| 21618 21650 21683 21715 21747 21779 21811 21844 6SO 21876 21908 21940 21972 22005 22037 22069 22101 22133 22166 690 22198 22230 22262 22294 22327 22359 22391 22423 22455 22488 700 22519 22551 22583 22615: 22648 22680 22712 22744 22776 22809 710 22841 22873 22905 22937 22970 23002 23034 23066 23098 23131 7 20 '23103 23194 23227 23259 23292 23324 23356 23388 23419 23453 730 234S4 23516 23548 23580 23613 23645 23677 23709 23741 23774 740 23806 23838 23S70 23902 23935 23967 23999 24031 24064 24096 750! 24128 241601 24192 24224 24257 24289 24321 24353 24386 24418 760' 24450 24482 24514 24546 24579 24611 24643 24675 24708 24740 770 ; 24771 24803 24835 24867 24900 249321 24964 24996 25029 25061 780 i 25093 25125 25157 25189 25222 25254 25286 25318 25350 25383 790 25415 25447 25479 25511 25544 25576 25608 25640 25671 25705 800 25730 25768 25800 25833 25865 25897 25929 25961 25993 26020 S10 1 26057 26089 26121 26154 26186 26218 26250 26282 26314 26347 820 26379 26411 26443 26476 26508 26540 26572 26604 26636 26669 830 26701 26733 26765 26798 26830 26862 26894 26926 26958 26991 840! 27023 ' 27055 27087 27120 27152 27184 27216 27248 27280 27313 850! 27344 27376 27408 27441 27473 27505 27537 27569 27601 27634 860 27666 27698 27730 27763 27795 27827 27859 27891 27923 27956 870, 27988 28020 28052 28085 28117 28149 28181 28213 28245 28278 ssO 28309 28341 28373 28406! 28438 28470 28502 28534 28566 28599 890 28031 28603 28695 28728 28760 28792 28824 28856 28888 28921 900 28953 .28985 29017 29050 29082 29114 29146 29178 29210 29243 910 29275 29307 29339 29372; 29404 29436 29468 29500 29532 29565 920129590 29628 29660 29693 29725 29757 29789 29821 29853 29886 930 29918 29950 29982 30015 30047 30079 30111 30143 30175 30208 940 30240 30272 30304 ! 30337 : 30369 30401 30433 30465 30497 30530 950 30561 30593 30625' 30658 30690 30722 30754 30786 30818 30851 960 30883 30915 30947 30980 31012 31043 31076 31108 31140 31173 970 31205 31237 31269 31302 31334 31305 31398i 31430 31462 31495 980 ! 31527 31559 31591 31624 31056 31687 31720 1 31752 31784 31817 990 : 31848 31880 31912 31945 31977 32009 32041 ; 32073 32105 32138 1000 32170 32202 32234 32207 32299 32331 32303 32395 32427 32400 112 ELEMENTS OF MECHANICS. 350. THE EARTH'S ATTRACTION ON ITS SURFACE. The attraction or weight of bodies on the earth's surface is slightly influenced by various causes, namely : 1st. The flatness of the poles makes the radius shortest in the direction of the earth's axis, which increases the force of attraction at the poles. The radius H of the earth in feet at any latitude L is about .#k 20887680(1 + 0.00164 cos. 2). 2d. The centrifugal force on the surface of the earth, which varies as the cosine for the latitude and acts in opposition to the force of attraction, is as follows : ^ __ A Rri 1 cos.L*^" 2933^5 ' in which A = unit of attraction at the radius R. n = revolutions per minute. (60x24) 2 2073600 3d. The centrifugal force of the earth's rotation around the sun influences the attraction as the sine of the sun's angle with the sur- face of the earth. This influence is so insignificant that it could not be detected in the most delicate scientific experiment yet known. 4th. The attraction of the sun and moon influences the attraction of bodies on the earth's surface directly as the sine of the sun's or moon's angle with the surface, and is demonstrated by the tidal wave, causing high and low water. {51. MASS OF THE EARTH. It is not necessary for the purpose of this treatise to observe the precision required in highly scientific investigations, for which reason we will limit ourselves within the supposition that the earth is a per- fect sphere of radius R = 20,887,680 feet, and is at rest. The delicate pendulum experiments of Cavendish approximated the mean density of the earth to be 5.44 times that of water. The weight of a cubic foot of distilled water of temperature 39 Fahr. is 62.388 pounds. Then the mass of the earth, expressed in matts., will be M ^*& 62.388x5.44 = 402^35^00,000,000,000,000,000 matts. 3 x 32.17 Log. M= 23.6050086. MASS OF TEE EARTH. 113 Let r denote the radius of the earth expressed by such a unit of length that the force F of attraction will be 82.17, which is the force of attraction between the earth and one matt, of matter on the surface of the earth. Then we have that radius of the earth to be ' '-li?' and r^lf, 111,886,700,000 units. Loff.r- 11.0487787. = 5356.59, and = 28,693,080. Log. = 3.7288886. Log. = 7.4577772. The force of attraction F in pounds between any two masses M and m, expressed in matts., and their distance apart D in feet, will be Example. Two bodies J^=450 and m = 360 matts. are held at a distance D = 2 feet apart. Required the force of attraction between them? ^=2i52. = a0014115ofa ^ d - When the masses of the attracting bodies are expressed by weights W and w in pounds, we have W= g M, and w = g m, W or M = , and m w . 9 9 y y 28693080 g 1 = 29,694,700,000, kg. = 10.4726794. Force of attraction, J-, __|^__. 10 H 114 ELEMENTS OF MECHANICS. 252. ATTRACTION OF A MOUNTAIN. The accompanying illustration represents a mountain of known quantity of matter M, at the side of which is hung on a string a b a mass m. The dotted line a c represents the vertical in which the body Ttt would hang if not attracted by the mountain, and a b is the resultant of the two forces of attraction from the earth and moun- tain. Let v denote the angle b, a, c formed by the pendulum, then the cosine for v represents the earth's attraction, and sin.v that of the mountain. W= weight of the body m, which is the force of attraction in the direction a, c, and F force of attraction of the mountain, we have Fig ' 89> F sin. v tan.v. W cos. v = W tan.v, and W^Fcot.v. To find the distance D between the body m and the centre of at- traction of the mountain is a rather complicated problem, which, however, can be approximated by knowing the form of the moun- tain. With the body m as a centre draw concentric circles equal dis- tances d apart through the mountain, representing spherical sections of areas e,f, g, h, etc., and call A = the sum of all the spherical sections. Reduce each area to a zone of the same radius, and multiply it with cosine for } the angle occupied by that zone ; the product will be the areas e,f, g, h, to be inserted in the following formulas : Then, and . f g h +> + * + - +,etc. \'l 4 9 16 The base-line of the mountain need not be referred to the level of the sea, but to the level of the ground surrounding it. Example 1. The quantity of matter in a mountain is estimated to INERTIA. 115 be M = 1,250,000,000,000 matte.; the distance between the mass m and centre of attraction of the mountain is approximated to D = 5000 feet ; and the weight of the body TF= 321.7 or the mass m = 10 inatts. Required the force of attraction between the body m and mountain M ? 1,250,000,000,000x10 == 28693080x5000- Example 2. In connection with the preceding example the body m is hung on a string 20 feet long. Required the angle v and the linear deviation a, b of the body ? Formula, tan.v = = - = 0.0000542, TV 321.7 which corresponds to an angle of v = 00' 11". The linear deviation a, b will be a, & = 20 xm.t; = 0.0011 feet = 0.0132 inches. An artificial horizon at the place of the body m would have the inclination. $ 53. INERTIA. * Inertia is the force of resistance which a body presents to any ex- ternal force tending to change the motion or rest of that body. An isolated rigid body is incapable within itself of changing its own state of motion or rest, which condition is called inert. A system of bodies by their own virtues of attraction may be capable of changing each other's motion or rest, as is the case with the heavenly bodies, which by their combined attractions regulate one another's motion in space. A falling body does not move simply by the earth's attraction alone, but by the combined virtues of attraction between the two bodies. If a 1 ody had no virtue of attraction, no other body could exercise any attraction on it, but there exists no ponderable matter without the virtue of universal attraction. If a body on the earth's surface could be deprived of its virtue of universal attraction, the earth's attraction would have no effect upon it, and that body would have no weight, but could be placed at rest anywhere in space without support. Two bodies of iron attract one another by their virtue of magnetic attraction, but a body of iron exercises no magnetic attraction on a, body of copper, because the latter possesses no such virtue of attraction..' 116 ELEMENTS OF MECHANICS. 54. FALLING BODIES. Falling bodies are drawn towards the earth by the force of gravity pr the earth's attraction, which force is equal to the weight of the *ody. F: M= V: T, But F = W, and M= , therefore W : = V : T of which V=g T. 9 That is to say, the velocity of a falling body in feet per second at the end of fall is equal to the time of fall multiplied by the accel- eratrix g = 32.17, which shows that the body gains a velocity of 32.17 feet per second for every second of its time of continued fall. The space fallen through will be equal to the time of fall, multi- y plied by the mean velocity . 2 Let /S denote the space in feet, we bave VT^g T* = V* = V* 2 "" 2 2g 64.34' By solving these formulas we have the time of fall, = == g V \ g = 4.01' Let u denote the space fallen through in the Tih second, we have - and T=-+ 55. FALL OF HEAVY AND LIGHT BODIES. In vacuum a heavy body does not fall faster than a light one, be- cause the weight of each body is equal to the force of gravity acting upon it ; but when a body falls in air or in a liquid, its force of gravity is diminished equal to the weight of the air or liquid displaced by the body ; and whilst the mass is constant, a smaller force has a heavier body to move, and the body will fall slower. A pound of lead dis- FALLING BODIES. 117 places less weight of air than does a pound of cork, for which reason the lead will fall faster than the cork in air. The force of gravity must also overcome the resistance of the air to the motion of the falling body, which is independent of the weight of air the body displaces. This resistance- increases as the square of the velocity and as the surface exposed to the motion. A pound of cork exposes more surface to the motion than does a pound of lead, for which reason the cork falls much slower. Formulas with these influences on falling bodies will be given far- ther on. 56. ILLUSTRATION FOR FALLING BODIES. The illustration represents the velocity V, space /S and time T of a body falling from a to b, with tihe corresponding numbers under the respective heads. In the second, the from T to 8 Fig.:90. 7 8 16 9 10 25 9 12 36 13 14 49 15 16 64 first body 1", a of 16.085 feet, which is represented by tiie first triangle, and which is the unit of the -system. The velocity attained in this period as F= 2x16.085 = 32. 17 feet per second. If the force of gravity ceased 'to act at the end of the first second, the body would continue to fall with a uniform velocity of 32.17 feet per second ; but as the force -of gravity acts constantly, the "body will attain an additional velocity of 32.17 feet for every second of its fall. Between 1" and 2" the body passes 3 triangles, which is the space 3 x 16.085 - 48.255 feet passed through in the second second ; and the. velocity at 2" is 4 x 16.085 = 64.34 feet per second. In the eighth second of its fall it passes 15 triangles, which corre- spond to 15 x 16.085 = 241.275 feet between 7" and 8" ; and the ve- locity at b is 16 x 16.085 = 257.36 feet per second. From a to b the body passed 64 triangles, which represents the linear space from a to b, or 64 x 16.085 = 1029.44 feet. 118 ELEMENTS OF MECHANICS. FALLING BODIES. V= velocity in feet per second at the end of fall. T= time in seconds of the fall. S= space fallen through in feet. V T S V T S V T 8 0.1 0.0031 .00015 5.1 0.1585 0.4042 11 0.3419 1.8804 0.2 0.0062 .00031 5.2 0.1616 0.4202 12 0.3730 2.2380 0.3 0.0093 0.0014 5.3 0.1647 0.4364 13 0.4041 2.6266 0.4 0.0124 0.0025 5.4 0.1678 0.4530 14 0.4352 3.0464 0.5 0.0155 0.0039 5.5 0.1709 0.4700 15 0.4663 3.4975 0.6 0.0186 0.0055 5.6 0.1740 0.4872 16 0.4973 3.9784 0.7 0.0217 0.0076 5.7 0.1771 0.5047 17 0.5284 4.4914 0.8 0.0248 0.0099 5.8 0.1802 0.5226 18 0.5595 5.0355 0.9 0.0279 0.0125 5.9 0.1833 0.5407 19 0.5906 5.6107 1. 0.0311 0.0155 6. 0.1865 0.5595 20 0.6217 6.2170 1.1 0.0342 0.0188 6.1 0.1896 0.5782 21 0.6527 6.8502 1.2 0.0373 0.0224 6.2 0.1927 0.5973 22 0.6838 7.5218 1.3 0.0404 0.0262 6.3 0.1958 0.6168 23 0.7149 8.2213 1.4 0.0435 0.0304 6.4 0.1989 0.6365 24 0.7460 8.9520 1.5 0.0446 0.0335 6.5 0.2020 0.6565 25 0.7771 9.7125 1.6 0,0477 0.0381 6.6 0.2051 0.6768 26 0.8082 10.566 1.7 0.0508 0.0432 6.7 0.2082 0.6975 27 0.8393 11.330 1.8 0.0539 0.0485 6.8 0.2113 0.7184 28 0.8704 12.185 1.9 0.0580 0.0551 6.9 0.2144 0.7397 29 0.9015 13.072 2. 0.0622 0.0622 7. 0.2176 0.7616 30 0.9325 13.987 2.1 0.0653 0.0685 7.1 0.2207 0.7835 31 0.9636 14.936 22 0.0684 0.0756 7.2 0.2238 0.8057 32 0.9947 15.915 3 0.0715 0.0822 7.3 0.2269 0.8282 33 1.0258 16.926 2.4 0.0746 0.0895 7.4 0.2300 0.8510 34 1.0569 17.967 2.5 0.0777 0.0971 7.5 0.2331 0.8741 35 1.0879 19.038 2.6 0.0808 0.1050 7.6 0.2362 0.8975 36 1.1190 20.142 2.7 0.0839 0.1135 7.7 0.2393 0.9213 37 1.1501 21.277 2.8 0.0870 0.1218 7.8 0.2424 0.9453 38 1.1812 22.443 2.9 0.0901 0.1305 7.9 0.2455 0.9697 39 1.2123 23.640 3. 0.0932 0.1398 8. 0.2487 0.9948 40 1.2434 24.868 3.1 0.0963 0.1492 8.1 0.2518 1.0168 41 1.2745 26.127 3.2 0.0994 0.1590 8.2 0.2549 1.0451 42 1.3056 27.417 3.3 0.1025 0.1691 8.3 0.2580 1.0707 43 1.3367 28.739 3.4 0.1054 0.1795 8.4 0.2611 1.0966 44 1.3678 29.407 3.5 0.1087 0.1886 8.5 0.2642 1.1228 45 1.3989 31.475 3.6 0.1118 0.2012 8.6 0.2673 1.1494 46 1.4300 32.890 3.7 0.1149 0.2125 8.7 0.2704 1.1762 47 1.4611 34.330 3.8 0.1170 0.2223 8.8 0.2735 1.2034 48 1.4922 35.813 3.9 0.1201 0.2355 8.9 0.2766 1.2259 49 1.5233 37.321 4. 0.1243 0.2486 9. 0.2797 1.2586 50 1.5544 38.830 4.1 0.1274 0.2611 9.1 0.2828 1.2867 51 1.5854 40.413 4.2 0.1305 0.2740 9.2 0.2859 1.3151 52 1.6165 42.029 4.3 0.1336 0.2872 9.3 0.2890 1.3438 53 1.6475 43.659 4.4 0.1367 0.2939 9.4 0.2921 1.3729 54 1.6786 45.322 4.5 0.1398 0.3145 9.5 0.2952 1.4022 55 1.7097 47.017 4.6 0.1429 0.3286 9.6 0.2983 1.4318 56 1.7407 48.740 4.7 0.1460 0.3431 9.7 0.3014 1.4618 57 1.7718 50.396 4.8 0.1491 0.3578 9.8 0.3045 1.4920 58 1.8029 52.284 4.9 0.1522 0.3729 9.9 | 0.3076 1.5226 59 1.8340 54.103 5. 0.1554 0.3885 10. 1 0.3108 1.5540 60 1.8651 55.953 FALLING BODIES. 119 FALLING BODIES. V= velocity in feet per second at the end of fall. T= time in seconds of the fall. 8= space fallen through in feet. V T S V T S V T 8 65 2.0206 65.669 530 16.478 4366.6 1030 32.027 16494 70 2.1769 76.260 540 16.788 4452.8 1040 32.338 16815 75 2.3314 87.427 550 17.099 4701.7 1050 32.649 17141 80 2.4868 97.472 560 17.409 4874.5 1060 32.950 17463 85 2.6422 112.29 570 17.720 5050.2 1070 33.261 17794 90 2.7976 125.89 580 18.030 5228.7 1080 33.572 18129 95 2.9530 140.27 590 18.341 5410.6 1090 33.883 18446 100 3.1085 155.42 600 18.651 5595.3 1100 34.194 18806 110 3.4194 188.07 610 18.961 5783.1 1110 34.504 19149 1-20 3.7302 223.81 620 19.271 5974.0 1120 34.815 19496 130 4.0411 262.67 630 19.582 6168.3 1130 35.126 19846 140 4.3519 304.63 640 19.893 6365.7 1140 35.436 20198 150 4.6627 349.70 650 20.204 6566.3 1150 35.747 20504 160 4.9736 397.88 660 20.515 6770.0 1160 36.058 20913 170 5.2844 449.18 670 20.826 6976.7 1170 36.369 21275 180 5.5953 503.36 680 21.137 7186.6 1180 36.680 21641 190 5.9061 561.08 690 21.448 7399.5 1190 36.991 22009 200 6.2170 621.70 700 21.759 7615.6 1200 37.302 22381 210 6.5279 689.43 710 22.070 7834.8 1210 37.613 22755 220 6.8387 752.26 720 22.380 8056.8 1220 37.924 23133 230 7.1496 822.20 730 22.691 8282.2 1230 38.235 23514 240 7.4604 895.25 740 23.002 8510.7 1240 38.546 23898 250 7.7713 971.41 750 23.313 8742.4 1250 38.857 24285 260 8.0821 1050.6 760 23.623 8976.7 1260 39.168 24676 270 8.3930 1133.1 770 23.934 9214.6 1270 39.479 25069 280 8.7038 1218.5 780 24.245 9455.5 1280 39.780 25459 290 9.0147 1308.2 790 24.556 9699.6 1290 40.090 25855 300 9.3255 1398.8 800 24.868 9947.2 1300 40.411 26267 310 9.6363 1493.7 810 25.179 10197 1310 40.722 26673 320 9.9472 1591.6 820 25.490 10451 1320 41.033 27081 330 10.258 1690.6 830 25.801 10707 1330 41.343 27493 340 10.569 1791.7 840 26.112 10967 1340 41.654 27908 350 10.879 1903.8 850 26.423 11230 1350 41.965 28326 360 11.190 2014.2 860 26.733 11495 1360 42.276 28747 370 11.501 2127.7 870 27.044 11764 1370 42.587 29172 380 11.812 2244.3 880 27.354 12035 1380 42.897 29599 390 12.123 2364.0 890 27.665 12311 1390 43.208 30029 400 12.434 2486.8 900 27.976 12589 1400 43.519 30463 410 12.745 2612.7 910 28.287 12871 1410 43.820 30893 420 13.055 2741.5 920 28.598 13155 1420 44.131 31333 430 13.366 2873.7 930 28.908 13442 1430 44.442 31776 440 13.677 3008.9 940 29.219 13733 1440 44.753 32222 450 13.989 3144.8 950 29.530 14027 1450 45.064 32671 460 14.300 3289.0 960 29.841 14323 1460 45.375 33123 470 14.611 3433.6 970 30.152 14623 1470 45.686 33579 480 14.922 3581.3 980 30.463 14927 1480 45.997 34037 490 15.233 3732.1 990 30.774 15233 1490 46.308 34499 500 15.545 3886.2 1000 31.085 15542 1500 46.631 34973 510 15.856 4043.3 1010 31.396 15855 1510 46.732 35082 520 16.167 4203.4 1020 31.707 16179 1520 47.043 35752 120 ELEMENTS OF MECHANICS. 57. FORMULAS FOR BODIES FALLING FREELY UNDER THE ACTION OF GRAVITY. Velocity in feet per second at the end of fall. Space in feet fallen through in the time T. Time of fall in seconds. Space in feet fallen through in the Tth second. . v. 8. T. U. V=gT. .1 8= 9 ?\ 5 T=. . 9 u = g(T-$). 13 2 9 V- 2 -j-> - 2 S -^' ' 6 T= . . 10 T=- + $. . 14 F=^.3 *"5' ' 7 Hy-- 11 o V* v^s F=8.02 1 /S..4 64.34 4.01 Example. A body is dropped from a height of $=80 feet. Ke- quired its time of fall, and with what velocity it reaches the ground? Time, Formula 11. T= 2x; 32.17 = 5 seconds. Velocity, Formula 4. F= 8.02j/80 = 71.73 feet per second. 58. ASCENDING BODIES. Ascending bodies are raised by external force superior to that of gravity. The start of a body from rest to a uniform ascending motion requires a greater force than that of gravity or weight of the body. The actual force which starts the body vertically upward is the dif- ference between the applied force and the weight of the body. The force required to maintain a uniform ascending motion is equal to the weight of the body. If the force suddenly ceases to act, the body will continue to ascend against the force of gravity with a retarded motion until it stops, when the force of gravity returns the body, which is then said to fall. The retarded motion of a body ascending against the force of gravity is inversely the same as the accelerated motion in its descent. Suppose the case of firing a ball vertically upward from a gun : ASCENDING BODIES. 121 F= the velocity given to the ball at the muzzle. y = the velocity at the time t from the time the ball left the muzzle of the gun. T= time the ball will ascend. t = any time less than T. S= height in feet to which the ball will ascend. s = the height the ball ascends in the time t. In the accompanying Fig. 91 a represents the muzzle of Fig. 91. a gun, at which the charge of powder has given an ascend- ing velocity V to the ball. The height to which the ball will reach at c is found by Formula 7, s-?-. The time T required for the ascent will be Formula 9, On reaching c the force of gravity will return the ball to the muzzle a in an equal time T and with the same velocity V, so that the ascending retarded motion is inversely the same as the accelerated descending motion, or at equal heights above the muzzle, the ball will have equal velocities of ascension or descension. If t denotes the time in which the body ascends from a to b, or descends from b to a, and T the time from a to - = - = 48 pounds. T 15 Example 3. What time T= ? is required to give a mass M = 580 matts. a velocity F=36 feet per second, the acting force being F= 480 pounds? Formula 4. T- - - 43.5 seconds. Example 4- A mass of unknown quantities is acted upon by a force of F*= 1680 pounds for a time of T= 45 seconds, in which it attained a velocity of V= 12 feet per second. Required the quantity of matter in the mass ? Formula 1. If- - 45 - 6300 matts., or about 90 tons, the answer. Example 5. A mass of M= 84 matts. has a velocity of 148 feet per second. What force is required to stop that mass in a time of T= 6 seconds ? Formula 2. F=~ ^= 84 X 14S = 2072 pounds. T 6 The weight of a body in pounds, multiplied by 0.0310849, will be the mass in matts. FALLING AND RISING OF BODIES. 129 W T hen the mass is expressed by ^ W-2-. ... 5 P fi weight the formulas will be v 9FT . 7 . 8 W gF' ' ' 63. FALLING AND RISING OF BODIES CONNECTED BY A ROPE OVER A PULLEY. The accompanying figure represents two weights TFand w hang- ing on a rope over a pulley. The heavy weight will fall, and draw up the lighter one with an equal velocity. The motive pi 9g force which sets the system in motion is equal to the dif- ference between the weights, or F W w. The mass of the two weights is The fundamental formula for dynamics of matter, as before described, is M: F=T: V. By inserting the values of .Fand M in this analogy, we have of which the dynamic momentums are \ TT i -,. Velocity, Time, -rr F 9 T(W- - j Tr +1O . g(W-w) Example. The large TT=36 pounds and w = BQ pounds. What velocity will the weights attain in a time T= 3 seconds from the time the weights start to move? , 32.17x3(36-30) , F- - * -- ' = 8.77 feet per second. 36 + 30 130 ELEMENTS OF MECHANICS. The space 8, which the weights will move from the start in the Ume T, will be =- ........ 6 2 The formulas will hold good when the experiment is made in vacuum, but when the weights W and w are of heavy materials, like that of metals, the difference of fall in air and in vacuum is very small and can be neglected for ordinary purposes. The weight of air dis- placed by solid iron is only 0.00015 of that of the iron. When the weights W and w are of light materials, like cork, the difference of fall in air and in vacuum is considerable. The weight of air displaced by cork is 0.005 of that of the cork. That is to say, a pound of cork displaces 33 times as much air as does a pound of iron. A piece of cork which we may suppose to weigh 1000 pounds Would displace 5 pounds of air, and would fall much slower in the air than in vacuum ; but let the cork be hung on the rope, Fig. 95, in place of the weight W, and hang another weight of whatever material weighing 5 pounds at w ; now let the system move in vacuum under the action of gravity, and the weight W will fall with the same ac- celeration of velocity as when it falls alone in air. 264. A FORCE ACTING TO MOVE A BODY ON A HORIZONTAL PLANE WITHOUT FRICTION. The body M is supposed to be at rest at a, and a force F is applied to move it from a to b, where the force ceases to act. If the time of action from a to b is known, then the velocity V at b is found by Formulas 3 and 7, 62. After the force J^has ceased to act, the body will continue from b with the uniform velocity F until it meets with some force of resistance, say at c, the force F' acting on the body in an opposite direction to the motion until it stops. The forces F and F' being constant, the velocity of the body is uniformly accelerated from a to b, and uniformly retarded from c to d, for which the mean velocity in both cases will be =- . FORCE ACTING TO MOVE A BODY. 131 We have learned ( 8) that space S is the product of time and ve- locity, in which Tis the time from a to b, and T' that from c to d. S = space from a to 5, and s = that from c to d. VT VT S = and *--g-, , ,. , T _ 2 2 a . oi which K = - = = - = . T T M M The product of either two .of these equal members will be alike, or 2 FT 2s F' T 28 FT 2 s F' T , T M T M TM T M ' of which F 8=F' s, or F ': F' =s : & That is to say, the force F, multiplied by its space of action, is equal to the force F f , multiplied by its space s, and the products are momentums of work, which will be explained hereafter. 3 .! f-. .2 Example 1. A force F= 300 pounds is applied on the body M from a to b, a space of 8=24. feet, after which the body meets with a re- sistance which stops it in a space of s = 0.5 feet. Required the force of resistance F 1 = ? Formula 2. F = = ?1^4 = ^^^ pounds> the answer. The forces F and F are independent of the weight or mass of the body M, for the reason that the spaces /S and s are inverse as the mass, as shown in the proof formulas. If the applied force F is equal to the weight of the body M, then the body will move on the horizontal plane with an equal accelerated velocity, as if it was falling vertically under the action of gravity ; therefore, in falling bodies the acting force is equal to the weight of 132 ELEMENTS OF MECHANICS. the body ; which is the same as to say that the weight of a body is the force of attraction between the earth and that body. The force of a falling body is equal to the resistance it meets with at the end of its fall. Example 2. A steam-hammer weighing 5 tons falls upon a red- hot mass of iron, in which it sinks s=1.5 inches = 0.125 feet; the height of fall to where the hammer stopped is $=6.25 feet, and the force F=b tons. Required the force of the blow of the hammer? Formula 1. F - = ^^ = 250 tons. a 0.125 It is a popular expression to say " the force of a falling or moving body," or " the force of the blow of a hammer," which is not correct. There is no more force in a moving body than in one at rest ; in fact, no force in either case. The force of inertia overcomes the resist- ance which the moving body strikes, and which resistance is the force which stops the motion of the body. 65. ON DRIVING A NAIL INTO A PIECE OF WOOD. Fig. 97. Fig. 97 represents a nail driven through a piece of wood by a weight W resting on the head of the nail. It is supposed that the resist- ance to the nail in the wood is equal to the weight W, so that the slightest additional force would cause the weight to drive the nail down to its head, as shown by the dotted lines. In driving a nail into a piece of wood the resistance is not uniform, for the deeper the nail is driven in the greater is the resistance ; but the mean force of resistance will always be as the following For- mula 2. Let s denote the space which the nail is driven into the wood by the weight. DRIVING A NAIL INTO WOOD. 133 Fig. 98. Let the same nail be driven into the same piece of wood by the aid of a lever, as represented by Fig. 98. The force F acting on the long lever L, presses on the nail equally to the weight W. The force of resistance F' to the nail in the wood, which is equal to the weight W, Fig. 97, acts on the short lever /. The fulcrum of the lever is at c. The force F with the lever L adjusted so that it balances the resist- ance F' , acting on the lever I. From the well-known analogy of levers we have F: F' = l:L. That is to say, the weight F, Fig. 98, is so much smaller than the weight W, Fig. 97, as the lever I is smaller than L. Let 8 represent the space which the weight falls in pressing down the nail in the wood, and 5 = the space the nail was driven in, which is the same as the space s, Fig. 97. It is well known in geometry that s:S-l:L, and as F : F = I : L, we have F : F' = s : S, and F S= F' s. F= 8 F8 - F FS These formulas are precisely the same as those in the preceding paragraph ; which proves that if the weight F is let to fall from an equal height S directly upon the head of the nail, the latter would be driven into the wood precisely the same distance 5 as by the aid of the lever and weight F, Fig. 98. The space of fall S must include the space s of penetration. 134 ELEMENTS OF MECHANICS. 66. PILE-DRIVER. The formulas for pile-drivers are the same as those for force of fall- ing bodies, the principles of action being the same. NOTATION OF LETTERS. Fig. 99. f7= weight of the ram in pounds. =set of the pile in feet by the blow of the ram. R = resistance in pounds to the pile in the ground. h = height in feet which the ram falls, in- cluding the set 8. V= velocity in feet per second with which the ram strikes the pile. Works Wh = RS. . . 1 Weight of ram W=^. . . 2 h Height of fall h = . . .3 Resistance to pile R = Set of pile . Striking velocity F=8 v /A-& . 6 The ram is raised, either by hand or steam, by a rope over a pulley at the top of the framing. In using the above formulas some allowance should be made for the work lost by the ram crushing the head of the pile, which may amount to some 25 per cent., or use only 75 per cent, of the actual weight of the ram for TTin the formulas. Set g- - 76 *. STEAM-HAMMERS. 135 67. STEAM-HAMMERS. The illustration, Fig. 100, represents a steam-hammer as formerly made by Henry G. Morris of Philadelphia, and is here intended only for illustrating the dynamics of forging. The steam-hammers now made by that gentleman are differently constructed. FORMULAS FOR STEAM-HAMMERS FALLING UNDER THE ACTION OF GRAVITY. -rr , ., \ elocity, Fig. 100. Work, Force Horse-power, pSn Time T= NOTATION OF W" weight of the ram in pounds. 8= stroke of the hammer in feet. s = depth of penetration by the blow in feet. V= velocity in feet per second of the blow. K= work in foot-pounds of the blow. n = number of blows per min- ute. EP = horse-power driving the hammer. LETTERS. F= mean force of the blow in pounds. P = steam-pressure in pounds on the top of the piston. when such is used. T= time of fall of the hammer in seconds. 2 = time in seconds required to raise the hammer, the stroke & p = steam-pressure in pounds on the under side of the piston. 136 ELEMENTS OF MECHANICS. Example. The weight of a steam-hammer is W= 5000 pounds, and is lifted $=4 feet; the penetration of the blow is s = 0.015 of a foot. Required the force of the blow ? - 1.338,333 pounds, or 600 tons, nearly, the force required. The steam-pressure p under the piston should be = 16 ** If it is required that the hammer should be lifted in the same length of time as that in which it falls, then make p = 2 W. Formulas for Steam-Hammers falling with aid of Steam on the Top of the Piston. Work K= ( W+ P) (8+ ) = F s. Force f J Horse-power Time 16(TP+P) P= Steam-pressure on the tops of the piston in pounds (not per square inch). |68. ON FORCE OF MOVING OR FALLING BODIES. This subject has been treated under different heads in the pre- ceding paragraphs namely, the Formulas 2 and 6, 62, give the force in regard to velocity and time, and the Formula 1, 65, in re- gard to space. It now remains to explain the subject in regard to velocity and space. The fundamental analogy of the four elements in dynamics of mat- ter, as before repeated, is M: F=T:V, in which Fis the velocity which the force F has given to the mass Min the time T. FORCE OF MOVING BODIES. 137 Space 8 is the product of velocity V and time T, but the velocity in the space 8, in which the force F started the mass M from rest to the velocity F, is only one-half of the final velocity V, or the uniform velocity of the mass after the force ceased to act. Therefore, 8 V T of which r= = 28 MV . M- MV* 28' 28 F V* ' MV* 2F' I2SF The formulas will hold equally good whether the force Fseis the mass at rest to motion, or brings a moving mass to rest. The mass M must be expressed in matts. of 32.17 pounds each. When the magnitude of the mass M or body W is expressed in pounds, the formulas will appear as follows : TFF 2 . 7 . 8 Example 1. A rifle-ball of W= 0.02 pounds was fired with a ve- locity F= 600 feet per second into a log of wood, in which it pene- trated $=0.9 of a foot. Required the force of resistance F<*? in the wood? Formula 5. F WV 1 0.02 x600 2 2x32.17x0.9 = 124.34 pounds. Example 2. A projectile of W= 300 pounds is fired from a gun S= 10 feet long with a muzzle velocity of V 800 feet per second. Required the mean force of the gunpowder ? O AA v O AA2 Formula5 ' ^^il^To' 298414 ^ 1 " 18 - 138 ELEMENTS OF MECHANICS. 69. FORCE OF MOVING BODIES MEASURED BY A SPRING. The illustration represents a body B moving with a velocity V between two equal spiral springs, the elasticity of which has been measured by experiments that when the spring is compressed the space s, its force of elasticity is f. The force of elasticity of spiral springs generally increases uniformly; that is, when the spring is compressed one-half s, the force will be one-half/, and so on. The height of the dotted triangle may represent the force / for the corresponding space s, from which we see that the mean force in the space s is equal to half the force /. Let S denote any space the spring may be compressed, and F' = force of the spring corresponding to the space S. Then S:8 = I*:f, 8-, MdJ*-^. The mean force in the space 8 will be M = mass of the body B expressed in matts. V= velocity of B in feet per second. Mean force, F= M V* 28 ' F= , 28 25 . 4 8= V* f MV* s' 8* ' FORCE OF A BODY FALLING UPON A SPRING. 139 Example 1. The mass of the body B is M = 2 matts. moving against the spring with a velocity F= 5 feet per second ; how much will it compress the spring ? when we know that f= 50 pounds will compress it s = 0.2 feet ? /O 2 x 2 Formula 5. 8 = 5-J^-^ = 0.447 feet. \ 50 After the spring is compressed by the blow it will push the body back with the same velocity to the other spring, where the same oper- ation would be repeated, and so the body would move between the springs for ever if there was no friction or other resistance to retard and finally stop the motion. It is not necessary that the springs should be alike, only that they should be perfectly elastic within the limit of compression. Example 2. A body ~B, weighing W= 50 pounds, moves against a spring with a velocity F=25 feet per second, and compresses it $=0.5 feet. Required the mean and ultimate forces of the spring? Force, 50x25 2 2x32.17x0.5 = 1000 pounds, the mean force of the spring, and the ultimate force will be double, or 2000 pounds. \ 70. FORCE OF A BODY FALLING FREELY UPON A SPRING. The weight W falling the space 8 will compress the spring the space s. After the weight is brought to rest the spring will throw it up again to the same height 8, and so the weight will continue to ascend and descend for ever if there is no other force of re- sistance to retard and finally stop the motion. The weight is acted upon by two opposite forces namely, the constant force of gravity, which is equal to the weight of the falling body, acts through the space 8, and the superior force of the spring acts in the oppo- site direction in the space 5. Let .F denote the mean force of the spring, and/ that at the greatest compression s. Then we have, as before proved, W: F=s:S, and WS=Fs. 2W:f=s:S, and2 TF=/s. 140 ELEMENTS OF MECHANICS. The greatest force of compression f=2F for ordinary spiral springs. TF= p= W_8 s a-**. S WS 8 = F ' -./ 28' . 2 WS 2W . 5 . 7 In these formulas the spaces S and s can be expressed by any unit of length, and also the weight W and force F by any unit of weight. Example 1. The weight TF=4 pounds falls a space $=50 inches upon a spring, which it compresses s = 3 inches. Required the mean force of the spring ? Mean force, F = - ^^ = 66.66 pounds. s 3 Example 2. It is known by experiments that the spiral spring is compressed s ' = 4.5 centimetres by a weight w = 6.3 kilogrammes resting upon it; from what height S must a weight W=3.5 kilo- grammes fall to compress the spring s = 12.5 centimetres ? Tins s : s' =f: w, of whichy= . insert this value for / in Formula 7, which will give the required space. w s' 6.3x12.5* 2 TTs' = 2x3.5x4.5 31.25 centimetres. \ 71. ELASTIC BODIES MOVING AGAINST HARD AND IMMOVABLE BODIES. We have now considered the moving body to be perfectly hard and moved against an elastic spring, but the body may also be perfectly elastic, like that of an india-rubber ball, and strike against a perfectly hard and immovable substance, when the motion of the body will be IMPACT OR COLLISION. 141 the same as when the non-elastic body moved against an elastic spring. A perfectly elastic body B falling the space 8 on a perfectly hard and solid base, will be flattened by the blow, or the Fig W3 product of the weight of the body multiplied by its height of fall 3, which is equal to the force F of elas- ticity multiplied by the space s of compression. The force of elasticity will restore the primitive form of the body and raise it to the same height S. The body will so continue to rise and fall for ever if there is no other force to disturb the operation. If the falling body is not perfectly elastic it will not rise to the same height from which it fell, but the height of rise is a measure of the grade of elasticity of the body. If half elastic, it will rise only half the space S, and so on. The elasticity of bodies can thus be found by similar experiments, but it is difficult to find a perfectly non-elastic base to experiment upon, because all bodies in nature are more or less elastic. A cast-iron ball falling upon a massive base of the same material will rebound some, which shows that cast-iron is partly elastic, and even a cobble-stone falling upon a solid rock will indicate some elasticity. IMPACT OR COLLISION. g 72. COLLISION OF PERFECTLY HARD AND NON-ELASTIC BODIES. A moving hard body striking an immovable one will stop at the col- lision. Practically, bodies are not perfectly hard nor non-elastic, and when a moving body stops suddenly against an immovable one, the momentum is destroyed by crushing and impressing the parts in contact. A moving hard body striking one at rest, but free to move, will set the one at rest in motion, so that both bodies will move with equal velocity after the collision. jjf = weight of a body of velocity V. m = weight of another body of velocity v. It is supposed that the direction of motion passes through the cen- tres of gravity of the operating bodies, and that no momentum is lost in the collision. Although the bodies are denoted by M and m, which means mass, they can be expressed by any unit of weight. 142 ELEMENTS OF MECHANICS. The sum of the momentums of motion of the bodies will be alike before and after the collision, and when the bodies are perfectly hard they will move with a common velocity v' after contact. The upper sign + is used when the bodies move in the same direc- tion, and - when in opposite directions. 273. ONLY ONE OF THE HARD BODIES IN MOTION. Fig. lot The body Jf moves with a velocity V, and strikes the body m at rest, or v = o, and conse- /' \%f quently m v = o. MV M+m M V-v 1 ' Example 1. A hard body 3f= 24 pounds moves with a velocity V= 42 feet per second, and strikes the body m = 16 pounds. Re- quired the common velocity of the bodies after collision ? Velocity, V - ^ = 25.2 feet per second. Example 2. A body m = 500 pounds, at rest but free to move, is to be set into motion of v' = 20 feet per second by a body M = 80 pounds. Required the velocity VI of the body M? Velocity, V= = 145 feet per second, the answer. Example 3. A body M= 250 pounds moves with a velocity V= 64 feet per second. What weight m ? must be put in the way so as to reduce the velocity of Jf from 64 to 5 feet per second? 250(64-5) , m ? '- = 34oO pounds. t) It is supposed that the body m = 3450 pounds is free to move after the collision. IMPACT OF HARD BODIES. 143 874. THE TWO HARD BODIES MOVE IN THE SAME DIRECTION. The body M moves fastest until it reaches Fi &- 105 - m, after which they will both move with the common velocity v' . v'(M+ m) , MV+mv v = . . . 1 M+m v _v'(M+m)-mv g M Example 4- The body M= 144 pounds moves with a velocity V 72 feet per second, in = 480 pounds arid v = 18. Required the common velocity v'? after contact ? 144x72+480x18 , Velocity. t>'= =30.15 feet per second. J 144 + 480 ? 75. THE HARD BODIES MOVE IN OPPOSITE DIRECTIONS. The body with the greatest momentum of Fi s- 106 - motion will gain on the other body, and re- turn it with the velocity v'. , MV-mv . I . 2 m(v' +v) M+ m v'(M+m)+mv V-v' ' M(V-v') M m v +v Example 5. The body m = 81 pounds, and moves with a velocity y = 36 feet per second against M= 27 pounds. What velocity must the body 3/have to stop the two bodies in the collision? v' =0. r 0(27 + 81) + 81x36 Velocity, F= L = 108 feet per second. 27 144 ELEMENTS OF MECHANICS. |76. COLLISION OF PERFECTLY ELASTIC BODIES. It has before been explained that when a perfectly elastic body strikes an immovable body, it will rebound with the same velocity as that with which it struck ; but if it strikes a movable body, some momentum will be lost by the striking body in giving motion to the one at rest ; and in all cases the sum of the momentums before and after collision will be alike, whether either or both bodies are non-, perfectly or partly elastic. F= velocity of JIf) v = velocity of in ) 877. ONLY ONE OF THE ELASTIC BODIES MOVES AND STRIKES THE ONE AT REST. Fig. 107. )= V(M-m). v'(M+m) = V V(M-m) 2V M * M+m ' t/(M-m) 2M 2V M V-V ' M(V- V) M(v'-2 V) . 5 . 6 . 7 . 8 Example 6. A body Jf=360 pounds, moving with an unknown velocity, strikes an elastic body m at rest and of unknown weight. After the collision it is found that the velocity of M is V = 42, and that of m is t/ = 96 feet per second. Required the weight of the body ml . 360(96-2x42) .. Formula 8. m = * = 90 pounds. 96 IMPACT OF ELASTIC BODIES. 145 g78. TWO PERFECTLY ELASTIC BODIES MOVING IN THE SAME DIRECTION. The body J/ moves fastest until it reaches the body m, after which it will move slower or perhaps return. Fig. 108. *%+)- V(M v'(M+m) = 2M V+v(m - M}. M+m M+m v'[V(M-m) + 2 2 MV+v(m-M) V'[2 M V+v(m-M)] V(M-m) ' ,_m(V' + V-2v) V -V = M(2V-v-v') v'+v M- m-M Example 7. The body m = 800 pounds, and moves with a velocity of v = 28 feet per second. M= 360 pounds. It is required to give such a velocity to the body M that when it strikes m all the momentum M V will be discharged into m ; so that M will be at rest after the collision, and m will move with the two momentums. mv' = MV+mv. Required the velocity Fof the body M? and v' of m? T . 0(360 + 800) -2x800x28 Formula 7. V= = 101.8 360 - 800 feet per second, the required velocity of the body M. MV+mv _8Mxl01.8 + 800x28 m 800 per second, the velocity of m after the collision. 13 K 146 ELEMENTS OF MECHANICS. |79. TWO PERFECTLY ELASTIC BODIES MOVE IN OPPOSITE DIRECTIONS. Fig. 109. V = V(M-m)-2m v M+m 2MV-v(m-M) M+m v'[V(M-m-)-2mv] 2MV-v(m-M) " v> _ V[2MV-v(m-]lf)] V(M-m)-2mv V'(M+m) = V(M-m)-2mv. 2 M V- v(m J/). 1 M ^m(2vv'+vfV- V'v} = 2VV+ V'v - V'v' ^M(2VV +V'v-v'V 2vv'+i/V- V'v v= v(mV'-2mv' -MV) ? 2MV'+mi/-Mv' m V'-2mv'-MV .8 Example 8. M= 24 kilogrammes, V= 16 metres per second, in = 10 kilogrammes, and v = 36 metres per second. Required the velocities V and v' after the collision? V = 16(24-10)-2xlQx36 = - 17.5 metres per second. 24+10 The negative sign means that the body M returns with that velocity. 2x24x16-36(10-24) v' = - * -- '- = +29.88 metres per second. 24 + 10 The positive sign + means that the body m continues its course with a velocity reduced from 36 to 29.88 metres per second. 80. COLLISION OF BODIES OF WHICH EITHER ONE OR BOTH ARE ELASTIC, NON-ELASTIC, OR PARTLY ELASTIC. Let E denote the grade of elasticity of the body M, and e = that of the body m. When a body is perfectly elastic, .Z^or eis equal to 1, and when per- fectly hard or non-elastic, E or e is zero or 0. Therefore, when a body is half elastic, E or e is 0.5, and so on. IMPACT OF ELASTIC BODIES. 147 2 81. ONE BODY IN MOTION AND ONE AT REST. v'(M+m) = > M+m VM(l+e v'(M-Em) M(\+e) , = V M(\ + e) V " M-Em ' Emv' Ev' M-Em ' er . 5 . 7 . 8 Example 9. The body Jf = 200 pounds, and elasticity ^=0.75, struck the body m = 150 pounds, and elasticity e = 0.25. After the collision it was found that the velocity of m was v' = 30 feet per second. Required the velocity V of the body Ml T7 , 30(200-0.75x160) , Formula 3. V - V,,., ^^ ' = 9.6 feet per second. 200(1+0.25) g 82. THE PARTLY ELASTIC BODIES MOVE IN ONE DIRECTION. Fig. 111. M+m -e M) M+ E)+v(m-eM) V(M-Em) + mr(\ v'v+V'e-V'V(I + E) V'v + v'VE-v'v(l + E) V'(M+m) M-Em + mv(l + E)' v'(M+m)-MV(l + E) m-eM 148 ELEMENTS OF MECHANICS. 83. PARTLY ELASTIC BODIES MOVING IN OPPOSITE DIREC- TIONS. Fig. 112. V(M+m) = V(M-Em)-mv(l v'(M+m) = M V(l + e) v(m e J V(M-Em)-mv(l + E) M+m , MV(l + E)-v(m-eM) Q E)-v(m-eM) f _ V'\M V(l + E) v(m- V(M-Em)-mv(l + ' 7=- Vv-v' E-v'v(l + E) V'(M+m) e M Example 10. A perfectly hard body M = 125 pounds is moving with a velocity of V= 54 feet per second against a perfectly elastic body w = 480 pounds, and w = 24 feet per second. Required the velocities V and v' after the collision ? In this case the elasticity of M is E=Q, and that of m is e = 1. Formula 1. V .54(125-0x480)-480x 2 4(l + 0) _ ^ ^ per second, the velocity with which the body M will return. 125x54(1 + 0)- 24(480- 1x125) _ n __ Formula2. */= 2 - -^ =- 2.92 feet per second, the returning velocity of the body m. Both the bodies will return in opposite directions after the collision. DYNAMICS OF MATTER. 149 DYNAMICS OF MATTER. 84. ON POWER AND WORK APPLIED TO MATTER. The three physical elements, force, motion and time, with their com- bination in space, power and work, have already been explained namely, Elements. Force =F -i - - '. . 1 Motion = V ' . ".'"" . 2 Time =T . .3 Functions. 8 = V T . -. 4 Power, P-FV . .5 Work, K=FVT . 6 In the application of this law upon matter, a fourth element, mass, enters into the combination which constitutes the dynamics of matter. The relation between these four elements has already been ex- plained in regard to force and motion, and it now remains to treat them in regard to power and work. The fundamental analogy in dynamics of matter is Multiply both the momentums by the velocity V, and we have the work K= M V* = F V T, which means that the work consumed in giving the mass M the velocity V is F V T, or the product of force, velocity and time. In the formula for work, K= F V T, V means the mean velocity of the force Fin the time T. When a constant force sets a body (at rest, but free to move) in motion, the velocity at the start is zero or 0, but is accelerated to V in the time T] and the mean velocity in that time is only one-half of the final velocity; therefore, if F means the final or uniform velocity of a moving body after the force has ceased to act upon it, the work stored in the body will be The space in which the body was set in motion is <&'-* VT, which, inserted in the formula for work, will be A body lifted vertically a space or height S against the force of 13* 150 ELEMENTS OF MECHANICS. gravity, but without regard to velocity and time, the work accom- plished is K=FS, but the force F is equal to the weight W of the body, and K= W 8. Now, let the body fall freely to the same spot from which it was lifted, and the force of gravity will perform an equal work, K= WS. In the falling work the velocity and time might be entirely differ- ent from that in the lifting work, but their product will be equal in both cases ; that is, the time multiplied by the velocity of the lifting work is equal to the time multiplied by the mean velocity of the fall- ing work. The space $= \ V T, in which V means the velocity the falling body has attained when striking the ground:, and Tthe time of fall. Insert this value of space in the formulas for work, and J5T- W V T. The time of fall is "Work K = M\ V*, which was to be proved. That is to say, the work stored in a moving body is equal to the mass multiplied by half the square of its velocity. $85. DYNAMICS OF MATTER REPRESENTED GEOMETRICALLY. The combination of the physical elements and functions with matter can be represented by the corresponding geometrical quantities, as explained in 2, and illustrated by Fig. 113. Let the arrow F represent the magnitude and direction of a con- stant force applied on a body M at rest, but free to move. The body M will then be started and moved with an accelerated velocity as long as the force acts. Let the line T represent the time of action of the force, in which the body moves from a to b, and let the vertical line F represent the velocity attained at b. The -force F ceases to act at b, and the body will continue its motion with the uniform velocity V until some other force is applied to change and stop its DYNAMICS OF MATTER. 151 motion. At b' the body meets a force F in opposite direction to its motion which will be stopped at a' in the time T . Whilst the body moved from b to b', the space S= VT", its momentum of motion was M V= F T= FT, or F : F = T : T. The illustration is so proportioned that F = 2 ^and T=2T'. Fig. 113. d > Draw the line a k, and the area of the triangle a, , & represents VT which the force F passed through in the time T. Draw from a the line a e at any convenient angle to represent the force F, and complete the wedge or prism a, b, c, d, e. Then the cubic contents of that prism will represent the work K consumed in giving the mass M the velocity V. T Complete the same operation at the other end of the illustration, and the prism a', b', % K CK = Jr ct. ana Jr = . 8t This expression - - can be inserted for the power P in any dynam- ct ical formula. Fig. 119 represents S = V T to be constant, and the force F var- iable. Fig. 119. #= - , which can be inserted for space in any dynamical formula. Work, K=FS, or 8 of which . 8s f\ jf -^- , can be inserted for force in any dynamical formula. C'S Either or all the elements F, V, T may be variable in the execu- tion of work. 3 89. POWER AND WORK WITH CONSTANT FORCE AND VARIABLE VELOCITY. Suppose a constant force F to be applied on a mass M free to move, Fig. 120, it will set the mass in motion with an accelerated velocity. When the force has acted for a time t, the velocity will be v and the power P= Fv. F: Jf= v:t, of which z!- , F and tit = M&v CONSTANT AND VARIABLE FORCES. 157 The differential work M-P&- FvMVv / MV* Work .fiT= I Jf v 8v = - , which is the well-known formula for work in a moving body. Fig. 120. The force F ceases to act at the time T, and the mass M will con- tinue to move with the uniform velocity V and generate the space S = V T". There is no work accomplished in the time T" ', but the mass will continue to move until some force F' is applied in opposite direction to retard and finally stop the body in the time T' . The work which stopped the body is equal to that which set it in motion, and which is represented by the volume of the two prisms, = , or FT- FT'. The two momentums of time F T and F' T' are alike, and each equal to the momentum of motion M V. I 90. FORCE VARIABLE WITH THE TIME OF ACTION. Fig. 121. When a force F, acting on a mass M, increases as the time of action, so that F= C T, C being a constant factor, the relation between time, ve- locity and space will be as follows : F-M= V- T 2 ' - - cm' 158 ELEMENTS OF MECHANICS. *-*% Ce& C M ^ S ** MS J J C ' 3 " C ' Time T _JzWs 1 Space *-fj 2 Velocity y_F_TjCT K/T 9 TL-f' jXL J>1_ 3 The velocity is as the square of the time, and the curve is therefore a parabola tangenting the time with its vertex at the start of motion. Example. Suppose it be known that the force f= 32 pounds after having acted on the mass Jf=50 matts., for a time of 2 = 4 seconds. Required the velocity and space when the force has acted for a time of T= 36 seconds ? f-Ct, and C= = = 8. Velocity F= ^^- = 103.7 feet per second. 2 x 50 Space 8- ^-^- = 2488.3 feet. 3x50 The area bounded within V ^represents the space passed through ; the area of the section / 1 represents the power in operation. The area of the base triangle represents the momentum of time ^ F T= M V, the momentum of motion, and the cubic content of the figure M V* represents the work done by the force F, which is K= - . The force /= C t. t* Mv Time VARIABLE FORCES. 159 The same time is obtained from Formula 2. When the mass is expressed by weight W, the formulas will be Time Velocity Space Time I3W8 91. FORCE VARIABLE INVERSELY AS THE TIME OF ACTION. A force F is applied to move a mass M toward o, and the force diminishes as the time of action ; so that in a time T the force / is reduced to o. Force/- C( T- ). Fig. 122. Velocity Space 160 ELEMENTS OF MECHANICS. When the formulas are integrated for the whole time T t we ri rpt Velocity F= 3 Time 8- BSJf \-c~ 4 CT* '3 M' These formulas are the same as those in the preceding paragraph. When the body M passes the centre o, the force / becomes negative and stops the motion at a ; which operation is accomplished in equal length of time T and space /S as that in which it was set in motion from rest to the velocity Fat o. Example. A force F= 240 pounds sets a body M= 15 matts. in motion, so that in a time T= 30 seconds of action the force is reduced to 0. Required the velocity of the body when arriving at 0, and what space it has passed through ? Formula 3. Velocity Formula 5. Space J /"Y ^iV o and G = = 80. T 30 F= ^p = 2400 feet per second. 3x15 The body arrived at with a velocity of 2400 feet per second, and it will continue to move with that velocity until some force is applied to change it; but if the force/ continue to act negatively in the same ratio, the body will be brought to rest at a, 1600 feet from 0. 92. FORCE VARIABLE AS THE SPACE OF ACTION. Assume F= CS, in which G is a constant factor. Fig. 123. F: M =V: T. MV^MV^ F CS Csct MVs Cs ' /'*-/ MS* rjn VARIABLE FORCES. 161 Time T-J^ hyp.kg.8. . 1 FT C8T CS8S Velocity F= = M M Velocity F-tf-r. . : ' - 2 ' s Example. The force F= 3 pounds at a distance s = 1 foot, and the mass M = 8 matts. Required the time and velocity of the body at a distance #=16 feet? Time, T= \F hyp-^0- 16 - S- 845 seconds. \ 3 Velocity, F= 16-J- = 9.8 feet per second. When the mass is expressed by weight W, we have Time, T=hyp.kg.S. , .4 Velocity, F=^^p . . . .5 I&J^-^JJP '. . . '.; 14* 162 ELEMENTS OF MECHANICS. \ 93. FORCE VARIABLE INVERSELY AS THE SPACE OF ACTION. A force F is applied to move a mass M toward o and diminishes a the space S, so that at o the force is reduced to nothing. Force /-<7 08V ). Fig. 124. F: M = V: T, Ftit _ C(S-s) M " M C = , then 8~~(- )8s. Velocity, When the velocity is integrated through the whole space S, we have * = S and the Velocity, V ' S \^ 2 The time of action through the space $ will be Time, T=^~ 3 That is to say, the time required to move the body to o is inde- pendent of the distance S as long as F= C 8. VARIABLE FORGES. 163 Example. A mass M= 8 matts. is acted upon by a force F*= 48 pounds at a distance $=16 feet. Required the time and velocity atO? ~ S " 16 ~ ' Velocity, V= 16-J- - 9.8 feet per second. \8 Time, T= -J- = 1.633 seconds. \3 When the mass is expressed by weight W, we have Velocity, V Time, T- I * * After the body has passed the centre o the force F is negative and stops the body at an equal distance - 8, 94. FORCE VARIABLE INVERSELY AS THE SQUARE OF THE DIS- TANCE FROM THE FORCE OF ACTION TO A GIVEN POINT. The force of attraction between bodies is inversely as the square of their distance apart. Let M and m denote the n *- 125 - in matts, and S their distance apart in feet. The force of attraction between them is Mm 18 SP = 28693080, the coefficient of attraction, 51. Suppose the mass M to re- main stationary and draw the mass m to it. s = distance moved by m. F: Mm 164 ELEMENTS OF MECHANICS. w ' 8T Time, I This is the time in which the mass m will move a space 8. Mm T M T vi rr M -- t then F- ' F' 9(8- 8? F8F- Velocity I 2M That is to say, the velocity is inversely as the square root of the distance between the attracting bodies. | 95. RESISTANCE OF AIR TO A MOVING BODY. A mass M rolling without friction on a level plane a b arrives with a velocity V at a, where it is left to work its own way against the re- sistance of the air. The force of resistance to a body moving in any perfect fluid is as the square of the velocity ; where- : - fore the resistance at a can be rep- resented by F C V*, of which C is a constant to be determined hereafter. tt 3 When the body has moved from a to b in the time t the resistance of the air has reduced the velocity Fig. 126. RESISTANCE OF AIR. 165 from Fto V , or V = V-v. The force of resistance of the air at b can therefore be represented by/= C( V-v) 2 . F:M=V: T, and T=. F Sl f r J < <7(F-t;) 2 o(F-t>) This formula gives the time in which the velocity V is reduced to F'=(F-v). The body's motion will be stopped when V = or whenv = F, which should be when M M M = 00 q F- F') C7 F C F That is to say, the resistance of the air requires an infinite length of time to stop the body, or, more correctly, the body will never be stopped by th*at means alone, but the velocity will be reduced so that no motion conld be perceptible in days or years. By solving the Formula 1 we find the velocity M+CtV' Call /8 r =the linear space the body would move through with the velocity Fin the time t without the resistance of air, or $= V t. 8' =the actual distance moved in the time t against the resistance of the air. s = the retarded distance in the time t, or s = /S/S'. From the Formula 1 we obtain CVtlt M+Ct' V r_ Vt-\hyp.log.(M+CV)-hyp.l<)g.M\. C [ 166 ELEMENTS OF MECHANICS. This linear space is represented by the area of the figure bounded within the lines c tv. The space S' which the body actually moves in the time t will be - TT| c L C Vt) - This linear space is represented by the area of the figure bounded within the lines V d V c. The linear space S which the body would have moved through with the constant velocity Fin the time Tis represented by the area of the rectangle Vt in the figure. The area of the base bounded within the lines Ft e represents the momentum of the resistance, which is equal to the momentum M v. The volume of the figure bounded within the lines fv t represents the work done by the resistance of the aif, and which is equal to the work | J/V. The time required for the body to move through the space S' is found as follows ; Call X=M+CVt. . . . . .6 . 7 8 The common logarithm multiplied by 2.30258509 is the hyperbolic logarithm. When the time t occupied by the body in passing through the space S' is correctly known, the initial velocity I 7 " will be at 96. Coefficient C. It now remains to find the coefficient C for the resistance of the air. The resistance is equal to the weight of a column of air with a base equal to the projecting area of the moving body, and a height equal to that from which a body falls and attains the same velocity as that of the resistance. A cubic foot of air of temperature 60 Fahr. and under a pressure of 30 inches of mercury may be assumed to weigh 530 grains. DYNAMICS Of MATTER. 167 A = area of resistance in square feet of the moving body. h = height in feet of the column of air. The force/ of the resistance of the air will then be in pounds. ,530,4 A The height A = 7000 F 2 . 10 . 7000x20 Of which the coefficient a**--. . .11 When the area is expressed in square inches a, we have the co- efficient 122367 12 A or a means a flat surface at rig"ht angle to the direction of motion. For a cylinder moving with its convex side to the motion the area of resistance is one-half of the projecting area. When the moving body is spherical, the area of resistance is one-quarter of the projecting area. D = diameter in feet, and d= diameter in inches of a moving sphere. ^4= 0.19635 D*, and a = 0.19635 d\ Example. A cast-iron ball of d = 8 inches in diameter, weighing W= 69.88 pounds or M = 2.1 matts, is fired from a gun with a ve- locity V= 1000 feet per second. Required the time t in which the ball will reach a target at 8' = 1500 feet horizontal distance from the muzzle of the gun ? a = 0.19635 x 8 J = 12.5664 square inches. 12.5664 1 Coefficient C= 122367 9737.65 HypJcg. 2.37 = 0.815253, and X- M- 0.27. 9737.65x0.27 , Time t = = 2.629 seconds. 168 ELEMENTS OF MECHANICS. In this time the ball would fall 32.17x2.629' ... = 1H.17 feet. Then, in order to hit the mark on the target, the centre line of the gun must be pointed to 111.17 feet above that mark, or the gun must be elevated to an angle =tanA 15'. 1500 97. RESISTANCE OF THE AIR WHEN THE BODY DESCRIBES A PARABOLA. When the gun is elevated to an angle z and the ball describes a parabolic curve a c d, the horizontal velocity v = V cos.z, which in- Fi g . 127. serted for Fin the preceding formulas, will make them answer for this case also. The vertical action of resist- ance in the ascent is counteracted in the descent, so that only the horizon- tal resistance of the air need be con- sidered in the operation. Without resistance of the air the ball would describe the parabola a ef, and make the horizontal range S=S r +s, but the resistance diminishes that space by s or to S' = S- s. MEAN FORCE. A variable force acting on a body free to move can be converted into a mean force. The mean force in time is the mean force of momentum. The mean force in space is the mean force of work. The mean force in time is at the centre of gravity of the momentum area of the base. The mean force in space is at the centre of gravity of the work volume. I 98. MEAN FORCE # IN TIME T. Figs. 121 and 122. When a force varies directly or inversely as the time, the mean force in the time T is $ = \F, and in the time t, $ = $(F+f). MEAN FORCE. 169 The mean force # acting on a body in the time Twill produce the same velocity as that of the variable force in the same time. The momentums of time, # T= % F T= M V, the momentum of motion. | 99. MEAN FORCE

IN SPACE ,S. Fig. 122. When the force varies inversely as the time, the mean force # in the space 8 will be as follows : Work, $S=%MV\ Insert Formula 5, 91, for 8 and Formula 3 for V. ^C T 3 Work, SM IN TIME T. Fig. 123. When a force varies directly as the space, the mean force in the time T is found as follows : Momentum of time # T= M V, momentum of motion. 15 170 ELEMENTS OF MECHANICS. Insert Formula 1, 91, for T and Formula 2 for V. Mean force, hyp.log.S \ 103. MIAN FORCE

\ Jjf \ 12 Example 2. Required the velocity t; at the compression s = 0.5 of the foot? Formula 6. v = x /5.477' - 360xQ - 5 * = 4. 75 feet per second. 12 x 1 The force of the spring at this velocity and compression is , Fs 360x0.5 . /= -- = - - = 180 pounds, and the power in operation at that moment is 855 p =/ v = 4.75 x 180 = 855 effects, or - = 1.55 horse-power. 550 Example 3. A mass M = 6 matts. moving against the spring with a velocity F"=4 feet per second. Required its velocity when the spring is compressed s =0.6 feet? Formula 6. v = -y/4* ; - 1 = - 2.37 feet per second. The negative sign proves that the mass was not able to compress the spring s = 0.6, for which is wanted an additional velocity of 2.37 feet per second to V] or F"=6.37 feet per second would just compress the spring that much but no more, and the velocity v would then be 0. 176 ELEMENTS OF MECHANICS. \ 108. THE MASS ACTED UPON BY TWO OPPOSING FORCES. The force .Facts on a mass M sliding on the base B with friction/. The mass is thus acted upon by two opposite forces, Fand/ of which Fig. 132. Fraust be greater than /before the mass can be moved, and the force which sets the mass in motion is Ff. The force F acts on the mass only in the space S, when the work of the force will be K-FS. 1 The work done by the friction will be k~fS. 2 The work utilized in giving the mass M the velocity V in the space 8 will be of which M:(F-f)=T:V. . MV 2S_ F-f V ' M "V M ' ' ' ' FRICTION FORCE. 177 of which T=. 8 s _T(F-f)_ MV* 2 M 2(F-f) . 10 MV , .11 The prism F V T represents the total work K of the force F and Formula 1, of which the work consumed by the friction is repre- sented by the light part of the prism fV Tand Formula 2. The shaded part of the prism represents the work utilized in giving the mass M the velocity V, which corresponds with Formula 3. The second prism fVt represents the work of friction consumed in bringing the mass to rest after the force F ceased to act. The cubic content of the prism fVt. is equal to that of the shaded part of the first prism which set the body in motion. Example 1. The force F= 160 pounds, Jf=64 matte., and the friction /= 100 pounds. What time is required to move the mass M a space $=20 feet? and what will be the velocity F? at the end of that space ? _, . _ T , 2x20(160-100) C10 , , , , Formula 7. V= */ = 6.125 feet per second, the required velocity at the end of the space S. Formula8 . r= _4 the time required to move the body S=2Q feet. M 178 ELEMENTS OF MECHANICS. I 109. ON FRICTION OF SCREW PROPELLERS WORKING IN WATER. The friction of screw propellers running in water is a considerable item of the propelling power, and is well worthy of attention when speed of the vessel and economy of fuel are desired. The solution of the problem furnishes a good example of the value of the calculus in dynamics. Fig. 133. Notation of Letters. P= pitch of the propeller in feet. R = radius of the propeller in feet. r = any radius less than R. 1 = length of a helix for one whole convolution in feet, at the radius r. A = area of the helicoidal surface for one whole convolution in square feet. jV= number of blades of the propeller. n = number of revolutions per minute of the propeller. v = velocity of the helix I in feet per second. h = horse-power of friction. f=* friction in pounds. = differential. The friction in pounds per square foot of surface of cast-iron and brass, of rough castings, and also of smooth surfaces, filed or ground, FRICTION OF SCREW-PROPELLERS. 179 but not polished, is approximated as follows when the surface moves with the veldrtty of one foot per second : Friction Surface. /' Rough cast-iron 0.0045 Smooth cast-iron : 0.0040 Brass, rough casting ; J T^5.0040 Brass, smooth' surface 0.0030 The friction for rough cast-iron will then be, in pounds, /= 0.0045 Av* 1 The differential area of the helicoidal surface for one convolution will be 8 A = 1 cr, . . . . .2 and the differential friction in pounds will be /= 0.0045 v*ldr 3 This force of friction multiplied by its velocity will be power in effects, and divided by 550 will be horse-power, when the differential horse-power will be a , 0.0045 v 3 lcr en = . . . .4 550 In , Pn* The velocity v = and ^ =*- . - . . 5 The differential horse-power will then be 0.0045 n^Er 550x60* -r 0.0045 w l n 3 Call X= = . 7 550 xGO 3 26,400,000,000 Then A = XJ 4 0r, .... 8 and k = xf?dr 9 But J- 1 /4r l + P a > and P = (4 ** r 2 + P 2 ) 2 . . 10 Insert Formula 10 in Formula 9, and we have . * . .11 4 . . .12 Then h = X T(16 ^ r* + 8 ^ r* P 2 + P 4 ) Cr. . .13 180 ELEMENTS OF MECHANICS. By integrating each term in the parenthesis we have 16 * 4 r 5 8 7T 2 7* * and = 26.319 r> P'. .15 Integrate the friction horse-power from the centre of the propeller to the periphery of radius P., then when r = 0, C= 0. Insert the Formula 7 for JTin Formula 14, with the values 15, and we have the frictional horse-power for one convolution of the screw and for one side of the helicoidal surface, A- L{/ /qi 1 71 7?*_u9AQ1Q 7?* > z -i- P 4N \ Ifi = | oil. 1 1 .fir + ^O.Oiy s J: -T J: ). .ID 26,400,000,000 The helicoidal Surface of screw propellers is generally cut up into small portions by a number of blades, each of a fraction of the pitch, and when the helicoidal surface is counted on both sides of the blade, the friction horse-power will be - _ P/+26.319 P'P'+P 4 ) .17 13,200,000,000 P^ This formula includes both the dragging and rotary friction horse- power of a propeller of rough cast-iron. Call/' = friction in pounds per square foot of surface moving with a velocity of one foot per second, and the friction horse-power will be Example. Required the friction horse-power of a propeller of the following dimensions: Diameter of propeller, 20 feet, or ............ P = 6 feet. Pitch ................................................ P= 18 feet. Length in the direction of axis ............... Z = 2.4 feet. Number of blades ................................ N= 4. Revolutions per minute ......................... n = 60. The horse-power consumed by friction will then be, for rough cast- iron surface, . 59,400,000x18 horse-power of friction. FRICTION OF SCREW-PROPELLERS. 181 In the year 1850 an experimental steamer was built in Kensington, Philadelphia, which was expected to make 20 to 30 miles per hour. The propeller was about 4 feet in diameter, with only one blade, extend- ing the whole convolution of the circle, and with a very fine pitch of about 6 inches. (The author does not remember the exact dimensions.) The propeller was expected to make 500 revolutions per minute. Example. Required the friction horse-power of the above-de- scribed propeller. 12 = 2 feet, P = 0.5 feet, n = 500, L = 0.5, N=l. Surface, rough cast-iron. 500- 2 2 6 59,400,000x0.5 horse-power, nearly. The power of the engine counted from the size of the steam boiler, did not amount to more than 50 horse-power, and the result was that when the trial trip came off the steamer could hardly crawl up against the tide. The building of the steamer was kept in the greatest secresy, and her performance was expected to astonish the world. There was another experimental steamer built in Kensington in the year 1864, in which several curious propellers were placed on each side of the vessel, which also turned out a failure on account of the friction of the propellers in the water being too great. A fine-pitched propeller has more friction-power than one with sharp pitch for equal speed of vessel. The proper pitch for propelling steamboats should be between two and two and a half times the diameter of the propeller. The sharpest vessel should have the sharpest pitch of propeller. For the proper proportions and construction of screw-propellers, see " Nystrom's Pocket-Book of Mechanics," thirteenth edition. 16 182 ELEMENTS OF MECHANICS. 110. GYRATION. Gyration means Circular Motion. The term is used in dynamics of matter in circular motion to designate the mean effect of all the particles in a revolving body or system of bodies. In motion of translation all the particles of a body move with equal velocity in straight and parallel lines, but in circular motion tie particles move with different velocities in different circles, of which one is called the circle of gyration. $ 111. DYNAMICS OF MATTER IN CIRCULAR MOTION. Fig. 134 represents a mass M free to move around the centre O in the circle of radius H. A constant force F is applied in the direc- Fig. 134 ** on ^ ^ e tangent of the circle to move the mass, the dynamics of which will be the same as if the mass moved in a straight line or -^ ^ I? rr> *** /> The velocity V of the mass M will be in the direction of the circle, which periphery is 2 * R. Circular velocity is generally expressed in revolutions per minute, and denoted by the letter n. When the radius JR is expressed in feet the circular velocity -mil be 2 - R n 60 F Example 1. A force F= 36 pounds is applied on the mass J/= 48 matts. for a time T= 9 seconds. Required the velocity F? and revo- lutions per minute ? with which the mass will continue to rotate after the force ceased to act ? The radius of the circle being R = 2 feet. of> q = 6.75 feet per second. 48 - - : = 32.6 revolutions per minute. i- X t>.14 X ^ GYRATION.. 183 The mass M will continue to rotate with this velocity until some force F' is applied in opposite direction to retard the motion and finally bring the mass to rest. Example 2. What force F' is required to stop the rotation of the mass Miu a time of jT=4 seconds? _, MV 48x6.75 F' = = = 80 pounds. The primitive work consumed in setting the body in rotation is equal to the realized work by which the rotation is stopped, like in straight linear motion, the 2 2 x 60 2 = 0.00548314 MJPri*, the work of rotation. Io.2.o77 This formula gives the primitive or realized work of a rotating mass, as illustrated by Fig. 134. When the mass is expressed by weight W, the formula will be = 32.17x182.377 = 5867.16 ' When the force F is applied on a radius r, and the centre of the mass M rotates with a radius , as represented by Fig. 136, the for- mulas will be as follows : F =^T" " <* M _Fr T ~ EV' v= Fr T = ME' T= 2 3 FT ' ^ *n ^ ,,, 60 rT ' no Fr T GO Fr T n = 6 7 ^ .< .. ^ 60 Fr Example 1. A mass J[f=96 matts. is to be put into a circular motion of n = 360 revolutions per minute around a radius R = 3 feet. The force F which sets the mass in rotation acts on a lever or crank of radius r = 0.5 of a foot. What force is required to give the mass that circular velocity in a time T=* 10 seconds ? , 96x2x3.14x3'x360 Formula? jF= = 6511.1 pounds. 60x0.5x10 184 ELEMENTS OF MECHANICS. 112. CENTRE OF GYRATION. Centre of gyration in a revolving body is a point in which, if all the matter were there contained, it would have the same dynamic effect as when distributed around that centre. The centre of gyration describes the circle of gyration. The centre of gyration is always outside of the centre of gravity of the revolving body. The less space the body occupies in the circle of gyration, and the greater the radius of revolution is, the nearer does the centre of gy- ration approach the centre of gravity of the body. |113. RADIUS OF GYRATION. The radius of gyration is the distance from the centre of rotation to the centre of gyration in a revolving body. The radius of gyration will herein be denoted by the letter X, to distinguish it from other radii. 114. MOMENT OF INERTIA. Moment of inertia is either weight, mass, volume or surface, multi- plied by the square of its radius of gyration. The moment of inertia will hereafter be denoted by the letter E, and may be expressed in either of the following units : E = W X* in square foot-pounds, or sq. ft. Ibs. E= MX* in square foot-matts., or sq. ft. mts. E= Q X* in square foot-volumes, or sq. ft. vol. E= X' in square foot-square, or sq. ft. sq. Moment of inertia is used for finding the radius of gyration. P5 135 Let Q denote the volume of any system of bodies, and X its radius of gyration rotating around the centre o, and A, , C, D, etc. represent the several volumes of respective radii of gyration v, x, y, z, i Then Q X**=A v t +x t + Cy^+JJz*, etc. The radius of gyration of the system will then be , etc. INERTIA. 185 We have learned that the work of a revolving body is 182.377 ' 5867.16 ' in which R means the radius of gyration, MR' 1 and W R* means the moment of inertia. When the moment of inertia E is given the work in the revolving body will be ^ = -lonor^' OI> ^ = 182.377' 5867.16' The moment of inertia is a constant quantity in a body or system of bodies rigid to an axis of rotation. Moment of inertia is E=MX'* ..... 1 Square radius of gyration is X* = . . . . . .2 E Radius of gyration is then X= * / . \ E E Mass of the body will be M= -t. . A. \ 116. FORCE OF INERTIA. The force of inertia is equal to any force applied to change the motion of a body. Let a constant force jPbe applied on a lever r to change the rotary motion of the mass M re- acting with its force of inertia ^Ton the lever or radius of gyration X. Then, F-.I = X:r, and the static momentums F r = I X. F r The force of inertia will be I . X Let I 7 denote the time in seconds, in which the velocity of the mass J/ is changed from V to v feet per second, or from N io n revolutions per minute. Then, I: M=(V-v): T, lo * 186 ELEMENTS OF MECHANICS. from which the force of inertia will be M(V-v) = Fr T ' X ' Static momentum Fr = IX= MX ( V ~ V \ ^ r , .__ . . A _ N F= -- -, v - -- , and ( V- v) = - (N- ). 60 60 60 v 9 JUT Y" s Static momentum ' F r = / X= - (TV- w). The time T in which the angular velocity is changed from V to v, or revolutions from ^Tto n, will be Fr 60 Fr The change of angular velocity in the time Twill be Fr T ... . &OFrT (F-)- , or (a--,)- The force jP required on the crank will be MX(V-v) Tr 60 Tr Example. A body weighing 2255 pounds, or mass M= 70 matts., revolving with a radius of gyration X= 5 feet, and making N= 120 revolutions per minute, is acted upon by a force F= 96 pounds on a lever r = 0.75 of a foot. What time Tis required to reduce the revo- lution to n = 60 per minute ? or 2 minutes 32.6 seconds, the answer. Example. A mass M= 360 matts. is revolving with a velocity of F= 30 feet per second in a circle of X= 12 feet radius of gyration, when a force ^=90 pounds acting on a crank r = 1.25 feet is applied RADIUS OF GYRATION. 187 in T= 180 seconds to increase the angular velocity. Required the increased velocity F? 90x1.25x180 , (F -' ) - S60xl2 -" F-v = 4.1, or F= 4.1 + 30 = 34.1 feet per second, the velocity required. The force of inertia of any revolving body in the circle of gyration will be Example. A fly-wheel weighing 5400 pounds, or M= 167.84 matts., is making N= 120 revolutions per minute, with, a radius of gyration JT=4.5 fe^t. The angular velocity of the wheel is to be reduced to n = 15 revolutions per minute in a time T= 90 seconds. What force I must be applied in the tangent of the circle of gyration to re- duce the revolutions from 120 to 15 in the time 90 seconds ? . 167.84 x2x 3.14 x 4.5. ir - 60^90 - ( " ^ " P Un This is the force of inertia under the conditions given in the example. 116. RADIUS OF GYRATION AND MOMENT OF INERTIA. Let two masses M and m rotate around a common centre C in cir- cles of gyration of radii E and r, and the two masses being rigid so as to rotate with a common angular velocity. Fig ia? Then the moment of inertia of each mass will be M R* and in r 2 , and of the two masses E = (M+ m)JP From this formula we have the radius of gyration to be x= J/+7W Let Z denote the distance of the centre of gravity of the two masses from the centre C, and we have the static momentum Z(M+m) = HR+mr, of which Z~ ?LI. M+m 188 ELEMENTS OF MECHANICS. It is to be proved that the centre of gyration is outside of the cen- tre of gravity of the two masses, or that X> Z. ^ Mtf+mr* z ,_IMR+mr\* M+m \ M+m j MlP+mr 1 (MR+mr? M+ m (M+ m)(M+ m) Reject the common denominator M+m. Multiply both members by M+m. (M+m)(MR"+m r t )>(MR+m r) 1 . M* R*+Mm R* + Mm S + m* r*>M* R* + 2MR m r + m* r\ Mm R* + Mm r>>2 MR m r. R 2 + r* > 2 R r ; which is true, because if R > r, the two squares R^+r 1 must be greater than two rectangles R r. HI"- GYRATION OF IRREGULAR BODIES. To find the radius of gyration and moment of inertia of an irregular body revolving around the centre C, Fig. 138. Draw concentric circles at equal distances r apart to represent cylindrical sections of the body concentric with the axis C. The first and last division should be only ^ r. The more divisions made the more correctly will the radius of gyration and moment of inertia be ascertained. The area of each cylindrical section a, b, c and d will then represent the mass of rotation at the respective section. Let denote the sum of all the areas of the sections, and X= radius of gyration of the body. The sum of the moments of inertia of the sections will be equal to X 1 . = a+b+c+d. Moments of inertia, X* = a r 2 +6(2 r) J + c(3 r) 2 +d(4 r) 1 . Rad. gyration, EQUILIBRIUM OF GYRATION. 189 |118. EQUILIBRIUM AND DISTURBANCE OF GYRATION. A body or system of bodies revolving around an axis c upon which it may be perfectly balanced in regard to gravity, but when rotating with a variable velocity, the equilibrium is disturbed in the axis of rotation. Two masses M and m revolving around their common centre of gravity o, will be in gyratic equilibrium only Fi tgg when their angular velocity is uniform, in \ which case the equilibrium is not disturbed in the axis of rotation ; but when the angular velocity is irregular, there is a tendency to change the axis of rotation, and the equilibrium ("^VT *" / ' of gyration is disturbed. / Let R and r denote the respective radii of gyration of the masses M and m, and X= radius of gyration of the system. Suppose the bodies to be balanced so that M:m=r:R, or M R = mr, in which case their dynamic momentum will be alike, or M V= m v ; but the work in each body will not be alike. From 112 we have the works in the revolving bodies to be M&rf mr^n* 182.377 ' ' " 182.377 ' Insert MR in the work k, and m r in the work K, which will be ,, m r R n K= . and = 182.377 ' 182.377 182.377 K 182.377 k mr R MRr KMRr^kmrR, or K:lc = m:M. That is to say, the work in the bodies are inversely as the masses ; therefore, the work which sets the unequal masses into rotation is unequally divided in the centre of rotation, which causes disturbance. When the masses rotate with a uniform velocity, there is no work transmitted to or from them through the centre of rotation, and there can consequently be no disturbance. Any change in the angular velocity will cause a disturbance in the journals, which disturbance cannot be avoided except by making M-R-^mr 2 ; but, then the rotating system will cause a disturbance during uniform velocity, because it is then not balanced for gravity. 190 ELEMENTS OF MECHANICS. 119. GYRATION TREATED BY THE CALCULUS. Any body or system of bodies may be considered to be composed of an infinite number of cylindrical sections concentric with the common axis C of rotation. Let denote the area of any such section rotating around the axis o, with the radius r. Q = cubic content of the body. Then cQ = 08r. Moment inertia Q X* = f Or'dr+C. Radius gyration ^ = \l I + ^' In order to bring the formulas into a practical shape, it is necessary to know the variation of the section in relation to r. ^120. GYRATION OF A STRAIGHT LINE OR ROD. Centre of Gyration and Moment of Inertia. Let a straight line or thin parallel rod L be attached with one p . H1 end, and at right angle to the axis of rotation o o. As the rod is parallel, its section at any dis- tance I is a constant quantity. -1---X *. Moment inertia Q X' 1 =J 0^ U = %0 L 3 sq. ft. vol. Q=OL, and = . L Moment inertia Q X* = \Q L*. IQ Tt IJTi Radius gyration X= -^1 = ^1 = L\/\. Radius gyration X= 0.5775 L. Example. A line or parallel rod is L = 12 feet long, and revolves around one of its ends. Required the radius of gyration ? X= 0.5775x12 = 6.93 feet. GYRATION. 191 121. GYRATION OF A BALANCED LINE OR PARALLEL ROD. The radius of gyration of a line or parallel rod revolving around an axis passing through the centre of gravity Fig 142 at right angle to the line will be o X=x 0.5775, L = 0.28875 L. ~ L \~ ' 122. GYRATION OF A STRAIGHT LINE NOT EXTENDING TO THE AXIS OF ROTATION. When the revolving line or parallel rod does not extend to the axis of rotation, the radius of mg ' gyration is found as follows : The radius of gyration of each elementary section of the line or rod, will be (a + ljfjl. Moment inertia L X* = f (a+W fil. = f (a+) 2 S*a+l LX*=\ (a 2 +2 a l+l*) fil = cf L+a Z?+%L 3 = L(o?-\ Radius gyration X= yd 1 + a L + \L'*. 123. CENTRE OF GYRATION REFERRED TO CENTRE OF GRAVITY. Having given the radius of gyration of a body or system of bodies when the axis of rotation passes through the centre of gravity, to find the radius of gyration of the same when the centre of gravity rotates with a radius R around an axis parallel with the former axis. Referring to the preceding figure, the radius R of the centre of gravity of the rod will be R = a + %L. a = R-\L, and a' 2 = 1? - R L + ^L*. The radius of gyration of the preceding figure is Insert the above values of a and a 2 , and we have x= /'R Radius gyration X This formula gives the centre of gyration for any parallel figure of length L. ""= 0.28875 L, 192 ELEMENTS OF MECHANICS. which is the radius of gyration of a line L rotating around its centre of gravity. Therefore, if r = radius of gyration of a body or system of bodies rotating around its centre of gravity, and R = radius of centre of gravity of the same system rotating around another axis parallel to the former, then Radius gyration Jf= j/.fif'-t-r 2 . This formula will hold good for any form of body or system of bodies. g 124. GYRATION OF A BODY REVOLVING OUTSIDE OF ITS AXIS OF ROTATION. The illustration represents the plan of the circles of rotation. The body A B is first supposed to revolve around its . centre of gravity o when its centre of gyration a de- scribes the dotted small circle of radius r. Now let the body revolve around the centre o', with a radius R from the centre of gravity o, so that the radii R anc? r are at right angles to one another. Then the radius of gyration will be X= o' a, which is the hypotenuse of the catheters R and r, and the centre of gyration will in this case be at b, describing the dotted large circle. This is an illustration of the formula in the preceding paragraph, The body A JB may revolve in any position in regard to the circles of gyration, with the condition that the two axes o and o' must be parallel. GYRATION OF A RECTANGULAR PLANE. The radius of gyration of any number of parallel lines rigid into a rectangular plane, which ends are parallel with the axis of ro- tation, is the same as that for a single line. R = radius of centre of gravity of the plane. r = radius of gyration of the plane when the axis of rotation passes through the centre of gravity and is parallel to the axis o, o. = area of the rectangular plane. Moment inertia E~ 0(R i + r 2 ) = 0(R* Radius gyration X= GYRATION. 193 I 126. GYRATION OF A PARALLELOPIPEDON. To find the radius of gyration and mo- FI s- 146 - ment of inertia of a parallelopipedon when the axis o o, of rotation passes through the centre of gravity and at right angles to either one of its sides. The radius of gyration of the dotted section passing through the axis and centre of gravity is ( 121) 0.28875 b. The radius gyration of any other parallel section at a distance I from the axis is 1/P+ (0.28875 b^ a b = area of one side of the parallelopipedon in the plane of rota- tion. Then the differential moment inertia will be ti(a b) X* = + (0.28875 bf[b til. Moment inertia a b X* =j[P+ (0.28875 &)*]& til. Integrate the moment inertia from I = o to I = \ a, and we have Radius gyration JT= -*/ Let the same parallelopipedon revolve around an axis o', o' parallel with o, o, and R = radius of the centre of gravity. Then the radius of gyration will be ( 123 and 124) 194 ELEMENTS OF MECHANICS. | 127. GYRATION OF A LATERAL TRIANGLE OR SOLID WEDGE ROTATING AROUND ITS VERTEX. Fig. 147. h = height and 6 = linear base of the triangle. a = any breadth parallel with b, but at a distance I from o, o. = ^ breadth, the area of the triangle. The differential moment of inertia will then be a : b = I : h, of which a = . h 60X 2 = 91. h Moment inertia, 4A Had. gyration, 0.70107 A. \ 128. GYRATION OF A LATERAL TRIANGLE OR SOLID WEDGE ROTATING AROUND AN AXIS PASSING THROUGH THE CENTRE OF GRAVITY. Fig. 148. but From the preceding section we know that when the body revolves with its vertex in the axis o' o', the rad. gyration is 0.70107A. We also know from 123 that when r = rad. gyration of a body when rotating around ita centre of gravity, X= T/tf + i*. of which r = i/X*-I?, - 3 0.2357A, which is the radius gyration of a triangle or wedge rotating around an axis passing through the centre of gravity. GYRATION. 195 |129. GYRATION OF A WEDGE ROTATING AROUND AN AXIS PASSING THROUGH THE MIDDLE OF ITS BASE. x - Fig - 149 - In this case R = \ k, and JR? = % h*. From the pre- ceding paragraph we have r = 0.2357A and r 2 = -fa A 2 . x = !/ A* + T V A 2 = Vi = 0.4082A. 130. CIRCULAR PLANE OR SOLID CYLINDER. Fig. 150 represents a circular plane or solid cylinder rotating around its centre. Let A represent the area of the circular end of the Fig. wo. cylinder, and r = any radius less than R. The circum- ference of the radius r is 2 x r, and the differential area ft A = 2 JT r dr. Differential mom. in., 6 A X* = 2x i*8r. X* but then r=. A 2x1? Radius gyration, X = Jf* = - \ '2 1 R 414 2" -0.7071 Radius of gyration, X= 0.7071 R. 131. ANNULAR RING. This example is the same as the foregoing, except that the formula is integrated from r to R instead of from o to R. s X* = 2 r* Fig. 151. 2^2- npw Rad. of gyrati The radius of gyration of a circle or a cylindrical surface rotat- ing around its centre or axis is equal to the radius of the circle or cylinder. 196 ELEMENTS OF MECHANICS. Fig. 154. \ 132. RADIUS OF GYRATION OF DIFFERENT FIGURES. Fig. 152. A circumference rotating around its diameter, and a circular plane or cylinder around its centre. r = radius of the circle. Radius gyration X= 0.7071 r. ~W~ Fig. 153. _l_ A Circular Plane Revolving around its Diameter. Radius gyration X= 0.5 r. r = radius of the circle. A Sphere Revolving around its Diameter. Spherical surface JT= 0.8165 r. A solid sphere X= 0.6324 r. A Cylinder Rotating around one of its Ends. v /4TT3? liadius gyration JL = -*J . \ 12 A Cylinder Rotating around its Middle. Radius gyration X" -*/ . A Cone Revolving around its Base X=. Vertis X= 12 A*+3 I? 20 GYRATION. 197 Fig. 158. A Cone Frustum Rotating around its Base. V / JT-J-I A "Wedge, or an Ann of a Wheel. = 0.204^/12^ A Cylinder whose Centre Line is parallel with, the axis of rotation. A Sphere Rotating with its Centre a distance a from the axis. Fig. 161. A Ring of Square Section, or a fly-wheel of very light arms. To find the Common Radius of gyration of a fly-wheel with arms of considerable weight. W= weight of the ring of square section. w = weight of all the arms. b = breadth of the arms in the direc- tion of rotation in feet. 17* 198 ELEMENTS OF MECHANICS. /6 ~\ Fig - 163 - \ 133. PARALLEL ARMS OF FLY-WHEELS. Fig. 163 represents a part of a fly-wheel with hub H, arm M, and ringZ It is required to find the radius of gyration of the arms. L = length of the parallel arm. A = area of cross section of the arm. r = radius of the hub. 1 = distance from the centre to any section A. Q = cubic content of the arm. y = radius of gyration of the arm. The momentum of radius of gyration of the consolidated element- ary sections A of the arm will be Q is the variable, and not y. f(# -i* L+r I? but Q = AL, and Eadius gyration y = |/r* -f r L + % L 9 FLY-WHEELS. 199 134. RADIUS OF GYRATION OF THE WHOLE FLY-WHEEL. x = Radius of gyration of the hub, to be calculated from 130. = weight in pounds of the hub. y = Radius gyration of the arms, to be calculated from 133. P= weight in pounds of all the arms. 2 = Radius gyration of the outer ring, to be calculated from 130. Q = weight in pounds of the ring. X= Radius gyration of the whole wheel. W= weight in pounds of the whole wheel. W-O+P+Q. Example. Required the radius of a cast-iron fly-wheel of the follow- ing dimensions ? C r = 6 inches, radius shaft in the hub. Hub, < R = 15 inches, outside radius of the hub. ( = 2798 pounds, weight of the hub. Arms { L = 10 ' 75 feet> Ien 8 th of the arms - 1 P= 10650 pounds, weight of six arms. f J?2 = 13 feet, outer radius of the ring. Km & { Q = 36500 pounds. Wheel, W= 49948 pounds, the weight of the whole wheel. Radius gyration of the hub will be, = 0.952 of a foot. 2 Radius gyration of the arms will be, y - y 1.25'+ 1.25 x 10.75 + x 10.75* = 7.32 feet. Radius gyration .of the ring, [TTo* 12.51 feet. 200 ELEMENTS OF MECHANICS. Radius gyration of the whole system of the fly-wheel will then be, /27 \ Q.952 1 +10650x7.32*+ 36500 xl2.5 ? f 49948 the radius gyration of the fly-wheel. This is 12- 11.217 =0.783 feet, or 9 inches less radius of gyration than the inner radius of the ring, which latter is generally taken in practice. In this case the radius gyration is only 0.934 of the inner radius, or 0.86 of the outer radius. \ 135. FLY-WHEELS. In fly-wheels of ordinary proportions the radius of gyration x can be assumed in practice to be the inner radius of the ring. TF= weight of a fly-wheel in pounds. X= radius of gyration in feet. n = revolutions per minute. K= work in the fly-wheel in foot-pounds. V= velocity of centre of gyration in feet per second. WV* 2 2x32.17' /9~ ~P~~, of which F s ^ 60 / 2x32.17x60' = 5867.16 ' This formula gives the work of a fly-wheel or of any rotating body ; that is to say, it requires that much work to bring a body from rest to a rotation of n revolutions per minute ; or if the body is rotating with an angular velocity of n revolutions per minute, it will re- generate that much work before it is brought to rest. 1 76.6 5867.16 ' 5867.16 K X' 1 n 1 ' n \W 76.6 4 Example 1. A fly-wheel of TF=2000 pounds, and radius of gy- ration X= 3 feet, is to be set in rotation by a weight F= 600 pounds, FLY-WHEELS. 201 Fig. 162. What angular velocity will the fly-wheel obtain by the weight falling 50 feet ? The work K= 600 x 50 = 3000 foot-pounds. hrfi A F6AAA Revolutions n = '\l - = 31.278 per minute. 3 \2000 In this case it makes no difference what radius the weight F is acting upon, because the weight multiplied by the fall will be the work stored in the fly-wheel ; and the same work will be re-utilized in bringing the fly-wheel to rest. Example 2. A fly-wheel weighing W= 16000 pounds, and radius of gyration X= 6 feet, is revolving at the rate of n 60 revolutions per minute. Required the work in the wheel ? v 16000 x6'x60 2 Q , Q/10 , , -^ = */. -,f> - = 3534 25 foot-pounds. 5867. lo The fly-wheel can wind up a weight of 353425 pounds one foot high, or 3534.25 pounds to a height of 100 feet. Whatever height divided into 353425 will be the weight the wheel can wind up to that height. 136. FLY-WHEELS IN REGARD TO TIME AND SPACE. In the preceding paragraph we treated fly-wheels in regard to work and angular velocity, without regard to the time and space which are constituent elements of work. We will now treat on the time required for storing or re-storing the work in a fly-wheel, and the space or total number of revolutions in the time required to bring the fly-wheel from rest to a uniform rotation or from a uniform rota- tion to rest. JV= total number of revolutions in the time T seconds. The space of uniformly accelerated or retarded motions is = 2 when V means the uniform velocity of centre of gyration. 2 + 60 202 ELEMENTS OF MECHANICS. W V* Wnrlr 7T =, - WXW 2? WX* 5867.16' ^120m j 2.454317^^* 5867.16 ^ 2.4543 WX*N* 1 ^ T } F 2 TP"^ \J\J -^A. -i-T-l ^^ . #- T ^ * 2.4543 X 1 N 9 ' Example 1. A fly-wheel at rest weighing W= 10000 pounds is to be put into n = 80 revolutions per minute in N= 6 revolutions of the wheel. Kequired the time T? The radius of gyration being X= 5 feet. ^ 10000x5'x80 l oaK , n , ,. . Work, K= = 295504 foot-pounds. 5867.16 Time, T= 1.5666 x 5 x 6 ^| 10000 _ 9 . 986 seconds. Example 2. The fly-wheel in the preceding example is to be stopped in T= 5 seconds. How many revolutions N will the wheel make in the action of stopping it ? Revolution, N= ^ J** = 3.3. 1.5666 \ 10000 CENTRING. 203 137. ON THE CENTRING OF REVOLVING BODIES. This subject has given much trouble to mechanical engineers, for the reason that a body perfectly balanced around its axis of rotation does not appear to be balanced when set into a high rotary velocity. Let the bodies IF and w be connected by an inflexible rod R r, and the system perfectly balanced on the line a b ; that is, R and r are distances from the fulcrum a b to the cen- tres of gravity of their respective bodies. Suppose the system to rotate in a plane at right angles to the axis a b, then the centrifugal force of each body will be 29335 F=J ^^_ 2 2933.5 But W R = w r, consequently the centrifugal forces of the bodies must be alike, and have no tendency to disturb the equilibrium in the fulcrum of rotation. The work stored in each body by setting the system in rotation wmbe 3 586 in which X and x are the respective radii of gyration, but for sim- plicity in the illustration we can (without much error) consider x = R and x = r y and we have the works, 5867.16 _ 5867.16 We have W R = wr, butTF-fi? is not equal to w r 2 , and conse- quently the works stored in each body are not alike. The small body w on the greater radius r has taken up more work than has the greater body W on the smaller radius R ; and it is this difference of 204 ELEMENTS OF MECHANICS. work which caused an action in the fulcrum whilst the bodies were set in rotation. This irregular distribution of work in revolving bodies has puzzled many good mechanics. Example. W= 150 pounds and JR = 0.5 feet. w = 20 pounds and r = 3.75 feet. n = 500 revolutions per minute. Required the centrifugal force? and work stored in each body? 150 x 5 x 500 2 Centrifugal force F= -- ^- - = 6391.7 pounds of each body. The works concentrated in the bodies will be respectively 150 x 0.5' x 500* . A = - - -- = 1598 foot-pounds. 5867 We see here that the work stored in the small body w is over seven times greater than that stored in the larger body TF; and it is this diiference which causes the revolving system to work irregular in the fulcrum. Suppose the two bodies W and w to be cast-iron balls, and find their diameter and radii of gyration ? The diameters are nearly Z> = 0.58 and d = 0.207 of a foot. r= \^ + ^ = \ a5M ^ =a533 feet - Radius gyration i J* i n 90Q 5 = 3.75 feet. 10 The radius of gyration of the large ball is 0.033 of a foot longer than the radius R ; but in the small ball there is no appreciable dif- ference between x and r. The real work stored in the large ball will then be ^ 150 x 0.533' x 500' JT - : - = 1815 foot-pounds. 5867 or 1815- 1598 = 217 foot-pounds more than in the first calculation. TRANSFORMATION OF WORK. 205 Any system of revolving bodies which is balanced when at rest will also be balanced in any uniform rotation ; but when the rotation is accelerated or retarded work is stored or re-stored, which causes an irregularity in the axis if the several radii of gyration are not balanced. 138. TRANSFORMATION OF STRAIGHT LINEAR MOTION TO ROTARY BY THE AID OF A CRANK. The illustration represents a steam-engine, of which the piston has a straight linear motion, which is transformed into rotary by the Fig. 165. crank E. It is supposed that the irregular motion of the piston ac- commodates itself to the uniform rotary motion of the crank. Let F denote the force acting constantly on the piston throughout the stroke ; then the force acting on the crank It in the direction of the tangent of the circle will be F sin.v. The volumes of the diagrams represent the works in the cylinder and crank respectively for a single stroke or half a revolution in the time T. The work in the cylinder is composed of a constant force by variable velocity, whilst that in the crank is composed of a variable force by constant velocity, but the two works or the products of the three simple FV Tare alike in both cases. The variable velocity of the piston and the variable force of rota- tion are represented by the ordinates in the respective semicircles. The time is represented in this case by the length of the semicircles. The force and velocity in one diagram take one another's place in the other diagram. 18 206 ELEMENTS OF MECHANICS. 139. ON THE REGULATION OF IRREGULAR WORK BY FLY-WHEELS. Fig. 166 represents a steam-engine and fly-wheel winding up a weight, which operation can be compared with any uniform work accomplished by an engine. F= constant force acting on the piston throughout the stroke 8. Fig. 166. Then F : W*= s : 8 for each single stroke or half a revolution of the wheel. TT/'ft TT Of TT7"o E* O T-. ^ -rr^ -r-f W S - ^ O rv " 6 -TO For simplicity in the illustration it is supposed that the connecting rod is infinitely long, that the steam-pressure is constant throughout the stroke of the piston, that all the work of the engine is transmitted through the crank-shaft and regulated by the fly-wheel to a uniform power in the realized work of rotation. When the force F is constant throughout the stroke S, the work in the cylinder can be represented by the area of a circle A a b B c d for one revolution of the engine. The area of the semicircle A b e B will then represent the work for a single stroke of the piston. The diameter A B is equal to the stroke S of the piston, and is sup- IRREGULAR WORK. 207 posed to be in the centre line of the engine, so that the crank-pin describes the circle and is on the centre at A and B. When the crank passes the centres A and the engine performs no work in the rotary motion, and when the power in the realized work is constant in all positions of the crank, the fly-wheel must per- form the full power of the engine when the crank passes the centres. The power of the engine varies as the sine for the angle of the crank to the centre line A &\ and the work performed in the time of one revolution of the crank is represented by the area of the circle de- scribed. Draw on A B the two rectangles A ef B and B g i A, equal to the area of the circle ; then the height A = A e represents the mean velocity of the piston throughout the stroke S. Suppose the engine to be running with a uniform rotation of the fly-wheel, the area of the semicircle will then represent the real work, and the rectangle the mean work on the piston for a single stroke. It will be seen that the circle projects over the rectangle with a segment a b, which is the work stored in the fly-wheel during that portion of the stroke, or whilst the crank-pin passes from a to b. The area of the corners bf B and B g c of the rectangles which project over the circle are equal to the area of the segment a b, or the work restored to the rotary motion by the fly-wheel, whilst the crank passes from b via centre B to c, and another segment of work c d is stored in the fly-wheel and restored to the work of rotation whilst the crank passes from d via centre A to a. The fly-wheel is thus regulating the irregular work in the steam cylinder to a uniform work of rotation. 3 140. TO FIND THE IRREGULARITY OF ROTATION OF A FLY-WHEEL. The alternate storing and re-storing of work by a fly-wheel causes a slight irregularity of rotation. The greater the fly-wheel is and the Fig . greater the velocity of rotation, the less will the irregularity be. Suppose the crank and fly-wheel to move in the direction indicated by the arrow, and the rotation will be fastest when the crank- pin passes b and d, and slowest at a and c. Let R denote the radius of the crank. The area or work of the semicircle A a b B will be -^ - IF. The area or mean work of the rectangle A e/B will be 2 R h. 208 ELEMENTS OF MECHANICS. Then | rr I? = 2 R h, of which A = = 0.7853 ^. That is to say, the mean velocity of the piston is equal to 0.7853 of the circular velocity of the crank-pin. The fraction 0.7853 is the sine for the angle v of the crank with the centre line of the engine when the rotary motion is a maximum at b and d, and minimum at a and c. The angle v = 51 45'. The work k stored and re-stored alternately by the fly-wheel or the area of the segment a b will be - 1? sin.v cos.v. 180 v = 51.75. sin.v = 0. 7853. cos.v = 0.61909. k - 8 ' 14 X 38 " 25 ^ _ Q.7853 x 0.61909 ^ = 0.18109^. This is the area of the segment a, 6 expressed in a fraction of a square radius ; but we want the segment of work to be expressed in a fraction of the semicircle, or of the work of a single stroke of the piston. The area of a semicircle = \K H?. Call .F#=the work of a single stroke, Then ^ That is to say, the irregular work stored and re-stored by a fly- wheel is equal to 0.1153 of the work of a single stroke of the piston. As the same proportion is constant for every stroke of the engine, it follows that the work of regulation by a fly-wheel is always 0.1153 of the work of the engine, when the steam-cylinder is double acting and the realized work is uniform. IRREGULAR WORK. 209 ON IRREGULARITY OF ROTATION OF A FLY-WHEEL. Notation of Letters (repeated). Fig. 168. B W= weight of fly-wheel in pounds. x = radius of gyration in feet. n = number of revolutions per minute. K= work in foot-pounds in the fly-wheel. /= irregularity in fraction of the revo- lutions n, or of the mean velocity of rotation. The mean velocity of the fly-wheel is when the crank passes the centres at A and , the greatest velocity at b and d, and the slowest at a and c. Whilst the crank-pin passes from A to a the work represented by the projecting corner A e a is given out from the fly-wheel, by which the velocity is reduced. The work of the corner A ea = x0.1153 F8= 0.0576 F8. W V 2 The work .ZT stored in the ball before striking is WV* 0.071 xlOOO* After penetration the work is = 1103.6-8.3 = 1095 foot-pounds. Velocity v after penetration will be ~Wq 11095x32.17 ,._,., , -\ F \ a07T~ =.4 feet per^cond. ^ The loss of velocity by penetration is 295.6 feet per second. The mean velocity through the door was 852 feet per second. The time of penetration was T= = 0.0000097844 of a second. The door is of white pine, 3 feet wide by 6 feet high, which will weigh 52 pounds. Suppose the door to move freely without friction in the hinges and without resistance of air. The ball penetrates the door in the centre, or at r = 1.5 feet from the line of the hinges. Radius of gyration of the door is X= 3x0.5775 = 1.7325 feet. In accordance with the assumed conditions the door will be set into rotation by the penetration of the ball. From 112 we have the number of revolutions per minute = 60 Fr 9 T = 60 x 100 x 1-5 x 0.0000097844 = 065851 ~ 2 TT W X" 1 6.28 x 0.071 x 1.7325 2 of one revolution, or it would require 15.1 minutes for the door to make only one revolution, which motion would hardly be perceptible, but the resistance of air and friction in the hinges would keep the door almost stationary. 222 ELEMENTS OF MECHANICS. 155. CENTRIFUGAL, AND CENTRIPETAL, FORCES. A mass M moving with a velocity V in the circle of radius R around the centre will have a tendency to move outward from the centre, for which a force equal to that tendency must be applied on the mass from the centre C, to keep it revolving in the circle. The tendency of the mass to move outward is called centrifugal force, and the force resisting that tendency is called centripetal force. These two forces are equal and in opposite directions ; neither one of them can exist without the other, in fact, their distinction is only action and re- action, the only condition under which force can be conceived or realized. Tie a stone at the end of a string and swing it round in the air, and a force is felt in the hand which you may call centrifugal or centripetal as you please. The tension of the string is the centri- fugal force, and the reaction by the hand is the centripetal force. 156. THEORY OF CENTRIFUGAL FORCE. Let a mass M, Fig. 180, move with a uniform velocity V in the direction a, b until it arrives at a, where a constant force F is applied on it at right angles to the motion, which causes the body to deviate from the straight line a, b ; but as that force acts at right angles to the motion, it has no effect on the velocity V of the mass. Suppose the direction of the force F to vary with the deviation of the motion from the straight line, so as to always be at right angles to the di- rection of motion ; it follows that the velocity of the mass will be constant. When the force F is so balanced as to cause the mass to move in a circle of radius JR around the centre C, the mass will then return to the same point a where the constant force was first applied. If the force F ceases to act at the moment it returns to a, the mass will continue with the same velocity in its original course a, b, which is the tangent to the circle. Therefore, when the centripetal force of a revolving body ceases to act, the body will fly off in the direction of the tangent to that point of the circle where it was let loose. Although the velocity of the mass was not changed in its circular motion, the force F actually stopped the mass at c, giving it a back- ward motion at d, stopped it at e and returned it into a forward motion again with the same velocity V at a, all in reference to the CENTRIFUGAL FORCE. 223 direction a, b ; which operation was accomplished in the space of the diameter of the circle. It follows that F is equal to a force which would give the mass M a velocity Fin a space equal to the diameter of the circle, and which is the centrifugal force of the mass. We have F: M = F: T, F VT , an M V Diameter 2 E = MV a aad T -v Centrifugal force F= F= force in pounds, JfeT=mass expressed in matts., V= velocity in feet per second of the mass in the circle of radius R in feet. {157. PROOF OF CENTRIFUGAL FORCE. Fig. 184. A body or mass M, Fig. 18, moving with a uniform velocity V in the direction A G, when arriving at A a force F is applied at right angles to the motion, which changes the course of the body. Let the direction of the force F be so changed as to always be at right angles to the motion of the mass M, and it will have no effect upon the velocity V in the path of motion. The force F can be of such mag- nitude as to cause the body to describe a circle A, , C, D, F of radius _K. Having given the mass M, velocity V and radius E, the problem is to find the magnitude of the force F. 224 ELEMENTS OF MECHANICS. When the body has arrived at B, the force F acting toward the centre H can be resolved into two forces a and b, of which a acts to stop the motion in the direction parallel to A, G, and b acts to set the body in motion in the direction parallel to A, H. a = F sin.x. and b = F cos.x. In the position C we have c = Fsin., and d=Fcos.' rn W^. x 2933.5 / 2933.5 tan ' X 51ieJ to7i< \ 54.16 oa.io^ 54.16 O These formulas give the number of revolutions per minute of the governor when the angle x, height h, radius r or length of the pen- dulum are given. The weight of the balls does not influence the angle of the governor, because the centrifugal force varies as the weight. The weights only serve to do the work of regulating the steam-valve when the velocity of rotation changes. Exampk. The pendulum arms of a governor are 1=2 feet from the upper joint to the centre of the balls. How many revolutions n ? must the governor make per minute to form an angle x = 45 ? 54.16 4o revolutions. |162. THE VARIABLE PENDULUM GOVERNOR. In this governor the balls are hung at a distance from the centre- line, by which the length of the pen- dulum varies with the angle x. The distance from the centre a of the ball to where the direction of the arm cuts the centre-line at b is the real length of the pendulum in that position of the governor. When the balls hang vertical, the length of the pendulum is infinite. d = right-angular distance from point of suspension to centre-line in Vfeet. e = distance from centre of ball to point of suspension. Then the length of the pendulum in any position of the balls will be in feet, l=d cosec.x -f e. The formulas and table for the ordinary centrifugal governor will also answer for this governor, only that the pendulum length must be measured from a to b. GOVERNORS. 231 Revolutions per minute of governors with different angles and lengths of the pendulum-arms. Length ANGLE OF PENDULUM IN DEGREES. I 2O 25 3O 35 4O 45 50 55 60 inches. n n n n n n n n n 1 193 197 202 207 214 223 234 248 265 2 137 140 143 147 152 158 166 175 187 3 112 114 116 120 124 129 135 143 153 4 97 99 102 104 108 112 117 124 133 5 87 88 90 93 96 100 105 111 119 6 79 81 83 85 88 91 96 101 109 7 73 75 76 78 81 84 89 94 100 8 68 70 71 73 76 79 83 88 94 9 64 66 67 69 71 74 78 82 88 10 61 62 64 66 68 70 74 78 84 11 58 59 61 63 65 67 70 75 80 12 56 57 58 60 62 64 67 72 77 15 50 51 52 53 55 58 60 64 69 18 45 46 47 49 51 53 55 58 63 21 42 43 44 45 47 49 51 54 58 24 39 40 41 42 44 46 48 50 54 27 37 38 39 40 41 43 45 48 51 30 35 36 37 38 39 41 43 45 48 33 34 35 36 37 38 39 41 43 46 36 32 33 34 35 36 37 39 41 44 39 30 31 32 33 34 35 37 39 41 42 30 31 32 33 34 35 36 38 41 45 29 30 31 32 33 34 35 37 40 48 28 29 30 31 32 33 34 36 39 Vertical height h in inches of the centre of suspension above the centre of the balls, of governors making u revolutions per minute. n h n h n h n h 20 88.00 32 34.38 51 13.53 114 2.71 21 79.82 33 32.32 54 12.07 120 2.44 22 72.73 34 30.45 57 10.83 126 2.22 23 66.54 35 28.74 60 9.78 132 2.02 24 61.07 36 27.16 66 8.08 138 1.85 25 56.32 37 25.71 72 6.79 144 1.69 26 52.07 38 24.38 78 5.78 150 1.56 27 48.29 39 23.14 84 4.54 162 1.34 28 44.90 40 22.00 90 4.34 174 1.16 29 41.86 42 18.18 96 3.82 186 1.01 30 39.11 45 17.38 102 3.38 i 198 0.88 31 36.63 48 15.27 108 3.02 i 200 0.87 232 ELEMENTS OF MECHANICS. Fig. 189. 163. ISOCHRONOUS GOVERNOR (Devised by the Author). The construction of this isochronous governor is readily understood by the illustration. It is perfectly bal- anced, and will work equally well in any position (horizontal, inclined, or vertical) it may be placed. It is independent of the force of gravity. W= weight in pounds of the four balls. F= centrifugal force. /= force in pounds on the spring in the direction of the spindle. Z> = diameter in feet of the circle de- scribed by the centre of the balls. d length of the two levers, as shown on the drawing. e = distance between the balls in the direction of the spindle. The centrifugal force of the four balls will be WDri* F- 5867 F:f-d:e, of which /= . WDerf and /" 5867 d ' The force/ of the spring ought to be so adjusted that the balls will form a square when the governor runs at the average speed. 164. CENTRIFUGAL FORCE OF BODIES MOVING ON CURVED ROADS. Fig. 190 represents a section of a circus ring, and a rider on a horse. It is well known that the faster the horse runs the more he leans to- ward the centre of the ring. Let the body B be suspended from a by the line B a, and swung around the circle in the same path and with the same velocity as that of the horse ; then the inclination of the line B a will be the same as that of the horse and the rider. From the formulas for the centrif- ugal governor we have the angle x as follows : Rn* MOMENTUM-CUR VE8. 233 R = radius in feet of the centre of the track. n = number of turns around the circle per minute. The rise of the track above horizon will be the same as the angle x. Let F denote the velocity in feet per second of the horse or ball. Kg. 190. a 60 n = ^0_F 2* R R 60' F' tan.x = 2933.5(2 * Ef 3600 F 2 tan.x 2933.5 x 4 x 9.86955 R 32.17 R This formula gives the angle x when the radius R and velocity F are given. When we have the velocity expressed in statute miles per . , ... ; Miles 2 hour, the angle x will be : tan.x = 6 14.956 R In railway curves the outer rail should be elevated an angle x above the inner rail. Call G = width of gauge in inches, h = eleva- tion of the outer rail in inches. h Miles 2 O Miles 2 7 and. o. n = -. G 14.956 R 15 R This formula is not strictly correct, because is sin.x, instead of G fan.z, but the difference is of no practical importance in the small angle of elevation of the outer rail in railroad curves. Example. A railroad train is to run at the- rate of 30 miles per hour on a curve of .# = 500 feet radius, anjfc- 'gauge (9 = 56.5 inches. Required the elevation of the outer rail ? . 56.5x30' h = = b.o inches. 15 x 500 With this elevation the wheels would bear equally on both rails. For slower speed on the same curve the wheels would bear heavier on the inner rail than on the outer one. 234 ELEMENTS OF MECHANICS. Fig. 191. d \ 165. TO MEASURE ANGULAR VELOCITY BY CENTRIFUGAL FORCE. The illustration, Fig. 191, represents a glass tube bent into the shape of a fork a, b, c, d, and filled with mercury to the line ef. The leg c d is set into rotation around the other leg a b as axis, and the centrifugal force of the mercury in the part b c will raise the mercury in the leg c d and lower it in a b to a difference h of level in the two legs. A = area of the cross-section of the tube in square inches. h = difference of height of the mercury in inches. R = radius of revolution of the leg c d in inches. n = number of revolutions per minute. W= weight in pounds of the column of mercury of the height A. w = weight of the mercury in the part b c. The weight of a cubic inch of mercury at the temperature of 60 Fahr. is 0.941 of a pound. The weights, W= 0.941 A h, and w = 0.941 A R. The centrifugal force of the mercury w will be F= E 70405' which must be equal to the weight W which acts as centrifugal force. Insert the values of W and w in the formulas, and we have 0.941 A l?n s 0.941 70405 241.44 241.44 70405' The differential height A of the mercury is independent of the sec- tional area A, which can be irregular. The parts a b and c d of the tube should be parallel, but the part b c need not be at right angles to the legs, nor need it be straight, but can be made of any curve or shape. Example 1. The radius between the centre lines of the legs is R = 3 inches, and makes n = 184 revolutions per minute. Required the differential height A ? 3* x 84 - = 4.327 inches, of which one-half, or 2.1635 inches, 70405 will fall in the leg a b and rise the other half in the leg c d. ROTATION GAUGE. 235 Fig. 192. Example 2. How many revolutions must the instrument make with a radius J? = 5 inches, to raise a differential column of h = 16 inches ? 241 44 n = 1/16 = 193.152 revolutions per minute, o When the areas of the cross-sections of the bore in the two legs are alike, the column of mercury will sink in the centre leg a b as much as it rises in the leg c d, or e e =//', and a graduated scale could be attached to the centre leg to indicate the number of revolutions of the instrument. It is not necessary to make the cross-sections of the legs alike, as will be explained in the following section. | 166. REVOLUTION INDICATOR. This instrument, Fig. 192, is based upon the principles described in the preceding section, and invented and patented by Edward Brown of Philadelphia. The centre tube a b is made of glass and con- tains the mercury which communicates with the iron tube c d. When the indicator stands still the mercury level is at/ e. The iron tube is fitted with a vessel at d, suf- ficiently large to contain the whole column e d of mercury, so that when the indicator revolves with its highest speed, the mercury will fall from e to e' in the glass tube, and rise to a small height in the vessel d. The part g is only a balance rod of solid iron. The vessel d is turned inward at the top to prevent the mercury from splashing out. The graduated scale is held in position by the framing, as shown by the illustration. The instrument is intended for indicating the rate revolutions of a steam-engine or other rotary machine, for which purpose it is not necessary that the indicator should make the same number of revolutions, but can be geared to run with any desired speed, and the scale is graduated to suit the gearing and indicate directly the number of revolutions of the engine. A = area of cross-section of the vessel d, and a = that of glass tube. h = difference of height of the mercury. x = depth to which the mercury sinks in the glass tube in inches. n = revolutions per minute of the indicator. R = radius of rotation of the iron tube c d. 236 ELEMENTS OF MECHANICS. 70405 - 1. x = - 0405/1 + -\ Example. The areas a : A = 1 : 16, and R = 4 inches, making n = 15 revolutions per minute. Required the sinkage x of mercury in the glass tube ? The indicator is geared, say 3 to 1 revolutions of the engine ; the latter will make 50 turns per minute when the mercury sinks 4.8125 inches in the glass tube. The divisions on the scale will be a ge- ometrical progression, or as the square of the revolutions. The scale can thus be graduated, but number the divisions so as to correspond with the revolutions of the engine. |167. CENTRIFUGAL FORCE OF A LIQUID ENCLOSED IN A ROTATING VESSEL. Fig. 193. Fig. 193 represents a cylindrical vessel of radius R and height _T, about half filled with any liquid, say to the line ef. When the system is set into rota- tion the centrifugal force will form the liquid into a complement paraboloid, or the section of the liquid will be a parabola, in accord- ance with the following formula : h 241.44 241.44 70405 n r h = abscissa and r = ordinate. The formulas give the form of the para- bola in the rotating vessel. The ribs b b are fastened inside to make the liquid rotate with the vessel. The angle z at any point of the parabola 2 A with the ordinate radius r will be tan.z = . r The same angle will be formed by the surface of the mercury in the vessel d, Fig. 192. The cubic content of a paraboloid is one-half of that of a cylinder of the same base and height ; the liquid in a rotating cylindrical ves- sel (Fig. 193) will therefore sink in the centre as much as it rides on the sides until the inverted vertex of the parabola reaches the bottom of the vessel. PENDULUM. 237 168. PENDULUM. A body freely suspended above its centre of gravity and made to swing is called a pendulum. Simple Pendulum is the combination of a small body hung on a light line and made to swing. Compound Pendulum is a rigid system of bodies suspended above the centre of gravity and free to swing. A body suspended at c by the line I will hang perpendicular under the point of suspension, and the combination is called a. plumb line. Draw the body aside from b to a, and leave it free to the action of gravity, which will draw the body back to b. The prim- itive work consumed in drawing the body aside is stored into it by the force of gravity which moves the body from a to b, where it arrives with a velocity equal to that due to an equal height of fall S. A body in motion cannot be stopped without restoring the work which has set it in motion, for which resistance is required, and as the body met with no resistance at b, its motion will be continued to c. Whilst moving from b to c the body meets the resistance of gravity, which has discharged the work when arriving at c, an equal height 8 above b. As the body is free to move in the arc of the circle, the force of gravity will draw it back to b, and the motion continue to a, where again it will be drawn back to b, and so it will continue to move fore and back for ever if no other force interferes with the ope- ration. This operation is called oscillation of a pendulum. U69. DYNAMICS OF THE PENDULUM. Let a body M, Fig. 195, be suspended by an inflexible rod without weight from the centre C, and free to revolve in the whole circle around that centre. J? = radius of the circle in feet. , which corresponds with that at the lowest position in the circle ; after which the forces F act opposite the motion and stop it at A, which corresponds with PENDULUM. 239 the highest position in the circle. The work accomplished by the forces F is represented by the area of the figure bounded within the curve B b D c A and the straight line A, B. The area B b D represents the work of drawing the body from via 1, 2, 3, etc. to 8, and which is equal to the weight W multiplied by the diameter of the circle, or in which V= velocity of the mass M when passing the lowest position at the division 8. This work is stored in the body, and cannot be taken out of it with- out realizing an equal amount of work, which is accomplished by the body continuing its motion in the circle against the action of the force of gravity, until it reaches its starting position at the highest point 0. The work accomplished by the force of gravity at any position of the mass M is the weight W multiplied by the vertical space, or sinus-versiis of the angle

, and R is the recoil of the gun. The recoil is generally partly coun- teracted by a force/ applied on MAS friction against the force F, as repre- sented by Fig. 210. . Fig. 214. (F-f) : M = V: T F:m = v : T F MV MV mv \F-f)~~F' m(F-f) 1 ~Y = M F MV ofwhlch r = of which S= 2 mv 2 MV* = +. 2(F-f) . 8 . 9 . 10 . 11 . 12 Recoil F M -+1 . 14 ORDNANCE DYNAMICS. 253 This is the recoil at the moment the mass or ball m reaches the space S, but the mass J/has then a velocity V, which must be stopped by the force f in an additional recoil r' . I MV* -, and the whole recoil r+i / = ^ + . . . 15 F M This formula reduces itself to [r , IFm(F-f) . 16 It is supposed in the above arguments that the mass or 'ball m moves through the bore without friction or other resistance, which cannot be the case ; and when that friction is taken into account the recoil will be diminished considerably. This friction and resistance to the ball in the bore acts to drag the gun with it, so that there may be ho recoil until the ball leaves the muzzle, as has been con- firmed by experiments with the ballistic pendulum. 'i 182. THE BALLISTIC PENDULUM. This pendulum, represented by Fig. 215, is designed for the pur- pose of measuring the velocity and work of a body striking it. It consists of a long rod suspended at one end, and with a block of wood or some other soft ma- terial at the other end, forming a pen- dulum. The point b is the centre of gravity of the whole pendulum, and c the centre of oscillation. A body, or say a rifle-ball J3, striking the wood block will move the pendulum an angle x and raise the centre of gravity a space s, which is the versed sine of x. The body B should strike the pendulum in a horizontal direction toward the centre of oscilla- tion c, in order to avoid jarring in the fulcrum a. The centre of oscillation should be determined by allowing the pendulum to swing and counting the number of single oscillations n per time t in seconds. The pendulum length Z/, from the fulcrum a to the centre of oscillation c, will be . 39.114 f* . 2j = , in inches. n 2 Fig. 215. _, 1 V I IS Aas. i \ ' j" !*li.- \ 254 ELEMENTS OF MECHANICS. The centre of gravity b is found by balancing the pendulum over a sharp edge or by suspending it in different positions, as described on 50. The work of raising the pendulum the vertical space s is equal to the work of the striking body B. W= weight of the pendulum in pounds. s = space in feet which the pendulum is lifted. B = weight of the striking body in pounds. V= striking velocity in feet per second. The work Ws = *9 The velocity The space s = jR ver.$in x. The angle x is measured by a graduated arc under the pendulum. Example. The weight of the pendulum is W=400 pounds, and the distance from the fulcrum to the centre of gravity JK = 10 feet. The weight of the striking body is _Z? = 0.08 of a pound, which moves the pendulum an angle ar = 36 53'. Required the striking velocity of the body B ? Space s = 10 x arer.sm.36 53' = 2 feet. Velocity F= J 2 * 32 - 17 * 400 * 2 = 02.3 feet per second. 0.08 . DYNAMICS OF HEAVY ORDNANCE. The force of ignited gunpowder enclosed in a gun varies with the quickness of the powder, and has been found to reach as high as 40 tons to the square inch. The work of gunpowder in heavy ordnance expressed by the ordinary unit foot-pound becomes a very high num- ber, for which a larger unit has been adopted by English ordnance officers namely, that of foot-ton, which means a work of lifting one ton of 2240 pounds one foot high. This unit reduces considerably the number which expresses the work of heavy ordnance. Let M denote the mass, and TFthe weight in pounds of a projectile of velocity V feet per second ; then the work stored in the projectile will be, as before proved, , MV* . K=* - = - in foot-pounds. ORDNANCE DYNAMICS. 255 W V 1 WV* K = = in toot-tons. . . 2 2^x2240 144121.6 144121.6 k Weight of projectile W=- pounds. . 3 Velocity of feet per second V= -*/ : . . . 4 The dynamic work of different kinds of gunpowder utilized in heavy ordnance varies between 60 and 90 foot-tons per pound of powder. The average may be taken to be 80 foot-tons. Let P denote the weight in pounds of powder in a charge, then the work W F 2 = 80P= Weight of charge Weight of projectile TF= 144121.6 IFF 2 11,500,000 11, 500,000 P v* TT, -. t .-i v /11,500,OOOP Q Velocity oi projectile l / = \/ ,'' " Example. A gun is loaded with a charge of P = 30 pounds of powder for a projectile weighing TF=200 pounds. Required the muzzle velocity of the projectile? . . 111,500,000x30 1Q1Q , . , . Formula 49. r "%/ --- -- = 1313 feet per second. Ans. * ^00 Quick powder in small firearms utilizes less work per weight of the explosive than does slow powder in heavy ordnance. The formulas from 47 to 49 inclusive are equally applicable for small firearms, in which the weights of the powder and projectile are expressed in grains. Coefficients, j 11 ' 500 ' 000 for heav ? ordnance ' i 8,000,000 for small firearms. 184. DYNAMIC DIAGRAMS OF HEAVY ORDNANCE. The following diagrams are deduced from English experiments with heavy ordnance made in the year 1870. The experiments were made by an 8-inch wrought-iron gun weighing 6 tons, and length 256 ELEMENTS OF MECHANICS. of bore 126 inches. The projectile was a cast-iron cylinder 15 inches long, turned to 7.99 inches in diameter, and weighing 180 pounds. The principal object of these experiments was to ascertain the maximum pressure of, and the effect produced by, different kinds of ignited gunpowder in the gun. The powders experimented with were of nearly the same compo- sition namely, saltpetre 75, charcoal 15, and sulphur 10, making 100 parts total. Result of the Experiments. Nature of Ponder. Charge of powder. Maximum pres- sure per square inch. Muzzle velocity per second. Rifle, large grain, R.L.G pounds. 30 tons. 29.8 feet. 1324 Russian prismatic 32 20.5 1366 Service pellet 30 17.4 1338 Pebble No. 5 35 15.4 1374 The rifle large grain is a quick powder, which gave the smallest velocity with the greatest maximum pressure. The Pebble, No. 5, ia a slow powder, which gave the greatest ve- locity with the smallest maximum pressure. The pressure of the ignited gunpowder was measured by Rodman's pressure-gauge. The time of operation in the gun was recorded by a chronoscope designed by Captain Andrew Noble, F. R. S., and from which the velocity of the projectile was calculated. (See London Engineer, Sept. 16, 1870.) The illustration, Fig. 216, represents a gun of the Rodman pattern, and not the one used in the English experiments. The drawn curves represent the performance with powder of rifle large grain, R. L. G., and the dotted curves that of the pebble powder, No. 5. The absciss axis of the diagram is divided into feet, and the ordi- nate axis into pressures in tons per square inch. The diagrams commence at the back end of the projectile w when ready to fire. The figure shows the position of the projectile with 35 pounds of pebble powder. When charged with 30 pounds of R. L. G. the pro- jectile is forced in farther until it reaches the charge. ORDNANCE DYNAMICS. 257 Fig. 216. Pressure Curves. The drawn line P' , P', P' represents the pressure curve with R. L. G., which reaches 30 tons to the square inch. The curve is not continued to that height on the diagram for want of space, but it shows that that maximum pressure is instantaneous. The ordi- nates measured from the base line to the curve show the pressure in tons per square inch at different positions of the projectile until it reaches the muzzle. The dotted line P, P, P is the pressure curve for a charge of 35 pounds of pebble powder. 258 ELEMENTS OF MECHANICS. The pressure curves approach the shape of an equilateral hyper- bola, but when the projectile attains a greater velocity in the bore the expansion of the gas does not follow up with its due pressure. Notation of Letters. Q = weight in grains of the powder in the charge. q = volume in cubic inches of the powder, including that, if any, between the charge and the projectile. Q = volume in cubic inches of the bore. q' =any volume of the bore enclosed by the projectile. JP = pressure in pounds per square inch of the ignited gas. /~>i = square of the semi-diameter of the hyperbola. 12 2.55 is a constant to be subtracted from the ordinate due to a regular hyperbola. The formula for the pressure curve P' , P' , P' will then be P-^-2.55. . . . .1 Example. Charge of powder R. L. G., 30 pounds x 700 = 210000 grains = G. Required the pressure of the ignited gunpowder when the projectile is at the first ordinate? The projectile being moved in 3 inches farther than shown on the drawing, which position is for pebble powder. The cross-area of the bore or projectile is 50 square inches. Volume, : 27 x 50 = 1350 cubic inches. 9 i noon Pressure, P' = - 2.55 = 10.41 tons. . 12x1350 The dotted pressure curve P, P, P is for a charge of 35 pounds of pebble powder, No. 5, which occupies 15 inches in the bore, or 750 cubic inches. This being a slow powder, drives the projectile ahead and increases the volume before all of it is ignited, and thus diminishes the max- imum pressure, which we find on the diagram to be when the projec- tile has moved about 6 inches, and when the volume is increased to q' = 50(15 + 6)1050 cubic inches. G = 35 x 7000 = 245000 grains. Pressure, P= ^ 2.55 = 16.89 tons. The diagram shows only 15.4 tons, being slow powder. ORDNANCE DYNAMICS. 259 The treatment of pressure, volume, temperature and work of ignited gunpowder belongs to dynamics of heat, from which the Formula 1 is borrowed. Work Curve. The work accomplished by the charge is represented by the area bounded within the pressure curve and abscissa to the ordinate of the pressure. This area, multiplied by the 50 square inches section of the bore, gives the work in foot-pounds accomplished by the ignited powder. The work has been calculated for each ordinate, and set off from the abscissa to the work curve K, K, K. The number on the scale, multiplied by 100, gives the work accom- plished in foot-tons. It is this work which is stored in the projectile when set in motion namely, wV* K= in foot-pounds ..... 2 _ZT= - in foot-tons. . . . .3 4480 g w = 180 pounds, weight of the projectile, V 1374 feet per second, the muzzle velocity of the projectile with pebble powder, No. 5. Work The diagram shows 2500 foot-tons accomplished by the charge, and 2500 - 2358 = 142 foot-tons, which must have been expended in fric- tion and leakage in the bore. The diagram can, however, not be very correct. Theoretical Work in Ordnance. S= length in feet of the bore of the gun. s = length occupied by the powder, s' = length passed through by the projectile in the bore. A = cross-area of the bore in square inches. The differential work will then be 12? Jr " 12 ? -2.55 . ... 4 260 ELEMENTS OF MECHANICS. As the bore is cylindrical, 8s' = 8q, and by integrating the formula from s to S, the work will be This work should be equal to that stored in the projectile when leaving the muzzle, or 2? Having given the charge G and the dimensions of the gun, the velocity of the projectile should be The velocity will be slightly less on account of friction and in the bore. In a gun of 8 inches in diameter of bore the cross-area is A = 50 square inches, and , 185. GUNPOWDER PILE-DRIVER. This pile-driver, invented by Thomas Shaw of Philadelphia, is worked by gunpowder as motive-power ; it consists of a gun a placed on a pile b, and a ram c with a plunger d fitting closely in the bore of the gun. The ram is guided by a high framing, as represented by Fig. 217. The explosion of the gunpowder in the gun, drives the plunger d with the ram c to a con- siderable height, and the recoil of the gun drives the pile into the ground. Whilst the ram is up a new charge is placed in the gun, and in the fall of the ram the plunger enters the bore and compresses the enclosed air to a high pressure and heat, which ignites the charge for another explo- sion for driving up the ram ; and so the ope- ration is, continued until the pile is driven home to its destination. The pile is also driven into the ground by the compression of the air in the gun, but the greatest drift is in the recoil. The ram can be held at any height by a friction-brake extending the whole height of the framing. At the top of the framing is a cushioning piston e, which fits closely in a bore in the ram, for the purpose of preventing the ram from striking or rising above the limited height. 264 ELEMENTS OF MECHANICS. The bores in the gun and ram are funnel-shaped at the top for admitting the plunger and piston with safety from striking the edges. A very slow gunpowder should be used in the charge. I 186. THEORY OF THE GUNPOWDER PILE-DRIVER. The recoil of the gun or set of the pile will be the same as that by the Formula 16, page 253. $=set of the pile in feet. L = length in feet of the bore in the gun passed through by the plunger, omitting the funnel. F= mean force in pounds of the gunpowder explosion. w = weight in pounds of the ram and plunger. W weight in pounds of the gun and pile. /= force of resistance in pounds to the pile in the ground, less the weight W. Call E = actual resistance to the pile, then E = W+f, and/= E - W. The efficiency of the pile-driver consists in obtaining great recoil, and the formula shows that the greater weight of ram and the greater length of bore the greater will be the set 8 on recoil ; but the bore should not be made longer than is necessary for the gases to raise the ram a sufficient height for igniting the charge in its fall. The force, of gunpowder cannot be correctly ascertained without experiments,; and different compositions of powder give different dia- grams of force and expansion. This subject belongs to dynamics of heat, from which the following three formulas are taken. They give an approximate value of the performance of gunpowder. G = weight in grains of powder in the charge. Q = volume in cubic inches of the gas of the exploded gunpowder. P = pressure in pounds per square inch of the gas. L = length of the bore of the gun in feet. I = length of bore occupied by the charge. F= mean force of the charge in pounds. Pressure P= . .... 2 Work JT-100.fo$r.(y\ . . 3 , 100 a , (L\ Mean force F= hyp. log. f I . .4 L-l \ I j GUN PILE-DRIVER. 265 Example. Assume the cross-area of the bore in the gun to be 40 square inches, and acted upon by the explosion of O = 700 grains of powder. Length of the bore L = 2 feet and 1= 0.08333 of a foot, which is one inch, the distance between the plunger and the bottom of the gun at the time of explosion. Required the pressure Pand P' per square inch in the gun at the time of explosion and when the piston leaves the muzzle ? The volumes of gas are Q = 1 x40 = 40 cubic inches at the time of explosion, and Q = 40 x 24 = 960 cubic inches, the volume of the bore. 100 x 700 Pressure P= = 1750 pounds per square inch. 100x700 , Pressure P = = 73 pounds per square inch. 960 Required the work done by the explosion ? Work K= 100 x 700 x hyp.log / ^ \- 222460 foot-pounds. Required the mean pressure jP? K 222460 Mean pressure F= j^ = 2 _ Q()833 = 116066 pounds. The work K of the explosion of the gunpowder performs two duties namely, to drive the pile into the ground and raise the ram. A = the height in feet to which the ram is driven, the work of which is w h. The work of driving the pile into the ground is S f. Then we have the work K=wh + Sf, of which/= , . . 5 and the resistance to the pile in the ground will be R^f+W^^-W. . 6 S The bearing capability of the pile should be calculated from For- mula 5, after the last blow or set S. The quality of the gunpowder can be ascertained experimentally by placing the gun on a solid foundation, so that there will be no recoil of the explosion. Then, F(L-t) = wh 7 and the mean force, F= 8 L I 23 266 ELEMENTS OF MECHANICS. 137. COMPRESSION OF AIR IN THE GUN. When the ram falls and drives the plunger into the bore of the gun, the air will be compressed to a small volume and heated to a high temperature. Knowing the weight of the ram and height of its fall, it is required to find the compression and temperature of the air in the gun. For this purpose we must borrow some formulas from dynamics of heat, which do not belong to this treatise : p = pressure of the air on the plunge-piston when entering the bore, which is 14.7 pounds to the square inch. P= pressure of the compressed air on the plunger. V= volume of air of atmospheric pressure in the bore. v = volume of the compressed air. Then, P=p( The bore of the gun is cylindrical, and the enclosed volume of air is therefore as the distance between the plunger and the bottom of the gun. Then, P= P(^T )" ..... 10 The work accomplished by compressing the air will be . . . .11 This work is wholly consumed by the generation of heat in the compressed air. In case the work of the falling of the ram w h is 7" 1.4 greater than p = -, the difference will be utilized in driving the pile into the ground. This set of the pile is not included in S, Form- ulas 5 and 6. P and p mean the actual pressure above vacuum of the enclosed air on the whole piston, from which must be subtracted the atmo- spheric pressure on the ram-plunger, which is 14.7 A, when A = area in square inches of the plunger piston. GUN PILE-DRIVER. 267 The work done by the ram in the compression of the air will then be .14 and if we include the work that may set the pile during the compres- sion of the air, which is/s, we have Theset s = -(w h- u.T A)(-^-+l-L\. . .16 188. TEMPERATURE OF THE COMPRESSED AIR IN THE GUN. T<= temperature of the compressed air. t = temperature of the air or gas of atmospheric pressure in the gun when the plunger enters the muzzle. The absolute zero = 461 below zero of Fahrenheit's scale. X /461+JY" 17 461+JY" 461 + * ) 161. . . . .18 It is supposed in these formulas that no air leaks out during the compression, and that no heat is conducted from the air by the metals enclosing it ; which cannot be the case, and for which reason a deduc- tion of at least 25 per cent, should be made in the length L. The temperature of ignition of slow gunpowder may be assumed to be T=600 Fahr., and assuming the temperature of the air or gas in the gun under atmospheric pressure to be t = 80, we have 461 + 600 = 1061, and 461 + 80 = 541, the absolute temperatures. Then = 5.208 Z, and I- 0.192 L, when ignition of the gunpowder takes place, but if 25 per cent, is deducted for leakage and conduction of heat, we have L = 6.944 I, and Z = 0.144 L. 268 ELEMENTS OF MECHANICS. The following data of performance of the gunpowder pile-driver has been tabulated by F. C. Prindle, C. E., U. S. N., from records of work done at League Island on the river Delaware : s 8 Driven. Diameter of Pile*. Inches. Distance Driren. No. of Bloirs per Weight of Powder per Pile. 5 | 1 * | Top. Bottom. Feet Pile. Pounds. II =| d - M M M x J Tn O * 1 3 a < a < 3 a < a a < a a < fc 811 10.5 13.1 13 6 S.7 -'2.5 14 19.4 19 3 5.2 1.3 i | 1300 6 i 9G6 15 10 12 11 6.5 s JO 21 24 ... 20 ... 2 1200 5* 457 19 9 12.2 14 7.5 8.7 36 20 29.2 8-5 11 H0.4 U u 3i 1700 6f 63 16 10 12.7 12 7.5 9 H 25 29.5 122 39 59.6 15 4 61 1200 5* 172 17.5 9 11.4 12 7 U Hi 26 29.2 30 6 12.7 4} 1 3.2 2170 7| Records Corresponding to the Numbers in the Table. Record 1. The first machine employed upon actual work. The framing was made of cast-iron, mounted on a scow and operated afloat at the landing wharf, Fig. 218. The piles of heavy yellow pine driven through mud containing clay to compact gravel. Record 2. The framing of wood and iron, mounted on land. Piles of hemlock, and firmly driven without pointing, through stiff clayey material, mixed with sand, to hard gravel and boulders. The piling was for the foundation for storehouses, etc. Number of blows and weight of powder approximate. Record 3. The same machine operating, in the same kind of ground as in Record 2. Foundation for iron-plating shop. Record 4- The same machine as in Records 2 and 3, but with lighter ram. Record 5. The new improved machine, with wrought-iron framing, completed the work. Fig. 218. This illustration, Fig. 218, represents Mr. Shaw's gunpowder pile-driver on a scow in the Delaware River at League Island Navy- Yard. DYNAMOMETER. 269 189. DYNAMOMETERS. Dynamometers are instruments for measuring force, power and work. The simplest form of dynamometer is that of a spring. 190. SPRING DYNAMOMETER. The construction of this dynamometer is readily understood hy Fig. 219. It is graduated by experiments in compressing the spring with known weights. Fig. 219. This dynamometer is, best adapted for measuring the force of pull- ing a load on a road, a boat on a canal, or of towing a ship. The force in pounds indicated by the dynamometer, multiplied by the velocity in feet per second, will be the power in effects, which divided by 550 will give the horse-power in operation. 2191. PRONY'S FRICTION DYNAMOMETER. Fig. 220. This dynamometer consists of a friction brake, as shown by the illustration. It is keyed on the shaft A, which transmits the power and work to be measured. The lever of the brake should be balanced at B before the weight W is put on the scale, and if it is not balanced, the weight of the lever and scale should be weighed at the scale and added to the weight W. The weight W on the scale is the force acting on the lever or radius R. It is supposed that all the power and work transmitted by the shaft is consumed by the friction in the brake. When the shaft is running with its average speed of n revolutions per minute, the strap is tightened up with the screws, so that the lever will barely lift the weight W, which is also adjusted to suit the motion. When the weight and friction are well balanced, count the revolutions per minute of the shaft. 23* 270 ELEMENTS OF MECHANICS. The power transmitted through the shaft is equal to the weight W multiplied by the velocity of the circumference of the radius It, mak- ing the same revolutions as the shaft. The velocity in feet per second is Power P = W F= i n e ff e cte f which divided by 550 give the WEn -p. . Horse-power IP = -- = 60x550 5253.2 The work Km foot-pounds consumed by the friction of the brake in the time T in seconds will be Work 60 9.55 All this work consumed by the friction is restored by generating heat, which makes the brake so hot that a constant stream of water must run on it to absorb the heat whilst the experiment is made, otherwise the wood in the brake would take fire. When convenient it is best to make the lever j= 10.5 feet, or 10 feet 6 inches, which will make the circumference 66 feet ; in which case, the horse-power will be g ^ 6&nW = 2nW ~ 550x60 " 1000 ' That is to say, the product of the revolutions per minute and weight W, multiplied by 2 and point off three places, will be the horse-power of the experiment. A lever of R = 5 feet 3 inches will make the circumference 33 feet, and the horse-power H.,^. 1000 DYNAMOMETER. 271 \ 192. BEVEL- WHEEL DYNAMOMETER. Fig. 221 represents a side elevation and Fig. 222 a plan of the dynamometer. Fig. 22L. It can be worked either by cranks c, c, or by pulleys a and J. Let c? be a shaft and pulley communicating motion by the dotted belt to the shaft and pulley e, and it is required to measure the power transmitted between the two shafts. Place the dynamometer so that it takes a belt/ from the pulley d on its pulley a, and another belt g from the pulley b to the pulley e. The pulley a and the bevel-wheel / are fastened on the crank-shaft. The bevel- wheel 3 is fast on the pulley 5, but both are loose on the crank-shaft. The bevel-wheels 2 and 4 are both loose on the arm h. Either one of the wheels 2 and 4 could be dispensed with, but the dynamometer works better with the two wheels. Now set the pulley and shaft c?in motion in the direction of the arrow, and all the power will be trans- mitted through the dynamometer to the pulley e. The problem is to measure the power transmitted. The arm h is fitted loose on the crank-shaft as a fulcrum, ai'ound which the arm is allowed a small angular motion, limited by the slot in the pillar i. 272 ELEMENTS OF MECHANICS. When power is transmitted the bevel-gear acts to lift the arm A, but a force F is applied to keep the arm in a horizontal position. If the pulley b with the wheel 3 is held stationary, and the arm be al- lowed to revolve, it would make one revolution whilst the crank- shaft makes two ; but when the pulley b and wheel 3 revolve the arm h is held horizontal by a force F. The force F is derived from a spiral spring enclosed in the drum^, which acts on the chain like that in a watch. A ratchet-wheel is fastened on the spring axis, and by the aid of the crank I the spring can be regulated to the exact force required to balance the arm A with the power transmitted through the dyna- mometer. The drum^ and ratchet-wheel k are both graduated to indicate the force F on the chain in pounds. The sum of the readings is the force on the chain. The weight W is for balancing the arm A. Notation of Letters. D = diameter in feet of each of the pulleys a and b. f= force of tension in pounds on the belts. R = radius of the arm A in feet. F= force in pounds on the chain. The static momentums of the combination will then be F:f=D:R, and F R=f D. FR -, and / When the dynamometer is worked by the cranks without the belts and pulleys, r = radius in feet of the crank c. f = force in pounds on the crank-pin. F:f'=2r:R, and FR-f'Zr. 2/'r ' FR J p = __, and /= __. Transmission of Power. y= velocity in feet per second of the belts. v = velocity of the end of the arm A if allowed to swing. n -= number of revolutions per minute of the crankshaft. D YNAMOMETER. 273 V:v = Z>: E, and VE = v D. F- and v= VR E ' D ' T . - D n , -En V = , and v = . 60 60 Power P =/ V= F v in effects. Power f j^.f^. The power transmitted through the dynamometer will then be P = *^L n = 0.05236 F E n. Fr.En FEn Horse-power Jtr = 60x550 10503.55 When the dynamometer is working we have given the force F from the spring graduation. The length E of the arm h is given and constant for each dynamometer. The number of revolutions per minute is obtained by counting the same. We have thus given all that is necessary for calculating the power transmitted through the dynamometer. In constructing a dynamometer of this kind it would be best to give such length to the lever E as to make an even number in the denominator of the formula; for instance, if .# = 5.251775 feet, we have the Horse-power IP = - . 2000 Allowing five per cent, for friction in the dynamometer, a five-feet lever would make the formula the same as above. Machinery very rarely transmits power uniformly from one locality to another ; which is particularly the case with the ordinary steam- engine, as has already been explained. The storage and delivery of work by a fly-wheel causes an irregu- larity in the power transmitted, which can be measured by the dynamometer. The oscillation of the arm h and spring-drum j indicates this irregu- larity, which can be read on the graduation, and thus enables us to determine with great precision the efficiency of a fly-wheel. Suppose the drum and ratchet-wheel to indicate a mean force F= 150 pounds 274 ELEMENTS OF MECHANICS. when the drum oscillates a difference 12 pounds ; the irregularity will then be , or 4 per cent. A dynamometer of this kind would be very useful in institutions where the subject of dynamics is taught. The students should be made to work the cranks of the dynamometer, and calculate their own force, power and work, and thus learn practically how to dis- tinguish the different elements and functions in dynamics, how they bear upon one another, and to conceive real magnitudes of such quantities. The dynamometer could easily be arranged with indicators by which to trace diagrams of the different elements and functions in- volved in the operation. 193. DYNAMOMETER AT THE ROYAL TECHNOLOGICAL INSTITUTE, STOCKHOLM. Fig. 223. The accompanying illustration represents an isometric perspective view of the dynamometer at the Royal Technological Institute, Stockholm. The power is applied on the crank a, and communicated through the pulley b, rope V, pulley c, wheel d, rope e, rope-pulley g, and is consumed by the fans///. The shafts n and o are in one line, but not connected between the pulleys b and c. The endless rope V Vis held tight in the grooves of the pulleys b and c by means of DYNAMOMETER. 275 two weights w and W, as will be understood by the drawing. If the two weights were alike, they could communicate no motion to the pulleys, but suppose w = 10 pounds, and W= 20 pounds, then there would be 10 pounds more weight on the sheave i, than on the sheave k, of which five pounds would pull on each pulley b and c. Let the radius of the crank a be equal to the radius of the pulleys, then it would require a force of five pounds to turn the crank in the direc- tion of the arrow. If the crank is turned with an irregular velocity, it would only raise or lower the weights, but a constant force of five pounds would always act on the pulley c to communicate motion to the fans. The power in operation will be equal to the force multi- plied by the velocity of the rope F, and the work accomplished will be equal to the power multiplied by the time of operation. Notation of Letters. E = radius of the crank in feet, which we have supposed to be equal to the radii of the pulleys i and c. F= force in pounds acting on the crank a. F= velocity in feet per second of the rope V, which, in the supposed case, will be equal to that of the crank-pin. 7 7 =time of operation in seconds. Wand w = weights on the pulleys b and c in pounds. P= power in dynamic effects, of which there are 550 per horse-power. K= work, in foot-pounds of work. n = number of revolutions per minute of the pulley c. Then we have The force F=\(W-'w\ and velocity F= 2 * R ". Power P=FV, and Example. Radius of the crank or pulleys, .#=1.25 feet, making n = 28 turns per minute. TF= 80, and w = 20 pounds. Required, the force F=?, velocity F=?, power P=?, and how much work, .5T=?, will be accomplished in one hour, or T= 3600 seconds ? Force F= |(80 - 20) = 30 pounds. .... 2x3.14x1.25x28 _ c . , , Velocity V- = 3.66 feet per second. 60 Power P= 30 x 3.66 = 109.9 effects. Work K= 109.9 x 3600 = 3956.4 foot-pounds. 276 ELEMENTS OF MECHANICS. The average power of a man working eight hours per day is 55 effects, which will be an accomplished work of K= 55 x 8 x 3600 = 1584000 foot-pounds in a day's work. In order to regulate the velocity to suit the power, the dyna- mometer has an arrangement by which to set the fans at any desired angle while in motion, which arrangement is not shown on the draw- ing. Students used to work the dynamometer in a spirit of emula- tion to outdo each other in power and work. Some could accomplish the greatest power, and work with less force and more velocity, whilst others preferred more force and less velocity. Arrangements could easily be made to register on the dynamometer the force, velocity, power, and work in the form'of diagrams. Dynamometers of this kind ought to be employed in all scientific institutions where dynamics are taught, for we have yet no better means by which to imbue the student with the real substance of dy- namics. Any student who has worked this instrument for a few hours will probably not commit the error of saying that work is in- dependent of time, or that time is included in power, which erroneous ideas are yet maintained in text-books. This dynamometer does not only teach the student the different properties of force, power and work, but it enables him to conceive and compare, with great precision, real magnitudes of those quantities, which is of great importance in designing machinery. In the year 1850 the author made a great many experiments with different kinds of screw-propellers, in which was employed a dyna- mometer of this description, made by Thomas Mason of Philadelphia, and which gave great satisfaction ; and he has always considered it the best form of dynamometer where it can be conveniently applied. SOUND. 277 194 DYNAMICS OF SOUND. Sound is work, consisting of the three simple elements F V T, of which F= force of the sound, F= velocity of vibration, and J"=time of continuance of the sound. The loudness of sound is Power P=FV. The pitch of the sound indicates the proportion between F and V. Two different sounds of different pitch may be of equal power or loudness, but the high-pitch sound is then produced by small force F and great velocity F", whilst the low-pitch sound is produced by greater force F' and smaller velocity V ', so that the products F V and F' V are alike in the two sounds. Let an elastic spring a b be drawn aside a space S where the force is F. Leave the spring to take its own course, f . 224 and it will move fore and back and set the surrounding air into vibration, which produces sound. The work expended in drawing the spring aside is K=F S, which work is restored by producing sound, and the spring will continue to sound until all the work F Sis consumed. The loudness or power P= F Vof the sound is greatest at the start of vibration, after which it will gradually diminish, and finally fade away to nothing. Differential work file = Pfit, but the power decreases as the time in- G creases that we can place P = , in which C is a constant factor. , Ctit T ^ rCdt = , K= \ = t J t T\ F 9 Time of sound, hyp.log. T= = - -. The force F of an elastic spring is as the space 8 within the limit of vibration, and the mean force in the space 8 is therefore \ F. The spring vibrates the same pitch of sound in any space 8, and the pro- portion between ^"and Vis therefore constant. The loudness of the sound or power jP= F V= OS' 1 , or the power of the sound, is as the .square of the space of vibration. The velocity V in the above for- mulas means that of the force producing the sound, and not the ve- locity of the sound from the sonorous body. 24 278 ELEMENTS OF MECHANICS. Noise in Machinery. All sound or noise in operating machinery represents so much work lost, and the loudness of the noise represents the power lost. The noise in a cotton-mill, rolling-mill, or railroads, etc. represents power and work lost. Noise of a Steam-hammer. A steam-hammer falling on its bare anvil will set the whole system, with the surrounding air, into vibration, and create a great noise, which represents the whole work of the hammer ; but the same ham- mer falling upon a puddle-ball, or upon white-hot iron, will create very little noise, and the greatest part of the work of the hammer is then utilized in forging the iron. Report of a G-un. The report of a gun represents the work lost of the total work due from the ignited powder. If the ignited gas in the gun could be allowed to propel the projectile until its force of expansion is reduced to the atmospheric pressure, there would be no loud report Sound of a Bell. The work of a clapper in striking its bell is represented by the sound produced. M = mass of the clapper and F= velocity of the strike. Then the MV* Work of the clapper = = work of the sound. Sound of Men and Animals. The sound of men or animals is produced by the heat in the body, which is work. A speaker, singer, or a musician playing a wind in- strument draws from his store of heat as when doing hand labor, and the louder he speaks, sing?, or plays, the greater power is drawn from his heat, and he ultimately becomes fatigued as when working a crank. Music of an Organ. The weight on the organ-bellows multiplied by the velocity with which it sinks whilst air is discharged and none enters is the power in effects, which, divided by 550, gives the horse-power of the organ. The same weight multiplied by the space it sinks is the work done by the organ in producing the music. Dynamics of Sound includes the science of acoustics, and is a very extensive and interesting subject, which requires a separate treatise. SOUND. 279 g 195. VELOCITY OF SOUND IN AIR. The velocity of sound in air Las been determined both by experi- ments and theory by Newton, Laplace, Dalton and various others; the summary of which is V= 1089.42^7 1 + 0.00208(!! - 32). D = 1089.42 TI/ 1 + 0.00208(2 - 32). V= velocity in feet per second of the sound. t = temperature Fahr. of the air. D = distance in feet travelled in the time T in seconds. 1089.42 = velocity of sound at 32, temperature of the air. 0.00208 = volume of expansion per degree Fahr. of the air. Distance in Feet which Sound Travels in Air at Different Temperatures Time see. , TEMPERA' 10 | 20 CURE 32 F THE 40 AIR, 50 FAHRI 60 INHEH 70 P SCAI 80 -E. ^ 100 1 1000 1064.2 1075.7 1089.4 1098.5 1109 1120 1131 1142 1153 1164 2 1985 i 2128 2151 2179 2197 2219 2241 2262 2285 2306 2328 3 2978 3193 3227 3268 3295 3328 3361 3393 3427 3459 3492 4 3971 4257 4303 4358 4394 4438 | 4482 4524 4570 4613 4656 5 4964 5321 5378 5447 5492 5548 5603 5655 5712 5766 5821 6 5956 6385 6454 6536 6591 6657 6723 6786 6855 6919 6984 7 6949 7449 7530 7626 7689 7767 7844 7917 7997 8072 8148 8 7962 8514 8606 8715 8788 8876 8964 9049 9140 9225 9312 9 ! 8934 9578 9681 9805 9886 9986 10085 10180 10282 10379 10476 10 ! 9927 10642 10757 10894 10985 11096 11306 11311 11425 11532 11640 11 110920 11706 11833 11983 12083112205 12326 12442 12567 12685 12804 12 11912 12770 12908 13073 13182 13315 13447 1357313710 13838 13968 13 12905 13835 13984 14162 14280 14424 145(57 14704 14852 14991 15132 14 13898 14899 15060 15252 15379 15534 15688 15835 15995116145 16296 15 14891 15963 16135 16341 16477 116644 16809 16966 1 17137 17298 17460 16 !l5883 i 17027 17211 17430 17576 17753 17929 18097118280 18451 18624 17 16876 17091 18287 18520 186741886319050 19228! 19422! 19604 19788 \ 18 1788919156 19363 19609 19773 19972 20170 20360 20565 20757 20952 19 20 18861 20220 19854 21284 20438 21514 20699 21788 20871 2108221291 21970 !22192i22412 21491 21707 21911 22622 122850 23064 22116 23280 21 '20847 '22348 22590 22877 23063 23301 23532 23753 1 23992 24217 24444 22 21839 23412 23665 23967 24167 124411:24653 24884 25135 25376 1 25608 23 22832 24477 24741 25056 25265 2552025773 26015 26277 26523 26772 24 23825 25541 25817 26146 26364 26630 26894 27146 27420 27677/27926 25 24818 26605 26892 27235 27462 27740 28015 28277 28562 28830 29100 280 ELEMENTS OF MECHANICS. 196. COUNTING BEATS OF SECONDS. When the occurrence of a distant sound is not anticipated, we are unprepared to record the exact moment, and before an appropriate timekeeper can be procured an uncertain time has elapsed. With some practice the beats of seconds can be counted in the mind with tolerable correctness without the aid of a timekeeper; which practice has been of great service to the author in astronomical obser- vations. Practice to count seconds by the aid of an oscillating second pendulum, or by the second-hand on a watch, until the counting agree with the timekeeper, without attention to the pendulum or second- hand. With good practice the counting should not differ more than one second per minute. When an unexpected distant sound is heard and its cause observed, we can always be ready to count seconds, and thus determine the distance. In astronomical observations at sea it is customary to keep a watch in the hand, or to station an assistant at the chronometer to note the time when the observer says " stop ;" but there are known cases when the captain has taken his observations without the aid of a watch or assistant, and walked slowly and comfortably to his cabin and noted the time of his observations from the chronometer, with no little amusement to other observers, who naturally supposed that the cap- tain's observations could not be very correct, but to their surprise were found to be as correct as their observations with ordinary precau- tions. The captain counted in his mind the beats of seconds, and de- ducted the sum from the time observed on the chronometer. The practice of counting seconds correctly is of great utility and service for estimating various movements. When the action is of very short duration, say less than 3 seconds, it is best to count half seconds, or even four times per second, and a short time may be de- termined with a correctness within a quarter of a second. ASTRONOMY. 281 ASTRONOMY. CREATION OF WORLDS. 197. MATTER in celestial space arranges itself into groups or nebulas by virtue of universal attraction, which by the aid of centri- fugal force are finally divided into definite bodies, of which the largest occupies the centre around which the smaller revolve, and the group is called a planetary system. Each fixed star is a central body of a planetary system like our sun. The rotary motion of each group, nebula or of each body around its axis has been caused by collisions of the matters constituting that group or body. The conditions under which a nebula can be formed into a plan- etary system are -first, that it must be set into a quick rotation; secondly, that it must consist of different kinds of matter ; and thirdly, that its ingredients must be of such proportions as to admit of divis- ion by the forces of attraction and centrifugal. Without the above conditions the nebula will remain permanent until it comes into collision with some other nebula or body. It appears that each kind of matter is derived from different parts of space, and in its course of travel meets and mingles with other kinds of matter, whereby nebulas of a variety of shapes are formed. The different forms of nebulas are caused by different dynamical and chemical actions of the mingling matters. The act of collision of nebulas is distinctly seen through powerful telescopes, and the mag- nitude of the collision is so enormous that a change in form is hardly perceptible during a lifetime of observations. The forms of the dif- ferent nebulas indicate their relative ages, which may be graduated from the form of a group of clouds to that of a permanently organized planetary system. The majority of the well-defined nebulas located within our scope of observation are spherical or egg-shaped, with a bright central spot indicating the act of forming a sun, about which the surrounding matter gradually divides itself into separate masses, forming at length a planetary system. Some nebulas are of the form of a ring, others consist of one or more spirals a configuration which indicates the relative motions of 282 ELEMENTS OF MECHANICS. its constituent elements. We also find nebulas of the form of a spindle, having a bright body at each end. The irregular nebulas, which are of the form of a group of clouds, may be classed as primary formations. The period of time in which a primary nebula is thus formed into a planetary system may be many millions of years. The operation of forming nebulas and the creation of worlds is going on all around us, even within our limit of observation through powerful telescopes, but the magnitude of that operation is too enor- mous for any human mind to conceive. The work is gradual and continued until all the matter in each part of space has assumed a definite form and motion. The matter in our part of space that is, the space occupied by our planetary system and the neighboring ones seems to be arranged into a definite form and motion ; but in other parts of space, where groups and bodies are yet forming, some matter or portions of bodies may be led astray by repulsive force in collisions, and by being overtaken by superior force of attraction from other systems is drawn toward a central body, as is the case with matter constantly flowing into our sun. When such stray matter is in the form of a solid body of perhaps thousands or millions of cubic miles in volume, it makes spots in the sun's photosphere ; but a great deal of such flowing matter is in the form of a gas or powder (which we sometimes see as zodiacal light when it passes near to the earth) which makes no visible spots in the sun. Stray matter is often taken up by planets, as experienced on our earth, arid we call it meteors. Falling meteors change the motion of the earth and disturb our chronology, but their mass is so very small compared with that of the earth that it requires many years of observation for us to appreciate any such change ; and as the meteors fall from all possible directions, some of them may counteract the action of their predecessors. The meteors which have fallen on the earth within our time of records and tradition have not changed our chronology more than, perhaps, a few seconds ; but there has evidently been a time when very large bodies have struck the earth and changed its rotation both around its axis and around the sun, before which time the present location of the poles might have been at or near the equator. The largest known meteor on our earth is lying on the pampa of Tucmnan, near Otumpa, in the Republic of Argentine, South America, weighing about 16 tons. A Comet is stray matter seeking a situation to supply other heavenly bodies with oxygen, hydrogen and nitrogen, and travels CREATION OF WORLDS. 283 from system to system, describing elliptical orbits around each central body, until its course is by chance directed near enough to some planet or sun to catch and retain the comet. The return of a comet cannot be calculated or predicted except when its orbit is limited within our system. When the orbit extends outside of our system, the comet will likely never return, but is over- taken alternately by other planetary systems or groups of matter. A planetary system may also have two suns, and subdivisions of groups or nebulas, in which the revolving bodies are called satellites, moving around a planet and forming a system within itself, but subject as one body to the main system. There are several planets in our system forming such a group, of which the earth and moon are one. The planets Saturn and Uranus have each eight satellites, Jupiter has four, and Neptune one. The condition under which the revolving bodies are maintained in their regular orbits is, that the force of attraction of the central body is equal to the centrifugal force of the revolving one. The orbits of periodical rotation are ellipses, in which the central body is in one of the foci. The orbit may accidentally be a circle, but there is no known planet or satellite which revolves in a perfect circle. A pendulum freely suspended and made to swing so as to describe a cone, the base of that cone would practically be an ellipse, for the reason that it is almost impossible to start the pendulum with such perfect velocity and direction in relation to its radius as to make it swing a perfect circle. Such is the case with the planetary orbits, in which the planets have not been started so as to describe perfect circles, and the orbits are also disturbed by the attraction between the planets. 2198. OUR PLANETARY SYSTEM. It will be observed in the following tables that the other worlds in our system are of a different nature from that of our earth, and that no two of them are alike. Their difference of density is a strik- ing feature ; some of them are light as wood, others heavy as rock ; and the planet Mercury must evidently be a chunk of precious metals. The density of the sun is given to be 1.128, compared with that of water, but the photosphere is included in the volume of that mean density. The core of the sun is evidently a very dense body. The column of density of the planets, however, indicates the tendency of the heavy materials to occupy the inner, and the lighter the outer, portion of a nebula or planetary system. The following tables contain the principal elements of our planetary system : 284 ELEMENTS OF MECHANICS. Elements of the Planetary System. The Mean distance from Sidereal peri od of | Rotat'n Eccentric of principal I the sun. one revolution. ! ar. axis. orb. in part of. planets. 55 Earth =1; Miles. Days. Yrs. mU. Hours. Sem. axis. Sun 60748 Mercury 6 0.3871 36,774,000 87.96926 2.93 24.05 0.2056179 Venus . 9 0.7233 68,613,000 224.7008 7.49 23.21 0.0068334 Earth . 1.000000 95,000,000 365.25637 1 00 24.00 0.01677046 Moon c Frt>m ' 237,360 27.32116 0.91 0.0 0.0635 Mars . 60* Centrifugal force, m lA-\ = ^ f orce o f attraction. \ oU / ^jooyoUou JJ' /9 TT W \ 2 Mass of the sun, M = 28693080 1?t J , in matts. The sidereal number of revolutions per minute of the earth around the sun will be 1 = 1 60x24x365.25 511350' log.b. 7087 182. 2 _ 1 "259679000000' %.11.4174364. %.5.9190422. ASTRONOMY. 291 D = 5280 E = 5280 x 95,000,000 = 501,600,000,000 feet, I? = 126,203,844,096,000,000,000,000,000,000,000,000 cubic feet. From paragraph 51 we know that the mass of the earth is 402,735,000,000,000,000,000,000 matte. ^.23.6050086. Then the mass of the sun compared with that of the earth will be 126,203,844,096,000,000 " 402,735^831002" %.5.5764616. This result is a little higher than that in the table, page 284. | 202. TO FIND THE DISTANCE FROM THE SUN TO ANY OF HIS PLANETS. Knowing that the centrifugal force of any planet revolving around the sun is FWi and that this centrifugal force is equal to the force of attraction, we have Z a nV ,, Ml 60 * * TV 4. n 'W 60 V Distance D Let t denote the time in minutes of one sidereal revolution of any planet around the sun. Then t = -, and 11 = -, which inserted in the formula for D, we have n t 292 ELEMENTS OF MECHANICS. Let T denote the time in days of one sidereal revolution of a planet around the sun, and we have *=rx 60x24 = 1440 T. n / JS//86400 T ^ \7V~2T- In this formula we have given the masses of the sun : Log. mass of the earth 23.6050086 Log. earth mass of sun add. 5.5800176 29.1850262 Log.

Let Jfand m denote the respective masses, expressed in matts. 8= distance apart in feet. T= time in seconds in which they would be drawn together. "Fand v = velocities in feet per second. R and r = the respective radii of the masses in feet, supposing them to be spherical. a and b = the respective distances in feet moved by the masses to the point of collision.

*. That is to say, the force of attraction between equal spheres in con- tact is as the fourth power of the diameter, or the force of attraction increases directly as the fourth power of the distance between the centres of attraction. 205. ELEMENTS OF THE EARTH AND MOON. The earth and the moon revolve around their common centre of gravity in periods of 29.53 days from new moon to new moon. M = mass of the earth, and m = that of the moon. r = radius of the earth, and JR = distance from the centre of the earth to the centre of the moon. It = 60 r. Let c denote the position of the common centre of gravity of the two bodies, a = distance from c to the centre of the earth, and b = dis- tance from c to the centre of the moon. Let d denote the position of the centre of attraction of the earth and moon. It is required to find the centres c and d? Centre of Gravity of the Earth and Moon. In regard to the centre of gravity c, we have the static momentums m b ra(60 r - a) '--^ I tf* ! a = - }\ Jt MM ^fcv! JHfa = w60r-wa; Jfa+ma = 60wr, or a( Jf + m) = 60 m r. 60mr Jf+m ASTRONOMY. 297 Assume the radius of the earth r = 1. Log.60 = 1.7781513 kg.m = 21.6619135 = 23.4400648 + = 23.6050086- %.0.684 = 9.8350562 That is, the distance of the centre of rotation of the earth and moon is at 0.684 r from the centre of the earth, or 0.316 of the earth's radius under the surface of the earth where the moon is in the zenith. The centres of the earth and moon thus describe elliptic orbits around the centre c as the common focus, whilst the centre c moves in an elliptic orbit around the sun. The radius of the earth is about 3956 miles, or 20887680 feet. a = 3956 x 0.684 = 2705.904 miles. b = 3956 x 60 - 2705.904 = 157534.096 miles. The velocity of a body moving in a circle is - . The earth and moon make one sidereal revolution in 27.32 solar days. 27.32 x 24 = 655.68 hours. 2x3.1472 R Velocity, 655.68 104.407' n ,1, 2705.904 nK Mt , ., Earth s centre, v - 25.917 miles per hour. 104.407 Moon's centre, "F= = 1508.8 miles per hour. 104.407 Centre of Attraction Between the Earth and Moon. In regard to the centre of attraction between the earth and moon, we have Force attraction, F= . I? It is required to find a point in the straight line between the earth and the moon where a body would be equally attracted from the two bodies, which is when m - , of which d= d* (R-d? ' M-m> 298 ELEMENTS OF MECHANICS. Assuming the mass of the moon as unit, or m = 1, then the mass of the earth will be M = 87.7. When E = 60 the distance d will be expressed in radii of the earth. [) = 54.2. d= 54.2x3956 = 214415 miles from the centre of the earth to the centre of attraction. A body placed at e, Fig. 228, would be equally attracted from the earth and moon. |206. ORBITS OF THE EARTH AND MOON. Fig. 229 represents the orbits of the earth and moon moving in the direction of the arrow, as seen from the North Star. The dotted line a represents the orbit described by the common centre of gravity of the earth and moon, which is an ellipse in which the sun is in one of the focii. The eccentricity of the ellipse is only 0.0168 of the major axis. Fig. 229. The drawn line b represents the orbit described by the centre of the moon, and the line c that of the earth. The line 0, 1, 2, 3 and 8 re- presents the radius-vectors from the sun. The illustration represents the orbits for one lunar month, or from new moon to new moon, which are accomplished in a time of 29 days, 31 minutes and 48 seconds. The mean velocity of the common centre of gravity in the dotted orbit a is 68,091 miles per hour. The following table shows the mean sidereal velocities in miles per hour of the earth and moon in their respective orbits at different positions of the moon. ASTRONOMY. 299 Sidereal Velocities of the Earth and Moon. No. Positions. Velocities in n Earth. ilcs per hoar. Moon. 1 2 3 4 5 6 7 8 68086 68110 68091 68072 68069 68072 68091 68110 68086 66582 67024 68091 69158 69600 69158 68091 67024 66582 Middle of first quarter Middle of second quarter Full moon Middle of third quarter Half moon Middle of fourth quarter New moon The greatest difference of velocity of the moon is 3013 miles per hour, a distance of about that between New York and Liverpool. The moon travels a distance of about 10 times the diameter of the earth per hour. A point on the equator describes a worm-line on the earth's orbit. 207. TO FIND THE MASS OF THE EARTH AND MOON. The notation of letters will be the same as in the preceding para- graph. The earth M and moon m revolve around their com- Fi &- m mon centre of gravity c. Having given the earth's radius and the moon's horizontal parallax, we obtain the distance R - 237360 miles, or P. = 1253260800 feet, between the centres of the two bodies. The force of attraction between the two bodies is F= in which y = 28693080. (2 7T \* / ecTr) =mb ( T=time in minutes of one sidereal revolution of the system, which is 27*, 7*, 43", IP, or T= 39343.183 minutes. Call -I-^ 2x3.14 60x39343.183 141150000000 log. 1 1.1496796. 300 ELEMENTS OF MECHANICS. The centrifugal forces are equal to the force of attraction, or Mm ,.. . , Mm , and -i. , m . J/ , = a o, and - = 00.

= 87000, and Jf=328. Volume P- 3 ' 14x8700 3 - 336,770,000,000,000 cubic miles. 6 1400850000000x828 336770000000000 The density of Jupiter is 1.36389 compared with that of water, or 0.252 of that of the earth. The astronomical data appear to be yet very incorrect, and it is therefore difficult to make them agree with the physical laws involved. ASTRONOMY. 303 g 209. TO FIND THE VELOCITY WITH WHICH A BODY SHOULD be started from the earth's surface in order to reach the Moon. Fig. 231. In the accompanying illustration M represents the earth and m, the moon. The point m' is the centre of attraction between the two bodies. A body at m' would be equally attracted by M and m ; therefore if the body m' is overtaken by the earth's attraction, it would arrive at b with a velocity equal to that with which it should be started from the earth's surface in order to reach the moon. M, m, and m' denote the masses of the respective bodies expressed in matte. S= distance in feet between the centres of the earth and moon. d = distance in feet of the body m' from the centre of the earth. F= force of attraction of the earth on the body m'. /= force of attraction of the moon. _ M m , . mm , and f-. The force acting on the body m' in its fall to the earth will be _ .. - f= - / - m \ J v\# (-<*)*/ Let v denote the velocity of the fall in feet per second. 7 7 = time of fall in seconds. (F-f) :M=v:T, of which (F-f) - ~. m! v m' I M m ~T~~ _T_lM__m_\ ^ 304 ELEMENTS OF MECHANICS. ri T T=- -, when d and v are variables, which inserted in Formula 1 ill be 8d /M v -- / --- / n 1 rlMfia mda \ v 8v = I) i. 1 Jemset Pendj. Panoh. Ngnn. Go. Pancha. Pat, 5 6 | Sittet. Shesh. Chha. Luck. Lock. Shash. Sex, 6 't \ Saybet. Helft. Sath. Tchut. Sytchi. Saptan. Ben, 7 S i Saymaniet. Hesht, Ath. Pbat. Hatchi. Ashta. Ott, 8 ! 9 | Tiset. Nuh. Nau. Geo. Ku. Navan. Net, 9 ,' 10 Eshret. Deh. Das. Shop. Dgiu. Dashan. Dis, ? 11 Ahedeshere. Yikadeh. Gyarah. Shopyat. Dgiuitchi. Aikadashan. Elr, "S 12 Ishnaueshere. Dudeh. Barah. Shopgaa. Dgiuni. D\vandashan. Ton,10 | NOMENCLATURE. 315 COMMENTS ON NOMENCLATURE. On account of the different pronunciations of letters and words in different languages, the true sound of a name cannot be conceived without a knowledge of the language in which it is written. The Japanese sound for 9 is written leu in the table, but for the English pronunciation it should be written koo. There are some letters of the alphabet which have nearly the same sound in all languages, and only such letters should be used in the coining of names for the figures and numbers in the duodenary system. The letters th, w, o, ur, ght in the English language, and also the letter C, which has two sounds in almost all languages, should not be used for the new names. The names given to the duodenary figures in the last column are clear and distinct sounds, -which would be well understood and pro- nounced alike in all languages. It would be useless to attempt to introduce the names of the figures and numbers in either of the languages above given as a universal nomenclature, for not only that they are net suited for more than the language in which they are writtea, but prejudices would be against them. The introduction of the French metrical system has been greatly retarded by reason merely of its cumbersome nomenclature. The best work on the etymology of -numbers known to the author is that of Professor S. Zehetmayr, published in Leipsic, 1854. In the establishment of a new and universal nomenclature of numbers we ought to select clear and distinct sounds, which can be understood and pronounced alike in all languages, without regard to the ety- mology of numbers. The Arabic notation of numbers is yet used only by about one-third of the population of the earth, and the other two-thirds use different kinds of irregular characters or hieroglyphics, which combinations are unfit for arithmetical calculations. The Roman notation was used in England up to the beginning of the seventeenth century, when the Arabic notation was gradually gaining ground against very strong opposition ; and at last caused the burning of the houses of Parliament. The Arabic notation was introduced into Germany in the twelfth century, and into Italy in the eleventh century. 316 ELEMENTS OF MECHANICS. Comparison of Numbers in the Duodenary and Decimal Systems, with the Corresponding New Names. New. Names. Old. New. Names. Old. o Zero 37 Tretoben 43 1 An. 1 38 Tretonott .. . 44 2 Do 39 Tret on ev 45 3 Tre 3 3T Tretondis 46 4 For 4 35 Tretonelv 47 5 Pat 5 40 Forton . 48 6 Sex 6 41 Fortonan 49 7 Ben 7 42 Fortondo 50 8 Ott 8 43 Fortontre 51 9 Nev... 9 44 Fortonfor 52 $ Dis 10 45 Fort on pat 53 8 Elv 11 46 Fortonsex 54 10 Ton 12 47 Fortoben 55 11 Tonan . 13 48 Fortonott 56 12 Tondo 14 49 Fortonev 57 13 Tontre 15 4$ Fortondis 58 14 Tonfor .. 16 45 Fortonelv .... 59 15 Tonpat 17 50 Paton . 60 16 Tonsex 18 51 Patonan 61 17 Toben 19 52 Patondo 62 18 Tonott 20 53 Patontre 63 19 Tonev 21 54 Patonfor 64 1$ Tondis 22 55 Patonpat 65 15 Tonelv 23 56 Patonsex ] 66 20 Doton . . 24 57 Patoben 67 21 Dotonan 25 58 Patonott 68 22 Dotondo 26 59 Patonev 69 23 Dotontre 27 5$ Patondis 70 24 Dotonfor 28 55 Paton elv 71 25 Doton pat 29 60 Sexton . 72 26 Dotonsex 30 61 Sextonan 73 27 Dotoben 31 62 Sextondo 74 28 Detonott 32 63 Sextontre 75 29 Detonev 33 64 Sextonfor 76 2$ Dotondis 34 65 Sextonpat 77 25 Dotonelv So 66 Sexton e ex 78 30 Treton .... 36 67 Sextoben 79 31 Tretonan 37 68 Sextonott 80 32 Tretondo 38 69 Sextonev 81 33 Tretontre 39 6$ Sextondis 82 34 Tretonfor 40 65 Sextonelv 83 35 Tretonpat 41 70 Benton 84 36 Tretonsex 42 71 Bentonan 85 NOMENCLATURE OF NUMBERS. 317 New. Names. Old. New. Names. Old. 72 Bentondo 86 51 Elvtonan 133 73 Bentontre 87 ft2 Elvtando 134 74 Bentonfor 88 ft3 Elvtontre 135 75 Bentonpat 89 ft4 El vtonfor 136 76 Bentonsex . . . 90 ft5 Elvtonpat . 137 77 78 Bentoben Bentonott 91 92 "86 ft7 Elvtoneex Elvtoben 138 139 79 Bentonev 93 #8 Elvtonott 140 7f Bentondis 94 ff9 Elvtonev 141 7ft Bentonelv 95 3f Elvtondis 142 80 Otton . . . 96 M Elvtonelv 143 81 Ottonan > 97 100 San. H4 82 Ottondo 98 148 San-fortonott 200 83 Ottontre 99 200 Dosan 288 84 Ottonfor 100 210 Dosan-ton 300 85 Ottonpat 101 300 Tresan 432 86 Ottonsex 102 358 Tresan-patonott 500 87 Ottoben 103 400 Forsan 576 88 Ottonott 104 420 Forsan-doton 600 89 Ottonev 105 500 Patsan 720 8? Ottondis 106 568 Patsan-sextonott 800 8ft Ottonelv 107 600 Sexan 864 90 Nevton 108 630 Sexan-treton 900 91 Nevtonan 109 700 Bensan 1008 92 Nevtondo 110 800 Ottsan 1152 93 Nevtoutre 111 900 Nevsan 1296 94 Nevtonfor 112 fOO Dissan 1440 95 Nevtonpat . 113 ftOO Elvsan 1584 96 Nevtonsex 114 1000 Tos 1728 97 Nevtoben 115 1100 Tossan 1872 98 Ne vtonott 116 1200 Tosdosam 2016 99 Nevtonev 117 1300 Tostresan 2160 9? Nevtondis 118 1400 Tosforsan 2304 9ft Nevtonelv 119 1500 Tospatsan 2448 $0 Diston 120 1600 Tossexan 2592 K Distonan 121 1700 Tosbensan 2736 f2 Distondo 122 1800 Tosottsan 2880 9?3 Distontre 123 1900 Tosnevsan 3024 $4 Distonfor . 124 IfOO Tosdissan 3168 $5 Distonpat 125 IftOO Toselvsan ... 3312 $6 Distonsex 126 2000 Dotos 3456 $7 Distoben 127 4000 Fortos 6912 $8 Distonott 128 6000 Sextos 10368 $9 Distonev 129 8000 Ottos 13724 $f Distondis 130 ?000 Distos 17180 . jguruiunjgj 1 HwPs $HIBRARY& HIBBAfiXfl^ -^HIBRAIT i.OKAl! fyKBAJNft-3\^ ^lOflFj> S^jflAiM-iv^ tfHIBRARYQ* %OJI1V3-JO^ ^HONV-SO' ^OfCAtfFO^ >AHVMain^ ^ijow-sm- I1V3-: UC SOUTHERN REGION J^ljjjjjjli^iiil,, AA 000777108 2 5 S ^UIBRARY^ -g^-UBRAR t fr-i | tymrn^ ^ojnvD-jo^ %OJITVD ^I0$i# ^OF-CAllFOM^ ^F-CMIF ^lAJMn^ yommn--^ ^o-mimi