SHOP MATHEMATICS BY EDWARD E. HOLTON HEAD OF DEPARTMENT OF MACHINE SHOP PRACTICE THE TECHNICAL HIGH SCHOOL SPRINGFIELD, MASSACHUSETTS THE HAYTOL SERIES OF TEXT BOOKS FOR INDUSTRIAL EDUCATION EDITED BY FRANK E. MATHEWSON THE TAYLOR-HOLDEN COMPANY SPRINGFIELD, MASS. 1910 Copyright, 1910 BY THE TAYLOR-HOLDEN COMPANY Springfield, Massachusetts Second Edition THE F. A. BASSETTE COMPANY PRINTERS SPRINGFIELD, MASSACHUSETTS PREFACE This book is the result of twelve years' experience as draftsman and shop foreman combined with an equal length of service as instructor in day and evening classes in technical and trades schools. The main feature of the book is the collection of practical shop problems, the most of them being either actual prob- lems which have arisen in the author's experience or those suggested by that experience, and were first collected for use in his own classes. The formulas given are those usually found in mechanics' hand books, and the author acknowledges his indebtedness to Wm. Kent and P. Lobbin for permission to use their formulas; also to Brown & Sharpe Mfg. Co. for courtesies extended, and especially to Mr. C. S. Bragdon of the Tech- nical High School, Springfield, Massachusetts, for assistance in collecting and arranging material. EDWARD E. HOLTON. May, 1910. 2066132 CONTENTS Introduction ....... vii Signs, Symbols, Abbreviations, Etc. . . ix Mechanics . . . . . . . 1 Moments of forces, work, horse power, energy, etc. 1 Mechanical Powers ...... 6 General law for machines .... 6 The six elements of machines . . 6 The lever .... . 6 Compound lever 9 Wheel and axle . 10 Combinations of wheels and axles . 12 The pulley . . 13 Fixed and movable pulleys . 14 Rule for pulleys . . 15 Lathe test indicator 18 The " gear " of a bicycle 21 Inclined plane . . 23 Screw . 24 Differential screw . 25 The wedge 26 Screw Threads 29 Spur Gears 32 Bevel Gears . 37 Worm Gears .... . 41 Spiral Gears . . 42 Pulleys . . 47 CONTENTS Friction .... Coefficient of friction Laws of friction . Angle of friction Leather Belting Manila Ropes Wire Cables .... Factor of Safety Chain Transmission Shafting . . . Journal Bearings Ball Bearings Machine Keys .... Linear Measuring Instruments Machines .... Lathes ..... The back gears Screw thread cutting The reversing feed gears Compound gearing Turning tapers Cutting speeds for various materials The metal planer Speeds and feeds The Universal Milling Machine The index head Compound indexing Differential indexing The spiral head Speeds and feeds of cutters Drill press Speeds and feeds . . . vi CONTENTS PAGE Hammer Blow . ..... 130 Horse Power of Machines . . . 132 The Prony brake 135 Fly Wheels 138 Power of the Steam Engine . . 140 Power of the locomotive . . . 143 Power of gasolene engines . . . 145 Engine Cylinders ...... 147 Indicator diagram ..... 149 Steam Boilers .. . . . . . . 149 The H. P. of boilers 150 To find heating surface of shell and tubes 151 To find water capacity . . . . 152 To find steam capacity ... 153 To find pressure carried on stay bolts . 153 To find the bursting strength . 154 The safety valve . . . . . 157 Hydraulics . . . . . . . 160 Pressure and head . . 160 Archimedes discovery . . . . 161 Specific gravity ... . . 161 Hydraulic machines . . . . . 163 Tables of Standard Units of Weights and Measures . 169 The Metric System . 173 Table of Decimal Equivalents . 177 Use of Formulas ... 178 Simple Trigonometric Functions . 186 Table of Natural Sines, Etc. . 189 Formulas ....... 199 Answers ........ 203 Index 221 INTRODUCTION CHARLES F. WARNER, Sc. D. Principal of the Technical High School and of the Evening School of Trades, Springfield, Mass. The loudest note sounded in all recent educational dis- cussions is the call for more practical methods in presenting the fundamental subjects of study. It is claimed that language, science, and mathematics have been taught with too little reference to their utility in the vocations and in the ordinary affairs of life. The methods commonly employed in presenting mathematical subjects have been especially open to criticism. This is true even in some schools designed to give vocational training. Mathematical books of the kind traditional in the older schools have been continued in use in the newer and more practical schools. These books were written almost entirely from the point of view of the teacher of pure mathematics with little reference to concrete problems of life and' having no reference whatever to the actual problems of the drafting-room and the shop. As a natural consequence the class room work in mathematics, in many of our most practical schools, has failed to utilize the material afforded by the shops and science laboratories to fix the knowledge of mathematical principles by concrete illustration and by practice. There is much truth in the criticism. But what are we going to do about it? It will not do to cast aside the old- time algebras and geometries unless something really better can be found to take their place. The effort to make the viii SHOP MATHEMATICS applications of mathematics more easily understood might lead to the substitution of a practical course in which the mathematical element is too much diluted. This would be folly. The real object should be to strengthen, not to weaken the teaching of mathematics in practical schools. What is needed is to purge the old books of useless material and put in place of it practical mathematical work distinctly planned to make up for the short-comings of the old methods when measured by the practical demands of modern times. The author of "Shop Mathematics" has had many years' experience in designing and making machine tools and in a wide range of practical shop work. In addition to this he has had a long experience as a teacher of drawing and of machine shop practice and tool-making in schools for boys and for adults. This has given him an unusual opportunity not only to find out what is needed, but to discover the facilities for supplying that need. The book contains a selection of problems that actually arise in shop practice. This is what is needed by the young mathematical student and for two reasons, first, that he may know what the shop problems are, and, second, that he may learn how to apply mathematical principles, rules, and formulas in the solution of such problems. No attempt is made in this book to teach mathematical theory or principles. That would be a need- less repetition of countless books already in existence. "Shop Mathematics" may be used to supplement the course in elementary algebra, geometry, and trigonometry, and if used in this way it will be found of great value in technical schools. But abundant rules and formulas are given under each subject so that the book will also find a place in brief practical courses which do not admit of the use of the ordinary mathematical text-book. SIGNS, SYMBOLS, AND ABBREVIATIONS USED IN THIS BOOK + , plus, is the sign of addition. , minus, is the sign of subtraction. Plus and minus are also used to indicate positive and negative quantities. X , times, or multiplied by, is the sign of multiplication. A dot . is sometimes used for X where the quantities are expressed by letters, but is usually omitted in algebraic formulas; thus a X b, or a . 6 is ordinarily written ab. H- divided by, is the sign of division. : without the dash between also indicates division, being used as the ratio sign; as, a : b, means the ratio of a to 6, or a divided by b. = is the sign of equality and indicates that the two quantities between which it is placed are of equal value. .'., therefore. >, greater than; as, 6>4, read 6 is greater than 4- <, less than; as, 4<6, read 4 is less than 6. Z , angle. J_, perpendicular to; as A_B, read A is perpendicular to B. ||, parallel to; as, C \\ D, read C is parallel to D. ., the decimal point; as, 0.2, read two tenths; or 0.004, four thousandths. x SHOP MATHEMATICS , the symbol for the degree in the measurement of angles. ' and ", the accents, denote minutes and seconds in the measurement of angles; as, 5 10' 15" , read 5 degrees 10 minutes 15 seconds. The above symbols, ' ", are also used for feet and inches in indicating dimensions; as, 3' 6" , read 3 feet 6 inches. The subscript is a small figure written at the lower right of a letter; as, A lt A 2 , A 3 , read A sub one, A sub two, A sub three. It is used to denote corresponding parts of related objects, or sometimes to avoid using too many different letters. \/, the radical sign, denotes the square root. Other roots are indicated by a small figure, called the index, written at the upper left of the radical sign; as, 3 5 n \/, \/, v 7 , read cube root, fifth root, nth root. The exponent is a small figure written at the upper right of a quantity to indicate a power, or the number of times the quantity is used as a factor; as, 5 2 means 5X5, 3* = 3X3X-3X3. ( ), parentheses, { } braces, [ ] brackets, vinculum, signify that the inclosed quantities are to be considered as one quantity. sin, sine. cos, cosine. tan, tangent. cot, cotangent. sec, secant. cosec, or esc., cosecant. TT, Pi, the ratio of the circumference of a circle. to its diameter, =3.1416. D, or dia., diameter of a circle. SHOP MATHEMATICS xi R, or r, radius of a circle. P or F, power or force. W, weight or resistance. H. P., horse power. R. P. M., revolutions per minute. F. P. M., feet per minute. /, coefficient of friction. ft.-lb., foot pound. ft., foot or feet. Z&., pound. in., inch or inches. pi., pitch. MECHANICS Mechanics treats of forces and of the effects of forces. Force is the action between two bodies tending to produce a change of position or shape; as when a horse pulls a load, a motor drives an electric car, elasticity causes the action of a steel spring, etc. The moment of a force. If a bar of uniform size is pivoted at its center and a weight placed on one end, the bar will rotate about the pivot. The numerical value of the im- portance of the force in producing motion about a pivot is called its moment and is equal to the product of the force by the distance from its line of action to the pivot. By reference to the accompanying sketch, the moment of the force FI, applied at point A, is seen to be 6X 10 = 60 ft.Jbs. of F 2 , 4% X10 = 45 ft.-lbs. of F 3 , 3 X 10 = 30 ft.-lbs. Diagram for Momenta of Force ,/>. When a force acting upon a body changes its position, work is done upon the body. The amount of work done 2 SHOP MATHEMATICS depends upon the force applied, also upon the distance through which it acts; that is, work is measured by the resistance overcome, multiplied by the distance through which it is overcome; as a 10 pound weight that is lifted to a height of 5 feet requires an amount of work equal to the product of 10 times 5 = 50 ft.-lbs. The unit of measure in the above example is expressed by the term foot-pound, that is, a force of one pound acting through a distance of one foot, or its equivalent; as 4 pounds acting through a distance of \ foot, or T V pound acting through 10 feet, etc. The unit can be expressed not only in ft-lbs. but in inch- pounds, foot-tons, centimeter-grams,* etc. The amount of work done in lifting a body a given distance is the same whether clone in 5 seconds or 5 minutes, but it is often necessary also to denote the rate of doing work. This is expressed in terms of horse power. An engine of one horse power (1 H. P.) means an engine capable of doing 33,000 ft.-lbs. of work in one minute. Therefore to find the horse power of any machine the formula used is, _ Ft.-lbs. of work done 33000 X time in minutes In electric power machines such as dynamos and motors one H. P. is equal to 746 watts; then the formula for power of electric machines is: volts X amperes ~r46~~ Energy is the capacity for doing work; as a coiled spring has the capacity to set in motion the mechanism of a clock *Note. For dimensions in the metric system see tables, pages 173 to 176. SHOP MATHEMATICS 3 or watch. The coiled spring of a watch possesses energy because at some previous time work has been performed upon it in the winding. The amount of energy in a body is measured in ft.-lbs., the unit used for measuring work. Mechanical energy is of two kinds, potential, and kinetic. Potential energy is due to the position of a body; as, for example, if a pile driver head weighing 50 pounds is suspended 20 feet above the ground , it has a potential energy of 20X50 = 1000 ft.-lhs. If now the weight is released and falls, the energy is of a different kind because of the motion of the falling weight. This is called kinetic energy. In either case the weight times the height equals the measure of the energy of the body or K = Wh. When the velocity of the falling body is given instead of the height from which it falls, Then by the laws of falling bodies ' Where v = velocity in feet per second g = 32.16 feet Wv 2 Then K = ^- 2g K = the kinetic energy W = weight of the body in pounds. PROBLEMS 1. A drop hammer weighing 400 Ibs. falls from a height of 36 in. What kind and how much energy will be exerted? 2. A weight of 500 Ibs. is used for breaking up old car wheels and is suspended 15 ft. above the anvil block. Cal- culate the kind and amount of energy. 4 SHOP MATHEMATICS 3. An elevator car weighing 2 tons requires how much energy to lift it a distance of 20 feet? 4. A tank contains 5,000 gals, of water, 50 ft. being the average height above the ground. What is the energy of the water at the ground? 5. A pile driver head weighing 175 Ibs. falls from a height of 18 ft. What is the energy at the end of the fall? 6. A man weighing 180 Ibs. jumps up to a platform 42 inches above the ground. How much energy was ex- pended? 7. If the man in problem 6 should climb a distance of 100 ft. above a certain point, how much energy would be exerted? 8. If a hammer that weighs 1 Ib. has a velocity of 22 ft. per sec., what is its energy? 9. If an iron ball weighs 100 Ibs., what is its energy when suspended at a height of 28 ft. from the ground? 10. A rock drill is operated by a 10 Ib. sledge hammer with a striking velocity of 30 ft. per sec. Find the energy used in the drilling. 11. W'hat is the H. P. of a pump that raises 100,000 gals, of water per hour to a reservoir 25 ft. above the level of the lake from which it is pumped? Note. One cu. ft. of water weighs approximately 62 Ibs. and contains about 1\ gals. 12. A motor is able to lift an elevator, which with load weighs 5 tons, to the top of a tower 500 ft. high in two min- utes. What H. P. is required to do the work? 13. What must be the H. P. of the engine required to raise a block of granite weighing 10 tons to the top of a wall 35 ft. from starting point when it takes 20 minutes to do the work? SHOP MATHEMATICS 5 14. If it takes 1^ hours to raise a weight of 20 tons 100 ft., what H. P. engine will be required? 15. If a workman carries 5 tons of pig iron up a flight of steps 14 ft. high in 10 hours, how much work does he accomplish expressed in ft.-lbs.? 16. A rope turns a pulley 48 in. dia. with a pull of 10 Ibs. at the rim at the rate of 2500 ft. per minute. How many ft.-lbs. of work are done in 5 hours consecutive movement? 17. A 12 Ib. ball falls 2500 ft. What is its velocity when it strikes and what is its kinetic energy? 18. A 500 volt electric motor runs 50 machines using 22 amperes of current. What is its H. P.? 19. What is the H. P. of a dynamo which will run 300 110-volt incandescent lamps if each lamp consumes ^ ampere of current? 20. An electric motor has a voltage of 250 and supplies 30 amperes to run a certain set of machines. Find the H . P. of motor. 21. What H. P. will be required to run a 250 lamp circuit, the lamps being the same as in problem 19? 22. A direct connected dynamo delivers power at 110 volts to a 1400 electric light circuit, also to 9 motors with normal amperage as follows: 1 at 144, 1 at 110, 1 at 80, 3 at 76, 1 at 57 and 2 at 20 amperes. What H. P. will be required to run the above equipment? MECHANICAL POWERS An appliance by which force can be used to do useful work is called a machine. The mechanical powers or elements of machines are six in number, as follows: 1. Lever. 2. Wheel and Axle. 3. Pulley. 4. Inclined plane. 5. Screw. 6. Wedge. The mechanical advantage of all kinds of machines, whether simple or compound, may be computed in accordance with the following GENERAL LAW The force multiplied by the distance through which it moves is equal to the resistance or weight multiplied by the distance through which it moves, or P X dP = W X dW. Each class of machines permits of a special statement of this law by substituting for the general terms the special terms used with that class of machines. 1. THE LEVER The lever is an inflexible bar or rod supported at some point, the bar being free to move about that point as a pivot. This pivotal point is the fulcrum, usually represented by F. The force applied to the lever is represented by P. SHOP MATHEMATICS The weight lifted, or the resistance to the force is repre- sented by W. The lever is classed according to the position of the three points P, F, W. A lever of the first class, Fig. 1, has F between P and W. A lever of the second class, Fig. 2, has W between F and P. A lever of the third class, Fig. 3, has P between F and W. The machinists' and the tinsmiths' pliers are -examples of levers of the first class. The nut cracker and lemon squeezer, are levers of the second class. The sheep shears and firm joint calipers are good examples of levers of the third class. The three classes of levers are operated and controlled by the following: LAW FOR LEVERS. The force multiplied by its distance from the fulcrum is equal to the weight multiplied by its distance from the fulcrum. SHOP MATHEMATICS Let P = force W = weight or resistance. Pa = distance from fulcrum to point where force is applied. Wa= distance from fulcrum to point where weight is applied. Then the law of Levers becomes PxPa = WxWa. From this equation the following are readily derived. WxWa W = PxPa WxWa PXPa Pa Wa P W Example. What force 18 in. from fulcrum will balance a weight of 870 Ibs. 3 in. from the fulcrum? Solution. By formula: WxWa 870X3 P = Pa 6 = 145 Ibs., Ans. The law for bent levers is the same as for straight levers but the lengths of arms are computed on lines from the fulcrum at right angles to -the direc- tions in which P and W act. The lengths of the arms of a bent lever are continually changing as the force and weight move into new positions. The case of pulling out a nail with the common claw hammer is a good illustration of the bent lever, Fig. 4. SHOP MATHEMATICS ^^^$5^^ The moving strut in Fig. 5 is a bent lever with one arm lacking. The force is applied at the same end at which the resistance is to be overcome. The resistance in this case is not the weight, W, but its resistance to being moved. The ratio between force and resistance changes as the angle A changes. Then force: resi stance = sin A: cos A or P X cos A = W X sin A WXsinA PXcos A and P = -. W = : ; cos A ; sin A The toggle joint is a double strut, Figs. 6 and 7. The statement is P : W = 2 sin A : cos A P _WX2 sin A cos A w _PXcos A 2 sin A A compound lever is a combi- nation of two or more levers which may be of the first, second or third classes. The calcula- tions for compound levers can be made by taking each lever ' separately by the formula for single levers. These separate operations, however, are usually condensed into one, in accordance with the following: 10 SHOP MATHEMATICS LAW OF COMPOUND LEVERS. The continued product of the force and all the force arms is equal to the continued product of the weight and all the weight arms. This law gives rise to the following formulas: P = W = PaXPa.XPa.,. PxPaXPa.XPa^ H. WHEEL AND AXLE The wheel and axle may be considered a continuous lever. By its use a continuous motion is obtained for raising a weight. The two arms of the lever are the diameters of the wheel and axle, or the radii of wheel and axle. If D is diameter of wheel: d= diameter of axle R = radius of wheel r = radius of axle C = circumf erence of wheel c= circumference of axle Then these formulas apply : P W = d : D P W = r :R P W = c:C Fry. 8, Fig. 8 shows sketch of simple windlass. Fig. 9 shows sketch of capstan, which is a wheel and axle SHOP MATHEMATICS 11 where the axle has a vertical position. It is used on ships for raising the anchor. The mechanical advantage of the wheel and axle may be increased by making the diameter of wheel larger or the diameter of axle smaller. The windlass is a modification of the wheel and axle where a crank is substituted for the wheel. It is used especially where the power is applied by hand. The differential windlass, Fig. 10, is a device for increasing the mechanical advantage of the axle by unwinding rope from a small axle and wind- ing it on to an axle of larger diameter; for with one turn of crank C, the section of rope supporting the weight will be shortened a distance equal to circumference of the large axle minus the circumference of the small axle and W will be raised half this distance. By the law of the wheel and axle, P : W = r :R then in the differential windlass (r.-r) 2 * r l = radius of large axle r = radius of small axle When motion is transmitted from one body to another by direct or by indirect contact the body that produces the motion is called the driver, the body that receives the motion is called the follower. In a combination of wheels and axles each pair may be 12 SHOP MATHEMATICS determined separately, but the shorter way is to use a formula similar to that used for compound levers, as follows : Where R, R 1} R 2 , are the radii of the wheels and r, i\, r 2 , are the radii of the axles. These problems may also be calculated by the following: RULE FOR DRIVERS AND FOLLOWERS. The speed of the first driver multiplied by the product of the sizes of all the drivers is equal to the speed of the last follower multiplied by the product of the sizes of all the followers. For the sizes of followers and drivers may be taken the circumferences, diameters, radii or number of teeth (if a gear wheel) ; only whatever dimension is taken for any driver the corresponding dimension must be taken for its follower. Thus for a train of wheels having four drivers and followers, the diameters can be taken for one pair, the radii for another, the circumferences for a third and number of teeth for the fourth, to solve the problem. Let N = th& R. P .M. of first driver n = the R. P. M . of last follower D, D lf D 2 =size of drivers F, F lt F 2 = size of followers Then by formula, N X D X D l X D 2 F ' X F l X F 2 The size of any driver or follower can be found by the formula : D_X_N *~~ n FXn, D ~ N where, NR, P. M. of the driver n = R. P. M. of the follower SHOP MATHEMATICS 13 Figures 12, 13, 14 and 15 are good illustrations of the above rule in shop practice. Fig. 12, shows a combination of the stud and change gears used on the engine lathe for thread cutting, and on the milling machine for cutting spiral flutes in cutting tools. Fig. 13, shows the belting from main line drive pulley through two countershafts to spindle pulley of machine. Fig. 14 shows sketch of chain and sprocket drive for bicycle or automobile, which form of drive is also being used extensively for positive drives in some kinds of machine construction. Fig. 15 is the bevel gear drive. III. THE PULLEY The pulley is a wheel over which a cord, band or chain is passed to transmit the force applied to the cord in another direction. The wheel is introduced to diminish the friction, 14 SHOP MATHEMATICS the band being the part that gives practical effect to the machine. There is no mechanical advantage gained with the fixed pulley as shown in Fig. 16, but as stated above it is of great use in changing the direction of the force. The fixed pulley when used in combi- nation with the movable pulley as shown at B in Figs. 17 and 18 has mechanical advantage, since the weight is carried by strands of the cord on either side o f the movable pulley. The usual arrangement of pulley blocks is shown in Figs. 17 and 18 where one or more wheels or sheaves (grooved pulleys) are placed in suitable pivots and bearings, and cords are passed over the pulleys connecting the force and weight. In Fig. 17, if there is a single pulley at A and B and a rope is passed over A and around B and fas- tened at C, then a pull of 1 pound at P will transmit a pull of 1 pound to the rope on the other side of pulley A ; this will transmit the same amount of force at B, which in turn will transmit the same force to the other side of pulley B. Thus 1 pound pull at P SHOP MATHEMATICS 15 will balance 1 pound at 1 and 2 or the combined puil of 1 and 2 at W, so that 1 pound at P will support 2 pounds at W. Again, if rope at P moves down 1 foot the rope must move up 1 foot on the other side of A: but the end of rope is fixed at C ' , so that when rope moves up at 1 and remains stationary at 2, one-half the motion will be given to the movable pulley between 1 and 2, which is in accordance with the general law of machines.'. W = 2 P. If there are two sheaves on each block A and B, each turning independently of the other in the bearings, the pull of 1 pound at P will be transmitted as in the first case and will also transmit the pull to the third and fourth strands, so that a pull of 1 pound at P will balance a weight of 4 pounds at W. Similarly it can be shown that 1 foot downward travel at P will give an upward travel of \ foot at W. From these prin- ciples is obtained the following: RULE FOR PULLEYS. The force multi- plied by the number of strands from the movable pulley will equal the weight that can be raised, or PXN=W, where N is the number of strands from the movable pulley. Whenever possible the pulleys should be arranged so as to pull in the direction in which the weight is to be moved, as shown in Fig. 18, for whereas in Fig. 17, N = 2, in Fig. 18, N = 3. The differential pulley, Fig. 19, is quite generally used in shop practice, the band for transmitting the force being an 16 SHOP MATHEMATICS endless-chain of iron links. The principle of the differential pulley is very similar to that of the differential windlass. The radius of the movable pulley F must be an exact mean between the radii of B and C, in order to have a uniform velocity ratio, so that chain links may fit the pockets in sprocket wheels. Let R = radius of pulley B, r = radius of pulley C. While the pull at P moves the chain between B and D up a distance R, the chain at C will move down a distance r, the loop around the pulley D will be shortened the distance R-r and W will be raised one-half this amount. From this is obtained the statement as follows: PXR=WX R ~ T WX 2 R-r 2 w= R PXR R-r MISCELLANEOUS PROBLEMS In the following problems the weight of the lever and friction of the bearings will not be considered in the cal- culations. 1. What force 36 in. from the fulcrum will balance a weight of 500 Ibs., 9 in. from the fulcrum in a lever of the first class? SHOP MATHEMATICS 17 2. If a pull of 75 Ibs. is made on the end of a bar 6 ft. long with the fulcrum 6 in. from the other end, what weight in a lever of the first class will just balance the pull? 3. If the lever of problem 2 had been of the second class and the weight 6 in. from the end, what weight would have been required? 4. When a weight of 324 Ibs. is balanced on the end of a lever of first class by a force of 62 Ibs. 27 in. from fulcrum, what distance is the weight from fulcrum? 5. When a weight of 685 Ibs. 2 in. from the fulcrum is balanced on a bar by a pull of 75 Ibs., what distance is the force from the fulcrum? 6. What weight 4 in. from the fulcrum will be balanced on a bar by a force of 96 Ibs. 48 in. from the fulcrum? 7. What force 10 ft. from the fulcrum will just raise a weight of 2465 Ibs. 8 in. from the fulcrum with a lever of the second class? 8. A bar of iron was placed with one end against the wall of a shop, the other end being placed against the leg of a machine that weighed 1500 Ibs. If the bar lay at an angle of 10 to the floor, what weight would be required on the end next to the machine to move the machine along the floor assuming that the resistance is equal to the weight of the machine? 9. A quartz crusher jaw requires a force of 10,000 Ibs. to crack a certain stone. If the angle through which toggle arms move is 5, find the force that must be put on the toggle arms to break the stone. 10. The power on the end of a strut is 124 Ibs., the angle is 5. What resistance could it overcome? 18 SHOP MATHEMATICS 11. A baling press requires a pressure of 6000 Ibs. to bale a bundle of steel chips. The platen is operated by a toggle joint with arms at an angle of 10. What force will be re- quired to operate the press? 12. The knife of a paper shear is operated by a toggle joint, the arms beginning the cut at an angle of 4^. If a pull of 650 Ibs. is required to start the arms to obtain a cut, what is the resistance on the knife blade? Fig. 20 shows a practical example of a lever of the first class. The long end of the pointer at C multiplies the error in the position of point A with respect to the axis of the lathe spindle, in proportion to the length of the arms A B and B C. If the distance from A to B is \ inch and from B to C is 8 inches the amount of motion of C will be in the ratio of 8 to or 16 tunes as much as A. 13. The pointer of a lathe test indicator is 15 in. long over all; if the short arm of lever is .375 in. and point c moves in a path ^ in. across, what amount of error in posi- tion of the point A is indicated? SHOP MATHEMATICS 19 14. The bar of a lathe test indicator is 13 in. from the end to a point in the work to be tested and the short arm is ^ in. How large a circle will be described at long end when short end is ^^^ in. out of line with the axis of lathe spindle? 15. The axis of a windlass is 6 in. dia. and the crank is 20 in. long. If a pull of 56 Ibs. is given at the end of the crank what weight will be raised? 16. Six men each exert a force of 125 Ibs. on the ends of 3 ft. capstan bars, the barrel of capstan is 16 in. and the bars enter 6 in. into the cap of capstan, which is 24 in. dia. What weight will be lifted? 17. The steering wheel of a ship is 6 ft. dia. and the drum is 12 in. dia. What resistance can be overcome on ropes from drum by a force of 200 Ibs. at the rim of the steering wheel? 18. The crank of a differential windlass is 30 in. long and the force applied at the end of the crank is 75 Ibs. When the diameters of the axles are 12 in. and 14 in., what weight will be raised? 19. A weight of 10,000 Ibs. is to be raised by a differential windlass with a force of 100 Ibs. What length of crank will be required with axles 12 in. and 13 in. diameters? 20. What is the force required on a crank 18 in. long on a differential windlass, with axles 18 in. and 20 in. diameters, to raise a weight of 8,000 Ibs? 21. A compound lever of first, second and third classes has levers each 4 ft. long, the short arm in each lever being 12 in. long. What weight can be lifted with a force of 10 Ibs.? 22. If the short arm of a lathe test indicator is 1 in. long, and the length of the pointer over all is 14 in., what is the 20 SHOP MATHEMATICS amount of motion at the end when indicated point is T ^Vv m - from axis of the lathe spindle? 23. A windlass is to raise a weight of 1000 Ibs. with axle 3 in. dia. What length of crank is required when a force of 120 Ibs. is to be used on end? 24. A windlass is to raise 1200 Ibs. with a 24 in. crank. If a force of 120 Ibs. is applied at the end of crank, what is the dia. of the axle required? 25. If a wheel and axle, 24 in. dia. of wheel and 3 in. dia. of axle, required 25 Ibs. force to raise a weight of 200 Ibs., what size wheel will be required to raise 500 Ibs. with an axle 2 in. dia. and a force of 36 Ibs.? 26. A windlass with an 18 in. crank and 3 in. axle will require how much force to raise a weight of 1500 Ibs.? 27. What dia. of axle will be required for a windlass with an 18 in. crank to raise a weight of 1000 Ibs. with a force of 35 Ibs.? 28. What dia. of axle will be required for a windlass to raise 1600 Ibs. with a 20 in. crank and a force of 45 Ibs.? 29. With a wheel 30 in. dia. and axle 5 in. dia., what force will be required to raise a weight of 500 Ibs.? 30. In a combination of three wheels and axles, when R is 10 in. for each wheel and r is 4 in. for each axle, what weight at W can be supported by a pull at P of 25 Ibs.? 31. If P moves a distance of 5 ft. in problem 30, how far will W move in the same time? 32. Find the ratio of speeds between P and W in prob- lem 31. 33. How many R. P. M. will a grinder spindle make that is driven through two counter shafts, Fig. 13, main line pulley 21 30 in. dia. and 150 R. P. M. with drivers on counters each 24 in. dia., followers 6 in. dia. and spindle pulley 3 in. dia.? 34. If it is desired to drive the spindle of problem 33 20,000 R. P. M., what size drive pulley will be required on main line if all the other pulleys remain the same? 35. A spindle is driven by a train of gears, as in Fig. 12; the first driver on shaft has 50 teeth, the two stud driving gears each 50 teeth, and the followers each 125 teeth with 150 teeth in gear on spindle. What is the number of revolu- tions of spindle with one turn of the shaft? 36. If one set of stud gears were thrown out of mesh in the train of problem 35 and the remaining gears moved into mesh, what is the number of revolutions of spindle for one turn of the shaft? 37. If 60 teeth gears were substituted in the place of the 50 teeth in problem 35, what would be the number of revo- lutions of the spindle for one turn of the shaft? 38. Find the "gear" of a bicycle, Fig. 14, having a rear wheel 28 in. dia. with 18 teeth in front and 7 teeth in rear sprocket. Note. The term gear as used in problem 38 means that the for- ward motion due to one revolution of the crank is the same as would be produced by one revolution of a wheel whose diameter in inches is equal to the "gear." 39. What is the "gear" with 17 teeth in front, 8 teeth in rear sprocket, with 30 in. dia. of wheel? 40. What is the "gear" with 30 teeth front, 10 teeth rear and 28 in. dia. of wheel? 41. W'hat is the "gear" with 16 teeth front, 7 teeth rear with 30 in. dia. of wheel? 42. Find the "gear" with 50 teeth front, 15 teeth rear and 28 in. wheel. 22 SHOP MATHEMATICS 43. Find the "gear" with 22 teeth front, 7 teeth rear and 26 in. wheel. 44. What is the "gear" with 32 teeth front, 12 teeth rear and 26 in. wheel? 45. What is the "gear" with 24 teeth front and 7 teeth rear sprocket and 24 in. dia. of wheel? 46. A man weighing 150 Ibs. has to raise a weight of 1,200 Ibs. How many sheaves must be placed in each pulley block as arranged in Fig. 17 to raise the weight? 47. What force would be required in problem 46 if a pulley is used as in Fig. 19, when sheave B is 8 in. dia. and C is 7 in. diameter? 48. With the pulley blocks as in Fig. 17, with 4 sheaves in both A and B, what weight can be raised with a 15 Ib. pull at P? 49. What weight can be lifted with a pair of 6 sheave pulley blocks, as in Fig. 17, with an 85 Ib. pull at P? 50. When the pull of 85 Ibs. is in the direction shown in Fig. 18, what weight can be raised with 6 sheave pulley blocks? 51. What is the ratio of the efficiency of the pulley blocks of problems 49 and 50? 52. What force will be required to lift a 1,200 Ibs. ma- chine with the pulley blocks of problem 48? 53. How many men weighing 150 Ibs. each can raise, by their own weight, a block of stone that weighs 10,000 Ibs. with a pulley block as in Fig. 19, when sheaves B and C are 10 in. and 9 in. dia. respectively? 54. Find the size of a differential pulley, to raise 6 tons with a force of 150 Ibs. when the difference of the diameters is inch. SHOP MATHEMATICS 23 IV. INCLINED PLANE The inclined plane is a flat surface sloping or inclined to the horizontal. With the inclined plane a weight can be raised by a force of less magnitude as the following illus- tration will show. Illustration. Suppose the weight W is to be raised from a horizontal AC to a point D, Fig. 21. If the weight is raised in a vertical line, as CD, then P and W act through the same dis- tance .'. by the general law of machines P = W. But if the weight is pushed up the incline AD, then the force acts through the distance AD while the weight is lifted only the distance CD; and the statement becomes P :W = CD :AD or P : W = height of plane : length of plane, from which are obtained the formulas WXH P- and W = L PXL H If the force acts along a line parallel to the base, AC then P : W = height : base from which is obtained WXH B w PxB and fF=_- 24 SHOP MATHEMATICS If the force acts at any angle to the plane as Y Fig. 21 A then P \W = sin x : cos Y WXsinx PXcos Y and P= ^ W = : cos Y sin x Example. A horse pulling a load on the level has only the friction to overcome, but the moment that it starts up an incline it has a part of the weight of the load added to the force required to overcome fric- tion. If the pull was 500 lbs. on the level and the load weighs 1,200 lbs., what extra force is required on an incline of 1 foot in 20 feet? Solution. By formula: P = f = '=60. Then 60+500 = 560 lbs. L 20 V. THE SCREW The screw is an inclined plane wrapped or wound around a cylinder. If the incline is long in comparison with the diameter of the cylinder, it may extend more than once around the cylinder forming the threads of the screw; the height of the incline in going once around is called the lead of the screw. The term pitch is used to designate either the number of threads per inch or the distance from the top of one thread to the top of the next; hence in a single thread screw the lead is the same as the pitch, but in a double or triple thread screw the lead is two or three times the pitch. When the screw is turned on its axis through one revolu- tion, the nut being stationary, the screw is raised or lowered a distance equal to the lead of the thread; the force moves in the same time a distance equal to the circumference of twice the length of the lever or bar that is used to turn the SHOP MATHEMATICS 25 screw on its axis. From this is obtained the following: RULE. The force multiplied by the circumference of the circle through which the force arm moves equals the weight multiplied by the lead of the screw. From this rule is obtained the statement: P :W=L :2irR shown in Fig. 23, the differential screw. This is made with two screws of different pitches, or leads of threads, either both right or both left hand threads. It will be seen from Fig. 23 that one turn of the large screw lifts the weight only the differ- ence between the leads of the large and small screws; then by the Gen- eral Law. where wrench screw then, L = lead of screw R = length of bar or used to operate the P = WXL L The screw can be com- pounded like the other ele- ments of machines, as Note. The nut N, Fig. 22, is the part that must be used with a screw to make it effective. 26 SHOP MATHEMATICS and P : W = Ll :2 WX(Ll) Ll Note. Owing to the amount of friction in the differential screw its practical use is limited. VI. WEDGE The wedge is a pair of inclined planes placed back to back. It is used in two ways; by being driven with a blow of the hammer, and by pressure which usually acts parallel to the base of the planes. The difficulty in calculating the effective- ness of the first kind is to determine the force of the hammer blow, otherwise, the statement and rule is the same for either kind. .-.P : W=T :L WXT or L PXL T where T= thickness. The rule is thus the same as for the inclined plane where force acts parallel to the base of incline. Example. A sliding wedge, Fig. 24, has a pressure of 100 Jfyq 24 Ibs., the width of back end is 4 in., the length is 20 in. What weight can be raised by its use? SHOP MATHEMATICS 27 Solution. By formula: TT7 PXL 100X20 W = = = = 500 Ibs. l 4 MISCELLANEOUS PROBLEMS Note. In the following problems friction of moving parts will not be considered. 1. An iron ball weighing 398 Ibs. rests on a surface which is inclined 16 45' to the horizontal. What force, acting at an angle of 14 30' to the incline, is required to hold the ball in position? 2. A weight of 3,500 Ibs. is to be drawn up an incline 640 ft. long, 85 ft. above the horizontal. What force acting parallel to incline will be required to keep the load on incline? 3. A cylinder of cast iron 24 in. dia., 30 in. long is to be rolled up an incline of 18 15'. What force, acting at 8 15' to the incline, will be required to hold the cylinder from rolling down? 4. What weight will be raised with a screw of in. lead when 100 Ibs. of force is applied at the end of a lever 18 in. from the center of screw? 5. When the lead of a screw is f in., R is 20 in. and a weight of 12,000 Ibs. is to be raised, what is the force required? 6. In Fig. 22, a screw of ^ in. lead is turned with a bar 1^ in. long, with 1 Ib. of force on end of R; what weight can be raised? 7. A sliding wedge, Fig. 24, is used to raise the knife on a shearing press that weighs 100 Ibs., the wedge is to move 18 in. and is 3 in. thick at back end. What force will be re- quired? 8. A truck loaded with an engine weighing 6 tons is to be drawn up an incline 12 ft. long and 5 ft. above the hori- 28 SHOP MATHEMATICS zontal. What force will be required when the pull is parallel to the incline? 9. What weight can be drawn up an incline 10 ft. long and 4 ft. high with a pull of 300 Ibs.? 10. Two men each pulling 125 Ibs. can pull what weight up an incline 8 ft. long and 6 ft. high? 11. What force will be required on a single thread screw having 3 threads per inch, with a bar 18 in. long from center of screw to raise a block of granite that weighs 5 tons? 12. Two jack screws are to be used to raise a block 10 ft. long, weighing 10,000 Ibs. One is a third more powerful than the other. Make sketch showing the position of screws to give the proportionate load on each. 13. How many jack screws with \ in. lead and having ft 16 in. long, will be required to raise a building weighing 50 tons, if the pull on each lever is 50 Ibs. ? 14. How many jacks will be required with screws of in. lead, and 12 in. levers, with 25 Ibs. pull on each lever to raise the building of problem 13? 15. A wedge 8 in. long, If in. thick at end will require how many Ibs. of a hammer blow to drive it into a log that has a resistance of 2,000 Ibs. against splitting? 16. When the screws of a differential are 8 and 12 pitch single threads respectively, with a pull of 5 Ibs., what length of lever will be required to raise a weight of 5, 000 Ibs.? 17. What length of bar will be required to raise a building of 100 tons weight with 10 Ibs. pull each on 100, \ in. lead jack screws? 18. A cylinder 25 in. dia. weighs 5,000 Ibs., Fig. 26. What force at P will hold the cvlinder on incline? SHOP MATHEMATICS 29 19. If the cylinder of problem 18 weighs 6,575 Ibs., the lever is 36? in. from P to fulcrum, 12| in. from weight to incline, and incline is 10 ft. long, 4 ft. high, what force on the end of lever P will prevent cylinder from rolling back? SCREW THREADS The formula for finding tap drill sizes is based on the depth from point to bottom of thread, as follows: Fig. 27 shows outline of the sharp, or V thread. The depth A of a thread of 1 inch pitch is found by trigonometry and is equal to .8660 inch. If then the thread is taken on both sides of a cylinder the double depth for 1 inch pitch = .8660X2 = 1. 732. 1 732 For any other pitch of thread the double depth = ' where N is the number of the threads per inch of the required screw. Then the formula for tap drill size is 1.732 Fiy.27. S=T N S = diameter of drill. T = outside diameter of bolt. Fig. 28 shows the outline of the United States Standard (U. S. S.) thread. Here one- eighth of the total depth of the sharp V is flattened on the points, and the same amount filled in at the bottom of the V, thus making the flattened parts of the U. S. S. thread each equal to one-eigth of the pitch so that the double depth for a 2^.28. 30 SHOP MATHEMATICS U. S. S. thread of 1 inch lead will be 1.732 2 X\ of 1.732 or 1.732 .432=1.3 1 3 For any other pitch the double depth will be -^=- where T = outside diameter of bolt. Then S is the size of tap drill to be used before tapping 1 3 threads in the nut. The formula is S= T ~ . jN Example. For a ^X 13 pi. U. S. S. bolt, what is the size of tap drill? Solution. By formula : s = T - '-= ~ - - - **> inch ta drm - It is sometimes necessary to use the British Standard thread, or Whitworth Standard, as it is usually called, after the name of the man who established it. In this thread the top is rounded over one-sixth of the depth and the bottom is filled in the same amount, so that one- third of the depth is subtracted for actual depth of thread, Fig. 29. I 28 The formula for tap drill is, S = T ^-. The included angle of the thread is 55 instead of 60 C as in the V and U. S. S. threads. PROBLEMS 1. Find the depth of a V thread of 12 pitch. 2. Find the depth of an 18 pi. V thread. 3. What is the depth of a U. S. S. thread of 12 pitch? SHOP MATHEMATICS 31 4. What is the double depth for a Whitworth thread of 11 pitch? 5. What is the tap drill size for a 1 in. by 8 pi. U. S. S. tap? 6. What is the tap drill size for ^ in. by 20 pi. V thread nut? 7. What size drill will be needed to allow a full thread on a in. by 12 pi. Whitworth tap? 8. Give the width of the flat on the top of a in. by 9 pi. U. S. S. thread. 9. Find tap drill size for 18 pi. double V thread nut ^ in. diameter. 10. What is the bottom dia. of a f in. by 11 pi. U. S. S. thread? 11. Find bottom dia. of a f in. by 10 pi. Whitworth thread. 12. Find bottom dia. of a in. by 13 pi. V thread. 13. What is the double depth for a f in. by 16 pi. U. S. S. bolt? 14. What is the double depth for a \ in. by 20 pi. V thread? 15. If there are 16 threads per inch on a 1 in. dia. bolt and the nut advances in. for each turn of the bolt, what is the pitch of the thread? 16. One turn of the feed screw on a milling machine moves the table ^ in., but the threads measure 8 per in. How should the threads be designated? Answer. % in. lead or 8 pitch double thread. 17. If there were 12 threads per in., and one turn of screw moves nut i in., how would the thread be designated? 32 SHOP MATHEMATICS TOOTHED GEAR WHEELS A toothed gear wheel is one with projections on its periphery ; these projections, or teeth, engage with the teeth of a similar wheel and the engagement or meshing of these teeth imparts a positive rotary motion from driver to follower. The ratio of the speeds of two gears that run together is called their velocity ratio, and is in inverse proportion to their size. Of two gears, if one revolves once while the other revolves twice, their velocity ratio is as 1 to 2=^, which indicates that the first gear is twice the size of the other. The pitch circle of a gear is the circle near the center be- tween top and bottom on the face of the teeth, such that if the teeth were to be made infinitely small the gear wheel would become a cylinder, Fig. 31. Fly. 31. The diameter of a gear wheel is always the diameter of the pitch circle, unless otherwise stated. The outside diameter is the diameter of the blank in which the teeth of a gear are cut. The pitch of the teeth of a gear, usually called the diametral pitch, is the number of teeth for each inch of the diameter of the gear, unless the circular pitch is stated, which is the distance SHOP MATHEMATICS 33 from center of one tooth to center of the next tooth measured on the pitch circle. When a pair of gears are running in mesh, the smaller of the pair is called the pinion. Let D = diameter of pitch circle of gear, and OD = diameter of blank. P = diametral pitch. N = number of teeth in large gear and n = number of teeth in pinion. C = circumference of pitch circle. CP= circular pitch. T = thickness of tooth on pitch circle, a == addendum. x = distance between centers of two gears in mesh with each other. \r \r 4- & TT TVimi P D OD~ CP N 2 D p-02> p C = CP 1.57 . T .151 Clearance 10 P 34 SHOP MATHEMATICS <2? Working depth = Whole depth = P 2.157 _ o The thickness of cast gears with cut teeth = -^ To find the diameters of two gears in mesh with each other when x and velocity ratio of the two gears are given: RULE. Divide the distance between the centers by the sum of the terms of the ratio; the pitch diameters will be twice the quotient multiplied by each term of the ratio. Example. Find the diameter of two gears with centers 10 in. apart and a velocity ratio of 2 to 5. Solution. By rule: 10 + 7 = 142857 1.42857X2X2 = 5.72 and 1 42857 X2X5 = 14.28 5.72 in. and 14-28 in. are the pitch diameters of gears. PROBLEMS 1. What is the circular pitch of an 8 pi. gear of 40 teeth? 2. What is the circular pitch of a 6 pi. gear of 108 teeth? 3. What is the pitch of a 75 tooth gear when the circular pitch is .7854? 4. A gear of 45 teeth has a circular pitch of .3142 in. Find the diametral pitch of gear. 5. What is the dia. of a gear blank that has 6 pi. teeth and 18 in. diameter? SHOP MATHEMATICS 35 6. A gear blank is 3| in. dia. and is to have 13 teeth. What is the pitch? 7. What is the number of teeth in a 6 pi. gear 18 in. diameter? 8. What is the number of teeth in a 4 pi. gear 19^ in. blank diameter? 9. W T hat is the thickness of tooth for a 75 tooth gear, .7854 in. circular pitch? 10. What is the thickness of tooth for a 4 pi. gear of 40 teeth? 11. Find the whole depth of tooth for a 4 pi. gear of 40 teeth. Xote. Unless otherwise stated the depth of tooth means the whole depth. 12. What is the depth of tooth for a 6 pi. gear with 30 teeth? 13. Find dia. of blank for a 16 pi. gear with 32 teeth. 14. What is the dia. of blank of a 10 pi. gear with 28 teeth? 15. What is the pitch of a 75 tooth gear 19^ in. outside diameter? 16. Find the number of teeth in a 16 pi. gear 8 in. diameter. 17. Find the number of teeth in a 10 pi. gear 64 in. diameter. 18. Find the number of teeth in a 20 pi. gear 10.1 in. outside diameter. 19. Find the sizes of two gears when the velocity ratio is 1 to 2 and the distance between centers is 12 inches. 20. Find the diameters of two gears 12 in. between cen- ters, with velocity ratio 2 to 4. 21. Find the diameters of two gears 28 in. between cen- ters, when the velocity ratio is 3 to 5. 36 SHOP MATHEMATICS 22. When the distance between centers of two gears is 27 in. and the velocity ratio is 4 to 5, find the diameters of the gears. 23. When velocity ratio of two gears is 4 to 5, and the distance between the centers is 36 in., find the diameters of the gears. 24. The velocity ratio of two gears is 2 to 3 and distance between the centers 30 in. Find the diameters of gears. 25. What is the clearance of a 4 pi. gear of 75 teeth? 26. Find the clearance of teeth for a 2 in. dia. gear with 32 teeth. 27. Find the dia. of a 6 pi. gear with 108 teeth. 28. What is the circumference of a gear blank with 32 teeth 16 pitch? 29. Two gears in mesh have 30 and 24 teeth respectively and are 6 pitch. W^hat is the distance between centers? 30. Two gears running together have 120 and 80 teeth respectively and are 10 pitch. Find the distance between the centers. Emery wheels are designed to run at 5,500 F. P. M. (surface feet per minute) , but the speeds of wheels are usually given in R. P. M.; F. P. M. = R. P. M. X irD, D = diameter of wheel in feet. F. P. M. R. P. M.= rD 31. A hand power grinding wheel, W, Fig. 32, is 6 in. dia. and is to run 5,500 F. P. M. What is the R. P. M. of crank C when A has 200 teeth, B 20 teeth, D 15 teeth, and E 250 teeth? SHOP MATHEMATICS 37 32. How many R. P. M. of W. in problem 31, when crank makes 25 R. P. Af .? 33. How many F. P. M. for x in problems 31 and 32 when C is 12 in. long? 34. If there is a resistance of 3 Ibs. on wheel W, problem 31, what force will be required at x to turn wheel 5,000 F. P. MJ 35. What is the velocity ratio between the R. P. M. of W and x in problem 31? Polishing wheels, with wood center and leather covered rim, are allowed to run 7,000 F. P. M. Cloth buffing wheels also run 7,000 F. P. M. Hair-brush cleaning wheels run 12,000 F. P. M. Ordinary grindstones (Ohio) should not run over 2,500 F. P. M. and Huron stones 3,500 F. P. M. 36. If a Huron stone of same size as in problem 31 was substituted for the emery wheel, what R. P. M. of C will give 3,500 F. P. M. of W? 37. If a brush wheel 4^ in. dia. was substituted for wheel of problem 31 to run 10,000 F. P. M., find R. P. M. of C. 38. What force will be required at x if there is 1 Ib. resistance at surface of W in problem 37? 39. What is the R. P. M. of C in problem 31 when A, B, D, and E are changed to friction wheels 9 in., 1\ in., 2 in., and 11 in. diameters respectively? BEVEL GEARS Positive rotary motion can be transmitted from driver to follower, when the shafts stand at an angle to each other, and in the same plane, by using a pair of bevel gears. 38 SHOP MATHEMATICS When the angle of the shafts is 90 and the velocity ratio is 1 to 1 , both gears are of the same size and are called miter gears. When the pair run at a velocity ratio other than 1 to 1, the smaller gear is called the pinion, the same as in spur gears. The pitch diameter of a bevel gear is the diameter of the base of the pitch cone, AB. All calculations for sizes, as in spur gears, are made on this base circle. The teeth of bevel gears vanish or become infinitely small at the apex of the pitch cone. O^AM is the normal cone, that is, one whose con- vex surface stands at right angles to the pitch cone 0AM. The shape and size of teeth are found on the normal cone. The addendum = - and P, D, N, C, CP, and T also clearance, whole and working depths of teeth are all found by the formulas on page 33 for spur gears. The calculations for sizes and shapes of blanks are found by the following formulas: HK-H D = diameter of gear. Dj = diameter of pinion. C = center angle of gear. C l = center angle of pinion, SHOP MATHEMATICS 39 AF then, tan C = --=, by (5) page 179 and C l = 90-C or tan C, = -^ (Jjb ~~P and AS l = a+ clearance. Angle FOS=C+AOS An* AS and tan AOb = r-^ AO Angle AOS is called the angle of increment, angle AOSi is the angle of decrement for the gear. AS, tan AOS^rj^. This angle is used in the setting of the blank at the proper angle for the depth of tooth. The OD for the bevel gear = D +2H and H = cos CxAS (by 4) page 179. Example. Two shafts at an angle of 90 to each other are connected by a pair of 16 pitch bevel gears with a velocity ratio of 3 to 4. If the pinion has 48 teeth find all the other dimensions of the gears. Solution. If the pinion has 48 teeth and the velocity ratio is 3 to 4 then 3 :4 = 4$ ' X which makes X=64, the number of teeth in the gear. D= = 4 inches. D. = = 3 inches. lo 16 AF 2 tn 10. The same as problem 8, except follower gear has 100 teeth and driver 10 teeth and crank 25 in. long. 11. The same as prob- lem 8, except W is 9,975 Ibs. to find P. 12. In Fig. 39, S is a screw of } in. lead, gear E, fast to S, has 100 teeth, D has 40 teeth, gears B and D fast to same stud, and B has 120 teeth, A has 30 teeth fast to crank C, 27 in. long. Find W when the force applied at x is 60 Ibs. 13. The same as problem 12, except gears D and B have 30 and 150 teeth respectively. 14. What power applied at x, problem 12, will raise a weight of 250,000 Ibs.? 15. In Fig. 40, C is a crank 12 in. long fast to pulley A 8 in. dia. A is belted to B which is 6 in. dia. D is a gear of 10 teeth fast to B, and D is in mesh with gear E having 50 teeth; screw S is ^ in. lead, fast to gear E. What weight can be moved up in- cline L, which is 12 in. long, 8 in. high, when a force of 75 Ibs. is applied at x? 52 SHOP MATHEMATICS 16. Find the force that will be required at x, Fig. 40, to move a weight of 75,000 Ibs. up L, when S is f in. lead, and C is 16 in., pulleys, gears and incline same as in problem 15. 17. If the positions of pulleys A and B, Fig. 40, were exchanged and C 24 in. long, otherwise same as in problem 16, what force will be required at x? 18. If a single thread worm was used in place of gear D and belted pulleys A and B of Fig. 40, and a 12 in. crank was fast to worm which meshes with a 50 tooth wheel 16 in. dia. at E, L and S, the same as in problem 15, what weight would be moved up L with force of 75 Ibs. at end of crank on worm? 19. If in Fig. 40, L is 8 ft. and H is 3 ft. and S is i in. lead double thread, gears E and D 15 in. and 6 in. diameters respectively, and A and B 15 in. and 12 in. diameters re- spectively, what force applied at x of a 12 in. crank will move 60,000 Ibs. up the incline at W? FRICTION Friction is the resistance offered a body moving on a surface. It is sliding friction when one surface slides on the other, and rolling friction when one body rolls on the other so that new surfaces are continually coming into contact. A wagon moving along a road illustrates rolling friction between the wheels and roadway, and sliding friction be- tween wheels and axles. Sliding friction varies greatly according to the materials used. For example, a sleigh drawn over bare ground has more friction than when drawn over ice. / is the symbol used to designate the coefficient of friction. It is the ratio between the force required to overcome the SHOP MATHEMATICS 53 resistance due to friction, and the weight or pressure of the moving body on a horizontal plane. From this is obtained /=|-, and P=/XW; W=~. Any pressure at right angles to the line of a moving body may be considered as part of its weight. Example. The slide valve of a steam chest may weigh only 15 Ibs., but a steam pressure of 100 Ibs. per sq. in. on the valve may bring a weight of 5,000 Ibs. on the valve seat to resist the sliding of the valve, so that pressure and weight may thus be equivalent terms. If, in this example, f=.10, then fX 5,000= .10 X 5,000 = 500 Ibs., which will be the force required to slide the valve on its seat. Example. If a block of granite weighs 6,000 Ibs. and /= .1666, what force will be required to slide the block along a platform? Solution. 6,000 X . 166 = 999 Ibs. Example. If the weight of a block is 300 Ibs. and the force necessary to slide it along is 50 Ibs., find/. Solution. f= 50 + 300 =.166. Axle friction is sliding friction, but the bearings are made very smooth and / is much smaller than for ordinary sliding friction. The value for / on well lubricated bearings is from .01 to .05, when not well lubricated /is usually taken at .075. The values for / on well lubricated flat slides have been established by experiment as .08 to .10, when not so well lubricated /is given as .16 to .20. The friction between a shaft bearing and its journal is axle friction. 54 SHOP MATHEMATICS Example. If a turning lathe with driving spindle 3 in. dia. making 50 R. P. M. has a pressure on bearings of 400 Ibs., what power will be required to run the lathe with a 24 in. drive pulley when/= .08? Solution. The resistance due to friction is 400 X. 08 = 32 Ibs., and the force necessary to , . ,. 32X1% overcome friction = = 4 Ibs. l& The travel of the force at the rim of a 24 in. dia. pulley is 50X2irr 12 and the power required is 314-16X4 1,256.64 = 50X6.2832 = 314.16 ft. per min. ft.-lbs. = .038 H. P. 33,000 33,000 Then the power necessary to overcome friction in fly wheels or other heavy axle bearings is found by taking the product of the frictional resistance and the distance of i r u- t r\ u r> ForceX travel . travel of this force. Or H . P. = ^7^7; The following 00,000 formula can be taken as a close approximation to find the H. P. absorbed by friction in heavy shaft bearings: p = WXfXNXvXd 33,000X12 The above formula becomes H.P. = W XfX NxdX .000008. W = load on bearing in pounds. d=dia. of shaft in inches. N=R. P. M. SHOP MATHEMATICS 55 PROBLEMS 1. A fly wheel for a steam engine weighing 6,000 Ibs. makes 74 R. P. M. Find power absorbed by friction if the fly wheel shaft is 10 in. dia. and/= .08. 2. A fly wheel weighs 12 tons, the shaft at bearing is 12 in. dia., makes 72 R. P. M. and f=.08. Find power ab- sorbed by friction. LAWS OF FRICTION 1. Friction is in direct proportion to the pressure with which bodies bear against each other. 2. Friction depends upon the quality of the surfaces in contact. 3. Velocity within ordinary limits has no influence on the value off. 4. The area of surfaces of contact does not affect the value of f within ordinary limits, but if the surfaces are unpropor- tionately large or small the friction will be increased, f for roll- ing friction such as that between the car wheels and rails, on railways, is .002, for iron tired wheels on hard roads .02, on soft roads, .06. ANGLE OF FRICTION OR REPOSE The angle which a plane makes with the horizontal when a body just begins to slide on that plane is called the angle of friction. It can be demonstrated that / is equal to the tan- gent of the angle of friction. FRICTION IN PULLEY BLOCKS Motion between two bodies in contact always produces friction, and in determining the efficiency of any appliance 56 SHOP MATHEMATICS or machine the friction must be taken into account. In a simple bar or lever there will be very little force lost in fric- tion, but in the pulley there is the friction of the rope bend- ing over the sheaves as well as the friction of the axles in the blocks. The amount of force required to overcome friction can be readily found by experiment. The following formula has been found to be right for the effective pull obtained by use of pulley blocks. When W is the weight that can be raised by blocks as arranged in Fig. 18, according to the number of sheaves in both blocks; and for pulleys as arranged in Fig. 17, The force required to overcome the friction of a body moving up an inclined plane and to balance the weight, by formula, page 23, will be ^ . ... F = f -- h friction. Li By trigonometry this becomes F= WXsin A + friction; but the friction is equal to the perpendicular pressure of the body on the inclined plane multiplied by the coefficient of friction. Let the pressure of a body on an inclined plane be repre- sented by line 1 3, Fig. 41, perpendicular to AB, while the whole weight of the body will be represented by the line 1 2 perpendicular to the base AC of the plane. SHOP MATHEMATICS 57 Draw 2 3 parallel to AB and perpendicular to 1 3, to complete the triangle. Then 2 3 represents the force re- quired to balance W on the plane. 23=WXsin A l3=WXcos A since triangle 123 is similar to triangle ABC .'. Friction = WXcos AXf, &ndF=WXcos Axf+WXsinA. Example. A weight of 300 Ibs. rests on a plane inclined at 30 to the horizontal; what force will be required to bal- ance the weight on plane? Solution. Friction not considered, F=WXsinA=300X.5 = 150 Ibs. Now the perpendicular pressure on AB, Fig. 41, is 300 X (cos A). 86603 =259. 8 Ibs., and the total force required to pull W up the incline if /= .15, is 259. 8 X. 15 +150 = 188. 97 Ibs. If W is moving down the incline then the force due to WXsin A will help to move it. If WXsin A is more than W X cos A Xf, the body will slide of itself. Thus in the above example it will require 150 Ibs. 38.97 Ibs. = 111. 03 Ibs. to keep the weight from sliding. When the force acts parallel to the base of the incline, as P lf Fig. 41, as it does in screw threads and helical cams, F=WXtanA and when both weight and friction are considered the formula is : . , . A \ * cos A (fXsin A) Example. When W = 300 Ibs., A =30, f= .15, find F. F-300X - ( ~ 238.5 Ibs. 58 SHOP MATHEMATICS The mechanical advantage for the screw has been given, as P : W = Lead :2-jrR, but the screw is an inclined plane of which the hypotenuse is the middle circumference of the thread, and the lead is the altitude of the triangle, with the force acting parallel to the base; hence when friction is considered the formula is : p _ w Lead + (fX^r) r * 2irr(fxLead) A R' where F = total force. W= weight. R = radius through which power acts. ' r= middle radius of screw thread. 2irr = middle circumference of screw thread. For V thread screw, the frictional resistance will be in- creased as - T = the secant of half the angle of thread, cos A which for U. S. S. thread = sec of 30 = 1.15. The formula for V threads is ._ r ^ 2-xr (LX/X1.16) R } where the letters have the same meaning as in formula above for square threads. MISCELLANEOUS PROBLEMS 1. What force will be required to raise a weight of 1 ton with a double sheave pulley as in Fig. 18? 2. What force will be required to raise 1,675 Ibs. with a double sheave pulley as arranged in Fig. 17? 3. How much extra weight would a 180 Ibs. man require to raise 2,000 Ibs. with a 3 sheave pulley as arranged in Fig. 17? SHOP MATHEMATICS 59 4. What is the value of / when the pull is 75 Ibs. and the weight is 600 Ibe.? 5. What is the value of / when the pull is 80 Ibs. and the weight is 1,000 Ibs.? 6. What is the pull when /is .15 and the weight is 800 Ibs.? 7. What force will be required to haul a machine weigh- ing 1,500 Ibs. up an incline with the horizontal length 12 ft. and height 5 ft. when/ is .2? 8. What force will be required to roll a cast iron cylinder weighing 1,150 Ibs. up an incline 12 ft. long and 6 ft. high when /is .01? 9. An engine fly wheel and shaft 10 in. dia. weighs 7 tons. What is the power required to move wheel when / is .08 and shaft makes 62 R. P. M.f 10. A truck loaded with a box of castings weighs 875 Ibs. If the two bearings of truck wheels l\ in. dia. have/=.15, and rims of wheels have /= .02 on floor of shop, what pull will be required to move truck? 11. A 5 in. pi. dia. worm and a 2 pi. worm wheel of 36 teeth are used to lift an elevator weighing, with the maximum load, 2 tons; the drum, carrying the load is 30 in. dia. and is fastened to worm wheel shaft which makes 10 R. P. M. The shaft bearings are 3 in. dia. with/= .05; the /for sliding of teeth of worm and wheel is .15 and for step bearing 3 in. dia. /= . 10. W T hat force will be required at the rim of an 18 in. dia. pulley on worm shaft to start the elevator? 12. If an elevator weighing 475 Ibs. is to be raised a dis- tance of 9 ft. with 4 sheave pulley blocks, what pull will be required, and what is the movement of the rope, with pulley arranged as in Fig. 17? 60 SHOP MATHEMATICS 13. How much force will be required to roll a barrel of sugar weighing 300 Ibs. up a plank 8 ft. long inclined at. 15 when /=. 005? 14. What force will be required at the end of a 26 in. lever to raise 20,000 Ibs. with a square thread screw 3 in. dia. and | in. lead when /= .15? 15. What weight can be lifted with a square thread screw of in. lead 2^ in. dia. with/=.15, when a force of 25 Ibs. is put on end of a 20 in. lever? 16. 25 tons weight is to be lifted with 50 jack screws of f in. lead, and 3 in. dia. of screw, with/=.10. What force will be required, distributed evenly at ends of 12 in. levers? 17. A hand screw press is used to punch holes in a sheet of iron, the pressure required being 1,250 Ibs. What force will be required at the end of a 16 in. lever on a triple square thread screw 2 in. dia. f in. lead and/=.25, to punch the holes? 18. A in. lead square thread screw 2 in. dia. with /= .10 is used to pull a broach through the in. square hole of an end wrench. A 12 in. pulley with 200 Ibs. pull at the rim was required to revolve the screw. What was the resistance of the pull? 19. A hand press with screw 2^ in. dia. and ^ in. lead V thread is used to force small pins into small discs. A pull of 40 Ibs. is used on the rim of a 16 in. hand wheel. If /=.15, what is the resistance against the pull? 20. If 100 ft. of shafting 2 in. dia. equipped with the usual number of drive pulleys and belting, at 100 R. P. M. requires 1 H. P. to keep it moving, what H. P. will be re- quired to move the same shaft at 170 R. P. M., when the power required increases directly as the speed? SHOP MATHEMATICS 61 21. If the power required to revolve a line of shafting increases as the cube of its dia., what power will be required to turn a line of shafting 3 in. dia., 150 ft. long when other requirements are the same as in problem 20? 22. What power is absorbed in friction in a lathe spindle 2^ in. dia. with 830 Ibs. weight on bearings due to belt pressure, etc., on a 10 in. dia. pulley at 100 R. P. M., when /is taken at .01? 23. What is the power absorbed in running a large lathe with a 10 in. dia. pulley making 20 F. P. M., if the weight on bearings is equal to 3,600 Ibs., the dia. of spindle is 5 in. and /equals .01? BELTING A common method for the transmission of rotary motion is with leather belting over pulleys. The adhesion of a leather belt to the surface of a pulley rim is given in terms of the amount of weight it is capable of lifting; the speed in feet per minute times the weight lifted equals the foot pounds of work it is capable of doing. Cooper, in his treatise on belting, considers a safe velocity for leather belts to be fifty square feet per minute ; for a belt one inch wide this would be equivalent to a linear velocity of 600 F. P. M. The breaking strain for leather belting is given as 3,200 pounds per square inch of cross section; using a factor of 3 200 safety of 11, -j = 290 Ibs. =the safe strain. Nowfor belting ^ inch thick and 1 inch wide, 290 X^55 Ibs. strain per inch. Double belts, that is, two single belts cemented together 62 SHOP MATHEMATICS may have about one-half more added for the allowable pull on the belt, or 82 pounds per inch of width. Modern shop practice allows less strain on the belt than that given above, as low as 30 pounds being given by some belt makers and users. This would correspond to a linear velocity of over 1,100 F. P. M. per H. P. for each inch of width, but about 800 F. P. M. per H. P. is a fair average for single belts, and 550 F. P. M. per H. P. for double belts. In selecting the size and speed for belts general shop conditions must be taken into account. Belts should not run faster than 6,000 F. P. M. on account of reducing the tension of the belt on the pulley by centrif- ugal force; between 3,000 and 4,000 F. P. M. is considered the better practice. The speed of a belt is found by multiplying the R. P. M. of driver by trD or by formula: F. P. M. = R. P. M.X*D. D is diameter of driver in feet. In the calculations for speeds of pulleys for belt transmis- sion, 2% may be allowed for slip, or creep, of belt on the pul- leys. The following rule is approximately correct for lengths of belts over driver and follower pulleys. RULE 1 . Add twice the distance between centers of shafts to half the sum of the diameters of the two pulleys multi- plied by TT, or by formula: I whole length of belt. L = distance between centers of pulleys. D = diameter of larger pulley. d= diameter of smaller pulley. SHOP MATHEMATICS 63 , v ----[--HA] *v The exact lengths for belts for two mating pulleys are found as follows: RULE 2. Find arc of contact of belt on driver and follower and to their sum add twice the dis- tance between points of contact on driver and follower. This distance may be found by the graphical method, or as follows: cos A R r . Fig. 42, on for arc of contact small pulley. L= distance between centers of shafts in inches. R = radius of large pulley. r radius of small pulley. A = arc of contact on small pulley. A = arc of contact on large pulley = 360 A . The arc of contact on cross belts, Fig. 43, is the same for both pulleys and formula is : /.. A\ R+r cos I 180 The length of arc of contact of belt on pulley is found by the formula T= AC 360' where L = length of arc of contact. 64 SHOP MATHEMATICS A = angle of arc. C = the circumference of pulley. The following approximate formula for width, of belt for a given H. P. may be taken as close enough for practical use for single belts. H. P. X 50, 000 . . H. P. X 50, 000 b= ; == -^= , by transposing a = = === -^ dxWxN bXWxN ,uu *uu- u IT/ H.P.X 50,000 b = width of belt m inches, W = -^ 7- ^ a /\ o /\ i v , ,. f H u AT- H.P.X50,000 a = diameter of pulley in inches, N = , , ' & X o X i' W = weight per square foot in ounces, ^ 50,000 N=R. P. M. 1 square foot of leather belt ^ inch thick, weighs 16 ounces. H. P. X 67, 500. For double belts, 6 = dxWxN PRESSURE ON BEARINGS Let P pressure on bearings from pull of belt. V= travel of belt in F. P. M. . , . D 3X33,OOOXH. P. Then approximately P= ~ PROBLEMS 1. A single belt ^ in. thick, 12 in. wide runs 5,000 F. P. M. If there is a strain of 55 Ibs. per in. in width, what H. P. can be transmitted by the belt? 2. A double belt 20 in. wide runs over a 4 ft. dia. pulley at 180 R. P. M. Find H . P. that the belt will transmit. SHOP MATHEMATICS 65 3. A single belt 24 in. wide runs on 14 ft. dia. pulley at 75 R. P. M. Find the H. P. it is capable of transmitting. 4. Two pulleys 60 in. and 24 in. diameters respectively are on separate shafts 12 ft. between centers. What is the approximate length of belt required? 5. What is the exact length of a cross belt running over two pulleys 60 in. and 24 in. diameters, when the shafts are 16 ft. between centers? 6. How many H. P. can be transmitted by a 3 in. belt at 55 Ibs. strain per in. of width at 3,500 F. P. M.? 7. How many H. P. can be transmitted by a 4 in. belt, 35 Ibs. strain, at 4,800 F. P. M.? 8. How many H. P. can be transmitted by a 24 in. double belt, over an 8 ft. driver at 115 R. P. M.? 9. How many H. P. can be transmitted by a 26 in. single belt at 33 Ibs. strain on a 14 ft. driver, at 80 R. P. MJ 10. If the belt in problem 9 is running over a 6 ft. fol- lower pulley, 16 ft. between centers, what is the weight for a belt ^ in. thick? 11. Find the length of a cross belt 16 ft. between the centers of pulleys that are 14 ft. and 6 ft. diameters respec- tively? 12. Find the exact length for an open belt 12 ft. be- tween centers of pulleys that are respectively 84 in. and 48 in. diameters. 13. What is the length of a cross belt 16 ft. between centers of pulleys that are respectively 7 ft. and 3 ft. diameters? 14. Find exact length for an open belt 18 ft. between the centers of pulleys that are respectively 30 in. and 70 in. diameters. 66 SHOP MATHEMATICS 15. What is the width of a double belt to transmit 112 H. P. on a 4 ft. diameter pulley at 180 R. P. M.f 16. What H . P. will a double belt 30 in. wide weighing 36 oz. per sq. ft., transmit on a 60 in. dia. pulley at 200 R. P. M.? 17. A 16 ft. dia. pulley runs at 75 R. P. M.; over this is running a 24 in. double belt that weighs 32 oz. per sq. ft. What H. P. will the belt transmit? 18. What is the pressure on bearings of a belt that is running 900 F. P. M. and transmits 60 H, PJ 19. What will be the pressure on the caps of lathe spindle bearings, when a 4 in. belt on 10 in. dia. pulley making 100 R. P. M. does 2 H. P. of useful work? 20. What is the pressure on the bearings for a 4 in. grinder belt at 3,000 F. P. M. transmitting H. P.? 21. What is the pressure on the bearings of a drop ham- mer pulley with 5 in. belt running at a velocity of 2,000 F. P. M. when 5 H. P. is to be transmitted? 22. What dia. of pulley would be required to increase the velocity of a belt from 900 F. P. M. over a pulley 20 in. dia. to 3,000 F. P. M. with the same R. P. M.f 23. Find the H. P. of a 24 in. double belt, 36 oz. weight per sq. ft. with 36 in. dia. drive pulley making 155 R. P. M. 24. Find H. P. of 40 in. double belt with 4 reinforcing strips each 4 in. wide. The strips weigh 16 oz. per sq. ft., the belt 32 oz. per sq. ft. and runs on a 20 ft. fly wheel pulley at 65 R. P. M. 25. What is the H. P. of a 16 in. single belt over an 8 ft. drive pulley at 110 R. P. M.f 26. What is the H. P. of a 24 in. double belt on 48 in. driver at 275 R. P. M. when the weight is 24 oz. per sq. ft.? SHOP MATHEMATICS 67 27. What width of double belt in. thick will be required for a 28 H. P. drive at 3,500 F. P. M.? 28. What width of double belt at 27 oz. per sq. ft. on a 24 in. dia. driver at 150 R. P. M. will be required for trans- mitting 62 i H. P. ? 29. Find b when w= 16, N= 125, d=20 in., H. P. = 50. 30. Find u> when 6 = 20 in., d = 36 in., N=2QO, H. P. = 75. 31. What weight of belt per sq. ft. will be required to transmit 250 H. P. over a 72 in. dia. pulley at 250 R. P. M., when belt is 36 in. wide? 32. What R. P. M. will a 34 in. dia. drive pulley make to transmit 100 H. P. with a 24 in. belt weighing 32 oz. per sq. ft.? 33. What R. P. M. will a 96 in. dia. drive pulley be re- quired to make in order to transmit 175 H. P. with a belt 36 in. wide and weighing 36 oz. per sq. ft.? 34. Find the dia. of a drive pulley making 125 R. P. M. for a 28 in. belt weighing 32 oz. per sq. ft. to transmit 185 H. P. 35. What is the dia. of pulley required at 250 R. P. M. for a belt 30 in. wide weighing 32 oz. per sq. ft. to transmit 185 H. P.? 36. What weight per sq. ft. will be required for a belt 12 in. wide running 3,600 F. P. M. on a 24 in. dia. pulley, to transmit 75 H. P.? 37. What weight per sq. ft. will be required for a 24 in. belt on a 46 in. dia. pulley making 225 R. P. M. to transmit 125 H. PJ 38. Find w when R. P. M. = 750, d=15 in., 6 = 7 in., H. P. = 25. 68 SHOP MATHEMATICS ROPES Rope drives are used for long distance transmissions and for drives leading off to several shafts at different angles to each other and not parallel to the driving shaft. The effectiveness of a rope drive depends upon the friction in the grooves of the pulleys. The velocity of rope transmission is from 1,500 to 5,000 F. P. M. and should not exceed 5,000 F. P. M. for then the loss due to centrifugal force will offset the gain in speed. The diameter of sheave should not be less than 40 times the diameter of rope in inches. The breaking strain for hemp rope is 6,000 pounds per square inch of cross section area; for manila rope 3,000 pounds per square inch. To find the breaking strength, multiply the cross section area of the rope by 6,000 for hemp and by 3,000 for manila rope. The horse power which a rope is capable of transmitting may be found by the formula: .. .. oUUO d= diameter of rope in inches. V=F. P. M. The weight of 1 inch diameter rope is T 3 ^ of a pound per foot in length, the weights of other sizes can be found by formula: W=d 2 X.3. PROBLEMS 1. Find the weight of 400 ft. of 1J in. dia. rope. 2. Find the weight per ft. of 1 in. dia. rope. 3. What weight will be required to break a 1| in. dia. manila rope? SHOP MATHEMATICS 69 4. What weight would be likely to break a f in. dia. hemp rope? 5. What weight would be apt to break a If in. dia. manila rope? 6. What H. P. will a \\ in. dia. manila rope transmit, running over two 8 ft. sheaves at 150 R. P. MJ 7. If the pulleys in problem 6 are 100 ft. apart on centers, what is the weight of the rope? 8. What strain will be put on a If in. manila rope run- ning at 4,500 F. P. M., to transmit 150 H. P.? 9. What H. P. will be transmitted with a If in. manila rope at 5,500 F. P. MJ 10. What is the minimum size of sheaves and R. P. M. of pulley for a If in. dia. manila rope at 3,500 F. P. MJ Also, what H. P. can be transmitted by this rope? 11. What weight would be required to break a f in. dia. manila rope? WIRE CABLE TRANSMISSION With a wire cable drive there should be no binding or wedg- ing of the cable in the groove of the sheave. The cable should run on wood or leather packing in the bottom of the groove, and the sheave for iron wire should not be less than 150 times the diameter of the cable. The cable should not run over 6,000 F. P. M. on account of centrifugal force, and 3,000 to 5,000 F. P. M. is preferable. The distance between centers of pulleys should not be less than 60 feet nor over 400 feet. Carrying pulleys are used when the spans are over 400 feet. 70 SHOP MATHEMATICS The H. P. transmitted by an iron wire cable is found by the formula: = d*XVX875 5,000 or H. P. = d 2 xVX.055, where the letters have the same significance as in the for- mulas for rope drives. A steel cable will transmit double the above amount. The breaking strain for iron cables is 40,000 pounds per square inch of cross section area, and for steel cables 80,000 pounds. A factor of safety of 10 is generally allowed for rope and cable transmission. A factor of safety is the number which expresses the ratio of the ultimate strength of a body to the working load. The weights of wire cables per foot in length are as follows : f in. dia. = .21 Ibs. f in. dia.= .57 Ibs. A in. dia. = .23 Ibs. f in. dia. = .92 Ibs. \ in. dia. = .31 Ibs. | in. dia. = 1.20 Ibs. 1 in. dia. = 1.50 Ibs. Cable Transmission Problems 1. What is the weight of a f in. dia. cable at 1.2 Ibs. per ft. 837 \ ft. long, and what H. P. can be transmitted at 100 R. P. M. over 12 ft. dia. sheaves? 2. What is the difference in H. P. of problem 1 when figured by the formula and by ft.-lbs. of work? 3. What is the weight of a f in. iron cable 700 ft. long, and what H. P. can be transmitted over 10 ft. dia. sheaves at 80 R. P.M. ? SHOP MATHEMATICS 71 4. What H . P. can be transmitted by a 1 in. cable at 95 R. P. M. over a 12 ft. dia. sheave? 5. The length of a 1 in. iron cable is 800 ft. and it weighs 2 Ibs. per ft. Find the F. P. M. required to transmit 200 H. P. 6. Find the dia. of sheave at 125 R. P. M. and length of a | in. steel cable to transmit 90 H. P. and be inside the safe limit. CHAIN TRANSMISSION One form of chain used for hoisting machinery is made of round wire links as short as possible for strength. Let d = diameter of wire, then the length of link inside = 2. 5 d; width of link inside = 1.5 d. The following table gives sizes of wire for hoists : Capacity in tons, i i i H 2 3 4 5 6 8 10 Diam. of wire in inches, A i A A f & i A I H Another form of chain used for giving a positive motion between two shafts by means of sprocket wheels, is the flat link and block chain; here the strain on the chain is limited by the shearing stress on the pivots. The sprockets should not have less than 7 teeth; the larger the sprocket the less will -. J be the strain on chain and consequently less / wear on the pivots of the rivets. The formula for sizes of sprockets is as follows : = 180 : ~N~' 72 SHOP MATHEMATICS __ sm x tan Y = B + cos x A sin N-= number of teeth in sprocket, A = distance between centers of rivets in link, B= distance between centers of rivets in block, 6 = diameter of end of block, for 1 inch pitch of chain 6 is usually .325 inch when A = .6 inch and B= A inch. The bottom diameter of sprocket wheel is the important dimension, therefore size b must be taken accurately. Then OD = PD+b and Bottom diameter =PD 6. PROBLEMS 1. What is the dia. of an 8 tooth sprocket wheel with a 1 in. pi. chain? 2. What is the OD and bottom dia. of a 20 tooth sprocket with 1 in. pi. chain? 3. What are the distances A, B, and dia. ofb, for a 1^ in. pi. chain, when proportionally the same as for 1 in. pitch? 4. What are the OD and PD for a 20 tooth sprocket, with 1 in. pi. chain? 5. Find the diameters for a 24 tooth sprocket for a 1 in- pi. chain. 6. Find A, B, and b, for If in. pi. when proportionally the same as for a 1 in. pi. chain. 7. Find the diameters for a 28 tooth sprocket with a If in. pi. chain. SHOP MATHEMATICS 73 8. What are the PD and OD for a sprocket of 28 teeth and 1 in. pi. chain? 9. Find the bottom dia. for a sprocket of 30 teeth and 1 in. pi. chain. SHAFTING The H. P. which a line shaft can impart to connected machinery is limited by the strength of the material of which it is made. The principal strain is in the twisting of the round bar when the pulleys are made to revolve carrying driving belts to the various machines. The twisting of the shaft is called its torsional strain, and the formula which determines the amount of torsion which a shaft will safely stand is , / rw = where d = the diameter of shaft in inches, w = the pull of the belt in pounds, r = the radius of the pulley in feet, C = the constant for breaking moment which is found by experiment for cold rolled steel to be 660 pounds. 10 is the factor of safety. Then the above formula may be written dss *IHX80 N where H = H.P. N = R. P. M. 74 SHOP MATHEMATICS PROBLEMS 1. Find values for H and N by transposing the formula for d, given above. 2. A 4 in. dia. shaft runs 150 R. P. M. What is the safe load in H. P. to put on the shaft? 3. Find the H . P. that can be transmitted by a shaft 1\ in. dia. at 200 R. P. M. 4. What H. P. can be transmitted by a 5 in. dia. shaft running at 275 R. P. MJ 5. What dia. shaft will be required to transmit 500 H. P. running at 150 R. P. MJ 6. What dia. of shaft will be required to transmit 250 H. P. running at 74 R. P. M.f 7. How fast should a shaft revolve that is to transmit 1,000 H. P. and is 12 in. diameter? 8. What is the dia. of a shaft that is to transmit 150 H.P.&H50R.P.MJ 9. Find the R. P. M. of a 5 in. shaft to transmit 300 H. P. 10. Find the dia. of a shaft to transmit 12} H. P. at 500 R. P. M. 11. Find the dia. of a shaft to transmit 400 H. P. at 200 R. R. M. 12. What dia. of shaft will be required to transmit 1,000 H. P. at 36 R. P. M.f 13. What dia. of shaft will be required to transmit 200 H.P.&tlSOR.P.M.f 14. Find the H. P. transmitted by an 8 in. dia. shaft at 115 R. P. M. SHOP MATHEMATICS 75 15. Find the H. P. transmitted by a 6 in. dia. shaft at 150 R. P. M. 16. What H. P. can be transmitted by an 18 in. dia. shaft at 50 R. P. M.f 17. How many R. P. M. should an 8 in. dia. shaft make to transmit 200H.PJ 18. How many R. P. M. should a 3 in. dia. shaft make to transmit 200 H. P.? 19. How many R. P. M. should a l}f in. dia. shaft make to transmit 28 H. P.? A jack shaft is a short shaft between the engine and main line of shafting, and is used to regulate the speed of the main line, and prevent the heavy strains on the main line shafting due to the engine drive belt. The bearings must be as near the pulleys as possible. The strain on shaft being greater, the constant is increased from 80 to 120. The formula is d=\ HX * the same \ N letters having the same meaning as in formula for shafting given on page 73. Idler shafts are used to change the direction in which the belts run and have only a bending or shearing strain. The distance between the centers of bearings for line shaft- ing varies in proportion to the size of shaft. Dia. of shaft li to If in. 2 to 1\ in. 2i to 4 in. Center of boxes 7 ft. 8 ft. 10 ft. 76 SHOP MATHEMATICS PROBLEMS 1. What dia. of jack shaft will be required at 225 R. P. M. to transmit 225 H. P.f 2. What dia. of jack shaft will be required at 500 R. P. M. to transmit 250 H. P.? 3. How many R. P. M. should a 6 in. dia. jack shaft make to transmit 500 H. P.f 4. How many R. P. M. should a 2 in. dia. jack shaft make to transmit 87 H. P.f 5. What H. P. will a 6 in. jack shaft transmit at 250 R. P. M.f 6. What H. P. will a 2 in. dia. jack shaft transmit at 250 R. P. M.f 7. Find H. P. transmitted by a 2 in. dia. jack shaft at 170 R. P. M. 8. Find dia. of jack shaft designed for a 250 H. P. engine, when the shaft is to make 84 R. P. M. 9. What should be the distance between the centers of hangers for a line shaft that will transmit 100 H. P. at 160 R. P. MJ 10. Find the distance between the centers of hangers on a line shaft to transmit 50 H. P. at 150 R. P. M. JOURNAL BEARINGS The lengths of journal bearings vary with the work to be done, lubrication of bearing, kind of materials used for the boxes, and diameter of journal. Line shaft bearings are designed from the formula : where L = length of bearing, d=dia. of shaft in inches. SHOP MATHEMATICS 77 The projected area of the journal is the size used in the calculations for the bearing surface; thus the area of bearing surface for a 4 inch diameter by 12 inch journal is 4 X 12 = 48 square inches. The length of bearings for machine spindles vary according to the service required, from 1 to 6 times the diameter of journal. For journals under 2 inches diameter having 100 to 1,000 R. P. M. use the formula: The allowable pressure on the ordinary shaft bearings is 40 pounds per square inch of projected area. BALL BEARINGS Ball bearings are used especially on high speed and light running machinery. The diameter of the ball race may be calculated from the size and number of balls as follows : In Fig. 45 by trigonometry AD let A0= __ sin AOD 8XAO=--D 2n n where n = number of balls in race D = dia. of circle at center of balls d=dia. of ball then AD = -^ when the disc on screw is turned through an arc of 3 degrees? 2. How far will a screw of ^g in. lead move a nut when turned through T ^ of a turn? 3. When the cross slide on a lathe is operated with a screw of in. lead, what is the movement of slide when a disc of 100 divisions on the screw is moved through an arc of one division? 4. Find the lead of feed screw on a slide that will move the slide ^^ in. when a disc with 100 divisions is turned through an arc of one division. 5. If the cross slide feed screw of a certain machine is ^ in. lead, what ratio of gears on an auxiliary crank rod geared to screw will be required on rod and screw to move the slide T tfW m -> when the graduated disc having 100 divisions on the crank moves through an arc of one division? 6. If the adjustable knee of a shaper is to be raised T ^o(T in. by turning crank through arc of one space of a disc fast to crank rod and having 100 divisions, and the screw that stands at right angles to crank rod has ^ in. lead, what will be the ratio between the number of teeth in the beveled gears running in mesh on crank rod and screw? 7. If the screw that moves the bed of a milling machine is in. lead, what number of spaces should be made on a disc to move the bed -r^W m - when the screw is moved through the arc of one space on the disc that is fast to screw? 8. If the cross feed screw of a lathe is i in. lead, how far will the movement of one space on a disc of 25 divisions that is fast to screw move the cross slide? SHOP MATHEMATICS 89 9. The screw that raises the knee on a milling machine is j% in. lead, and is turned by a crank rod with a pair of bevel gears on rod and screw. On crank rod is a disc having 100 divisions. What ratio of gears will be needed to raise the knee T^Vtr m - when the disc fast to the crank is turned through an arc of one division? 10. The cross feed screw for a milling machine bed is 20 pitch. Into how many divisions should the disc that is fast to the screw be divided to move the bed T oW m - when the disc is moved through the arc of one division? 11. If the table of a horizontal boring mill is moved with an 8 pitch single thread screw, into how many divisions should a disc fast to screw be divided to give ^ 3X3 3X10 2X20 30 J,O_ 35 = 7X5 = 7X10 X 5X20 ~ 70 WO Therefore gears 30 and 40 will be used as drivers and 70 and 100 as followers. 100 SHOP MATHEMATICS These principles are summarized in the following: RULE FOR COMPOUND GEARING. 1. Find the ratio between the pitch of the lead screw and the pitch of the screw to be cut. 2. Separate each term of this ratio into two factors and express the result as the product of two fractions. 3. Multiply both numerator and denominator of each fraction by such a number as will make each product repre- sent the number of teeth found hi one of the gears furnished with the lathe. The numerators of these two fractions will denote the driving gears, and the denominators the follower gears. PROBLEMS IN COMPOUND GEARING 1. The gears furnished with a 14 in. Reed lathe have 24, 32, 40, 44, 48, 52, 56, 60, 64, 72, and 110 teeth, and the lead screw is 6 pitch. Find the compound gears required for cutting a 3^ pitch thread. 2. With the same set of gears and same pitch lead screw as given in problem 1, find compound gearing required for cutting a 25 pitch screw. A 28 pitch screw. A 7% pitch screw. A screw of f in. lead. 3. The change gears furnished with a 13 in. Reed lathe have 25, 30, 35, 40, 45, 50, 55, 60, 65, 69, 70, 80, 90, 100, 110 and 120 teeth, and lead screw has 5 threads to the inch. Find the compound gearing required for cutting an 1 1 pitch pipe thread. 4. With the same lathe and same set of gears, find the compound gearing required for cutting a 40 pitch screw. A 23 pitch screw. A 27 pitch screw. A screw of ^ inch lead. A screw of ^ inch lead. SHOP MATHEMATICS 101 TURNING TAPERS Lathes are also used to turn tapers on round stock, and although there are several attachments in use for taper turning, a common method is that of offsetting the tail or foot stock center so that it is out of line with driving center the necessary amount to produce the required taper. When the taper is to be turned the whole length of the bar, the dead center is offset one-half the difference between the diameters at each end of piece. The formula is, _ x = offset of center. D = diameter at large end of taper. d = diameter at small end of taper. When the taper runs only a part of the length of piece, the amount of offset is determined by the length of piece between the centers, and the diameters at each end of the tapered part. The formula becomes x = L ( I where x = offset of center. D = diameter at large end of taper. d = diameter at small end of taper. L = whole length of piece. 1 = length of the part tapered. Example. What is the offset of dead center, when a bar 36 inches long is to be turned 9 inches of its length, tapered from 2 inch dia. at end to 4 inch diameter? Solution. Bv formula 102 SHOP MATHEMATICS The taper per inch in a turned piece is found by subtracting the small diameter from the large and dividing the remainder by the length; as a piece 8 inches long with diameters at ends 1 and 1\ inches is tapered ^ inch per inch in length or Y = ^ = J- = iXi = *V When it is desired to find the taper per foot, the formula is = 12(D d) IJ ~ I where y = the taper per foot and other values same as given above. Example. What is the taper per ft. in the example given above? Solution. Bv formula: which s. the L tJ 7 taper per foot. PROBLEMS 1. A bar of steel is tapered 1^ in. per ft. and diam- eters at end of taper are 1 in. and If in. Find length of taper. 2. A bar is tapered 16 in. of its length and diameters at ends of taper are 1 in. and 2^ in. Find taper per foot. 3. A bar of steel is tapered 7 in. of its length, the diam- eters at ends of taper are 2J in. and 3f in. What is the taper per foot? 4. A bar is turned on a taper 24 in. of its length. The diameters at each end of taper are f in. and ^ in. What is the taper per foot? 5. A shank mill has a tapered end 5^ in. long. The taper is ^ in. and ^f in. diameters at the ends. What is the taper per foot? SHOP MATHEMATICS 103 6. A small cutter has shank tapered 3 in. long, the dia. at small end is .365 in., at large end the taper is .525 in. dia. What is the taper per foot? 7. The taper shank on a cutter is 3^ in. long, the small dia. is .573 in., and the large dia. is .749 in. What is the taper per foot? 8. The taper on a reamer is 8^ in. long, the small dia. is 2^ in., the large dia., is 2^ in. What is the taper per foot? 9. Find the taper per ft. of a pin 2 in. long, and in. and f in. diameters respectively at ends. 10. What is the offset of dead center for turning a taper 16 in. long on the end of bar 40 in. long, when the diameters at ends of the taper are 1J in. and 4 inches? 11. What is the offset of center for turning a taper 19 in. long on a bar 25 in. long, if the diameters at ends of taper are 1 in. and If inches? 12. A bar is 39 in. long; a taper 11 in. long turned on one end is 2| in. dia. at small end and 3j% in. dia. at large end. What is the offset of the center for turning the taper? 13. Find the offset of the center for turning a taper 30 in. long on a bar 34 in. long, when the diameters of taper are \\ in. and If inches. 14. Find the offset of the center for turning a taper 33 in. long on a bar 44 in. long, with the diameters 3 in. and 4\ in. on ends of taper. 15. What is the offset for a lathe center to turn a taper 17 in. on a bar 19 in. long, if the diameters of taper are 1 in. and 1 1 in. on ends? 16. How much should the dead center be moved over to turn u taper bar 4 in. long, when one end of piece is .779 in. dia. and the other end is .982 in. diameter? 104 SHOP MATHEMATICS 17. What is the taper per ft. in problem 16? Note. The formula for offset of lathe center for turning tapers, is very helpful. It is, however, only a close approximation as the calculation is only exact between the ends of centers, whereas, the stock to be turned is supported a little way up the countersunk part on each end. CUTTING SPEEDS The speed at which any piece of work should revolve in taking a cut on the lathe depends on the following conditions: (1) The kind and condition of the material of which the work is made. (2} The kind and condition of the cutting tool. (3) The lubrication of the tool while making the cut. (4) The size of the chip. Therefore any rule for speeds will be only approximate, but from experience with the carbon steels which are used most in the toolmaking department for the finishing cuts, the following speeds have been found practical as a basis for trial: Soft brass 80 F. P. M. Gray iron castings Jfi F. P. M. Machinery steel 30 F. P. M. Annealed tool steel 20 F. P. M. Note. The high speed steel cutting tools can be used at about double the above speed. The F. P. M. can be found for lathe turning from the formula : irRd = 12 C= cutting speed in F. P. M. R=R.P. M. of the work. d= diameter of work in inches. SHOP MATHEMATICS 105 The approximate rule for finding the F. P. M. is to count the revolutions for a quarter of a minute and multiply this number by the diameter of the work in inches. This rule is based on the fact that if in formula just given d=l inch, then - = foot approximately. *.# "T Then approximately, Rd "T and R^~ PROBLEMS In the following problems the approximate formula for R will be used unless otherwise stated. 1. A bar of 4 in. dia. cold rolled machinery steel is to be turned in lathe. What approximate R. P. M. should it revolve to turn a chip at 30 F. P. M.? 2. What is the approximate R. P. M. for turning a chip on a tool steel reamer 1 in. diameter? 3. A cast iron pulley 10 in. dia. is to have the rim face turned in the lathe. How many R. P. M. should it make? 4. How many R. P. M. should a brass rod f in. dia. make in taking a cut in the lathe? 5. The cast iron cylinder of a gasolene motor 5 in. dia. is to be bored in lathe. How many R. P. M. should a boring bar make to do the work? 6. If a ring gauge of tool steel is to be bored in a chuck to 2 in. dia., what is the approximate R. P. M. that it should make? 106 SHOP MATHEMATICS 7. The hub of a cast iron pulley is to be bored for finish- ing with a 3^ in. dia. reamer. How many R. P. M. should it make approximately in the chuck? 8. A brass nut is to be threaded in the lathe for a \\ in. bolt. What would be the maximum speed in R. P. M.? 9. A machinery steel collar is to be shrunk onto a H in. arbor. What speed should it revolve when boring to size? 10. A crank pin 4 in. dia. is to be shrunk into place. AVhat approximate R. P. M. should it make in turning to size, if made of annealed tool steel? 11. A cast iron pulley 16 in. dia. should turn approxi- mately at what R. P. M. for turning the rim? 12. A l^f in. cold rolled machinery steel shaft is to be turned one chip in depth near the end to fit a flange coupling. How many R. P. M. should it make? 13. A crank shaft made of cold rolled machinery steel is to have the bearings finished 7} in. dia. What speed in R. P. M. should it have? MISCELLANEOUS PROBLEMS 14. When a main line of shafting runs at 200 R. P. M. with a 10 in. dia. driver belted with a 14 in. pulley on coun- tershaft, what R. P. M. will a lathe spindle make that has a 9 in. dia. cone pulley belted to a 6 in. pulley on counter? 15. When main line shafting runs 150 R. P. M. with a 16 in. driver belted to a 12 in. follower on countershaft, how many R. P. M. will the spindle make with a 3 in. dia. pulley belted to a 7 in. pulley on counter? 16. When a main line runs 150 R. P. M. from a 14 in. driver belted to 10 in. dia. pulley on counter, cone pulley on SHOP MATHEMATICS 107 counter 8^ in. belted to 6J in. dia. on lathe spindle, find R. P. M. of spindle when gear fast to cone has 28 teeth in mesh with 88 teeth on back gear and 36 teeth on back gear in mesh with 96 teeth gear fast to spindle? 17. If 2% is allowed for slip on each belt in problem 16, what R. P. M. will spindle make? 18. What dia. of pulley should be put on a countershaft to make 175 R. P. M. when the pulley is to be belted to a 15 in. dia. on main line making 160 R. P. M.? 19. A lathe has cone pulley 9 in. dia. belted to a counter- shaft cone 12 in. dia. with back gears 39 and 88 teeth in mesh with 65 teeth on spindle and 27 teeth on cone; the lathe is to bore a 4 in. dia. cast iron disc at 40 F. P. M. When the dia. of drive pulley on main line is 15 inches, what dia. of follower pulley on the countershaft will be re- quired if main line runs 150 R. P. M .? 20. An axle lathe is to turn 4^ in. dia. crucible steel axles at 9 F. P. M.; the back gears have 66 and 22 teeth in mesh with 26 and 58 teeth on cone and spindle respectively. When main line runs 160 R. P. M. and dia. of driver on main line is 10 in., what dia. of follower on counter will be required, if a 12 in. dia. pulley on lathe spindle is belted to 11 in. dia. pulley on countershaft? 21. If 2% were to be allowed for slip in each of the two belts of problem 20, how much larger dia. of drive pulley will be required on main line? 22. A crank pin is to have a taper seat in crank 4^ in. long. The diameters at each end of taper are 4 in. and 4^ in. What offset will be required for dead center to turn the taper when whole length of pin is 10 j% inches? 23. The reamer to finish seat for the pin in crank of problem 22, was 8 in. long. What was the offset of the dead 108 SHOP MATHEMATICS center to turn the taper on a reamer when the small dia. of taper was 4^\ in. dia. and taper was 6^ in. long? 24. An axle turning lathe was equipped with high speed tools for turning at 100 F. P. M. What R. P. M. should the spindle revolve for turning an axle 3 in. diameter? 25. A lathe is fitted to turn heavy cast iron pulleys at 60 F. P. M. What approximate R. P. M. will spindle make for a 30 in. dia. pulley? METAL PLANER A sketch of the driving mechanism for a Powell planer is shown in Fig. 50. Two belts connect the countershaft with Main ttne l6Dia-IOaT the driving shaft, the cross belt driving the platen forward for the cut and the straight belt causing the return. The motion is transmitted from the driving shaft A to the platen SHOP MATHEMATICS 109 by means of a system of gearing as shown in the cut, the last gear meshing with a rack on the under side of the platen. The cutting speed for planer tools is about the same as for lathe tools but heavier cuts can be taken on account of the rigid support of work on the platen. One great dis- advantage with doing work on a planer is that the return stroke is time lost for the tool in the cut, although belted to return about three times as fast as the cutting stroke. The calculations for the planer are to determine sizes of pulleys to give required travel of work in F. P. M. The formula is essentially the one for drivers and followers. Example. What is the travel of platen B, Fig. 50, for one revolution of shaft A ? Solution. By formula: DriversXR of 1st driver W r _ , _ !L _ Followers n = revolution of last follower, 1X15X13 13 The travel = R. P. M. X * X diameter of last follower or, .024X3.1416X18.75 = 1.4137 inches. PROBLEMS 1. Find the F. P. M. for the cut on a planer when the 16 in. dia. pulley on main line runs 170 R. P. M. and is belted to a 12 in. follower on countershaft on which is another 12 in. dia. pulley belted to an 18 in. dia. pulley on shaft A, the gearing from A to platen being the same as given in example illustrated above. 110 SHOP MATHEMATICS 2. What is the return stroke in F. P. M. when the pulley on countershaft is changed from 12 in. to 20 in. dia. belted to an 1 H in. pulley on shaft A in place of the 18 in. dia. pulley of problem 1, with the gearing from A to platen remaining the same. 3. When the drive pulley on main line is changed from 16 in. dia. to 20 in. dia., otherwise the same as problem 1, what is the F. P. M. of the cut? 4. If the belt from driver on main line to countershaft has a slip of 2% and the one from countershaft to shaft A has a creep of 4%, what is the speed of problem 1 in F. P. M.f 5. What is the time required to take 150 strokes on a plate 24 in. long when the travel in F. P. M. is the same as in problems 1 and 2, and 3 in. travel is allowed at the begin- ning and end of stroke for shipping the belt? Figure 51 shows an outline of the gearing for feeding the tool carrier across the plane of the work. The tool carrier is moved by a inch lead square thread screw S, to which is fastened a 19 tooth gear C in mesh J^ra. 5J. w ^h a gear B, having 78 teeth. The ratchet wheel A with 80 teeth is connected by a rod to the adjustable cam that imparts a rotary motion to it at each stroke of the planer; fast to gear B is the ratchet P which holds the gear B from moving back with the rod when it returns back at end of stroke for the next move- ment for the feed. The feed motion is calculated by the for- mula used for gear trains, page 12. For one tooth movement , . 78X.250in.Xl of ratchet wheel A the feed of carrier = jy x oU .01283 inch. SHOP MATHEMATICS 111 7 '8X. 250X2 lor two teeth movement, the leed = = iy x oO .02566 inch. 6. When a planer mechanism is arranged as in problems 1 and 2 and the feed is for a two teeth movement of ratchet as shown in Fig. 51, what is the time required to plane a plate 12 in. square when the time allowance for shipping is the same as in problem 5? 7. What is the time required to finish a surface 24 in. long by 18 in. wide when the gearing is the same as in prob- lems 1 and 2 with 8 tooth movement of A ? 8. What gears should be used at B and C to give a .01 in. feed to the tool for 1 tooth movement of A ? 9. What changes will be required in the sizes of two driving pulleys on main line and countershaft if pulleys are in the same ratio as in problem 1, to give 40 F. P. M. for planing cast iron, when the other mechanism remains the same? UNIVERSAL MILLING MACHINE The universal milling machine is used with various attach- ments for taking special cuts on work, and more especially for cutting the teeth of mills, cutters, gears, reamers, drills, etc. The most important attachment is the index or spiral head which is used with a dead center in tail stock, both fastened to a bed or platen that can be moved back or forward in a horizontal plane under a revolving cutter carried by the main driving spindle of the machine. The feeding mechanism for the platen is shown in Fig. 52. A feed belt connects the main driving spindle with an intermediate shaft on which is a stepped cone belted to a second cone pulley in 112 SHOP MATHEMATICS reverse order of steps allowing for changes in speeds of the feed gearing; from the second cone, the feed is connected 63T- Fig. 52. to the bed by a system of shafts and bevel gears. This shows a simple system of connecting the feeding of the bed from the driving spindle but it is done in a variety of ways. The calculations for the speeds and feeds of the milling machine are almost identical with those for the lathe, therefore the mathematical work of the miller will be confined to the index head attachment, in connection with such parts of the feed- ing mechanism as are related to it. THE INDEX HEAD The index head is used to divide the periphery of a piece of work into &ny number of equal parts and to hold the work at these positions to allow cuts to be taken at equal intervals, SHOP MATHEMATICS 113 as in gear cutting, etc. The mechanism for indexing is shown in Fig. 53, with the body and supporting parts removed. The work which is to be divided into equal divisions is sup- ported and revolved on centers in much the same manner as work is turned in a lathe. A latch pin 5 is held by the tension of a spring in one of the holes of the index plate 6; when the pin is pulled out of the plate, crank 4, can be turned, transmitting motion through worm shaft 3, and its worm, 3^ to worm wheel 2 fast to spindle 1, in which is the live center that supports the work to be cut. The worm wheel 2 has 40 teeth driven by a single thread worm 3 lt so that one revolution of worm 3 l} which is also the same as one revolution of index crank 4, will move the spindle 1 carrying the work through ^ of a revolution; then forty revolutions of crank 4 will give the work one 114 SHOP MATHEMATICS complete revolution. If then 40 teeth, or divisions, are required on the work, one revolution of the crank will be made be- tween each cut as the work is moved under the cutter on the main driving spindle. If 20 teeth, or divisions, are required, the crank will be turned twice around for each cut since 20 is one-half of 40, 10 divisions will require four turns of the crank; 8 divisions five turns; 5 divisions eight turns, etc. From this is obtained the formula for any required number of divisions. Let N= number of divisions required. R number of turns of the crank for each cut. Then R = ^. Example. Find the indexing required for cutting a gear having 60 teeth. Solution. By formula: ^=T7 = =; therefore any 1\ bO plate having the number of holes in a row divisible by 3 may be used. In this case take the 39 hole index and for each one of the 60 teeth cut move the index pin 5 around 26 holes, since 26 = f of 39. The length of the crank 4 is adjustable to any size of the plate 6, but to avoid using a plate of a large diameter, three detachable plates are supplied with each machine so that they can be changed easily to allow any simple indexing to be made. The following list gives the usual number of circles of holes on the three plates. Plate Number of holes in rows: 1 = 151617181920 2 = 2 123 27 29 3 133 3 = 373941434749 SHOP MATHEMATICS 115 A sector 7 is adjustable so that its -two arms can be moved to take the space between any two holes of a series, thus avoiding the necessity of counting the number of holes for each division as it is spaced on the work. PROBLEMS 1. Find the indexing for 3 divisions. 2. If a mill is to have 32 teeth, what is the indexing required? 3. If a tap is to have 4 flutes cut in the universal miller, what indexing can be used? 4. A gear is to have 24 teeth. What is the required indexing? 5. A mill is to have 35 teeth cut on miller. What is the indexing required? 6. Find the indexing required for 45 divisions. 7. Find the indexing required for 72 divisions. 8. Find the indexing required for 68 divisions. 9. If 82 teeth were required to be cut on a gear in the universal miller, what indexing would be made? 10. A gear is to have 108 teeth. Find the indexing re- quired. 11. An index plate with 116 holes is to be drilled on the milling machine with horizontal drilling attachment. What is the indexing required? 12. A ratchet wheel is to have 148 teeth to be cut on its periphery. Find the required indexing. 13. A circular plate is to be marked on the circumference with 164 divisions. What indexing will allow the work to be done on the universal miller? 14. A knurl is to have teeth ^ in. circular pitch at bot- tom of cut on the circumference; if the blank is 3.3706 in. 116 SHOP MATHEMATICS dia. and the teeth are ^ m - deep, what indexing will be necessary for cutting the teeth in the universal miller? 15. A thin disc is to have 165 notches cut on its edge at equal intervals around its circumference. If the work is done in universal miller, what indexing will be necessary? 16. A worm wheel blank 3.6330 in. dia. is to mesh with a T V in. lead V thread screw. What indexing will allow the wheel to be cut on miller? 17. An indexing disc blank 4.1253 in. dia. for the planer centers has V cuts in the circumference .06 in. wide from point to point. Find the indexing to cut the teeth in a uni- versal miller. 18. A disc 31.2 in. in periphery has lines T V in. apart on the circumference. What indexing will allow the mark- ing to be done in miller? 19. A disc 1 in. in periphery has lines at % in. intervals around the circumference. Find the indexing necessary to do the work in the universal milling machine. COMPOUND INDEXING When simple indexing will not give the required spacing compound indexing may be used as follows: Example. Find the indexing required to cut 69 teeth in a gear. Solution. (1) 69 = 23X3 (2) 3323=10 = (3) (4) .40 = 2X2X2X5 . (5) 33=11X3 (6) 23=23X1 SHOP MATHEMATICS 117 (1) Set down the number of divisions required and re- solve into factors. (2) Choose an index plate as 33 and 23 holes and set their difference under the required number and factor as in (1). (3) Draw a line under the two sets. (4) Set down the number representing teeth in worm wheel and factor as in (1). (5) Set down the number of the larger index and factor as in (1). (6) Set down the number of the smaller index and factor as in (1). (7) Cancel factors above the line with the factors below and if all the factors above the line cancel, then the right index plate has been chosen. If the factors above the line do not all cancel, other numbers representing sets of holes in the index plates must be chosen until a number is found that will permit cancellation of all factors above the line. The product of the factors below the line remaining uncancelled, 2X2X11 = 44, will be the number required to give the right indexing fraction; that is, turning the 33 hole index 44 holes in one direction and the 23 hole index 44 holes in the other direction will make the movement of the index IHH ofatum - If any whole number is subtracted from both of these fractions the resulting fraction is unchanged. In this case subtracting one turn from each fraction the indexing will be |^ of a turn in one direction and ^ of a turn in the 118 SHOP MATHEMATICS opposite direction. It does not matter in which direction the indexing is done, as it will simply turn the work in either direction throughout the cut. It will be seen that pin 16 is not adjustable and is in line only with the outside row of holes in index plates, therefore the outside row of holes of one of the three plates must be chosen for the larger number for the compounding. PROBLEMS 20. Find compound indexing required for 77 divisions. 21. What indexing will be required to cut a gear of 91 teeth in the universal milling machine? 22. A ratchet wheel is required to have 96 teeth spaced at equal intervals on its circumference. What indexing will be required to do the work in the universal milling machine? Note. In problem 22 it is possible to use simple indexing for 48 divisions and cut around once, then turn the work one-half the distance between two cuts and cut around again. 23. A gear with 99 teeth is to be cut in milling machine. Find the required indexing. 24. What compound indexing can be made for 147 divisions? 25. A gear is to be cut in the milling machine to have 154 teeth. What indexing will be required to do the work? 26. What indexing in the milling machine will be re- quired to do the cutting of 174 equal divisions? 27. Find the indexing required for 182 equal spaces on the circumference of a disc. 28. What indexing will cut 186 teeth in a gear? SHOP MATHEMATICS 119 DIFFERENTIAL INDEXING The method of compound indexing has been to a large extent displaced by the differential system of indexing that is now a special feature of all the Brown and Sharpe Manu- facturing Company's universal milling machines. With the differential indexing any number of divisions from 1 to 382 may be obtained in the same manner as for plain indexing except that properly selected change gears are used. In plain indexing the index plate is held firmly in position by means of pin 16, Fig. 53, so that ^ of a turn of the index crank moves the work on centers of ? V = T ^ of a turn. In differential indexing the index plate is connected with the work spindle through a train of gears. Every movement of the index crank causes a movement of the index plate, and this motion may be either forward or backward as desired. The effects of these motions may be seen from the follow- ing illustrations: Fig. A represents the method of plain in- dexing, with index plate fixed. \ of a turn of the index crank from M to between the arms of the sector produces ^ of -G = T;VV f a turn of the work on work spindle 1, Fig. 53. Fig. B represents the result when index plate is geared to move forward. While the index crank is being moved through of a turn, from 120 SHOP MATHEMATICS N to of sector, the index plate is moving forward the dis- tance N M so that the work is actually turned through more than ^ of 4*0 of a turn. Fig. C represents the result when the index plate is geared to move backward. While the index crank is being moved through 3 of a turn as indicated by the distance N be- tween the arms of the sector, the index plate is being moved in the opposite direction, so that the movement of the work on the work spindle centers is thus less than y^ of a turn. The amount of rotation of the index plate may be regulated by the difference in the velocity ratios of the change gears. The gears furnished with the Brown and Sharpe Mfg. Co. milling machine for differential indexing and spiral milling cuts are as follows: 3 gears of 24 teeth and one each with 28, 32, 40, 44, 48, 56, 64, 72, 86, and 100 teeth respectively. Example. Find the indexing required for 53 divisions. Solution. By simple indexing the index crank would be revolved through | of a turn for each division, but as there is no index plate having 53 holes this spacing is im- possible, therefore another fraction is selected whose value is near , say M = |, then the 21 or 49 hole index plate can be used, as 15 holes in the 21 row or 35 holes in the 49 row gives the required fraction. Indexing in this way for 53 divisions gives 5 wnen the dia. of cylinder is 7| in., length 00,000 of stroke 15f in., making 200 R. P. M., when a brake test of the engine shows 8 H. P. 14. A gas engine does 25 horse power in a test with a 6 in. dia. cylinder, 12 in. stroke, at 600 R. P. M. What is the M. E. P.? 15. A single cylinder motorcycle 2| in. dia. and 3 in. length of stroke makes 2,500 R. P. M. What is the H. P. whenM. E. P. islOlbs.? SHOP MATHEMATICS 147 16. If the engine in problem 15 has two cylinders, and dimensions the same, what is the H. P.f 17. Find the H. P. by formula, H. P. = S f ^ '**, of a two cylinder gasolene engine, 5 in. stroke, 4 in. dia., 1,000 R. P. M. 18. Find the H. P. by formula, H. P. = P ^^ N , of a four 1,000 cylinder automobile engine making 750 R. P. M., 4 in. dia. and stroke 4f in. long with M. E. P. 87 Ibs. 19. What is the H. P. of a six cylinder automobile gaso- lene motor if D = 4f in. and L = 4 \ inches. 20. Find the H. P. of a four cylinder gasolene engine 3f in. diameter. 21. What is the H. P. of a six cylinder automobile engine with cylinder 4f in. dia. and length of stroke 4f inches? 22. What is the H. P. of a single cylinder gasolene motor 3 1 in. dia., 5f in. length of stroke? ENGINE CYLINDERS The size of the cylinder can be found by transposing the formula for H. P. = - and getting the diameter 33,000 from the area. PROBLEMS 23. Find the H. P. of an engine when the cylinder is 18 in. dia., 26 in. stroke, R. P. M. 200, and M. E. P. of 48 Ibs. 24. What is the H. P. of an engine per Ib. M. E. P. when piston is 20 in. dia., 36 in. stroke, 175 R. P. M.f 148 SHOP MATHEMATICS 25. A locomotive's cylinders are 16 in. dia., 26 in. stroke, drivers 6 ft. dia. What is the tractive power when boiler pressure is 200 Ibs., M. E. P. taken at 50% of boiler pressure, as shown by the steam gauge? 26. The same as problem 25 except boiler pressure is 175 Ibs. and M . E. P. taken at 60%. 27. What size cylinder will be required for an engine to develop 250 H. P. when the boiler pressure is 100 Ibs., and M. E. P. taken at 50%, 24 in. crank, at 74 R. P. M.f 28. What is the H. P. of an engine per Ib. M. E. P. when the piston is 18 in. dia., 30 in. stroke, crank making 112 R. P. M.f 29. Find the size of a cylinder to develop 75 H. P. with 18 in. crank at 115 R. P. M., boiler pressure of 85 Ibs. and M. E. P. taken at 45% of the boiler pressure. 30. What dia. of cylinder will be required for a 150 H. P. engine with 24 in. crank, at 85 R. P. M., 80 Ibs. boiler pres- sure if M. E. P. is taken at 40%? 31. Find the size of a cylinder required for a 75 H. P. engine with 16 in. crank at 150 R. P. M. and M. E. P. 37 Ibs. 32. Find the length of an engine crank when R. P. M. is 85, dia. of cylinder 16 in., M. E. P. 37 Ibs. when the en- gine is to develop 62 H . P. 33. What is the length and R. P. M. of a crank for a 50 H. P. engine, when the piston is 12 in. dia., and the length of crank is of the dia. of piston and the M. E. P. is 30 Ibs.? 34. Find the H. P. of an engine with a 20 in. crank at 140 R. P. M. with cylinder 20 in. dia. and M. E. P. 49 Ibs. 35. Find the H. P. per Ib. M . E. P. with an 18 in. crank at 130 R. P. M. and dia. of cylinder 16 inches. SHOP MATHEMATICS 149 INDICATOR DIAGRAM The M. E. P. on the piston of an engine can be found accurately with an indi- cator. By the mechanism of the instrument the exact pressure on the piston during the entire length of the stroke is obtained in diagram on a A card. A sketch of an indicator card diagram from one end of cylinder is shown at Fig. 59. To find the M. E. P. from the diagram, the line AB, which is the atmospheric line, is drawn by the pencil of the indicator when the connections with the engine are closed and each side of the piston is open to the atmosphere. Divide the length of a base line as AB into any even number of equal parts, say 10, setting off half a part from ends A and B with 9 parts between and from these points of division erect per- pendiculars to the base line AB crossing the diagram at top and bottom; add together the lengths of these lines cut off by the diagram and divide by the number of lines. This will give the mean height, which multiplied by the scale of the spring used to get the diagram will give the M . E. P. The area can also be found by Simpsons' rule, or more accurately by using a planimeter. STEAM BOILER The unit of H. P. for the steam boiler is derived from the number of pounds of water evaporated or converted into steam in one hour. The American Society of Mechanical 150 SHOP MATHEMATICS Engineers gives the following rule which has been adopted as the standard unit for H. P. of boilers. 1 H. P. = The evaporation per hour of SO pounds of water from a feed water temperature of 100 F. into steam at 70 pounds gauge pressure, or 84% pounds per hour from and at 212. From this standard the capacity of a boiler, fired with good anthracite coal, to give the above evaporation per H. P. with maximum economy, is given by Kent as follows: Proportions of Grate and Heating Surface 1. The heating surface per H. P. = 11% square feet. 2. The grate surface per H. P. = % square foot. 3. The ratio of heating to grate surf ace =34% to 1. 4. The water evaporated per square foot of heating surface from 212F. = 3 pounds per hour. 5. The combustible burned per H. P. per hour =3 pounds. 6. The combustible burned per square foot of grate sur- face per hour = 9 pounds. 7. The water evaporated at a temperature of 212F. per pound of combustible = 11% pounds. Example. What heating and grate surface will be re- quired for a 200 H. P. boiler? Solution. The heating surface by (1) is 11% sq. ft. per H. P., then for 200 H. P. the heating surface will be 11% X200 = 2,300 sq.ft. If % sq. ft. of grate surface is required per H. P. then for 200 H. P. it will require 200 X i sq. ft= 66% sq. ft. of grate surface. Example. How many Ibs. of good coal will be required to run a 200 H. P. boiler for 10 hours? Solution. If 3 Ibs. of coal are required per H. P. for 1 hr., for 200 H. P. it will require 200X3=600 Ibs. per hr.; and if 600 Ibs. of coal are required for 1 hr., for lOhrs. it will require 600X10 =6, 000 Ibs. SHOP MATHEMATICS 151 TO FIND HEATING SURFACE FOR VERTICAL TUBULAR BOILER RULE. Multiply the circumference of fire box in inches by the height above the grate; multiply the combined circum- ference of all the tubes by the length of one tube, in inches, and to the sum of these two products add the area of the lower tube sheet, first subtracting the area of all the tubes in inches ; divide the sum of these products by 144 to find the square feet of heating surface. Example. What is the heating surface of a vertical tubular boiler having twenty-four 3 in. dia. and 8 ft. length of tubes, when the fire box is 16 in. high and boiler 30 in. diameter? Solution: By rule cXh = 30X^X16 = 1508 sq. in. and 3% X * X 24 X 96 = 25,334 sq. in. Area = 30X30X .7854 = 707 sq. in. then 1508 +25, 334 +707 = 27, 549 sq. in. and 7,549 + 144 = 191.3 sq. ft. TO FIND HEATING SURFACE FOR HORIZONTAL TUB- ULAR BOILER RULE. Take all dimensions hi inches. Multiply two- thirds of the circumference of the shell by its length ; multiply the combined circumference of all the tubes by the lengths of one tube, to the sum of these two products add two-thirds of the area of both tube sheets and subtract twice the com- bined area of all the tubes; divide this remainder by 144 to find the square feet of heating surface. 152 SHOP MATHEMATICS Example. What is the heating surface of a horizontal tubular boiler 5 ft. dia. and 16 ft. long with fifty 2 in. dia. tubes? Solution: , 2XCXI 2X*X60X192 By rule = = 24127.5 sq. in. o 3 then 2%X50X-*X192 = 75398.4 sq. in. | area of heads minus twice the area of all the tubes or X * xa ? X - 7M *-Htx>X(t-X.7W- t> 3769.92490.875 = 3279 sq. in. 24127.5+75,398.4+3279 __ fot Then = 713.9 square feet. 144 TO FIND THE HEATING SURFACE FOR ANY NUMBER OF TUBES RULE. Multiply the number of tubes by the diameter of one tube in inches, this product by its length hi feet, and this product by .2618. The final product will give the square feet of heating surface. Example. Find the heating surface of fifty-six 3 in. dia. tubes, 18 ft. long. Solution. By the rule, 56X3X18X. 2618 =791. 68 square feet. TO FIND THE WATER CAPACITY OF A BOILER RULE. Multiply two-thirds of the area of the head in square inches, by the length of the boiler hi niches, and sub- tract the volume of the tubes hi cubic inches ; divide the re- mainder by 23 1 , to find the capacity in gallons. SHOP MATHEMATICS 153 Example. What is the water capacity of a horizontal tubular boiler 72 in. dia. and 18 ft. long with 66 tubes 3f in. outside diameter? Solution: By rule V = A X L volume of tubes. Then 586,297157,457 = 428,840 cubic inches. then ^j~ = 1856. 4 gallons. TO FIND THE STEAM CAPACITY OF A BOILER RULE. Multiply one-third area of head in square inches by the length of the boiler in inches, and divide this product by 1728 to find capacity in cubic feet. Example. What is the steam capacity of a horizontal tubular boiler 6 ft. dia., 18 ft. long with sixty-six 3 in. dia. tubes? Solution: A y 7 By rule ^+1,7*8, then +i.rn--mji ,. ft. TO FIND THE PRESSURE CARRIED BY STAY BOLTS RULE. Find the area of the space be- > tween any set of adjacent bolts as A, B, C, D, Fig. 60, and multiply this area in square inches by the pressure of steam gauge. This gives the pressure carried by one bolt; c, .1) to find the pressure for any number of bolts, ^, . ^ multiply by that number. <-/ The area of bolt must be subtracted for accurate results. 154 SHOP MATHEMATICS TO FIND THE BURSTING STRENGTH OF A BOILER B = bursting strength of boiler. d= diameter of boiler in inches. t = thickness of plate. T = tensile strength. c = coefficient of strength of the riveted joint and is the ratio of the strength of the joint to the solid plate. For double riveted joints c= .7. For single riveted joints c= .5. The safe working pressure for a boiler may be taken as follows : 2tTc "~W P = safe working pressure. /= factor of safety usually taken as 6. The horizontal seams in a boiler are double riveted unless otherwise stated. Example. Find the bursting strength and safe working pressure of a 78 in. dia. boiler 18 ft. long with f in. thickness of plates tested at 62,000 Ibs. Solution. By formula: 2t T c 2 X f X 62,000 X .7 54,250 B =^T ~78- -^- = 695.5 Ibs. and dividing by 6 for the safe working pressure 695.5 , f , 07 , = 115.9 Ibs. b The length of a boiler is approximately 3\ times its diameter. SHOP MATHEMATICS 155 PROBLEMS 1. Find the cost for coal at $3.50 per ton to run a 200 H. P. boiler for 20 days of 10 hrs. at the average coal supply for a well equipped boiler. 2. What is the cost of water and coal for running an economically equipped power plant of 150 H. P. for a year of 300 days of 10 hrs. each, when coal is $3.50 per ton and water is 8c per 100 gallons? 3. Find H. P. of a vertical tubular boiler 48 in. dia. having thirty 2 in. dia. tubes, fire box 20 in. high and tubes 8 ft. long. 4. What is the H. P. of a horizontal tubular boiler 72 in. dia., 18 ft. long with one hundred 3^ in. dia. tubes? 5. What is the heating surface of a horizontal tubular boiler 20 ft. long, 5 ft. 6 in dia., with ninety 3 in. dia\ tubes? 6. Find the heating surface of a horizontal tubular boiler, 18^ ft. long, 5 ft. 3 in. dia., having eighty 3 in. tubes. 7. What will be the heating surface of a horizontal tubular boiler, 18^ ft. long, 5 ft. 2 in. dia., with seventy-two 3 in. tubes? 8. Find the steam capacity for the boiler in problem 4. 9. Find the heating surface of a horizontal tubular boiler 18 ft. long, 5 ft. dia., with sixty-six 3 in. tubes. 10. What is the grate surface in sq. ft. of problem 6? 11. Find the sq. ft. of grate surface of problem 7. 12. How much more was the cost of coal per week of 56 hrs. at 4 dollars per ton of 2,000 Ibs. for running a 100 H. P. boiler with 4,800 Ibs. of coal per day of 10 hrs. than if the boiler had been operated on the economical basis? 13. If the grate area had been 24 sq. ft. in problem 12, how much should it be increased to bring it up to the econom- ical measurement? 156 SHOP MATHEMATICS 14. What is the water capacity for a boiler 18 ft. long, 72 in. dia., with ninety 3 in. tubes? 15. What is the water capacity for a boiler 21^ ft. long, 6 ft. dia., with ninety-six 3^ in. tubes? 16. Find the steam capacity for a boiler 16^ ft. long with one hundred 3 in. tubes when boiler is 6 ft. diameter. 17. Find the steam space for a boiler 21 ft. long by 78 in. dia. with one hundred 3 in. tubes. 18. What is the bursting strain for a boiler 48 in. dia., single riveted, with a \ in. thickness of plate tested at 60,000 Ibs. per square inches? 19. Find the bursting pressure of a 72 in. dia. boiler for ^ in. boiler plate tested at 55,000 Ibs. 20. What should be the thickness of plates for a 66 in. dia. boiler to carry 125 Ibs. steam pressure with 6 for factor of safety and plates to be tested at 60,000 Ibs.? 21. What is the heating surface and bursting strength of a horizontal tubular boiler 18 ft. long, 66 in. dia. with seventy-two 3 in. tubes, when made of in. plates tested at 65,000 Ibs.? 22. Find the bursting strength of a boiler 66 in. dia. with in. plates tested at 65,000 Ibs. 23. What is the bursting strength of a boiler 10 ft. dia. with H in. thickness of steel plates at 80,000 Ibs. tensile strength? 24. What is the allowable pressure on a boiler 7 ft. dia. made of 1 in. steel plates tested at 65,000 Ibs. tensile strength? 25. W T hat thickness of plate with a test of 65,000 Ibs. per sq. in. should be used for a 6 ft. dia. boiler, if 6 is the fac- tor of safety and 100 Ibs. per in. boiler pressure is to be used? SHOP MATHEMATICS 157 26. Find the thickness of the plate to use for a boiler 66 in. dia. to carry 80 Ibs. gauge pressure when plates are tested at 65,000 Ibs. Note. In all calculations for strength of boilers it is assumed that the riveting is of right proportions and good workmanship throughout. A boiler plate is supposed to be tested at not over one-third the breaking strain of the plate. The tensile strength of a stay bolt is not over 6,000 Ibs. per square in. and area is taken at the bottom dia. of thread. Example. What pressure is carried on forty f in. stay bolts that are spaced 4 in. on centers, when steam gauge shows 75 Ibs.? Solution: Area of pressure = 4X4 = 16 sq. in. Pressure on one bolt = 16X75 = 1,200 Ibs. Pressure on 40 bolts =1,300X40 = 48, 000 Ibs. Area of bolt=.6875X .6875X.7854= .371 sq. in. Area of 40 bolts = .371X40=14.84 sq. in. 14.84X75 = 1,113 Ibs. Actual total pressure on 40 bolts = 48, 000 1,113 = 46,887 Ibs. PROBLEMS 27. What is the total pressure carried on thirty-six in. stay bolts spaced 4 in. between centers, (see Fig. 60,) when the steam gauge shows 87 Ibs. pressure? 28. Find the pressure on forty-eight in. X 10 pi. stay bolts with U. S. S. thread when boiler pressure is 95 Ibs. and area between a set of bolts is 12 square inches. SAFETY VALVE The lever safety valve, Fig. 61, is a lever of the third class, and calculations for lengths of arms and weights required for given boiler pressure are made from the formulas of the lever except that weights of lever and valve must be taken into account. 158 SHOP MATHEMATICS When the lever and valve connected to it will just balance over a knife edge, this point is called the center of gravity of the lever; the fulcrum is at the center of pivot on which the lever swings. Then let g = the distance between center of gravity and fulcrum. w = weight of ball in pounds. VL = weight of valve and lever in pounds. A = area of safety valve in square inches. a = distance between ball and fulcrum in inches. b = distance between center of valve and ful- crum in inches. P = pressure per square inch on steam gauge. The formula for weight to balance the pressure is and g, A, P, a, etc., can be found by transposing terms and solving by algebra; the above formula can be written w _APb-VLg a Example. What distance from the center of fulcrum will a weight be placed, if the steam gauge shows 95 Ibs., weight is 15 lbs v area of valve 3 square in., and valve and lever weigh 18 Ibs., cen- ter of valve 2 in. from fulcrum, 0= 12 inches? SHOP MATHEMATICS 159 Solution. Substituting the values given in the formula 3X95X2$ 18X12 10 = r - 3X96X8} 18X18 then a = - =33.1 in. 15 PROBLEMS 1. From the law of levers show how the formula for W is derived. 2. What is the weight for a ball on a lever safety valve 5 in. dia., if the ball is placed 30 in. from center of fulcrum, the center of gravity is 12 in. from fulcrum, valve and lever weigh 20 Ibs., steam gauge registers 80 Ibs., and centre of valve is 3 in. from fulcrum? 3. What is the dia. of a cast iron ball for safety valve, when the center of gravity is 12 in. from fulcrum, lever and valve weigh 14 Ibs., distance from center of ball to fulcrum is 27 in., center of 2 in. dia. valve is 3 in. from fulcrum, and the steam gauge shows 85 Ibs. pressure? 4. When the center of a safety valve 3.75 in. dia. is 2 in. from fulcrum, what distance from the fulcrum must a 6^ in. iron ball be placed, if the valve and lever weigh 16 Ibs., boiler pressure is 70 Ibs., and center of gravity is 16 in. from fulcrum? 5. What weight placed 30 in. from the fulcrum of a safety valve 4^ in. dia. will just balance 80 Ibs. boiler pres- sure, when the valve and lever weigh 12 Ibs., the center of gravity is 13 in. from fulcrum, and the center of valve is 4J in. from fulcrum? 160 SHOP MATHEMATICS HYDRAULICS Hydraulics treats of liquids in motion, especially of the action of water in canals, pipes, machinery for raising water and the use of water as a source of power. Pressure varies directly as the depth from the free surface. This depth from the free surface is called the head. If the weight of a cubic foot of water is taken as 62 pounds, then the weight of a column of water 1 foot high and 1 square inch in cross section = 62. 5 -r- 144= -4$4 pound. Therefore the pressure per square inch at any point in a body of water = the depth below the surface, or the head, X.W, then let P = pressure per square inch. Then P = HX.434 and H To find the head when pressure is given RULE. Divide the pressure by .434 or multiply the pres- sure by 2.302. The folio whig laws apply to liquids : The pressure does not depend upon the size or shape of the vessel. The pressure increases with the depth below the free surface. At any point in a liquid the upward, downward and lateral pressures are equal. The pressure which a body of liquid exerts on the contain- ing vessel such as the walls of a tank, is subject to the follow- ing: SHOP MATHEMATICS 161 RULE. Pressure is equal to the product of the head, the area of the surface on which the liquid presses and the weight of a cubic foot of the contained liquid. From this rule is obtained the formula: P=H A W Let W = 62.5 pounds = weight of a cubic foot of water. .-1 = area in square feet of surface in contact with liquid. H = the head, which is the distance from the free surface of the liquid to the center of the sur- face in contact with the liquid. Archimedes' discovery, that a solid body immersed in a liquid displaces the same volume as itself, furnishes an ex- cellent method of finding the volume of any irregular shaped body, by immersing it in water and measuring the volume of the water displaced. The weight of the body can then be found by multiplying the volume of water displaced by the weight of a cubic inch of the substance of which the body is composed. Example. An iron casting when immersed displaces half a gallon of water. Find the weight of casting. Solution. A cu. in. of cast iron weighs .2604 lb. and there are 115.5 cu. in. in a half gal., therefore 115.5 X .3604 = 30.076 Ibs. The specific gravity of a substance is its weight as com- pared with the weight of an equal volume of water. PROBLEMS 1. Find the specific gravity of lead when a cu. in. weighs .4106 Ib. 2. Find the specific gravity of coal when a cu. ft. weighs 57 Ibs. 162 SHOP MATHEMATICS 3. Find the specific gravity of southern pine timber when a cu. ft. weighs 60 Ibs. 4. A cu. in. of a certain composition metal weighs .358 Ib. What is its specific gravity? 5. What pressure will be shown on the water gauge of a boiler if the supply tank has a head of 200 feet? 6. What head will show 120 Ibs. pressure on the water gauge of a receiving tank? 7. A tank that is 3 ft. square is filled to a depth of 3 ft. with water. What is the pressure of the water on the bottom of the tank? 8. What is the pressure of water on one side of tank of problem 7? 9. A tank is 12 ft. square on its base and 5 ft. high. What is the pressure of the water on one side, if the tank is filled to a depth of 5 feet? 10. When the tank of problem 9 is a closed tank and a pipe 5 in. dia. runs 5 ft. above top of tank, what is the pres- sure on top surface of tank when the water is filled to the top of pipe? 11. A 5 in. dia. pipe closed on lower end is sunk in a ver- tical position in a tank of water to a depth of 30 in. What is the upward pressure on the closed end of the pipe? 12. A sluice gate is 6 ft. high and 4 ft. wide. When closed the water is up to the top on one side and 2 ft. high on the other. What is the pressure on the gate? 13. When a cast iron ball displaces 1 gal. of water, what is the weight of the ball? 14. If an iron casting displaces 3 qts. of water, how much does it weigh? SHOP MATHEMATICS 163 15. A steel bar was sunk in a tank full of water and it was found that 15 qts. of water ran out to allow space for the bar. What was the weight of the bar? 16. An irregular shaped plate of steel was immersed in a tank of water and found to displace 6 gal. What was the weight of the plate? 17. A steel forging was found to displace 6 qts. of water. How much did it weigh? 18. How many feet of lead pipe weighing 10 Ibs. per ft. in length, will displace 10 gal. of water? HYDRAULIC MACHINES Elevators and pumps are the most common examples of hydraulic machines used in shop practice. The hydraulic elevator is based on the principle known as Pascal's law to the effect that pressure exerted upon a liquid in a containing vessel is transmitted equally and undimin- ished throughout the liquid. If pipe a, Fig. 62, has an area of 1 sq. in. and the pipe 6 has an area of 100 sq. in. and water is pumped into 6 through pipe a at a pressure of 200 Ibs. per sq. in., then for each sq. in. of surface on piston B there will be a pressure of 200 Ibs., and on the whole surface of B there will be a pressure of 100X200 Ibs. or 20,000 Ibs. Example. A supply pipe for a 12 in. plunger hydraulic elevator piston is 1 in. area, and the pressure into supply pipe is pumped up to 150 Ibs. per sq. in. What is the total pressure on bottom of piston? Fij .62. 164 SHOP MATHEMATICS Solution, Since pressure on a liquid is transmitted un- diminished in all directions, a pressure of n Ibs. per sq. in. on small pipe will produce the same pressure per sq. in. on large pipe at the piston, therefore 150 : x = 1 in. : 113.10 in. x = 16,965 Ibs. Pumps are of several kinds and are operated by hand or power. The formula for lifting or forcing water either under pressure or head is as follows : P = H A W, where H is the distance from the level of the source of sup- ply to the point of discharge. Example. What is the pull on a pump rod, when dia. of bucket is 5 in. and water is raised 24 feet? Solution. By formula: P = H AW= &4Xr X 62.5=204-45 Ibs. 144 which is the pull on rod necessary to operate the pump; to this must be added the amount of power required to overcome friction in the moving parts. The steam pump is often used to supply the feed water to the boiler. The pump in such cases is usually made with a steam piston at one end of a connecting rod and the water piston at the other end. In this case the steam piston must be enough larger in diameter than the water piston to over- come the friction of the mechanism and leakage in the valves, besides the steam pressure in boiler, against which the pump is working. From 5% to 40% is allowed according to the condition of the mechanism of pump. SHOP MATHEMATICS 165 TO FIND THE CAPACITY OF A PUMP PER HOUR RULE. Find the cubical contents of the water cylinder per stroke in cubic inches, multiply by number of strokes per hour and divide the product by 231 to find the number of gallons or by 1,728 to find the capacity hi cubic feet. TO FIND THE H. P. REQUIRED TO PUMP WATER TO A GIVEN HEIGHT RULE. Multiply the weight in pounds of water to be raised per minute by the height hi feet and divide by 33,000; the quotient will be the H. P. required. The formula is *. P. Example. What is the capacity per hour of a pump with water piston 6 in. dia. and 8 in. stroke, when the piston makes 75 strokes per minute? Solution. The contents of water cylinder, if cylinder is filled at each stroke is AxL = 28.274X8 = 226.2 cu. in. At 75 strokes per minute there will be 75X60 = 4,500 strokes per hour. If the piston pumps 226.2 cu. in. per stroke then for one hour it will pump 226.2X4,500 = 1,017,900 cu. in., and 1,017,900 + 331 = 4,406.4 gal- per hr. Example. Find the H. P. required to pump the water of above example to a height of 50 ft. above source of supply. Solution. If a pump will raise 4,406.33 gal. of water per hour, it will raise 4, 406. 33 + 60 =7 3.. $8 gal. per minute and 73.438 gal. weighs 73. 438X8$ = 61 1.983 Ibs. This weight of 166 SHOP MATHEMATICS water is to be pumped 50 ft. high per minute; then by formula = . = .^ 33,000 33,000 660 PROBLEMS 1. Find the number of gal. of water delivered per hour by a single action pump 6 in. dia. with a 12 in. stroke of the water piston at 100 strokes per minute. 2. What H. P. would be required to operate the pump of problem 1 if the discharge is 80 ft. above the source of supply? 3. How many gals, per hour will be delivered by a 6 in. dia. water piston with 10 in. length of stroke and making one stroke per second, if the cylinder only fills three-fourths full at each stroke? 4. If the plunger of a hydraulic elevator is 16 in. dia. and the pressure of supply pipe is 120 Ibs. per sq. in., what weight can be lifted if car and plunger weigh 400 Ibs. and the friction of moving parts takes 5% of the power? 5. How many cu. ft. of water per hour will a single action pump 12 in. dia., 15 in. stroke deliver at 50 double strokes per minute, when water cylinder is three-fourths full per stroke? 6. What H. P. will be required to raise the water of problem 5, 40 ft. above source to a delivery tank? 7. Find the amount of water in gals, that will be pumped with a 4 in. dia. single action pump with water piston moving at 100 F. P. M. and cylinder drawing only seven-eighths full per stroke. SHOP MATHEMATICS 167 8. What power will be required to pump water 200 ft. above source with a 12 in. dia. double action water cylinder moving 100 F. P. M.f 9. Find the power required to pump from a river to a tank at the top of a factory building 67 ft. from river to top of tank with an 8 in. dia. single action water piston moving 100 F. P. M.f 10. What steam pressure will be required on the steam piston of a direct acting pump to deliver water 70 ft. above source, when the steam and water pistons are both 8 in. dia. if 25% is allowed for leakage and friction in the mech- anism? 11. A tank 10 ft. dia. and 10 ft. high is set on the top of a building 70 ft. above the level of a water supply from which an 8 in. dia. single action piston pump moving 100 ft. per minute is delivering to tank at 200 F. P. M. How long will it take to fill the tank if 30% is allowed for loss by leakage, etc.? Note. The velocity of water through a pipe at 200 F. P. M. requires a pipe to be of diameter .J5V gal. per minute. Between 100 and 200 F. P. M., the pipe should be of a diameter I gal, per minute \ velocity in F. P. M. 12. What size pipe will be required for delivery pipe of problem 11? 13. When water is pumped through a delivery pipe at 150 F. P. M., what dia. of pipe will be required from a 12 in. dia. by 18 in. stroke piston of a single action pump making 60 strokes per minute? 14. When the piston travel is 125 F. P. M. what is the number of cu. ft. of water pumped per minute by a double action pump with 10 in. dia. of piston? 168 SHOP MATHEMATICS 15. What is the H. P. required to run the pump of prob- lem 14, when it delivers to a tank 72 ft. above the source? 16. Find the size of a delivery pipe required for problem 15 if it were to discharge into the tank at 150 F. P. M. 17. When the bucket of a hand suction pump is 3 in. dia. and the supply is drawn from a depth of 30 ft., what pressure is required on the handle 26 in. from the fulcrum when the center of bucket rod is 5 in. from the fulcrum? 18. When the piston travel of a double action steam pump is 150 F. P. M. and 15 in. dia., find the H. P. required to fill a supply tank 62 ft. above the source. 19. What dia. of pipe will be required to deliver the water of problem 18 at a speed of 200 F. P. M.? STANDARD UNITS OF WEIGHTS AND MEASURES Measure is that by which extension, capacity, force, duration, or value is estimated or determined. The weight of a body is the measure of the force of the earth's attraction for that body, commonly called the force of gravity. Linear or Long measure is used in measuring distances. Table of Linear Measure 12 inches =1 foot (ft.). 3 feet =1 yard (yd.). 5* yds. or 16* ft.= 1 rod (rd.). 320 rods = 1 mile. 1 mile =320 rds.= 1,760 yds. = 5,280 ft.= 63,360 in. Surface measure is used in measuring in two dimensions, length and breadth. Table of Square Measure 144 square inches = 1 square foot. 9 square feet = 1 square yard. 30^ sq. yds. or 272 sq. ft. = 1 square rod. 160 square rods = 1 acre. 640 acres = 1 square mile. 1 sq. rd. =30^ sq. yds. = 272^ sq. ft. = 39,204 sq. in. Pressures of liquids and gases are usually given in pounds per square inch or square foot. 170 SHOP MATHEMATICS Cubic measure is used in measuring the volume of solids. Table of Solid or Cubic Measure 1,728 cubic inches = 1 cubic foot. 27 cubic feet = 1 cubic yard. 1 cu. yd. = 27 cu. ft. = 46,656 cu. in. The measure of volume for liquids. Table of Liquid Measure 4 gills = 1 pint. 2 pints = 1 quart. 4 quarts = 1 gallon. 31^ gallons = 1 barrel. Liquids are sometimes measured with cubic measure, the U. S. gal. contains 231 cu. in. so that a cu. ft. of water con- tains approximately 1\ gallons. One gallon of water taken at the temperature of maximum density, 39.1F., weighs 8.3389 Ibs., avoirdupois, which is approximately 8J Ibs. A cubic foot of water weighs 62.355 Ibs., approximately 62 Ibs. Water freezes at 32 F. or C., and boils at 212 F (Fahrenheit) or 100 C. (Centigrade). The standard unit for weight in shop practice is the avoirdupois pound. Table of Avoirdupois Weight 16 ounces = 1 pound. 100 pounds = 1 hundred weight (cwt.). 20 cwt. = 1 ton. 1 ton = 20 cwt. = 2,000 Ibs. = 32,000 oz. SHOP MATHEMATICS 171 Time is the measure of duration. Table of Measure for Time 60 seconds = 1 minute. 60 minutes = 1 hour. 24 hours = 1 day. 7 days = 1 week. 365 days = 1 year. 366 days on leap years. Angular measure is used in measuring angles. Table of Angular Measure 60 seconds (60") = 1 minute (I'). 60 minutes (60') = 1 degree (1). 360 degrees (360) = 1 circumference. Money is the measure for values. It is used as a medium of exchange. Table of U. S. Money 10 mills = 1 cent. 10 cents = 1 dime. 10 dimes = 1 dollar. 10 dollars = 1 eagle. 1 eagle = $ 1 = 100 dimes = 1 ,000 cents = 10,000 mills. MISCELLANEOUS MEASURES Table 12 units = 1 dozen. 12 dozen = 1 gross. 12 gross = 1 great gross. 1 gross = 12 doz.= 144 units. 172 SHOP MATHEMATICS Weights of Materials 1 cu. in. of cast iron weighs .2604 Ib. melts at 2,500 F. 1 cu. in. of wr't. iron weighs .2779 Ib. melts at 3,100 F. 1 cu. in. of steel weighs .2833 Ib. melts at 3,000 F. 1 cu. in. of copper weighs .3195 Ib. melts at 1 ,930 F. 1 cu. in. of brass ] . [ weighs .3029 Ib. melts at ( zinc 35 ) 1 cu. in. of lead weighs .4106 Ib. melts at 625 F. 1 cu. in. of aluminum weighs .0963 Ib. melts at 1,160 F. The approximate weight of round bar iron or steel can be found by the formula T _(dX4) 2 6 ' where L = weight of one foot in length of the bar and d = diameter of the bar in inches. Horse Power The horse power is an expression for foot-pounds of useful work accomplished in a specified time. 1 H. P. = 33, 000 pounds raised 1 foot in one minute, or 1 pound raised 33,000 feet in one minute. Electric Units The volt is the unit of electrical pressure. The ampere is the unit of current strength or rate of flow. The ohm is the unit of resistance. The watt is the unit of power. The ohm, ampere and volt are defined in terms of one an- other as follows: Ohm, the resistance of a conductor through which a current of one ampere will pass when the electro-motive force is one volt. SHOP MATHEMATICS 173 Ampere, the quantity of current which will flow through a resistance of one ohm when the electro-motive force is one volt. Volt, the electro-motive force required to cause a current of one ampere to flow through a resistance of one ohm. The relation which these quantities bear to one another is expressed by Ohm's Law. E M F in volts Current in amperes = -=: : : j . Resistance in ohms E Then C-I-. H where E = the electro-motive force in volts, R = resistance in ohms, C= current in amperes. The electric H. P. = 746 watts. METRIC SYSTEM OF MEASURES The metric system is used in almost all the European countries and is authorized by law in the United States and is in general use for all scientific treatises. Metric Linear Measure 10 millimeters (mm.) = l centimeter (cm.). 10 centimeters =1 decimeter (dm.). 10 decimeters = 1 meter (m.). 10 meters =1 dekameter (Dm.). 10 dekameters = 1 hektometer (Hm.). 10 hektometers = 1 kilometer (Km.). 10 kilometers = 1 myriameter (Mm.) . i Mm. = 10 Km. = 100 Hm. = 1,000 Dm. = 10,000 meters. 1 m.= 10 dm. = 100 cm. = 1,000 mm. = 39.37 inches. The unit for metric linear measure is the meter. 174 SHOP MATHEMATICS Metric Square Measure 100 sq. millimeters = 1 sq. centimeter (sq. cm.). 100 sq. centimeters = 1 sq. decimeter (sq. dm.) 100 sq. decimeters =1 sq. meter (sq. m.). 100 sq. meters = 1 sq. dekameter (sq. Dm.). 100 sq. dekameters = 1 sq. hektometer (sq. Hm.). 100 sq. hektometers = 1 sq. kilometer (sq. Km.). The unit for metric square measure is the square meter for small surfaces, and the are for land measurements. 100 centares= 1 are 1 sq. meter = 1.196 sq. yard 100 ares= 1 hectare Metric Cubic Measure 1,000 cu. millimeters = 1 cu. centimeter (cu. cm.). 1,000 cu. centimeters = 1 cu. decimeter (cu. dm.). 1,000 cu. decimeters = 1 cu. meter (cu. M.). The unit for cubic measure is the cubic meter and 1 cu. m. = 1.308 cu. yard. Metric Measure of Capacity 10 milliliters = 1 centiliter (cl.). 10 centiliters =1 deciliter (dl.). 10 deciliters =1 liter (1.). 10 liters = 1 dekaliter (DL). 10 dekaliters = 1 hektoliter (HI). 10 hektoliters = 1 kiloliter (Kl.). The unit for capacity for both liquid and dry measure is the liter. 1 liter = .908 of a dry quart. 1 liter = 1.0567 of a liquid quart. SHOP MATHEMATICS 175 Metric Measure of Weight 10 milligrams =1 centigram (eg.). 10 centigrams =1 decigram (dg.). 10 decigrams = 1 gram (g.). 10 grams =1 dekagram (Dg.). 10 dekagrams = 1 hektogram (Hg.). 10 hektograms = 1 kilogram (Kg.). 10 kilograms =1 myriagram (Mg.). 10 myriagrams =1 quintal (Q.). 10 quintals = 1 tonneau (T.). The unit of weight is the gram. 1 gram =.03527 oz. Avoirdupois. EQUIVALENTS OF COMMON AND METRIC SYSTEMS Linear Measure 1 inch = 2.54 cm. 1 foot =0.3048 m. 1 yard = 0.9 144m. 1 rod = 5.029 m. 1 mile =1.6093 Km. 1 cm. = 0.3937 in. 1 dm. =0.328 ft. 1m. =1.0936 yds. 1 Dm.= 1.9884rds. 1 Km. = 0.62137 mile Square Measure 1 sq. in. = 6.452 sq. cm. 1 sq. ft. = 0.0929 sq. m. 1 sq. yd. = 0.8361 sq. m. 1 sq. rd. =25.293 sq. m. 1 acre =40.47 ares 1 sq. mile = 259 hectares 1 sq. cm. =0.155 sq. in. 1 sq. dm. = 0.1076 sq. ft. 1 sq. m. = 1.196 sq. yds. 1 are =3.954 sq. rds. 1 hectare =2.471 acres 1 sq. Km. = 0.3861 sq. mile 176 SHOP MATHEMATICS Cubic Measure 1 cu. in. = 16.387 cu. cm. 1 cu. ft. =28.317 cu. dm. 1 cu. yd. =0.7645 cu. m. 1 cord = 3.624 steres 1 cu. cm. =0.061 cu. in. 1 cu. dm. =0.0353 cu. ft. 1 cu. m. = 1.308 cu. yds. 1 stere =0.2759 cord Measures of Capacity 1 liquid qt. = 0.9463 liters 1 dry qt. =1.101 liters 1 liquid gal. = 0.3785 Dl. 1 peck = 0.881 Dl. 1 bushel =0.3524 HI. 1 liter = 1.0567 liquid qts. 1 liter = 0.908 dry qts. 1 Dl. = 2.6417 liquid gals. 1 Dl. =1.135 pecks 1 HI. = 2.8375 bushels Measures of Weight 1 grain Troy= 0.0648 gram. 1 oz. Avoir. = 28.35 g. 1 oz. Troy =31.104g. 1 Ib. Avoir. = 0.4536 Kg. 1 Ib. Troy = 0.3732 Kg. 1 ton (2000 lbs.)= 0.9072 tonneau, 1 tonneau =1.1023 tons (2000 Ibs.) 1 g. = 0.03527 oz. Avoir. 1 g. = 0.03215 oz. Troy 1 g. = 15.432 grains Troy 1 Kg. = 2.2046 Ibs. Avoir. 1 Kg.= 2.679 Ibs. Troy SHOP MATHEMATICS 177 DECIMAL EQUIVALENTS OF PARTS OF AN INCH A .01563 A .03125 A .04688 A .0625 A .07813 A .09375 A .10938 i .125 A .14063 A .15625 Ji .17188 A .1875 H .20313 A .21875 if .23438 J .25 H .26563 A .28125 H .29688 A -3125 ft .32813 H .34375 ft .35938 I .375 H .39063 Jf .40625 tt .42188 A -4375 H .45313 H .46875 ft .48438 * .5 H .51563 J} .53125 H .54688 A -5625 ft .57813 H .59375 H .60938 .625 H .64063 ft .65625 If .67188 ft .6875 H .70313 H .71875 tf .73438 I .75 tt .76563 H .78125 ft .79688 H .8125 ft .82813 Ji .84375 if .85938 t .875 H .89063 H .90625 H .92188 tt .9375 tt .95313 H .96875 ft .98438 1 1.00000 USE OF FORMULAS A formula, as used in technical books and papers, may be defined as a rule in which letters and symbols are used in place of numbers or words. A formula, then, is a short way of expressing a rule, and is more convenient, as it shows at a glance the operations to be performed. For example, the volume of a rectangular solid is equal to the product of its three dimensions, length, width, and height. The same statement would be expressed by the formula, V = LWH. The signs of addition, subtraction, multiplication and division are used in formulas in the same way as in arithmetic except that the sign of multiplication between two letters is usually omitted ; as V = LWH means V = L X W X H. In working out or solving a formula, the numerical values of all but one of the letters must usually be known. These values are substituted for their respective letters and the value of the remaining letter may then be easily found. Example. In the formula S=% gt 2 . Let g = 32, t=5. Find S. Solution. By formula, S=$ of 32X5* = 16X25 = 400. Every formula is in reality an algebraic equation, and a knowledge of some of the simpler principles of equations will be of great assistance in working out formulas. The equation may be illustrated by reference to a scale pan with equal arms. If equal weights, say five pounds, are put in each pan, they will just balance. If two pounds is SHOP MATHEMATICS 179 t 3 added to one pan, two pounds must be added to the other pan, and if three pounds is taken from one pan, three pounds must also be taken from the other or the pans will not balance. The same is true of the equation; whatever oper- ation is performed on one side, the same operation must be performed on the other side. The following operations may be performed on an equation without changing its value. I The same number may be added to each side. II The same number may be subtracted from each side. III Each side may be multiplied by the same number. IV Each side may be divided by the same number. V The same root may be taken of each side. VI Each side may be raised to the same power. TRANSPOSING TERMS Example. Five added to five times a certain number is equal to 15. What is the number? Solution. Let x = the unknown number. Then the state- ment of the equation will be 5x-\-5 = 15. But since by (II) 5 may be taken from each side, the state- ment then becomes, 5x = 10. By (IV) x=2. In the above example the statement: 180 SHOP MATHEMATICS 5x-\- 5 = 15 could be changed to read as follows: 5x = 15 5 from which could be obtained the following. RULE : Any term of an equation can be moved from one side to the other by changing its sign. This is called transposing a term. CLEARING OF FRACTIONS Example. The sum of the half and the fifth parts of a number is 7. What is the number? Solution. Let z = the unknown number. Then the statement will be as follows: -+- = 7 2^5 ' By (III) it is possible to multiply both sides of the equa- tion by the same number, in this case 10, which is the L.C.M . of 2 and 5. Then the equation becomes: 5x+2x = 70 7x = 70 By (IV) x = 10 To prove that the result is correct, the value found for x when substituted in the original equation should make both sides identical. Then 10 .10 y+y = 7, or 7 = 7. This is called checking the problem. A coefficient is any factor of the rest of the term; as, in ab, a is the coefficient of 6 and 6 is the coefficient of a, but in an algebraic formula the coefficient is usually understood to SHOP MATHEMATICS 181 mean the numerical coefficient of the given term; as, 5 is the coefficient of x in the expression 5x. Terms that differ only in their numerical coefficients are called similar and may be added or subtracted by finding the sum or difference of then* coefficients and affixing to the result the common letters. Examples. 7a+3a+6a = 16a. 2xm-\-12xm8xm = 6xm. 8abc Sabc = 5abc. The law for precedence of signs. When the signs +, , X, and -v- occur in the same formula, the operations of multipli- cation and division must be performed before those of addi- tion and subtraction; as, 9X36 + 2 = 27-3 = 24. The law of signs for addition. In adding similar terms having + and signs, subtract the less number from the greater (arithmetically) and prefix the sign of the greater number; as, 5x+( 8x) = 3x. The law of signs for subtraction. Change the sign of every term in the subtrahend and then proceed as in addition; as, 5x- (-8x) = 5x+(+8x) = ISx. The law of signs for multiplication. The product of two terms of like signs has the plus sign; the product of two terms of unlike signs has the minus sign; as, (+a)X(+6)=a6. (-a)X(-b)=ab. An exponent denotes the number of times a quantity is used as a factor; as, aXaXa = a 3 . 182 SHOP MATHEMATICS Then multiplying a 2 by a 3 means, (aXa)X(aXaXa) or a used as a factor five times. The result is written a 5 . From the above follows: The law for exponents in multiplication. The exponent of a letter in the product is equal to the sum of its exponents in the factors. The law of signs for division. If the numbers to be divided have the same sign, the sign of the quotient will be plus; if the numbers to be divided have unlike signs the sign of the quotient will be minus; as, The law for exponents in division. The exponent of a letter in the quotient is equal to its exponent in the dividend minus its exponent in the divisor. Example. Divide a 4 by a 2 . c, , .. a 4 aXaXaXa rp f c ^ Solution. = . Two factors of the a- aXa dividend will be canceled by the two factors of the divisor and the quotient = a 2 , or a 4 -r-a 2 = a 4 ~ 2 =a 2 . SHOP MATHEMATICS 183 PROBLEMS INVOLVING USE OF FORMULAS Example. Find the value of V in the formula V=LWH. When L=20, W=15 and H = 5. Solution. Substituting the known values for the letters L, W, and H in the formula: Then V= 20X1 5X5 = 1500. Example. What is the value for H in the formula V= L W H, when V=SOOO, L = 45 and W=20? Solution. Substituting the known values for the letters V, L, and W in the formula: Then 3000 = 45X20XH. The quantities L, W and H are factors of the quantity V. Therefore the values of one of the factors of V must be the quotient obtained by dividing V by the product of the other factors; Then by transposing, B- In like manner any other term can be found, as, V WH LH E 1. G = ~. Find E when C = 10 and R = 5. ti 2. j^ = ~- FindF when D = 10 and L=#0. 3. 7 = f X. Find when V=816 and H=^. o 4. =4 Find C when P = 6, W = 96 and L = 7. 184 SHOP MATHEMATICS 5. PXa = WXb. Find the value of each letter in terms of the other letters. a c,. a. Find a in terms of c and sin A. Also find 6. SmA= . , , . . c cm terms of a and sin A. 7. Tan A =- Find tan A when a = 8 and b = 6. o 8. C = ^-(F-32}. Find F in terms of C. t7 Note. In the solution of this problem the result could be obtained by first taking f of T'' and from this product sub- tracting f of 32; but the parenthesis indicates that the enclosed terms are to be treated as one; 32 should first be subtracted from ^ and then f of the difference should be found. g (pg2} From the formula C = ^ ^ - a formula for F may be derived as follows: Clearing of fractions, 9C = 5 (F - 32) Removing parenthesis, 9C = 5F - 160 Transposing, 5F=9C+160 Dividing by the coefficient ) no_9C . _- of the unknown term, ( 5 The operation of removing the parenthesis from a formula should be performed first and the resulting number treated according to the signs of operation that affect the whole parenthesis; thus, | (3+5) 2 = f (5) 2 = Again 6a-(3b-2c+l}=6a-Sb+2cl. Also, 6a+(3b-2c+l)=6a+Sb-2c+l. From the above it follows: that SHOP MATHEMATICS 185 To remove a parenthesis. When the plus sign precedes it, the signs of all the terms within the parenthesis remain un- changed. But if the minus sign precedes, the sign of every term within the parenthesis must be changed; this is but an illustration of the law of signs in subtraction where the signs of all the terms in the subtrahend are changed before adding. When a parenthesis or other sign of union occurs within a parenthesis, the inner ones are usually removed first; as, 5a+8a+4a+lla+9a= 9. Change 77 F to the equivalent Centigrade reading. 10. Change 18 F to the equivalent Centigrade reading. 11. Change 10 C to the equivalent Fahrenheit reading. 12. Change 10 C to the equivalent Fahrenheit reading. 186 SHOP MATHEMATICS SIMPLE TRIGONOMETRIC FUNCTIONS The right triangle ABC has six parts or elements; three sides a, b, c, and three angles A, B, and (7. When three of these elements are given one of the known parts being a side the three unknown parts can be determined by computation. The process of finding the unknown parts is called solving the triangle. The radios of the sides of the right triangle are called trigonometric functions and are named as follows: The sine of an angle is the ratio of the side opposite the angle to the hypotenuse, as sin A= , sin B = . C C The cosine of an angle is the ratio of the adjacent side to the hypotenuse, as cos A = , cos B = . C C The tangent of an angle is the ratio of the side opposite to the side adjacent, as tan A -J-, tan B = . o a The cotangent of an angle is the ratio of the adjacent side to the opposite side, as cot A = , cot B=-^~. a o The secant of an angle is the ratio of the hypotenuse to C C the adjacent side, as sec A = -j--, sec B = . The cosecant of an angle is the ratio of the hypotenuse C C to the opposite side, as esc A = , esc 5 = -=-. a b The following arrangement of the functions of an angle will be found convenient, and with the tables of natural SHOP MATHEMATICS 187 functions given on pagi necessary for the solutioi (1). Sine (2). Side opposite (3) . Cosine (4). Side adjacent (5). Tangent (6). Side opposite (7) . Cotangent (8). Side adjacent (9) . Hypotenuse ( 1 rf\ TTTrrt on 11 ae> 3s 189 to 193 furnish all the data i of right triangles, side opposite hypotenuse = hypotenuse X sine side adjacent hypotenuse = hypotenuse X cosine side opposite side adjacent = side adjacent X tangent side adjacent side opposite = cotangent X side opposite side opposite sine side adjacent How to Use the Tables Example. If a = 47 ft., and c = 63 ft., find angle A. Solution. By proposition (1) sin A = = = .74603. Re- C oo fer to page 193 in column marked sines at bottom on right, find .74606 just above the figures 48 in the column marked D for degrees and on the same line with figures 15 in column marked M for minutes. It is evident that the angle A is very close to 4$ 15', as the next smaller angle given in the table is 48 0' for which the sine is .74314 and as the machinists' protractor has quarter degrees for the smallest graduations 188 SHOP MATHEMATICS the nearest approximation is taken to the size of the angle as found in the table, that is 48 15' . Example. Find cosine of angle A when A 35 30' . Solution. On page 192 in column headed D on left side, follow down the column until the figure 35 is reached, in the next column headed M on the same line is o, under this is 15 and under 15 is 30 which is the angle required, that is, 35 30' . On this line at the right under column headed cosines (at top) will be found the decimal .81412 which is the cosine of angle 35 3Cf . Note. For angles between and 45 the names of the functions are found at the top of the page and the angles at the left; for all angles between 45 and 90 the names of the functions are found at the bottom of the page and the angles at the right. SHOP MATHEMATICS 189 NATURAL TRIGONOMETRICAL FUNCTIONS D M Sines Cosines Tangents Cotangents Secants Cosecants .00000 1.0000 .00000 Infinite I. 0000 Infinite 90 15 .00436 .99999 .00436 229 . 182 1.0000 229 . 18 45 30 .00873 .99996 .00873 114.589 1.0000 114.59 30 45 .01309 .99991 .01309 76.3900 1.0001 76.397 15 1 .01745 .99985 .01746 57.2900 1.0001 57.299 89 15 .02181 .99976 .02182 45.8294 1.0002 45.840 45 30 .02618 .99966 .02619 38.1885 1.0003 38.202 30 45 .03054 . 99953 .03055 32.7303 1.0005 32 . 746 15 2 .03490 .99939 .03492 28.6363 1.0006 28.654 88 15 .03926 .99923 .03929 25.4517 1.0008 25.471 45 30 .04362 .99905 .04366 22.9038 1.0009 22.926 30 45 .04798 .99885 .04803 20.8188 1.0011 20.843 15 3 .05234 .99863 .05241 19.0811 1.0014 19.107 87 15 .05669 .99839 .05678 17.6106 1.0016 17.639 45 30 .06105 .99813 .06116 16.3499 1.0019 16.380 30 45 .06540 .99786 .06554 15.2571 1.0021 15.290 15 4 .06976 .99756 .06993 14.3007 1.0024 14.336 86 15 .07411 .99725 .07431 13.4566 1.0028 13.494 45 30 .07846 .99692 .07870 12.7062 1.0031 12.745 30 45 .08281 .99657 .08309 12.0346 1.0034 12.076 15 5 .08716 .99619 .08749 11.4301 1.0038 11.474 85 15 .09150 .99580 .09189 10.8829 1.0042 10.929 45 30 .09585 .99540 .09629 10.3854 ! 1.0046 10.433 30 45 .10019 .99497 . 10069 9.93101 1.0051 9.9812 15 6 . 10453 .99452 .10510 9.51436 1.0055 9.5668 84 15 . 10887 .99406 . 10952 9.13093! 1.0060 9 . 1855 45 30 .11320 .99357 .11394 8.77689 1.0065 8.8337 30 45 .11754 .99307 .11836 8.44896J 1.0070 8.5079 15 7 .12187 .99255 . 12278 8.14435; 1.0075 8 . 2055 83 15 .12620 .99200 .12722 7.86064: 1.0081 7.9240 45 30 . 13053 .99144 .13165 7.59575! 1.0086 7.6613 30 45 . 13485 .99087 . 13609 7. 34786 i 1.0092 7.4156 15 8 .13917 .99027 . 14054 7. 11537 | 1.0098 7.1853 82 15 . 14349 .98965 . 14499 6.89688; 1.0105 6.9690 45 30 . 14781 .98902 . 14945 6.69116; 1.0111 6 . 7655 30 45 .15212 .98836 . 15391 6. 49710 i 1.0118 6.5736 15 9 .15643 .98769 .15838 6.31375! 1.0125 6.3924 81 15 . 16074 .98700 . 16286 6.14023, 1.0132 6.2211 45 30 .16505 .98629 .16734 5.975761 1.0139 6.0589 30 45 . 16935 .98556 .17183 5.81966 1.0147 5.9049 15 10 .17365 .98481 .17633 5.67128! 1.0154 5 . 7588 80 Cosines Sines Cotangents Tangents Cosecants Secants D M From 80 to 90 read from bottom of table upwards. 190 SHOP MATHEMATICS NATURAL TRIGONOMETRICAL FUNCTIONS D M Sines Cosines Tangents Cotangents Secants Cos ecants 10 .17365 .98481 .17633 5.67128 1.0154 5.7588 80 15 .17794 . 98404 . 18083 5.53007 1.0162 5.6198 I 45 30 .18224 .98325 . 18534 5.39552 1.0170 5.4874 30 45 . 18652 .98245 . 18986 5.26715 1.0179 5.3612 15 11 . 19081 .98163 .19438 5.14455 1.0187 5.2408 79 15 . 19509 .98079 .19891 5.02734 1.0196 5.1258 45 30 . 19937 .97992 . 20345 4.91516 1.0205 5.0158 30 45 .20364 .97905 . 20800 4.80769 1.0214 4.9106 15 12 .20791 .97815 .21256 4.70463 1.0223 4.8097 78 15 .21218 .97723 .21712 4.60572 1.0233 4.7130 45 30 .21644 .97630 .22169 4.51071 1.0243 4.6202 30 45 .22070 .97534 . 22628 4.41936 1.0253 4.5311 15 13 . 22495 .97437 . 23087 4.33148 1.0263 4.4454 77 15 .22920 .97338 .23547 4.24685 1.0273 4.3630 45 30 . 23345 .97237 .24008 4.16530 1.0284 4.2837 30 45 .23769 .97134 .24470 4.08666 1.0295 4.2072 15 14 .24192 .97030 . 24933 4.01078 1.0306 4.1336 76 15 .24615 .96923 .25397 3.93751 1.0317 4.0625 45 30 .25038 .96815 . 25862 3.86671 1.0329 3.9939 30 45 .25460 .96705 .26328 3.79827 1.0341 3.9277 15 15 . 25882 .96593 .26795 3.73205 1.0353 3.8637 75 15 .26303 .96479 .27263 3.66796 1.0365 3.8018 45 30 .26724 .96363 . 27732 3 . 60588 1.0377 3 . 7420 30 45 .27144 .96246 .28203 3.54573 1.0390 3.6840 15 16 . 27564 .96126 . 28675 3.48741 1.0403 3.6280 74 15 .27983 .96005 .29147 3.43084 1.0416 3.5736 45 30 . 28402 .95882 .29621 3.37594 1.0429 3.5209 30 45 .28820 .95757 .30097 3.32264 1.0443 3.4699 15 17 .29237 .95630 .30573 3.27085 1.0457 3.4203 73 15 . 29654 .95502 .31051 3 . 22053 1.0471 3.3722 45 30 .30071 . 95372 .31530 3.17159 1.0485 3.3255 30 45 .30486 .95240 .32010 3.12400 1.0500 3.2801 15 18 .30902 .95106 .32492 3.07768 1.0515 3.2361 72 15 .31316 .94970 .32975 3.03260 1.0530 3 . 1932 45 30 .31730 .94832 .33460 2.99868 1.0545 3.1515 30 45 .32144 .94693 .33945 2.94591 1.0560 3.1110 15 19 .32557 .94552 .34433 2.90421 1.0576 3.0715 71 15 .32969 .94409 .34922 2 . 86356 1.0592 3.0331 45 30 .33381 .94264 .35412 2.82391 1.0608 2.9957 30 45 .33792 .94118 .35904 2.78523 1.0625 2.9593 15 20 .34202 .93969 .36397 2.74748 1.0642 2.9238 70 Cosines Sines Cotangents Tangents Cosecants Secants D M From 70 to 80 read from bottom of table upwards. SHOP MATHEMATICS 191 NATURAL TRIGONOMETRICAL FUNCTIONS D M Sines Cosines Tangents Cotangents Secants Cosecants 20 .34202 .93969 .36397 2.74748 1.0642 2.9238 70 15 .34612 .93819 .36892 2.71062 1.0659 2.8892 45 30 .35021 .93667 .37388 2.67462 1.0676 2.8554 30 45 .35429 .93514 .37887 2.63945 1.0694 2.8225 15 21 .35837 .93358 .38386 2.60509 1.0711 2.7904 69 15 .36244 .93201 .38888 2.57150 1.0729 2.7591 45 30 .36650 .93042 .39391 2.53865 1.0748 2.7285 30 45 .37056 .92881 .39896 2.50652 1.0766 2.6986 15 22 .37461 .92718 .40403 2.47509 1.0785 2.6695 68 15 .37865 .92554 .40911 2.44433 1.0804 2.6410 45 30 .38268 .92388 .41421 2.41421 1.0824 2.6131 30 45 .38671 .92220 .41933 2.38473 1.0844 2.5859 15 23 .39073 .92050 .42447 2.35585 1.0864 2.5593 67 15 .39474 .91879 .42963 2.32756 1.0884 2.5333 45 30 .39875 .91706 .43481 2.29984 1.0904 2.5078 30 45 .40275 .91531 .44001 2.27267 1.0925 2.4829 15 24 .40674 .91355 .44523 2.24604 1.0946 2.4586 66 15 .41072 .91176 .45047 2.21992 1.0968 2.4348 45 30 .41469 .90996 .45573 2.19430 1.0989 2.4114 30 45 .41866 .90814 .46101 2.16917 1.1011 2.3886 15 25 .42262 .90631 .46631 2.14451 1 . 1034 2.3662 65 15 .42657 .90446 .47163 2.12030 1 . 1056 2.3443 45 30 .43051 .90259 .47698 2.09654 1 . 1079 2.3228 30 45 .43445 .90070 .48234 2.07321 1.1102 2.3018 15 26 .43837 .89879 .48773 2.05030 .1126 2.2812 64 15 .44229 .89687 .49315 2.02780 .1150 2.2610 45 30 .44620 .89493 .49858 2.00569 .1174 2.2412 30 45 .45010 .89298 .50404 1.98396 .1198 2.2217 15 27 .45399 .89101 . 50953 1.96261 .1223 2 . 2027 63 15 .45787 .88902 .51503 1.94162 .1248 2.1840 45 30 .46175 .88701 .52057 1.92098 .1274 2.1657 30 45 .46561 .88499 .52613 1.90069 .1300 2.1477 15 28 .46947 .88295 .53171 1.88073 .1326 2 . 1300 62 15 .47332 .88089 .53732 1.86109 1 . 1352 2.1127 45 30 .47716 .87882 .54296 1.84177 1.1379 2.0957 30 45 .48099 .87673 . 54862 1.82276 1 . 1406 2.0790 15 29 .48481 .87462 .55431 1.80405 1 . 1433 2.0627 61 15 .48862 .87250 .56003 1.78563 1.1461 2.0466 45 30 .49242 .87036 .56577 1.76749 1 . 1490 2.0308 30 45 .49622 .86820 .57155 1 . 74964 1.1518 2.0152 15 30 .50000 .86603 .57735 1.73205 1 . 1547 2.0000 60 Cosines Sines Cotangents Tangents Cosecants Secants D M From 60 to 70 read from bottom of table upwards. 192 SHOP MATHEMATICS NATURAL TRIGONOMETRICAL FUNCTIONS D M Sines Cosines Tangents Cotangents Secants Cosecants 30 .50000 .86603 .57735 .73205 1.1547 2.0000 60 15 .50377 .86384 .58318 .71473 1.1576 1.9850 45 30 .50754 .86163 .58904 .69766 1.1606 1.9703 30 45 .51129 .85941 .59494 .68085 1.1636 1.9558 15 31 .51504 .85717 .60086 .66428 1.1666 1.9416 59 15 .51877 .85491 .60681 .64795 1.1697 1.9276 45 30 .52250 .85264 .61280 .63185 1.1728 1.9139 30 45 .52621 .85035 .61882 .61598 1.1760 1.9004 15 32 .52992 .84805 .62487 .60033 1.1792 1.8871 58 15 .53361 .84573 .63095 1.58490 1 . 1824 1.8740 45 30 .53730 .84339 .63707 1.56969 1.1857 1.8612 30 45 .54097 .84104 .64322 1.55467 1.1890 1.8485 15 33 .54464 .83867 .64941 1.53986 1.1924 1.8361 57 15 .54829 .83629 .65563 1.52525 1.1958 1.8238 45 30 .55194 .83389 .66188 1.51084 1.1992 1.8118 30 45 .55557 .83147 .66818 1.49661 1.2027 1.7999 15 34 .55919 .82904 .67451 1.48256 1.2062 1.7883 56 15 . 562SO .82659 .68087 1.46870 1.2098 1.7768 45 30 .56641 .82413 .68728 1.45501 1.2134 1.7655 30 45 .57000 .82165 .69372 1.44149 1.2171 1.7544 15 35 .57358 .81915 .70021 1.42815 1.2208 1 . 7434 55 15 .57715 .81664 .70673 1.41497 1.2245 1.7327 45 30 .58070 .81412 .71329 1.40195 1.2283 1.7220 30 45 .58425 .81157 .71990 1.38909 1.2322 1.7116 15 36 .58779 .80902 .72654 1 .37638 1.2361 1.7013 54 15 .59131 .80644 .73323 1.36383 1.2400 1.6912 45 30 .59482 .80386 .73996 1.35142 1.2440 1.6812 30 45 .59832 .80125 .74673 1.33916 1.2480 1.6713 15 37 .60181 .79864 .75355 1.32704 1.2521 1.6616 53 15 .60529 .79600 .76042 1.31507 1.2563 1.6521 45 30 .60876 .79335 .76733 1.30323 1.2605 1.6427 30 45 .61222 .79069 .77428 1.29152 1.2647 1.6334 15 38 .61566 .78801 .78129 1.27994 1.2690 1.6243 52 15 .61909 .78532 .78834 1.26849 1.2734 1.6153 45 30 .62251 .78261 .79543 1.25717 1.2778 1.6064 30 45 .62592 . 77;-'8< .80258 1.24597 1.2822 1.5976 15 39 .62932 .77715 .80978 1.23490 1.2868 1.5890 51 15 .63271 .77439 .81703 1.22394 1.2913 1.5805 45 30 .63608 .77162 .82434 1.21310 1.2960 1.5721 30 45 . 63944 .76884 .83169 1.20237 1.3007 1.5639 15 40 .64279 .76604 .83910 1 . 19175 1.3054 1.5557 .50 Cosines Sines Cotangents Tangents Cosecants Secants D M From 50 to 60 read from bottom of table upwards. SHOP MATHEMATICS 193 NATURAL TRIGONOMETRICAL FUNCTIONS M Sines Cosines Tangents Cotangents Secants Cosecants 40 .64279 . 76604 .83910 1.19175 1.3054 1.5557 50 15 .64612 .76323 .84656 1.18125 1.3102 1.5477 45 30 .64945 . 76041 .85408 1 . 17085 1.3151 1.5398 30 45 .65278 .75756 .86165 1.16056 1.3200 1.5320 15 41 .65606 . 75471 .86929 1 . 15037 1.3250 1.5242 49 15 . 65935 .75184 .87698 1 . 14028 1.3301 1.5166 45 30 .66262 . 74896 .88472 1.13029 1.3352 1.5092 30 45 .66588 .74606 .89253 1.12041 1.3404 1.5018 15 42 .66913 .74314 .90040 1.11061 1.3456 1.4945 48 15 .67237 .74022 .90834 1 . 10091 1.3509 1.4873 45 30 . 67559 . 73728 .91633 1.09131 | 1.3563 1.4802 30 45 .67880 .73432 .92439 1.08179 1.3618 1.4732 15 43 .68200 : .73135 .93251 1.07237 1.3673 1.4663 47 15 .68518 .72837 .94071 1.06303 1.3729 1.4595 45 30 .68835 j .72537 .94896 1.05378 1.3786 1.4527 30 45 .69151 . 72236 .95729 1.04461 1.3843 1.4461 15 44 . 69466 .71934 .96569 1.03553 1.3902 1.4396 46 15 .69779 .71630 .97416 1.02653 1.3961 1.4331 45 30 .70091 .71325 .98270 1.01761 1.4020 1.4267 30 45 .70401 .71019 .99131 1.00876 1.4081 1.4204 15 45 .70711 .70711 1.00000 1.00000 1.4142 1.4142 45 Cosines Sines Cotangents Tangents Cosecants Secants D M From 45 to 50 read from bottom of table upwards. FORMULAS volts X amperes _. , . . H.P. = - - . Electric horse power. Wv 2 K = . Kinetic energy ......... 3 PXPa=WxWa. Law of lever . . . , . . 8 PXcos A = WXsin A. The moving strut. ... 9 PXcos A = WX sin A. The toggle joint. ... 9 ti P= - -. The compound lever. . 10 PaXP^XPa^ PXR=WXr. The wheel and axle ...... 10 -. The differential windlass ... 11 P= 75 ^ 5-^. The law for trains of wheels and K X -ft/! X /v 2 axles ............. 12 t, _, . . , . , . .. n = ' -. The law for driver and follower r X-T j pulleys and gears .......... 12 = PXN. The pulley block ........ 15 P= . The differential pulley ..... 16 K Px L W 77 . The inclined plane when P is parallel to n plane ............. 23 SHOP MATHEMATICS 195 Page W = 77. The inclined plane when P is parallel to n base. . ........... 23 P X cos Y W ' = - : - . The inclined plane when P is at an sin a: angle to incline .......... 24 rR . The screw. ....... 25 j Li , The wedge .......... 26 S=T-~. Tap size drill for V threads. ... 29 N S = 7 1 ^f. Tap size drill for U. S. S. threads. 30 N CXR. P. M.Xb _ H. P. = . The horse power for pulleys. 48 oOO p /= . The coefficient of friction ....... 53 " p -= u ^ x/x .v x d x .ooooos. Axle friction ........... 54 W = F(1 +0.9 +0.9 2 +0.9 3 +0.9 4 ). The efficiency of the pulley ............ 56 sin A+(fXcosA) . .. p = W X - tt , TT The inclined plane parallel cos A (fXsin A) to base. ........... 57 g- The S( 5 uare thread R screw ............. 58 thread so,.. 58 196 SHOP MATHEMATICS . Length of belt. ' ...... 62 <& , H. P. X 50,000 . b = j 7 . Width of belt ....... 64 dXwXN D 3X33,OOOXH.P _ r = y . Pressure on bearings. ... 64 H. P. = d 2 xVx.0025. Horse power transmitted by ropes ........... * . 68 H. P. = d z x Vx.055. Horse power transmitted by wire cables ............ 70 A P D = =. Diameter of chain sprockets. 72 sin 1 . Diameter of shafting 73 L=6\/d. Length of journal bearings. ..... 76 & = . t = . Size of machine keys 82 4 3 x = L X -T?- The micrometer screw. 87 N ( - I . Offset of footstock center for I turning tapers 101 C = . The cutting speed in F. P. M. for lathe turning. 104 R = j-. Simple indexing, milling machine. . . . 114 R F = . Gearing for spiral head 124 SHOP MATHEMATICS 197 x R.P.M. = jr. Speeds for drills. . . . . 129 WH F=^. Hammer blow. . . .130 WXPXR. P. M.X*d f -' ^ 33,000X12 ' machines ......... ... 132 H. P. = CxW. Horse power of machines. . . . 132 H. P. = . H. P. measurement by the dyna- 00,000 mometer ............ 136 F=WXRXN 2 X. 000341 . Centrifugal force in fly wheels ............. 138 5 = W XRXN 2 X. 00005427. The tensile strain in fly wheels. 139 LJ H H B ir H P H n vN iOV . 141 . 145 . 154 . 158 . 160 . 161 . 165 1A7 oo ,000 D D * N ^ * <4.t? 2t T c a a p .434 = H A W Water pressure in tanks. WxH P- QQ ooo' P um P ln g watci to a given height. V in F. P. M. ANSWERS Mechanics, Page 3 1. 1,200 ft.-lbs. 2. 7,500 ft.-lbs. 3. 80,000 ft.-lbs. 4. 2,083,350 ft.-lbs. 5. 3,150 ft.-lbs. 6. 630 ft.-lbs. 7. 18,000 ft.-lbs. 8. 1\ ft.-lbs. 9. 2,800 ft.-lbs. 10. 139.9 ft.-lbs. 11. 10.52 H. P. 12. 75.76 H. P. 17. 13. 1.06 H. P. 14. 1.35 H. P. 15. 140,000 ft.-lbs. 16. 7,500,000 ft.-lbs. V=400ft. per sec. K = 30,000 ft.-lbs. 18. 14.75 H. P. 19. 22.12 H. P. 20. 10.05 H. P. 21. 18.43 H. P. 22. 200.38 H. P. Miscellaneous Problems, Page 16 1. 125 Ibs. 2. 825 Ibs. 3. 900 Ibs. 4. 5 in. 5. 25.117 in. 6. 1,024 Ibs. 7. 164^ Ibs. 8. 264.49 Ibs. 9. 1,925.86 Ibs. 10. 1,417.25 Ibs. 11. 2,115.94 Ibs. 12. 8,258.96 Ibs. 13. 0.0004 in. 14. 0.0442 in. 15. 373lbs. 16. 3,937^ Ibs. 17. 1,200 Ibs. 18. 4,500 Ibs. 19. 25 in. 20. 222f Ibs. 21. 30 Ibs. 22. 0.026 dia. circle 23. 12^ in. 24. 4.8 in. 25. 27| in. 26. 125 Ibs. SHOP MATHEMATICS 199 27. 1.26 in. 28. 1| in. 29. SS^lbs. 30. 390| Ibs. 31. A ft. 32. 15f 33. 24,000 R. P. M. 34. 25 in. dia. 35. ft rev - 36. T V rev. 37. AW rev. 38. 72 39. 63f 40. 84 41. 68$ 42. 93 43. 81f 44. 69 45. 82f 46. 4 sheaves 47. 75 Ibs. 48. 120 Ibs. 49. 1,020 Ibs. 50. 1,105 Ibs. 51. 92.2% 52. 150 Ibs. 53. 4 men 54. 10 in. and 9f in. Problems, Page 27 1. 118.477 Ibs. 2. 464.84 Ibs. 3. 1,118.3 Ibs. 4. 45,239.04 Ibs. 5. 71.62 Ibs. 6. 150.8 Ibs. 7. 16$ Ibs. 8. 5,000 Ibs. 9. 750 Ibs. 10. 333^ Ibs. 12. 11. 29.47 Ibs. 2} ft. ) 2f ft. / 13. 5 14. 9 15. 437 \ Ibs. 16. 6.631 in. 17. 7.957 in. 18. 644.33 Ibs. 19. 900.68 Ibs. 3 to 4 Screw Threads, Page 30 1. 0.0722 in. 5. 0.8375 in. 2. 0.0481 in. 6. 0.1634 in. 3. 0.0542 in. 7. 0.3933 in. 4. 0.1164 in. 8. 0.0139 in. 200 SHOP MATHEMATICS 9. 0.5913 in. 10. 0.5068 in. 11. 0.622 in. 12. 0.3668 in. 13. 0.0813 in. 14. 0.0866 in. 15. 16 pitch double 17. 12 pitch triple Gears, Page 34 1. 0.3927 in. 2. 0.5236 in. 3. 4 pitch 4. 10 pitch 5. 18^ in. 6. 4 pitch 7. 108 teeth 8. 75 teeth 9. 0.3927 in. 10. 0.3925 in. 11. 0.5392 in. 12. 0.3595 in. 13. 2i in. 14. 3 in. 15. 4 pitch 16. 128 teeth 17. 640 teeth 18. 200 teeth 8 in. 16 in. 8 in. 16 in. 21 in. 35 in. 19. 20. 21. 22. 23. 24. 24 in. 30 in. 32 in. 40 in. 24 in. 36 in. 25. 0.0393 in. 26. 0.0098 in. 27. 18 in. 28. 6.6759 in. 29. 4 in. 30. 10 in. 31. 21.008 R. P. M. 32. 4166J R. P. M. 132 F. P. M. . 157.08 F. P. M. 34. 125 Ibs. 35. 166j 36. 13.368 R. P. M. 37. 50.93 38. 3H Ibs. 39. 176.84 R, P. M. 33. SHOP MATHEMATICS 201 Bevel Gears, Page 40 D O D 40 60 6| in. 10 in. 6.944 in. . 10.185 in. Cutting Angles ( 31 51' , Index head ( 54 29' !33 41' 56 19' Increment 1 35' Decrement 1 50' Center face (70 32' angles of blank \ 115 48' N 40 D4in. O D 4.1414 in. Cutting angle to f set head I Angle C 45 Center face angle of blank Increment 2 1' Decrement 2 15' 45' 94 2' X D 96 120 12 15 Cutting Angles to ( 37 49' set index head \ 50 29' 2Q , , P Angles G! and C < 5 Increment 45' Decrement 51' Center face (77 4' angles of blank ) 104 30' o 54 5 in. 6in. 28 43' 56 47' f 4.234 in. ' \ 6.795 in. Cutting angles 4 {30 58' 59 2' Center face f 65 37' angles blank 1 121 45' Decrement 2 15' Increment 1 50V 202 SHOP MATHEMATICS X D O D 5. 30 40 5 in. 6| in. 5.267 in. 6.867 in. '34 7' 50 23' 36 52' 53 8' Center face f 78 20' angles of blank 1 110 52' Increment 2 18' Decrement 2 45' D O D Cutting angles Angles G! and C 6. f NJ 12 1 150 15 in. 18f in. 15.195 in. 18.906 in. 38 0' 50 40' 38 40' 51 20' Center face f 78 30' angles of blank \ 103 50' Increment 35' Decrement 40' Cutting angles Angles G! and C 1. 18,000 Ibs. 2. 11 in. 3. 4f in. Worm Wheels, Page 42 4. 47,250 Ibs. 5. 189,000 Ibs. Helical Gears, Page 46 1. 2, 3. D 3.677 in. O D 4.077 in. D 1.414 in. D 1.614 in. D 2.121 in. O D 2.371 in. /Centers 2.357 in. \O D 2.524 in. 5. 1.414 in. 6. No teeth 11 and 22 D for both gears 1.754 Pulleys, Page 49 1. 2,040 R. P. M. 2. 12,600 R. P. M. 3. 0.167 in. 4. 25f in. 5. 40 in. 6. 3.825 in. 7. 8.325 in. 8. 11.156 in. SHOP MATHEMATICS 203 ft 0.205 in. 'N6 9. J c 0.125 in. h 4.332 in. [T 0.535 in. e 1.733 in. 10. 3.287 in. 11 . < t 0.5625 in. fb 13.82 in. T 1.375 in. 11. J c 0.250 in. D t 8 in. [s 0.375 in. .L 10.667 Miscellaneous Problems, Page 49 flead 11.107 in. 7 . 75,398.4 Ibs. 1. J D 3.535 in. 8. 150,796.8 Ibs. lO D 3.635 in. 9 . 75,398.4 Ibs. 2. flead 13.329 in. J D 4.243 in. [ O D 4.443 in. 10 11 12 13 . 628,320 Ibs. . 6.615 Ibs. . 203,575.68 Ibs. . 339,292.8 Ibs. flead 8.650 in. 14 . 73.68 Ibs. 3. J D 2.946 in. 15 . 63,617.4 Ibs. [0 D 3.113 in. 16 . 49.7 Ibs. 4. 16,000 Ibs. 17 . 18.65 Ibs. 5. 20,000 Ibs. 18 . 848,232 Ibs. 6. 64,000 Ibs. 19 . 37.30 Ibs. Friction, Page 55 1. 2.842 H. P. | 2 . 13.271 H. P. Miscellaneous Problems, Page 58 1. 488.39 Ibs. 5 . .08 2. 541.18 Ibs. 6 . 120 Ibs. 3. 474.27 Ibs. 7 . 853.85 Ibs. 4. 0.125 8 . 584.96 Ibs. 204 SHOP MATHEMATICS 9. 5.6 H. P. 10. 148f Ibs. 11. 542.87 Ibs. 12. 92.67 Ibs. 13. 79.09 Ibs. 14. 209.5 Ibs. 15. 2,282.95 Ibs. 16. 16.76 Ibs. 17. 33.36 Ibs. 18. 8,948.02 Ibs. 19. 1,300.52 Ibs. 20. 1.7 H. P. 21. 8.606 H. P. 22. 543.235 ft.-lbs. per min. 23. 360 ft.-lbs. per min. Belting, Page 64 1. 100 H. P. 2. 81.92 H. P. 3. 96.77 H. P. 4. 34.996 ft. 5. 43.764 ft. 6. 17^ H. P. 7. 20.36 H. P. 8. 125.61 H. P. 9. 91.48 H. P. 10. 139.58 Ibs. 11. 69.9ft. 12. 41.5 ft. 13. 49.3 ft. 14. 49.25 15. 24.3 in. 16. 192 H. P. 17. 163.84 H. P. 18. 6,600 Ibs. 19. 756.3 Ibs. 20. 8.25 Ibs. 21. 247.5 Ibs. 22. 66f in. 23. 71.42 H. P. 24. 408.23 H. P. 25. 54.1 H. P. 26. 112.64 H. P. 27. 4.4 in. 28. 43.4 in. 29. 62 \ in. 30. 20+ oz. 31. 26+ oz. 32. 258 R. P. M. 33. 95 R. P. M. 34. 11 1.5 in. dia. 35. 52 in. 36. 30 oz. 37. 34 oz. 38. 15.8 oz. SHOP MATHEMATICS 205 1. 187ilbs. 2. 0.675 Ibs. 3. 5,301.45 Ibs. 4. 2,650.725 Ibs. 5. 7,215.86 Ibs. 6. 14.73 H. P. Rope Drives, Page 68 7. 105.53 Ibs. 8. 1,100 Ibs. 9. 42.11 H. P. f 23.11 H. P. 10. J 205 + R. P. M. [ 65 in. 11. 1,804 Ibs. 1. 1,005 Ibs. 158.34 H. 2. 43.53 H. P. 399 Ibs. 54 H. P. 'I 3 Wire Cables, Page 70 P. 4. 205.19 H. P. 5. 2,873.15 F. P. M. 7.41 ft. dia. 1,110ft. 6. 1. 2. 3. 4. o. P D 2.565 in. D 2.890 in. O D 6.699 in. B D 6.049 in. A 0.9 in. B 0.6 in. b 0.4875 in. 9.561 in. 10.049 in. B D 10.974 in. P D 11.461 in. O D 11.949 in. fpD \0 D Chains, Page 72 I A 1.05 in. 6. < B 0.7 in. [b 0.569 in. IB D 15.038 in. P D 15.607 in. O D 16.176 in. ( P D 8.917 in. I O D 9.242 in. 9. 9.229 in. 206 SHOP MATHEMATICS Shafting, Page 74 d 3 N HX80 10. It in. 80 ' cl 3 11. 5.4 in. 2. 120 H. P. 12. 13.05 in. 3. 39 H. P. 13. 4iin. 4. 429.7 H. P. 14. 736 H. P. 5. 6.4 in. 15. 405 H. P. 6. 6i in. 16. 3,645 H. P. 7. 46+ R. P. M. 17. 31+ R. P. M. 8. 4.3 in. 18. 592 + R. P. M 9. 192+ R. P. M. 19. 308 + R. P. M Jack Shaft, Page 76 1. 4.93 in. 6. 16| H. P. 2. 3.92 in. 7. 22.14 H. P. 3. 277.8 R. P. M. 8. 7.1 in. 4. 1,312.5 R. P. M. 9. 10ft. 5. 450 H. P. 10. 10ft. Journal Bearings, Page 79 1. 1.226 in. 12. 13.991 in. 2. 1.462 in. 13. 15.223 in. 3. 1.087 in. 14. 7.111 in. 4. 7.306 in. 15. 2.507 in. 5. 1.515 in. 16. 11.567 in. 6. 2.915 in. 17. 12.219 in. 7. 9.367 in. 18. 3 pulleys 8. 8.352 in. 19. 10.727 in. 9. 10.283 in. 20. 16.121 in. 10. 11. 124 in. 21. 5 pulleys 11. 4 in. 22. 8.67 in. SHOP MATHEMATICS 207 23. 20 in. 24. 14.29 in. 25. 1.82 in. 26. 0.17 in. 27. iin. 28. 10 balls 1. 29. 1.633 in. 30. 8 balls 31. .126 in. 32. iin. 33. .343 in. 34. .109 in. Machine Keys, Page 82 ,b 1 & 1 in. 1 1 1 & .833 in. i b % in. ' t .333 in. 3. Micrometer, Page 88 1. &in. 6. | 2. TsW in- 7. 250 divisions 3. stfcr in. . 8. T $ v in. 4. i in. 9. T 5 5. f 10. 50 divisions 11. 75 divisions Lathe Work, Page 92 1. 97+R. P. M. 2. 173+ R. P. M. 3. 103+ R. P. M. 4. 571 + R. P. M. 5. 82+ R. P. M. 6. 529 + R. P. M. Back Gears, Page 94 1. 12+ R. P. M. 2. 32+ R. P. M. 3. 27 + R. P. M. 4. 15+ R. P. M. 5. 13 + R, P. M. 6. 19+ R. P. M. 208 SHOP MATHEMATICS 1. Ratio of 2. 60 teeth 3. 25 teeth 4. 72 teeth Screw Cutting, Page 97 5. 66 teeth 6. 48 teeth 7. 24 teeth 8. 64 teeth 10. 9. Ratio of f Ratio of 1. 7| in. 2. 1 in. per ft. 3. 2^ in. per ft. 4. s\ in. per ft. 5. I'j- in. per ft. 6. 0.64 in. per ft. 7. 0.603 in. per ft. 8. 0.706 in. per ft. Taper Turning, Page 102 9. 0.6 in. per ft. 10. 3i in. 11. 0.411 in. 12. 1.44 in. 13. 0.283 in. 14. 0.833 in. 15. 0.419 in. 16. 0.1(32 in. 17. 0.609 in. per ft. Miscellaneous Lathe 1. 30 R. P. M. 2. 80 R. P. M. 3. 16 R. P. M. 4. 426 + R. P. M. 5. 32 R. P. M. 6. 40 R. P. M. 7. 46+ R. P. M. 8. 256 R. P. M. 9. 80 R. P. M. Problems, Page 105 10. 19+ R. P. M. 11. 10 R. P. M. 12. 61+ R. P. M. 13. 16 R. P. M. 14. 95+ R. P. M. 15. 466 + R. P. M. 16. 34 R. P. M. 17. 32 R. P. M. 18. 13.716+ in. SHOP MATHEMATICS 209 19. ( 16.266 in. ( 15.533 in. approx. 22. 0.513 in. 23. 0.389 in. 20. ( 26.297 in. 24. 114+ R. P. M. ( 25.112 in. approx. 25. 8 R. P. M. 21. 10.412 in. Planer, Page 109 1. 17.864 F. P. M. 6. 54+ min. 2. 46.584 F. P. M. 7. 34+ min. 3. 22.317 F. P. M. 8. V 4. 16.788 F. P. M. n ( 24" in. dia. 5. 29 + min. 9 \ (18 in. dia. Milling Machine, Page 115 Simple Indexing 1. 13 1 turns 10. i^turn 2. H H. H 3. 10 12. i? 4. If 13. H 5. Ij 14. i 6. 15. A 7. $ 16. i 8. H 17. A .18. A Compound Indexing, Page 118 OfJ 30 30 r\r 94. 13 3 r>r \). Jy ^^ Ul ^t. ^T 4^ Ol A+A H A 21. A+H 25. &&OT 22. A+A or H M t A 26. ii A 23. A+A or 27. H-T& 28. 210 SHOP MATHEMATICS Differential Indexing So many combinations possible that answers* are omitted. 1. Ratio of J 2. Ratio of y 3. Ratio of | 4. Ratio of V 5. Ratio of V 6. 9.524 in. 7. 36 in. 8. 2i in. 9. 21 26' 10. 27 38' 34 30' 18 36' 13 15' 14. 16.327 in. 15. 31.416 in. 11. 12. 13. 16. The Spiral Head, Page 125 , 18 58' 6 min. 17. 41 + R. P. M. 18. 21 + R. P. M. 19. 40 R. P. M. 20. 80 R. P. M. 21. 192 R. P. M. 22. 128 R. P. M. 23. 80 R. P. M. 24. 86 + R. P. M. 25. 18.54 F. P. M. 50.4 F. P. M. 35+ R. P. M. 27. 0.617 in. 28. 0.278 in. 29. 1.867 in. 26. 1. 6. Drill Press, Page 129 293 R. P. M. 41 sec. 2. 0.64 in. 3. 57 sec. 4. 47 sec. 52+ sec. 125+ R. P. M. 5. 1 min. 22 sec. 220 R. P. M. 7. 4 min. 12 sec. 8. 4 min. 4 sec. 9. 4 min. SHOP MATHEMATICS 211 1. 288,000 Ibs. [329,143 Ibs. 2. 1 384,000 Ibs. [460,800 Ibs. 3. 276,000 Ibs. Hammer Blow, Page 131 4. 326,400 Ibs. 5. 20,000 Ibs. 6. 288,000 Ibs. 7. 10,000 Ibs. 8. 20,000 Ibs. 13,125 Ibs. Horse Power of Machines, Page 133 1. 0.589 H. P. 10. 0.551 H. P. 2. 0.929 H. P. 11. 0.506 H. P. 3. 0.524 H. P. 12. 1.875 H. P. 4. 0.220 H. P. 13. 2.142 H. P. 5. 0.23 H. P. 14. 2.076 H. P. 6. 1.805 H. P. 15. .785 H. P. 7. 1.106 H. P. 16. .880 H. P. 8. 0.413 H. P. 17. 1.276 H. P. 9. 0.743 H. P. 18. 2.856 H. P. 1. 281 +R. P. M. 2. 18.75 H. P. Dynamometers, Page 136 3. 25 H. P. 4. 2.801 ft. 5. 50 Ibs. Fly Wheels, Page 139 1. 19,432.7 Ibs. 2. 414.6 Ibs. 3. 13,063 + Ibs. 4. 7,124+ Ibs. 5. 14,563 + Ibs. 6. 9,167+ Ibs. 7. 77,676 + Ibs. 8. 3,607 + Ibs. 9. 9,770 + Ibs. 10. 210+ R. P. M. 11. 480+ R. P. M. 12. 136+ R. P. M. 13. 5,654.88 F. P. M. 14. 940.368 F. P. M. 212 SHOP MATHEMATICS Horse Power of 1. 2(5:5.2 H. P. 2. 215.0 H. P. :;. 177.4 H. P. 4. 719.7 II. P. 5. 0.476 H. P. porlb. M. E. P. 0. 26.00 H. P. porlb. M. E. P. 7. 1.37 H. P. perlb. M. E. P. 8. O.S675 II . P. pcrlb. M. E. P. 9. 423.1 H. P. 10. 415.9 H. P. 11. 15,402 Ibs. 15,360 lb/. 552.7 1}. P. at 1 mile [ per In in. 13. 10.95 1'bs. M. E. P. 14. 24.3 Ibs. M. E. P. 15. 2.67+ H. P. 16. 5 H. P. 17. 1HH. P. Engines, Page 142 18. 33.47 H. P. 19. 52+ H. P. 20. 22 H. P. 21. 25.9 H. P. 22. 6+ H. P. 23. 320.8 H. P. 24. 9.996 H. P. per lb. M. E. P. 25. 18,488.89 Ibs. 26. 19,413.33 Ibs. 27. 18.838 in. 28. 4.318 H. P. per 11 . M. E. P. 29. 10.927 in. 30. 17.019 in. 31. 10.249 in. 32. 9.577 in. ^ 8 in. crank l>182+ R. P. M. 34. 435.4 H. P. 35. 4.752 H. P. per lb. M. E. P. 1. $210 2. $3,852.90 ::. 10.4S H. P. 4. Hi5.2 II. P. 5. 1899.4 sq. ft. Steam Boiler, Page 155 6. 1386.8 sq. ft. 7. 1267.2 sq. ft. 8. 169.65 cu. ft, 9. 1141.2 sq. ft. 10. 40.2 H. P. SHOP MATHEMATICS 213 11. 36.7 H. P. 20. .589 in. 12. 13. 14. 15. 16. $20.16 9J sq. ft. 1,728 gals. 2,000 gals. 174.36 cu. ft. ( 861+ ibs. ( 1283 + sq. ft. 22. 689 Ibs. 23. 1,167 Ibs. 24. 168 Ibs. 17. 232.3 25. .474 in. 18. 312^ Ibs. 26. .348 in. 19. 00 1.5 Ibs. 28. 53, : 27. 62,261 Ibs. ,42 Ibs. Safety Valve, Page 159 2. 3. 149.08 Ibs. 4. 45 in. 6.652+ in. 5. 175-V Ibs. Hydraulic Machines, Page 161 1. 11.38 10. 44,957.4 Ibs. j. .914 11. 2 1.3 Ibs. 4. .962 9.92 12. 4,000 Ibs. 13. 90.228 Ibs. 5. 86.8 Ibs. 14. 14.114 Ibs. 6. 27(.5 ft. 15. 245.41 Ibs. 7. 1,687.5 Ibs. 16. 392.65 Ibs. 8. 9. S43| Ibs. 9,375 Ibs. 17. 98.16 Ibs. 18. 94.85 ft. Steam Pumps, Page 166 1. 8,812.68 gals. 2. 1.48 H. P. 3. 3,304.8 gals. 4. 22,521 Ibs. 5. 2 ; 208.9 cu. ft. 6. 2.79 H. P. 214 SHOP MATHEMATICS 7. 3 ,427.2 13. 6.632 in. 8. 29.75 H. P. 14. 68.177 cu. ft. 9. 2 .22 H. P. 15. 9.29 H. P. 10. 37.975 Ibs. per sq. in. 16. 10 in. nearly 11. 64 min. 37 sec. 17. 17.7 Ibs. 12. 3 .346 in. 18. 21.8 H. P. 19. 12.988 in. Formulas, Page 183 1. 50 c a 2. 20 V. \~> . A 1 " sm A 3. 108 4. 112 7. or .5 WXb ,, WXb 9. 25 C -p-; F a 10. -7f C 5. , PXa w PXa 11. 50 F W ' ~b~~ 12. 14 F INDEX Abbreviations ix Addendum 32 Adhesion of belts 61 Angle to set milling ma- chine table 126 Angle of friction 55 Archimedes' discovery. . 161 Arithmetical signs ix Avoirdupois pound 170 Axle friction 53 B Bar iron, rule for weight of 172 Ball bearings 77 Bearings, belt pressure. . 64 Bearings, journal 76 Bearings, shaft 76 Bearings, weight on. ... 77 Belts, leather 61 Belts, H. P. of 64 Belts, length of 62 Belts, centrifugal force in 62 Bevel gears 37 Boilers, steam 149 Boiler, heating surface of 151 Boilers, strength of 154 Boilers, water capacity. . 152 Breaking strain of man- ila rope 68 Breaking strain, wire rope 70 Capstan 10 Centrifugal force in pul- ley rims 138 Chain transmission 71 Chain link 71 Coefficient of friction ... 52 Combination of wheels, gears, examples 49 Combination of screws, pulleys, etc., examples 51 Compound indexing on milling machine 1 16 Compound lever, mech. power 9 216 INDEX Cutting speeds for vari- ous materials 104 D Decimal equivalents. . . . 177 Dedendum 32 Depth of gear teeth. ... 34 Diametral pitch 32 Differential indexing milling machine 119 Differential pulley 15 Differential screw 25 Differential windlass. ... 11 Drill press 129 Driver and follower, rules for 12 Dynamometer 135 E Electric H. P 2 Emery wheel, speed of . . 36 Energy, potential and kinetic 3 Engine details 147 Engine fly wheel 138 Engines, gas and gaso- lene 145 Engine lathe 91 F Factor of safety 70 Feather kevs. . 82 PAGE Feeds, lathe carriage ... 90 Feed, lead screw 95 Fixed pulley 14 ; Fly wheel 138 ; Force 1 Formulas, use of 178 Friction 52 Friction, laws for 55 Friction in bearings .... 54 Friction in mechanical powers - 55 G Gallon, cubic contents. . 170 Gas unit of measure. ... 160 Gasolene engines . 145 Gear wheel definition ... 32 Gears, rules for pitch of. 33 Gear, spiral 42 Gear, worm 41 General law of machines 6 Grinder, speeds for 37 Grindstone speed 37 H Hammer blow 130 Head of water 160 Helical gears 42 Horse power, definition . 2 Horsepower in belts. ... 62 in boilers 150 in electrical machinery 2 in the locomotive. . 144 INDEX 217 in machines 132 in pulleys 48 in manila ropes 68 in shafting 73 in steam engines 141 in pumps 165 in wire cables 70 Hydraulics 160 Hydraulic machines. . 163 Inclined plane mechani- cal power 23 Inclined plane, friction in 56 Inclined plane, angle of friction for 55 Index head, milling machine 112 Indexing, compound ... 116 Indexing, differential. .. llg Jack shaft 75 Journal for ball bearings 77 Journal bearing allow- able pressure on 77 Journal bearing for ma- chines 77 Journal for shafting. . 76 Keys, feather 82 Keys, machine 82 Lathe, description of . . . 91 Lathe, carriage feed for. 90 Lathe, change gears for. 95 Lathe, compound gears for ! 98 Lathe, back gears for. . . 93 Lathe, reversing gear for 97 Lathe, lead screw for. . . 95 Lathe, feed rod for 91 Lathe, speeds for cuts on 104 Lathe test indicator. ... 18 Laws for friction 55 Leather belts 61 Leather belts, lengths of 62 Length, unit of 169 Lever, three kinds 7 Lever, bent 8 Lever, calculations for. . 8 Liquid measure 170 Locomotive, power of . . . 143 If Machines 90 Machines, general law for. . 6 , Machine keys 82 218 INDEX Machines, general 90 Manila rope 68 Materials, weights and melting points 172 Mathematical signs ix Measurement, units of . . 169 Measurement by micrometer 83 Measurement of liquid . . 170 Mechanical powers 6 Melting points of metals 172 Meter 173 Metric tables 173 Micrometer explained. . . 83 Mile in feet 169 Milling machine Ill Milling cutters 127 Moment of forces 1 Movable pulley 14 N Natural sines, etc., table 181 O Open belt, lengths of. ... 63 Offsetting the center for taper turning 101 P Planer ' 108 Power and speed, Gene- ral law 6 Powers, mechanical .... 6 Pressure in Ibs. per sq. in 169 I Pressure or head of water 160 Pressure on bearings by belts 64 Pressure on bearings by weight 54 Pressure, unit of meas- ure 169 Pulley, mechanical power 13 Pulley design 47 Pulley safe speed 47 Pumps 163 I R Reversing gears for lathe 97 Roll bearings 77 Rolling friction 52 Rope, breaking strain of 68 Rope, manila 68 Rope transmission 68 Rule for trains of wheels or pulleys 12 Rules for driver and fol- lower 12 S Safe speed for pulleys. . . 47 Safety* valves 157 Screw cutting, lathe. ... 95 Screw, differential 25 INDEX 219 Screw, friction in 58 Screw, mechanical power 24 Screw, lead or pitch of . . 24 Screw threads 29 Shafting .. 73 Sliding friction 52 Speeds of pulleys, safe . . 47 Speeds of emery wheels . 36 Speeds of grindstones. . . 37 Speeds of lathe cuts 104 Speeds of drills 129 Speeds of milling cutters 127 Speeds of polishing wheels 37 Speed and power, gen- eral law 6 Spiral gears 42 Spiral head, milling ma- chine 123 Sprockets 71 Spur gears 32 Square measure 169 Steam boiler 149 Steam engine formulas. . 141 Steam engine details. . . . 147 Strut, moving 9 Symbols, mathematical . ix T Table of decimal equiva- lents 177 Table, natural sines, etc . 181 Tapers per foot 102 Tapers per inch 102 Taper turning 101 Tension in pulleys 139 Trigonometric functions 186 Thread cutting, lathe ... 95 Threads, U. S., V, Whit- worth 29 Toggle joint 9 U Units of length, etc 169 Use of formulas 178 V Velocity of pulleys 47 Velocity ratio 32 Vernier caliper 85 V-threads 29 W Water pressure 160 Wedge, mechanical power 26 Weight of water 170 Weight of metals 172 Windlass, simple 10 Windlass, differential. . . 11 Wheel and axle 10 Wire rope, H. P., etc ... 70 Wire rope sheaves 69 Whitworth threads 30 Work 1 Worm gears 41 A 000 038 664 9