PLEA DO NOT REMOVE THIS BOOK CARD! University Research Library i i This book is DUB on the last date stamped below WAY 1 9 1952! 3 1 1952 REC '-'"iURL JAN 2 9 1975 'JAN 2 1 19 Form L-9-15m-8,'26 INDUSTRIAL SERIES SHOP MATHEMATICS PART I SHOP ARITHMETIC PREPARED IN THE EXTENSION DIVISION OF THE UNIVERSITY OF WISCONSIN BY EARLE B. NORRIS, M. E. ASSOCIATE PROFESSOR OF MECHANICAL ENGINEERING, IN CHARGE OF MECHANICAL ENGINEERING COURSES IX THE UNIVERSITY EXTENSION DIVISION AND KENNETH G. SMITH, A. B., B. S. ASSOCIATE PROFESSOR OF MECHANICAL ENGINEERING, DISTRICT REPRESENTATIVE IN CHARGE OF FIRST DISTRICT, THE UNIVERSITY EXTENSION DIVISION, MILWAUKEE FIRST EDITION SECOND IMPRESSION McGRAW-HILL BOOK COMPANY 239 WEST 39TH STREET, NEW YORK 6 BOUVERIE STREET, LONDON, E. C. 1912 $883 COPYRIGHT, 1912, BY THE BOOK COMPANY 7113 THE. MAPLE. PRESS- YORK. PA N n \r PREFACE The aim of this book is to teach the fundamental principles of mathematics to shop men, using familiar terms and processes, and giving such applications to shop problems as will maintain the interest of the student and develop in him an ability to apply the mathematical and scientific principles to his every day problems of the shop. The problems and applications relate largely to the metal working trades. It has, however, been the aim in preparing this volume not to apply the work to these particular trades so closely but that it shall be of interest and value to men in other lines of industry. This volume presents the first half of the instruction papers in Shop Mathematics as developed and used by the Extension Division of the University of Wisconsin. As here offered, it embodies the point of view obtained through apprenticeship and shop experience as well as the experience gained through its use during the past four years as a text for both correspondence and class room instruction. It is believed that the book will be found suitable for home study and for use as a text in trade, industrial, and continuation schools. The instruction in arithmetic ends with Chapter XII. The remaining chapters are introduced to give further practice in calculation and to develop an ability to handle simple formulas, as well as to impart a knowledge of the principles of machines. The second volume will take up more fully the use of formulas and will teach the principles of geometry and trigonometry as applied to shop work. The authors are indebted to Mr. F. D. Crawshaw, Professor of Manual Arts in The University of Wisconsin, for a careful reading of the proof and for valuable criticisms and suggestions. E. B. N. THE UNIVERSITY OF WISCONSIN, MADISON, Wis. June 1, 1912. CONTENTS CHAPTER I COMMON FRACTIONS ART. PAGE 1. Why we use fractions 1 2. Definition of a fraction 2 3. The denominator 2 4. The numerator 2 5. Writing and reading fractions 2 6. Proper fractions 3 7. Improper fractions 3 8. Mixed numbers 3 9. Reduction of fractions 3 10. Reduction to higher terms 4 11. Reduction to lower terms 4 12. Reduction of improper fractions 5 13. Reduction of mixed numbers 6 CHAPTER II ADDITION AND SUBTRACTION OF FRACTIONS 14. Common denominators 9 15. To find the L. C. D 10 16. To reduce to the L. C. D 10 17. Addition of fractions 11 18. Subtraction of fractions 12 CHAPTER III MULTIPLICATION AND DIVISION OF FRACTIONS 19. A whole number times a fraction 17 20. "Of" means times 17 21. A fraction times a fraction 18 22. Multiplying mixed numbers 18 23. Cancellation 19 24. Division the reverse of multiplication 21 25. Compound fractions 21 26. How to analyze practical problems 22 CHAPTER IV MONEY AND WAGES 27. U. S. money 24 28. Addition 25 vii viii CONTENTS ART. PAOE 29. Subtraction '. 26 30. Multiplication , 26 31. Division . 26 32. Reducing dollars to cents '27 33. Reducing cents to dollars 28 34. The mill 28 35. Wage calculations 29 CHAPTER V DECIMAL FRACTIONS 36. What are decimals? 33 37. Addition and subtraction 35 38. Multiplication 36 39. Short cuts 37 40. Division 37 41. Reducing common fractions to decimals 39 42. Complex decimals 40 43. The micrometer 40 CHAPTER VI PEKCENTAGE 44. Explanation 44 45. The uses of percentage 46 46. Efficiencies 47 47. Discount . .' 47 48. Classes of problems 48 CHAPTER VII CIRCUMFERENCES OF CIRCLES: CUTTING AND GRINDING SPEEDS 49. Shop uses 51 50. Circles 51 51. Formulas 52 52. Circumferential speeds 54 53. Grindstones and emery wheels 55 54. Cutting speeds 57 55. Pulleys and belts 58 CHAPTER VIII RATIO AND PROPORTION 56. Ratios 59 57. Proportion '.'... 60 58. Speeds and diameters of pulleys 63 59. Gear ratios 64 ' CONTENTS ix CHAPTER IX PULLEY AND GEAR TRAINS; CHANGE GEARS Am. J'ACib; 60. Direct and inverse proportions 65 61. Gear trains 60 62. Compound gear and pulley trains 68 63. Screw cutting 72 CHAPTER X AREAS AND VOLUMES OF SIMPLE FIGURES 64. Squares 75 65. Square root 75 66. Cubes and higher powers 75 67. Square measure 76 68. Area of a circle 77 69. The rectangle 79 70. The cube 80 71. Volumes of straight bars 80 72. Weights of metals 82 73. Short rule for plates 83 74. Weight of casting from pattern 83 CHAPTER XI SQUARE ROOT 75. The meaning of square root 85 76. Extracting the square root 86 77. Square roots of mixed numbers 87 78. Square roots of decimals 87 79. Rules for square root 88 80. The law of right triangles 89 81. Dimensions of squares and circles 91 82. Dimensions of rectangles 91 83. Cube root 92 CHAPTER XII MATHEMATICAL TABLES (CIRCLES, POWERS, AND ROOTS) 84. The value of tables 94 85. Explanation of the tables 95 86. Interpolation 96 87. Roots of numbers greater than 1000 5)7 88. Cube roots of decimals ! 98 89. Square root by the table 99 x CONTENTS CHAPTER XIII LEVERS ART. I'A<;K 90. Types of machines 115 91. The lever 115 92. Three classes of levers 117 93. Compound levers 118 94. Mechanical Advantage 119 95. The wheel and axle 120 CHAPTER XIV TACKLE BLOCKS 96. Types of blocks . . . , 123 97. Differential pulleys 126 CHAPTER XV THE INCLINED PLANE AND SCREW 98. The use of inclined planes 130 99. Theory of the inclined plane 130 100. The wedge 132 101. The jack screw 133 102. Efficiencies 134 CHAPTER XVI WORK POWER AND ENERGY; HORSE-POWER OF BELTING 103. Work 137 104. Unit of work . . . 137 105. Power 138 106. Horse-power of belting 139 107. Widths of belts 140 108. Rules for belting 141 CHAPTER XVII HORSE-POWER or ENGINES 109. Steam engines 145 110. Gas engines 147 111. Air compressors 149 112. Brake horse-power 11!) 113. Frictional horsc-pbwer 151 114. Mechanical efficiency 151 CONTENTS xi CHAPTER XVIII MECHANICS OF FLUIDS ART. PAGE 115. Fluids 153 116. Specific gravity 153 117. Transmission of pressure through fluids 154 118. The hydraulic jack 155 119. Hydraulic machinery 158 120. Hydraulic heads 158 121. Steam and air 159 CHAPTER XIX HEAT 122. Nature of heat 163 123. Temperatures 164 124. Expansion and contraction 168 125. Allowances for shrink fits 171 HAPTER XX STRENGTH OF MATERIALS 126. Stresses 173 127. Ultimate stresses 174 128. Safe working stresses 174 129. Strengths of bolts 175 130. Strengths of hemp ropes 177 131. Wire ropes and cables 177 132. Strengths of chains 177 133. Columns . .178 SHOP ARITHMETIC CHAPTER I COMMOX FRACTIONS 1. Why We Use Fractions. When we find it necessary to deal with things that are less than one unit, we must use fractions. A machinist cannot do all his work in full inches because it is generally impossible to have all measurements in exact inches. Consequently, for measurements less than 1 in., he uses fractions of an inch; he also makes use of fractions for measurements between one whole number of inches and the next whole number. If a bolt is wanted longer than 4 in. but shorter than 5 in., it would be 4 in. and a fraction of an inch. This fraction of an inch might be nearly a whole inch or it might be a very small part of an inch. The system used to designate parts of a unit is FIG. 1. easily seen by looking at a machinist's scale or at a foot-rule of any sort. Each inch on the scale is divided into a number of equal parts. A wooden foot-rule usually has eight or sixteen parts to each inch, while a machinist's steel scale has much, finer divisions. Now, if we want to measure a piece of steel which is not an inch long, we hold a scale against it, as dn Fig. 1, and find out how many of these divisions of an inch it takes to equal the length of the piece. The scale in Fig. 1 is 3 in. long and each inch is divided into eight parts. We see that this piece 'is as 1 2 SHOP ARITHMETIC long as five of these eight parts of an inch, or we say that it is "five-eighths " of an inch long. 2. Definition of a Fraction. A Fraction is one or more of the equal parts into which anything may be divided. Every fraction must contain two numbers, a numerator and a denominator. These are called the terms of a fraction. 3. The Denominator. The Denominator tells into how many equal parts the unit is divided. In the case shown in Fig. 1, 1 in. was the unit and it was divided into eight equal parts. The denominator in this case was eight. 4. The Numerator. The Numerator shows how many of these parts are taken. In giving the length of the piece of steel in Fig. 1, we divided the inch into eight parts and took five of them for the length. Five is the numerator and eight is the denominator. 5. Writing and Reading Fractions. In writing fractions, the numerator is placed over the denominator and either a slanting line, as in 5/8, or a horizontal line, as in f , drawn between them. The horizontal line is the better form to use, as mistakes are easily made when a whole number and a fraction with a slanting line are written close together. -j is read one-fourth or one-quarter. -= is read one-half. 2 is read three-fourths or three-quarters. -= is read five-eighths. o 3 = is read three-sevenths. We can have fractions of all sorts of things besides inches. An hour of time is divided into sixty equal parts called minutes. A minute is merely -fa of an hour. Likewise, 20 minutes is f $ of an hour. In the same way, 1 second is -^ of a minute. In the early days, before we had the unit called the inch, the foot was the common unit for measuring lengths. When it was necessary to measure lengths less than 1 ft., fractions of a foot COMMON FRACTIONS 3 were used. This got to be too troublesome, so one-twelfth of a foot was given the name of inch to avoid using so many fractions. For instance, where formerly one said f\ of a foot, we can now say 5 in. This shows how the use of a smaller unit reduces the use of fractions. In Europe, a unit called the millimeter is used in nearly all shop work. This is so small, being only about T T of an inch, that it is seldom necessary in shop work to use fractions of a millimeter. 6. Proper Fractions. If the numerator and denominator of a fraction are equal, the value of the fraction is 1, because there are just as many parts taken as there are parts in one unit. 4_! 8 10 4~ 8~ 10" In each of these cases, the numerator shows that we nave taken the full number of parts into which the unit has been divided. Consequently, each of the fractions equals a full unit, or 1. A Proper Fraction is one whose numerator is less than the denominator. The value of a proper fraction, therefore, is always less than 1. 3 5 7 27 A' T' Q' QO are a ^ P r P er fractions. 4 ID o ou 7. Improper Fractions. An Improper Fraction is one whose numerator is equal to or larger than the denominator. There- fore, an improper fraction is equal to, or more than 1. 24 14 17 64 To' IP T' 7 are a ^ im proper fractions. 1 _. o It) C)4 8. Mixed Numbers. A Mixed Number is a whole number and a fraction written together: for example, 4 is a mixed number. 4 is read four and one-half and means four whole units and one-half a unit more. 9. Reduction of Fractions. Quite often we find it desirable to change the form of a fraction in order to make certain calcula- tions; but, of course, the real value of the fraction must not be changed. The operation of changing a fraction from one form to another without changing its value is called Reduction. By referring to the scale in Fig. 2 it will be seen that, if we take the first inch and divide it into 8 parts, each \ in. will con- SHOP ARITHMETIC tain 4 of these parts. Hence, ^ in. = in. In this case, we make the denominator of the fraction 4 times as large, by making 4 times as many parts in the whole. It then takes a numerator 4 times as large to represent the same fractional part of an inch. This relation holds whether we are dealing with inches or with any other thing as a unit. FIG. 2. 10. Reduction to Higher Terms. When we raise a fraction to higher terms, we increase the number of parts in the whole, as just shown, and this likewise increases the number of parts taken. Therefore, the numerator and denominator both become larger numbers. 1 . 4. - m. = in. A 16 10. = 32 m ' A fraction is raised to higher terms by multiplying both numer- ator and denominator by the same number. Examples : 1X4 Similarly, -!= 5X2 = 10 16 16X2 32 2 Suppose we want to change y^ of an inch to 64ths. To get 64 for the denominator, we must multiply 16 by 4 and, therefore, must multiply 3 by the same number. 3 = 3X4 ^12 lo = 16X4 64 11. Reduction to Lower Terms. When we reduce a fraction to lower terms, we reduce the number of parts into which the whole unit is divided. This likewise reduces the number of parts which are taken. 4 . 1 . 2 . 1 . A fraction is reduced to lower terms by dividing both numer- ator and denominator by the same number. When there is no COMMON FRACTIONS 5 number -which will exactly divide both numerator and denomina- tor, the fraction is already in its lowest terms. Example : Reduce ._ to its lowest terms. 36 36 4-2^18-4-2^ 9 128 128-5-2 64-J-2 32 There is no number that will exactly divide both 9 and 32 and, therefore, the fraction is reduced to its lowest terms. 12. Reduction of Improper Fractions. When the numerator of a fraction is just equal to the denominator, we know that the value of the fraction is 1 (see Art. 6) : 8 = 1 64 = 1 !? = i 8 64 10 In each of these cases we have taken the full number of parts into which we have divided the unit. Consequently, each of these fractions is one whole unit, or 1. When the numerator is greater than the denominator, the value of the fraction is one or more units, plus a proper fraction, or a whole plus some part of a whole. Examples : 12 8.4 .4 ,1 -8- = 8 + 8 = 1 8 rl 2 47 = 36 11 11 12 12 + 12 12 From these examples we may see that to reduce an improper fraction to a whole or mixed number the simplest way is as follows: Divide the numerator by the denominator. The quotient will be the number of whole units. If there is anything left over, or a remainder, write this remainder over the denominator since it represents the number of parts left in addition to the whole units. We now have a mixed number, or an exact whole number, in place of the improper fraction. Examples : 27 _ 27 . 7 _Q 6 7)27(3 y =27 ^ 7=3 7 21 6 45 3 1 6)45(7 -6 =45 ^ 6 = 7 6 2 f 2 3 6 SHOP ARITHMETIC These show that a fraction represents unperformed division. In fact, division is often indicated in the form of a fraction. The numerator is the dividend and the denominator is the divisor. 24 24 -f- 3 can be written o 2 2-7-8 can be written = 13. Reduction of Mixed Numbers. It is often necessary or desirable to change mixed numbers to improper fractions. The method of doing this may be seen from the following examples. Examples : Reduce 5- to an improper fraction. 4-+J In one unit there are two halves. Therefore, , 5X2 10 1 = 10 1 = 11 2 22 2 If 7 ^ were to be reduced to an improper fraction we would say: "Since there are 4 fourths in 1, in 7 there are 4X7, or 28 fourths. 28 fourths plus 1 fourth equals 29 fourths." 1 = 28 1 = 29 4~ 4 4~ 4 The rule which this gives us is very simple: Multiply the whole number by the denominator of the fraction and write the prod- uct over the denominator. This reduces the whole number to a fraction. Add to this the fractional part of the mixed number. The sum is the desired improper fraction. In working problems like the above, the work should be arranged as in the following example. Example : Reduce 5 5 in. to eighths of an inch, o _ 1 40 1 41 5 H = ~Q~ + fi = ^~' Answer - o o o o COMMON FRACTIONS 7 USEFUL TABLES Measures of Length 12 inches (in.) = 1 foot (ft.) 3 ft. or 36 in. = 1 yard (yd.) 5J yd. or 164 ft. = 1 rod (rd.) 320 rd. or 5280 ft. = 1 mile (mi.) Measures of Time 60 seconds (sec.) =1 minute (min.) 60 minutes =1 hour (hr.) 24 hours =1 day (da.) 7 days =1 week (wk.) 365J days =; 1 average year (yr.) 100 years = 1 century Note. Thirty days are generally considered as one month, though the number of days differs for different months. Miscellaneous Units 12 things =1 -dozen (doz.) 12 dozen or 144 things = 1 gross (gr.) 12 gross = 1 great gross 20 things = 1 score QUESTIONS AND PROBLEMS 1. What is a fraction? 2. Name some fractions of an inch commonly used. 3. Write the following as fractions or mixed numbers. Five-sixteenths Nine thirty-seconds Twenty and one-eighth Twenty-one eighths Three and three-fourths 4. Write out in words the following: ol 5 7 8 5 1 21 3 6' 8' 16' 8' 24' 2 ~4' 4 6. Indicate the proper fractions, the improper fractions, and the mixed numbers among the following: 3 _1 16 21 5 7 9 4' 6 ~&' 16' 16' ^e' 8' 16 6. Change yg of an inch to eighths of an inch. 4 Change of an inch to fourths of an inch. 8 SHOP ARITHMETIC g 7. How many sixteenths of an inch in j in.? 3 How many thirty-seconds of an inch in -r in.? 13 7 33 8. Which is greater, -r^ in. or n in.? 77j or To ? 9. Reduce the following mixed numbers to improper fractions: ,1 ^1 o3 ! ,1 V V V V '16 10. Reduce the following improper fractions to whole or mixed numbers: 21 8 24 7 121 16' 8* 3' 2' 12 11. I want to mix up a pound of solder to be made of 5 parts zinc, 2 parts tin, and 1 part lead. What fraction of a pound of each metal zinc, tin and lead must I have? 12. If a train is running at the rate of a mile a minute, how many feet does it go in 1 second? 7 13. An apprentice who is drilling and tapping a cylinder for ^ in. studs, 3 1 tries a -7 in. drill but the tap binds, so he decides to use a drill TTT in. larger. W 7 hat size drill does he ask for? 14. The tubes in a certain boiler are 15 ft. 11 in. long. How many inches long are they? 15. How many seconds in an hour? 40 seconds is what fraction of an hour? 16. An 8-ft. bar of steel is cut up into 16 in. lengths. AVhat fraction of the whole bar is one of the pieces? 17. When a man runs 100 yd. in 10 seconds, how many feet does he go in 1 second? 18. Wood screws are generally put up in boxes containing one gross. If 36 screws are taken from a full box for use on a certain job, what fraction of the gross is used on this job and what fraction is left in the box? Reduce both fractions to their lowest terms. 19. A steel plate 2 ft. 6 in. wide is to be sheared into four strips of equal width. How wide will each strip be in inches? 20. In one plant all drawings are dimensioned in inches, while in another all dimensions above 2 ft. are given in feet and inches. If a dimension is given as 89 in. in the first plant, how would the same dimension be stated in the other plant? CHAPTER II ADDITION AND SUBTRACTION OF FRACTIONS 14. Common Denominators. Fractions cannot be added unless they contain the same kind of parts, or, in other words, have the same denominator. When fractions having different denominators are to be added, they must first be reduced to frac- tions having a common denominator. A number of fractions are said to have a common denominator when they all have the same number for their denominators, f and cannot be added as they stand, any more than can 3 bolts and 5 washers. Both the fractions must be of the same kind, that is, must have the same denominator, may be changed to f . By making this change, the fractions are given a common denominator and can now be added. 6 eighths plus 5 eighths equals 1 1 eighths, in just the same manner as 6 inches plus 5 inches equals 11 inches. The work of this example would be written as follows: 4X2~8 6 5 11 ,3 + = ~~ =1 > Answer - 8 is called the Least Common Denominator (L. C. D.) of f and f , because it is the smallest number that can be used as a common denominator for these two fractions. In this case, the least common denominator is apparent at a glance; in many other cases it is more difficult to find, especially if there are several fractions to be added. In the case just given, the denominator of one fraction can be used for the common denominator. When we have two denominators like 5 and 8, neither of them is an exact multiple of the other number, and so neither can be the common denominator. In such a case, the product of the two numbers can be used as a common denominator. 9 10 SHOP ARITHMETIC Example : 5X8 = 40, theL. C. D. 3^3X8^24 5 5X8 40 5^5X5^25 8~8X5~40 24 25 49 9 We can always be sure that the product of the denominators will be a common denominator, to which all the fractions can be reduced, but it will not always be the least common denominator. For example, if we wish to add T 5 ^- and -j^-, we can use 12X16 = 192 for the common denominator, but we readily see that 48 will serve just as well and not make the fractions so cumbersome. In this case 48 is the least common denominator. 15. To Find the L. C. D. If the L. C. D. cannot be easily seen by examining the denominators, it may be found as follows: Suppose we are to find the L. C. D. of \, f , f , and T \- First place the denominators in a row, separating them by commas. 2)4, 3, 9, 16 2)2, 3, 9, 8 3)1,3,9, 4 1, 1, 3, 4 L. C. D. =2X2X3X3X4 = 144 Select the smallest number (other than 1) that will exactly divide two or more of the denominators. In this case, 2 will exactly divide 4 and 16. Divide it into all the numbers that are exactly divisible by it, that is, may be divided by it without leaving a remainder. When writing the quotients below, also bring down any numbers which are not divisible by the divisor and write them with the quotients. Now proceed as before, again using the smallest number that will divide two or more of the numbers just obtained. Continue this process until no number (except 1) will exactly divide more than one of the remaining numbers. The product of all the divisors and all the numbers (except 1's) left in the last line of quotients is the Least Common Denominator. 16. To Reduce to the L. C. D. Having found the least common denominator of two or more fractions, the next step is to reduce ADDITION AND SUBTRACTION OF FRACTIONS 11 the given fractions to fractions having this least common denominator. Let us take $, f, , and We first find the L. C. D., which turns out to be 120. We next proceed to reduce the fractions to fractions having the L. C. D. Divide the com- mon denominator by the denominator of the first fraction. Multiply both numerator and denominator of the fraction by the quotient thus obtained. Do this for each fraction, as illus- trated here. 2)3, 5, 4, 8 2)3, 5, 2, 4 3, 5, 1, 2 L.C.D. =2X2X3X5X2 = 120 1 1X40 40 3 3X40 120 120-i-3 = 120.5 = 24 rsx24 - 120 120.4 = 30 i= 1X3 30 120.8 = 4 4X30 120 3 = 3X15^ 45^ 8~~8X15~ 120 17. Addition of Fractions. Addition of fractions is very simple after the fractions have been reduced to fractions with a common denominator. Having done this it is only necessary to add the numerators and place this sum over the common denominator. The sum should always be reduced to lowest terms and if it turns out to be an improper fraction it should be reduced to a mixed number. Example : Find the sum of _5^ 3 9 7 16 4 32 32 Common denominator =32 . 16~16X2~32 3 3X8 = 24 4 4X8 32 10 24 _9_ 7 = 50 :u 32 32 32 32 50 25 .9 If there are mixed numbers and whole numbers, add the whole numbers and fractions separately. If the sum of the fractions 12 SHOP ARITHMETIC is an improper fraction, reduce it to a mixed number and add this to the sum of the whole numbers. Example : How long a steel bar is needed from which to shear one piece each of the following lengths: 1731 7 S in., 5^ in., 4 7 in., 65 in. ? 2 o 4 o 7 1 4 ? 2 8 5 7 7 Explanation: The sum of the whole numbers is 22. ,3 6 18 4 7 o The sum of the fractions is -^-t which reduces to 2-r- o 4 A* 1 Adding this to the sum of the whole numbers (22), 8 8 1 gives 24 -r as the sum of the mixed numbers. Hence 22 18 ,-.1 4 == 2 ~~ 1 2\ 84 we mus t have a bar 24- inches long. 4 24-r' Answer. 4 18. Subtraction of Fractions. Just as in addition, the fractions must first be reduced to a common denominator. Then we can subtract the numerators and write the result over the common denominator. Example : Subtract - from -=- o 16 Common denominator = 16 5 = 10 8 16 15 10 5 -T^=TS' Answer. 16 16 16 In subtracting mixed numbers, subtract the fractions first and then the whole numbers. Example : in. long? 1 How much must be cut from a 15 in. bolt to make it L. C. D. =16 7^' Answer. Ib ADDITION AND SUBTRACTION OF FRACTIONS 13 Sometimes, in subtracting mixed numbers, we find that the fraction in the subtrahend (the number to be taken away) is larger than the fraction in the minuend (the number from which the subtrahend is to be taken) . In this case, we borrow 1 from the whole number of the minuend and add it to the fraction of the minuend. This makes an improper fraction of the fraction in the minuend and we can now subtract the other fraction from it. Example : 3 i Take 9 7 from 12 5 4 8 19 3 12 5 = 11 5 Explanation: -7 cannot be substracted from o o 4 1 / 8\ Q 3 = 6 -, so we borrow 1 (or^) from 12 and write the 4 _8 9 3 minuend 11~- 2 5 > Answer. o ( or^j If the minuend happens to be a whole number, borrow 1 from it and write it as a fractional part of the minuend. Then sub- tract as before. Example : 10-4-T -r^> Answer. ID PROBLEMS 53 9 10 24 9 21. Reduce to the L. C. D. jg j> and ^H- Answer, ^ oo Q-X- 22. Reduce to the L. C. D. | |. ^> and ~- 4 o 1U lo , 10 11 , 15 109 _13 23. Add 25, y^ and ^- Answer, -^ = 2^- 24> + ++" 7 14 SHOP ARITHMETIC 25. Add 2g> 5^ and 7^- 15 3 26. Find the sum of 8, 3o 4g> and JTT- 4 7 27. Subtract from - 3 1 29. Find the difference between 13j and 202" 30. 15-ll|=? o 31. The weights of a number of castings are: 412^ lb., 270^ lb., 1020 lb., 75^ lb., 68^ lb. What is their total weight? 32. Four studs are required: 2- in., lg in., 2-r^ in., and loo in. long; how 3 long a piece of steel will be required from which to cut them allowing -7 in. altogether for cutting off and finishing their ends? 33. Monday morning an engineer bought 48n gallons of cylinder oil; 2 on Monday, Tuesday, and Wednesday he usedj gallon per day; on Thurs- 7 1 day he used g gallon; and on Friday H gallon. How much oil had he left on Saturday? 94-- _ 2 i-l"_ "8 Fia. 3. 34. Find the total length of the roll shown in the sketch in Fig. 3. 35. A piece of work on a lathe is 1 ft. in diameter; it is turned down in 3 five cuts; in the first step the tool takes off 09 in- from the diameter; then jg in. ; then oo m - > then ^o m -> an< ^ the fifth time gj in. What is the diam- eter of the finished piece? 36. How long must a machine shop be to accommodate the following machines installed in a single line: lathe, 8?> ft. long; planer, 14^ ft. long; 17 1 milling machine, 4g ft. long; engine, 7o ft. long; tool room, 12g ft. long? Allow 3j ft. between a wall and a machine, and 3^ ft. between two machines. The tool room is to be placed at the end of the shop. 37. In doing a certain piece of work one man puts in lq hours, a second man ?> hour, a third works 2~- hours, and a fourth man works lj hours. How long would it take one man to do the work? 38. By mistake, the draftsman omitted the thickness of the flange on the drawing of a gas engine cylinder in Fig. 4. From the other dimensions given, calculate the thickness of the flange. -J- fr* 1C Fia. 4. 39. A millwright has to rig up temporarily a 6 in. belt to be 367g in. long. In looking over the stock of old belting he finds the following pieces of the 17 3 right width; one piece 126^ in. long, one 142g in. long, and one 133g in. long. 16 SHOP ARITHMETIC How many inches must be cut from one of the pieces so that these pieces can be laced together to give the right length? 40. The time cards for a certain piece of work show 2 hours and 15 minutes lathe work, 3 hours and 10 minutes milling, 1 hour and 10 minutes planing, and 1 hour and 15 minutes bench work; what is the total number of hours to be charged to the job? CHAPTER III MULTIPLICATION AND DIVISION OF FRACTIONS 19. A Whole Number Times a Fraction. In the study of multiplication, we learn that multiplying is only a short way of adding. 4x7 is the same as four 7's added together. Either 4X7, or 7+7+7+7 will give 28. If we apply this same principle to the multiplying of fractions, we see that 4 X | is the same as four of these fractions added together. 4X 7 = V + V = 28 A 8 8^8^8^8 8 This shows that multiplying a fraction by a whole number is performed by multiplying the numerator by the whole number and placing the product over the denominator of the fraction. In other words, the size of the parts is not changed, but the number of parts is increased by the multiplication. After multiplying, the product should be reduced to lowest terms and, if an improper fraction, should be reduced to a whole or mixed number. Example : What would be the total weight of 12 brass castings each weighing f of a pound? f\ *3f\ 12X~=-r=9 lb.. Answer. 4 4 20. "Of" Means "Times." The word "of" is often seen in problems in fractions, as for instance, "What is of 5 in.?" In such a case, we work the problem by multiplying, so we say that " of" means " times." You can see that this is so by taking a piece of wood 5 in. long and cutting it into four equal parts and then taking three of these parts. These three parts will be | of 5 in., and by actual measurement will be 3J in long, so we know that of 5 = 3f . Now see what times 5 is' 3 vr 15 3 TXo = - r =3j 4 44 which is the same value. Therefore, we see that the word "of" in such a case signifies multiplication. 2 17 18 SHOP ARITHMETIC 21. A Fraction Times a Fraction. To multiply two or more fractions together, multiply the numerators together for the numerator of the product and multiply the denominators together for the denominator of the product. Example : Multiply gxf- o o 7 2_14_ 7 Explanation: The numerator of the product is 8 3~24~T2 obtained from multiplying the numerators to- gether: 7X2 = 14. The denominator of the product, in the same manner, is 8 X 3 = 24. This 14 7 gives the product,, - which can be reduced to j-_- 24 12 Let us see what multiplication of fractions really means, and why the work is done as just shown. Suppose we are to find | of | in. This means that of an inch is to be divided into 4 equal parts and 3 of these parts are wanted. If we divide | in. into 4 equal parts, each part will be one-fourth as large as | in. and, therefore, can be considered as being made up of 7 parts, each one-fourth as large as | in. Then \ of f =-jV- Three of these parts will naturally contain three times as many thirty-seconds, or - 8 ^- = f^. Therefore: ? 7 3 7 3X7 21 4 8~4 X 8~4X8~32' 22. Multiplying Mixed Numbers. This is one of the most difficult operations in the study of fractions, unless one adopts a fixed rule and follows it in all cases. The student will have no trouble if he will first reduce the mixed numbers to improper fractions, and then multiply these like any other fractions. Example : Find the product of 3^ and 2^- 1 13 1 5 113 513X565 1 ~~ ~*' * To multiply a mixed number by a whole number, we can reduce the mixed number to an improper fraction and then multiply it; or we can multiply the fractional part and the whole number part separately by the number and then add the products. MULTIPLICATION AND DIVISION OF FRACTIONS 19 Example : 3 What would be the cost of ten J in. by 6 in. machine bolts at 1 Q o cents a piece? First method: 10x1^ = 10 X-Q- =- 8 -= 13x = 13 r cents. Second Method: 10 3 Explanation: First multiply 10 by 5- This gives 8 30 6 3 . or 3 6 . or 3.- Set this down. Then multiply 6 o o 4 i 3 4 lObyl. This gives 10, and we add this to the 3 l.'V* cents, Answer, giving a total of 13^> 23. Cancellation. Very often the work of multiplying frac- tions may be lessened by cancellation, as it avoids the necessity of reducing the product to lowest terms. To get an idea of cancellation we must first understand what a "factor" is. A Factor of a number is a number which will exactly divide it. Thus, 2 is a factor of 8, 3 is a factor of 27, 5 is a factor of 35, etc. When the same number will exactly divide two or more numbers it is called a common factor of those numbers. Thus, 2 is a common factor of 8 and 12, because it will divide both 8 and 12 without leaving a remainder. 4 is also a common factor of 8 and 12. Similarly, 7 is a common factor of 14 and 21. This idea of common factors we have already used in reducing fractions to lowest terms. Thus, when we have -fa we divide both 8 and 12 by 4 and get 8 = 8-h4 = 2 12 12+4 3 Cancellation is a process of shortening the work of reduction by removing or cancelling the equal factors from the numerator and denominator. Example : Suppose we have several fractions to multiply together, as ?x 2 x 3 x 21 - 4*3 X 14 X 32 ,, . . . 3X2X 3 X21 378 The,r product is -- 20 SHOP ARITHMETIC This is not in its lowest terms so we divide both numerator and denominator by 2, 3, and 7, and get 378-^-2 189-i-3 63-r-7 9 5376^2 2688*3 ~ 896 -h 7 ~ 128 Now, if we had struck out the common factors from the numerator and denominator before multiplying the fractions, we would have shortened the work and our answer would have been in its lowest terms without reducing. Thus: 1 1 3 9 2X1X2X32 128 2 1 2 Explanation: First the 3 in the numerator is cancelled with the 3 in the denominator. This merely divides the numerator and denominator by 3 at the outset instead of waiting until the terms are all multiplied together; and, as 3 -f-3 = 1, we cancel a 3 from both numerator and denomin- ator and place 1's in their stead. Next we divide both terms by 2. The gives 1 in place of the 2 in the numerator and 2 in place of the 4 in this denominator. Next we see that 7 is a common factor of the numerator and denominator, so we divide the 21 and 14 each by 7 and place 3 and 2 in their places. There are no more common factors; so we multiply together 9 the numbers we now have and get -=-== iZo Another Example : 1 2465 = 120 " Explanation: First we cancel 250 out of 500 and 250; and then 9 out of 36 and 63; then 7 out of 42 and 7; then 10 out of 50 and 20; and finally 2 out of 2 and 2. This removes all the common factors and we get 120 for the answer. PROBLEMS IN MULTIPLICATION 41. 7X^ = ? 42. 8 X g = ? 23 5 43. Multiply ^ by Tg' 3 5 44. Find the product of j and 45. What is % of 2^? o o 46. What is | of | of ^ of |? MULTIPLICATION AND DIVISION OF FRACTIONS 21 24. Division The Reverse of Multiplication. Division is just the opposite of multiplication and this fact gives us the cue to a very simple method of dividing fractions. To divide one fraction by another, invert the divisor and then multiply. To invert means to turn upside down. Invert f and we get f ; invert and we get . Example : TV -J 27 U 3 Divide 09 by-- 9 27^3 = ?J * = 9 J 8 3 4 Explanation: The divisor is -: Inverting this gives - In multiplying, 9 1 we make use of cancellation to simplify the work, and we get - or 1 for o o the result. Suppose we have a fraction to divide by a whole number; as -h2 = ? Therefore, 14 2 1 4- 2 = ? 2 is the same as n If we invert this we get ? lu 1 2i 7 14 2 Ml 7 - 25. Compound Fractions. Sometimes we see a fraction which has a fraction for the numerator and another fraction for the denominator. This is called a Compound Fraction. If we remember that a fraction indicates the division of the numerator by the denominator, we will see that a compound fraction can be simplified by performing this division. Example : 27 W 1 i;it i s iii ? T 4 27 3 This means the same as --*- and, therefore, would be solved as follows: ' 4 27 9 1 27 * x 8 1 32 27 3 27 * 9 ,1 --- x - 1 Answer - 22 SHOP ARITHMETIC PROBLEMS IN DIVISION 47. Divide 175 by |- 15 7 48. Divide yg by ITT- 49. Divide s| by 5g- 60 27-A_ ? &0< 32^10"' 7 61. Find the quotient of 21 -J-K- 62. - = ? 26. How to Analyze Practical Problems. The chief trouble that students have in working practical problems is in analyzing the problems to find out just what operations they should use to work them. Problems in multiplication or division of fractions will fall in one of the three following cases: 1. Given a whole; to find a part (multiply). 2. Given a part; to find the whole (divide). 3. To find what part one number is of another (divide). Example of Case I : The total weight of a shipment of steel bars is 3425 Ib. ^ of this consists &O 3 1 of j in. round bars and the balance is ?> in. round. What weight is there of each size? 137 Explanation : In this example we 7 3>f2S _., .3. , have the whole (3425 Ib.) ; to find a ^ X ~ =959 Ib. of j in. bars. / 7 \ * part \-~-A If the whole is 3425, then 3425 -959 = 2466 Ib. of ~ in. bars. 7 V ' ~o{ 3425 = 959 Ib. The balance, which or ^o - 137 1 1 8 3423 1 consists of in. bars, will be 3425 959, ^ X^^ =2466 Ib. of 5 in. bars. rv 1 * ., M1 , 1 7 ^o 7 lo ,,, or it will be l-25=25-25 = 25 ofthe 18 whole, g of 3425 = 2466 Ib. Example of Case 2 : 3 The base of a dynamo weighs 270 Ib.; the base is yr of the total weight; find the total weight. MULTIPLICATION AND DIVISION OF FRACTIONS 23 4- of the whole = 270 lb. Explanation: Here we have a part given to find the whole, -jy whole = 270 H- jj- of the whole is 270 j b Thig means 90 that if the whole machine were 3 270 11 divided into eleven equal parts, 270 -5-jj = -y- X-^ =990 lb., Answer. three of these parts together would weigh 270 lb. Then one part would weigh 270 H- 3 =90 lb. Since there are 11 of these equal parts, the whole machine weighs HX90 = 9901b. This is the same as dividing 270 by g the fraction vy Example of Case 3 : A molder who is on piece work sets up 91 flasks, but the castings from 7 of them are defective. What fractional part of his work does he get paid for? 91 7 = 84 sound castings. ' Explanation: The problem is: What part 84 84-*-? 12 of 91 is 84? There are 91 parts in his whole = ~ = TT Answer. g4 work and he gets paid for 84 parts, or Q-T of 12 the whole. This can be reduced to-pr; which is the answer. PROBLEMS 2 53. A gallon is about JF of a cubic foot. If a cubic foot of water weighs 62^ lb., how much does a gallon of water weigh? 2 4 64. Aluminum is 2^ times as heavy as water; and copper is 8v times as heavy as water. Copper is how many times as heavy as aluminum? 66. If a certain sized steel bar weighs 2-= lb. to the foot, how long must a piece be to weigh 8j lb.? 66. What is the cost of a casting weighing 387 lb. at 4j cents a pound? 67. How many steel pins to finish lg in. long can be cut from an 8 ft. 3 rod if we allow -jg in. to each pin for cutting off and finishing? 68. A certain piece for a machine can be made of steel or of cast iron. 3 1 If drop forged from steel it would weigh 7j lb. and would cost 6g cents per pound. If made of cast iron, it would have to be made much hetvier 24 SHOP ARITHMETIC and would weigh 14 Ib. and cost 2g cents per pound. Which would he the cheaper and how much? 69. I want to measure out 2-r gallons of water, but I have no measure at hand. However, there are some scales handy and I proceed to weigh out 7 the proper amount in a pail that weighs ITT Ib. What should be the total weight of the pail and the water, if one gallon of water weighs 8^ Ib. ? 60. I want to cut 300 pieces of steel, each 112 in. long for wagon tires. I have in stock a sufficient number of bars of the same size, but they are 120 in. long; and I also have a sufficient number 235 in. long. Which length should I use in order to waste the least material? Calculate the total num- ber of inches of stock that would be wasted in each case. CHAPTER IV MONEY AND WAGES 27. U. S. Money. Nearly every country has a money system of its own. The unit of money in the United States is the dollar. To represent parts of a dollar, we use the cent, which is yfo of a dollar. Fifty cents is $ of a dollar; it is also one-half dollar (y 5 ^ = J) . Likewise, twenty-five cents is T 2 ^ dollar, or one-quarter dollar. In writing United States money, the dollar sign ($) is written before the number; a period called the decimal point, is placed after the number of dollars; following this decimal point is placed the number of cents. Two dollars and seventy cents is written $ 2 . 70 Fifteen dollars and seven cents is written $ 15.07 One Hundred twenty-five dollars is written $125.00 One dollar and twenty-five cents is written $ 1 . 25 Thirty-five cents is written $ . 35 Eight cents is written $ . 08 Since one cent is T ^ dollar, it follows that the figures to the right of the decimal point represent a fraction of a dollar. These figures are the numerator, and the denominator is 100. $ 2 . 70 is the same as $ 2- i y - $15.07 is the same as $15 r J- - $ . 08 is the same as $ MONEY AND WAGES 25 The first figure following the decimal point can be said to indicate the number of dimes, because 1 dime = 10 cents, and this figure indicates the number of tens of cents. Also this number represents tenths of a dollar, because 1 dime = 10 cents = i l column and not in the cents column. $1 can be $5.00, Answer. written $1.00, if desired, to avoid any danger of a mistake. 26 SHOP ARITHMETIC 29. Subtraction. The same rules should be followed in sub- tracting. If any figure in the subtrahend is larger than the corresponding figure in the minuend, we can borrow 1 from the figure next to the left, just as in ordinary subtraction. Example : A man draws $24.75 on pay day and immediately pays bills amounting to $8.86. How much does he have left to put in the bank? Explanation: Having set down the numbers properly, we subtract j ust as if we were subtract- $24.75 ing whole numbers. The only difference is that, 8.86 after subtracting, we put the decimal point in $15 gg Answer. tn remainder directly below the other decimal points, to separate the dollars and cents in the remainder. 30. Multiplication. In multiplying an amount of money by any number, the process is the same as in simple multiplication, remembering, however, to keep the decimal point to separate the dollars from the cents. The reasoning for this is just the same as in addition. Multiplying cents, gives cents; multiplying dimes gives dimes, and multiplying dollars gives dollars. The figures left over are carried forward just as in plain multiplication, because 10 cents = 1 dime, and 10 dimes = 1 dollar. Examples : 1. What should a machinist receive for finishing 48 gas engines pistons at 25 cents each? $25 Explanation: The 25 cents is put down as $.25. The multiplication of 25 by 48 is performed as 200 usual. Then the decimal point is placed in the 100 product to separate the dollars and cents by leaving <12 00 Answer. the * wo places, counting from the right, to repre- sent cents. 2. If you owe 2$ weeks board at $3.25 a week, how much money will it take to settle the bill? Explanation: First, we put down the numbers $3.25 and then multiply as if we were simply multi- _2i plying 325 X2$. The product is 812$, but since 162$ we are dealing with dollars and cents, we must 650 put the decimal point in the product to show or two-tenths. 37 .37 is y^:> or thirty-seven one-hundredths. 526 . 526 is > or five hundred twenty-six one-thousandths. 2749 . 2749 is > or two thousand seven hundred forty -nine ten-thousandths. 42 .042 is > or forty-two one-thousandths. 1000 The last case (.042) presents an interesting problem. Here we have a numerator so small in respect to the denominator that it is necessary to have a cipher, or zero (0) between it and the decimal point, in order that the denominator can be indicated correctly. Let us see how we would go about writing such a common fraction into a decimal. Take -nnnr- If we merely wrote .5 that would be -^ and would, therefore, not be right. From the rule for finding denominators of decimals we see that DECIMAL FRACTIONS 35 there must be as many figures after the decimal point as there are ciphers in the denominator. In this case the denominator (1000) has 3 ciphers, so we must have three figures in our decimal. We, therefore, put two ciphers to the left of the 5 and then put down the decimal point. We now have .005, which can be easily seen to be -nmr- One thing that must be carefully borne in mind is that adding ciphers after a decimal does not change the value of the fraction. .5 is the same in value as .50 or .500 because T 5 T is the same in value as T 5 ^ or -nnnr- O n tne other hand, ciphers immediately following the decimal point do affect the value of the fraction, as has just been shown. Mixed numbers are especially easy to handle by decimals, because the whole number and the fraction can be written out in a horizontal line with the decimal point between them. We read mixed decimals just as we would any mixed number; first the whole number, then the numerator, and lastly, the denominator. Example : 42137.24697 In this example, 42137 is the whole number, and .24697 is the fraction. The number reads "forty-two thousand one hundred thirty-seven and twenty-four thousand six hundred ninety-seven hundred-thousandths. The names and places to the right and left of the decimal point are as follows: 5 i ^ ^ <5 7654321. 123456 37. Addition and Subtraction. Knowing that all figures to the right of the decimal point are decimal parts of 1 thing and that all figures to the left are whole numbers and represent whole things, it will be seen readily that in addition and subtraction the figures must be so placed that the decimal points come under 36 SHOP ARITHMETIC each other. As was shown under U. S. Money, the operations can then be carried out just as if we were dealing with whole numbers. Examples : Addition Subtraction 783.5 22.7180 21.473 1.7042 804.973 21.0138 Pay no attention to the number of figures in the decimal. Place the decimal points in line vertically. You can, if you desire, add ciphers to make the number of decimal places equal in the two numbers. Remember, however, that the ciphers must be added to the right of the figures in the decimal. Proceed as in ordinary addition and subtraction, carrying the tens forward in addition and borrowing, where necessary, in subtraction just as with whole numbers. 38. Multiplication. In multiplication forget all about the decimal point until the work is finished ; multiply as usual with whole numbers. Then point off in the product as many decimal places, counting from the right, as there are decimal places in the multiplier and multiplicand together. Example : 6 . 685 Multiplicand (3 places) 5.2 Multiplier (1 place) 13370 33425 34.7620 Product (4 places) Since there are three decimal places in one number, and one in the other, we count off in the product four (3 + 1) places from the right and place the point between the 7 and the 4. The last can be dropped after pointing off the product, giving the result 34.762 (or 34-,^). The reason for this can be seen from the following: The whole numbers are 6 and 5. The result must be a little more than 6X5 = 30, and less than 7X6 = 42, since the numbers are more than 6 and 5, and less than 7 and 6. The actual result is 34.762. The position of the decimal point can be reasoned out in this way for any example, but the quickest way is to point off from the right a number of decimal places equal to the sum of the numbers of decimal places in the multiplier and multiplicand. DECIMAL FRACTIONS 37 Examples : (a) .0045X2.7 (b) .000402x4.26 . 0045 (4 places) . 000402 (6 places) 2.7(1 place) 4 . 26 (2 places) :U5 2112 90 804 .01215 (5 places) 1608 .00171252 (8 places) In the above examples it was necessary to put ciphers before the product in order to get the required number of decimal places. To see the reason for this take a simple example such as .2X-3 The product is .06 or -j-f-g-, as can be readily seen if they are multiplied as common fractions (^X^\ = ^^i). This checks with the rule of adding the number of decimal places in the two numbers to get the number in their product. The product of two proper fractions is always less than either of the fractions, because it is part of a part. 39. Short Cuts. If we want to multiply or divide a decimal by 10, 100, 1000, or any similar number, the process is very simple. Suppose we had a decimal .145 and then moved the decimal point one place to the right and made it 1.45. The number would then be 1-j 4 ^ or {%% instead of iV 4 Tsir5 so we see that mov- ing the decimal point one place to the right has multiplied the original number by 10. Therefore, we see that: To multiply by 10 move the decimal point one place to the right. To multiply by 100 move the decimal point two places to the rigJiT. For other' similar multipliers move the decimal point one place to' the right for each cipher in the multiplier. This process is reversed in division, the rules being: To divide by 10 move the decimal point one place to the left. To divide by 100 move the decimal point two places to the left, etc. Example : Reduce 10275 cents to dollars. 10275-^-100 = $102. 75 (Decimal point moved two places to left). 40. Division. The division of decimals is just as easy as the multiplication of them after one learns to forget the decimal point entirely until the operation of dividing is finished. Divide 38 SHOP ARITHMETIC as in simple numbers. Then point off from the right as many decimal places in the quotient as the number of decimal places in the dividend exceeds that in the divisor. In other words, we subtract the number of decimal places in the divisor from the number in the dividend and point this number off from the right in the quotient. Example : Explanation: The number of Divide 105.587 bv .93 places in the dividend is 3, and ooMnc KC7/-i"iQ K_L A c in the divisor 2. Hence, when .93) 105. 587(113.5 + , Answer. thfi division ^ complet ' ed) we _ point off 3-2 = 1 place in the quotient. The + sign after the Dividend, 3 places quotient means that the division 328 Divisor, 2 places ^ not come out even> but that 279 Quotient, 1 place there was a remainder and that 497 the quotient given is not com- 465 plete. If desired, ciphers could ~32 Remainder. , e PJ. ac ^ after the dividend and the division carried farther, giv- ing more decimal places in the quotient. It makes no difference if the divisor is larger than the dividend, as in the following example. In such a case the quotient will be entirely a decimal. Example : 22.762-5-84.25 = ? 84.25)22.762000(.2701+ or .2702, Answer. 16 850 5 9120 5 8975 14500 8425 6075 Explanation: The divisor being larger than the dividend, the quotient turns out to be an entire decimal. In this case we will presume that we wanted the answer to four decimal places. We have, therefore, added ciphers to the dividend until we have six decimal places. When these have all been used in the division, we have 6 2 = 4 places in the quotient. The remainder is more than half of the divisor, showing that if we had carried the division to another place, the next figure would have been more than 5. We, therefore, raise the last figure (1) of the quotient to 2, because this is nearer the exact quantity. In stopping any division this way, if the next figure of the quotient would be less than 5, let the quotient stand as it is, but, if the next figure would be 5 or more, as in the example just worked, raise the last figure of the quotient to the next higher figure. DECIMAL FRACTIONS 39 Sometimes the decimal places are equal in dividend and divisor, as for instance, if we divide .28 by .07. .07). 28 4 As the numbers of decimal places in the dividend and divisor are the same, the difference between them is zero, and there are no decimal places in the quotient. The answer is simply 4. The decimal point would come after the 4 where it would, of course, be useless. If there are more decimal places in the divisor than in the dividend, add ciphers at the right of the decimal part of the dividend as far as necessary. In counting the decimal places, be sure to count only the ciphers actually used. Examples : l.H-.025 = ? 4.2 + 38.25 = ? . 025) 1 . 000(40, Answer. 38 . 25)4 . 200000( . 1098 + , Answer. 1 00 3 825 37500 34425 30750 30600 150 41. Reducing Common Fractions to Decimals. Common fractions are easily reduced to decimals by dividing the numer- ator by the denominator. In the case of , we divide 1.0 by 2 and get .5 All that is necessary is to take the numerator and place a decimal point after it, adding as many ciphers to the right of the decimal point as are likely to be needed, four being a common number to add, as four decimal places (ten thousandths) are accurate enough for almost any calculations. If -g^- is to be reduced to a decimal, the work is simply an example in long division, the placing of the point being the main thing to consider. Simply divide 1.00000 by 32. This gives .03125 or 3125 one hundred-thousandths. 32)1.00000(.03125 96 40 32_ 80 64_ 16,0 160 40 SHOP ARITHMETIC 42. Complex Decimals. A complex decimal is a decimal with a common fraction after it, such as .12J, .0312^, etc. The frac- tion is not counted in determining the number of places in the decimal. .12 is read "twelve and one-half hundredths." .0312 is read "three hundred twelve and one-half ten-thou- sandths." To change a complex decimal to a straight decimal, reduce the common fraction to a decimal and write it directly after the other decimal, leaving out any decimal point between them. Examples : .06^=. 065 .8|=.875 .03g= .03125 43. The Micrometer. The micrometer is a device to measure to the thousandth of an inch and is best known to shop men in the form of the micrometer caliper shown in Fig. 6. The whole principle of the micrometer, as generally made, can be said to depend on the fact that ^V of TV~TtW- The micrometer, as shown in Fig. 6, is made up of the frame or yoke 6, the anvil c, the screw or spindle a, the barrel d, and the thimble e. The 1 2 3 4 50 1 - 20 e 15 a inliniiiiHu ,. niiliiii ) r- 1 FIQ. 6. spindle a is threaded inside of d. The thimble e is attached to the end of the spindle a. The piece to be measured is inserted between c and a, and the caliper closed on it by screwing a against it. The screw on a has 40 threads to the inch, so if it is open one turn, it is open ^ in., or T^, or .025. Along the barrel d are marks to indicate the number of turns or the number of fortieths inch that the caliper is open. Four of these divisions (A) wn l represent one-tenth of an inch, so the tenths of an inch are marked by marking every fourth division on the barrel. 41 Around the thimble e are 25 equal divisions to indicate parts of a turn. One of these divisions on e will, therefore, indicate -^ of a turn, and the distance represented will be -^ of ^ = T^ns m - To read a micrometer, first set down the number of tenths inch as shown by the last number exposed on the barrel. Count the number of small divisions on the barrel which are exposed be- tween this point and the edge of the barrel. Multiply this number by .025 and add to the number of tenths. Then observe how far the thimble has been turned from the zero point on its edge. Write this number as thousandths of an inch and add to the reading already obtained. The result is the reading in thousandths of an inch. Example : Let us read the micrometer shown in Fig. 6. . 7 Explanation: First we find the figure 7 exposed on .025 the barrel, indicating that we have over ^ in. This .018 we put down as a decimal. In addition, there is one 7743 in. Answer. f * ne smaller divisions uncovered. This is .025 in more. And on the thimble, we find that it is 3 divisions beyond the 15 mark toward the 20 mark. This would be 18, and indicates .018 in. more. Adding the three, .7 + .025 + .018 = .743 in., Answer. This can perhaps be better under- stood as being 7 thousandths less than ^ in. Lots of men locate a decimal in their minds by its being just so far from some common fraction. Most micrometers have stamped in the frame the decimal equivalents of the common fractions of an inch by sixty-fourths from -fa in. to 1 in. A table of these decimal equivalents is given in this chapter, and will be found very useful. Everyone should know by heart the decimal equivalents of the eighths, quarters, and one-half, or, at least, that one-eighth is .125. Then f = 5 X. 125 = .625; and $ = 7 X. 125 = .875, etc. Also, if possible, learn that -^- = .062^, or .0625. To get the decimal equivalent of a number of sixteenths, add .062^ to the decimal equivalent of the eighths next below the desired sixteenths. Example : 13 What is the decimal equivalent of -^ in.? || = |+^=. 750+. 062^=. 812 Jin., or .8125 in. To set a micrometer to a certain decimal, first unscrew the thimble until the number is uncovered on the barrel correspond- ing to the number of tenths in the decimal. Divide the remainder by .025. The quotient will be the additional number of the 42 SHOP ARITHMETIC divisions to be uncovered on the barrel and the remainder will give the number of divisions that the thimble should be turned from zero. Example : Calculate the setting for ft in. (= .4375 or .437$). First unscrew the micrometer until the 4. is uncovered on the barrel. Then divide the remainder .0375 by .025. This gives 1 and leaves a re- mainder of .0125. The thimble should, therefore, be unscrewed one full turn or 1 division beyond 4 on the barrel, plus 12.5 divisions on the thimble. TABLE OF DECIMAL EQUIVALENTS FROM & TO 1. Fraction Decimal Equivalent Fraction Decimal Equivalent &.... i .015625 .03125 ||.... 4J. . .515625 .53125 , .046875 .0625 f|.... 9 .546875 .5625 16 5 751- A. .078125 .09375 10 II...- 1|. . .578125 . 59375 A---- .109375 .125 5 .609375 .625 5 A. . 140625 . 15625 8 iiv SI .640625 . 65625 ||.... 3 .171875 .1875 11 .671875 .6875 10 i, 04 ' A. .203125 .21875 10 .... .703125 .71875 H,... 1 .234375 .25 if...; 3 .734375 .75 4 J7 84 ' A. .265625 .28125 4 I!-. 64 . 765625 78125 19 b4 ' ' ' ' 5 .296875 .3125 -... 13 . 796875 .8125 16 S1 6 4 ' u. . .328125 .34375 16 If..-. .828125 84375 I!.... 3 .359375 .375 If---. 7 .859375 875 8 -... n .390625 .40625 8 .... n. . .890625 90625 7 ;; .421875 .4375 69 ??. 15 .921875 9375 16 J9 ft . . . . .453125 .46875 16 :::: H .953125 96875 , .484375 .5 &.... 1 .984375 1 00000 2 DECIMAL FRACTIONS 43 PROBLEMS 71. Write the following as decimals: One and twenty-five one-hundredths. Three hundred seventy-five one-thousandths. Three hundred and seventy-five one-thousandths. Sixty-two and one-half one-thousandths. Seven hundred sixty-five and five one-thousandths. 72. Read the following decimals and write them out in words: .075 .137 100.037 .121 1.09375 73. Find the sum of .2143, 783.5, 138.72, and 10.0041. 74. From 241. 70 take 215.875. 76. a. Find the product of 78.8763 X .462. b. Multiply 21.3 by .071. 76. a. Divide 187.2421 by 123.42. b. Divide 25 by .0025. 77. Reduce in. to a decimal and compare with the table. 78. Reduce or, in. to a decimal and compare with the table. 79. Calculate the decimal equivalent of HT in. 80. Write .8125 as a common fraction and reduce it to the lowest terms. 81. If an alloy is .67 copper and .33 zinc, how many pounds of each metal x would there be in a casting weighing 75 lb.? v 82. A steam pump delivers 2.35 gallons of water per stroke and runs 48 strokes per minute; how many gallons will it deliver in one hour? 83. The diameter of No. 8 B. W. G. wire is .165 in. and of No. 12 wire is .109 in. What is the difference in diameter of the two wires? What do the letters B. W. G. stand for? I 84. A machinist whose rate is 27.5 cents per hour puts in a full day of 10 hours and also 3 hours overtime. If he is paid "time and a half" for overtime, how much should he be paid altogether? 3 85. The depth of a thread on a v in- bolt with U. S. Standard threads is .065 in. What is the diameter at the bottom of the threads? 86. I want 5000 ft. of $ in. Q (square) steel bars. I find from a table that this size weighs 1.914 lb. per foot of length. How many pounds must I order and what will it cost at $1.85 per 100 lb.? 2 i 87. Explain how you would set a micrometer for JQQQ in. over g in. \ 44 SHOP ARITHMETIC 88. A 28-tooth 7-pitch gear has an outside diameter of 4.286 in. The diameter at the bottom of the teeth is 3.67 in. How deep are the teeth cut? 89. A 2 in. pipe has an actual inside diameter of 2.067 in. The metal of the pipe is .154 in. thick. What is the outside diameter of the pipe? 90. Read the micrometer shown below in Fig. 7. Fio. 7. CHAPTER VI PERCENTAGE 44. Explanation. Percentage is merely another kind of fractions or, rather, a particular kind of decimal fractions, of which the denominator is always 100. Instead of writing the denominator, we use the term "per cent" to indicate that the denominator is 100. When we speak of "6 per cent" we mean T | 7 or .06. These all mean the same thing; namely, six parts out of one hundred. Instead of writing out the words "per cent" we more often use the sign % after the number, as, for instance, 6%, which means "6 per cent." Since per cent means hun- dredths of a thing, then the whole of anything is 100% of itself, meaning ]---, or the whole. If a man is getting 40 cents an hour and gets an increase of 10%, this increase will be 10% (or -j^V or .10) of 40 cents and this is easily seen to be 4 cents, so his new rate is 44 cents. Another way of working this would be to say that his old rate is 100% of itself and his increase is 10% of the old rate, so that altogether he is to get 110% of the old rate. Now 110% is the same as 1.10 and 1.10x40 = 44 cents, the new rate. PERCENTAGE 45 Any decimal fraction may be easily changed to per cent. 07 K 875-^-87.5%. Here we first change the decimal to a common fraction having 100 for a denominator. Then we drop this denominator and use, instead, the per cent sign (%) written after the numerator. This sign indicates, in this case, 87.5 parts out of 100, or 87.5 100 The change from a decimal to percentage can be made without changing to a common fraction as was just done. Having a decimal, move the decimal point two places to the right and write per cent after the new number. .625=62.5% .06=6% 1.10=110% If it is desired to use a certain number of per cent in calcula- tions, it is usually expressed as a decimal first and then the calculations are made. For example, when figuring the interest on $1250 at the rate of 6%, we would first change 6% to .06 and multiply $1250 by .06 which gives $75.00. $1250 .06 $75.00 A common fraction is reduced to per cent by first reducing it to a decimal and then changing the decimal to per cent. Example : The force in a shop is cut down from 85 men to 62. What per cent of the original number of men are retained? no 62 is | of 85. oo ' AO ^=.729 = 72.9% 62 is 72. 9% of 85. Therefore, the number of men retained is 72.9% or nearly 73% of the original number of men. If we want to reduce the fraction | to per cent, we first get = .125 and then, changing this decimal to per cent, we have .125 = 12.5%. Then | of anything is the same as 12% of it, because 46 SHOP ARITHMETIC The following table gives a number of different per cents with the corresponding decimals and common fractions: Per cent Decimal Fraction Per cent Decimal Fraction 1% .01 1 100 25% .25 25 1 100 4 2% .02 2 1 100 " 50 33J% .88| 33^ 1 100 3 2*% .025 2i .. 1 100 40 37}% .375 37J 3 100 ~ 8 5% .05 5 1 100 20 50% .50 50 1 100 2 61% .0625 6i _ 1 100 16 75% .75 75 3 100 ~ 4 10% .10 10 1 100 10 90% .90 90 9 100 10 12J% .125 12^ 1 100 ~ 8 100% 1.00 100 100 ~ 16f% -16| 163 l 100 6 200% 2.00 200 100~ * 45. The Uses of Percentage. In shop work, the chief use of percentage is to express loss or gain in certain quantities or to state portions or quantities that are used or unused, good or bad, finished or unfinished, etc. Very often we hear expressions like: "two out of five of those castings are bad;" or "nine out of ten of those cutters should be replaced." If, in the first illustration, we wanted to talk on the basis of a hundred castings instead of five, we would say "40 per cent of those castings are bad," because "two out of five" is the same as f, = T \\> =40%. And in the second case: "90 per cent of those cutters should be replaced." Here, "nine out of ten" =^,=^,=90%. If a piece of work is said to be 60% completed, it means that, if we divide the whole work on the job into 100 qual parts, we have already done 60 of these parts or -$ of the whole. If a shop is running with 50% of its full force, it means that T'W or \ f the full force is working. If the full force of men is 1300, then the present force is 50% of 1300 = .50X1300 = 650. If the full force were 700 men, then the 50% would be 350. Another very common use of percentage is in stating the por- tions or quantities of the ingredients going to make up a whole. We often see formulas for brasses, bronzes, and other alloys in PERCENTAGE 47 which the proportions of the different metals used are indicated by per cents. For example, brass usually contains about 65% copper and 35% zinc. Then, in 100 Ib. of brass, .there would be 65 Ib. of copper and 35 Ib. of zinc. Suppose, however, that instead of 100 Ib. we wanted to mix a smaller amount, say 8 Ib. The amount of copper needed would be 65 % or .65 of 8 Ib. .65x8 = 5.20 Ib., or 5 T \ Ib., the copper needed. .35X8 = 2.80 Ib., or 2 T 8 T Ib., the zinc needed. Sometimes, in dealing with very small per cents, we see a decimal per cent such as found in the specifications for boiler steel, where it is stated that the sulphur in the steel shall not exceed .04%. Now this is not 4%; neither is it .04; but it is .04%, meaning four one-hundredths per cent, or four one- hundredths of one one-hundredth. This is -j^-g- of T ^- = y o o o o > so if we write this .04% as a decimal, it will be .0004. It is a very common mistake to misunderstand these decimal per cents, and the student should be very careful in reading them. Likewise, be careful in changing a decimal into per cent that the decimal point is shifted two places to the right. 46. Efficiencies. Another common use of percentage is in stating the efficiencies of engines or machinery. The efficiency of a machine is that part of the power supplied to it, that the machine delivers up. This is generally stated in per cent, meaning so many out of each hundred units. If it requires 100 horse-power to drive a dynamo and the dynamo only generates 92 horse-power of electricity, then the efficiency of the dynamo is i\\ or 92%. If the engine driving a machine shop delivers 250 horse-power to the lineshaft, but the lineshaft only delivers 200 horse-power to the machines, then the efficiency of the lineshaft is ff = .80 = 80%. The other 50 horse-power, or 20%, is lost in the friction of the shaft in its bearings and in the slipping of the belts. The efficiencies of all machinery should be kept as high as possible because the difference between 100% and the effi- ciency means money lost. The large amount of power that is often lost in line shafting can be readily appreciated when we try to turn a shaft by hand and try to imagine the power that would be required to turn it two or three hundred times a minute. 47. Discount. In selling bolts, screws, rivets, and a great many other similar articles, the manufacturers have a standard list of prices for the different sizes and lengths and they give their 48 SHOP ARITHMETIC customers discounts from these list prices. These discounts or reductions in price are always given in per cent. Sometimes they -are very complicated, containing several per cents to be deducted one after another. Each discount, in such a case, is figured on the basis of what is left after the preceding per cents have been deducted. Example : The list price of i in. by 1} i Q - stove bolts is $1 per hundred. If a firm gets a quotation of 75, 10 and 10% discount from list price, what would they pay for the bolts per hundred? 100 X . 75 = 75 cents Explanation: 75, 10 and 10% discount means 100 75 = 25 cents 75% deducted from the list price, then 10% de- 1 ducted from that remainder, then 10% taken from = 2 ce the second remainder. 1 1 Starting with 100 cents, the list price, we de- 25-2^ = 22- cents duct the first discount of 75%. This leaves 25 1 , cents. The next discount of 10% means 10% off 22- X . 10 = 2- cents from this balance. Deducting this leaves 22* cents. Next, we take 10% from this, leaving 20\ cents 22! _ 2! = 20- cents per hundred as the actual cost of these stove bolts. 24 4 48. Classes of Problems. Nearly all problems in percentage can be divided into three classes on the same basis as explained in Article 26. There are three items in almost any percentage problem: namely, the whole, the part, and the per cent. For example, suppose we have a question like this: "If 35% of the belts in a shop are worn out and need replacing, and there are 220 belts altogether, how many belts are worn out?" In this case, the whole is the number of belts in the shop, 220. The part is the number of belts to be replaced, which is the number to be calculated. The per cent is given as 35%. Any two of these items may be given and we can calculate the missing one. We thus have the three cases: 1. Given the whole and the per cent, to find the part. 2. Given the part and the per cent that it is of the whole, to find the whole. 3. Given the whole and the part, to find what per cent the, part is of the whole. The principles taught under common fractions will apply equally well in working problems under these cases, the only difference being that here a per cent is used instead of a common fraction. In working problems, the per cent should always be changed to a decimal. One difficulty in working percentage problems is in deciding PERCENTAGE 49 just what number is the whole (or the base, as it is often called). The following illustration shows the importance of this. If I offer a man $2000 for his house, but he holds out for $3000, then his price is 50% greater than my offer, while my offer is 33^% less than his price. The difference is $1000 either way but, if we take my offer as the base, it would be necessary for me to raise it i, or 50%, to meet his price. On the other hand, for him to meet my bid, he would only have to cut his price J, or Examples of the three types of problems, before mentioned, may help somewhat in getting an understanding of the processes to be used. Example of Case i : How many pounds of nickel are there in 1 ton of nickel-steel containing 2.85% nickel? Explanation: Here we have the whole 1 ton = 2000 Ib. ( 200 lb -) and th e per cent (2.85%) to 2. 85% =.0285 find the part. After changing the 2.85% .0285X2000 = 57 lb., Answer. to a decimal fraction, the problem be- comes a simple problem in multiplica- tion of decimals. Example of Case 2 : The machines in a small pattern shop require altogether 12 horse-power and are to be driven from a lineshaft by a single electric motor. If we assume that 20% of the power of the motor wfll be lost in the line shaft and belting, what size motor must we install? 100% 20% = 80% Explanation: Here the per cent given (20%) 80%= .80, or .8 is based on the horse-power of the motor, which 12 -T-. 8 = 15, Answer. is, as yet, unknown. The horse-power of the motor is 100% of itself and, if 20% is lost, then the machines will receive 80%, or .8 of the power of the motor. This is 12 horse- power. The whole will be 12 -f-. 8 = 15 horse- power. This is the size of motor to install. Example of Case 3 : If the force in a shop is increased from 160 to 200 men, what per cent is the capacity of the shop increased? 200-5-160 = 1.25 Explanation: The present force is 1} or 1.25 = 125% 1.25 of the old force, because 200 is :';; of 125% -100% = 25%, Answer. 160, and fgg = l}. This is the same as 125%. The increase is, therefore, 25% of the former force. Note. To reduce the present force back to the old number would require a reduction of only 20%, because now the base is different on which to figure the per cent. 40 + 200= .20, or 20%. PROBLEMS 91. Write 4% as a decimal and as a common fraction. 92. Write 25% as a common fraction and reduce the fraction to its lowest terms. 4 50 SHOP ARITHMETIC 93. What per cent of an inch is ^ in.? 94. If there are 240 men working in a shop and 30% of them are laid off, how many men will be laid off and how many will remain at work? 96. Out of one lot of 342 brass castings, 21 were spoiled and out of an- other lot of 547, 32 were spoiled. Which lot had the larger per cent of spoiled castings? 96. 500 Ib. of bronze bearings are to be made; the mixture is 77% copper, 8% tin, and 15% lead. How many pounds of copper, tin, and lead are required? Note. This is a standard bearing mixture used by the Pa. R. R. and by some steam turbine manufacturers. 97. The boss pattern maker is given a raise of 25% on Christmas, after which he finds that he is receiving $130 a month. How much did he get per month before Christmas? 98. In testing a shop drive it was found that the machines driven by one motor required horse-power as follows: 60 in. mill 3.31 horse-power 20 in. lathe 75 horse-power 48 in. lathe 2.42 horse-power 42 in. by 42 in. by 12 ft. planer 4.82 horse-power 16 in. shaper 33 horse-power The total power delivered by the motor was 13.65 horse-power. What per cent of the total power was used in belting and lineshaft? What per cent by the machines? 99. A man who has been drawing $2.50 a day gets his pay cut 10% on May 1, and the following September he is given an increase of 10% of his rate at this time. How much will he get per day after September? 100. The following weights of metals are melted to make up a solder: 18 Ib. of tin, 75 Ib. of bismuth, 37.5 Ib. of lead, and 19.5 Ib. of cadmium. What per cent of the total weight is there of each metal? CHAPTER VII CIRCUMFERENCES OF CIRCLES; CUTTING AND GRINDING SPEEDS 49. Shop Uses. In the running of almost any machine, judg- ment must be used in order to determine the speed which will give the best results. Lathes, milling machines, boring mills, etc., are provided with means for changing the speed, according to the -judgment of the operator. Emery wheels and grind- stones, however, are often set up and run at any speed which the pulleys happen to give, regardless of the diameter. If an emery wheel of large size is put on a spindle that has been belted to drive a smaller wheel, the speed may be too great for the larger wheel and, if the difference is considerable, the large wheel may fly to pieces. Every mechanic should know how to calculate the proper sizes of pulleys to use for emery wheels or grindstones, the correct speed at which to run the work in his lathe, or the most economical speeds to use for belts and pulleys. A little data on this subject may be useful and will afford applications for arithmetical principles. 50. Circles. To understand what has just been mentioned, it is necessary to get a knowledge of circles and their properties. The distance across a circle, measured straight through the center, is called the Diameter. Circles are generally designated by their diameters. Thus a 6 in. circle means a circle 6 in. in diameter. Sometimes the radius is used. The Radius is the distance from the center to the edge or circumference and is, therefore, just half the diameter. If a circle is designated by the radius, we should be careful to say so. Thus, there would be no misunder- standing if we said " a circle of 5-in. radius" ; but unless the word radius is used, we always understand that the measurement given is the diameter. The Circumference is the name given to the distance around the circle, as indicated in Fig. 8. The circumference of any circle is always 3.1416 times the diameter. In other words, if we measure the diameter with a string and lay this off around the circle, it will go a little over three times. 5 51 52 SHOP ARITHMETIC This number 3.1416 is, without doubt, the most used in practical work of any figure in mathematics. In writing formulas, it is quite common to represent this decimal by the Greek letter TT (pronounced "pi")> instead of writing out the whole number. For this reason, the number 3.1416 is given the name "pi." FIG. 8. Where it is more convenient and extreme accuracy is not quired, the fractio exact value 3.1416. 22 required, the fraction -- may be used for n instead of the more = 3 = 3.1429 It, therefore, gives values of the circumference slightly too large, but in many cases it is sufficiently accurate and saves time. Examples : 1. What length of steel sheet would be needed to roll into a drum 32 in. in diameter? When rolled up, the length of the sheet will become the circumference of a 32-in. circle. The circumference must be x times 32. 3.1416X32 = 100.5+ in. The length of the sheet must, therefore, be 100 in. and, if it is to be lapped and riveted, we would have to add a suitable allowance of 1 in. or so for making the joint. 2. A circular steel tank measures 37 ft. 8$ in. in circumference. What is its diameter? If the circumference of a circle is 3.1416 times the diameter, then the diameter can be obtained by dividing the circumference by 3.1416. 37 ft. 8^ in. = 37^ ft. = 37.7+ ft. 2i 1 37.7 -4- 3.1416 = 12 ft., Answer. 61. Formulas. A formula, in mathematics, is a rule in which mathematical signs and letters have been used to take the place of words. We say that "the circumference of a circle equals CIRCUMFERENCES OF CIRCLES 53 3.1416 times the diameter." This is a rule. But suppose we merely write This is the same rule expressed as a formula. We have used C instead of the words "the circumference of a circle;" the sign = replaces the word "equals;" the symbol it is used instead of the number 3.1416; X stands for "times;" and D stands for "the diameter." We found in the second example under Article 50 that, when the circumference is given, we can obtain the diameter by divid- ing the circumference by n. As a formula this would be written D = - or D = C + K 7T This arrangement is useful when we want to get \he diameters of trees, chimneys, tanks, and other large objects. We can easily measure their circumferences and, by dividing by 3.1416, we get the diameters. Formulas do not save much, if any space, because it is necessary usually to explain what the letters stand for. They have, however, the great advantage that intricate mathematical opera- tions can be shown much more clearly than if they were written out in a long sentence or statement. One can usually see in one glance at a formula just what is to be done, with the numbers that are given in the problem, to find the quantity that is unknown. Example : What is the circumference in feet of a 16-in. emery wheel? C = 3.1416 X 16 = 50.2656 in. 50.2656 in. -T- 12 = 4.1888 ft., Answer. Explanation: We have the diameter given and want to get the circum- ference. We, therefore, use the formula which says that C = nXD. x is always 3.1416 and D in this case is 16 in. Then C comes out 50.2656 in. But the problem calls for the circumference in feet. This is T ' ? of the number of inches, or it is the number of inches divided by 12. In the work of this chapter, the circumferences of circles are always used in feet, and, consequently, should always be calculated in feet. If we use D in feet, we will get C in feet, while, if D is in inches, C will come out in inches. If the diameter can be reduced to exact feet, it is easiest to use the diameter in feet when multiplying by TT, rather than to reduce to feet after multiplying. 54 SHOP ARITHMETIC Example : What is the circumference of a 48-in. fly wheel? 48 in. -5- 12 = 4 ft., the diameter. C = nXD (7 = 3.1416X4 = 12.5664 ft., Answer. This is much shorter than it would be to multiply 3.1416 by 48 and then divide the product by 12. 52. Circumferential Speeds. When a fly wheel or emery wheel or any circular object makes one complete revolution, each point on the circumference travels once around the circum- ference and returns to its starting-point. When the wheel turns ten times, the point will have travelled a distance of ten times the circumference. In one minute, it will travel a distance equal to the product of the circumference times the number of revolu- tions per minute. The distance, in feet per minute, travelled by a point on the circumference of a wheel is called its Circum- ferential Speed, Rim Speed, or Surface Speed. It is also some- times called Peripheral Speed, because the circumference is sometimes given the name of periphery. It is the surface speed by which we determine how to run our fly wheels, belts, emery wheels, and grindstones, and what speeds to use in cutting materials in a machine. Written as a formula: S = CXN where : S is the surface speed C is the circumference N is the number of revolutions per minute (R.P. M.). Expressed in words this formula states that the surface speed of any wheel is equal to the circumference of the wheel multiplied by the number of revolutions per minute. Example : What would be the rim speed of a 7 ft. fly wheel when running at 210 revolutions per minute? C = xXD Explanation: First we find the circum- ^_22 ference of the wheel, by multiplying the ~^f diameter by x. Here is a case where it is S = CXN much easier to use *f- for n than to use the 5=^2x210 = 4620 decimal 3.1416, and the result is sufficiently 4620 ft. per min., Answer. accurate for our purposes. We get 22 ft. for the circumference. We can now get the rim speed, which is equal to the product of the circumference times the number of revolutions per minute; orS = CxN. C being 22 ft. and N being 210 revolutions per minute, we find that S is 4620 ft. per min. Hence, the rim of this fly wheel travels at a speed of 4620 ft. per minute. CIRCUMFERENCES OF CIRCLES 55 If we have given a certain speed which is wanted and have the circumference of the wheel, then the R. P. M. (revolutions per minute) will be obtained by dividing the desired speed by the circumference. In the example just worked, if we want to give the fly wheel a rim speed of 5280 ft. per minute, it requires no argument to show that the wheel will have to run at 5280-^-22 = 240 revolutions per minute. In such a case, we would use our formula in the form A7 S N= c This formula expresses the same relation as S = NxC, but now it is rearranged to enable us to find the R. P. M. when the rim speed and the circumference are given. Sometimes, especially with emery wheels, we know the proper surface speed and we have an arbor belted to run a certain number of R. P. M. The problem then is to find the proper size of stone to order. The desired speed divided by the number of R. P. M. will give the circumference, and from this we can figure the diameter of the stone. c- s C ~N Here again we have merely rearranged the formula S = CXN so as to be in more suitable form for finding the circumference when the surface speed and the R. P. M. are given. 53. Grindstones and Emery Wheels. Makers of emery wheels and grindstones usually give the proper speed for the stones in feet per minute. This refers to the distance that a point on the circumference of the stone should travel in 1 minute and is called the "surface speed" or the "grinding speed." The proper speed at which to run grindstones depends on the kind of grinding to be done and the strength of the stones. For heavy grinding they can be run quite fast. For. grinding edge tools they must be run much slower to get smooth sur- faces and to prevent heating the fine edges of the tools. The following surface speeds may be taken as representing good practice: Grindstones: For machinists' tools, 800 to 1000 ft. per minute. For carpenters' tools, 550 to 600 ft. per minute. 56 SHOP ARITHMETIC Grindstones for very rapid grinding: Coarse Ohio stones, 2500 ft. per minute. Fine Huron stones, 3000 to 3400 ft. per minute. Sometimes the rule is given for grindstones as follows: "Run at such a speed that the water just begins to fly." This is a speed of about 800 ft. per minute and would be a good average speed for sharpening all kinds of tools. Examples : 1. A 36-in. grindstone, used for sharpening carpenters' and pattern- makers' tools, is run at 60 R. P. M. Is this speed correct? We must first find the circumference and then the surface speed to see if it falls between the allowable limits. 36 in. -i- 12 = 3 ft., the diameter Explanation: First we find the C = 7tXD circumference, which comes out C = 3. 1416X3 = 9.4248 ft. 9.4248ft. Using this and the R. P. S = CxN M., we find S to be 565 F. P. M. 5 = 9.4248X60 = 565.488 (feet per minute). As this lies be- S = 565.488 ft. per minute. tween the allowed limits (550 to 600 F. P. M.) the speed of the stone is correct. 2. At what R. P. M. should a 50-in. Huron stone be run if it is to be used for rough grinding? C nXD Explanation: First we find the C = 3. 1416X50 = 157.08 in. circumference of the stone in feet, C = 157.08 in. = 13.09 ft. which turns out to be a little over 13 ~~To ft. The proper speed is given as g 3000 to 3400 F. P. M. Trying 3200 N=Y? .we find that N comes out 246 R. TVI Q Qonn v P M P. M. The stone should, therefore, Take S = 320C F. P. M. be be]ted to fun about 24Q Qr 25Q AT = 3200 = 246.+ R.P.M. R.P.M. lo Emery wheels are usually run at a speed of about 5500 ft. per minute. A good, ready rule, easy to remember, is a speed of a mile a minute. Most emery wheel arbors are fitted with two pulleys of different diameters. When the wheel is new, the larger pulley on the arbor should be used and, when the wheel becomes worn down sufficiently, the belt should be shifted to the smaller pulley. Never shift the belt on an emery wheel, however, with- out first calculating the effect on the surface speed of the wheel. Many serious accidents have been caused by emery wheels bursting as a result of being driven at too great a speed. Before cutting a new wheel on an arbor the resultant surface speed CIRCUMFERENCES OF CIRCLES 57 should be calculated, to see if the R. P. M. is suitable for the size of the wheel. Example : What size wheel should be ordered to go on a spindle running 1700 R. P.M.? -I 3 - 106 D = 12 in. wheel, Answer. Note. A wheel of exactly 12 in. diameter would, at 1700 R. P. M., have* a surface speed of 5340 F. P. M. (1700 XT: = 5340). 54. Cutting Speeds. Cutting speeds on lathe and boring mill work may be calculated in the same way that grinding speeds are calculated. The life of a lathe tool depends on the rate at which it cuts the metal. This cutting speed is the speed with which the work revolves past the tool and is, therefore, obtained by multiplying the circumference of the work by the revolutions per minute. The same formulas are used as in the calculations for emery wheels and grindstones but, of course, the allowable speeds are much different. Tables of proper cutting speeds are given in many handbooks in feet per minute. To find the necessary R. P. M., divide the cutting speed by the circumference of the work. The cutting speeds used in shops have increased considerably with the advent of the high speed steels. No exact figures can be given for the best speeds at which to cut different metals. The proper speed depends on the nature of the cut, whether finishing or roughing, on the size of the work and its ability to stand heavy cuts, the rigidity and power of the lathe, the nature of the metal being cut, and the kind of tool used. If the work is not very rigid it is, of course, best to take a light cut and run at rather high speed. On the other hand, it is generally agreed that more metal can be removed in the same time if a moderate speed is used and a heavy cut taken. As nearly as any general rules can be given, the following table gives about the average cutting speeds. 58 SHOP ARITHMETIC CUTTING SPEEDS IN FEET PER MINUTE Kind of tool Material Carbon steel High speed steel Cast iron 30 to 40 60 to 80 Steel or wrought iron 25 to 30 50 to 60 Tool steel. . . 20 to 25 40 to 50 Brass. . . . . . 80 to 100 160 to 200 Cutting speed per minute (in feet) revolutions per minute. ^Circumference of work (in feet) Example : A casting is 30 in. in diameter. Find the number of R. P. M. necessary for a cutting speed of 40 ft. per minute. 94.248 _ _ ^,^ = 7.804 ft., circumference. \.Zi N = ~ =^ ^7 = 5.09 R. P. M., Answer. / . oo4 The same principles apply to milling and drilling, except that in these cases the tool is turning instead of the work. Conse- quently, the cutting speeds are obtained from the product of the circumference of the tool times its R. P. M. In calculating the cutting speed of a drill, take the speed of the outer end of the lip or, in other words, the speed of the drill circumference. Example : A ^-in. drill is making 300 revolutions per minute; what is the cutting speed? 3.1416X^ = 1.5708, circumference in inches a =.131 ft. (nearly) .131X300 = 39.3 ft. per minute, cutting speed. 55. Pulleys and Belts. If the rim of a pulley is run at too great a speed, the pulley may burst. The rim speeds of pulleys are calculated in the same manner as are grinding and cutting speeds. A general rule for cast iron pulleys is that they should not have a rim speed of over a mile a minute (5280 ft. per minute) . RATIO AND PROPORTION 59 This speed may be exceeded somewhat if care is taken that the pulley is well balanced and is sound and of good design. The proper speeds for belts is taken up fully in a later chapter under the general subject of belting. It is well, however, to point out now that the speed at which any belt is travelling through the air is practically the same as that of the rim of either of the pulleys over which the belt runs; and, if we neglect the small amount of slipping which usually occurs between a belt and its pulleys, we can say that the speed of a belt is the same as the rim speed of the pulleys. It will be seen from this that if two pulleys are connected by a belt, their rim speeds are practically the same. PROBLEMS 101. A stack is measured with a tape line and its circumference found to be 88 in. What is the diameter of the stack? 102. An emery wheel 16 in. in diameter runs 1300 R. P. M. Find the sur- face speed. 103. The Bridgeport Safety Emery Wheel Co., Bridgeport, Conn., build an emery wheel 36 in. in diameter and recommend a speed of 425 450 revolutions. Calculate the surface speeds at 425 and at 450 revolutions. 104. An emery wheel runs 1000 R. P. M. What should be its diameter to give a surface speed of 5500 ft.? 105. A grindstone 3 \ ft. in diameter is to be used for grinding carpenters' tools; how many R. P. M. should it run? 106. Calculate the belt speed on a high-speed automatic engine carrying a 48 in. pulley and running at 250 R. P. M. 107. How many revolutions will a locomotive driving wheel, 72 in. in diameter, make in going 1 mile? 108. What would be the rim speed in feet per minute of a fly wheel 14 ft. in diameter running 80 R. P. M.? 109. At how many R. P. M. should an 8 in. shaft be driven in a lathe to give a cutting speed of 60 ft. per minute? 110. At what R. P. M. should a 1^ in. high speed drill be run to give a cutting speed of 80 ft. per minute? If the drill is fed .01 in. per revolution, how long will it take to drill through 2 in. of metal? CHAPTER VIII RATIO AND PROPORTION 66. Ratios. In comparing the relative sizes of two quantities, we refer to one as being a multiple or a fraction of the other. If one casting weighs 600 lb., and another weighs 200 lb., we say that the first one is three times as heavy as the second, or that GO SHOP ARITHMETIC the second is one-third as heavy as the first. This relation between two quantities of the same kind is called a Ratio. In comparing the speeds of two pulleys, one of which funs 40 revolutions per minute and the other one 160 revolutions per minute, we say that their speeds are " as 40 is to 160," or " as 1 is to 4." In this sentence, "40 is to 160" is a ratio, and so also is "1 is to 4" a ratio. Ratios may be written in three ways. For example, the ratio of (or relation between) the diameters of two pulleys which are 12 in. and 16 in. in diameter can be written as a fraction, -J-f ; or, since a fraction means division, it can be written 12-7-16; or, again, the line in the division sign is sometimes left out and it becomes 12:16. The last method, 12:16, is the one most used and will be followed here. It is read "twelve is to sixteen." A ratio may be reduced to lower terms the same as a fraction, without changing the value of the ratio. If one bin in the stock room contains 1000 washers, while another bin contains 3000, then the ratio of the contents of the first bin to the contents of the second is " as 1000 is to 3000." The relation of 1000 to 3000 can be reduced by dividing both by 1000. This leaves the ratio 1 to 3. 1000 -s- 1000 1 Hence, the ratio between the contents of the bins is also as 1 is to 3. Likewise, the ratio 24:60 can be reduced to 2:5 by dividing both terms by 12. If we write it as a fraction we can easily see that 24 = 24--12_2 60~60-f-12~5 Therefore, 24:60 = 2:5. The ratio of the 1000 washers to the 3000 washers is 1000:3000 or 1:3. The ratio of 8 in. to 12 in. is 8:12 or 2:3. The ratio of $1 to $1.50 is 1:1 J or 2:3. The ratio of 30 castings to 24 castings is 30:24 or 5:4. 57. Proportion. When two ratios are equal, the four terms are said to be in proportion. The two ratios 2:4 and 8:16 are clearly equal, because we can reduce 8:16 to 2:4 and we can RATIO AND PROPORTION 61 therefore write 2:4 = 8:16. When written thus, these four numbers form a Proportion. Likewise, we can say that the numbers 6, 8, 15, and 20 form a proportion because the ratio 6:8 is equal to the ratio of 15:20. 6:8 = 15:20 Now, it will be noticed that, if the first and fourth terms of this proportion be multiplied together, their product will be equal to the product of the second and third terms: /First\ /SecondN /Third\ /Fourth\ Vterm/ V term ) \ term / \ term / 6:8 15 , : 20 6X20 = 120 8X15 = 120 Therefore, 6X 20 = 8X15 This is true of any proportion and forms the basis for an easy way of working practical examples, where we do not know one term of the proportion, but know the other three. The first and fourth terms are called the Extremes, and the second and third are called the Means. Then we have the rule: " The product of the means is equal to the product of the extremes." This relation can be very nicely and simply expressed as a formula. Let a, &, c, and d represent the four terms of any proportion so that a : b=c : d Then, according to our rule, we have aXd=bXc Let us now see of what practical use this is. We will take this example: If it requires 137 Ib. of metal to make 19 castings, how many pounds will it take to make 13 castings from the same pattern? Now very clearly the ratio between the number of castings 19:13 is the same as the ratio of the weights, but one of the weights we do not know. Writing the proportion out and put- ting the word " answer" for the number which we are to find, we have 19:13 = 137 :Answer 02 SHOP ARITHMETIC From our rule which says the product of the means equals the product of the extremes: 13 X 137 1781, product of " means." This must equal the product of the extremes which would be 19 X Answer. Then: 19 X Answer = 1781 Answer = 1781-*- 19 Answer = 93. 7+ Ib. In using proportion keep the following things in mind: (1) Make the number which is the same kind of thing as the required answer the third term. Make the answer the fourth term. (2) See whether the answer will be greater or less than the third term; if less, place the less of the other two numbers for the second term; if greater, place the greater of the other two numbers for the second term. (3) Solve by knowing that the product of the means equals the product of the extremes, or by this rule: Multiply the means together and divide by the given extreme; the result will be the other extreme or answer. Let us see how these rules would be applied to a practical example. Example : A countershaft for a grinder is to be driven at 450 R. P. M. by a lineshaft that runs 200 R. P. M. If the pulley on the countershaft is 8 in. in diameter, what size pulley should be put on the lineshaft? A proportion can be formed of the pulley diameters and their revolutions per minute. Applying the rules of proportion, we get the following analysis and solution to the problem. (1) The diameter of the lineshaft pulley is the unknown answer. The other number of the same kind is the diameter of the countershaft pulley (8 in.). So we have the ratio. 8 :Answer (2) If the countershaft pulley is to run faster, its diameter must be smaller than the other one. Therefore, the answer is greater than 8. Hence, the greater revolutions (450) will be placed as the second term and the other R. P. M. (200) will be the first term. Therefore, we have the completed proportion: 200:450 = 8 : Answer (3) Solving this we get: 450X8 = 3600, product of means. 3600-^-200 = 18, Answer. Hence, an 18-in. pulley should be put on the lineshaft to give the desired speed to the countershaft. RATIO AND PROPORTION 03 Sometimes the letter X is used to represent the unknown number whose value is sought. The following is an example of such a case. 6:40 = 5:X. Find what number X stands for. 40 X 5 = 200, product of the means. Hence, 6XX = 200 200 X=~ =33.3+, Answer. 58. Speeds and Diameters of Pulleys. As shown in an example previously worked, if two pulleys are belted together, their diameters and revolutions per minute can be written in a pro- portion having diameters in one ratio and R. P. M. in the other ratio of the proportion. It will be noticed from the example which was worked, that the numbers which form the means apply to the same pulley, while the extremes both refer to the other pul- ley. Then, since the product of the means equals the product of the extremes, we obtain the following simple relation for pulleys belted together: The product of the diameter and revolu- tions of one pulley equals the product of the diameter and revolu- tions of the other. This gives us the following simple rule for working pulley problems. Rule for Finding the Speeds or Diameters of Pulleys. Take the pulley of which we know both the diameter and the R. P. M., and multiply these two numbers together. Then divide this product by the number that is known of the other pulley. The result is the desired number. Examples : 1. A 36-in. pulley running 240 R. P. M. is belted to a 15-in. pulley. Find the R. P. M. of the 15-in. pulley. 36 X 240 = 8640, the product of the known diameter and revolutions. 8640 -=-15 = 576, the R. P. M. of the 15-in. pulley, Answer. 2. A 36-in. grindstone is to be driven at a speed of 800 R. P. M. from a 6-in. pulley on the lineshaft which is running 225 R. P. M. \Vhat size pulley must be put on the grindstone arbor? S Explanation: First we must find C the R. P. M. for the grindstone as 800 explained in Chapter VII. To get = Q . .-- ~ =85 R. P. M., nearly, the required surface speed we find v99* ,o- n 85 R. P. M. necessary. , Now we have the R. P. M. and the size of the lineshaft pulley. Use a 16-m. pulley on the arbor. The product of these two nu J jnber8 is 1350. Dividing this by the R. P. M. of the grindstone arbor gives 16 in. as the nearest even size of pulley, so we will use that size. 64 SHOP ARITHMETIC 69. Gear Ratios. The same principles as are applied to pulleys can be applied to gears. If we have two gears running together as shown in Fig. 9, the product of the diameter 'and R. P. M. of one gear will be equal to the product of the diameter and R. P. M. of the other. In studying gearing, we do not deal with the diam- eters so much as we do with the numbers of teeth. We find that gears are generally designated by the numbers of teeth. For example, we talk of 16 tooth gears and 24 tooth gears, etc., but we seldom talk about gears of certain diameters. In making these calculations for gears, we can use the numbers of teeth instead of the diameters. When a gear is revolving, the number of teeth that pass a certain point in one minute will be the product of the number of teeth times the R. P. M. of the gear. FIG. 9. If this gear is driving another one, as in Fig. 9, each tooth on the one gear will shove along one tooth on the other one. Conse- quently, the product of the number of teeth times R. P. M. of the second gear will be the same as for the first gear. This gives us our rule for the relation of the speeds and numbers of teeth of gears. Rule for Finding the Speeds or Numbers of Teeth of Gears. Take the gear of which we know both the R. P. M. and the num- ber of teeth and multiply these two numbers together. Divide their product by the number that is known about the other gear. The quotient will be the unknown number. PULLEY AND GEAR TRAINS 65 Example : A 38 tooth gear running 360 R. P. M. is to drive another gear at 190 R. P. M. What must be the number of teeth on the other gear? 38 X 360 = 13,680, the product of the number of teeth and revolutions of one gear. 1 90 X Answer = 13, 680 13,680 Answer - 190 Answer = 72 teeth. PROBLEMS 111. (a) If you draw $33.00 on pay day and another man draws $22.00, what is the ratio of your pay to hist (b) What is the ratio of his pay to yours? 112. The speeds of two pulleys are in the ratio of 1:4. If the faster one goes 260 R. P. M., how fast does the slower one go? 113. Two castings are weighed and the ratio of their weights is 5:2. If the lighter one weighs 80 lb., what does the heavier one weigh? 114. Find the unknown number in each of the following proportions: (a) 2:10 = 5:Answer (b) 6:42 = 5:Answer (c) 7:35 = 10:X (d) 6:72 = 8 iX" 115. If it takes 72 lb. of metal to make 14 castings, how many pounds are required to make 9 castings? 116. A 14 tooth gear is driving a 26 tooth gear. If the 14 tooth gear runs 225 revolutions per minute, what is the speed of the 26 tooth gear? 117. A 12 in. lineshaft pulley runs 280 revolutions and is belted to a machine running 70 revolutions. What must be the size of the pulley on the machine? 118. A lineshaft runs 250 R. P. M. A grinder with a 6 in. pulley is to run 1550 R. P. M. Determine size of pulley to put on the lineshaft to run the grinder at the desired speed. 119. An apprentice was given 100 bolts to thread. He completed three- fifths of this number in 45 minutes and then the order was increased so that it took him 2 hours for the entire lot. How many bolts did he thread? 120. A 42 in. planer has a cutting speed of 30 ft. per minute and the ratio of cutting speed to return speed of the table is 1 : 2.8. What is the return speed in feet per minute? CHAPTER IX PULLEY AND GEAR TRAINS CHANGE GEARS 60. Direct and Inverse Proportions. A proportion formed of numbers of castings and the weights of metal required to make them is a direct proportion, because the amount of metal required increases directly as the number of castings increases. 66 SHOP ARITHMETIC When two pulleys (or gears) are running together, one driving the other, the larger of the two is the one that runs the slower. The proportion formed from their diameters and revolutions is, therefore, called an Inverse Proportion, because the larger pulley runs at the slower speed. The number of revolutions of one pulley is said to vary inversely as its diameter, since the greater the diameter, the less the number of revolutions it will make. In every pair of gears one of them is driving the other, so the one can be called the driving gear, or the driver, and the other the driven gear, or the follower. These names are in quite general use to designate the gears and to assist in keeping the propor- tions in the right order. Accordingly, we have the proportion: R. P. M. of driven /R. P. M. of\ = /No. of teethN /No. of teethN ' y driver / \ on driver / ' \ on driven / This is an inverse proportion because the driver and the driven are in the reverse order in the second ratio from what they are in the first ratio. Perhaps this can be seen better if the ratios are written as fractions. R. P. M. of driven _ No. of teeth on driver R. P. M. of driver No. of teeth on driven Fio. 10. Here the reason for the name "inverse proportion" is easily seen. The second fraction has the driver and the driven inverted from what they are in the first fraction. This method of writing proportions as fractions is much used in solving problems in gears or pulleys. 61. Gear Trains. A gear train consists of any number of gears used to transmit motion from one point to another. Fig. PULLEY AND GEAR TRAINS 67 10 shows the simplest form of gear train, having but two gears. Fig. 1 1 shows the same gears A and J3, as in Fig. 10, but with a third gear, usually called an intermediate gear, between them. The intermediate gear C can be used for either of two reasons: 1. To connect A and B and thus permit of a greater distance between the centers of A and B without increasing the size of the gears; or 2. To reverse the direction of rotation of either A or B. If A turns in a clockwise direction, as shown in both Figs. 10 and 11, B in Fig. 10 will turn in the opposite, or counter-clockwise direction, but in Fig. 11, B will turn in the same direction as A. FIG. 11. The introduction of the intermediate gear C has no effect on the speed ratio of A to B. If A has 48 teeth and B 96 teeth, the speed ratio of A to B will be 2 to 1 in either Fig. 10 or Fig. 11. In Fig. 10 suppose A to be the driver. R. P. M. of driver _ No. of teeth on driven R. P. M. of driven" No. of teeth on driver R. P. M. of driver _ 96 _2 R. P. M. of driven" 48~I Hence, the speed ratio of A to B is 2 to 1. In the case shown in Fig. 11 when A moves a distance of one tooth, the same amount of motion will be given to C, and C must at the same time move B one tooth. To move B 96 teeth, or one revolution, will require a motion of 96 teeth on A, or two revolutions of A. Hence, A will turn twice to each one turn of 6 68 SHOP ARITHMETIC B, or the speed ratio of A to B is 2 to 1, just as in the case of Fig. 10. 62. Compound Gear and Pulley Trains. Quite often it is desired to make such a great change in speed that it is practically necessary to use two or more pairs of gears or pulleys to accom- plish it! If a great increase or reduction of speed is made by a single pair of gears or pulleys, it means that the difference in the diameters will have to be very great. The belt drive of a lathe is an example of a compound train of pulleys, though here the train is used chiefly for other reasons. In the first step, the pulley on the lineshaft drives a pulley on the countershaft; then another pulley on the countershaft drives the lathe. The back gearing on a lathe is an example of compound gearing, two pairs of gears being used to make the speed reduction from the cone pulley to the spindle and face plate. FIG. 12. Fig. 12 shows a common arrangement of compound gearing. Here A drives B and causes a certain reduction of speed. B and C are fastened together and therefore travel at the same speed. A further reduction in speed is made by the two gears C and D. A and C are the driving gears of the two pairs and B and D are the driven gears. In making calculations dealing with compound gear or pulley trains, we might make the calculations for each pair as explained in Chapter VIII and then proceed to the next pair, etc., but this can be shortened to form a much simpler process. The speed ratio for a pulley or gear train is equal to the product of the ratios of all the separate pairs of pulleys or gears making up the train. PULLEY AND GEAR TRAINS 69 In using this principle for calculations, the ratios are written as fractions and Ave have the following formula: R. P. M. of last driven gear Product of Nos. of teeth of all drivers K. P. M. of first driver 1 roduct of Nos. of teeth of all driven gears Or, if we want the ratio stated the other way around R. P. M. of first driver Product of Nos. of teeth of all driven gears R. P. M. of last driven gear" Product of Xos. of teeth of all drivers Example : Let us calculate the speed ratio for the train of gears in Fig. 12. This would be the ratio of the speed of A to the speed of D. A is the first driver and D the last driven gear, and the ratio of their speeds is the ratio for the whole train. Speed of A ^ Teeth on Bx Teeth on D Speed of~Z> "" Teeth on A X Teeth on C 2 5 I 10 Hence, Speed of A: Speed of D = 10:l In other words, A revolves 10 times as fast as D. Problems in getting the speed ratios of pulley trains are solved in the same way except that diameters are used instead of num- bers of teeth. Speed of last driven pulley _ Product of diameters of all driving pulleys Speed of first driving pulley Product of diameters of all driven pulleys Example : Let us take the pulley train of Fig. 13 and calculate the ratio of the speeds of pulleys A and D. A and C are the drivers and B and D are the driven pulleys. Speed of A _ Diameter of B X Diameter of D Speed of D Diameter of A X Diameter of C 5 _ 5 __ i_ ~36 7.2 3 12 Hence, Speed of A : Speed of D = 1 : 7.2 Trains are frequently used having combinations of pulleys and gears. In nearly all machine tools, we will find both pulleys and gears between the lineshaft and the work. In wood-working 70 SHOP ARITHMETIC machinery, on the other hand, we usually find only pulleys and belts, on account of the high speeds at which the machines are run. In calculating the speed ratios of these combined trains, we can use the diameters of the pulleys and the numbers of teeth of the gears in the same formula. /R. P. M. of\ /Product of diameters\ x /Product of teeth oA \first driver/ _ \of all driven pulleys/ \ all driven gears / /R. P. M. of\ = /Product of diametersX /Product of teeth of\ yast driven j \of all driving pulleys j x \^ all driving gears ) Fio. 13. If a problem calls for the calculation of the size of one pulley or gear in a train, all the others and the speed ratio being known, start at one or both ends of the train and work toward the gear or pulley in question until you get a proportion which will give the desired quantity. Example: The punch shown m Fig. 14 is to be set up so that it will make 20 strokes per minute. (The 80 tooth gear must, therefore, run 20 R. P. M.) The punch is to be driven from a countershaft and we want to calculate the size of the pulley to put on the countershaft, to drive the punch at the desired speed. We find that the main lineshaft runs 240 R. P. M. and carries a 16-in. pulley which drives a 24-in. pulley on the countershaft. Working from the lineshaft: 24 in.xR. P. M. of countershaft = 16X240. R. P. M. of countershaft=^p = 160 R. P. M. 24 PULLEY AND GEAR TRAINS Working from the punch: 20XR. P. M. of 20 T gear = 80X20. R. P. M. of 20 T gear = -^-- = 80 71 This is also the R. P. M. of the 24-in. pulley and this pulley is driven by the unknown pulley on the countershaft, which we have found runs 160 R. P. M. IGOxDiam. of pulley = 80X24 1920 Diam. of pulley = -^r- = 12 in., Answer. Fio. 1 1. Another way to solve this would be to write put an equation for the entire train, using X to represent the pulley whose size we want to find. 20 = 16 XXX 20 240 "24X24X80 ? 1 == 12 6X24 6 24 XI IX 1 Now, if ^ X is to equal ^-.y then must be which would be when A" 12 24 is 12. Hence, X 12 in., Answer. 72 SHOP ARITHMETIC 63. Screw Cutting. Most lathes are equipped with a small plate giving the necessary gears to use for cutting different threads, but every good machinist should know how to calculate the proper gear setting for such work. This is a simple problem in gear trains and should cause no difficulty for the man who understands the principles of gear trains. The lathe carriage and tool are moved by a "lead screw" having usually 2, 4, 6, or 8 threads per inch. If a lathe has a lead screw having 6 threads per inch, each revolution of the lead screw will move the carriage in. ; a 4 pitch screw would move the carriage \ in. for each revolution of the screw. Then, if the spindle of the lathe and the lead screw turn at the same speed, the lathe will cut a thread of the same pitch as that on the lead screw. If a finer thread is wanted than that on the lead screw, the spindle should make more turns than does the lead screw. Suppose we want to cut 24 threads per inch and have a 6 thread per inch lead screw. It will require 6 turns of the lead screw to move the carriage 1 inch. Meanwhile, the work should revolve 24 times. Then the ratio of spindle speed to lead screw speed should be 4:1 Speed of spindle Threads per inch to be cut Speed of lead screw Threads per inch on lead screw The first driving gear is that on the spindle, while the last driven gear is that on the end of the lead screw. Hence, Threads per inch to be cut _ Product of Nos. of teeth on driven gears Threads per inch on lead screw Product of Nos. of teeth on driving gears PROBLEMS 121. In Fig. 10, if we removed the 48 tooth gear and put a 64 tooth gear in its place, what would be the speed ratio of A to B? 122. In Fig. 11, if B makes 6 revolutions, how many turns will C make and how many will A make? 123. What would be the speed ratio of the train of Fig. 12 if we put a 32T gear on at C and a 48T gear at D? 124. The lineshaft in Fig. 15 runs 250 R. P. M. Determine the size of lineshaft pulley to run the grinder at 1550 R. P. M. using the countershaft as shown in the figure. 126. A machinist wishes to thread a pipe on a lathe having 2 threads per inch on the lead screw. There are to be 11^ threads per inch on the pipe. What is the ratio of the speeds of the spindle and the lead screw? 126. Two gears are to have a speed ratio of 4.6 to 1. If the smaller gear has 15 teeth, what must be the number of teeth on the larger gear? PULLEY AND GEAR TRAINS 73 COUNT CJ?SHAFT GRINDLR FIQ. 15. FIG. IS. 74 SHOP ARITHMETIC 127. It has been decided to equip the punch in Fig. 14 with a motor drive by replacing the fly wheel with a large gear to be driven by a small pinion on the motor. If the motor runs 800 R. P. M., and has a 16 tooth pinion, what must be the number of teeth on the other gear? Speed of the punch to be 20 strokes per minute. 128. A street car is driven through a single pair of gears, a large gear on the axle being driven by a smaller one on the motor shaft. If a car has 33-in. wheels and a gear ratio of 1 :4, how fast would the car go when the motor is running 1200 R. P. M.? 129. Fig. 16 shows the head stock for a lathe. The cone pulley carries with it the cone pinion A, which drives the back gear B. B is connected solidly with the back pinion C which drives the face gear D. If the gears have the following numbers of teeth, determine the back gear ratio (speed of A : speed of D) : Teeth on cone pinion, A 28 Teeth on back gear, B 82 Teeth on back pinion, C 25 Teeth on face gear, D 74 130. If you were to cut a 20 pitch thread on a lathe having a 4 pitch lead screw, what would be the ratio of the speeds of the spindle and the lead screw? CHAPTER X AREAS AND VOLUMES OF SIMPLE FIGURES 64. Squares. In taking up the calculation of areas of surfaces and the volumes and weights of objects, the expressions "square" and "square root" will be met and must be understood. To one unfamiliar with these names and the corresponding operations the signs and operations themselves seem difficult. They are in reality very simple. The square of a number is simply the prod- uct of the number multiplied by itself; the square of 2 is 2 X2 = 4; the square of 5 is 5X5 = 25; the square of 12.5 is 12.5X12.5 = 156.25. Instead of writing 2x2 or 5X5, it is customary to write 2 2 and 5 2 . These are read "2 squared" and "5 squared." 12.5 2 =12.5 squared, and so on. The little 2 at the upper right hand corner is called the Exponent. 65. Square Root. The square root of a given number is simply another number which, when multiplied by itself (or squared), produces the given number. Thus, the square root of 4 is 2, since 2 multiplied by itself (2X2) gives 4. The square root of 9 is 3, since 3X3=3 2 = 9. Square root is the reverse of square, so if the square of 5 is 25 the square root of 25 is 5. The mathematical sign of square root, called the radical sign, is \/7 Then V9 = 3; V 25 = 5. These expressions are read "the square root of 9=3"; "the square root of 25 = 5". Square roots of larger numbers can usually be found in handbooks and the actual process of calculating them, which is somewhat compli- cated, will be taken up later on. 66. Cubes and Higher Powers. In the same way that 2 2 (2 squared) =2X2 =4, 2 3 (2 cubed) =2x2x2 = 8. The exponent simply indicates how many times the number is used as a factor, or how many times it is multiplied together. 4 s = 4x4x4 = 64. 3 3 =3X3X3 = 27. Just as square root is the reverse of square, so cube root is the reverse of cube. The sign for cube root is V7 So if 3 8 = 3X3X3=27, then ^27 = 3. Sometimes a factor is repeated more than 3 times, in which case, the exponent indicates the 7 75 76 SHOP ARITHMETIC number of times. 2* means 2x2x2x2 and is read "2 to the fourth power." 2 s = 2X2X2X2X2 and is read "2 to the fifth power," and so on.'" The roots are indicated in the same way. ^16 = fourth root of 16 = 2; 5 4 = 5X5X5X5 = 625, etc. 67. Square Measure. Before going further, it will be well to get clearly in mind just what the term "Square" means in terms of the things we see. Areas of figures are measured in terms of the "square" unit. For instance, if the dimensions of the base of a milling machine are 3 ft. by 5 ft., the floor space covered by this base is 15 square feet. In this case the area is 5'- FIQ. 17. measured by the unit known as the square foot. A Square Foot is a surface bounded by a square having each side 1 ft. in length. In case of the milling machine base represented in Fig. 17, there are by actual count 15 sq. ft. in this surface and this is readily seen to be the product of the length and the breadth of the base, since 3X5 = 15. The Square Inch is another common unit of area. This is much smaller than the square foot, being only one-twelfth as great each way. If a square foot is divided into square inches it will be seen to contain 12X12 or 144 sq. in. (see Fig. 18). It will be readily seen that the area of any square is equal to the product of the side of the square by itself. In other words, the area of a square equals the side "squared" (referring to the process explained in Article 64). Looking at it the other way around, the square of any number can be represented by the area of a square figure, one side of which represents the number itself. The actual things which the number represents makes no difference whatever. If the side of a square is 5 in., the area is AREAS AND VOLUMES OF SIMPLE FIGURES 77 25 sq. in.; if the side is 5 ft., the area is 25 sq. ft. If we simply have the number 5, its square is 25, no matter what kind of things the 5 may refer to. As mentioned before, 1 sq. ft. is the area of a square 1 ft. on each side and, if divided into square inches, will be found to contain 12 2 or 144 sq. in. Likewise, a square yard is 3 ft. on each side and, therefore, contains 3* = 9 sq. ft. The following table gives the relation between the units ordinarily used in measuring areas: FIG. 18. MEASURES OF AREA (SQUARE MEASURE) 144 square inches (sq. in.) = 1 square foot (sq. ft.) 9 square feet=l square yard (sq. yd.) 30 J square yards = 1 square rod (sq. rd.) 160 square rods=l acre (A) 640 acres = 1 square mile (sq. mi.) 68. Area of a Circle. If a circle is drawn in a square as shown in Fig. 19, it is easily seen that it has a smaller area than the square because the corners are cut off. The area of the circle is always a definite part of the area of the square drawn on its diameter, the area of the circle being always .7854 times the area of the square. This number .7854 happens to be just one- fourth of the number 3.1416 given in Chapter VII. Just why this is so will be shown later on. If the diameter of the circle = 10 in., as in Fig. 19, the area of the square is 100 sq. in. and the area of the circle is .7854x100 = 78.54 sq. in. You can prove this to your own satisfaction in the following manner. Cut a 78 SHOP ARITHMETIC square of carHboard of any size, and from the center describe a circle as shown just touching on all four sides. Weigh the square, and then cut out the circle and weigh it. The circle will weigh . 7854 X weight of the square. A pair of balances such as are found in a drug store are the best for this experiment. 10"- FIQ. 19. Rule for Area of Circle. The area of any circle is obtained by squaring the diameter and then multiplying this result .by .7854. If written as a formula this rule would read where A = area of a circle of which D = the diameter. Example : Find the area of a circle 3 in. in diameter. A= .7854 XD 2 A= .7854 X3 2 = .7854X9 _ = 7.0686 sq. in., Answer. If you think a little you will see that, if the diameter is doubled, the area is increased four times. This can also be seen from Fig. 20. The diameter of the large circle is twice that of one of the small circles, but its area is four times that of one of the small circles. This is a very important and useful law and may be stated as follows: "The areas of similar figures are to each other as the squares of their like dimensions." A 2 in. circle contains 2 2 X. 7854 = 3. 1416 sq. in., while a 6 in. circle contains 6 2 X .7854 = 28.2744 sq. in., or nine times as much. This we can find AREAS AND VOLUMES OF SIMPLE FIGURES 79 by saying the 6 in. circle has three times the diameter of the 2 in. circle and, therefore, the area is 3 2 , or nine times as great. A piece of steel plate 6 in. in diameter weighs nine times as much as a piece 2 in. in diameter of the same thickness. Likewise a 10 in. square has four times the area of a 5 in. square. If we Fio. 20. let A represent the area of the larger circle, a the area of the smaller circle, D the diameter of the larger circle, and d the diam- eter of the smaller circle, then we have the direct proportion: 69. The Rectangle. When a four-sided figure has square corners it is a Rectangle. Each side of a brick is a rectangle. 3" Fio. 21. A Square is a special kind of rectangle having all the sides equal. The area of a rectangle is obtained by multiplying the length by the breadth. In Fig. 21 the area is 2X3 = 6 sq. in., as can be seen by counting the 1-in. squares, which each contain 1 sq. in. 80 SHOP ARITHMETIC 70. The Cube. Just, as the square of a number is represented by the area of a square, one side of which represents the number, so the cube of a number is represented by the volume of a cubical block, each edge of which represents the number. The volume of a cube which is 10 in. on each edge is 10 X 10 X 10 - 1000 cubic inches, and since. this is obtained by "cubing" 10 (10 3 = 10X10X10 = 1000), we can see that the cube of a number FIQ 22 can be represented by the volume of a cube, the edge of which represents the number. If the edge of the cube is one-half as long, that is 5 in., the volume is 5X5X5 = 1 25 cubic inches, or only the volume of the 10 in. cube. ^ s> ^> I , "p\ 1 1 \ ^ Fia. 23. MEASURES OF VOLUME (CUBICAL MEASURE) 1728 cubic inches (cu. in.) = 1 cubic foot (cu. ft.) 27 cubic feet = l cubic yard (cu. yd.) (Larger units than cubic yards are seldom, if ever, used.) 71. Volumes of Straight Bars. A piece 1 in. long cut from a bar will naturally contain just as many cu. in. as there are sq. in. on the end of the bar. In the billet shown in Fig. 24, there are 3x4 = 12 sq. in. on the end of the bar, and a piece 1 in. long contains 12 cu. in. The entire billet, contains 10 slices just like this one, so there are 12X10 = 120 cu. in. in the entire billet. AREAS AND VOLUMES OF SIMPLE FIGURES 81 Therefore, we see that to find the number of cu. in. in any straight bar we proceed as follows: Calculate the area of one end of the bar in square inches; then multiply this result by the length of the bar in inches; the result will be the number of cubic inches in the bar. For bars of square or rectangular section, the volume is the product of the three dimensions, length, breadth, and thickness. If L, B, and T represent the length, breadth, and thickness, and V stands for the volume, then V=LXBXT L 10"- Fio. 24. Example : How many cubic inches of steel in a bar 2 in. square and 4 ft. long? 4 ft. =48 in. V=LXBXT = 48X2X2 = 192 cu. in., Answer. For round bars, the area on the end is .7854 times the square of the diameter, and this, multiplied by the length, gives the volume. Then, if D represents the diameter of the bar and'L its length, the volume V will be 7=.7854X 2 XL This will apply equally well to thin circular plates or to long bars or shafting. With thin plates, we would naturally speak of thickness (T) instead of length (L). Fig. 25 shows that the two objects have the same shape except that their proportions are different. Examples : 1. How many cubic inches of steel in a shaft 2 in. in diameter and 12 ft. long? 12 ft. = 12X12 = 144 in., length of bar. F = .7854XD 2 XL V = . 7854 X2 2 X 144 V = .7854 X 4 X 144 =452.39 cu. in., Answer. 82 2. How many cubic inches in a blank for a boiler head 60 in. in diameter and TTT in. thick? 7 = .7854X60 2 X ^ F = . 7854 X 3600 XTF = 1237 cu. in., Answer. Fio. 25. 72. Weights of Metals. The chief uses in the shop for calcu- lations of volume are in finding the amount of material needed to make some object; in finding the weight of some object that cannot be conveniently weighed; or in finding the capacity of some bin or other receptacle. Having obtained the volume of an object, it is only necessary to multiply the volume by the known weight of a unit volume of the material to get the weight of the object. In the case of the shaft of which we just got the volume, 1 cu. in. will weigh about .283 lb., so the total weight of the shaft will be 452 . 39 X . 283 = 128 . + pounds. The weight of the boiler head will be 1237 X . 283 = 350 + pounds. The following table gives the weights per cubic inch and per cubic foot for the most common metals and also for water: Material . 1 cu. in. 1 cu. ft. Cast iron .260 Ib. 450 Ib. Wrought iron .278 Ib. 480 Ib. Steel .283 Ib. 489 Ib. Brass . 301 Ib. 520 Ib. Copper. . .3181b. 550 Ib. Lead .411 Ib. 711 Ib. Aluminum . 094 Ib. 162 Ib. Water .036 Ib. 62.4 Ib. 73. Short Rule for Plates. A flat wrought iron plate in. thick and 1 ft. square will weigh 5 Ib., since 12X12x^ = 18 cu. in., and 18 X. 278 = 5 Ib. The rule obtained from this is very easy to remember and is very useful for plates that have their dimensions in exact feet. Rule. Weight of flat iron plates = area in square feetX number of eighths of an inch in thickness X 5. This rule can also be used for steel plates by adding 2 per cent, to the result calculated from the above rule. Example : 3 Find the weight of a steel plate 30 in. X96 in. X s in. o 1 3 30 in. = 2^ ft., 96 in. =- 8 ft., g in. = 3 eighths. 2^X8X3X5 = 300 (weight if it were of wrought iron). 2% of 300 = 6 Ib. 300 + 6 = 306 Ib.; weight of steel plate, Answer. If this weight is calculated by first getting the cubic inches of steel, we get : Q 30 X 96 X 77 = 1080 cu. in. o 1080 X. 283 = 305.64 Ib. weight of steel plate, Answer. We see that the results check as closely as could be expected and, in fact, different plates of supposedly the same size would differ as much as this because of differences in rolling. 74. Weight of Casting from Pattern. In foundry work, it is often desired to get the approximate weight of a casting in order to calculate the amount of metal needed to make it. The prob- 84 SHOP ARITHMETIC able weight of the casting can be obtained closely enough by .weighing the pattern and multiplying this weight by the proper number from the following- table. In case the pattern contains core prints, the weight of these prints should be calculated and subtracted from the pattern weight before multiplying; or else the total pattern weight can be multiplied first and then the weight of metal which would occupy the same volume as the core print be subtracted from it. PROPORTIONATE WEIGHT OF CASTINGS TO WEIGHT OF WOOD PATTERNS For each 1 Ib. weight of pattern when made of (less weight of core prints) Casting will weigh if made of Cast iron Copper or bronze Brass Aluminum White pine 16. 12. 10.2 10.6 0.84 2.6 0.95 19.6 14.7 12.5 13. 1. 3.2 1.3 18.5 14. 11.7 12.3 0.95 3.1 1.2 5.9 4.5 3.8 3.9 0.32 0.95 0.38 Mahogany Pear wood Birch Brass Aluminum Cast iron PROBLEMS 131. Find the weight of a piece of steel shafting 2 in. in diameter and 20 ft. long. 132 . What is the weight of a billet of wrought iron 4 in. square and 2 ft. 8 in. long? 133. What would a steel boiler plate 36 in. by 108 in. by \ in. weigh? 134. A cast steel cylinder is 42 in. inside diameter, 4 ft. 6 in. long and 1J in. thick. Find its weight. 135. A steam engine cylinder 4 in. inside diameter has the cylinder head held on by four studs. When the pressure in the cylinder is 125 Ib. per square inch, what is the total pressure on the cylinder head and what is the pull in each stud? 136. 50 studs 1\ in. long and 1 in. in diameter are to be cut from cold rolled steel. Find the length and weight of bar necessary, allowing \ in. per stud for cutting off. 137. What would be the weight of a \ in. by 3 in. wagon tire for a 40 in. wheel? (Length of stock = circumference of & 39 J in. circle.) SQUARE ROOT 85 138. A copper billet 2 in. by 8 in. by 24 in. is rolled out into a plate of No. 10 B. & S. gage. The thickness of this gage is .1019 in. What would be the probable area of this plate in square feet? 139. The steel link shown in Fig. 26 is made of f in. round steel (round steel f in. in diameter). Find the length of bar necessary to make it and then find the weight of the link. 140. A steel piece is to be finished as shown in the sketch below (Fig. 27). The only stock available from which to make it is 4 in. in diameter. Com- pute the length of the 4 in. stock which must be upset to make the piece and have -5*5 extra stock all over for finishing. T Fio. 26. Fio. 27. CHAPTER XI SQUARE ROOT 76. The Meaning of Square Root. The previous chapter showed the usefulness of squares in finding areas and of cubes in finding volumes. Problems often arise in which it is necessary to find one edge of a square or cube of which only the area or , volume is given. For instance, what must be the side of a square so that its area will be 9 sq. in.? The length of the side must be such that when multiplied by itself it will give 9 sq. in. A moment's thought shows that 3X3 = 9, or 3 2 = 9. Therefore, 3 is the necessary side of the square. Finding such a value is called Extracting the Square Root, and is represented by the sign V called the square root sign or radical sign. Thus v9 = 3 ; \/16 = 4. To make clear the idea of extracting square roots, the student should consider it as the reverse or "the undoing" 86 SHOP ARITHMETIC of squaring, just as division is the reverse of multiplication or as subtraction is the reverse of addition. 5 2 = 25, and its reverse is: \/25 = 5. The square roots of some numbers, like 4, 9, 16, 25, 36, 49, 64, 81, etc.', are easily seen, but we must have some method that will apply to any number. There are several methods of finding square root, of which two are open to the student of shop arith- metic: (1) by actual calculation; (2) by the use of a table of squares or square roots. A third method which uses logarithms will be explained in the chapters on logarithms. In many handbooks will be found tables giving the square roots of numbers, but we must learn some method that can be used when a table is not available and the method that will now be explained should be used throughout the work in this chapter. 76. Extracting the Square Root. The first step in finding the square root of any number is to find how many figures there are in the root. This is done by pointing off the number into periods or groups of two figures each, beginning at the decimal point and working each way. 12 = 1 10 2 =1'00 100 2 =1'00'00 From these it is evident that the number of periods indicates the number of figures in the root. Thus the square root of 103684 contains 3 figures because this number (10'36'84) con- tains three periods. Also the square root of 6'50'25 contains three figures since there are three periods. (The extreme left hand period may have 1 or 2 figures in it.) We must not forget that, for any number not containing a decimal, a decimal point may be placed at the extreme right of the number. Thus the decimal point for 62025 would be placed at the right of the number (as 62025.) The method of finding the square root of a number can best be explained by working some examples and explaining the work as we go along. The student should take a pencil and a piece of paper and go through the work, one step at a time, as he reads the explanation. Example : Find the square root of 186624. Point off into periods of two figures each (18'66'24) and it will be seen that there are 3 figures in the root. The work is arranged very similarly to division. SQUARE ROOT 87 18'66'24(432 Explanation: First find the largest number whose square is equal to or 2X40 = 80 JJ 83 2X430 = 860 2 862 9~fiS~ l ess ^an I**, *ke ^ rst P ef iod. This is 4, since 5 2 is more than 18. Write the 4 to the right for the first figure of the root just as the quotient is put down 17 24 in long division. The first figure of the root is 4. Square the 4 and write 17 24 its square (16) under the first period (18) and subtract, leaving 2. Bring down the next period (66) and annex it to the remainder, giving 266 for what is called the dividend. Annex a cipher to the part of the root already found (4) giving 40; then multiply this by 2, making 80, which is called the trial divisor. Set this off to the left. Divide the dividend (266) by the trial divisor (80). We obtain 3, which is probably the next figure of the root. Write this 3 in the root as the second figure and also add it to the trial divisor, giving 83, which is the final divisor. Multiply this by the figure of the root just found (3) giving 249. Subtract this from the dividend (266) leaving 17. Bring down the next period (24) and annex to the 17, giving a new dividend 1724. Repeat the preceding process as follows: Annex a cipher to the part of the root already found (43) giving 430; and multiply by 2, giving 860, the. trial divisor. Divide the dividend by this divisor and ob- tain 2 as the next figure of the root. Put this down as the third figure of the root and also add it to the trial divisor, giving 862 as the final divisor. Multiply this by the 2 and obtain 1724, which leaves no remainder when subtracted from the dividend. As there are no more periods in the original number, the root is complete. 77. Square Roots of Mixed Numbers. If it is required to find the square root of a number composed of a whole number and a decimal, begin at the decimal point and point off periods to right and left. Then find the root as before. Example : Find the square root of 257.8623 2' 5 7.86' q 23'00(16.058 + , Answer. 20 _6 26 _1 to or less than 2, the first period. Proceeding as I 57 before, we get 6 for the second figure. After subtracting the second time (at a) we find that , eg the trial divisor 320 is larger than the dividend 186. In this case, we place a cipher in the root, annex another cipher to 320 making 3200, annex the next period, 23, to the dividend and then 1 60 25 proceed as before. If the root proves, as in this 25 98 00 case, to be an interminable decimal (one that does not end) continue for two or three decimal 5 320.5 32100 8 32108 25 68 64 places and put a + sign after the root as in divi- sion. In this example the decimal point comes 29 36 after 16, because there must be two figures in the whole number part of the root since there are two periods in the whole number part of our original number. 78. Square Roots of Decimals. Sometimes, in the case of a decimal, one or more periods are composed entirely of ciphers. The root will then contain one cipher following the decimal point for each full period of ciphers in the number. 88 SHOP ARITHMETIC Example : Take .0007856 as an example. Beginning at the decimal point and pointing off into periods of two figures each, we have .00'07'85'60. Hence, the first figure of the root must be a cipher. To obtain the rest of the root we proceed as before. .00'07'85'60(.0280 + , Answer. 4 40 _8 48 385 384 560 | 1 60 It will be noticed that the square root of a decimal will always be a decimal. If we square a fraction, we will get a smaller fraction for its square, () 2 = ^; or as a decimal, .25 2 = .0625. Therefore, the opposite is true; that, if we take the square root of a number entirely a decimal, will get a decimal, but it will be larger than the one of which it is the square root. Notice the example just given: .0007856 is less than its square root .028. 79. Rules for Square Root. From the preceding examples the following rules may be deduced: 1. Beginning at the decimal point separate the number into periods of two figures each. If there is no decimal point begin with the figure farthest to the right. 2. Find the greatest whole number whose square is contained in the first or left-hand period. Write this number as the first figure in the root; subtract the square of this number from the first period, and annex the second period to the remainder. 3. Annex a cipher to the part of the root already found and multiply by 2; this gives the trial divisor. Divide the dividend by the trial divisor for the second figure of the root and add this figure to the trial divisor for the complete divisor. Multiply the complete divisor by the second figure in the root and sub- tract this result from the dividend. (If this result is larger than the dividend, a smaller number must be tried for the second figure of the root.) Bring down the third period and annex it to the last remainder for the new dividend. 4. Repeat rule 3 until the last period is used, after which, if any additional decimal places are required, annex cipher periods and continue as before. If the last period in the decimal should contain but one figure, annex a cipher to make a full period. SQUARE ROOT 89 5. If at any time the trial divisor is not contained in the dividend, place a cipher in the root, annex a cipher to the trial divisor and bring down another period. 6. To locate the decimal point, remember that there will be as many figures in the root to the left of the decimal point as there were periods to the left of the decimal point in our original number. 80. The Law of Right Triangles. One of the most useful laws of geometry is that relating to the sides of a right angled triangle. Fig. 28 shows a right angled triangle, or "right triangle," so FIQ. 29. called because one of its angles (the one at C) is a right angle, or 90. The longest side (c) is called the hypotenuse. "In any right triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides." Written as a formula this would read This can be illustrated by drawing squares on each side, as in Fig. 29, and noting that the area of the square on the hypotenuse is equal to the sum of the areas of the other two. In using this rule, however, we do not care anything about these 90 SHOP ARITHMETIC areas and seldom think of them except as being the squares of numbers. It is used to find one side of such a triangle when the other two are known. Examples : 1. If the trolley pole in Fig. 30 is 24 ft. high, and the guy wire is anchored 7 ft. from the base of the pole, what is the length of the guy wire? The guy wire is the hypotenuse of a right triangle whose sides are 24 ft. and 7 ft. c 2 = b' + a 2 c 2 = 24-< + 7 2 = 576 + 49 = 625 c = v X 625 = 25 ft., Answer. 2. If the triangle of Fig. 31 is a right triangle having the hypotenuse c = 13 in. and the side a = 5 in., what is the length of the side 6? Hence, 2 = 132-52 = 169-25 = 144 b =\/144 = 12 in., Answer. This property of right triangles is also useful in laying out right angles on a large scale more accurately than it can be done with a square. This is done by using three strings, wires, or chains of such lengths that when stretched they form a right triangle. A useful set of numbers that will give this are 3, 4, and 5, since 3 2 + 4 2 = 5 2 (9 + 16-25). Any three other numbers having the same ratios as 3, 4 ? and 5 can be used if desired. 6, 8, and 10; 9, 12, and 15; 12, 16 ; and 20; 15, 20, and 25; any of these sets of numbers can be used. SQUARE ROOT 91 A surveyor will often use lengths of 15 ft., 20 ft., and 25 ft. on his chain to lay out a square corner; this method can also be used in aligning engines, shafting, etc. 81. Dimensions of Squares and Circles. Square Root must be used in getting the dimensions of a square or a circle to have a given area. If the area of a square is given, the length of one side can be obtained by extracting the square root of the area. If we wish to know the diameter of a circle which shall have a certain area, we can find it by the following process: The area is . 7854 X the square of the diameter or, briefly, If we divide the given area by .7854, we will get the area of the square constructed around the circle (see Fig. 19). One side of this square is the same as the diameter of the circle and is equal to the square root of the area of the square. Then, to find the diameter of a circle to have a given area: Divide the given area by . 7854 and extract the square root of the quotient. n- P 1 " = \77854~ 82. Dimensions of Rectangles. Occasionally one encounters a problem in which he wants a rectangle of a certain area and knows only that the two dimensions must be in some ratio. It may be that a factory building is to cover, say 40,000 sq. ft. of ground and is to be four times as long as it is wide, or some problem of a similar nature. Suppose we take the case of this factory and see how we would proceed to find the dimensions of the building. Example : Wanted a factory building to cover 40,000 sq. ft. of ground. Ratio of length to breadth, 4:1. Find the dimensions. I 1 1 1 1 1 40000 -s-4 b Fio. 32. = 10000 = 10000 = 100 = 4X100 = 400. 92 SHOP ARITHMETIC Explanation: If we divide the total area by 4, we get 10,000 as the area of a square having the breadth 6 on each side. From this we find the breadth or width 6 to be the square root of 10,000 or 100 ft. If the length is four times as great it will be 400 ft. and the dimensions of the building will be 400 by 100. 83. Cube Root. The Cube Root of a given number is another number which, when cubed, produces the given number. In other words, the cube root is one of the three equal factors of a number. The cube root of 8 is 2, because 2 3 = 2x2x2 = 8; also the cube root of 27 is 3 (since 3 3 = 27) and the cube root of 64 is 4 (since 4 3 = 64). The sign of cube root is V placed over the number of which we want the root. Thus we would write 4 ^1000= 10 If we consider the number of which we want the cube root as representing the volume of a cubical block, then the cube root of the number will represent the length of one edge of the cube. The cube root of 1728 is 12 and a cube containing 1728 cu. in. will measure 12 in. on each edge. There are four ways of getting cube roots: (1) by actual calculation, (2) by reference to a table of cubes or cube roots, (3) by the use of logarithms, and (4) by the use of some calcu- lating device like the slide rule. The use of a table is the simplest way of finding cube roots, but its value and accuracy is limited by the size of the table. Tables of cubes or cube roots are to be found in many handbooks and catalogues and should be used whenever they give the desired root with sufficient accuracy. Logarithms give us an easy way of getting cube roots, but here also a table is necessary and the accuracy is limited by the size of the table of logarithms. 'The use of logarithms will be explained in a later chapter. The ordinary pocket slide rule will give the first three figures of a cube root and for many calculations this is sufficiently accurate. The method of actually calculating cube roots is very complicated and is used so seldom that one can never remember it when he needs it. Consequently, if it is necessary to hunt up a book to find how to extract the cube root, one might just as well look up a table of cube roots or a logarithm table, either of which will give the root much quicker. The next chapter contains tables of cube roots and a chapter further on explains the use of logarithms. SQUARE ROOT PROBLEMS 93 141. Extract the square roots of (a) 64516 Answer 254 (6)198.1369 Answer 14.076 + (c) .571428 Answer .7559 + Note. These answers are given so that the student can see if he understands the operations of square root before proceeding further. 142. The two sides of a right triangle are 36 and 48 ft. ; what is the length of the hypotenuse? 143. A square nut for a 2 in. bolt is 3 in. on each side. What is the length of the diagonal, or distance across the corners? 144. A steel stack 75 ft. high is to be supported by 4 guy wires fastened to a ring two-thirds of the way up the stack and having the other ends anchored at a distance of 50 ft. from the base and on a level with the base. How many feet of wire are necessary, allowing 20 ft. extra for fastening the ends? 145. What would be the diameter of a circular brass plate having an area of 100 sq. in.? FIG. 33. 146. A lineshaft and the motor which drives it are located in separate rooms as shown in Fig. 33. Calculate the exact distance between the centers of the two shafts. 147. I want to cut a rectangular sheet of drawing paper to have an area of 235 sq. in. and to be one and one-half times as long as it is wide. What would be the dimensions of the sheet? 148. A 6 in. pipe and an 8 in. pipe both discharge into a single header. Find the diameter of the header so that it will have an area equal to that of both the pipes. 149. What would be the diameter of a 1 Ib. circular cast iron weight J in. thick? 94 SHOP ARITHMETIC 150. How long must be the boom in Fig. 34 to land the load on the 12 ft. pedestal, allowing 4 ft. clearance at the end for ropes, pulleys, etc.? FIG. 34. CHAPTER XII MATHEMATICAL TABLES (CIRCLES, POWERS, AND ROOTS) 84. The Value of Tables. There are certain calculations that are made thousands of times a day by different people in different parts of the world. For example, the circumferences of circles of different diameters are being calculated every day by hundreds and thousands of men. To save much of the time that is thus wasted in useless repetition, many of the common operations and their results have been "tabulated," that is, arranged in tables in the same way as are our multiplication tables in arith- metics. These tables are not learned, however, as were the multiplication tables, but are consulted each time that we have need for their assistance. Just what tables one needs most, depends on his occupation. The machinist has use for tables of the decimal equivalents of common fractions, tables of cutting speeds, tables of change gears to use for screw cutting, etc. The draftsman would use tables of strengths and weights of different materials, safe loads for bolts, beams, etc., tables of proportions of standard machine parts of different sizes, etc. The engineer uses tables of the prop- erties of steam, and of the horse-power of engines, boilers, etc, MATHEMATICAL TABLES 95 There are certain mathematical tables that are of value to nearly everyone. Among these are the tables given in this chapter: Tables of Circumferences and Areas of Circles; Tables of Squares, Cubes, Square Roots, and Cube Roots of Numbers. 85. Explanation of the Tables. The first table is to save the necessity of always multiplying the diameter by 3.1416 when we want the circumference of a circle, or of squaring the diameter and multiplying by .7854 when the area of a circle is wanted. To find the circumference of a circle: Find, in the diameter column, the number which is the given diameter; directly across, in the next column to the right, will be found the corre- sponding circumference. Examples : Diameter 1J Circumference 3.9270 Diameter 27, Circumference 84.823 Diameter 90J, Circumference 284.314 To find the area of a circle: Find, in the diameter column, the number which is the given diameter; directly across, in the second column to the right (the column headed "Area") will be found the area. Examples : Diameter 66, Area 3421.2 Diameter 17, Area 226.98 Diameter f , Area 0.3067 If the area or circumference is known and we want to get the diameter, we find the given number in the area or circumference column and read the diameter in the corresponding diameter column to the left. Examples : Area 78.54, Diameter 10 Area 706.86, Diameter 30 Circumference 281. Diameter 89 J The second table, that of squares, cubes, square roots, and cube roots, is especially valuable in avoiding the tedious process of extracting square or cube roots. The table is read the same as the other one. Find the given number in the first column; on a level with it, in the other columns, will be found the corre- sponding powers and roots, as indicated in the headings at the tops of the columns. 96 SHOP ARITHMETIC Examples : ^25 = 2.924 \/260 - 16.1245 295 3 = 25,672,375 865 2 = 748,225 86. Interpolation. This is a name given to the process of finding values between those given in the tables. For example, suppose we want the circumference of a 30 \ in. circle. The table gives 30 and 30^ and, since 30J is half way between these, its circumference will be half way between that of a 30 in. and a 30J in. circle. Circumference of 30 \ in. circle = 95. 8 19 Circumference of 30 in. circle = 94 . 248 The Difference = 1.571 Then the circumference of the 30 \ in. circle is just half this difference more than that of the 30-in. circle, 94.248 + ^ of 1.571 = 95.034 A This method enables us to increase greatly the value of tables. For most purposes the interpolation can be done quickly, and while it requires some calculating, is much shorter than the complete calculation would be. This is especially true in finding square or cube roots. Example : Find from the table the cube root of 736.4 3/737 = 9.0328 = 9.0287 Difference =41 .4 X the difference = .4X41 = 16.4 9.0287 16 ^736>4= 9.0303, Answer. Explanation: The root of 736.4 will be between that of 736 and that of 737, and will be .4 of the difference greater than that of 736. In making this correction for the .4, we forget, for the minute, that the difference is a decimal and write it as 41 merely to save time. We then multiply it by .4, and, dropping the decimal part, add the 16 to the 90287. This gives 9.0303 as the cube root of 736.4 Hence, MATHEMATICAL TABLES 97 87. Roots of Numbers Greater than 1000. For getting the cube roots of numbers greater than 1000, the easiest and most accurate way is to look in the third column headed "cubes" for a number as near as possible to our given number. Now, we know that the numbers in the first column are the cube roots of these numbers in the third column. If we can find our number in the third column, there is nothing further to do because its cube root will be directly opposite it in the first column. Examples : 3/1728 = 12 3/6967871 = 191 3/166375 =55 Likewise, the numbers in the first column are the square roots of the numbers in the second column. But suppose we want the cube root of a number which is not found in the third column, but lies somewhere between two consecutive numbers in that column. In this case we pursue the method shown in the following example: Example : Find 3/621723 In the column headed " Cube" find two consecutive numbers, one larger and one smaller than 621723. These numbers are 636056 whose cube root is 86 and 614125 whose cube root is 85 Hence, the cube root of 621723 is more than 85 and less than 86; that is, it is 85 and a decimal, or 85 + . The decimal part is found as follows: Subtract the lesser of the two num- bers found in the table from the greater and call the result the First Difference. 636056-614125 = 21931, First difference. Then subtract the smaller of the two numbers in the table from the given number and call the result the Second Difference. 621723-614125 = 7598, Second Difference Now the first difference, 21931, is the amount that the number increases when its cube root changes from 85 to 86. Our given number is only 7598 more than the cube of 85, so its cube root will be approximately 85 zWVr- We do not want a fraction like this, so we reduce it to a decimal as follows: Divide the second difference by the first difference and annex the quotient to 85. This will give us the cube root of our number, approximately. Second Difference _ 7598 _ ., . First Difference " 2l93l " This is the decimal part of the root sought and the whole root is 85.346 + . Hence 3/621723 = 85.346 + . This method is not exact and the third decimal place will usually be slightly off, so it is best to drop the third decimal if less than 5, or raise it to 10, if more than 5. In this case we will call the root 85.35. 98 SHOP ARITHMETIC 88. Cube Roots of Decimals. In getting the cube root of either a number entirely decimal, or a mixed decimal number, it is best to move the decimal point a number of periods, that is, 3, 6, 9, or 12 decimal places, sufficient to make a whole number out of the decimal. After finding the cube root, shift the decimal point in the root back to the left as many places as the number of periods that we moved the decimal point in our original number. For example, suppose that we had .621723 of which to find the cube root. Moving the decimal point two periods (of three places each) to the right gives us 621723, of which we just found the cube root to be 85.35. We moved the decimal point of our original number two periods to the right, so we must move the decimal point back two places to the left in the root; we then have ^.621723 = .8535. The following illustrations will show the principle: 4/621723 =85.35 V62 1.723 = 8.535 V. 621723 .8535 10 298116 162771336 23.3666 8.1733 591 349281 206425071 24.3105 8.3919 547 299209 163667323 23.3880 8.1783 592 350464 207474688 24.3311 8.3967 548 300304 164566592 23.4094 8.1833 593 351649 208527857 24.3516 8.4014 549 301401 165469149 23.4307 8.1882 594 352836 209584584 24.3721 8.4061 550 302500 166375000 23.4521 8.1932 595 354025 210644875 24.3926 8.4108 551 303601 167284151 23.4734 8.1982 596 355216 211708736 24.4131 8.4155 552 304704 168196608 23.4947 8.2031 597 356409 212776173 24.4336 8.4202 553 305809 169112377 23.5160 8.2081 598 357604 213847192 24.4540 8.4249 554 306916 170031464 23.5372 8.2130 599 358801 214921799 24.4745 8.4296 555 308025 170953875 23.5584 8.2180 600 360000 216000000 24.4949 8.4343 556 309136 171879616 23.5797 8.2229 601 361201 217081801 24.5153 8.4390 557 310249 172808693 23.6008 8.2278 602 362404 218167208 24.5357 8.4437 558 311364 173741112 23.6220 8.2327 603 363609 219256227 24.5561 8.4484 559 312481 174676879 23.6432 8.2377 604 364816 220348864 24.5764 8.4530 560 313600 175616000 23.6643 8.2426 605 366025 221445125 24.5967 8.4577 561 314721 176558481 23.6854 8.2475 606 367236 222545016 24.6171 8.4623 562 315844 177504328 23.7065 8.2524 607 368449 223648543 24.6374 8.4670 563 316969 178453547 23.7276 8.2573 608 369664 224755712 24.6577 8.4716 564 318096 179406144 23.7487 8.2621 609 370881 225866529 24.6779 8.4763 56.') 319225 180362125 23.7697 8.2670 610 372100 226981000 24.6982 8.4809 566 320356 181321496 23.7908 8.2719 611 373321 228099131 24.7184 8.4856 567 321489 182284263 23.8118 8.2768 612 374544 229220928 24.7386 8.4902 568 322624 183250432 23.8328 8.2816 613 375769 230346397 24.7588 8.4948 569 323761 ] 184220009 23.8537 8.2865 614 376996 231475544 24.7790 8.4994 570 324900 185193000 23.8747 8.2913 615 378225 232608375 24.7992 8.5040 571 326041 186169411 23.8956 8.2962 616 379456 233744896 24.8193 8.5086 672 327184 187149248 23.9165 8.3010 617 380689 234885113 24.8395 8.5132 573 328329 188132517 23.9374 8.3059 618 381924 236029032 24.8596 8.5178 571 329476, 189119224 23.9583 8.3107 619 383161 237176659 24.8797 8.5224 575 330625 190109375 23.9792 8.3155 620 384400 238328000 24.8998 8.5270 576 331776 191102976 24. 8.3203 621 385641 239483061 24.9199 8.5316 577 332929 192100033 24.0208 8.3251 622 386884 240641848 24.9399 8.5362 578 334084 193100552 24.0416 8.3300 623 388129 241804367 24.9600 8.5408 579 335241 194104539 24.0624 8.3348 624 389376 242970624 24.9800 8.5453 580 336400 195112000 24.0832 8.3396 625 390625 244140625 25. 8.5499 581 337561 196122941 24.1039 8.3443 626 391876 245314376 25.0200 8.5544 582 338724 197137368 24.1247 8.3491 627 393129 246491883 25.0400 8.5590 583 339889; 198155287 24.1454 8.3539 628 394384 247673152 25.0599 8.5635 584 341056 199176704 24.1661 8.3587 629 395641 248858189 25.0799 8.5681 585 342225 200201625 24.1868 8.3634 630 396900 250047000 25.0998 8.5726 110 SHOP ARITHMETIC SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF NUMBERS. Continued No. Square Cube Square root Cube root No. Square Cube Square root Cube root 631 398161 251239591 25.1197 8.5772 676 456976 308915776 26. 8.7764 632 399424 252435968 25.1396 8.5817 677 458329 310288733 26.0192 8.7807 633 400689 253636137 25.1595 8.5862 678 459684 311665752 26.0384 8.7850 634 401956 254840104 25.1794 8.5907 679 461041 313046839 26.0576 8.7893 635 403225 256047875 25.1992 8.5952 680 462400 314432000 26.0768 8.7937 636 404496 257259456 25.2190 8.5997 681 463761 315821241 26.0960 8.7980 637 405769 258474853 25.2389 8.6043 682 465124 317214568 26.1151 8.8023 638 407044 259694072 25.2587 8.6088 683 466489 318611987 26.1343 8.8066 639 408321 260917119 25.2784 8.6132 684 467856 320013504 26.1534 8.8109 640 409600 262144000 25.2982 8.6177 685 469225 321419125 26.1725 8.8152 641 410881 263374721 25.3180 8.6222 686 470596 322828856 26.1916 8.8194 642 412164 264609288 25.3377 8.6267 687 471969 324242703 26.2107 8.8237 643 413449 265847707 25.3574 8.6312 688 473344 325660672 26.2298 8.8280 644 414736 267089984 25.3772 8.6357 689 474721 327082769 26.2488 8.8323 645 416025 268336125 25.3969 8.6401 690 476100 328509000 26.2679 8.8366 646 417316 269586136 25.4165 8.6446 691 477481 329929371 26.2869 8.8408 647 418609 270840023 25.4362 8.6490 692 478864 331373888 26.3059 8.8451 648 419904 272097792 25.4558 8.6535 693 480249 332812557 26.3249 8.8493 649 421201 273359449 25.4755 8.6579 694 481636 334255384 26.3439 8.8536 650 422500 274625000 25.4951 8.6624 695 483025 335702375 26.3629 8.8578 651 423801 275894451 25.5147 8.6668 696 484416 337153536 26.3818 8.8621 652 425104 277167808 25.5343 8.6713 697 485809 338608873 26.4008 8.8663 653 426409 278445077 25.5539 8.6757 698 487204 340068392 26.4197 8.8706 654 427716 279726264 25.5734 8.6801 699 488601 341532099 26.4386 8.8748 655 429025 2S1011375 25.5930 8.6845 700 490000 343000000 26.4575 8.8790 656 430336 282300416 25.6125 8.6890 701 491401 344472101 26.4764 8.8833 657 431649 283593393 25.6320 8.6934 702 492804 345948408 '26.4953 8.8875 658 432964 284890312 25.6515 8.6978 703 494209 347428927 26.5141 8.8917 659 434281 286191179 25.6710 8.7022 704 495616 348913664 26.5330 8.8959 660 435600 287496000 25.6905 8.7066 705 497025 350402625 26.5518 8.9001 661 436921 288804781 25.7099 8.7110 706 498436 351895816 26.5707 8.9043 662 438244 290117528 25.7294 8.7154 707 499849 353393243 26.5895 8.9085 663 439569 291434247 25.7488 8.7198 708 501264 354894912 26.6083 8.9127 664 440896 292754944 25.7682 8.7241 709 502681 356400829 26.6271 8.9169 665 442225 294079625 25.7876 8.7285 710 504100 357911000 26.6458 8.9211 666 443556 295408296 25.8070 8.7329 711 505521 359425431 26.6646 8.9253 667 444889 296740963 25.8263 8.7373 712 506944 360944128 26.6833 8.9295 668 446224 298077632 25.8457 8.7416 713 508369 362467097 26.7021 8.9337 669 447561 299418309 25.8650 8.7460 714 509796 363994344 26.7208 8.9378 670 448900 300763000 25.8844 8.7503 715 511225 365525875 26.7395 8.9420 671 450241 302111711 25.9037 8.7547 716 512656 367061696 26.7582 8.9462 672 451584 303464448 25.9230 8.7590 717 514089 368601813 26 . 7769 8.9503 673 452929 304821217 25.9422 8.7634 718 515524 370146232 26.7955 8.9545 674 454276 306182024 25.9615 8.7677 719 516961 371694959 26.8142 8.9587 675 455625 307546875 25.9808 8.7721 720 518400 373248000 26.8328 8.9628 MATHEMATICAL TABLES 111 SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF NUMBERS. Continued No. Square Cube Square root Cube root No. Square Cube Square root Cube root 721 519841 374805361 26.8514 8.9670 766 586756 449455096 27.6767 9.1498 722 521284 376367048 26.8701 8.9711 767 588289 451217663 27.6948 9.1537 723 522729 377933067 26.8887 8.9752 768 589824 452984832 27.7128 9.1577 724 524176 379503424 26.9072 8.9794 769 591361 454756609 27.7308 9.1617 725 525625 381078125 26.9258 8.9835 770 592900 456533000 27.7489 9.1657 726 527076 382657176 26.9444 8.9876 771 594441 458314011 27.7669 9.1696 727 528529 384240583 26.9629 8.9918 772 595984 460099648 27.7849 9.1736 728 529984 385828352 26.9815 8.9959 773 597529 461889917 27.8029 9.1775 729 531441 387420489 27. 9. 774 599076 463684824 27.8209 9.1815 730 532900 389017000 27.0185 9.0041 775 600625 465484375 27.8388 9.1855 731 534361 390617891 27.0370 9.0082 776 602176 467288576 27.8568 9.1894 732 535824 392223168J 27.0555 9.0123 777 603729 469097433 27.8747 9.1933 733 537289 393832837 27.0740 9.0164 778 605284 470910952 27.8927 9.1973 734 538756 395446904 27.0924 9.0205 779 606841 472729139 27.9106 9.2012 735 540225 397065375 27.1109 9.0246 780 608400 474552000 27.9285 9.2052 736 541696 398688256 27.1293 9.0287 781 609961 476379541 27.9464 9.2091 737 543169 400315553 27.1477 9.0328 782 611524 478211768 27.9643 9.2130 738 544644 401947272 27.1662 9.0369 783 613089 480048687; 27.9821 9.2170 739 546121 403583419 27.1846 9.0410 784 614656 481890304 28. 9.2209 740 547600 405224000 27.2029 9.0450 785 616225 483736625 28.0179 9.2248 741 549081 406869021 27.2213 9.0491 786 617796 485587656 28.0357 9.2287 742 5505641 408518488 27.2397 9.0532 787 619369 487443403 28.0535 9.2326 743 552049 410172407 27.2580 9.0572 788 620944 489303872 28.0713 9.2365 744 553536 411830784 27.2764 9.0613 789 622521; 491169069 28.0891 9.2404 745 555025 413493625 27.2.947 9.0654 790 624100 493039000 28.1069 9.2443 746 556516 415160936 27.3130 9.0694 791 625681 494913671 28.1247 9.2482 747 558009 416832723 27.3313 9.0735 792 627264 496793088 28.1425 9.2521 748 559504 418508992 27.3496 9.0775 793 628849 498677257 28.1603 9.2560 749 561001 420189749 27.3679 9.0816 794 630436 500566184 28.1780 9.2599 750 562500 421875000 27.3861 9.0856 795 632025 502459875 28.1957 9.2638 751 564001 423564751 27.4044 9.0896 796 633616 504358336 28.2135 9.2677 752 565504 425259008 27.4226 9.0937 797 635209 506261573 28.2312 9.2716 753 567009 426957777 27.4408 9.0977 798 636804 508169592 28.2489 9.2754 754 568516 428661064 27.4591 9.1017 799 638401 510082399 28.2666 9.2793 755 570025 430368875 27.4773 9.1057 800 640000 512000000 28.2843 9.2832 756 571536 432081216 27.4955 9.1098 801 641601 513922401 28.3019 9.2870 757 573049 433798093 27.5136 9.1138 802 643204 5158496081 28.3196 9.2909 758 574564 435519512 27.5318 9.1178 803: 644809 517781627 28.3373 9.2948 759 576081 437245479 27.5500 9.1218 804 646416 519718464 28.3549 9.2986 760 577600 438976000 27.5681 9.1258 805 648025 521660125 28.3725 9.3025 761 579121 440711081 27.5862 9.1298 806 649636 523606616 28.3901 9.3063 762 580644 442450728 27.6043 9.1338 807 651249; 525557943 28.4077 9.3102 763 -582169 444194947 27.6225 9.1378 808 652864 527514112 28.4253 9.3140 764 583696; 445943744 27.6405 9.1418 809 654481 529475129 28.4429 9.3179 765 585225 447697125 27.6586 9.1458 810 656100 531441000 28.4605 9.3217 112 SHOP ARITHMETIC SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF N UM BERS . Continued No. Square Cube Square root Cube root No. Square Cube Square root Cube root 811 657721 533411731 28.4781 9.3255 856 732736 627222016 29.2575 9.4949 812 659344 535387328 28.4956 9.3294 857 734449 629422793 29.2746 9.4986 813 660969 537367797 28.5132 9.3332 858 736164 631628712 29.2916 9.5023 814 662596 539353144 28.5307 9.3370 859 737881' 633839779 29.3087 9.5060 815 664225 541343375 28.5482 9.3408 860 739600 636056000 29.3258 9.5097 816 665856 543338496 28.5657 9.3447 861 741321 638277381 29.3428 9.5134 817 667489 545338513 28.5832 9.3485 862 743044 640503928 29.3598 9.5171 818 669124 547343432 28.6007 9.3523 863 744769 642735647 29.3769 9.5207 819 670761 549353259 28.6182 9.3561 864 746496 644972544 29.3939 9.5244 820 672400 551368000 28.6356 9.3599 865 748225 647214625 29.4109 9.5281 821 674041 553387661 28.6531 9.3637 866 749956 649461896 29.4279 9.5317 822! 675684 555412248 28.6705 9.3675 867 751689 T 651714363 29.4449 9.5354 823 677329 557441767 28.6880 9.3713 868 753424 653972032 29.4618 9.5391 824 678976 559476224 28.7054 9.3751 869 755161 656234909 29.4788 9.5427 825 680625 561515625 28.7228 9.3789 870 756900 658503000 29.4958 9.5464 826 682276 563559976 28.7402 9.3827 871 758641 660776311 29.5127 9.5501 827 683929 565609283 28.7576 9.3865 872 760384 663054848 29.5296 9.5537 828 685584 567663552 28.7750 9.3902 873 762129 665338617 29.5466 9.5574 829 687241 569722789 28.7924 9.3940 874 763876 667627624 29.5635 9.5610 830 688900 571787000 28.8097 9.3978 875 765625 669921875 29.5804 9.5647 831 690561 573856191 28.8271 9.4016 876 767376 672221376 29.5973 9.5683 832 692224 575930368 28.8444 9.4053 877 769129 674526133 29.6142 9.5719 833 693889 578009537 28.8617 9.4091 878 770884 676836152 29.6311 9.5756 834 695556 580093704 28.8791 9.4129 879 772641 679151439 29.6479 9.5792 835 697225 582182875 28.8964 9.4166 880 774400 681472000 29.6648 9.5828 836 698896 584277056 28.9137 9.4204 881 776161 683797841 29.6816 9.5865 837 700569 586376253 28.9310 9.4241 882 777924 686128968 29.6985 9.5901 838 702244 588480472 28.9482 9.4279 883 779689 688465387 29.7153 9.5937 839 703921 590589719 28.9655 9.4316 884 781456 690807104 29.7321 9.5973 840 705600 592704000 28.9828 9.4354 885 783225 693154125 29.7489 9.6010 841 707281 594823321 29. 9.4391 886 784996 695506456 29.7658 9.6046 842 708964 596947688 29.0172 9.4429 887 786769 697864103 29 . 7825 9.6082 843 710649 599077107 29.0345 9.4466 888 788544 700227072 29.7993 9.6118 844 712336 601211584 29.0517 9.4503 889 790321 702595369 29.8161 9.6154 845 714025 603351125 29.0689 9.4541 890 792100 704969000 29.8329 9.6190 846 715716 605495736 29.0861 9.4578 891 793881 707347971 29.8496 9.6226 847 717409 607645423 29.1033 9.4615 892 795664 709732288 29.8664 9.6262 848 719104 609800192 29.1204 9.4652 893 797449 712121957 29.8831 9.6298 849 720801 611960049 29 . 1376 9.4690 894 799236 714516984 29.8998 9.6334 850 722500 614125000 29.1548 9.4727 895 801025 716917375 29.9166 9.6370 851 724201 616295051 29.1719 9.4764 896 802816 719323136 29.9333 9.6406 852 725904 618470208 29.1890 9.4801 897 804609 721734273 29.9500 9.6142 853 727609 6206.50477 29.2062 9.4838 898 806 10 i 724150792 29.9666 9.6477 854 729316 622835864 29.2233 9.4875 899 808201 726572699 29.9833 9.6513 855 731025 625026375 29.2404 9.4912 900 810000- 729000000 30. 9.6549 MATHEMATICAL TABLES 113 SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF NUMBERS. Continued No. Square Cube Square root Cube root No. Square Cube Square root Cube root 901 811801 731432701 30.0167 9.6585 946 894916 846590536 30.7571 9.8167 902 813604 733870808 30.0333 9.6620 947 896809 849278123 30.7734 9.8201 903 815409 736314327 30.0500 9.6656 948 898704 851971392 30.7896 9.8236 901 817216 738763264 30.0666 9.6692 949 900601 854670349 30.8058 9.8270 903 819025 741217625 30.0832 9.6727 950 902500 857375000 30.8221 9.8305 906 820836 743677416 30.0998 9.6763 951 904401 860085351 30.8383 9.8339 907 822619, 746142643 30.1164 9.6799 952 906304 862801408 30.8545 9.8374 908 824461 748613312 30.1330 9.6834 953 908209 865523177 30.8707 9.8408 909 826281 751089429 30.1496 9.6870 954 910116 86S250664 30.8869 9.8443 910 828100 753571000 30.1662 9.6905 955 912025 870983875 30.9031 9.8477 911 829921 756058031 30.1828 9.6941 956 913936 873722816 30.9192 9.8511 912 831744 758550528 30.1993 9.6976 957 915849 876467493 30.9354 9.8546 913 833569 761048497 30.2159 9.7012 958 917764| 879217912 30.9516 9.8580 914 835396 763551944 30.2324 9.7047 959 919681 881974079 30.9677 9.8614 Fia. 84. is borne by the threads (which are the inclined planes). The threads are advanced and the weight is raised by a pull (which we will call P) on the end of the rod whose length is marked R. The distance the weight W is moved for one revolution of the screw equals the lead of the screw expressed as a fraction of an Inch. The lead of a screw is the distance it advances lengthwise in one turn or revolution. The force P moves through a distance 134 SHOP ARITHMETIC equal to the circumference of a circle whose radius is the length of the handle; if we represent the length of the handle or lever by R, then the distance traversed by P in one revolution is TT X 2 X R. If we let the letter L represent the lead of the screw then we will have the work accomplished in one revolution of the screw = WxL. Meanwhile, the work expended in doing it = PXKX2XR. Assuming that there is no friction in the screw, we have Distance P moves 7TX2X.R or Mech. Adv. =^ - ^r~- -- = r= - Distance W is raised L If stated in words, these formulas would read: "The force multiplied by the circumference of the circle through which it moves equals the weight multiplied by the lead of the screw." "The mechanical advantage of a jack screw equals the cir- cumference of the circle through which the force moves divided by the lead of the screw" (the amount the screw advances in one turn) . Example : With a 1$ in. jack screw having 3 threads per inch and a pull of 50 Ib. at a radius of 18 in., calculate: (a) The theoretical load that can be lifted by the screw; (6) The actual load if the efficiency of the screw is 18%. (a) Mech. Adv. = .R = 18 in. and L 1 . = 3 in. Hence, 3.1416X2X18 ,. , . , Mech. Adv. = = 3 _ OOJ7 _! 3 W = 339X50 = 16050 Ib., Answer. (b) 18% of 16950 = 3051 Ib., Answer. Explanation: If there are 3 threads per inch, the lead is $ in., and if the radius is 18 in., we have the Mech. Adv. =339.3. In theory then we should be able to lift 50 Ib. X 339 = 16950 Ib. with this screw. But a screw has con- siderable friction and for this reason only 18% of the energy expended in this case is effective, the remaining 82% being all lost in friction. The actual weight lifted is, therefore, only 18% of 16950 Ib. or 3051 Ib. 102. Efficiencies. In explaining the machines of this chapter and of Chapters XIII and XIV, it was assumed that no work is lost in friction within the machines. In a properly mounted lever there is little energy lost. In a tackle block the loss depends on THE INCLINED PLANE AND SCREW 135 the size of the pulleys as compared with the size of the rope and on the nature of the pulley bearings. The efficiency may vary from 60 to 95 %. The more pulleys there are, the lower will be the efficiency, because each bend in the rope and each pulley means a loss in friction. With inclined planes, the efficiency may vary all the way from to nearly 100%. It will be lowest if the weight is merely slid on the plane and will be much higher if wheels or rollers are* used. In any machine, if the weight will start back of its own accord when the force is removed, the friction is less than 50% and the efficiency is greater than 50%. If the weight will not start back, the efficiency is less than 50%. This can be shown as follows: Of the force applied to a machine, part of it is absorbed in over- coming the friction within the machine. The balance goes through the machine and is effective in accomplishing the work to be done. Of the whole force applied, the per cent which this effective force represents is called the Efficiency. If the efficiency is less than 50%, it shows that the friction absorbs more than half of the total force and, therefore, that the friction is greater than the effective force. Now, suppose we had a simple machine such as a jack-screw, being used to raise a weight. If the applied force is removed, the friction will remain the same, but will now act to hold the weight from 'going back. If the friction is suffi- cient to hold the weight, it must at least equal the effective or theoretical force required to raise the weight. Therefore, if a machine does not run backward when the force is removed, the friction must be more than one-half of the total force required to raise the weight, and the effective force must be less than one- half of this total force. Hence, the efficiency in such a case is less than 50%. A jack-screw will not go down of its own accord when the force is removed and therefore its efficiency is less than 50%. In reality, for the usual dimensions of screws, it has been found to be only from 15 to 20%. Mr. Wilfred Lewis has derived, from experiment, a simple formula which gives the average efficiency for a jack-screw under ordinary conditions. L in which E is the efficiency, as a decimal, where L is the lead of the screw, and D is the diameter of the screw. 12 136 SHOP ARITHMETIC Example : Find the probable efficiency of the screw given in the example under Article 101. T l A n i 1 - jL/=77 in., and D = IH in. o ~ .33 .33 + 1.5 .# = 18%, Answer. One can get an approximate idea of the efficiency of any machine by observing, as before explained, whether or not it will run backward of its own accord when the force is removed. This will tell whether the efficiency is above or below 50%. If it is above 50% and a considerable force is required to keep the weight from going back, then the efficiency is high. If, however, a very slight pull will hold it from going back, then the efficiency is not very much above 50%. If we find the efficiency to be under 50% but find that only a very small pull will start the weight down, then the efficiency is not far under 50%. On the other hand, if it seems as if almost as great a force is required to lower the weight as to raise it, this signifies that the efficiency of the machine is extremely low. PROBLEMS 176. An engine weighing 5 tons is to be loaded onto a car, the floor of which is 6 ft. from the ground. If 16 ft. timbers are used for the runway, find the pull necessary to draw the engine up the slope, neglecting friction. 177. How many pounds must a locomotive exert to pull a train of 50 cars, each weighing 50 tons, up a grade of 3 in. in 100 ft.? 178. A building is to be raised by means of 4 jack-screws; the screws are 2 in. in diameter, with 4 threads to the inch. The lever is 20 in. long and a 30 Ib. force is required on each handle. Calculate th6 theoretical weight which the four screws should lift under these conditions. 179. Calculate the probable efficiency of these jack-screws from Lewis' formula and estimate the probable weight of the building. FIG. 65. 180. A windlass with an axle 8 in. in diameter and crank 18 in. long is used in connection with an inclined plane 20 ft. long and 5 ft. high, as shown in Fig. 65. Neglecting friction, what weight can be pulled up the slope with a force of 150 Ib. on the crank? CHAPTER XVI WORK, POWER, AND ENERGY; HORSE-POWER OF BELTING 103. Work. Whenever a force causes a body to move, work is done. Unless the body is moved, no work is accomplished. A man may push against a heavy casting for hours and, unless he moves it, he does no work, no matter how tired he may feel at the end of the time. It is evident that there are two factors to be considered in measuring work force and distance. In the study of levers, tackle blocks, and inclined planes we dealt with the problem of work. In any of these machines the work accomplished in lifting a weight is measured by the product of the weight and the distance it is moved. The work expended or put into the machine to accomplish this is the product of the force exerted times the distance through which this force must act. We found that, if we neglect the work lost in friction, the work put into a machine is equal to the work accomplished by it. The actual difference between the work put in and the work accomplished is the amount that is lost in friction. The follow- ing expressions may make these relations clearer: Work lost in Friction = Work put in Work got out. _~ . Work got out Efficiency = ^ r-& Work put in 104. Unit of Work. The unit by which work is measured is called the Foot-pound. This is the work done in overcoming a resistance of one pound through a distance of 1 ft.; that is, if a weight of 1 Ib. is lifted 1 ft., the work done is equal to 1 foot- pound. All work is measured by this standard. The work in foot-pounds is the product of the force in pounds and the distance in feet through which it acts. In lifting a weight vertically, the resistance, and hence, the force that must be exerted, is equal to the weight itself in pounds. The work done is the product of the weight times the vertical distance that it is raised. If a weight of 80 Ib. is lifted a distance of 4 ft., the work done is 80X4 or 320 foot-pounds. It would require this same amount of work to lift 40 Ib. 8 ft., or to lift 20 Ib. 16 ft. 13 137 138 SHOP ARITHMETIC When a body s moved horizontally, the only resistance to be overcome is the friction. When a team of horses pulls a loaded wagon, the only resistances which it must overcome are the friction between the wheels and the axles, and the resistance on the tires caused by the unevenness of the road. The work necessary to pump a certain amount of water is the weight of the water times the height through which it is lifted or pumped (plus, of course, the work lost in friction in the pipes). The work necessary to hoist a casting is the weight of the casting times the height to which it is lifted. The work done by a belt is the effective pull of the belt times the distance in feet which the belt travels. The work done in hoisting an elevator is the weight of the cage and of the load it carries times the height of the lift. Numerous other illustrations of work will suggest themselves to the student. 105. Power. Power is the rate of doing work; that is, in cal- culating power the time required to do a certain number of foot- pounds of work is considered. If 10,000 Ib. are lifted 7 ft. the work done is 70,000 foot-pounds, regardless of how long it takes. But, if one of two machines can do this in one-half the time that the other machine requires, then the first machine has twice the power of the second. The engineer's standard of power is the Horse-power, which may be defined as the ability to do 33, 000 foot-pounds of work per minute. The horse-power required to perform a certain amount of work is found by dividing the foot-pounds done per minute by 33,000. If an engine can do 1,980,000 foot-pounds in a minute, its horse- power would be 1,980,000^-33,000 = 60. An engine that can raise 66,000 Ib. to a height of 10 ft. in 1 minute will do 66,000 Ib. XlO ft. =660,000 foot-pounds per minute, and this will equal a ff$=20 horse-power. If another engine takes 4 minutes to do this same amount of work, it is only one-fourth as powerful; the work done per minute will be &&S>AQ. = 165,000 foot-pounds per minute; and its horse-power is - L H7nnj- = 5 horse-power. Example : An electric crane lifts a casting weighing 3 tons to a height of 20 ft. from the floor in 30 seconds; what is the horse-power used? 3 tons = 6000 Ibs. 6000 Ib. X20 ft. = 120,000 foot-pounds done. 120,000 foot-pounds done in 30 seconds = 240,000 foot-pounds per 1 minute. = 7.27 horse-power used. WORK, POWER, AND ENERGY 139 106. Horse -power of Belting. A belt is an apparatus for the transmission of power from one shaft to another. The driving pulley exerts a certain pull in the belt and this pull is transmitted by the belt and exerted on the rim of the driven pulley. The power transmitted by any belt depends on two things the effective pull of the belt tending to turn the wheel, and the speed with which the belt travels. From the preceding pages, it is easily seen that these include the three items necessary to measure power. The pull of the belt is the force. The speed, given in feet per minute, includes both distance and time. Force, distance and time are the three items necessary for the measure- ment of power. The total pull that a belt will stand depends on its width and thickness. It should be wide enough and heavy enough to stand for a reasonable time the greatest tension put upon it. This is, of course, the tension on the driving side. This tension, however, does not represent the force tending to turn the pulley. The force tending to turn the pulley (or the Effective Pull, as it is called) is. the difference in tension between the tight and the slack sides of the belt. The effective pull that can be allowed in a belt depends prima- rily on the width, thickness, and strength of the leather, or what- ever material the belt is made of. Besides, we must consider that every time a belt causes trouble from breaking or becoming loose, it means a considerable loss in time of the machine, of the men who are using it, and of the men required to make the repairs and, therefore, it should not be loaded as heavily as might otherwise be allowed. Leather belts are called "single/' "double," "triple," or "quadruple," according to whether they are made of one, two, three, or four thicknesses of leather. Good practice allows an effective pull of 35 Ib. in a single leather belt per inch of width. In a double belt a pull of 70 Ib. per inch of width may be allowed. The pull times the width gives the total effective pull or the force transmitted by the belt. The force times the velocity, or speed, of the belt in feet per minute will give the foot-pounds transmitted by it in 1 minute. One horse-power is a rate of 33,000 foot-pounds per minute; hence, the horse-power of a belt is obtained by dividing the foot-pounds transmitted by it per minute by 33,000. The velocity of the belt is calculated from the diameter and revolu- tions per minute of either one of the pulleys over which the belt 140 SHOP ARITHMETIC travels, as explained in Chapter VII. From these considerations, the formula for the horse-power that a belt will transmit may be written PXWXV H = 33000 where H = horse-power P = effective pull allowed per inch of width W = width in inches V = velocity in feet per minute Stated in words, this formula would read as follows: "The horse- power that may be transmitted by a belt is found by multiplying together the allowable pull per inch of width of the belt, the width of the belt in inches, and the velocity of the belt in feet per minute and then dividing this product by 33,000. Example : Find the horse-power that should be carried by a 12-in. double leather belt, if one of the pulleys is 14 in. in diameter and runs 1100 R. P. M. Explanation: To get the horse- P = 70 Ib. power, we must first find the values W = 12 in. O f Pt W> and F. We will take P v ,N/^ vunn as 70 Ib. since this is a double belt. r 71 A Vr * * AUU Tr , . . , . T , .-, | ., 12 W is given, 12 in. V, the velocity, = 4032 ft. per min. is obtained by multiplying the cir- TT_PXWX V cumference of the pulley by the R. 33000 P- M., which gives us 4032. Multi- 70X12X4032 plying these three together gives = - oonnn 3,386,880 foot-pounds per minute, -102+ horse-Dower Answer and divi ding by 33,000 we have >e-power, Answer. 102 + as the horse-power that this belt might be required to carry. 107. Widths of Belts. It is possible, also, to develop a formula with which to calculate the width of belt required to transmit a certain horse-power at a given velocity. One horse-power is 33,000 foot-pounds per minute. Then the given number of horse-power multiplied by 33,000 gives the number of foot-pounds to be transmitted per minute. Foot-pounds per minute = 33000 X H If we know the velocity in feet per minute, we can divide the foot-pounds per minute by the velocity; the quotient will be the force or the effective pull in the belt. Force ^ X \J V/V> * y- WORK, POWER, AND ENERGY 141 Now the force can be divided by the allowable pull per inch of width of belt. The result will be the necessary width. , 33000 XH Stated in words, this formula would read: "To obtain the width of belt necessary for a certain horse-power; multiply the horse- power by 33,000 and divide by the product of the allowable pull per inch of width of belt times the velocity of the belt in feet per minute." Example : Find the width of a single belt to transmit 10 horse-power at a speed of 2000 ft. per minute. II = 10 Explanation: We have given the horse- V = 2000 power and the velocity, and we know that P = 35 for a single belt a pull of 35 Ib. per inch is al- 33000 XH lowable. This data is all that is needed to PXV calculate the width, which comes out 4 in. The next larger standard width is 5 in., so 2 that is the size that would be used. 33000 X^ 33 . ~ ??X#W> = 7 7 Use 5 in. belt, Answer. 108. Rules for Belting. 1. Belt Thickness. It is generally advisable to use single belting in all cases where one or both pulleys are under 12 in. in diameter, and double belting on pulleys 12 in. or larger. Triple and quadruple belts are used only for main drives where considerable power is to be transmitted and where a single or double belt would have an excessive width. A triple belt should not be run on a pulley less than 20 in. in diam- eter, nor a quadruple belt on a pulley less than 30 in. in diameter. 2. Tension per Inch of Width. An effective pull of 35 Ib. per inch of width of belt is allowable for single belts. For double belts an effective pull of 70 Ib. per inch is allowable unless the belt is used over a pulley less than 12 in. in diameter, in which case only 50 Ib. per inch should be allowed. A prominent manu- facturer of rubber belting recommends 33 Ib. per inch of width of belt for 4-ply belts and 43 Ib. for 6-ply rubber belts. 3. Belt Speeds. The most efficient speed for belts to run is from 4000 to 4500 ft. per minute. Belts will not hug the pulley and therefore will slip badly if run at a speed of over one mile per minute. These figures are seldom reached in machine shops. 142 SHOP ARITHMETIC Belts for machine tool drives run from 1000 to 2000 ft. per min- ute, while main driving belts for line shafts are more often run about 3000 ft. per minute. On wood-working tools we find higher speeds, usually 4000 ft. per minute or over. 4. Distance between Centers. The best distance to have be- tween the centers of shafts to be connected by belting is 20 to 25 ft. For narrow belts and small pulleys this distance should be reduced. 5. Arrangement of Pulleys. It is desirable that the angle of the belt with the floor should not exceed 45 degrees; that is, the belt should be nearer horizontal than vertical. Fig. 66 shows Driver Fio. 67. the effect of having a belt nearly vertical. Any sag in the belt causes it to drop away from the lower pulley and lose its grip on it. Fig. 67 shows the best arrangement. Have the belt some- where near horizontal and have the tight side of the belt under- neath, if possible. This will increase the wrap of the belt around the pulleys. If the lower side is the loose side, the wrap will be decreased by the sag. It is also desirable, whenever possible, to arrange the shafting and machinery so that the belts will run in opposite directions from the shaft, as shown in Fig. 68. This arrangement balances somewhat the belt pulls, and reduces the friction and wear in the bearings. WORK, POWER, AND ENERGY 143 For belts which are to be shifted, the pulley faces should be flat; all other pulleys should have the faces crowned (high in the center) about T 8 in. per foot of width. I. ! FIG. 68. 6. Grain and Flesh Sides. The grain side of the leather is the side from which the hair is removed. It is the smoothest but weakest side of the leather, and should run next to the pulley surface. It will wrap closer to the pulley surface and thus get a better grip on the pulley. Furthermore, the flesh side, being stronger, is better able to stand the stretching which must occur in the outside of the belt in bending around a pulley. oursioc. Fio. 69. 7. Belt Joints. Whenever possible, the ends of belts should be fastened together by splicing and cementing. Never run a wide cemented belt onto the pulleys as one side is liable to be stretched out of true. Rather lift one shaft out of the bearings, 144 SHOP ARITHMETIC place the belt on the pulleys, and force the shaft back into place. Of other methods of fastening belts, the leather lacing is un- doubtedly the best when properly done. In lacing a belt, begin at the center and lace both ways with equal tension. Fig. 69 shows an excellent method of lacing belts. The lacing should be crossed on the outside of the belt. On the inside, the lacing should lie in line with the belt. Holes should be about 1 in. apart and their edges should be at least $ in. from the ends of the belt. The holes should be punched, preferably with an oval punch, the long dimension of the oval running lengthwise of the belt so as not to weaken the belt too much. PROBLEMS 181. A casting weighs 300 Ib. How much work is required to place it on a planer bed 3 ft. 5 in. above the floor? 182. How much work is required to pump 5000 gallons of water into a tank 150 ft. above the pump? 183. Find the horse-power that may be transmitted per inch of width by a single belt running at 2500 ft. per minute. How does this compare with a double belt running at the same speed? 184. A 6 in. double belt is carried by a 48 in. pulley running 250 R. P. M. Find the horse-power that may be transmitted. 185. A shop requires 50 horse-power to run it. The main shaft runs 250 R. P. M. Select a main driving pulley and determine width of double belt to run the shop. 186. A foundry fan runs 3145 R. P. M., and requires 24 horse-power to run it. There are two single belts on the blower running over pulleys 7 in. in diameter. Determine the necessary width of belt. Note. (Each belt should be wide enough to drive the fan so that in case one breaks, the other will carry the load.) 187. A belt is carried by a 36 in. pulley running at 150 R. P. M. The effective pull in the belt is 240 Ib. Find the horse-power. 188. A pumping engine lifts 92,500 gallons of water every hour to a height of 150 ft. What is the horse-power of the engine? 189. If a freight elevator and its load weigh 5000 Ib., what horse-power must be exerted to raise the elevator at a rate of 2 ft. per second? 190. A touring car is travelling on a level road at a rate of 45 miles an hour. If it is shown by actual test that a force of 200 Ib. is required to maintain this rate of speed, what horse-power must the engine deliver at the wheels? CHAPTER XVII HORSE-POWER OF ENGINES 109. Steam Engines. In the last chapter, the meaning of the term horse-power was explained and its application to belting was discussed. We will now take up the calculations of the horse-powers of steam and gas engines. One horse-power was given as the ability to do 33,000 foot- pounds of work in 1 minute. From this we see that the best way to get the horse-power of any engine is to find out how many foot-pounds of work it does in 1 minute and then to divide the number of foot-pounds delivered in a minute by 33,000. Let us study the action of the steam in the cylinder of the ordinary double-acting steam engine. In Fig. 70 is shown a FIQ. 70. section of a very simple boiler and engine. We find that steam enters one end of the cylinder behind the piston and pushes the piston toward the other end of the cylinder. Meanwhile, the valve is moved to the other end of the valve chest. The opera- tion is then reversed and the piston is pushed back to the starting- point. It has thus made two strokes, or one revolution. The steam pressure on the piston is not the same at all points in the stroke, but varies according to the action of the valve in cutting off the admission of steam into the cylinder. However, it 145 146 SHOP ARITHMETIC is possible to obtain the average or "mean effective pres- sure" per square inch during a stroke, and, if we multiply this by the piston area in square inches, we will have the average total pressure or force exerted during one stroke. Now reduce the length of the stroke to feet and multiply this by the total pressure just found, and we have the number of foot-pounds of work done during one stroke. This result, when multiplied by the number of working strokes per minute, gives the foot-pounds per minute and this divided by 33,000 gives the horse-power. The following are the symbols generally used : H. P. Horse-power. P = Mean pressure in pounds per square inch. A = Area of piston in square inches. L = Length of stroke in feet . N = Number of working strokes per minute. PX A = Total pressure on piston. PX A xL = it. Ib. of work done per stroke. PXA XLX N it. Ib. of work done per minute, and hence PXAXLXN = 33000 " or ' as usuall y wntten > PXLXAXN - . In the latter form, the letters in the numerator 33000 spell the word Plan and the formula is thus easily remembered. In the common steam engine, there are two working strokes for every revolution of the engine, that is, the engine is \vhat is called double acting, and N is twice the revolutions per minute. A few steam engines, like the vertical Westinghouse engine, are single acting and, hence, have only one working stroke of each piston per revolution. Unless otherwise stated, it will be as- sumed in working problems that a steam engine is double acting. Example : Find the horse-power of a 32 in. by 54-in. steam engine running at 94 R. P. M. with an M. E. P. (Mean Effective Pressure) of 60 Ib. Note. In giving the dimensions of an engine cylinder, the first number represents the diameter and the second number the stroke. P = 60 Ib. L = 54 in. =4^ ft. A =Area of 32 in. piston = 804. 25 sq. in. N = Number of strokes per minute = 94 X 2 = 188 H p _-PxLxAxJV 33000 60X4.5X804.25X188 1237+ Jiorse-power, Answer. HORSE-POWER OF ENGINES 147 Notice particularly that the area of the piston is expressed in square inches, because the pressure is given in pounds per square inch; but that the stroke is reduced to feet because we measure work in foot-pounds and, consequently, must express in feet the distance which the piston moves. If an engine has more than one cylinder, the horse-power of each can be calculated and the results added; or, if the cylinders are arranged to do equal amounts of work, we can find the horse- power of one cylinder and multiply this by the number of cylinders. The mean effective pressure can be obtained for any engine by the use of a device called an "indicator," which draws a diagram showing just what the pressure is in the cylinder at each point in the stroke. From this diagram, we can calculate the average or mean effective pressure for the stroke. This pressure must not be confused with the boiler pressure or the pressure in the steam pipe. For instance, when the steam comes from the boiler to the engine at 100 Ib. pressure, the mean pressure in the cylinder will not be 100 Ib., as it would be very wasteful to use steam from the boiler for the full stroke. Instead, the M. E. P. (Mean Effective Pressure) will be from 20% to 85% of the boiler pressure depending on the type of the engine and the load it is carrying. Horse-power calculated as explained here is called Indicated Horse-power because an indicator is used to determine it. The indicated horse-power represents the power delivered to the piston by the steam. 110. Gas Engines. The most common type of gas or gasoline engine works on what is called the four stroke cycle. Such an engine is called a four-cycle engine. Fig. 71 shows in four views the operation of such an engine. Four strokes, or two revolutions, are required for each explosion that occurs in the cylinder. Consequently, in calculating the horse-power of a single cylinder gas engine, the number of working strokes (or N in the horse-power formula) is one-half of the R. P. M. There is another type of gasoline engine called the two-cycle engine. A single cylinder two-cycle engine has one working stroke for each revolution of the crank shaft and N is therefore the same as the number of R. P. M. The mean effective pressure of a gas engine is from 40 to 100 Ib. per square inch, depending chiefly on the fuel used. For 148 SHOP ARITHMETIC gasoline or natural gas or illuminating gas it is usually between 80 and 90 Ib. per square inch. 2. Compression 4-. Exttausr Fio. 71 Example : What horse-power could be delivered by a single cylinder 5 in. by 8 in. four-cycle gasoline engine running 450 R. P. M.? Note. Use a value of P = 80 Ib. per square inch. P = 80 A = . 7854 X5 2 = 19.6 sq. in. Then, H. P. 33000 80x|xl9.6x225 O " 33000 - = 7.13 horse-power, Answer. HORSE-POWER OF ENGINES 149 111. Air Compressors. An air compressor is like a double acting steam engine in appearance; but, instead of delivering up power, it requires power from some other source to run it. This power is stored in the air and later is recovered when the air is used. An air compressor takes air into the cylinder, raises its pressure by compressing it, and then forces it into the air line or the storage tank. In calculating the horse-power of a com- pressor, the same formula can be used as for a steam engine. The value of P to use is not the pressure to which the air is raised, but is the average or mean pressure during the stroke. It is usually somewhat less than half the final air pressure; for ex- ample, when an air compressor is delivering air at 80 Ib. pressure, the mean pressure on the piston is about 33 Ib. Most air compressors are double acting, though there are many small single acting ones. Example : A double acting 12 in. by 14 in. air compressor is running 150 R. P. M. It is supplying air at 100 Ib. and the mean pressure in the cylinder is 37 Ib. per square inch. Calculate the horse-power necessary to run it. P = 37 Ib. A = Area of 12 in. piston = 113.1 sq. in. N = Strokes = 150 X 2 = 300 per minute. ThPn ff P Then, H.P. - 6 110 In this case 12 appears in the denominator in order to reduce the 14 inches to feet. 112. Brake Horse-power. The Brake Horse-power of an engine is the power actually available for outside use. It, therefore, is equal to the indicated horse-power minus the power lost in friction in the engine. Brake horse-power can be readily determined by putting a brake on the rim of the fly-wheel and thus absorbing and measuring the power actually delivered. Fig. 72 shows such a brake arranged for use. This form is known as the "Prony Brake." It consists of a steel or leather band carrying a number of wooden blocks. By tightening the bolt at A, the friction between the blocks and the rim of the wheel can be varied at will. The corresponding pull which this friction gives at a distance R ft. from the shaft is weighed by a 150 SHOP ARITHMETIC platform scale or spring balance. From the scale reading must be deducted the weight due to the unbalanced weight of the brake arms, which can be determined by reading the scales when the brake is loose and the engine is not running. If an engine is capable of maintaining a certain net pressure W on the scale, and meanwhile maintains a speed of N revolutions per minute, we can readily see that this is equivalent to an effective belt pull of W pounds on a pulley of radius R running at N revolutions; or it can be considered as being equivalent to raising a weight FIG. 72. equal to W by means of a rope wound around a pulley of radius R turning at N revolutions per minute. This weight would be lifted at the rate of 3.1416 X2XRXN ft. per minute and the brake horse-power will be '

<6* p Explanation: As the piston rests on the water, the pressure f the water on the bottom of 1000 the piston must be sufficient to -5^-== 35.4 Ib. per sq. in., Answer, support the weight. The area of the bottom is 28.27 sq. in., 1000 and the pressure on each sq. in. will be ^nvr or 35.4 Ib. persquare inch. Zo.Z I This pressure is transmitted throughout the water and is exerted by it with equal force in all directions. 118. The Hydraulic Jack. This property of water of trans- mitting pressure in any direction is made use of in many ways. The same property is, of course, common to other fluids such as FIG. 74. oil, air, etc. Wherever we find a powerful, slow-moving force required in a shop, we usually find some hydraulic machine. (The word " hydraulic " refers to the use of water but it is often applied to machines using any liquid water, oil, or alcohol.) Fig. 74 shows the principle of all these hydraulic machines. 156 SHOP ARITHMETIC A small force P is exerted on a small piston and this produces a certain pressure in the water. This pressure is transmitted to the larger cylinder where the same pressure per square inch is exerted on the under side of the large piston. If the large piston has 100 times the area of the small piston, the weight supported (W) will be 100 times P. If the water pressure produced by P is 100 Ib. per square inch and the large piston has an area of 100 sq. in., then the weight W that can be raised will be 10,000 Ib. Like the lever, the jackscrew, and the pulley, this increase in force is obtained only by a decrease in the distance the weight is moved. The work done on the small piston is theoretically the same as the work obtained from the large piston. For example, suppose that the large piston has 100 times the area of the small one and we shove the small piston down 1 in. ; the water that is thus pushed out of the small cylinder will have to spread out over the entire area of the large piston; the large piston will, therefore, be raised only one one-hundredth of the distance that the small piston was moved. Thus, we have, here also, an applica- tion of the law that the work put into a machine is equal, neglecting friction, to the work done by it. Force X distance moved = weight X distance raised. The Mechanical Advantage of such a machine will be seen to be the ratio of the areas of the pistons. In the case just men- tioned, the ratio of the areas of the pistons was 100:1; hence, the mechanical advantage was 100. In Fig. 74, the motion that can be given to W is very limited, but by using a pump with valves, instead of the simple plunger P, we can continue to force water into the large cylinder and thus secure a considerable motion to W. Fig. 75 shows a common form of hydraulic jack which operates on this principle. The top part contains a reservoir for the liquid, and also has a small pump operated by a hand lever on the outside of the jack. By working the lever, the liquid is pumped into the lower part of the jack between the plunger and the casing, thus raising the load. The load may be lowered by slacking the lowering screw Y. This opens a passage to the reservoir, and the load on the jack forces the liquid to flow back through this passage to the reservoir. In calculating the mechanical advantage of a hydraulic jack, we must consider the mechanical advantage of the lever which MECHANICS OF FLUIDS 157 operates the pump as well as the advantage due to the relative sizes of the pump and the ram. Fia. 75. Example : If the ram of Fig. 75 is 3 in. in diameter and the pump is 1 in. in diameter, while the lever is 15 in. long and is connected to the pump at a distance of 1 J in. from the fulcrum, what is the mechanical advantage of the entire jack? Explanation: The areas of the ram and pumps are as 9:1, hence their mechanical advantage is 9. The lever has a mechanical advantage of 10. Hence, that of the whole jack is 9X10 = 90, and a force applied at the end of the lever would be multiplied 90 times. This force would, however, move through a distance 90 times as great as the distance the load would be raised. 7854X3 2 _9 .7854 Xl 1 ""!" 9X10 = 90, Answer. 158 SHOP ARITHMETIC The hydraulic jack has usually an efficiency of over 70% and is, therefore, a much more efficient lifting device than the jack screw. A mixture containing one-third alcohol and two-thirds water should be used in jacks. The alcohol is added to prevent freezing. 119. Hydraulic Machinery. In the shop, we often find water pressure used to operate presses, punches, shears, riveters, hoists, and sometimes elevators. These machines are seldom operated by hand power but have water supplied under pressure from a central pumping plant. The admission of the water and the consequent motion of the machine is controlled by hand operated valves. Most of these hydraulic machines are used where tremendous forces are required. Therefore, very high water pressures are used, occasionally as high as 3000 Ib. per square inch. 1500 Ib. per square inch is a common working pressure for hydraulic machines. FIG. 76. FIG. 77. Fig. 76 shows a press operated by hydraulic pressure. It will be noticed that the movable head is connected to two pistons a large one for doing the work on the down stroke, and a smaller one above, used only for the idle or return stroke of the press. 120. Hydraulic Heads. Quite often we use a high tower or MECHANICS OF FLUIDS 159 tank to secure a water pressure, or we make use of some natural source of water which is at some elevation. This is most often seen in the water supplies for towns and cities. Water tanks are put upon the roofs of some factories for the same purpose. The higher the tank, the greater will be the pressure which it will maintain in the system. Let Fig. 77 represent such a system. The water at the bottom has the weight of a column of water h ft. high to support and, consequently, will be under a pressure equal to the weight of this column of water. A volume of water 1 in. square and 1 ft. high weighs .434 lb., so the pressure per square inch at the base of the column in Fig. 77 will be .434 X h. Notice particularly that the shape and size of the tank has no influence on the pressure, it being used merely for storage and to keep the pressure from falling too fast if the water is drawn off. The water in the tank on either side of the outlet is supported by the bottom of the tank and has no effect on the pressure in the pipe. The pressure at the bottom of the pipe would be the same if the pipe alone extended up to the height h without the tank. Also the size of the pipe has no effect on the pressure per square inch. The water in a large pipe will weigh more than in a small pipe, but the pressure will be spread over a larger area and if, the heights are the same, the pressure per square inch will be the same. In pumping water to an elevated tank or reservoir, the pressure required per square inch is also determined in the same manner and is .434 times the height to which the water is raised, plus an allowance for friction in the pipes. Thus, to pump water to a height of 100 ft. requires a pressure somewhat greater than .434 X 100 = 43.4 lb. per square inch. 121. Steam and Air. Steam and air are likewise used to pro- duce pressures in shop machinery. Being more elastic than water, they are preferred where the machines are to be operated quickly. Devices using air are called "pneumatic appliances," among the most common of which are pneumatic drills, hammers, and hoists. The air for operating these is supplied by air com- pressors located in the power house. These take the air from out of doors and compress it into a smaller volume. The resistance of the air to this compression causes it, in its effort to escape, to exert a pressure on the walls of the tank or pipe containing it. The more the air is compressed, the greater is the pressure exerted by it. The air pressure used in shop work is usually about 160 SHOP ARITHMETIC 80 Ib. per square inch. The air is conducted through pipes and hose to the point where it is to be used and there allowed to exert its pressure on the piston of the appliance which is to be driven. PROBLEMS 201. The specific gravity of Lignum Vitse (a hard wood) is 1.328. this wood float or will it sink in water? Will TTT fit FIG. 78. FIG. 79. 202. What weight on the small piston of Fig. 74 would support a weight of 30,000 Ib. on the large piston if the small piston is 1 in. in diameter and the large one 12 in. in diameter? 203. If a hydraulic press works with a water pressure of 1500 Ib. per square inch, what must be the diameter of the ram if a total pressure of 75,000 Ib. is to be produced? 204. If the air hoist of Fig. 78 has a cylinder 10 in. in diameter inside, and the piston rod is 1J in. in diameter, and an air pressure of 80 Ib. per square inch is exerted on the bottom of the piston, what weight can be lifted by the hoist? 205. If a city wishes to maintain a water pressure of 80 Ib. per square inch at their hydrants, how high above the streets must be the water level in the stand pipe? MECHANICS OF FLUIDS 161 206. Fig. 79 shows the principle of one form of hydraulic elevator, the car being fastened directly to a long ram which is raised by water pressure. The weight of the ram and car are partially balanced by a counterweight. If this elevator is operated with water from the city mains at 80 Ib. pressure per square inch and the ram has a diameter of 10 in., what load can be lifted allowing 30% for losses in friction, etc.? FIQ. 80. Fia. SI. 207. If a steam pump, such as shown in Fig. 80, has a 12 in. steam piston and an 8 in. water piston, what water pressure can be produced with a steam pressure of 90 Ib. per square inch? How many gallons would be pumped per minute when the pump is running at 100 strokes per minute, the stroke of the pump being 12 in.? (1 gallon = 231 cu. in.) 208. What water pressure must a pump be capable of producing in order to force the water to a reservoir at an elevation of 300 ft. above the pump? 15 162 SHOP ARITHMETIC 209. A gravity oiling system has an oil tank placed 10 ft. above the bearings to be lubricated. The tank is connected by small tubes to the various bearings. If the specific gravity of the oil is .88, what pressure will the oil have at the bearings? 210. In Fig. 81 we have a hoist operated by a hydraulic ram in the top of the crane post. The motion of the ram is multiplied by the system of pulleys shown in the figure. What size must the ram be that a load of 10,000 Ib. can be lifted with a water pressure of 72 Ib. per square inch, the efficiency of the whole apparatus being 70%? CHAPTER XIX HEAT 122. Nature of Heat. Some of the effects of heat are very- useful in shop work and every mechanic should know something of the nature of heat and of the laws which govern its applica- tions to shop work. Heat is a form of energy; that is, it is capable of doing work. This we see amply illustrated in the steam engine and the gas engine, where heat is used in producing work. The steam engine uses heat which has been imparted to the steam in the boiler. Part of the heat of the steam is changed to work in the engine and the rest is rejected in the exhaust. Heat is not a substance as was formerly supposed it cannot be weighed and cannot exist by itself. It is always found in some substances. We generally get heat by burning some fuel such as coal, wood, gas, or oil. In burning, the fuel unites with oxygen, one of the constituents of air, and this process, called combustion, generates the heat. We cannot get heat by this process, therefore, without air. No fuel will burn without a supply of air, and as soon as we shut off the air from a fire, combustion stops and no more heat is generated. A fire may continue to give off heat for some time after the air is cut off, but this heat comes from the cooling of the hot fuel in the fire. Of the heat generated during combustion, some of it goes through the furnace walls to the surrounding air; some goes to heat up the bed of coals and any object that may be placed in the fire to be heated; but the greater part of the heat goes off in the gases that are formed by the union of the fuel with the air. It is to save this heat that we sometimes see steam boilers set up in connection with the furnaces of large forge shops. There are other ways of generating heat besides that of com- bustion. One method, that is coming into considerable use and which is especially interesting to shop men because of the ease with which it can be controlled, is by the use of electricity. We now have electric annealing and hardening furnaces for use in 16 163 164 SHOP ARITHMETIC tool rooms, where a close regulation of the heat is very desirable. Then there are the electric furnaces by which aluminum and carborundum are produced. We also have electric welding as an example of the production of heat from electricity. Another method of heat generation that is frequently encoun- tered in shops, often where it is not desired, is the production of heat from work. We have seen how heat is turned into work, but here we have work returned into heat. One common case of this is in bearings, where heat is produced from the work that is spent in overcoming the friction. Another example is seen in the heating of a lathe tool when it is taking a heavy cut, or in the heating of the tool when it is being ground. In either event, the work spent in removing the metal goes into heat. 123. Temperatures. Temperature is the indication of the height or intensity of the heat in a body. Lowering the tem- perature means a removal of heat from a body, and raising the temperature means the addition of more heat. The common method of measuring temperature is by means of an instrument known as a thermometer, which usually consists of a glass tube which is partly filled with mercury and which has the air ex- hausted from the other part of it. The mercury expands and contracts as the temperature rises or falls and, therefore, the height of the column of mercury is a measure of the temperature. Alcohol is often used instead of mercury for outdoor thermome- ters where the mercury might freeze. There are two kinds of thermometer scales in common use the Centigrade (abbreviated C.) and the Fahrenheit (abbreviated Fahr. or F.) . On the Centigrade thermometer the space between the freezing-point of water and the boiling-point at atmospheric pressure is divided into 100 equal parts called Degrees (represented by ) the freezing-point being marked zero (0) and the boiling- point 100. The balance of the scale is then divided into spaces of equal length below zero and above 100 in order that tem- peratures higher than 100 and lower than zero may be read. On the Fahrenheit scale the freezing-point of water is marked 32 and the boiling-point 212, so the space between is divided into 180 (212 -32 = 180). This scale is also marked with divisions below 32 and above 212 in order to make the thermom- eter read through a wider range. The Fahrenheit thermometer is used more commonly in the United States than the Centigrade, which is used extensively in Europe. The Centigrade scale is, HEAT 165 however, used in this country for most scientific work and is becoming so common that it is desirable to understand the rela- tions between the two scales. Fig. 82 shows the relation of the two scales up to 212 F. or 100 C. Since the same interval of temperature is divided into 100 parts in the Centigrade scale, and 180 parts in the Fahrenheit scale, each Centigrade degree is }$, or f Fahrenheit degrees. Similarly, one Fahrenheit degree is f of a Centigrade degree. A change of 30 in tempera- ture on the Centigrade scale would equal f of 30, or 54 change on a Fahrenheit thermometer. Likewise, when the mercury moves 27 on a Fahrenheit thermometer, it would move only J- of 27 = 15 on a Centigrade scale. In changing a reading on one thermom- eter scale to the corresponding reading on the other, it is necessary to remem- ber that the zero points are not the same. The Centigrade zero is at 32 F. In other words, the two zeros are 32 Fahrenheit degrees apart. To change a reading on the Centigrade scale to the corresponding Fahrenheit reading: First multiply the degrees C. by f . This gives an equivalent number of degrees on the F. scale. To this add 32, in order to have the reading from the F. zero. To change a reading on the Fahrenheit scale to the corresponding Centigrade reading: First subtract 32. This gives the number of F. de- grees above freezing (which is the C. zero) . Multiply the result by f , thus obtaining the desired C. reading. Examples : 1. Change 30 C. to the corresponding Fahrenheit reading. 30 X r = 54, the equivalent number of F. degrees. 54+32-86 F., the reading on a F. thermometer. 2. What would a Centigrade thermometer read when a Fahrenheit ther- mometer stood at 72? 210 200 190 180 170 160 150 140 130 120 110 100 so- so 70 60 BO 40 F ^0= 20 10 -10 -20 -30 -40 1 212F = 100C BOILING i 100 - 90 - 80 - 70 60 50 1 40 | 30 - - 20 ; 10 32 F. 0C 1 FREEZING o C \ 10 \ 20 1 30 - 1G6 SHOP ARITHMETIC 72 32 = 40, the number of F. degrees above freezing. 5 2 40 X = 22 C., Answer. These rules or relations are often expressed by the following formulas, in which C stands for a reading on the Centigrade thermometer and F for a reading on the Fahrenheit thermometer. The parenthesis ( ) when used as above means that the work indicated inside of it is to be done first and then the result multiplied by f . In the second formula, the C is first to be multiplied by f and then the 32 is added to the product. It is always to be understood that multiplications and divisions are to be performed before additions and subtractions unless the reverse is indicated, as was done in the first of these formulas, by the use of the parenthesis ( ) . When it comes to measuring the temperatures in furnaces, as is often desirable in fine tool work, a thermometer is clearly out of the question. As the mercury thermometer is ordinarily made, it should not be used for temperatures above 500, but, by filling the glass tube above the mercury with nitrogen gas under pressure, a thermometer can be made that may be read up to 900. For higher temperatures, devices called Pyrometers are used. There are numerous kinds of pyrometers, but the one most used in the shops for furnace temperatures is what is called the Le Chatelier pyrometer. In this pyrometer, one end of a porcelain tube about f in. in diameter and from 12 in. to 40 in. long is thrust into the furnace and held there or, if frequent readings are to be taken, it may be placed there permanently. Inside this tube are some wires of special composition that generate an electric current when they get hot. From the other end of the tube a couple of wires run to a small box containing a "galvanometer," that is, a device for indicating the strength of the electric current generated. This has a needle swinging over a dial and the dial is usually laid off in degrees so the temperature is read direct. Most of these pyrometers have centigrade gradu- ations, but one should be sure which scale a pyrometer has before he uses it. HEAT 167 For example, suppose we wanted to get 1000 F. and by mis- take had 1000 C. instead. ? X 1000 + 32= 1832 F. o 1000 C. = 1832 F. This shows that it would be pretty serious to use the w r rong scale. It has for years been the practice of the older shop men to tell the temperature of steel or iron by its color. This method has its disadvantages, however, as so much depends on the sensitive- ness of the man's eye and on whether the work is being done in bright sunlight or in a dark corner of the shop. A bar will show red in the dark when it would still be black in the sunlight. For the lower range of temperatures (those used in tempering tools) we can judge the temperature by the color which will appear on a polished steel surface when heated in the air. These tempering colors and their uses for carbon tool steels are about as follows: 430 F. Very pale yellow Scrapers Hammer faces Lathe, shaper, and planer tools 460 Straw yellow Milling cutters Taps and dies 480 Dark straw color Punches and dies Knives Reamers f>00 Brownish-yellow Stone cutting tools Twist drills .">20 Yellow tinged with purple Drift pins 530 Light purple Augers Cold chisels for steel oo() Dark purple Hatchets Cold chisels for iron Screw drivers Springs 570 Dark blue Saws for wood 610 Pale blue 630 Blue tinged with green More uniform results can be obtained if the steel is heated in a bath of sand or of oil, the bath being maintained at the desired temperature and a pyrometer being used to observe the temperature. For higher temperatures, molten lead or mineral 1G8 SHOP ARITHMETIC salts such as common salt, barium chloride, potassium chloride, and potassium cyanide are used. When steel and -iron are heated to higher temperatures, they successively become red, orange, and white. These colors and the corresponding temperatures are about as follows: 957 F. First signs of red 1290 Dull red 1470 Dark cherry 1655 Cherry red 1830 Bright cherry 2010 Dull orange 2190 Bright orange 2370 White heat 2550 Bright white welding heat 2730 ^ 2910 t0 } Dazzling white. 124. Expansion and Contraction. Nearly all substances expand when heat is applied to them and contract when heat is removed. This phenomenon is greatest in gases and least in solids, but even in solids it is of enough moment to be extremely useful at times or to cause considerable trouble when allowance is not made for it. There are a few metal alloys which, within certain limits, do not change their volumes with changes of temperature, and there are also some which between certain temperatures will even expand when cooled and contract when heated. A nickel steel containing 36% nickel has practically no expansion or contrac- tion with changes in temperature and is, therefore, used in some cases for accurate measurements where expansion of the measur- ing instruments would introduce serious errors. When a solid body is heated it expands in all directions, if free to do so, but as a rule we are concerned only with the change of one dimension and not with the change in volume. Thus, in the case of a steam pipe we do not care about the change in thickness or in diameter, but we are concerned with the change in length. On the other hand, when a bearing gets hot and seizes, it is the change in diameter that causes the trouble. There are few machinists who have not had the experience, in boring a sleeve to fit a certain shaft, of having a free fit when tested just after taking a cut through the sleeve, and then later of finding that the sleeve fitted so tightly that it had to be driven off the shaft. Of course, the explanation is that the sleeve becomes HEAT 169 warm when being bored in the lathe, while the shaft is much cooler. When the sleeve cools to the temperature of the .shaft, it contracts and seizes or "freezes" to the shaft. In accurate tool work the effect of differences in temperature between the measur- ing instruments and the work may become serious. For this reason many gages are provided with rubber or wooden handles which do not conduct heat readily. They thus prevent the heat of the hand from getting into the gages and expanding them. But this is enough to give some idea of the troubles caused by this property of materials; let us now see of what benefit it is. We have already seen the use that is made of the expansion of mercury in thermometers. There are numerous heat regulating devices" (called thermostats) which depend on the expansion or contraction of a bar to perform the desired operations. We find these used for regulating house heating boilers and furnaces, incubators, anti. other devices where uniform temperatures are required. Probably the greatest shop use of expansion and contraction is in making shrink fits. When we want to fasten securely and permanently one piece of metal around another, we generally shrink the first onto the second. This process is used for attaching all sorts of bands and collars to shafts, cylinders, and the like, for putting tires on locomotive wheels, and for similar work. The erecting engineer uses it to put in the links in a sectional fly-wheel rim or to draw up bolts in the hub or in any other place where he wants to make a rigid permanent joint. The amount of linear expansion which a body undergoes depends upon the kind of material of which the body is made, upon the amount of the temperature change and, of course, upon the original length. The coefficient of linear expansion of a substance is that part of its original length which a body will expand for each degree change in temperature. Coefficients for different metals have been determined for our use by careful experiments, and can be found in hand books or tables under the head of "Coefficients of Expansion." The values given in different books do not al- ways agree. In fact, the exact compositions of the metals used in the tests were undoubtedly different for the different tests that are on record. Hence, different tables give slightly different rates of expansion. The following values are taken from the most reliable authorities and are sufficiently accurate for most purposes. 170 SHOP ARITHMETIC COEFFICIENTS OF EXPANSION Metal Coefficient Aluminum . 00001234 Brass . 00001 Cast iron .0000055 to Wrought iron and machine steel . 36% nickel steel .000006 .0000065 . 0000003 The above values are based on a temperature rise of 1 F. For one Centigrade degree change in temperature the coefficients would be f of those just given. The student is not expected to memorize these values. Remember that if the length is given in feet the expansion calculated will be in feet, and if the length is in inches the expansion calculated will be in inches. To get the actual expansion per degree for any certain length, multiply the coefficient of expansion by the length. If the temperature change is 100, the expansion will be 100 times that for 1. Example : The head of a gas engine piston in operation has a temperature of about 400 higher than the cylinder in which it is running. What allowance must be made for this expansion in a 12 in. piston? (The piston is made of cast iron.) .000006X400 = .0024 in. expansion per inch .0024X12 =.0288 in. expansion in 12 in., Answer. The head of the piston must, therefore, be turned at least .0288 in. small to allow for the expansion to take place without the piston seizing in the cylinder. The law of expansion and contraction may be expressed by a formula as follows: where E = TxCxL E is the change in length T is the change in temperature C is the coefficient of linear expansion L is the original length of the body. HEAT 171 Example : What will be the expansion in a steam pipe 200 ft. long when subjected to a temperature of 300, if erected when the temperature was 60? 7 7 = 300-60 = 240; C = .0000065; L = 200 E=TXCXL = 240 X. 0000065X200 = .312 ft., Answer. Notice particularly that here we use L in feet and, consequently, the expan- sion E comes out in feet. This can be reduced to inches if desired, giving 12 X. 312 = 3. 744 in. or 3| in. nearly. 125. Allowances for Shrink Fits. In making a shrink fit, the collar or band, or whatever is to be shrunk on, is bored slightly smaller than the outside diameter of the part on which it is to be shrunk. It is then heated and thus expanded until it can be slipped into place. When it cools, it cannot return to its original size but is in a stretched condition. It, therefore, exerts a power- ful grip on the article over which is has been shrunk. Practice differs considerably in the allowances that are made for shrink fits. A rule which has been widely and successfully used is to allow oVu" m< f r each inch of diameter. According to this rule, if we were shrinking a crank on a 6 in. shaft, the crank should be bored . 006 in. small or else the shaft turned . 006 in. oversize and the crank bored exactly 6 in. For a 10-in. shaft we would allow .010 in, and so on for other sizes. This could be expressed by the following formulas: A = .OOlXl>, or, since -001=77)7)7), this could be written D 1000 where A stands for " allowance " and D for the diameter Assuming that an allowance of .001 XD is made, let us see what temperature is necessary in order to give the necessary expansion so that a steel tire can be put over a locomotive driving wheel. For each degree that the tire is heated, it will expand .0000065 in. per inch of diameter. We must have an expansion of at least .001 in. The number of de'grees necessary to get this will be .001 . 0000065 17 = 154 172 SHOP ARITHMETIC It would look as if a difference of 154 would be sufficient. However, a greater difference is necessary in practice. There must be sufficient clearance so that the tire can be slipped quickly into place before it has time to cool off or to warm the wheel. Once in place, the tire will grip the wheel when a temperaUire difference of 154 exists. PROBLEMS 211. In testing direct current generators, it is customary to specify that under full load the temperature of the armature shall not rise more than 40 Centigrade above a room temperature of 25 C. ; that is, the temperature of the armature under these conditions should not exceed 65 C. In making a test a Fahrenheit thermometer was used. The room was at a temperature of 77 F. and the temperature of the armature at the end of the run was 180 F. Did the generator meet the specifications? What was the temperature change, Centigrade? 212. In erecting a long steam line that will have a variation in tempera- ture of 320, how far apart should the expansion joints be placed if each joint can take care of a motion of 3 inches? 213. If a brass bushing measures 2 in. just after boring, when its tem- perature is 95 F., what will it caliper when it has cooled to 65 F.? 214. A steel link 2 ft. long is made ^ in. too short for the slot in the fly- wheel rim into which it is to be shrunk. How hot must the link be before it will go in? 215. If a hub bolt is heated until it just begins to show red and is im- mediately screwed up snug and allowed to cool, what shrinkage allowance per inch of length would we be allowing by such a plan? 216. If we wished to maintain a tempering bath at a temperature of 500 F., what should be the reading on a Centigrade pyrometer? 217. If the brass bearings for a 2 in. steel crank shaft are given a running clearance of .002 in. at a temperature of 60 F., what would be the clearance when running at a temperature of 100 F.? 218. A horizontal steam turbine and dynamo are to be direct-connected, their shaft centers being 3 ft. above the bed plate. If the bearings are lined up at a temperature of 70, how much will they be out of alignment under running conditions when the temperature of the dynamo frame is 80 F. and that of the turbine is 215 F., both frames being of cast iron? CHAPTER XX STRENGTH OF MATERIALS 126. Stresses. When a load is put upon any piece of material, it tends to change the shape of the piece. The material naturally resists this and, therefore, exerts a force opposite to the load. If the load is not too heavy, the material may be able to exert a sufficient force to hold it, but often the strength of the material is exceeded and the piece breaks. The resistance which is set up when a piece of material is loaded is called the Stress. For instance, if a casting weighing 3 tons or 6000 Ib. is suspended by a single rope, the stress in the rope will be 6000 Ib. x\\\\\\ \v\\\\ W tension \\\\\\\\\\\VC compression FIG. 83. shear There are three different kinds of stresses that can be produced, depending on the way the load is applied (1) Tensile stress (pulling stress). (2) Compressive stress (crushing or pushing stress) . (3) Shearing stress (cutting stress). Fig. 83 shows how these different stresses are produced. We sometimes recognize two other kinds of stresses, but these are really special cases of the three just given. These two others are: 173 174 SHOP ARITHMETIC (a) Bending Stress (really a combination of tension on one side and compression on the other). (b) Torsional or Twisting Stress (a form of shearing stress) . 127. Ultimate Strengths. By taking specimens of the different materials and loading them until they break, it has been possible to find out just what each kind of material will stand. The load to which each square inch of cross-section must be subjected in order to break it, is called the Ultimate strength of the material. The strength of most materials differs for the different methods of loading shown in Fig. 83. The Tensile Strength of a material is the resistance offered by its fibers to being pulled apart. The Compressive Strength of a material is the resistance offered by its fibers to being crushed. The Shearing Strength of a material is the resistance offered by its fibers to being cut off. The following table gives the average values for the most used materials. ULTIMATE STRENGTHS POUNDS PER SQUARE INCH Material Tension Compression Shear Timber 10000 8000 3000 (across grain) Cast iron 20000 90000 20000 Wrought iron Machine steel 50000 65000 50000 65000 40000 50000 128. Safe Working Stresses. Having found how great a stress is required to break one square inch of material, we naturally would not allow anywhere near this stress to come on a piece of material in actual service. The Ultimate Strength is usually divided by some number, known as the Factor of Safety, and the quotient is used as the Safe Working Stress. For example, if 60,000 Ib. per square inch will break a piece of soft steel and we use a factor of safety of 5, this would give: (\C\OC\O Safe working stress = = = 12000 Ib. per square inch. o The following table gives the Safe Working Stresses of the most used materials. STRENGTH OF MATERIALS 175 SAFE WORKING STRESSES POUNDS PER SQUARE INCH Material 8 t Tension S c Compression s, Shear Timber 700 700 500 Cast iron .... 3- 4000 15-18000 3- 4000 Wrought iron .... Machine steel. . . . 8-10000 10-16000 8-10000 10-16000 7- 9000 8--12000 Instead of writing " safe loads in pounds per square inch " for tension, compression, or shear, the symbols S t , S c , and S 8 are used. So if A = area in square inches, then the load W which can be carried safely = Area X safe load per square inch or AxS t = W (tension) A X S c = W (compression) AxS a = W (shear) Or, in general, for all stresses AXS = W Perhaps more often we would want to find the area necessary in order to support a certain weight or load. In this case, we would want a formula which would give A. If we divide the total load by the safe stress, we will get the necessary area; or W A = S This simply says, area of metal necessary = total weight to be carried divided by safe load in pounds per square inch. From the area of a bolt or rod, its diameter can be easily found. 129. Strengths of Bolts. There is a well-known saying that " a chain is only as strong as its weakest link." This means, in general, that any mechanism must be so designed that its weakest part will be strong enough to stand the greatest load that may come on it. In figuring the size of a bolt to hold a certain load, we would not calculate the full diameter of the bolt and make the area there just sufficient, but we must see to it that the bolt has a cross-sectional area at the root of the threads large enough to support the load. Then the body of the bolt will have a surplus of strength. 176 SHOP ARITHMETIC Example : ' What size of steel eyebolt will support a weight of 5000 lb.? Take 12,000 lb. as the safe load in tension. W 5000 Then ' A= T = l2ooo 5 A = 2 sq. in. = .416 sq. in. .416 sq. in. is then the necessary area to support the weight. Of course, the example could be completed by saying .7854 D 2 = .416 sq. in., where D = diameter at the root of the thread. By then solving for D we would get the diameter at the root of the threads. But the Bolt Tables afford an easier method than this. In the following table, .4193 is given as the area of a | in. bolt at the root of the thread. Therefore, a | in. eyebolt would probably be used. In figuring the allowable loads for steel bolts, it is best not to allow over 12,000 lb. stress per square inch and 10,000 lb. is perhaps even more usual on account of the sharp root of the threads, which makes a bolt liable to develop cracks at this point. BOLT TABLE. U. S. S. THREADS Diam. Threads to inch Diam. at bottom of thread Area of bolt Area at bottom of thread tin. 20 .1850 .0491 .0269 fs in. 18 .2403 .0767 .0454 1 in. 16 .2938 .1104 .0678 T 7 T in. 14 .3447 .1503 .0933 i in. 13 .4001 .1963 .1257 -, 9 ,T in. 12 .4542 .2485 .1621 fin. 11 .5069 .3068 .2018 J in. 10 .6201 .4418 .3020 I in- 9 .7307 .6013 .4193 1 in. 8 .8376 .7854 .5510 IJin. 7 .9394 .9940 .6931 Hin. 7 1 . 0644 1 . 2272 .8899 If in. 6 1.1585 1.4849 1.0541 liin. 6 1.2835 1.7671 1.2938 If in. 5i 1.3888 2.0739 1,5149 If in. 5 1.4902 2.4053 1.7441 2 in. 4i 1.7113 3.1416 2.3001 2\ in. 4i 1.9613 3.9761 3.0213 2iin. 4 2.1752 4.9087 3.7163 2| in. 4 2.4252 5.9396 4.6196 3 in. 34 2.6288 7.0686 5.4277 STRENGTH OF MATERIALS 177 130. Strength of Hemp Ropes. It is quite common in calcu- lating the strength of ropes and cables to assume that the section of the rope is a solid circle. Of course, the strands of the rope do not completely fill the circle but, if we find by test the allow- able safe strength per square inch on this basis, it will be perfectly safe to make calculations for other sizes of ropes on the same basis. The safe working stress based on the full area of the circle is 1420 Ib. per square inch. The Nominal Area (as the area of the full circle by which the rope is designated is called) is A = .7854X.D 2 . The safe stress is 1420 Ib. per square inch and, consequently, the weight that can be supported by a rope of diameter D is W = SxA = 1420 X. 7854 X# 2 Here we have two constant numbers (1420 and .7854) that would be used every time we were to calculate the safe strength of a rope. If this were to be done often we would not want to multiply these together every time, so we can combine them now, once and for all. 1420 X. 7854 = 1120, approximately Hence * TF = 1120XZ) 2 Example : Find the safe load on a hemp rope of in. diameter. = 1120X4 = 280 Ib., Answer. 131. Wire Ropes and Cables. For wire ropes made of crucible steel, a safe working load of 15,000 Ib. per square inch of nominal area is allowable. For cables of Swedish iron but half this value should be used. 132. Strength of Chains. It has been demonstrated by re- peated tests that a welded joint cannot be safely loaded as heavily as a solid piece of material. Of course, there are often welds that are practically as strong as the stock, but it is not safe to depend on them. For this reason, the safe working load per 178 SHOP ARITHMETIC square inch for chain links is often given as 9000 lb., which is just f of 12,000 lb. If D = the diameter of the rod of which the links are made A=2X.7854X> 2 W = S t XA F = 9000X2X.7854X> 2 Combining the constant numbers, this can be simplified into W = 14,000 X> 2 This is used in the same way as the formula for a rope. 133. Columns. The previous examples were cases of tension. The size of a rod or timber subjected to compression is computed in the same way unless it is long in comparison with its thickness. When a bar under compression has a length greater than ten times its least thickness, it is called a Column and must be con- sidered by the use of complicated formulas which take account of its length. It can be seen by taking a yardstick, or similar piece, that it is much easier to break than a piece of shorter length but otherwise of the same dimensions. A long piece, when compressed, will buckle in the center and break under a light thrust or compression. An example of this can be found in the piston rod on a steam engine, where, on account of the length of the rod, it is necessary to use much lower stresses than those given in the tables. The compressive stress allowed in piston rods varies with the judgment of different designers but is generally about 5000 lb. per square inch, using a pressure on the piston of 125 lb. per square inch. Example : Find the size of rod for a 30 in. by 52 in. Corliss engine with 125 lb . steam pressure. 30 in. is the diameter of the cylinder and 52 in. is the stroke, which is not considered in the problem except in that it has reduced the allowable stress in the rod. .7854 X30 2 = 706.86 sq. in., area of piston. 706.86X125 = 88357.5 lb., total pressure on piston. Using 5000 lb. per square inch, allowable stress in the rod. 88358-7-5000 = 17.67 sq. in. sectional area of rod, From the table of areas of circles, it is seen that this is the area of a circle nearly 4f in. in diameter, so we would use a 4 in. rod. PROBLEMS Note. In all examples involving screw threads, to get areas at root of thread, use the table given in this chapter. Give sizes of bolts always as diameters. STRENGTH OF MATERIALS 179 219. If the generator frame shown in Fig. 84 weighs 3000 lb., what size steel eyebolt should be used for lifting it, allowing a stress of 10,000 lb. at the root of the thread? 220. What would be the safe load for a J in. chain? 221. What size hemp rope would be necessary to lift a load of 4000 lb.? Fio. 84. 222. What force would be necessary to shear off a bar of machinery steel 2 in. in diameter? 223. A certain manufacturer of jack screws states that a 2J in. screw is capable of raising 28 tons. If the diameter of the screw at the base of the threads is 1.82 in., what is the stress per square inch at the bottom of the threads when carrying 28 tons? 224. A soft steel test bar having a diameter of .8 in. is pulled in two by a load of 31,500 lb. What was the breaking tensile stress per square inch? Fio. 85. 225. The cylinder head of a small steam engine (Fig. 85) having a cylinder diameter of 7 in. is held on by 6 studs of $ in. diameter. When there is a steam pressure of 125 lb. per square inch in the cylinder, what will be the pull on each stud? And what will be the stress per square inch in each stud, due to the steam pressure? 226. With a cylinder diameter of 10 in. and an air pressure of 100 lb. per square inch, find the greatest weight that can be lifted by the air hoist, 180 SHOP ARITHMETIC shown in Fig. 86. Also find the size of piston rod necessary, assuming that it is screwed into the piston. Notice that this rod is subject only to tension and, therefore, a greater stress is allowable than in steam engine piston rods. 227. Work out a formula for the strength of crucible steel cables on the same plan as that given for hemp rope. 228. What is the greatest load that should be lifted with a pair of tackle blocks having 3 pulley in the movable block and 2 in the fixed block, and having a f in. rope. m m FIG. 86. INDEX Addition of decimals, 35 of fractions, 1 1 Air compressors, 149 steam and, 159 Allowances for shrink fits, 171 Analyzing practical problems in fractions, 22 Area of a circle, 77 Areas of circles, table of, 101 Arrangement of pulleys, 142 Belting, horse power of, 139 rules for, 141 Belt joints, 143 Belts, grain and flesh sides, 143 speeds of, 141 tension per inch of width, 141 thickness of, 141 width of, 140 Blocks, types of tackle, 123 Bolts, strength of, 175 Bolt table, U. S. S. threads, 170 Cables, strength of wire rope and, 177 Cancellation, 19 Casting, weight of, 83 Centigrade thermometers, 164 Chains, strength of, 177 Circle, area of, 77 circumference of, 51 diameter of, 51 radius of, 51 Circumference of a circle, 51 Circumferences of circles, table of, 101 Circumferential speeds, 54 Classes of levers, 117 Coefficient of linear expansion, 169 Columns, 178 Common denominator, 9 fractions reduced to decimals, 39 Complex decimals, 40 Compound fractions, 21 gear and pulley trains, 68 Contraction, expansion and, 168 19 181 182 INDEX Cube, the, 80 root, 75, 92 roots of decimals, 98 of numbers greater than 1000, 97 table, 103 Cubes, 75 and higher powers, 75 of numbers, table of, 103 Cubical measure, units of, 80 Cutting speeds, 57 Decimal equivalents, table of, 42 fractions, 33 Decimals, addition of, 35 complex, 40 cube roots of, 98 division of, 37 multiplication of, 36 short cuts, 37 subtraction of, 35 Denominator, common, 9 least common, 10 of a fraction, 2 Diameter of a circle, 51 from area, calculation of, 91 Differential pulleys, 126 pulley, mechanical advantage of, 127 Dimensions of circles, 91 rectangles, 91 squares, 91 Direct and inverse proportions, 65 Division of decimals, 37 fractions, 21 Efficiencies, 134 Efficiency of engines, mechanical, 151 of hydraulic jack, 158 of jack screw, 135 Emery wheels, 55 Expansion and contraction, 168 Extracting square root, 86 cube root, 96 Fahrenheit thermometers, 164 Fluids, 153 transmission of pressure through, 154 Foot pound, 137 Formulas, 52 INDEX 183 Fractions, addition of, 11 common fractions, reducing to decimals, 39 compound, 21 decimal, 33 definition of, 2 denominators of, 2 division of, 21 improper, 3 multiplication of, 17 numerators of, 2 proper, 3 reduction of, 3 subtraction of, 12 whole number times a fraction, 17 writing and reading of, 2 Frictional horse power, 151 Gas engines, 147 horse power of, 147 Gear ratios, 64 Gears, relation of sizes and speeds of, 64 Gear trains, 66 compound, 68 Grindstones and emery wheels, 55 Heat, nature of, 163 Horse power, 138 brake, 149 frictional, 151 of belting, 139 of gas engines, 147 of steam engines, 145 Hydraulic heads, 158 jack, 155 mechanical advantage of, 156 efficiency of, 158 Hypotenuse, 89 Improper fractions, 3 reduction of, 5 Inclined planes, mechanical advantage of, 131 theory of', 130 use of, 130 Interpolation, 96 Inverse proportion, 65 Iron and steel, colors at different temperatures, 167 Jack screws, 133 efficiency of, 135 mechanical advantage of, 134 184 INDEX Law of right triangles, 89 Least common denominator, 10 Levers, 115 classes of, 117 compound, 118 mechanical advantage of, 119 Linear expansion, coefficient of, 169 Machines, types of, 115 Mean effective pressure, 146 Measure, cubical, 80 square, 76 Measures of length, 7 of time, 7 of volume, 80 Mechanical advantage of differential pulley, 127 hydraulic jack, 156 inclined plane, 131 jack screw, 134 lever, 119 tackle blocks, 125 wedge, 132 efficiency of an engine, 151 Metals, weights of, 82 Micrometer, the, 40 Mill, the, 28 Mixed numbers, 3 multiplication of, 18 reduction of, 6 Money, U. S., 24 addition, 25 division, 26 multiplication, 26 reducing cents to dollars, 28 reducing dollars to cents, 27 subtraction, 26 table of, 28 the "mill," 28 Multiplication of decimals, 36 fractions, 17 mixed numbers, 18 whole numbers and fractions, 17 Nature of heat, 163 Numbers, mixed, 3 multiplication of, 18 Numerator of a fraction, 2 Percentage, 44 INDEX 185 Percentage, classes of problems under, 48 uses of, 46 Peripheral speed, 54 Periphery, 54 Planes, inclined, 130 Plates, short rule for weights of, 83 Power, 138 Powers, cubes and higher, 75 Pressure through fluids, transmission of, 154 Proper fractions, 3 Proportion, 59 direct and inverse, 65 Pulleys and belts, 58 Pulleys, arrangement of, 142 diameters of, 63 distance between centers of, 142 speeds of, 63 Pulley trains, compound, 68 Pyrometers, 166 Radius of a circle, 51 Ratio and proportion, 59 Rectangle, the, 79 Rectangles, dimensions of, 91 Reduction of fractions, 3 of improper fractions, 5 of mixed numbers, 6 Right triangles, 89 Rim speed, 54 Roots of numbers, by table, square, 99 Roots of numbers, cube, 75, 92 square, 75 table of, 103 Ropes, strengths of, 177 Rules for area of a circle, 78 belting, 141 gears, 64 pulleys, 63 square root, 88 weights of plates, 83 Safe working stresses, 174 Screw cutting, 72 Shrink fits, allowances for, 171 Specific gravity, 153 Speeds of pulleys, 63 circumferential, 54 cutting, 57 peripheral, 54 186 INDEX Speeds, rim, 54 surface, 54 Square measure, 76 table of, 77 root, 75 by table, 99 extracting, 86 meaning of, 85 rules for, 88 roots of numbers, table of, 103 Squares of numbers, table of, 103 Steam and air, 159 engines, 145 horse power of, 145 Strengths of bolts, 175 of cables and wire ropes, 177 of chains, 177 of hemp ropes, 177 of wire ropes and cables, 177 Stresses, definition of kinds of, 173 safe working, 174 Subtraction of decimals, 35 of fractions, 12 Surface speed, 54 Tables, areas of circles, 101 bolt table^-U. S. standard thread, 176 circumferences and areas of circles, 10 1 coefficients of expansion, 170 cube roots of numbers, 103 cubes of numbers, 103 cutting speeds, 58 decimal equivalents, 42 explanation of, 95 measures of length, 7 of time, 7 miscellaneous units, 7 square measure, 77 roots of numbers, 103 squares of numbers, 103 U. S. money, 28 weights of castings from patterns, 84 of materials, 83 Tackle blocks, mechanical advantage of, 125 types of, 123 Temperatures, 164 of iron and steel by color, 167 Thermometers, Centigrade, 164 Fahrenheit, 164 INDEX 187 Thermometers, relation of Centigrade to Fahrenheit, 165 for temperatures above 500, 166 Thermostats, 169 Threads, cutting of, 72 Triangles, right, 89 Types of machines, 115 of tackle blocks, 123 Ultimate strengths, 174 Unit of work, 137 U. S. money, 24 Volume, measures of, 80 Volumes of straight bars, 80 Wage calculations, 29 Wedge, 132 mechanical advantage of, 132 Weights of castings from patterns, 83 of materials, 83 of metals, 82 Wheel and axle, 120 Widths of belts, 140 Work, unit of, 137 2883