THE WILEY TECHNICAL SERIES FOR VOCATIONAL AND INDUSTRIAL SCHOOLS EDITED BY JOSEPH M. JAMESON GIRARD COLLEGE PREPAEATOEY MATHEMATICS FOR USE IN TECHNICAL SCHOOLS THE WILEY TECHNICAL SERIES EDITED BY JOSEPH M. JAMESON MATHEMATICS TEXTS Mathematics for Technical and Vocational Schools. By Samuel Slade, B.S., C.E., and Louis Margolis, A.B., C.E. 491 pages. 5i by 8. 353 figures. Cloth. Mathematics for Machinists. By H. VV. BuRNHAM, M.A. 229 pages. 5 by 7. 175 figures. Cloth. Arithmetic for Carpenters and Builders. By R. BuRDETTE Dale, M.E. 231 pages. 5 by 7. 109 figures. Cloth. Practical Shop Mechanics and Mathematics. By James F. Johnson. 130 pages. 5 by 7. 81 figures. Cloth. CASS TECHNICAL HIGH SCHOOL SERIES Mathematics for Shop and Drawing Students. By H. M. Keai, and C J. Leo.nard. 213 pages. 4Jby7. 188 figures. Cloth. Mathematics for Electricel Students. By II. M. Keai, and C. J. Leonard. 230 pages. 4 J by 7. lti.j figures. Cloth. Preparatory Mathematics for Use in Technical Schools. By Harold B. Ray and .Arnold V. Doub. 68 pages. 4 J by 7. 70 figures. Cloth. Preparatory Mathematics for the Building Trades. By Harold B. Ray, .Arnold V. Doub, and O. Frank Carpenter. 05 pages. 41 by 7. 58 figures. Cloth. Technical Mathematics. Vol. I. By II. M. Keal, N. S. Phelps and C. J. Leonard. 231 pages. 45 by 7. 145 figures. Cloth. Technical Mathematics. Vol. II. 271 pages. 4i by 7. 306 figures. Cloth. Technical Mathematics. Vol. III. 138 pages. 4 5 by 7. 136 figures. Cloth. Tables for Technical Mathematics. By H. M. Keal, N. S. Phelps and C. J. Leonard. 85 pages. 4 J by 7. Cloth. PREPARATORY MATHEMATICS FOR USE IN TECHNICAL SCHOOLS BY HAROLD B. RAY AND ARNOLD V. DOUB INSTRUCTORS IN MATHEMATICS, CASS TECHNICAL HIGH SCHOOL DETROIT NEW YORK JOHN WILEY & SONS, Inc. London: CHAPMAN & H.ALL, Limited Copyright, 1921 BY HAROLD B. RAY and ARNOLD V. DOUB All Rights Reserved This book or any pari thereof must not be reproduced in any form without the written permission of the pxiblisher. Printed in U. S. A. 9/30 PRESS OF BRAUNWORTH & CO,, INC. BOOK MANUFACTUREHa BROOKLYN, NEW YORK Kq PREFACE ^ A LAEGE number of continuation and evening school ^ students, who wish to take up technical courses, are deficient ^ in the knowledge of certain fundamentals of elementary 3 mathematics. Most important of these are the rules and ^ methods used in solving problems involving common frac- ^ tions, decimals, and square root. A short, yet thorough j^ review of these topics is necessary before such students can ^ do satisfactory work, either in the shop and drafting room, ^ or in the accompanying more advanced mathematics, t*- Applications of these basic operations in solving prob- lems of mensuration, threads, gears, the micrometer, and ^— percentage, suggest and anticipate the problems of the ^ trade courses. Lu The material and methods of this brief text have been "^ selected after careful study, in the class room, of the needs of several thousand students taking this preparatory course in Cass Technical High School. Each discussion and exercise has passed the test of "selling itself" to the student doing individual work. H. B. R. A. V. D. Detroit, Mich., 1921. ^p:if^29 TABLE OF CONTENTS CHAPTER I PAGE Fractions '- Definition. Reduction of Fractions. Equivalent Fractions. Addition. Least Common Denominator. Subtraction. Mul- tiplication. Areas. Division. Applied Problems. Screw Threads. Gears. 30 CHAPTER II Decimals Definition. Reading. Addition. Subtraction. Multipli- cation. Division. Decimal Equivalents. Percentage. Efficiency. The Micrometer. Taper. Circles. Cutting Speed. Volume of Solids. CHAPTER III Powers and Roots ^° Powers. Roots. Square Root. Right Triangles. Review Problems. Appendix "•'^ Index 67 MULTIPLICATION TABLE 1 2 3 4 5 6 7 8 9 10 11 1 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 2 4 6 8 10 12 14 16 18 20 22 24 3 6 9 12 15 18 21 24 27 30 33 36 4 8 12 16 20 24 28 32 36 40 44 48 5 10 15 20 25 30 35 40 45 50 55 60 6 12 18 24 30 36 42 48 54 60 66 72 7 14 21 28 35 42 49 56 63 70 77 84 8 16 24 32 40 48 56 64 72 80 88 96 9 18 27 36 45 54 63 72 81 90 99 108 10 20 30 40 50 60 70 80 90 100 110 120 11 22 33 44 55 66 77 88 99 110 121 132 12 24 36 48 60 72 i 84 96 108 120 132 144 PREPARATORY MATHEMATICS CHAPTER I FRACTIONS 1. Definitions. — The line from A to B, or simply the line AB, Fig. 1, is four inches in length. It is divided at C into two equal parts, AC and CB, each two inches in length. Each of these parts is one-half of AB, or two inches < 4 „ r c Fig. 1. is one-half of four inches. AC is divided at D into two equal parts, AD and DC, and each part is one inch in length. CB is divided similarly at E. There are thus four equal parts or divisions of AB, each one inch in length, and each part is one fourth of AB. That is, AZ) = i of AB; DC = l of AB; CE^loiAB; EB = loiAB. 2 FRACTIONS Then, AD-\-DC = l of AB-\-l of AB, or, AC = ioiAB. Also, AE = i o( AB, and Ai5 = t of AB. Each of the numbers, |, j, f, f, and t, is a fraction. The number above the line, as 1, 2, 3, is the numerator of the fraction. The number below the line, 4, is the denominator of the fraction. The numerator and denomi- nator are the terms of the fraction. 2. Reduction of Fractions. — A fraction is reduced to its lowest terms when the numerator and denominator taken together can be divided by no number except 1. In Section 1 it is stated that AC = \ of AB; also that AC^ioi AB. Thenf of A5 = iof ^5, orf = i If both terms of the fraction f are divided by 2, the resulting fraction is \. 2-^2_l 4--2~2' A factor of a number is a divisor of the number. One is the only common factor of 1 and 2, therefore the fraction f , when reduced to ^, is expressed in its lowest terms. Similarly : 6^32 4^41 15-^5 3. 24h-2 ^12, 12-^3 ^4 9^'"3' 8T4~2' 25-^5~5' 30^2~15' 15^3~5" Sometimes the common divisor of the two terms is written before the fraction, thus 3 | t2 =f . REDUCTION OF FRACTIONS EXERCISE 1 Reduce the following fractions to their lowest terms: 5 6 1. H 2. -A 3 12^ • T8- T 4. 20.^ T3"- ? 4 /'^ 6 4- :?^ 7 2.8 /^ 9 5 6 8. ^^A 3 3"- 10 48 11. The line AB, Fig. 2, is divided into sixteen equal parts. How many eighths in four of these parts? 16^ ^8 iii. I M I I I I I I Fig. 2. How many eighths in twelve parts? How many fourths in four parts? In twelve parts? 12. Fig. 3. The circle of Fig. 3 is divided into twelve equal parts. How many fourths of the circle in nine of these parts? How many thirds in eight parts? How many sixths in ten parts? 3. Equivalent Fractions. — If both terms of the fraction \ are multiplied by 2, the result is f . 1X2^2 2X2 4* Similarly, any fraction can be changed to an equivalent fraction by multiplying both numerator and denominator by the same number. 2X3 _6 4><7_28. 3X8 _24 3X3~9' 5X7~35' 11X8~88' Examples.- FRACTIONS It is often necessary to change a fraction into an equiva- lent fraction having a specific number for the denominator. This is done as in the following example : Example. — Change f to thirty-seconds; or f=3^. Divide 32 by 4, the result is 8. Multiply 3 by 8, the result is 24, the numerator of the new fraction. Therefore f =tI- EXERCISE 2 1. 4-32- 3. 2 _ ? K 4 _ ? 5 — 40- "• 7— "2 8' 7. 5 _ ? T2— T8"- 2. 1 _ ? 4. 7 ■' e 1 1 _ ■? S — 6T- O. T6—T2- 8. 3 _ ? 8 — 3^ I I 1 1 I 1 1 I I 1 ?_. 4 — 16, 10. Fig. 4. The line AB, Fig. 4, is di- vided into sixteen equal parts. How many sixteenths in j of the line? ITS, 5 __L-. S — 1 6; 7 _ ? . 1 _ ? 8 — TB"j S — Tff- The circle, Fig. 5, is divided into 12 equal parts^ How many twelfths in | of |fce circle? i? In I? In f ? In Fig. 5. 4. Addition of Fractions. — In addition of fractions, the fractions to be added must first be changed to fractions having the sa7ne number for a denominator. This number is called the Common Denominator. After the common denominator is selected, the next step is similar to the work done in Exercise 2. ADDITION 5 Example 1. — Add ^ and ^. The common denominator must be a number divisible by the two denominators. 6 and 12 are two numbers divisible by both 3 and 2. It is advisable to use the smallest number possible; this number is called the Least Common Denominator. (Abbreviated L. C. D.) After selecting 6 for the L. C. D., change ^ and ^ to sixths. • 1 — 2. 3 ~ 6 > 1 _ 3.. 2 — 6 , f+t=-|- Ans. The last step is performed by adding the numerators and writing the sum over the common denominator. When the denominators are small numbers, the L. C. D. is easily obtained bj^ reference to the multiplication table. If this is not satisfactory, the product of the denominators may be used as the common denominator, although this method will not always give the least common denominator. Instructions will be given later for finding the L. C. D. when the denominators' are large numbers. Example 2.— i + f +f = ? In the table of 6's (see table facing page 1), 72 is the smallest number in the table that is exactly divisible by both 8 and 9. Then it is the least common denom- inator for the three fractions. Arrange the solution as follows : 3 _ 27 S ~ 72 2 _ JJ8. ■ff — 72 The sum of the numerators (12+27+16) is 55; the sum of the fractions is rl. Ans. FRACTIONS Example 3.— 1+^-|-^ = ? 2 _ 28 3 =¥2 1 ^ |i. The product of the denominators (3X2X7) iL— 2j4 is 42. The sum of the numerators is 73; the 7 — 42 ' sum of the fractions is if • 7|=lfi Ans. A fraction whose numerator is larger than the denomi- nator is an Improper Fraction. Such a fraction should be reduced by dividing the numerator by the denominator. 734-42=1, with a remainder of 31. The result is written, Ifi. This is a mixed number. A Mixed Number is a number made up of a whole number and a fraction. 5. Addition of Mixed Numbers. — If mixed numbers, or mixed numbers and fractions, are to be added, add the whole numbers and fractions separately, and combine the results. Example.— 81+ t+12f = ? 82 _ O 2.0. 3 — 3 1= U The L. C. D. is 30. 12f=mf ^3lf 6 3 _ 9_3_ _ 9 1_ . 3~0 — •^ 3 — ^ITU • 20+2Tio=22TV. Ans. EXERCISE 3 1. i+| = ? 3. RH? 5. i+f = ? 7. 2f+8f = ? 2. i + | = ? 4. i^h = '^ 6. U+24 = ? 8. 4|+Gf = ? ADDITION 9. Find the over-all dimension of Fig. 6. ,// _ r II < p< 2\ "^ '10 ^ 1 i is I 7 1 ^1 I Fig. 6 10. Find the over-all dimension of Fig. 7. Fig. 7. 11. Measure the length of the lines indicated in Fig. 8, add, and check by measuring the over-all length. Note. — In problems 11 and 12 the student should use a machinist's scale or a rule with yj inch divisions. 8 FRACTIONS 12. Measure the lines of Fig. 9 and check as in Prob. 11. I ■ ■. I ■ r I Fig. 8. 6. To Find the Least Common Denominator. — In some cases, the least common denominator can not be found easily by the methods explained in Section 4. Fia. 9. LEAST COMMON DENOMINATOR 9 Example.— Find the L. C. D. of t^, -h, and i^. The following method is generally used. 2 )12 21 14 7 ) 6 21 7 3) 6 3 1 2 1 1 2X7X3X2X1X1 = 84,L.C.D. Rule. — Divide the denominators by any number that will divide exactly two or more of them, bringing down the denomi- nators that cannot be exactly divided by the divisor. Repeat the operation until no two numbers left are exactly divisible by the same number. The product of the remaining numbers multi- plied by the divisors will be the Least Common Denominator. EXERCISE 4 Add: 1. 6i +71 + 32^^. Ans. 17,^. 2. 4i +56^4+ 6f. Ans. 16M. 3. 7f +2f + 9^. Ans. 19iff. 4. -V-+ ¥+ tV. Ans. 12i. 5. n +4TV+18f. Ans. 33 iV- 6. 4f +7i + 9i. Ans. 21f . 7. 8| +32^4+ 8i^. Ans. 20^. 8. 2TV+lf + BiV. Ans. 7^. 9. How many rods of fencing will it take to inclose a lot the sides of which measure 10| rods, 17f rods, 12t^ rods, and 8f rods? Ans. 493^^ rods. 10. A machinist wishes to cut three lengths from a stock. The first length must be If", the second 2^6", and the third 96V"- How long must the stock be to furnish the required lengths? Ans. 12|f". 10 FRACTIONS 11. Find the missing dimensions in Figs. 10 and 11. Mi Fig. 10. if^lfe'^ 1 1 ] 1 1 1 i 1 i „ i rf ? ^t* -? >1 Fig. 11. 12. Find the total length of a generator shaft if 12f inches are allowed for the armature, 4t inches for the bear- ings, SiTT inches for the commutator, and 1^ inches are clear. Ans. 211^ inches. 7. Subtraction of Fractions. — In the subtraction of frac- tions, as in addition, the given fractions must be changed to equivalent fractions having a common denominator, then the difference of the numerators can be found. Example 1. Example 2. l_i — ? 12^— ^i — ? The L. C. D. is 6. The L. C. D. is 24. 2 _ 1 Oc — Oo 2T Ans. '^T- Ans. -k SUBTRACTION EXERCISE 5 Subtract the following: 1. 2f 3. 31t^ 5. 91 Si 7. 1631 97f 2. 131 4. ^ *• 3 3 8 6. 3 If 4i 8. 24f 19f 11 9. If 2f inch screws are used to fasten porcelain insulators on a wall, and the insulators are lye inches thick, how far will the screws penetrate the wall? 10. Find the missing dimensions in Figs, 12 and 13. Fig. 12. Fig. 13. 11. Find the lengths indicated in Fig. 14, (a), (6), (c), and (d) by subtraction, and check each by measuring. Examples.— 5f- If = ? The L. C. D. is 8. 5t = 5|=4|+|=4i^ Check: 13 = lf=.lf =lf Does, lf+3| 31. Ans. ■5f? 12 FRACTIONS Example 4. — Subtract 9f from 17f . 17| = 17H=16ii+-H=16H Qi. = qi2 _ 012. 7H Ans. Fig. 14. In Example 3, f cannot be subtracted from f. To increase f so that the subtraction is possible borrow one from the whole number 5, leaving the whole number 4, change 1 to eighths (1 = f ), and add to |. The upper number becomes 4-V-, from which If can be subtracted. In Example 4, the same plan is used, except that after 1 is borrowed from 17, it is changed to fifteenths. In each problem change the 1 borrowed to a fraction having the L. C. D. for numerator and denominator. Observe that the correctness of the answer can be checked by going over the solution carefully, or by adding the answer to the number subtracted, as suggested in Example 3. The student should form the habit of checking each solution. SUBTRACTION 13 ^ Subtract: 1. Ill 21 EXERCISE 6 2. 4i 3. 11t^ 41 5. 15i-6T^ = ? 6. 14 -511= ? 7. 8i+7i-3i = ? 8. 9|-2| + li = ? 9. Find the missing dimension, Fig. 15. 4. 83i Ans. 8i 7 ^ . 23 ->J 1 "' 1 -lef Fig. 15. 10. Find the missing dimension, Fig. 16. Fig. 16. 14 FRACTIONS 11. -1^ l'-^->\<- Find the missing di- mension, Fig. 17. -?.i- ->H- -7f- Fig. 17. 12. Find the missing dimension, Fig. 18. Fig. 18. 13. Find the lengths indicated in Fig. 19 (a), (b) (c) and (d), and check by measuring. '\-^ 1 H "h — ?-H 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 II I I M (a) MM M ITT ir) Fig. 19. MULTIPLICATION 15 8. Multiplication of Common Fractions. — The method of multiplying one fraction by another or by a whole or mixed number is shown in the following examples: Example 1.— Multiply f by f . f X f = ii- Example 2.— Multiply f by 9. Example 3.— Multiply 2| by f. ^sX^-^X^ 14 ^T¥- 2 Example 4.— Multiply ^ by 4. 3fX4 = -V-XT = ^=15f. Example 5.— Multiply 4f by 1^. 12 7 ';4 P,o ft4 45X134- ^X^^ 17 ^• 17 Example 6. — Square 3^. oiv'^i — i:^Vi:^ — 169 — 10' Rule. — To multiply fractions, cancel the common factors; the required result will he the product of the remaining numer- ators over the product of the remaining denominators. Observe that in Example 2, the whole number 9 is expressed as a fraction with 1 as the denominator. In problem 3 the mixed number, 2|, is changed to an improper fraction, %^-. 16 FRACTIONS In problem 6, to square any number, multiply it by itself. EXERCISE 7 1. Multiply i by f . 2. Multiply t by StV- 3. Multiply 3| by 6i 4. Multiply f X 2f X Y X |. 5. Find the product of 9 and 144 1. 6. ^X7|X6ixH = ? 7. Find the product of A and 7. . 8. l|x4iXi = ? 9. Mensuration. — The area of a rectangle equals the product of the base and the altitude. The perimeter is the distance around a figure. EXERCISE 8 1. Find the area and perimeter of the rectangle of Fig. 20. -6^ — base- Fig. 20. AREAS 17 2. -Hi"- Find the area of the rectangle of Fig. 21. Fig. 21. 3. Find the area and perimeter of Fig. 22, Ans. 28| sq. in. 23^ in. 51 -=r— 2— -^ J i- Fig. 22. The area of a square is equal to the square of one side. EXERCISE 9 Find the area of the square, Fig. 23. Ans. IQeT sq. in. Fig. 23. 18 FRACTIONS 2. ~^ Find the area and perimeter of Fig. 24. Ans. 7i^ sq. in.; 11 in. O .^ Find the area of the shaded part of the square, Fig. 25. Ans. 15tV sq. in. Fig. 25. 4. Measure to the nearest ye inch the sides of the two rectangles in Fig. 26 and find the area of the shaded portion. Fig. 26. Measure to the nearest YE inch the sides of the two squares in Fig. 27 and find the area of the shaded portion. ^^ Fig. 27. AREAS 19 The area of a triangle equals one-half the product of the base and altitude. EXERCISE 10 1. Find the area of ~ the triangle shown in Fig. 28. Ans. 9f sq. in. Fig. 28. 2. Find the area of Fig. 29. Ans. 12if sq. in. Fig. 29. 3. Find the area of the triangle in Fig. 30. Ans. 16irr sq. in. Fig. 30. 20 FRACTIONS 10. Division of Fractions and Mixed Numbers. — Example 1. — Divide 4^ by 4f . ^2 ^^ 12 • 8 U^f> 15- Rule. — To divide fractions, invert the divisor and follow exactly the rule for multiplication of fractions. 1. 2. 3. 4. 5. 11. 12. 13. weight. EXERCISE 11 l-i-? 6. 32-^3f = ? K4 = ? 7. 9f-^8| = ? 8i^-l0 = ? 8. 6f^lf=? 20^21 = ? 9. 16f^2i = ? f^6 = ? 10. 86f-^4|=? How many eighths in 3? (3-^| = ?) Divide 101 into 56f . Six castings 5 weigh 118 1 pounds. Fint Find the average 14. Fig. 31. Find the base of the rectangle. Fig. 31. Ans. 27i". Note. — We have learned that the area of a rectangle = base X alti- tude. Therefore, base = area divided by altitude, and altitude = area divided by base. 15. r 16. Fig. 32. 21 Find the base of the rectangle, Fig. 32. oi _ Find the altitude of Fig. 33. Ans. 9i". Fig. 33. Fig. 34. Find the altitude of Fig. 34. Note— The area of a triangle = ^ base X altitude. Therefore, the base of a triangle = 2 Xarea divided by the altitude, and the altitude = 2Xarea divided by the base. 18. Find the base of the triangle, Fig. 35. Ans. 12f ". Fig. 35. 22 FRACTIONS 19. Find the base of the triangle of Fig. 36, Ans. IV'. Fig. 36. 20. Find the altitude of the triangle of Fig. 37. Ans. 6' Fig. 37. 21. Find the altitude of the triangle of Fig. 38. Ans. 51''. Fig. 38. APPLIED PROBLEMS 23 22. Find the altitude of the triangle of Fig. 39. Ans. 12". Fig. 39. EXERCISE 12 Applied Problems. Common Fractions. 1. Find the diameter of the smaller circle, Fig. 40. Fig. 40. 24 FRACTIONS 3. Find the diameter of each circle, Fig. 41. Find the thickness cf the ring, Fig. 42. Outer diameter 2f ". Inner diameter, 2tg". Fig. 42. 4. Find the distance between centers, Fig. 43. Fig. 43. THREADS 25 Diameter of large circle is 5|". Diameter of small circle is 3]". 5. A mechanic earns $6.00 in 8^ hours. What does he earn per hour? 6. Find the length of the rod from which 15 pieces, each 4f " in length, can be cut? 7. Which is the larger, f or xV? How much? 8. In a certain drawing |" represents 1'. What length of line will represent 14 feet? Ans. If". 9. If the scale is -^" = V Q", what distance in the drawing represents 200 feet? Ans. 12|". 10. If a planer cuts a strip -^'^ wide, how many cuts are necessary to plane a piece 3^" wide? Ans. 50. Note. — The last cut may not be as wide, but it is the same length as the other cuts. 11. If a sheet of metal is y&" thick, how many sheets are required to make a pile one foot thick? Ans. 28. 12. An armature core is built up of sheets of steel -^ inch thick. How many sheets are required to build an armature 9| inches thick? Ans. 52. 11. Lead and Pitch of Screw Threads. — The distance from any point on a screw thread to the corresponding point on the next thread is the Pitch of the thread (see Fig. 44). Rule. To find the pitch v)hen the number of threads per inch is given, divide 1 inch by the number of threads per inch. 26 FRACTIONS In Fig 44, there are six threads per inch; l"-^6 = i", the pitch. ABC Fig. 44.— Single V Thread. If the thread of Fig. 44 is traced around the screw from A, it will reappear at B, and on the next turn at C. The so-called threads of this screw are parts of one thread, or it is single-threaded. If the screw is turned in a nut, it will advance through the nut | of an inch in each revolu- tion. This advance is called the Lead. On a single-threaded screw the pitch and lead are equal. In Fig. 45, if the thread starting at A is traced around A B C D Fig. 45.— Double V Thread. THREADS 27 the screw, it reappears at C; starting at B, it reappears at D. Evidently there are two threads on the screw, or it is double-threaded. The lead on a double-threaded screiv is twice the pitch. Similarly a screw may be triple-threaded, four-threaded, or more. To obtain the lead of such a screw, multiply the pitch by 3, 4, etc. The feed of a lathe is the distance the cutting tool advances for each revolution of the stock. The feed of a drill is the depth of cut in each revolu- tion of the drill. EXERCISE 13. APPLIED PROBLEMS X. What is the pitch of a single-threaded screw that has 8 threads to the inch? 10 threads? 2f threads? What is the lead for each screw? 2. Find the pitch and lead of double-threaded screws having 3, 5|, and 12 threads to the inch, respectively. 3. How many threads per inch on a screw if the pitch is I of an inch? If f of an inch? {Suggestion: Divide 1 inch by the pitch.) 4. A drill cuts 1 inch in 40 revolutions. What is the feed? 5. A drill will cut through a certain steel plate 1 inch thick in 64 revolutions. What is the feed? How many revolutions are necessary to drill 5 holes? 6. How many minutes will be required to turn a stock 3 J feet long, if the feed is -gV", and 60 revolutions are made per minute? Ans. 20f. 7. A piece 56 inches in length, making 80 revolutions per minute, was turned in 35 minutes. What was the feed? Ans. sV". 28 FRACTIONS 8. Find the time required for drilling 8 holes in a steel plate W thick, feed ^", R. P. M. 120. Allow 2^ minutes per hole for adjusting the machine. Ans. 22x1 minutes. 12. Gears. — A wheel with teeth on its rim is a gear. When gears are in mesh, the number of revolutions of the driven gear is found by dividing the product of revolutions of the driving gear and the number of its teeth by the number of the teeth of the driven gear. The above rule may be expressed in this manner: Revolutions of the driven gear = Revolutions of the driving gearX Teeth of driving gear Teeth of driven gear Or when: 7^ = Number of revolutions of driving gear. r = Number of revolutions of driven gear. !r= Number of teeth of driving gear. ^ = Number of teeth of driven gear. The above statement becomes, RXT r= t , RXT _, rXt T_r> O T3 -2 •a c3 a (V 3 00 O .2 1 p n H K H H ffi 000 .000000 30 ADDITION AND SUBTRACTION 31 EXERCISE 1 Read the following decimals: 1. .0076 3. 23.9026 5. 1674.01674 2. 7.986 4. 275.275 6. 27.0467 Write the following in figures: 7. Eight hundred ninety-five ten thousandths. 8. Two hundred ninety seven hundred thousandths. 9. Four hundred thirty-one and sixty-two thousandths. 10. One hundred and twenty-seven millionths. 15. Addition of Decimals : Example.— Add 6.749, 23.0764, .00072, 100.0000702. 6.749 23.0764 .00072 100.0000702 129.8261902 Rule. — Place the numbers to be added so that the decimal points form, a straight line, and proceed as in the addition of whole numbers. 16. Subtraction of Decimals : Example.— Subtract 10.0846 from 15.00001. 15.00001 10.08460 4.91541 32 DECIMAL FRACTIONS 17. Multiplication of Decimals : Example.— Multiply 1003.21 by .064. 1003.21 .064 401284 601926 64.20544 Rule. — Multiply as in whole numbers and point off as many decimal places in the product as there are in the factors taken together. EXERCISE 2 1. Add 3.985, 4.06, 98.4763, .0356, and 100.728. 2. Add 13.8505, 909.73, .00007, and 148.561. 3. Add 978.001, 100.999, 47.0878, and 79.38. 4. Add 59.86, .13, 9.9, .008, and 749. 5. Subtract 19.496 from 97.382. 6. Subtract .987634 from 1.00175. 7. Subtract 123.4567 from 987.8321. 8. Subtract .3594 from 1. Find the product in the following: 9. 3.945X8.6. 10. 578.01X49.7. 11. .003X7.8. 12. 79.05 X. 46. 13. 98.8 X. 069. 14. 3.1416X.25. 15. 6.321X100. DIVISION 33 18. Division of Decimals. — -Placing the decimal point in the quotient is the greatest difficulty the student has in mastering division of decimal fractions. A careful study of the following examples will aid the student in overcoming the difficulty. Example 1.— Divide .0093 by 4. ■ 002325 Quotient Divisor 4). 009300 Dividend Divide as in whole numbers, annexing ciphers if necessary (two were annexed in this problem). The decimal point of the quotient is placed directly above the decimal point of the dividend; when the divisor is a whole number there are as many decimal places in the answer as in the dividend. In the discussion of common fractions it was shown that both numerator and denominator of a fraction may be multiplied or divided by the same number without changing the value of the fraction. TVinc 3. 6 _15_30_ 3000 „J.p J-ilLlb 8 ~ 1 6 — 40 — 80 — 8000J t^l^*^' Example 2.— Divide 3354 by .078. 3354 This problem may be written as a fraction, . If the numerator and denominator are multiplied by 1000, it gives, 3354 1000 3354000 .078^1000" 78 ' this fraction can be reduced by division. Ordinarily the problem is solved as follows: 34 DECIMAL FRACTIONS 430 00. Quotient Divisor .078x) 3354.000^ Dividend 312 234 234 000 If the decimal points of the dividend and the divisor are shifted three places to the right, the result is the same as was obtained by multiplying both by 1000. The new positions of the decimal points are indicated by x- Place the first figure of the quotient above the last figure of the first product; 4 in the quotient is written above 2 in 312. Example 3.— Divide X383. 594 by 48.7. 14.0368 Quotient Divisor 48. 7x) 683. 5x9400 Dividend 487 1965 1948 1794 1461 3330 2922 4080 3896 184 Remainder, DIVISION 35 When the dividend and the divisor are multipUed by 10, the divisor has no decimal places. Place x in the dividend and divisor one place to the right of the decimal point. The first figure and decimal point of the quotient are placed as suggested in Example 2. When the third subtraction is made there is a remainder, 333. Further division is possible if one or more ciphers are annexed to the dividend. The final remainder indi- cates that the quotient obtained is not exact. Note. — Pointing off to the iiearest thousandth : Since .0368 is nearer to .0370 than to .0360 in value, the quotient may be written 14.037. Division to the fourth place is necessary to determine this result. EXERCISE 3 Divide the following: 1. .085^4 7. 5771.402-7.34 2. 98.07^3 8. 5.698^154 3. .7656^8 9. .01-^8 4. 126^.07 10. 1.296-^1.8 ^. 1833^3.9 11. 3.4592^73.6 6. 4.56^.16 12. 22.002 H- .057 13. Divide to the nearest .001. 70.90742^14.6 14. Divide 28.52806 by 8.6. 15. Divide $90 by 24. 19. Reduction of Common Fractions to Decimals. — A common fraction may be reduced to a decimal by dividing the numerator by the denominator. 36 DECIMAL FRACTIONS Example 1.- — Reduce xe to a decimal, 1 . 1875 16)3.0000 1 6 1 40 1 28 120 112 80 80 Example 2. — Reduce rs to a decimal. 1.0666 Ans. .1875. 15)1.0000 Ans. .067. 90 100 90 100 90 10 FXERCISE 4 1. Reduce the following fractions to decimals: -^o? tt> A i 2. _A_ 5. 7) 3) 9) 11) 6- 2. Divide 6f by Ig- (use fractions). 3. Change the fractions in problem 2 to decimals and divide. PERCENTAGE 37 Compare the two answers. 4. Change to decimals and add: 3.1 5 I 1 I 3 I 9 15 8ll6~32r64ll6l8. Ans. 1.953125. 5. Decimal equivalents of fractions, ^ inch to 1 inch; solve for the missing values. 17 64 9 32 19 64 TS = ? A =.078125 A =.09375 6T=. 109375 i = .1250 ^=.140625 ^f=. 15625 i^=.1875 if =.203 125 7 . T2- 1 5 . ST- i=? .265625 .28125 .296875 .3125 .328125 .34375 .359375 = '? .390625 .40625 .421875 .4375 .453125 .46875 .484375 .5000 If = .515625 H = .53125 tf = . 546875 _9__? 16 — • fi=. 578125 if =.59375 If =.609375 I = .6250 il=. 640625 M=? M=. 671875 ii=.6875 6T~ f = .7500 || = . 765625 |f = ;78125 If = .796875 if =.828 125 M=. 84375 tf=. 859375 I = .8750 64 — ■ ft =.90625 lt=.9375 If =.953 125 3J._ 9 32 — • If =.984375 1 = 1 20. Percentage. — Percent means by the hundred, and is written " %." Examples.— 50 per cent of 72 = t%°oX72, =.50X72 = 36. 12%of 45 = tWX45=. 12X45 = 5.4. Observe that the symbol " % " is used instead of the denominator 100, or two decimal places. In the solution of problems, the decimal form is used most frequently. ar.^!ir^r>^ 38 DECIMAL FRACTIONS EXERCISE 5 1. Find 15% of 60; 87% of 387; 37|% of 472 ft; 33|% of $72; i% (i of 1%) of 640"; |% of 1728 cubic inches; 7^% of 1585 pounds; 125% of 48; 12^% of 48; 682% of 956. 2. Aluminum bronze is 10% aluminum and 90% copper. How many pounds of each metal in 125 pounds of the alloy? Ans. 12i lbs., 112^ lbs. 3. If a lathe costing $1350 depreciates (decreases in value), 12^% in one year, what is it worth at the end of the year? Ans. $1181.25. 4. Bonds of the Fourth Liberty Loan bear 4j% interest, payable semiannually. How much interest is due April 15, and October 15, on bonds having a par value of $650? Ans. $13.81. 21. Relation of Numbers Expressed as a Percent. — To express one number as a percent of another numl^er, divide the first by the second to the nearest .01, and write the result as percent. Examples. — 20 — .15— 15%; -V-- 5.00 = 500%; f=. 2857 = 28.57%. Note. — Where greater accuracy is desired, decimal places after the second may be retained in decimal form as in the last example. EFFICIENCY 39 EXERCISE 6 1. A gear blank in the rough weighed 9 pounds. The finished piece weighed 7 pounds. How many pounds were wasted? What percent of the blank was wasted? Ans. 22%. 2. In 250 pounds of type metal there are 200 pounds of lead and 50 pounds of antimony. What is the percentage of each metal? Ans. 80%, 20%. 3. Gold coin is 900 parts gold, 75 parts copper, and 25 parts silver. What is the percentage of each metal in the alloy? Suggestion: What is the total number of parts? Ans. 90%, 7i%, 2i%. 4. An article which cost $375 was sold for $425. The profit was what percent of the cost? Ans. 13.3%. 22. Efficiency. — The efficiency of a machine is the ratio of useful work the machine does to the energy it receives. It may be stated thus : Output Efficiency = Input * If a power plant converted all the fuel into useful power, output would equal input, and the power plant efficiency would be 100%. If a steam engine, using fuel that contains 100 horse power of energy, delivers 12 horse power to the belt or drive wheel, its thermal efficiency is T^ or 12%. 40 DECIMAL FRACTIONS EXERCISE 7 Compute the efficiency of the following: INPUT. OUTPUT. EFFICIENCY. Gasoline Motor 150 66 ? Diesel Engine 200 154 ? Electric Motor 45 37.4 ? 23. The Micrometer. — -The micrometer is used to measm'e to one thousandth of an inch. The parts are named as shown in Fig. 48. The anvil and sleeve are attached Set Screw Spindle of Screw \ Anvil / Ratchet Thimble Frame of Yoke Fig. 48. to the frame, and are not movable. The sjyindle turns in the sleeve. It is a single threaded screw with 40 threads to the inch. Since the pitch of the thread is tV, or .025 in., one turn of the screw changes the width of the opening .025 in. This is the distance the thimble moves on the sleeve; and its position at the end of each revolution is indicated by the marks on the sleeve. Every fourth mark is longer; the distance between these marks is 4X.025 in. or .100 in. THE MICROMETER 41 The thimble turns with the spindle. The beveled edge has 25 divisions. When the thimble is turned 2V of one revolution, the screw moves -it of .025 in. or .001 in. There are four steps in reading the micrometer: 1. Read the tenths of an inch from the numbered marks on the sleeve. 2. Multiply the number of divisions to the right of the last numbered mark on the sleeve by .025. 3. Read the thousandths on the thimble. 4. Add the three results. / / \ ■10 yo 1 2 . h X W Fig. 49. 4 5 6 Fig. 51. 42 DECIMAL FRACTIONS In Fig. 49, the reading .257, is obtained as follows: 2 X. 100 = .200 in. 2 X. 025 = .050 in. 7X. 001 = .007 in. .257 in. EXERCISE 8 1. What is the reading in Fig. 50? 2. What is the reading in Fig. 51? 3. What is the reading in Fig. 52? EXERCISE 9. REVIEW PROBLEMS 1. Tin weighs 459 pounds per cubic foot. "What is the weight of one cubic inch? (1728 cubic inches =1 cubic ■foot.) Ans. .2656 lb. 2. If cast iron weighs .261 pound per cubic inch, how many cubic inches in an iron casting weighing 845 pounds? Ans. 3237.5 cubic inches. How much more would the same volume of steel weigh? (Steel weighs .283 pound per cubic inch.) Ans. 71.22 pounds. 3. What is the cost of 1250 feet of wrought iron, weigh- ing 1.83 pounds per foot, and costing 95 dollars a ton? Ans. $108.65. 4. The weight of one cubic foot of water is 62.3 pounds, and the weight of the same volume of rolled brass is 524 pounds. How many times as heavy as water is rolled brass? How many times as heavy as rolled brass is water? Ans. 8.41; .1189. TAPER 43 5. Number 6, U. S. Standard Plate Iron and Steel is .203125 inch thick and weighs about 4.96 pounds per square foot. How many sheets are in a stack 6.5 inches high? What is the weight of this stack if each sheet con- tains 9 square feet? Ans. 32 sheets; 1428.48 pounds. 24. Taper — Taper is the decrease in the diameter of a piece of work in a unit of length, usually one foot. In some problems the taper per inch is given ; in others it is convenient Fig. 53. to find the taper per inch. In Fig. 53 the taper per foot is I in. (1" — I") or .500 in. Distance between centers rather than the actual length of the stock must be consid- ered in computing taper. In the shop, detailed instructions are given for adjusting machines for taper cutting. EXERCISE 10 1. The diameters of the ends of a piece one foot long are 1 inch and .400 inch, respectively. What is the taper? 2. Find the taper of the finished piece of work 6 inches in length, if the diameter of the large end is 1.5 inches and of the smaller end 1.25 inches. Ans. .5 in. per foot. 3. The taper of a certain piece of work 7 inches long is .600 inch per foot. The diameter of the large end is 1| inches. Find the diameter of the small end? Ans. .775 inch. 44 DECIMAL FRACTIONS 25. The Circle. Fig. 54. The radius of a circle is one-half of the diameter. The circumference is 3.1416 times the diameter. This rule may be stated as a formula, C = tD, or C = 2wR. The letter w, called " pi," has the value 3.1416. If C is given, D and R may be found by using the formulas; D = C -£ EXERCISE 11 1. (a) Find C when /) = 7.15 inches. (b) when /? = 4.54 inches. Ans. (a) 22.46244 inches; (h) 28.5257 inches. 2. Find D and R when C= 11.781. Ans. D = 3.75 inches; /?= 1.875 inches. 3. A fly wheel is 42 inches in diameter, and makes 250 R. P. M. At what speed (feet per minute) does it drive a belt? Ans. 2748.9 feet. 26. Cutting Speed. — The rate at which a point on the surface of a piece of stock turns past the tool in a lathe is the Cutting Speed. The distance traveled in one revolu- tion is the circumference of a circle having a diameter equal to the diameter of the piece of stock. This distance multi- plied by the number of revolutions per minute (R.P.M.) CUTTING SPEED 45 gives the Cutting Speed. The answer should be expressed in feet. Example 1. — ^Find the cutting speed when a l|-inch rod is revolving 95 times per minute. 1^X3.1416 = 4.7124, the circumference in inches. 4.7124X95 12 = 37.3, cutting speed in feet. EXERCISE 12 1. What is the cutting speed when 2|-inch stock makes 175 R.P.M.? Ans. 97. 2. At how many R.P.M. should If-inch stock be turned in a lathe to give a cutting speed of 40 feet per minute? Ans. 87. 27. Areas. — The area of a circle is found by multiplying the square of the diameter by .7854. The formula is, A = .7854^^-. Example. — Find the area of a circle whose diameter is 5 inches. A = 5X5X. 7854= 19.635, area in square inches. EXERCISE 13 1. The diameter of the piston of a gas engine is 3^ inches. Find the area. Ans. 9.621 square inches. 2. The diameter of a circle is one inch. Find the area. Ans. .7854 square inch. 3. The circumference of a circle is 14.1372 inches. What is the area? Ans. 15.9 square inches. 46 DECIMAL FRACTIONS 28. Volumes.— jT/ie volume of a rectangular solid or cylinder is equal to the product of the area of he base and the altitude. Fig. 55. Fig. Sf). Fig. 58. In Fig. 55, 8"X5" = 40 square inches, area of base. 40 square inchesXl2"=480 cubic inches, volume of solid. In Fig. 57 or 58, 3" X 3" X 3. 1416 = 28.2744 square inches, area of base. 28.2744 square inches X 12" = 339.292 cubic inches, volume of solid. VOLUME OF SOLIDS 47 EXERCISE 14 1. Find the volume of the sohd shown in Fig. 56. 2. Find the volume of a rectangular solid of steel whose edges are 6 inches, 5 inches, and 3 inches respectively. What is its weight if steel weighs 490 pounds per cubic foot? Ans. 25.52 pounds. 3. How many solids, each containing .875 cubic inch, can be taken from a bar of steel the area of the base being six square inches, and the length 12 feet? Allow 18% for waste. Ans. 809. 4. A cylindrical Portland cement pillar is 12 inches in diameter, and 15 feet high. If Portland cement weighs 90 pounds per cubic foot, what is the weight of the pillar? Ans. 1060.29 pounds. 5. Find the weight of a lead pipe 10 feet long, if it weighs 5.75 pounds per foot. Compare this result with the result obtained when the pipe is considered as the difference between the volumes of two right circular cylinders, whose lengths are 10 feet each, and whose diameters are 1.75 and 1.25 inches, respectively. (Lead weighs .4105 pound per cubic inch.) Ans. 58.03 pounds. CHAPTER III POWERS AND ROOTS 29. Powers. — If a number is used two or more times as a factor, the product is a power of the number. 4X4=16, the second power or square of 4. 4X4X4 = 64, the third power or cube of 4. 4X4X4X4 = 256, the fourth power of 4. The writing of the factors may be shortened by using the common factor or base once, and putting above and to the right, a small number, showing how many times the base is used as a factor. This number is an Exponent. 4^ means 4X4, 2 being the exponent. Similarly, 5'^ means 5X5X5; 9-i means 9X9X9X9. EXERCISE 1 1. Find the value of 6^, 15^, 8^ 5^. 2. Square 3. Compare the square of the result with the fourth power of 3. 3. Which is the larger, 6-* or IP? 4. Add the square of 12 to the square of 16. Compare this with the square of 20. 5. What is the value of (h)^ (2|)2, (f)^? 6. Square 3.5; 11.25. 7. To find the area of a square, square the length of one side. Find the area of a square 7^ inches on a side. Of one 16.2 inches on a side. Of one 3| inches on a side. 48 SQUARE ROOT 49 30. Roots. — A root of a number is one of the equal factors of the number. Since 4X4=16, 4 is a root of 16. Four is the square root of 16, because 4 squared equals 16. Likewise, since 4"^ = 64, 4 is the cube root of 64. The radical sign V is placed before a number to indicate that a root of the number is to be found. A line is drawn over the number to show how much of what follows is affected by the radical; thus V729. If there is no number in the opening of the radical sign, the square root is to be found. 31. Square Root. — It was learned in the multipli^tion table that 3X3 = 9; therefore, V9 = 3 . Similarly, V25 = 5 ; V49 = 7. EXERCISE 2 Find by reference to the multiplication table or by experi- ment, the square root of the following: 1. 100 3. 1 5. 144 7. 16 9. 225 2. 4 4. 64 6. 121 8. 400 10. 81 32. Computing Square Roots. — The multiplication table (facing page 1) does not include products larger than 144 and only a few of the products in the table are squares. For finding the square roots of other numbers and all numbers larger than 144, another method is commonly used. Example. — Find the square root of 3844. 6 2. 38 '44. 36 122)244 Ans. 62. 244 50 POWERS AND ROOTS Beginning at the decimal point, point off the number into periods of two figures each. In the period at the left, 38, find the largest square. This is 36, and the square root of 36 is 6. Write 6 above the first period and write 36 under 38. Subtract 36 from 38 and bring down the next period which gives 244. Multiply 6 by 2, writing the product, 12, at the left. This number, 12, is called a trial divisor. Divide 12 into 24, which gives the quotient 2. Annex 2 to 12, making the complete divisor 122. Multiply this number by 2 and write 244 in the proper place. Write 2 in the answer, making 62 the square root of the number. Example 2. 3 8. 7 14'49.'32'49 9 68)540 544 7607 53249 Ans. 38.07. 53249 Point off as in Example 1, beginning at the decimal point. Division by the trial divisor 6, gives 9. If 69 is multi- plied by 9, the result is 621, which is larger than 549. Use 8 as the last figure of the complete divisor; multiplying, the result, 544, which can be subtracted from 549, is obtained. In the next step, 53 is not divisible by 76. In such case write as the next part of the answer, and annex to the SQUARE ROOT 51 trial divisor. Then bring down the next period. Divide 5324 by 760; the result is 7, the last figure of the answer. Point off one decimal place, counting from the right, in the answer, for each decimal period in the given number. Example 3. — Find the square root of 764 to the nearest .001. 2 7. 6 4 5 7'64.'00'00'00' 00 4 47) 364 3 29 35 00 546) 32 76 2 24 00 5524) 2 20 96 3 04 00 00 KK^