FRAGILE DOES NOT CIRCULATE QA ■73 Cornell University Library ^ The original of tiiis bool< is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924002978561 The Mannheim & Polyphase Slide Rules A Self Teaching Manual with tables of settings, equivalents and gauge points By William E. Breckenridge, A. M. Associate in Mathematics Columbia University New York City Published by KEUFFEL & ESSER Co. - ilEWyORKJ2ZFuhonSt. GenenlQ£Rcaan3FMiorles,HQBOKEN,N.U^. • CHICAGO STLOUIS SANFKANCISCO MONTREAL. 46-20 S-DearbomSt. SlTLocustSl. 30-34 Seami St SNoireBajneSfW ftawii^Materials *>1afliemaHcalandSurvcyin|iishnimerfs*hlE^^ S Cornell University Library QA 73.B82 The Mannheim and polyphase slide rules; c 3 1924 002 978 561 DATE DUE WW*»««*n4j 13L^. CAVLORD PRINTED tMU.S A. K&.E MANNHEIM SLIDE RULE. itJ^l»| . ^Mlj. i |f|!i:jjiSi^.;„,,jVi,^ M'l.v.jil V ri ii ((ii i i i i|f4}iito|i!if'*[* fa 140S~ uo< 18000, a. 190!D,ss:: 200l!0-_ 204- o w -143 145 -CO Diagram Illustrating the reading of the graduations of the rule. li-r- The Slide Rule in its present form has become an indispensable aid not only to the engineer and scientist, but also to the manufacturer, the merchant, accountant and all others whose occupation or business invol- ves calculations. "We manufacture slide rales and devote to them a separate department of our factory which is thoroughly equipped with the most improved special machinery. Several of our improvements are protected by patents, and are, there- fore, not embodied in other Rules. Cross section of K & E Mannheim Slide Rule showing slide adjustment. » Polyphase Slide Rule No. 4058-3. Magnifier for K & E Mannheim Slide Rule. ^hw^^^^^^'^^^'^ ~' — ~ — ^^ — ■^g^.tjafej^^y^^s?!! A|^^^j|«^^^^i[ KEUFFEL & ESSER CO., NEW YORK ll^^^^^O^^^^ [^^i^fe.^ KS*- — -"^-" V -'--^ tt>4WUS-X!ii^. JSi\. 1 The Mannheliti & Polyphase Slide Rules COMPLETE MANUAL with tables of settings, equivalents and gauge points ■By William E. Breckenridge, A. M. Associate in Mathematics Columbia University New York City Published By KEUFFEL CESSER CO. NEW YORK aA73 Copyright 1922, by KEUFFEL & ESSER CO. O / ^ f ^ ^7 — 1 ^^^^c^ KEUFFEL & ESSER CO.. NEW YORK 1 ^^^fep^®!**®^®^ THE SLIDE RULE. PKEFACE. This manual is designed to meet the needs of all who desire to learn the use of the slide rule. Chapter 1. is self-teaching by means of numerous cuts and examples simply explained. For some persons a few lessons in this chapter will teach all that is desired. It is suggested that everyone learning t(?use the slide rule begin by working the examples in Chapter 1. In Chapter 11., a simple explanation of the theory of the slide rule is followed by the advanced subjects of Cubes, Cube Root, Sines, Cosines, Tan- gents and Logarithms. Special work for technical men and typical problems from various occupa- tions are presented in Chapter III. Who should use the Slide Rule? I. Teachers in the following types of schools: 1. Elementary Schools as early as the eighth grade. 2. Junior High Schools for part of their practical mathematics. 3. High Schools in connection with logarithms, practical mathematics, or trigonometry. 4. Colleges in their courses in algebra or trigonometry. Many colleges have already made the slide rule a part of the trigonometry course. 5. Evening schools will hold the students better while teaching the slide rule than any other subject. 6. Engineering and Trade Schools find the rule indispensable. II. Engineers, Mechanics, Chemists, and Architects have long understood its value. III. Private Secretaries can check reports by the slide rule in a small fraction of the time required by ordinary calculation. IV. Estimators, Accountants and Surveyors can make approximate calcu- lations rapidly and with sufficient accuracy to check gross errors. By means of the slide rule, all manner of problems involving multiplication, division and proportion can be correctly solved without mental strain and in a small fraction of the time required to work them out by the.^ usual "figuring." 2 — For instance, rapid calculation is made possible in such problems as the following, which are of every-day occurence in office and shop: Estimating, discounts, simple and compound interest, converting feet into meters, pounds into kilograms, foreign money into U. S. money, the taking of a series of dis- counts from list prices, adding profits to costs; while dozens of equivalents are instantly shown, such as cubic inches or feet in gallons, and vice versa; centi- meters in inches, inches in yards or feet; kilometers in miles, square centimeters in square inches, liters in cubic feet, kilograms in pounds; pounds in gallons; feet per second in miles per hour; circumferences and diameters of circles. Uow mucb education is necessary? Anyone who has a knowledge of decimal fractions can learn to use the slide rule. How mucli time will It take? The simplest operations may be learned in a few minutes, but it is recom- mended that at least the examples in Chapter I. be worked thoroughly and checked by the answers, in order to gain accuracy and speed. This will take from one to ten hours, according to the previous training of the student. How accurate is the Slide Kule? With the ten-inch slide rule results can be obtained correct to three and sometimes four significant figures. With the twenty-inch rule, results are correct to four figures. How to use this manual. For the man who desires to perform the simplest operations of multiplica- tion and division, the first few lessons in Chapter 1 will be sufficient. Work the illustrative examples and as many others for practice as seem necessary to obtain accuracy and speed. For educational use, Chapter II. furnishes the necessary theory and history of the rule, while Chapter I. provides additional examples for practice. Chapter III. may be used for advanced work. It Is recommended that the users of the Polyphase Slide Rule carefully read page 73, etc. before beginning the general study of the slide rule outlined In this book. — 3 — CHAPTER I. ESSENTIAIiS SIMPLY EXPLAIXED. The slide rule is an instrument that may be used for saving time and labor in most of the calculations that occur in the practical problems of the business man, mechanic, draftsman, engineer, or estimator. There are four scales on the rule, A,B,C, and D, as shown in Fig. 2. Scales A andB are graduated from 1 to 100, the left half of the rule extending from 1 to 10. Scales C and D are graduated from 1 to 10. In order that you may see how the rule is used on simple] problems where you know the answers, let us take the following: 1. 2X3. Set 1 on scale C opposite 2 on scale D. Then move the glass runner to 3 on scale O. Directly below this 3 you will find 6, the answer. Fig. 2. 2 X 3 = 6, or 6 -^ 3 = 2. 2 2X2. 3.' 2 X 4.' (See Fig. 3). |p | l|l |l |l |l | l | l |l|!i|^ i' i' i | il'i | i' i '/|i ij i i ' '|t ii|f| i yi | i ij Fig. 3. 2X4 =8 or 8-^4= 2. 4. 3X3. (See Fig. 4). ^llftl il ^lfl#plirt l ff1^ ' 'l 'l 1^^> l 1^^ ^ l^i'l1^f1ft ^^^ Fig. 4. 3 X 3 =^or 9 -r- 3 =3. 5. 6 -r 3. (See Fig. 2). Opposite 6 on scale D, set 3 on scale C. Look along C to the left, till you come to 1 at the end of the slide. Under this 1 you will find 2, the answer, on ''=^^^^- 6. 8■^4. (See Fig. 3). 7. 9-^3. (See Fig. 4). It will be noted that the cuts shown'are not in the same scale. This arrangement is lor the purpose of illustrating various lengths of the rule. — 4 — r— - 1_ 1I-- _JB- T,-.^ j -T ^T«aT^_i_irri: — rr- ^^S j^^^^^KBUrPE. . .SS.. CO. NEW YORK SQUARES AND SQUAKE BOOTS You will remember that to square a number means to multiply that number by itself; e. g., 3^ means 3X3=9. 8. On the slide rule this would be done as follows: Set the hair line of the glass runner to 3 on scale D. Above, on scale A, opposite the hair line, you will find 9. (Fig. 5). Fig. 5. 32 = 9 and 2* = 4. 9. In the same way find 2^ (See Fig. 5). To find square roots simply do the work in the reverse order. 10. Find the square root of 9; i. e., find the number which multiplied by itself will give 9. The square root of 9 is indicated thus: -\/9. Set the hair line of the runner to 9 on scale A, being careful to use the 9 on the left-hand half of the rule, because the other 9 is really 90. Below, on scale D, find 3, the answer. (Fig. 5). 11. 1/4. Set the runner to 4 on A. Opposite the hair line on scale Z>, find 2, the answer. (Fig. 5). We shall now proceed to apply the same methods to numbers of two or more figures. 12. 2 X 1.5. Set 1 on C to 2 on D. Move the runner to 1.5 on C. This will be half way between 1 and 2. Under the hair line of the runner, find 3 on D. V" ti i i |) | 'lii i t |i | n i | ii \ ^ KEUFFELaES^ERC! N.Y. Fig. 6. 2 X 1.5 = 3, and 2 X 1.8 = 8.6. — 5 — ej^>;a»^Ega»gg^^iiUj=^jrB\i_»i_^ui iinJj-.i_ugBl^»sg^!ai^^«.S&^fi^sgR8^:a^S^ll m^^^m KEUFFEL 8= ESSER CO., NEW YORK 1 ^^IP^O^i^i^tSM 1 1 fe4^fe£^^fe^^r.,S^r^ 1 ^Sm^j^^^H^^^ri^W 13. 2 X 1.8. Using Fig. 6, see if you can make it 3.6. The tenths on D are the large subdivisions. 14. 1.5 X 2.5. Set 1 on C to 1.5 on D. Move the runner to 2.5 on C. Below 2.5, find 3.75, the answer, on D. Note that this answer is half way between 3.7 and 3.8, which makes it 3.75. (Fig. 7). Fig. 7. 1.5 X 2.5 = 3.75. 15. 2 X 15. This is worked on the rule exactly like example 12, but you can see by looking at the problem that the answer is 30 and not 3. All of these problems are worked like Example 12. As far as the slide rule is concerned, we multiply 2 by 1.5 and get 3. Then we place the decimal point by inspec- tion. From arithmetic we remember that in multiplying decimals we first multiply as though there were no deci- mal point, then point off as many decimal places in the answer as there are decimal places in both numbers to be multiplied together. Thus, in Problem 22, there are two decimal places in .02 and three in .015. So in the answer, 30, we must have 2 4- 3, or 5 places, making the result .00030. Of course the at the right does not count and the final result is .0003. From these examples it is evident that the decimal point is not considered in operating the slide rule. After the work of the rule has been done, the decimal point is placed by looking at the problem and making a rough cal- culation. 23. 23 X 32. 16. 20 X 15. 17. 200 X 15. 18. 20 X 150. 19. 2 X .15. 20. 2 X .015. 21. .2X 15. 22. .02 X .015 Fig. 8. 23 X 32 = 736. Set 1 on C to 2.3 on D. Move the runner to 3.2 on C. Under the runner line on D find 736, On the rule the hair line comes a little beyond 735, which is evidently 736. By looking at the example, you see that the units figure is 6 because 2X3=6. If we had to depend entirely on the rule, the third figure might be 5, 6, or 7 So when we read numbers on the slide rule, we can be sure of the first two figures, and the third will not be more than one point away from the correct result. PLACING THE DECIMAli POINT. Since 23 X 32 is roughly 20 X 30, or 600, place the decimal point in 736 so that the result will be as near 600 as possible. Evidently the answer is 736. (Fig. 8). Ill |i |gj i' lw |i'i m i> . i )ii^i i l i il !i |i.l! i' Fig. 9. 18 X 34 = 612. 24. 18 X 34. On the rule the result is a little over 61. Looking at the problem, we see that 4 X 8 = 32; hence 2 will be in the third place, making the order of figures 612. By a rough calculation the problem is about equal to 20 X 30 = 600. Hence we make 612 look like 600 by placing the decimal point after the 2. The answer is 612. 25. 16 X 24. (Answer, 384). 26. 14 X 26. (Answer, 364). 27. Fill in the blanks in the following multiplication table, using the slide rule; 31 32 33 34 85 21 23 23 24 25 36 ~ 37 28 29 30 — 7 — A^^^M^^^SpJ KEUFFEL & ESSER CO_ NEW YORK ]|S^ WHICH INDEX TO USE. We shall now call the 1 at the extreme left of the scale C the left-hand index, and the 1 at the extreme right of scale C the right-hand index. In multiplying 30 by 34, if we use the left index of C, 34 comes beyond the right-hand end of the rule. In this case use the right-hand index, setting 10 on C to 30 on D. The right or left-hand index must be used in accordance with the following rule: When the first digits of the factors are such that their product gives a number greater than ten, use the right index on C; otherwise use the left. Take as examples the following: Example 1. 2.13 X 3.33, 3X2=6. Use the left index. Example 2. 7.23 X 4.71, 7 X 4 = 28. Use the right index. Example 3. .131 X 4.6, -1X4=4. Use the left mdex. If this very simple rule be kept in mind, much time will be saved. It will be a little difficult to make a decision in certain cases like the following: 3.13 X 3.31. By the rule, the left-hand scale should be used, considering only the first digit of each of the factors; a closer inspection will show however, that this is not correct, and that the right index should have been used. The operator should not try to memorize these rules, but he should appreciate their sig- nificance. 28. Suppose you are earning 56 cents per hour and you are given an increase of 8 cents. What per cent increase do you receive ? Of course you will divide 8 by 56. To divide one number by another on the slide rule we simply reverse the order of the work we have been doing in multiplication. Set the runner to 8 on scale D. Move the slide so as to set 56 on C to the hair line of the runner. Fig. 10. 8 -=- 56 = .14. Under 1 on C we find 14 and a little over. But the result is nearer 14 than 15. Hence the correct result to two figures is 14. By inspection the decimal point must be placed before the number, making the answer .14 or 14 per cent. 29. Before the war a man earned 35 cents per hour. While recovering from wounds, he learned a new trade which increased his earning power to 67 cents per hour. What per cent increase did he receive? His increase is 32 cents per hour. The per cent of increase is found by dividing 32 by 35. Set the runner to 32 on D. Set 35 on C to the runner line. The result would be found under the 1 at the extreme left of scale C, but this 1 projects beyond scale D. So we use the 1 on the extreme right of scale C. Under this 1, find 91 on scale D. (Fig. 11). pwi' i fiwi||iwji^ ^ ^ ^ ^^^^ ^ Fig. 11. 32 -=- 35 = .91. In the same way, for practice, try the following: 30. What per cent of 91 is 45? (Divide 45 by 91.) 31. What per cent of 73 is 24? 32. What per cent of 67 is 61? 33. What per cent of 53 is 31? 84. What per cent of 82 is 13? 35. What per cent of 42 is 9? If you have a long report to make out in which a large number of per cents are to be calculated, why not use the slide rule? A secretary to the president of a big corporation recently said: "The slide rule does my work in one-third of the time that would be required other- But suppose you had to get per cents in a problem like the following: 36. A baseball player made 57 hits out of 286 times at bat. What his percentage? T| KEUFrci.aEsJtRCSN.Y. ^ ' Fig. 12. 57 -T- 286 = .199. Opposite 57 on D set 286 on C. When we look for 286 we observe that between 2.8 and 2.9 there are five spaces on the rule. Then every space counts one-fifth of .1, which is .02. Since we want six points for the third figure, we have to use three spaces, every one worth .02. 3 X .02 = .06. Under the left-hand index find 199. When we read the result, we see that it comes on the rule between 1.9 and 2.0. There are ten small spaces between 1.9 and 2.0. Hence eyery space counts one point. We see that just a little practice in decimals enables us to read the rule to three figures. Now we must 57 60 1 place the decimal point. A rough calculation shows that — is nearly — , or -. 286 300 6 Then the decimal point must be placed so as to make the result some- where near one-fifth or .2. Evidently the result is .199. This may be read 19Vio Psr cent or 19Vio hundredths, or 199 thousandths. '37. If your income is $2,600 per year and you save $451, what per cent do you save? Fig. 13. 451 -7- 2,500 = .180. Opposite 451 on D set 25 on C. Under the index find 180 on D. Hence the answer is .180, or 18 per cent. We note that when we look for the 1 in 451 on the rule, there are only two spaces between 45 and 46. Then each space counts one-half of a hundredth or one-half of .01, which is .005 or five points for the third figure. We estimate one-fifth of the small space to obtain .001. 38. If your salary is $57.50 per week, and you are given an increase of $12.40, what per cent increase do you receive? l fl l |lll l [l ll h l |l|y[V[! immm iJTilililil TppI^ M trrrmr mf KeuffeI Fig. 14. 124 -i- 575 = .216. Opposite 124 on D set 575 on C. This means that between 5 and 6 on C we must take 7 of the large divisions and one of the small divisions. Under the right-hand index read 216 on D. Hence the answer is 21^% per cent. ■ 39. 5.42 ^ 2.42. - 40. 7.35 ^ 3.14. 41. 6.13 -;- 4.61. 42. 9.56 -H 7.26. 43. 10 -H 3.14. For 10, use either the right or the left-hand index. To place the decimal point these problems are roughly: 44. 16.5 -r .245. 16 -=- .2 = 80. 45. .00655 ^ .00034. .0060 -=- .0003 = 20. 46. .00156 -r 32.8. .0015 -t- 30 = .00005. 47. .375 -7- .065. .36 -=- .06 = 6. 48. .0385 -=- .0014. .038 h- .001 = 38. Here is another method of placing the decimal point in division. Work the problem as though both dividend and divisor were integers (i. e., not decimals), pointing off as usual. Move the decimal point to the left as many places as 10 — there are decimal places in the dividend. Then move it to the right as many places as there are decimal places in the divisor. For example in problem 44, 165 -;- 245 gives .673. Move the point one place to the left because there is one decimal place in the dividend, giving .0673. Then move it three places to the right because there are three places in the divisor, giving as a result 67.3. Try both methods and see which one you like the better. Let one check the other. Suppose we have more than three figures, as in the following example: 49. Find the circumference of a wheel 28 inches in diameter. Here we must multiply 28 by 3.1416. But the slide rule only reads to three figures. So cut off the fourth and fifth figures in 3.1416 and call it 3.14, since the number is nearer 3.14 than 3.15. 50. Multiply 26 by 3.145. In this case call 3.145 equal to 3.15. 61. If bell metal is made 25 parts of copper to 11 parts of tin, find the weight of tin in a bell weighing 404 pounds. 11 X 404 The tin is evidently eleven thirty-sixths of 404, or - 36 f$ill^. jll|![ l [ l i ' | 'i lll i l l jl!l||li!ll | l|| | ftp iiiiiiii TTITir iii| | ^i ^([ iiii|iili [ji i iiii| i j! ii i ii |j i ik|i i ' :|IMI|MN{MII{lllllill|lll KcurrcLaEs^ERCSN-Y. 11 X 404 Fig. 15. = 123. 36 Opposite 11 on D set 36 on C. (Fig. 15). Move the runner to 404 on C. Under the runner line read 128 on D. To place the decimal point, make a rough calculation as follows: The 10 X 400 example is roughly equal to = 100. So make 123 look as nearly like 40 100 as possible by placing the point after 3. The answer is 123 pounds of tin. 14 X 525 24.5 X 43.4 1.35 X 3.16 52. . 53. . 54. . 47 3620 (See Fig. 16). Fig. 16. — 11 — 1 te^.^^;^^^^^^-"'^"'''^'^ "■ -T-n >^^^^,^,&fe,asga8g^^Ssg || I ^^^^^^^l^^^^'§[ KEUFFEL & ESSER CO. NEW YORK J^^^MJ Opposite 1.35 on D, set 6.2 on C. If we try to move the runner to 316 on C, it is impossible because 316 lies beyond the extremity of D. In such a ease proceed as follows: Move the runner to the right-hand index of C. (Fig. 17.) Fig. 17. Then move the slide, setting the left-hand index of C to the runner line. Fig. 18. Now we can move the runner to 316 on C. (Fig. 19.) nm i i( ii l i( #l| ii iiiW^ ^^ ^ n ii | i Nl|i i ii|y4i i ii|4» i i i "Aii i A KeurrEL«Es^iiCSN.Y. ' Fig. 19. and read the result under the hair line on D. The answer is 688. 1X3 3 A rough calculation for the decimal point gives us = - ,or .5. Mak- '^ 6 ing 688 look as much as possible like .5, we have .688. 2.28 X .0125 .55. . 4.36 2 X .012 The rough calculation for the decimal point might be = .006. , 56. . 57. 7.63 X 2.34 . 24.3 2.56 X 1.78 58. 82.5 X 9.3 59. 32.6 X 22.1 7.4 56.5 9.25 60. If an aeroplane flying 100 miles an hour travels 86 miles in a given time, how far will an automobile traveling 22 miles an hour go in the same time? Writing this in the form of a proportion: 100 : 86 = 22 : x which means that 100 is to 86 as 22 is to the answer. The work on the rule is as follows: 12 — Fig. 20. 100 : 86 = 22 : 18.9. Opposite 86 on D, set 100 on C. (Use right index for 100). Move the runner to 22 on C and under the hair line on D read the answer, 18.9. An easy method of remembering this is: C D C D 100 : 86 = 22 : 18.9. In placing the decimal point, note that 100 has the same relation to 86 that 22 has to the answer. Since 86 is about nine-tenths of 100, we must place the decimal point in 189 so that the answer shall be about nine-tenths of 22. Hence the answer is 18.9. In the same way solve the following proportions: - 61. '24 : 31 = 15.2 : x. ■ 62. 1.4 : 2.5 = 12 : x. ■ 63. 3.71 : 2.4 = 51.2 : x. . 64. 2.54 : 4.72 = 7.48 : x. ■ 65. 12.3 : 15.2 = 8.5 : x. Problems from Camp Taylor (Field Artillery Training School): • 66. If a post 13.2 feet high casts a shadow 27.2 feet long, how high is a tower which casts a shadow 116.8 feet long? ^' 27.2 Fig. 21. 27.2 : 27.2 : 13.2 = 116.8 Solve like example 60. 67. At 2,400 yards an increase of 1 mil in the elevation of a gun increases the range 25.0 yards. What change in elevation will increase the range 40 yards? The mil is the unit of angle in the artillery. It is equal to ^^ of 360.° 68. The effects of wind on a shell are approximately proportion^ to the velocity of the wind. At 3,000 yards for a 3-inch gun, a rear wind of 10 miles per hour increases the range 30.1 yards. (a) What wind will increase the range 42.8 yards? (b) What wind will decrease the range 68.5 yards ? Answer: (a) Rear wind of 14.2 miles per hour, (b) Head wind of 22.8 miles per hour. — 13 — ^^§^^^^^^^§1 KEUFFEL & ESSER CO., NEW YORK 3i^^^^^^^^ .i^a^^iCJa^ ,^^ SQUARBiS. 69. Find the area of a square plot of ground measuring 128 yards on a side. Fig. 22. Set the runner to 128 on D. Directly above on A find the square required, 164. To place the decimal point, make a rough calculation. (128)2 is roughly (130)« or 16900. Then make 164 look like 16900 by plac- ing the point as follows: 16400. The result is only correct to three figures. The complete result is 16384. Case 1. — When the given number lies between 1 and 10, if the runner is set to this number on D, the square will be found opposite the runner line on A. EXAMPLE: 70. Square 6.5. The runner set to 6.5 on D indicates the square 42.25 on A. Notice that the A scale runs from 1 to 100 with 10 in the middle of the rule. Case 2. — ^When the given number does not lie between 1 and 10, square as in case 1, and then place the decimal point by a rough calculation. 71. Example: Square 652. Set the runner to 6.52 on D reading the square 425 on A. Roughly (652)'' = (600)2 = 360,000. To make 425 look as much as possible like 360,000, the decimal point must be placed as follows: 425,000. Notice that here the arithmetic square would be 425104, but on the slide rule we can only get the first three figures, 425. This, however, is close enough for most practical purposes, such as estimating on contract work. Find the square of 8.65 78. .66 81. .0244 276. 79. .0625 82. 2240. 34.2 80. .0057 a circular plot of ground measuring 14.5 feet in .7854 d^, which means that the area of the circle is equal to .7854 multiplied by the square of the diameter. Set the runner to 145 on D. The square is found directly above on A, but need not be read. Set the right-hand index of the slide to the runner line. Move the runner to .7854 on B. Opposite the runner line on A find the result, 165 sq. ft. 72. 3.2 75. 73. 4.65 76. 74. 1.12 77. 83. Find the area of diameter Use the formula A = — 14 — SQUARE ROOTS. 84. How long must one side of a square garden bed be made in order that it shall contain 8 square yards? Here we have to find the square root of 8. Pig. 23. -v/8 = 2.83. Set the runner to 8 on scale A. Note that scale A runs from 1 to 100, so that 8 is found on the left-hand half of the rule. Now opposite the runner line on scale D, find 2.83, the square root. Then the result is 2.83 yards. 85. Find VS. Set the runner to 3 on A. Opposite the runner line find 1.73 on D. 86. Find \/30- Set the runner to 30 on A, being careful to notice that 30 is indicated by 3 on the right-hand halt of the rule. Opposite the runner line on D, find 5.4 8. Fig. 24. VSO = 5.48. 87. Find V300. Since scale A only runs from 1 to 100, 300 is not found on the rule. Move the decimal point an even number of places in order to obtain a number that is on the r ule. This can be done by moving the point two places to the left, giving \/3-00. Find the ^/Z, which is 1.73. Then move the decimal point half as many places as it was moved in the first place, but in the opposite direction. In this case, move the point in 1.73 one place to the right, giving 17.3. — 15 — 1 %a^^ca!i'^is»«as'B*s»->."->i>^ 88. Find V-30. Move the point two places to the right, obtaining 30. Find VSO = 5.48. Move the point one place to the left, obtaining .548 for the result. 89. Find V-OS. Move the decimal point two places to the right, obtaining -\/3. Find VS = 1.73. Move the point one place to the left, obtaining .173. 90. Find \/^3. _ Move the point four places to the right, obtaining -v/SO. Find VSO = 5.48. Move the point two places to the left, obtaining .0548. Find the square roots of 91. 1.42. 94. .142. 96. 85.4. 98. .31416. .92. 14.2. 95. 2.43. 97. .365. 99. 1450. 93. 142. 100. Make a list of square roots of whole numbers between 110 and 130. 101. On a baseball field, find the distance from home plate to second base, measured in a straight line. (The distance between the bases is 90 feet). 102. Water is conducted into a tank through two lead pipes havmg diameters of 5^ and 1% inches, respectively. Find the size of the lead waste pipe that will allow the water to run out as fast as it runs in. Use 5^ and 1 J^ in the decimal form. Find. V(-625)2 + (1.75)2. NOTE:— Perform the addition by arithmetic. The slide cannot be used to advantage in addition. 103. Two branch iron sewer pipes, each 6 inches in diameter, empty into a third pipe. What should be the diameter of the third pipe in order to carry ofif the sewage? WHEN MORE THAN TWO NUMBERS ARE TO BE MULTIPLIED TOGETHER. 104. Find the product of 3.2 X 52 X .24. Find 3.2 X 52 as usual, obtaining 166. Set the left-hand index to the run- ner line. Move the runner to 24 on C. Opposite the runner on D, read the answer, 39.9. 105. Find the product of 7.2 X 3.2 X .25 X 5.4. 7.2 X 32 gives 230. Set the left index to the runner. Move the runner to 25 on C. Opposite the runner on D, read 576. The product of the first three numbers is 57.6. Set the right index to the runner line. Move the runner to 5.4 on C. On D opposite the runner line, read 311. The decimal point is placed by a rough calculation. The problem: 7.2 X 3.2 X .25 X 5.4 is roughly equal to 10 X 3 X .2 X 5 = 30. Then 311 must be made to look as much as possible like 30. In order to do this the answer is evidently 31.1. •106. Find the product of .75 X 1.1 X 6.5 X 8.65. Answer, 46.4. • 107. Find the product of 8.2 X .45 X 6.4 X 16. Answer, 378. . 108. Test for the simple operations of the slide rule. — 16 — ^^^K s^V^ KEUFFEL a ESSER CO.. NEW YORK M TO THE STUDENT. Read carefully the following instructions: 1. Copy the test on your paper in the form given below. 2. Work the problems straight through, setting down the answers in the column at the extreme right. 3. Fold these answers underneath the paper. 4. Work the problems through again, setting down the answers in the other column. 5. Compare the two sets of answers. 6. If the answers to any problem do not agree (within one point in the third place), work the problem again. 7. The correct results are given on page 71. . TEST. Credits Answers Answers Second First Time Time 20 1. 1.28 X 2.46. 20 2. 84 -^ 59.5. 20 3. 58.5 X 15.2. 78 20 4. 6.25 : 24.2 = 9.5 X 20 5. ^182. CHAPTER II. THE THEORY OF THE SLIOE KUIiE. This chapter includes a brief historical note on the Slide Rule, the theory of the operations described in Chapter 1, and additional subjects, including Cubes, Cube Roots, Sines, Tangents and Logarithms. HISTORICAL NOTE. The earliest instrument that can be called a slide nde was invented by Wm. Oughtred (1574-1660), an English mathematician, and was described in publications of his pupil, Wm. Forster, in 1632 and 1653. The present form of the rule was invented in 1859 by Lieutenant Am§dee Manheim.of the French Artillery. He had the good fortune to have his rule made by a firm of national reputation and to have it adopted by the French Artillery. In the United States the rule was little known until 1881, when Edwin Thacher invented the cylindrical form. Modern improvements have been introduced by Keuffel & Esser, such as celluloid facing, devices for smooth and even working of the slide in the stock, allowing for shrinking and warping, the addition of scales for cube root, sines, tangents and logarithms, inverted scales, the Duplex Form, and the 20-inch length. — 17 — ^^B ^!^^^iPS^^^3§| KEUFFEL & ESSER CO., NEW YORK For a complete history of the Logarithmic Slide Rule, the student is referred to "A History of the Logarithmic Slide Rule," by Florian Cajori, published by the Engineering News Publishing Company, New York City. This book traces the growth of the various forms of the rule from the time of its invention to 1909. ACCURACY. The accuracy of a result depends upon o), accuracy of the observed data, 6), accuracy of mathematical constants, c), accuracy of physical constants. ACCURACY OF THE OBSERVED DATA. The precision of a measurement is evidently limited by the nature of the instrument and by the care used by the observer. Example 1. If a distance is measured by a scale whose smallest sub- division is a milliineter, and the result recorded 134.8 mm., evidently the result is correct to 134, but the .8 is estimated. Hence it is known that the actual measurement lies between 134 and 135 and is estimated to be 134.8. The result 134.8 is said to be "correct to four significant figures." If the result were desired correct to only three figures, it would be recorded 185, since 134.8 is nearer 135.0 than 134.0. This result is said to be "correct to three significant figures." Example 2. If the distance is measured by a rule whose smallest sub- division is .1 inch, and found to be exactly 8. inches, the result would be recorded 8.00 inches. The zeros record the fact that there are no tenths and no hundredths, but the distance is exactly 8 inches. The result, 8.00 inches, is said to be "correct to three significant figures." Example 3. If an object is weighed on a balance capable of weighing to .01 gram, then .001 gram can be estimated. Suppose several objects are weighed with the following results: 1. Seven grams recorded 7.000 grams. 2. Seven and a half grams " 7.500 " 3. Seven and 9/100 grams " 7.090 " 4. Seven and 6/1000 grams " 7.006 " 5. 4/100 and 2/1000 grams " .042 " Note that readings with the same instrument should show the same number of "places filled in to the right of the decimal point, even if zero occurs in one or all of these places. In number 5, the result, .042 grams is said to be "correct to two significant figures." The first significant figure is 4 and the- second is 2. Example 4. When we say that light travels 186,000 miles per second, we mean that the velocity of light is nearer 186,000 miles than it is 185,000 miles, or 187,000 miles. The result is said to be "correct to three significant figures." Summarizing the preceding examples: Example 1. 134.8 is correct to four significant figures. Example 2. 8.00 is correct to three significant figures. Example 3. .042 is correct to two significant figures. Example 4. 186,000. is correct to three significant figures. Counting from the left, the first significant figure is the first figure that is not zero. — 18 — After the first significant figure, zero may count as a significant figure, as in Example 2, where it represents an observed value, or it may not so count, as in Example 4, where the zeros merely serve to place the decimal point cor- rectly, the number 186,000. being correct only to the nearest thousand miles. ACCURACY OF MATHEMATICAL CONSTANTS. A mathematical constant may be carried to any desired degree of accuracy, e. g., the value of tt usually given as 3.14159 has been calculated to 707 places. For ordinary calculations 3.14 or 3f is sufficiently accurate. ACCURACY OF PHYSICAL CONSTANTS. Most physical constants are only correct to three significant figures and some only to two figures. e. g., The weight of a cu. ft. of water is 62.5 lb. The weight of a cu. in. of cast iron is .26 lb. LIMITS OF ACCURACY HOLMAN'S RULE. If numbers are to be multiplied or divided, a given percentage error in one of them will produce the same percentage error in the result. In other words, a chain is no stronger than its weakest link. Since physical constants are not usually correct beyond three significant figures, and the observed data in an experiment is rarely reliable beyond this point, the slide rule reading to three figures gives results sufficiently accurate for most kinds of practical work. PKRCENTAGE OF ERROR. If a result is correct to three significant figures, the ratio of the error to the result is less than 1:100. Suppose, for example, the result is 3527.6, which is known to be correct to three significant figures. Then the figures 352 are known to be correct and the figures 7.6 are doubtful. Since 7.6 is less than 10 and 3527.6 is greater than 1000, the error must be less than 10:1000 or 1:100. 7.6 10 10 ■^ Q KO'T e "^ 1 nn A »^^ " 3527.6 ^ 3527.6 ^ 1000 ' 100 A result read on the 10-inch slide rule to four significant figures is 1324, which is correct to three figures, 132, while the fourth figure, 4, is a dose estimate not more than one point away from the correct reading. The error here is less than ^„„. , which is less than:^^ . Hence the error in this reading is less than one-tenth of one per cent. By a similar method it may be shown that in a result correct to three significant figures, the error is less than one per cent. It will be found that readings on the left half of the slide rule may be obtained to four significant figures, or correct to one-tenth of one per cent. — 19 — On the right half of the rule where the spaces are smaller, readings may be obtained to three significant figures or correct to one per cent. If greater accuracy is desired, a 20-inch rule will give one more significant figure and a Thacher Cylindrical Rule will give readings correct to five and sometimes six figures. LOGARITHMS. 102 = 100. Another form of making this statement is: The logarithm of 100 is 2. In the same way, 10' = 1,000. or the logarithm of 1,000 is 3. From these examples it is evident that the logarithm is the exponent which is given to 10. EXERCISE. Fill out the blanks in the following table: 1. 10* = 10,000 Log 10,000 = 2. 10= = 100,000 Log 100,000 = 3. 10' = 10 Log 10 4. 10° = 1 Log 1 = LAW OF MULTIPLICATION. W = 100. 10' = 1,000. 102 X 10» = 100 X 1,000. 10' = 100,000. Log 100,000 is 5. Since 5 is the sum of 2 and 3, log 100,000 = 2 + 3 = log 100 + log 1,000, or The logarithm of a product is the sum of the logarithms of the multiplicand and the multiplier. Hence to multiply one number by another, add their logarithms. The construction of the rule allows this addition to be done easily. The scales are divided proportionally to the logarithms of the numbers. If the 10-inch rule is considered as divided into 1,000 units, then any number; 1, 2, 3, etc., is placed on the rule so that its distance from the left index is proportional to its logarithm. Since log 1 = 0, 1 is found at the extreme left. " log 2 = .301, 2 is found 301 units from the left. " log 3 = .477, 3 " " 477 " " log 9 = .954, 9 " " 954 " " log 10 = 1000, 10 " " 1000 " On the scale the number 8 is placed three times as far from the left index as 2, because the logarithm of 8 is three times the logarithm of 2. MULTIPLICATION. When we multiply 2 by 4, we set the left index of the slide to 2 on scale D, and under 4 on scale C find the product, 8 on scale D. This is equivalent to adding log 2 to log 4 and finding log 8 (Fig. 25). — 20 — , C 1 4 1 D |2 1 8 -LOG 2- -LOG 4- Fig. 25. Example 1. Multiply 2.45 by 3.52. Set 1 on C to 2.45 on D, and under 3.52 on C find 862 on D. Roughly calculating, 2.45 X 3.52 =2X4=8. Hence, we place the decimal point to make the result as near 8 as possible, and the result is 8.62. Example 2. Multiply 24.5 by 35.2. Working this like the preceding example, without regard to the decimal point, we obtain 862. Roughly calculating, 24.5 X 35.2 = 25 X 36 = 900. Placing the decimal point to make 862 as near 900 as possible, we obtain 862. Example 3. Multiply 6.234 by 142.6. Taking each number correct to three significant figures we multiply 6.23 by 143. Set 1 on C to 623 on D. Under 143 on C find 885 on D. Roughly calculating, 6 X 140 = 840. Therefore the result is 885. EXAMPLE 4. Multiply 2.46 by 7.82. When the product of the first significant figures of each number is greater than 10, the sum of their logarithms will exceed the length of the rule. Hence if we set the left index of the slide to 246 on D, the other number 782 on C projects beyond the rule. In this case, think of the projection as wrapped around and inserted in the groove at the left, which would be the case in a circular slide rule. Now the right and left-hand indices coincide. Hence set the right index of the slide to 246 on D. Under 782 on C find 192 on D. Roughly calculating, 2 X 8 = 16. Hence the result is 19.2 EXAMPLE 5. Multiply .00146 by .0465. Set 1 on C to 146 on D. Under 465 on C, find 679 on D. Roughly calculating, .001 X .05 = .00005. Hence the result is .0000679. Find the value of 1. 2.34 X 3.16. 4. 8.54 X 6.85. 7. .023 X 2.35. EXERCISE. 2. 3.76 X 5.14. 5. 34.2 X 7.55. 8. .00515 X .324 3. 1.82 X4.15. 6. 4.371 X 62.47. 9. .00523 X .0174. 21 ^ 10. Find the circumferences of circles having diameters of 4 ft., 6.5 ft., 14 ft. Set 1 on B, to jr on A. Above 4, 6.5, and 14 read the circumferences on A. DIVISION. In division, reversing the operation of multiplication, 8 -r 4 = 2. We subtract log 4 from log 8 and obtain log 2. PROPORTION. Problems in proportion are special cases of multiplication and division. Example 1. Solve 16 : 27 = 17.5:a;. 27 X 17.5 Then multiply by Example 2. 16 Solve X : 24 C X ; To 18 on D, set 11 on C. 16 Solve by the method of Example 51, Chapter I. First divide 27 by 16, by setting 16 on C to 27 on D. 17.5 by moving the runner to 17.5 on C. On D, opposite the runner line, will be found the result, 295. To place the decimal point, note that 16 has the same relation to 27 that 17.5 has to X. Since 27 is not quite twice 16, x will be not quite twice 17.5. Hence the decimal point must be placed so that the answer is 29.5. The method of working a proportion is easily remembered as follows: CD CD. : 27 = 17.5 : x. = 11 : 18. D C D. ; 24 = 11 : 18. Opposite 24 on D find x on C. X = 147. To place the decimal point, note that since 11 is a little more than half of 18, X will be a little more than half of 24, or 14.7. SQUARES AND SQUARE ROOTS. {wy = w X 103. = 10". Since 6=2X3, Log (10') = =2 X log 10'. Hence, to square a number, multiply its logarithm by 2. Scale A is graduated from 1 to 100, while scale D runs from 1 to 10. Since the space given to each number on scale D is twice that on scale A, we multiply the log of a number by 2 if we look for it on scale D. As an example, suppose we wish to square 3. This can be done by doubling the space given to 3 on scale A and finding 9, or looking for 3 on scale D and finding its square above it on scale A. Reversing the operation gives the square root. As an example, find the square root of 9. Look for 9 on scale A , and directly below it on D find 3, its square root. — 22 — ^H ;^ ^^ KEUFFEL a ESSER CO.. NEW YORK ||§^ CUBES. Set right index on the slide to 5 on D. Fig. 26. 2' = 8. 1. Find Z\ Set the left index of the slide to 2 on D. Opposite 2 on 5 read 8 on A. Note that we have taken 2 X log 2 + log 2 = 3 log 2, or log 8 Scale D, Scale B Scale A where log 2 on B has been used as the unit. In the same way, show that 2. 3. 4. 5. 6. 7. 8. 9. IP = 1,331. Roughly calculating, we know 11' is a little larger than 10', or 1,000. We also can see that the units figure will be the cube of 1, or 1. The mark on the rule gives 133. Hence the total result is 1,331. Show that 10. 12' = 1,728. Find the cubes of the following numbers correct to 4 significant figures. Problem 11. 13. Problem 16. 18. 17. 19. 18. 20. 19. 21. Find the cubes of the following numbers correct to 3 significant figures. 3' = 27. 43 = 64. 5' = 125. 6' = 216. 73 = 343. 8' = 512. 93 = 729. 12. 14. 13. 15. 14. 16. 15. 17. 20. 22. 21. 31. 22. 46. 23. 47. (The complete answer is 10,648, but on the slide rule we get 10,600 correct to 3 significant figures. The error is less than one-halt of one per cent. (Set the right-hand index on the slide to 47 on D. 24. 53. •23 — — ^-— 77;=^ " . ^«T<>.^<.jWj.,R^^j^a»^i■^^S 1 ^§( KEUFFEL & ESSER CO, NEW YORK || ^^^^^M^^^Sljl 25. 64. 26. 758. 31. .342. 27. 232. 32. .057. 28. 425.6* 33. .0068. 29. 87.9 34. 1.03. 30. 139. 35. 2.12. *ln Example 28, 425.6 is approximately 426. Roughly approximating the result 400' = 64,000,000. The rule gives us 773. Hence the result is 77,300,000 correct to three significant figures. The complete result is 77,308,776. Example 36. How many gallons will a cubical tank hold that measures 26 hiches in depth. (1 gal. = 231 cu. in.). CUBE ROOT. Example 1. Find the cube root of 9. Set the runner to 9 on scale A. We know that the first figure of the root is 2, since the greatest cube in 9 is ^(ag.^fe^^^i^«»^^ ' ■ -mT ■■^!^,ftfe,6^^5Sg^^SiJ>ll i^^^^sr^ KEUFFEL a ESSER CO.. NEW YORK 1%^^^^^^^ {-^m^i^m. = "'■' "^-^ " ^ __„„ ^.- _ . ,25a®^£^^,^fflSt FIRST METHOD. —Work the example without regard to the square root, then find the square root of the result: SECOND METHOD.— Using scales A andB: A B B D 21.4 X 3.45 X 640 = X. 4.15 X .75 X B B ,08 B Intermediate on A. 516, opposite index. 1. To 21.4 on A set 4.15 on B. Be careful to use 21.4 on the right half of A and not 2.14 on the left half, since the square root of 21.4 has different significant figures from the square root of 2.14. For the same reason use 4.15 on the left half of B. 2. Runner to 3.45 on B (left half of rule) 178, opposite runner. 3. Move slide setting .75 (right half) to runner. 237, opposite index. 4. Runner to 6.4 (left half) on B. 152, opposite runner. Change 640 to 6.4, by moving the decimal point an even number of places , in order not to change the square root. 5. Move slide setting 8 (left half) onB to runner. 190, opposite index. 6. Opposite right index of B find 436 on D. Without taking the square root the result is 190 on A. Dropping down to D gives the square root, 436. Rough Calculation 4 '20 X 3 X 600 = V90000 = 300. 4 X 1 X .1 Placing the decimal point so as to make 436 as near as possible to 300, the result is 436. BXEKCisi:. Find 1 the value of: 3.26 X .0235 4.22 6.75 X 1.35 14.4 ■3. 26.4 X 4.8 X 7.12 . 4. 6.2 X 28 X .35 X 5.4 , 5. .65 X 24 X 7.5 X 9.5 , 6.45 4.55 X .0276 1 2.66 X .75 X 1.42 3 2.54 X 7.45 — 30 — 1 S5^^'W^se^'»>^;g;g»j»^,Sfefi^ggSa?^S8gll KEUFFEL & ESSCR CO, NEW YORK j^^^al Set 2 on CJ to the runner. Move the ranner to 10 on CI. Set 8 on CJ to the runner. Opposite 1 on CI, find 96 on D. Example 3. Find the value of 1.25 X 1.42 X 6.14 by the inverted slide. Set the runner to 125 on D. Set 142 on CI to the runner. Runner to 1 on CI. Set 614 on CI to the runner. Under 10 on CI, find 109 on D. Placing the decimal point by inspection, the result is 10.9. EXERCISE. Find the value of 1. 6.1 X 24 X .32. 2. .53 X 42 X 6.5. 3. 14.3 X 126 X .238. 4. 6.25 X .014 X 58. 5. 7.4 X 42.5 X .85. DIVISION. Example 1. Divide 6 by 2, using the inverted slide. Fig. 30. To 6 on D, set 1 on CI. Under 2 on CI, find 3 on D. EXPLANATION. We have subtracted log 2 on CI from log 6 on D, and obtained log 3 on £). Example 2. Divide 3.72 by 14.5. To 372 on D, set 1 on CI. Under 145 on CI, find 256 on D. Roughly calculating, 3.72-S-14.5 = 3/15 = .2. Hence, the result is .256. Example 3. Divide 12.5 by 6.8. If we set 1 on CI to 125 on D, 6.8 is beyond the left end of the rule. We therefore use the other index. To 125 on D, set 10 on CI. Under 68 on CI, find 184 on D. Placing the decimal point by inspection, the result is 1.84. — 35- EXERCISE. Find the value of the examples on Page 9, using the inverted slide. SINES. FIRST METHOD. Remove the slide from the groove, turn it over so that the face that was underneath is now uppermost and insert it in the groove with the indices coinciding as in Fig. 31. Fig. 31. The scale marked S is a scale of sines. Angles are given on scale S, opposite their sines on scale A. Example 1. Find sine 20°- Opposite 20 on scale S is found its sine on scale A. This reads 342. To place the decimal point, a number read on the right half of scale A has the first significant figure in the first decimal place, except sine 90, which is 1; a number read on the left half of scale A has the first significant figure in the second decimal place. Hence sine 20° = .3420. Example 2. Find sine 2°. The significant figures are 349. The reading is on the left of scale A, hence the result is .0349. SECOND METHOD. Without reversing the slide, set the given angle on scale S to the mark opposite the right index on the under side of the rule; then under the right index of scale A read the sine on scale B. COSINES. Since the cosine of an angle is equal to the sine of the complement of the angle, the cosine may be found on the slide rule. Example. Find cos 30°. Cos 30° = sin (90°— 30°). = sin 60°. = .866. Example 3. Find sin 5° 40' X 35. METHOD I. — ^With the slide having scales B and C uppermost, set sin 5° 40' on S to the mark in the hollow at the right end of the rule. Under 35 on A, read the product 3.46 on B. EXPIiANATION. Evidently we have added log sin 5° 40' to log 35, thei sum being counted on scale B. Or sin. 5° 40' X 35 = a; may be written as a proportion using scales A and B. A BAB 1 : sin 5° 40' =35 : x. — 36 — fti^^W'ssiasZi^fe*^**— ■ 'J^s> ' ~iri Tirr i im— .^j,^ ^^^1 J^§^^M^^^gi( KEUFFEL & ESSER CO. NEW YORK |^ j^#4(ffii^.aSBS^6^. -1 , — , 1^. — i„..,.55bI Under 1 on A, set sin 5° 40' on B. Under 35 on A, find 3.46 onB. METHOD II. — With the slide having scales iS and T uppermost, A : to 35 : Find 3.46 S : Set Right Index : Over 5° 40' EXPLiAXATIOSr. Log 35 is added to log sin 5° 40', the sum being counted on scale A. 35 Example 4. Find . sin 5° 40' METHOD 1. — ^With the slide having scales A and B uppermost, set sin ' 40' to the mark in the hollow at the right end of the rule. Over 35 onB, read the quotient 354 on A. METHOD II. With scale S uppermost A : To 35 : Find 354. S : Set 5° 40' : Over left index. To place the decimal point, note that sin 5° 40' is a trifle less than .1. JHence dividing 35 by .1 we have 350 for the rough calculation. EXPIiANATION. METHOD 1. We have solved the proportion: sin 5° 40' : 1 = 35 : x. B ABA. METHOD 11. — We have taken the proportion by alternation securing, sin 5° 40' : 35 = 1 : x. S ASA. EXERCISE. Find the sines of: 1. 90°. 2. 45°. . 3. 30°. 4. 3°. 5. 40'. cosme of: 11. 80°. 12. 65°. 13. 42°. 14. 14°- 15. 12°. 21. Sin 25° X 45. 22. Cos 56° X 27 18 23. sin 12° 30' 6. 15° 20' 7. 1°30' 8. 8° 30' 9. 2° 15' 10. 21° 30' 16. 75° 30' 17. 54° 10'. 18. 20° 30'. 19. 81° 45'. 20. 88° 25'. 24. 21.5 sin 42° 10' 37 — fe^a|C!M?^^^^i^^ite^^:#^**^=a^S^^S*^5^^^*=%a=ss*^i^5S,fJ,S&^^SS;^^^^!#|| NEW YORK J^^^») 1 Jfe=^ll^S.^^WKf= 1 ■ ■■"t - - ■ 25. 26. A. =32°. c = 65 Find a ■A. = 70° 30'. a = 15.4. Findc. Fig. 83. 28. Holes A and C are to be drilled on the milling machine. After drilling C, in order to drill A, how much movement of the table will there be in each direction? The table moves from C to B, then from B to A. BC = 5 X cos 20°. BA = 5 X sin 20°. Fig. 32 27. A disk is 21 inches in diameter. Find the distance necessary to set a pair of dividers in order to space off a), 7 sides; 6), 8 sides; e), 10 sides; d), 13 sides. The angle DAB = 1/7 of 360° = 51° 26» (to the nearest minute)./ ' The angle DAC = 1/2 of 51° 26' = 25° 43'. CD = sine angle DAC. A D CD = AD X sine angle DAC. BD = 2 X CD. Fig. 34. TANGENTS. Fig. 35. With the slide in position for reading sines, scale T gives readings for angles whose tangents are found opposite on scale D. The first significant figure comes in the first decimal place for all values found on the rule. Example 1. Find tan 30°. — 38 — llte^^ca!y^Ba»gaaiMt^ =^^ ■ Tct ,-;s,,«^^^^^ ^ ^^^^^ffi KEUFFEL & ESSER CO, NEW YORK |^^2^^ METHOD I. Opposite 30 on scale T, find 577 on D. Pointing oflf, we have tan 30° = .5770, which is correct to three significant figures, the result correct to four figures being .5774. ^'method 11. With the slide in the usual position showing scales B and C, set 30 on the T scale to the mark in the notch at the left end of the rule and opposite 1 on D read 577 on C. Example 2. Find the value of tan 18° 30' X 175. METHOD 1.— Find tan 18° 30' by Example 1, Method II. Above 175 on D, find 586 on C. Since tan 18° 30' is .3346, the product must be roughly 1/3 of 175, making the result 58.6. ' The scale gives tangents only as far as 45°. For larger angles, use the formula: 1 tan A = . Example. Fmd the tan of 75°. tan (90°— A) tan (90 — 75] 1 tan 15° 1 .2679 = 3.73. e3. Find the value of 565 -=- tan 65'. 5fi5 -r tan. 65° 1 = 565 -^ . tan 25° = 565 X tan 25°. = 263. Example 4. Find the value of 256 -^ tan 10° 30'. . METHOD 1.— To 256 on D set 10° 30' on T. Under the left index on T, find 138 on D. By a rough calculation, remembering tan 45° = 1, 256 256 -— = 172. tan 10° 30' .18 Making 138 look as much as possible like 172, we have 138. METHOD 11.— With the C and D scales uppermost, set 10° 30' on T to the mark in the groove at the left end of the rule. Under 256 on C, find 138 on D. Example 5. Find the value of 256 -^ tan 40° 10'. By Method 11, setting 40° 10' on T to the mark in the groove at the left end of the rule, we find 256 on C comes beyond the left end of the rule. — 39 Set the runner to 10 on C, then set 1 on C to the runner. Now, under 256 on C, find 303 on D. 256 256 Rough] V calculating: — = = 320. tan 40° 10' 8 Hence, the result is 303. The tangent of an angle less than 5° 43' cannot be obtained directly from the ordinary 10 in. rule, but the sine may be used in place of the tangent, since the sine and the tangent of any of these angles are identical to three significant figures. tan 1° 30' = sin 1° 80' = .0262. COTANGENTS. The cotangent may be found as follows: Cot A = tan (90° — A). Example: Find cot 65°. Cot 65° = tan (90° — 65°). = tan 25°. = .466. SECANT AND COSECANT. The secant and cosecant may be found by the formulas: 1 sec A = ■ . esc A cos A. 1 sin A. EXERCISE. Find the tangents of the following angles 1. 25°. 7. 75° 10'. 2. 14° 80'. 8. 20° 10'. 3. 35° 30'. 9. 15° 5'. 4. 26° 20'. 10. 6° 25'. 5. 18° 80'. 11. 1° 45'. 6. 55° 20'. 12. 42° 20'. 13. Tan. 15° X 18 18.2 14. Tan 65° 30' X 13.2 = tan. 24° 30' 5.62 15. tan 10° 8.5 16. = 8.5 X tan 19° 40 tan 70° 20' Fig. 36. ■40- 17. To find BE, the height of a building, a transit is set up at A, a level line AC is sighted on a rod held atE. CE is found to be 5.2 ft. EF, which is equal to CA, is measured and found to be 138 ft. The angle CAB is taken by the transit and found to be 28° 30'. Find BE, the height of the building. BE = BC + CE. BC = CA X tan A. BE = CA X tan A + CE. = 138 X tan 28° 30' + 5.2. = 74.9 + 5.2. = 80.1 ft. Fig. 37. 18. To find CB, the width of a river. A transit is set up at C and a right angle, BCA is laid off. CA is measured and found to be 235 ft. Then the transit is set up at A and the angle A found to be 75° 30'. Find CB, the width of the river. BC = CA X tan A. = 235 X tan 75° 30' 235 tan 15° 30' = 847 ft. LOGARITHMS. Between the scale of sines and the scale of tangents is a scale of equal parts by which the logarithm of a number may be found. Example 1. Find log 50. With the slide in its usual position having the scale of equal parts under- neath, set the left index of the slide to 5 on D. On the scale of equal parts opposite the right index on the under side of the rule, read log 5 = 699. Placing the decimal point and prefixing the characteristic, as usual in working with logarithms, log 50 = 1.699. (The characteristic is found by taking one less than the number of figures at the left of the decimal point.) — 41 Example 2. Find (2.36)5 = X. Log a; = 5 X log 2.36. = 5 X .373. = 1.865. X = 73.2. Example 3. Findv/l87 = X. Log 187 = 2.272. Log 1^187 = 1/5 of 2.272 = .454. a; = 2.84. EXEKCISE. Find the logarithms of 1. 1.24. 2. 54.5. 3. .312. 4. .067. 5. 735. Find the value of the following to three significant figures 6. (3.2)^ 7. (425)*. 8. ^SA6. 9. ■^286. 10. -v/1430. CHAPTER III. TYPIOAIi PROBLEMS FROM VARIOUS OCCUPATIONS. Secretarial Work. A secretary in checking a traveling man's expense account for one week found items as follows: Raihroad fares $27.50 Hotel bills 56.00 Total 83.50 Find what per cent of the total expense was used in hotel bills. Solution: 56 -^ 83.50 = 67 per cent. ESTIMATING PLASTERING. As a detail in estimating the cost of plastering a house, find the number of square yards in the walls and ceiling of a room 24 ft. long, 16 ft. wide, and 9 ft. high without any allowance for openings. Ceiling, 24 + 16 = 384 by the slide rule. Walls, 2 X (24 X 16) X 9 = 720 Total 1104 EXCAVATING. What will be the cost of excavating rock for a cellar measuring 43 ft X 28 ft. to an average depth of 6.5 ft. at $2.50 per cubic yard? — 42 — ^^jC^ff^^^aifl^'^ -r—t.^.r^irm .1 TB-T . ^ , ^^H |^^^|M^^^gg[ KEUFFEL & ESSER CO.. NEW VORK 43 X 28 X 6.5 X 2.5 27 = $725. Correct to the nearest dollar. BUSINESS PER CENT OF PROFIT. A merchant purchased a bill of goods for $318 and sold the same for $360. Find the per cent of profit reckoned,' a. On the cost. 6. On the selling price. Solution: Profit = $360 — $318 = $42. 42 Per cent of profit reckoned on the cost = = 13.2 per cent. 318 42 Per cent of profit reckoned on the selling price = = 11.7 per cent. 360 DISCOUNT. Goods marked $7.25 are sold at a discount of 353^ per cent. Find the net price. Solution: The net price is 100 per cent — 35 J^ per cent or 64 J/g per cent. .645 X 7.25 = 4.68. COMPOUND INTEREST. How many years will it take a sum of money to double itself if deposited in a savings bank paying 4 per cent interest, compounded semi-annually. Using the formula A = P (1 + r)u , where A is the amount, P the principal, r the interest of $1. for 6 months, and n the number of half years, if we take $1. as P, we have: n 2 = (1+.02). log 2 and n = . log 1.02 .301 On the .0086 = 35 half years, or 17 J/^ years. NOTE. — It is advisable to use a 20-inch rule for this problem. 10-inch rule the result can only be found very roughly. PHYSICS. In a photometer a 16 c. p. lamp is used as a standard. The following distance readings are obtained in testing a nitrogen filled lamp. 317 mm. 304 mm. 322 mm. 248 mm. 683 mm. 696 mm. 678 mm. 702 mm. By experiment 1 2 3 — 43 — ^Si^^:p^;^^^S^3SfQ^'^!^J^='f-i^^^^-^^-^^*-!- ^'ea^J--^^-Jt;^tg^.fi^;^:. Allowing J inch for weld- ing, we have 16.90 + .25 = 17.15 inches. — 45 — %^B»afg^iB^8a»<'%^^Ji-M^!J«tJ-^^^-^'" ^Jilj^=!--!Jg»g^gg?V^feg&S^PgBSg^a^^^ ll ^ra^Kfc KEUFFEL & ESSER CO.. NEW YORK' l^^^^}3jli..aa a ^Tgg^ a bs» a ^^Hl KEUFFEL a ESSER CO.. NEW YORK 49 = Pounds per cubic foot. 785 = Kilogs per cubic meter. 27 = Pounds per cubic yard. 16 = Kilogs per cubic meter. 89 = Cubic feet per minute. 42 = Liters per second. 700 = Imperial gallons per minute. 53 = Liters per second. 840 = U. S. gallons per minute. 53 = Liters per second. 38 = Weight of fresh water. 39 = Weight of sea water. 5 = Cubic feet of water. 312 = Weight in pounds. 1 = Imperial gallons of water. 10 = Weight in pounds. 3 = U. S. gallons of water. 25 = Weight in pounds. 50 = Pounds per U. S. gallon. 6 = Kilogs per liter. 10 = Pounds per Imperial gallon. 1 = Kilogs per liter. 30 = Pounds per U. S. gallon. 25 = Pounds per Imperial gallon. 3 = Cubic feet of water. 85 = Weight in kilogs. 46 = Imperial gallons of water. 209 = Weight in kilogs. 14 = U. S. gallons of water. 53 = Weight in kilogs. — 53 — %ii^^T/'^^»tf*-«5ft?¥**^^ ^ — ^ " — ' '"^f^ ^^ j^^|^^^^^^S>§| KEUFFEL & ESSER CO., NEW YORK ISC 44 = Feet per second. 30 = Miles per hour. 88 = Yards per minute. 3 = Miles per hour. 41 = Feet per second. 750 = Meters per minute 82 = Feet per minute. 25 = Meters per minute. 340 = Footpounds. 47 = Kilogrammeters. 72 = British horse power. 73 = French horse power. 3700 = One cubic foot of water per minute under one foot of head. 7 = British horse, power. 75 = One liter of water per second under one meter of head. 1 = French horse power. In no case does the departure, in these equivalents, from the exact ratio attain one per thousand. EXAMPLES. What is the pressure in pounds per square inch equivalent to a head of 34 feet of water? C D Set 60 To 26 Under 34 Find 14.75 pounds — Answer. What head of water. In feet, is equivalent to a pressure of 18 pounds per square inch? C D Set 26 To 60 Under 18 Find 41.5 feet — Answer. How many horse power will 50 cubic feet of water per minute give under a head of 400 feet? C D Set 3700 To 7 Runner to 4001 1 to R Under 50 Find 37.8 H. P.— Answer. 54 — METHODS OF WORKING OTJT MECHAlS^ICAIi AND OTHEB FORMUIiAS. Diameters and Areas of Circles. A = .7854 DK A To 205 A Toll B Set 161 C Find Areas. B or C Set 6 Find Areas in square feet. D Above Diameters D Above Diameter in inches. 2. To Calculate Selling Prices of Goods, with percentage of profit on Cost Price. C D Set 100 To 100 plus percentage of profit Below cost price. Find selling price. 3. To Calculate Selling Prices of Goods, with percentages of profit on Selling Price. c Set 100 less percentage of profit Below cost price D To 100 Find selling price Example: If goods cost 45 cents a yard, at what price must they be sold to realize 15 per cent profit on the selling price? C D Set 85 (=100—15) Below 45 D To 100 4. To find the Area of a Ring. A = Set sum of the two diameters Find area. Find 53 — Answer (D + d) X (D — d) 1.2732 To 1.273 Above difference of the two diameters 5. Compound Interest. Set the left index of C, to 100 plus the rate of interest, on D, then take the corresponding number on the scale of Equal Parts, and multiply it by the number of years. Set this product on the scale of E. P. to the index on the under side of the Rule, then on D will be found the amount of any coinciding sum on C for the given years at the given rate. Example: Find the amount of $150 at 5 per cent at the end of 10 years. C D Set 100 To 105 E. P. =21.2X10 =212 Under side of Rule and 212 to 1 Slide D Below 150 Find $244.35— Ans. We thus obtain on D, below 1 on C, a gauge-point for 10 years at 5 per cent, and can obtain in like manner similar ones for any other number of years and rate of interest. — 5B — te^^ca7^BjE^<5B*^fe^=^*s**=-=^=^T-. ^=**=%«=^s«,^^rf|ft&^ ^M i^^^i^^te^SgJ KEUFFEL & ESSER CO., NEW YORK li^^^ [^^@^^^WS5;f~. ,,;S;^„„_;__S^ -—^S^r^—^iS^'^^ D 6. Levers. Set distance from fulnrum to power or weight applied To distance from fulcrum to power or weight transmitted Find power or weight transmitted. Above power or weight applied. 7. Diameter of Pulleys or Teeth of Wheels Inv. or, C D Set diameter or teeth of driving To revolutions of driving Set diameter or teeth of driving Diameter or teeth of driven Revolutions of driven Revolutions of driven To diameter or teeth of driven Revolutions of driving 8. Diameter of two Wheels to work at given Velocities. Set distance between their centers Find diameter. To half sum of their revolutions Above revolutions of each 9. Strength of Teeth of Wheels. P=- VH 0.6V A To H. P. to be transmitted B C Set gauge point 0.6 Rtol Velocity in ft. per second to R Under 1 D Pitch in inches D X x 10. Diameter and Pitch of Wheels. N=- C II Set 226 R to 1 Pitch to R Under diameter of pitch circle D 1 To 710 Find number of teeth 11. Strength of Wrought Iron Sliafting, \ N for crank shafts and prime movers. 165 H for ordinary shafting A \ N To 83 or 65 B C Set revolutions per min. R to Ind. H. P. Number coinciding with R = diameter Under 1 D Same coinciding number = diameter 56 — NOTE. — In this, as in other cases, the coefficients (83 and 65) may be altered to suit individual opinions, without in any way altering the methods of solution. 12. To find the Change Wheel in a Screw-Cutting Lathe. sxw n=tmxp w^n Mxp T = S where N = Number of threads per inch to be cut. T = M = W = P = S = on traverse screw, teeth in wheel on mandril. " stud wheel (gearing in M). " stud pinion (gearing in S). " wheel on traverse screw. C IISetT RtoP StoR Under M D |toN Find No. of teeth in W or stud wheel 13. Rules for Good Leather Belting, W = 600 or 375 H. P. V ft. per min. Set 600 D I To velocity in feet per min. C II Set 375 Find width in inches D II To velocity in feet per min. Above actual H. P. Find width in inches -for Single Belts Above actual H. P. 14. Best Manila Rope Driving, -for Double Belts A To velocity in feet per min. Find Actual Horse Power. B C Set 307 Above diameter in inches D A To 4 Find Strength in Tons, B C Setl Above diameter in inches D A To 107 Find Working Tension in Pounds. B C Setl Above diameter in inches D A To 0.28 Find Weight per Foot in Pounds. B C Setl Above diameter in inches D — 57 — jj^^^^g^^^g^^la.^i^a-I.^J^^J^ ' " ' '"'"^ as^^t^^j^-jg^^gj^y-i^^t'? ll ESSER CO. NEW YORK rii~i "iTTLj'* 'Ii^i jj^^^a 15. Weight of Iron Bars in Pounds per Foot Length. A Tol Weight of Square Bars. B C Sets Above width of side in inches D A To 55 Weight of Round Bars. B C Set 21 Above diameter in inches D , c D Set 0.3 Breadth in inches Below thickness in inches Weight of Flat Bars. 16. Weight of Iron Plates in Pounds per Square Foot. D Set 32 To 40 Below thickness in thirty-seconds of an inch Find weight in pounds per square foot. 17. Weights of other Metals. C D Setl To weight in iron Below G. P. for other metals Find weight in other metals. G. P Weight. . Gauge-points of other metals, and weight per cubic foot. Cast Steel Cast W. 1. C. I. Steel. Plates. Copper. Brass. Lead. Zinc . 1 .93 1.02 1.04 1.15 1.09 1.47 .92 . 480 450 490 500 550 525 710 440 pounds. Example: What is the weight of a bar of copper, 1 foot long, 3 inches broad and 2 inches thick? Set 0.3 R to 2 inches thick 1 to R Below G. P. 1.15 D To 3 inches broad I Find 23 pounds — Ans. 18. Weight of Cast Iron Pipes. Set .4075 Below Difference of inside and outside diameters in inches D To Sum of inside and outside diameters in inches Find weight in pounds per lineal foot. G. P. for other metals Brass. Copper. Lead. W.Iron .355 .333 .259 .38 ■58 19. Safe Load on Chains. A Safe load in tons. B C Set 36 Above 1 D To diameter in sixteenths of an inch 20. Gravity. Set 1 D To seconds Below 32.2 Velocity in feet per second. A Space fallen through in feet Set 1 Under 8 D Velocity in feet per second. A Space fallen through in ft. B C Setl Above 16.1 D To seconds 21. Oscillations of Pendulums. A B C Set length pendulum in in. Below^l D To 875 VNumber oscillations per minute. 22. Comparison of Thermometers. C Set 5 Degrees Centigrade. D To 9 Degrees+32= Fahrenheit. C Set 4 Degrees Reaumur D To 9 Degrees + 32 = Fahrenheit. C D Set 4 I Degrees Reaumur To 5 I Degrees Centigrade. — 59 — \^l^m^^^^ ^B. ^^^^s KEUFFEL a ESSER CO.. NEW YORK 1^^^^ — >^^,. -sj ss-*— .-2i~^-n;>2^^ 23. Force of Wind. B C Set 10 Find pressure in pounds per square foot D To 66 Velocity in Feet per second. A B C Set 10 Rnd pressure in pounds per square foot D To 45 Velocity in Miles per hour. 24. Discharge from Pumps. A Gallons delivered per stroke. B C Set 294 Stroke in inches D To diameter in inches 25. Diameter of Single-acting Pumps. A To 294 B C Set length of stroke in inches. R to gallons to be delivered per min. No. strokes per min. to R Below 1 D Diam. pump in inches. D 26. Horse Power required for Pumps. Set G. P. To cubic feet or gallons to be raised per minute Height in feet to which the water is to be raised. Horse power required. Gauge Points with different percentages of allowance. Percent None 10 20 30 40 50 60 70 80 For Gallons Imp. 3300 3000 2750 2540 2360 2200 2060 1940 1835 " C Feet 528 480 440 406 377 352 330 311 294 " U S Gallons 3960 3600 3300 3050 2830 2640 2470 2330 2200 A Head in feet B C Setl Under 8 D Velocity in feet per second. 60 1 ^g^^^^^^g^a^^^ ^^^^^fe^ff KEUFFEL a ESSER CO.. NEW YORK l^^^^^^^^^J) 28. Theoretical Discharge from an Orifice 1 inch Square. A B C Set 1 Under head in feet D To G. P. 3.34 Discharge in cubic feet per minute. If the hole is round and one inch dia. ' the G. P. is 2.62. 29. Real Discharge from Orifice in a Tank 1 inch Square. Setl Under head in feet. To 2.1 G. P. Discharge in cubic feet per minute with coefficient .63. A B C D Gauge Points for other coefficients. Coefficient 60 .66.69 .72 .75 .78 .81 .84 If the hole is round and 1 inch diam., the G. P. is 1.65. .87 .90 .93 .96 G. P. Square. . " Round. . 2. 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3. 3.1 3.2 .1.57 1.73 1.80 1.88 1.96 2.04 2.12 2.20 2.28 2.36 2.44 2.52 30. Discharge from Pipes when renl Telocity is known. A C B D Discharge in cubic feet per min. Inverted Velocity in feet per second Above 1.75 Diameter in inches 31. Delivery of Water from Pipes. .71^: D' H Set 1 D I To diameter in inches EPX5=X itol Under side of Rule. W=4 Eytelwein's Rule. {Con. to next line.) A ToZ B C Rtol Length in feet to R Under head in feet Setl Under 4.71 D FindZ Cu.ft.per min. This is worked out similarly to Formula 5, which is explained in full. — 61 — ^^ ^^S^ml iiTrjiL'j^^ii'^i'zisrjiS Inverted D 32. Gauging Water with a Weir. Depth in inches Depth in inches Under 4.3 Discharge in cubic feet per minute from each foot width of sill. 33. Discharge of a Turbine. Vhxv 0.3 = D Inverted A C B D Head in feet Under 0.3 Square inches water vented Cubic feet discharged per minute, 34. Revolutions of a Turbine. A To head in feet B C Set diameter in inches R to 1840 Ito R Under rate of peripheral velocity D Find revolutions per min. 35. Horse Power of a Turbine. c Set 530 R to discharge per c. ft. per min. ItoR Percentage useful effect D Head in ft. Horse power. or, A Under head in ft. B C Setl Rto head in ft. 158 to R Rto vent in sq. in. to R Under useful effect D Horse power 36. Horse Power of a Steam Engine, D Set 21,000 To diam. in inches R to diam. 1 toR Rto stroke in ft. 1 toR R to rev. per min 1 toR Mean pressure per sq. inch Horse power. — 62 — ^^B ^^^^^^^^^^§1 KEUFFEL a ESSER CO.. NEW YORK A H. P. B C Set 21,000 R to stroke in ft. 1 toR Rto revolutions ItoR Mean pressure D To diam. in inches 37. Dynamometer to Estimate the H. P. indicated by- Set 5252 R to length of lever in feet from center of shaft lUnderrev. of 1 to R|shaftperniin. D weight applied at end of lever in pounds, including weight of scale. actual horse power. 63 1 9'i^t?a?aj&^'^'Wf' ■ — r -iTT^^FIFF TrT, ■^^-, ,,p~ — , ,-- -^ -\ j^^^^ KHUPP.L .SS.R CO.. -Jg^^j^^^^ SUPPLEMENT THE SLIDE RULE PLANE TRIGOI^^OMETRY. J. M. WILLiARD Department of Mathematics School of Liiberal Artti State College of PennaylTanlo The Slide Rule is commonly used by engineers to solve the numerical problem a : b = c : X but it does not seem to be generally known that practically every problem of plane trigonometry, including the more difficult solutions of oblique triangles, can be made viiih one setting of the ordinary Slide Rule. It is proposed here to take up the different cases usually treated in plane trigonometry, giving an illustration of each. 1. Given degrees, to find the corresponding radian measure of an angle, or the reverse. A on X Read any radians B C Set 180 Read any degrees D 64 — arc sec (V) A To 10 Under 7 S T Set 1 (sin 90°) Read (90°— answer) 45°.5 D PLANE RIGHT TRIANGLES. 3. Given two legs. Using the usual notation A,B, C, for angles; a, h, c, for sides opposite Let a = 0.0227, 6 = 0.084, C = 90° A S T Set 1 (tan 45°) Read A D Over 84 Over 227 A = 15° 7' B = 90° — (15° 7') = 74° 53'. To find e A Under 227 Read answer c = 0.087 S T Set 15° 7' Over 1 (sin 90°) D follows; If either angle is less than 6° the sine may be used for the tangent as 4. Let a = 64.4 h = 5.99. To 64.4 Under 5.99 Set 1 (sin 90°) Read answer B =5° 19' B = 5° 19' A = 90° — (5° 19') = 84° 41' — 65 ^M ^ ^^^^^^^^3p\ KEUFFEL & ESSER CO.. NEW YORK 5. Given a leg and the hypotenuse. Let c = 7.46 6 = 0.647 A Under 746 Under 647 Read a S T Set 1 (sin 90°) ReadB Over 85° 1' D 5=4° 59' A = 90°— (4° 59') = 85° 1' a = 7.43 6. If a only is required in the last example it may be better to use the formula a =v' (c+ 6) (c — 6)" A To 8.107 B C Set'l Under difference 6.813 D Read answer a = 7.43 7. When 6 is nearly equal to c, we can find A more accurately by the formula A c—b Sin' — = 2 2e Let 6 = 7.4169 c = 7.4451 A Toe— 6=0.0282 Index to 2.295 B C Set 2c = 14.8902 Read answer 4 = 2° 29' D Read 2.295 8. Given a leg and an acute angle.- Let 6 = 0.084 A = 15° 9' whence B = 74° 51' — 66 A To 0.084 Read c= 0.087 Read a =0.0227 S T Set 74° 51' Over 1 (sin 90°) Over 15° 9' D If only a is desired we read more accurately A D Set 1 (tan 45°) Over 0.084 Under 15° 9' Read a = 0.0227 9. Given the hypotenuse and an acute angle. Let c = 426 A = 34° 15' whence B = 55° 45' A On 426 Read a = 240 Read 6 = 353 S T Set 1 (sin 90°) Over 34° 15' Over 55° 45' D 10. Latitude and departure of plane surveying is a special case of right triangles: A On "distance" Read "departure" Read "latitude" S T Set 1 (sin 90°) Over "course" Over (90°— course) D 11. Projecting a line (or stress) on another line. A On length Read projection S T Set 1 (sin 90°) Over (90°— angle) D • — 67 — ■ In^^E KEUFFEL & ESSER CO. NEW YORK a^^^s 12. Let a force of 5 lbs. act at an angle of 59° to another line. Deter- mine its resolute. A On 5 Read answer = 2.58 S T Setl (sin 90°) Over 31° D PLANE OBLIQUE TRIANGLES. 13. Given a side and two adjacent angles. Using the formula a : Sin A = b : Sin B — e : sin C. Let a = 25, B = 50°, C = 60°; whence A = 70° A On 25 Read b =20.6 Read c =23.2 S T Set 70° Over 50° Over 60° D 14. Given two sides and an angle opposite one of them (case of double solution). The minutes of arc cannot be read correctly to the unit with the slide rule. Let A = 48° 34', a = 46.24, 6 = 60.02 A To 46.24 Under 60.02 Read c = 50.36 Read r^ =29.08 S T Set 48° 34' Reads B2=180°— (76°42') Over 54° 44' B,=76°42',Ci=54°44' 0^ er 28°8' B2 = 103° 18' C2 =28° 8' D ' 15. Sometimes in the double solution case we have only one solution. Let A = 32° 43', a = 988, 6 = 672. Here A is acute and o> 6, hence we have only one solution as follows: A On 988 Under 672 Read c = 1484 S T Set 32° 43' Reads Over 54° 17' D B = 21° 34' C = 180° —(A+B) = 180° — (54° 17') — 68 — i-gi-,Tj--^E^rt-c t^;^j___:reiLt,^^.,_rFE — nrr ttt 9 ^^ A^^^^M^^^gil KEUFFEL & ESSER CO.. NEW YORK 16. Given two sides and the included angle. Use the formula Tan B— C ctn _2 b + c b — c - The fact that the tangent scale runs only to 45° makes two cases. Case I. -^ greater than 45°- Let 6 =6.24, c =2.35, A =110° 32', whence -4" = 55° 16'. ^^^ =90—4' ^ dt ^ 34° 44', 6 + c = 8.59. 6 — c = 3.89. A S T Set 34° 44' Read ^7^- 17° 26' D On 8.59 Over 3.89 Adding and subtracting — ^ — , — ^ — B = 52° 10' C = 17° 18' Case II. -g- less than 45°. A S T Set 1 (tan 45°) Runner to y Read ^7^ D On 6 +c Index to runner Over 6 — c Let 6 =226.2, c =138.7, A =59° 13', whence 6 +c= 364.9, 6— c =87.5, ^ = 29° 36', ^?-i^ = 60° 24' A S T Set 1 (tan 45°) Runner to 29° 36' Read ^7^ = 22° 54' D On 364.9 Index to runner Over 87.5 Adding and subtracting B =83° 18' C =37° 30' B+C B—C — 69 17. Given the three sides. The common solution requires first the finding of the radius of the in- scribed circle, by the formula r' = (s — g) (s — 6) (s — c) a +b + e and then the angles, by the formulas A r tan r . B r . , tan -H- = r> etc. s — a 2 s — 6 This is the only problem of plane triangles which requires several settings of the slide rule for its solution. But after r is found one setting gives any angle, as follows; Let 0=260, 6=280, c = 300, whence s=420, s— a = 160, s— 6 = 140, s— c = 120. A On 160 1 to runner B C Set 420 Runner to 140 Under 120 D Read r = 7.99 Invert the slide. A S T Setl Read ^ = 26° 34' D On 160 Over r ( = 7.99) A = 53° 08' This completes the solution of all cases of plane triangles. ANSWERS. CHAPTER I. (Answers given with the problems are not given below.) 16. 300. 17. 3000. 18. 3000. 19. .3 27. 20. 21. .03 31 32 33 34 35 21 651 672 693 714 735 22 682 704 726 748 770 23 713 736 759 782 805 24 744 768 792 816 840 25 775 800 826 860 876 26 806 832 858 884 910 27 837 864 868 896 891 918 945 28 924 962 980 29 899 928 957 986 1015 30 930 960 990 1020 1060 30. 49% 31. 33% 32. 91% 33. 58% 34. 16% 35. 21% 39. 2.24 40. 2.34 41. 1.33 42. 1.32 43. 3.18 44. 67.3 45. 19.3 46. .0000476 47. 5.77 48. 27.5 49. 87.9 In. 50. 81.9 52. 156. 53. .294 55. .00654 56. .735 57. .615 58. 13.6 59. 77.9 61. 19.6 62. 21.4 63. 33.1 64. 13.9 65. 10.5 66. 56.7 67. 1.6 mils. 72. 10.2 73. 21.6 74. 1.25 75. 74.8 76. 76200. 77. 1170. 78. .436 79. .0039 80. .0000325 81. .000595 82. 5020000. 91. 1.19 92. 3.76 93. 11.9 94. .377 95. 1.56 96. 9.24 97. .604 98. .660 99. 38.2 100. Square roots of numbers from 110 to 130. Number Square Roots 110. 10.5 111. 10.5 112. 10.6 113. 10.6 114. 10.7 116. 10.7 116. 10.8 117. 10.8 118. 10.9 119. 10.9 120. 11.0 71 121. 11.0 122. 110 123. 11.1 124. 11.1 125. 11.2 126. 11.2 127. 11.3 128. 11.3 129. 11.4 130. 11.4 103. 8.5 inches (Use a 9-in. pipe, the nearest standard Answers Page 16. to test problems on 101. 127 feet, 3 inches. 102. 1.9 inches. Use a 2-in. pipe, the nearest standard size. 3.15 1.41 11.4 36.8 13.5 1. 7.39 2. 19.3 3. 7.55 4. 58.5 5. 258. 11. 2197. 12. 2744. 13. 3375. 14. 4096. 15. 4913. 16. 5832. 17. 6859. 18. 8000. 19. 9261. (Th ree significant figures) 20. 10600. 21. 29800. 22. 97300. 23. 104000. 1. 1.44 2. 3.107 3. 6.69 4. .669 5. .3107 6. .144 7. 13.77 8. 3.628 9. .922 10. 35.59 11. 3.68 Multiplication 1 92.4 2. 114.7 3. 17,490,000 . ANSWERS. Chapter II Multiplication. 6. 7. 8. 9. 10. 273. .0541 .00167 .0000910 12.56, 20.4 44.0. Cubes. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. Cube Roots. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 4. 5. 149000. 262000. 436,000,000. 12,500,000. 77,300,000. 679,000. 2,690,000. .04 .000185 .000,000,314 1.09 9.53 76.1 gal. 1.94 .832 6.22 15.66 37.34 .2535 .211 1.012 47.7 20.4 1.309 56.1 — 72 — K ^^-x. - :^rT~ J«^J^=^J^tJ^ifig&<^^giiSg4J^S8gll ^§1 KEUFFEL a ESSER CO.. NEW YORK f i^P^^^i^^^BJ Combined M ultiplication and Division. 1. .01815 8. .1585 2. .633 9. 4.58 3. 902. 10. 1.69 4. 328. 11. .0886 5. nil. 12. .0786 6. 51.4 13. 1.073 7. .353 Miscellaneous Calculations. 1. 32.3 7. 57,300,000, 2. 1.91 8. 1.234 3. .516 9. .408 4. .45 10. .00642 5. 1627. 11. 81.4 6. 35.8 12. 6.4 Sines and Cosines , 1. 1. 16. .250 2. .707 17. .585 3. .5 18. .937 4. .0523 19. .1435 5. .0116 20. .0276 6. .264 21. 19. 7. .0262 22. 15.1 8. .1478 23. 83.2 9. .0393 24. 32.0 10. .3665 25. 34.5 11. .1736 26. 16.3 12. .423 27. 0=9.11, 7) =8.04, c=6.49, 13. .743 d = 5.03. 14 .970 28. BC =4.70, .8^=1.71 15, .978 Tangentt i. 1. .466 9. .270 2. .259 10. .1125 3. .713 11. .0306 4. .495 12. .911 5. .335 13. 4.82 6. 1.446 14. 29.0 7. 3.78 15. 31.9 8. .367 16. 30.4 Logarithms. 1. .093 6. 33.6 2. 1.736 7. 32,600,000,000. 3. 1.494, or 9.494-10. 8. 1.512 4. 2.8260, or 8.8260-10 9. 3.10 5. 2.866 10. 2.82 73 THE POLYPHASE SLIDE RULE. No. 4053-3. The Polyphase Slide Rule is especially devised for the solution of prob- lems containing combinations of three factors and problems involving squares, square roots, cubes, cube roots, and many of the higher powers' and roots. Furthermore, it offers in many cases a more ready and accurate means of per- forming the calculations which are possible with the ordinary Mannheim Slide Rule and it is especially adapted for electrical and hydraulic work. Description. The Polyphase Slide Rule has in addition to the regular scales of the Mannheim, a scale of cubes (K) on the vertical edge of the rule, and an inverted scale (CI) on the face of the slide. The cube or "K" scale consists of three consecutive and identical log- arithmic scales, so placed that the extreme right and left indexes coincide with the right and left indexes of the "D" scale. Each of these three scales is graduated like the "C" and "D" scales and is one-third of their length. The Inverted Scale (CI) consists of a complete logarithmic scale placed in reversed order, that is, prime 1 is at the right index, the scale progressing towards the left. How to use the Manual with a Polyphase Rule. Work all problems in Chapters 1, 2 and 3 according to directions for the use of the Mannheim Rule, with the exception of problems involving Recipro- cals, Product of three factors, Cubes, Cube Roots, and Inverted Slide. Operations in these specific cases are performed more readily by the use of the special scales CI and K of the Polyphase Rule. Use of Scales CI and K. RECIPROCALS. The reciprocal of 'any number on the CI scale will be found directly opposite on the C scale. For examples see page 33. MULTIPLICATION. Two Factors: Set one of the factors on the CI scale to the other factor on the D scale. Under the index on C find the product on D. Example: 25x36=900. Set 25 on CI 'to 36 on D. Under the right index of C find the product, 900, on D. Three Factors: One of the many useful advantages of the Polyphase Rule is that it enables an operator to find the product of three factors with only one setting of the slide. Example: 65x24X125=195,000. Set 65 on CI to 24 on D. Under 125 on C read 195,000 on D. Compare with Example 2 on page 33, and note the advantage of the Poly- phase over the Mannheim Rule. Work the problems on page 34. -74- n.-;;;^jr--^E-^^qi._U?. - ■ r^^f^ 1 i^Ti J-T irjiTt, __i_j ^^H i^i^KI'^^^K^ NEW YORK CUBES AND CUBE ROOTS. Cubes. Since scale D is three times the length of each of the three scales which comprise the K scale, the values of the readings on the K scale are cubes of the coinciding values on the D scale, and conversely, the values on the D scale are cube roots of the coinciding values on the K scale. Therefore, to find the cube of a number, set the runner at the given number on D, when the coinciding number on K will be its cube. Work the problems beginning on page 22, using the K scale and placing the decimal point by a rough calculation. It will be found that the method of using the K scale for cubes and cube roots is more rapid, while the method given on page 22 involving scales A, B, C, and D is more accurate. Hence, even on the Polyphase Rule, when greater accuracy is required it may be worth while to take time to use scales A, B, C, and D. Cube Roots. To find the cube root of a number: When the number lies between 1 and 10, use the left band scale of K. When the number lies between 10 and 100, use the middle scale of K. When the number lies between 100 and 1000, use the right hand scale of K. When the number does not lie between 1 and 1000, follow the method of page 24. Set the mark on the under side of the runner to the given number on K, selecting the proper one of the three scales as directed above. Opposite the runner line on D find the cube root. Example 1. Find the cube root of 8. Set the runner to 8 on the left hand scale of K. Opposite the runner line on D find 2, the cube root. Example 2. Find the cube root of 90. Set the runner to 90 on the middle scale of K. Opposite the runner line on D find 4.48, the cube root. Example 3. Find the cube root of 900. Set the runner to 900 on the right hand scale of K. Opposite the runner line on D find 9.66, the cube root. Work the problem on page 24 beginning with Example 6, using scale K. IIIGnEB POWERS AND ROOTS. The fourth root of a number is obtained by finding the square root of the square root. The sixth root is obtained by finding the square root of the cube root. The eighth root is the square root of the fourth root. Expressions Which May Be Read Directly By Means Of The Indicator, Without Setting The Slide: 1. X = a", set Runner to a on D, read'X on A. 2. X = / fls', set as on B to a on K, over a on D, read X on C. 38. X = B , — , set a on B to OS on K, over a on D, read X on CI. y as' 39. X = >/«'•, set a on CI to a on K, under a on B, read X on D. SETTINGS FOR TWO FACTORS. 40. X = ah, set 1 to o on D, under b on C, read X on D. 41. X = ~r, set as on CI to b on D, over 1 on D, read X on C. a 42. X = ~T", set 6 on C to as on D, under 1 on C, read X on D. 43. X = — , set * on C to a on D, over 1 on D, read X on C. 44. X = as J*, set 1 to a on A, over b on C, read X on A. 45. X = —Ti, set b on CI to a on A, under 1 on A, read X on B. a 46. X := "T^, set S on C to a on A, over 1 on B, read X on A. a* 47. X = "7" > set a on C to S on A, under 1 on A, read X on B. 48. X = aH^, set 1 on C to a on D, over * on C, read X on A. 49. X = ajs, set as on CI to b on D, under 1 on A, read X on B. as* 50. X = "p", set a on C to J on D, under 1 on A, read X on B. 51. X = a l/T7 set 1 to a on D, under b on B, read X on D. 52. X = — -;^ set a on CI to b on A, over 1 on D, read X on C. ay b 53. X = ^ " , set S on C to a on A, under 1 on C, read X on D. a 54. X = , — , set i on B to as on D, under 1 on C, read X on D. y * 55. X ^ d^ |/ J , set as on CI to a on D, under b on B, read X on D. 56. X ^ . — , set S on B to as on D, under a on C, read X on D. V b 57. X = ^ " , set * on C to a on A, under b on CI, read X on D. b' 58. X = ab'. Set 1 on C to as on K, under b on C, read X on K. a 59. X = "Tj", set J on C to d on K, under 1 on C, read X on K. a' 60. X = -p", set a on CI to as on A, over b on CI, read X on A. ■77 — 61. X = "jj". set J on B to a on D, over b on CI, read X on A. = a'b', set S on CI to a on D, over b on B, read X on A. = a f^~b^ , set 1 on C to J on K, under a on C, read X on D. = 1 -f- ffl ^ J .set fls"on CI to b on K, over 1 on D, read X on C. = an- F* , set a on C to i on K, over 1 on D, read X on C. = ■^ a T- J, set i on C to ffi on K, under 1 on C, read X on D. = a'J', set 1 on C to J on D, under a on C, read X on K. = "jT . set 6 on C to a on D, under 1 on C read X on K. = o6*, set b on CI to a on A, over S on C, read X on A. = &• \/a^, set 6 on CI to o on A, under 6 on C, read X on K. 62. X 63. X 64. X 65. X 66. X 67. X 68. X 69. X 70. X 78 INDEX. Accuracy of Slide Rule, Chemistry, . . Cubes Cube Root, Cutting Speed, Decimal Point, Division, Fonnulas, Gearing Gauge Points, Historical Note, Index, which to use, Inverted Slide, Excavating, Discount Compound Interest, . . Cosines Logarithms, Multiplication, two numbers, . . Multiplication, more than two numbers. Multiplication, theory, Multiplication and Division Combined, Patterns, Per Cents, Physics, . . Plastering, Polyphase Slide Rule, Proportion Secant and Cosecant, Significant Figures, Sines, Squares, Square Roots, ... Speeds of Pulleys, Surveying, . .^ Tangents ^d Cotangents, .... Test, Triangles, Solution by Slide Rule, . . . FAOB . 2, 17, 18 . . . .43 . . . .22 . 23, 24, 25 . . . .43 6,11 . . .9,21 . . 46, etc. . . . .44 . . 48, etc. .... 16 . . . . 7 .... 32 . . . .41 . . . .42 . . . .42 . . . .35 . . 19, 40 . . .3—6 . .15,26 . .19 . .10,26 . . . .44 . .7,8,9 . . . .42 . . .41 . . . 73 11, 12, 21 . . . .39 . .17,18 . . . .35 . 4, 13, 21 . 4, 14, 21 . . . .43 . . . .45 37, 38, 39 .... 16 ■ . 64, etc. SLIDE RULES OF ALL KINDS. We are the largest manufacturers of slide rules in America and have the largest assortment. Some of the best known of our slide rules are shown below. MANNHEIM Mannheim Slide Rules can be used to advantage in every line of busi- ness. They have our patent adjustment for obtaining any desired friction of the slide. POLYPHASE The Polyphase Slide Rule is of the Mannheim type, but has an inverted C scale and a scale of cubes in addition to the regular Mannheim scales. This arrangement facilitates the solution of many problems involving three factors, as well as many powers and roots. DUPLEX Duplex Slide Rules are practically two rules in one, with all the scales on the exterior faces, where they may be readily seen and utilized by merely turning the rule over. POLYPHASE DUPLEX i | ii) ijtii|iiiii!i ;l!;i: |ii;| i 'i"'li' i |' AoiluXifni^iurtHillii ' ■^■■" i «| i ii i |»i|»| iiii M n | i | i f i| | iWi i »iilW i i The Polyphase Duplex Slide Rules are a combination of the Polyphase and the Duplex Rules, with the addition of several special scales. K & E STADIA SLIDE RULE r r T T Ji j»n|l.i,|,„i|,!.|.,|,.,^ j;,x,ri,ih , ij , h , iii4i:i^igfii^4i» ^ This form of Stadia Slide Rule is remarkable for its simplicity. By one setting of the slide, the horizontal distance and vertical height can be obtained at once when the stadia rod reading and the angle of elevation or depression of the telescope are known. SLIDE RULES FOR MANY PURPOSES. In addition to our extended line of regular Mannheim and Duplex Slide Rules, we manufacture a variety, of rules for special purposes, of which we name a few below: THE ROYLANCE ELECTRICAL SLIDE RULE This rule is a modification of our regular Slide Bale No. 4035 and can be used for all the calculations made with the ordinary Slide Rule. In addition to the usual scales, it carries a series of scales or gauge marks by means of which the different properties of cop- per wire, such as size, conductivity, weight, etc., may be determined without the use of tables. SURVEYOR'S DUPLEX SLIDE RULE liW-s ti jM'i'j pii H * *i^' ! "^ w ifw^ ^ - liiiiiii 1iit'niP ii 'i i iT'ni i inimrB i ' ii '!T.'i'L lu ' aiitWtf. tWs nIl!|ilTii(iiI::::i;r; •.vi 'i' J ■ ' i 'ii C" f , iB'!ia-fi i iiiftrr i ff-.1-tf'!-i » Wi' The fact that aU astronomical data essential to surveying, such as azimuth, time, latitude, etc., can be ascertained b^ means of the usual tj^e of Transit with vertical circle but without solar attachment, while generally known, is rather seldom utilized in this country. The main reason for this surprising condition is the difficulty of computing, in the field, by spherical trigonometry, the resiilts of observations. The new K &E Surveyor's Slide Enle entirely eliminates this difficulty by reducing the hitherto complicated calculations to mere mechanical operations, thereby rendering the method of field astronomy with the regular Engineer's Ti-ansit extremely simple and practical. MERCHANT'S (CALCULATING) SLIDE RULE 1 ljl !li{iijlj# l i}! i |t| i | i ;i | ! i i;| l ft w 4^ 1 ^ . W■ lll |..M il ^l ' ■^l■ ill |^j^^^^^ ' ^^!"M'^l^3 ' ■3""i ' "!i il*: i::::>iiM}w W ^^S j««^ |l ' W ' |'l ' | i | 'l' [i ll ! 1 f ! ■ ! ! r ! f Especially designed for the merchant, importer, exporter, accountant, manager, mechanic, foreman, etc. By means of it, all manner of problems involving multiplication, division and proportion can be correctly solved without mental strain and in a small frac- tion of the time required to work them out by the usual "figuring". THE CHEMIST'S DUPLEX SLIDE RULE For the rapid solution of problems in Stoichiometry, such as Gravimetric and Volu- metric Analysis, Equivalents, Percentage Composition, Conversion Factors and many others. ' EVERY REQUISITE OF THE ENGINEER FOR THE FIELD or OFFICE K & E Surveying In- struments, Transits, Levels, etc., are used on nearly every important Government, Municipal and Private engineering work. They excel in de- sign, workmanship and accuracy and embody the latest improvements. We carry the largest and most complete assort- ment of Drawing Papers, Tracing Cloths and Papers, Blueprint and Brownprint papers. Write lor General Catalogue Engineer's Expedition TracBit No. 6079A. We manufacture the celebrated K & E Measuring Tapes, Steel, Woven, Flat Wire Tapes, Bandchains, etc. Accurate. Excellent quality. Large assortment. The K & E Ready Reading Tapes represent the latest improvements in the method of numbering tapes, resulting in a saving of time and reducing the possibility of errors in reading. FICHED:_mi 3253. as, x, kk. hM.