Manual / Guide · 1938
K&E's full manual for its Mannheim rules with the 4041 scale set (A, B, C, D; S, L, T on the back of the slide) — the same scale set as the 4058C Beginner's rule. Covers reading, multiplication, division, proportion, squares and roots, the trig and log scales, and tables of settings and equivalents. 78 pp.
%S 7725 FF MANNHEIM The | : : Slide Rule a A Self Teaching Manual ot j tables of settings, equivalents and gauge points | ae BY BY WILLIAM E. BRECKENRIDGE, A. M. Ya Associate in Mathematics Columbia University New York City PuBLIsHED By : KEUFFEL & ESSER CO. S| NEW YORK, 127 Fulton Street, _ General Office and Factories, HOBOKEN, N. J. CHICAGO ST. Louis DETROIT 516-20 S$. Dearborn St. 817 Locust St. Concourse, Gen’l. Motors Bldg. ‘SAN FRANCISCO LOS ANGELES MONTREAL 30-34 Second St, 730 S. Flower St. 17-9 Notre Dame St., W Drawing Materials, Mathematical and Surveying Instruments, Measuring Tapes. | o PD aa ee ‘ K& E MANNHEIM SLIDE RULE. REG: U.S. PAT: OFF, FEUER A ESEA CO } n 2 " RA ee : — i ’ 2 , — fe el> Lula aa eT ee 3 | om i stata | , a a sub Vs ee $3 . eS eee oS ee natal olo. wp oo gress a= TTT TITTY TTTT TTTTTTTTTT TTT il & | ibtltity Luu filifitltitulventl fei lirieletitieee —— s75—1— — DT es | ge ee FOI - Diagram illustrating the reading of the graduations of the rule. = Se Baath Soh fs th Me Oath a al, dh A Ta De Delhi hah Dh H tu MADE IHUS & Par-1,934.232 No. N4041. 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 in- volves calculations. We manufacture slide rules; and devote to them a separate department of our factory, which is thorougbly equipped with the most improved special machinery. Several of our improvements are protected by patents, and are, there- fore, not embodied in other Rules. NG Cross section of K & E Mannheim Slide Rule showing slide adjustment. + Mannheim Slide Rule No. 4031. 4035. A041. 4045. 4051. 4055. 4056. 4058. ribo iRnON. SAMARAS inna, a sts. Magnifier for K& E Mannheim Slide Rule. Note: We manufacture a complete line of Slide Rules for all uses, and publish a separate book of instructions for each type. Write for complete information on our Polyphase, Polyphase Du- plex, Log Log Duplex, L L Trig Duplex, Mer- chants, and other special types of Slide Rules. The MANNHEIM Slide Rule 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 i PUBLISHED By KEUFFEL & ESSER CO. NEW YORK, 127 Fulton Street, General Office and Factories, HOBOKEN. N. J. Zp CHICAGO . ST.LOUIS DETROIT re 516-20 $. Dearborn St. 617 Locust St, Concourse, Gen'l. Motors Bldg SAN FRANCISCO LOS ANGELES MONTREAL 30-34 Second St. 130 S. Flower St. 7-9 Notre Oame St., W Drawing Materials, Mathematical and Surveying Instruments, Measuring Tapes. COPYRIGHT, 1938, BY KEUFFEI, & ESSER CO. PRINTED IN U, S, A. peer or aT. 934 > wrk wae Lie t duabats Llu dadretstsnaboac He Bip ethane eet Fetes tit Lstiht iow 1isSustan”, «8 tt hr “3 ptt i ala an a A rama Hoatanlittanttadialtuduult taal Be J latin! Laid cre ee ee ae eS . THE MANNHEIM SLIDE RULE PREFACE. This manual is designed to meet the needs of al] who desire to learn the use of this slide rule. Chapter 1, through the use of numerous cuts and examples simply explained, is self-teaching. Some persons will learn all that they require from a few lessons in this chapter. It is suggested that everyone fearning to use the slide rule begin by working the problems in Chapter I. In Chapters II, III, IV, and V, a simple explanation of the theory of the slide rule is followed by the advanced subjects of Cubes, Cube Root, Sines, Cosines, Tangents, Logarithms, and the Solution of Triangles. Special work for technical men and typical problems from various occupa- tions are presented in Chapters VI, VIJ, and VIII. WHO SHOULD USE THE SLIDE RULE? I. Teachers in the following types of schools: 1. Elementary Schools in the higher grades. 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. Most colleges have already made the slide rule a part of the trigonometry course. 5. Evening schools; since no subject holds the students so well as the teaching of the use of the slide rule. 6. Engineering and Trade Schools find the rule indispensable. 1]. Engineers, Mechanics, Chemists, and Architects who have long understood its value. Ill. Private Secretaries to check reports by the slide rule in a small fraction of the time required by ordinary calculation. Ivy. Estimators, Accountants and Surveyors to 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.” a For instance, rapid calculation is made possible in the following everyday problems of office and shop: estimating; discounts; simple and compound interest; the conversion of feet into meters, pounds into kilograms and foreign money into U.S. money; the taking of a series of discounts from list prices; and adding profits to costs. Dozens of equivalents are instantly found, such as cubic inches or feet in gallons, and vice versa; centimeters 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. How much education is necessary? Anyone who has a knowledge of decimal fractions can learn to use the slide rule. How much time will it take? The simplest operations may be learned in a few minutes, but it is recom- mended that at least the problems in Chapter 1 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 Rule ? The accuracy of the slide rule is about proportional to the unit length of the scales used. The 10 inch scale gives results correct to within about 1 part in 1000, or one tenth of one per cent. The 20 inch scale gives results correct to within one part in about 2000. The Thacher Cylindrical slide rule gives an accuracy of about 1 part in 10000. 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 problems for practice as seem necessary to obtain accuracy and speed. For educational use, Chapter II furnishes the necessary theory and history af the rule, while Chapter I provides additional examples for practice. Chapters Ill, [V, and V may be used for advanced work. Foard ot OE em — 5. —, CHAPTER I ESSENTIALS OF THE SLIDE RULE SIMPLY EXPLAINED ; 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. On scales C and D if 1 at extreme left is taken as unity, then 1 at the extreme right of these scales is 10. ’ On scales A and Bif 1 at the extreme left is taken as unity, then 1 in the middle of the scale is 10 and 1 at the extreme right is 100. In order that you may see how the rule is used on simple problems where you know the answers, let us take the following: Example: 2 X 3. (See Fig. 1) Opposite 2 onscale Dset 1 onscale C. Then move the indicator or glass runner so that the hair line is over 3 on scale C. Directly below this 3 you will find 6, the answer. Indicator A. Tt fh i aula a 2 a ‘ - 6 I" P18 f ‘. | Dy . Keurree & ES: ee co a ew Fig. 1. 23 =6,0r6 +3 = 2. Example: 2 > 4. (See Fig. 2) @ Lit Tl Re ' Neuevexin radi C2 NY "Ti Fig.2. 2X4=8o0r8+4 =2. —_lL— Example: 3X3 (See Fig. 3) _| ‘ 3 1 al Ra a i‘ peel greeny i ; i -" " 1 a 2 is at 7 fe. 1, -_ | Jette fontaul I" Pu 1 — Fig. 8. 3X3 =9,0r9+3 =3. —6— Example: 6 +3. (See Fig. 1) 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 scale D. Example: In the same way find 8+ 4. (See Fig. 2) Example: ae ss « 9+ 3. (See Fig. 3) It will be noted that the cuts shown are not in the same scale. This arrangement is for the purpose of iliustrating various lengths of the rule, SQUARES AND SQUARE ROOTS Example: You will remember that to square a number means to multiply that number by itself; e. g., 3? means 8 X 3 = 9. On the slide rule this is done as follows: set the indicator to 8 on scale D. Above, on scale A, using the face of the rule under the indicator you will find 9, the answer. ‘Fig. 4). 1 | ' , i uit | t i q luton tllaalia UA HARLUMLGLUS UE WL HL q KeurFera etsy {aC N.Y. Lf a as Fig. 4. 3% =9 and 2% = 4. Example: In the same way find 2%. (See Fig. 4) To find square roots simply do the work in the reverse order. To find the square root of 9, find the number which multiplied by itself will give 9. The square root of 9 is indicated thus: /9. Set the indicator 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. 4). Example: Find V4. Set the indicator to4on A. Under the indicator on scale D, find 2, the answer. (Fig. 4). —7— MULTIPLICATION OF TWO OR MORE FIGURES Example: Find the value of 2 X 1.5. Opposite 2 on D set 1 on C. Move the indicator to 1.bonC. This will be between 1 and 2 at the division numbered 5; since the numbered divisions between 1 and 2 on C and D are the tenths. Under the indicator, find 3 on D, (Fig. 5) Fig. 5. 2X15=8. Example: 2 X 1.8. Using Fig. 5, see if you can make it 3.6. Example: 1.5 X 2.5. Opposite 1.5 on Dset 1 onC. Move the indicator to 2.5 on C. Below 2.5, find 3.75, the answer, on D. Note that this answer is halfway between 8.7 and 8.8, which makes it 3.75. (Fig. 6). 4 a 8 A PTE LTTE i q 4 by 2 3 $ 9 a H ailadade sliced li 4] HO Lon OL Ti] ] the DU ED 1 Keurreca Esser C2 NY. Fig. 6. 1.5 X 2.5 = 3,75. HOW TO READ THE SCALES Graduations on the slide rule are not measures of length, but represent figures. On the 8” and 10” slide rules, scales C and D consist of nine prime spaces of unequal length; the first line of each space is numbered, respectively, 1 (calied left index), 2, 3, 4, 5, 6, 7, 8, 9, ; the last line is numbered 1, and is called the right index. The spaces 1-2, 2-3, 3-4, etc., decrease in length, the space from 1 to 2 being the longest; and every succeeding space being shorter than the one preceding it. Each of these prime spaces is divided into ten (secondary) spaces, also decreasing in length, the nine lines between prime 1 and prime 2 being num- bered 1,2, 3, 4, 5, 6, 7, 8, 9, in smaller figures than those of the prime gradu- ations. Space does not permit the numbering of the other secondary lines. Each of the spaces between these secondary lines is again subdivided. Thus, each secondary space between prime 1 and prime 2 is divided into ten (unequal) parts. The secondary spaces between prime 2 and prime 4 are subdivided into five (unequal) spaces, =P The secondary spaces from 4 to the end are subdivided into two (unequal) parts by one line between the two secondary lines. To find a number, always read the first figure to the left on the prime line, the second figure of the number on the secondary line to the right thereof, and the third figure on the subdivision; thus, to read 435 (say four, three, five, not four hundred and thirty-five) find prime 4, secondary 3 and sub. 5. PLACING THE DECIMAL POINT Example: 2 X 15. This is worked on the rule exactly like the above examples, but you can see by looking at the problem that the answer is 30 and not 3. All of these problems are worked like the above. ESE? As far as the slide rule is concerned we multiply 2 by 1.5 1. 20% 15. and get 3. Then we place the decimal point by inspec- 2200 X 15. tion. From arithmetic we remember that in multiplying 3, 20 x 150. decimals we first multiply as though there were no deci- 4. 2X AS. mal points, then point off as many decimal places in the 5. 2X .015, answer as there are total decimal places in the two 6 2x 15. numbers which were multiplied together. Thus, in 7. 02x O15. Problem 7, there are two decimal places in .02 and three in .015. Soin the answer, 30, we must have 2 + 8, or 5 places, making the result .00030. Of course the 0 at the right does not count and the final result is .0003. From the above explanation it is evident that the decimal point is not con- sidered in operating the slide rule. After the work of the rule has been done, the decimal point can usually be placed by inspection; i. e. through a mental survey of the influence of the involved factors upon the result. Where this is not feasable, a rough arithmetical calculation will serve to properly locate the decimal point. Fig. 7. 2.3 X 8.2 = 7.36. Example: 2.3 X 3.2. Opposite 2.30n Dset1onC. Move the indicator to3.20nC. Under the hair line on D find 736. That the unit figure is 6 is further confirmed by observing that the product of the unit figures 3 and 2 in the example is 6. Since 2.3 X 3.2 is roughly 2 X 3, or 6, place the decimal point in 736 so that the result will be as near 6 as possible. Evidently the answer is 7.36. (Fig. 7) BUA SS ER i uj t ( “or Keurreva £sser C2 N.Y. Fig. 8. 18 X34 = 61.2 ‘ Example: 18 x 3.4. Using the same method as in the previous example, the slide rule gives 612. By a rough calculation the problem is about equal to 20 x 3 = 60. “Hence we make 612 look like 60 by placing the decimal point after the 1. The answer is 61.2. Example: 16 xX 2.4. Answer 38.4. Example: 1.4 X 2.6. Answer 3.64. Problem 8. Fillin the blanks in the following multiplication table, using the slide rule; 21 22 23 24 25 26 27 28 29 81 32 33 84 Set left index of C to3l on D. Note that the factors 21 to 29 can be taken without resetting the slide. WHICH INDEX TO USE If we attempt to multiply 30 by 45, using the preceding methods of setting the 1 on the left hand end of C to 30 on D, we shall find it impossible to move the indicator to 45, since 45 on scale C lies beyond the right hand end of seale D. In such a case, begin the work on the rule by setting the 1 on the right hand end of C to 30 onseale D, It is then possible to set the indicator to 45 on C. Opposite the 45 on C find 135 on D. Placing the decimal point by inspection, the result is 1350. We will now define the left hand 1 on scale C as the left index and the right hand 1 on seale C as the right index. In most examples, the following rule will be found useful in determining which index to use: Lf the product of the first figures of the given numbers is less than 10, use the left index, if this product ts greater than 10, use the right index. Example 1. 2.13 x 3.388, 3X2 =6. Use the left index, Example 2. 7.238 X 4.71, 7 4 = 28. Use the right index. Example 8. .181 x 4.6, 1*4=4. Use the left index. i ae [Ge An exception to this rule will be found in such a case as 3.12 X 3.81. According to the rule the left index should be used. It will be found, however, that it is necessary to use the right index. This is due to the fact that while the product of the first figures of the two numbers is less than 10, the product of the complete numbers is greater than 10. In most cases, the use of the above rule will save time, PER CENT Example: 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 indicator to 8 on scale D. Move the slide so as to set 56 onC to the hair line of the indicator. 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. Example: A man earned 35 cents per hour. He learned a new trade which increased his earning power toe 67 cents per hour. What per cent increase did he receive? His increase is 82 cents per hour. The per cent of increase is found by dividing 32 by 33. Set the indicator to 32 on D. Set 35 onC to the indicator, The result cannot be found under the left index Ze. the 1 at the extreme left of scale C, since this projects beyond scale D. So we use the right index of C. Under this index, find 91 on scale D. (Fig. 10). a | y } i i A i i fl | ai f int i Esser C2 N.Y. 6 Cl) Fig. 10, 32 + 85 = 91. In the same way, for practice, try the following, obtaining the result correct to two figures: Problem 9. What per cent of 91 is 45? (Divide 45 by 91) 10. What per cent of 73 is 24? 11. What per cent of 67 is 61? 12, What per cent of 53 is 31? 13. What per cent of 82 is 13? 14. 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- wise.” READING TO THREE FIGURES Suppose you had to get per cents in a problem like the following: Example: A baseball player made 57 hits out of 286 times at bat. What is his percentage? : t a tyre n z 3 ‘ 5 fe ee pluslutuutae y W f} 4 KeurrevaEssea Ce WY. Fig. 11. 57+ 286= 199. | J 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. Hence 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 index of C look for the resulton D. When we read this 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 every space counts one point. The index is close to the ninth of these divisions. Hence the reading is 199. Now we must 57 . 60 1 aa6 8 nearly 300° TE Hence 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 19°/1) per cent or 19%/1. hundredths, or 199 thousandths. Example: If your income is $2,500 per year and you save $451. what per eent do you save? place the decimal point. A rough calculation shows that ! PTLANUNL EOD SPOONING TTLATE Typ TTT Keurrec a Esser C2 N.Y. Fig. 12, 451 + 2,500 = .180. Opposite 451 on Dset 250nC. Under the index find 1800n D. Hence the answer is .180, or 18 per cent. We note that when we look for the 1 in 451 on the rule, we find only two spaces between 45 and 46. Hence 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. (Fig. 12) Example: If your salary is $57.50 per week, and you are given an increase of $12.40, what per cent increase do you receive? Lule RAAT a Pee KEUFFE Fig. 18. 12.4 + 57.5 = .216. Opposite 124 on D set 575 onC. 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. Problem 15. 5.42 + 2.42. 16. 7.85 + 3.14. 17. 6.13 + 4.61. 18. 9.56 + 7.26. 19. 10 + 3.14. For 10, use either the right or ieft index. In the following problems the location of the decimal point is determined by working the problems in round numbers. Problem 20. 16.5 + .245 is approximately 16 +.2 = 80. 21. .00655 + .00034 “ s .0060 + .0003 = 20. 22. .00156 + 32.8 ee ff .0015 + 380 = .00005. 23. .875 + .065 ee ss 36 + .06 = 6. 24, .03885 + .0014 “ “ 038 = 001 = 38. There is another method of placing the decimal! point in division. Work the problem as though both dividend and divisor were integers (7. e., not decimals), pointing off as usual. Move the decimal point to the left as many places as 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 20, 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. MORE THAN THREE FIGURES IN A FACTOR Suppose we have more than three figures, as in the following example: Problem 25. Find the circumference of a wheel 28 inches in diameter. Here we must multiply 28 by 3.1416. But the 10” 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. It is, however, somewhat more convenient to work this problem on the A and B scales, where (w) 3.1416 ig accurately marked. Use A in place of D, and B in place of C. Problem 26. Multiply 26 by 8.149. Call 8.149 equal to 8.15. —13— COMBINED MULTIPLICATION AND DIVISION Example: If bell metal is made 25 parts of copper to 11 parts of tin, find the weight of tin in a bell weighing 402 pounds. 11 x 402 The tin is evidently eleven thirty-sixths of 402, or 36 CO Fig. 14. ux =123, Opposite 11 on D set 836 on C. (Fig. 14) Move the indicator to 402 on C. Opposite 402 on C read 123 on D. To place the decimal foint, ne a rough calculation as follows: The ‘0 of Os 100. So make 123 look as nearly like 100 as possible by placing the point after 3. The answer is 123 pounds of tin. example is roughly equal to 14 X 525 Problem 27. 47 24.5 X 43.4 Problem 28. 3620 Eximple: oes (See Fig. 15) (' i wTytt re HUI i i 1 | ‘ My $ | | AAAAAAAAARAAN ARRARARARR ANAT ARRAAANTN J i] TUTTE { UF ey — aa KEUFFEL & 1.35 6.2 Opposite 1.85 on D, set 6.2 on C. If we try to move the indicator to 316 on C, it is impossible because 316 lies beyond the extremity of D. In such a case proceed as follows: Move the indicator to the right-hand index of C. (See Fig. 16) { ; ce Be me b K reo eucabants Fig. 16. | Fig. 15. —14— Then move the slide, setting the left-hand index of C to the indicator. (Fig. 17.) T By) Kourrece : i ce NY. a Fig’ 17. Now we can move the indicator to 316 on C and under 316 on C read the answer 688 on D. (Fig. 18.) ily i it i i ny i H Sa ale : : Keurrec afssea Ce N.Y. Fig. 18. as ae A rough calculation for the decimal point gives us : * 2. - or .5. Mak- ing 688 look as much as possible like .5, we have .688, ee 2.28 X .0125 . : : . 2X .012 The rough calculation for the decimal point might be —— = .006, The answer is .00654. 7.63 X 2.34 Problem 29. 2.43. Problem 30. a6 x E78 82.5 X 9.3 Problem 31. — 565 82.6 X 22.1 Problem 32. 925 — PROPORTION Example: 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? 100 : 22 =86 : x which means that 100 is to 22 as 86 is to the answer, The work on the rule is as follows: adler Aber @en J’ ee bru ) Terry " rrr 7 mene | fet une vee A Futana Jos bulalundued sLutstidal PRUE BUC UL Fig. 19. 100 : 22=86 : 189. — i Gia Opposite 22 on D, set 100 on C. (Use right index for 100). Opposite 86 on C read the answer, 18.9 on D. An easy method of remembering this is: C D Cc D 100 ; 22 =86 ; 18.9 In placing the decimal point, note that 100 has the same relation to 22 that 86 has to the answer. Since 22 is about one-fifth of 100, we must place the decimal point in 189 so that the answer shall be about one-fifth of 8€. Hence, the answer is 18.9. In the same way solve the following proportions, Problem 33. 24 : 81 =152 : x. 34. 14 : 25=12 : x. 35. 3.71 : 24 =512 : x _Problem 36. 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? 27.2 116.8 Fig. 20. 27.2 : 13.2 = 116.8 : h. Problem 37. 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 sty, of 360°. Example: The effects of wind on a shell are approximately proportional 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. SQUARES Example: Find the area of a square plot of ground measuring 128 yards on a side. turtle th i ulti fT emia ! rr Fatlusltrtulatuctuuls SDUDOHAE INAV STETONTOHA(TI CUPTTOATITAA reagent pret HU a 2 3 + ‘ KeuFFEL& e| Fig. 21 Set the indicator to 128 on D. Directly above on A find the square re- quired, 164. To place the decimal point, make a rough calculation. — 16 — (128)? 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. If greater accuracy is desired, a number may be squared by the use of the longer scales C and D. Example: Find the square of 128. Regard this as an example in multiplication equivalent to: Find 128 x 128. To 128 on D set left index. Opposite 128 on C read 1638 on D. Placing the decimal point by a rough calculation, the result is 16380. Example: Square 652. Set the indicator to 652 on D reading the square 425 on A. Notice that here the arithmetic square would be 425104, but on the slide rule we can get only the first three figures, 425. This, however, is close enough for most practical purposes, such as estimating on contract work. ; To place the decimal point, 652? > 600? = 360000. . < 700? = 490000. since the value is between these limits the result is 425000. Find the squares of the following numbers: Problem 38. 3.2 Problem 42. 276, Problem 46, .0057 39, 4.65 43. 34.2 47. .0244 40. 1.12 44, 66 48. 2240. 41. 8.65 45. .0625 _ Example: Find the area of a circular plot of ground measuring 14.5 feet in diameter. _ Use the formula A = .7854 d?, which means that the area of the circle is equal to .7854 multiplied by the square of the diameter. Set the indicator to 1450n D. The square is found directly above on A, but need not be read. Set the right-hand index of the slide to the indicator. Move the indicator to the constant, .7854 on B, and opposite find the result, 165 sq. ft. on A. ‘This constant, .7854, is so frequently used that it has been marked by a special line on the right-hand half of the A and 2 scales. SQUARE ROOTS ; Example: 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. § 9 Pei q | Rs 1 Ste Co N.Y Fig. 22. «/8 =2.93. Set the indicator to 8 on scale A. Assume that scale runs from 1 to 100, so that 8 is found on the left-hand half of the rule. Now under the hair line on scale D, find 2.88, the square root. Then the result is 2.83 yards. Example: Find 4/3. Set the indicator to 3.on A. Under the hair line find 1.73 on D. —~17— Example: Find +/30. Set the indicator to 30 on A, being careful to notice that 30 is indicated by 3 on the right-hand half of therule. Opposite the indicator on D, find 5.48. 8 __ Fig. 28. +/80 = 5.48. Example: Find +/300. Move the decimal point an even number of places in order to obtain a number that is between 1 and 100. This can be done by moving the point two places to the ieft, giving +/3.00. Find the +/3, 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. Example: Find +/.30. Move the point two places to the right, obtaining 30. Find 4/30 = 5.48. Move the point one place to the left, obtaining .548 for the result. Example: Find \/.03. _ _ Move the decimal point two places to the right, obtaining +/3. Find 3 = 1.78. Move the point one place to the left, obtaining .173. Example: Find +/.003. Move the point four places to the right, obtaining +/30. Find 1/30 = 5.48. Move the point two places to the left, obtaining .0548. Find the square roots of the following numbers: Problem 49. 1.42 Problem 52. .142 Problem 55. .365 50. 14.2 53. 2.43 56. .31416 51. 142 54. 85.4 57. 1450 Problem 58. Make alist of square roots of whole numbers between 110 and 130. Problem 59. Ona baseball field, find the distance from home plate tosecond base, measured in a straight line. (The distance between the bases is 90 feet). Problem 60. Water is conducted into a tank through two lead pipes having diameters of 54 and 134 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 54 and 134 in the decimal form. __ Find -4/(.625)? + (1.75)?. NOTE:—Perform the addition by arithmetic. The slide rule cannot be used to advantage in addition. Problem 61. 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 garry off the sewage? — 18 — TEST PROBLEMS Read carefully the following instructions: a. Copy the test on your paper in the form given below. s, Work the problems straight through, setting down the answers in the column at the extreme right. ¢. Fold these answers underneath the paper. d. Work the problems through again, setting down the answers in the other column. é. Compare the two sets of answers. J. If the answers to any problem do not agree (within one point in the third place), work the problem again. g. The correct results are given on page 71. TEST Answers Answers Second First Credits Time Time Problem 62. 1.28 « 2.46 20 " 63. 84 + 59.5 20 7 58.5 X 15.2 64, ag 20 i 65. 6.25 : 242 59.5 3 x 20 a 66. +/182 20 — 19 — CHAPTER II THEORY OF ‘THE SLIDE RULE HISTORICAL NOTE In 1614 John Napier, of Merchiston, Scotland, first published his ‘Canon of Logarithms.” Napier concisely sets forth his purpose in presenting to the world his system of Logarithms as follows: ‘Seeing there is nothing (right well beloved Students of Mathematics) that is so troublesome to mathematical practice, nor doth more molest and hinder calculators, than the multiplications, divisions, square and cubical extractions of great numbers, which besides the tedious expense of time are for the most part subject to many slippery errors, I began therefore to consider in my mind by what certain and ready art I might remove those hindrances.” Napier builded better than he knew. His invention of logarithms made possible the modern slide rule, the fruition of his early conception of the importance of abbreviating mathematical calculations. In 1620 Gunter invented the straight logarithmic scale, and effected calculation with it by the aid of compasses. In 1630 Wm, Oughtred arranged two Gunter logarithmic scales adapted to slide along each other and kept together by hand. He thus invented the first instrument that could be called a slide rule. In 1675 Newton solved the cubic equation by means of three parallel logarithmic scales, and made the first suggestion toward the use of an indicator. In 1722 Warner used square and cube scales. In 1755 Everard inverted the logarithmic scale and adapted the slide rule to gauging. In 1815 Roget invented the log-log scale. In 1859 Lieutenant Amédée Mannheim, of the French Artillery, invented the present form of the rule that bears his name. In 1881 Edwin Thacher invented the cylindrical form which bears his name. In 1891 Wm. Cox patented the Duplex Slide Rule. The sole rights to this type of rule were then acquired by Keuffel & Esser Co. For a complete history of the Logarithmic Slide Rule, the student is referred to “‘A History of the Logarithmie 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 (a), accuracy of the observed data; (6), accuracy of mathematical constants; (c), accuracy of physical constants; (d), precision of the computation. ACCURACY OF THE ORSERVED DATA The precision of a measurement is evidently limited by the nature of the instrument, and the care taken by the observer. Example 1. If a distance is measured by a scale whose smallest sub- division is a millimeter, 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 135, 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 — 20 — 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 & 7090 “ 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 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 8. .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. 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. Similarly in results derived from calculation, zero counts as a significant figure if it represents a definite value, ¢. g. 25 X 36 = 900. Both zeros in 900 are significant figures. On the other hand, zero is not a significant figure if it does not represent a definite value, but merely serves to place the decimal point. Find the cube of 234. The complete result is 12,812,904. On the slide rule only the first three significant figures can be found, and the result is 12,800,000. Here 128 are significant figures and the five zeros following are not significant, since they do not represent definite values, but merely serve to place the decimal point. As far as calculation on the slide rule can determine, each of these five zeros might be any one of the numbers from 0 to 9. Arithmetical ealeulation shows that they are really, 12,904. ACCURACY OF MATHEMATICAL CONSTANTS A mathematica! constant may be carried to any desired degree of accuracy, e. g., the value of x usually given as 3.14159 has been calculated to 707 decimal places. For ordinary calculations 3.14 or 3} is sufficiently accurate. ACCURACY OF PHYSICAL CONSTANTS Most physical constants are only correct to three significant figures and some only to two figures. —21— €. g., The weight of a cu. ft. of water is 62.5 Ib. The weight of a cu. in. of cast iron is .26 Ib. LIMITS OF ACCURACY Holman’s rule states that 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 are rarely reliable beyond this point, the slide rule reading to three figures gives results sufficiently accurate for most kinds of practical work. PERCENTAGE 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 1 3627.6 ~ 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 close estimate not more than one point away from the correct reading. The error here is less than — i which is less than Hence the 1324 error in this reading is less than one-tenth of one per cent. It is evident that the per cent of error holds throughout the length of the slide rule, since the first significant figure increases from 1 to 10 as spaces decrease. e. g., On the right end of the rule, a result read 998 might be really 999 making an error of 1 in 999 or approximately 00 or 0 of 1%. ait 1000° If greater accuracy is desired, a twenty-inch Tule will give results correct to within one part in two thousand; while the Thacher Cylindrical Rule will give results correct to within one part in ten thousand. LOGARITHMS 10? = 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. Fill out the blanks in the following table: 10 = 10,000 Log 10,000 = 105 = 100,000 Log 100,000 = 10! = 10 Log 10 = 10° = 1 Log 1 = — 23 — LAW OF MULTIPLICATION 10? = 100. 10° = 1,000. 10? 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 scale is considered as divided into 1,000 units, then any number—1, 2, 8, 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, 2isfound 301 units from the left. ae log 3 os ATT, 3 ii “ce 4977 ee se #6 ae log 3 = 954, ] sé iii 954 “f “ss its “ log 10 =1.000, 10“ “* 1000 “* “ “ On the scale the number 8 is placed three times as far from the left i