The complete book of slide rule use

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The Complete Book of Slide Rule Use

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Ira Rirow, currently a consulting engineer for Airborne Instruments Laboratory Division of Cutler-Hammer, Inc., has been a practicing electrical engineer since 1944. After obtaining his bachelor’s and master’s degrees in electrical engineering, he did graduate work at New York University and Polytechnic Institute of Brooklyn. He has taught, lectured, and written extensively in the fields of applied mathematics and engineering and has had considerable experience in the design of automatic control, computing, and radar systems. He is the author of a Dolphin Handbook Original, Capsule Calculus.

THE COMPLETE BOOK OF SLIDE RULE USE

BY

Ira Ritow CONSULTANT, AIRBORNE DIVISION

INSTRUMENTS

LABORATORY

OF CUTLER-HAMMER,

INC.

Doubleday & Company, Inc. Garden City, New York

1963

Portions of this material previously appeared in Electrical Manufacturing (now Electro-Technology) magazine under the title, “Slide Rule Mathematics,” Copyright 1958 by the Gage Publishing Company. It is reproduced here with the kind consent of the Gage Publishing Company (now C-M Technical Publications).

Library of Congress Catalog Card Number 63-8766

Copyright © 1963 by Ira Ritow All Rights Reserved Printed in the United States of America

4

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PREFACE

This book is designed to teach anyone with a knowledge of arithmetic to learn the how and why of slide rule use as quickly as possible. There are practice problems with diagrammed solutions, mathematics review sections, a flexible arrangement of chapters and data on the construction of do-it-yourself slide rules to help you learn. To get the most out of this book: 1. Borrow, buy or make a slide rule in order to follow along the examples in the book. 2. Do each example on your own slide rule and work out enough practice problems in each set of practice problems to develop a feeling of confidence in solving the type of problem covered in each set. (Two correct answers in a row are a good indication of understanding.) 3. Skip over paragraphs marked “Math Facts” that cover familiar material. 4. After finishing Chapter I, go directly to those chapters of greatest interest to you. The table of contents indicates the topics and slide rule scales covered in each chapter and the background reading for each chapter. 5. Study for many short time periods rather than a few long ones.



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CONTENTS

Preface

CHAPTER E The Basic Slide Rule The slide rule Math facts — Exponents Multiplication by exponent addition Development of the slide rule Multiplication on the slide rule Division and reciprocals Repeated and mixed processes Locating the decimal point Negative numbers Math facts — Logarithms Slide rule as a logarithm scale Logarithms by slide rule

II

Antilogarithms Inverted scales Advanced Scales and

Operations Displaced scales Proportions Squares and square roots

PREREQUISITE READING

SLIDE

RULE

SCALES INTRODUCED

L

CI, DI CF, DF, CIF A, B, R1, R2, Sql, Sq2

Cubes and cube roots Math facts — Powers and

roots by logarithms

K

*

Contents PREREQUISITE READING

CHAPTER

SLIDE RULE SCALES INTRODUCED

Use of L scale to find roots and powers

54

Pythagorean scale Til

Trigonometry Calculations

PAGE NUMBER

ds I

Math facts — Trigonometry terms Sine Arc sine Cosine Are cosine

54 59 59

SRE

SSE

Tangent

T

Are tangent

64 68 69 70

71 75

Multiple values of inverse

functions

76

Functions of angles over ninety degrees

77 77

Reciprocal functions Minute and Second conversions

IV

The Log-Log Scales

78

I

79

Introduction to log-log

scales Powers of e

L11, LL2, LL3

Raising a number to a power

79 81 82

The LLO scales

LLO, LLOO,

83

LL000, LLO01, LL02, LLO3, Ln0, Ln1, Ln2, Ln3, Ln—0, Ln—1, Ln—2 Ln-—3

Fractional powers or roots

V

Odd logarithms Exponential equations Extended values Vector Diagram Problems

87

Teh

87 89 90 92

Math facts — Vectors

92

Vector diagrams

92

Vector problems

93

Contents °

CHAPTER VI Hyperbolic Functions Math facts — Hyperbolic functions Hyperbolic sine (sinh)

PREREQUISITE READING EATIOL,

Hyperbolic tangent (tanh) Hyperbolic cosine (cosh)

Hyperbolic functions by exponential formulas vil Complex Functions I, ii, fV; Vi Introduction Raising e to an imaginary power Raising e to a complex power Evaluation of complex functions VIII. _—_—Circular Slide Rules I Appendices A. Summary B. How to Make Your Own Slide Rule C. Answers to Practice Problems Index ,

SLIDE RULE SCALES INTRODUCED

SH1, SH2 TH

~

am bsSee

©

LIST OF ILLUSTRATIONS

. Typical slide rule 2. Slide rule diagram . Conversion chart for obtaining corresponding power of ten for any number . Conversion chart of Figure 3 distorted to obtain straight line . Elements of slide rule multiplication . Scale reading exercises . Slide rule scales used in the examples in this book

po Ae = S(t - >) =) = C—)] ake = BIC yo (-—-) = ——)] Sees = Sle >) -P (==) = (——)] co an = 1S yp -— =) = I i toe Lol = 0) 4 (=) =] pe f= 216) = C=) = ae ee UOC = ==) = (-——)] ae ee 0 es) 4) = (| eee LS 2 U5 (Eee) =] -4X2X 7.5 X 25 = 1500 . (12 + 8) + 75 = 0.02

. {{2 + 30) x 9] + 6} X 5 = 3.0 . Development of logarithm scale

. Logi 50 = 0.699 + 1 = 1.699 (—- —-); antilogi 0.5 = 3.16 (- + +)

pret One 0a ete) = (— 2/6) X 3 = OS (e-) PE ==) =" )] . Equal distances on inverse scales

ee

= BIG

eee

ee ==) ==

1.0 = 00

IC)

(-——)}

ae) Ps

0) en (>)

. Displaced scales . Displaced scales on slide rule

. (0.4/7) X 7.5 = 0.955 [(- - -) + (--)

75 X0AX «= 9.42[¢-=) + E— ers 19 (Ce +) + =>) = . Layout of slide with CIF scale Selon" 1.00, oo. 0

xili

= (—)]

+)

List of Illustrations

xiv = 7.5 -) + C--) =

44

39. Conversion chart for root scales

45

43. R1 & R2 are double length square root scales 44. CI scale square root method

48 50 51 51 53 55 55 57 57 60 61 62 64 65 66 67 68 69 7a 72 73

38. V2 = 1.414 (---) ; (2.5)? = 6.25 (----)

45

40. (4 + V35) X7.5 = 3[¢ +) — C-) + 6-) = 41. Incorrect setup for (4 + 25) X 7.5 = 3 42, (2! + 84) X 45 = 20[(- + -) — C--) + *-") = I) 45. Use of C scale for CI scale 46. Cube roots of 8, 80 and 800 47. Cube roots with A-B scales 48. Log 16 = 0.204 + 1 = 1.204 49. Antilog 1.806 = 64 50. Right triangle 51. Pythagorean scale use 52. Periodic functions 53. Exponential functions 54. Basic right triangle 55. Tangent x = sin z/cosz = a/b 56. Slide rule coverage of sine function 57. Sine scale layout 58. Sin 2° = 0.035 (-—-), sin 30° = 0.5 (- « -) 59. (8 + sin 10°) X sin 8° = 0.905[(- - -) — (--) + ©:-:) = (— 60.3 +6 =0.5[(- - -) — (--) = (—)] [are sin 0.5 = 30°] 61. Slide rule coverage of tangent function 62. Tangent scale layout 63. Tangents of various angles

46 46 47

64. 15 X tan 30° = 8.66 [(- - -) + (---) = (—)] 65. 15 X tan 60° = 26 [(- - -) + (--) = (— )]

74 74

66. Csc 50° = 1.30 ; cot 65° = 0.466 67. Log-log scale arrangement for numbers greater than one

77 80

68. ¢9-977 = 1.08, 1.3 = ¢-262, el5 = 4.48

81

69. (5)*5 = 56

82

70. 1.024 = 2800 [(Ln 1.02) X 400 = Ln 2800]

[(e) ==)

=

(——-)]

83

71. Log-log scale arrangement for numbers less than one

72. e©5 = 0.6065 (-—-); e-9-%% = 0.9656 (- - +); 0.98

=

e-0-0202 (

y)e0.93

=

e-0-0725 (- a

+)

73. (0.8)? = 0.64 [(Ln 0.8) X 2 = Ln (0.64)] [(- - -) + (---) = (J 74, (0.98) = 0.403 [(Ln 0.98) X 45 = Ln 0.403]

(- +) +C--) 75. 0.975 = 1/1.0256

= (—)

84 85

85

86

List of Illustrations

76. ~/10,000 = 10 77. Logarithms to the base 4.5

78. 3¥ = 100;y= 4.19[(Ln 3) X y = Ln 100][(-- -) + (---) = (—)] 79. (0.8)5-*/%9 = 0.96 [(Ln 0.8) X 5.5 + 30 = Ln 0.96]

+

C-2 = C=) =

80. Addition of vectors 81. Development of vector components

82. 83. 84. 85. 86.

3 + tan“1(3/7) = 7.61 Evaluating components Hyperbolic functions Hyperbolic sine scale arrangement Sinh 0.65 = 0.697 ( ); sinh 2.64 = 6.97 (—_); sinh! 4 = 2.1(- - -) 87. [3 + (sinh 2)] X (sinh 0.7) = 0.627

(+) —G-+6--) =)

88. Hyperbolic tangent scale arrangement

89. (2 + 3) X (tanh 0.7) = 0.403 [(- - -) —- (--) + C--:) 90. 2 (cosh 0.7) = [(sinh 0.7) + (tanh 0.7)] X 2 = 2.51

ee Sp = 91. 92. 93. 94. 95.

Representation of e7* Complex term evaluation Representation of sinh (x + jy) Circular slide rule (5 X 1.5) + 1.94 = 3.86

96. Practice slide rules 97. Construction of practice slide rule 98. Parts for practice slide rule 99. Assembly of practice slide rule

= (— )]

Chapter I

THE

BASIC

SLIDE

RULE

Tur Siipe Rue. The slide rule is a computing device. As a computing device, it has such dissimilar relatives as: a pencil and paper, an abacus, a desk calculator and an electronic computer.

The user of a computing device must consider accuracy, speed, cost, and convenience in selecting his weapon. The slide rule is fast, versatile, and cheap. Its accuracy of +4 per cent or better on most problems is adequate for most practical calculations. It is portable and relatively foolproof. The modern slide rule is the result of the cascaded efforts of many men, from John Napier (1550-1617) to Herman Ritow (the author’s father), whose combined ingenuity over a period of over three centuries has resulted in the modern slide rule. BODY

DI

(1) (9) 4

f

p {0.99 5)

(8)

(7)

SLIDE

(6)

(0.99)

INDICATOR INDICATOR Fig. 1. Typical Slide Rule 1

ASSEMBLY

2

The Complete Book of Slide Rule Use

Figure 1 shows a typical slide rule and the names of its parts. The body of the rule is marked with scales usually referred to by letter designation such as “D scale,” “P scale,” “L scale,” ete. The slide also has scales and is free to slide in grooves cut in the body of the slide rule. An indicator assembly slides left and right over the body of the slide rule. The indicator assembly is transparent and has a fine black line (indicator) that runs from top to bottom. Since the purpose of the indicator assembly is only to support the indicator and since the body and slide only serve to support fixed and sliding scales, the illustrative diagrams in this book show only the indicator and the fixed and sliding scales.

pi

(1) (9) (8)

p (0.995)

(7) (0.99)

(6)

(5)

(4)

(0.98)

(3) (0.95)

(2) 0.9)

1) (0.8)

(0.7)

(0.2)

py p

Fig. 2. Slide Rule Diagram

Figure 2 shows a diagrammatic representation of the slide rule of Figure 1. In both Figure 1 and Figure 2 the slide and indicator are set so that the indicator is on 4 on the D scale and 3.33 on the C scale. Marx Facts — Exponrents. Exponents are a shorthand for writing amounts that are formed by multiplying numbers by themselves. For example, 3? means 1 X 3 X 8 or one times three multiplied twice so that a= or

LX Sx= 9

oO

In the above example, 3 is called the main* number and 2 is called *Base number is also used for main number. confusion with the use of base in logarithms.

Main is used here to avoid

The Basic Slide Rule

3

the exponent of 3. 3? can be put into words by saying ‘three raised to the two power.” Similarly, 4=1xX4xX4x4x4xX4 = 1024 In this example, four is raised to the fifth power and five is the exponent. The multiplication by unity (one) in each of the above examples is not necessary here but will be needed later. Because they occur very often, ‘‘to the two power” is often abbreviated to “squared” and “to the three power” is abbreviated to “cubed.” Therefore, 3? = ‘“‘three squared” and 43 = “four cubed.” Numbers that are not whole numbers can be raised to a power. For example,

(1.3)? = 1 X (1.8) X G3) = 1.69 The main use of exponents in slide rule theory is in the multiplication of numbers written as the same main number raised to various powers. For example, 45 X 4 = 46)

= 45 = 1024

As long as the exponents are powers of a common main number (4 in this case), numbers can be multiplied together by adding their exponents. As a final example of exponent addition,

10? X 10¢ = 10°*® = 10° = 1,000,000 Any main number raised to unity (one) power equals the main number itself. Thus W=1XKW=$17 or

1S

=1%15=1.5

and

10' = 1 X 10 = 10.

Exponents can be negative numbers. As might be expected, since raising a main number to a positive power means multiplication, then raising a main number to a negative power means division. Thus,

43=12+4+44+4=1X4+XiXj = 1/64

4

The Complete Book of Slide Rule Use

Four raised to the minus three power can also be interpreted as:

1 4 3=]1+

1

43 Le 43 =

64

Therefore, 4-3 = 1/4%® so that a number raised to a negative exponent is the reciprocal* of the number raised to the same but positive power. Multiplication of numbers with negative and positive exponents can be combined as long as the main number is the same in all cases. Therefore, 7X75

XK 7-8 = 74-8)

= Tl = 7

An example will demonstrate the meaning of raising a number to the zero power. V+ P= eK 4% = 40

but therefore,

V+Vv=16+16=1 4=

1]

Similarly, any number raised to the zero power equals one. Exponents need not be whole numbers. 9-°, for example, can be evaluated by realizing that 9° K 9° = 965+) = 9! = 9 so that 94 is the “square root of 9,” that is, the number which when multiplied by itself equals 9. By trial and error one can find out that 9° = 3 since 3 X 3 = 9. Fractional exponents can be written in either decimal or fractional form so that 9° = 9!/2. Just as the .5 power is called “the square root” of a number, the .8333 or 1/3 power of a number is called ‘‘the cube root.” With few exceptions, there are no direct methods to compute the values of numbers raised to fractional exponents. Such problems can be solved only with the use of special tables of numbers or slide rules. MULTIPLICATION BY Exponent ApprTion. The basic slide rule operates by a scheme that solves multiplication problems by adding distances on a scale. This scheme will now be explained. ; 1 1 *q” and “b” are reciprocals if a = 5 orb =-— {Except zero to the zero power which is indeterminate.

The Basic Slide Rule

5

Suppose it is desired to multiply 12 x 4. (12 and 4 are called “factors.”?) The usual method of multiplication involves repeated addition, such as: 12 _ ftimes

12 12

48 The word “times” for multiplication is a clue to the process of adding one factor into a sum a certain “number of times’’ corresponding to the second factor. Desk calculators operate on the “times”’ principle. The slide rule multiplies by a completely different principle based on exponents. 4 X 12 can be multiplied as follows: 4 SK 12

=

10 © -602) x

194-078)

=

1) © -602+1.078)

=

1900-88)

=

48

In this method each factor is rewritten as 10 raised to some exponential power, the exponents are then added and the resultant power of ten gives the desired answer.

Se

Oaant0 CORRESPONDING POWER OF TEN10s

25200

30

SEO

50) 260m NUMBER

Fic. 3. Conversion

Chart for Obtaining

07080

Corresponding

Any Number

Power

90.

100

of Ten for

6

The Complete Book of Slide Rule Use

DEVELOPMENT OF THE SuIpE Rutz. As a first step toward making a slide rule, a graph of ten raised to various exponential powers is shown in Figure 3. The following steps are necessary to use the graph of Figure 3 to solve 4 X 12: 1. Convert 4 and 12 to exponential powers of ten. This is done by finding points b and d on the graph. Dashed lines ab and cd represent the exponents of ten that correspond to 4 and 12 respectively. 2. Add the exponents. To add the exponents, it is only necessary to draw line segment bd’ as shown so that bd’ has the same length as line segment cd and note the sum, line segment ad’.

TEN OF POWER CORRESPONDING

Or

On2

Ose

HO

Pa)

5

OSe2.0)

50

100

Fig. 4. Conversion Chart of Figure 3 Distorted to Obtain Straight Line

The Basic Slide Rule

7

3. Convert the exponential sum into a number. Line segment ad’ now represents the sum of the exponents of ten that in turn represented the original factors, 4 and 12. Converting distance ad’ to a number by moving point d’ to the right to intersect the curve at point e gives the product desired, 48. The above scheme using Figure 3 has a drawback in that the numbers close to zero are difficult to read. Asa second step toward developing a slide rule, Figure 4 is a redrawn version of Figure 3 with the horizontal scale distorted so that the graph is a straight line. Again line segment ab represents 4, cd represents 12 and cd’ represents the product (4 X 12). Notice, however, that since the graph is a straight line, segments a,b; and cid; could have been added to give ad’; which is read directly as 48. Therefore, the horizontal scale of Figure 4 by itself will permit multiplication by

adding distances! Figure 5 shows the final step in the development of the slide rule. The distorted number scale of Figure 4 has been drawn twice in Figure 5, as scale C and scale D. Distances representing 4 and 12 are added as before to give the product, 48. Notice that it would be very convenient to have scale C movable left and right so that any set of distances could be conveniently added. The C scale is movable in a slide rule. Notice also that the scale from 10 to 100 has the same layout as the scale from 1 to 10. Asa result, y has the same relative position on the C scale as xz and using distance wy instead of wz in the multiplication problem converts the problem to 4 X 1.2 from 4 X 12. The answer to 4 X 1.2 appears at z on scale D; x = 4.8. In order to obtain maximum accuracy in a calculation, slide rules are made with only one “cycle” (ie., 1 — 10) in both C and D scales and decimal places are shifted as needed to make the answer appear on the scale. As a result, the slide rule does not give decimal places.* (How to find where to put the decimal point in a problem answer will be discussed later in this chapter. The first slide rule problems and examples shown are simple enough to make the decimal point location obvious.) *Methods have been devised to keep track of the decimal place during slide rule manipulation, but although ingenious, they are easily remembered wrong and few people use them.

The Complete Book of Slide Rule Use en

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The Basic Slide Rule

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Answers:



a= 720, k = 463,

q

b6=254, 1 = 133,

»b

c=118, m= 280,

o

d= 513, n = 840,

Luh

8

e= 904, 0 = 2112,

y

Z

}

S

g=1015, gq = 2012

Vv

“9 “DIY SOSIOIOX YTB SUIPBVOY

f= 1525, p= 556,

9

{



h =1735,

¢

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7 = 627,

L

7 = 1105,

6

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10

The Complete Book of Slide Rule Use

In solving problems with the slide rule, the user must insert the decimal point in the answer himself. Another problem that the slide rule user faces is reading the scales on the slide rule. This takes more practice than most people will admit. Because of the uneven scale spacing, the smallest division may be 1/5, 1/2, or 1/10 of a major division, depending on the position on the scale. Readings just beyond 1.00 are also troublesome. As exercise, read the pointer settings in Figure 6 as rapidly as possible. The correct answers are shown upside down at the bottom of the figure. As a further practice exercise, borrow a slide rule and, looking only at the answers listed in Figure 6, set the indicator at the stated number on the D scale. Then check your settings with the illustrations in Figure 6. From this point on, it is very important that the reader do the examples and practice problems himself on a bought, borrowed, or self-built slide rule. The best choice is to buy a slide rule. For the benefit of those who build their own practice slide rules by following the instructions in Appendix B, Figure 7 shows the four slide rules that are used for all the examples in this book. The number of the slide rule used is indicated in every example. The scales of the rules of Figure 7 are designed to simplify and speed learning and do not represent the best choice in scale layout or markings. Similarly the problems and examples used are designed to teach the principles of slide rule use. As a result most of the problems in this book can be done at a glance. The problems have been designed to avoid tricky scale reading and slide manipulations until the reader has had practice on simple problems. MULTIPLICATION ON THE SLIDE Rute. Figure 8 shows simple multiplication, 2X 4 = 8[(: - +) + (~--) = ( )], using the C and D scales of the slide rule. The dots and dashes after the equation tie the slide rule movement to the problem and show the distance addition that corresponds to the multiplication. In Figure 8, the dotted line shows the 2 distance, the dashed line shows the 4 distance and the solid line shows the 8 distance. This coding shows the distance addition used to solve the problem. The vertical lines across the slide rule show the positions of the indicator. The dashed indicator shows the temporary setting used

The Basic Slide Rule

Pp (0.995)

(0.99)

11

(0.98)

1000

2.0

(0.97)

(0.96)

2°(88°) 15°(75°)

=20°(70°

10000

LL2

1.10

LLI

(0.95)

3°(87°) 30(60°)

LL3

2.5e

4°(86°) 5°(85°) 40°(50°

RULE # 4 Fig. 7. Slide Rule Scales Used in the Examples in This Book

The Complete Book of Slide Rule Use

12

to line up 2 on the D scale with 1 on the C scale, and the solid indicator shows the final setting for reading the answer which is

Fia. 8. (Rule No. 1)

2X4=8

[(---)+(C--)

=)

circled. All examples and problem answers will use the coding form of Figure 8 so that the method used to solve each example is apparent.

Fig. 9. (Rule No. 1)

4X2=8

[((-: -) + C--) =

Figure 9 shows another correct way of setting up the same example since multiplication can be done in any order. PRAcTICE PROBLEM Set No. 1

The following is the first set of practice problems. It is suggested that as a minimum the reader do as many problems of each set on his slide rule as he needs to do to get two problems correct. The answers to the problems are diagrammed in Appendix C. The

The Basic Slide Rule

13

problems are so numbered that the answers to practice problems from the same set appear on different pages in Appendix C. Problem No.

Problem

1-1

a0 hm

2-1 3-1 4-1

es ee i ae4 A> = 27 o 2.5 =a

Se

5-1 6-1

Ry ee eM! § 2% 3.5 =?

Although most people manipulate the slide rule to leave the answer on the D scale, it is perfectly proper to have it end up on the C scale as shown in Fig. 10 for the same example,

2X4=8

Fia. 10. (Rule No. 1)

[+ -)+C- =

2x4=8

[((-: -)+C--)

=]

Practice PRoBLEM Set No. 2

In each of the following, the answer is to appear on the C scale. Problem No. 1-2 2-2 3-2 4-2 5-2 6-2

Problem Bx 2.508 ¢ 25 X22 ae a ii = Ta Ae o>Gece tar |S aa Mid 4X15=?

The Complete Book of Slide Rule Use

14

Sometimes the answer to a problem is literally “off the end of the rule.”? Such a case is shown in Figure 11 where

27:5 se"

(C6

) Pea)

is carried out.

Fig. 11. (Rule No. 1)

[Ca sdnicle (===)

2X 7.5 = 15

ad]

If the D scale were continued to the right another cycle, as shown dashed in Figure 11, all would be well, but this is not possible. The way out is to shift the C scale to the left, instead, as shown in Figure 12.

D py

Fra. 12. (Rule No. 1)

27.5

= 15

| (1) (9)

(+++)

Saak (8)

KS

+ C--) = (—)

The reader should compare Figures 11 and 12 to prove to himself that they are equivalent. In fact, if the C and D scales repeated themselves several times, such as is done in part in Figure 5, there would be one point in each cycle where the hairline could be located to read out the answer. In Figure 12, the shift of the slide by a whole cycle puts the right

The Basic Slide Rule ’

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The Complete Book of Slide Rule Use

16

hand “1” of the C scale in the former place of the left hand “1” of the C scale. This process is called “interchanging indices.” “One” on the C or D scale is called an index because it is the starting place for distance measurements. The left index corresponds to the vertical line at 1.0 in Figure 4. Practice

ProspuemM

Set

No.

3

In these problems the answer is to appear on the D scale. Problem No. 1-3 2-3 3-3 4-3 5-3 6-3

Problem 5X8=? Ux oer 2 1 ee Vie ats Pe 5X9 = 7 OP ar

DIvIsIon AND ReEcrprocats.

by subtracting distances.

Division is done on the slide rule

This can be seen from:

10°/10° = 10¢-» In words, the quotient equals ten raised to the difference between _the exponents, when (102) dividend and (10°) divisor are both written as ten raised to various powers.

L

%

0.1

0.2

ie

ti

4

5

EE

O13 ra

CF

0.4 in

0.5

|: Pak

0.6

0.7

0.8

ae

re ee ltet Co hanes Wa

0.9

2

5

Saye

Cc

(es

2

eat b 3

10g I

an

DF 2

4

5

Cpt

1 pa 3 a SEC emt tat WO Upasasadentanant cahansanaalanl-cdeta:)—-G¢--) =C—)

Notice the dashed intermediate indicator setting.

The Basic Slide Rule

17

Pratcicr PropieM Set No. 4 In these problems the answer is to appear on the D scale. Problem No. 14

pi

1) (9)

Problem Beto 2 =?

2-4

15~

3-4 4-4 5-4 6-4

(ee ei 66> 3g=? 5+ = ? 9.5 + 4.756 =?

(8)

p 0.995)

(7)

(6)

(0.99)

Fia. 14. (Rule No. 1)

(5)

(4)

(0.98)

4+8=0.5

25=?

(3)

(0.95)

(1)

(0.9)

10.8) (0.7)

(0.2)

DI

p

sens). — Aare)

An example of division is shown in Figure 14 in which the answer appears off in space under the left hand C scale index at the end of the solid line. Because the C and D scales can be repeated endlessly, the same number value appears under the right hand index of the C scale as would be found under the left hand index of the C scale and the problem answer is read under whichever index is more convenient. PRAcTICE PRoBLEM Set No. 5

In these problems the answer is to appear on the D scale. Problem No. 1-5 2-5 3-5 4-5 5-5

6-5

Problem 2+ a1 15+75=7 35+ 5 = 72 3+ =a Q=— 5=?

20.

=?

The Complete Book of Slide Rule Use

18

Fig. 15. (Rule No. 1)

4+8=0.5

[((- »:) — (---) = ——

As in multiplication, the roles of the C and D scales can be reversed in division. Figure 15 shows the division problem of Figure 14 solved with the previous scale roles reversed. In this case, the right hand D scale index substitutes for the left hand D index. Practice ProspieM Set No. 6 The answers to these problems are to appear on the C scale. Problem No.

Problem

1-6 2-6 3-6 4-6 5-6

2+ 8+ 2:2).

(0.7) (0.8)

(0.9) a o =

Zz

> Oo a a onl

FOLD DOWN

Fia. 96. Practice Slide Rules

(A)

Practice Slide Rule No. 1

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How to Make Your Own Slide Rule

121

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(B)

Practice Slide Rule No. 2

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123

How to Make Your Own Slide Rule

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{¢6°0) (96°0)

(C)

Practice Slide Rule No. 3

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125 How to Make Your Own Slide Rule ,

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How to Make Your Own Slide Rule STAPLE

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TAPE TO FORM SLIDE

STAPLE OR SCOTCH

UNBENT PAPER CLIP. TO ACT AS INDICATOR

127

TAPE IN POSITION

PASTE FORMED BODY TO CARDBOARD

BASE

Fig. 97. Construction of Practice Slide Rule

Do-it-yourself slide rule No. 3: Cut out pieces of cardboard and celluloid as indicated in Figure 98. The quantities shown in Figure 98 will make one practice slide rule.

The cardboard should be about 42” thick.

Either the card-

board used to back writing pads or that used by laundries to ironed shirts in shape is suitable. Assemble the cardboard celluloid parts as shown in Figure 99 using slide rule scales cut Figure 96A. For a full set of practice rules repeat using the rule scales of Figures 96B, C, and D.

hold and from slide

“pit 34!x 5° — MAKE FOUR OF CARDBOARD

"G"— 1%'x 5 — MAKE ONE OF CARDBOARD

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niin trees 219/35

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MAKE ONE OF CARDBOARD AND OF CELLULOID WITH INDICATOR LINE AS SHOWN BY DASHES.

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“pr qx 5" — MAKE ONE OF CARDBOARD

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TWO

OF CARDBOARD

Fig. 98. Parts for Practice Slide Rule

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B (CARDBOARD)

B (CELLULOID)

de SCALES Fst

LINE UP OUTER EDGES AND GLUE TOGETHER

G

INDICATOR

ASSEMBLY

LINE UP OUTER EDGES AND

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GLUE TOGETHER

E

BODY ASSEMBLY

PRINTED

CENTER PARTS AND GLUE TOGETHER

SLIDE ASSEMBLY Frc. 99. Assembly of Practice Slide Rule (A) Indicator Assembly (B) Body Assembly (C) Slide Assembly

|

APPENDIX ANSWERS

TO PRACTICE

Problem 1-1. (Rule No. 1)

UG Bied

PROBLEMS

2.5% 3 = 7.5

eeotcea’ |pect Gomereed)

) (9) (8) Pa p95)

Problem 1-2. Bue No. 1)

Ca

C

cae

(7) (0.99)

Sat ei ey OS

ie earsJe 2sGere | 131

(0.98)

The Complete Book of Slide Rule Use

132

Problem 1-3. (Rule No. 1)

Nee

5X8

es =)

Problem 1-4. (Rule No. 1)

= 40

= Caney

8+2=4

(++) = G--) = —

Di ft) (9) (8) (7) p (0.995) (0.99)

(6)

Problem 1-5. (Rule No. 1)

8) (0.f 8)

(1)

(0.8) (0.7)

2+ 8 = 225

((-)-C- =

(0.2)

DI

p

Answers to.Practice Problems

Problem 1-6. (Rule No. 1)

133

2+8

(---)-C- =

p95)

(0.99)

(0.98)

Problem 1-7. (Rule No. 1)

ie)

Problem 1-8A.

= 0.25

1 + 3 = 0.333

l=) = Cj

(Rule No. 1)

(ee) hese

5 X 6 = 30

=

The Complete Book of Slide Rule Use

134

Problem 1-8B. (Rule No. 1)

30 X 1.5 = 45

)=) (6 +E-3

Problem 1-9. (Rule No. 1)

(5-6) x G75

(-)-C-34+6---) =

DI p (0.995)

(0.99)

Problem 1-10.

(0.98)

(Rule No. 1)

logis 700 = 0.845 + 2 = 2.845

Answers to Practice Problems

135

D pi)

(9) (8)

p 0.995)

(7)

(6)

10.99)

Problem 1-11.

GB)

@

(0.98)

~ J

(Rule No. 1)

(1)

(0.95)

(0.9)

(0.8) (0.7)

antilogio (3.4) = 10°4

= 10 X 10°* = 1000 X 2.51 = 2,510

Problem 1-12.

Problem 1-13.

(Rule No. 1)

(Rule No. 1)

on

ee

1/7 = 0.1428

(1/5) X 30 = 6

ee)

(0.2)

py

p

The Complete Book of Slide Rule Use

136

Problem 1-14. (Rule No. 1)

3X2=6

Problem 1-15. (Rule No. 3)

3 X 2.5 X 8 = 60

Problem 1-16. (Rule No. 1)

(6/7) X 3°] 407

J+ 6-9 =(—I

( )+E-J+6+-) =o

()+6-J =)

Answers to Practice Problems

137

Problem 1-17. (Rule No. 1)

(4 + 2.5) X w = 5.027

Problem 1-18.

(2.5 +15)

() FC“ 4+6--) =

(Rule No. 1)

xX9=15

(+) -—G-J+6°-) =

138

Problem 1-20. (Rule No. 1)

Problem 1-21. (Rule No.2)

The Complete Book of Slide Rule Use

C0 ela

— 0.036 = 0.1897

Problem 1-22. (Rule No.2) 9 X V3 = 5.20 RGR eat wt Gem2)

x=5

Answers to Practice Problems

Problem 1-28. a

No. 2)

Nah

139

60 + 5? = 2.4

eae Bee poe)

Problem 1-24. (Rule No. 2)

0.09 = 0.3

Problem 1-25. (Rule No.2)

—_-~/27,000 = 30

140

The Complete Book of Slide Rule Use

Problem 1-26. (Rule No.2) 27,000 = 30 (30)? X 80 = 27,000 kl as Ha eer

:

1

us

bp Na pal NS Naar a p 0.995)

(0.99)

2

3

eC a es Ba (0.98)

4

4

ee (0.95)

5

(2) ee

(0.9)

6

rf

Sian

Pia

ry

ee (0.8) (0.7) (0,2)

p

i 251+ = 125, Problem 1-27. (Rule No. 1) log 25 = 1.398 (-—-), antilog 2.097 = 125 (- - :)

Problem 1-28. (Rule No. 4)

sin 45° = 0.707

Answers to Practice Problems

141

Problem 1-29.

(-)-G-+6+--) =O

(Rule No. 4)

(4 + sin 40°) X sin 20° = 2.11

Problem 1-80.

(Rule No. 4)

sin"! 0.707 = 45°

Figure 1-31. (Rule No. 4)

(+)

2(cos 63°) = 0.908

+C-3 =)

142

The Complete Book of Slide Rule Use

Problem 1-32. (Rule No. 4)

cos! 0.423 = 65°

°(8 7°)

Problem 1-34. (Rule No. 4)

Kes i

9(865}

5°(85°

4 + (tan 3°) = 76.3

bl |eg) le

Answers to Practice Problems

10°%|(80°}

bpes (|) A (9)A(8) (7) |e

(6)

143

15°(75°)—-20°(70°}

(5)

(4)

2

30°(60°)

(3) 3

(3) i

Problem 1-36. (Rule No. 3)

e

)