The Use of the 
 
 Slide Rule 
 
 By ALLAN R. CULLIMORE 
 
 Director, Newark Technical School 
 NEWARK, NEW JERSEY 
 
 Copyright, 1915, 1920 by 
 KEUFFEL & ESSER CO. 
 
 PUBLISHED BY 
 
 • KEUFFEL & ESSER Co. » 
 
 NEW YORK J2Z Fulton St. General Office *ndF AC tori C s,HOBOKEN,N. */. 
 
 CHICAGO ST.LOUIS SAN FRANCISCO AVONTREAL, 
 
 516-20 S.DcarbornSt. 81T Locust Si. 30-34 Second St 5NotrcDameStV 
 
 Drawin£Materials * Mathematical andSurveyin^iistruments * Measurin&Tapes 
 
* r * « * r * 
 
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 4- a .v . 
 
 PREFACE TO THE FIRST EDITION. 
 
 The need for a small book of this type arose in the work of teaching 
 the use of the slide rule to engineering and industrial students, and this Manual 
 is a direct result of sets of notes issued to classes consisting of engineering 
 students and men of more or less practical experience. The book is not a 
 treatise in any sense, its aim being to develop the ideas of the operator rather 
 than to give empirical rules. Those rules that have been given are for the 
 purpose of training the student in the formulation of processes, and it is not 
 intended that they shall ever be committed to memory. 
 
 The examples have been taken largely from Hydraulics and Mechanics, 
 and while the actual field covered by specific problems is narrow, the idea has 
 been to make them fundamental. It is hoped that these examples will serve 
 the purpose of development better than the more specialized problems arising 
 in different branches of engineering. 
 
 PREFACE TO THE SECOND EDITION. 
 
 The second edition is made necessary by the increasing popularity of 
 the Polyphase type of rule, a popularity rightly based on the combined ef- 
 ficiency and simplicity of this type. For students in engineering and vocational 
 schools the Polyphase slide rule is to be strongly recommended. While the 
 general principles outlined in the first edition apply equally to the Mannheim 
 and the Polyphase types of rule, it seems advisable to make additions to certain 
 chapters explaining the use of the Polyphase rule when the use of this type 
 makes for more efficient calculation. 
 
THE USE OF THE SLIDE TITTLE. 
 
 I. ACCURACY AND SIGNIFICANT FIGURES. 
 
 It is absolutely necessary for a proper and efficient use of the slide 
 rule that the operator should have a clear and definite idea of the conditions 
 under which the rule may be used to advantage. Even among engineers, the 
 idea is often expressed that the slide rule is inaccurate, because, in the hands 
 of a reasonably expert operator, the rule will give results accurate only to 
 1/10 of 1%. It should be borne in mind that the rule is inaccurate in exactly 
 the same way that a four place table of logarithms is inaccurate. 
 
 In a very large class of engineering calculations, however, results which 
 are well within the allowable error may be computed by the slide rule. The 
 question most often arising is whether the rule is adapted to a given problem or 
 not. To answer this, requires a knowledge of the proper use of significant 
 figures and a close inspection of the data of the problem, together with a 
 knowledge of the means employed in obtaining these data. 
 
 By a significant figure we mean any figure which is significant in that 
 it gives some real information as regards the quantity which is represented. 
 Thus: 
 
 18700.1 has six significant figures. 
 
 13.7303 " six 
 
 0.0032 " two 
 
 13000 " two or five significant figures. 
 
 Notice that in the last case an ambiguity arises and that any one or all 
 of the zeros may or may not be significant. 
 
 Take the problem of finding the area of a circle whose radius is 4.67 feet, 
 measured with a steel tape. It is easily seen that the recorder of the data 
 meant that he was unable to state the distance closer than 1/100 of a foot. In 
 short, that he felt sure that the distance was nearer .67 of a foot, than either 
 .66 or .68. He, therefore, recorded the result 4.67 feet, using three significant 
 figures with an accuracy of one part in 467, or not quite 1/5 of one per cent. It 
 should be borne in mind that the number of significant figures expresses accuracy, 
 while the number of decimal places may or may not do so. Thus 761 millimeters 
 is identical with .761 meters, and the decimal place shows nothing. The 
 
 4CyjJt ; - 
 

 engineer recording or computing data should force himself to express, by the 
 number of significant figures in the result, the accuracy of that result. The 
 frequent habit of carrying results to a greater number of significant figures 
 than the data warrant comes perilously near to lying with figures; it certainly 
 creates a wrong impression as to the accuracy of the result. Certain mathe- 
 matical constants can, however, be computed to any number of significant 
 figures; for instance ir may be expressed as 3.142 or 3.141592654. On the 
 other hand, certain physical constants are very uncertain, even in the third 
 place. Take, for example, the weight of a cubic foot of water generally given 
 as 62.5 lbs.; conditions of temperature and solution may easily alter the last 
 figure of the three. With recorded data, however, the number of significant 
 figures, as well as their character, gives very definite information. If we say 
 that light travels 186000 miles per second, we do not mean that it covers 
 186000 miles to within the smallest fraction of an inch in one second of time, but 
 that the distance covered is nearer 186000 miles than it is 185000 or 187000, and 
 the accuracy 1/186 is expressed by three significant figures. In this particular 
 instance, it will be noticed that the three zeros to the right of the six, may or 
 may not be significant figures. To prevent this ambiguity it has been suggested 
 that results like the above be written 186 X (10) 3 which obviates the difficulty. 
 
 If, therefore, the slide rule will consistently give results to within 1/10 
 of one per cent, or to one part in a thousand, we have a right to use it where 
 three significant figures are warranted in the result. The following rule given 
 by Holman should be rigidly observed in all cases: "If numbers are to be mul- 
 tiplied or divided, a given percentage error in one of them will produce the same 
 percentage error in the result." This amounts to saying that all problems, 
 involving data correct to three significant figures only, can be computed 
 advantageously by means of the Slide Rule. The answer is not only near 
 enough, but is as accurate as the data warrant. Of course, in cases when the 
 slide rule is used as a more or less rough check on logarithmic or other calcual- 
 tions, these questions of accuracy do not apply. The consideration of a very 
 simple example will serve to illustrate the rule as stated above. 
 
 Suppose we wish to compute the cubical contents of a prism of earth. 
 Consider that the horizontal distances have been measured with a tape to the 
 nearest 1/100 of a foot, and that the heights have been measured by a level to 
 the nearest 1/10 of a foot, and the following dimensions recorded: 
 
 Multiplying: 
 
 Length 101.13 ft. Breadth 7.34 ft. Depth 9.3 ft. 
 
 101.13 
 7.34 
 
 742.2942 
 9.3 
 
 6903.33606 
 
 Now, if the answer be as indicated, we know the contents to within 1/100000 
 of a cubic foot, or better, with an accuracy of 1/7000000 of one per cent. 
 
— 5 — 
 
 which is, of course, ridiculous. Suppose the depth, somewhere between 9.25 and 
 9.34: The correct height might be 9.25, or 9.34 but if tenths alone were ex- 
 pressed, it would be recorded in both cases as 9.3. We see, therefore, that the 
 correct result lies between the following: 
 
 742.2942 
 9.25 
 
 6866.221350 
 
 742.2942 
 9.34 
 
 6933.027828 
 
 We see, then, that actually we know, only, that the result surely liesbeetwen 
 6866.221350 and 6933.027828, but we certainly know nothing more definite 
 than this. If we express the answer first found as 6903.33606, we know nothing 
 about the last seven figures. We are sure of only the first two and the result 
 should have been written 6900. In the light of this we now perform the same 
 multiplication as follows: 
 
 101.13 
 7.34 
 
 742.0 
 6900. 
 
 giving the answer 6900, which is as near the true value as we can know by 
 the data recorded. 
 
 It will be readily seen that a knowledge of the proper number 
 of significant figures saves an immense amount of time in calculation. This 
 is true no matter what means are used in calculating, whether it be multiplica- 
 tion, logarithms or the slide rule. The operator should accustom himself 
 first to examine the data of a problem and mentally calculate the desired 
 accuracy of the result, as well as the approximate numerical value of that result. 
 
 II. DESCRIPTION OF THE RULE. 
 
 The usual type of engineer's slide rule (K. & E. Mannheim or Polyphase) 
 is of wood, faced with a white composition upon which the units are graduated 
 in black, and is about ten inches long. Along the center line of the rule, a slide 
 of wood moves easily in a longitudinal groove. Each rule is provided with an 
 indicator or runner of glass, marked with a hair line, which serves as a reference 
 line in calculating. There are four distinct scales on the face of the Mannheim 
 and five on the Polyphase rule, and for the sake of convenience we will always 
 call the scales A, B, C, D, beginning at the top and reading down. If the 
 slide be inverted (that is, turned upside down in its groove) or if it be reversed 
 (exposing the back of the slide), the second scale from the top will always be 
 referred to asB, the third as C, etc. As the two top scales are double, we will 
 speak of the right or left A, as the case may be. If the slide be reversed, three 
 scales are seen on the back of the slide; a scale of sines marked S, a scale of 
 tangents marked T, and a scale of equal parts in the center. Notice on the 
 
— 6 — 
 
 back of the rule proper a piece of transparent material set in the end of the 
 rule with a line marked upon it by which the scales can be read. In operating 
 the rule the slide will be used in four positions: direct, inverted, reversed direct, 
 and reversed inverted. The terms for these positions are explained above. 
 
 III. PRINCIPLE OF THE RULE. 
 
 The rule is based on the principle that the addition of the logarithms of 
 two numbers gives the logarithm of the product of the two numbers, and that 
 the subtraction of the logarithm of one number from the logarithm of another 
 number gives the logarithm of the quotient obtained by dividing the second 
 number by the first. If then we add a length on the rule to another length, and 
 these two lengths are proportional to the logarithms of certain numbers, then 
 the length which represents the sum of these two lengths will be proportional 
 to the product of the two numbers. This can be shown quite simply on the 
 rule. On the back of the slide we find in the center a scale of 500 equal spaces. 
 Suppose we reverse and invert the slide, and bring the extreme left-hand 
 graduations into coincidence. Set the runner to 2 on scale D, and read under 
 the runner on the middle scale of equal spaces. We read 301. Reading 
 on AD in the same way we have 602, and on SD — 903. The distance between 
 1 and 2 is therefore 301 units, and between 1 and 4 is 602 units; adding these 
 we would have 903, which, as we have seen, corresponds to 8. Division is, 
 of course, the reverse of this process. 
 
 In describing different settings, LA and RB will be used for the left-hand 
 scale of A and the right-hand scale of B respectively. R alone, refers to the 
 runner. LLA would mean the left-hand index on scale A. The operation, R 
 to 3LA, would consist in moving the runner until the hair line on it concided 
 with the 3 on the left-hand scale of A. SC to ALA would mean placing oC in" 
 such a position that 3C and ALA are the same distance from the end of the 
 rule; that is, both would be brought into the same straight line. This is best 
 done by placing the runner so that the line on it is on the number on the fixed 
 scale, and then moving the slide until the number on the slide is under the 
 line on the runner. 
 
 IV. DECIMAL POINT AND READING THE SCALE. 
 
 Success as an operator depends upon a quick and accurate reading 
 of the graduations, and this can only be acquired by faithful practice. It 
 should be constantly borne in mind that there is no way to distinguish by direct 
 reading alone the position of the decimal point. The reading on the rule 
 would be the same for each of these numbers 1751, .1751, 17.51. The rule 
 gives simply a succession of figures in their proper order, but without determin- 
 ing the decimal point. This determination always must be made independently 
 of the actual solution of the problem in exactly the same way that the char- 
 acteristics of logarithms are independently computed. Suppose the problem 
 

 7 — 
 
 is to set the hair-line on the runner to 478 on Scale D (R to 478D). We 
 see that between 4 and 5 (care should be taken not to confuse this 4 and 5 with 
 1.4 and 1.5 which occur further to the left), there are 20 divisions, a very long 
 division marking 4.5, 45, or 450, whichever we please to call it. Between 
 450 and 500 there are 4 long marks and 5 short ones; the first long one following 
 being 460, the next 470, the next 480, etc. Our 478 lies, then, between the 
 second and third long mark to the right of 450; that is, between 470 and 480. 
 Between 470 and 480 we see a smaller division marking 475, and there is no 
 division between 475 and 480. 478 lies in this blank space and lies 3/5 or 
 6/10 of the distance between the marks to the right of 475. In the same 
 way, to set on 1673 on Scale D (R to 1673D), we see 1 at the extreme left fol- 
 lowed by a smaller 1 marking 1.1D or HOOD, then a small 2 marking 1200D, 
 etc. 1600 is easily found, and between 1600 and 1700 are ten divisions, 1650 
 being marked by a long one. The next short division to the right of 1650 is 
 1660, the next 1670, and our number lies between 1670 and 1680; 3/10 of the 
 distance from 1670 to 1680, to the right of 1670, lies 1673D. This procedure 
 should be gone through over all parts of the scale until the operator is familiar 
 with the different graduations and can set on numbers quickly and accurately. 
 It will be seen that the value of the distance between graduations changes on 
 different parts of the rule. Three significant figures should be read on all 
 parts of D and C; on the left, four may be obtained. On A and B, two and 
 sometimes three are obtainable. 
 
 In regard to the fixing of the decimal point, this is, of course, fixed for 
 any given set of data. General rules might be given for fixing the decimal 
 point, but it seems best in a vast majority of cases to, fix the decimal point 
 independently of the rule. This is simple in many cases and gives the added 
 advantage of a rough check on the work. Examples will be taken up under 
 each class of computation. It is sufficient to consider here a very simple case: 
 Find the square-root of 1873. This evidently has a square root lying between 
 10 and 100; pointing off two places to the left of the decimal point is, therefore, 
 correct. 
 
 V. EQUIVALENT RATIOS. 
 
 One of the most important uses of the slide rule is in converting quantities 
 from one kind of units to another. Printed on the back of the rule are a series 
 of ratios existing between different systems of units. The time saved by using 
 these ratios is very great, especially when a considerable number of quantities 
 are changed. 
 
 Suppose we wish to change ounces to grams and we have 63^, 7 %, 9}4> 
 \\%, ounces to be expressed as grams. We see on the back of the rule that 
 6 ounces equals 170 grams. One ounce will equal approximately 30 grams, 
 and this knowledge allows us to fix the decimal point easily in any case. Set 
 the runner to 6D and put 170C to the runner; we then read 6 ounces equals 
 170 grams, reading ounces on D and grams immediately above on C. Notice 
 after this setting is made that one ounce equals 28.3 grams, and that one 
 gram equals .0353 ounces. Reading directly over 6.5, 7.75, 9.25, on D, the 
 equivalent grams are shown on C as follows: 184, 220, 262. We cannot read 
 
— 8 — 
 
 over 11.67, as no scale is there, so we shift the slide to the left, putting the right 
 index of C where the left one stood previously and then on C, by means of the 
 line on the runner, we read 331 grams. 
 
 Consider this example: Find the intensity of water pressure in pounds 
 per square inch at depths of 14.33, 17.84, 19.83, 9.76 feet below the surface. On 
 the back of the rule we find feet of water : pounds per square inch :: 60 : 26, 
 or roughly, 2 ft. of water gives a pressure of one pound per square inch; the 
 fixing of the decimal point is, therefore, easy. Set 26C to 60D (by means of 
 runner). Notice that one pound per square inch pressure corresponds to head 
 of 2.3 ft., and that 1 ft. head gives a pressure of .434 lbs. per square inch; and 
 corresponding to 14.33 ft., 17.84 ft., 19.83 ft., and 9.76 ft., we have 6.21 lbs., 
 7.73 lbs., 8.60 lbs., 4.23 lbs., noticing that the slide must be shifted on the first 
 and last reading. In this connection, it is well to notice that these settings 
 for conversion can be made on scales A and B, and then no shifting of the slide 
 is necessary, but the loss in accuracy in most cases prohibits the use of these 
 scales. Instead of setting 26C to 60D we might have set 60C to 26D and read 
 feet on C to pounds pressure per square inch on D. 
 
 Ratios in constant use in a special class of calculations are sometimes 
 more conveniently expressed as round numbers. We see on the back of the 
 rule the ratio of the diameter of a circle to its circumference given as 113 to 355 
 instead of 1 to 3.1416. The first setting is somewhat easier and in all classes 
 of work the change in ratio is easily effected. Instead of memorizing a large 
 number of these equivalents, it is advisable to work them out when necessary , 
 using simply the few given on the back of the rule. Suppose we wish the 
 equivalent of U. S. gallons of water expressed in kilograms. We get the 
 
 , . , J _, , . , ,, . U. S. Gallons 3 Pounds 75. 
 
 following from the back of the rule: Pounds of water - W . ~^^ - j; 
 
 One gallon = -5- lbs. One pound = == Kilos. Therefore, One gallon = 
 
 kilos = q kilos, or 9 gallons = 34 kilos. Therefore, -=== = — , which is 
 
 the required ratio. Consider finding the equivalent of chains in feet, by the 
 
 865 
 43 
 
 rule, Chains: Meters :: 43 : 865; Yards : Meters :: 82 : 75; 1 Chain = 
 
 Meters; 1 Meter 
 865 X 82 X 3 
 
 82 
 75 
 
 Yards; 1 Yard = 3 Feet. Therefore, 1 Chain = 
 
 Feet; Chains : Feet :: 215 : 14200, or as 1 is to 66, which is the 
 4o X 75 
 
 correct answer. These examples will serve to show how any desired equivalent 
 may be obtained. They may easily be solved by following the instructions 
 which will be given for continued multiplication and division. 
 
 The slide rule also furnishes a ready means of converting inches into 
 decimals of a foot, and decimals of an inch into any desired fractional part 
 thereof; it facilitates inverse operations. To reduce the number of inches and 
 fractions, given to the nearest \ in., to decimals of a foot, setlC to 12D and over 
 
9 — 
 
 inches on D, find feet and tenths on C. For example to find the decimal of a foot 
 corresponding to 3 M inches, read above 3.25D and get .2708 on C. This setting 
 can be conveniently used when the distances are given to the nearest }4 or 
 perhaps \i in. We must, of course, remember the decimals of an inch correspond- 
 ing to each eighth of an inch as follows: .125, .250, .375, .500, .625, .750, .875. 
 The reduction to decimals from sixteenths is simple. Set \C to 16D and over 
 the number of sixteenths on D find the decimal on C. As can readily be seen, 
 this method is applicable to any fractional part, such as a sixty-fourth. 
 
 Take the following example: Express 7 9/64 in. as a decimal of a foot. 
 Put \C to 64D, using the left-hand index on C. Reading on C over 9 on D, 
 we find .141. Set 1C to 12D and read .595 on C, over 7.141 on D. These 
 examples are really special cases of a general rule. To reduce fractions to deci- 
 mals, set numerator on C to denominator on D. Read on C over ID to find 
 equivalent decimal, and to reduce decimals to fractions, set decimal on C to ID, 
 and corresponding numerators will be found over their denominators. 
 
 .31 
 
 Example: Find the decimal corresponding to — . Set 31C to 47D, 
 
 47 
 60 198 
 
 and over ID read .66 either — or . We are enabled to find reciprocals in 
 
 91 300 
 much the same way. Example: Find the reciprocal of 3.17. Set 3.17C over 
 ID, and read on D under \C. Answer is .316. 
 
 On the back of the rule will be noticed gauge points for different metals, 
 with the weight of one cubic foot of the metal given. In using these ratios, it 
 is convenient to reduce all weights to equivalent weight of iron, and then 
 reduce by means of ratios given. Suppose we wish to make a table of weights 
 of brass plates 12 in. on a side, varying by 32nds of an inch in thickness. 
 
 480 
 Dividing = 40, which is the weight of 1 square foot of iron, 1 in. thick; set 
 
 12 
 32C to 40D, and multiply in each case by 1.09, the gauge point for brass. ] The 
 best method for this setting is toset32C to 40D, R to 1.09C. Answer 1.36D. 
 Then Rto2C, ICto R,R to 1.09C. Answer 2.72 D, etc. ^ = 1-36; T V = 2.72; 
 T 3 5 = 8.16, etc. Or we might have done as follows: 
 
 525 
 
 = 43.7. Set 32C to 43.7D and we have as before 1.36, 2.72, 8.16. 
 
 12 
 This same idea is applicable to any section of bar, plate or structural shape. 
 (See page 10). 
 
 VI. SQUARES AND SQUARE ROOTS. 
 
 The numbers on A are the squares of those on D. We have then a 
 table of squares and square roots ready at hand. To find the square of any 
 number set the runner to the number on D and read the answer on A . Example : 
 to find the square of 4.17, set the runner to 417D and find under runner on A 
 17.4. The position of the decimal point is very evident as shown. Finding 
 the square root is the reverse of this process. Set the runner to any number 
 
— 10 — 
 
 
 
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 on A and under the runner find the square root of the given number. Care must 
 be taken to choose the proper scale on A. In finding the square root of 5, if 
 we set the runner on 5LA, we find, under the runner on D, 2.24, and if we use 
 5RA, we find 7.07. It is evident that 2.24 is the square root of 5, and 7.07 the 
 square root of 50. We can use this rule : If a number has an uneven number of 
 digits to the left of the decimal point, use the left-hand scale; if an even number, 
 the right-hand scale on A. Thus we see that under the left-hand 5 we would find 
 the square roots of 5, 500, 50000, which would be 2.24, 22.4, 224. It is perhaps 
 better in most cases to estimate the result mentally and set on the scale so as 
 to approximate this result. Estimating that the square root of 5 would be 2, 
 we can quickly see that the left-hand scale must be used to obtain an answer 
 in the neighborhood of 2. 
 
 We have 
 
 Suppose we wish to find th 
 525 X 22 X d 2 
 
 weight in lbs. per lineal foot of brass rods. 
 51 Xrf 2 
 
 lbs. which is equal (almost) to 
 
 18 
 
 We can solve 
 
 144 X 7 X 4 
 
 then very easily in the following manner: Set 1SB to 51A, read answer on A 
 over diameter in inches on C. In setting 1SB to 51A, the operation has been 
 performed with the exception of multiplying by the variable d 2 . Now d 2 on 
 B corresponds to d on C, so multiplying by d on C is equivalent to using d 2 on 
 B, hence the setting as given. 
 
 VII. MULTIPLICATION AND DIVISION. 
 
 Notice that if we place 1C to 2D, every number on D is twice the number 
 immediately above it on C; that is, we have formed a series of proportions, 
 using the ratio 1 to 2. All numbers on C have this ratio to the numbers im- 
 mediately below on D. Suppose that we wished to solve the proportion 3.8 : 
 2.18 :: x : 4.53. Set 3.8C to 2.18D (always use the runner to make this class 
 of setting) and over 4.53D find 7.9C. In making this setting, place the runner 
 first on 2.18D and then slide C along until 3.8 comes under the runner; then slide 
 the runner to 4.53D and read 7.9 on C, which is the answer. This gives us a 
 method of conveniently performing the operations of multiplication and division. 
 These operations can be performed, using the two top or the two bottom scales, 
 as determined by the percentage accuracy required in the result. To multiply 
 on the bottom scale, set \C at a factor on D; under the other factor on C 
 read the answer on D. Setting is made as follows: Slide C until 1 is over a 
 factor on D, slide the runner to the other factor on C, and read the answer on D 
 under the runner. Example: 3.26X2.83. The answer is estimated as 
 approximately 9.0. Set 1C to 3.26D and under 2.83C read 9.22 on D, which is 
 the answer. Example: 3.91 X 7.33. The answer is estimated to be about 
 28. Set 1C to 3.9LD, under 7.33C-find 28.7, answer. In this second example, it 
 will be noticed that we were forced to use the right-hand \C instead of the index 
 to the left. A very considerable amount of time will be saved if the operator 
 decides, before trying, which index he will have to use. The following example 
 will serve to illustrate: 13.14 X 16.93. The answer will lie between 100 and 
 
— 11 — 
 
 300. As to using the right-hand or left-hand scale, we suggest 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: 
 
 2.13 X 3.33, 2X3=6. Use the left index. 
 7.23 X 4.71, 7 X 4 = 28. " " right " 
 .131 X 4.6, 1X4=4. " " left " 
 
 If this very simple rule be kept in mind much time will be saved. It will 
 be a little difficult in certain cases, like the following, to make a decision: 
 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 significance. The one just given, if appreciated, will save an immense 
 amount of time. If the process of multiplication be carried out between scales 
 A and B, and if we always use the middle index on B, the second factor will 
 never fall outside the scale of A. 
 
 Division is the reverse of the process of multiplication. Set the divisor 
 on C to the dividend on D, read the answer on D under 1C (The runner should 
 be used in setting the divisor on the dividend.) Example: Divide 313 by 8.9; 
 the answer is estimated to lie between 30 and 40. Set 8.9C to 313D and read 
 under \C the answer 35.2. Notice that in the case of division only one index 
 can appear on the scale, so that no ambiguity can possibly arise. 
 
 Attention should be called at this point to a very simple setting for 
 finding the areas of circles. Suppose that we use scales A and B. Required 
 the area of a circle whose diameter is 11.6 inches. We use the formula A = 
 
 = .7854 d 2 . (Notice that .7854 is marked on scales RA and RB by a longer 
 
 4 
 line in the same way that -n- is marked on the same scales.) We might solve 
 this by simple division, but the following will be found easier. R to 11. 6D (see 
 the square under the runner on A), set IB to R, and over .78545 read the answer 
 on A. It is seen to be 105.7. Or simpler still is the following: Reading on the 
 back of the rule: Diameter circle : side of equal square :: 79 : 70. Set 
 79C to 70D, R to 11.6C, and under runner read the answer on A, 105.7. The 
 advantage of this last method is apparent, for after the first setting of 79C to 
 70D the area of any circle may be found by one setting of the runner. The 
 slide rule is primarily a time-saving device and the idea always should be to 
 choose the setting giving the quickest solution. In this special case, the last 
 setting will always prove the most efficient when the number of areas is greater 
 than two or three. 
 
 In engineering work, formulae very frequently take the form of a fraction 
 in which the denominator and the numerator are made up of several factors. 
 
 142 X 15 X 15 X 45000 
 
 Consider the solution of the following example: ; first 
 
 4 X 16 X 16 
 
 determine the decimal point mentally. The two 16s cancel the two 15s 
 
— 12 — 
 
 approximately, 4 into 142 goes 35 times, 45 X 35 = 1000, add three ciphers and 
 the answer will be approximately 1600000. This process of determining the 
 decimal point should always be followed, no matter what the problem. 
 
 It is evident that we might multiply all the factors of the numerator first 
 and divide the result by each of the factors of the denominator in turn. Or 
 we might alternately multiply and divide. The second method is much simpler 
 and shorter, and an actual comparison of the two methods in this problem will 
 show the gain in speed of the second method over the first. The first method : 
 Set 1C to 1.42D, R to 15C, 1C to R, R to 15C, 1C to R, R to 45C (note here a 
 change of index), AC to R, R to 1C, 16C to R, R to 1C, 16C to R, read under \C 
 the answer 1404000 on D. The second method: Set 16C to 142D, R to 15C; 
 16C to R, R to 15C, AC to R; read under the 45C the answer 1404000 on D. It 
 will be seen that the work of the first method is almost double that of the 
 second. The success of this method lies in the correct choosing of divisors. 
 Suppose that the example had been solved as follows: AC to 142D, R to 15C, 
 16C to R, R to 15C, 16C to R, R to 1C, 1C to R, read under 45C 1404000 on D. 
 This process is evidently more laborious. It is not always possible to arrange 
 
 45X130X101 
 
 divisors so as to prevent the shifting of indices. Take for example, :- 
 
 72X84 
 
 mentally, 45 into 84 twice, 2 into 130, 65 times, 72 into 6500 about 90, which is 
 the approximate answer. Probably the best solution is 1C to 45D, R to 13C, 
 72C to R, R to 1C, 84C to R, R to 1C, 1C to R, read the answer on D under 
 101C as 97.7 ; another solution is 1C to 45D, R to 13C, 72C to R, R to 1C, 1C to R, 
 R to 101C, 84C to R, read under 1C the answer 97.7 on D. 
 
 The solution of the following types is almost as simple as that of simple 
 
 division: 
 
 b' b 2 ' b 2 ' ^b"' 
 
 VjL 1-1 
 b ' \b 
 
 Int—, R to aD, bB to R. Answer on A over LB. 
 
 In -f^> R to aD, bC to R. Answer on A over LB. 
 b 2 
 
 In ■§;, R to aA, bC to R. Answer on A over LB. 
 
 '^ 
 
 
 In — =•» R to aD, bB to R. Answer on D under 1C 
 
 In y a R to aA, bB to R. Answer on D under 1C. 
 b ' 
 
 sl 
 
 In I—, R to aA, bB to R. Answer on D under 1C 
 
 The thorough mastery of the principles involved in these settings will 
 suggest others as simple. 
 
— 13 — 
 
 VIII. THE USE OF THE RULE WITH INVERTED SLIDE. 
 
 Invert the slide so that the former scale C is adjacent to A; scale C now 
 becomes B, and B becomes C, A and D remaining the same. If we make the 
 indices of the slide to coincide with the indices of the stationary scale, we have, 
 reading from scale D to scale B, a table of reciprocals. Set R to AD and read 
 scaleU under R, and we see .25, the reciprocal of 4. The position of the decimal 
 point is easily determined by inspection. Find the reciprocal of 31.5. 100/32 
 is about 3, so the approximate answer will be .03. Set R to 31.5D, and reading 
 under the runner onB, we find 318, or as we know it to be, .0318. This gives 
 ready means for converting fractions of feet or inches into decimals, provided 
 the numerator is unity. Set R to 64D, reading onB under the runner .01562. 
 Or 1/64 of an inch = .01562 in. 
 
 Multiplication and division can readily be performed on the rule with 
 the slide inverted, the process being in each case the reverse of the process 
 with the slide direct. Invert the slide and set R to AD, and 2B to R and under 
 LB find 8 on D. In general, then, to multiply, set the runner to one factor on 
 D and bring the other factor onB under the runner, and look under IB on D for 
 the answer. With slide inverted, set R to AD, and LB to R, and under 2B 
 find 2 on D; to divide, therefore, set the runner to dividend on D, set IB to the 
 runner and find the answer on D under the divisor. It will be easily seen that 
 multiplication with the slide inverted is simpler than the same process with the 
 slide direct, and all problems involving only the product of two or more quantities 
 should be solved with the slide inverted. In multiplying with the slide direct, 
 we found that the question arose as to which of the indices to use. This 
 question cannot arise with the slide inverted. The only disadvantage con- 
 nected with the inverted slide is that scales B and D are separated, but this is 
 more than offset by the absence of ambiguity in choosing indices, and by the 
 fact that both right and left indices on the slide coincide. As an example take: 
 137 X 16.3 X 8.13; by inspection the answer will be near 21000. Solve as 
 follows with the slide inverted:' R to 137D, 163B to R, R to LB, 8135 to R, and 
 under LB find the result on D as 18150. 
 
 With the slide direct we would have the following: \C to 137D, R to 
 163C, \C to R, note here a change of index, R to 813C, and read result on D under 
 R, as 18150. The advantage of the first method is obvious. Both multiplica- 
 tion and division with the slide inverted are necessary to the speedy solution 
 of some of the more complicated formulae which will be shown later, and they 
 should, therefore, be mastered. 
 
 The inverted scale is also useful in cases of equal products of the general 
 type ax = be, where x is an unknown quantity. With the slide inverted, set 
 LB to 8Z> and, with the runner set at random, the products of the numbers under 
 the runner on B and D will always equal 8. Consider the following example : 
 A 16 in. pulley makes 137 R. P. M.; required the diameters of pulleys driven 
 by the specified pulley so as to give speeds of 150-175-200-250-300 R. P. M. 
 With inverted slide, set 16B to 137D, and read the required diameters onB over 
 the speeds on D. We see corresponding to a speed of 150 a diameter of 14.6 
 inches, to 175 — 12.5, etc. This problem depends upon the fact that the speed 
 of a pulley (belted or meshing) in revolutions per minute multiplied by the 
 
— 14 — 
 
 diameter of the pulley is a constant for a constant rim speed. We have found 
 this constant and have divided it by the speeds as given and found the respective 
 diameters. Notice that the indices must be interchanged between speeds of 
 200 and 250 R. P. M. 
 
 Consider the reduction of inches and decimals to inches and sixty-fourths 
 with the scale inverted. Required to express 14.6 in. as inches and sixty- 
 fourths. Set R to 6D, 64B to R, the result will be found on D under LB, and 38 
 will be seen to be the nearest whole number. So the answer is 14-38/64 inches. 
 Now set R to 5D, 64JB to R, and under LB find 32, as it should be. This setting 
 is rather inconvenient when a great many decimals are to be changed and the 
 following is preferable. Set R to 64D, 6B to R, under LB find 38 as before. Set 
 f>B to R, find 32, etc. It will be seen that we really divide the decimal by the 
 reciprocal of 64; this second method will prove most efficient when more than 
 two decimals are to be converted. As a further example of equal products, 
 consider this example: Two vertical pipes are connected by a tube. The 
 cross sectional area of pipe A is 3.42 square inches, and the cross sectional 
 area of B is 13.61 square inches. If these pipes contain water, and the surface 
 in A is depressed 8.19 inch, what will be the increase in height in B? We have 
 a case of equal products as in the previous example. We might solve by con- 
 
 3.42 X 8.19 
 
 sidering x = and solving with the slide direct. The answer is 
 
 13.61 
 approximately 2 by inspection. With slide direct, set 1361C to 342D, R to 1C, 
 \C to R, and under 8.19C find 2.06 on D. Notice in this setting that the runner 
 must be used to make the first and last setting and that in addition the indices 
 must be shifted, making in all five operations. With slide inverted, set 819B 
 to 342D, R to LB, shift indices and under 136LB, find 2.06. If conditions of 
 accuracy permit, we may use scales A and C when the slide is inverted; this is, 
 of course, shorter in that the indices are not shifted. Set 819C to 342A, R to 
 1361C, read answer on A under R, as 2.06. We might have set R to 136LA and 
 read answer on C, giving the same result, 2.06. 
 
 THE POLYPHASE SLIDE RULE. 
 
 Multiplication, division and continued multiplication and division are 
 very much simplified by the use of the Polyphase rule. In general 50% saving 
 in time can be made if the relations of C, CI and D are understood. The red 
 scale CI obviates the necessity of inverting the slide and all problems calling 
 for inverted slide can be readily solved with slide direct by using scale CI. 
 The study of a typical problem in detail will serve to illustrate the method and 
 advantage of using CI. 12.8 X 6.14 X 183.2 X 62.5. Solving : First using 
 C and D, we have, R to 12.8D, L1C to R, R to 6.14C, R1C to R, R to 183.2C, 
 L1C to R, R to 62. 5C. Answer on D under R = 90000. 
 
 This requires seven settings. Solving : Now using C, CI. and D we 
 have R to 12.8D, 6.14CL to R, R to 1.83C, 62.5CJ to R. Answer on D 
 under \C = 90000. This requires four settings. With no possible ambiguity 
 as to right or left index. Notice R to 12.8D, 6.14CL to R gives a multiplication 
 
15 — 
 
 answer under \C on D, to multiply this by 1.83 we need only to put runner to 
 1.83, answer is then under R on D. Setting 62. 5CI to R multiplied by 62.5 
 and the answer is on D under 1C. In general for simple multiplication it is 
 advantageous to set two factors together and read the answer, rather than to 
 set the index on one and the runner to the other on the slide. It is, of course, 
 very simple to remember that the processes of multiplication and division be- 
 tween Ci" and D are just the reverse of those between C and D. In this type 
 of rule all the advantages of both direct and inverted slide are available for 
 any particular problem. 
 
 326 
 
 Run through the following: R to 32.6D, 18.44C to 
 
 18.44 X 1.68 X 47 
 
 R, notice here why CI. can be used to advantage, and the next setting is R to 
 47C.1 and 16.8C to R. Answer on D, under 1C = .224. The student should 
 not leave this subject until he sees clearly the interrelation between C D and CI. 
 
 It frequently happens that we wish to successively divide a number by 
 various other numbers. Suppose we time 85 revolutions of a current meter in 
 47.3, 48.1, 46.4 seconds, respectively, and wish to reduce to revolutions per 
 second. Scale D with CI should be used. Set 1C to 85D and read under 
 47.3, 48.1, 46.4 and we have 1.80, 1.77, 1.83 revolutions per second, respectively. 
 
 IX. THE FORM XY = B AND X 2 Y = B. 
 
 Let us now consider problems of the general type ab 2 = x; that is, to 
 find the product of two factors, one of which is the square of a given number. 
 Find (3) 2 X 9. Set 1C to 3D, read answer on A over 9B, which is 81. It is 
 evident that we first square 3 and multiply the answer by 9 on scales A and B. 
 The reverse of this problem occurs in the designing of beams, the problem of 
 design being to determine the dimensions of a section of a beam to withstand 
 a given moment. As this problem is typical of the use of the slide rule in a wide 
 variety of problems, it will be taken up more in detail than the problem itself 
 might seem to warrant. Consider the design of a rectangular wooden beam 
 to withstand a moment of 50000 in. pounds. We have from statics M = 1/6 
 fbh 2 , where/ = allowable stress in cross bending in pounds per square inch, b the 
 width, and h the depth of the beam in inches. Let 1200 lbs. equal /, and we 
 have 50000 = 200 bh 2 or 250 = bh 2 , and we wish to determine b and h so that 
 their product shall be 250 or very slightly in excess of that figure. It is, of 
 course, the exact reverse of the problem above. The problem as given is evi- 
 dently indeterminate, as for every valve of b assumed we can find a corresponding 
 value for h. There are, as a matter of fact, certain limitations. We desire a 
 beam of minimum material, and as a commercial size must be used, we desire 
 a beam whose theoretically desired section shall vary as little as possible from 
 the commercial sizes. For the purposes of this problem let us say we can 
 obtain beams 2 X 6, 2 X 8, 2 X10, 4 X6, 4 X 8, 4 X 10, 6 X 6, 6 X 8, 6 X 10. 
 
 Set R to 250A (the left-hand 250). The reason for this choice will be 
 pointed out later. Slide left LB to the runner. We then see that (15.8) 2 X 1 = 
 250, or a beam 1 in. in width would require a depth of 15.8 in. Put 2B under 
 the runner, h then would have to be greater than 11 in., but 2 X 10 is our 
 
— 16 
 
 deepest commercial size. Put 4B under the runner and we see by the rule that 
 h = 7.9, that, therefore, a 4 X 8 would be sufficient with a waste of .1 of an inch 
 in height X 4 in. in width or .4 square ins. Put QB to runner, /i=6.47, or we 
 would require a 6 in. X 8 in., as the smallest commercial size available with 
 a width of 6 in. We have seen that a 4 X 8 is sufficient; the 6X8 then is 
 much too large. Notice that for each assumed width the rule shows the waste 
 between the best theoretic section and the commercial section of the same 
 width. Now as to the choice of the right or left-hand scale on A. As a matter 
 of fact, either may be used if the proper scale on B is used. Suppose we had 
 used LA with RB, for b = 4 we would have found h = 25. Obviously (25) 2 
 X 4 is not equal to 250, .but (2.5) 2 X 40 = 250 and such a choice would theo- 
 retically fulfill the conditions of our problem. No ambiguity will arise if we 
 roughly check our choice once for each problem. 
 
 Let us consider one more example: 1225 = bh 2 . Set R to 1225ZA, 
 set 8LB to R; we find h equals approximately 40. Now (40) 2 X 8 does not 
 equal 1225. So use 8RB and we find h = approx. 12.5 and (12.5) 2 X 8 = 
 approx. 1225. Rules might be given for this class of calculation, but the above 
 method is simpler and more satisfactory. Use either scale on A and determine 
 by trial the scale to be used on B. 
 
 Consider another problem of the same type but simpler in 'principle. 
 We desire to determine the number, width, and thickness of steel plates to be 
 used in the flange of a plate girder. The plates must have a gross sectional 
 area of 33.6 square inches. They may vary in width from 12 to 16 by half 
 inches (that is, we may use 12, 12^2, 13, 13 H, etc.) and the plates may vary 
 in thickness from % in. to % in. by 16ths of an inch (that is, we may use 
 H, 7 /i6> M> etc., up to %). Required the number of plates, their width and 
 thickness, so that their total sectional area will be as little in excess of 33.6 in. 
 as possible. The plates must, of course, be all the same width, but each plate 
 may vary in thickness between the limits specified; further it is desired to have 
 the plates of as nearly uniform thickness as possible. Set R to 33.6D, set 12C 
 to R. Read under 16C, 44.8 on D. This means that with a width of 12 in. the 
 
 44.8 
 
 plates must total , but we consider only full 16ths according to the con- 
 
 16 
 ditions of the problem, so 45/16 are required. Tabulating waste of each width 
 from 12 to 16, sliding the runner to 12^, 13, etc., to determine 16ths required, 
 we have : 
 
 12 12J4 13 13^ 14 14% 15 15}^ 16 
 .25 .7 .2 .6 9 2 3 4 
 
 So 12 Yl shows least waste, and putting 1232 under runner we have 43/16 as 
 total thickness. Now two whole numbers the product of which is nearest 43 
 are 7 and 6, so use five plates Vie X 12 Y% and one plate Yz X 12.5. 
 
 5 X 12.5X 7 
 Checking: - 
 
 16 
 
 12.5 
 = 27.3 and = 6.25, and 
 
 27.3 
 6.25 
 
 33.55 chk. 
 
— 17 — 
 
 THE POLYPHASE SLIDE RULE. 
 
 With the Polyphase rule cases of xy = B are sometimes simpler of solution 
 using CI. in conjunction with D. Consider the problem as given above. 
 Set 16CI to 33.6 D. Run to 12CI, read under R on D, 44.8, which is the 
 number of 16ths required in the total thickness of plates with a wastage shown. 
 Shift R to 12.5CI and we see 43, with no wastage. R to 13 CI under R 41.3 
 with .7 wastage, so 12 % width plates should be used. 
 
 X. CUBES. 
 
 Cubing a number (a) is, of course, the same as multiplying a by b 2 when 
 (a) and (b) are equal. To cube 2.17, set R to 2.17D. Set IB to R and over 
 2.17J5 read answer 10.2 on A. It is frequently of advantage to cube a number 
 with the slide inverted. This setting is in reality simpler than with the slide 
 direct and should always be used in connection with problems in which for 
 any reason the slide is in the inverted position. To cube 2.17, set R to 2.17P, 
 set 2.17C to R, read over IB the answer 10.2 on A. Notice that this operation 
 consists in squaring the number in the regular way, and then multiplying the 
 result by the number itself; using scales A and C. It is interesting to notice that 
 if we set R to 2.Y1D, 2.17B to R and read over 2.17C on A, we have the answer 
 on A, 10,2. 
 
 The application of these last settings is clearly seen when we wish 
 to raise a number to the 3/2 power. This can be performed very simply with 
 the slide direct, but the solution with slide inverted is in many ways more 
 satisfactory. Let us consider the first problem with the slide direct. To raise 
 3.163 to the 3/2 power. Set \C to 3.163D and read the answer on D under 
 3.163B, which is 5.63. Notice if we had used the right-hand index instead of 
 the left we would have read 1780. Suppose we perform the same operation 
 with the slide inverted. Set R to 3.163D and 3.163C to R, read answer on B 
 under \A, 5.63, or on D under 1C. We will notice that two readings under (1) 
 are possible; one reading being 5.63 and the other 1780. There will, of course, 
 never be any doubt as to which of these two values is correct in a particular 
 example, if the operator forms the habit of estimating the result. In this case, 
 (3.16) 3 is approximately 30, and V30 ^ es between 5 and 6. As to the value 
 1780, it is, of course, (31.63) |; the two values arising in finding the square root 
 of 31.63 and 3.163 cubed. In making this setting, notice that under the runner 
 on scale C is the number itself, on A is the square of 3.163, on B is \/(3.163) 3 . 
 Under one of the C indices on D read v / (3.163) 3an( i under the other v 3.163. 
 
 THE POLYPHASE SLIDE RULE. 
 
 Below D on the edge of the polyphase will be found a scale of cubes. Set 
 R on any number on D and read the cube of that number on the edge of rule 
 below D. It should be noticed that the accuracy obtainable by using this 
 scale is not as great as in the method given above, but it is sufficient for many 
 purposes, being approximately ^ of 1%. 
 
— 18 — 
 
 
 
 
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 l^^^kl^^^^W^t KEUFFEL & ESSER CO., NEW YORK 
 
 XL CUBE ROOTS. 
 
 In finding the cube roots of numbers, the slide may be used either direct 
 or inverted, and in either case the process is the reverse of finding the cube with 
 the corresponding position of the slide. We must in general recognize three 
 cases. Take for example the problem of finding the cube roots of the following: 
 270, 27.0 and 2.70. These will serve to outline the process. To find y/270 
 with the scale direct: R to 270i?A, slide B and C to the left, using thereby 
 the right-hand scale onJB until the same number onB appears under the runner 
 that is seen on D under the right-hand index on C. When 6B is under R, we 
 read 6.7 on D, and when 6.46S is under the runner, we read 6.46 on D under 1C. 
 Therefore, 6.46 is v^TOT To find ^27^0, set R to 2.70LA, and using the same 
 method with RB, we read 3 on B under R and also 3 on D under \C. Using 
 RA and LB, we arrive at the same result*. To find \/2X, set R to 2.7LA and 
 move slide to the right so as to use LB. 1.39 then appears as the v / 2.7 
 
 Tabulating, the following are evident: 
 
 tf270 =6.46 Using right-hand scales. 
 
 \/2n^ =3.00 " " A with left B or left A with right B. 
 
 ^2770=1.39 " left-hand scales. 
 
 This general rule can then be formulated: If a number is greater than 
 unity, point off periods of three places to the left of the decimal point; and if 
 there are no digits to the left of the last period, use the two right-hand scales; 
 if one digit remains to the left, use the two left-hand scales ; if two digits remain 
 to the left, use the right or left with the left or right. With a number greater 
 than unity, in most cases it is easier roughly to approximate the result and use 
 the combinations of scales giving this result. To find the cube root of 31700 
 by rule, point off 31 '700 and we have two digits to the left, and the rule calls 
 for left with right or right with left. Using the former we get 31.8. The cube 
 roots of fractions presents greater difficulty. Making a table as above we have: 
 
 ^2.7 = 1.39 Using the two left-hand scales. 
 
 #27 = .646 " " " right " 
 
 ^\027 = .30 " " right with the left or vice versa. 
 
 #0027 = .139 " " two left -hand scales. 
 
 This leads to a rule for fractions as follows: Point off the number of 
 zeros immediately to the right of the decimal point in groups of three, counting 
 the decimal point as one zero; if there are no zeros immediately to the right of 
 the last group, use the two left-hand scales; if one zero remains, use the two 
 right-hand scales; if two zeros remain, use right with left or vice versa. To 
 find the cube root of .000000738, pointing off .00'000'0738, we have one zero 
 remaining; so, by the rule, we use two rights and read 904, and as we have two 
 groups, .00904 is the answer. 
 
 It should be distinctly understood that these rules are not intended to 
 be memorized, but a study of them will simplify cube root computations. As 
 already suggested, if the number is greater than unity, a method of approxima- 
 tion and trial is most satisfactory. 
 
— 19 — 
 
 With fractions, if a rule be desired, the following is simple: Count off 
 the zeros between the decimal point and the first significant figure of the number 
 in the following way (counting the decimal point as one zero): Right, either, 
 left, right, either, left, etc., the scales to be used being determined by the 
 word falling on the last zero. As, for instance, in the following: .000386. 
 Counting as indicated above, we use the two right-hand scales, and read 
 .0728 as the cube root desired. 
 
 In finding the cube root with the slide inverted, the rules given above 
 regarding the proper reading of the scales hold. Suppose we wish to find the 
 cube root of 1734; evidently the result lies between ten and twenty. Set R 
 to 1734LA and place the middle index on C to R. Find a value on C such that 
 the number on the same point on D is identical. Reading on RC, we find im- 
 mediately above 2585 on D, 2585 on C. Obviously, this cannot be the cube root 
 sought, so reading now LC we read 1202, so 12.02 is the root sought. Again 
 taking the same example, we might have applied the rules as given. Pointing 
 off 1'734, one digit remains, and the two left-hand scales are called for. These 
 are the ones which we have used and the result will be as before. Notice if 
 the number had been 17340, we would have used, as we did in the first trial, the 
 LA with the RC and found 25.85, which would have been correct had the number 
 been 17340. 
 
 Take for another example the cube root of .000892 ; counting according 
 to the rule given, we end on the word "right" and we find .0963 as the correct 
 answer. In finding the cube roots with the slide inverted, the runner should be 
 used in finding the coincidences, as it helps the eye materially and is, of course, 
 not needed at the center index of C after the placing of that index. 
 
 We can now solve problems of the type f (84) 2 . With the slide direct, 
 set R to 84D and move the slide until the same number onB is under R, that is 
 seen on D under the left index of C; in this case, using the left scale onB, we 
 read 413. Notice that 80 2 = 6400 and that 41.3 cannot, therefore, be the number 
 sought; however, if we use RB we find 192 and 19.2 is the correct answer. The 
 question of which slide to use, direct or inverted, is largely a personal one. The 
 operator should, however, master both methods. 
 
 THE POLYPHASE SLIDE RULE. 
 
 With the Polyphase rule cube roots may be found to an accuracy of 
 14 of 1 % by the use of the scale on the edge of the rule. It will be noticed that 
 the scale is divided into three parts. The left-hand scale being for numbers of 
 one digit to the left of decimal point, the middle scale for numbers of two 
 digits to the left and the right-hand scale for numbers of three digits to the 
 left of the decimal point. If the number of digits be greater than three, sub- 
 tract any multiple of three and use the scales as indicated above; thus, for the 
 cube root of 1,860,000, subtract 6 from 7 and find cube root of 1.86 on scale 
 to left, setting runner to number on the cube scale and reading the answer 
 on D as 1.23. 
 
 It will be seen that the cube scale makes problems of the type At very 
 simple. Set R to A on the cube scale and read the answer on A under R. 
 10s = 4.65. Set R to 10 on cube scale and read 4.65 on A under R. 
 
— 20 — 
 
 XII. LOGARITHMS. 
 
 Logarithms of numbers may be found on the slide rule in two ways. 
 First, set LLC to the number on D and obtain the mantissa of the logarithm 
 on the center scale of equal parts to be found on the back of the slide, by 
 reversing the rule and reading under the line etched on the xylonite on the left 
 end of the back of the rule. To find the logarithm of 265, set 1C to 265D and 
 reading on the center scale on the back of the rule under the line, we find 423. 
 The desired logarithm is then 2.423. It should be remembered that the number 
 found on the rule gives only the mantissa of the logarithm and that the 
 characteristic must in every case be supplied in order to complete the logarithm. 
 In the second method, we reverse and invert the slide and read the mantissa 
 on the center scale directly over the number on D; the runner being, of course, 
 used to do this. Thus we find the logarithm of 265 to be 2.423 as before. 
 
 In finding the numbers corresponding to logarithms, the reverse of 
 either of these two methods is employed; the mantissa alone being set upon 
 the scale. Example: to find the number corresponding to the logarithm 
 3.697: With slide direct, set 697 on center scale on back of slide to line on 
 xylonite on back of rule, and read answer on D under 1C, 4980; or reverse and 
 invert the slide, put the indices in coincidence and set runner to 697 on the 
 scale in center, and read under runner 4980, the answer, on D. 
 
 XI11. GENERAL INVOLUTION AND EVOLUTION. 
 
 Logarithms are used in slide rule work mainly to raise numbers to 
 fractional powers. In raising numbers to integral powers, we can use either the 
 logarithmic method or a combination of squaring and cubing with direct 
 multiplication. Thus: (317) 7 = [(3.17) 2 ] 2 (3.17) 3 or (3.17) 2 X (3.17) 2 X (3.17)3. 
 This method is applicable to any power or root. Suppose we solve the problem 
 first by the method outlined above, with this change (3.17 X 3.17) 2 X 
 (3.17)3 = (3.17) 7 . Slide direct; 1C to 3.17D, R to 3.17B, LB to R, R to 10LB. 
 Read answer under Ron A, 3210. The logarithmic method depends, of course, 
 upon the fact that if we multiply the logarithm of the number by the exponent, 
 .the answer is the logarithm of the required number. Set \C to 3 . 1 ID. Read on 
 center scale on back of slide .501, the characteristic being 0. Set 1C to 5011), 
 R to 1C. Read on D under R 3.505. Set 505 on center scale on back to mark 
 on xylonite and read under 1C, 3210 on D, which is the answer. It may be 
 easily seen that the first method is the more accurate, while for the seventh 
 power, at least, the second is much the shorter. The logarithmic method is 
 especially adapted to problems involving fractional powers or roots, except 
 where the desired root or power is a multiple of two or three. For instance, 
 suppose we wish to find (717)1, or as it might be stated, the fifth root of 717 
 cubed. We may cube 717 and extract the fifth root. Cubing 717 gives 
 370000000, and the logarithm of this number is by the rule 8.568. Dividing 
 by 5 gives 1.713 and the number corresponding to the mantissa .713 is 516, so 
 the correct answer will be 51.6. An easier solution is this: Divide 3 by 5 
 giving .6; the logarithm of 717 is 2.855; 2.855 X .6 = 1.713 and the number 
 
— 21 — 
 
 corresponding as before to the mantissa .713 is 51.6. The problem as outlined 
 above is actually worked out as follows : Mentally — = .6; set \C to 717D, 
 
 o 
 
 read under line on back of rule 855, and with characteristic 2 the logarithm is 
 2.855; set IRC to 2.855D and read on D under 6C the number 1.713; set 713 
 on the center scale on the back of the slide to the line, and read on D under 
 1C, 516, and as the characteristic was 1 the answer will be 51.6. Remember 
 that the rule in every case gives only the mantissas and that the characteristic 
 must be supplied independently. 
 
 The problem of finding the root of numbers, the index of the root being 
 fractional, is slightly more complicated. Solve •'v / L~9 as an example: This 
 is equivalent to (1.9).oV The value of 1/.03, determined by the method of 
 finding reciprocals as previously explained, is 33.3. The logarithm of 1.9 =0.279; 
 0.279 X 33.3 = 9.3; number corresponding to mantissa .3 is 2. So the answer 
 will be 2 000 000 000. 
 
 As another type, take (.00273) - 232 , which represents the most difficult 
 class of problems in involution. Find logarithm of .00273 to be 7.436 — 10 or 
 — 3.436, remembering that the characteristic is negative and the mantissa 
 positive. -3.0 X .232 = -.696 
 .436 X .232 =.101 
 
 -.595 adding and subtracting 10, the logarithm is 
 9.405-10; and the number corresponding to mantissa .405 is 254. Therefore, 
 (.00273)- 232 . = .254 
 
 Let us now check this result as follows: 
 (.00273)* 232 = .254 if the work is correct. 
 
 (.00273)7x3= .254, for we set 232C to ID and under 3C find 13D. 
 (.00273) 3 should equal, if the work performed is correct, (.254) 13 . 
 (.00273) 3 = .0000000203. 
 
 = -ir(.254) 2 ]^ 3 X .254. 
 = .0646. 
 = .00416. 
 
 (.254) 13 
 (.254) 2 
 (.0646) 2 
 (.00416) 3 = .000000072. 
 .000000072 X .254 = .0000000183c/ifc. 
 
 THE POLYPHASE SLIDE RULE 
 
 The methods of squaring and cubing having been explained, it is easy 
 to see how to raise a number directly to the 4th, 5th and 6th power by the use 
 of CI scale. 
 
 a 4 , aCl to aD. Answer: A over LB. 
 
 a 5 , aCl to aD. Answer: A over aB. 
 
 a 6 , aCl to aD. Answer, on cube scale under 1C 
 
 XIV. EQUATIONS INVOLVING FRACTIONAL EXPONENTS. 
 
 Equations of the type X« =6 may be solved by the methods already 
 
 given, for X N = b is equivalent to X = 6n . Now let us consider a few examples 
 
 occurring in engineering work. The volume of the compression space in the 
 
 cylinder of a gas engine is \i of the piston displacement. What is the efficiency 
 
 (Vd\k-\ 
 = \Ve) 
 
 of the engine? The formula to be used is E 
 
 where Vd = \ Vc and 
 
— 22 
 
 k = 1.405. Substituting these values in the formula, E = (.25). 405 . Set \C 
 to 25D and read under line on back of the rule on the center scale 398, so the 
 logarithm required is -1.398, -1 X .405 = -.405 
 
 .398 X .405 = .1609, IRC to .405D, read 
 
 under 398C. Next subtract 
 
 -.244 
 -.244 from 10-10, and we have, as the logarithm of the answer, 9.756-10. 
 Set 756 on the logarithmic scale on the back of the rule and read on D under 
 1C, 574; E = 1-.574 or about 43% 
 
 Take as another example, Hodgkinson's formula for the strength of 
 cast-iron columns the lengths of which exceed thirty times their diameters. 
 
 J)3-55 
 
 The formula is W = 99318 ; where W equals the breaking load; D the 
 
 £1-7 
 
 diameter of the column in inches; andB its length in feet. Let D = 3"; and 
 
 3.3-55 
 
 B = 11.5'; W = X 99318. The example may be solved by finding 
 
 11.5»'» 
 the numerator and the denominator of the fractions separately and then 
 dividing them. 
 
 Log. 3 = .477 Log. 3 355 = .477 X 3.55 = 1.695 Number corres. = 49.5 
 Log. 11.5 = 1.061 Log. 11.5 1 " =1.061 X 1.7 = 1.805 " " = 63.8 
 
 49.5 
 So W = 99318 X = 76900. 
 
 63.8 
 
 This method is quite simple, but the following in many cases will prove 
 
 to be more satisfactory: 
 
 33.55 
 
 We may write the formula: W = 99318 . Since 3* = 11.5, 
 
 (32-22)1.7 
 
 Log. 11.5 1.061 3 3 ' 58 X99318 
 
 and x = = = 2.22; then we have = W => 
 
 Log. 3 .477 3 377 
 
 33-55 1 
 
 99318 X = 99318 X ; and since Log. 3* 22 = .477 X .22 = .105 and 
 
 33.77 322 
 
 99318 
 
 the number corresponding = 1.27, we have W = = 77000. This will 
 
 1.27 
 
 cover most of the cases which arise in involution and evolution. 
 
 XV. TRIGONOMETRIC COMPUTATIONS. 
 
 Trigonometric functions may be found on the rule either with slide 
 direct or reversed; sines and tangents being given directly by the readings on 
 the rule. On the reverse of the slide will be found two scales, one at the top 
 marked S giving sines, and the lower marked T giving tangents. To find 
 the sine of an angle with the slide direct, set the angle on scale S to the line on 
 xylonite on the back of the rule, and read the sine of the angle on B under the 
 index on scale A. For example, to find the sine of 15°, we set 15 on scale S, 
 to the line and under middle or right index of A we read .259; if we wish the 
 tangent of 15°, we set 15° on scale marked T'to the line on the back of the rule 
 
— 23 
 
 and read on C under index on D, .268. Or the slide may be reversed — the 
 left-hand indices brought into coincidence, and the sines and tangents read 
 directly, using scale of sines with A and scale of tangents with D. Reading 
 on A over 15° onB we have .259, and reading on D under 15° on C we find .268. 
 Notice that if the sine occurs on the scale LA, one zero stands between the 
 decimal point and the first significant figure of the desired sine; the sine of 2° 
 is .0349 and the sine of 20° 26' is .349. This should be kept clearly in mind. In 
 the same way, all the values of the tangent to be read on scale D will lie between 
 1. and .1. The sines and tangents of small angles are practically identical, and 
 may be found in one of two ways :- either by using the gauge point on the 
 rule, or by an application of the proposition that the sines and tangents of very 
 small angles are proportional to the angles themselves. Using the first method, 
 let us find the sine of 4' and 56". Set the gauge point which is seen just before 
 the 2° mark on the scale i? to 4A and read on A over LB, 1162. Then set gauge 
 point which is near 1° 10' to 56A and read on A over LB, 2715. Remembering 
 that the sine of 1" is .000005 and that the sine of 1' is .0003, we will write our 
 first reading .001162 and add to that our other reading 0.002715 and obtain 
 ,0014335, which is either the sine or tangent of 4' 56". 
 
 The second method is as follows: The sine of 4' 56" is equal to 1/10 the 
 sine of 40' 560" = 1/10 sine 49.33'; and by the rule we find sine 49' 33" = 
 .0143 and 1/10 of this value is .00143. Suppose we multiply 4' 56" by 20, we 
 have sine 4' 56" = 1/20 sine 1° 38.66'. We find sine 4' 56" = 1/20 = .0286 
 = .00143; the latter method is preferable in almost all cases. As will be 
 noticed, we cannot find directly the sine of angles between 90° and 360°, nor 
 tangents of angles greater than 45°. All that is necessary is, of course, to reduce 
 the functions as expressed to functions of an angle less than 90°. Thus the sine 
 of 169° = sine 11° by the familiar rule of trigonometry, and the tangent of 
 
 75° = i-r-n* In cases where we have to deal with other functions than the 
 
 tan 15 
 
 sine or tangent we have to transform the function by trigonometry until the 
 required function is expressed in terms of the sine or the tangent. Cosine A = 
 sine (90°-A); that is, to find the cosine of 49° 13' we find the sine of 40° 47', 
 which by the rule is .653. This method enables us to find all the functions 
 
 very simply: cos A = sin (90°-A), cot A = r t-> sec A = T esc A = 
 
 J * J v tan A sin A 
 
 -,or in finding the cosine of A we might divide the sine of A by the tangent 
 
 sin A 
 
 of A. Thus to find the cos 30° 15', find .503 as the sine of 30° 15' on scale 
 
 A, set R to 503D, 30° 15' C to R and read under 1C .864 on D, which is the 
 
 cosine required. In general, the first method is easier when the ang e lies 
 
 between 45° and 90°, and the second, when the angle lies between 0° and 45°. 
 
 Let us consider the example of finding the cotangent of 75°; evidently cot 
 
 75° = tan 15°. Notice that with the slide reversed we can multiply directly 
 
 by the sine or tangent of the angle without first determining the actual numerical 
 
 value of the function; needless to say, the process of division is equally simple. 
 
 217 3 
 Solve — , '•' , ., ', set 13° 14'C to 217.3D and reading under 1C find on scale 
 tan 13 14 
 
 D 923.0 as the answer. Solve 653 X sin 36° 41'; set LB to 653A, R to 36° 41'.B 
 
 and read answer 390 on A under R. 
 
— 24 — 
 
 1.36 
 
 Required the solution of the right triangle as given above. With scale direct 
 divide 2.17 by 4.36 and find .497 as the result. Let the runner rest at .497 and 
 reverse the slide, set indices together and read on C under R, 26° 25', which is 
 the value of the angle opposite the shorter leg; the other angle will be 63° 35'. 
 Set R to 2.17 A, 26° 25'B to R and read on A over IB 4.86, which is the length 
 of the hypothenuse. Check this result by squaring 2.17 = 4.71, and squaring 
 4.36 = 19, adding the two = 23.71, extracting the square root = 4.86 chk. 
 
 We may solve an oblique triangle using the law of sines very simply, in 
 any case where that law applies. 
 
 Solve the following oblique triangle: 
 
 4.32 
 
 34° 40' 
 
 With the slide reversed, set 34° 40' B to 3.19A and read the answer onB under 
 4.32A, as 50° 35', which is the value of the angle opposite the side 4.32. The 
 other angle will, of course, be 94° 45' and since we know sine 94° 45' = sine 
 85° 15', we can easily find the side opposite, by a simple proportion as above. 
 
 In general, it can be stated that any formula adapted to logarithmic 
 computations is adapted to computations with the slide rule. If a particular 
 problem calls for logarithmic functions, they can be very simply found by finding 
 the logarithm of the function. Suppose log tan 38° is desired. Reverse the 
 slide and set the runner on 38° C, then invert the slide reversed as it is, and 
 read on the scale of logarithms under the runner .893, which is the required 
 mantissa, the complete logarithm being 9.893-10. In finding the log sine use 
 the same general process, but multiply the mantissa as obtained on the scale 
 by 2, and take the fractional remainder as the required mantissa. Find the 
 log sin 20°. With the slide reversed and the indices coinciding, set R to 20° B, 
 invert the slide and read on the logarithmic scale under R, 767. Multiplying 
 by 2 gives 1.534 and .534 is the mantissa required. The logarithmic functions 
 are often useful in checking logarithmic work. 
 
— 25 — 
 
 XVI. VARIATION AS THE SQUARE. 
 
 The following example is typical of a wide variety of problems to which 
 the slide rule offers a simple solution. The example as given is very simple, but 
 from it others may be deduced. Given the parabola as shown: 
 
 OD ° 5ft. 
 OL ' 18ft. 
 
 to find the distance OA which corresponds to a distance of AB = 3.37 feet. 
 Subtracting 3.37 from 5 gives 1.63 = CK. To solve, set 18C to 5A, R to 1.63A, 
 and read answer under R on C as 10.29. This may be checked by finding the 
 
 324 
 
 equation of the parabola, which is found to be y 2 = x; when x = CK and 
 
 5 
 
 y = OA, we have y = 
 
 V 
 
 324 X 1.63 
 
 = 10.29. Problems of this type are very 
 
 useful in structural design, where the moment curve is a parabola, or in rail- 
 roads where a vertical curve is parabolic in form. The setting, as given, is 
 deduced from the fact that on a parabola the abcissae are proportional to the 
 squares of the ordinates. Let us work one simple problem in detail. 
 
 Given a beam uniformly loaded with a load of 1000 lbs. per foot ; the 
 beam being 17 feet long, to find the moments at one foot intervals over the span: 
 in this case the moment curve is a parabola with vertical axis. Center distance 
 WL* 1000 X17 2 17 
 
 = = ; SB to 17D, read the answer on A over LB as 36100; set — or 
 
 8 8 2 
 
 8.5C to 3.61A and corresponding to the following distances on the beam, as 
 measured from the center, we have as per the table: 
 
 8.5 7.5 6.5 5.5 4.5 3.5 2.5 1.5 0.5 
 36100 28200 21200 15100 10100 6130 3130 1120 125 
 
 These values are obtained by setting as above and reading the numbers 
 on A corresponding to distances on C. That is, 8.5C corresponds to 36100A, 
 7.5 to 28200A, etc. To find the moment subtract each number as found from 
 36100 and we have the actual moments at corresponding points. 
 
 End 123 45 678 
 
 7900 14900 21000 26000 29970 32970 34980 35980 
 
 The distances being expressed now as distances from the end of the beam 
 instead of from the center. 
 
— 26 
 
 XVII. SOLUTIONS OF SPECIAL FORMS. 
 The following solutions of general formulae are instructive. 
 
 (1). The type — ; which occurs in hydraulic computations. 
 gx 
 
 Let v = 31, g = 32, x = 8. 
 
 Mental approximation; 31 divided by 8 is approx. = 4. 
 
 Set 32B to 31D, R to IB, 8B to R, read answer on A over IB, 3.75. 
 
 (2). The type 
 
 J- 1 
 
 is solved in the same way except that the answer 
 
 is read on D instead of A. Care should be taken, however, in all problems 
 involving roots to use the proper scales, when using A andB together. 
 
 m v 2 
 
 (3). The type ; same as (1), with a coefficient. 
 
 2g 
 Let v = 8.5, m = .93, 2g = 64.4 
 Mental approximation: 80/64 = approx. 1. 
 
 Set 64.4J5 to 8.5D, and over .935 find 1.04 answer on A. 
 
 (4). The type ^ 2gh ; 
 a 
 
 Let 2g = 64.4, h = 12.5, a = 5. 
 
 Mental approximation; — - — = approx. 5.0. 
 
 5 
 
 Invert the slide; set 64.4C to 12. 5A, under 5B find answer on D as 5.67. 
 
 Care should be taken to use the proper scales; notice that with the wrong 
 
 choice 179 is read under 5B. 
 
 (5). The type b ; 
 c 
 
 Find the square root of b, divide by c and multiply by a. This can be 
 
 done either direct or inverted. Direct: set cC to bA, and under aC find the 
 
 answer on D. 
 
 (6). The type a 
 
 bXc 
 
 ;take for an example of this type the formula 
 
 \ d 
 for the flow through a nozzle as expressed by the formula: g = .7854d 2 
 
 V 
 
 2gh 
 
 ^-^prv and let us consider its solution in detail. First find the value 
 
 of .7854d 2 , which is the area of a circle with diameter d. 
 Let d =1| in., then .7854d 2 = 2.4 sq. in. 
 Let D = Sy 2 in., 2g = 64.4, c = .97, h = 130. 
 
 (^) 2 = 1.06; (^) 2 = .25 [(1) 2 -(|) 2 ] = .81. The formula has 
 
 now been reduced to 2.4 
 
 -JL- and this formula is of the type^ Xc . Mental 
 
 ^1 .81 "V d 
 
 approximation: 65X150 = 10000,1/10000 = 100, 100 X 2.4 = 240. 
 
the process 
 
 Considering this formula, it is evident that when we have completed 
 
 4 
 
 IbXc 
 
 the result should be indicated upon scale A in order to 
 
 facilitate the finding of the square root; this being found, we multiply the result 
 by a. Any method of multiplication which gives the result on A can be used. 
 Theie is no reason, therefore, why the scale should not be used direct. 
 
 Set SIB to 64.4A, R to 130.B, IB to R, under 2.4C read the answer on D as 
 244 cubic feet per second. Again we must be careful in selecting the scales to 
 use on A and B. Estimating the value of the expression under the radical and 
 mentally extracting the square root, it is seen to be about 100, therefore, the re- 
 sult of the quantity under the radical should appear on the left-hand scale of A. 
 This simple approximation should always be made where a root is involved. 
 
 m \/H* 
 
 (7). The type — or . The analysis shows that we should 
 
 b b 
 
 cube H so that the result falls on A, transfer to D and divide by b. 
 
 Let H = 1.79, b = 3.32. Notice that H 3 = approx. 8, so we wish 
 the result of this step to fall on LA. 
 
 Mental approximation: 3/3.3 = .9. 
 
 Slide direct: set 1C to 179D, R to 1795, 332C to R, read answer on D 
 under 1C, as .722. With slide inverted, set 179C to 179D, and under 332E find 
 answer on D as .722. As the second method requires a change of index, in 
 this particular case it is little, if any, shorter. In most cases, however, the second 
 method is the quicker of the two, and is, therefore, to be recommended. 
 
 (8). The type a X bi ; an example of this type is the formula Q = 3.336/tl 
 due to Francis. The quantity Q is the discharge over a trapezoidal weir in 
 cubic feet per second, b the length on the crest in feet, and h the head of the 
 weir. The formula may be written &/.3 X fcf; as is readily seen, the inverted 
 slide offers the best solution. 
 
 Let h = 1.43, and b = 4.51. 
 
 Mentally approximate: (l^) 3 = 3.5, V3.5 = 2, 2 X 5 X 3 =30. Ans- 
 
 Solution with slide inverted: Set 1.43C to 1.43D, R to 1C, 4.5LB to R, 
 read answer on D under 3J5 as 25.7. Notice that the cube of 1.43 is approxi- 
 mately 3.5, so that the result of the process of cubing should appear on the 
 left scale of A. If it is made to appear on the right, the answer is found to 
 be 813, which is impossible. 
 
 VaX b 
 (9). The type is sometimes given. This is, of course, equiva- 
 
 lent to the expression 
 
 x 
 
 «X6 2 
 
 and in this form, it is of more common occurrence. 
 
 c- 
 
 Its solution requires us to find the ratio — , square it and multiply by a. 
 
 c 
 
 Let a = 9.18, b = 3.12, c = 4.26. 
 
 Mental approximation: 81/16 = 5. 
 
 Use the slide direct; set 4.26C to 3.12D, read answer on A over 9185 
 as 4.93. These types shown are few in number, but they serve to illustrate the 
 method of finding a setting for a given formula. 
 
- 28 
 
 XVIII. GENERAL CONDITIONS. 
 
 Remember in general that all formulae should be expressed in the form 
 of products, as the first step in their solution by means of the slide rule. The 
 expression for the cosine of any angle of a triangle in terms of its sides is a 
 
 familiar example of this process. Cos 3^A 
 
 ^ 
 
 s (s-a) 
 
 be 
 
 where s 
 
 the 
 
 sum of the sides a, b, c. 
 simple transformation. 
 
 This expression is derived from the law of cosines by a 
 As is readily seen, the form above is adapted to slide 
 
 rule computation, while the law of cosines is not. That is, Cos A = 
 
 b 2 +c 2 
 
 26c 
 
 cannot be readily solved by either the slide rule or by logarithms. 
 
 It is sometimes helpful to add and subtract numbers on the rule when 
 paper and pencil are not at hand. This may be done by reversing and inverting 
 the slide; suppose we have the following: Add 3.17 
 
 5.36 
 -3.81 
 -7.42 
 6.89 
 —4.19 
 With the slide reversed and inverted, bring indices into coincidence, 
 R to 3.17 log scale, log scale to R, R to 5.36 log scale, 3.81 log scale to R, R to 
 log scale, 7.42 log scale to R, R to log scale, invert slide and set right-hand 
 indices into coincidence. Read the distance on log scale between runner and 
 6.89. Read in this way: 3 1.1 .09 or 4.19. The operator will easily grasp 
 the principle of this setting and can readily see in what cases its application 
 will be of advantage. 
 
— 29 — 
 
 PROBLEMS. 
 
 EQUIVALENT RATIOS. 
 
 '(1). Find the areas of circles the diameters of which are respectively! 
 7.93. 9.76, .01345, 64.8, and 33.71 in. 
 
 Ansicers: 49.4, 74.8, .0001421, 3300, 893. 
 
 f (2). Find the number of cubic feet corresponding to the following 
 numbers of U. S. gallons: 8.0, 9.78, 351, 1003. 
 
 Answers: 1.070, 1.307, 46.9, 134.1. 
 
 (3). Transform the following volumes expressed in cubic feet to 
 litres: 33.9, 67.8, 85.4. 
 
 Answers: 960, 1921, 2420. 
 
 * (4). Find decimals of inches corresponding to the following fractions 
 of inches: 1/8, 3/32, 5/16, 9/16, 17/64, 19/64. 
 
 Answers: .125, .0938, .313, .563, .266, .297. 
 
 (5). Find decimals of a foot corresponding to the following numbers 
 of inches: 3X, 5H, 7%, 9*A, HM, &H, *H. 
 
 Answers: .260, .438, .646, .792, .979, .698, .375. 
 
 . (6). Find the pressures in tons per square foot due to the following 
 depths of water: 6J4 9, 17, 31.5, and 76 feet. 
 
 Hint: Determine first the ratio between pounds per square inch and 
 tons per square foot. 
 
 Answers: .203, .281, .531, .984, 2.38. 
 
 (7). Express to the nearest even sixteenth of an inch the following 
 decimals of a foot: .938, .642, .733, .657. 
 
 Answers: 11 \i, 7 ,, / i6 , 8"/i6, 7|. 
 
 (8). Transform the following pressures given in inches of mercury 
 to equivalent pressures expressed as atmospheres: 35.7, 33.4, 31.6, 27.3. 
 
 Answers: 1.194, 1.117, 1.057, .910. 
 
 (9). Find the weight of copper strips 1 in. wide, 1 ft. long, and of the 
 following thicknesses: 3 /i6, \i, 5 A,^-% in. 
 
 Answers: .716, .953, 2.38, 4.28. 
 
 (10). Find the weight in grams of a steel plate circular in shape with 
 diameter .1125 in. and thickness u / 32 of an inch. 
 
 Answer: .759. 
 
30 — 
 
 SQUARES AND SQUARE ROOTS. 
 
 (1). Find the squares of the following: 10.17, .979, 126.5, .0326, 
 4.77, 87.9. 
 
 Answers: 103.4, .958, 16000, .001063, 22.75, 7730. 
 | (2). Find the square roots of the following: .0735, .520, 1391, 600, 3.17. 
 Answers: .271, .721, 37.3, 24.5, 1.78. 
 
 (3). Find the squares of the following: 43/52, 47/64, 89/71. 
 Hint: 52C to 43D, read answer on A over IB. 
 
 24 8.9 ' 27 34.5 110 100.5 
 
 Answers: .683 or — or — ; .540 or — or ; 1.57 or or . 
 
 35 13 50 64 70 64 
 
 (4). Find the square roots of the following expressed as even sixty- 
 fourths: 9/64, 19/32, 3/8, 7/16, 1/32. 
 
 Answers: 24/64, 49/64, 39/64, 42/64, 11/64. 
 
 (5). Find the length of the side of a square to the nearest 64th of an 
 inch, the square containing 9.81 square inches. 
 
 200 
 Answer: — or 2>Y%. 
 64 
 ■ (6). Find the diameter of a circle to the nearest thirty-second of an inch, 
 the circle containing 7.97 square inches. 
 
 Hint: Use the ratio found on the reverse of the rule 79: 70. 
 
 204 
 Answer: — or 3 3 /i6. 
 64 
 
 MULTIPLICATION AND DIVISION. 
 
 (1). Solve 16.16 X 3.1416 X 18.13. Answer: 921. 
 
 39.8 
 Solve . Answer: .852. 
 
 (2). 
 
 (3). 
 
 (4). 
 
 (5). 
 (6). 
 
 46.7 
 
 83.7 X 1163 
 Solve . Answer: 12.54. 
 
 77.6 
 
 100.3 X 69.3 X 33.7 
 Solve . Answer: 2.75. 
 
 88.2 X 54.6 X 17.70 
 1131 X 94.8 
 Solve . Answer: .1750. 
 
 79.9 X 104.4 X 73.5 
 Solve 24 
 
 47 
 18 
 51 
 1 
 19 
 
 Answer: 0.0761. 
 
— 31 — 
 
 \J (7). Solve the following proportion, stating the answer to the nearest 
 
 17i 9| 54 22 11 
 
 thirty-second: = — . Answer: x = — or 1 — or 1— . 
 
 3£ x 32 32 16 
 
 (3.18) 2 x 
 
 (8). Solve the following: = . Answer: x = 3.27. 
 
 (10.17) 2 33.4 
 
 (9). Give the general setting for the solution of problems of the type 
 
 a 2 x 
 of (8) — = — . Ansiver: Set bC to aD, answer on A over cB. 
 £ 2 c 
 
 (10). Find the answer to this problem in kilogram meters when the 
 quantities are expressed in feet and pounds: 
 
 3.1416 X (5.07) 2 X 189 X 2.73 X 550 
 
 _ m Answer: 791000. 
 
 4 
 
 Hint: Deduce from the table of equivalents a factor to satisfy this case 
 and multiply the expression by it. 
 
 INVERTED SLIDE. 
 
 (1). Find the reciprocals of: 389, 40.9, 7.79, 1019, 6.59. 
 
 Answers: .00257, .0244, .1284, .000981, .1517. 
 
 (2). Multiply 13.14 X 9.63 X 7.46 X 349. Answer: 329000. 
 
 Multiply 159.6 X .834 X 76.7 X 9.13. Answer: 93200. 
 (3). Solve this example with the slide inverted and also with the 
 
 3.43 X 79.6 X 928 X .1119 X .06888 
 
 slide direct: . Answer: .0001471. 
 
 19.15 X 47.7 X 344 X 42.2 
 
 (4). Find a simple method for solving the following and determine 
 
 1111 
 
 their numerical values: , , 
 
 (3.29) 2 (18.72) 2 (4.13) 2 (1.251) 2 
 
 Answer: Invert slide and read over 3.29, 18.72, etc., onJ5, the answer 
 on A as: .0924, .00285, .0586, .638. 
 
 (5). Find a simple setting for problems of this class and determine 
 numerical results with the data as given. 
 
 A lever of the first class has the fulcrum 6 in. from the end where rests 
 a weight of 380 lbs. At what distances from the fulcrum will the weights given 
 be placed, one weight acting at a time, in order to insure equilibrium? The 
 weights are 50, 73.4, 16.93, and 217 lbs. 
 
 Answer: Invert slide, set 5B to 380D and under the number of pounds 
 in B find the answer in feet on D as follows: 3.80, 2.59, 11.22, .875. 
 
 (6). Find the supporting forces on the two ends of a beam 22.3 feet 
 long under the action of a movable load of 625 pounds. Reactions are desired 
 when the load is concentrated at distances of 2, 4, 6, 8, 10, 12 feet from the left- 
 hand end of the beam. Answers to be given to the nearest pound. Answers 
 55-570, 112-513, 168-457, 224-401, etc. 
 
— 32 
 
 (7). Find the number of amperes required to run a motor developing 
 15 horsepower at voltages of 110, 220, 550, and 750; using the formula H. P. — 
 
 VXA 
 
 . Answer: 102, 51, 20.4, 14.9. 
 
 746 
 
 (8). At a pressure of two atmospheres a certain quantity of a eartain 
 gas has a volume of 22 cubic feet. Assuming Boyle's law to hold, that is; 
 PRESSURE X VOLUME = CONSTANT, find the volume of the gas at 
 pressures of 33, 37, 40, 45, 46 pounds per square inch. 
 
 Answers: 19.6, 17.5, 16.2, 14.4, 14.1. 
 
 (9). If a pump A of capacity 47 cubic feet per minute can empty an 
 excavation in 12 l /i days at a cost of $3.45 per day, and if a pumpB, capacity 
 38 cubic feet per minute, which runs at a cost of $2.83, is also available, which 
 will be the cheaper to use ? Answer: A. 
 
 (10). State the simplest setting for (9). 
 
 Answer: Invert slide. Set 47B to 1225D, R to 38B, 283B to R, R to 
 1225D. If the number on B under R is greater than 345, A is the cheaper. 
 
 THE FORMS XY = B AND X 2 Y = B. 
 
 (1). Find the dimensions of a rectangle, its sides being determined to 
 the nearest inch only, so that the rectangle shall contain as nearly as possible 
 1317 square inches. The short side of the rectangle must be over 25 in long. 
 
 Answer: 28 X 47 in. 
 
 (2). Determine the dimensions of a right triangle so it will contain, at 
 least, and as near 317 square inches as possible. The sides are to be measured 
 to the nearest J^ in. The short leg of the triangle must be 15 in. or over. 
 
 Answer: 29 J^ X 21 J^ in. 
 
 (3). A sewer must have a cross sectional area of at least 716 square 
 inches and is elliptical in section. Find the lengths of the major and minor 
 axes to the nearest Inch. The minor axis must be between 20 and 25 inches 
 long. Area of ellipse is .7854 a X 6, where a is the major and b the minor 
 
 axis. Answer: 24 X 38. 
 
 (4). Determine the number, width and thickness of steel plates for 
 the flange of a girder, so that the gross cross-sectional area of the plates shall 
 exceed by an amount as small as possible, 23.7 square inches. The widths of 
 the plates can vary by half inches from 10 to 12 inches, and in thickness from 
 % to Ys by sixteenths. The plates must all be the same width and should vary 
 in thickness as little as possible. 
 
 Answer: 3 PI. 1VA X V% in., 1 PL 11 K X Vie in. 
 
 (5). Determine plates required under conditions which are identical 
 with problem (4), except that the gross area must be 43.6 and the plates may 
 vary in width by half inches from 14 to 18 in. 
 
 Answer: 5 PL 14 X Ys in., or 4 PL 17 ^ X Ys in. 
 
— 33 — 
 
 (6). Design the section of a rectangular wooden beam using the 
 
 fbh* 
 
 formula M = , where M = 31250 inch pounds, and / = 980 lbs. per 
 
 6 
 square inch. The minimum value of 6 is two inches. Both b and h must be 
 multiples of two. Answer: 2 X 10 in. 
 
 (7). Design a beam under the same conditions as problem (6) where 
 M = 72500 inch pounds and / = 1000 lbs. per square inch. Minimum value 
 for 6 is 6 in. Answer: 6 X 10 in. 
 
 (8). Determine b and h for a rectangular reinforced concrete beam 
 where M = 125 b h 2 . M = 1500000 inch pounds, h must be at least double 
 b, and b at least 13 in. Determine the dimensions to the nearest J^ in. Answer: 
 13^x30 in. 
 
 CUBES. 
 
 (1). Cube the following numbers; 728, 8.07, 55.9, 10.17. 
 Answers: 386000000, 526, 174700, 1052. 
 
 (2). Cube the following with the slide inverted: 9.77, 100.5, .246, 
 .0133. Answers: 933, 1015000, .01489, .000002353. 
 
 (3). Solve the following with the slide direct: (15.1)?, (2.34)1, (.0327)1. 
 ( 21 / 3 i)I. Answers: 58.7, 3.58, .00591, 17 /32 or .532. 
 
 (4). Deduce the setting for xf with the slide inverted, when x lies 
 between 1 and 10. Answer: xRC to xD, under IRC read answer on D, or 
 xB to xLA, under IRC read answer on D. 
 
 (5). With the scale inverted find the following; expressing the answer 
 to the nearest thirty-second of an inch. (3.71)1, ( 7 A 6 )I, (1V 8 )I, (47.5)i. 
 
 229 9 38 10480 
 
 Answers , — , — , . 
 
 32 32 32 32 
 
 CUBE ROOTS. 
 
 (1). Find the cube roots of the following with the slide direct: 2.613, 
 17.8 .00321, 1793, .0488. Answers: 1.377, 2.61, .1475, 12.15, .3655. 
 
 (2). Find the cube roots of the following with the slide inverted; 
 3.93, 14.16, .000444, .01331, 571. Answers: 1.578, 2.42, .0763, .237, 8.30. 
 
 (3). Solve the following: f/{8.93) 2 , f / (71.7) 2 ^(IA2S)\ f / (.0321)K 
 Answers: 4.30, 17.26, 1.265, .1010. 
 
 (4). Deduce the general setting for the type Hi, with the slide inverted. 
 Answer: 1C to HD. Look for coincidences between A and B, or C and D. 
 The numbers coinciding will give the answer. 
 
 (5). Solve the equation x| = 18.17. Answer: x = 77.5. 
 
 (6). Deduce the setting for the solution of the type shown in (5). 
 Answer: Set right-hand member on A to right-hand member on B and 
 read x on D under 1C. Slide inverted. 
 
— 34 
 
 (7). Solve the equation x% = 9.81. Answer: x = 4.58. 
 
 (8). Deduce the setting for the solution of the type shown in (7) with 
 the slide direct. Answer: R to right-hand member on D, move the slide until 
 the same number on B is under R, that appears on D under 1C. This number 
 will be the required value of x. 
 
 (9). A right circular cone contains 347 cubic inches and the radius 
 of the base is 7{ in. What will be the volume of a similar cone, the radius of 
 the base being 8.19 in. ? Answer: 500 cu. in. 
 
 (10). If a sphere of (a) inches radius contains 400 gallons, what will be 
 the radius of a sphere containing 600 gallons. Answer: 1.144 a. 
 
 (11). Find in problem (10) the radii of the two spheres in feet. 
 Answer: 2.33 ft. and 2.67 ft. 
 
 LOGARITHMS, INVOLUTION AND EVOLUTION. 
 
 (1). Find the logarithms of the following numbers to three places 
 in the mantissa: 2181, 131.7, .983, .0436, 7.94. 
 
 Ansivers: 3.339, 2.120, 9.993-10, 8.639-10, 0.900. 
 
 (2). Find the numbers corresponding to the following logarithms; 
 2.736, 7.1316-10, .4717, 6.979-10. Answers: 544.5, .001354, .2963, .000953. 
 
 (3). Find the fifth power of 3.72 by two distinct methods. 
 Answer: 712. 
 
 (4). Solve (1.217)1 by the use of logarithms. Answer: 1.387. 
 
 (5). Solve (8.44) 127 and (.00511) 311 . Answers: 15.01 and .0000000747. 
 
 (6). Solve (.0132)-" 05 and - 31 \/1729. Answers: .173 and 27.8 billions. 
 
 EQUATIONS INVOLVING FRACTIONAL POWERS. 
 
 wx 
 We f 
 
 (1). Solve the equation S = , when W is 10 tons, x 
 
 f 
 
 100 feet, 
 
 w = .27 pounds per cubic inch, / = 10000 lbs. per square inch. Value for e 
 
 will be found on the reverse of the rule. Express the answer in square inches. 
 
 Answer: 2.066. 
 
 99318 X D 3 " 55 
 
 (2). Using Hodgkinson's formula W = , find W when D 
 
 L 17 
 
 = 2.5 andL = 17.8. Answer: 19230 lbs. per sq. in. 
 
 (3). Solve the equation x 218 = 3.43. Answer: x = 1.760. 
 
 14 X 3.1416 X (5)J 
 
 (4). Solve for t: t = 
 
 15 
 
 —V 64.4 X 12 
 4 
 
 Answer: t = 23.6. 
 
 (5). Solve for q in the equation: q = 3.01 X 16.7 X (2.32) 1 . 53 
 Answer: 182.2 
 
— 35 
 
 
 *S^> 
 
 TRIGONOMETRIC COMPUTATIONS. 
 
 (1). Find sin 32° 14', sin 85° 38', tan 19° 14', tan 22° 41'. 
 Answers: .533, .997, .349, .418. 
 
 (2). Find cos 29° 17', cos 73° 04', cot 89° 19', cot 12° 51'. 
 Answers: .872, .291, .01193, 4.38. 
 
 (3). Find sec 45° 09', sec 73° 33', esc 11° 13', esc 36° 28'. 
 Answers: 1.419, 3.53, 5.14, 1.685. 
 
 (4). Find sin 00° 18', tan 00° 07', sin 00° 41', tan 01° 03'. 
 Answers: .00524, .00204, .01193, .01833. 
 
 (5). Perform the following operations: 
 
 sin 41° 03' X 386 tan 20° 10' X 31.3 
 36.1 X sin 30° 35', 
 
 sin 10° 17' tan 43° 16' 
 
 Answers: 18.4, 1420, 12.21. 
 
 (6). Find log sin 39° 19', log sin 15° 52', log tan 40° 40', log cot 49° 35', 
 log sec 86° 17', log esc 4° 44'. 
 
 Answers: 9.802-10, 9.437-10, 9.934-10, 9.930-10, 1.188, 1.083. 
 (7). Solve the triangle as given. 
 
 31.3 
 
 104.6 
 
 Answers: A = 16°39',B = 73° 21', c = 109.2. 
 (8). Solve the triangle as given. 
 
 65°40' 
 
 Answers: A = 24° 20', b = 108.5, a = 49.1. 
 (9). Solve the triangle as given. 
 
 30°03: 
 
 Answers: A = 22° 14', B = 127° 43', b = 12.86. 
 
— 36 — 
 
 (10. Solve the triangle as given. 
 17° 20' 
 
 12° 13' 
 
 Answers; B = 150° 27', b = 7.76, a = 4.70. 
 
 VARIATION AS THE SQUARE. 
 
 (1). A circle has an area of 34.71 square inches. What will be the 
 diameter of a circle whose area is 3 34 times as large? Answer: 12 in. 
 
 (2). An equilateral triangle has its sides 3.17 in. long and has an 
 area of 4.33 sq. in. What is the area of an equilateral triangle whose side is 
 3.47 in.; one whose side is 7.62; one whose side is 17.1 ? 
 
 Answers: 5.18, 25.01, 125.9 in. 
 
 (3). A bar of steel 1 ft. long and \\i in. in diameter weighs 4.17 lbs. 
 What will be the weight of bars 1 ft. long and with the following diameters: 
 134 1% and 2 ^ in. ? Answers: 3.38, 8.18, 16.6 lbs. 
 
 (4). In problem (1) if e is the area of the given circle, 6 the ratio of 
 the area of the required to the given circle, find the general setting for problems 
 of this type. Answer: Set 1RB to eRA, R to bLB, 79C to 70D, answer under 
 runner on C. 
 
 (5). A beam 28 feet long is loaded with a uniform load such that the 
 moment at the center is 7070000 inch pounds. Find the moments at 2 foot 
 intervals over the beam. 
 
 Answers in tens of thousands of inch pounds. 
 
 2 4 6 8 10 12 14 16 etc. 
 
 188 346 476 577 649 693 707 693 
 
 (6). With the parabola as given with the vertex at (a) find the distance 
 (b) so that c is 21.5 feet. Answer: 61.4 ft. 
 
 34.5 
 
 < lOO 
 
> 
 
 - . • 
 
 — 37 — . 
 
 (7). A parabola as given with the vertex at a; dc = 238', ad = 440'. 
 Determine perpendicular offsets to the parabola from points 40 feet apart 
 
 on ad Answer: 
 
 
 
 
 
 
 
 
 
 a 40 80 
 
 120 
 
 160 
 
 200 
 
 240 
 
 280 
 
 320 
 
 360 
 
 400 
 
 1.9 7.9 
 
 17.7 
 
 31.5 
 
 49.3 
 
 71.0 
 
 96.8 
 
 126.0 
 
 160 
 
 197 
 
 SPECIAL FORMS. 
 
 (1 
 (2 
 (3 
 (4 
 (5 
 
 (6 
 
 (7 
 
 (8 
 (9 
 
 (10 
 
 (8.72) 2 
 
 Compute . Answer: .289. 
 
 32.2 X 8.17 
 
 (91.5) 2 
 Compute A /— — . Answer: 3.90. 
 
 4 
 
 32.2 X 17.17 
 
 .925 X (13.11) 2 
 
 Compute . Answer: 2.47. 
 
 64.4 
 
 \/MA X >/73.8 
 
 Compute . Answer: 15.81. 
 
 4.36 
 
 13.16 X \/.036 
 
 Compute . Answer: .1290. 
 
 19.37 
 
 12.718 X 3.4 
 
 Compute 8.88 A . Answer: 3.14. 
 
 73.7 
 
 4 
 
 \/(3.13) 3 
 
 Compute . Answer: .420. 
 
 13.19 
 
 Compute 8.9 (.032)f. Answer: .0510. 
 
 \/3.18 X 9.72 
 
 Compute . Answer: 13.02. 
 
 1.331 
 
 132.9 X (1.313) 2 
 
 Compute .» . Answer: 7.71. 
 
 (.977) 2 
 
 4 
 
— 38 
 
 TABLE OF CONTENTS. 
 
 Page 
 Preface 2 
 
 Accuracy and Significant Figures . . 3 
 
 Description of the Rule 5 
 
 Decimal Point and Reading the Scale 6 
 
 Polyphase Slide Rule 14 
 
 Principle of the Rule .6 
 
 Equivalent Ratios 7 
 
 Squares and Square Roots . 9 
 
 Multiplication and Division 10 
 
 Inverted Slide 13 
 
 Forms xy = B, x 2 y = B . 15 
 
 Cubes 17 
 
 Cube Roots 18 
 
 Logarithms . . 20 
 
 Involution and Evolution (General) 20 
 
 Equations Involving Fractional Exponents .... 21 
 
 Trigonometric Computations 22 
 
 Variation as the Square 25 
 
 Special Forms . 26 
 
 General Conditions 28 
 
 Problems . 29 
 
 
» • * 
 
 c ■ ■ 
 
 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 
 
 fLM4 LfakJAU; 
 
 **t* 
 
 T ^ r ,^, fV ^r l ,.,,.,,,,l 1 . l |{.,. 1 ^ r p (l . Mj ..j-^.i 
 
 | r| i i-V-|U-i- l V|,| ,M. l , l i l ' . ' |,,,ViiiriyiVi ',,;V r i '' iV i: H " "l '*P 
 
 i 'i tt ( t'^^'TnrrT 
 
 ,1,1,1,1, f.T,l.l.l.l.l.[.I.I.I.Bl r l.l.lJ.[.i.l.i.fi.i-l,i,t,M.l.l.fi.il.l.l.l.i.l,l.a.l.:.lJjJjJ 
 
 TT 
 
 ■ 1 \ t ■ ■ jJ 1 1 jli i t J A t ij t ,^^ ■ - f^t 1 1 ^ .t r ^^ . j^u-'^ u ^L i ^ ;,.. < j! M 1 i fc^ . -i , r . -7f . i' . -? ; -^ . -^^- . a : , -^T I ? ; .^ i : i :.^ t ; . .'. i- .' j-V : :-" J • - ! j;J.h j-0.- . t : ^ 
 
 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 
 
 anan 
 
 lnnl« Mliii.li»,tn,li,nt,i)i.>Ii lilil ,lil,l,l:l,l,IJIIililJ,l l !,l,lllili!,'lli l il,J,M.liiiiTiinli l Ji-,'' 
 
 » 1' >" i ■ ■ i ■>>. . . i. < .ii" ..j.^irrapnSfi^vi,! 1 1 m i j < i f j< : ' i ' i ■ i H* [yfin" J» w: ■V^i'W'iTi'i'rti' i ...... . 4 . .- 
 
 11 tu^iiki3f'^i,n)ri,l,iJrt,lJilili>,l,ljl,t.i„mi,l,i,i,l>i,l,i,l,*,l,l,lilililil,lil<f<litihlil,lihliliV 
 c liii,linl!lii Imil.iul? X,i,.mI?hI»X XhkSUJiaJiiiiIiiiiIiiiiIiiiiIiiiiL 
 
 
 :;;!:■ ■'■irllnUI|,?il4'lil,l,l l,li''l.l,!! l .l. l i'Ul,lAli ! Jiiiilm*'iiili'iMnilj"iiiili 
 
 ^Tiprijwrp^ 
 
 JiHimfalwiiiMuulii !.,,,ij,iu,„',.,;.,t .1. ,i„,)„.l:~Y.„i, „J,, ,,!„,',, „l,,,Xi, ill, -I 
 
 •V T " r ' '''' T r"'"?T^.* '-■ ,' ? .' ;'■'■' '■'■' ' • ' ■ ' L ^' ' ■ ' >r 1 1 ■ I ■» ■ I . t -i ^-i i . ri ■ i.i.'.rT.r.-.i.i.i. i .i.!.,?;,; i.i.nr i - 1 iT.r i.i i.j ■ __j^ 
 
 t*.q'|iii|l|i|Mlinn l INI|M ; ''|'lf|i'^.'H'|i|'ii|.riWI'|i|'[ 
 
 hrimffiwii'rtii'tfifiij 89* 
 
 j _ , i j i I ,py 
 
 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 
 
 a-t^—a' 
 
 "|i i i a H^|3J!ii^ i | i | i|i | i | i |i|i[ i | i | i |i| i|i 
 
 ClF biofuitdnuLlulmiliiiJim 
 
 ■ ' .. j . ,.| ? , | . ..! " . . 
 
 m .f.Vn'i i'j-i^u.1i^,A„,i,„?i 1„„ 
 
 JiiiiJJi.^iJlJoiil iiIiimLiiIiim1i.iLiM.iU[i 
 
 ■ J ' Mfl ' i^: J ' J ' l ' KJW ' fl^tKfW^ i ■ ■ ' I" "[ ' j "I 1 m[|;n|m ' j^H'm|'Tj < [ i ii i £ i ifr 
 
 jj. 
 
 ^TT 
 
 £# 
 
 UjJ,IJiliJiMililit.l. 
 
 i 'lhI ii .; ,!!,,,, . j.: ; 
 
 ^—■r—^ 
 
 
 
 iii ftfl ii finr| i j i jJ i j ii | i r i jfjiJifm i T ! . ' r ' {jf 
 
 i.i.U* 
 
 
 T 
 
 T>/t*fiOTiii^' 
 
 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 
 
 10 80 JO *- ^o tip 
 
 liiiilini l,inN«*Ji..,!iuH.,»i.!ilii.'ilililihl l l,liiiliMl l W,Wili,'d',ai|ii,il,iii ]„il, ■!. ,!■■ J -.1 ., . iV. UimlnJutoU il 
 
 •^■i 
 
 ■ ■' I ■' ■« "J . i.i riiii'ui'uiiliujUi^M iM iilii.i : iliUn 
 
 "ii"i l .-> , ao"WiK , * , " - «" , "S'"Tir r aini | i,' 
 
 U£iKlUU£JCalU^U k SIN 2a 
 
 _r f r t i r f>\ f , , y, ■ . tf , iftny.Ei/.t^Mi.Miy liafj i; 
 
 J p- 
 
 4oa soe tuc 000 1LJ 
 
 ulaJ:mJii.ii,...l....ii.'.i.ii l,U 
 
 EOS?! 
 
 .■i,;ff;iii.|iiiit|JiiViliij™iui|.'iiiij|j 
 
 This form of Stadia Slide Rule is remarkable for its simplicity. When 
 the stadia rod reading and the angle of elevation or depression of the telescope 
 are known, the horizontal distance and vertical height can be obtained at 
 once by one setting of slide. 
 
 Write for our Slide Rule Booklet, fully describing the foregoing, as well 
 as all other Slide Rules made by us. 
 
EVERY REQUISITE of the ENGINEER 
 for the FIELD or OFFICE 
 
 K & E Surveying Instru- 
 ments, Transits, Levels, etc., are 
 used on nearly every important 
 Government, Municipal and 
 Private engineering work. They 
 excel in design, workmanship 
 and accuracy and embody the 
 latest improvements. 
 
 We carry the largest and most complete assortment of Draw- 
 ing Papers, Tracing Cloths and Papers, Blueprint and Brownprint 
 Papers. 
 
 K & E Ready Reading Tapes represent the latest improve- 
 ments in the method of numbering tapes, resulting in a saving of 
 time and reducing the possibility of errors in reading. 
 
 Write for our 566 pa£e General Catalogue. 
 
 • KEUFFEL & ESSER Co. • 
 
 NEW YORK J27 Fulton St. GewrelOmcc *ndF&rtori cs ,HOBOKEN,N.</. 
 
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 General Library 
 
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 UNIVERSITY OF CALIFORNIA LIBRARY