# Of the Usage of the Sliding Rule

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1 Of the Usage of the Sliding Rule Laurent GRÉGOIRE August 2007 ontents 1 Introduction 2 2 Mathematical principles 2 3 Multiplying Principle and? Example Out of range product Mantissa and exponent ivisions First method Second method Reciprocal haining multiplications and divisions General method Multiplying 3 factors Proportions & conversions 9 7 Squares & square roots Squares Square roots ubes and cube roots 11 9 Usual calculations onverting degree/minute Volume of a cylinder π-shifted scales

2 10 Trigonometry Logarithms, exponential, powers ecimal logarithms ecimal powers Powers Natural logarithms, exponential Squares Introduction From the nineteenth century up to the 70 s, slide rules have been the precious auxiliaries of engineers, architects and technicians worldwide. Since then, they have been outdated by electronic pocket calculators. Slide rules allow one to quickly perform a lot of numerical computations, such as multiplying, dividing, computing squares, cubes, extracting square and cube roots, trigonometry, etc. Figure 1: Graphoplex slide rule, Polyphase type. In this article, we are going to study in details the principles of a classical slide rule ( Mannheim or Polyphase and derivatives), through the analysis of the underlying mathematical relations and several numerical examples. An high-school level in mathematics should be sufficient enough to read this article. 2 Mathematical principles All sliding rules are based upon the usage of logarithms. The logarithm is a quite handy mathematical function which allows us to transform a multiplication into an addition: log(a b) = log(a) + log(b) The logarithm of the product is the sum of the logarithms. That way we transform a multiplication, which is quite complex to perform, into an addition, which is much easier. The adding of two values is easy to achieve by the juxtaposition of two physical lengths: to multiply a and b, we add their logarithms. The value c which logarithm is equal to this sum is then equal to the product of a by b. Following the same idea, we use the equation below to divide two numbers: log(a/b) = log(a) log(b) 2

3 A division is tranformed into a substraction. The same principle is used for manual computation, but by using numerical logarithmic tables. 3 Multiplying 3.1 Principle Below is a logarithmic scale, on which each graduation a is located at a length of l a = log(a) from the origin 1 (figure 2). l a = log(a) a Figure 2: Logarithmic scale. This logarithmic scale has the noteworthy property that if you translate a cursor on this scale by some length k starting from any number a, you will find this number multiplied by k (k = log(k )) (figure 3). k k a 1 a 1 k a 2 a 2 k Figure 3: Adding a distance to multiply. In order to multiply two numbers a and b, we use two logarithmic scales and. The length l a on between the graduation 1 and the graduation corresponding to the number a is l a = log(a). In the same way, the graduation of b on the scale will give us the length l b = log(b). Thus the length l c = l a + l b is the logarithm of c = a b. More precisely, we place the two scales and side by side, and by translating of a length l a we align the base of scale (the 1) with the graduation of a on scale. Then, we report length l b after l a (starting from scale base) up to the b graduation on scale. The length l c from the base of scale to this graduation, is l c = l a + l b. We can then directly read on scale the product a b (figure 4). l c = l a + l b b 1 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1, l a a l b a b Figure 4: Multiplying a by b. 3

6 l b c = l a l b l b 1 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1,9 2 3 l a a a /b Figure 8: ivision, first method. 8.3, / /10 Figure 9: ividing 3.95 by Second method We use here a second scale I, inverted from (I Inverted, in red on figure 10). The base of scale I (graduation 1 on the right hand-side) should be aligned here with the graduation on scale of dividend a. Going back to the left (that is, by substracting from l a the length l b ), the graduation on scale aligned with b on scale I gives directly the result c = a /b. l b c = l a l b l b ,9 1,8 1,7 1,6 1,5 1,4 1,3 1,2 1,1 1 I l a a a /b Figure 10: ividing, second method. If the result of the division is not in the interval [1..10], align the left base (graduation 10) of I, compute the quotient 10 and divide the result by 10. Figure 11 show the computation of 2.14/ I / /10 Figure 11: ividing 2.14 by The main advantage of this method, apparently more complex because needing a new type of scale (I), is clearly revealed in the section Reciprocal The use of the two scales and I allows the direct computation of the reciprocal 1/a of a number a. 6

7 Indeed, log(1/a) = log(a), and log(10 1/a) = log(10) log(a), from where: log(10 1/a) = 1 log(a) Figure 12 explicits the computation log(a) = log(10 1 ) a a I ,9 1,8 1,7 1,6 1,5 1,4 1,3 1,2 1,1 1 I log(a) a Figure 12: omputation of 1/a. As in multiplication, it is easy to compute the reciprocal of a number outside the interval [1..10], by writing the number in scientific notation and by using the formula 1/10 b = 10 b. 5 haining multiplications and divisions 5.1 General method To chain operations, like a quotient of products such as: a 1 a 2 a 3 b 1 b 2 it is possible to compute the product a = a 1 a 2 a 3, keep the result a, then compute b = b 1 b 2, keep the result b, then finally divide c = a/b. But we need to keep two intermediate results. There is another method, way simpler, which prevents us of writing down any intermediate results. If we use the second method to divide, the result of the division is on the scale, ready to be re-used for the next computation stage. Thus, by alternating multiplications and divisions, we use the previous intermediate result as a base for the next computation. The method is the following: ompute a 1 /b 1, Multiply this result by a 2, ivide by b 2, Multiply then by a 3, which gives us the final result. The first operand of stage n + 1 is in all cases the result of stage n. There is no intermediate result to keep. For example we will compute

8 Only four moves of the rule are needed to obtain the result. ompute 7.1/ (figure 13) ,9 1,8 1,7 1,6 1,5 1,4 1,3 1,2 1,1 1 I 1,3 1,4 1,5 1,6 1,7 1,8 1, / Figure 13: hained computation, first step. Align the 10 of the multiplication scale on the result which was computed, that is Multiplying by 0.51 gives 14.5 (figure 14) ,9 1,8 1,7 1,6 1,5 1,4 1,3 1,2 1,1 1 I Figure 14: hained computation, second step. Align the 10 of the division scale I on the result. ividing 14.5 by 61.5 gives (figure 15) I ,9 1,8 1,7 1,6 1,5 1,4 1, 1 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1, / Figure 15: hained computation, third step. Then, align the 1 of the multiplication scale on the result, and multiply by 22800, which gives us as a result 5380 (figure 16) I ,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1, Figure 16: hained computation, fourth and last step. The exact result equals 5369, 678.., which gives us an error of roughly 0, 2%. 8

9 5.2 Multiplying three factors There is a quick method to multiply 3 factors a b c, using only one slide rule translation. Using the I inverted scale, we can multiply two numbers a and b by aligning them on the I and scale. The product a b is then on the scale, aligned with the base of I scale (1). This intermediate product can then be easily reused as the first operand of a classical multiplication with the scale, using c as the second operand. The graduation of c on shows us directly the product a b c on (figure 17). l c l a + l b l 10 l 10 l a a c I ,9 1,8 1,7 1,6 1,5 1 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1, l b a b/10 b a b c/10 l a + l b + l c l 10 Figure 17: Multiplying 3 factors a b c. 6 Proportions & conversions Proportion problems are encountered frequently, such as converting from one unit to another (inch to millimeters, knots to km/h for example), or to apply a proportion rule (how much weight 3.73 litres of a product, when 1.27 litre weight kg). These computations are equivalent to the computation of a 1 with a 1 b 1 = a 2 b 2 knowing a 2 /b 2 and b 1. This proportion rule can be written as a 1 /a 2 = b 1 /b 2, that is log(a 1 /a 2 ) = log(b 1 /b 2 ). Knowing that log(x/y) = log(x) log(y), the last equality can be rewritten as log(a 1 ) log(a 2 ) = log(b 1 ) log(b 2 ). Let us pose l a = log(a 1 ) log(a 2 ) and l b = log(b 1 ) log(b 2 ). We can remark (figure 18) that by aligning factors a 2 et b 2 on respectively and scales, a 1 will be directly read on scale aligned with b 1 on. b 2 l b b 1 1 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1, a 2 l a a 1 Figure 18: Proportions In other words, the distance from b 1 to b 2 on scale is the same as the distance from a 1 to a 2 on scale. Once the scales setup, one can convert how many values he wants, without modifying the sliding rule ratio. 9

10 When converting out of range values, shift the scale of a length l 10 to the left, to bring back the 10 index in place of the 1 index, without forgetting to adjust the computed result of a factor 10. Figure 19 gives an example of converting with a ratio of 1.74/ ,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1, / Figure 19: 3.70 is to 2.64 what 2.44 is to Squares & square roots 7.1 Squares We use a scale, called A, graduated according to the logarithm of the square root. The length from the origin l of a graduation a is given by: l = log( a) To compute the square of a number a, we slide the cursor over the graduation for a on the scale. The b value aligned on scale A equals to this square: b = a 2. We have l a = l b, l a = log(a) and l b = log( b), so log(a) = log( b), thus b = a 2 (figure 20). l b = log( b) b = a 2 A A l a = log(a) a Figure 20: Square computation. In order to compute the square of a number out of the range [1..10], by writing this number on scientific form a = a m.10 ae, we have a 2 = a 2 m.10 ae2, or a 2 = a 2 m.10 2.ae. The mantissa of the square is the square of the mantissa, the exponent of the square is twice the exponent. See the example of on figure = A A = Figure 21: Square of

11 7.2 Square roots To extract the square root of a number, we could believe the opposite process would do it. The problem being that scale A has two intervals, respectively [1..10] and [ ]. Which one are we going to use to compute the square root of a number out of the range [1..100]? We have to write the number a on which we want to compute the square, under the form a = a m.10 2.ae, with a m in the interval [1..100] and a e integer. The square of a equals to: a = am.10 2.ae thus a = am.10 ae To compute the square root of , write = , compute , we obtain then = ubes and cube roots The method is based on the same principle as computing squares and square roots. The scale used here, called K, is graduated upon: l = log( 3 a) To compute the cube root, write the number a on which we want to compute the root under the form a = a m.10 3.ae, with a m in the interval [ ] and a e integer. The remaining of the procedure is straightforward. 9 Usual calculations 9.1 onverting degree/minute We found, on the and scale of most slide rules, some constants: ρ, ρ et ρ (figure 22). This constants allows, by aligning the base of a scale on one of them, to transform angles expressed using the MS system (egree, Minute, Second) onto angle in radian (arc length), or the reverse. ρ = π 3437, 747, converting an angle in minutes, ρ = 60 ρ , converting an angle in sexagesimal seconds, ρ = 100 ρ , converting an angle in decimal seconds. 11

12 ρ ρ ρ Figure 22: Usual conversion constants. 9.2 Volume of a cylinder The constant = 4 1, π allows the calculation of the volume V of a cylinder of diameter d and height h: V = πd2 4 h This equation can also be written under the form: V = d h Align the mark of the constant on scale with d on scale. Slide the cursor on h on scale B, the volum V can then be directly read on scale A. Figure 23 visually explains the computation. log(d/. h) = log( V ) A A B log(d/) log( h) h 1 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1, log(d) d V Figure 23: ylinder s volume. 9.3 π-shifted scales On some slide rules we found scales F, F, even IF. Theses scale are identical to respectively, and I, but shifted by a π factor, on the right by a length l π = log(π) With theses scale one can perform all classical computations, but multiplied/divided with a π factor. For example, to compute π , place the origin of scale aligned to 2.3 on, then read the result on F: Without using the shifted scales, we can use the π constant on scales A, B, and to easily perform theses computations. The 3-factor multiplication method ( 5.2) could be used in that case. 12

13 10 Trigonometry Scale S (figure 24), graduated according to the law l = log(10. sin(α)), allows the computation of the sine of an angle from 5.74 to 90 (sin(5.74 ) 0.1, sin(90 ) = 1, that is, the interval of the logarithmic scale divided by 10). To compute the sine of an angle, directly read on the logarithmic scale the result and divide it by 10. To compute the angle for which the sine is known, proceed in the same manner but interverting the two scales. log(10. sin(α)) S S α 10. sin(α) Figure 24: Sine scale from 5.7 to 90. The computation of the tangent and its inverse is made using the same principle on a scale T, graduated upon the law l = log(10. tan(α)) (figure 25). log(10. tan(α)) T T α 10. tan(α) Figure 25: Tangent scale from 5.7 to 45. To compute the sine or the tangent of an angle in the range and 5.7 (sin(0.573 ) tan(0.573 ) 0.01), we use in both cases a unique scale called ST (figure 26), graduated according to the law l = log(100.(sin(α) + tan(α))/2). We can use one rule for both because for small angles, sine and tangent are nearly the same. The error, ǫ, between the sine and the tangent for a given angle α, is: tan(α) sin(α) ǫ = 2. tan(α) + sin(α) which gives 0.496% for α = 5.7 ; and only 0.005% for α = The error is in the same order of magnitude as the reading error, and greatly less in many cases, so is negligible. Lastly, for angles lower than 0, 573, the sine and the tangent of an angle can be directely approximated by the arc length (the angle in radians). The error between the arc length and the sine for an angle α is: ǫ = α sin(α) α equals to % for α = 0.57, negligible in all the cases compared to the general precision. 13

14 log(100. tan(α)) log(100. sin(α)) α ST 1 1,5 2 2,5 3 3,5 4 4,5 5 5,5 ST 100. tan(α) 100. sin(α) Figure 26: Sine and tangent scale from 0.57 to 5.7. The computation of the cosine, sine or tangent of an angle outside the interval [0..90 ] is directly deduced from the sine or tangent of an angle from this interval by the usual trigonometric formulae: cos(α) = sin( π 2 + α) sin( α) = sin(π + α) = sin(α) tan( π 2 + α) = 1 tan(α) tan( α) = tan(π α) = tan(α) 11 Logarithms, exponential, powers 11.1 ecimal logarithms The computation of a logarithm is straightforward when one has a logarithmic scale. The logarithm of a number a on the scale is equals to the length l a : the scale L is thus simply linear (figure 27). l a = log(a) a L 0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 L log(a) log(a) Figure 27: Logarithms. To compute logarithms of numbers greater than 10 or smaller than 1, we write a as: a = a m.10 ae, then log(a) = log(a m.10 ae ) = log(a m ) + log(10 ae ), knowing that log(10 x ) = x, we obtain: log(a) = log(a m ) + a e If we write the number under a scientific form, the logarithm is equals to the exponent a e added to the logarithm of the mantissa a m. 14

15 11.2 ecimal powers To compute 10 a, we have to split the integer part a e (also called characteristic) from the decimal part a f (also called mantissa) of a: a = a e + a f. We can remark that 10 a = 10 ae+a f can also be written as 10 a = 10 ae.10 a f. It is easy to compute 10 a f using the linear scale L and the logarithmic scale (figure 28), it is the mantissa of the result written in scientific form. 10 ae is straightforward to compute and is the exponent part of the same result. log(10 a f ) = a f 10 a f L 0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 L a f a f Figure 28: Power of 10. For example: = Using the cursor with on scale L we found on scale. Thus, , which is approximatively Powers The objective here is to calculate c = a b. First method We compute the logarithm of the two members of the expression: log(c) = log(a b ). We have log(a b ) = b. log(a), so: The procedure is thus the following: c = 10 b.log(a) ompute the logarithm of a (see 11.1), Multiply this logarithm by b, ompute the power of 10 of the last result (see 11.2). This method, though indirect, is using only scales, and L, and works with all numbers a and b (Except for the necessary adjustments). Second method We use a scale LL (logarithm of the logarithm, or log-log), graduated according to the law l = log(k. ln(a)). Indeed, log(k. ln(a b )) = log(k.b. ln(a)), so log(k. ln(a b )) = log(b) + log(k. ln(a)) Number a on scale LL gives us a length l a, on which we add the distance l b given by b on log scale. The value c on scale LL aligned with b on scale is the result of a to the power b: a b (figure 29). 15

16 LL3 l c = log(k. ln(a b )) a LL3 1 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1,9 2 3 a b l a = log(k. ln(a)) l b = log(b) b Figure 29: alculating a b. LL scales are not relatives but absolutes: they are graduated with the logarithm of a logarithm, thus the rule log(10.a) = 1 + log(a) cannot be applied. We then have to use several scales according to the interval we should use. Usual LL scales are: LL3: l 3 = log(ln(a)) (interval [e..e 10 ]), LL2: l 2 = log(10. ln(a)) (interval [e 0,1..e]), LL1: l 1 = log(100. ln(a)) (interval [e 0,01..e 0,1 ]). The end of scale LL1 corresponds to the beginning of scale LL2, the end of LL2 to the beginning of LL3 (e being the base of natural logarithms). To calculate (figure 30), we align the origin of scale on of scale LL2. The result, 1.808, can be read using the cursor on LL2 aligned with 4.32 on scale. LL ,12 1,14 1,16 1,18 1,2 1,25 1,3 1,35 1,4 1,45 1,5 1,6 1,7 1,8 1,9 2 2,2 2,4 2,6 LL11,01 1,012 1,014 1,016 1,018 1,02 1,025 1,03 1,035 1,04 1,045 1,05 1,06 1,07 1,08 1,09 1,1 LL1 1 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1, Figure 30: alculating LL Natural logarithms, exponential We know that ln(a) = ln(10). log(a), so: ln(a) log(a) This is equivalent to compute a decimal logarithm which can then be multiplied by a constant. To compute an exponential without using a LL scale, we use the first method of computing a power ( 11.3). log(e a ) = a. log(e), we should then remind us the constant log(e) We only need to multiply a with this constant, 0.434, and read the result on scale L, with all the necessary adjustments. 16

17 11.5 Squares To compute c = b a, we can remark that b a = a 1/b. The calculation is the same as a power, with the exponent being the reciprocal of the square base. 17

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