The slide rule

By combining the rules for multiplication and division, we can readily evaluate expressions of the form a
b × c
d × e
f × g
h = x. The simplest case, a × c
b can be solved by one setting of the slide.^[3] Take as an example, 14·45 × 60
8·5 = 102. Setting 8·5 on C to 14·45 on D, we can, if desired, read 1·7 on D under 1 on C, as the quotient. However, we are not concerned with this, but require its multiplication by 60, and the slide being already set for this operation, we at once read under 60 on C the result, 102, on D. The figures in the answer are obvious.

When there are more factors to take into account, we place the cursor over 102 on D, bring the next divisor on C to the cursor, move the cursor to the next multiplier on C, bring the next divisor on C to the cursor, and so on, until all the factors have been dealt with. Note that only the first factor and the result are read on D; also that the cursor is moved for multiplying and the slide for dividing.

Number of Digits in Result in Combined Multiplication and Division.—For those who use rules the author’s method of determining the decimal point in combined multiplication and division may be used. Each time multiplication is performed with the slide projecting to the right, make a − mark; each time division is effected with the slide to the right, make a | mark; but allow the | marks to cancel the − marks as far as they will. Subtract the sum of the digits in the denominator from the sum of digits in the numerator, and to this difference add any uncancelled memo-marks, if of | character, or subtract them if of − character.

Ex.—43·5 × 29·4 × 51 × 32
27 × 3·83 × 10·5 × 1·31 = 1468.

Set 27 on C to 43·5 on D, and as with this division the slide is to the right, make the first ⵏ mark. Bring cursor to 29·4 on C, and as in this multiplication the slide is to the right, make the first − mark, cancelling as shown. Setting 3·83 on C to the cursor, requires the second ⵏ mark, which, however, is cancelled in turn by the multiplication by 51. The division by 10·5 requires the third ⵏ mark, and after multiplying by 32 (requiring no mark) the final division by 1·31 requires the fourth ⵏ mark. Then, as there are 8 numerator digits, 6 denominator, and 2 uncancelled memo-marks (which, being 1, are additive) we have

Had the uncancelled marks been − in character, the number of digits would have been 8 − 6 − 2 = 0.

For quantities less than 0·1 the digit place numbers will be negative. The troublesome addition of these may be avoided by transferring them to the opposite side and treating them as positive.

The first numerator, 0·00356, has −2 digits. Note this by placing 2 below the lower line as shown. 27·1 has 2 digits; place 2 over it. 0·08375 has −1 digit; hence place 1 below the lower line. The first denominator has no digits; the second, 9·85, has 1 digit; hence place 1 under it. 0·00002 has −4 digits; place 4 above the upper line. The sum of the top series is 2 + 4 = 6; of the bottom series 2 + 1 + 1 = 4. Subtracting the bottom from the top, we have 6 − 4 = 2 digits, to which 1 has to be added for an uncancelled memo-mark, and the result is read as 288.

Moving the decimal point often facilitates matters. Thus, 32·4 × 0·98 × 432 × 0·0217
4·71 × 0·175 × 0·00000621 × 412000 is much more conveniently dealt with when re-arranged as 32·4 × 9·8 × 432 × 2·17
4·71 × 17·5 × 6·21 × 4·12 = 141.

To determine the number of figures in the result by rough cancelling and mental calculation, we note that 4·71 enters 432 about 100 times; 9·8 enters 17·5 about 2; 6·21 into 32·4 about 5; and 2·17 into 4·12 about 2. This gives 500
4 = 125, showing that the result contains 3 digits. From the slide rule we read 141, which is therefore the result sought.

The occasional traversing of the slide through the rule, to interchange the indices—a contingency which the use of the C and D scales always involves—may often be avoided by a very simple expedient. Such an example as 6·19 × 31·2 × 422
1120 × 8·86 × 2.09 = 3·93 is sometimes cited as a particularly difficult case. Working through the expression as given, two traversings of the slide are necessary; but by taking the factors in the slightly different order, 6·19 × 31·2 × 422
8·86 × 2·09 × 1120, so that the significant figures of each pair are more nearly alike, we not only avoid any traversing the slide, but we also reduce the extent to which the slide is moved to effect the several divisions.

Such cases as a × b
c × d × e × f × g or a × b × c × d × e
f × g really resolve themselves into a × b × 1 × 1 × 1
c × d × e × f × g and a × b × c × d × e
f × g × 1 × 1 × 1, but, of course, if rules are used to locate the decimal point, the 1’s so (mentally) introduced are not to be counted as additional figures in the factors.