MULTIPLICATION.
In the preliminary notes it was shown that by mechanically adding two lengths representing the logarithms of two numbers, we can obtain the product of these numbers; while by subtracting one log. length from another, the number represented by the latter is divided by the number represented by the former. Hence, using the C and D scales, we have the
Rule for Multiplication.—Set the index of the C scale to one of the factors on D, and under the other factor on C, find the product on D.
Fig. 10.
Thus, to find the product of 2 × 4, the slide is moved to the right until the left index (1) of C is brought over 2 on D, when under the other factor (4) on C, is found the required product (8) on D. Following along the slide, to the right, we find that beyond 5 on C (giving 10 on D), we have no scale below the projecting slide (Fig. 10). If we imagine the D scale prolonged to the right, we should have a repetition of the earlier portion, but, as with the two parts of the A scales, the repeated portion would be of tenfold value, and 10 on C would agree with 20 on the prolonged D scale. We turn this fact to account by moving the slide to the left until 10 on C agrees with 2 on D, and we can then read off such results as 2 × 6 = 12; 2 × 8 = 16, etc., remembering that as the scale is now of tenfold value, there will be two figures in the result. Hence, for those who prefer rules, we have the
Rule for the Number of Digits in a Product.—If the product is read with the slide projecting to the LEFT, ADD THE NUMBER OF THE DIGITS IN THE TWO FACTORS; if read with the slide to the RIGHT, deduct 1 from this sum.
Ex.—25 × 70 = 1750.
The product is found with the slide projecting to the left, so the number of digits in the product = 2 + 2 = 4.
Ex.—3·6 × 25 = 90.
The slide projects to the right, and the number of digits in the product is therefore 1 + 2 − 1 = 2.
Ex.—0·025 × 0·7 = 0·0175.
The product is obtained with the slide projecting to the left, and the number of digits is therefore −1 + 0 = −1.
Ex.—0·000184 × 0·005 = 0·00000092.
The sum of the number of digits in the two factors = −3 + (−2) = −5, but as the slide projects to the right, the number of digits will be −5 − 1 = −6.
From the last two examples it will be seen that when the first significant figure of a decimal factor does not immediately follow the decimal point, the minus sign is to be prefixed to the number of digits, counting as many digits minus as there are 0’s following the decimal point. Thus, 0·03 has −1 digit, 0·0035 has −2 digits, and so on. Some little care is necessary to ensure these minus values being correctly taken into account in determining the number of digits in the answer. For this reason many prefer to treat decimal factors as whole numbers, and to locate the decimal point according to the usual rules for the multiplication of decimals. Thus, in the last example we take 184 × 5 = 920, but as by the usual rule the product must contain 6 + 3 = 9 decimal places, we prefix six cyphers, obtaining 0·00000092. When both factors consist of integers as well as decimals, the number of digits in the product, and therefore the position of the decimal point, will be determined by the usual rule for whole numbers.
Another method of determining the number of digits in a product deserves mention, which, not being dependent upon the position of the slide, is applicable to all calculating instruments.
General Rule for Number of Digits in a Product.—When the first significant figure in the product is smaller than in EITHER of the factors, the number of digits in the product is equal to the SUM of the digits in the two factors. When the contrary is the case, the number of digits is 1 LESS than the sum of the digits in the two factors. When the first figures are the same, those following must be compared.
Estimation of the Figures in a Product.—We have given rules for those who prefer to decide the number of figures by this means, but experience will show that to make the best use of the instrument, the result, as read on the rule, should be regarded merely as the significant figures of the answer, the position of the decimal point, if not obvious, being decided by a very rough mental calculation. In very many instances, the magnitude of the result will be evident from the conditions of the problem—e.g., whether the answer should be 0·3 in., 3 in., or 30 in.; or 10 tons, 0·1 ton, 100 tons, etc. In those cases where the magnitude of the answer cannot be estimated, and the factors contain many figures, or have a number of 0’s following the decimal point, the use of notation by powers of 10 (page 8) is of considerable assistance; but more usually it will be found, that a very rough calculation will settle the point with comparatively little trouble. Considerable practice is needed to work rapidly and with certainty, when using rules. Moreover, the experience thus acquired is confined to slide-rule work. The same time spent in practising the “rough approximation” method will enable reliable results to be obtained rapidly, with the advantage that the method is applicable to calculations generally. However, the choice of methods is a matter of personal preference. Both methods will be given, but whichever plan is followed, the student is strongly advised to cultivate the habit of forming an idea of the magnitude of the result.
Ex.—33·6 × 236 = 7930.
Setting 1 on C to 33·6 on D, we read under 236 on D and find 793 on D, as the significant figures of the answer. A rough calculation, as 30 × 200 = 6000, indicates that the result will consist of 4 figures, and is therefore to be read as 7930.
Ex.—17,300 × 3780 = 65,400,000.
By factorising with powers of 10
Setting 1 on C to 1·73 on D, we read, under 3·78 on C, the result of the simple multiplication, as 6·54. Multiplying by 107 moves the decimal point 7 places to the right, and the answer is 65,400,000.
If it is required to find a series of products of which one of the factors is constant, set 1 on C to the constant factor on D and read the several products on D, under the respective variable factors.
If the factors are required which will give a constant product (really a case of division), set the cursor to the constant product on D. Then obviously, as the slide is moved along, any pair of factors found simultaneously under the cursor line on C, and on D under index of C, will give the product. A better method of working will be explained when we deal with the inversion of the slide.
It is sometimes useful to remember that although we usually set the slide to the rule, we can obtain the result equally well by setting the rule to the slide. Thus, bringing 1 (or 10) on D to 2 on C, we find on C, over any other factor, n on D, the product of 2 × n. But note that the slide and rule have now changed places, and if we use rules for the number of digits in the result, we must now deduct 1 from the sum of the digits in the factors, when the rule projects to the right of the slide.
With the ordinary 10 in. rule it will be found in general that the extent to which the C and D scales are subdivided is such as to enable not more than three figures in either factor being dealt with. For the same reason it is impossible to directly read more than the first three figures of any product, although it is often possible—by mentally dividing the smallest space involved in the reading—to correctly determine the fourth figure of a product. Necessarily this method is only reliable when used in the earlier parts of the C and D scales. However, the last numeral of a three-figure, and in some cases the last of a four-figure, product can be readily ascertained by an inspection of the factors.
Ex.—19 × 27 = 513. Placing the L.H. index of C to 19 on D, we find opposite 27 on C, the product, which lies between 510 and 515. A glance at the factors, however, is sufficient to decide that the third figure must be 3, since the product of 9 and 7 is 63, and the last figure of this product must be the last figure in the answer.
Ex.—79 × 91 = 7189.
In this case the division line 91 on C indicates on D that the answer lies between 7180 and 7190. As the last figure must be 9, it is at once inferred that the last two figures are 89.
When there are more than three figures in either or both of the factors, the fourth and following figures to the right must be neglected. It is well to note, however, that if the first neglected figure is 5, or greater than 5, it will generally be advisable to increase by 1 the third figure of the factor employed. Generally it will suffice to make this increase in one of the two factors only, but it is obvious that in some cases greater accuracy will be obtained by increasing both factors in this way.
Continued Multiplication.—To find the product of more than two factors, we make use of the cursor to mark the position of successive products (the value of which does not concern us) as the several factors are taken into the calculation. Setting the index of C to the 1st factor on D, we bring the line of the cursor to the 2nd factor on C, then the index of C to the cursor, the cursor to the 3rd factor, index of C to cursor, and so on, reading the final product on D under the last factor on C. (Note that the 1st factor and the result are read on D; all intermediate readings are taken on C.)
If the rule for the number of digits in a product is used, it is necessary to note the number of times multiplication is effected with the slide projecting to the right. This number, deducted from the sum of the digits of the several factors, gives the number of digits in the product. Ingenious devices have been adopted to record the number of times the slide projects to the right, but some of these are very inconvenient. The author’s method is to record each time the slide so projects, by a minus mark, thus −. These can be noted down in any convenient manner, and the sum of the marks so obtained deducted from the sum of the digits in the several factors, gives the number of digits in the product as before explained.
Ex.—42 × 71 × 1·5 × 0·32 × 121 = 173,200.
The product given, which is that read on the rule, is obtained as follows:—Set R.H. index of C to 42 on D, and bring the cursor to 71 on C. Next bring the L.H. index of C to the cursor, and the latter to 1·5 on C. This multiplication is effected with the slide to the right, and a memorandum of this fact is kept by making a mark −. Bring the R.H. index of C to the cursor and the latter to 0·32 on C. Then set the L.H. index of C to the cursor and read the result, 1732, on D under 121 on C, while as a slide again projects to the right, a second − memo-mark is recorded. There are 2 + 2 + 1 + 0 + 3 = 8 digits in the factors, and as there were 2 − marks recorded during the operation, there will be 8 − 2 = 6 digits in the product, which will therefore read 173,200 (173,194·56).
For a very rough evaluation of the result, we note that 1·5 × 0·3 is about 0·5; hence, as a clue to the number of figures we have