Chapter 2 – Multiplication and Division (C and D Scale)
2.1 Adding with Uniform Scales
Fig. 2-1 shows how we could construct a simple calculating device to add numbers (e.g. two ordinary rules).
Fig 2-1
To calculate 2 + 3:
If we wish to find 2 + 7 we would follow the first two steps as above. Then for the third step go along the upper scale to 7 and read off 9 on the lower scale as the answer.
The Slide Rule is designed to add or subtract lengths. In using the C and D scales, the lengths we add or subtract are logarithms of the numbers marked on the graduations. Thus when lengths are added the numbers are multiplied.
Fig 2-2
Example 1: 2 x 3 = 6 (Fig. 2-2)
Note: The hair line on the cursor may be used in the following ways for this method of multiplication.
Example 2: 1.14 x 1.32 = 1.505 (Fig. 2-3).
Exercise 2(a)
Use the three steps as shown in the examples above to calculate the following:
Note: In parts (v) and (vi) of Exercise 2(a) it would be better to do the multiplication in the reverse order. (e.g. 1.25 x 7.6 instead of 7.6 x 1.25). This would mean much less movement of the slide, thus speeding up the calculation.
Fig 2-3
Consider the product 9 x 8. If we follow the steps as used in the examples above, the 8 on the C scale falls hopelessly off the end of the D scale. The problem is overcome, by placing the right index of the C scale on the 9, instead of the left index.
Fig 2-4
Example 1: 9 x 8 = 72 (Fig 2-4).
Note:
Example 2: 6.62 x 6.4 = 42.4
Exercise 2(b)
2.4 Locating the Decimal Point
From the last set of exercises we can readily see that for multiplication on the C and D scales the Slide Rule does not give us the position of the decimal point. This we have to decide for ourselves. Do not try to learn a rule for locating the decimal point as this has been proved to be inadequate and confusing.
The Best method is to make a quick estimate of the answer. This will always give you the position of the decimal point and checked the magnitude of your answer too.
Example 1:
(i.e. approx. 4 x 125 = 500)
therefore the answer is 493.0
Example 2:
488 x 0.283 = ‘1381’
(i.e. approx. 500 x .3 = 150)
therefore the answer is 138.1
Note: The other possibilities are 13.81 and 1,381 which are a long way off 150. So even if the approximation is fairly rough, it will not spoil this method.
Example 3:
0.46 x 72.3 x 52.2 = ‘1735’
(i.e. 0.5 x 70 x 50 = 0.5 x 3500 = 1750)
therefore the answer is 1735.0
When very large or very small numbers are involved, the above method can be streamlined by using standard form (or scientific notation). For example, we can express 2900 as 2.9 x 103 and 0.0012 as 1.2 x 10-3. Thus 2900 could be approximated by 3 x 103 and by 1 x 10-3. Always approximate 2.5, 2.6 etc. by 3 and 2.4, 2.3 etc. by 2.
Example 4:
640 x 0.00024 = ‘1538’
(i.e. approx. 6 x 102 X 2 x 10-4 = 12 x 10-2 = .12)
therefore the answer is 0.1538
Example 5:
0.024 x 36 x 430 x 0.057 = ‘1538’
(i.e. approx. 2 x 10-2 X 4 X 10 X 4 x 102 X 6 x 10-2 = 192 x 10-1 X = 19.2)
therefore the answer is 21.2
Exercise 2(c)
Locate the decimal point in the following:
To Multiply three or more numbers together:
Example: 2 x 15 x 88 = 2640
Step 1
Step 2
To locate decimal point:
2 x 15 x 88 ≈ 2 x 20 x 90 ≈ 360
there for the answer is 2640
(≈ stands for "approximately equals")
Exercise 2(d)