Slide Rule Triple Multiplication

published 2026-04-02

by Christopher Howard

One question with a basic slide rule is what is the most efficient way to complete a long chain of multiplications. E.g.,

a × b × c × d × e = y

One approach that is straightforward is to put the C index above 'a' on the D scale, move the cursor to 'b' on the C scale, put the C index on the new D scale position, move the cursor to 'c' on the C scale, and so on. This requires two actions per number — a slide movement and a cursor movement.

Another approach allows you to multiple three numbers at a time. Consider a three factor problem:

a × b × c = y

This can be reformatted as this ratio problem:

a ∶ (1 ∕ b) = y ∶ c

I prefer to see it like this, where the left side of the ratio, i.e., the numerator of the fraction, corresponds with the CI scale, and the denominator corresponds with the D scale:

(1 ∕ b) ∶ a = c ∶ y

Now, we can solve this problem with this technique: set the cursor to 'a' on the D scale, and move the slide so that the cursor lines up with 'b' on the CI scale. Now, move the cursor to 'c' on the C scale. The product then will be at the cursor on the D scale.

This technique required three actions for three numbers, i.e., one action per number, rather than two actions per number with the previous technique. Also, our cursor is set up so that the product on the D scale is set up to be the first number on another three factor problem, meaning we can multiply another two factors into the problem with only one more slide movement and one more cursor movement, maintaining our average of one action per number. This works cleanly for any problem involving an odd number of factors: 3, 5, 7... In the case of an even number of factors, like 4, we could use C index instead of our usual last number, which is to say the final factor is 1. Since the C index is touching the D index, a final cursor movement is not necessary, so that we don't have to do that final wasted action.

Copyright

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