Slide Rule Addition

published 2026-04-18

by Christopher Howard

Through an accident, I own a copy of the instruction manual for the Hemmi 266 slide rule, but not the slide rule itself. This is an electronics slide rule produced by a Japanese company. The manual is in English, though a few ESL artifacts are included, such as the word "sylinder" instead of "cylinder".

Browsing through this manual, I noticed that it has two interesting sets of scales, one for solving the parallel resistance problem (or series capacitor problem) and another for solving series impedance problems. The first set of scales is the r1 and r2 scales — not to be confused with the r1 and r2 square scales on other slide rules. The second set is the P and Q scales. You can observe them on this photograph:

Hemmi_266_Electronics_dcQE_May1966.jpg (JPEG Image, 2400 × 1010 pixels) — Scaled (64%)

A parallel resistance problem has this fundamental form:

1/r1 + 1/r2 = 1/R

Only two actions are required: move ∞ on the r2 scale to r1 on the r1 scale, then set the hairline to r2 on the r2 scale. The value R is then read under the hairline on the R1 scale.

At first, I was confused at how this was done, trying to imagine how some log rule transformation could be used here. But I chatted with somebody about it on IRC, and they pointed out that it is very simple: the values shown on the r1 and r2 are placed at a fractional distance from ∞ and the index (1), corresponding to 1/x. You can see that 2 is placed 1⁄2 the distance between ∞ and 1; 3 is at 1⁄3 the distance, and 0.5 is at 1/0.5 or 2 times the distance. So, all you are doing is adding the two distance distances, and then using one of the scales to convert the distance back from 1/R to R. The ∞ label comes from 1/∞ approaches zero.

Regarding the P and Q scales: a series impedance problem has this form:

p² + q² = P²

where p and q are the resistance and reactance. To solve this in another way using the usual methods, you would need to either: square those values, add them together, and then take the root; or you could the angle using tan(ϑ) = q/p, and then use sin(ϑ) × q. But with the P and Q scales, all you have to do is set 0 on the Q scale to p on the P scale, then read P about q on the Q scale.

How this works is that the numbers on the P and Q scales correspond to a distance that is equal to a square of that number, the distance being referenced to the distance between 0 and the 1 mark, which is not labeled. You can further confirm this by observing that the marker for 2, corresponding to 2² = 4, is ¼ the distance from the 0 mark as the marker for 4, which corresponds to 4² = 16. I.e., 4⁄16 = ¼. So, we are simply adding together those distances and then using a scale to convert the new distance back from P² to P.

For both these problems, this is the simplest method I've seen of solving the problems, except of course entering the values into a computer program that does the squaring, etc. behind the scenes. Impedance problems can be solved fairly easily with a protractor and graph paper, but that of course requires making a large drawing on graph paper, and there are a few extra steps required for measurement of the line lengths.

Now, you can use this method if one of your numbers is very large, and one is very small, such as a difference of two orders of magnitude. But in these applications, at least, you wouldn't be doing that anyway. With resistance, putting a large resistance in parallel with an existing small one is not likely to make any practical difference in your circuit, especially taking into account component tolerances. In regards to impedance problems, the general practice is that if your resistance or your reactance is at least ten times larger than the other value, you just assume your impedance is equal to the larger value.

I imagine that this basic idea could be generalized to create scales for other problems of form

f(a) + f(b) = f(c)

but I'm not sure what restrictions exactly must be imposed on the function f.

Copyright

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