Square Roots and Squares

Most slide rules sport A and B scales, and with them you can work some wonderful square root and square magic. The idea behind this really drives home the elegance of what Napier wrought. In particular:

log(sqrt(x)) = 1/2 * log(x)

Examining your slide rule, you'll note that the C and D scales are laid out logarithmically. But the A and B scales are measured half as long, per the right side of the expression above. Hence, they represent square roots when compared to numbers on the C and D scales. Because of that 1/2 factor, there are two copies along A and B.

Or, if you prefer to think in reverse, the C and D scales are twice as long as the spans on the A and B scales, and hence represent squares:

log(x2) = 2 * log(x)

To find a square root, then, all one need do is locate the desired argument on the A scale, and read its length along the D scale. (The B and C scales on the slide will be used in chained arguments, to be discussed later).

Now it might occur to you: since there are two copies of the log scale on A, which one should be used? That's easy to sort out. The first half of A represents arguments of 1.0 to 10.0, and hence square roots of 1.0 to 10.0 are found on the D scale.

The right half of the A scale represents arguments of 10.0 to 100.0, with their square roots found on D as well, but obviously larger. (You might want to note that 10.0 in the middle of A corresponds to sqrt(10) ≈ 3.16 on D.)

From then on, thanks to how decimal squaring behaves, the two halves of A keep repeating. The left half corresponds to arguments whose whole number part consists of an odd number of digits, while the right half maps to those with an even number of digits. For example, you would locate the following arguments on the left half of A:

2, 2.25, 222, 222.5, 22222.5, etc.

and these on the right half:

22, 22.5, 2222.5, etc.

For emphasis, when finding square roots, simply count the digits of the whole number part of the argument; if odd, locate the argument on the left half of the A scale, but if even, find it on the right half.

Pretty clearly, we can work this in reverse to compute squares of numbers, too. Just pinpoint the argument on the D scale, and the corresponding value on the A scale, remembering how to interpret the odd/even number of digits. Some examples should make this all clear.

Finding Square Roots


Example 1.1: Find sqrt(3.8).
  1. Set the cursor at 3.8 on the A scale. Because there is an odd number of digits in the whole part of the argument (one), use the left half of A.
  2. Read the result under the hairline on the D scale: 1.95.
Here's how she looks in pictures. Remember, you can always click a photo to magnify it for closer inspection.


Cursor to A:3.8, result at D:1.95

Notice that the settings of the slide rule would be identical were you searching for sqrt(380); only your interpretation of the order of magnitude would differ at the very end.

Example 1.2: Find sqrt(38).

Same digits here, but note there are now two in the whole number part, and so we must use the right half of A.
  1. Set the cursor at 38 (3.8) on the right half of the A scale.
  2. Read the result under the hairline on the D scale: 6.16.
In summary:


Cursor to A:38, result at D:6.16

Finding Squares


Working backwards to find squares of numbers is easy if you remember what we learned about counting the whole number digits.

Exercise 2.1: Find 27.52.
  1. Set the cursor at 27.5 (2.75) on the D scale.
  2. Read the result under the hairline on the A scale: 760.
When I did it on my Pickett 1006-ES, it looked like this:


Cursor to D:27.5, result at A:760

Since the hairline wound up in the left half of the A scale, the answer is 7.60, or 760, or 76,000 and so on. Obviously, 760 is the required result.

Exercise 2.2: Find 9202.
  1. Set the cursor at 920 (9.2) on the D scale.
  2. Read the result under the hairline on the A scale: 850,000.
Pictorially:


Cursor to D:920, result at A:850,000

Since we started with something close to 9 (920) and ended up on the right half of the A scale, we expect something close to 81, a two-digit result. Factoring in the 1002, we ended up with 85,000. Or if you wish, simply estimate the order of magnitude by reckoning what 9002 must be.

I've provided a whole lot more examples of square roots and squares in Multiplication by Square Roots and Multiplication by Squares, should you want to jump ahead a bit.

Next installment: Cube Roots and Cubes