Multiplication by Squares

In the previous section, we worked from the principle that the B and A scales were half the length of C and D, hence must represent square roots. Now we'll just change our perspective and note that the C and D scales are twice the length of the B and A scales, so must represent squares.

There are really just two main methods for products involving squares (that I've found!). So let's dive in with a problem and see them in action.

Exercise 2.1: Find the product of 3.7 and 1.5^2.

Method #1: As usual, let's commence with the direct approach (Method #1 from Basic Multiplication) bearing in mind it might lead to some index-swapping. Note that action begins on the A scale, but the square is found on the C scale, with the final result returning to the A scale. In other words, we're really working this problem on the half-length scale, realizing that any distance measured along the full-length scale must designate the square of a number.
  1. Set the cursor at 3.7 on the A scale.
  2. Align the left index with the hairline.
  3. Move the cursor to 1.5 on the C scale.
  4. Read the result under the hairline on the A scale: 8.3.
Here's what it looks like. (Reminder: click on any image to enlarge it).

Left index on A:3.7, cursor on C:1.5, result at A:8.3

Something pleasant to note: when taking a square root with respect to the C or D scales, you need to be careful to choose the left or the right half of the A scale, according to how many digits appear in the argument. But here we're working in reverse (thinking of A as the point of reference, and C twice as long), so that no longer matters. In short, in the example above, we've could have just as properly used the 3.7 which appears on the right half of A and everything would work out. Give it a try now and convince yourself.

Having said that, I'll mention that I continue to use the "left is odd number of digits" and "right is even number of digits" scheme even in situations where it doesn't matter. This is simply to keep beating the protocol into my mind for when it does make a difference. So, in Example 2.1, above, I used the left-hand 3.7. Moreover, if this was part of a chained operation (say involving a scale-transfer to K), then again choosing the number in the correct span will matter.

Method #2: Realizing that in multiplication with the "basic" method you might go off-scale, let's try this again using the old divide-by-the-reciprocal ploy (which is guaranteed to keep you on-scale one way or another).
  1. Set the cursor at 3.7 on the A scale.
  2. Align 1.5 on the CI scale with the hairline.
  3. Move cursor to the right index.
  4. Read the result under the hairline on the A scale at 8.3.
On a typical Rietz type rule, you can do all four steps pronto. With the Pickett 1006-ES I'm using here, a duplex-flip is required, since CI is on the reverse side relative to A. So, here are three pics showing it all:

Align cursor with A:3.7, then duplex-flip

Align CI:1.5 with the hairline, then duplex-flip

Align cursor with index, result at A:8.3

One final thing. I mentioned in the section concerning products with square roots that using commutativity of multiplication to reorder the arguments is sometimes helpful. That's certainly possible here, too, should you find it easier to think about a square times an ordinary number. As an exercise, go ahead and try the problem for above as 1.5^2 times 3.7. It's actually kind of pretty how you end up at the same result!

"What about index-swapping?," I hear you ask. Try the next problem and see.

Exercise 2.2: Find the product of 42 and 75.6^2.

Let's try Method #1 from above. Whether you use the leftmost 42 (well 4.2, actually, at the next order of magnitude) on A, or the rightmost version, you'll note that 75.6 (again, 7.56) is off-scale. With this method, we have no option but to do an index-swap at the outset. With practice, you'll see it coming...
  1. Set the cursor at 4.2 on the A scale, (either one is okay).
  2. Align the right index with the hairline.
  3. Move the cursor to 7.56 on the C scale.
  4. Read the result under the hairline on the A scale: 2.4.
Here's what you'll see:

Right index at A:4.2, cursor on C:7.56, result at A:2.4

Estimation, or consideration of the orders of magnitude involved, pegs this at 240,000.

Let's try this one again, using Method #2 in which we divide by a reciprocal.
  1. Set the cursor at 4.2 on the A scale, (either one is okay).
  2. Align 7.56 of the CI scale with the hairline.
  3. Move the cursor to the left index.
  4. Read the result under the hairline on the A scale: 2.4.
Remembering that the CI scale is on the reverse side of the rule I'm using here, the steps look like:

Align cursor with A:4.2, then duplex-flip
Align 7.56 with hairline, then duplex-flip
Move cursor to left index, read result at A:2.4

Wrap-Up: As usual, the CI approach gets you quickly from the start to the end with no hesitation at the outset on which index to use. And with a Rietz rule, there's no duplex-flipping involved which is even more fleet. So, my suggestion would be to continue with the "divide by a reciprocal" approach, unless something else in a chained computation suggests otherwise.

And remember, with squares, you can use either the left or the right half of A with impunity, but only in simple problems like the above. But for consistency with how you handle square roots or in chained operations, you should stick with the "left side for odd-digit numbers and the right side for even-digit numbers" protocol.

Next installment: Multiplication by Cube Roots