Quintics

Yes, you really can compute quintics (fifth powers) on something as simple as a Rietz type rule, and with just one movement of the slide! Truly, this is cool.

Here's the deal. Just think of x5 in the following way:

x5 = (x * x)2 * x

That first product is easily handled on C and D, the squaring on A, and the final product with the help of B.

Let's go through the motions with 25, which is easy to keep track of.

You would first calculate the product of 2 and 2. When finding quintics, it's best to use the old "multiply by a reciprocal" routine. So, align 2 of CI with 2 on D. The result 4 is under the right index on D, but ignore it. Instead, look to A and you'll find 16 is also under the right index. So far, so good; that's 24.

To bring in the concluding factor, just slide the cursor to 2 on the B scale (thus multiplying 24 by 2) and read the result under the cursor: 32. Pretty slick, don't you think? The slide rule ever startles me with its embedded patterns and logarithmic beauty.

For emphasis, just one setting of the slide is all it takes to compute quintics. Let's try a couple problems.

Exercise 1.1: Compute 5.75.

I'll be using my trusty Pickett 1006-ES for this, and so will need a duplex-flip, since CI is on the reverse side of A and B. If you're using a Rietz type rule, you won't need to do this, since everything required is staring you in the face.
  1. Set the cursor to 5.7 on the D scale.
  2. Align 5.7 of the CI scale with the hairline.
  3. Duplex-flip, if needed.
  4. Move the cursor to 5.7 on the B scale.
  5. Read the result under the hairline on A: 6000.
When I did it, it looked like this:

Cursor to D:5,7, CI:5.7 to hairline

Duplex-flip, cursor to B:5.7, result at A:6000

Of course, you'll need to estimate the order of magnitude (which is morally good for a person anyway). This isn't too onerous of a task here. Just think of 52 = 25, then 252 = 625, then multiplying by 5 again gives something greater than 3000.

Exercise 1.2: Compute 255.

Of course, we'll just reckon this as 2.5, and worry about the order of magnitude at the end.
  1. Set the cursor at 25 (2.5) on the D scale.
  2. Align 25 (2.5) of the CI scale with the hairline.
  3. Duplex-flip, if needed.
  4. Move the cursor to 25 (2.5) on the B scale.
  5. Read the result under the hairline on the A scale: 9,750,000.
She goes like this:

Cursor to D:25, CI:25 to hairline
Duplex-flip, cursor to B:25, result at A:9,750,000

When I worked the problem, I estimated the result by thinking of the argument as being 2.5 * 101. Thus, five places and/or zeros were bound to stack up at the end, in addition to whatever else the mantissa raised to the fifth contributed.

Using the log-log scales on fancier slide rules might seem more direct, since the order of magnitude comes along more or less automatically. However, if you're willing to count some zeros and estimate, notice that the method just described with just D, C, A and B is good for arguments of any size.

What electricity is to the Amish, the log-log scales are to me. I'm getting a kick out of seeing what can be accomplished with just the simplest of slide rules.

Next installment: Sextics