Slides rules are actually astonishingly powerful things. The simple slide rule does multiplication and division using the C and D scales; strictly speaking, you can have a basic rule with nothing but C and D. But you almost never see a rule that simple. (The only one I've ever seen with only the two scales was a circular rule used as a promotional giveaway.)
The other scales are where things get a bit complicated; but it's a lot of fun to figure out how they work, and to see how much you can actually do with, basically, two attached rulers with a bunch of different markings.
As an example, I'm going to walk through how you do exponents and roots using a slide rule. They're interesting, because they're the hardest things to do on the slide rule; so if you can get how to do them, there's pretty much nothing that you can do with a slide rule that you won't be able to figure out by understanding the scale. I'm using the same virtual slide rule program as yesterday, but this time, the images are of a Pickett N3-T, which is a log-log duplex trig scientific rule. We'll need the extra scales of a log-log for today.
One quick note of clarification: a slang term for a slide rule is a "slipstick", or just "stick". I'm actually used to calling it a stick; that's what my father used to call it when he taught me to use it. So if I slip up and say "stick", you'll know what I'm talking about.
Simple Powers: Squares, Cubes, and their Roots
The simplest way of doing squares and square roots is to just use the D and A scales together. The A and B scales are half-scale relative to the C and D scales, which means that the number markings are in a square relationship. To, to square a number, you slide the cursor to that number on D, and then look at the corresponding number of A. So, for example, let's try 212. We slide the cursor to 2.1 on A:
When we follow the cursor up to A, it's on roughly 4.4. 20 squared is about 400, so with powers of ten, we'd say 21 squared is 440. (It's actually 441.)
For cubes, do the same thing, but use the K scale.
Roots are harder - because powers of 10 matter. The square root of 2 is 1.414...; the square root of 200 is 14.14..., but the square root of 20 is 4.472. (sqrt(2) * sqrt(10)). If you look at the topmost scale on the rule, you'll see that it's labeled with a square-root radical; and next to it is something about "odd digit/zero nos". That double-scale is set up to be used for taking square roots. If the number is greater than one, and the total number of digits plus trailing zeros is odd, then you read the top of the scale. If the number of digits plus trailing zeros is even, then you read the bottom of the scale. If the number is *less than* one, then you look at the number of *leading* zeros after the decimal point; if it's odd, you use the top, even you use the bottom.
So let's look at the square roots of 20 and 200. 20 is two digits, which is even, so we'll use the bottom half of the square root scale. 200 is three digits, odd, so it'll be the top of the scale. We slide the cursor to 2.0 on the A scale, and then look up to the bottom markings on the square root scale:
So before adjusting powers of ten, the square root of 20 is 4.47; and the square root of 200 is 1.41. For 200, we need to adjust the power of ten, making it 14.1; for 20, the power of ten is correct.
You can also use the A scale for doing roots; it's often less convenient, but it can do the job. The A scale repeats the 1 to 10 range twice (remember, it's half-scale of C/D.) So the left half of A is equivalent to the top of the square root scale; the right half of A is equivalent to the bottom of the square root scale.
There's also a cube root scale on the very bottom of the rule, with three sets of markings. You use the top if the number of digits is 1 mod 3; the middle for 2 mod 3; and the bottom for 0 mod 3.
General Exponents and Roots: Log and LogLog scales
Where things start to get really interesting (and complicated), we move to the log scales. Almost all rules have at least one log scale; on the Pickett N3 stick that I'm getting my images from, it's on the back, labeled L. The L scale is the base 10 logarithm of the C/D scales. There's also usually a Ln scale, which is the natural log.
So let's do a quick log. What's are the base 10 and natural logs of 2? We just flip the rule so that we can see the L and Ln scales, slide the cursor to 2 on D, and look:
Log base 10 of 2, we read "0.3" on the "L" scale; natural log, we read "0.69". By calculator log10(2)=0.3010, and ln(2)=0.6931. The catch, as usual, is dealing with the powers of 10. For the base ten log, it's easy. On the stick, we're always working in scientific notation; take the power of ten from the exponent part of the number, and just *add it* to the base ten logarithm. So, for example, suppose we wanted to base-10 log of 2000 instead of 2. 2000 is 2.0×103, so we'd take the log of 2.0 (which we saw just a moment ago was 0.3), and we'dd add 3 to it. The base-10 log of 2000 is 3.30.
For the natural log, its similar, except of course that the powers of ten aren't so convenient. We need to do some work to deal with the powers of 10 in natural log. What we need to do with the natural log is *multiply* the natural log of 10 (which we can easily see is 2.3 from the LN scale) by the 10s exponent, and then add that to the result. So, for example, to do the natural log of 2000, we'd start with the natural log of 2 (0.69). Then we'd take the natural log of 10 (2.3) and multiply it by 3, getting 6.9; then add that to the natural log of 2, getting 7.6. And of course, we can use our slide rule to do that multiplication.
The most general way of doing exponents and roots is to just work with the logarithms. We can use the basic logarithmic formula: xy = aloga*y. So, for example, to take the fourth root of 24, we can take the base-10 logarithm of 24 (0.38 + 1 = 1.38); divide it by four (0.344); and then look for 0.344 on the log scale, and translate back, which gives you about 2.21.
So, what about the LL scales? "LL" scales are "log-log" scales. On the Pickett rule that I'm using for images, the LL scales are on the back of the rule. There are four of them: LL0, LL1, LL2, and LL3. The way that the LL scales are set up is that LLN scaled is one power of ten larger than LL(N-1). One *really* important thing to remember about the LL scale is: they do do powers of 10. So the range for which the LL scales is useful is very limited.
So the if you pick a value on LL1, and use the cursor to find the same point on LL2 will be the LL1 value to the tenth power. So, for example, we can look at 1.015 on LL1; the corresponding value on LL2 is 1.16:
What's 1.015^10? About 1.1605. We can do tenth roots by going the other way on the LL scales; from LL2 to LL1 is the tenth root.
The LL scale also work for negative powers; there's a second set of numbers of each of the LL scales; they're called the -L scales when we use the negative numbering. LL1, LL2, and LL3 all have negative scales. The negative scales are important precisely because LL scales keep their powers of 10. So if the LL0 starts at 1, we can't do anything smaller than 1 without using one of the negative LL scales.
That's all fine. But we aren't really interested in just raising things to the 10th power, or taking 10th roots. What we'd like to be able to do is do arbitrary roots and exponents. How do we do that using the LL scales?
It's pretty much the same thing we did above with taking the log and multiplying. Suppose we want to compute 18 to the power of 2.7? Well, we find 18 on the LL scales. It's on LL3. We slide the C scale so that the 1 is directly above 18 on LL3 (using the cursor to line it up):
Then we slide to cursor up to 2.7 on C, and read the result on LL3, which gives us about 2400. (Actually, about 2450 is correct.) Not very precise in this range; but it's a reasonable approximation, and it's a lot faster than doing the log and exponent steps separately.
But that's a very easy version of using the LL scales. Most of the time, you need to "cross scales", because the answer can't be done on a single LL scale. For example, what if we wanted to raise 2.5 to the 3.4? We can find 2.5 on LL2; but when we go to multiply by 3.4, it's off the right edge of the stick. And since LL scales do use powers of 10, we can't just switch to the 1 on the other side, because that would give the wrong answer. In fact, if we used the one on the right-hand side of C, the answer would be 2.50.34 instead of 2.53.4. And hidden in that fact is our solution!
Once again, we're going to pull out another mathematical identity involving exponents which we're going to use to make things work. ab×c = abc. Using this, we can restate the problem as 2.50.34×10, or 2.50.3410. So, we move the cursor to 2.5 on LL2; then slide the right hand one on C to line up with the cursor:
And now slide the cursor down to 3.4 on C:
The cursor is now right around 1.36 on LL2. But that's not our answer; we still need to take its tenth power. And if you look upwards just a bit you'll see that whatever is on LL3 at the same place is the value on LL2 to the tenth power. So the value on LL3 is the answer. LL3 reads 22.5. The right answer is around 22.54.
That's pretty much it for the slide rule. It's really not hard to use, as you can see. But it's based on really fascinating math. The structure of the rule is built on the principle of logarithms; and the ways we do calculations are based on understanding how to apply a series of basic mathematical identities in order to help the rule give us the results we want.
It's really a brilliant device. Like I said before, the idea that having what's basically just a pair of sliding rulers, if you write numbers on them in the right places, then you can use them to do all sorts of math - that is really amazing.
There's also a bit of an aesthetic aspect to slide rules. They are, in their own idiosyncratic way, really beautiful things. They need to be very precise, so good rules are remarkably well crafted things; and the layout of the scales on the rules to make them easy to use is an aesthetic choice, particular to each manufacturer.
As you can probably tell, I really love these things. Silly, but as a person who finds math incredibly fun, using a slipstick is just fun in a way that using a calculator can't be. Because with a slipstick, you're really using your own understanding of math at every step.
If, by chance, reading this has interested you in getting hold of a slide rule of your own, they're not too hard to find. The two best places are Ebay (where you can find very good used slide-rules dirt cheap); and Sphere's Slide Rule Universe, where they have a ton of brand new, unopened slide rules. (They're all thirty-odd years old, but they're still untouched in the original packaging. There were entire warehouses full of slide rules when the electronic calculator suddenly appeared, and many of them have still never been sold.)
I think you have a typo in the sentences round the first picture:
"We slide the cursor to 2.1 on A:" then "When we follow the cursor up to A".
Same typo in the roots example:"...slide the cursor to 2.0 on the A scale..." - should be the D scale
Well I can't reproduce it directly like you did in the comments, but a^(b*c) NOT= a^(b^c).
(eg look at a=2, b=2, c=3).
However it is true (a^b)^c = a^(b*c), and this is why exponentiation is defined to left associate, I think. Just something that caught my eye, and an example of a function that isn't associative in the way we normally think, and that always gets me messed up for some reason.
I am playing with a borrowed ACU-Math 400, with S/K/A, B/C1/C, D/L/T. S looks like arcsin(A).
I am just young enough to have a physics teacher who would only let us use calculators if we showed him we could multiply on the slide rule, which is as much as I ever knew, and really had forgotten. I borrowed this a few years ago, meaning to learn, but...
And so these lessons were fun and a nice intro. I am going to practice, return this one, and buy my own.
Thank you for a clear explanation and pleasant writing. I agree with you that the slipstick, as a mechanical extension of our mind, is in a separate class from more automatic devices. There's a certain satisfaction in knowing the answer is correct because we've worked through the underlying math every step of the way.