Squares and Square Roots

Nov 13, 2022 by @turgon@dosgame.club

Let's do the unthinkable and skip division for now. We'll go straight for squares and square roots!

Remember, slide rules love numbers in normalized scientific notation, that is, a coefficient between 1 and 10, times ten to some power. The exponent tells us exactly where the decimal goes. We'll start without the slide rule and see what happens when we square numbers in that form, say $x\cdot {10}^m$:

${(x\cdot {10}^m)}^2$ which rearranges nicely into:

$x^2\cdot {10}^{2m}$ thanks again to commutativity and exponent rules.

Since squaring is a special case of multiplication, it isn't surprising to see that $x^2$ is also bounded between $1$ and $\mathrm{100,}$ and we have the same two cases as before: either $x^2$ is less than 10, or between $10$ and $100.$ When $x^2$ is less than $10$ it directly gives us the normalized scientific notation coefficient of the answer and the answer's exponent is simply $2\cdot m$, an even value.

But when $x^2$ is greater than $\mathrm{10,}$ we can wrestle the result back into normal form by factoring out a $10$ from the coefficient and incrementing the exponent: $\frac{x^2}{10}\cdot {10}^{(2\cdot m+1)}$. Note in that case the exponent is an odd value.

Because the A scale is logarithmic, $10$ appears exactly in the center of the A scale, since the log of the square root of $10$ is $\mathrm{0.5.}$ The entire left half of the A scale accounts for all squares with a normalized scientific notation exponent that is even, and the right half covers the odd exponents. Take a look at the A scale on your slide rule and notice its left half and right half are labeled very much alike. We'll come back to this when we take square roots.

photo of my Frederick Post 1447 with B's left margin aligned to A=10, showing that the left half of B and the right half of A are the same

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When we did multiplication we had to mess about with the slide, but squaring is a lookup that's indexed from the C or D scale to the A scale. For the rest of the post I'll assume it's on D, but you can always figure it out because lookups are always body to body or slide to slide. So if your slide rule's A scale is on the body, its index will be too. To find a square, simply move the cursor to the operand on the D scale, and read the result from the A scale.

Finding the exponent is pretty easy thanks to the visual cue of the split scale on A. If your square lands on the left side of the A scale, just double the exponent. And if it's on the right side, double it and add $1$ instead.

For example, the square of $200$ or $2.0\cdot {10}^2$ is $4.0\cdot {10}^{(2\cdot 2)}$ or $4.0\cdot {10}^4$ or $\mathrm{40,000}$

And the square of $\mathrm{5,000}$ or $5.0\cdot {10}^4$ is $2.5\cdot {10}^{(2\cdot 4+1)}$ or $2.5\cdot {10}^9$ or $\mathrm{25,000,000}$

photo of my Frederick Post 1447, showing 5,000 squared is 2.5 ⋅ 10^(9) on the right side of the A scale

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There's a photo of the latter calculation above. Did you notice how $5^2=25$ gave up a factor of ten to the exponential term and became $2.5$, making it a normalized coefficient?

Wonderful!

This observation about the "sidedness" (not an official math term) of the A scale also helps when taking a square root. To take the square root of $y\cdot {10}^n$ you must first determine which side of the A scale to work with. But we now know this is as easy as checking whether the exponent $n$ is even or odd.

If the exponent is odd, move the cursor to the $y$ value on the right side of the A scale, and read the result from D. The exponent of the answer will be $\frac{(n-1)}{2}$.

Likewise if the exponent is even, move the cursor to the y value on the left side of the A scale, and read the result from D. The exponent of the answer in this case is simply $\frac{n}{2}$.

If you prefer, rather than remember the expressions for both cases for the exponent of a square root, you can divide $n$ by $2$ and take the $floor$. You can fiddle with the expressions to show that's true. You don't need to worry about the coefficient since the slide rule provides it in normal form either way.

Let's do a quick example and take the square root of Avogadro's constant, $6.02\cdot {10}^{23}$

photo of my Frederick Post 1447, showing that the square root of Avogadro's contstant $6.02\cdot {10}^{23}$ is about $7.8\cdot {10}^{11}$

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The exponent of $23$ is odd, so we'll be on the right side of the A scale. On my Post 1447 pictured above, the labels go by tens (annoying, but a sign of how different brands evolved), which means this particular slide rule has left the factor of $10$ in its coefficients. Weird, but fine! We need to find $60.2$ with the cursor, and then we can read the result, which appears to be around $7.8$. We'll divide the exponent in half and take the floor: $floor\left(\frac{23}{2}\right)=floor\left(11.5\right)=11$. The square root of Avogadro's constant is about $7.8\cdot {10}^{11}$

Many slide rules provide a copy of the A scale called the B scale, typically with one on the body and the other on the slide. That's exciting for two reasons. First, it makes it easier for us to see that there's a gap in the A scale between 10 and 11. If you'd like to see what I mean, try sliding the left margin of the B scale to align with $10$ on A. You should see that the entire left half of B and the right half of A are metrically identical, although they may differ slightly in labels. There's no $10.5$ because $10.5$ is actually $1.05\cdot {10}^1$!

It felt very strange to me at first to see what appeared to be a discontinuity on the A scale. It helped me to think of the A scale as two distinct scales that cover the same squared normalized coefficients, but the right side version carries with it an extra square root of 10 and pairs up with different square root coefficients. I think of the left side as the A0 scale and right as the A1 scale, where the 0 and 1 represent the squared exponent's value modulo $2$.

The other reason having a B scale is great is that it allows you to chain other operations with squares and square roots. We haven't discussed chaining operations yet, but a big part of the craft of calculation with a slide rule involves thoughtfully ordering your operations to minimize error. In the previous post I mentioned that most slide rule operations provide three digits of precision with a 2% error bound. In that light, it's pragmatic to take care to minimize error over multiple operations. I'll cover that in more detail in time.

Alright, I hope that's enough for now!

Next post: cubes and cube roots