Right! Except that you can remember it and do it without calculus.
Newton’s method is more general, of course.
When I first came up with that method, incidentally (and I don’t claim that I’m the only person to ever come up with it, just that I did so independently), I briefly considered modifications to bring the mediant closer to the arithmetic mean (for instance, you could multiply the numerator and denominator each by some number, to bring the denominators closer together). But then I realized that trying to get closer to the arithmetic mean isn’t really what we want, anyway: What you really want for the in-between-number is something that approximates the geometric mean (the geometric mean itself would be perfect, of course, but calculating it requires the very same square root we’re looking for).
That is neat. Kinda surprising that it works no matter which side of the approximation x and y start on.
I poked around a little with higher-order iterations, and came up with this:
x’ = -1/(2x[sup]3[/sup]) + 3/(2x) + 3x/8
That gives you triple the number of digits per round instead of double. But it looks like a bit net loss with all those multiplies. The x[sup]3[/sup] in particular kills you.
Computers of the human female model were also extensively employed in the Manhattan Project. Of course, they had rather sophisticated mechanical calculators to work with.
There were college-level classes where the “human computer” profession was taught, just as there have always been secretarial classes that taught typing, filing, and stuff like that. Only computing was certainly much more technical and must certainly have required a solid knowledge of basic math at least.
I used to have a collection of old math textbooks that I found in various used-book stores and thrift shops. One of them was a textbook for such a class. There were specific lessons for things like how to enter the first number of an addition or multiplication into the machine, how to enter the second number, and so forth. Entering the second number wasn’t the same as entering the first number. Thus, they had the terminology like “addend” and “augend”, “multiplier” and “multiplicand”, that one hardly ever hears any more. Do today’s kiddos even learn “minuend” and “subtrahend”?
ETA: I had a trig textbook from 1914. I had, and still have, a college algebra text from 1947 or so. I always though those books were vastly better than the cruft we see for textbooks today – although the selection of topics to cover was somewhat dated, and not entirely the same as topics covered today. There was an entire chapter on working with approximate numbers. Today, students don’t even learn “linear interpolation” any more.
I pointed out to my Calculus teacher that the old 1941 trig text gave reductions formulas for cot, sec, and csc to put any angle into the first quadrant, which she had never heard of before.
The paper (this one, not the one you linked to) just shows a fragment of such a table. The section suggests that one might compute a large table, and then find the rows such that the difference is small (in practice, either -2 or 1). m and n would not be the same, but you don’t know in advance what the values are, so you have to compute all the values for each one.
The table is relatively easy to construct because of finite differences:
120[sup]2[/sup] + 241 = 121[sup]2[/sup]
121[sup]2[/sup] + 243 = 122[sup]2[/sup]
122[sup]2[/sup] + 245 = 123[sup]2[/sup]
You can see that the 241, 243, 245, … sequence just differs by 2 each time, so the table is easy to make through basic addition.
My father, in the middle of the 20th century, was required to memorise some of those tables, to get his professional licence. Not the whole table, but at spaced intervals, so that he could linearly interpolate between the points.
I took Regents Math in high school (circa 1980). Because we were the smart kids our math teacher liked us so one day he showed us this method for finding square roots (somewhat similar to writing out long division). He didn’t spend any more time on it because pocket calculators were common at this point, and because even us smart kids found it rather confusing (so did the teacher*!) I think he just did it for fun, showing us young uns’ how he had to learn it!* 
From Feynman’s memoirs - the Manhattan project used IBM card machines to do large numbers, since the requirements included doing large matrix calculations in sequence. (Essentially simulations with time steps, I think) I think it got to the point by then where the machines could add and multiply, the results had to be transcribed.
He mentions occasionally they had typos (keypunch errors) or other problems, which meant throwing the latest result out and backing up to the last good step and redoing it. He then pointed out they could repair the problem by simply redoing the column and row in error, as one number only affected a that part of the result. they could do the correction on different color cards and give them priority to get caught back up. He said he came back the next morning, and found a rainbow of cards as they were using his technique to repair numerous errors at once.
I also recall reading about a persistent deep error in some published logarithmic tables - someone made a typo in the early 1800’s in some manual, and it followed academia around for a century, until computers (the electronic ones) came along.
there was also a product being hawked in the early 90’s or so as “turn your 386 into a 486”. The 486 had the math coprocessor on the chip (usually). Reading between the lines, this software replaced the floating point software math library DLL for a 386 with something that only gave about 3 digits accuracy instead of 8 - significantly cutting the calculation time. When most games had a resolution of 600x800, that was all the accuracy needed when speed was important.
When I was quite young, my father showed me that method too. It superficially resembles doing a long-division problem.
You can do a similar technique for finding cube roots too. Instead of grouping the digits in groups of two, you start by grouping the digits in groups of three.
When I first read Thudlow Boink’s post I skipped over the linked pdf file. Mistake! When Indistinguishable linked back to it I took another look. It addresses–and gives as much of an answer as we are likely to find–my exact questions. So it looks like the method adding numerators and denominators may well be the answer.
Hm, interesting, I hadn’t actually read that link either.