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Another important bit of geometry to remember (which is not widely taught):
The angular size of a distant object is just the linear physical size across the line of sight divided by distance. The result is an angle as a pure ratio (a.k.a. "in radians") and can be converted to more familiar minutes or arc by multiplying by 3438 (this "magic number" is just 360·60/2pi):
  (angular size) = 3438 · (physical size) / (distance).
Next time you see a substantial ship approaching the coast, give that a try. Find out how big the ship is. Estimate the size "across" the line of sight. Divide by distance and multiply by 3438. Then get out your sextant (or some simpler angular estimator) and measure it. Notice that you can easily turn this math around to measure distance from observed angular size. Angles are ratios: the ratio of size to distance.

This can be considered an approximation (though this is not a necessary point of view), and if so, it should be counted as excellent for angles smaller than 5°, which covers anything you'll see in the night sky and also covers the angular sizes of ships at sea, distant lighthouses, etc. 

So if I am standing on the Moon, looking at the Earth in the sky, what is its angular SD? The linear size here is just the radius of the Earth. The distance is the Earth-Moon distance. Divide 4000 miles by 240,000 miles, and you get 1/60. Multiply by 3438, and you find that the Earth's SD (which is essentially identical to the tabulated HP) is about 57'. Now turn it around. You're standing on the Earth looking at the Moon. The radius of the Moon is 27.24% of the Earth's radius. Call it 1000 miles for a round number. Divide 1000 by 240,000 miles (same distance!) and then multiply by 3438 and you'll find about 15' for the Moon SD. If we worked the numbers more carefully, the ratio of Moon SD to Earth SD (or HP) would have to be 0.2724 because the distance is the same, whether Earth-to-Moon or Moon-to-Earth.

What happens as distance changes? Suppose the Earth-Moon distance is 230,000 miles. Try it out. What do you get for Earth SD (which is HP)? And what do you get for Moon SD? What is the ratio of the two? And try it again for a distance of 250,000 miles. See what happens in every case??

Notice that the tabulated HP in the almanacs and other resources is simply a convenient surrogate for the distance to the Moon. We could just as easily tabulate the Moon's distance in miles for every hour of every day. Then we would get the HP by doing some simple math: HP = 3438(Earth_Radius / Moon_Dist). By tabulating HP directly, we are spared from that repetitive arithmetic.

Frank Reed

PS1: That 3438 is only necessary for "practical math". There are variants for different angular outcomes. If you take an angle as a pure ratio and mutliply by 57.30, you get the angle in degrees. If instead you multiply by 206265, you get the angle in seconds of arc. These three numbers have the same origin, and they have a simple ratio relationship: 57.30-to-3438-to-206265 is 1:60:3600. In old astronomy books, especially from the 19th century, you'll often see another variant of this factor. It's the inverse of 206265, but it's usually written as sin1" (the sine of one arcsecond). It was almost never the case that they wanted the actual sine of one arc second. This was just a simple shorthand for (2pi)/(3600·360) which is the exact conversion from angles as pure ratios to angles in seconds of arc, and which is almost exactly 1/206265, and which is almost exactly sin1".

PS2: The radius of the Earth in nautical miles is also 3438. Freaky?? Well, if you think it through it all makes sense. But supposed I have the distance to the Moon in nautical miles, and I want to get the HP. From above, that's HP = 3438(Earth_Radius / Moon_Dist). But the Earth's radius in n.m. is 3438... So then
  HP = 3438² / Moon_Dist,
which is one of those "weird but true" things.