Re: (TFT) Conical hexes

[pvk@oz.net]

My first shot would be like this:

1. Approximate Earth radius to say 6,360,000 meters.

2. Observe that with conical hexes, hex length will be proportional to
circumference of the sphere, which is the same as the circumference of a
circle of radius R. Circumference = Radius x 2 x pi. So hex length is
proportional to altitude above the center of the earth.

3. So hex length (L) = altitdue above sea level (ASL) + 6,360,000 meters /
6,360,000 meters.

4. So at 0 ASL. L = 1 standard hex length. At 80,000m ASL, L = (6,360,000
+ 80,000) / 6,360,000, = 1.0126 x standard hex length.

At that point, I think I'd decide it was close enough and not worry about
adding hexes as I ascended.

Another approach might be to make a standard BM 10 hexes x 10 hexes at sea
level, and then calculate when it would want to become 11 hexes x 11
hexes, or the point when you descended to where a BM would be 9 x 9 hexes
(say, the point when L = 1.05, or L = 0.95):

1.1 = (A + 6,360,000) / 6,360,000
6,360,000 x 1.1 = A + 6,360,000
A = 6,360,000 x 1.1 - 6,360,000
A = 6,360,000 x 0.1
A = 636,000

So intervals where one BM scale rounds to the correct size are 636,000m
deep in altitude, so you'd need to sink or elevate half that distance
(318,000m) before you'd want to add or subtract a hex from your BM.

Again, I can't imagine wanting to care unless I was going to hand it over
to a computer program (which I would, but that's another story...).

PvK
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