On Jan 6, 7:09 pm, marcelo <[email protected]> wrote:
>
> Your example, (0,0; 87,0; 0,180), puts all three points on the same
> great circle and therefore is not a spherical triangle.

Isn't it? I'm not sure who your "No" was directed to.

There *are* three great circles there, I think: part of the Prime
Meridian (0,0 to 87,0), part of the Equator (0,0 to 0,180), and part
of an oblique great circle connecting 87,0 to 0,180. Using
http://maps.forum.nu/gm_flight_path.html and points 85,0 to 0,180
which lie on the map, the line goes around the Pole, on the opposite
side of the earth from the US [ie across Asia]. It doesn't follow the
Meridian. That does describe a spherical triangle.

Even if the northernmost point of the three was actually at the Pole,
there would still be two great circles involved, the Meridian and the
Equator, and a bilinear segment of a quarter of the Earth would
result. The "centre of gravity" of those points on the surface
wouldn't be at 30,60 though, which does show that the simplification
of the arithmetic mean is a simplification which breaks down in
boundary cases. (In fact because you only have two great circles
involved, you can divide by 2 and get a point at 45,90, which does
seem reasonable for the 3D solid surface).

Andrew
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