Next message: Dave Weilacher: "Re: The circumnavigator's paradox. was: Benetnasch and Alkaid revisited"
> Mark three points around the equator. Point A is at 0deg West, point B is
> at 120deg West, point C is at 240deg West. Join A to B, B to C, and C to A
> again, with great circles, going Westerly each time. Then you have three
> vertexes, each subtending a 180deg angle. The resulting triangle divides
> the circle into two equal halves
>
> You might object that with such a 180deg angle, then each vertex has become
> a straight line, not a corner. That's true, because it's a limiting case.
> Think about it, if you prefer, when the angle at A, B, and C is not quite
> (but as near as dammit) equal to 180, so there's a VERY obtuse angle at
> each corner. Then increase these angles, very slightly.
George,
See your point(s). Had not considered that case, C to A as the smallest
GCD; but rather joined my three arbitrary points (C to A) with the greater
GCD.
>
> This business of the "spherical excess" is quite new to me, but it seems to
> work. If we add the three "angles", each 180, we get 540 deg. The spherical
> excess over a tiny triangle, in which the angles always sum to 180deg, is
> therefore 360 deg, or 2pi radians. Multiply this by r squared, and we get 2
> pi r-squared, which is indeed the area of the half-sphere that the
> "triangle" embraces. Today, I've learned something new...
Ain't life grand? Been doing dumb human tricks with spherical trig today.
Once I got got the bigger picture, it occurred to me if swap out AP Lat and
GP Lat (declination) and vise versa in the Z formula(s) I could obtain the
AP, GP, pole angle. Add those up (-180d, pi radians) and I can calculate
area in the triangle. I am dumbfounded as to how to use the area inside the
triangle--other than to impress sailing friends whose eyes glaze over when I
mention cel nav with the number of square miles of water included in the
triangle--but betting one the list mentors has a practical application.
Bill
