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Re: spherical geometry

Regarding Carl's rearrangement of his formula:

First let me simplify my formula to:

tan(B) = sin(T) * cot(P-P0)

Are you happier with this? It allows negative values, although of
course it still can't distinguish quadrants I/IV from II/III.

At least twice as happy as before. Actually, the ambiguity your
result still now has is *supposed* to be there. This is because
there are 2 geodesic paths connecting two arbitrary fixed points
on a sphere going in opposite directions. Your formula now gives
both the short way *and* the long way around the sphere from the
same formula.

If you want another spherical geometry exercise,
rederive Clairaut's formula.

Heck, I'm never going to get anything else done, but it's fun.
Equation (3) divided by (2) in my PDF gives:

tan(A) = tan(T)/sin(P-P0)

Solve this for (P-P0) and substitute into my bearing formula above
to get:

cos(T) * cos(B) = cos(A) = constant

which is Clairaut's formula, and certainly looks like a nice way to
satisfy Hugh's request for a simple method of correcting course
enroute. Swap cos(B) for sin(H) for Brian's desire to get headings
relative to north. -Carl

Good work!

David Bowman