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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.
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
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