Re: spherical geometry

["Carl E. Mungan" <mungan@USNA.EDU>]

> Carl's result is nice but the formula for B above is manifestly
> positive because of the sqrt() function.  This means it won't work
> correctly for heading angles that are obtuse (i.e. westerly oriented
> headings that have a negative cos(B) value).  Also it doesn't
> discriminate between northerly headings (positive B) and southerly
> headings (negative B) because cos(B) = cos(-B).

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.
>    http://williams.best.vwh.net/avform.htm#Clairaut

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
--
Carl E. Mungan, Asst. Prof. of Physics  410-293-6680 (O) -3729 (F)
      U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5040
mailto:mungan@usna.edu       http://usna.edu/Users/physics/mungan/