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Re: spherical geometry

I *agree* with Carl's answer for the Guantanamo Bay location.

Great. Here then is my method. Let the two points on the earth's
surface have (latitude,longitude) of (T1,P1) and (T2,P2) with
north/east positive (ie. T=theta, P=phi). They lie on a great circle
like in my PDF document *except* that this circle no longer crosses
the equator at P=0. Instead it is shifted over to say P=P0. But if we
replace P in my document by deltaP=P-P0, then all the equations will
work as before. In particular, Eq. (3) divided by (2) gives the
constant tilt angle A in terms of T and deltaP. Write this ratio down
twice, once for each set of coordinates, and equate them. That gives
an equation for P0, which can easily be solved to find for the Gitmo

P0 = 73.625 or -106.375 degrees.

Now that we've found the great circle on which the two points lie, we
simply need to find the angle (call it B say) between any point
(T,deltaP) on this great circle and east. An easy way to do this is
to note that cos(B) = dot product between a unit vector tangent to
the great circle and a unit vector pointing east. Both of these unit
vectors lie in the plane perpendicular to (x,y,z) given by Eqs.
(1)-(3) in my PDF. The east unit vector is therefore
(-y,x,0)/sqrt(x^2+y^2). The tangential unit vector is seen from the
diagram in my PDF to be (-sin G, cos G cos A, cos G sin A) where
G=gamma and A=alpha are the "great circle" coordinates of the point.
After some minor cleanup, this becomes:


Plug in the given values of T1 and P1 with the above value of P0 to
get the desired B. Carl

ps: Thanks to John Denker for going to the trouble to very nicely
draw and re-explain the concept of transporting a vector around an
area on a sphere.
Carl E. Mungan, Asst. Prof. of Physics 410-293-6680 (O) -3729 (F)
U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5040