Re: Equal Area Spherical Triangles Apex Locus on Sphere

From: Oscar Lanzi III (ol3_at_webtv.net)
Date: 12/04/04 

Date: Sat, 4 Dec 2004 16:27:33 -0600

OK, I'll start that.

Let LtA be the latitude of A, north positive, LoA be the longitude of A,
east positive, with similar nomenclature for points B and C. Set N to
the North Pole. Draw Triangle ANB. From the Law of Cosines and known
quantities we have:

cos(AB) = sin(LtA)sin(LtB)+cos(LtA)cos(LtB)cos(LoA-LoB)

Likewise draw Triangles NAC and NBC and get analogous expressions for
cos(AC) and cos(BC). With the sides of Triangle ABC thus characterized,
obtain the angles in that triangle by using the Law of Cosines solved
for a vertex angle; for example:

cos C = (cos(AB)-cos(AC)cos(BC))/(sin(AC)sin(BC))

where angle C is taken to be within Triangle ABC (no longer involving
N). To get the sines in the denominator you must take square roots (sin
x = (+/-) sqrt(1-cos^2 x)); give the square roots positive signs because
the arcs measure 180 degrees or less. Take the inverse cosines of your
angles, add them up and set the sum to the desired value.

The locus may be a simple closed curve, but its algebra is not very
simple!

--OL

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