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# Re: spherical geometry

• From: Carl Mungan <mungan@USNA.EDU>
• Date: Sun, 15 Aug 2004 00:35:39 -0400

David Bowman wrote:

Suppose we straightened out the 2nd leg of the path so *all three*
legs of the path are geodesically straight and the length of the 2nd
leg is the same as the length of the 1st and 3rd legs. The whole
closed path is now an equilateral triangle as inscribed onto the
spherical surface. The problem is to find a formula for the measure
of the interior angle of such an equilateral triangle as a function
of the length s of the sides of the triangle (conveniently in units
of the sphere's radius). A few hints are that 1) the value of the
formula must boil down to 60 deg in the limit of s becoming a
zeroth fraction of the sphere's radius, 2) the value of the formula
becomes 90 deg when s is 1/4 of the circumference of the sphere,
3) the maximum size triangle occurs for a great circle with 3
equally-spaced vertices (120 deg apart from each other) on it with
the interior angle at each vertex being 180 deg across the vertex and
each side having a length s of 1/3 of the sphere's circumference, and
4) the messy intermediate math eventually simplifies to a relatively
simplified formula in the general case.

Resisting the temptation to look up any references on this challenge,
I have come up with what I believe to be the solution (where A =
interior angle of interest):

cos(A) = tan(s/2)/tan(s)

This satisfied hints 1 through 4, so I suspect it's correct. I don't
remember having seen this formula before, so it was a neat problem. I
set up the triangle with its base on the equator and its top vertex
on the Greenwich meridian, to make the derivation simpler.

For a lot of extra credit points you can also find the proper formula
for the *area* of this spherical equilateral triangle in terms of the
length s of the sides of the triangle (making sure that the formula
boils down to all the correct values for the variously known special
cases).

I find the following integral formula but I don't have a symbolic
math program at home to see what it evaluates to:

area = 2 * integral from 0 to MAX of {arcsin[tan(X)/tan(A)] * sin(X) * dX}

where MAX = arcsec[cos(s/2)/cos(s)]

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/