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Recalling that cos(A)=tan(s/2)/tan(s), I find the following in the
small s limit:
cos(A) =~ (s/2)/(s) = 1/2 => A = 60 degrees agreed
(Did I say something that made you think otherwise?)
That integral almost looks like an elliptic integral, except for
the sign in front of the trig fn. It has the correct (trivial)
limiting values for A = 0 and 90 degrees. Carl
However, your comments about the area made me take a second look at
my result and I am *super* embarrassed to discover that I mixed up
the latitude and colatitude. The corrected integral now becomes:
area = 2*integral from 0 to s/2 of {dY/sqrt(1+cot^2(A)*csc^2(Y)}
Check out the small s limit:
area =~ s/sqrt[1+(1/3)*(4/s)^2] =~ s/sqrt[(1/3)*(4/s)^2]
which wouldn't you know it is *exactly* your quoted result above!!
But the really important thing: Unlike the messages from you and
John, my method for finding the above expressions for the angle and
area is *very* simple. If you give me another hour, I will write
them up in PDF and post them for you to inspect. 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/