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Re: spherical geometry

Regarding Carl's more recent post:

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

Good. I knew you knew that your formula for A was correct.

(Did I say something that made you think otherwise?)

Well, sorta. In your previous post you wrote:

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

This quote mentions A = 0 as a limit. I kinda figured it was
probably a typo. But being the obnoxious lowlife sort of person I am
I just had to highlight it anyway.

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!!

Yes, it gets the s --> 0 limit correct, and it is also correct for
s = [pi]/2, A = [pi]/2 case as well. However for the largest
possible triangle case s = 2*[pi]/3, A = [pi] the correct area is
supposed to be 2*[pi] but your integral gives the area --> 0.

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

Oh, BTW Carl, I was wondering about the undergrad curriculum at the
USNA. How much spherical geometry are the cadets taught nowadays?
I suppose that the GPS system probably makes lots of the old
fashioned navigational arts obsolete.

David Bowman