Re: spherical geometry

["Carl E. Mungan" <mungan@USNA.EDU>]

David Bowman wrote:

>  >area = s - 2*integral from 0 to s/2 of {dY/sqrt(1+tan^2(A)*sin^2(Y)}
>
> Sorry, but your formula *doesn't* have the correct limiting value for
> small s.  In the small s limit the expression above approaches
>
> area = (s^3)/8
>
> (which is dimensionally impossible).  But the correct limiting
> formula for the small s area is
>
> area = (sqrt(3)/4)*s^2 .
>
> Also, when the side length s is small A *does* not go to zero; it
> goes to 60 deg (whose squared tangent is 3).

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?)

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/