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*From*: "Carl E. Mungan" <mungan@USNA.EDU>*Date*: Tue, 17 Aug 2004 11:55:31 -0400

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/

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