Finding Area of a Spherical Triangle
Date: 07/08/2004 at 20:32:13 From: Denise Subject: spherical triangles Assuming the earth is a sphere, and given three (latitude, longitude) coordinates, what's the easiest way to calculate the area of the spherical triangle formed by those three points? For example: What is the area between coordinates 34, -105; 34.1, -105.5; 34.2, -105.9 ? I have found various formulas to calcuate spherical triangles, such as Girard's theorem, but they assume I know the angles. I don't know how to go from coordinates to angles and take into account all the other stuff like great circles. I tried finding the distances between the 3 points and then using 1/4*square root (p*(p-2a)(p-2b)(p-2c)), where a,b, and c are the lengths of the sides and p is the perimeter, but of course this doesn't take into consideration that the Earth is not flat.
Date: 07/09/2004 at 14:07:56 From: Doctor Rick Subject: Re: spherical triangles Hi, Denise. My CRC Standard Mathematical Tables contain the basic formula for the area of a spherical triangle: Area = pi*R^2*E/180 where R = radius of sphere E = spherical excess of triangle, E = A + B + C - 180 A, B, C = angles of spherical triangle in degrees This is the formula you say isn't helpful because you don't know the angles, right? Well, the tables also have the following formula for the spherical excess E: tan(E/4) = sqrt(tan(s/2)*tan((s-a)/2)*tan((s-b)/2)*tan((s-c)/2)) where a, b, c = sides of spherical triangle s = (a + b + c)/2 You can find the sides using either the cosine formula or the haversine formula, found on the following pages in the Dr. Math archives: Distance using Latitude and Longitude http://mathforum.org/library/drmath/view/54680.html Deriving the Haversine Formula http://mathforum.org/library/drmath/view/51879.html Good luck! Write back if you are still stuck and show me what you've been able to do. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
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