Spherical Polygon AreaDate: 08/09/99 at 14:56:51 From: Min Chen Subject: Spherical polygon area Would you please explain to me what the meaning of "theta" is in the formula of spherical polygon area below? S = sum(theta) - (n-2)Pi Date: 08/09/99 at 16:19:25 From: Doctor Tom Subject: Re: Spherical polygon area The thetas are the internal angles of the polygon in question. By the way, the formula above is for spherical polygons on a unit sphere - a sphere of radius 1. For example, suppose the earth has radius 1, and you want to find the area of the triangle connecting the North Pole and two points on the equator 90 degrees apart. The triangle covers 1/8 of the surface area of the earth. The total area should be (4 pi)/8 = pi/2. The triangle has three angles, so n = 3. All three angles are 90 degrees, or pi/2. The sum of the angles is (3 pi)/2. The surface area is thus (3 pi)/2 - (3-2)pi = pi/2, just as it should. - Doctor Tom, The Math Forum http://mathforum.org/dr.math/ Date: 08/09/99 at 16:23:16 From: Min Chen Subject: Re: Spherical polygon area Is it possible for the area for a convex spherical polygon to be negative? That happens in my program. -Min Date: 08/09/99 at 16:27:40 From: Doctor Tom Subject: Re: Spherical polygon area Hi Min, There's something wrong with your program. Every triangle has more than 180 degrees (more than pi radians), so when you subdivide your polygon into triangles, each will have more than pi radians. So when you subtract pi from it, the area will be bigger than zero. Are you sure you're just looking at the interior angles of the polygon? If you look at exterior angles the formula is nonsense. - Doctor Tom, The Math Forum http://mathforum.org/dr.math/ Date: 08/09/99 at 16:32:35 From: Min Chen Subject: Re: Spherical polygon area Sorry for bothering you again. Given the coordinates of polygon vertices and the center of the sphere, how do you compute the interior angle of the polygon? Just want to check where my program got wrong. Thanks for your help. -Min Date: 08/09/99 at 17:20:30 From: Doctor Tom Subject: RE: Spherical polygon area Hi Min, What I would do is find the equations for the two planes passing through the center and the two pairs of points. Normalize the plane equations to the form Ax + By + Cz = D, where A^2 + B^2 + C^2 = 1, and then (A,B,C) will represent a vector of unit length perpendicular to the plane. Take the dot product of the two vectors for the two intersecting planes and that will be the cosine of the angle between the planes (and hence of the spherical polygon edges). - Doctor Tom, The Math Forum http://mathforum.org/dr.math/ Date: 08/09/99 at 18:08:14 From: Min Chen Subject: Re: Spherical polygon area But which direction of the normal vector should I choose? You know, (A,B,C) and (-A,-B,-C) are both perpendicular to the plane. -Min Date: 08/09/99 at 18:52:33 From: Doctor Tom Subject: RE: Spherical polygon area Okay, try this. Suppose the three points on the sphere are A, B, and C, and you want to get the angle made by the path A to B to C along the surface. Let O be the center of the sphere. OA cross OB is a vector perpendicular to the plane OAB, and OB cross OC is another perpendicular to plane OBC. Normalize these vectors to length 1, and take their dot product. This is the cosine of the angle between the planes with a consistent orientation. - Doctor Tom, The Math Forum http://mathforum.org/dr.math/ |
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