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Spherical Polygon Area


Date: 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/   
    
Associated Topics:
College Higher-Dimensional Geometry

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