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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.

-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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