# Re: area enclosed by a polygon on a sphere

• Subject: Re: area enclosed by a polygon on a sphere
• From: Ronn Kling <ronn(at)rlkling.com>
• Date: Wed, 04 Aug 1999 00:58:00 GMT
• Newsgroups: comp.lang.idl-pvwave
• Organization: Deja.com - Share what you know. Learn what you don't.
• References: <7o4m5g\$uvv\$1@nnrp1.deja.com> <7o74n7\$aa7\$1@news.lth.se>
• Xref: news.doit.wisc.edu comp.lang.idl-pvwave:15964

```In article <7o74n7\$aa7\$1@news.lth.se>,
Struan Gray <struan.gray@sljus.lu.se> wrote:

>     No code, but an idea which essentially uses Green's Theorem.
>
>     1) convert lat/lon to cartesian coords
>     2) use them to make an IDLgrPolygon object
>     3) use the IDLgrTessellator object to turn that
>         into a set of triangles
>     4) for each triangle work out the solid angle it
>        subtends from the centre of the earth
>     5) add up the solid angles and convert to an
>        area.
>
>     Working out the solid angle subtended by an arbitrary
> triangle of points on the surface of a sphere is left as an
> exercise for the reader (watch out for triplets of points on
> the same great circle :-).
>
> Struan
>

I found out how to do this today.  For a spherical triangle the solid
angle subtended is the sum (in radians) of the interior angles  - Pi.
This can be generalized to any polygon as

Area = (sum of all interior angles) - (n-2)*Pi

Where n is the number of points.

-Ronn

--
Ronn Kling
Ronn Kling Consulting
www.rlkling.com

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```