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MathGroup Archive 1995

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Re: finding irregular areas

  • To: mathgroup at christensen.cybernetics.net
  • Subject: [mg1105] Re: finding irregular areas
  • From: rubin at msu.edu (Paul A. Rubin)
  • Date: Sun, 14 May 1995 23:31:38 -0400
  • Organization: Michigan State University

In article <3os827$r7q at news0.cybernetics.net>,
   Richard Mercer <richard at seuss.math.wright.edu> wrote:
->Here is a formula for the area of a polygon with vertices 
->
->{(xk,yk): k = 1,...,n}:
->
->Area = 1/2 [(x1*y2 - x2*y1) + (x2*y3 - x3*y2) + ... + (xn*y1 - x1*yn)].

Minor caveat:  I believe this requires the vertices to numbered 
consecutively (and counterclockwise).  It's derived by finding the areas of 
the triangles spanned by the origin and each consecutive pair of vertices.

->This formula appears in an Article by Gil Strang of MIT on p. 253 of the 
March  
->1993 issue of The American Mathematical Monthly, with the note that it is 
 
->"known, but not well known". There is also a very brief discussion of 
proofs  
->and other references, including an article by Bart Braden of Northern 
Kentucky  
->U., a known Mathematica enthusiast.
->
->If your friend is successful in digitizing enough points to define the 
region,  
->this formula will calculate the area. Being a polygon is not a practical 
 
->limitation, you just might have to use more points!

This is probably better than my suggestion about Delaunay triangles, as 
this works properly with nonconvex polygons (assuming you order the 
vertices correctly, which should be easy enough).

Paul

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* Paul A. Rubin                                  Phone: (517) 432-3509   *
* Department of Management                       Fax:   (517) 432-1111   *
* Eli Broad Graduate School of Management        Net:   RUBIN at MSU.EDU    *
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they translate it into their own language, and at once it is something
entirely different.                                    J. W. v. GOETHE


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