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MathGroup Archive 1995

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Re: finding irregular areas

  • To: mathgroup at christensen.cybernetics.net
  • Subject: [mg1084] Re: finding irregular areas
  • From: Richard Mercer <richard at seuss.math.wright.edu>
  • Date: Sun, 14 May 1995 20:01:26 -0400

>  Greetings:
>  

>  I have a colleague who is looking for a full or partial
>  Mathematica solution to the following situation.
>  

>  He has some aerial photographs of agricultural fields.
>  These fields are broken up by fences.  The fields are
>  not of any nice shape being defined by rivers, roads,
>  etc.  He was to be able to scan the photos into his
>  computer and then find the areas of the various fenced
>  areas.
>  

>  He will need to be able to define the fenced areas with
>  a mouse or similar device and then feed the numbers to
>  a program to find the area.
>  

>  Any ideas on how to handle all or part of this kind of
>  problem?
>  

>  [I did suggest that he just cut the photos into the
>  various areas and weight each and compare to a known
>  area's weight, but that was not technical enough.  :-)
>  ]
>  

>  Thanks.
>  

>  Steve Christensen
>  


Steve,

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)].

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!

Richard Mercer


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