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- From: newberry@kepler.as.arizona.edu (Mike Newberry)
- Newsgroups: comp.graphics
- Subject: Re: Calculating polygon areas
- Message-ID: <1992Dec18.204704.24387@organpipe.uug.arizona.edu>
- Date: 18 Dec 92 20:47:04 GMT
- References: <1992Dec14.163751.24617@sophia.smith.edu> <1992Dec16.173613.5737@medusa.prime.com> <1992Dec16.192007.29022@sophia.smith.edu>
- Sender: news@organpipe.uug.arizona.edu
- Organization: University of Arizona, Tucson, AZ
- Lines: 22
-
- In article <1992Dec16.192007.29022@sophia.smith.edu> orourke@sophia.smith.edu (Joseph O'Rourke) writes:
- >
- > In article <1992Dec16.173613.5737@medusa.prime.com> mrj@CIS.Prime.COM writes:
- > >In article 24617@sophia.smith.edu, orourke@sophia.smith.edu (Joseph O'Rourke) writes:
- > >> So if the coordinates of vertex v_i are x_i and y_i,
- > >> twice the area of a polygon is given by
- > >> 2 \A( P ) = \sum_{i=0}^{n-1} (x_i y_{i+1} - y_i x_{i+1})
- > >
- > >There is a more general form of the above:
- > >
- > > Area = 0.5 \sum_{i=0}^{n} (P_i X P_{i+1})
- > >
- > > s.t. i+1 = (i+1) mod n i.e. when i = n then P_{i+1} = P_0
- > >Where X denotes a vector cross product.
- >
- >These are equivalent formulae: I just unraveled the cross product
- >into x and y coordinates.
- >
- > >The proof is simple. If anyone is interested I'll post it to the net.
- >
- It's just a straight application of Green's Theorem from calculus.
- .
-