Jeff Massung wrote:
Michael,
I assume you are talking about 2D polygons...
Just take a 3D cross product of the first two vector (P3 - P2) x (P2 - P1).
If the resultant vector is coming out of the screen (Z < 0) then the polygon
is counter-clockwise. If it is going into the screen (Z > 0) then the
polygon is clockwise.
http://chortle.ccsu.edu/VectorLessons/vch12/vch12_1.html
That page may be of help if you don't know what the cross product is. Note:
to do the 3D cross product, just use Z=0 for all the points.
Well, I confess I don't properly remember what the cross-product is, so
I will go read that page.
But even without doing so, I am sure that this can't be a correct
solution; the following two polygons have opposite flow, but P1, P2 and
P3 are the same
0,0; 100,0; 100,100; 0,0 (i.e. a counter-clockwise triangle)
0,0; 100,0; 100,100; 200,100; 200,-10; 0,-10; 0,0 (i.e. a clockwise
L-shape)
I'll post a couple of suggested solutions shortly (once I've got the
stack running properly)
-- Alex.
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