ksbooth@watcgl.waterloo.edu (Kelly Booth) (04/30/89)
P.S. The algorithm presented in article <5397@cs.utexas.edu> by atc@cs.utexas.edu (Alvin T. Campbell III) is typical of most of these "solutions". It walks the entire polygon looking for the "top" vertex (incorrectly, as it turns out, but making it right takes even more code), when Newell's formula would perform a fairly simple (but floating-point, to be sure) calculation at each vertex and be done with it. For quadrilaterals and triangles (by far the most common cases), I believe the total computation is less for the Newell's formula. As the number of vertices gets large, Newell's formula may lose, but in these cases the chance of "bad" data increases, so Newell's formula is even more appropriate. (And if you only want the orientation, you don't have to compute all three coefficients, just c).
rhbartels@watcgl.waterloo.edu (Richard Bartels) (04/30/89)
In article <9458@watcgl.waterloo.edu> ksbooth@watcgl.waterloo.edu (Kelly Booth) writes: > >(And if you only want the orientation, you don't have to compute all three >coefficients, just c). > The context of the midterm question, which Kelly forgot to mention, is one in which the lines of sight are all parallel to the z axis. This is why c alone gives the (back/front facing) orientation. -Richard