Thanks much! It seems so obvious now that you pointed it out... anyways, thanks again!
Tyler On Wed, Feb 9, 2011 at 3:47 PM, Mathieu Bouchard <ma...@artengine.ca> wrote: > On Wed, 9 Feb 2011, Tyler Leavitt wrote: > > At the moment, no, this isn't allowed =) I had thought of that, and then >> with all the difficulty I've had without taking this into account, I'd >> figured one step at a time. >> > > You replied in private. (why ?) > > Anyway, I reply back to the list. > > suppose you have 4 points a,b,c,d with polygon edges ab, bc, cd. They have > diagonals ac, bd. They form triangles abc and bcd that intersect. The > intersection is bce where e is the single point at the intersection of ac > and bd. > > e is a + p*(c-a) where p is some number between 0 and 1, which is the > fraction of the travel from a that you have to do to reach e, when going in > a straight line towards c. > > e is also b + q*(d-b). > > e = a + p*(c-a) = b + q*(d-b). > > I don't quite remember how to continue the reasoning from there. > > However I found this out http://en.wikipedia.org/wiki/Line_intersectionwhich > I can translate to : > > e = (det(a,c)*(b-d) - det(b,d)*(a-c)) / det(a-c,b-d) > > after that, it's easy, as the area of the bce triangle is det(b-e,c-e)/2, > as you'd do for any triangle. This part can be computed using this > abstraction : http://gridflow.ca/help/%23polygon_area-help.html > > With your example, you have to do the above calculation 5 times on > different combinations of points, to get 5 different inner points and 5 > different areas. Then you combine the inner points to get the areas of 5 > more triangle. Area #1, in the middle, is not a triangle, but you can > compute its area with [#polygon_area] anyway. > > > _______________________________________________________________________ > | Mathieu Bouchard ---- tél: +1.514.383.3801 ---- Villeray, Montréal, QC >
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