Quote:Original post by badmoonQuote: Original Post by Dmytry
it must work with non-convex polyhedrons, you will have nonzero resulting mass only inside polyhedron - "negative" masses will do the job.
Do you mean the first method will work for non-convex polyhedrons? Because in this case some of your pyramids could partly be inside the mesh and partly not, so if you look at just their positive volume (i.e. the part in the mesh) it's no longer a pyramid but one with 'bits' missing from it - and that just makes another polyhedron so we haven't simplified the problem.
If you meant the 2nd method will work for non-convex polyhedrons, then can you elaborate more on this, because I still can't quite see how this would work
Both should work with non-convex polyhedrons, i think. It's why we need negative masses. And if you'll draw thing, you'll see that for every "wrong" positive volume you have "negative" volume that removes wrong part from positive volume.
For 2d case you can do a drawing, it's simple to see that.
For 3d case, if it will work when P is placed outside of convex polyhedron,it will work with non-convex polyhedrons.
If divide non-convex polyhedron into convex using some polygons, each such polygon is added to our summ twice, with opposite signs, so it's same as if you don't divide.
So it's needed to only prove that it will work if P is placed outside of convex polygon. Can be proven using equations(the only more-or-less reliable way to prove things), but probably it's waay too long.
My English is not good enough to prove that "in words".
