General Formula to Calculate Polyhedron Volume

Question

Given a list of vertices (v), and a list of edges connecting the vertices (e), and a list of surfaces that connect the edges (s), how to calculate the volume of the Polyhedron?

Answer 1

Similarly with a polygon where we can split it into triangles and sum the areas,
you could split a polyhedron into pyramids and sum their volumes. But I'm not sure how hard is to implement an algorithm for that.

(I believe there is a mathematical way/formula, like using vectors and matrices.
I suggest to post your question also on http://mathoverflow.net)

Answer 2

First, break every face into triangles by drawing in new edges.

Now look at one triangle, and suppose it's on the "upper" surface (some of these details will turn out to be unimportant later). Look at the volume below the triangle, down to some horizontal plane below the poyhedron. If {h1, h2, h3} are the heights of the three points, and A is the area of the base, then the volume of the solid will be A(h1+h2+h3)/3. Now we have to add up the volumes of these solids for the upper faces, and subtract them for the lower faces to get the volume of the poyhedron.

Play with the algebra and you'll see that the height of the polyhedron above the horizontal plane doesn't matter. The plane can be above the polyhedron, or pass through it, and the result will still be correct.

So what we need is (1) a way to calculate the area of the base, and (2) a way to tell an "upper" face from a "lower" one. The first is easy if you have the cartesian coordinates of the points, the second is easy if the points are ordered, and you can combine them and kill two birds with one stone. Suppose for each dace you have a list of its corners, in counter-clockwise order. Then the projection of those points on the x-y plane will be counterclockwise for a an upper face and clockwise for a lower one. If you use this method to calculate the area of the base, it will come up positive for an upper face and negative for a lower one, so you can add them all together and have the answer.

So how do you get the ordered lists of corners? Start with one triangle, pick an ordering, and for each edge the neighbor that shares that edge should list those two points in the opposite order. Move from neighbor to neighbor until you have a list for every triangle. If the volume of the polyhedron comes up negative, just multiply by -1 (it means you chose the wrong ordering for that first triangle, and the polyhedron was inside-out).

EDIT: I forgot the best part! If you check the algebra for adding up these volumes, you'll see that a lot of terms cancel out, especially when combining triangles back into the original faces. I haven't worked this out in detail, but it looks as if the final result could be a surprisingly simple function.

Answer 3

I have done this before, but the surface mesh I used always had triangular facets. If your mesh has non triangular facets, you can easily break them up into triangular facets first. Then I fed it to TetGen to obtain a tetrahedralization of the interior. Finally, I added up all the volumes of the tetrahedra. TetGen is reasonably easy to use, and is the only library other than CGAL I know of that can handle complicated meshes. CGAL is pretty easy to use if you don't mind installing a gigantic library and use templates like crazy.