Computing the volume of intersection between a tetrahedron and a hexahedron.

Question

I am trying to compute the volume of two intersecting objects in 3D. One is a tetrahedron and the other is a hexahedron. I just need the volume but it looks like I will have to find the polyhedron that intersects these two objects and then compute it's volume. This seems very complicated.

In terms of more detail: I know that the two objects intersect and am trying to compute the volume of the hexahedron that is in the tetrahedron. I represent both objects with vertices (in 3D space). I can compute the faces and their normals and also have the flexibility to transform them if that would help.

Is there an easier approach to just the volume or do I have to get the intersecting polyhedron and then compute the volume for it?

Here is an image.

Answer 1

If you could accept non-exact value for the intersection volume, then you can try the famous Monte Carlo Method.

• Surround your two polyhedrons by an iso-oriented cube and generate uniformly distributed (in this cube) sample points - for each point you can quickly verify if it's inside the intersection or not.
• Calculate two numbers - the number of all samples and the number of samples, lying inside the intersection, then divide the second one over the first one... you understand, I think.

The more the number of sample points the better the approximation.

UPDATE:

There are many strategies to select sample points. You can even distribute them regularly (without randomness) inside the cube, or inside the first polyhedron. Any other way, which guarantees the uniform distribution of sample points, will work in this case.