0
$\begingroup$

Given two polyhedra(with triangulated faces), P1 and P2. I want to create a polyhedron P, which is the result of joining the two at their region of contact. With join I do not mean union but physically, it is more like welding P1 and P2, so if there is say a cylinder(P1) and a cube(P2) aligned on same axis with P1 partially out of P2, union operation will result in disappearance of the part of cylinder inside the cuboid but I want to preserve that region in P. In other words, P1 and P2 should be concatenated at their region of contact.

I would like to know if there is any practical algorithm for this? I think that it is more like computing overlay of P1 and P2 but I do not find any approach for the same in 3D.

$\endgroup$
3
  • $\begingroup$ Maybe something like this $\endgroup$
    – adrianN
    Commented Sep 20, 2016 at 10:49
  • 1
    $\begingroup$ Can you be more precise about how the polyhedra are represented? $\endgroup$
    – D.W.
    Commented Sep 20, 2016 at 11:22
  • $\begingroup$ @D.W. Please see, I have edited accordingly. $\endgroup$
    – Pranav
    Commented Sep 20, 2016 at 11:28

1 Answer 1

1
$\begingroup$

I was recently referred to an article 'Mesh arrangements for solid geometry' by Qingnan Zhou et. al. [pdf] and the corresponding code by Dr. Campen which exactly solves this problem.

This algorithm inserts new vertices and edges in the region of intersection between $P1$ and $P2$ thereby a generating new polyhedron $P$ and evaluates boolean operators on $P$ by selecting/rejecting vertices and edges depending on the operator. For my problem, $P$ forms the solution.

However, out of curiosity I would like to know if anyone has an alternative to this approach.

Update: I have tested the tool referred above against my original problem, it successfully prevents self-intersection by inserting new vertices and edges at the region of contact between $P1$ and $P2$.

$\endgroup$
1
  • $\begingroup$ I have replaced the link with that of ACM repository. $\endgroup$
    – Pranav
    Commented Oct 6, 2016 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.