Re: Question on concave to convexs converting alorithm
- From: "Dave Eberly" <dNOSPAMeberly@xxxxxxxxxxxxxxx>
- Date: Mon, 26 Sep 2005 13:23:48 GMT
"kiplring" <kiplring@xxxxxxxxxxx> wrote in message
news:1127723323.471129.9040@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> How can I get optimized convexs from an 3D concave?
>
> This is what I'm planning.
<snip>
If only it were that easy. What you are describing sounds similar
to "ear clipping" for triangulating a simple polygon in 2D. In the
2D case, you locate a "convex" vertex V0 (the interior angle at
the vertex has angle smaller than pi radians). Let Vm and Vp
be its adjacent vertices. If the triangle <Vm,V0,Vp> does not
contain any other vertices of the polygon, then this triangle is
part of the triangulation and is removed from the current polygon.
The new polygon is simple and has fewer vertices, so you can
repeat this process until the last triangle is removed.
Your 3D description sounds like you are attempting to find
a "convex" vertex of the 3D polyhedron. If that vertex is
shared by three triangular faces, you could test if the remaining
polyhedron vertices are outside the tetrahedron formed by
the vertex and three faces. If so, remove the tetrahedron and
repeat the process. However, if that vertex is shared by four
or more faces, it is not apparent that you can test/remove the
polyhedron formed by the vertex and the faces. In particular,
the edges formed by those faces and adjacent to the given
vertex do not necessarily form a planar polygon.
What you want is generally a hard problem to solve. You will
need to split faces and/or add more points to the data set. One
method that comes to mind is decomposition with BSP (binary
space partitioning) trees. The leaf nodes represent the convex
subregions of your original polyhedron.
--
Dave Eberly
http://www.geometrictools.com
.
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