Re: Calculating principal radii of curvature ?

On Aug 4, 1:50 am, "Dave Eberly" <dNOSPAMebe...@xxxxxxxxxxxxxxx>
wrote:
"broli" <Brol...@xxxxxxxxx> wrote in message

news:7420ae98-22af-4064-90d8-cf7ca0f1604a@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
The easiest answer: The principal curvatures at any triangle
interior point are zero. After all, the triangle is flat :)

The algorithm I mentioned is in the realm of "discrete differential
geometry". You are thinking of the continuous case, where your
triangle mesh is an approximation to a C^2 surface. Without
any additional information, you can try to fit a surface to the
triangle using the information you know at its three vertices
(positions, normals, principal curvatures, principal directions).

You could treat the ray-triangle intersection as another vertex
and subdivide your triangle based on that intersection, then
apply the method I mentioned in my PDF. The normal vector
at the intersection could just be the triangle normal. Or you
could attempt to interpolate the 3 vertex normals.

You could also just use barycentric coordinates of the ray-triangle
intersection to compute a weighted average of the principal
curvatures. Whether or not the results are acceptable depends
on your application...


Your answer that the principal radii of curvature in interior of
triangle is zero has somewhat disappointed me :)

Is it possible to use the principal radii of curvature at the vertices
and multiple them with the barycentric coordinates of point of
intersection to find the principal radii of curvature at the point ? I
also found a thesis which finds the per face curvature (on page 3) :

http://www.cs.princeton.edu/gfx/pubs/_2004_ECA/curvpaper.pdf

It is easy to build a 3d orthognal system by taking one direction (u)
as direction vector along an edge and then u X N where N is surface
normal gives v.

Can you please describe the steps in your thesis please ?

I understood the whole thing until finding vertex normals but I could
not understand the math behind other steps.


.

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