"Tim Nikias" <tim### [at] gmxde> wrote in message
news:3d70e2d7$1@news.povray.org...
>
> Any links?
>

No links, but I can see a fairly straightforward way to accomplish this.
Three closest points *will not* work. No amount of time will ever make it
work, so patience isn't even an issue. Point B may be the closest point to
Point A, but Point A is not *necessarily* the closest point to PointB. Here
is an example:
A=========B==C

This is how I would accomplish what you are trying to. I don't know any
math, so I hope that I can communicate what I mean with words. Make a circle
around each point which is a little larger than half the maximum distance
around the closest points and make a polygons out of the circle
intersections. Then, make a polygon around any vertex for which the circle
method did not work out of all surrounding intersections or vertices.
Triangulate the polygons, and then you will be done, but with extra
triangles.

I had a better idea while I was typing the first. Just add planes making
sure all vertices are on one side of the plane until all vertices are
included and then  figure out the  plane intersections.

I haven't tested this, but it seems to make sense as I type.

 -Shay

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