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Subject: 
Re: Sphere tesselation
Newsgroups: 
lugnet.cad.dev
Date: 
Tue, 12 Sep 2000 19:06:39 GMT
Viewed: 
644 times
  
"Steve Bliss" <steve.bliss@home.com> wrote in message
news:f5csrssq9rkotteig88soujitik82bkj48@4ax.com...
In lugnet.cad.dev, Travis Cobbs wrote:

"Steve Bliss" <steve.bliss@home.com> wrote in message
news:ppnprssksi9anl3hi45qgq5kkbarndid1n@4ax.com...
In lugnet.cad.dev, Travis Cobbs wrote:

At some point, I'd like to replace the sphere p-files with something
better-looking.  Something like the 48\1-8sphe.dat I posted (see
<http://news.lugnet.com/cad/dat/parts/primitives/?n=39>), but with only • 5
vertices on each edge, to fit with the standard curved p-files.

How did you generate this file?  It certainly looks like it was
algorithmically generated.  Don't tell me you figured out all those • points
BY HAND?

It was a fractal-thingy--it started with a 10-vertex triangular mesh (the
three points on the axes, two points along each edge, and the point at the
center of the surface of the hemi-demi-semi-sphere).  Then it divided each
triangle in the mesh into 4 subtriangles.  To figure the new points, it
found the midpoint of each edge, and projected that point onto the surface
of the unit sphere.  These divisions were repeated for a specified number
of iterations.

Thanks for the info.  The next thing I'm going to try will be similar, but
iterative instead of recursive in order to allow for a set of points that's
not a power of 2 (which may or may not work).  You had to make your starting
case overly complicated in order to handle 12, as explained below.


I'm not sure why I started with the 10-mesh that I did.  It would have • been
a lot less code to start with just the 3 points on the axes.  Maybe I
wanted to have that center-point?

You had to, because 12 (the number of subdivisions along each edge) is not a
power of two, and since you could only cut the edge of each triangle in half
(subdividing it into 4 new triangles), you had to start with nine triangles
which could then be subdivided twice each to yield 9*n^4 total triangles,
where n is the number of subdivisions of those triangles, or 9*2^4=144
triangles.

--Travis Cobbs (tcobbs@san.REMOVE.rr.com)



Message has 1 Reply:
  Re: Sphere tesselation
 
(...) I wonder if it's possible to push the (initial?) vertices toward the center. I was playing around with fractaling from a simplest case (1 triangle in the mesh). I noticed the center area was made up of one type of triangle (nearly (...) (24 years ago, 12-Sep-00, to lugnet.cad.dev)

Message is in Reply To:
  Re: Sphere tesselation
 
(...) It was a fractal-thingy--it started with a 10-vertex triangular mesh (the three points on the axes, two points along each edge, and the point at the center of the surface of the hemi-demi-semi-sphere). Then it divided each triangle in the mesh (...) (24 years ago, 12-Sep-00, to lugnet.cad.dev)

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