- To: mathgroup at smc.vnet.net
- Subject: [mg26264] Re: ListContourPlot3D
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Sun, 10 Dec 2000 00:19:38 -0500 (EST)
- Organization: Universitaet Leipzig
- References: <firstname.lastname@example.org>
- Sender: owner-wri-mathgroup at wolfram.com
1. Just set EdgeForm in the ContourStyle option
2. you must generate a 3d interpolation and make a
ContourPlot3D on the interpolation
You may look at
the OpenGL Viewer has now a buil-in iso-surface command
with interactive adjustment of the level. Clearly
MathGL3d is much faster computing 3d contour surfaces,
50000 polygons/second can be found and rendered with
MathGL3d. It has an automatic polygon reduction that
will also create nice triangulations for fine resolved
data and functions.
David Djajaputra wrote:
> I'm using ListContourPlot3D to generate a surface based on numerical
> data points on an 8x8x8 cubic grid.
> 1. Is it possible to remove the lines drawn on the surface so that I
> have just a nice surface?
> 2. Is it possible to "smooth" this surface using some kind of smooth
> interpolation in Mathematica? Or do I really have to compute my data
> points on a finer grid to get a smoother surface? The surface that I'm
> dealing with is pretty much well behaved and slowly varying.
> Any pointer on how to do this smoothing business in Mathematica will be
> greatly appreciated.
Prev by Date:
Re: transposing an equation
Next by Date:
RE: Easy way to simplify expr to a+bi?
Previous by thread:
Next by thread:
transposing an equation