Re: To color surfaces using z values with ContourPlot3D

• To: mathgroup at smc.vnet.net
• Subject: [mg23937] Re: To color surfaces using z values with ContourPlot3D
• From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
• Date: Fri, 16 Jun 2000 00:57:22 -0400 (EDT)
• Organization: Universitaet Leipzig
• References: <8i9ov0\$2dm@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Hi,

what is with

poly = ContourPlot3D[Evaluate[eq], {x, -1, 3/2}, {y, -1, 3/2}, {z, -1,
3/2},
PlotPoints -> 35, Axes -> Truel,DisplayFunction->Identity];

Show[poly /.
Polygon[pnts_] :> {Hue[(Last[Plus @@ #/Length[#] &[pnts]] + 1)/2.5],
Polygon[pnts]}, Lighting ->
False,DisplayFunction->\$DisplayFunction]

Regards
Jens

Valeria Olivati wrote:
>
> Hallo eveybody,
>
> I'm a student of Mathematics at University of Pavia, Italy.
>
> I start using Mathematica 4 two months ago, and I need to plot a cubic
> function and color it without using LightSources. I use ContourPlot3D to
> plot the function but I don't know how to color the surface using the z
> values, as you can do using Plot3D and ColorFunction->Hue.
> I cannot use Plot3D because the function is quite complicated:

```

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