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: