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To color surfaces using z values with ContourPlot3D
*To*: mathgroup at smc.vnet.net
*Subject*: [mg23881] To color surfaces using z values with ContourPlot3D
*From*: "Valeria Olivati" <olivaleNNN at tin.it>
*Date*: Thu, 15 Jun 2000 00:51:18 -0400 (EDT)
*Organization*: TIN
*Sender*: owner-wri-mathgroup at wolfram.com
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:
eq = 81*(x^3 + y^3 + z^3) - 189*(x^2*y + x^2*z + y^2*x + y^2*z + z^2*x +
z^2*y) + 54*x*y*z + 126*(x*y + x*z + y*z) - 9*(x^2 + y^2 + z^2) - 9*(x+y+z)
+ 1
Now I plot this funtion in this way:
sup = ContourPlot3D[eq, {x, -1, 3/2}, {y, -1, 3/2}, {z, -1, 3/2},
PlotPoints->{5,7}, Axes->True, ViewPoint->{0.056, -3.382, -0.076},
ContourStyle->{Hue[1], EdgeForm[GrayLevel[0.6]]}, Lighitng->False]
but with only one color It's not a clear shape.
If I use the following LightSources instead of ConturStyle
LightSources->{{{0,0,0.7}, RGBColor[0,1,0]}, {{1,0,1}, RGBColor[1,1,0]},
{{-1,0,1}, RGBColor[0,0,1]}, {{-1,0,0}, GrayLevel[1]}}
I don't like the colors I obtain, they are not linked to the surface but
they are the best LightSources I found after a big number of changes. Of
course my problem with LightSources is that I haven't understood how they
work....
I hope someone has suggestions for me
Thanks a lot
Best Regards
Olivati Valeria
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