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How to do that with version 6?

  • To: mathgroup at
  • Subject: [mg90457] How to do that with version 6?
  • From: vasil michev <michev at>
  • Date: Thu, 10 Jul 2008 06:33:31 -0400 (EDT)

I have a 3D Plot, say

p = Plot3D[n[\[Mu], t], {\[Mu], -44, 0}, {t, .0001, 0.2},
  PlotPoints -> 100, PlotRange -> All]

In previous versions I was able to do this:

bla1 = Show[ContourGraphics[p], ColorFunction -> Hue,
  Contours -> {0.25, 0.3, 0.35, 0.4, 0.45, 0.5, 0.55, 0.6, 0.65, 0.7,
    0.75, 0.8, 0.85, 0.9, 0.95, 1., 1.05, 1.1, 1.15, 1.2, 1.25, 1.3,
    1.35, 1.4, 1.45, 1.5, 1.55, 1.6, 1.65, 1.7, 1.75, 1.8, 1.85, 1.9,
    1.95, 1.999}, PlotRange -> All, AspectRatio -> .6]

and get the points for the above contour lines. However, the same
procedure doesnt work with version 6. It returns ContourGraphics
object, but im unable to visualize it, and looking at its input form,
its no way near what its supposed to be, it only has an
ContourGraphics added to the output of the Plot3D command:


I tried using the legacy packages (<< Version5`Graphics`, <<
Graphics`Graphics`; << Graphics`Graphics3D`), but it doesnt work as
expected again. I guess the problem is with ContourGraphics expecting
a SurfaceGraphics object, but im not really sure, and I dont really
know how to convert it to SurfaceGraphics in v6.

So if anyone can help me with this, or tell me a way to do this in v6
I would be very thankfull. (Especially if its possible without
reevaluating the plots, I have lots of them so it will take a lot of
time. Also, i will most likely need more contours, so just doing
ContourPlot's is much more time consuming. I do have the 3D plots
evaluated in v6, with GraphicsComplex and all)

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