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Re: ContourPlot3D appearance

  • To: mathgroup at smc.vnet.net
  • Subject: [mg89236] Re: ContourPlot3D appearance
  • From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
  • Date: Sun, 1 Jun 2008 03:34:34 -0400 (EDT)
  • References: <g1o8g2$hdp$1@smc.vnet.net>

Hi,

this is a bug in Mathematica. Mathematica create total
nonsense for the surface normals.

Try

maxh = 3000; maxv = 500;
n = 50; r0 =
  n*(n - 1); m = (((n (n - 1))^(n - 1)) Exp[-(n -
         1)])^2; ContourPlot3D[((x^2 + y^2 + z^2)^(n -
        1)) Exp[-2 ((x^2 + y^2 + z^2)^0.5)/
       n] ((x^2 + y^2)/(x^2 + y^2 + z^2))^(n - 1) == m/2, {x, -maxh,
   maxh}, {y, -maxh, maxh}, {z, -maxv, maxv}, BoxRatios -> Automatic,
  Mesh -> None, NormalsFunction -> None]

this give some messages and no smooth lighting,
  but the surface is visible.

Regards
   Jens

Adrian Lupascu wrote:
> Hi everybody,
> 
> 
> Here are the lines of a notebook where I use ContourPlot3D:
> 
> 
> maxh = 3000; maxv = 500;
> 
> n = 50; r0 =
>   n*(n - 1); m = (((n (n - 1))^(n - 1))
>      Exp[-(n - 1)])^2; ContourPlot3D[((x^2 + y^2 + z^2)^(n - 1))
>     Exp[-2 ((x^2 + y^2 + z^2)^0.5)/
>        n] ((x^2 + y^2)/(x^2 + y^2 + z^2))^(n - 1) == m/2, {x, -maxh,
>    maxh}, {y, -maxh, maxh}, {z, -maxv, maxv}, BoxRatios -> Automatic,
>   Mesh -> None]
> 
> ContourPlot3D[(Sqrt[x^2 + y^2] - 2450)^2 + z^2 == 300^2, {x, -maxh,
>    maxh}, {y, -maxh, maxh}, {z, -maxv, maxv}, BoxRatios -> Automatic,
>   Mesh -> None]
> 
> 
> 
> The contour for the first plot should be a torus, ans this is what I get
> indeed. Then I plot using also ContourPlot3D the contour of another
> function, which should correspond to about the same thing. I do get
> again my torus. The problem is that even though the options are the
> same, the first and second plot are very different in appearance (colour 
> and light). I prefer the second plot.
> Any help is greatly appreciated.
> 
> Thank you,
> Adi
> 
> 
> 
> 


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