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
>
>
>
>