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 > > > >