Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2005
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2005

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Re: 3d plots in mathematica 5.0

  • To: mathgroup at smc.vnet.net
  • Subject: [mg59127] Re: [mg59124] Re: 3d plots in mathematica 5.0
  • From: "David Annetts" <davidannetts at aapt.net.au>
  • Date: Sat, 30 Jul 2005 01:25:01 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

 Hi Fiz,
> 
> Here is the function
> 
> w=exp(-2(x^2+y^2/r0^2))*exp(-2(z^2/z0^2))
> 
> r0=1.4 and z0=1.4

The function is Exp, not exp, and it needs square brackets not parentheses.

r0 = 1.4;
z0 = 1.4;
w = Exp[(-2(x^2 + y^2/r0^2))]*Exp[(-2(z^2/z0^2))] // FullSimplify

The function you're after is ContourPlot3D, and you use it like

Needs["Graphics`ContourPlot3D`"]
ContourPlot3D[w, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, Contours -> {.75}, \
PlotRange -> All, Axes -> True, BoxRatios -> {1, 1, 1}];

This produces "something", but it's not (IMHO) particularly useful or
informative.

You might get better value from either animating the 3D plot or another
approach.

Dave.


  • Prev by Date: Re: Add terms surrounded by zero together in matrix
  • Next by Date: Re: Mathematica 5.2 and mathssh
  • Previous by thread: Re: Re: 3d plots in mathematica 5.0
  • Next by thread: Re: Re: 3d plots in mathematica 5.0