Re: Re: Re: graphic
- To: mathgroup at smc.vnet.net
- Subject: [mg104618] Re: [mg104583] Re: [mg104518] Re: graphic
- From: "David Park" <djmpark at comcast.net>
- Date: Thu, 5 Nov 2009 03:48:27 -0500 (EST)
- References: <hcl3qc$btt$1@smc.vnet.net> <200911030752.CAA01048@smc.vnet.net> <27376704.1257318514372.JavaMail.root@n11>
Now that I see a nice example:
Needs["Presentations`Master`"]
f1[x_, y_] = x^2 + y^2 - 2;
f2[x_, y_] = x^2;
Draw3DItems[
{ContourDraw[f1[x, y] == 0, {x, -2, 2}, {y, -2, 2}] //
RaiseTo3D[f2[#1, #2] &]},
NiceRotation]
David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/
From: DrMajorBob [mailto:btreat1 at austin.rr.com]
Or, more generally,
f1[x_, y_] = x^2 + y^2 - 2;
f2[x_, y_] = x^2;
cp = ContourPlot[f1[x, y] == 0, {x, -2, 2}, {y, -2, 2}];
Graphics3D[
Replace[First[cp],
GraphicsComplex[points_, data_] :>
GraphicsComplex[{#1, #2, f2[#1, #2]} & @@@ points, data]]]
Correct?
Bobby
On Tue, 03 Nov 2009 01:52:03 -0600, Szabolcs Horv=E1t <szhorvat at gmail.com>
wrote:
> On 2009.11.01. 23:58, Nodar wrote:
>> Dear All,
>>
>> May I build 3D curve graphic z=F2(x,y), where x and y are the
>> solutions of equation F1(x,y)==0 ?
>> May I receive list of points from a "ImplicitPlot" graphic ?
>
> The simplest way would be to solve the F1(x,y)==0 equation. Here's a
> hackish solution for the case when this is not possible:
>
> Suppose F1(x,y) = x^2+y^2-2 and F2(x,y) = x^2
>
> cp = ContourPlot[x^2 + y^2 - 2 == 0, {x, -2, 2}, {y, -2, 2}]
>
> Graphics3D[
> Replace[First[cp],
> GraphicsComplex[points_, data_] :>
> GraphicsComplex[{#1, #2, #1^2} & @@@ points, data]]]
>
>
> This will work in Mathematica 6 or later.
>
--
DrMajorBob at yahoo.com
- References:
- Re: graphic
- From: Szabolcs Horvát <szhorvat@gmail.com>
- Re: graphic