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MathGroup Archive 2005

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Re: 3D graphics domain

  • To: mathgroup at smc.vnet.net
  • Subject: [mg55759] Re: 3D graphics domain
  • From: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
  • Date: Tue, 5 Apr 2005 05:45:09 -0400 (EDT)
  • Organization: Uni Leipzig
  • References: <d2tesa$qj2$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi,

a quick and dirty solution may be

f[x_, y_] := -64*x + 320*(x^2) - 512*(x^3) + 
256*(x^4) + 20*y - 64*x*y +
64*(x^2)*y - 4*(y^2)

gg = Graphics3D[
Plot3D[f[x, y], {x, -4, 5}, {y, 1, -100}, 
PlotPoints -> 256,
Mesh -> False, DisplayFunction -> Identity]] /.
Polygon[pnts_] /; Length[pnts] > 3 :>
Polygon /@ Partition[pnts, 3, 2, {1, 1}];

simpleTest[Polygon[pnts_]] := Module[{cntr},
cntr = Plus @@ pnts/Length[pnts];
upper[cntr] > 0 && lower[cntr] > 0
]

poly = Select[Cases[gg, _Polygon, Infinity], 
simpleTest];

Show[Graphics3D[{EdgeForm[], poly}, PlotRange -> 
All, BoxRatios -> {1, 1, 1}]]

This draw only polygons with the center is inside 
of the domain. A longer version would select all 
polygons that intersect the boundary and compute 
the intersection polygon that lies inside of the 
boundary. This gives you a cleaner boundary curve.

Regards

  Jens


"Richard Bedient" <rbedient at hamilton.edu> schrieb 
im Newsbeitrag news:d2tesa$qj2$1 at smc.vnet.net...
> Thanks to Bob and Dan for helping me get this 
> far. Again, I've exhausted
> my Mathematica knowledge along with anything I 
> can find in the Help
> files.  I now need to take the function they 
> found for me and graph it
> in 3D over a restricted domain. Here's the 
> problem:
>
> Graph the function
>
> f(x,y) = -64*x + 320*(x^2) - 512*(x^3) + 
> 256*(x^4) + 20*y - 64*x*y +
> 64*(x^2)*y - 4*(y^2)
>
> over the domain:
>
> y <= 4*x*(1-x)
> y >= 4*x*(1 - 2x)
> y >= 4*(x - 1)*(1 - 2x)
>
> Thanks for any help.
>
> Dick
> 



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