- Do you want a grid of (x,y) locations where the z value is a parabola? So it's like you have a bowl sitting above (or below) the x-y plane? Because if that's so, you don't need z. The I value is the z value, which is the distance from the x-y plane.
- Or do you want to do some kind of Delaunay triangulation of a set of (x,y,z) points to form a surface with a bunch of triangles on it?
Create a 3D mesh of a paraboloidal region
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I have a paraboloid region. I would like to plot this region and build a tetrahedron or Hexahedron mesh, for this I use:
a = 40;
d = 1;
x=-40:step:40;
y=-40:step:40;
z=-40:step:40;
I = 16*a^2.*y.^4+a.*y.^3.*(-32*a^2-8*d^2-8.*x.^2-8.*z.^2)+a.*y.*(32*a^2*d^2+8*d^4-8*a^2.*x.^2+2*d^2.*x.^2-10.*x.^4-8*a^2.*z.^2+2*d^2.*z.^2-20.*x.^2.*z.^2- 10.*z.^4)+y.^2.*(16*a^4-8*a^2*d^2+d^4+32*a^2.*x.^2-2*d^2.*x.^2+x.^4+32*a^2.*z.^2-2*d^2.*z.^2+2.*x.^2.*z.^2+z.^4)-(16.*a.^4.*d^2+8*a^2*d^4+d^6+20*a^2*d^2.*x.^2- 3*d^4.*x.^2-a^2.*x.^4+3*d^2.*x.^4-x.^6+20*a^2*d^2.*z.^2-3*d^4.*z.^2-2*a^2.*x.^2.*z.^2+6*d^2.*x.^2.*z.^2-3.*x.^4.*z.^2-a^2.*z.^4+3*d^2.*z.^4-3.*x.^2.*z.^4-z.^6)<= 0;
The following step is use:
[X,Y,Z] = meshgrid(x,y,z);
but the usage the memory is so large (+70 gb).
Is there another way to create geometry and meshing for this region?
1 commentaire
Image Analyst
le 17 Jan 2020
Modifié(e) : Image Analyst
le 17 Jan 2020
Not sure what you're looking for. I'm not sure what a "a tetrahedron or Hexahedron mesh" is. Do you have some kind of picture of it that you can post? Some guesses:
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