plot a 3-d figure (rhombus inside a cricle)

Hello everybody.

I have an assignment to plot shape inside other shape. The assignment is related to electromagnetic fields, I need to describe the potential behavior from low potential to high potential. anyway all of this doesn't matter. because if someone will help with the code of the plot, I'll handle the rest.

The figure I need is a rhombus inside a circle a 3-d plot of course. I added an example of square inside a circle to illustrate my meaning.

http://img200.imageshack.us/i/unleddm.jpg/

THANK YOU VERY MUCH!

Answer 1

[Teja Muppirala]

Just solve Laplace's equation: Uxx + Uyy = 0.

This makes the plot you have shown. Doing it with a rhombus instead of a square should be pretty straight forward.

N = 25;
NN = 2*N+1;
boxh = 10;

[xgrid,ygrid] = meshgrid(-N:N);
r = xgrid.^2 + ygrid.^2;

maskouter = find(r >= N^2);
maskinner = find(abs(xgrid) <= boxh &  abs(ygrid) <= boxh);

L = speye(NN^2);

for n = setdiff(1:NN^2,[maskinner(:) ; maskouter(:)])
    L(n,n + [-NN -1 0 1 NN]) = [1 1 -4 1 1]; 
end
RHS = zeros(NN^2,1);
RHS(maskinner) = 1;

U = reshape(L\RHS,NN,NN);
surf(U);

Answer 2

[Sean de Wolski]

Pseudocode (I don't want to do your whole assignment for you):

1. Use the formula for a circle to create a logical circle the size you want (both circle size and matrix size) http://matlab.wikia.com/wiki/FAQ#How_do_I_create_a_circle.3F . (M in the code below)
2. Create a new variable that is the double precision equivalent of the circle. (C in the code below)
3. Set the square in the middle of the circle (the logical circle) to false (M)
4. Find the row/col coordinates of both sets of points in the map (true and false) ([rhave chave] and [rwant cwant])
5. Use griddata and some fun logical indexing expressions to overwrite those values. I'll provide the code for that.
6. Use mesh() to visualize

C(M) = griddata(rhave,chave,C(~M),rwant,cwant,'linear');

And to prove that I'm not talking out of my ass:

You can scale it as you wish at this point because the minimum value is zero and the square is 1; any multiplication will be a linear scale.

Good Luck!

[Matt Fig]

I like your graphic, Sean de. It is basically what I did when I saw this post. However I think there is more to the story, as the image posted by the OP looks like it was derived from more elaborate constraints. If you will notice, the OP image looks like a square plate punched through an elastic circular membrane. Thus it almost looks to me like the solution to some sort of an Elastic Lagrange minimization problem.

[Sean de Wolski]

Matt you're absolutely right! His obviously non-convex while mine is. There was nothing in the problem statement about that though; he said he'd take care of it. (This was a fun Monday morning wake-up challenge)

[Andrew Newell]

Actually, the problem statement did say that it is a potential, as in Teja's answer.

Answer 3

[Andrew Newell]

Suppose your potential function is z = V(x,y). Define V so that it can input matrices for x and y, e.g.,

V = @(x,y) 1./(x.^2+y.^2+1);

then create the matrices for your problem area, e.g.,

x = -40:1:40; y = -40:1:40;
[x,y] = meshgrid(x,y);
z = V(x,y);

Now you can plot it:

surf(x,y,z)

See surf,surfc for ways to adjust the appearance of the plot.

[Sean de Wolski]

or do that inside bsxfun, i.e.:

z = bsxfun(@V,-40:40,(-40:40).');

Which means in theory (a detailed enough V function), this entire plot could be one-lined!

Answer 4

[eli]

Thank you all guys. In a matter of fact I eventually succeeded by myself. All I had to do is to think better of the part which sets 100v in the matrix(stubbornness is sin :)). I appreciate your response. If someone interested I can upload the entire code.

[Andrew Newell]

For the benefit of other people reading this question, I recommend you accept Teja's response.