Meshgrid multi-valued function data

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Ahmed Zankoor
Ahmed Zankoor on 12 Jun 2018
Edited: Ahmed Zankoor on 12 Jun 2018
If I have data z = f(x,y) which have at some (or all) (x,y) coordinates more than a value for z. How can I represent the corresponding z value for the X and Y mesh grid, where [X Y] = meshgrid(x,y). I also wonder how this is handled in sphere function. For example: >> [x y z] = sphere(50) , surf(x,y,z) Since x,y,z meshgrids are not in the normal format of mesh grids?
  2 Comments
Rik
Rik on 12 Jun 2018
You can best think of meshgrid as returning coordinate grids. That means that you can use it as an input to a function that returns an non-scalar, but you'll have to use cellfun or a similar function.
sphere does not use meshgrid internally. The reason the vector outputs of sphere are not in a meshgrid format seems simple: you'dd have way too many points that are useless. You can have a peek by typing edit sphere, although I'dd urge you never to edit internal functions.
Ahmed Zankoor
Ahmed Zankoor on 12 Jun 2018
Edited: Ahmed Zankoor on 12 Jun 2018
Thank you Rik, I got your point for sphere. What I can not understand is the meshgrid part, how can we express (x1,y1,z1) and (x1,x2,z2) using the same mishgrid X,Y matrices. Take this for example:
{radius = [3];
x = linspace(-radius,radius,100);
y = linspace(-radius,radius,100);
[X Y] = meshgrid(x,y);
Z = - sqrt(radius^2 - X.^2 - Y.^2);
Z(imag(Z)~= 0) = 0 ;
surf(X,Y,Z)
Z(imag(Z)~= 0) = 0 ;
surf(X,Y,Z)
end}
What if I want to surf both halves of the sphere ??

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Accepted Answer

Rik
Rik on 12 Jun 2018
You should use hold on to plot multiple parts. The code below is a fix to your code.
radius=3;
x = linspace(-radius,radius,100);
y = linspace(-radius,radius,100);
[X,Y] = meshgrid(x,y);
Z = sqrt(radius^2 - X.^2 - Y.^2);
Z(imag(Z)~= 0) = 0 ;
%set coordinates outside of the sphere to the edge
L=(X.^2+Y.^2)>(radius^2);
phi=atan2(Y,X);
X(L)=radius*cos(phi(L));
Y(L)=radius*sin(phi(L));
%put the two halves together
Z = [Z,-Z];
X = [X,X];
Y = [Y,Y];
figure(1),clf(1)
surf(X,Y,Z)
  1 Comment
Ahmed Zankoor
Ahmed Zankoor on 12 Jun 2018
Edited: Ahmed Zankoor on 12 Jun 2018
Thank you so much Rik. So this is my point, meshgrid 2D matrices must be single-valued (single value of z corresponding to each (x,y)), otherwise, we should use multiple surfaces (different Z matrices for the same X and Y).

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