Translating shape functions into isosurfaces, a new question

Question

Brian 2012-10-29

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/52214-translating-shape-functions-into-isosurfaces-a-new-question

Hey, all, thanks for reading this,

I am trying to create a set of isosurfaces defined by shape functions (kind of like CAD curves). My code for such is here: http://pastebin.com/80ZBXC4u.

What I can do is take a shape function for a sphere, for instance, and plot it as a isosurface pretty easily. When I try one of the equations for my example, however, it doesn't work. I am thinking it could possibly be a result of me not translating it right into MATLAB.

The line:

% our_function = @(x,y,z) (sqrt(x.^2+y.^2+z.^2)<10); % Example

is the sample code for the sphere. I then have the example:

    % our_function = @(x,y,z) (z.*15*15*15+2*(x.^2+2.25*y.^2)*(x.^2+2.25*y.^2)<0); % Shape Function 1 Simplified

which is a bit more convoluted. When I try to do this one, I get the error:

Error using * Inputs must be 2-D, or at least one input must be scalar. To compute elementwise TIMES, use TIMES (.*) instead.

which means somewhere I am multiplying a vector element instead of a scalar element (what I think, at least).

I then have the more advanced line, which I haven't even started yet:

    % our_function = @(x,y,z) (z.*15*15*15+2*(x.^2+2.25*y.^2)*(x.^2+2.25*y.^2)<0)&&(x.^2>0.01*15*15)&&(z>-2.3*15)
    % Shape Function 1
    % Extended

which is the last equation plus some constraints for x,y and z.

What this should plot is a 2D surface in 3 dimensions, meaning its a sheet of sorts. However, I cannot MATLAB well, so somewhere along the line I think I am inputting this wrong.

Thanks for all the advice you can give!

Actually, scratch this, I managed to solve it myself. I needed .*, and just one & for the constraints since it was a vector value.