How to preserve meshgrid compatibility after rotation
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Dear All,
I have generated a vector field in 3D. the coordinates are created by meshgrid. Now I need to rotate the domain 90° about y-axis. When I do this and try to plot the isosurface of some values, I get the following error:
Input grid is not a valid MESHGRID.
How can I solve this problem?
Thanks,
Ahmad
3 comentarios
Matt J
el 29 de Oct. de 2012
Editada: Matt J
el 29 de Oct. de 2012
How are you performing the rotation? A 90 degree rotation about y should transform (x,y,z) to (z,y,-x). Are you doing this formula directly, or are you using trigonometric functions to generate rotation matrices? If you are using trig functions are you using COS & SIN or are you using COSD & SIND ?
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Matt J
el 29 de Oct. de 2012
You can't perform a coordinate transformation on a meshgrid and still expect it to be a meshgrid. Only certain transforms like translations preserve a mesh. Use griddata or TriScatterdInterp to obtain gridded samples.
3 comentarios
Walter Roberson
el 29 de Oct. de 2012
isosurface() requires that the X and Y be monotonic, not that they be monotonic increasing.
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