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How to preserve meshgrid compatibility after rotation

3 ビュー (過去 30 日間)
AP 2012 年 10 月 28 日
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?
  3 件のコメント
Matt J
Matt J 2012 年 10 月 29 日
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 ?


回答 (1 件)

Matt J
Matt J 2012 年 10 月 29 日
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 件のコメント
Walter Roberson
Walter Roberson 2012 年 10 月 29 日
isosurface() requires that the X and Y be monotonic, not that they be monotonic increasing.


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