> Well, the rotation matrix above seems like the way to go. > Why is this not a "nice" result? > Can you provide an example? > Can you use interp1 (or one of its variations) > to get your "even" spacing? > > > > > /JS
Thanks for your reply!
Yes, for now I have had to resort to rotating the grid to be "upright" and interpolate the rotated grid onto a perfect (even-spaced) one with GRIDDATA. Given the small discrepancies, this interpolation does not change much, so it is acceptable for now.
An example would be this. Imagine a grid with unit spacing, rotated 10 degrees clockwise thus forming a diamond with the lower left corner set in (0,0). If I position myself in this left corner (0,0) and I wish to move to the next point along the lower edge of the grid (to read off the Z coordinate associated with that point), which in the "upright" version of the grid would be (1,0), how do I find that point easily in the rotated grid? Since I know the spacing to be 1, and the angle to be 10 degrees, I can of course predict the coordinates where the point should be in the "upright" system and then look for it in the list. But this just appears to be a clumsy solution. This is the same as rotating the the entire grid up and handling the uneven coordinates with some rounding scheme.
But I have now thought about it for a while and I guess there is no "correct", elegant way of doing this, that is, defining my own coordinate system, with axes parallel to the grid's edges.