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JS
Posts:
49
Registered:
8/4/08


Re: Navigating in a rotated grid?
Posted:
Nov 16, 2010 8:20 AM


On Nov 15, 9:08 pm, JS <joe778...@gmail.com> wrote: > Hi NG, > > I have now struggled with a problem for a while so I finally resort to > asking here in the hope somebody can help me, or at least give a few > pointers. The problem is very simple, so I hope there is a simple > solution also, that I just haven't seen. Here goes: > > I have a list of coordinates (X,Y,Z) which I know form a rectangular > grid with all datapoints within that rectangle existing in the list. > However, the grid isrotatedand translated, relative to a "normal" > coordinate system. That is, if I do scatter(X,Y), I see the rectangle > as a diamond shape (i.e.,rotated) and shifted away from the origin > (i.e., translated). The points are equally spaced. > > What I would like is a matrix with the surface formed by the > coordinates (X,Y,Z) so that row 1 in the matrix corresponds to all the > Z values found along the lower edge of the rectangle, row 2 > corresponds to the next row up, etc. In effect, I am thereby defining > my own coordinate system like this, with axes parallel to the sides of > the rectangle. Simple as that. Had the data not beenrotated, this > would of course be trivial, but since it is, I can't "travel" along, > e.g., a row in therotatedgrid and copy the Z values over to the > right spot in a matrix, since a row is not a row in the "straight" > coordinate system. > > I of course know the angle and offset so I have tried to simply shift > all points so that the lower left corner has (0,0) and then rotate the > points with: > > xCoordsRot = (xCoords.*cosd(rotAngle))(yCoords.*sind(rotAngle)); > yCoordsRot = (xCoords.*sind(rotAngle))+(yCoords.*cosd(rotAngle)); > > This works, however, this is not a very "nice" result, in that there > are rounding issues causing the point spacing that I know to be an > even number (500) no longer to be so. This, in turn makes is difficult > to find the correct coordinates in my matrix for a given Z value. > > So, how can I move around along a row in arotatedgrid in a simple, > efficient manner? Any suggestions? > > Thanks very much in advance! > > /JS
Noone? I should think it would be relevant to some imageanalysis applications with a rotated raster image needs to be manipulated?
/JS


Date

Subject

Author

11/15/10


JS

11/16/10


JS

11/16/10


Curious

11/17/10


JS


