David Young

Hi Gomathy. Your code is almost OK. One problem is that you have chosen very small rotations to test, so it is hard to see if the results are right. Instead of fractions of a degree, try using many degrees, so that you can tell if there's a big error or just a small inaccuracy.

The second problem is that your conversion from the y-coordinate of the correlation peak to an angle is not correct. Remember that the number of wedges is just a full circle, so in your case a change of 500 in the y-coordinate corresponds to 360 degrees or 2*pi radians. If you use this, then the code works.

Here is my test code, based on yours, but simplified to make it easier to analyse.

 Iref = imread('cameraman.tif');
 deg = 73;
 I = imrotate(Iref, deg, 'bilinear', 'crop');

 rmin = 5;
 rmax = 125;
 nw = 500;
 xc = size(Iref, 2)/2;
 yc = size(Iref, 1)/2;

 lp_Iref = logsample(Iref, rmin, rmax, xc, yc, [], nw);
 lp_I = logsample(I, rmin, rmax, xc, yc, [], nw);

 cc = normxcorr2(lp_I, lp_Iref);
 [max_cc, imax] = max(cc(:));
 [ypeak, xpeak] = ind2sub(size(cc), imax)

 drot = 360 * (ypeak - nw) / nw % prints 72.72 for initial deg=73

The choice of the parameters is up to you - you seem to have chosen good ones for this problem.

Yeoh cs: There are no general rules for setting those parameters. It depends on the original image and the purpose of doing the transform.

rmax determines the "field of view" of the transform. That is, it gives the radius of the region of the original image that is sampled.

nw determines the resolution at the outside ring. If you want to capture the full detail of the original image, with one transform pixel for each original pixel, you need to make nw equal to 2*pi*rmax. However, this results in massive oversampling nearer the centre, so you may want a smaller value.

I normally set only one of rmin and nr, and allow the other to default, so that the pixels in the sampled image have an aspect ratio of 1. There is always a "hole" at the centre of a log-polar image, and rmin determines how big this hole is. Near the centre of the log-polar image the original image is grossly oversampled, and the smaller rmin is, the worse this is.

In the end though, you have to look at what you are using the log-polar image for, and use either theoretical arguments or empirical tests to decide the parameters.

Gracias David Young por este algoritmo, me ayudo mucho para detectar comprimidos en un blister,

Hi David
Thanks for your submission. I have been following your answers and submissions and they all are very useful. Greatly appreciate your work.

I am trying to use imrotate for checking the rotation between two image frames (1024x1024 int16) coming from HH and HV polarized antennas of a marine radar. They are 16-bit data (comes as Matlab matrices) and the HV is rotated w.r.t HH. I tried to use the above piece of code but first I tried with the rotation of HH w.r.t. itself but drot always comes as 0. Is it due to the data? Should this valid for int8 only? Here is the modified code I used. Greatly appreciate your reply.

HHdble=double(HH);
HVdble=double(HV);

Iref = HHdble;
deg = 10;

 I = imrotate(Iref, deg, 'bilinear', 'crop');
 rmin = 5;
 rmax = 512; %%125;
 nw = 512;
 xc = size(Iref, 2)/2;
 yc = size(Iref, 1)/2;
 lp_Iref = logsample(Iref, rmin, rmax, xc, yc, [], nw);
 lp_I = logsample(I, rmin, rmax, xc, yc, [], nw);
 cc = normxcorr2(lp_I, lp_Iref);
 [max_cc, imax] = max(cc(:));
 [ypeak, xpeak] = ind2sub(size(cc), imax)
 drot = 360 * (ypeak - nw) / nw %% comes as '0' .

Hi Gomathy. Measuring small rotations is difficult, but you could try my affine optic flow estimator at http://www.mathworks.co.uk/matlabcentral/fileexchange/27093-affine-optic-flow If you have difficulty with it, please let me know.