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Re: How to calculate the angle between two images?
Posted:
Apr 6, 2009 4:54 AM
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"khoo" <jim_khoo@hotmail.com> wrote in message <grce8v$8g$1@fred.mathworks.com>...
> > Why on earth are you trying to convert an image into a 1 dimensional vector? you are correlating 2 histograms - histograms are essentially graphs - in your case angle along the x axis, number of pixels of that angle in y direction. If you have say 256 angles, then each bin represents (2*pi)/256 radians. Thus the histogram will be a vector of 256 numbers, each number representing the number of pixels of that angle.
> >
> > I suggest you carefully read help hist(), doesn't that default to 10 bins - you will need to have much better resolution than that.
> >
> > I have had a look at your images, and would have thought they were ideally suited to this methodology.
> >
> > Regards
> >
> > Dave Robinson
>
> Bro Dave, this is my code currently...and i ady change the hist not to default 10 binsand set to 256..
>
>
> clear all;clc
>
> im1=imread('image1.jpg');
> im11=rgb2gray(im1);
> im111=im2double(im11);
> [dx,dy]=gradient(im111);
> p1=atan2(dy,dx);
> im1111=edge(p1,'canny');
> bw1 = bwmorph(im1111,'dilate');
> im11111=p1.*bw1;
> im6=hist(im11111(:),256);
> im66=fft(im6);
>
> im2=imread('image2.jpg');
> im22=rgb2gray(im2);
> im222=im2double(im22);
> im333=zeros(size(im111));
> im333(1:size(im2,1),1:size(im2,2))=im222;
> [dx,dy]=gradient(im333);
> p2=atan2(dy,dx);
> im2222=edge(p2,'canny');
> bw2 = bwmorph(im2222,'dilate');
> im22222=p2.*bw2;
> im7=hist(im22222(:),256);
> im77=conj(fft(im7));
> im99=im66.*im77;
> im8=real(ifft(im66.*im77));
> XXX=hist(im8(:),256);
> stem(1:256,XXX, 'Marker', 'none');
>
> and this is my new image2..
> http://i278.photobucket.com/albums/kk119/khoo011/image2-1.jpg
>
> just now i check all the histogram for the im11111 and im22222..all is correct..but once i fft the both histograms..when i inverse the FFT..the histogram cannot inverse and still stay in FFt mode...can you help me check my code got any problem ? and teach me how to fix the problem..so far i find the peak is at left side which is after FFT..
Why are you histogramming im8? which should be the correlation that you are looking for - just plot im8 and if everything else is correct then you should have one dimensional distribution reaching a peak at the position corresponding to the relative rotation angle of the 2 images.
Regards
Dave Robinson
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