Re: where is area?

[Przemek Klosowski]

On 09/08/2011 05:11 PM, Luiz Portella wrote:
> Hi,
>
> I´m working in project for computed tomography.
> I get matrix 5x5, rotate it with bilinear interpolation (it now is
> 7x7) and sum elements of columns and get for 0, 45, 90, 135 and 180
....
>  For 0, 90 and 180 the sum of this numbers are 4, but not for 45 and
> 1350! What? I dont understand.... All for cell with 1 (there are 4) is
> inside matrix 5x5, away from border.

It's unrealistic to expect that the image rotation algorithms preserve 
the integrated intensity very well. After all, cropping the rotated 
image deletes significant chunks, and introduces non-existent segments.
Take a look at this:

x=1:180; a = zeros(7,7); a(3,3) = a(5,3) = a(5,4) = a(5,5) = 1;
for i=x
 s.bilinear(i)=sum(sum(imrotate(a,i,"bilinear", "crop", 0)));
 s.bicubic(i)=sum(sum(imrotate(a,i,"bicubic","crop",0)));
 s.nearest(i)=sum(sum(imrotate(a,i,"nearest","crop",0)));
 s.fourier(i)=sum(sum(imrotate(a,i,"fourier","crop",0)));
end
plot(x,s.bilinear,x,s.bicubic,x,s.nearest,x,s.fourier)

the integrated intensity varies by as much as +75% for
"nearest". Aparently bicubic has least variance: the integrated 
intensity varies from 3.95 to 4.25.

Interestingly, there may be a problem with 'fourier' as indicated by its 
unreasonable behavior for small rotations; see my separate posting.