CT #s corresponding to spatial coordinates
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I have a CT image data(transverse slices),CT of 512*512*30. I know how to find CT # for any slice k corresponding to ith row and jth column, CT#=CT(i,j,k).I now trace a ray,starting from source point(0,0,100) then goes through CT image data and finally strikes to detector point(say 10,25,-50). Spatial coordinates for source and detector points are determined with isocenter taken at (0,0,0).Using linspace I can find spatial coordinates(i.e.,x,y,z)of points between source point and detector point. So, I have,say 105 intermediate points and I know their x,y,z coordinates.Say, 15 points lie outside of CT image data(i.e.in air) before passing through it and another 15 points also lie outside of CT image data(i.e.in air) after exiting through it. For remaining 75 points(assuming a ray goes through each slice), how can I assign CT # to each points(with their known spatial coordinates). Basically it means how can I get CT#=CT(x,y,z,15)for a point in 15th slice with its spatial coordinates x,y,z. I tried with, c = improfile(I,xi,yi,n)but got error message when tried to extend it, c = improfile(I,xi,yi,zi,n). Any help will be appreciated. Thanks!!
6 Comments
Naz
on 27 Dec 2011
I understood that you have some number of rays and you want to assign to each coordinate a number of the ray that passes through it. Is that right?
It depends on how you design your system. If your x-ray is a point source, than unlimited number of rays pass through each coordinate. If you have an array source of parallel rays, then you can assume some width of the ray and consequently each ray will pass through a limited set of coordinates around the line of propagation (no scattering).
Naz
on 27 Dec 2011
In other words you can imagine each ray as a bar rather than a line.
Bibek
on 27 Dec 2011
Naz
on 27 Dec 2011
And what is CT#?
Image Analyst
on 28 Dec 2011
Those are the digital numbers the Computed Tomography unit gives, called Hounsfield Units: http://en.wikipedia.org/wiki/Hounsfield_scale.
Bibek, how did my suggestion work out for you?
Bibek
on 28 Dec 2011
Answers (1)
Image Analyst
on 27 Dec 2011
0 votes
Why not just use trilinear interpolation? I don't think there's a built-in function for it, but it would be easy enough to program up yourself.
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