CT #s corresponding to spatial coordinates

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Bibek
Bibek on 27 Dec 2011
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
Image Analyst
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
Bibek on 28 Dec 2011
In matlab help I found the code, VI = interp3(X,Y,Z,V,XI,YI,ZI). So, in my case V=CT, it is of size 512*512*30,X=x_coordinates, Y=y_coordinates and Z=z_coordinates. All these X,Y,Z are matrices I created myself and are also of size 512*512*30. Since I am trying to find CT # corresponding to a point with x,y,z coordinates(1.5,25,15), I kept XI=1.5,YI=25,ZI=15. But I got a error message:
vi=interp3(x_Coordinates,y_Coordinates,z_Coordinates,CT,1.5,25,15);
??? Error using ==> interp3 at 138
X, Y and Z must be matrices produced by MESHGRID. Use
TriScatteredInterp instead
of INTERP3 for scattered data.
In my case I create x,y,z coordinates as
X9=1:1:512;
x_Coordinates=repmat(X9,[512,1,75]);% first row corresponds to 712 z-coordinates(all zeros) for 1st image and so on
Y9=1:1:512;
y_Coordinates=repmat(Y9,[512,1,75]);
Z9=0:2.5:185;
for i=1:75;
z_Coordinates(:,:,i)=repmat(Z9(1,i),[512,512,1]);
end
Since I don't know anything about meshgrid, I am not sure I will be able to create x,y,z coordinates like above using meshgrid.
When I tried with TriScatteredInterp I got this error message:
vi=TriScatteredInterp(x_Coordinates,y_Coordinates,z_Coordinates,CT,1.5,25,15);
??? Error using ==> TriScatteredInterp
Incorrect number of arguments for TriScatteredInterp.
I don't know where I did a mistake. Probably I was wrong in defining X,Y,Z and XI,YI,ZI. Any help will be appreciated.

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Answers (1)

Image Analyst
Image Analyst on 27 Dec 2011
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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