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Iam working with CTA volume having dimensions 512*512*304. How I can convert the volume into isotropic form having axial size of 256*256.I have gone through Converting non-isotropic to isotropic voxel but it didnt help me .

Iam not sure how to proceed but still i have read necessary parameters (may be required for above problem)from dicom including spacing between slices=0.45 AND pixel spacing is 0.4 for both x & y.

One more question I want to ask is about transformation into world coordinates. When I am working with original volume(512*512*304), The extracted tube is identical in shape to the reference data , but in terms of coordinates position, it shows a notable difference. My colleague asked me that reference data is in world coordinate.

Many thanks in advance;

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M Jawaid
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    Your question is unclear to me: what does "isotropic form" mean? I suggest that you add some relevant tags to the question, since knowledge of MATLAB alone is insufficient for your question. Also, leave only one question per question. – Andras Deak -- Слава Україні Jan 29 '16 at 18:30
  • An element of an array in MATLAB can map to any physical-world volume element you like. What exactly do you need to do? – Carl Witthoft Jan 29 '16 at 18:56
  • Andras, Actually Iam following paper at http://link.springer.com/article/10.1007/s10554-011-9894-2/fulltext.html In preprocessing they have written that for computational efficiency they interpolated and downsampled given volume of 512*512*304 to acquire isotropic volume with the axial image size equal to 256*256 pixels.. – M Jawaid Jan 29 '16 at 19:07
  • Dear Carl, Iam extracting the arterial centreline. Based on original volume <512*512*304> I have obtained centreline containg the respective positions representing true artery; However when Iam comparing it with reference centreline provided by Rotterdam coronary challenge organizers, The shape is identical that means i have extracted true artery but the coordinate points of reference centreline are much smaller/scaled down. I contacted with organizers & they responded me that the reference is given in world coordinates...thats what is making me confused. – M Jawaid Jan 29 '16 at 19:16

1 Answers1

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It sounds like you are trying to redefine your coordinate grid to one with isotropic (i.e. cube, i.e. equal height, width, length) voxels.

Perhaps something like this would do it: (I'm using matlabs sample mri data for an example)

load mri D
D=double(squeeze(D));       % Remove singleton dimension and convert from int to double
szD_a=size(D);              % calculate size of original image stack
vox_a = [.4,.4,.45];         % define original voxel size
vox_b = [.25,.25,.25];       % set target voxel size
szD_b=ceil((szD_a-1).*vox_a./vox_b)+1; % calculate size of target image stack
                                       % plus and minus 1 account for 
                                       % pixel (1,1,1) being at coordinate [0,0,0]
% Get x,y,z coordinates of each voxel in original image stack
[Xa,Ya,Za]=meshgrid(...
    [0:szD_a(1)-1]*vox_a(1),...
    [0:szD_a(2)-1]*vox_a(2),...
    [0:szD_a(3)-1]*vox_a(3));

% Define x,y,z coordinates of each voxel in target image stack
[Xb,Yb,Zb]=meshgrid(...
    [0:szD_b(1)-1]*vox_b(1),...
    [0:szD_b(2)-1]*vox_b(2),...
    [0:szD_b(3)-1]*vox_b(3));

% interpolate imagestack from original to target coordinates
D_target=interp3(Xa,Ya,Za,double(D),Xb,Yb,Zb);
figure
subplot(1,2,1)
imagesc(D(:,:,1))
subplot(1,2,2)
imagesc(D_target(:,:,1))
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