Triangulating fiducial marker position in c-arm image (DICOM) using stereoParameters

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Joel Poder 2018-3-18

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评论： Matt J 2018-6-4

Hi,

I have 2 DICOM images exported from a c-arm imaging device. The images are 60 degrees apart in rotation and there is no translation between the two images. There is a marker in each image of which I would like to return the 3-D coordinates in world coordinate system.

The source-imager distance, pixel size, coordinates of the marker in each image, and angle between the two images are all known.

I'm attempting to use stereoParameters and triangulate to achieve this however I am unable to change the FocalLength parameter manually.

Is anyone able to assist me with this? Or recommend an alternative method to achieve my aim?

Regards, Joel

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Matt J 2018-6-4

Yan Li commented:

Hi,Joel I am a graduate student now and interested in your C_arm image because I wanna study the parameters of them, could you upload them and share, I will appreciate it ~ THANKS ：） Leanne

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Matt J 2018-3-18

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编辑：Matt J 2018-4-6

triangulateRays.m

If you have the 3x4 projection matrices for the two gantry positions, then you can use the attached routine to do the triangulation. It isn't the optimal triangulation algorithm, but if your 2D localization errors are small, it serves most practical purposes.

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Matt J 2018-4-5

编辑：Matt J 2018-4-6

在 MATLAB Online 中打开

Hi Joel,

The coordinates are up to you, but they must be consistent. In other words all distances must be measured in the same units (either mm or pixels) in both the input xy coordinates and the intrinsic parameters.

Below, I illustrate the use of the code when all of the data you mentioned is converted to millimeters, but you can test for yourself that you get the same result when both the intrinsics and xy are re-expressed in pixels. I had to make some assumptions (see the A matrix) about how your 2D image axes are oriented relative to the 3D world coordinates.

pix=0.388; %pixel size
K = [1650, 0, 0; 0, 1650, 0; 512*pix, 384*pix, 1].' ;%intrinsic
C=[0;-1000;0]; %source location
A=[1 0 0 ; 0 0 -1; 0 1 0]; %assumed world coordinates of camera axes in position 1
R=[0, -1, 0; 1, 0, 0; 0, 0, 1]; %90 deg rotation
E1=A*[eye(3),-C ]; %Extrinsics position 1
E2=(A*R.')*[eye(3), -R*C]; %Extrinsics position 2
P(:,:,2)=K*E2; %projection matrix stack
P(:,:,1)=K*E1;
xy=pix*[524.7754, 372.8574; 392.1953, 375.5078].'; %observed 2D positions
xyz=triangulateRays(P,xy), %3D location in mm

This gives the result

as the triangulated 3D location.

Matt J 2018-4-9

Yes, it should be a 2x3x2 matrix, but xy(1,:,:) should be the x-coordinates and xy(2,:,:) should be the y-coordinates.

Joel Poder 2018-4-11

Brilliant.

Thanks so much Matt

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Triangulating fiducial marker position in c-arm image (DICOM) using stereoParameters

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