[Insight-users] RaycastInterpolateImageFunction

Luis Ibanez luis.ibanez at kitware.com
Sat Jun 28 18:51:44 EDT 2008


Hi Jef,

What were the coordinates of the origin of the output image ?

Please keep in mind that normally those are the coordinates
of the first pixel in the image, which means that they relate
to a corner (not the center) of the image.

Please let us know,


    Thanks


       Luis


----------------
jef vdmb wrote:
> Hi, 
> 
> I am trying to make a digitally reconstructed radiograph, using  the RaycastInterpolateImageFunction. I started with  the example DigitallyReconstructedRadiograph1.cxx, but would like to define the geometry differently.  I define the focal point and the origin of the output image. Sofar so good. The input image is placed in this physical space just by using it's origin (which was correctly set in a previous step). However, it doesn't seem to be in the right place, ie the point which should be projected in the middle of the output image, isn't there. When I calculate focalpoint and output origin with respect to the middle of the input volume (as you would do if the origin of the physical space were to be defined in the center of this volume) the projection is as expected. It seems that by default the  center of the input volume is the origin of physical coordinate space. The image origin specified in the header of the image is just added to this one. Does this make any sens
e? 
> 
> This is how I define the projection geometry:
> 
> 
>    focalpoint[0]= mIsoCenter[0] + mSourceToAxis;
>   focalpoint[1]= mIsoCenter[1];
>   focalpoint[2]= mIsoCenter[2];
>   interpolator->SetFocalPoint(focalpoint);
> 
> //projection in the YZ plane
>   originOutput[0] = mIsoCenter[0]- (mSourceToScreen - mSourceToAxis); 
>   originOutput[1] = mIsoCenter[1]-(((double)sizeOuput[1]*spacingOutput[1])/2.);
>   originOutput[2] = mIsoCenter[2]-((double)sizeOuput[2]*spacingOutput[2]/2.); 
> 
>  
> a point in the image whith coordinates of the isocenter should be projected in the middle of output image right?
> 
> 
> Thank you in advance for your help,
> Jef
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