[Rtk-users] Lateral blur in a FDK reconstructed volume

[Vincent Libertiaux]

On 11.05.22 15:20, Vincent Libertiaux wrote:
> On 11.05.22 15:15, Simon Rit wrote:
>> Hi,
>> Yes, I think it's correct. To be sure you correctly understand it, 
>> you can always do test cases with the source and detector positions, 
>> u v vectors in the coordinate system of your object.
>> http://www.openrtk.org/Doxygen/classrtk_1_1ThreeDCircularProjectionGeometry.html#a0fb1475ed76a28cde24fac85eae18e1e
>> and then check the resulting angles and distances.
>> Simon
>>
>> On Wed, May 11, 2022 at 2:15 PM Vincent Libertiaux <vl at xris.eu> wrote:
>>
>>     On 10.05.22 22:54, Simon Rit wrote:
>>     > Hi Vincent,
>>     > RTK can parametrize any orientation of the detector with the three
>>     > angles GantryAngle, InPlaneAngle, OutOfPlaneAngle. 0.025° seems
>>     very
>>     > small indeed! I don't know how much you know about software B
>>     but the
>>     > easiest would be to have either the projection matrix or the
>>     source
>>     > position, detector position, u axis and v axis in patient/object
>>     > coordinates to derive the RTK parameters.
>>     > Good luck with this!
>>     > Simon
>>
>>     Hi Simon !
>>
>>     Unfortunately, I don't have access to B projection matrices.
>>
>>     As for the detector orientation in RTK, I have made this picture
>>     to make
>>     sure I understand properly how to use the gantry angle to achieve my
>>     desired geometry:
>>
>>     https://ibb.co/J3H8z9M
>>
>>     The cyan detector is the default configuration with a 0° gantry
>>     angle.
>>     The blue detector is at a gantry angle of alpha (largely
>>     exaggerated for
>>     the sake of clarity).  So in order to simulate an out-of-plane
>>     rotation
>>     of the detector around its vertical axis, I should translate this
>>     blue
>>     detector so that its center matches the coordinates of the cyan
>>     one, and
>>     translate the source accordingly (along the black vectors on the
>>     picture) ?  I assume that proj_iso_x/y and source_x/y are
>>     expressed in
>>     the gantry system of coordinates (local) ?
>>
>>
>>     Thank you again for your feedback,
>>
>>     kindest regards,
>>
>>     V.
>>
> Thanks Simon,
>
> I'll investigate more and let you know.  Hopefully, it might be useful 
> to someone else one day !
>
> V.
>
Hi Simon,

I finally got some time to investigate further this issue this week.  I 
managed to get sharp edges everywhere now and it was indeed the detector 
out-of-plane angle colinear with the gantry angle that was the cause.  
The value given by the other software seems to have been in rad rather 
than degrees; the angle I found was 1.15°.  This makes me wonder what 
were the assumptions under which no effect was found for angles below 
2°.  If you know the title of the seminal paper, I'd be interested to 
read it.


As for the mean to include this angle in the geometry, no extra code was 
indeed needed.  If we call this extra angle "c", the following 
modifications have to be made in rtksimulatedgeometry:

- first angle = c

- sdd = sdd_0 * cos(c)

- sid = sid_0 * cos(c)

- source_x = source_x0 - sid*sin(c)

- proj_iso_x = proj_iso_x0 + (sdd-sid)*sin(c)

I can't really promise I'll find time to do it, but if it is the case, 
I'll submit a PR to include that in the matrices computation.

Hopefully, it will help others on the list who encountered a similar issue.

Best regards,

Vincent
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