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

[Vincent Libertiaux]

On 19.07.22 18:23, Simon Rit wrote:
> Hi Vincent,
> Thanks for the report. I don't believe that there is need for a PR. It 
> comes down to using a different parameterization which I think you can 
> always go around with one of the different versions of AddProjection.
> Did I mention that the out of plane angle has no effect below 2°? If 
> yes, I'm not sure you can trust this information... as I don't know 
> where it comes from.
> Best regards,
> Simon
>
> On Tue, Jul 19, 2022 at 11:34 AM Vincent Libertiaux <vl at xris.eu> wrote:
>
>     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
>
Hi Simon,

you didn't mention that the out of plane angle has no effect below 2°.  
I have read that in several papers about CT geometric calibration.  To 
be honest, I am glad that no PR is needed, I have a lot on my plate at 
work these days :).

Best regards,

V.
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