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

[Simon Rit]

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
>
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