On 19.07.22 19:21, Jerome Lesaint wrote:
Hello all,
The question of neglecting out-of-plane offsets if less than 2 degrees
is discussed in Yang et al, Med. Phys., 2006, section III.A.
Regards,
Jerome
Le mar. 19 juil. 2022 à 18:23, Simon Rit
<[email protected]> a écrit :
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 <[email protected]>
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
<[email protected]> 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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Thank you very much for the reference Jérôme !
Best regards,
Vincent
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