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 <v...@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 <v...@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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