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 >
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