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. > >
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