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<div class="moz-cite-prefix">On 11.05.22 15:15, Simon Rit wrote:<br>
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<blockquote type="cite"
cite="mid:CAF0oig34ofV8Qz4MQw2Kbqt2LCk+Jq0duO_u3Mp=kdVyCJjROg@mail.gmail.com">
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<div>Hi,</div>
<div>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.</div>
<div><a
href="http://www.openrtk.org/Doxygen/classrtk_1_1ThreeDCircularProjectionGeometry.html#a0fb1475ed76a28cde24fac85eae18e1e"
moz-do-not-send="true" class="moz-txt-link-freetext">http://www.openrtk.org/Doxygen/classrtk_1_1ThreeDCircularProjectionGeometry.html#a0fb1475ed76a28cde24fac85eae18e1e</a></div>
<div>and then check the resulting angles and distances.<br>
</div>
<div>Simon<br>
</div>
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<br>
<div class="gmail_quote">
<div dir="ltr" class="gmail_attr">On Wed, May 11, 2022 at 2:15
PM Vincent Libertiaux <<a href="mailto:vl@xris.eu"
moz-do-not-send="true" class="moz-txt-link-freetext">vl@xris.eu</a>>
wrote:<br>
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<blockquote class="gmail_quote" style="margin:0px 0px 0px
0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">On
10.05.22 22:54, Simon Rit wrote:<br>
> Hi Vincent,<br>
> RTK can parametrize any orientation of the detector with
the three <br>
> angles GantryAngle, InPlaneAngle, OutOfPlaneAngle. 0.025°
seems very <br>
> small indeed! I don't know how much you know about
software B but the <br>
> easiest would be to have either the projection matrix or
the source <br>
> position, detector position, u axis and v axis in
patient/object <br>
> coordinates to derive the RTK parameters.<br>
> Good luck with this!<br>
> Simon<br>
<br>
Hi Simon !<br>
<br>
Unfortunately, I don't have access to B projection matrices.<br>
<br>
As for the detector orientation in RTK, I have made this
picture to make <br>
sure I understand properly how to use the gantry angle to
achieve my <br>
desired geometry:<br>
<br>
<a href="https://ibb.co/J3H8z9M" rel="noreferrer"
target="_blank" moz-do-not-send="true"
class="moz-txt-link-freetext">https://ibb.co/J3H8z9M</a><br>
<br>
The cyan detector is the default configuration with a 0°
gantry angle. <br>
The blue detector is at a gantry angle of alpha (largely
exaggerated for <br>
the sake of clarity). So in order to simulate an out-of-plane
rotation <br>
of the detector around its vertical axis, I should translate
this blue <br>
detector so that its center matches the coordinates of the
cyan one, and <br>
translate the source accordingly (along the black vectors on
the <br>
picture) ? I assume that proj_iso_x/y and source_x/y are
expressed in <br>
the gantry system of coordinates (local) ?<br>
<br>
<br>
Thank you again for your feedback,<br>
<br>
kindest regards,<br>
<br>
V.<br>
<br>
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<p>Thanks Simon,</p>
<p>I'll investigate more and let you know. Hopefully, it might be
useful to someone else one day !</p>
<p>V.<br>
</p>
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