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</o:shapelayout></xml><![endif]--></head><body lang=DE link=blue vlink=purple><div class=WordSection1><p class=MsoNormal><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Hello,<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>thanks for the link to the paper but I dont have access to it. Aside from how the trajectory is interpreted within in RTK. My actual question was<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>if any of those two trajectories would need another reconstruction filter than the FDK Filter. From my point of understanding a specific rotation around<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>the object is necessary for fbp/fdk (like c-arm bow, standard circular cone-beam trajectory). That’s why I asked If the first trajectory needs some other reconstruction <o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>filter because the object itself doesn’t rotate around itself. It actually gets translated on a circular path. So I was more expecting a “yes” or “no” to the fdk filter <o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>or a hint to another filter (except iterative reconstructions) I should use for these trajectories.<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>To trajectory 2: I think the projections are different. The object rotates and each projection shows a different view.<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Kind regards,<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Robert C.<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><b><span lang=EN-US style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'>Von:</span></b><span lang=EN-US style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'> simon.rit@gmail.com [mailto:simon.rit@gmail.com] <b>Im Auftrag von </b>Simon Rit<br><b>Gesendet:</b> Dienstag, 1</span><span style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'>0. Oktober 2017 20:29<br><b>An:</b> Robert Calliess<br><b>Cc:</b> Cyril Mory; rtk-users@public.kitware.com<br><b>Betreff:</b> Re: [Rtk-users] FDK for planar ct<o:p></o:p></span></p><p class=MsoNormal><o:p> </o:p></p><div><div><div><div><p class=MsoNormal>Hi,<o:p></o:p></p></div><p class=MsoNormal>Let me try to clarify what I mean by "source trajectory wrt the object." In tomography, you need to determine the source trajectory in the object coordinate system, we don't really care about the source trajectory in the room coordinate system. For example, rotating the source on a circular trajectory or rotating the object makes no difference for the reconstruction algorithm. That's why we call diagnostic scanners "helical scanners".<o:p></o:p></p></div><p class=MsoNormal>So for trajectory 1, it seems that the source trajectory (again, wrt to the object) is a circle but the object is offset. This is somewhat similar to <a href="https://doi.org/10.1109/TNS.2006.880977">https://doi.org/10.1109/TNS.2006.880977</a> except that the detector is not tilted so FDK would be the only FBP algorithm I could think of. But the situation is really not good, data are missing and iterative reconstruction should give better results.<o:p></o:p></p></div><div><p class=MsoNormal>Trajectory 2: what I said in my previous email is true, it's useless I believe, all projections are similar up to a 2D transform of the projection.<o:p></o:p></p></div><div><p class=MsoNormal>Simon<o:p></o:p></p></div></div><div><p class=MsoNormal><o:p> </o:p></p><div><p class=MsoNormal>On Tue, Oct 10, 2017 at 8:07 PM, Robert Calliess <<a href="mailto:robert.calliess@gmx.de" target="_blank">robert.calliess@gmx.de</a>> wrote:<o:p></o:p></p><div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Hello,</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>I try to clarify the both trajectories.</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'> </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Trajectory 1:</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>No, i dont move the source on two circles. The xray source is fixed. Only the object and the detector moves. Both move on a circular path so that the iso-ray</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>always passes through the pcb centre and the detector centre. There is one orthogonal view and the others are the ones moving on the circular path. </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>(Object is not rotating around its own axis).</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'> </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Trajectory 2:</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Yes, the xray source lies in the rotation axis and only the object rotates around its z-axis. Detector and xray source are fixed and the detector is tilted.</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>It’s almost like this trajectory here <a href="https://www.ikeda-shoponline.com/engctsoft/wp-content/uploads/2016/06/Oblique-View-CT1.jpg" target="_blank">https://www.ikeda-shoponline.com/engctsoft/wp-content/uploads/2016/06/Oblique-View-CT1.jpg</a></span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>except that the xray source lies on the rotation axis.</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'> </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>I hope this helps to understand the trajectories I have to deal with.</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'> </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Kind regards, </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Robert</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'> </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'> </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><b><span lang=EN-US style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'>Von:</span></b><span lang=EN-US style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'> <a href="mailto:simon.rit@gmail.com" target="_blank">simon.rit@gmail.com</a> [mailto:<a href="mailto:simon.rit@gmail.com" target="_blank">simon.rit@gmail.com</a>] <b>Im Auftrag von </b>Simon Rit<br><b>Gese</b></span><b><span style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'>ndet:</span></b><span style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'> Dienstag, 10. Oktober 2017 19:06<br><b>An:</b> Robert Calließ<br><b>Cc:</b> Cyril Mory; <a href="mailto:rtk-users@public.kitware.com" target="_blank">rtk-users@public.kitware.com</a></span><o:p></o:p></p><div><div><p class=MsoNormal><br><b>Betreff:</b> Re: [Rtk-users] FDK for planar ct<o:p></o:p></p></div></div><div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'> <o:p></o:p></p><div><div><div><div><div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'>Hi,<o:p></o:p></p></div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'>It's still not clear to me but what is helpful is to think in terms of source trajectory wrt the object.<o:p></o:p></p></div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'>Trajectory 1: if I understand, you move the source on two circles plus one point. I don't know of a FBP algorithm to reconstruct this, but there might be one. I would consider iterative reconstruction first.<o:p></o:p></p></div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'>Trajectory 2: your trajectory is a point, the source does not move with respect ot the object since it lies on the rotation axis. So each projection contains exactly the same information up to a simple 2D projection deformation. So it's hopeless to reconstruct from one projection only.<o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'>To create the correct geometry, I would suggest using the function <a href="https://github.com/SimonRit/RTK/blob/master/code/rtkThreeDCircularProjectionGeometry.h#L92" target="_blank">AddProjection</a> for which you provide the source and detector positions plus the 3D coordinates of the two axes of the coordinate system of the projection.<o:p></o:p></p></div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'>I hope this helps<o:p></o:p></p></div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'>Simon<o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'> <o:p></o:p></p><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'>On Tue, Oct 10, 2017 at 5:43 PM, "Robert Calließ" <<a href="mailto:Robert.Calliess@gmx.de" target="_blank">Robert.Calliess@gmx.de</a>> wrote:<o:p></o:p></p><div><div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>Hello,</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>thank you for the fast reply.</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>To answer your questions first.</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>In this case the abbrevation pcb stands for printed circuit board.</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>Next point is the trajectory we are currently handling with.</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>Please see the attached image "trajectory.png". There are two schematics showing the side view and top view for trajectory type 1</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>and a side-view for trajectory type 2.</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'> </span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>For type 1:</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>The xray source is fixed. The pcb is clamped within a transport, so the pcb and the detector are moveable with in the xy plane.</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>As you can see at the image, the pcb moves along a circular path but the pcb itself is not rotating. And let's assume that the iso ray</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>always passes through the centre of the pcb and the centre of the detector.</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'> </span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>For type 2:</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>The xray source is fixed and the detector is tilted. The pcb lies centred in the middle of a table. So that the pcb rotates around its centre</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>around the z-axis.</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'> </span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'> </span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>I hope this makes clear what trajectory i'm dealing with. Thank you.</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'> </span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>Kind regards,</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>Robert C.</span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'> </span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'> </span><o:p></o:p></p></div><div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'> </span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'> </span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'> </span><o:p></o:p></p></div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'> </span><o:p></o:p></p></div><div style='border:none;border-left:solid #C3D9E5 1.5pt;padding:0cm 0cm 0cm 8.0pt;margin-left:7.5pt;margin-top:7.5pt;margin-right:3.75pt;margin-bottom:3.75pt;word-wrap:break-word' name=quote><div style='margin-bottom:7.5pt'><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><b><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>Gesendet:</span></b><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'> Dienstag, 10. Oktober 2017 um 15:31 Uhr<br><b>Von:</b> "Cyril Mory" <<a href="mailto:cyril.mory@creatis.insa-lyon.fr" target="_blank">cyril.mory@creatis.insa-lyon.fr</a>><br><b>An:</b> "Robert Calliess" <<a href="mailto:robert.calliess@gmx.de" target="_blank">robert.calliess@gmx.de</a>>, <a href="mailto:rtk-users@public.kitware.com" target="_blank">rtk-users@public.kitware.com</a><br><b>Betreff:</b> Re: [Rtk-users] FDK for planar ct</span><o:p></o:p></p></div><div><div><div name=quoted-content><div><p style='background:white'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>Dear Robert,</span><o:p></o:p></p><p style='background:white'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>Your description of the trajectory is very obscure to me. Maybe you have a very unusual X-ray system. Could you make the following points clear :</span><o:p></o:p></p><p style='background:white'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>- what is a PCB ?</span><o:p></o:p></p><p style='background:white'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>- what is fixed/moving in your system (we need this information for the object, the source and the detector), and what kind of trajectories have the moving parts ?</span><o:p></o:p></p><p style='background:white'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>- can you re-draw your sketch with just 2 or 3 positions (ideally, on similar but separate drawings), each one with the object, the source and the detector ?</span><o:p></o:p></p><p style='background:white'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>If you do that, we should have a clear understanding of how your acquisition goes, and be able to give you appropriate advice.</span><o:p></o:p></p><p style='background:white'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>Best regards,</span><o:p></o:p></p><p style='background:white'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>Cyril</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'> </span><o:p></o:p></p><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'>On 10/10/2017 15:02, Robert Calliess wrote:</span><o:p></o:p></p></div><blockquote style='margin-top:5.0pt;margin-bottom:5.0pt'><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'>Hello rtk users,<o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'>I have question to the RTK FDK Filter. As far as I understand from to the fourier slice theorem the object to be reconstructed needs a circular trajectory and needs to rotate its own centre.<o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'>Please have a look at the attached sketch. With this planar trajectory (Object, a PCB, is moved on a circle trajectpry “in-plane”, PCB itself is not rotating) do I need<o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'>a special filtering if I want to use FDK for planar CT with respect to the sketched trajectory ? I tried a circular in-plane trajectory where the PCB is centred and rotates<o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'>around its centre point. And with 100 projections I get good results. But with the trajectory I described (sketch, attached image) the results are not so good.<o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'>Because of the row-wise ramp filter It looks like there is a directional dependency. My assumption is, and with respect to fourier slice theorem, that the missing object<o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'>rotation (rotation around itself) causes there directional effects.<o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'> <o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'>So my questions to the experts are. Do I need to apply a special filtering before backprojecting with FDK or is it just the wrong<o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'>algorithm for this kind of trajectory ?<o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'> <o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'>kind regards,<o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'>Robert C.<o:p></o:p></p></div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;background:white'><span style='font-size:9.0pt;font-family:"Verdana","sans-serif"'> </span><o:p></o:p></p><pre style='background:white'>_______________________________________________<o:p></o:p></pre><pre style='background:white'>Rtk-users mailing list<o:p></o:p></pre><pre style='background:white'><a href="mailto:Rtk-users@public.kitware.com" target="_blank">Rtk-users@public.kitware.com</a><o:p></o:p></pre><pre style='background:white'><a href="http://public.kitware.com/mailman/listinfo/rtk-users" target="_blank">http://public.kitware.com/mailman/listinfo/rtk-users</a><o:p></o:p></pre></blockquote></div></div></div></div></div></div></div></div><p class=MsoNormal style='mso-margin-top-alt:auto;margin-bottom:12.0pt'><br>_______________________________________________<br>Rtk-users mailing list<br><a href="mailto:Rtk-users@public.kitware.com" target="_blank">Rtk-users@public.kitware.com</a><br><a href="http://public.kitware.com/mailman/listinfo/rtk-users" target="_blank">http://public.kitware.com/mailman/listinfo/rtk-users</a><o:p></o:p></p></div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'> <o:p></o:p></p></div></div></div></div></div></div><p class=MsoNormal><o:p> </o:p></p></div></div></body></html>