<div dir="ltr"><div><div>Hi,<br></div>I see one drawing only, not two. And the object does not seem to be moving on your drawing, is it? If not and the source is also static (as it seem), this is equivalent to one large projection.<br></div>Simon<br></div><div class="gmail_extra"><br><div class="gmail_quote">On Tue, Oct 24, 2017 at 3:58 PM, "Robert Calließ" <span dir="ltr"><<a href="mailto:Robert.Calliess@gmx.de" target="_blank">Robert.Calliess@gmx.de</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div><div style="font-family:Verdana;font-size:12.0px"><div>Hello,</div>
<div>I suppose there are still misunderstandings with respect to the trajectory.</div>
<div>Attached you can find the two difference trajectories. I also had a closer look to</div>
<div>the off centered fdk ( the paper you suggested). But I don't think it is in my case.</div>
<div>The iso ray passes object center and detector center at each view. Off centered fdk</div>
<div>has a different preweighting scheme.</div>
<div> </div>
<div>You said that the RTK ramp filter is along the u axis (orthogonal to rotaion axis). For planar_ct_1 trajectory that</div>
<div>should fit. As you can see at the picture, the object is moving on a circular path but not rotating around the</div>
<div>center point (red cross in the image).</div>
<div> </div>
<div> </div>
<div>Kind regards,</div>
<div>Robert C.</div>
<div>
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<div style="margin:0 0 10px 0"><b>Gesendet:</b> Donnerstag, 12. Oktober 2017 um 07:14 Uhr<br>
<b>Von:</b> "Simon Rit" <<a href="mailto:simon.rit@creatis.insa-lyon.fr" target="_blank">simon.rit@creatis.insa-lyon.<wbr>fr</a>><br>
<b>An:</b> "Robert Calliess" <<a href="mailto:robert.calliess@gmx.de" target="_blank">robert.calliess@gmx.de</a>><br>
<b>Cc:</b> "<a href="mailto:rtk-users@public.kitware.com" target="_blank">rtk-users@public.kitware.<wbr>com</a>" <<a href="mailto:rtk-users@public.kitware.com" target="_blank">rtk-users@public.kitware.com</a>><div><div class="h5"><br>
<b>Betreff:</b> Re: [Rtk-users] FDK for planar ct</div></div></div><div><div class="h5">
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<div>Hello,</div>
No. The filter should be orthogonal to the rotation axis. The RTK ramp filter is along the u axis of the projection.</div>
Trajectory 2: if you take photos by rotating the cameras, they are photographies of the same point-of-view. This is what I meant.</div>
Cheers,</div>
Simon</div>
<div class="gmail_extra">
<div class="gmail_quote">On Wed, Oct 11, 2017 at 8:58 PM, Robert Calliess <span><<a href="mailto:robert.calliess@gmx.de" target="_blank">robert.calliess@gmx.de</a>></span> wrote:
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<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">Hello,</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">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</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">if any of those two trajectories would need another reconstruction filter than the FDK Filter. From my point of understanding a specific rotation around</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">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 </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">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 </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">or a hint to another filter (except iterative reconstructions) I should use for these trajectories.</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)"> </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">To trajectory 2: I think the projections are different. The object rotates and each projection shows a different view.</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)"> </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)"> </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">Kind regards,</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">Robert C.</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)"> </span></p>
<p class="MsoNormal"><b><span style="font-size:10.0pt;font-family:Tahoma,sans-serif">Von:</span></b><span 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>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</span></p>
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<div class="m_-55466648088813648h5"><br>
<b>Cc:</b> Cyril Mory; <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</div>
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<p class="MsoNormal">Hi,</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".</p>
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<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" target="_blank">https://doi.org/10.1109/TNS.<wbr>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.</p>
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<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.</p>
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<p class="MsoNormal">Simon</p>
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<p class="MsoNormal"> </p>
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<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:</p>
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<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">Hello,</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">I try to clarify the both trajectories.</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)"> </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">Trajectory 1:</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">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></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">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></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">(Object is not rotating around its own axis).</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)"> </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">Trajectory 2:</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">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></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">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.<wbr>com/engctsoft/wp-content/<wbr>uploads/2016/06/Oblique-View-<wbr>CT1.jpg</a></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">except that the xray source lies on the rotation axis.</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)"> </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">I hope this helps to understand the trajectories I have to deal with.</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)"> </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">Kind regards, </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)">Robert</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)"> </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)"> </span></p>
<p class="MsoNormal"><b><span style="font-size:10.0pt;font-family:Tahoma,sans-serif">Von:</span></b><span 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></p>
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<p class="MsoNormal"><br>
<b>Betreff:</b> Re: [Rtk-users] FDK for planar ct</p>
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<p class="MsoNormal">Hi,</p>
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<p class="MsoNormal">It's still not clear to me but what is helpful is to think in terms of source trajectory wrt the object.</p>
</div>
<p class="MsoNormal">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.</p>
</div>
<p class="MsoNormal">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.</p>
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<div>
<p class="MsoNormal">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.</p>
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<p class="MsoNormal">I hope this helps</p>
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<p class="MsoNormal">Simon</p>
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<p class="MsoNormal"> </p>
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<p class="MsoNormal">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:</p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">Hello,</span></p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">thank you for the fast reply.</span></p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">To answer your questions first.</span></p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">In this case the abbrevation pcb stands for printed circuit board.</span></p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">Next point is the trajectory we are currently handling with.</span></p>
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<p class="MsoNormal"><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></p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">and a side-view for trajectory type 2.</span></p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif"> </span></p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">For type 1:</span></p>
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<p class="MsoNormal"><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></p>
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<p class="MsoNormal"><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></p>
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<p class="MsoNormal"><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></p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif"> </span></p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">For type 2:</span></p>
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<p class="MsoNormal"><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></p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">around the z-axis.</span></p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif"> </span></p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif"> </span></p>
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<p class="MsoNormal"><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></p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif"> </span></p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">Kind regards,</span></p>
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<p class="MsoNormal"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">Robert C.</span></p>
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<p class="MsoNormal"><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.<wbr>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></p>
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<p style="background:white"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">Dear Robert,</span></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></p>
<p style="background:white"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">- what is a PCB ?</span></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></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></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></p>
<p style="background:white"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">Best regards,</span></p>
<p style="background:white"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">Cyril</span></p>
<p class="MsoNormal" style="background:white"><span style="font-size:9.0pt;font-family:Verdana,sans-serif"> </span></p>
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<p class="MsoNormal" style="background:white"><span style="font-size:9.0pt;font-family:Verdana,sans-serif">On 10/10/2017 15:02, Robert Calliess wrote:</span></p>
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<blockquote style="margin-top:5.0pt;margin-bottom:5.0pt">
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<p class="MsoNormal" style="background:white">Hello rtk users,</p>
<p class="MsoNormal" style="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.</p>
<p class="MsoNormal" style="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</p>
<p class="MsoNormal" style="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</p>
<p class="MsoNormal" style="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.</p>
<p class="MsoNormal" style="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</p>
<p class="MsoNormal" style="background:white">rotation (rotation around itself) causes there directional effects.</p>
<p class="MsoNormal" style="background:white"> </p>
<p class="MsoNormal" style="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</p>
<p class="MsoNormal" style="background:white">algorithm for this kind of trajectory ?</p>
<p class="MsoNormal" style="background:white"> </p>
<p class="MsoNormal" style="background:white">kind regards,</p>
<p class="MsoNormal" style="background:white">Robert C.</p>
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<p class="MsoNormal" style="background:white"><span style="font-size:9.0pt;font-family:Verdana,sans-serif"> </span></p>
<pre style="background:white">______________________________<wbr>_________________</pre>
<pre style="background:white">Rtk-users mailing list</pre>
<pre style="background:white"><a href="mailto:Rtk-users@public.kitware.com" target="_blank">Rtk-users@public.kitware.com</a></pre>
<pre style="background:white"><a href="http://public.kitware.com/mailman/listinfo/rtk-users" target="_blank">http://public.kitware.com/<wbr>mailman/listinfo/rtk-users</a></pre>
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______________________________<wbr>_________________<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/<wbr>mailman/listinfo/rtk-users</a></p>
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