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<div style="line-break:after-white-space">Dear Timothy and Noémie,
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<div>I did spend some time figuring out what may have gone wrong with my previous equations. I must have made a stupid copy/paste mistake as my own Mathematica notebook had a different result for PF than the writeup - in agreement with Timothy’s notebook.</div>
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<div>So I set out to re-derive all equations, and at the same time I requested (and received) updated geometric and density numbers from Chris Keith. So here is the new version of my dilution factor note, with BOTH fixes implemented. As you can see, some of
the numerical constants have changed a bit, which may perhaps influence the final result for the PF and the DF. Please check and let me know what you find.</div>
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<div>Thanks and apologies - Sebastian</div>
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<div>On Oct 8, 2024, at 6:44 PM, Timothy Hayward <haywardt@mit.edu> wrote:</div>
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<div style="line-break:after-white-space">Hi Sebastian,
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<div>Thanks for double checking versus your notebook. I did use data for all five target types and checked the epX, epi+X and epi-X channels (they were all consistent) as well as checked for any kinematic dependence (which was not observed). 0.527 does sound
nicer though…</div>
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<div>Thanks again,</div>
<div>Timothy<br id="x_lineBreakAtBeginningOfMessage">
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<div>On Oct 8, 2024, at 6:10 PM, Sebastian Kuhn <kuhn@jlab.org> wrote:</div>
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<div class="x_PlainText">Dear Timothy,<br>
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I took a look at your Mathematica notebook and found your 2nd-to-last equation:<br>
(0.565345 xC xCH xf (NMT xA - 1. NA xHe)) / (xA (1. NMT xC xCH xf + 0.0159627 Nf xC xCH xHe - 1.25501 NCH xC xf xHe + 0.23905 NC xCH xf xHe))<br>
agrees with my own Mathematica notebook. I haven’t figured out yet why the equation 10 in my note is so different - I may have copied something wrong.
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If you get 0.45 for the PF from this above formula (do you have data for all of the ingredients?), there is one thing to keep in mind - this number would have to be multiplied with the entire length of the bath (5.86 cm) for an equivalent length of l_A = 2.64
cm. Normalizing just to the length of the target cell, I get a PF of 0.527. This is still less than I would have expected, but in the right direction.
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One check would be that this number should be entirely independent of the channel or the kinematics of the data that you use as input. If that’s not the case, there is still something wrong. I can’t promise when I will have time to take another look, but you
can obviously do the programming (and debugging) in Mathematica yourself.<br>
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- Sebastian<br>
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> On Oct 7, 2024, at 9:29 AM, Timothy Hayward <haywardt@mit.edu> wrote:<br>
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> Dear Sebastian, <br>
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> If you have a chance — could you double check the packing fraction calculation in your note? Specifically the transition from eq. 8 to eq. 10? I am unable to reproduce eq. 10 like I was for the DF jump from eq. 7 to eq. 9 (c.f. the first two //FullSimplify’s
from the attached notebook). Using the function that I I derived in Mathematica I get a packing fraction of about 0.45 (which Noémie and I are hoping is too low) but the eq. 10 in the note seems to give me a value greater than 1 for the PF.<br>
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> Very possible that I’m doing something simple incorrectly. <br>
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> Thank you for your help,<br>
> Timothy<br>
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