[G14_run] Locations of the Kel-F windows

[Andy Sandorfi]

Hi Alessandra,
A few comments might help:

  1.  I constructed that table which Tsuneo forwarded in order to define the separation of the beam-entrance and beam-exit kel-F widows, for the sake of comparing with and calibrating the vertex reconstruction.
  2.  The dimensions in that table do not define the effective HD length. That is determined by the amount of gas that was condensed. The silver (21a) target was about 4 cm long, and not a right cylinder, due to problems when the target was formed. The other two had smooth injections and appear to be right-cylinders of 5 cm in length. These conclusions come from a comparison between the vertex image and a calculation of the amount of HD gas that was condensed.
  3.  The aluminum wire diameter and density are well known (yes, I measured them), but the lengths are only approximately known. (Imagine trying to position a thousand hairs, one by one.)

Andy



From: G14_run <g14_run-bounces at jlab.org> on behalf of Alessandra FIlippi via G14_run <g14_run at jlab.org>
Date: Friday, May 5, 2023 at 10:23 AM
To: Tsuneo Kageya <kageya at jlab.org>
Cc: g14_run at jlab.org <g14_run at jlab.org>, Tsuneo Kageya <kageya at jlab.org>
Subject: [G14_run] [EXTERNAL] Re: Locations of the Kel-F windows
Hi Tsuneo,

>
> Ciao Alessandra,
>
>
>  here is an expanded view of Fig. 28 (attached to this email).
>
>  I would like to suggest to show z range -15 < z < -5 cm for Fig. 28.
>
>  I don't think we need the range of z > -5 cm.

ok, I just put there what I had ready, I can zoom it a little bit.

>
>  We can see the following things from the figure.
>

aaaah, now I understand... the "Aluminum wire" is that sort of messy hank
inside the cell! I believed the wires were tensioned along the walls of
the cylinder. So, they make an overall contribution in the central part
which could potentially be relevant, it makes sense.
We know the density of this wire, has anybody ever weighted the total
amount of Al mass in this region? its length and its cross section, to be
more quantitative?
I'll correct the text accordingly. Sorry my bad, I never understood this
was the real target structure.

>
>  (1) Relative positions of three Kel-F windows for Gold2 (19b) and Empty
> (21a).
>
>  I am attaching the info. about target cell 19b and 21a.
>
>
>  (a) The positions of downstream Kel-F windows for Gold2 and Empty are
> close at Fig. 28.
>
>
>  (b) As can see L1 on the table of the attacment, the difference of 96.21
> - 95.74 = 0.47 mm

ok. Do we care about what happens in the very upstream region, L4?
I assumed not.

>
> is seen at the exit windows at Fig. 28.
>
>
>  (c) L4 difference shown on the table as 36.90 - 34.10 = 2.8 mm can be
> seen at the entrande windows at Fig.28.
>
>

ok

>  (2) Small diffence of downstream Kel-F positions between Silver (21a) and
> Empty (21a) can be observed as 1 mm or so
>
> which could be negligible.
>

yes, I believe the important information is how long is the "HD" volume,
that is l = L1-L3-L4

>
>  In order to understand eq. 35, can you define, l_empty, l_gold, Z
> ^gold_KelF1 and Z^empty_KelF1 ?
>
> What are their equivalent numbers at the table ?

l_{cell} = L1-L3-L3 - they differ for about 1 mm (61.63-59.20 mm), you can
see them in the table. This means that the HD length is slightly different
in the two cases.
Z^whatever_KelF1 is the z I get from a gaussian fit of the entrance window
peak for each cell. I need this and the HD length (see above) to write the
transformation for the z coordinate, to "align" the targets, ie bring the
gold2 entrance window to level up with the empty target entrance. These
are the values reported on the table.
Cheers,
    Alessandra



>
>  Ciao, Tsuneo Kageya.
>
>
>


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