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Hi Tegan,<br>
<br>
Thanks for the explanation. It is a nice summary of what Andrei did
and how this can help us in collecting hits into a clusters.<br>
<br>
Cheers, Elton.<br>
<pre class="moz-signature" cols="72">Elton Smith
Jefferson Lab MS 12H3
12000 Jefferson Ave STE 4
Newport News, VA 23606
(757) 269-7625
(757) 269-6331 fax </pre>
<div class="moz-cite-prefix">On 8/6/15 11:08 AM, beattite wrote:<br>
</div>
<blockquote cite="mid:7da2b0a5b8950a2bda297d2ea3612cbd@uregina.ca"
type="cite">Re-sending. Some people missed it. I'll shove it on
DocDB at some point as well, I suppose.
<br>
<br>
<br>
-------- Original Message --------
<br>
Subject: Re: Curvature
<br>
Date: 13.07.2015 14:51
<br>
From: beattite <a class="moz-txt-link-rfc2396E" href="mailto:beattite@uregina.ca"><beattite@uregina.ca></a>
<br>
To: <a class="moz-txt-link-rfc2396E" href="mailto:halld-cal@jlab.org"><halld-cal@jlab.org></a>
<br>
<br>
Accidentally sent the e-mail before attaching things. Here you
are.
<br>
<br>
<br>
On 13.07.2015 14:48, beattite wrote:
<br>
<blockquote type="cite">Hello.
<br>
<br>
Last week, Mark asked for some sort of plot showing the effects
of
<br>
the shower curvature for why we might want to be doing a
correction to
<br>
the IU shower code, since it's been a while since Andrei's
original
<br>
work and we all want to be on the same page.
<br>
<br>
The attached picture (15degrees.jpg) is point r vs. point z for
a simulated 5 GeV
<br>
photon gun at 15 degrees. The picture shows all reconstructed
points
<br>
in the run. It gives the point z distribution for an incident
photon
<br>
at 15 degrees (I chose such a large energy, 5 GeV, so that there
are
<br>
enough points in layer 4 to talk about).
<br>
<br>
If we expect the centroid of the point z distribution to lie
along
<br>
the trajectory of the photon, we would expect the centroid in
the four
<br>
layers to be r/tan(15):
<br>
Layer 1 - 242 cm
<br>
Layer 2 - 256 cm
<br>
Layer 3 - 274 cm
<br>
Layer 4 - 302 cm
<br>
<br>
What we actually see are centroids in the four layers of:
<br>
Layer 1 - ~253 cm
<br>
Layer 2 - ~257 cm
<br>
Layer 3 - ~260 cm
<br>
Layer 4 - ~265 cm
<br>
<br>
This is similar to what Andrei's plots from a lot time ago show,
as
<br>
in, the centroid in layer one is shifted in the positive
z-direction
<br>
from the photon trajectory while the centroids in layers three
and
<br>
four are shifted in the negative z-direction from the photon
<br>
trajectory. The centroid of layer two lines up nicely with the
<br>
trajectory.
<br>
<br>
The IU code uses the Moliere radius around the trajectory to
include
<br>
points. At low angles like 15 degrees, the spread of the cone
in z is
<br>
something like +\- 75 cm. That is to say, the code is looking
for
<br>
points in the shower between z = 230 and z = 370 or so in layer
four.
<br>
While this does actually cover the shower including the
curvature (in
<br>
this case), it also has a huge acceptance in the positive
z-direction
<br>
that is unneeded and which possibly catches points which belong
to a
<br>
different shower. So, the idea is to take the initial energy
estimate
<br>
from the IU shower along with the angle estimate of the IU
shower
<br>
(which is based off points in layers 1, 2, and 3, and so should
be
<br>
close to the angle of the actual incident particle), then use
the
<br>
curvature tables provided by Andrei which simulated in more
detail the
<br>
energy deposition of an incident photon in the BCAL at different
<br>
angles and energies. The tables give energy deposition
centroids in
<br>
each layer which we can then use to fill the showers with points
<br>
rather than the Moliere radius that the IU code used.
<br>
<br>
The effect is less pronounced at, say, 50 degrees
(50degrees.jpg).
<br>
Here, the centroid in layer four should be around 68, but it
lies
<br>
more around 65, for instance.
<br>
</blockquote>
<br>
<fieldset class="mimeAttachmentHeader"></fieldset>
<br>
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