Difference between revisions of "Determining wire-theta correspondence"

From New IAC Wiki
Jump to navigation Jump to search
Line 18: Line 18:
  
  
<center>{| border=1
+
<center>
 +
{| border=1
 
   |+ Constraints for 1st Guard Plane
 
   |+ Constraints for 1st Guard Plane
 
|-
 
|-
Line 29: Line 30:
 
   | 228.078 cm
 
   | 228.078 cm
 
   | 4.694 degrees
 
   | 4.694 degrees
|}</center>
+
|}
 +
</center>
  
  

Revision as of 20:39, 15 May 2017

[math]\textbf{\underline{Navigation}}[/math]

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]


Determing Wire-Theta Correspondence

To associate the hits with the Moller scattering angle theta, the occupancy plots of the drift chamber hits by means of wire numbers and layer must be translated using the physical constraints of the detector. Using the data released for the DC:

DC: Drift Chambers(specs)
DC Geometry(geom)


Constraints for 1st Guard Plane
Variable Measurement
dist2target th_min
228.078 cm 4.694 degrees


At 25 degrees above the beam line direction, we can create a line that is perpendicular to the plane of the wires, which are at 65 degrees above the reverse beam line direction. This line will be a perpendicular line with respect to all parallel layers of wires with respect to the initial guard wire plane. We can use this geometry and dist2target, which is the distance from the target to the first guard wire plane along the normal of said plane, to calculate other distances.


GuardWireLocation.png
[math]\frac{228.078\ cm}{cos\ 25^{\circ}}=251.6562\ cm[/math]


The parallel planes of wires are shown in the image below.


Wires parallel all types.png


The angle th_min is defined as the polar angle to the first guard wire’s (in the first guard wire plane) midpoint where the wire mid-point is the intersection of the wire with the chamber mid-plane. The first guard wire can be found at 4.694 degrees above the beam line. This position can be taken as the starting position for the wire normal plane.


1st position.png

Using the law of sines


[math]\frac{251.6562}{sin\ 110.306^{\circ}}=\frac{d}{sin\ 4.694^{\circ}}\Rightarrow d=21.9587\ cm[/math]

Making a right triangle,


[math]21.9587\ cm\ sin\ 65^{\circ}=19.914\ cm\ \hat{x}[/math]


[math]21.9587\ cm\ cos\ 65^{\circ}=9.2802\ cm[/math]


[math]251.6562-9.2802=242.376 cm\ \hat{z}[/math]


Examining the geometry file, we can see that each plane, not just the plane of the sense wires, is separated by a distance of


[math]D=.3861\ cm[/math]


We can use the geometry of the wire placements to find the separation distance between two adjacent sense wires on the same plane


[math]d=4(.3861\ cm)\cos(30^{\circ})=1.337\ cm[/math]


DC geometry wires copy.png


Tilt position.png

The starting position for sense wires with respect to the first guard wire at midplane, can be found to be at:

[math]2.3166\ cm\ \cos\ 5^{\circ}=2.3078\ cm\ \hat{x}[/math]


[math]2.3166\ cm\ \sin\ 5^{\circ}=.2019\ cm\ \hat{z}[/math]



Since the separation between adjacent sense wires is uniform and at a set angle of 25 degrees with respect to the beam line, we can use this fact to determine the angle theta each wire makes when measured from the vertex.


DC geom.png


[math]\Delta\ x=1.337\cos(25^{\circ})cm=1.212cm[/math]


[math]\Delta\ z=-1.337\sin(25^{\circ})cm=-0.5652cm[/math]


The new x and z coordinates for wire 2 can be found using the change in the components

[math]x=x_1+\Delta x[/math]
[math]z=z_1+\Delta z[/math]


This can be extended to any point along the same wire plane, starting from the coordinates for wire 1 minus the difference in position to the next wire crossing on the plane.

[math]x_0=x_1-\Delta x[/math]
[math]z_0=z_1-\Delta z[/math]
[math]x=x_0+\Delta x\times\ n[/math]
[math]z=z_0+\Delta z\times\ n[/math]

The angle theta that the wire makes with the vertex is given by

[math]\tan{\theta}=\frac{x_0+\Delta x}{z_0+\Delta z}\Rightarrow \theta=\arctan{\frac{x_0+\Delta x}{z_0+\Delta z}}[/math]


For the values found earlier for the starting position:

[math]\frac{19.9014\ cm\ +2.3078\ cm}{242.3760\ cm+.2019\ cm}=\frac{22.2092\ cm}{242.5779\ cm}=.09155=\tan\ \theta[/math]


[math]\theta(wire\ 1)=\arctan\ .09155=5.2311^{\circ}[/math]


[math]x_0=22.2092-1.212=20.9972\ cm\ \hat{x}[/math]
[math]z_0=242.5779+0.5652=243.1431\ cm\ \hat{z}[/math]



[math]x(n)=20.9972+1.212\ n[/math]


[math]z(n)=243.1431-0.5652\ n[/math]
[math]\tan{\theta}=\frac{20.9972+n1.212}{243.1431-n0.5652}\Rightarrow \tan{ \theta}=\frac{-959.637}{n - 430.189} - 2.14437 \Rightarrow 430.637 - n = \frac{959.637}{\tan{ \theta}+2.14437}[/math]


[math]\Rightarrow n = \frac{-959.189}{\tan{ \theta}+2.14437}+430.189[/math]




[math]\textbf{\underline{Navigation}}[/math]

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]