Difference between revisions of "DV MollerTrackRecon"

From New IAC Wiki
Jump to navigation Jump to search
Line 12: Line 12:
  
 
[[File:Event29.png]]
 
[[File:Event29.png]]
 +
 +
 +
We take the phi angle from the Simulated Event momentum.
 +
 +
 +
To find the change in phi we compare it to the final position from Timer Based Tracking.
 +
 +
The final position is the same as the
  
  

Revision as of 18:52, 22 October 2015

Moller events WITH Solenoid

LUND file with Moller events (with origin of coordinates occurring at each event)

2       1       1       1       1       0       0.000563654     3.53715 0       6.2002
1 -1 1 11 0 0 0.69 -2.4999 10993.7998 10993.80 0.000511 0 0 0
2 -1 1 11 0 0 -0.69 2.4999 6.5852 7.08 0.000511 0 0 0


From a GEMC run WITH the Solenoid.

Event29.png


We take the phi angle from the Simulated Event momentum.


To find the change in phi we compare it to the final position from Timer Based Tracking.

The final position is the same as the


Detector position.png

EulerAngles.png

Transformation Matrix

[math]\left( \begin{array}{ccc} \cos (\theta ) & 0 & -\sin (\theta ) \\ 0 & 1 & 0 \\ \sin (\theta ) & 0 & \cos (\theta ) \\ \end{array} \right).\left( \begin{array}{c} x \\ y \\ z \\ \end{array} \right)[/math]


[math]=\left( \begin{array}{c} x \cos (\theta )-z \sin (\theta ) \\ y \\ z \cos (\theta )+x \sin (\theta ) \\ \end{array} \right)[/math]


For event #29, in sector 3, the location of the first interaction is given by

Conversions.png


Converting -25 degrees to radians, [math]\theta =-0.436332[/math]


[math]\left( \begin{array}{ccc} \cos (\theta ) & 0 & -\sin (\theta ) \\ 0 & 1 & 0 \\ \sin (\theta ) & 0 & \cos (\theta ) \\ \end{array} \right).\left( \begin{array}{c} -15.76 \\ 0 \\ 237.43 \\ \end{array} \right)[/math]

[math]=\left( \begin{array}{c} 86.0588 \\ 0. \\ 221.845 \\ \end{array} \right)[/math]

Finding [math]\phi =\frac{120\ 2 \pi }{360};[/math] since "sector -1" =3-1=2*60=120 degrees

[math]\left( \begin{array}{ccc} \cos (\phi ) & -\sin (\phi ) & 0 \\ \sin (\phi ) & \cos (\phi ) & 0 \\ 0 & 0 & 1 \\ \end{array} \right).\left( \begin{array}{c} 86.0588 \\ 0. \\ 221.845 \\ \end{array} \right)[/math]

[math]\left( \begin{array}{c} -43.0294 \\ 74.5291 \\ 221.845 \\ \end{array} \right)[/math]


DV_RunGroupC_Moller#Moller_Track_Reconstruction