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| | [[File:beam_up_down5.png]]<br> | | [[File:beam_up_down5.png]]<br> |
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| − | =Vacuum pipe location (only the kicker angle)= | + | =Vacuum pipe location (<math> \Theta_c/2</math>)= |
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| | ==collimator location== | | ==collimator location== |
Revision as of 17:50, 25 May 2010
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Some measurements of 90 experimental degree exit port
Critical angle and displacement calculations
[math]\Theta = \frac{m_ec^2}{E_{beam}} = \frac{0.511\ MeV}{44\ MeV} = 0.67\ ^o[/math]
Kicker angle and displacement calculations
accelerator's side wall
[math]\Delta = 286\ cm\ *\ \tan(0.67^o) = 3.34\ cm[/math]
[math]x^2+x^2 = 3.34^2\ cm \ \ \Rightarrow\ \ x = 2.36\ cm[/math]
[math]\Delta = 2.36\ cm \ \ \Rightarrow\ \ \tan^{-1}\left(\frac{2.36}{286}\right) = 0.47\ ^o[/math]
detector's side wall
[math]\Delta = (286\ cm + 183\ cm)\ *\ \tan(0.67^o) = 5.48\ cm[/math]
[math]\Delta = (286\ cm + 183\ cm)\ *\ \tan(0.47^o) = 3.85\ cm[/math]
Off-axis collimation geometry
Vacuum pipe location ([math] \Theta_c/2[/math])
collimator location
1) center position
[math]286\ cm \cdot \tan (0.47) = 2.35\ cm[/math] (wall 1)
[math](286 + 183)\ cm \cdot \tan (0.47) = 3.85\ cm[/math] (wall 2)
2) assume collimator diameter is [math]\Theta_c/2 = 0.67^o/2 = 0.335^o[/math]
[math]286\ cm \cdot \tan (0.335) = 1.67\ cm[/math] (wall 1)
[math](286 + 183)\ cm \cdot \tan (0.335) = 2.74\ cm[/math] (wall 2)
collimator critical angle
[math] AB = AC - BD/2 = (2.35 - 1.67/2)\ cm = 1.52\ cm [/math]
[math] A_1D_1 = A_1C_1 + B_1D_1/2 = (3.85 + 2.74/2)\ cm = 5.22\ cm [/math]
[math] ED_1 = A_1D_1 - A_1E = (5.22 - 1.52)\ cm = 3.70\ cm [/math]
from triangle [math]BED_1[/math]:
[math] \tan (\alpha) = \frac{3.70\ cm}{183\ cm} \Rightarrow \alpha = 1.16^o[/math]
minimal distance from the wall
from triangle FAB:
[math] FA = \frac{AB}{\tan (1.16^o)} = \frac{1.52\ cm}{\tan (1.16^o)} = 75\ cm [/math]
collimator and pipe geometry
Vacuum pipe location (kicker + multiple scattering angle angles)
1) take multiple scattering angle [math] \Theta = 0.27^o[/math]
[math] 0.67^o \longrightarrow (0.67^o - 0.27^o) = 0.40^o[/math]
[math] OR = 211 cm\cdot \frac{tan (1.16^o)}{tan (1.16^o) + tan (0.40^o)} = 157\ cm[/math]
[math] RQ = OQ - RQ = (211-157)\ cm = 54\ cm [/math]
[math] PR = 157\cdot \tan (0.40^o) = 1.09\ cm[/math]
4) minimal distance:
[math] OR = 157\ cm\ \ (vacuum\ pipe\ length) [/math]
[math] RA = OA - OR = (286 - 157)\ cm = 129\ cm\ \ (from\ the\ wall\ to\ the\ pipe) [/math]
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