Difference between revisions of "Geometry (44 MeV LINAC exit port)"

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   <math>(286 + 183)\ cm \cdot \tan (0.47) = 3.85\ cm</math>  (wall 2)
 
   <math>(286 + 183)\ cm \cdot \tan (0.47) = 3.85\ cm</math>  (wall 2)
  
2) assume diameter is <math>\ \ \theta_C/2 = 0.67^o/2 = 0.335^o</math>
+
2) assume diameter is <math>\Theta_crit/2 = 0.67^o/2 = 0.335^o</math>
  
  

Revision as of 05:09, 25 May 2010

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Some measurements of 90 experimental degree exit port

Exit port1.png


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

1 foot = 30.48 cm

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

Beam up down5.png


Vacuum pipe

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 diameter is [math]\Theta_crit/2 = 0.67^o/2 = 0.335^o[/math]



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