Difference between revisions of "Geometry (25 MeV LINAC exit port)"

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[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]
 
[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]
  
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=Minimum accelerator energy to run experiment=
  
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=Critical angle=
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=25 MeV Critical angle=
  
 
<math>\Theta_C = \frac{m_ec^2}{E_{beam}} = \frac{0.511\ MeV}{25\ MeV} = 1.17\ ^o</math><br>
 
<math>\Theta_C = \frac{m_ec^2}{E_{beam}} = \frac{0.511\ MeV}{25\ MeV} = 1.17\ ^o</math><br>
  
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=Kicker angle=
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=25 MeV Kicker angle=
  
 
   <math>\Delta_1 = 286\ cm\ *\ \tan(1.17^o) = 5.84\ cm</math> <br>
 
   <math>\Delta_1 = 286\ cm\ *\ \tan(1.17^o) = 5.84\ cm</math> <br>
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   <math>\Delta_2 = 4.13\ cm \ \ \Rightarrow\ \ \tan^{-1}\left(\frac{4.13}{286}\right) = 0.827\ ^o</math><br>
 
   <math>\Delta_2 = 4.13\ cm \ \ \Rightarrow\ \ \tan^{-1}\left(\frac{4.13}{286}\right) = 0.827\ ^o</math><br>
  
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=Collimator geometry (<math> \Theta_c/2</math>)=
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=25 MeV Collimator geometry (<math> \Theta_c/2</math>)=
  
 
==center position==
 
==center position==
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   <math>\bigtriangleup BED_1  \Rightarrow \tan (\alpha) = \frac{(9.165 - 2.67)\ cm}{183\ cm} \Rightarrow \alpha = 2.033^o</math>:
 
   <math>\bigtriangleup BED_1  \Rightarrow \tan (\alpha) = \frac{(9.165 - 2.67)\ cm}{183\ cm} \Rightarrow \alpha = 2.033^o</math>:
  
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=minimal distance from the wall (<math> \Theta_c/2</math>)=
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=25 MeV minimal distance from the wall (<math> \Theta_c/2</math>)=
  
 
   <math>\bigtriangleup FAB  \Rightarrow FA = \frac{AB}{\tan (2.033^o)} = \frac{2.67\ cm}{\tan (2.033^o)} = 75\ cm </math>
 
   <math>\bigtriangleup FAB  \Rightarrow FA = \frac{AB}{\tan (2.033^o)} = \frac{2.67\ cm}{\tan (2.033^o)} = 75\ cm </math>
  
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=Collimator geometry (<math> \Theta_c/4</math>)=
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=25 MeV Collimator geometry (<math> \Theta_c/4</math>)=
  
 
==center position==
 
==center position==
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   <math>\bigtriangleup BED_1  \Rightarrow \tan (\alpha) = \frac{(7.965 - 3.4)\ cm}{183\ cm} \Rightarrow \alpha = 1.429^o</math>:
 
   <math>\bigtriangleup BED_1  \Rightarrow \tan (\alpha) = \frac{(7.965 - 3.4)\ cm}{183\ cm} \Rightarrow \alpha = 1.429^o</math>:
  
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=minimal distance from the wall (<math> \Theta_c/4</math>)=
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=25 MeV minimal distance from the wall (<math> \Theta_c/4</math>)=
  
 
   <math>\bigtriangleup FAB  \Rightarrow FA = \frac{AB}{\tan (1.429^o)} = \frac{3.4\ cm}{\tan (1.429^o)} = 136\ cm </math>
 
   <math>\bigtriangleup FAB  \Rightarrow FA = \frac{AB}{\tan (1.429^o)} = \frac{3.4\ cm}{\tan (1.429^o)} = 136\ cm </math>
  
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=Funny pictures...=
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=25 MeV Funny pictures...=
  
 
==how it looks 1 (<math> \Theta_c/2</math>, box 3"x4" and then pipe 4")==
 
==how it looks 1 (<math> \Theta_c/2</math>, box 3"x4" and then pipe 4")==

Revision as of 20:56, 11 June 2010

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Minimum accelerator energy to run experiment

25 MeV Critical angle

[math]\Theta_C = \frac{m_ec^2}{E_{beam}} = \frac{0.511\ MeV}{25\ MeV} = 1.17\ ^o[/math]

25 MeV Kicker angle

  [math]\Delta_1 = 286\ cm\ *\ \tan(1.17^o) = 5.84\ cm[/math] 

  [math]x^2+x^2 = 5.84^2\ cm \ \ \Rightarrow\ \  x = 4.13\ cm[/math] 

  [math]\Delta_2 = 4.13\ cm \ \ \Rightarrow\ \ \tan^{-1}\left(\frac{4.13}{286}\right) = 0.827\ ^o[/math]

25 MeV Collimator geometry ([math] \Theta_c/2[/math])

center position

  [math]286\ cm \cdot \tan (0.827) = 4.13\ cm[/math]  (wall 1)

  [math](286 + 183)\ cm \cdot \tan (0.827) = 6.77\ cm[/math]  (wall 2)

collimator diameter

  [math]286\ cm \cdot \tan (1.17/2) = 2.92\ cm[/math]  (wall 1)

  [math](286 + 183)\ cm \cdot \tan (1.17/2) = 4.79\ cm[/math]  (wall 2)

collimator critical angle

  [math] AB = AC - BD/2 = (4.13 - 2.92/2)\ cm = 2.67\ cm [/math]

  [math] A_1D_1 = A_1C_1 + B_1D_1/2 = (6.77 + 4.79/2)\ cm = 9.165\ cm [/math]

  [math]\bigtriangleup BED_1  \Rightarrow \tan (\alpha) = \frac{(9.165 - 2.67)\ cm}{183\ cm} \Rightarrow \alpha = 2.033^o[/math]:

25 MeV minimal distance from the wall ([math] \Theta_c/2[/math])

  [math]\bigtriangleup FAB  \Rightarrow FA = \frac{AB}{\tan (2.033^o)} = \frac{2.67\ cm}{\tan (2.033^o)} = 75\ cm [/math]

25 MeV Collimator geometry ([math] \Theta_c/4[/math])

center position

  [math]286\ cm \cdot \tan (0.827) = 4.13\ cm[/math]  (wall 1)

  [math](286 + 183)\ cm \cdot \tan (0.827) = 6.77\ cm[/math]  (wall 2)

collimator diameter

  [math]286\ cm \cdot \tan (1.17/4) = 1.46\ cm[/math]  (wall 1)

  [math](286 + 183)\ cm \cdot \tan (1.17/4) = 2.39\ cm[/math]  (wall 2)

collimator critical angle

  [math] AB = AC - BD/2 = (4.13 - 1.46/2)\ cm = 3.4\ cm [/math]

  [math] A_1D_1 = A_1C_1 + B_1D_1/2 = (6.77 + 2.39/2)\ cm = 7.965\ cm [/math]

  [math]\bigtriangleup BED_1  \Rightarrow \tan (\alpha) = \frac{(7.965 - 3.4)\ cm}{183\ cm} \Rightarrow \alpha = 1.429^o[/math]:

25 MeV minimal distance from the wall ([math] \Theta_c/4[/math])

  [math]\bigtriangleup FAB  \Rightarrow FA = \frac{AB}{\tan (1.429^o)} = \frac{3.4\ cm}{\tan (1.429^o)} = 136\ cm [/math]

25 MeV Funny pictures...

how it looks 1 ([math] \Theta_c/2[/math], box 3"x4" and then pipe 4")

File:Vacuum pipe collimator .png

how it looks 2 ([math] \Theta_c/4[/math], box 3"x4" and then pipe 4")

File:Vacuum pipe collimator .png


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