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

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[[File:min_energy.png|1000px]]
 
[[File:min_energy.png|1000px]]
  
  <math> \Theta_C = \frac{0.511\ MeV}{E\ MeV}\frac{180}{\pi} </math> <br>  
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  <math> \Theta_C = </math> <br>  
  <math>\Delta = 286\ \tan(\Theta_C) \Rightarrow  x = \frac{\Delta}{\sqrt{2}}</math><br>
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  <math>\Delta = \Rightarrow  </math><br>
  <math> r = \frac{1}{2}\ 286\ \tan{\frac{\Theta_C}{2}} </math>
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  <math> r =  </math>
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 +
 
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The minimum energy of accelerators defined by fitting  collimator size (<math>r_2</math>) into hole (<math>R_2</math>)
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<math>x_2 + r_2 = R</math>
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<math>x = \frac{286}{\sqrt{2}}\ \tan(\frac{0.511}{E}\frac{180}{\pi}) +
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          \frac{286}{2}\ \tan{\frac{\frac{1}{2}\frac{0.511}{E}\frac{180}{\pi}}   </math>
  
 
=25 MeV geometry=
 
=25 MeV geometry=

Revision as of 22:59, 11 June 2010

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Minimum accelerator energy to run experiment

Min energy.png

[math] \Theta_C =  [/math] 
[math]\Delta = \Rightarrow [/math]
[math] r = [/math]


The minimum energy of accelerators defined by fitting  collimator size ([math]r_2[/math]) into hole ([math]R_2[/math])

[math]x_2 + r_2 = R[/math]
[math]x = \frac{286}{\sqrt{2}}\ \tan(\frac{0.511}{E}\frac{180}{\pi}) +
           \frac{286}{2}\ \tan{\frac{\frac{1}{2}\frac{0.511}{E}\frac{180}{\pi}}   [/math]

25 MeV geometry

critical angle

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

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]

geometry ([math] \Theta_c/2[/math])

collimator 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]:

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]

geometry ([math] \Theta_c/4[/math])

collimator 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]:

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]

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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