Geometry (25 MeV LINAC exit port)

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Critical angle

$\Theta_C = \frac{m_ec^2}{E_{beam}} = \frac{0.511\ MeV}{25\ MeV} = 1.17\ ^o$

Kicker angle and displacements on the wall

accelerator's side wall

  [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{2.36}{286}\right) = 0.827\ ^o[/math]

detector's side wall

  [math]\Delta_1 = (286\ cm + 183\ cm)\ *\ \tan(1.17^o) = 9.58\ cm[/math] 

  [math]\Delta_2 = (286\ cm + 183\ cm)\ *\ \tan(0.827^o) = 6.77\ cm[/math]

Vacuum pipe location ($\Theta_c/2$)

collimator location

1) center position:

  [math]286\ cm \cdot \tan (1.17) = 5.84\ cm[/math]  (wall 1)

  [math](286 + 183)\ cm \cdot \tan (1.17) = 9.58\ cm[/math]  (wall 2)

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 = (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 $BED_1$:

  [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]

Vacuum pipe location ($\Theta_c/4$)

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) collimator diameter:

  [math]\Theta_c/4 = 0.67^o/4 = 0.168^o[/math]

  [math]286\ cm \cdot \tan (0.168) = 0.84\ cm[/math]  (wall 1)

  [math](286 + 183)\ cm \cdot \tan (0.168) = 1.38\ cm[/math]  (wall 2)

collimator critical angle

  [math] AB = AC - BD/2 = (2.35 - 0.84/2)\ cm = 1.93\ cm [/math]

  [math] A_1D_1 = A_1C_1 + B_1D_1/2 = (3.85 + 1.38/2)\ cm = 4.54\ cm [/math]

  [math] ED_1 = A_1D_1 - AB = (4.54 - 1.93)\ cm = 2.61\ cm [/math]
 

from triangle $BED_1$:

  [math] \tan (\alpha) = \frac{2.61\ cm}{183\ cm} \Rightarrow \alpha = 0.82^o[/math]

minimal distance from the wall

from triangle FAB:

  [math] FA = \frac{AB}{\tan (0.82^o)} = \frac{1.93\ cm}{\tan (0.82^o)} = 135\ cm [/math]

Funny pictures...

how it looks ($\Theta_c/2$, pipe 3")

how it looks 1 ($\Theta_c/4$, pipe 3")

how it looks 2 ($\Theta_c/4$, pipe 3")

how it looks 4 ($\Theta_c/2$, pipe (2 1/2)" and then pipe 4")

need to adjust to converter position

how it looks 5 ($\Theta_c/2$, box 3"x4" and then pipe 4")

need to adjust to converter position

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