Geometry (25 MeV LINAC exit port)

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Minimum accelerator energy to run experiment

The minimum energy of accelerator (MeV) is limited by fitting the collimator ([math]r_2[/math]) into the hole ([math]R[/math])

Assuming the collimator diameter is [math]\Theta_C/2[/math] and the hole radius is R = 8.73 cm:

[math]x_2 + r_2 = R[/math]

[math]\frac{1}{\sqrt{2}}\ (286+183)\ \tan\left(\frac{0.511}{E_{min}}\right) +
           \frac{1}{2}\ (286+183)\ \tan\left(\frac{1}{2}\ \frac{0.511}{E_{min}}\right) = 8.73 \Rightarrow E_{min} = 26.3\ MeV  [/math]

Solving for [math]E_{min}[/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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