Geometry (44 MeV LINAC exit port)

From New IAC Wiki
Jump to navigation Jump to search

Go Back

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

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/2 = 0.67^o/2 = 0.335^o[/math]
  [math]286\ cm \cdot \tan (0.335) = 1.67\ cm[/math]  (wall 1)
[math](286 + 183)\ cm \cdot \tan (0.335) = 2.74\ 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 [math]BED_1[/math]:

  [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]

how it looks ([math] \Theta_c/2[/math])

Vacuum pipe collimator 0.335 2.png

Vacuum pipe location ([math] \Theta_c/4[/math])

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 [math]BED_1[/math]:

  [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]

how it looks 1 ([math] \Theta_c/4[/math])

Vacuum pipe collimator 0.168 2.png

how it looks 2 ([math] \Theta_c/4[/math])

File:Vacuum pipe collimator 168 3.png


Go Back