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

From New IAC Wiki
Jump to navigation Jump to search
Line 71: Line 71:
 
=Funny pictures...=
 
=Funny pictures...=
  
==how it looks (<math> \Theta_c/2</math>, pipe 3")==
+
==how it looks 1 (<math> \Theta_c/2</math>, box 3"x4" and then pipe 4")==
  
[[File:vacuum_pipe_collimator_0.335_2.png]]<br>
+
[[File:vacuum_pipe_collimator.png]]<br>
  
==how it looks 1 (<math> \Theta_c/4</math>, pipe 3")==
+
==how it looks 2 (<math> \Theta_c/4</math>, box 3"x4" and then pipe 4")==
  
[[File:vacuum_pipe_collimator_0.168_2.png]]<br>
+
[[File:vacuum_pipe_collimator.png]]<br>
 
 
==how it looks 2 (<math> \Theta_c/4</math>, pipe 3")==
 
 
 
[[File:vacuum_pipe_collimator_168_1.png]]<br>
 
 
 
==how it looks 4 (<math> \Theta_c/2</math>, pipe (2 1/2)" and then pipe 4")==
 
 
 
need to adjust to converter position
 
 
 
[[File:vacuum_pipe_collimator_335_4.png]]<br>
 
 
 
 
 
==how it looks 5 (<math> \Theta_c/2</math>, box 3"x4" and then pipe 4")==
 
 
 
need to adjust to converter position
 
 
 
[[File:vacuum_pipe_collimator_335_5.png]]<br>
 
  
  
  
 
[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]

Revision as of 05:54, 11 June 2010

Go Back


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

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

collimator location

1) 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)

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

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]

Funny pictures...

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

Vacuum pipe collimator.png

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

Vacuum pipe collimator.png


Go Back