Difference between revisions of "Geometry (44 MeV LINAC exit port)"

From New IAC Wiki
Jump to navigation Jump to search
Line 42: Line 42:
 
   <math>(286 + 183)\ cm \cdot \tan (0.47) = 3.85\ cm</math>  (wall 2)
 
   <math>(286 + 183)\ cm \cdot \tan (0.47) = 3.85\ cm</math>  (wall 2)
  
2) assume diameter is <math>\Theta_c/2 = 0.67^o/2 = 0.335^o</math>
+
2) assume collimator diameter is <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)<br>
 
   <math>286\ cm \cdot \tan (0.335) = 1.67\ cm</math>  (wall 1)<br>

Revision as of 14:17, 25 May 2010

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 (only the kicker angle)

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) assume collimator diameter is [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 - AB = (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

1) from triangle QAB:

  [math] QA = \frac{AB}{\tan (0.67^o)} = \frac{1.52\ cm}{\tan (0.67^o)} = 75\ cm [/math]

3) from triangles OPR and QPR:

  [math] OQ = OA - QA = (286 - 75)\ cm = 211\ cm [/math]
  [math] OR\cdot \tan (0.67^o)) = (211 - OR)\cdot \tan (1.16^o) \Rightarrow[/math]
[math] OR = 211 cm\cdot \frac{tan (1.16^o)}{tan (1.16^o) + tan (0.67^o)} = 134\ cm[/math]
  [math] RQ = OQ - RQ = (211-134)\ cm = 77\ cm [/math]
  
  [math] PR = 134\cdot \tan (0.67^o) = 1.57\ cm[/math]

4) minimal distance:

  [math] OR = 134\ cm\ \ (vacuum\ pipe\ length) [/math]
  [math] RA = OA - OR = (286 - 134)\ cm = 152\ cm\ \ (from\ the\ wall\ to\ the\ pipe) [/math]

collimator and pipe geometry

Vacuum pipe collimator2.png

Vacuum pipe location (kicker + multiple scattering angle angles)

1) take multiple scattering angle [math] \Theta = 0.27^o[/math]

  [math] 0.67^o \longrightarrow (0.67^o - 0.27^o) = 0.40^o[/math]
  [math] OR = 211 cm\cdot \frac{tan (1.16^o)}{tan (1.16^o) + tan (0.40^o)} = 157\ cm[/math]
  [math] RQ = OQ - RQ = (211-157)\ cm = 54\ cm [/math]
  
  [math] PR = 157\cdot \tan (0.40^o) = 1.09\ cm[/math]

4) minimal distance:

  [math] OR = 157\ cm\ \ (vacuum\ pipe\ length) [/math]
  [math] RA = OA - OR = (286 - 157)\ cm = 129\ cm\ \ (from\ the\ wall\ to\ the\ pipe) [/math]


Go Back