Difference between revisions of "Geometry (44 MeV LINAC exit port)"
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<math> QA = \frac{AB}{\tan (1.16^o)} = \frac{1.52\ cm}{\tan (1.16^o)} = 75\ cm </math> | <math> QA = \frac{AB}{\tan (1.16^o)} = \frac{1.52\ cm}{\tan (1.16^o)} = 75\ cm </math> | ||
+ | 2) OQ = OA - QA = (286 - 75) cm = 211 cm | ||
+ | |||
+ | 3) from triangles OPR and QPR: | ||
+ | |||
+ | <math> PR = QR\cdot \tan (1.16^o) = (211 - QR) \tan (0.47^o) \Rightarrow</math> | ||
+ | <math> PR = QR\cdot \tan (1.16^o) = (211 - QR) \tan (0.47^o) \Rightarrow</math> | ||
[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 05:59, 25 May 2010
Some measurements of 90 experimental degree exit port
Critical angle and displacement calculations
Kicker angle and displacement calculations
1 foot = 30.48 cm
accelerator's side wall
detector's side wall
Off-axis collimation geometry
Vacuum pipe
collimator location
1) center position
(wall 1)
(wall 2)
2) assume diameter is
(wall 1)
(wall 2)
collimator critical angle
AB = AC - BD/2 = (2.35 - 1.67/2) cm = 1.52 cm
A1D1 = A1C1 + B1D1/2 = (3.85 + 2.74/2) cm = 5.22 cm
ED1 = A1D1 - AB = (5.22 - 1.52) cm = 3.70 cm
from triangle
:
minimal distance from the wall
1) from triangle QAB:
2) OQ = OA - QA = (286 - 75) cm = 211 cm
3) from triangles OPR and QPR: