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> | ||
− | + | 3) from triangles OPR and QPR: | |
+ | |||
+ | OQ = OA - QA = (286 - 75) cm = 211 cm | ||
+ | |||
+ | <math> RQ\cdot \tan (1.16^o) = (211 - RQ)\cdot \tan (0.47^o) \Rightarrow</math><br> | ||
+ | <math> RQ = 211\cdot \frac{tan (0.47^o)}{tan (0.47^o) + tan (1.16^o)} = 61\ cm</math> | ||
+ | |||
+ | 4) minimal distance: | ||
− | + | RA = RQ + QA = (61 + 75) cm = 136 cm (from the wall) | |
− | + | OR = OA + RA = (286 + 136) cm = 150 cm (from the wall) | |
− | |||
[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 06:10, 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:
3) from triangles OPR and QPR:
OQ = OA - QA = (286 - 75) cm = 211 cm
4) minimal distance:
RA = RQ + QA = (61 + 75) cm = 136 cm (from the wall)
OR = OA + RA = (286 + 136) cm = 150 cm (from the wall)