Minimum accelerator energy to run experiment: Difference between revisions
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<math>A_1D_1(E,\ \Theta_C/m) = \frac{469}{2}\tan\left(\frac{0.511}{E}\right) | <math>A_1D_1(E,\ \Theta_C/m) = \frac{469}{2}\tan\left(\frac{0.511}{E}\right) | ||
+ \frac{469}{\sqrt{2}}\tan\left(\frac{1}{m}\frac{0.511}{E}\right)</math> | + \frac{469}{\sqrt{2}}\tan\left(\frac{1}{m}\frac{0.511}{E}\right)</math> | ||
To fit the collimator size into the hole through the concrete wall with radius R = 8.73 cm we need to solve equation: | To fit the collimator size into the hole through the concrete wall with radius R = 8.73 cm we need to solve equation: | ||
<math>A_1D_1(E,\ \Theta_C/m) = 8.73\ cm</math> | <math>A_1D_1(E,\ \Theta_C/m) = 8.73\ cm</math> | ||
1) some solutions of this equation for different collimator size m | 1) some solutions of this equation for different collimator size m | ||
Revision as of 22:27, 14 June 2010
general setup
fitting the collimator size into the hole through the concrete wall
I can express the distance [math]\displaystyle{ A_1D_1 }[/math] as function of collimator size [math]\displaystyle{ \Theta_C/m }[/math] and electron beam energy E:
[math]\displaystyle{ A_1D_1(E,\ \Theta_C/m) = \frac{469}{2}\tan\left(\frac{0.511}{E}\right)
+ \frac{469}{\sqrt{2}}\tan\left(\frac{1}{m}\frac{0.511}{E}\right) }[/math]
To fit the collimator size into the hole through the concrete wall with radius R = 8.73 cm we need to solve equation:
[math]\displaystyle{ A_1D_1(E,\ \Theta_C/m) = 8.73\ cm }[/math]
1) some solutions of this equation for different collimator size m
[math]\displaystyle{ m = 1 \Rightarrow E_{min} = 33.1\ MeV }[/math]
[math]\displaystyle{ m = 2 \Rightarrow E_{min} = 26.3\ MeV }[/math]
[math]\displaystyle{ m = 4 \Rightarrow E_{min} = 22.8\ MeV }[/math]
2) in general for arbitrary collimator size m the solutions are:
All energy under this line is good to run experiment for condition above
GH = 5.08 cm condition
Also I can express the distance [math]\displaystyle{ GH }[/math] as function of collimator size [math]\displaystyle{ \Theta_C/m }[/math] and electron beam energy E:
It's pretty long formula to show here
If I would like that nothing hitting the 4" box to go through the collimator I need to solve equation:
[math]\displaystyle{ GH(E,\ \Theta_C/m) = 5.08\ cm }[/math]
1) some solutions of this equation for different collimator size m
[math]\displaystyle{ m = 1 \Rightarrow E_{min} = 33.1\ MeV }[/math]
[math]\displaystyle{ m = 2 \Rightarrow E_{min} = 26.3\ MeV }[/math]
[math]\displaystyle{ m = 4 \Rightarrow E_{min} = 22.8\ MeV }[/math]
2) in general for arbitrary collimator size m the solutions are:
All energy under this line is good to run experiment for condition above
1) assuming the collimator diameter is [math]\displaystyle{ \Theta_C }[/math]
[math]\displaystyle{ E_{min} = 73.7\ MeV }[/math]
2) assuming the collimator diameter is [math]\displaystyle{ \Theta_C/2 }[/math]
[math]\displaystyle{ E_{min} = 36.9\ MeV }[/math]
3) assuming the collimator diameter is [math]\displaystyle{ \Theta_C/4 }[/math]
[math]\displaystyle{ E_{min} = 18.4\ MeV }[/math]
4) for arbitrary collimator size [math]\displaystyle{ \Theta_C/m }[/math]:
All energy under this line is good to run experiment for condition above
both conditions above are together
All energy under this linse is good to run experiment for both conditions above


