Geometry (25 MeV LINAC exit port)

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Minimum accelerator energy to run experiment

The minimum energy of accelerator (MeV) is limited by fitting the collimator ($r_2$) into the hole ($R = 8.73\ cm$)

[math]x_2 + r_2 = R[/math]

1) Assuming the collimator diameter is $\Theta_C$:

[math]\frac{1}{\sqrt{2}}\ (286+183)\ \tan\left(\frac{0.511}{E_{min}}\right) +
       \frac{1}{2}\ (286+183)\ \tan\left(\frac{0.511}{E_{min}}\right) = 8.73 \Rightarrow E_{min} = 33.1\ MeV  [/math]

2) Assuming the collimator diameter is $\Theta_C/2$:

[math]\frac{1}{\sqrt{2}}\ (286+183)\ \tan\left(\frac{0.511}{E_{min}}\right) +
       \frac{1}{2}\ (286+183)\ \tan\left(\frac{1}{2}\ \frac{0.511}{E_{min}}\right) = 8.73 \Rightarrow E_{min} = 26.3\ MeV  [/math]

3) Assuming the collimator diameter is $\Theta_C/4$:

[math]\frac{1}{\sqrt{2}}\ (286+183)\ \tan\left(\frac{0.511}{E_{min}}\right) +
       \frac{1}{2}\ (286+183)\ \tan\left(\frac{1}{4}\ \frac{0.511}{E_{min}}\right) = 8.73 \Rightarrow E_{min} = 22.8\ MeV  [/math]

4) In general:

25 MeV geometry

geometry calculation

collimator diameter	$\Theta_{critical}$	$\Theta_{kicker}$	$\alpha_{collimator}$	$AC$	$A_1C_1$	$BD$	$B_1D_1$
$\frac{\Theta_{critical}}{2}$	$1.17^o$	$0.83^o$	$2.03^o$	4.13	6.78	2.92	4.79
$\frac{\Theta_{critical}}{4}$	$1.17^o$	$0.82^o$	$1.43^o$	4.13	6.78	1.46	2.40

geometry pictures

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

File:Vacuum pipe collimator .png

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

File:Vacuum pipe collimator .png

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