Geometry (25 MeV LINAC exit port)

From New IAC Wiki
Jump to navigation Jump to search

Go Back

Minimum accelerator energy to run experiment

Min energy.png

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

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

1) Assuming the collimator diameter is [math]\Theta_C[/math]:

[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 [math]\Theta_C/2[/math]:

[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 [math]\Theta_C/4[/math]:

[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:

Plot energy collimatorsize.jpeg

25 MeV geometry

geometry calculation

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

geometry pictures

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

File:Vacuum pipe collimator .png

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

File:Vacuum pipe collimator .png


Go Back