Difference between revisions of "Minimum accelerator energy to run experiment"

From New IAC Wiki
Jump to navigation Jump to search
(Created page with '[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back] =Minimum accelerator energy to run experiment= ==condition 1: fitting the collimator s...')
 
Line 1: Line 1:
 
[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]
  
=Minimum accelerator energy to run experiment=
+
=condition 1: fitting the collimator size into the hole=
 
 
==condition 1: fitting the collimator size into the hole==
 
 
 
[[File:min_energy.png|800px]]
 
 
 
The minimum energy of accelerator (MeV) is limited by fitting the collimator size <math>r_2</math> into the hole R = 8.73 cm:
 
 
 
<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) for arbitrary collimator size <math>\Theta_C/2</math>:
 
 
 
[[File:plot_energy_collimatorsize.jpeg]]
 
 
 
All energy above this line is good to run experiment
 
 
 
==condition 2: F1A = 286 cm==
 
 
 
1) assuming the collimator diameter is <math>\Theta_C</math>
 
 
 
<math> E_{min} = 73.7\ MeV  </math>
 
 
 
2) assuming the collimator diameter is <math>\Theta_C/2</math>
 
 
 
<math> E_{min} = 36.9\ MeV  </math>
 
 
 
3) assuming the collimator diameter is <math>\Theta_C/4</math>
 
 
 
<math> E_{min} = 18.4\ MeV  </math>
 
 
 
4) for arbitrary collimator size <math>\Theta_C/m</math>:
 
 
 
[[File:plot_energy_F1A.jpeg]]
 
 
 
All energy above this line is good to run experiment
 
 
 
==both conditions above are together==
 
 
 
[[File:plot_energy_bothcondition.jpeg]]
 
 
 
All energy above this lines is good to run experiment
 
 
 
[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]
 
 
 
=Minimum accelerator energy to run experiment=
 
 
 
==condition 1: fitting the collimator size into the hole==
 
  
 
[[File:min_energy.png|800px]]
 
[[File:min_energy.png|800px]]
Line 91: Line 30:
 
All energy above this line is good to run experiment
 
All energy above this line is good to run experiment
  
==condition 2: F1A = 286 cm==
+
=condition 2: F1A = 286 cm=
  
 
1) assuming the collimator diameter is <math>\Theta_C</math>
 
1) assuming the collimator diameter is <math>\Theta_C</math>
Line 111: Line 50:
 
All energy above this line is good to run experiment
 
All energy above this line is good to run experiment
  
==both conditions above are together==
+
=both conditions above are together=
  
 
[[File:plot_energy_bothcondition.jpeg]]
 
[[File:plot_energy_bothcondition.jpeg]]

Revision as of 19:03, 14 June 2010

Go Back

condition 1: fitting the collimator size into the hole

Min energy.png

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

[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) for arbitrary collimator size [math]\Theta_C/2[/math]:

Plot energy collimatorsize.jpeg

All energy above this line is good to run experiment

condition 2: F1A = 286 cm

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

[math] E_{min} = 73.7\ MeV  [/math]

2) assuming the collimator diameter is [math]\Theta_C/2[/math]

[math] E_{min} = 36.9\ MeV  [/math]

3) assuming the collimator diameter is [math]\Theta_C/4[/math]

[math] E_{min} = 18.4\ MeV  [/math]

4) for arbitrary collimator size [math]\Theta_C/m[/math]:

Plot energy F1A.jpeg

All energy above this line is good to run experiment

both conditions above are together

Plot energy bothcondition.jpeg

All energy above this lines is good to run experiment


Go Back