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=Minimum accelerator energy to run experiment=
=general setup=


==condition 1: fitting the collimator size into the hole==
[[File:minimum_energy_condition.png]]


[[File:min_energy.png|800px]]
= fitting the collimator size into the hole through the concrete wall=


The minimum energy of accelerator (MeV) is limited by fitting the collimator size <math>r_2</math> into the hole R = 8.73 cm:
I can express the distance <math>A_1D_1</math> as function of collimator size <math>\Theta_C/m</math> and electron beam energy E:
[[File:formula_cond11.png]]


<math>x_2 + r_2 = R</math>
To fit the collimator size into the hole through the concrete wall with radius R = 8.73 cm we need to solve equation:


1) Assuming the collimator diameter is <math>\Theta_C</math>:
<math>A_1D_1(E,\ \Theta_C/m) = 8.73\ cm</math>


<math>\frac{1}{\sqrt{2}}\ (286+183)\ \tan\left(\frac{0.511}{E_{min}}\right) +
1) some solutions of this equation for different collimator sizes m are:
      \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>m = 1 \Rightarrow E_{min} = 33.1\ MeV  </math><br>
<math>m = 2 \Rightarrow E_{min} = 26.3\ MeV  </math><br>
<math>m = 4 \Rightarrow E_{min} = 22.8\ MeV  </math><br>


<math>\frac{1}{\sqrt{2}}\ (286+183)\ \tan\left(\frac{0.511}{E_{min}}\right) +
2) in general for arbitrary collimator size m the solutions are:
      \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>:
[[File:energy_condition1.jpeg]]


<math>\frac{1}{\sqrt{2}}\ (286+183)\ \tan\left(\frac{0.511}{E_{min}}\right) +
All energies above this line is good to run experiment for condition above
      \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>:
=critical collimator line condition=


[[File:plot_energy_collimatorsize.jpeg]]
Also I can express the distance <math>GH</math> as function of collimator size <math>\Theta_C/m</math> and electron beam energy E:
[[File:formula_cond21.png]]
If I would like that nothing hitting the 4" box to go through the collimator I need to solve equation:


All energy above this line is good to run experiment
<math>GH(E,\ \Theta_C/m) = 5.08\ cm</math>


==condition 2: F1A = 286 cm==
1) some solutions of this equation for different collimator sizes m are:


1) assuming the collimator diameter is <math>\Theta_C</math>
<math>m = 1 \Rightarrow E_{min} = 73.7\ MeV  </math><br>
<math>m = 2 \Rightarrow E_{min} = 36.9\ MeV  </math><br>
<math>m = 4 \Rightarrow E_{min} = 18.4\ MeV  </math><br>


<math> E_{min} = 73.7\ MeV  </math>
2) in general for arbitrary collimator size m the solutions are:


2) assuming the collimator diameter is <math>\Theta_C/2</math>
[[File:energy_condition2.jpeg]]


<math> E_{min} = 36.9\ MeV  </math>
All energies above this line is good to run experiment for condition above


3) assuming the collimator diameter is <math>\Theta_C/4</math>
=both solutions are together=


<math> E_{min} = 18.4\ MeV  </math>
[[File:energy_condition12.jpeg]]


4) for arbitrary collimator size <math>\Theta_C/m</math>:
All energies above this lines is good to run experiment for both conditions above


[[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]]
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]
[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]

Latest revision as of 19:03, 24 May 2012

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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:

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 sizes m are: 

[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 energies above this line is good to run experiment for condition above

critical collimator line 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:

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 sizes m are: 

[math]\displaystyle{ m = 1 \Rightarrow E_{min} = 73.7\ MeV   }[/math]

[math]\displaystyle{ m = 2 \Rightarrow E_{min} = 36.9\ MeV   }[/math]

[math]\displaystyle{ m = 4 \Rightarrow E_{min} = 18.4\ MeV   }[/math]

2) in general for arbitrary collimator size m the solutions are:

All energies above this line is good to run experiment for condition above

both solutions are together

All energies above this lines is good to run experiment for both conditions above


Go Back

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