Minimum accelerator energy to run experiment: Difference between revisions

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=general setup=
=general setup=


[[File:minimum_energy_condition.png|600px]]
[[File:minimum_energy_condition.png]]


= fitting the collimator size into the hole in the concrete wall=
= fitting the collimator size into the hole through the concrete wall=


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:
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:
   
   
  <math>A_1D_1(E,\ \Theta_C/m) = \frac{469}{2}\tan(\frac{0.511}{E}) + \frac{469}{sqrt{2}}\tan(\frac{1}{m}\frac{0.511}{E})<math>
  [[File:formula_cond11.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:
To fit the collimator size into the hole through the concrete wall with radius R = 8.73 cm we need to solve equation:


<math>x_2 + r_2 = R</math>
<math>A_1D_1(E,\ \Theta_C/m) = 8.73\ cm</math>


1) Assuming the collimator diameter is <math>\Theta_C</math>:
1) some solutions of this equation for different collimator sizes m are:


  <math>\frac{1}{\sqrt{2}}\ (286+183)\ \tan\left(\frac{0.511}{E_{min}}\right) +
  <math>m = 1 \Rightarrow E_{min} = 33.1\ MeV  </math><br>
      \frac{1}{2}\ (286+183)\ \tan\left(\frac{0.511}{E_{min}}\right) = 8.73 \Rightarrow E_{min} = 33.1\ MeV  </math>
<math>m = 2 \Rightarrow E_{min} = 26.3\ MeV  </math><br>
<math>m = 4 \Rightarrow E_{min} = 22.8\ MeV  </math><br>


2) Assuming the collimator diameter is <math>\Theta_C/2</math>:
2) in general for arbitrary collimator size m the solutions are:


<math>\frac{1}{\sqrt{2}}\ (286+183)\ \tan\left(\frac{0.511}{E_{min}}\right) +
[[File:energy_condition1.jpeg]]
      \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>:
All energies above this line is good to run experiment for condition above


<math>\frac{1}{\sqrt{2}}\ (286+183)\ \tan\left(\frac{0.511}{E_{min}}\right) +
=critical collimator line condition=
      \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>:
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:plot_energy_collimatorsize.jpeg]]
[[File:formula_cond21.png]]
 
All energy under this line is good to run experiment for condition above
If I would like that nothing hitting the 4" box to go through the collimator I need to solve equation:
 
=GH = 5.08 cm condition=
 
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>
  <math>GH(E,\ \Theta_C/m) = 5.08\ cm</math>


3) assuming the collimator diameter is <math>\Theta_C/4</math>
1) some solutions of this equation for different collimator sizes m are:


  <math> E_{min} = 18.4\ MeV  </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>


4) for arbitrary collimator size <math>\Theta_C/m</math>:
2) in general for arbitrary collimator size m the solutions are:


[[File:plot_energy_F1A.jpeg]]
[[File:energy_condition2.jpeg]]


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


=both conditions above are together=
=both solutions are together=


[[File:plot_energy_bothcondition.jpeg]]
[[File:energy_condition12.jpeg]]


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






[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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