Difference between revisions of "Determining wire-phi correspondance"

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<center>[[File:DC_stereo.png]]</center>
 
  
 
<center>[[File:DC_Geomentry_Sideview.png]]</center>
 
 
 
 
xdist - distance between the line of intersection of the two end-plate planes and the target position =8.298 cm th_min=4.694
 
 
Using Mathematica, we can produce a 3D rendering of how the sectors for Level 1 would have to interact with a steady angle theta with respect to the beam line, as angle phi is rotated through 360 degrees.
 
<center>[[File:PhiCone.png]]</center>
 
 
 
 
Looking just at sector 1, we can see that the intersection of level 1 and the cone of constant angle theta forms a conic section.
 
<center>[[File:Projection_side_view.png]]</center>
 
 
 
<center>[[File:Projection_Rear_view.png]]</center>
 
 
 
Following the rules of conic sections we know that the eccentricity of the conic is given by:
 
 
<center><math>e=\frac{\sin \beta}{\sin\alpha}</math></center>
 
 
 
Where β is the angle of the plane, which in our case is 25 degrees for the sectors with respect to the beam line (axis of rotation of the cone).
 
 
This leaves α is the slant of the cone, which is the angle theta that the particle must be traveling with respect to the beamline.
 
 
If the conic is an circle, e=0
 
 
If the conic is an parabola, e=1
 
 
If the conic is an ellipse, <math>e=\sqrt{1-\frac{b^2}{a^2}}</math>
 
 
<center>[[File:Conic_section.png]]</center>
 
 
 
For ellipses centered at (h,k):
 
 
<center><math>\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1</math></center>
 
 
where
 
 
<center><math>a = major\ radius\ (= \frac{1}{2}\ length\ major\ axis)</math></center>
 
 
 
 
<center><math>b = minor\ radius\ (= \frac{1}{2}\ length\ minor\ axis)</math></center>
 
 
<center>[[File:Ellipse.png]]</center>
 
 
Where
 
<center><math>c^2=a^2-b^2</math></center>
 
 
 
 
For a parabola:
 
 
<center><math>4px = y^2</math></center>
 
 
where
 
 
p = distance from vertex to focus (or directrix)
 

Latest revision as of 04:14, 4 March 2017