Determining wire-phi correspondance: Difference between revisions

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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>
<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_side_view.png]]</center>


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<center>[[File:Projection_Rear_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, and α is the slant of the cone.
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.png]]</center>
<center>[[File:Conic_section.png.png]]</center>

Revision as of 05:18, 3 January 2017


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.


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.



Following the rules of conic sections we know that the eccentricity of the conic is given by:

[math]\displaystyle{ \e=\frac{\sin [\beta]}{\sin[\alpha]} }[/math]


Where β is the angle of the plane, and α is the slant of the cone.

If the conic is an circle, e=0

If the conic is an parabola, e=1

If the conic is an ellipse, [math]\displaystyle{ e=\sqrt{1-\frac{b^2}{a^2}} }[/math]

File:Conic section.png.png