Difference between revisions of "Determining wire-phi correspondance"
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![DC stereo.png](/./images/d/d4/DC_stereo.png)
![PhiCone.png](/./images/5/50/PhiCone.png)
![Projection side view.png](/./images/8/8c/Projection_side_view.png)
![Projection Rear view.png](/./images/4/48/Projection_Rear_view.png)
File:Conic section.png.png
Line 17: | Line 17: | ||
Following the rules of conic sections we know that the eccentricity of the conic is given by: | Following the rules of conic sections we know that the eccentricity of the conic is given by: | ||
− | <center><math> | + | <center><math>e=\frac{\sin [\beta]}{\sin[\alpha]}</math></center> |
Revision as of 05:19, 3 January 2017
![DC stereo.png](/./images/d/d4/DC_stereo.png)
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.
![PhiCone.png](/./images/5/50/PhiCone.png)
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.
![Projection side view.png](/./images/8/8c/Projection_side_view.png)
![Projection Rear view.png](/./images/4/48/Projection_Rear_view.png)
Following the rules of conic sections we know that the eccentricity of the conic is given by:
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,