Difference between revisions of "Determining wire-phi correspondance"
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If the conic is an ellipse, <math>e=\sqrt{1-\frac{b^2}{a^2}}</math> | If the conic is an ellipse, <math>e=\sqrt{1-\frac{b^2}{a^2}}</math> | ||
− | <center>[[File:Conic_section | + | <center>[[File:Conic_section.png]]</center> |
+ | |||
+ | |||
+ | For ellipses centered at (j,k): | ||
+ | |||
+ | <center><math>\frac{(x-j)^2}{a^2}+\frac{(y-k)^2}{b^2}=1</math></center> | ||
+ | |||
+ | where | ||
+ | |||
+ | a = major radius (= 1/2 length major axis) | ||
+ | b = minor radius (= 1/2 length minor axis) | ||
+ | |||
+ | |||
+ | For a parabola: | ||
+ | |||
+ | <center><math>4px = y^2</math></center> | ||
+ | |||
+ | where | ||
+ | |||
+ | p = distance from vertex to focus (or directrix) |
Revision as of 05:33, 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:
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,
For ellipses centered at (j,k):
where
a = major radius (= 1/2 length major axis) b = minor radius (= 1/2 length minor axis)
For a parabola:
where
p = distance from vertex to focus (or directrix)