Determining wire-phi correspondance: Difference between revisions
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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, [math]\displaystyle{ e=\sqrt{1-\frac{b^2}{a^2}} }[/math]

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)