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.png.png]]</center>
+
<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

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


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


Projection Rear view.png


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

[math]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]e=\sqrt{1-\frac{b^2}{a^2}}[/math]

Conic section.png


For ellipses centered at (j,k):

[math]\frac{(x-j)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/math]

where

a = major radius (= 1/2 length major axis) b = minor radius (= 1/2 length minor axis)


For a parabola:

[math]4px = y^2[/math]

where

p = distance from vertex to focus (or directrix)