Determining wire-phi correspondance

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DC stereo.png


DC Geomentry Sideview.png


xdist - distance between the line of intersection of the two end-plate planes and the target position =8.298 cm th_min=4.694

dist2tgt - distance from the target to the first guard wire plane along the normal of said plane


Hex.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, which in our case is 25 degrees for the sectors with respect to the beam line (axis of rotation of the cone).

This leaves α is the slant of the cone, which is the angle theta that the particle must be traveling with respect to the beamline.

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 (h,k):

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

where

[math]a = major\ radius\ (= \frac{1}{2}\ length\ major\ axis)[/math]


[math]b = minor\ radius\ (= \frac{1}{2}\ length\ minor\ axis)[/math]
Ellipse.png

Where

[math]c^2=a^2-b^2[/math]


For a parabola:

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

where

p = distance from vertex to focus (or directrix)