[[File:Plane_side_view.png|thumb|center|600px|alt=Cone of constant Theta for varying Phi|'''Figure 5:''' A cone of constant Theta with varying Phi.]]
Here we must take note of the fact that the center of the cone and the center of the ellipse do not coincide as can be shown visually by extending the cone length.
[[File:Ellipse_Cone_overlay_full.png|thumb|center|600px|alt=Cone of constant Theta for varying Phi|'''Figure 5:''' A cone of constant Theta with varying Phi.]]
Here we will make use of Dandelin spheres to help determine the ellipse components with reference to the cone created at a constant theta. The basic construction of using Dandelin spheres is that within the cone, there are two spheres which are tangent to the plane as well as tangent to the cone as shown below in the xz plane. The Dandelin method picks an arbitrary point,P, on the intersection of the cone and the ellipse's plane. Drawing a line from the cone vertex through the point P, the directrix lines are intersected as shown below at D1 and D2. From the geometry as shown in the xz plane, it is shown that the line segments F1P=D1P and F2P=D2P since they share a common line segment, all three angles, and both have a the respective Dandelin radius as parts of the triangles they form.
[[File:Dandelin_spheres2.png|thumb|center|600px|alt=Cone of constant Theta for varying Phi|'''Figure 5:''' A cone of constant Theta with varying Phi.]]
We can use the law of sines to find a portion of the ellipse
[[File:Law_of_sines_conic_section.png|thumb|center|600px|alt=Cone of constant Theta for varying Phi|'''Figure 5:''' A cone of constant Theta with varying Phi.]]
[[File:Law_of_sines_conic_section2.png|thumb|center|600px|alt=Cone of constant Theta for varying Phi|'''Figure 5:''' A cone of constant Theta with varying Phi.]]
Since these lines meet at the intersection of each other, and they both originate at the vertex positions for the ellipse, we can state that the semi major axis is:
<center><math>2a\equiv a_1+a_2</math></center>
[[File:Dandelin_spheres.png|thumb|center|600px|alt=Cone of constant Theta for varying Phi|'''Figure 5:''' A cone of constant Theta with varying Phi.]]
Using the fact that half the semi-major axis will give use the center of the ellipse
| style="background: gray" | <math>a=\frac{2.53sin(\theta)}{2}\left (csc(65^{\circ}-\theta)+csc(115^{\circ}-\theta) \right )</math>
|}
[[File:Directrix_delta_z.png|thumb|center|600px|alt=Cone of constant Theta for varying Phi|'''Figure 5:''' A cone of constant Theta with varying Phi.]]
| style="background: gray" | <math>\Delta a=\frac{2.53sin(\theta)}{2}\left (csc(65^{\circ}-\theta)-csc(115^{\circ}-\theta) \right )</math>
|}
<center><math>\Delta z=\Delta a \cdot cos(65^{\circ})</math></center>
<center><math>\Delta x=\Delta a \cdot sin(65^{\circ})</math></center>
Since the vertex position is at (0,0,0) this shows the center of the ellipse to be at
<center><math>c_{ellipse} \equiv (-\Delta a sin(65^{\circ}), 0,z+\Delta z)</math></center>
<center><math>c_{ellipse} \equiv (-\Delta a sin(65^{\circ}), 0,2.53+\Delta z)</math></center>
The sum of the distance from both focal points to a point on the surface of the ellipse is a constant
[[File:Ellipse_sum_rule1.png|thumb|center|600px|alt=Cone of constant Theta for varying Phi|'''Figure 5:''' A cone of constant Theta with varying Phi.]]
<center><math>PF1+PF2\equiv 2a</math></center>
A property of an ellipse is that the distance from the center of the ellipse to a focal point is given by
[[File:Ellipse_sum_rule2.png|thumb|center|600px|alt=Cone of constant Theta for varying Phi|'''Figure 5:''' A cone of constant Theta with varying Phi.]]
Additionally, using the fact that the distance from the center to a focal point is ae
[[File:Dandelin_spheres_zoom.png|thumb|center|600px|alt=Cone of constant Theta for varying Phi|'''Figure 5:''' A cone of constant Theta with varying Phi.]]
<center><math>R_{Lower\ Dandelin}=(ae-\Delta a) tan(65^{\circ})</math></center>
<center><math>r_{D1}=R_{Lower\ Dandelin}sec(\theta)=(ae-\Delta a) tan(65^{\circ})cos(\theta)</math></center>
By symmetry,
<center><math>R_{Upper\ Dandelin}=(ae+\Delta a) tan(65^{\circ})</math></center>
<center><math>r_{D2}=R_{Lower\ Dandelin}sec(\theta)=(ae+\Delta a) tan(65^{\circ})cos(\theta)</math></center>
The D1 and D2 components can be obtained by finding the x and y components of the two directrix circles, which will be a function of their respective radii and the angle phi around the z axis.
Using these two points, which will be at the same angle phi around the z axis, a line can be created
[math]\displaystyle{ \vec d=r_{D2}cos(\phi)-r_{D1}cos(\phi)\ \hat x+r_{D2}sin(\phi)-r_{D1}sin(\phi)\ \hat y+r_{D2}cot(\theta)-r_{D1}cot(\theta)\ \hat z }[/math]
Since this line is composed of the difference in the cartesian components of the two points it acts as a direction vector pointing from one point to the next. We can use the fact that a line in space is simply the intersection of two planes which in this situation results in the direction vector. Finding the parametric form equation of this line
Since this condition implies that the 3 equations listed above for all conditions, we can solve for the [math]\displaystyle{ \phi=0 }[/math] situation which must then hold for all other instances of [math]\displaystyle{ \phi }[/math]
We can substitute for z to find the point of intersection for the plane and cone at a given [math]\displaystyle{ \theta }[/math] and [math]\displaystyle{ \phi }[/math]
We can use the geometry of the sector to impose limits on the allowed x and y for this plane. The angle which the two sides of the triangle that make up the sector is given as [math]\displaystyle{ 59^{\circ} }[/math]. Each focii will have a ray that creates a circle around the point that is equal to the length found from the D1P and D2P lines. These circles will overlap at a common y component, which can be used to determine if the point is within the detector limits.
Using the physical constraints of the distance the interior walls of the side walls intersect each other from the beam line, as well as the angle the lowest guard wire at the plane midpoint.
where b' is the location on the y' axis where the two interior side walls of the sector intersect.
[math]\displaystyle{ b'=\frac{0.08298\ m}{sin(65^{\circ})}=.09156\ m }[/math]
The quantity [math]\displaystyle{ x_0^' }[/math] is the corresponding x' distance from the beam line and detector plane intersection.
[math]\displaystyle{ x_0^'=\frac{sin(4.694^{\circ})2.53\ m}{sin(110.306^{\circ})}=.0872(2.53)=.22070/ m }[/math]
Since the interior sides of the sector make an angle of [math]\displaystyle{ 59^{\circ} }[/math] in the plane of the wires, this can be represented by equations
[math]\displaystyle{ x'=\pm cot(29.5^{\circ})y^'+.09156\ m }[/math]
Using the properties of an ellipse, we can work in the coordinates of the ellipse, in the plane of the ellipse, where the vertex is located at the crossing of the ellipse and the cone. Since the Dandelin construction was used, we know that
Test for [math]\displaystyle{ \theta=20 }[/math] and [math]\displaystyle{ \phi=1 }[/math]
All previous quantities where calculated for [math]\displaystyle{ \theta=20^{\circ} }[/math] and do not depend on the angle [math]\displaystyle{ \phi }[/math]. The quantities that do change
The two possible answers denote shifting to the left on right on the y axis. We take the direction of positive and negative to be the same as the sign convention for the angle phi starting on the x axis and shifting positive clockwise. A shift of 1 degree in phi at theta equal to 20 degrees only results in a small change in the x and y. This changes depending on the angles.
Function for the change in x' in the detector frame for change in [math]\displaystyle{ \phi }[/math] and constant [math]\displaystyle{ \theta }[/math] in the lab frame
Function for the wire number in the detector frame for change in [math]\displaystyle{ \phi }[/math] and constant [math]\displaystyle{ \theta }[/math] in the lab frame
Using the expression for wire number n in terms of [math]\displaystyle{ \theta }[/math] for the detector mid-plane where [math]\displaystyle{ \phi=0 }[/math]:
[math]\displaystyle{ n = \frac{-957.412}{\tan(\theta)+2.14437}+430.626 }[/math]
We can use the inverse of this function to find the neighboring wire's corresponding angle theta
[math]\displaystyle{ r_{D1}=R_{Lower\ Dandelin}cos(\theta)=(ae-\Delta a) tan(65^{\circ})cos(\theta)\qquad \qquad r_{D2}=R_{Lower\ Dandelin}cos(\theta)=(ae+\Delta a) tan(65^{\circ})cos(\theta) }[/math]
We can take this point to be the x axis intercept and use the fact that each wire is titled by 6 degrees to the horizontal in the plane of the detector to create an equation
where the initial wire and x' position at the given theta is represented by [math]\displaystyle{ n_0 }[/math]
This equation can be solved for a hypothetical wire 0, which will allow the wire number to be the multiplicative factor for the change from the starting position.
