Wire Number Function

Function for the wire number in the detector frame for change in $\phi$ and constant $\theta$ in the lab frame

Using the expression for wire number n in terms of $\theta$ for the detector mid-plane where $\phi=0$:

$n = \frac{-957.412}{\tan(\theta)+2.14437}+430.626$

We can use the inverse of this function to find the neighboring wire's corresponding angle theta

$\theta\equiv 4.49876 +0.293001 n+0.000679074 n^2-3.57132\times 10^{-6} n^3$

$\theta(n \pm 1)\equiv 4.49876 +0.293001 (n \pm 1)+0.000679074 (n \pm 1)^2-3.57132\times 10^{-6} (n \pm 1)^3$

We also know what the x' function must follow dependent on phi in the detector plane

$x_1^'=\frac{((x_{D2}-x_P)^2+(y_{D2}-y_P)^2+(z_{D2}-z_P)^2)-((x_P-x_{D1})^2+(y_P-y_{D1})^2+(z_P-z_{D1})^2)}{4ae}-ae$

$x_1^'=\frac{(r_{D2}^2)-(r_{D1}^2)+cot^2(\theta)(r_{D2}^2-r_{D1}^2)-2x_P(x_{D2}-x_{D1})-2y_P(y_{D2}-y_{D1})-2z_P(z_{D2}-z_{D1})}{4ae}-ae$

$x_{D1}=r_{D1}\ cos(\phi)\qquad y_{D1}=r_{D1}cos(\phi)\qquad z_{D1}=r_{D1} cot(\theta)$

$x_{D2}=r_{D2} cos(\phi)\qquad y_{D2}=r_{D2} sin(\phi)\qquad z_{D2}=r_{D2} cot(\theta)$

$x_P=\frac{2.53cos(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}$

$y_P=\frac{2.53sin(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}$

$z_P=\frac{2.53cot(\theta)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}$

$r_{D1}=R_{Lower\ Dandelin}cos(\theta)=(ae-\Delta a) tan(65^{\circ})cos(\theta)\qquad \qquad r_{D2}=R_{Lower\ Dandelin}cos(\theta)=(ae+\Delta a) tan(65^{\circ})cos(\theta)$

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

$x_{wire\ n}'=tan(6^{\circ})y'+x_{n_0}$

where the initial wire and x' position at the given theta is represented by $n_0$

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.

$\frac{x_{n=1}}{sin 5.20988^{\circ}}=\frac{2.52934}{sin 109.79^{\circ}} \Rightarrow x_{n=1}=.2441$

Since each wire is separated by .01337 meters

$.2441-.01337=.2307=x_{n=0}$

Each wire becomes an equation of the form,

$x_{wire\ n}'=tan(6^{\circ})y'+.2307+.01337\ n$

This agrees with CED simulation