Determining wire-phi correspondance: Difference between revisions

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=Function for the change in x' in the detector frame for change in <math>\phi</math> and constant <math>\theta</math> in the lab frame=
<center><math>D2P=\sqrt{(x_{D2}-x_P)^2+(y_{D2}-y_P)^2+(z_{D2}-z_P)^2}</math></center>
<center><math>D1P=\sqrt{(x_P-x_{D1})^2+(y_P-y_{D1})^2+(z_P-z_{D1})^2}</math></center>
<center><math>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</math></center>
{| class="wikitable" align="center"
| style="background: gray"      | <math>x_{D1}=r_{D1}\ cos(\phi)\qquad y_{D1}=r_{D1}cos(\phi)\qquad z_{D1}=r_{D1} cot(\theta)</math>
|}
{| class="wikitable" align="center"
| style="background: gray"      | <math>x_{D2}=r_{D2} cos(\phi)\qquad y_{D2}=r_{D2} sin(\phi)\qquad z_{D2}=r_{D2} cot(\theta)</math>
|}
{| class="wikitable" align="center"
| style="background: gray"      | <math>x_P=\frac{2.53cos(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}</math>
|}
{| class="wikitable" align="center"
| style="background: gray"      | <math>y_P=\frac{2.53sin(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}</math>
|}
{| class="wikitable" align="center"
| style="background: gray"      | <math>z_P=\frac{2.53cot(\theta)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}</math>
|}
<center><math>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</math></center>
<center><math>x_1^'=\frac{x_{D2}^2-2x_Px_{D2}+x_P^2+y_{D2}^2-2y_Py_{D2}+y_P^2+z_{D2}^2-2z_Pz_{D2}+z_P^2-x_P^2+2x_Px_{D1}-x_{D1}^2-y_P^2+2y_Py_{D1}-y_{D1}^2-z_P^2+2z_Pz_{D1}-z_{D1}^2}{4ae}-ae</math></center>
<center><math>x_1^'=\frac{(x_{D2}^2+y_{D2}^2)-(x_{D1}^2+y_{D1}^2)+z_{D2}^2-z_{D1}^2-2x_P(x_{D2}-x_{D1})-2y_P(y_{D2}-y_{D1})-2z_P(z_{D2}-z_{D1})}{4ae}-ae</math></center>
<center><math>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</math></center>
Expressing this as functions of <math>\phi</math> and non-differentiable constants
<center><math>x_1^'=\frac{c_1+c_2-2x_P(\phi)x_{D2}(\phi)+2x_P(\phi)x_{D1}(\phi)-2y_P(\phi)y_{D2}(\phi)+2y_P(\phi)y_{D1}(\phi)-2z_P(\phi)c_3}{4c_4}-c_4</math></center>
Differentiating with respect to <math>\phi</math>
{| class="wikitable" align="center"
| style="background: gray"      | <math>x_{D1}=r_{D1} cos(\phi)\Rightarrow \dot x_{D1}=-r_{D1} sin(\phi)</math>
|}
{| class="wikitable" align="center"
| style="background: gray"      | <math>y_{D1}=r_{D1}sin(\phi)\Rightarrow \dot y_{D1}=r_{D1}cos(\phi)</math>
|}
{| class="wikitable" align="center"
| style="background: gray"      | <math>x_{D2}=r_{D2} cos(\phi)\Rightarrow \dot x_{D2}=-r_{D2} sin(\phi)</math>
|}
{| class="wikitable" align="center"
| style="background: gray"      | <math>y_{D2}=r_{D2}sin(\phi)\Rightarrow \dot y_{D2}=r_{D2}cos(\phi)</math>
|}
{| class="wikitable" align="center"
| style="background: gray"      | <math>x_P=\frac{2.52934271645cos(\phi)}{cot(\theta)+cos(\phi)cot(65^{\circ})}\Rightarrow \dot x_P=\frac{-2.52934271645cot(\theta)sin(\phi)}{(cos(\phi)cot(65^{\circ}+cot(\theta))^2}</math>
|}
{| class="wikitable" align="center"
| style="background: gray"      | <math>y_P=\frac{2.52934271645sin(\phi)}{cot(\theta)+cos(\phi)cot(65^{\circ})}\Rightarrow \dot y_P=\frac{-1.7206+2.52934271645 cos(\phi) cot(\theta)}{(cos(\phi) cot(65^{\circ}) + cot(\theta))^2}</math>
|}
{| class="wikitable" align="center"
| style="background: gray"      | <math>z_P=\frac{2.52934271645cot(\theta)}{cot(\theta)+cos(\phi)cot(65^{\circ})}\Rightarrow \dot z_P=\frac{-1.7206 cot(\theta)sin(\phi))}{(cos(\phi) cot(65) + cot(\theta))^2}</math>
|}
<center><math>\frac{dx_1^1}{d\phi}=\frac{-2}{4c_4}\frac{d}{d\phi}(x_P(\phi)x_{D2}(\phi))+\frac{2}{4c_4}\frac{d}{d\phi}(x_P(\phi)x_{D1}(\phi))-\frac{2}{4c_4}\frac{d}{d\phi}(y_P(\phi)y_{D2}(\phi))+\frac{2}{4c_4}\frac{d}{d\phi}(y_P(\phi)y_{D1}(\phi))-\frac{2c_3}{4c_4}\frac{d}{d\phi}z_P(\phi)</math></center>
<center><math>\frac{dx_1^1}{d\phi}=\frac{-2}{4c_4} \left ( (\dot x_P(\phi)x_{D2}(\phi)+x_P(\phi)\dot x_{D2}(\phi))-(\dot x_P(\phi)x_{D1}(\phi)+x_P(\phi)\dot x_{D1}(\phi))+(\dot y_P(\phi)y_{D2}(\phi)+y_P(\phi)\dot y_{D2}(\phi))-(\dot y_P(\phi)y_{D1}(\phi)+y_P(\phi)\dot y_{D1}(\phi))+c_3\dot z_P(\phi) \right )</math></center>


===Function for the wire number in the detector frame for change in <math>\phi</math> and constant <math>\theta</math> in the lab frame===
===Function for the wire number in the detector frame for change in <math>\phi</math> and constant <math>\theta</math> in the lab frame===

Revision as of 03:49, 4 March 2017




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{ \theta\equiv 4.49876 +0.293001 n+0.000679074 n^2-3.57132\times 10^{-6} n^3 }[/math]


[math]\displaystyle{ \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 }[/math]


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

[math]\displaystyle{ 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 }[/math]


[math]\displaystyle{ 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 }[/math]


[math]\displaystyle{ x_{D1}=r_{D1}\ cos(\phi)\qquad y_{D1}=r_{D1}cos(\phi)\qquad z_{D1}=r_{D1} cot(\theta) }[/math]


[math]\displaystyle{ x_{D2}=r_{D2} cos(\phi)\qquad y_{D2}=r_{D2} sin(\phi)\qquad z_{D2}=r_{D2} cot(\theta) }[/math]


[math]\displaystyle{ x_P=\frac{2.53cos(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})} }[/math]


[math]\displaystyle{ y_P=\frac{2.53sin(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})} }[/math]


[math]\displaystyle{ z_P=\frac{2.53cot(\theta)}{(cot(\theta)+cos(\phi)cot(65^{\circ})} }[/math]
[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

[math]\displaystyle{ x_{wire\ n}'=tan(6^{\circ})y'+x_{n_0} }[/math]

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.


[math]\displaystyle{ \frac{x_{n=1}}{sin 4.79^{\circ}}=\frac{2.52934}{sin 110.21^{\circ}} \Rightarrow x_{n=1}=.2252 }[/math]


Since each wire is separated by .01337 meters


[math]\displaystyle{ .2252-.01337=.2117=x_{n=0} }[/math]


Each wire becomes an equation of the form,

[math]\displaystyle{ x_{wire\ n}'=tan(6^{\circ})y'+.2117+.01337\ n }[/math]

This agrees with CED simulation



Setting up Mathematica for DC Theta-Phi Isotropic Cone

File:FirstPass.pdf


File:SecondPass.pdf