Determining wire-phi correspondance

Test for [math]\displaystyle{ \theta=20 }[/math] and [math]\displaystyle{ \phi=0 }[/math]

Substituting in the values found earlier for the case of [math]\displaystyle{ \theta=20^{\circ} }[/math] and [math]\displaystyle{ \phi=0 }[/math]

[math]\displaystyle{ a=\frac{a_1+a_2}{2}=1.0459 }[/math]

[math]\displaystyle{ e\equiv \frac{sin(25^{\circ})}{cos(\theta)}= \frac{sin(25^{\circ})}{cos(20^{\circ})}=.4497 }[/math]

[math]\displaystyle{ x_1^'=\frac{r_2^{'2}-r_1^{'2}}{4ae}-ae=\frac{r_2^{'2}-r_1^{'2}}{4(1.0459)(.4497)}-(1.0459)(.4497) }[/math]

Since

[math]\displaystyle{ \lVert \overrightarrow{D1P} \rVert \equiv \lVert \overrightarrow{F1P} \rVert \qquad \qquad \lVert \overrightarrow{D2P} \rVert \equiv \lVert \overrightarrow{F2P} \rVert }[/math]

[math]\displaystyle{ D1P \approx .576\ \text{m}\qquad D2P \approx 1.516\ \text{m} }[/math]

The [math]\displaystyle{ x_1^' }[/math] distance from focal point 1 is:

[math]\displaystyle{ x_1^'=\frac{1.516^2-.576^2}{4(1.0459)(.4497)}-(1.0459)(.4497)=\frac{1.97}{1.88}-.4703=.576\ \text{m}=r_1 }[/math]

This is the radius from focal point 1, which is to be expected since the y component is equal to zero for [math]\displaystyle{ \phi=0 }[/math]

The focii are located at

[math]\displaystyle{ F1\equiv (-\Delta a+ae,0)=(-.1775+1.0459(.4497),0)=(.292,0)\ \text{m}\qquad F2\equiv (-\Delta a-ae,0)=(-.1775-1.0450(.4497),0)=(-6.478,0)\ \text{m} }[/math]

This implies that with respect to the origin, x', we find

[math]\displaystyle{ x'=x_1'+\Delta a+ae=.576+.292=.868\ \text{m} }[/math]

This is verified with CED

Since the x' dimension is the hypotenuse in a right triangle of 65 degrees

[math]\displaystyle{ x'=\frac{x_{lab}}{sin 65^{ \circ}}=.868=x' }[/math]

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

[math]\displaystyle{ x_{D1}=r_{D1}\ cos(\phi)=.5901cos(1^{\circ}))=.5900\text {m}\qquad y_{D1}=r_{D1}cos(\phi)=.5901 sin(1^{\circ}))=.0103\qquad z_{D1}=r_{D1} cot(\theta)=.5901cot(20)=1.6212\ \text{m} }[/math]

[math]\displaystyle{ x_{D2}=r_{D2} cos(\phi)=1.3055cos(1^{\circ}))=1.3053\text {m}\qquad y_{D2}=r_{D2} sin(\phi)=1.3055sin(1^{\circ}))=.0228\qquad z_{D2}=r_{D2} cot(\theta)=1.3055cot(20)=3.5868\ \text{m} }[/math]

[math]\displaystyle{ x_P=\frac{2.53cos(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}=\frac{2.53cos(1^{\circ}))}{(cot(20^{\circ})+cos(1^{\circ}))cot(65^{\circ})}=0.7869 }[/math]

[math]\displaystyle{ y_P=\frac{2.53sin(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}=\frac{2.53sin(1^{\circ}))}{(cot(20^{\circ})+cos(1^{\circ}))cot(65^{\circ})}=.0137 }[/math]

[math]\displaystyle{ z_P=\frac{2.53cot(\theta)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}=\frac{2.53cot(20^{\circ})}{(cot(20^{\circ})+cos(1^{\circ}))cot(65^{\circ})}=2.1624 }[/math]

[math]\displaystyle{ D2P=\sqrt{(x_{D2}-x_P)^2+(y_{D2}-y_P)^2+(z_{D2}-z_P)^2}=\sqrt{(1.3053-0.7869)^2+(.0228-.0137)^2+(3.5868-2.1624)^2}=\sqrt{(.5184)^2+(.0091)^2+(1.4244)^2}=1.51582872713\ \text{m} }[/math]

[math]\displaystyle{ D1P=\sqrt{(x_P-x_{D1})^2+(y_P-y_{D1})^2+(z_P-z_{D1})^2}=\sqrt{(0.7871-.5900)^2+(.0137-.0103)^2+(2.1624-1.6212)^2}=\sqrt{(.1971)^2+(.0034)^2+(.5412)^2}=.575983862621\ \text{m} }[/math]

[math]\displaystyle{ x_1^'=\frac{r_2^{'2}-r_1^{'2}}{4ae}-ae=\frac{1.5158^{'2}-.5758^{'2}}{4(1.0459)(.4497)}-(1.0459)(.4497)=.575\ \text{m} }[/math]

Using the pythagorean theorem

[math]\displaystyle{ y'=\sqrt{.576^2-.575^2}=.03\ \text{m} }[/math]

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

[math]\displaystyle{ D2P=\sqrt{(x_{D2}-x_P)^2+(y_{D2}-y_P)^2+(z_{D2}-z_P)^2} }[/math]

[math]\displaystyle{ D1P=\sqrt{(x_P-x_{D1})^2+(y_P-y_{D1})^2+(z_P-z_{D1})^2} }[/math]

[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_{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{ 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{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]

[math]\displaystyle{ 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]

[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]

Expressing this as functions of [math]\displaystyle{ \phi }[/math] and non-differentiable constants

[math]\displaystyle{ 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]

Differentiating with respect to [math]\displaystyle{ \phi }[/math]

[math]\displaystyle{ x_{D1}=r_{D1} cos(\phi)\Rightarrow \dot x_{D1}=-r_{D1} sin(\phi) }[/math]

[math]\displaystyle{ y_{D1}=r_{D1}sin(\phi)\Rightarrow \dot y_{D1}=r_{D1}cos(\phi) }[/math]

[math]\displaystyle{ x_{D2}=r_{D2} cos(\phi)\Rightarrow \dot x_{D2}=-r_{D2} sin(\phi) }[/math]

[math]\displaystyle{ y_{D2}=r_{D2}sin(\phi)\Rightarrow \dot y_{D2}=r_{D2}cos(\phi) }[/math]

[math]\displaystyle{ 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]

[math]\displaystyle{ 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]

[math]\displaystyle{ 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]

[math]\displaystyle{ \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]

[math]\displaystyle{ \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]

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