|
|
| (One intermediate revision by the same user not shown) |
| Line 1: |
Line 1: |
| | | | |
| − |
| |
| − |
| |
| − |
| |
| − |
| |
| − |
| |
| − | ===Function for the wire number in the detector frame for change in <math>\phi</math> and constant <math>\theta</math> in the lab frame===
| |
| − |
| |
| − | Using the expression for wire number n in terms of <math>\theta</math> for the detector mid-plane where <math>\phi=0</math>:
| |
| − |
| |
| − | <center><math>n = \frac{-957.412}{\tan(\theta)+2.14437}+430.626</math></center>
| |
| − |
| |
| − | We can use the inverse of this function to find the neighboring wire's corresponding angle theta
| |
| − |
| |
| − | <center><math>\theta\equiv 4.49876 +0.293001 n+0.000679074 n^2-3.57132\times 10^{-6} n^3</math></center>
| |
| − |
| |
| − |
| |
| − |
| |
| − | <center><math>\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></center>
| |
| − |
| |
| − |
| |
| − | We also know what the x' function must follow dependent on phi in the detector plane
| |
| − |
| |
| − | <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{(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>
| |
| − |
| |
| − |
| |
| − | {| 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>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></center>
| |
| − |
| |
| − | 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
| |
| − |
| |
| − | <center><math>x_{wire\ n}'=tan(6^{\circ})y'+x_{n_0}</math></center>
| |
| − |
| |
| − | where the initial wire and x' position at the given theta is represented by <math>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.
| |
| − |
| |
| − |
| |
| − | <center><math>\frac{x_{n=1}}{sin 4.79^{\circ}}=\frac{2.52934}{sin 110.21^{\circ}} \Rightarrow x_{n=1}=.2252</math></center>
| |
| − |
| |
| − |
| |
| − | Since each wire is separated by .01337 meters
| |
| − |
| |
| − |
| |
| − |
| |
| − | <center><math>.2252-.01337=.2117=x_{n=0}</math></center>
| |
| − |
| |
| − |
| |
| − | Each wire becomes an equation of the form,
| |
| − |
| |
| − | <center><math>x_{wire\ n}'=tan(6^{\circ})y'+.2117+.01337\ n</math></center>
| |
| − |
| |
| − | This agrees with CED simulation
| |
| − |
| |
| − |
| |
| − |
| |
| − |
| |
| − | <center>[[File:DC_geom.png]]</center>
| |
| − |
| |
| − | ==Setting up Mathematica for DC Theta-Phi Isotropic Cone==
| |
| − |
| |
| − | <center>[[File:FirstPass.pdf]]</center>
| |
| − |
| |
| − |
| |
| − |
| |
| − | <center>[[File:SecondPass.pdf]]</center>
| |