Difference between revisions of "Right Hand Wall"

From New IAC Wiki
Jump to navigation Jump to search
Line 1: Line 1:
 
 
This same process can be applied to the side walls for the detector.  For the sidewalls, we have approximated them as lines following the equation
 
This same process can be applied to the side walls for the detector.  For the sidewalls, we have approximated them as lines following the equation
  
Line 58: Line 57:
 
\end{bmatrix}=
 
\end{bmatrix}=
 
\begin{bmatrix}
 
\begin{bmatrix}
0.09156\ cos\ 6^{\circ}+\t cos\ 6 ^{\circ}cos\ 29.5^{\circ}-t\ sin\ 6 ^{\circ}sin\ 29.5^{\circ} \\
+
0.09156\ cos\ 6^{\circ}+t\ cos\ 6 ^{\circ}cos\ 29.5^{\circ}-t\ sin\ 6 ^{\circ}sin\ 29.5^{\circ} \\
 
t\ cos\ 6 ^{\circ}sin\ 29.5^{\circ}+0.09156\ sin\ 6^{\circ}+t\ cos\ 29.5^{\circ}sin\ 6^{\circ} \\
 
t\ cos\ 6 ^{\circ}sin\ 29.5^{\circ}+0.09156\ sin\ 6^{\circ}+t\ cos\ 29.5^{\circ}sin\ 6^{\circ} \\
 
0
 
0

Revision as of 03:28, 28 April 2017

This same process can be applied to the side walls for the detector. For the sidewalls, we have approximated them as lines following the equation

[math]x=cot\ 29.5^{\circ}\ y + 0.09156[/math]

Parameterizing this

[math]r \mapsto {y\ cot\ 29.5^{\circ} + 0.09156, y, 0}[/math]


[math]t \mapsto {t\ cos\ 29.5^{\circ} + 0.09156, t\ sin\ 29.5^{\circ} , 0}[/math]


[math]\begin{bmatrix} x'' \\ y'' \\ z'' \end{bmatrix}= \begin{bmatrix} cos\ 6^{\circ} & -sin\ 6^{\circ} & 0 \\ sin\ 6^{\circ} & cos\ 6^{\circ}& 0 \\ 0 & 0 & 1 \end{bmatrix}\cdot \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix}[/math]



[math]\begin{bmatrix} x'' \\ y'' \\ z'' \end{bmatrix}= \begin{bmatrix} cos\ 6^{\circ} & -sin\ 6^{\circ} & 0 \\ sin\ 6^{\circ} & cos\ 6^{\circ}& 0 \\ 0 & 0 & 1 \end{bmatrix}\cdot \begin{bmatrix} t\ cos\ 29.5^{\circ}+0.09156 \\ t sin 29.5^{\circ}\\ 0 \end{bmatrix}[/math]



[math]\begin{bmatrix} x'' \\ y'' \\ z'' \end{bmatrix}= \begin{bmatrix} 0.09156\ cos\ 6^{\circ}+t\ cos\ 6 ^{\circ}cos\ 29.5^{\circ}-t\ sin\ 6 ^{\circ}sin\ 29.5^{\circ} \\ t\ cos\ 6 ^{\circ}sin\ 29.5^{\circ}+0.09156\ sin\ 6^{\circ}+t\ cos\ 29.5^{\circ}sin\ 6^{\circ} \\ 0 \end{bmatrix}[/math]

(x y z

)= (0.09156cos 6 \[Degree]+t (cos 6 \[Degree]cos 29.5\[Degree]- sin 6 \[Degree]sin 29.5\[Degree]) 0.09156 sin 6 \[Degree]+t (sin 6 \[Degree] cos 29.5\[Degree]+cos 6 \[Degree]sin 29.5\[Degree]) 0

)

Using the equation for y we can solve for t

[math]y''=0.09156\ sin\ 6^{\circ}+t (sin 6 \[Degree] cos 29.5^{\circ}+cos 6 ^{\circ}sin 29.5^{\circ}) -\gt t=(y''-0.09156 sin 6 ^{\circ})/(sin 6^{\circ} cos 29.5^{\circ}+cos 6^{\circ}sin 29.5^{\circ})[/math]

Substituting this into the expression for x

[math]x''=0.09156cos 6^{\circ}+t (cos 6^{\circ}cos 29.5^{\circ}- sin 6^{\circ} sin 29.5^{\circ})=0.09156cos 6 ^{\circ}+\frac{y''-0.09156 sin 6^{\circ}}{sin 6^{\circ} cos 29.5^{\circ}+cos 6^{\circ}sin 29.5^{\circ}} (cos 6^{\circ}cos 29.5^{\circ}- sin 6^{\circ} sin 29.5^{\circ})[/math]


[math]x''=0.09156\ cos\ 6^{\circ}+\frac{y''-0.09156\ sin\ 6^{\circ}}{sin\ 6^{\circ} cos\ 29.5^{\circ}+cos\ 6 ^{\circ}sin\ 29.5^{\circ}} (cos\ 6 ^{\circ}cos\ 29.5^{\circ}- sin\ 6^{\circ}sin\ 29.5^{\circ})[/math]


[math]x''=(0.994522)0.09156+\frac{y''-0.09156 (0.104528) }{0.0909769+.489726} (0.865588- 0.051472)[/math]


[math]x''=(0.091058)+\frac{y''-.0095706 }{0.580703} (.814116)[/math]


[math]x''=(0.091058)+(y''-.0095706 ) (1.401949)[/math]


[math]x''=1.401949\ y''-.013417+.091058[/math]


[math]x''=1.401949\ y''+.077641[/math]


rightRotated = 
  ContourPlot[x2 == 1.401949 y + 0.077641, {y, -1, 1}, {x2, 0, 1.8}, 
   Frame -> {True, True, False, False}, 
       PlotLabel -> 
    "Right side limit of DC as a function of X and Y", 
   FrameLabel -> {"y (meters)", "x (meters)"}, 
   ContourStyle -> Black, 
       PlotLegends -> Automatic];