Difference between revisions of "Right Hand Wall"

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\end{bmatrix}</math></center>
 
\end{bmatrix}</math></center>
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<center><math>\begin{bmatrix}
 
<center><math>\begin{bmatrix}
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Using the equation for y'' we can solve for t
 
Using the equation for y'' we can solve for t
  
<center><math>y''=0.09156\  sin\ 6^{\circ}+t (sin 6 \[Degree] cos 29.5^{\circ}+cos 6 ^{\circ}sin 29.5^{\circ}) -> t=(y''-0.09156  sin 6 ^{\circ})/(sin 6^{\circ} cos 29.5^{\circ}+cos 6^{\circ}sin 29.5^{\circ})</math></center>
+
<center><math>y''=0.09156\  sin\ 6^{\circ}+t (sin\ 6^{\circ} cos\ 29.5^{\circ}+cos\ 6 ^{\circ}sin\ 29.5^{\circ}) \Rightarrow t=\frac{y''-0.09156\ sin\ 6 ^{\circ}}{sin\ 6^{\circ} cos\ 29.5^{\circ}+cos\ 6^{\circ}sin\ 29.5^{\circ}}</math></center>
  
 
Substituting this into the expression for x''
 
Substituting this into the expression for x''

Revision as of 03:34, 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]


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

Using the equation for y we can solve for t

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