Difference between revisions of "Right Hand Wall"

From New IAC Wiki
Jump to navigation Jump to search
 
(14 intermediate revisions by the same user not shown)
Line 1: Line 1:
 +
<center><math>\underline{\textbf{Navigation}}</math>
 +
 +
[[The_Wires|<math>\vartriangleleft </math>]]
 +
[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]]
 +
[[Left_Hand_Wall|<math>\vartriangleright </math>]]
 +
 +
</center>
  
  
Line 7: Line 14:
 
Parameterizing this  
 
Parameterizing this  
  
−
<center><math>r \mapsto {y\ cot 29.5^{\circ} + 0.09156, y, 0}</math></center>
+
<center><math>r \mapsto {y\ cot\ 29.5^{\circ} + 0.09156, y, 0}</math></center>
 +
 
 +
 
 +
<center><math>t \mapsto {t\ cos\ 29.5^{\circ} + 0.09156, t\ sin\ 29.5^{\circ} , 0}</math></center>
  
  
−
<center><math>t \mapsto {t\ cos 29.5^{\circ} + 0.09156, t\ sin\ 29.5^{\circ} , 0}</math></center>
 
  
 
<center><math>\begin{bmatrix}
 
<center><math>\begin{bmatrix}
Line 29: Line 38:
  
  
−
(x''
 
−
y''
 
−
z''
 
  
−
)=(cos 6\[Degree] -sin 6\[Degree] 0
 
−
sin 6\[Degree] cos 6\[Degree] 0
 
−
0 0 1
 
  
−
) . (t cos 29.5\[Degree]+0.09156
+
<center><math>\begin{bmatrix}
−
t sin 29.5\[Degree]
+
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
 
0
 +
\end{bmatrix}</math></center>
 +
  
−
)
 
  
−
(x''
 
−
y''
 
−
z''
 
  
−
)= (0.09156cos 6 \[Degree]+t cos 6 \[Degree]cos 29.5\[Degree]-t sin 6 \[Degree]sin 29.5\[Degree]
+
 
−
t cos 6 \[Degree]sin 29.5\[Degree]+0.09156 sin 6 \[Degree]+t cos 29.5\[Degree]sin 6 \[Degree]
+
<center><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
 
0
 +
\end{bmatrix}</math></center>
  
−
)
 
  
−
(x''
 
−
y''
 
−
z''
 
  
−
)= (0.09156cos 6 \[Degree]+t (cos 6 \[Degree]cos 29.5\[Degree]- sin 6 \[Degree]sin 29.5\[Degree])
+
<center><math>\begin{bmatrix}
−
0.09156  sin 6 \[Degree]+t (sin 6 \[Degree] cos 29.5\[Degree]+cos 6 \[Degree]sin 29.5\[Degree])
+
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
 
0
−
 
+
\end{bmatrix}</math></center>
−
)
 
  
 
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''
  
−
<center><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></center>
+
<center><math>x''=0.09156\ cos\ 6^{\circ}+t\ (cos\ 6^{\circ}cos\ 29.5^{\circ}- sin\ 6^{\circ} sin\ 29.5^{\circ})</math></center>
 +
 
 +
 
 +
<center><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></center>
  
  
Line 90: Line 114:
  
  
−
<pre>
+
[[File:rwall.png]]
−
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)"},
+
<center><math>\underline{\textbf{Navigation}}</math>
−
  ContourStyle -> Black,
+
 
−
      PlotLegends -> Automatic];
+
[[The_Wires|<math>\vartriangleleft </math>]]
−
</pre>
+
[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]]
 +
[[Left_Hand_Wall|<math>\vartriangleright </math>]]
 +
 
 +
</center>

Latest revision as of 20:33, 15 May 2018

[math]\underline{\textbf{Navigation}}[/math]

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]


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''=0.09156\ cos\ 6^{\circ}+t\ (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''=0.09156\ 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]


Rwall.png



[math]\underline{\textbf{Navigation}}[/math]

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]

Retrieved from "https://wiki.iac.isu.edu/index.php?title=Right_Hand_Wall&oldid=122816"