Difference between revisions of "Left Hand Wall"

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<center><math>\underline{\textbf{Navigation}}</math>
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 +
[[Right_Hand_Wall|<math>\vartriangleleft </math>]]
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[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]]
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[[The_Ellipse|<math>\vartriangleright </math>]]
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</center>
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<center><math>x=-y\ cot\ 29.5^{\circ}+0.09156</math></center>
 
<center><math>x=-y\ cot\ 29.5^{\circ}+0.09156</math></center>
  
 
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>
  
  
 
where the negative sign is applied to the sine function by the even odd relationships of cosine and sine, i.e. ( sin(-t)=-sin(t), cos(-t)=cos(t)) and the fact that the y'' component is in the 4th quadrant.   
 
where the negative sign is applied to the sine function by the even odd relationships of cosine and sine, i.e. ( sin(-t)=-sin(t), cos(-t)=cos(t)) and the fact that the y'' component is in the 4th quadrant.   
  
−
(x''
+
<center><math>
−
y''
+
\begin{bmatrix}
 +
x'' \\
 +
y'' \\
 
z''
 
z''
−
 
+
\end{bmatrix}=
−
)=(cos 6\[Degree] -sin 6\[Degree] 0
+
\begin{bmatrix}
−
sin 6\[Degree] cos 6\[Degree] 0
+
cos\ 6^{\circ} & -sin\ 6^{\circ} & 0 \\
−
0 0 1
+
sin\ 6^{\circ} & cos\ 6^{\circ} & 0 \\
−
 
+
0 & 0 & 1
−
) . (x'
+
\end{bmatrix} \cdot
−
y'
+
\begin{bmatrix}
 +
x' \\
 +
y' \\
 
z'
 
z'
 +
\end{bmatrix}</math></center>
 +
  
−
)
+
<center><math>
−
 
+
\begin{bmatrix}
−
(x''
+
x'' \\
−
y''
+
y'' \\
 
z''
 
z''
−
 
+
\end{bmatrix}=
−
)=(cos 6\[Degree] -sin 6\[Degree] 0
+
\begin{bmatrix}
−
sin 6\[Degree] cos 6\[Degree] 0
+
cos\ 6^{\circ} & -sin\ 6^{\circ} & 0 \\
−
0 0 1
+
sin\ 6^{\circ} & cos\ 6^{\circ} & 0 \\
−
 
+
0 & 0 & 1
−
) . (t cos (29.5\[Degree])+0.09156
+
\end{bmatrix} \cdot
−
-t sin (29.5\[Degree])
+
\begin{bmatrix}
 +
t\ cos\ 29.5^{\circ}+0.09156 \\
 +
-t\ sin\ 29.5^{\circ} \\
 
0
 
0
 +
\end{bmatrix}</math></center>
  
−
)
 
  
−
(x''
+
<center><math>
−
y''
+
\begin{bmatrix}
 +
x'' \\
 +
y'' \\
 
z''
 
z''
−
 
+
\end{bmatrix}=
−
)= (0.09156cos 6 \[Degree]+t cos 6 \[Degree]cos (29.5\[Degree])+t sin 6 \[Degree]sin (29.5\[Degree])
+
\begin{bmatrix}
−
-t cos 6 \[Degree]sin (29.5\[Degree])+0.09156 sin 6 \[Degree]+t cos (29.5\[Degree])sin 6 \[Degree]
+
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''
+
<center><math>
−
y''
+
\begin{bmatrix}
 +
x'' \\
 +
y'' \\
 
z''
 
z''
−
 
+
\end{bmatrix}=
−
)= (0.09156cos 6 \[Degree]+t (cos 6 \[Degree]cos(29.5\[Degree])+ sin 6 \[Degree]sin (29.5\[Degree]))
+
\begin{bmatrix}
−
0.09156  sin 6 \[Degree]-t (cos 6 \[Degree]sin (29.5\[Degree])-sin 6 \[Degree] cos (29.5\[Degree]))
+
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\ (cos\ 6^{\circ}sin\ 29.5^{\circ}-sin\ 6^{\circ} cos\ 29.5^{\circ}) \\
 
0
 
0
 +
\end{bmatrix}</math></center>
  
−
)
 
  
−
(x''
+
<center><math>
−
y''
+
\begin{bmatrix}
 +
x'' \\
 +
y'' \\
 
z''
 
z''
−
 
+
\end{bmatrix}=
−
)= (0.09156cos 6 \[Degree]+t cos (6\[Degree] -29.5\[Degree])
+
\begin{bmatrix}
−
0.09156  sin 6 \[Degree]+t sin (6 \[Degree]-29.5\[Degree])
+
0.09156\ cos\ 6^{\circ}+t\ cos\ (6^{\circ} -29.5^{\circ}) \\
 +
0.09156\ sin\ 6^{\circ}+t\ sin\ (6^{\circ}-29.5^{\circ}) \\
 
0
 
0
 +
\end{bmatrix}</math></center>
  
−
)
 
  
−
(x''
+
<center><math>
−
y''
+
\begin{bmatrix}
 +
x'' \\
 +
y'' \\
 
z''
 
z''
−
 
+
\end{bmatrix}=
−
)= (0.09156cos 6 \[Degree]+t cos (-23.5\[Degree])
+
\begin{bmatrix}
−
0.09156  sin 6 \[Degree]+t sin (-23.5\[Degree])
+
0.09156\ cos\ 6^{\circ}+t\ cos\ (-23.5^{\circ}) \\
 +
0.09156\ sin\ 6^{\circ}+t\ sin\ (-23.5^{\circ}) \\
 
0
 
0
 +
\end{bmatrix}</math></center>
  
−
)
 
  
−
(x''
+
<center><math>
−
y''
+
\begin{bmatrix}
 +
x'' \\
 +
y'' \\
 
z''
 
z''
−
 
+
\end{bmatrix}=
−
)= (0.09156cos 6 \[Degree]+t cos (23.5\[Degree])
+
\begin{bmatrix}
−
0.09156  sin 6 \[Degree]-t sin (-23.5\[Degree])
+
0.09156\ cos\ 6^{\circ}+t\ cos\ 23.5^{\circ} \\
 +
0.09156\ sin\ 6^{\circ}-t\ sin\ 23.5^{\circ} \\
 
0
 
0
−
 
+
\end{bmatrix}</math></center>
−
)
 
  
  
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<center><math>x''=-2.299843\ y''+.113069</math></center>
 
<center><math>x''=-2.299843\ y''+.113069</math></center>
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 +
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[[File:leftwall.png]]
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----
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 +
 +
<center><math>\underline{\textbf{Navigation}}</math>
 +
 +
[[Right_Hand_Wall|<math>\vartriangleleft </math>]]
 +
[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]]
 +
[[The_Ellipse|<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]


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


where the negative sign is applied to the sine function by the even odd relationships of cosine and sine, i.e. ( sin(-t)=-sin(t), cos(-t)=cos(t)) and the fact that the y component is in the 4th quadrant.

[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\ (cos\ 6^{\circ}sin\ 29.5^{\circ}-sin\ 6^{\circ} cos\ 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} -29.5^{\circ}) \\ 0.09156\ sin\ 6^{\circ}+t\ sin\ (6^{\circ}-29.5^{\circ}) \\ 0 \end{bmatrix}[/math]


[math] \begin{bmatrix} x'' \\ y'' \\ z'' \end{bmatrix}= \begin{bmatrix} 0.09156\ cos\ 6^{\circ}+t\ cos\ (-23.5^{\circ}) \\ 0.09156\ sin\ 6^{\circ}+t\ sin\ (-23.5^{\circ}) \\ 0 \end{bmatrix}[/math]


[math] \begin{bmatrix} x'' \\ y'' \\ z'' \end{bmatrix}= \begin{bmatrix} 0.09156\ cos\ 6^{\circ}+t\ cos\ 23.5^{\circ} \\ 0.09156\ sin\ 6^{\circ}-t\ sin\ 23.5^{\circ} \\ 0 \end{bmatrix}[/math]


Using the equation for y we can solve for t

[math]y''=0.09156\ sin\ 6^{\circ}-t\ sin\ 23.5^{\circ} \Rightarrow t=\frac{-(y''-0.09156 sin 6^{\circ})}{sin 23.5^{\circ}}[/math]


Substituting this into the expression for x

[math]x''=0.09156\ cos\ 6^{\circ}+t\ cos\ 23.5^{\circ}[/math]


[math]x''=0.09156\ cos\ 6 ^{\circ}+\frac{-(y''-0.09156 sin 6 ^{\circ})}{sin 23.5^{\circ}}(cos 23.5^{\circ})[/math]


[math]x''=0.091058+\frac{y''-.0095706 }{-0.398749} (.917060)[/math]


[math]x''=0.091058+(y''-.0095706 ) (-2.299843)[/math]


[math]x''=-2.299843\ y''+.022011+.091058[/math]


[math]x''=-2.299843\ y''+.113069[/math]


Leftwall.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=Left_Hand_Wall&oldid=122817"