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
  
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<center><math>r\mapsto {-y\ cot\ 29.5^{\circ}+0.09156,y,0}</math></center>
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<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>
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<center><math>t\mapsto \{t\ cos\ 29.5^{\circ}+0.09156,-t\ sin\ 29.5^{\circ},0 \}</math></center>
  
  
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t\ cos\ 29.5^{\circ}+0.09156  \\
 
t\ cos\ 29.5^{\circ}+0.09156  \\
 
-t\ sin\ 29.5^{\circ} \\
 
-t\ sin\ 29.5^{\circ} \\
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z'
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0
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\end{bmatrix}</math></center>
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<center><math>
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\begin{bmatrix}
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x'' \\
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y'' \\
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z''
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\end{bmatrix}=
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\begin{bmatrix}
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0.09156\ cos\ 6^{\circ}+t\ cos\ 6^{\circ}cos\ 29.5^{\circ}+t\ sin\ 6^{\circ}sin\ 29.5^{\circ}  \\
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-t\ cos\ 6^{\circ}sin\ 29.5^{\circ}+0.09156\ sin\ 6^{\circ}+t\ cos\ 29.5^{\circ}sin\ 6^{\circ} \\
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0
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\end{bmatrix}</math></center>
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 +
 
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<center><math>
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\begin{bmatrix}
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x'' \\
 +
y'' \\
 +
z''
 +
\end{bmatrix}=
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\begin{bmatrix}
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0.09156\ cos\ 6^{\circ}+t\ (cos\ 6^{\circ}cos\ 29.5^{\circ}+ sin\ 6^{\circ}sin\ 29.5^{\circ})  \\
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0.09156\  sin\ 6^{\circ}-t\ (cos\ 6^{\circ}sin\ 29.5^{\circ}-sin\ 6^{\circ} cos\ 29.5^{\circ}) \\
 +
0
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\end{bmatrix}</math></center>
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 +
 
 +
<center><math>
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\begin{bmatrix}
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x'' \\
 +
y'' \\
 +
z''
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\end{bmatrix}=
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\begin{bmatrix}
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0.09156\ cos\ 6^{\circ}+t\ cos\ (6^{\circ} -29.5^{\circ}) \\
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0.09156\  sin\ 6^{\circ}+t\ sin\ (6^{\circ}-29.5^{\circ}) \\
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0
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\end{bmatrix}</math></center>
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 +
 
 +
<center><math>
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\begin{bmatrix}
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x'' \\
 +
y'' \\
 +
z''
 +
\end{bmatrix}=
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\begin{bmatrix}
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0.09156\ cos\ 6^{\circ}+t\ cos\ (-23.5^{\circ}) \\
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0.09156\  sin\ 6^{\circ}+t\ sin\ (-23.5^{\circ}) \\
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0
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\end{bmatrix}</math></center>
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 +
 
 +
<center><math>
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\begin{bmatrix}
 +
x'' \\
 +
y'' \\
 +
z''
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\end{bmatrix}=
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\begin{bmatrix}
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0.09156\ cos\ 6^{\circ}+t\ cos\ 23.5^{\circ} \\
 +
0.09156\  sin\ 6^{\circ}-t\ sin\ 23.5^{\circ} \\
 +
0
 
\end{bmatrix}</math></center>
 
\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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[[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"