Difference between revisions of "The Wires"

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<center><math>\underline{\textbf{Navigation}}</math>
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[[Points_of_Intersection|<math>\vartriangleleft </math>]]
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[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]]
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[[Right_Hand_Wall|<math>\vartriangleright </math>]]
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</center>
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We can parametrize the equations for the wires and wire midpoints to express the equation in vector form.  In the y'-x' plane the general equation follows the relationship:
 
We can parametrize the equations for the wires and wire midpoints to express the equation in vector form.  In the y'-x' plane the general equation follows the relationship:
  
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<center><math>x' = y\ sin\ 6^{\circ}+x_0</math></center>
 
<center><math>x' = y\ sin\ 6^{\circ}+x_0</math></center>
 +
 +
 
<center><math>y' = y\ cos\ 6^{\circ}</math></center>
 
<center><math>y' = y\ cos\ 6^{\circ}</math></center>
 
 
 
The parameterization has reduced two equations with two variables, to two equations which depend on one variable.  Working in the y-x plane, we will undergo a positive rotation,
 
The parameterization has reduced two equations with two variables, to two equations which depend on one variable.  Working in the y-x plane, we will undergo a positive rotation,
  
−
R(Subscript[\[Theta], yx])=(cos 6\[Degree] -sin 6\[Degree] 0
 
−
sin 6\[Degree] cos 6\[Degree] 0
 
−
0 0 1
 
  
−
)
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<center><math>R(\theta_{yx})=\begin{bmatrix}
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cos\ 6^{\circ} &-sin\ 6^{\circ} & 0 \\
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sin\ 6^{\circ} & cos\ 6^{\circ} &0 \\
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0 &0 & 1
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\end{bmatrix}</math></center>
  
  
−
(Components of
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<center><math>\begin{bmatrix}
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same vector
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Components\ of \\
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in new system
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same\ vector \\
−
 
+
in\ new\ system
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)=(Passive
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\end{bmatrix}
−
transformation
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=\begin{bmatrix}
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Passive \\
 +
transformation \\
 
matrix
 
matrix
 +
\end{bmatrix}\cdot
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\begin{bmatrix}
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Components\ of \\
 +
vector\ in \\
 +
original\ system
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\end{bmatrix}</math></center>
  
−
) . (Components of
 
−
vector in
 
−
original system
 
  
−
) (New
 
−
basis
 
−
vectors
 
  
−
)=(Active
 
−
transformation
 
−
matrix
 
  
−
) . (original
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<center><math>
−
basis
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\begin{bmatrix}
−
vectors
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x'' \\
−
 
+
y'' \\
−
)
 
−
 
−
 
−
(x''
 
−
y''
 
 
z''
 
z''
−
 
+
\end{bmatrix}=
−
)=(cos 6\[Degree] -sin 6\[Degree] 0
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\begin{bmatrix}
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sin 6\[Degree] cos 6\[Degree] 0
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cos\ 6^{\circ} &-sin\ 6^{\circ} & 0 \\
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0 0 1
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sin\ 6^{\circ} & cos\ 6^{\circ} &0 \\
−
 
+
0 &0 & 1
−
) . (x'
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\end{bmatrix}\cdot
−
y'
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\begin{bmatrix}
 +
x' \\
 +
y' \\
 
z'
 
z'
 +
\end{bmatrix}</math></center>
 +
  
−
) (x'
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<center><math>
−
y'
+
\begin{bmatrix}
−
z'
+
x'' \\
 +
y'' \\
 +
z''
 +
\end{bmatrix}=
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\begin{bmatrix}
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cos\ 6^{\circ} &-sin\ 6^{\circ} & 0 \\
 +
sin\ 6^{\circ} & cos\ 6^{\circ} &0 \\
 +
0 &0 & 1
 +
\end{bmatrix}\cdot
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\begin{bmatrix}
 +
y'\ sin\ 6^{\circ}+x_0 \\
 +
y'\ cos\ 6^{\circ} \\
 +
0
 +
\end{bmatrix}</math></center>
  
−
)=(cos 6\[Degree] sin 6\[Degree] 0
 
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-sin 6\[Degree] cos 6 \[Degree] 0
 
−
0 0 1
 
  
−
) . (x''
 
−
y''
 
−
z''
 
  
−
)
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<center><math>
−
+
\begin{bmatrix}
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(x''
+
x'' \\
−
y''
+
y'' \\
 
z''
 
z''
 +
\end{bmatrix}=
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\begin{bmatrix}
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-y'\ cos\ 6^{\circ}sin\ 6^{\circ}+x_0\ cos\ 6^{\circ} +y'\ cos\ 6^{\circ}sin\ 6^{\circ}\\
 +
y'\ cos^2 6^{\circ}+x_0\sin\ 6^{\circ}+y sin^2 6^{\circ} \\
 +
0
 +
\end{bmatrix}</math></center>
 +
  
−
)=(cos 6\[Degree] -sin 6\[Degree] 0
 
−
sin 6\[Degree] cos 6\[Degree] 0
 
−
0 0 1
 
  
−
) . ( y sin 6\[Degree]+Subscript[x, 0]
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<center><math>
−
y cos 6\[Degree]
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\begin{bmatrix}
 +
x'' \\
 +
y'' \\
 +
z''
 +
\end{bmatrix}=
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\begin{bmatrix}
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x_0\ cos\ 6^{\circ}\\
 +
y'\ +x_0\sin\ 6^{\circ} \\
 
0
 
0
 +
\end{bmatrix}</math></center>
  
−
)
 
−
 
−
(x''
 
−
y''
 
−
z''
 
  
−
)= (-y cos 6 \[Degree] sin 6 \[Degree]+Subscript[x, 0]cos 6 \[Degree] +y cos 6 \[Degree]sin 6 \[Degree]
+
This relationship shows us that x'' is a constant in this frame while y'' can have any value, which is the horizontal line with respect to the y axis as expected.
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y cos^2 6 \[Degree]+Subscript[x, 0]sin 6 \[Degree]+y sin^2 6 \[Degree]
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−
0
 
  
−
) (x'
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----
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y'
 
−
z'
 
  
−
)= (x'' cos 6\[Degree]+y " sin 6\[Degree]
 
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-x'' sin 6 \[Degree]+y " cos 6\[Degree]
 
−
0
 
  
−
)
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<center><math>\underline{\textbf{Navigation}}</math>
−
 
−
  (x''
 
−
y''
 
−
z''
 
  
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)= (Subscript[x, 0]cos 6 \[Degree]  
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[[Points_of_Intersection|<math>\vartriangleleft </math>]]
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y +Subscript[x, 0]sin 6 \[Degree]
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[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]]
−
0
+
[[Right_Hand_Wall|<math>\vartriangleright </math>]]
  
−
)
+
</center>

Latest revision as of 20:32, 15 May 2018

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

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


We can parametrize the equations for the wires and wire midpoints to express the equation in vector form. In the y'-x' plane the general equation follows the relationship:

[math]x'=y'\ tan\ 6^{\circ}+x_0[/math]

where [math]x_0[/math] is the point where the line crosses the x axis.

[math]y' \Rightarrow {y\ tan\ 6^{\circ}+x_0, y, 0}[/math]


In this form we can easily see that the components of x and y , in the y'-x' plane are

[math]x' = y\ sin\ 6^{\circ}+x_0[/math]


[math]y' = y\ cos\ 6^{\circ}[/math]

The parameterization has reduced two equations with two variables, to two equations which depend on one variable. Working in the y-x plane, we will undergo a positive rotation,


[math]R(\theta_{yx})=\begin{bmatrix} cos\ 6^{\circ} &-sin\ 6^{\circ} & 0 \\ sin\ 6^{\circ} & cos\ 6^{\circ} &0 \\ 0 &0 & 1 \end{bmatrix}[/math]


[math]\begin{bmatrix} Components\ of \\ same\ vector \\ in\ new\ system \end{bmatrix} =\begin{bmatrix} Passive \\ transformation \\ matrix \end{bmatrix}\cdot \begin{bmatrix} Components\ of \\ vector\ in \\ original\ system \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} 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} y'\ sin\ 6^{\circ}+x_0 \\ y'\ cos\ 6^{\circ} \\ 0 \end{bmatrix}[/math]


[math] \begin{bmatrix} x'' \\ y'' \\ z'' \end{bmatrix}= \begin{bmatrix} -y'\ cos\ 6^{\circ}sin\ 6^{\circ}+x_0\ cos\ 6^{\circ} +y'\ cos\ 6^{\circ}sin\ 6^{\circ}\\ y'\ cos^2 6^{\circ}+x_0\sin\ 6^{\circ}+y sin^2 6^{\circ} \\ 0 \end{bmatrix}[/math]


[math] \begin{bmatrix} x'' \\ y'' \\ z'' \end{bmatrix}= \begin{bmatrix} x_0\ cos\ 6^{\circ}\\ y'\ +x_0\sin\ 6^{\circ} \\ 0 \end{bmatrix}[/math]


This relationship shows us that x is a constant in this frame while y can have any value, which is the horizontal line with respect to the y axis as expected.



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

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

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