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Revision as of 16:22, 1 May 2017
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]\displaystyle{ x'=y'\ tan\ 6^{\circ}+x_0 }[/math]
where [math]\displaystyle{ x_0 }[/math] is the point where the line crosses the x axis.
[math]\displaystyle{ 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]\displaystyle{ x' = y\ sin\ 6^{\circ}+x_0 }[/math]
[math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ \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]\displaystyle{
\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]\displaystyle{
\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]\displaystyle{
\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]\displaystyle{
\begin{bmatrix}
x'' \\
y'' \\
z''
\end{bmatrix}=
\begin{bmatrix}
x_0\ cos\ 6^{\circ}\\
y'\ cos^2 6^{\circ}+x_0\sin\ 6^{\circ}+y sin^2 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.