The Wires

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

[math]\displaystyle{ \vartriangleleft }[/math] [math]\displaystyle{ \triangle }[/math] [math]\displaystyle{ \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]\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'\ +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]\displaystyle{ \underline{\textbf{Navigation}} }[/math]

[math]\displaystyle{ \vartriangleleft }[/math] [math]\displaystyle{ \triangle }[/math] [math]\displaystyle{ \vartriangleright }[/math]