The Wires: Difference between revisions

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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


)
<center><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></center>




(Components of
\begin{bmatrix}
same vector
Components of \\
same vector \\
in new system
in new system
 
\end{bmatrix}<
)=(Passive
)=\begin{bmatrix}
transformation
Passive \\
transformation \\
matrix
matrix
 
\end{bmatrix}\cdot
) . (Components of
\begin{bmatrix}
vector in
Components of \\
vector in \\
original system
original system
\end{bmatrix}</math></center>


) (New
basis
vectors


)=(Active
transformation
matrix


) . (original
basis
vectors


)
<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
) . (x'
\end{bmatrix}\cdot
y'
\begin{bmatrix}
x' \\
y' \\
z'
z'
\end{bmatrix}</math></center>


) (x'
<center><math>
y'
\begin{bmatrix}
z'
x'' \\
 
y'' \\
)=(cos 6\[Degree] sin 6\[Degree] 0
-sin 6\[Degree] cos 6 \[Degree] 0
0 0 1
 
) . (x''
y''
z''
 
)
(x''
y''
z''
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></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]
y cos 6\[Degree]
0


)
<center><math>
\begin{bmatrix}
(x''
x'' \\
y''
y'' \\
z''
z''
 
\end{bmatrix}=
)= (-y cos 6 \[Degree] sin 6 \[Degree]+Subscript[x, 0]cos 6 \[Degree] +y cos 6 \[Degree]sin 6 \[Degree]
\begin{bmatrix}
y cos^2 6 \[Degree]+Subscript[x, 0]sin 6 \[Degree]+y sin^2 6 \[Degree]
-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
0
\end{bmatrix}</math></center>


) (x'
y'
z'


)= (x'' cos 6\[Degree]+y " sin 6\[Degree]
<center><math>
-x'' sin 6 \[Degree]+y " cos 6\[Degree]
\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
0
\end{bmatrix}</math></center>


)
  (x''
y''
z''
)= (Subscript[x, 0]cos 6 \[Degree]
y +Subscript[x, 0]sin 6 \[Degree]
0


)
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.

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]


\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.