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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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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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.
 
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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----
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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>

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]