Difference between revisions of "The Ellipse"

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<center><math>\underline{\textbf{Navigation}}</math>
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[[Left_Hand_Wall|<math>\vartriangleleft </math>]]
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[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]]
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[[Plotting_Different_Frames|<math>\vartriangleright </math>]]
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</center>
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Viewing the conic section <math>\phi</math> maps out on the DC sector plane, we know that it follows an elliptical path centered on it's x axis.  Performing a passive rotation on points in the DC section plane does not physically change the position in space, i.e.  passive rotations only give the components in a new coordinate system.  Once such a rotation has been performed, the equation describing these points must be done within that plane.  
 
Viewing the conic section <math>\phi</math> maps out on the DC sector plane, we know that it follows an elliptical path centered on it's x axis.  Performing a passive rotation on points in the DC section plane does not physically change the position in space, i.e.  passive rotations only give the components in a new coordinate system.  Once such a rotation has been performed, the equation describing these points must be done within that plane.  
  
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\end{bmatrix}</math></center>
 
\end{bmatrix}</math></center>
 
   
 
   
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<center><math>\begin{bmatrix}
 
<center><math>\begin{bmatrix}
 
x'' \\
 
x'' \\
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0
 
0
 
\end{bmatrix}</math></center>
 
\end{bmatrix}</math></center>
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<center><math>\Rightarrow ( x ", y'', z'' ) =( \Delta a\ cos\ 6^{\circ}, \Delta a\ sin\ 6^{\circ}, 0 ) = (h'', k'', 0  )</math></center>
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<center><math>(x'', y'', z'')_{center} = (\Delta a\ cos\ 6^{\circ} , \Delta a\ sin\ 6^{\circ} , 0 )= (h'', k'', 0) </math></center>
  
Performing an active rotation, we will rotate the equation for an ellipse in the frame of the DC to the frame of the wires .  In the frame of the DC, the ellipse is centered on the x' axis, with the intersection points not having a uniform spacing in the ellipse parameter.  In the frame of the wires, the ellipse is tilted <math>6^{\circ}</math> counterclockwise from the x'' axis, with the intersection points having a uniform spacing in the ellipse parameter.
 
  
  
(x''
 
y''
 
z''
 
  
)=(cos 6\[Degree] sin 6\[Degree] 0
+
Performing an active rotation, we will rotate the equation for an ellipse in the frame of the DC to the frame of the wires .  In the frame of the DC, the ellipse is centered on the x' axis, with the intersection points not having a uniform spacing in the ellipse parameter.  In the frame of the wires, the ellipse is tilted <math>6^{\circ}</math> counterclockwise from the x'' axis, with the intersection points having a uniform spacing in the x'' component.
-sin 6\[Degree] cos 6\[Degree] 0
+
 
0 0 1
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) . (x'
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<center><math>\begin{bmatrix}
y'
+
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 \\
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0 & 0 & 1
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\end{bmatrix}\cdot
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\begin{bmatrix}
 +
x' \\
 +
y' \\
 
z'
 
z'
 
+
\end{bmatrix}</math></center>
)
 
 
 
 
 
(x''
+
<center><math>\begin{bmatrix}
y''
+
x'' \\
 +
y'' \\
 
z''
 
z''
 
+
\end{bmatrix}=
)=(x'cos 6\[Degree]+y' sin 6\[Degree]
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\begin{bmatrix}
-x'sin 6\[Degree]+y'cos 6\[Degree]
+
x'\ cos\ 6^{\circ}+y'\ sin\ 6^{\circ} \\
 +
-x'\ sin\ 6^{\circ}+y'\ cos\ 6^{\circ} \\
 
0
 
0
 
+
\end{bmatrix}</math></center>
)
 
 
 
Substituting this into the equation for an ellipse in the frame of the wires,
+
Substituting this into the equation for an ellipse in the frame of the wires, that is parallel to the frame of reference x and y axis,
 
 
(x''+h'')^2/a^2+(y''+k'')^2/b^2=1
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<center><math>\frac{(x''+h'')^2}{a^2}+\frac{(y''+k'')^2}{b^2}=1</math></center>
 
 
 +
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<center><math>\frac{((x'+\Delta a\ cos\ 6^{\circ})cos\ 6^{\circ}+(y'+\Delta a\ sin\ 6^{\circ})sin\ 6^{\circ})^2}{a^2}+\frac{(-(x'+\Delta a\ cos\ 6^{\circ})sin\ 6^{\circ}+(y'+\Delta a\ sin\ 6^{\circ})cos\ 6^{\circ})^2}{b^2}=1</math></center>
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((x'+\[CapitalDelta]a Cos[6 \[Degree]])Cos[6 \[Degree]]+(y'+\[CapitalDelta]a Sin[6 \[Degree]])Sin[6 \[Degree]])^2/a^2+(-(x'+\[CapitalDelta]a Cos[6 \[Degree]])Sin[6 \[Degree]]+(y'+\[CapitalDelta]a Sin[6 \[Degree]])Cos[6 \[Degree]])^2/b^2=1
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<center><math>\frac{(x'\ cos\ 6^{\circ}+\Delta a\ cos^2 6^{\circ}+y'\ sin\ 6^{\circ}+\Delta a\ sin^2 6^{\circ})^2}{a^2}+\frac{(-x'\ sin\ 6^{\circ}-\Delta a\ cos\ 6^{\circ}sin\ 6^{\circ}+y'\ cos\ 6^{\circ}+\Delta a\ sin\ 6^{\circ}cos\ 6^{\circ})^2}{b^2}=1</math></center>
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(x'Cos[6 \[Degree]]+\[CapitalDelta]a (Cos^2)[6 \[Degree]]+y'Sin[6 \[Degree]]+\[CapitalDelta]a (Sin^2)[6 \[Degree]])^2/a^2+(-x'Sin[6 \[Degree]]-\[CapitalDelta]a Cos[6 \[Degree]]Sin[6 \[Degree]]+y'Cos[6 \[Degree]]+\[CapitalDelta]a Sin[6 \[Degree]]Cos[6 \[Degree]])^2/b^2=1
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 +
<center><math>\frac{(x'\ cos\ 6^{\circ}+y'\ sin\ 6^{\circ}+\Delta a (cos^2 6^{\circ}+ sin^2 6^{\circ}) )^2}{a^2}+\frac{(-x'\ sin\ 6^{\circ}+y'\ cos\ 6^{\circ}+\Delta a (cos\ 6^{\circ}sin\ 6^{\circ}-cos\ 6^{\circ}sin\ 6^{\circ}))^2}{b^2}=1</math></center>
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 +
 
 +
<center><math>\frac{(x'\ cos\ 6^{\circ}+y'\ sin\ 6^{\circ}+\Delta a )^2}{a^2}+\frac{(-x'\ sin\ 6^{\circ}+y''\ cos\ 6 ^{\circ})^2}{b^2}=1</math></center>
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 +
This shows the preservation of the vector length of the offset from the origin on the x axis in the frame of the DC section.  The equation shows the ellipse rotated so that the vertex and co-vertex are no longer parallel to the frame of reference x and y axis.
 +
 
 +
 
 +
----
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(x'Cos[6 \[Degree]]+y'Sin[6 \[Degree]]+\[CapitalDelta]a (Cos^2)[6 \[Degree]]+\[CapitalDelta]a (Sin^2)[6 \[Degree]])^2/a^2+(-x'Sin[6 \[Degree]]+y'Cos[6 \[Degree]]+\[CapitalDelta]a Cos[6 \[Degree]]Sin[6 \[Degree]]-\[CapitalDelta]a Cos[6 \[Degree]]Sin[6 \[Degree]])^2/b^2=1
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<center><math>\underline{\textbf{Navigation}}</math>
  
(x'Cos[6 \[Degree]]+y'Sin[6 \[Degree]]+\[CapitalDelta]a ((Cos^2)[6 \[Degree]]+ (Sin^2)[6 \[Degree]]))^2/a^2+(-x'Sin[6 \[Degree]]+y'Cos[6 \[Degree]]+\[CapitalDelta]a (Cos[6 \[Degree]]Sin[6 \[Degree]]-Cos[6 \[Degree]]Sin[6 \[Degree]]))^2/b^2=1
+
[[Left_Hand_Wall|<math>\vartriangleleft </math>]]
 +
[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]]
 +
[[Plotting_Different_Frames|<math>\vartriangleright </math>]]
  
(x'Cos[6 \[Degree]]+y'Sin[6 \[Degree]]+\[CapitalDelta]a )^2/a^2+(-x'Sin[6 \[Degree]]+y'Cos[6 \[Degree]])^2/b^2=1
+
</center>

Latest revision as of 20:33, 15 May 2018

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

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


Viewing the conic section [math]\phi[/math] maps out on the DC sector plane, we know that it follows an elliptical path centered on it's x axis. Performing a passive rotation on points in the DC section plane does not physically change the position in space, i.e. passive rotations only give the components in a new coordinate system. Once such a rotation has been performed, the equation describing these points must be done within that plane.

An ellipse centered at the origin can be expressed in the form

[math]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/math]

For an ellipse not centered on the origin, but instead the point (h',k'), this expression becomes


[math]\frac{(x+h')^2}{a^2}+\frac{(y+k')^2}{b^2}=1[/math]


In the plane of the DC sector, this equation becomes


[math]\frac{(x'+\Delta a)^2}{a^2}+\frac{(y')^2}{b^2}=1[/math]

where the center of the ellipse is found at [math]\{\Delta a, 0\}[/math].


Switching to the frame of the wires, the ellipse is still centered at [math]\{\Delta a,0\}[/math] in the DC sector, with the semi-major axis lying on the x' axis. For a rotation in the y-x plane, this corresponds to a positive angle [math]\theta[/math], with the rotation matrix [math]R(\theta_{yx})[/math]. In the frame of the wires, this center point falls at


[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} \Delta a \\ 0 \\ 0 \end{bmatrix}[/math]


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


[math](x'', y'', z'')_{center} = (\Delta a\ cos\ 6^{\circ} , \Delta a\ sin\ 6^{\circ} , 0 )= (h'', k'', 0) [/math]



Performing an active rotation, we will rotate the equation for an ellipse in the frame of the DC to the frame of the wires . In the frame of the DC, the ellipse is centered on the x' axis, with the intersection points not having a uniform spacing in the ellipse parameter. In the frame of the wires, the ellipse is tilted [math]6^{\circ}[/math] counterclockwise from the x axis, with the intersection points having a uniform spacing in the x component.


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

Substituting this into the equation for an ellipse in the frame of the wires, that is parallel to the frame of reference x and y axis,

[math]\frac{(x''+h'')^2}{a^2}+\frac{(y''+k'')^2}{b^2}=1[/math]


[math]\frac{((x'+\Delta a\ cos\ 6^{\circ})cos\ 6^{\circ}+(y'+\Delta a\ sin\ 6^{\circ})sin\ 6^{\circ})^2}{a^2}+\frac{(-(x'+\Delta a\ cos\ 6^{\circ})sin\ 6^{\circ}+(y'+\Delta a\ sin\ 6^{\circ})cos\ 6^{\circ})^2}{b^2}=1[/math]


[math]\frac{(x'\ cos\ 6^{\circ}+\Delta a\ cos^2 6^{\circ}+y'\ sin\ 6^{\circ}+\Delta a\ sin^2 6^{\circ})^2}{a^2}+\frac{(-x'\ sin\ 6^{\circ}-\Delta a\ cos\ 6^{\circ}sin\ 6^{\circ}+y'\ cos\ 6^{\circ}+\Delta a\ sin\ 6^{\circ}cos\ 6^{\circ})^2}{b^2}=1[/math]


[math]\frac{(x'\ cos\ 6^{\circ}+y'\ sin\ 6^{\circ}+\Delta a (cos^2 6^{\circ}+ sin^2 6^{\circ}) )^2}{a^2}+\frac{(-x'\ sin\ 6^{\circ}+y'\ cos\ 6^{\circ}+\Delta a (cos\ 6^{\circ}sin\ 6^{\circ}-cos\ 6^{\circ}sin\ 6^{\circ}))^2}{b^2}=1[/math]


[math]\frac{(x'\ cos\ 6^{\circ}+y'\ sin\ 6^{\circ}+\Delta a )^2}{a^2}+\frac{(-x'\ sin\ 6^{\circ}+y''\ cos\ 6 ^{\circ})^2}{b^2}=1[/math]

This shows the preservation of the vector length of the offset from the origin on the x axis in the frame of the DC section. The equation shows the ellipse rotated so that the vertex and co-vertex are no longer parallel to the frame of reference x and y axis.




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

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