Difference between revisions of "The Ellipse"

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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.  
+
<center><math>\underline{\textbf{Navigation}}</math>
 +
 
 +
[[Left_Hand_Wall|<math>\vartriangleleft </math>]]
 +
[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]]
 +
[[Plotting_Different_Frames|<math>\vartriangleright </math>]]
 +
 
 +
</center>
 +
 
 +
 
 +
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
 
An ellipse centered at the origin can be expressed in the form
Line 19: Line 28:
  
  
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
+
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
  
 
 
 
 
Line 41: Line 50:
 
 
 
   
 
   
  (x''
+
<center><math>\begin{bmatrix}
y''
+
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}
 +
\Delta a \\
 +
0 \\
 +
0
 +
\end{bmatrix}</math></center>
 +
 
  
)=(cos (6\[Degree]) -sin (6\[Degree]) 0
 
sin (6\[Degree]) cos (6\[Degree]) 0
 
0 0 1
 
  
) . (\[CapitalDelta]a
+
<center><math>\begin{bmatrix}
0
+
x'' \\
 +
y'' \\
 +
z''
 +
\end{bmatrix}=
 +
\begin{bmatrix}
 +
\Delta a\ cos\ 6^{\circ} \\
 +
\Delta a\ sin\ 6^{\circ} \\
 
0
 
0
 +
\end{bmatrix}</math></center>
  
)
 
 
   
 
   
  (x''
+
<center><math>(x'', y'', z'')_{center} = (\Delta a\ cos\ 6^{\circ} , \Delta a\ sin\ 6^{\circ} , 0 )= (h'', k'', 0)  </math></center>
y''
 
z''
 
  
)=(\[CapitalDelta]a cos (6\[Degree])
 
\[CapitalDelta]a sin (6\[Degree])
 
0
 
  
)
 
 
 
  -> (x ", y'', z''Subscript[), Center] = (\[CapitalDelta]a cos 6\[Degree], \[CapitalDelta]a sin 6\[Degree], 0) = (h'', k'', 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 6\[Degree] counterclockwise from the x'' axis, with the intersection points having a uniform spacing in the ellipse parameter.
 
  
 +
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.
  
(x''
 
y''
 
z''
 
  
)=(cos 6\[Degree] sin 6\[Degree] 0
 
-sin 6\[Degree] cos 6\[Degree] 0
 
0 0 1
 
  
) . (x'
+
<center><math>\begin{bmatrix}
y'
+
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'
 
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]
+
\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
+
<center><math>\frac{(x''+h'')^2}{a^2}+\frac{(y''+k'')^2}{b^2}=1</math></center>
 
 
 +
 +
<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>
 +
 
 
((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
+
<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>
+
 
(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
+
 
 +
<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>
 +
 
 +
 
 +
<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>
 +
 
 +
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.
 +
 
 +
 
 +
----
 +
 
  
(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
+
<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]