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>
 
 
Substituting this into the equation for an ellipse in the frame of the wires,
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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,
 
 
 
<center><math>\frac{(x''+h'')^2}{a^2}+\frac{(y''+k'')^2}{b^2}=1</math></center>
 
<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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<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>
 
<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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<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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<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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----
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<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>]]
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</center>

Latest revision as of 20:33, 15 May 2018

Navigation_


Viewing the conic section ϕ 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

x2a2+y2b2=1

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


(x+h)2a2+(y+k)2b2=1


In the plane of the DC sector, this equation becomes


(x+Δa)2a2+(y)2b2=1

where the center of the ellipse is found at {Δa,0}.


Switching to the frame of the wires, the ellipse is still centered at {Δa,0} 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 θ, with the rotation matrix R(θyx). In the frame of the wires, this center point falls at


[xyz]=[cos 6sin 60sin 6cos 60001][xyz]



[xyz]=[cos 6sin 60sin 6cos 60001][Δa00]


[xyz]=[Δa cos 6Δa sin 60]


(x,y,z)center=(Δa cos 6,Δa sin 6,0)=(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 counterclockwise from the x axis, with the intersection points having a uniform spacing in the x component.


[xyz]=[cos 6sin 60sin 6cos 60001][xyz]


[xyz]=[x cos 6+y sin 6x sin 6+y cos 60]

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)2a2+(y+k)2b2=1


((x+Δa cos 6)cos 6+(y+Δa sin 6)sin 6)2a2+((x+Δa cos 6)sin 6+(y+Δa sin 6)cos 6)2b2=1


(x cos 6+Δa cos26+y sin 6+Δa sin26)2a2+(x sin 6Δa cos 6sin 6+y cos 6+Δa sin 6cos 6)2b2=1


(x cos 6+y sin 6+Δa(cos26+sin26))2a2+(x sin 6+y cos 6+Δa(cos 6sin 6cos 6sin 6))2b2=1


(x cos 6+y sin 6+Δa)2a2+(x sin 6+y cos 6)2b2=1

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.




Navigation_