Difference between revisions of "The Ellipse"

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Revision as of 17:58, 15 May 2017

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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.