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Re: ellipse, mean distance from point on perimeter to a focus ?
Posted:
Feb 12, 2010 2:45 PM
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In article <ut8bn5hhqceuqnjnnmk261gcrc14setco2@4ax.com>,
Cary <cary@domain.invalid> wrote:
>On Fri, 12 Feb 2010 10:22:33 +0000 (UTC), rob@trash.whim.org (Rob
>Johnson) wrote:
>
>>In article <20100211111323.200$wv@newsreader.com>,
>>David W. Cantrell <DWCantrell@sigmaxi.net> wrote:
>>>rob@trash.whim.org (Rob Johnson) wrote:
>>>> In article
>>>> <a36f3aaf-77d5-499b-bd30-f603613dcc8d@j6g2000vbd.googlegroups.com>, G
>>>> Patel <gaya.patel@gmail.com> wrote:
>>>> >Is the mean distance from a point on perimeter to a focus equal to the
>>>> >semi major axis length?
>>>>
>>>> Yes.
>[snip]
>
>>>But I initially (before seeing Achava's post) calculated the mean distance
>>>with respect to theta, thinking of the ellipse in polar coordinates as
>>>given by
>>>
>>>r = a (1 - e^2) / (1 - e cos(theta))
>>>
>>>Doing that, we find instead
>>>
>>>mean distance = a sqrt(1 - e^2) = b, the semi-minor axis length.
>>
>>I get that as well. Furthermore, averaging with respect to time,
>>using equal area in equal time, I get a mean of a(1+e^2/2).
>
>
>This paper may be of interest to those following this thread.
><http://faculty.matcmadison.edu/alehnen/kepler/kepler.htm>
Thanks for the reference; that paper goes the distance. It covers
the means that David Cantrell (True Anomaly) and I (Time) discussed.
I note that they also discuss average over arc-length, and since arc
length is symmetric with respect to the foci, that average is a.
It may not be obvious at first, but eccentric anomaly is also
symmetric with respect to the foci. The eccentric anomaly is the
angle from the center of the ellipse after the ellipse has been
scaled in the direction of either the major or the minor axis to a
circle. Thus, the mean with respect to the eccentric anomaly is
also a.
Rob Johnson <rob@trash.whim.org>
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