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Re: ellipse, mean distance from point on perimeter to a focus ?
Posted:
Feb 12, 2010 5:22 AM
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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.
>
>Well, as Achava correctly pointed out, it depends on the variable with
>respect to which the mean is calculated.
Indeed. I realized that soon after I posted. As soon as I read the
original post, the picture of the "string and pins" construction
popped into my head and I went with it. Any weight symmetric with
respect to the foci will give the semi-major axis as the mean.
>> Recall the "string and pins" method of drawing an ellipse, which
>> depends on the fact that the sum of the distances from a point on the
>> ellipse to the two foci of that ellipse is the major axis of that
>> ellipse. Since the sum of those two distances is always the major
>> axis, the sum of their means is the major axis. By symmetry, the
>> means of these distances are the same. Therefore, each is equal to
>> the semi-major axis.
>
>That's a simple and excellent argument that _a_ mean is the semi-major axis
>length. Perhaps in some sense, that is the most useful mean.
>
>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).
Rob Johnson <rob@trash.whim.org>
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