In article <email@example.com>, David W. Cantrell <DWCantrell@sigmaxi.net> wrote: >firstname.lastname@example.org (Rob Johnson) wrote: >> In article >> <email@example.com>, G >> Patel <firstname.lastname@example.org> 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).
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