Re: A Better "Great-Circle" Arcradius

Er, any comments...or should I be posting this some place else? P=|

On Dec 4, 4:41 pm, Kaimbridge <kaimbri...@gmail.com> wrote:
> The current default radius used in Google Maps distance calculations
> is the WGS-84 equatorial radius, "a", 6378.137 km (6378137 m).
> While "a" is ideal along the equator, that is only one extreme:  The
> other is along a north-south meridian, which approximately equals "[.5*
> (a^2+b^2)]^.5", the "elliptical quadratic mean".  For WGS-84, this
> works out to 6367.454, with the actual meridional value being about
> 6367.449 km.
> If you average the two extremes, you get the "ellipsoidal quadratic
> mean", "Qr":
>
>         / equatorial^2 + meridional^2 \^.5
>   Qr = (-------------------------------),
>         \              2              /
>
>         / a^2 + .5*(a^2+b^2) \^.5
>      = (----------------------),
>         \          2         /
>
>         / 3*a^2 + b^2 \^.5
>      = (---------------);
>         \      4      /
>
> So, for the WGS-84 spheroid model, where a = 6378.137 and
> b = 6356.752314, Qr ~=~ 6372.7975559, 6372.797 or——given all of the
> different "flavors" discussed in the "Talk:Earth" link given in the
> link below—— even just rounded to "6372.8"!  See:
>
> http://groups.google.com/group/comp.infosystems.gis/msg/5897cf05c924a569
>
> for a fuller explanation why, as well as the Wikipedia link to an even
> much more thorough explanation.
>
>      ~Kaimbridge~
>
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>
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