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Topic: Meridional curvature
Replies: 1   Last Post: Jan 31, 2013 1:59 PM

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Kaimbridge M. GoldChild

Posts: 79
From: 42.57°N/70.89°W; FN42nn (North Shore, Massachusetts, USA)
Registered: 3/28/05
Re: Meridional curvature
Posted: Jan 31, 2013 1:59 PM
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On Jan 30, 6:19 pm, Taurio <sci.math@googlegroups.com> wrote:
> A meridian's radius of curvature at the equator equals b^2/a, an
> ellipse's /semi-latus rectum/.
> Does the polar radius of curvature, which equals a^2/b, have a
> similar elliptical property equivalency?
> Besides the equator, isn't the meridional curvature's radius the
> only curvature radius that is geocentric?


Any ?radius of curvature? equals an equivalent linear radius.
With a circle or sphere, curvature is constant, therefore radius of
curvature is constant, equaling the linear, geocentric radius.
Now consider an ellipse.
The curvature at the x-axis/?equator? is curled inward, while at the y-
axis/?pole? it?s flattened, meaning, if you took an infinitesimal snip
of equatorial arc and made enough copies to form a circle, and did the
same with a snip of polar arc, the composite polar circle would be
bigger than the equatorial one: If you make an oblatum (i.e., an
oblate spheroid) with the same equatorial and polar radii, the
composite equatorial circle would be smaller than the equator, b^2/a,
while the polar circle would be larger, a^2/b. Along the equator,
which is a great circle, the horizontal radius of curvature equals the
geocentric, equatorial radius, a.
There are two angular extremes of curvature radii??the vertical, 0°,
*Meridional*, M(Lat), and the perpendicular horizontal, 90°, *Normal*,
N(Lat).
The meridional equals an ellipse?s radii of curvature, thus M(0) = b^2/
a, while M(90°) = N(90°) = a^2/b.
Now look at the radii graph for Earth:

http://upload.wikimedia.org/wikipedia/en/c/c5/EarthEllipRadii.jpg

Only M(?45.072°) and N(0) equal their respective geocentric radius,
meaning, vertically/meridionally, only (geodetic/geographic) latitude
45.072° has a geocentric radius of curvature.
If you look at fig.2 on pg.3 of Rod E. Deakin?s ?The Great Elliptic
Arc on an Ellipsoid?,

http://user.gs.rmit.edu.au/rod/files/publications/Great%20Elliptic%20Arc.pdf

you will see that the linear radius equivalent of N(P) extends below
the geocenter, O, to H: N(P) = PH.
As for the polar radius of curvature, a^2/b, I?m not sure if NH
represents the linear radius equivalent.
Of coarse, there is a *geodetic/geographic* radius of curvature in any
direction at a given point, G, *IN THE GREAT ELLIPTIC PLANE*, which is
the arcradius (not to be confused with the more recognized and cited
Euler's radius of curvature in the normal section).

~Kaimbridge~

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