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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 <> 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*,
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:

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?,

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).


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