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



Kaimbridge M. GoldChild
Posts:
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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


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 /semilatus 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 xaxis/?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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