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There are several curvatures depending on the ellipsoid used.

Along the meridians, the local radius of curvature at is at φ is

M(φ)=a(1-e^2 )/(1-e^2 〖sin〗^2 φ)^(3⁄2)

The curvature of the meridian at φ is the inverse of M(φ), e.g. 1/M(φ).

As you go north to the pole, the radius of the ellipsoid get smaller, the curvature gets a little larger.

The parallels do something similar based on the length of the normal is dependent on φ, and is:

N(φ)=a/√(1-e^2 〖sin〗^2⁡φ )

The radius of curvature of a parallel at latitude is:

ρ(φ)=N(φ) cos⁡φ

Again the curvature is the inverse 1/ρ(φ).

Both M(φ) and ρ(φ) can for an determined arc (in radians) and converted to meters (arc)*(radius of curvature) and use a integral of the function to get lengths of just about anything.. The idea is from what is called Riemannian Geometry. The Wikipedia article on Metric tensor lays out the math fairly easily.

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Anonymous

5y ago

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