Talk:Reference ellipsoid

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[edit] Coordinate section

Coordinates: "These formulae have a closed-form inverse, though the algebra is rather involved. It can be shown that"

Yes there is a closed-form inverse, but this is not it! This is just the initial guess in Bowring's iterative method. While quite accurate at near terrestial points, it is not good for large h (e.g.: satellite positions).

Yup, fixed! P=)  ~Kaimbridge~ 01:43, 1 November 2006 (UTC)

[edit] Notation

Should (for example) sin(ae/2)^2 be (sin(ae/2))^2 -- the notation used here is not clear... —The preceding unsigned comment was added by 65.95.168.33 (talk) 01:22, 22 January 2007 (UTC).

[edit] Puzzlement

Referring to the formula

N=N(\phi)=\frac{a}{\sqrt{1-(\sin(\phi)\sin(o\!\varepsilon))^2}}\,\!

is the radius of curvature in the prime vertical.

What is the behaviour at the equator and the poles? Clearly, at the equator the latitude is zero so sin(phi) is zero and the result is then N = a, which is clear: a circle of radius a would be tangent to the ellipse at the equator. Notionally interchanging x and y suggests that at the pole the radius of curvature should be b, but the formula does not yield that so far as I can see. The latitude at the poles is 90 degrees so sin(phi) is 1, and the formula becomes

N=N(90)=\frac{a}{\sqrt{1-(sin(o\!\varepsilon))^2}}\,\!

and given that

o\!\varepsilon=\arccos\left(\frac{b}{a}\right) or \cos(o\!\varepsilon)=\frac{b}{a} from the definitions

and (sin(o\!\varepsilon))^2=1 - (\cos(o\!\varepsilon))^2

then N=N(90)=\frac{a}{\sqrt{1-(1-(\frac{b}{a})^2)}}\,\!
which is N=N(90)=\frac{a}{\frac{b}{a}}\,\!

or

N=N(90)=\frac{a^2}{b}

Which is not b. So I'm puzzled. Rewriting the formula in terms of eccentricity,

N=N(\phi)=\frac{a}{\sqrt{1-(e\sin(\phi))^2}}\,\!

has the same problem if e^2 = \frac{a^2 - b^2}{a^2} but not if the second eccentricity is used instead, e^2 = \frac{a^2 - b^2}{b^2}

So, how should I adjust my ideas? NickyMcLean 04:33, 11 May 2007 (UTC)

You're looking at it wrong——change in radius of curvature goes in the opposite direction to that of the axial radius underneath it: If you take (theoretically) a (small) oblate ellipsoid, slice it in half at the equator, lay it flat on a piece of paper and trace the circumference you will have a circle, meaning the radius of curvature——here, N—— equals the radius. Now put the ellipsoid back together, then slice it down the middle, lay it flat and trace that circumference and you will see that you have an ellipse, the radius of curvature equalling M. Now look and compare the arcs at the equator and poles with the corresponding axes: While the equatorial axis is greater than the polar, the equatorial arc "pinches", while the polar arcs flatten, hence the radius of equatorial arc/curvature is smaller than the axial radius underneath it,{a\to\frac{b^2}{a}}\,\!, and the polar radius of arc/curvature is greater than its axial radius, {b\to\frac{a^2}{b}}\,\!.  ~Kaimbridge~14:33, 11 May 2007 (UTC)
After I posted, and while walking home after having dozed on the train ride, I realised that the ellipse at its pointy end is more sharply curved than the circle tangent to it there of radius a and likewise at the flattened pole it is less curved than a circle of radius b tangent there. I was failing to distinguish horizontal and vertical slices of an ellipsoid, as you describe: "slice it in half along the equator" meaning a horizontal slice through the ellipsoid, which are always circles of some radius out from the polar axis, while the slice down the middle is at right angles to the equator and is an ellipse. Thanks. NickyMcLean 03:13, 12 May 2007 (UTC)