Talk:Reference ellipsoid

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)

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).

Yes it should be. The double angle formulas are incorrect unless the square (^2) applies to the sine operator, not the operand. the correct way to write this is something like sin^2(ae/2). TelecomNut ^(talk) 21:22, 20 August 2008 (UTC)

If you have any doubts as to the validity, just think of it as derivative notation:

$(\sin(x))'=\sin '(x)\neq \sin(x)'.\,$

So

$(\sin(x))^{2}=\sin ^{2}(x),\;\mathbf {not} \;\sin(x)^{2}.\,$

I think a good part of the problem is that "sin(x)^2" is the way it is usually presented in both regular ASCII text (such as in Usenet) and in computer programming. JMHO!

74.10.197.201 (talk) 14:08, 23 August 2008 (UTC)

Undefined Symbols

In the formulas:

$X_{t}=[N+h]\cos(\phi )\cos(\lambda );\,\!$

$Y_{t}=[N+h]\cos(\phi )\sin(\lambda );\,\!$

$Z_{t}=[\cos ^{2}(\alpha )N+h]\sin(\phi );\,\!$

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

Alpha appears to be undefined. It may be defined by mathematical or geographical convention, but without the appropriate background, I have no idea what values to plug in for a given ellipsoid. 70.89.118.42 (talk) 00:18, 11 January 2012 (UTC)

Apologies. I made an edit recently in which I removed the definition of alpha from the first section and I forgot to go back and remove all the other instances of alpha. Basically

sin\alpha =e

cos\alpha= \sqrt(1-e^2)

If you feel up to it it would be good if you could make these changes. Have a look at the appendices of

A guide to coordinate systems in Great Britain.

[1]]

There you will find the usual notation. Peter Mercator (talk) 16:21, 11 January 2012 (UTC)

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)

Future of this page

I have restored the conventional form of the equations for the direct transformation from geodetic to Cartesian coordinates. This was how they appeared on this page until one editor tried to convert every Wiki appearance of eccentricity (e) into angular eccentricity (alpha). I have given references and links to other wiki pages for the inverse transformation.

This whole area of Wiki is a mess, with much duplication of topics. For example this page Reference ellipsoids overlaps with geodetic system, geographic coordinate system , Latitude, Figure of the Earth, Earth Ellipsoid and probably many others. There could be some short term rationalising but perhaps a more thorough appraisal is need. Comments?  Peter Mercator (talk) 21:07, 11 January 2012 (UTC)

Principia quotation and oblate ellipsoid/spheroid

The statement in the article

In 1687 Isaac Newton published the Principia in which he included a proof that a rotating self-gravitating fluid body in equilibrium takes the form of an oblate ellipsoid of revolution which he termed an oblate spheroid.

is not true, because in the quoted section of the Principia, Newton simply proves that the shape is oblate, but does not prove that it is a spheroid or ellipsoid. --98.234.249.218 (talk) 12:49, 18 February 2012 (UTC)

It isn't spelled out in that paragraph but within two or three pages Newton describes the shape as an ellipse rotated about its minor axis. He characterises the shape by only the ratio of the axes (b/a) and therefore it must be an ellipsoid. I.e. the simplest deviation from a sphere is an elllipsoid. Any other shape requires more parameters for its description. Please reconsider your edit.  Peter Mercator (talk) 09:42, 20 February 2012 (UTC)

He may describe it as an ellipsoid, but he does not present any proof of it. He may have only computed the ratio of the axes, but any number of oval-like shapes have one axis shorter than the other. It seems circular for you to say that since he only computed those parameters (because he assumed it was an ellipse), then it must be an ellipse because any other shape would require more parameters. Perhaps the true shape is something else that requires more parameters, and he just didn't realize it. --98.234.249.218 (talk) 06:48, 21 February 2012 (UTC)

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