It's pretty difficult to Google for the meaning of a formula.

This is the equation, it has to do with ellipses and GIS coordinates.

$$\nu =\frac{ a} {\sqrt{(1 - (e^2 \cdot \sin(\varphi))^2)}}$$

$a$ is an ellipsoid axis.

$\varphi$ is geodetic coord latitude in radians.

$e^2$ is eccentricity squared.

I see it all over the code I'm porting and would like to separate it out but I can't figure out what to call the function!

Edit: On page 38 of this PDF the equation and its context is described.


You link has broken in the last six years, but something like that expression now appears in Annex B (page 47) of https://www.ordnancesurvey.co.uk/docs/support/guide-coordinate-systems-great-britain.pdf which has $$\nu =\frac{ a} {\sqrt{1 - e^2 \sin^2(\varphi) }}$$

If the world was spherical with radius $r$ then a point with latitude $\varphi$ and longitude $\lambda$ would have $(X,Y,Z)$ co-ordinates

  • $X=r \cos(\varphi) \cos(\lambda)$
  • $Y=r \cos(\varphi) \sin(\lambda)$
  • $Z=r \sin(\varphi)$

but with an ellipsoid, the coordinates of a point on the ellipsoid (ignoring height above the ellipsoid) would be

  • $X=\nu \cos(\varphi) \cos(\lambda)$
  • $Y=\nu \cos(\varphi) \sin(\lambda)$
  • $Z=(1-e^2)\nu \sin(\varphi)$

with $\nu$ defined as above. Note that $\nu$ varies with latitude $\varphi$

So $\nu$ is close to the idea of a radius: because of the $(1-e^2)$ term, it is not quite the distance from the centre of the ellipsoid to a point of latitude $\varphi$, but it is not far away

Personally I would just call it nu or perhaps something like ellipsoid_nu

  • $\begingroup$ Thanks. What made you revisit this question after 6 years!? $\endgroup$ – James Apr 23 '18 at 7:24
  • $\begingroup$ @James - I was looking at something else with the geodesy tag and came across it $\endgroup$ – Henry Apr 23 '18 at 7:28

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