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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.

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    http://en.wikipedia.org/wiki/Ellipse#Polar_form_relative_to_center2012-09-23

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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

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    @James - I was looking at something else with the `geodesy` tag and came across it2018-04-23