What property does this equation calculate? 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.
 A: 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
