Allow simulation on lat/lon locations

[lrntct]

$\varphi$, $\lambda$ are latitude and longitude in radiant, respectively.

Surface area

https://onlinelibrary.wiley.com/doi/full/10.1111/tgis.12636

Distances / cell sizes

$f = \frac{a-b}{a}$
$e^2 = f(2 - f)$

where $a$ is the semi-major axis, $b$ the semi-minor axis, and $f$ the flattening.
For the WGS84 ellipsoid:
$a = 6378137$
$b = 6356752.314245$

Meridional radius:

$M(\varphi)=\frac{1-e^2}{a^2} N(\varphi)^3$

Prime vertical radius:

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

Distances:

$\Delta X = \Delta\lambda N(\varphi\prime)\cos(\varphi\prime)$
$\Delta Y = \Delta\varphi M(\varphi\prime)$
Where $\varphi\prime$ is the mean for the given range (in our case, the center of the given cell)

Reference

Compare results with Karneys's solution, for both distances and areas.
For the tiny distances here (likely a few hundred meters max), it should be accurate to under a millimeter.