Find Latitude x miles north of starting latitude using ellipsoid earth model Let us say I am given a starting (latitude, longitude)=(lat,lon) coordinate in degrees.  The objective is to compute the new latitude, lat' when moving d miles north along a fixed longitude, lon.  
More specifically, I want a formula for lat' given an initial (lat, lon) and a distance in miles (or kilometers), d on how long one will move North across the fix starting longitude lon.
Ideally, we'd take into account the fact that the earth is an ellipsoid.  A rough approximation will also work.
Also for simplification let us assume that one is not too close to the poles (100+ miles away from the poles) and that the distance d is less than 20 miles north - this will avoid the need to handle edge cases.
Thanks!
 A: If your distances are fairly short like this, we can use the formula for the meridian arc $M(\phi_o)$ to determine the local meridian radius of curvature at the starting latitude $\phi_o$, and then use it as a basis for the calculation of the latitude of destination $\phi_d$. 
$$M(\phi_o) = \frac{(ab)^2}{\left(\left(a \text{ cos } \phi_o \right)+ \left(b \text{ sin } \phi_o\right)^2\right)^\frac{3}{2}}$$
$$\phi_d = \phi_o + d/M(\phi_o)$$
Where $a$ is the semi major-axis and $b$ is the semi minor-axis of the ellipsoid. All angles are in radians.
Finally, calculate the midpoint latitude (average of $\phi_o$ and $\phi_d$) and recalculate the radius of curvature with this new value, then use this result in the second formula instead of the old $M(\phi_o)$ to get the final answer. 
With this method, for a distance of 20 miles, the resulting destination latitude will always be within 1 millimeter of the real answer.
Of course, an much simpler but less accurate method would be to assume that the Earth is spherical, and use a constant Earth radius: $\phi_d = \phi_o + d/r$ This can generate a maximum error of about 0.5% of the distance. 
This Wikipedia article explains the problem in more detail and how to solve it more accurately with integrals and iterative solutions.
