Manually reversing the rotation should do the trick; there should be a formula for rotating spherical coordinate systems somewhere, but since I can't find it, here's the derivation ( ' marks the rotated coordinate system; normal geographic coordinates use plain symbols):
First convert the data in the second dataset from spherical (lon', lat') to (x',y',z') using:
x' = cos(lon')*cos(lat')
y' = sin(lon')*cos(lat')
z' = sin(lat')
Then use two rotation matrices to rotate the second coordinate system so that it coincides with the first 'normal' one. We'll be rotating the coordinate axes, so we can use the axis rotation matrices. We need to reverse the sign in the ϑ matrix to match the rotation sense used in the ECMWF definition, which seems to be different from the standard positive direction.
Since we're undoing the rotation described in the definition of the coordinate system, we first rotate by ϑ = -(90 + lat0) = -55 degrees around the y' axis (along the rotated Greenwich meridian) and then by φ = -lon0 = +15 degrees around the z axis):
x ( cos(φ), sin(φ), 0) ( cos(ϑ), 0, sin(ϑ)) (x')
y = (-sin(φ), cos(φ), 0).( 0 , 1, 0 ).(y')
z ( 0 , 0 , 1) ( -sin(ϑ), 0, cos(ϑ)) (z')
Expanded, this becomes:
x = cos(ϑ) cos(φ) x' + sin(φ) y' + sin(ϑ) cos(φ) z'
y = -cos(ϑ) sin(φ) x' + cos(φ) y' - sin(ϑ) sin(φ) z'
z = -sin(ϑ) x' + cos(ϑ) z'
Then convert back to 'normal' (lat,lon) using
lat = arcsin(z)
lon = atan2(y, x)
If you don't have atan2, you can implement it yourself by using atan(y/x) and examining the signs of x and y
Make sure that you convert all angles to radians before using the trigonometric functions, or you'll get weird results; convert back to degrees in the end if that's what you prefer...
Example (rotated sphere coordinates ==> standard geographic coordinates):
southern pole of the rotated CS is (lat0, lon0)
(-90°, *) ==> (-35°, -15°)
prime meridian of the rotated CS is the -15° meridian in geographic (rotated 55° towards north)
(0°, 0°) ==> (55°, -15°)
symmetry requires that both equators intersect at 90°/-90° in the new CS, or 75°/-105° in geographic coordinates
(0°, 90°) ==> (0°, 75°)
(0°, -90°) ==> (0°,-105°)
EDIT: Rewritten the answer thanks to very constructive comment by whuber: the matrices and the expansion are now in sync, using proper signs for the rotation parameters; added reference to the definition of the matrices; removed atan(y/x) from the answer; added examples of conversion.
EDIT 2: It is possible to derive expressions for the same result without explicit tranformation into cartesian space. The x, y, z in the result can be substituted with their corresponding expressions, and the same can be repeated for x', y' and z'. After applying some trigonometric identities, the following single-step expressions emerge:
lat = arcsin(cos(ϑ) sin(lat') - cos(lon') sin(ϑ) cos(lat'))
lon = atan2(sin(lon'), tan(lat') sin(ϑ) + cos(lon') cos(ϑ)) - φ
If you are trying to calculate correct distance by projecting geographic coordinates (lon-lat) to map coordinates (E-N) and then using plane coordinate geometry, then you
- either need to project using an appropriate equidistant map projection -- probably not a good idea if you have points at arbitrary locations and distances in arbitrary directions
- or need to use a conformal projection -- which you are doing (Mercator) -- and apply the appropriate scale factor.
For long lines and accurate results, you may need to apply two or three scale factors. See Calculating distance scale factor by latitude for Mercator. At latitude 47°, the SF is 1 / cosine (47°) = approx 1 / 0.68 which, applied to your distorted-by-Mercator distance of 89NM, is now approx 60NM, i.e., a correct distance!
Best Answer
There are many more possible representations of our Earth on a map. You might want to learn about map projections. A good starting point is Carlos A. Furuti's Cartographical Map Projections page.