How do I find out (calculate somehow) the central meridian for a UTM zone for which I know the zone number?
Coordinate System – Finding Central Meridian of UTM Zone
coordinate systemutm
Related Solutions
UTM uses a transverse Mercator projection with a scale factor of 0.9996 at the central meridian. In the Mercator, the distance scale factor is the secant of the latitude (one source: http://en.wikipedia.org/wiki/Mercator_projection), whence the area scale factor is the square of this scale factor (because it applies in all directions, the Mercator being conformal). Understanding the latitude as the spherical distance to the equator, and approximating the ellipsoid with a sphere, we can apply this formula to any aspect of the Mercator projection. Thus:
The scale factor is 0.9996 times the secant of the (angular) distance to the central meridian. The area scale factor is the square of this quantity.
To find this distance, consider the spherical triangle formed by traveling along a geodesic from an arbitrary point at (lon, lat) = (lambda, phi) straight towards the central meridian at longitude mu, along that meridian to the nearest pole, and then back along the lambda meridian to the original point. The first turn is a right angle and the second one is an angle of lambda-mu. The amount traveled along the last portion is 90-phi degrees. The Spherical Law of Sines applied to this triangle states
sin(lambda-mu) / sin(distance) = sin(90 degrees) / sin(90-phi)
with solution
distance = ArcSin(sin(lambda-mu) * cos(phi)).
This distance is given as an angle, which is convenient for computing the secant.
Example
Consider UTM zone 17, with central meridian at -183 + 17*6 = -81 degrees. Let the outlying location be at longitude -90 degrees, latitude 50 degrees. Then
Step 1: Spherical distance from (-90, 50) to the -81 degree meridian equals ArcSin(sin(9 degrees) * cos(50 degrees)) = 0.1007244 radians.
Step 2: The area distortion equals (0.9996 * sec(0.1007244 radians))^2 = 1.009406.
(Numerical calculations with the GRS 80 ellipsoid give the value as 1.009435, showing that the answer we computed is 0.3% too low: that's the same order of magnitude as the flattening of the ellipsoid, indicating the error is due to the spherical approximation.)
Approximations
To get a feel for how the area changes, we can use some trig identities to simplify the overall expression and expand that as a Taylor series in lambda-mu (the displacement between the point's longitude and the longitude of the UTM central meridian). It works out to
Area scale factor ~ 0.9992 * (1 + cos(phi)^2 * (lambda-mu)^2).
As with all such expansions, the angle lambda-mu must be measured in radians. The error is less than 0.9992 * cos(phi)^4 * (lambda-mu)^4, which is close to the square of the difference between the approximation and 1--that is, the square of the value after the decimal point.
In the example with phi = 50 degrees (with a cosine of 0.642788) and lambda-mu = -9 degrees = -0.15708 radians, the approximation gives 0.9992 * (1 + 0.642788^2 * (-0.15708)^2) = 1.009387. Looking past the decimal point and squaring, we deduce (even without knowing the correct value) that its error cannot be greater than (0.009387)^2 = less than 0.0001 (and in fact the error is only one-fifth that size).
From this analysis it is evident that at high latitudes (where cos(phi) is small), scale errors will always be small; and at lower latitudes, area scale errors will behave like the square of the difference in longitudes.
I guess you stumbled over this note on page 53:
subsequent civilian adoption of the systems usually ignores the zone prefix to easting. Where this is the case the formulas below do not apply: use the standard TM formula separately for each zone
The formulas on page 54 should only be used if the zone number is written before the false Easting, which is not the case here. So I took only the formulas from p 50ff into Libre Office and got these results:
M0 0
e1 0.0016649922
mue1 0.4772587465
M1 3038553.85538554
T1 0.2701106629
C1 0.0059292372
esq 0.0066822021
D 0.0183352698
phi1 0.4793002154
nue1 6381782.36093598
rho1 6348382.77011411
phirad 0.4792124095 phi 27.4568485565
lamrad 1.4867384817 lam 85.1838402406
which returns the same lat and lon on the Everest ellipsoid as
cs2cs +proj=tmerc +lon_0=84 +lat_0=0 +k=0.9999 +a=6377276.345 +b=6356075.41314024 +x_0=500000 +units=m +towgs84=282,726,254 +no_defs +to +proj=longlat +a=6377276.345 +b=6356075.41314024 +towgs84=282,726,254 +no_defs -f "%%.6f"<MUTM88.txt >WGS84out.txt
85.183840 27.456849 1230.000000
Transforming to WGS84 returns:
85.181615 27.457128 1188.383496
which is still not exactly what you expected, but more close than your result.
Best Answer
The central meridian of a 6 degree wide UTM zone will be given by: