latitude longitudemercator
Say I'm looking at a portion of a map that's displayed using Mercator projection with the view bounded at 40°N and 20°N. How do I compute the latitude of the point that's in the centre of that view?
I figured it's not 30° and can compute it by firstly projecting the bounding latitudes onto a square using Mercator, computing the middle point, and then projecting back onto a sphere. But is there a simpler way of achieving the same without having to project to and from Mercator?
This question was answered by @acraig5075 here.
After correcting my mistake and adjusting southern latitude I get: Fixed link: http://jsfiddle.net/an3BZ/145/
By definition, the scale is the amount by which (infinitesimal) distances are multiplied by the projection. Whenever a tiny displacement of d meters on the earth is associated with a displacement of d/s meters on the map, the scale is written as 1:s. It may depend on the direction of the displacement.
The scale factor compares the scale at points on the map to the map's nominal scale. For instance, if the nominal scale is 1:25000 but the scale at a point (in a particular direction) actually is 1:50000, then the scale factor at that point is (1/50000) divided by (1/25000), equal to 1/2.
For many projections, scale depends on location and bearing. For conformal projections like the Mercator, though, scale is constant at all bearings (at any given location) and depends only on position. (This is the defining characteristic of a conformal projection.) Moreover, with cylindrical projections of surfaces of revolution (and the Mercator is one of them, too), scale does not depend on longitude. Therefore, the Mercator scale depends only on the latitude.
For precise work, the scale and scale factors depend a little bit on the model of the earth's surface.
For a spherical model, the scale factor of the Mercator projection is proportional to the secant of the latitude (Snyder, equation 7-3).
(This is intuitively obvious because any cylindrical projection, with its vertical meridian lines, allows for no horizontal adjustment in distances on the map. Therefore the horizontal scale at any latitude must be inversely proportional to lengths on the earth's surface in east-west directions. These will shrink in proportion to the circumferences of the circles of latitude. Equivalently, they are proportional to the radii of those circles, which--by definition of the cosine--vary with the cosine of the latitude. The secant merely is the reciprocal of the cosine.)
For example, at a latitude of 60 degrees the secant equals 2 whereas at the Equator (0 degrees) the secant equals 1. Therefore sizes of features are magnified by 2/1 at 60 degrees latitude compared to at the Equator.
The graph of the secant function rises infinitely high as the latitude approaches 90 degrees.
For an ellipsoidal model with eccentricity e, the scale factor is sqrt(1 - e^2 sin^2(f)) * sec(f) at latitude f (Snyder, equation 7-8). The function sqrt(1 - e^2 sin^2(f)) multiplies the secant of f to correct for the eccentricity. For the (common) WGS 84 model, for instance, e is approximately 0.006694379990141317. This multiplier therefore ranges from 1 at the Equator down to sqrt(1 - e^2) at the poles, amounting to a shrinkage of 22.41 parts per million compared to the secant formula.
This plot shows the multiplicative correction to the simple secant formula that should be applied for the WGS 84 ellipsoid. The correction is practically indistinguishable from this for any ellipsoid in standard use.
An algebraically equivalent expression for the scale factor is Sqrt(((Sec(f))^2*(1-e^2) + e^2); this requires only one trigonometric calculation rather than two. The values 1-e^2 and e^2 are constants that can be precomputed. An excellent approximation (accurate to 5E-11 everywhere) which avoids the square root is (1 + c2*(cos(2*f) - 1)) / cos(f) where c2 = 0.00001120378. (It is based on the first two terms of the Fourier series for the correction factor.)
In either case the scale factor for the Mercator projection ranges from a low of 1 at the Equator up to arbitrarily large values near the poles.
John P. Snyder, Map Projections--A Working Manual. USGS Professional Paper 1395, 1987. Available on the Web at http://pubs.er.usgs.gov/publication/pp1395.
Best Answer
Assuming you know the map coordinates of the center -- from those of the top and bottom of the map window -- the problem is called inverse (or reverse) map projection.
From wikipedia: Mercator projection, The spherical model:
latitude = 2 arctan ( exp [y / R]) - pi/2
where y = map coordinates of the center
and R = earth radius