Are you assuming the Earth is a sphere? The WGS84 ellipsoid is used in most calculations. Brent On 6/6/2014 5:57 PM, Henry Baker wrote:
Thanks very much, Gene & Dan!
If the projection had been simply "central cylindrical", then great circles would be coplanar with the center of the sphere, and also coplanar with the ellipse which is cut out of the cylinder by this plane.
Thus in central cylindrical projections, great circles do indeed map to true sinusoids.
However, the Mercator project is NOT a central cylindrical projection, regardless of what we might have been taught in graded school or by TV news readers.
Now if we roll up our Mercator map so that it is a cylinder again, we see that our "great circles" are not planar, hence not even circles (in 3D), but do seem to form a ruled surface whose edge is the track of the "great circle" on the Mercator-projection map.
It might be fun to 3D print a cylindrical Mercator globe that comes apart at this "great circle surface" -- e.g. for a great circle route from L.A. to London -- to show just how non-sinusoidal these curves really are.
BTW, satellite "ground tracks" are further distorted from sinusoids by the rotation of the earth while the satellite is travelling, as well as by the eccentricity of the satellite orbit.
The ground tracks of the extremely eccentric Molniya orbits are utterly amazing.
http://en.wikipedia.org/wiki/Molniya_orbit
At 04:15 PM 6/6/2014, Dan Asimov wrote:
Very nice and clear explanation, Gene -- thanks.
Reading a bit about the history of the Mercator projection, it was interesting to learn that Mercator never described his method; he just published (in 1569) maps that used it.
Its property of straight lines on the map corresponding to lines of constant bearing (i.e., angle to longitude lines), a.k.a. rhumb lines, made these maps very popular with mariners.
The first rigorous description of the Mercator projection was apparently given by Edwin Wright in 1599 -- it's a lovely geometric description I hadn't heard before:
Imagine a vertical cylinder tangent to the globe at the equator.
Now inflate (uniformly expand) a perfectly spherical balloon that initially coincides with the globe, stopping each point at the moment it reaches the cylinder.
The correspondence between a point on the globe and the point it reaches on the cylinder is the Mercator projection, and can readily be seen to be conformal.
--Dan
P.S. For many years in the late 20th century, the Encylopaedia Britannica misstated the Mercator projection as just being the central projection of the globe to the vertical cylinder tangent at the equator.
This mistake was copied in innumerable other places, alas.
On Jun 6, 2014, at 3:50 PM, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
First, what is the Mercator projection? Conformally map the unit sphere with latitude λ and longitude φ onto the unit cylinder, (λ,φ) → (z,φ), and unroll the cylinder.
The arc of length cosλ dφ on the sphere maps to an arc of length dφ on the cylinder, so the scale factor is secλ.
The arc of length dλ on the sphere maps to an arc of length dz on the cylinder.
For the map to be conformal, dz/dλ = secλ.
The clever trick is to take the integral as
z = asinh tanλ.
Second what is the equation of a great circle? Let the normal to the plane of the great circle lie on the international date line, longitude φ = π, at latitude μ.
This normal has coordinates
n = (-cosμ, 0, sinμ). The general point on the sphere has coordinates r = (cosλ cosφ, cosλ sinφ, sinλ). The points on the great circle satisfy n.r = 0, which gives the relation tanλ = cotμ cosλ. Putting these together, the map of a great circle is z = asinh (cotμ cosλ). The map is approximately sinusoidal when cotμ is small, i.e. when the great circle is near the equator.
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