I should have added that if, in the demonstration, you put one point on the equator and move the other point above and below the equator, the curve alternates between concave and convex. Therefore, it must at some point (i.e. on the equator) be a straight line. So "always sinusoidal", no. Ever sinusoidal, or how to give a general description of the projected great circles, I don't know. Jeff On Fri, Jun 6, 2014 at 1:08 PM, Jeff Caldwell <jeffrey.d.caldwell@gmail.com> wrote:

You might be interested in the Wolfram Demonstraton Project "Great Circles on Mercator's Chart".

http://demonstrations.wolfram.com/GreatCirclesOnMercatorsChart/

Jeff

On Fri, Jun 6, 2014 at 12:17 PM, Henry Baker <hbaker1@pipeline.com> wrote:

...

Yes, I know this question is 400 years old, but what is the shape of the image of a great circle on a Mercator projection map?

I've seen it described as "sinusoid" shaped, but that can't be right, can it? (Yes, we've all seen those satellite location maps on TV, and they look pretty sinusoidal, but are they really?)

I'm not trying to measure lengths or angles, but am simply interested in the mathematical shapes of these great circle on the flat Mercator map.

I'm also not interested in the images of non-great circles.

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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.

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.

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

My first impression was the same as yours, Gareth. But I don't think points follow straight lines in this scheme. With Wright's description, any point on the (unit) sphere will first reach the cylinder when its latitude circle has expanded to radius = 1. By uniform expansion, the local magnification factor in the vertical direction will have to be the same, i.e. 1/cos(lat) = sec(lat). The local equality of magnification factors occurs if and only if angles are preserved. Just considering what happens to a longitude arc A starting from the equator and parametrized by angle t from the horizontal: This arc is mapped by call it t |-> L(t) to the generator of the cylinder meeting the equator at the same point as A, parametrized by height z. By what we know of the magnification, L'(t) = sec(t) (as Gene mentioned), with initial condition L(0) = 0. Thus L(t) = Integral_{0 to t} sec(u) du = ln(sec(t) + tan(t)) Which some people prefer to write as L(t) = ln(tan(t/2 + pi/4). So evidently, contrary to intuition, as the balloon expands and the points reaching the cylinder stop, it moves away from the equator faster than it would from expansion alone. --Dan On Jun 6, 2014, at 6:01 PM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:

On 07/06/2014 00:15, Dan Asimov wrote:

...

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.

I'm not sure I understand this.

The simplest interpretation of "uniformly expand" would seem to be that each point moves radially outwards, but that gives exactly the interpretation you said the Encyclopaedia Britannica wrongly gave for years: central projection from the sphere to the cylinder.

So maybe you intend some fancier (more physically realistic?) notion of uniform expansion, but I can't tell what.

What am I missing?

-- g

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun