Eisenlohr’s optimal conformal map of the world

[PeteD]

If you understand conformal radii better than I do, or at least know a source I missed, feel free to check the proof.

I definitely don't!

As far as I understand, your conjecture is that the optimal conformal projection is circular and has minimum scale at its centre. This seems reasonable and I'm willing to go along with it. However, for the Lagrange projection to be optimal, you would have to show that it's the only possible circular projection with minimum scale at its centre. I don't see why that should be the case.

[Milo]

As far as I understand, your conjecture is that the optimal conformal projection is circular and has minimum scale at its centre. This seems reasonable and I'm willing to go along with it.

No, what I say is that if a circular projection with minimum scale at the center exists, then it is optimal. Note that it is possible in some cases for such a projection to not exist, in which case the optimal projection would necessarily be something else. Conversely, the conjecture theoretically shows that if multiple such projections exist, then they must all be tied for optimal (but this probably cannot happen: see below).

However, for the Lagrange projection to be optimal, you would have to show that it's the only possible circular projection with minimum scale at its centre. I don't see why that should be the case.

http://en.wikipedia.org/wiki/Riemann_mapping_theorem:

Henri Poincaré proved that the map f is essentially unique: if z₀ is an element of U and φ is an arbitrary angle, then there exists precisely one f as above such that f(z₀) = 0 and such that the argument of the derivative of f at the point z₀ is equal to φ. This is an easy consequence of the Schwarz lemma.

Picking apart the math, what this is saying is that there exists only one conformal projection of any region onto a circle with a given point at its center (the part about the angle can be ignored, since a circle is rotationally symmetric). I guess that, hypothetically, this doesn't rule out the existence of two different choices of "center point" that both result in that point having minimum scale. I think I've proven that this won't happen, but please check my proof for errors:

Let projection A be centered on point z_A and projection B be centered on point z_B, both conformal projections of the same region onto the unit disc. Let f be the function that maps A to B, and g be the function that maps B to A (these will be inverses of each other, so f(g(z)) = z).

Then:
The conformal radius of A at z_A is 1 (per definition).
The conformal radius of B at z_B is 1 (per definition).
The conformal radius of A at z_B is 1/f'(g(0)) = g'(0) ≤ 1.
The conformal radius of B at z_A is 1/g'(f(0)) = f'(0) ≤ 1.
See the inverse function rule for the latter equalities, and for the inequalities, note that the conformal radius of the unit disc at a point at distance r from the center is equal to 1−r² ≤ 1 (the relevant function is the Möbius transformation h(x) = (x−r)/(1−r⋅x)).

Suppose that A has minimum scale at z_A, i.e., s_A(z) / s_A(z_A) ≥ 1 everywhere. In particular s_A(z_B) / s_A(z_A) ≥ 1, or equivalently, s_A(z_A) / s_A(z_B) ≤ 1.

Observe that s_A(z) / s_B(z) = g'(B(z)) = 1/f'(A(z)).

Then s_B(z_A) / s_B(z_B) = s_A(z_A) / s_A(z_B) ⋅ f'(0) ⋅ g'(0). This is the product of three values that have each been shown to be smaller than or equal to 1, meaning that B does not have minimum scale at z_B unless A and B are the same projection.

Note that this applies to projections of any arbitrary Euclidean or non-Euclidean geometry, not just spherical!

[PeteD]

Picking apart the math, what this is saying is that there exists only one conformal projection of any region onto a circle with a given point at its center

OK, so if I understand correctly, this is a special property of circles and isn't true for any arbitrary shape?

(the part about the angle can be ignored, since a circle is rotationally symmetric).

Sorry to be pedantic, but the angle can be ignored because a circle is circularly symmetric. Rotational symmetry of order n wouldn't be sufficient to ignore the angle for finite n.

[Milo]

Picking apart the math, what this is saying is that there exists only one conformal projection of any region onto a circle with a given point at its center

OK, so if I understand correctly, this is a special property of circles and isn't true for any arbitrary shape?

For shapes that aren't rotationally circularly symmetric, you also need to account for different choices of angle at the central point. But with that extra qualifier, conformal projections are still unique.

For example, the Adams and Guyou projections are two different ways of conformally mapping the same source region (an equatorial hemisphere) onto the same destination region (a square), with the same centering (the center of the hemisphere is mapped onto the center of the square), but with different rotation (in Adams, "north" at the center points towards a corner of the square, while in Guyou, "north" at the center points towards an edge center of the square). But the Adams projection is the only conformal projection with all of these properties, including the direction of north.

The exact information needed to uniquely specify a conformal projection is: (A) the source shape (for example, the meridian-interrupted sphere for Lagrange, Eisenlohr, Mercator, etc. - technically, different choices of interruption meridian produce different projections, although these are often thought of as different "aspects" of the same named projection), (B) the destination shape (for example, a circle or a square), (C) a pairing of one anchor point in the source shape and one anchor point in the destination shape which are identified with one another (usually you'll pick the centers for this, but you don't have to, and an asymmetric shapes may not have an obvious "center"), and (D) a pairing of a tangent line through the source anchor point and a tangent line through the destination anchor point which are also identified with each other. (Note that there is more than one way to specify the same projection, since you can choose any point as the anchor point. However, naturally, once you decide on either the source anchor point or the destination anchor point, that also determines the other one, since that's the point of projections. Likewise for tangent lines.)

Another way to put requirements (C) and (D) is that you know both the value and the derivative of the projection at some predefined point.

The Riemann mapping theorem gives additional conditions under which a conformal projection satisfying any arbitrary choice of (A)-(D) is guaranteed to exist, in addition to being unique. When these additional conditions are not satisfied, there will still be at most one such projection, but there might be none. (Well, except when the region to be projected consists of multiple disjoint components. In that case, you need a pair of anchor points and tangent lines for each component to uniquely determine the projection.) The most important case where the additional conditions are not satisfied is when the source shape is the sphere interrupted at only one point, in which case the destination shape may only be the whole infinite plane (i.e., the stereographic projection).

This is not true of non-conformal projections! For example, the Mollweide and Hammer projections project the same source shape (meridian-interrupted sphere) onto the same destination shape (2:1 ellipse) with the same anchor points (center-to-center) and tangent lines (at the center, north-south is the ellipse's minor axis and east-west is its major axis), and are both equal-area, yet they're different projections. But there is only one conformal projection with the same properties. On the other hand, creating equal-area projections onto an ellipse with a different aspect ratio is a trivial matter of rescaling, whereas creating conformal projections onto an ellipse with a different aspect ratio requires more fundamental changes to the projection.

[PeteD]

OK, I understand it better now. Thanks for taking the time to explain.

[daan]

A mere year after submitting the manuscript, the paper is finally published. Author’s manuscript is available here. Many thanks again to Milo and PeteD for their help.

Cheers,
— daan

[PeteD]

Glad I could help, however small my contribution.

[PeteD]

The green zones denote where the Lagrange projection has lower area distortion than the Eisenlohr projection at the same point (when normalized such that the scale at the center is 1). The yellow zones denote where the Lagrange projection has worse area distortion than the Eisenlohr projection at the same point. The red zones denote where the Lagrange projection has worse area distortion than the Eisenlohr projection at any point. The chart itself is in Lagrange projection.

eisenlohr-lagrange-compare.png (716.69 KiB) Viewed 35117 times

Revisiting this, it occurred to me that a similar comparison of the Lagrange and Mercator projections could be very interesting. The Mercator should win over Australia, New Zealand, the west coast of North America and the east coast of Asia, and it could give the Lagrange a run for its money.

[Milo]

Let's see if I still remember how my old code works well enough to tweak it.

Here we go. Assuming that I didn't accidentally compare a different metric than the one I used last time (since I didn't double-check all my old code):

mercator-lagrange-compare.png (558 KiB) Viewed 35112 times

mercator-lagrange-compare-sinusoidal.png (266.14 KiB) Viewed 35112 times

So, not just the west coast!

No red zones because the Mercator projection reaches infinite distortion at the poles, but the yellow zones (where Mercator is less inflated than Lagrange) are pretty large.

The important caveat is that both these and my previous comparisons are based on a normalization of equal scale at the center.

Also note that the hourglass shape in this chart is not identical to the one in the previous chart, even though they look very similar.

South America looks nearly the same in the Lagrange and Mercator projections, but it makes sense that it's still slightly less distorted in the Mercator projection. After all, the Mercator projection has minimal scale along the entire equator, Lagrange only at the center. It's just that Mercator mangles Antarctica so badly that it drives up the average over the whole map. But it's not like Lagrange handles Antarctica well...

Lagrange definitely wins over Svalbard, but it's hard to argue that Svalbard is the center of the world.

So is Mercator just the best conformal projection after all, overinflated Greenland and inability to show the entire world in finite space and all?

(Mercator is already favored for navigational maps either way, since it has the unique property of preserving compass directions.)

And what metric might we devise that actually formalizes these observations? I guess that for conformal maps, what's important isn't so much absolute distortion (which is arbitrary and depends on normalization anyway) but rather relative distortion... and even more to the point, relative distortion between nearby points on the map. The exaggeration of Greenland compared to Africa isn't that important because, even if they're on the same meridian, they're not that close to each other. The exaggeration of Greenland compared to Newfoundland is a bigger deal because it affects local shapes. So I think some sort of metric based on the second derivative of a conformal projection, measuring the rate of change of flation over the map, might be called for.

[daan]

And what metric might we devise that actually formalizes these observations? I guess that for conformal maps, what's important isn't so much absolute distortion (which is arbitrary and depends on normalization anyway) but rather relative distortion... and even more to the point, relative distortion between nearby points on the map. The exaggeration of Greenland compared to Africa isn't that important because, even if they're on the same meridian, they're not that close to each other. The exaggeration of Greenland compared to Newfoundland is a bigger deal because it affects local shapes. So I think some sort of metric based on the second derivative of a conformal projection, measuring the rate of change of flation over the map, might be called for.

My thoughts exactly. This is why areal and angular distortion are not naturally comparable. You can talk about points of no distortion on equal-area maps as centers away from which distortion increases, but that does not work for conformal maps. The most natural analog would seem to be where the rate of change in inflation is least (inflation, rather than flation, because you would always take the “uphill” direction for a positive measure). Most useful conformal maps have at least a point where that rate is zero.

I haven’t looked into it in any rigorous way, and I’m not aware of any study that has, but I could imagine directly comparing rate of inflation to some simple transformation of angular deformation as a way of meaningfully comparing conformal and equal-area maps. More generally, you could combine these two measures for any map, and compare combined measures. I suspect this would result in something a lot more useful than arbitrary combinations of the Tissot metric. Since angular deformation α has range [0..180]°, maybe something as simple as tan α/2 could be usefully compared against rate of inflation s, with range [0..∞]. I suspect, though, that the most natural way to compare would be a little more complicated: inflation and deformation tend to increase exponentially away from minima.

— daan