Eisenlohr’s optimal conformal map of the world

[Milo]

I've done some math to determine the extreme points of the Eisenlohr projection.

According to my calculations:
longitude = +pi, latitude = 0 -> x = sqrt(8) - ln(3+sqrt(8)), y = 0
longitude = +pi, latitude = +pi/3 -> x = sqrt(3) - ln(2+sqrt(3)), y = sqrt(3) - pi/3

Or in hyperbolic notation:
longitude = +pi, latitude = 0 -> x = sqrt(8) - acosh(3), y = 0
longitude = +pi, latitude = +pi/3 -> x = sqrt(3) - acosh(2), y = sqrt(3) - acos(1/2)

Other notations for the first value include x = 2 * (sqrt(2) - ln(1+sqrt(2))) and x = 2 * (sqrt(2) - asinh(1)).

(All of these are in the coordinates you're using above, so without the 3 + √8 scaling factor.)

Thus, the overall aspect ratio of the projection is (sqrt(8) - acosh(3)) / (sqrt(3) - pi/3), or approximately 1.556. This seems to match the aspect ratio I'm seeing from screenshotting existing examples of the projection, so I think I did my math right.

[Milo]

And here's my implementation!

I'm still missing this part:

My paper describes a reasonably simple test for determining whether z is outside the range, which I omit here.

...so what you get is a peek of what an indefinitely-extended Eisenlohr projection looks like. There seems to be some fractal behavior near the vertical axis, although using a real map rather than some abstract pattern obscures the details.

eisenlohr.png (722.15 KiB) Viewed 26038 times

Incidentally, I didn't need to do anything about the sign in (1 ± 2 sin β cos β) (in the computation of ψ). I just used a consistent (1 - sin(2β)) and it works fine. In fact, using a plus sign gives nonsensical results that zoom in on a distorted Africa and ignore everything else.

[daan]

I've done some math to determine the extreme points of the Eisenlohr projection.

According to my calculations:
longitude = +pi, latitude = 0 -> x = sqrt(8) - ln(3+sqrt(8)), y = 0
longitude = +pi/3, latitude = +pi -> x = sqrt(3) - ln(2+sqrt(3)), y = sqrt(3) - pi/3

I’m going to say you’ve reversed longitude and latitude in the second case. Presuming so, your calculations match mine.

— daan

[Milo]

I’m going to say you’ve reversed longitude and latitude in the second case. Presuming so, your calculations match mine.

Uh, d'oh. Fixed.

Meanwhile:

s = |[cos λ + i sin λ] / [(sin φ + 1) ∙ (√(2ψ) + ψ + 1)²]|

I think I have derived a version of this that's written purely in terms of ψ:
s = (|ψ| + 1/|ψ|) / |√2 + √ψ + 1/√ψ|²
I still need to double-check this, but it seems more convenient than obtaining λ = 𝕀(ln(ψ)) only to immediately plug it back into e ^{i λ} = cos λ + i sin λ.

[daan]

s = |[cos λ + i sin λ] / [(sin φ + 1) ∙ (√(2ψ) + ψ + 1)²]|

I think I have derived a version of this that's written purely in terms of ψ:
s = (|ψ| + 1/|ψ|) / |√2 + √ψ + 1/√ψ|²
I still need to double-check this, but it seems more convenient than obtaining λ = 𝕀(ln(ψ)) only to immediately plug it back into e ^{i λ} = cos λ + i sin λ.

I think in most cases you would start from λ, φ and so you have them available. Still:
cos λ + i sin λ = ψ / |ψ| = sgn(ψ), and
sin φ + 1 = (1 – |ψ|²)/(1 + |ψ|²)
so it is clear that you can easily compute purely in ψ. I should note that in the paper. Thanks.

— daan

[Milo]

cos λ + i sin λ = ψ / |ψ| = sgn(ψ)

Come to think of it, isn't the cos λ + i sin λ part totally redundant? Absolute values are multiplicative (|a b| = |a| |b|), and so since you take the absolute value in the end anyway, cos λ + i sin λ can simply be reduced to 1.

sin φ + 1 = (1 – |ψ|²)/(1 + |ψ|²)

This doesn't seem to be correct. Just looking at positive real values of psi, where you don't even need to take the absolute value, gnuplot gives me this:

gnuplot1.png (13.95 KiB) Viewed 26026 times

By contrast, when I plug in sin φ + 1 = 2/(1 + |ψ|²), I get a perfect match:

gnuplot2.png (12 KiB) Viewed 26026 times

Speaking of that factor 2 in the latter formula, when I plug in ψ = 1 (i.e., λ = φ = 0) in your formula for s, I get s = 1 / (√2 + 2)² = 1 / (√2 + 2)², a factor 2 off from the expected 1 / (3 + √8).

[Milo]

And finally, what I've been after all this time.

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 (562.75 KiB) Viewed 26021 times

Eisenlohr wins out over Europe, and some icy places nobody cares about. Lagrange wins out over North America, South America, Asia, and Australia. Africa is tied. The red zones are entirely limited to the Arctic Ocean and Antarctica.

Of course, since it just so happens to be Europe that's exaggerated in the Lagrange projection, the Gall-Peters crowd are still going to be decrying it as imperialist.

[PeteD]

Very interesting. Would you be able to project your chart onto an equal-area projection so that we could see the relative sizes of the regions where each of the Lagrange and Eisenlohr projections is better?

[Milo]

Very interesting. Would you be able to project your chart onto an equal-area projection so that we could see the relative sizes of the regions where each of the Lagrange and Eisenlohr projections is better?

In theory, yes. In practice, I just thought I was being clever using complex math all the way, which of course is optimal for making comformal projections easy and is more of a hassle for equal-area ones.

The reason I was talking earlier about computing scale purely from ψ is because that's exactly what I'm doing.

But I managed to wrangle a sinusoidal projection for you. This works by first converting the sinusoidal coordinates to longitude-latitude, then converting that to Mercator coordinates, and then using conformal math from there. And because I'm lazy, forgetting the longitude-latitude coordinates and recomputing them from the Mercator coordinates later when I need them again to reference the source image. (I did check it against a sinusoidal projection made using a completely different algorithm to confirm I did it right.)

sinusoidal.png (265.41 KiB) Viewed 26015 times

An interesting note is that the red zones look convex above, but concave here.

[PeteD]

Thank you! That makes it much more obvious that Lagrange is the clear winner over most of the globe.