Eisenlohr’s optimal conformal map of the world

[Milo]

This has got me thinking: instead of trying to come up with a good measure for areal distortion and an analogous measure for angular distortion and then combining them to use as an overall distortion metric, maybe it's better to come up with a single measure that serves as a measure of areal distortion in conformal projections, a measure of angular distortion in equal-area projections and a measure of both in compromise projections.

That's kind of what I suggested above:

My line of thinking is to try to come up with a metric that works well for conformal projections, try to come up with a metric that works well for equal-area projections, and then try to come up with a generalized metric that simplifies to either of the preceding two when restricted to projections in that category, but is also applicable to other projections.

Again, resolution-efficiency meets the definition you give: it simplifies to either an areal distortion metric or an angular distortion metric as appropriate, but it isn't simply obtained from an arbitrary weighted combination of two separate distortion metrics. Though I originally invented it as a way of judging compromise projections, and its applicability to conformal and equal-area projections was a somewhat-coincidence.

Could skewness be that measure? It certainly serves as a measure of areal distortion in conformal projections.

Maybe. How does it behave for equal-area projections?

It seems to me that it would still be measuring the derivative of angular distortion, not angular distortion itself, which does not seem overly useful.

This question can ultimately only be answered using a global distortion metric. While it's true that we may never agree on one, that's no reason not to try.

I agree, I was just pointing out that there are multiple options and no obvious reason to favor one over another.

Isn't this desirable? If you're asking the question "What's the best projection for this map?", then the answer will depend very strongly on the region that you're mapping, and an overall distortion metric should reflect this.

That part is gotten when you average the local distortion over the mapped region (using whatever type of average you want). However, it's useful if you still have a local distortion metric that's defined independently and relates in a logical way to your global distortion metric. Especially for world maps, it's often of interest which parts of the map have the highest or lowest distortion (many world maps achieve seemingly-good shapes by pushing most of their distortion to the Pacific ocean and the poles).

Non-Mercator cylindrical projections, for example, have their standard parallel (location of zero angular distortion) remain the same regardless of whether you're using them to project the whole world or a small region. The standard parallel could even fall outside of the mapped region entirely, though that's not in general a good projection.

We're proposing a method by which "location of zero areal distortion" could be defined in a similarly map-extent-independent way (and which you would logically want to keep somewhere inside the region being projected, thus explaining why Mercator is not favorable for mapping regions that do not include the equator).

So 2. and 3. are always the same, while 1. differs only by a constant factor of 2/π, so it doesn't matter which option we take.

Caveat: this is true for the derivative of flation.

If you want to take the derivative of linear scale (with, as you suggest, the convention for non-conformal projections to, for each direction, measure the scale in that direction and then take the derivative in the same direction, and then average those directions), then this isn't a gradient anymore, and so that doesn't apply.

For conformal projections, of course, the distinction doesn't matter.

The integral of the square of these metrics doesn't converge for the equirectangular or Mercator projections, and I dare say for many others, so the root mean square is out. The number of arbitrary choices is dropping like flies.

Good point.

Another observation: the mean of a continuous function is essentially its integral (and, for two-dimensional functions, a surface integral). So what we're doing is taking the second derivative of the projection (first to obtain the Tissot ellipses and thus flation, second to obtain the gradient of logarithmic flation), and then taking a double integral of that (since it's a surface integral)... theoretically putting us back at the same "level" of differentiation/integration as we started! However, it doesn't just give you the original projection function back, since the derivatives are taken in different directions than the integrals, and we took the logarithm between the first and second derivative while doing no such thing for the integral. Furthermore, if you wanted to generalize this concept to a measure of hypervolume distortion on projections of spheres in arbitrary dimension, you would still always take the second derivative (exactly), but the number of integrals would match the dimension of the sphere.

Still, it's a vague and handwavy argument in favor of taking the integral in the "same dimension" (i.e., not squaring) as the derivative.

[PeteD]

Maybe. How does it behave for equal-area projections?

Here's a plot of global skewness S versus global angular deformation E_a for a few equal-area projections, where E_a² = ⟨ε_a²⟩, where ε_a = ln a/b.

S vs Ea.png (168.38 KiB) Viewed 11387 times

There's definitely a correlation, but let's have a closer look at which projections fare well and which projections fare poorly according to each metric.

According to both metrics, the three worst projections are Lambert CEA, sinusoidal and Gall-Peters, which sounds right.

According to E_a, the top 10 projections are Francula V and XIV, Wagner VII, Eckert IV, Canters W31-W34, Strebe 1995 and equal Earth, which again sounds right. Of the remaining projections, Eckert VI beats Mollweide, which beats Hammer. All good.

On the other hand, according to S, Hammer beats 9 of those top 10 projections, and Briesemeister beats all 10 of them! There's something wrong here. And Hammer beats Mollweide, which beats Eckert VI, which also doesn't seem right to me.

It seems to me that it would still be measuring the derivative of angular distortion, not angular distortion itself, which does not seem overly useful.

Yes, I think that's what it's doing, which is obviously not quite what we want. It's not a complete disaster, but E_a does seem to be the better angular distortion metric. I don't think skewness is the metric we're looking for.

[quadibloc]

In case it may be helpful, on my web page, at
http://www.quadibloc.com/maps/mcf0702.htm
I give Snyder's formula for the Eisenlohr, and I also have drawings of both the alpha component and the beta component.

[daan]

I don’t know of any others, although Snyder’s GS50 is nearly such.

Over the area it's optimized for, or when pathologically extended to the whole globe?

It falls apart quickly outside its region of optimality; you don’t have to expand the range literally to the whole globe to see gargantuan inflation and deflation.

— daan

[mapnerd2022]

I don’t know of any others, although Snyder’s GS50 is nearly such.

Over the area it's optimized for, or when pathologically extended to the whole globe?

It falls apart quickly outside its region of optimality; you don’t have to expand the range literally to the whole globe to see gargantuan inflation and deflation.

— daan

Precisely. "As the distance from the projection center increases,meridians, parallels and shorelines begin to exhibit loops, overlapping and other undesirable curves." Also, John Snyder refers to a "world" map on the GS50 as "nearly illegible" "with meridians and parallels intertwined like wild vines."

[PeteD]

I can't think of a definition of angular distortion that's analogous to the derivative of flation in the same way that E_p and E_a are analogous to each other.

Just to briefly elaborate on this, it's not so much that there isn't an obvious counterpart to the derivative of flation as that this obvious counterpart isn't a measure of angular distortion.

Our measure of areal distortion based on the derivative of flation is given by:

abs(derivative of ln ab)

or equivalently:

abs(1/(ab) * derivative of ab),

and its obvious counterpart is:

abs(derivative of ln a/b)

or equivalently:

abs(b/a * derivative of a/b).

Since ln a/b is already a good measure of angular distortion, there doesn't seem much point in taking its derivative. On the other hand, if we consider the equivalent abs(b/a * derivative of a/b) definition and use some wishful thinking, then we might be able to convince ourselves for a minute that it's essentially just b/a and that multiplying by the derivative of a/b is just another way to ensure that inflations and deflations by the same factor are treated equally.

We could go on to tell ourselves that, just as with E_p and E_a, there's a symmetry between the above counterparts: the first is equal to 2 abs(tan ϕ) for the Mercator projection, abs(tan ϕ) for equirectangular projections and zero for cylindrical equal-area projections, while the second is equal to zero for the Mercator projection, abs(tan ϕ) for equirectangular projections and 2 abs(tan ϕ) for cylindrical equal-area projections.

Hold on, equirectangular projections? Cylindrical equal-area projections? Surely an angular distortion measure should have different values for equirectangular projections and cylindrical equal-area projections with different standard parallels?

Here the illusion shatters – of course it should. The second of the above counterparts has a value of zero along the equator of any equirectangular projection and any cylindrical equal-area projection, even though angular distortion is equal to zero along the equator of these projections only if the standard parallel is set to the equator. The second of the above counterparts therefore can't be a measure of angular distortion, which was kind of obvious all along when you consider the abs(derivative of ln a/b) definition.

[Milo]

Okay. Let's try to get some real results.

To be completely unambiguous...

The measure of local areal distortion that will be used in this post is: the norm of the gradient of half the logarithm of flation. (The "half" factor is somewhat arbitrary, but it compensates for flation being a two-dimensional measure.)

The measure of global areal distortion that will be used in this post is: the mean of local areal distortion over the whole sphere / region being mapped (i.e., the surface integral of local areal distortion over the region divided by the area of that region).

Then the Mercator projection has:
local areal distortion = |tan(latitude)|
global areal distortion = 1

Now for the Lagrange projection. I worked out the formula for flation of the Lagrange projection back when I wrote the program to make my comparison images, but that was a while ago, so I don't remember exactly how I did it and will need to look over my old work carefully to make sure I'm using it right.
...Okay, here we go.
If we define the complex value "mercator = asinh(tan(latitude)) + longitude*i", then:
linear scale = square root of flation = cosh(ℜ(mercator)) / |1 + cosh(mercator/2)| * 2 = 1/cos(latitude) / |1 + cosh(mercator/2)| * 2 = (bunch of annoying math) = 2/cos(latitude) / sqrt(1 + sqrt((cos(longitude)+1)*(cos(latitude)+1)/cos(latitude)) + (cos(longitude)*cos(latitude)+1)/(2*cos(latitude)))
I can't manage to simplify it more than that.
Now, taking the gradient of that takes caution, since we want to take the gradient on the sphere: sqrt((∂ln(scale)/∂longitude/cos(latitude))² + (∂ln(scale)/∂latitude)²)
...Which is really messy and I've given up on trying to simplify it, which is going to prevent me from computing an integral. Still, using Maxima to assist in differentiation and not bothering to check the result, I at least get a formula which I can plug in to output another comparison chart, which gives:

mercator-lagrange-compare.png (265.95 KiB) Viewed 48575 times

...That's a lot more favorable to Lagrange than the last chart. Did I make a mistake somewhere? Or does Lagrange win after all, even over Alaska?
Through an even more handwavy hackjob of numeric integration, I get a global areal distortion for Lagrange of 0.854.

[PeteD]

Thanks for the calculations!

Along the equator, where both projections fare well, the Mercator fares better, whereas towards the poles, where both projections fare badly, the Mercator fares worse, so I suppose it's not too surprising that the Mercator has a greater flation gradient over most of the globe.

[daan]

Did I make a mistake somewhere?

Not that I can find; the scale measure matches my own calculations. Here’s a “simplification” of the gradient; I don’t think you’ll dig a closed-form integral out of that: [Edited for clarity]

(machine form)
sqrt(((1 + (-4*cos(lambda)^3 - 72*cos(lambda)^2 - 192*cos(lambda) - 128)*cos(phi)^5 + (-40*cos(lambda)^2 - 160*cos(lambda) - 128)*cos(phi)^4 + (3*cos(lambda)^3 + 64*cos(lambda)^2 + 164*cos(lambda) + 104)*cos(phi)^3 + (35*cos(lambda)^2 + 160*cos(lambda) + 136)*cos(phi)^2 + (25*cos(lambda) + 32)*cos(phi))*sqrt((cos(lambda) + 1)*(cos(phi) + 1)/cos(phi)) - 24*(cos(phi) + 1)*(-1/3 + (cos(lambda) + 4/3)*(cos(lambda) + 4)*cos(phi)^4 + (-cos(lambda)^2/6 + (4*cos(lambda))/3 + 8/3)*cos(phi)^3 + (-(2*cos(lambda)^2)/3 - (14*cos(lambda))/3 - 5)*cos(phi)^2 + (-(5*cos(lambda))/3 - 19/6)*cos(phi))*(cos(lambda) + 1))/(cos(phi)^2*((1 + (cos(lambda)^2 + 16*cos(lambda) + 16)*cos(phi)^2 + (14*cos(lambda) + 16)*cos(phi))*(1 + (cos(lambda) + 2)*cos(phi))*sqrt((cos(lambda) + 1)*(cos(phi) + 1)/cos(phi)) + 6*(3 + (cos(lambda) + 4)*cos(phi))*(cos(phi) + 1)*(1/3 + (cos(lambda) + 4/3)*cos(phi))*(cos(lambda) + 1))))/2

(better human readability)
½ sec 𝜑 ∙
√(
(
(
1 +
[−4 cos³𝜆 − 72 cos²𝜆 − 192 cos 𝜆 − 128] ∙ cos⁵𝜑 +
[−40 cos²𝜆 − 160 cos 𝜆 − 128] ∙ cos⁴𝜑 +
[3 cos³𝜆 + 64 cos²𝜆 + 164 cos 𝜆 + 104] ∙ cos³𝜑 +
[35 cos²𝜆 + 160 cos 𝜆 + 136] ∙ cos²𝜑 +
[25 cos 𝜆 + 32] ∙ cos 𝜑
) ∙
√(
[cos 𝜆 + 1] ∙ [cos 𝜑 + 1]/cos 𝜑
)

−

24 ∙
(cos 𝜑 + 1) ∙
(cos 𝜆 + 1) ∙
(
−1/3 +
[cos 𝜆 + 4/3] ∙ [cos 𝜆 + 4] ∙ cos⁴𝜑 +
[−cos²𝜆/6 + (4 cos 𝜆)/3 + 8/3] ∙ cos³𝜑 +
[−(2 cos²𝜆)/3 − (14 cos 𝜆)/3 − 5] ∙ cos²𝜑 +
[−(5 cos 𝜆)/3 − 19/6] ∙ cos 𝜑
)
)

/

(
(
1 +
[cos²𝜆 + 16 cos 𝜆 + 16] ∙ cos²𝜑 +
[14 cos 𝜆 + 16] ∙ cos 𝜑
) ∙
(
1 +
[cos 𝜆 + 2] ∙ cos 𝜑
) ∙
√(
[cos 𝜆 + 1] ∙ [cos 𝜑 + 1]/cos 𝜑
)

+

6 ∙
(
3 + [cos 𝜆 + 4] ∙ cos 𝜑
) ∙
(
1/3 + [cos 𝜆 + 4/3] ∙ cos 𝜑
) ∙
(cos 𝜑 + 1) ∙
(cos 𝜆 + 1)
)
)

— daan

[quadibloc]

I see it was from a paper of yours (of Daniel Strebe's) that I recently learned that Chebyshev first hypothesized the condition of uniform scale around the boundary for an optimal conformal projection. Oddly, though, you refer to Pafnuty Chebyshev, as if he wasn't the Chebyshev, who hardly needs to be described further, but simply refer to the one who then proved that conjecture simpy as Grave.
When I added this tidbit to my page on August's conformal and the Eisenlohr, I made a quick Google search, and I believe it was Dimitri Grave to whom you were referring.