Eisenlohr’s optimal conformal map of the world

[daan]

If you parameterize Lagrange such that it has the same width as August and the same scale at the center, then the Lagrange looks similar but has less area than the August. That means that, on average, points have less areal inflation.

I suppose I should illustrate that. The overlay in red coastlines is Lagrange, with the equator as the center, straight parallel, and about 132°48′ as the circular meridian (I did not calculate the exact value). The poles show that Lagrange has greater scale factors there (actually by many multiples), but you can see that the Lagrange is less inflated over the rest of the map. So, unless you’re interested specifically in the poles (in which case you should, perhaps, consider some other projection), it does look like August has no particular merits over alternatives.

Lagrange overlaid on August

AugustLagrange.png (314.62 KiB) Viewed 21851 times

Cheers,
— daan

[Milo]

The overlay in red coastlines is Lagrange, with the equator as the center, straight parallel, and about 132°48′ as the circular meridian (I did not calculate the exact value).

Which Lagrange projection are you using here? The optimality we discussed before is specifically for the circular Lagrange projection, where the Mercator projection is multiplied by a factor of exactly 0.5, while your picture looks like a member of the generalized Lagrange family where the Mercator projection is multiplied by a different (shallower/larger) factor. Which I guess makes it meaningful that even this other, less-optimal variant of the Lagrange projection still looks better than the August projection, but even so I doubt I'd ever actually use it over the circular Lagrange projection.

Technically speaking, I think the circular form of the projection is due to Lambert, while Lagrange's only innovation was trying other scaling factors. So if the circular form is the only one I care about, it could be argued I should be calling it the Lambert projection. However, Lambert already has way too many map projections named after him: Lambert azimuthal equal-area (quite common and valuable, but also straightforward enough that anyone with a basic knowledge of mathematics could reinvent it), Lambert cylindrical equal-area (which isn't even that good, basically any other standard parallel would produce a better projection), and then just for a change, Lambert conformal conic (both the Lambert azimuthal and Lambert cylindrical in fact being a special case of the Albers conic). Aside from being boringly repetitive, it also means you can't call anything a "Lambert projection" without a wordy disambiguation. Lagrange, while also well-known for contributions in other fields of math and physics, only ever had one map projection to his name, making for a much snappier name. Which is why I prefer calling it that even when I only really care about the circular version.

Over here, I formulated a different way of generalizing the (circular) Lagrange projection that actually is optimal within a certain range of parameters, although it's still not that practical compared to the standard Lagrange projection.

I suppose the one advantage the August and Eisenlohr have over the Lagrange (all variations) is they they don't have singularities at the poles. But like you say, if you care about the poles, none of these projections are particularly advisable.

[daan]

Which Lagrange projection are you using here?

The one that Snyder reports as Lagrange’s projection which, at a glance, appears to be what Lambert reports in his treatise. I had forgotten the details of the optimality conversation; sorry about that. Yes, of course the circular degeneration is the one with the least area. It’s also the one with the most massive distortion at the poles. It’s reasonable, then, to say that the August is a fair alternative to one of these parameterized Lambert’s. Obviously any conformal map is optimal for something by Chebyshev’s criterion; you just need to find the one that’s optimal for the region you want.

And I’m not particularly fussy about singularities.

Cheers,
— daan

[mapnerd2022]

Yes, of course, Lagrange even used a factor that was meant for a map of Europe. He used it to minimize scale distortion near Berlin.

[mapnerd2022]

So yes, it was created by Lambert, but it is called the Lagrange projection because Lagrange studied it's properties thoroughly and developed it's elipsoidal form.

[mapnerd2022]

It's named after not the developed, but a promotor, just how the Bonne projection which is named after Bonne is only so named because he used it considerably for a map of the coast.

[mapnerd2022]

It's not named after the one who developed it, but a promotor, just how the Bonne projection which is named after Bonne is only so named because he used it considerably for a map of the coast, since the Bonne projection had been in use before his birth.

[quadibloc]

Why use August instead of Lagrange?
Obviously, if no one knew about the optimality of Lagrange because that was discovered here on this forum, that's one reason.
But the other thing is that, even if August's conformal projection is imperfect, it is much simpler than the Eisenlohr, and it shares one important property with the Eisenlohr: unlike any form of the Lagrange projection, it is conformal everywhere, including the poles, because the poles are cusps so that angles can remain correct even at them.
Many people, from the behavior of the Stereoscopic and Mercator projections, even if they've also seen the circular case of the Lagrange projection, or the Adams projection, will think that it's impossible for a conformal projection that brings the whole globe into a finite area not to have some points where the projection is not conformal.
And so people have used the August conformal, not even considering the more extended forms of the Lagrange, because conformality everywhere was a property very important to the goals they had in mind for their map.

[PeteD]

Obviously, if no one knew about the optimality of Lagrange because that was discovered here on this forum, that's one reason.

It's been established on this forum that the Lagrange has lower overall areal distortion than the August and the Eisenlohr, but has it been shown to be optimal, i.e. that no conformal projection exists with even lower overall areal distortion (without additional interruption)?

And so people have used the August conformal, not even considering the more extended forms of the Lagrange, because conformality everywhere was a property very important to the goals they had in mind for their map.

I'm struggling to think of an application for which a lack of conformality at two infinitesimally small points would present a serious problem. What goals did you have in mind?

[Milo]

It's been established on this forum that the Lagrange has lower overall areal distortion than the August and the Eisenlohr, but has it been shown to be optimal, i.e. that no conformal projection exists with even lower overall areal distortion (without additional interruption)?

See the last post on the previous page:

As far as I know, the optimality of the Lagrange projection was only noticed in this forum. It was first suggested for this purpose by dummy_index, then expanded on by me, but still without a formally-rigorous proof (even though I'm satisfied by it).

So it's not quite conclusively proven, because I made one assumption that I couldn't find a reliable source for, but it's probably true. If you understand conformal radii better than I do, or at least know a source I missed, feel free to check the proof.