Tobler’s hyperelliptical projection

[Atarimaster]

I don’t think that I’ve ever found a configuration that I liked better than Tobler’s preferred one (α = 0, γ = 1.183136, k = 2.5).
I have not tried many configurations, though.


(Edit:)
Tobler’s fig. 6 with α = 0.25, γ = 1.194, k = 2 is quite nice, too.

[daan]

Does that mean we won’t get the “Projection parameters are not compatible with each other” message anymore on the hyperelliptical projection?

Yes. It will be awhile before I put out an update, though.

— daan

[daan]

Speaking of which, which is your favourite configuration of the Hyperelliptical projection?

I like Tobler’s preferred configuration. If I have to go with a pseudocylindric projection, it will do fine because it gives much of the advantages of a projection that stretches the poles into lines without actually stretching the poles into lines.

— daan

[Milo]

Turns out I don't actually use the same parameters Tobler does. I'm looking at Wikipedia's definition, and I'm barely making any sense of this. It's ridiculous to require "γ ≈ 1.27323394" to reproduce something as simple as the Mollweide projection. And the definition the article gives for hyperellipses is clearly actually for hypercircles: how are you supposed to vary the aspect ratio? Would, say, Bromley still have an α of 0 but a different γ? Okay, I've figured it out. At least when α = 0, the aspect ratio is actually equal to π/γ, and the 1.27323394 value on Wikipedia is for the Bromley projection. For the conventional 2:1 Mollweide projection (or any other 2:1 projection), you'd use γ = π/2 ≈ 1.57079632. Meanwhile, α values other than 0 appear to perform a crude blending with the Lambert cylindrical equal-area projection (and always Lambert, not some other aspect ratio / standard parallel, regardless of the value of γ), resulting in projections that are no longer actually true hyperellipses in shape?

Instead, my implementation asks for map width, map height, x exponent, and y exponent. The "official" Tobler projection as defined by Tobler himself always has both exponents equal to each other, but trying other combinations is a trivial generalization, and some interesting things can happen if you do. If the x exponent is larger than the y exponent, you can get a projection that looks quite similar to a flat-pole projection like Eckert IV, without actually having a pole line, no α necessary.

(α is also not necessary to produce an actual cylindrical equal-area projection, since that is the limit as the exponents go to infinity.)

Compare, a hyperelliptical projection with x exponent 4 and y exponent 2:

tobler42+154.png (661.58 KiB) Viewed 422 times

And the Eckert IV projection:

eckertIV+154.png (668.53 KiB) Viewed 422 times

You can probably get even closer to Eckert IV with finer tweaking of the exponents, but it's not like making an exact copy is the point here.

Of course, the plain Mollweide projection is pretty nice on its own, but even if it's technically a special case of the Tobler projection, naming it feels like cheating

[mapnerd2022]

Very interesting information! I have searched the hyperellipse a while ago and found that it is also called Lamé curve since it was devised by Gabriel Lamé.

[daan]

Meanwhile, α values other than 0 appear to perform a crude blending with the Lambert cylindrical equal-area projection (and always Lambert, not some other aspect ratio / standard parallel, regardless of the value of γ)

It’s pretty hokey, but it came for free, pretty much. I’d call it obsolete after my homotopy, letting you blend any two projections. If both have straight parallels, so will the result. If both are equal-area, so will be the result.

— daan

[daan]

Instead, my implementation asks for map width, map height, x exponent, and y exponent. The "official" Tobler projection as defined by Tobler himself always has both exponents equal to each other, but trying other combinations is a trivial generalization, and some interesting things can happen if you do. If the x exponent is larger than the y exponent, you can get a projection that looks quite similar to a flat-pole projection line Eckert IV, without actually having a pole line, no α necessary.

I like the Eckert IV recreation lacking the pole-line. I think we have the mathematical and computational tools now to eliminate pole-lines while keeping the much of their benefit.

— daan

[daan]

Does that mean we won’t get the “Projection parameters are not compatible with each other” message anymore on the hyperelliptical projection?

Yes.

— daan

Not true. In any circumstance where…

1 > α γ – (α – 1) γ √π 2^-2/k Γ(1 + 1/k) / Γ(1/2 + 1/k)

…the projection can’t fulfill its own requirements. So, for example, γ = 1.12, k = 2.5 does not yield a mathematically coherent projection because the right side of the inequality evalutes to 0.946…. In fact, Tobler’s recommended γ = 1.183136 is very near the lower limit for γ at k = 2.5; the right side evaluates to 1.00002… . At least in my update, parameter combinations that work mathematically will also work computationally.

— daan

[daan]

Of course, the plain Mollweide projection is pretty nice on its own, but even if it's technically a special case of the Tobler projection, naming it feels like cheating

The whole world is wondering what γ must be for the hyperelliptic to result in the Mollweide, since Tobler only gave k = 2 and α = 0, ignoring γ, when he claimed that the Mollweide was a degenerate case of the hyperelliptic.

The solution is γ = 4/π.

— daan

[Milo]

I’d call it obsolete after my homotopy, letting you blend any two projections.

I've read about that one. It's pretty clever, but I do find it a little inelegant due to its lack of symmetry, which prevents a 50% blend of two projections from actually feeling like a proper "midpoint".

That said, it's interesting to note that there is at least one important projection whose usual definition is technically a Strebe homotopy: the Lagrange projection. (This, incidentally, demonstrates that "needing infinite space" is not a property preserved by the process, since both of the input projections have it but the blended one doesn't. Blending them in the opposite direction, however, would preserve the property.)

I'm looking at Wikipedia's definition, and I'm barely making any sense of this.

Can someone who has access to Tobler's original work check if Wikipedia's formulae are correct? Because something just isn't adding up.

Wikipedia says:
αy = sin φ + (α − 1) / γ ⋅ ∫₀^y (γ^k − z^k)^1/k dz

Plugging in α = 0, this can be rearranged to:
sin φ = 1/γ ⋅ ∫₀^y (γ^k − z^k)^1/k dz
Then, taking the derivative of both sides with regard to y gives:
d sin φ / dy = 1/γ ⋅ (γ^k − y^k)^1/k
Plugging in y = 0, this cancels out to:
d sin φ / dy = 1

Likewise, plugging in y = 0 to the formula for x gives x = λ.

Taken at face value, this means that the scale at the equator in both the x and the y directions is independent of the value of γ, which is clearly nonsensical if γ is supposed to control the aspect ratio. And just looking at the formula for x on its own makes it clear that it does.

(Incidentally, may I note that the symbols for γ and y look annoyingly similar to each other?)

1 > α γ – (α – 1) γ √π 2^-2/k Γ(1 + 1/k) / Γ(1/2 + 1/k)

What.

I mean, this is a valid mathematical formula and I'm technically capable of parsing it, but wow is that complicated.

So, for example, γ = 1.12, k = 2.5 does not yield a mathematically coherent projection because the right side of the inequality evalutes to 0.946....

Again, this doesn't make sense if my interpretation of γ as a way of controlling the aspect ratio is correct.

The example you give would correspond to an aspect ratio of π/1.12 ≈ 2.80499344, and my implementation is totally capable of handling that alongside k = 2.5:

impossible.png (159.46 KiB) Viewed 5308 times

(Rounding to the nearest pixel means this image technically has γ = π ⋅ 285 / 800 ≈ 1.11919238.)

The whole world is wondering what γ must be for the hyperelliptic to result in the Mollweide, since Tobler only gave k = 2 and α = 0, ignoring γ, when he claimed that the Mollweide was a degenerate case of the hyperelliptic.

The solution is γ = 4/π.

Again, this clashes with my own interpretation, which gave γ = π/2.

The only way this makes sense is if either one of us is seriously misreading something, or the formulae on Wikipedia are wrong.