Tobler’s hyperelliptical projection

[Milo]

I don’t care to develop new pseudocylindric projections; that’s a broadly plowed field, and I don’t really care for pseudocylindric projections in general. In fact, I don’t think I’ve ever intentionally¹ developed one. I view them as tools on the way to more interesting projections.

Can you give an example of a pseudocylindrical projection (that isn't just cylindrical) that is actually used as part of the construction of another, non-pseudocylindrical projection?

Now, since you accidentally bungled the integrand, you will think that 𝜃(𝑦(𝜑)) is 𝑦(𝜑), unaware that it’s mangled,

That is the part I disagree with.

What you believe 𝑦 is depends on the integral, not just on 𝜃.

For simplicity, let's consider the case where 𝑘 = 1, 𝛾 = 2, giving the Collignon projection. However, it's mangled by 𝜃(𝑦) = (𝑦−1)³+1, just to come up with an arbitrary function that has fixed points at 0 and 2 as well as the correct overall integral over that interval.

You will then compute:
sin 𝜑 = 1/2 ∙ ∫₀^𝑦(1 − (𝑧−1)³) d𝑧
sin 𝜑 = 𝑦/8 ∙ (8 − 6𝑦 + 4𝑦² − 𝑦³)

Whereas the correct value is given by:
sin 𝜑 = 𝑦 - (𝑦/2)²
And this same formula distorted by 𝜃 is:
sin 𝜑 = 𝑦/4 ∙ (12 − 21𝑦 + 22𝑦² − 15𝑦³ + 6𝑦⁴ − y⁵)

It is obvious from the degree alone that these are not the same polynomial.

[daan]

Now, since you accidentally bungled the integrand, you will think that 𝜃(𝑦(𝜑)) is 𝑦(𝜑), unaware that it’s mangled,

That is the part I disagree with.

What you believe 𝑦 is depends on the integral, not just on 𝜃.

For simplicity, let's consider the case where 𝑘 = 1, 𝛾 = 2, giving the Collignon projection. However, it's mangled by 𝜃(𝑦) = (𝑦−1)³+1, just to come up with an arbitrary function that has fixed points at 0 and 2 as well as the correct overall integral over that interval.

You will then compute:
sin 𝜑 = 1/2 ∙ ∫₀^𝑦(1 − (𝑧−1)³) d𝑧
sin 𝜑 = 𝑦/8 ∙ (8 − 6𝑦 + 4𝑦² − 𝑦³)

Whereas the correct value is given by:
sin 𝜑 = 𝑦 - (𝑦/2)²
And this same formula distorted by 𝜃 is:
sin 𝜑 = 𝑦/4 ∙ (12 − 21𝑦 + 22𝑦² − 15𝑦³ + 6𝑦⁴ − y⁵)

It is obvious from the degree alone that these are not the same polynomial.

I’m not clear on what you disagree about. My assertions do not depend on those two polynomials being equivalent. Forgetting, for the moment, any self-delusions of variety, “you will think that 𝜃(𝑦(𝜑)) is 𝑦(𝜑)”, do you agree that the procedure I detail results in an equal-area projection?

If you agree with that, then are you disputing that a person would mistakenly use 𝜃(𝑦) as 𝑦? The reason I say they would is because (a) I did—because my calculations were inaccurate; and (b) In the forward form of the projection, you have no other means of obtaining 𝑦, and so even if you mess that up with 𝜃, that’s all you’ve got.

— daan

[Milo]

do you agree that the procedure I detail results in an equal-area projection?

No.

If I attempt it, I come up with a mangled Collignon projection where 𝑦 is computed from 𝜑 according to:
sin 𝜑 = 𝑦/8 ∙ (8 − 6𝑦 + 4𝑦² − 𝑦³)
And then 𝑥 is computed from 𝜆 and the computed 𝑦 according to:
𝑥 = 𝜆 ∙ (1 - 𝑦/2)

Then:
d𝑦/d𝜑 = 2 cos 𝜑 / (1 − (𝑦 − 1)³)
d𝑥/d𝜆 = (1 - 𝑦/2)
d𝑥/d𝜆 ∙ d𝑦/d𝜑 / cos 𝜑 = 1 / (𝑦² − 𝑦 + 1) ≠ 1
(unless 𝑦 = 0 or 𝑦 = 1, which happens when 𝜑 = 0 or 𝜑 = asin(5/8))

This mangled Collignon projection will be pseudocylindrical and have the same boundary shape as the true Collignon projection, but will not be equal-area.

Is this not the same projection that your procedure would generate, given the same parameters including 𝜃?

[daan]

Then:
d𝑦/d𝜑 = 2 cos 𝜑 / (1 − (𝑦 − 1)³)

(I think your integrand should be (2 − (𝑦 − 1)³) here because 𝛾 = 2, but that’s not material to the conclusions.)
This is how I compute d𝑦/d𝜑:
d/d𝜑 (∫₀^𝑦 1 − (1 − 𝑧)³ d𝑧) = d/d𝜑 (2 sin 𝜑)
[d/d𝜑 𝑦(𝜑)] (𝑦³(𝜑) − 3𝑦²(𝜑) + 3𝑦(𝜑)) = 2 cos 𝜑
[d/d𝜑 𝑦(𝜑)] = 2 cos 𝜑 / (𝑦³(𝜑) − 3𝑦²(𝜑) + 3𝑦(𝜑)), so, discarding functional notation since we aren’t doing any more calculus,
d𝑦/d𝜑 = 2 cos 𝜑 / (𝑦³ − 3𝑦² + 3𝑦) = 2 cos 𝜑 / (1 − (1 − 𝑦)³)

This differs from your result. Meanwhile, what I am stating is to use the mangled function of 𝑦 in 𝑥:
𝑥 = 𝜆 (1 − (1 − 𝑦)³) / 2

I think that is where we have been talking past each other. You state 𝑥 = 𝜆 (1 − 𝑦/2). The reason I don’t use that is because, presumably, the same forces that resulted in 𝜃 within the integral calculation to compute the integrand would also be in play in computing 𝑥, since 𝑥 is a function of the integrand. And that is, in fact, what happened to me.

Of course, we have not solved for 𝑦 yet at this point; it’s an annoying quartic equation, so I won’t bother, but continuing,
d𝑥/d𝜆 = (1 − (1 − 𝑦)³) / 2

and so
d𝑦/d𝜑 ∙ d𝑥/d𝜆 = cos 𝜑

as required. To generalize, this generates equal-area projections:
𝑥 = 𝜆∙𝑓(𝑦)
sin 𝜑 = ∫₀^𝑦𝑓(𝑧) d𝑧

— daan

[daan]

I don’t care to develop new pseudocylindric projections; that’s a broadly plowed field, and I don’t really care for pseudocylindric projections in general. In fact, I don’t think I’ve ever intentionally¹ developed one. I view them as tools on the way to more interesting projections.

Can you give an example of a pseudocylindrical projection (that isn't just cylindrical) that is actually used as part of the construction of another, non-pseudocylindrical projection?

My 1995 projection, for example.

Cheers,
— daan

[Milo]

d/d𝜑 (∫₀^𝑦 1 − (1 − 𝑧)³ d𝑧) = d/d𝜑 (2 sin 𝜑)

You made a sign error here, I said (𝑧 − 1) not (1 − 𝑧).

Our respective final values for d𝑦/d𝜑 differ only in that exact sign difference.

To really spell things out pedantically:
I defined: 𝜃(𝑦) = (𝑦−1)³+1
The raw integral of interest (verbatim from your post) is: ∫₀^𝑦(𝛾^𝑘 − 𝜃(𝑧))^1/𝑘 d𝑧
Substituting the above 𝜃 gives: ∫₀^𝑦(𝛾^𝑘 − (z−1)³ − 1)^1/𝑘 d𝑧
Substituting 𝛾 = 2 and 𝑘 = 1 gives: ∫₀^𝑦(2 − (z−1)³ − 1)^1/𝑘 d𝑧
Simple elimination of terms gives ∫₀^𝑦(1 − (z−1)³)^1/𝑘 d𝑧

(Incidentally, I just now notice that your formula neglects to raise 𝜃(𝑧) to the power of 𝑘, but that is irrelevant in our case where 𝑘 = 1, and any desired exponentiation could just be subsumed into 𝜃 anyway.)

Meanwhile, what I am stating is to use the mangled function of 𝑦 in 𝑥:
𝑥 = 𝜆 (1 − (1 − 𝑦)³) / 2

I think that is where we have been talking past each other. You state 𝑥 = 𝜆 (1 − 𝑦/2). The reason I don’t use that is because, presumably, the same forces that resulted in 𝜃 within the integral calculation to compute the integrand would also be in play in computing 𝑥, since 𝑥 is a function of the integrand.

This is indeed the part I'm not following.

In the "real" scenario I'm modelling, you wouldn't actually know the behavior of 𝜃. You're trying to calculate ∫ 𝑓(𝑦) d𝑦, but because of inaccuracies in your numerical integration algorithm, you're instead calculating ∫ 𝑓(𝜃(𝑦)) d𝑦. Because these inaccuracies are inherent to your integration algorithm, they cannot be separated from each other. You can calculate 𝑓(𝑦), or you can calculate ∫ 𝑓(𝜃(𝑦)) d𝑦 (and presumably you can also calculate the inverses of those expressions, through numerical solution if nothing else, if you're trying to figure out the value of 𝑦), but you cannot calculate 𝑓(𝜃(𝑦)) or ∫ 𝑓(𝑦) d𝑦 (or their respective inverses).

The scenario you're describing would only make sense if the numerical inaccuracies take place inside the calculation of 𝑓, rather than in the integration algorithm.

The equation describing 𝑥 in terms of 𝑦 is extremely simple and does not involve integration at all, and therefore you're unlikely to miscalculate that equation itself to any significant degree, even if you're plugging in the wrong value for 𝑦. Therefore, I don't see how you're getting your "two wrongs make a right" result.

To generalize, this results in equal-area projections:
𝑥 = 𝜆∙𝑓(𝑦)
sin 𝜑 = ∫₀^𝑦𝑓(𝑧) d𝑧

Yes, I agree with that part. This is indeed a valid pseudocylindrical equal-area projection, for any reasonable value of 𝑓. (Reasonable = non-negative over its domain, and the integral eventually reaches 1 for high enough 𝑦. Although without that, it might still be a valid regional projection, just not extensible to a full world map.) Indeed, that exact formulation is very close to how my implementation calculates those projections (except that instead of specifying 𝑓 mathematically, it's read from an image file). However, it only works if you are able to compute the integral to sufficient accuracy (which I do through pixel-counting).

[Atarimaster]

It now surprises me that Siemon and Wagner did not make it all the way to the general homotopy method that I described.

Well, maybe Wagner thought of it.
I was just browsing through "Kartographische Netzentwürfe" looking for something else when a certain paragraph caught my eye. I must have read before but at that time, I apparently didn’t really think about it and forgot it right away. It’s in the final chapter “Concluding Remarks”, right after Wagner covered blended projections (Winkel Tripel and Eckert III to VI).
Here it is (translation by DeepL with a few corrections):

Im Grunde genommen sind solche Mischungen auch nichts anderes als Netztransformationen in der Art der vorher aufgeführten Umbezifferungen. Diese ihrerseits sind vom mathematischen Standpunkt aus auch nur Spezialfälle, die in viel allgemeineren Umwandlungen bereits enthalten sind. Um diesem Problem auf den Grund gehen zu können, bedarf es jedoch eines beachtlichen Maßes an mathematischem Rüstzeug (Differentialgeometrie bis zu partiellen Differentialgleichungen), so daß wir es uns hier versagen müssen, auch nur andeutungsweise darauf einzugehen.
----
Basically, such blends are also nothing more than transformations in the manner of the previously listed Umbeziffern. From a mathematical point of view, these in turn are also only special cases that are already contained in much more general transformations. In order to get to the bottom of this problem, however, a considerable amount of mathematical equipment is required (differential geometry up to partial differential equations), so that we must refrain here from going into it even slightly.

I have no idea if the homotopy uses differential geometry and partial differential equations – I wouldn’t know them if they hit me in the face –, but is it possible that Wagner referred to solutions like this?

Kind regards,
Tobias

[daan]

Nice find, Tobias.

I think that that passage indicates that Wagner did not conceive of the homotopy, nor even of “substitute deprojection”, but instead believed Umbeziffern to be a closed-form special case of something a lot more complicated. I think this because calculus has no role in the homotopy technique or for substitute deprojection (such as used in my 1995 projection). Differential calculus might be useful in constructing a formal proof of those techniques, but even then, I doubt it’s needed. (I did not offer formal proofs in my papers.)

Interesting.

Thanks.
— daan

[Atarimaster]

Thank you for clearing that up!

It’d be interesting to learn what “something a lot more complicated” Wagner had in mind. But I guess unless he elaborated on that in a later paper I don’t know about, we’ll never know.

[Milo]

Having interesting ideas and then failing to write them down seems to be a common failing of mathematicians.