Experimental projections

[mapnerd2022]

I'm responding to Mr.Milo's opinion on the Van der Grinten I, sorry if it isn't clear.

[PeteD]

In case anyone else is as confused as I was, most of the post in question is a quotation of Milo's comment from 16th March, and the last paragraph, starting from the bit about Alois Bludau, is mapnerd's response thereto.

[daan]

In case anyone else is as confused as I was, most of the post in question is a quotation of Milo's comment from 16th March, and the last paragraph, starting from the bit about Alois Bludau, is mapnerd's response thereto.

Thanks, PeteD. I updated the posting in question to use the quote mechanism.

— daan

[mapnerd2022]

Thank you, Mr Strebe. I'm sorry if my quotation was ambiguous in any way.

[PeteD]

Open book projection:

open book.png (292.76 KiB) Viewed 10306 times

[daan]

Now that’s cute. I can imagine ways of generating that. What is your way?

— daan

[Atarimaster]

Open book projection

I like it, too!
I wouldn’t use it as a map in an atlas but, you know… to show a map in an atlas.

[mapnerd2022]

Really nice projection! Since I absolutely love reading, it's would be the perfect projection for me!

[PeteD]

What is your way?

Step 1

In the first step, the parallels are shifted upwards along most of their length by a height h, but either side of the central meridian, they become quarter-circles of radius h instead. The meridians are equally spaced along the parallels:

var h = 0.45;

if (abs(lambda) < h * halfPi) {
var x = sign(lambda) * h * (1 - cos(lambda/h)),
y = sqrt2 * phi + h * sin(abs(lambda/h));
} else {
var x = lambda + sign(lambda) * h * (1 - halfPi),
y = sqrt2 * phi + h;
}

open book step 1.png (158.06 KiB) Viewed 10293 times

Step 2

In the second step, the height by which the parallels are shifted upwards is reduced by a factor of cos²(lambda/2). Unfortunately, this messes up the equal spacing of the meridians along the parallels. I'm by no means claiming that this is the best way of doing things -- it's just a simple way that I thought looked good enough:

var h = 0.45;

if (abs(lambda) < h * halfPi) {
var x = sign(lambda) * h * (1 - cos(lambda/h)),
y = sqrt2 * phi + h * sin(abs(lambda/h)) * pow(cos(lambda/2), 2);
} else {
var x = lambda + sign(lambda) * h * (1 - halfPi),
y = sqrt2 * phi + h * pow(cos(lambda/2), 2);
}

open book step 2.png (199.76 KiB) Viewed 10293 times

Step 3

In the final step, the width of the projection is reduced by a factor of 1 + (w-1)/pi * (phi + pi/2), where w is the factor by which the width is reduced at the north pole. This is supposed to give the appearance of perspective. Of course, in that case, there should also be some foreshortening, but again, I'm by no means claiming that this is the best way of doing things:

var h = 0.45,
w = 0.95;

if (abs(lambda) < h * halfPi) {
var x = sign(lambda) * h * (1 - cos(lambda/h)) * (1 + (w-1)/pi * (phi + halfPi)),
y = sqrt2 * phi + h * sin(abs(lambda/h)) * pow(cos(lambda/2), 2);
} else {
var x = (lambda + sign(lambda) * h * (1 - halfPi)) * (1 + (w-1)/pi * (phi + halfPi)),
y = sqrt2 * phi + h * pow(cos(lambda/2), 2);
}

open book.png (292.76 KiB) Viewed 10293 times

I can imagine ways of generating that.

I'd be interested to know how you would have done it.

[PeteD]

I recently came across the Winkel-Snyder projection, and it's now one of my favourite pseudocylindricals. It's the average of the equirectangular and Mollweide projections and was apparently invented by accident when Snyder got confused about how Winkel II was derived. Note that before averaging with the equirectangular, the Mollweide is scaled up to the size of Apian II.

I mentioned the Winkel-Snyder to Tobias, who was already aware of it and had already thought of the obvious next step -- if the Winkel-Snyder is essentially Winkel II but with a modified spacing of the parallels and is derived by replacing Apian II with the Mollweide in the derivation of Winkel II, then replacing the Aitoff with the Hammer in the derivation of the Winkel tripel will give a projection that's essentially the Winkel tripel but with a modified spacing of the parallels:

0 Winkel-Hammer.png (165.96 KiB) Viewed 10198 times

Again, before averaging with the equirectangular, the Hammer is scaled up to the size of the Aitoff.

However, Tobias and I agreed that unlike the Winkel-Snyder, this projection, which we've provisionally been calling the Winkel-Hammer, doesn't actually look very good. I was wondering why this might be and thought that maybe it was because the spacing of the meridians is also modified, with everything getting pushed outwards to regions of higher distortion and the Americas in particular looking rather bent.

It occurred to me that you could modify the spacing of the parallels in the Winkel tripel without changing the spacing of the meridians if you stick with Winkel's average of the equirectangular and Aitoff projections for the x values but take the average of the equirectangular and (scaled up) Hammer projections for the y values. Here's the result, both with Winkel's original standard parallel of arccos(2/pi) for the parent equirectangular projection and with Bartholomew's standard parallel of 40°:

1a Winkel tripel lower areal distortion.png (164.19 KiB) Viewed 10198 times

1b Winkel tripel lower areal distortion 40 deg.png (170.11 KiB) Viewed 10198 times

Africa becomes a bit skinny in the cos phi0 = 2/pi version, but overall the projection is better than the Winkel-Hammer, particularly the Americas. Compared to the original Winkel tripel, it has lower areal distortion at the expense of higher angular distortion.

In the interest of clarity and conciseness, I'll refer to this as projection 1 and number the projections in the subsequent posts accordingly.