Experimental projections

[PeteD]

Regarding the interior of the map, I like your last example best although I’m not so sure that the left “hemisphere” has a rounded pole and the right a spiky one.

I agree, so I've come up with something that improves on the shape of Africa in the non-umbeziffert azimuthal equidistant projection with "unequal hemispheres" without making the polar regions narrow and spiky -- an Umbeziffern gradient.

As I mentioned before, the right "hemisphere" has a width of 226.8° at the equator and needs to be umbeziffert with a bounding meridian of 142.9° for this width to correspond to 180° of the parent azimuthal equidistant projection. However, as we move away from the equator, the width of the right "hemisphere" gradually decreases to 180° at the poles (this can best be seen in my beer glass map), where no Umbeziffern is required for the width to correspond to 180° of the parent azimuthal equidistant projection. We can therefore apply an Umbeziffern transformation with a different bounding meridian for each parallel such that the length of every parallel in the right "hemisphere" corresponds to 180° of the parent azimuthal equidistant projection:

unequal_hemispheres_azimuthal_equidistant_umbeziffern_gradient.png (88.73 KiB) Viewed 2340 times

As far as I'm aware, no such Umbeziffern gradient has ever been applied to a parent projection before (please correct me if I'm wrong), so I'd say this constitutes a new projection rather than merely a new way of interrupting an existing projection. It's my favourite of all the "unequal hemisphere" maps that I've posted above, and I'm quite proud of it.

[PeteD]

I’ve found some which I think are not too bad: Two equal-area projections, Strebe-Mollweide and Hufnagel 10 (the latter again with a slight horizontal stretch), and a compromise projection, Apian II.

Not bad at all. Are the contour intervals 10° for angular deformation and 0.5 for areal distortion?

[Atarimaster]

Are the contour intervals 10° for angular deformation and 0.5 for areal distortion?

No, this kind of visualization works differently.
Here are the last ones again with isolines at max. angular deformation of 10°, 20°, 30°, 40°, 50°, 60° and 80° (yes, no 70°) and areal inflation at 1.0, 1.2, 1.5, 2.0, 2,5, 3.0 and 3.5.

pointed-poles-2.png (203.1 KiB) Viewed 2337 times

Apparently, there’s a bug when you use the isolines in conjunction with custom boundaries that result in a “less than a hemisphere” map (or something like that)…

As far as I'm aware, no such Umbeziffern gradient has ever been applied to a parent projection before (please correct me if I'm wrong)

I have never heard of it, and I assume it has never been applied, but of course I only know a very small part of the map projection literature.
I like this example, too!

Kind regards,
Tobias

[quadibloc]

The troubles you are having with Greenland and Iceland have reminded me of a map projection I have been looking for.
I tried to create it myself, but with uninspiring results.
And that is: a projection that stands between two Stereographic hemispheres, and the August two-cusped epicycloidal projection, so that the Eastern and Western hemispheres are joined by a line that is shorter than a full meridian, say 90 degrees or less.

[Milo]

In theory, according to the Riemann mapping theorem, for any shape you choose to draw, there exists a unique conformal mapping onto that shape. Hence, it should be possible to just draw a rough sketch of what you think your map should look like and then use numerical methods to compute that map. However, I do not know of any software that would make such computation convenient.

dummy_index has done some preliminary work or calculating something similar to what you're asking for (based on Eisenlohr rather than August), also using numeric methods, but it only shows graticules and not an actual projection. I think you could recreate it if you look at http://github.com/dummy-index/gp-mapprojections/blob/master/world.plt and search for "ApproxEisenlohr90". There's no explanation of how he arrived at those formulas (except maybe if you can read Japanese), but they should be usable with a little effort.

[daan]

Hi, I'm new here.

Welcome to the forums, brsr! Thanks for your posting.

I've been working on a variation of the Chamberlin Trimetric projection.

I read your paper a while ago; sorry that I did not get back to you about it. It’s a good paper. Do you intend to publish?

For the most part it just looks like the Chamberlin Trimetric, but when the control points are pushed out to an entire hemisphere, it maps the hemisphere to a triangle. Maybe too much distortion to be practically useful, but it would make a good logo for something.

hemisphere.png

It looks much better on the use cases that the Chamberlin Trimetric is useful for. If anyone is interested, I have a writeup on the new projection on GitHub.

I might implement this in Geocart if you don’t mind. It probably won’t be soon; I’ve been completely swamped lately. But someday?

Cheers,
— daan

[PeteD]

The troubles you are having with Greenland and Iceland have reminded me of a map projection I have been looking for.
I tried to create it myself, but with uninspiring results.
And that is: a projection that stands between two Stereographic hemispheres, and the August two-cusped epicycloidal projection, so that the Eastern and Western hemispheres are joined by a line that is shorter than a full meridian, say 90 degrees or less.

I'd be interested in seeing that if you ever get good results!

If I understand correctly, the usual two-hemisphere problem of cutting through landmasses would be solved by the fact that the join allows the central interruption to be moved eastwards, presumably passing between Iceland and Ireland, without cutting through Africa. A join of just 60° should be sufficient for this. However, if the two hemispheres are still equal, then eastern Siberia will still get cut off and moved to the other side of the map. On the other hand, if you move the interruptions further eastwards such that the outer interruption passes through the Bering Strait, then you'll need a join of about 130° along the central interruption in order to avoid cutting through Europe.

[PeteD]

Apparently, there’s a bug when you use the isolines in conjunction with custom boundaries that result in a “less than a hemisphere” map (or something like that)…

It's not nearly as bad as the mess that I get when I try to plot contour lines for two-hemisphere (or two-"unequal hemisphere") maps!

By the way, I've slightly improved on my previous attempts. I've straightened the central interruption (on the globe, not the map, i.e. I've reduced the extent to which it deviates from a meridian). This means that I can keep each "hemisphere" symmetric without moving more of the Pacific than necessary to the right "hemisphere". As you can see, Hawai'i is now firmly in the left "hemisphere":

unequal_hemispheres_azimuthal_equidistant_2.png (89.2 KiB) Viewed 2287 times

This is now very similar to the first map that Tobias posted on this subject, with the only real difference being the fact that Iceland appears in the left "hemisphere" and Greenland doesn't need an extension to avoid getting cut up.

Since the right "hemisphere" isn't quite as wide at the equator as it was in my previous attempts, less Umbeziffern (i.e. a greater limiting meridian) is required for the width to correspond to 180° of the parent azimuthal equidistant projection:

unequal_hemispheres_azimuthal_equidistant_umbeziffern_gradient_2.png (87.55 KiB) Viewed 2287 times

I've used the Umbeziffern gradient as before, but this time, instead of applying the same Umbeziffern transformation to the left "hemisphere" or just leaving it un-umbeziffert, I've applied the same principle to the left "hemisphere", i.e. just enough Umbeziffern for the width to correspond to 180° of the parent azimuthal equidistant projection, which results in a very similar curvature of the boundary in the polar regions of both "hemispheres". Since the width of the left "hemisphere" is less than 180°, this requires a limiting meridian greater than 180°, which is OK in this case because it's not applied over the whole globe.

[dummy_index]

Earth can be divided equally, not by great circle, similar to my Baseball Projection (unintroduced yet).
I search some cutting line avoid cutting any continent and any island, except Fox Islands.
Cutting line is sine curve plotted on Mercator map and rotated by Euler angle.
I found some cutting lines, whether cutting Canary Islands and how cutting Fiji. Below shows two of them.
Cyan: curve A: nearest to great circle.
Yellow: curve E: Canary Islands is not divided.

CurlyHemisphere_CurveA_CurveE.png (93.93 KiB) Viewed 2270 times

Then on making two-hemisphere configuration, I use a little bit of Umbeziffern to make same scale for top/bottom points and left/right points. Once put major axis of each oval region onto equator and scale longitude to make quasi- small circle.
Upper unclipped lines are some of Fox Islands.

TwoCurlyHemisphereA.png (77.75 KiB) Viewed 2270 times

[PeteD]

Excellent work. Love it!

Does the point on the map where the two hemispheres touch correspond to the same point on the globe in each hemisphere?