A Modified Azimuthal Equidistant/Stereographic Projection

[PeteD]

Very nice! I like it almost as much as the Hellerick boreal triaxial projection:

Hellerick_triaxial_boreal_projection.jpg (68.79 KiB) Viewed 1476 times

(Image taken from the Wikipedia article on Modern flat Earth beliefs.)

Apart from Antarctica, it depicts landmasses with very low distortion.

The Pierce quincuncial would be a good alternative if you prefer a square world with a rim that you can fall off (or an infinitely repeating tiled world) to a circular world with an all-encompassing ice wall.

[Atarimaster]

Here's another projection to consider.

This projection is azimuthal equidistant until the equator, and then pseudoazimuthal equidistant-equal-area (analogous to the sinusoidal projection, but in azimuthal rather than cylindrical form: parallels are at the same spacing as in an equidistant projection, but rescaled to make the projection equal-area) beyond the equator.

That’s not unlike John Bartholomew’s “Tetrahedral” projection.
See its decription in the web archive of progonos.com, or an image here.

[PeteD]

That’s not unlike John Bartholomew’s “Tetrahedral” projection.
See its decription in the web archive of progonos.com, or an image here.

From the information in your link, it looks like the fundamental difference with respect to the Bartholomew tetrahedral projection is how it solves the problem that "parallel lengths do not coincide at the boundary latitude on the pristine Lambert [or azimuthal equidistant] and Bonne/Werner projections ... thus parallels must be lengthened in the lobes ..."

As far as I understand, in Milo's projection, the spacing of the parallels in the lobes is "rescaled to make the projection equal-area beyond the equator" at the expense of correct scale along the central meridians of the lobes, whereas in Bartholomew's projection, the spacing of the parallels in the lobes is unmodified for correct scale along the central meridians at the expense of equal-area lobes.

Personally, I prefer a third solution to this trilemma: replace the azimuthal equidistant with the conic equidistant and set the outer standard parallel to the boundary latitude so that it has the correct length and the parallels in the lobes don't need to be lengthened in the first place, as in the Bartholomew regional projection. That way, you can eat your cake and still have it: equal-area lobes and correct scale along their central meridians!

Here's my own variant of the Bartholomew regional projection with five lobes and the inner standard parallel moved to the North Pole:

bartholomew_regional_5_lobes.png (102.43 KiB) Viewed 1445 times

Please ignore the lines connecting the different parts of Antarctica.

Filling in the gaps between the lobes with extra ocean as Milo did and replacing Antarctica with the ice wall of the azimuthal equidistant gives my attempt at a flat earth projection:

bartholomew_regional_flat_earth.png (114.34 KiB) Viewed 1445 times

Apart from Antarctica, landmasses are less distorted than in the projections in Tobias's link, and the region from Iran to India is better than in the Hellerick boreal triaxial projection that I mentioned in my previous post, though I admit this projection wouldn't stand up to the scrutiny of a trip to the Bering Strait. Distortion in the northern hemisphere could be further reduced by reintroducing the pole line of the Bartholomew regional projection, but I didn't want to open up the Arctic even further.

[Milo]

As far as I understand, in Milo's projection, the spacing of the parallels in the lobes is "rescaled to make the projection equal-area beyond the equator" at the expense of correct scale along the central meridians of the lobes,

Correction: scale is correct along the central meridian of each lobe, just not the other meridians.

Mind you, it's not conformal along the central meridians (if I did everything right, then angular distortion along the central meridians of the lobes of the outer hemisphere should be equal to angular distortion along the hemisphere-boundary of the equidistant projection used for the inner hemispere), but both the semimajor and semiminor Tissot axes should remain constant, just not equal to each other.

Personally, I prefer a third solution to this trilemma: replace the azimuthal equidistant with the conic equidistant

This doesn't work well at the Arctic Sea, where it introduces an interruption.

Admittedly, the Arctic Sea is not the most inhabited area, and probably not travelled that much by flat-earthers. But interrupting the Bering Strait to put Alaska and Chukchi far apart from each other is going to stand out to anyone who's even vaguely familiar with the local geography, or, you know, the history of the Ice Age.

though I admit this projection wouldn't stand up to the scrutiny of a trip to the Bering Strait.

Oh, you're already aware of that.

I'll also note that both your and my projection drastically exaggerate the distance between Australia and Brazil. Those are both relatively large and wealthy countries, and someone is bound to have noticed that travel between them doesn't typically entail a trip over the North Pole.

[PeteD]

Correction: scale is correct along the central meridian of each lobe, just not the other meridians.

Now I'm confused. The hemisphere boundary of the azimuthal equidistant is stretched east–west, so if the lobes are to be equal-area, then surely their central meridians must be compressed in order to compensate for this east–west stretching, right? So if scale is correct along the central meridians, then the lobes can't be equal-area, can they?

I'll also note that both your and my projection drastically exaggerate the distance between Australia and Brazil. Those are both relatively large and wealthy countries, and someone is bound to have noticed that travel between them doesn't typically entail a trip over the North Pole.

Yes, this is indeed a serious flaw.

[Milo]

Now I'm confused. The hemisphere boundary of the azimuthal equidistant is stretched east–west, so if the lobes are to be equal-area, then surely their central meridians must be compressed in order to compensate for this east–west stretching, right? So if scale is correct along the central meridians, then the lobes can't be equal-area, can they?

Stretching is irrelevant to the equal-area property. You can take a cylindrical equal-area projection, stretch it east-west by a factor of ten while keeping its north-south scale the same, and it'll still be a cylindrical equal-area projection.

[PeteD]

Well, strictly speaking, you can:

1. take a cylindrical equal-area projection, stretch it east-west by a factor of ten while keeping its north-south scale the same; and
2. shrink the projection to one-tenth its new area while maintaining its new aspect ratio

and it'll still be a cylindrical equal-area projection.

Or, equivalently,

1. take a cylindrical equal-area projection, stretch it east-west by a factor of ten while keeping its north-south scale the same; and
2. adjust the nominal scale by a factor of sqrt(10)

and it'll still be a cylindrical equal-area projection.

Step 2 normally isn't very important because the size of the projection relative to its nominal scale doesn't affect the appearance of the projection. However, if you're claiming two different properties -- equal-area and correct scale -- that are each defined relative to the nominal scale, then it is important that they're both defined relative to the same nominal scale.

Yes, a cylindrical equal-area projection of any aspect ratio is equal-area, but only the Lambert cylindrical equal-area projection has correct scale along the equator.

Claiming equal-area lobes and correct scale along their central meridians is the equivalent of claiming correct scale along the equator of the Gall-Peters projection. Yes, you could enlarge the Gall-Peters projection relative to its nominal scale such that it has correct scale along the equator, all while maintaining its aspect ratio, but then the whole thing would be too big to be equal-area. On the other hand, if you set its nominal scale such that the projection is equal-area (and if no nominal scale is explicit, then you'd usually assume this), then the equator is too short.

[Milo]

Yes, a cylindrical equal-area projection of any aspect ratio is equal-area, but only the Lambert cylindrical equal-area projection has correct scale along the equator.

I think we're using different definitions of "scale".

All cylindrical projections "preserve scale on the equator" in the sense that if points A, B, C, and D are all on the equator, then the ratio of the disance between A and B and the distance between C and D will be correct. More generally cylindrical projections preserve scale along every parallel, but not between parallels (if A and B are on one parallel, and C and D are on another parallel, then the ratio of the distance between A and B and between C and D will be incorrect).

In the special case of cylindrical equidistant projections, scale is also preserved both along and between meridians, and between meridians and one specific parallel only (the standard parallel). This is not just the plate carree projection: other equidistant projections will depict meridians differently, but will still correctly allow distances along them to be measured.

Pseudocylindrical projections still preserve scale along each parallel (and also usually not between parallels, but as a special case, the sinusoidal projection does), but do not preserve scale either along or between meridians (the sinusoidal projection does on the central meridian only, but most don't manage even that). An equatorial-aspect orthographic projection looks superficially like a pseudocylindrical projection, in that it has straight parallels, but unlike a true pseudocylindrical projection it does not preserve scale along those parallels.

You're not going to have truly correct scale on the equator unless your map is 40000 kilometers long. When talking about map accuracy I always implicitly take "correct" to mean "proportionally correct" - while setting your units such that the constant of proportionality is 1 (in your chosen units) may make the formulae more convenient, it's still just an arbitrary convention.

[PeteD]

I think we're using different definitions of "scale".

Yes, it looks like we are. What you call "preserving scale", I would call "constant scale". My usage of "correct scale" (or "true scale") versus "constant scale" is consistent with the usage in the directory of map projections on this website. For example, under "cylindric equal-area", it says:

Scale
Correct along one parallel and the parallel of opposite sign; constant along each parallel.

When talking about map accuracy I always implicitly take "correct" to mean "proportionally correct" - while setting your units such that the constant of proportionality is 1 (in your chosen units) may make the formulae more convenient, it's still just an arbitrary convention.

If you take an equal-area projection and enlarge it such that it's the same size as the earth's surface, then whether distances along a given line are really equal to, or merely proportional to, the corresponding distances on the earth is more than just an arbitrary distinction. Similarly, if you enlarge an equal-area projection such that distances along a given line are equal to the corresponding distances on the earth, then whether the projection is the same size as the earth's surface or not is also more than just an arbitrary distinction.

Anyway, it seems I misunderstood your projection. Is the scale along the central meridians of the lobes the same as the scale along the meridians of the azimuthal equidistant part?

[Milo]

If you take an equal-area projection and enlarge it such that it's the same size as the earth's surface, then whether distances along a given line are really equal to, or merely proportional to, the corresponding distances on the earth is more than just an arbitrary distinction.

It matters in an indirect sense: if you know the answer to that question, then you can use it to compute the angle distortion on the line. However, what you really care about here is the angle distortion. If angle distortion does not specifically interest you, then the details of scale (which, on any map projection except a globe, is going to be wrong almost everywhere anyway) also aren't that important. You're better off asking about angle distortion directly.

Anyway, it seems I misunderstood your projection. Is the scale along the central meridians of the lobes the same as the scale along the meridians of the azimuthal equidistant part?

Yes.