Snyder's equal-area projection

[Milo]

the Shirley-Chiu transformation

...The what?

I did a search on "Shirley-Chiu" and found this paper. The chart on page 3 is instructive. The "squircle" and "elliptical" columns look interesting, though they're not equal-area yet. The "elliptical" one, in particular, has very smooth patterns for both D_A (area distortion) and D_I (angle distortion, dunno why the authors called it "I" apparently it stands for "isotropy"), so is probably a good compromise projection. No, it works great inside one disc/square (no kinks at the diagonals, like Shirley has), but it still doesn't behave right at the edges, causing problems if you want to tape two hemispheres together. The problem is that the radial lines on the disc, when projected, aren't perpendicular to the edges of the square. (Note that by the theory of Tissot indicatrices, at any point where a projection is continuously-differentiable, there will always be some pair of lines through that point that are perpendicular in both the original and the projection. So you don't need conformality for this. You just need to make sure either the semiminor or semimajor axis of the Tissot indicatrix is parallel to the edge.)

That paper also contains explicit formulae for the conformal projection (disc-to-square only, not general polygons). Predictably, they're way more complicated than anything else in the paper and take up half a page, but they're described neatly and readably enough that you could make a program to implement this, if you wanted (or download the author's example code).

If you want to find a differentiable equal-area polygon projection (or prove that it doesn't exist), maybe start with the Shirley-Chiu transformation so you don't have the change from one geometry to another complicating things.

Well, maybe not the Shirley mapping specifically, but yeah, if you want to project a hemisphere/dihedron, the spherical-face-to-flat-disc part is easy, so you just need to focus on getting the disc-to-polygon part right.

For more complicated polyhedra, projecting a face into a disc isn't so simple (if you just apply an azimuthal projection to an arbitrary spherical polygon, you'll get an awkward Euclidean shape that isn't so easy to work with). But I'm hoping that if we can get the dihedral case right, it'll provide inspiration for how to tweak the formula for the more general case.

[Milo]

Some further observations.

Consider the simplest way of attempting a "polygonal" projection: projecting all meridians as straight lines, as in an azimuthal projection, just with their lengths adjusted. One octant of a square would look like this:

polygon_azimuths.png (14.45 KiB) Viewed 1622 times

Immediately you will observe two problems:

The first problem is that the areas are wrong. The area to the left of a particular meridian on the sphere is A_s = longitude - asin(sin(longitude)*cos(r)). The area to the left of a particular meridian on the plane is tan(longitude)*r^2/2, or, when scaled so that the total area is correct, A_p = tan(longitude) / tan(max_longitude) * (max_longitude - asin(sin(max_longitude)*cos(r))). For the simplest case, a square dihedron (r = pi/2, max_longitude = pi/4), these simplify to A_s = longitude and A_p = tan(longitude) * pi/4. Graphic these makes it obvious that A_p < A_s for 0 < longitude < max_longitude.

That is, these straight-line meridians will always split the polygon in a way that leaves too little room on the left, and too much room on the right. To remedy this, we would need to curve the meridians so that, at least somewhere in the projection, they extend further right than the lines depicted above. In turn this means that, at least somewhere along their extent, the meridians have an angle that is further-right / more-horizontal (in the above image's orientation) than the straight-line meridians.

The second problem is that the meridians aren't perpendicular to the far edge, so multiple polygons in a polyhedron won't line up correctly. To remedy this, we would need to curve the meridians to the left, since they need to end up totally vertical.

Meanwhile, the projection needs to be conformal at the center of the polygon.

Therefore, an equal-area polyhedral projection would need to have sigmoid meridians: they start out having the correct angle at the center of the polygon, then curve further right to grab some more area, then curve left again to allow them to end vertical. Something like this:

polygon_sigmoid.png (2.87 KiB) Viewed 1622 times

Indeed, when looking at a pole-centered Gringorten projection, even though it does not appear to perfectly satisfy the perpendicularity condition above, it does have such sigmoid meridians.

This is an obvious problem, because angle distortion is likely to be pretty bad at the inflection point. Then again, currently we don't have a projection that even meets the conditions at all, even with severe angle distortion, so maybe that's jumping the gun a little!

[daan]

Meanwhile, the projection needs to be conformal at the center of the polygon.

Is that true? I think singularities at a finite number of points are “allowed”; even conformal projections have them. It would mean that opposite meridians would kink at the pole, which is merely not nice.

Much more difficult is the constraint that the polygon edges must all be conformal, right? Otherwise, under coordinate transformation you will get kinks along them. That’s because if the Tissot indicatricies are squashed, most lines that enter the edge can’t exit the far side of the edge without kinking. And since conformal in an equal-area projection means no distortion, that’s the constraint the edges must have. In other words, the edges must have correct scale. And yet that is impossible because the left and right edges in your diagram are different lengths for meridians of the same length.

I think I’ve just demonstrated that there’s no equal-area solution without kinking for general polygons.

— daan

[Milo]

Is that true? I think singularities at a finite number of points are “allowed”; even conformal projections have them. It would mean that opposite meridians would kink at the pole, which is merely not nice.

Well, if a projection is differentiable at the center (meaning it has elliptical Tissot indicatrices), and it also has threefold or greater rotational symmetry at that point (as you'd expect of any regular polygon), then it must be conformal, because the only ellipse with threefold or greater rotational symmetry is a circle.

This reasoning does leave some loopholes though. You could, as you suggest, allow the projection to be non-differentiable at the center, and the reasoning also doesn't apply to irregular polygons.

I would like to prefer to avoid singularities at the face centers, because that violates the letter of my "differentiable everywhere except the vertices" condition, and the general idea that projections should try to minimize distortion at the "center", because that's where you try to put the stuff that interests you most. Distortion (and even singularity) at the vertices of a polyhedron is unavoidable, because you're going to cut them open to unfold the polyhedron into a net. Distortion at the center of the face is avoidable, because that's a smooth part of the polyhedron far away from any cuts, and so it's desirable to avoid it.

If I wanted a projection that has singularities at both the vertices and the face centers of a cube, I'd look for a projection of the tetrakis cube, not the normal cube. (Since that uses irregular polygons without rotational symmetry, I'm not 100% sure if conformality at the center is actually necessary or even possible, but it does seem like something at least worth trying to aim for.)

You mention that even conformal projections have singularities, but they do try to push them to the edge of the projection. For example, the Lagrange projection has singularities at the poles (the endpoints of the interruption meridian), but nowhere else.

Much more difficult is the constraint that the polygon edges must all be conformal, right?

They don't have to be conformal, just perpendicular.

Note that at any point where a projection is differentiable, there will always be some pair of lines which are perpendicular to each other in both the original and the projection (specifically, the minor and major axes of the Tissot indicatrix at that point), even if all other perpendicular lines at the point get distorted. Therefore, what I'm asking is that this one pair of perpendicular lines is arranged so that one of them is the edge of the polygon.

(More generally, it's true that for any arbitrary angle, there will exist two pairs of lines that preserve that angle - except for angle pi/2, where the two pairs are equal, and angle 0, where any line trivially satisfies the condition. For example, under the transformation (x,y) -> (2*x,y), the vectors (±3683,±9297) and (±7838,±6210) have angle approximately pi/6 in both the original and the projection.)

I don't think actual conformality at the edges is achievable. We've been over that here. (For sufficiently small polygons - those with small area relative to their circumference - it would be achievable by projecting them onto a non-polygonal shape like a Reuleaux triangle, but for polygon-to-polygon projections, which we're after here, it just don't work.)

I think I’ve just demonstrated that there’s no equal-area solution without kinking for general polygons.

Sorry, but no.

You just need the Tissot indicatrices to have the right orientation. (The indicatrices' orientation is a subject that gets discussed far less than their shape! You considered something similar here, though, calling it "torsion" in the context of a since-disproven conjecture.)

[daan]

Much more difficult is the constraint that the polygon edges must all be conformal, right?

They don't have to be conformal, just perpendicular.

…
Note that at any point where a projection is differentiable, there will always be some pair of lines which are perpendicular to each other in both the original and the projection (specifically, the minor and major axes of the Tissot indicatrix at that point), even if all other perpendicular lines at the point get distorted. Therefore, what I'm asking is that this one pair of perpendicular lines is arranged so that one of them is the edge of the polygon.

Since that doesn’t avoid kinking of arbitrary lines that cross the edge, it doesn’t make a good tiling. It’s a little nicer if the graticule is arranged in a way that its lines cross smoothly, and presumably that’s possible, but if the coastline at the same place kinks, I’m not sure we’ve achieved much.

I think I’m addressing what you wrote, and I think my reasoning is correct here, but let me know if I’m not following or have reasoned incorrectly.

Cheers,
— daan

[Milo]

Since that doesn’t avoid kinking of arbitrary lines that cross the edge,

Yes it does. Their angle relative to the edge might get distorted, but it gets distorted by the same amount on both sides, so they'll still line up.

The theory of Tissot indicatrices really make this very simple. It guarantees that, so long as a projection is differentiable at a point and therefore has an elliptical Tissot indicatrix, the projected angles at that point depend only on the Tissot indicatrix, and nothing else. Therefore, if you glue two projections together such that their Tissot indicatrices at the seams are identical, then the result won't have any kinks at the seams. (It may still violate higher-order smoothness requirements, such as second-differentiability, but we're not worried about that right now.)

For our polyhedral case, the two projections being glued together are mirror images of each other across the edge, which requires that the Tissot indicatrices have reflection symmetry across that edge. An ellipse with a correctly-oriented major axis has this reflection symmetry and therefore satisfies the condition.

Here is an example with a cylindrical equal-area projection up to 45° being glued to a plate carree projection beyond 45°. The standard parallel for the former was chosen as 32.7651°, so that the Tissot indicatrices at the seams align. No kinks I can see.

glued.png (551.14 KiB) Viewed 1600 times

[daan]

You are absolutely correct. Apologies for the hasty thoughts.

Cheers,
— daan

[Milo]

No worries. I could probably have been clearer.

I worked this out a while ago, so it's obvious to me now, but it wasn't originally.