What is shear in map projections?

[daan]

Shear has a lot of meanings, depending on the domain. Most of those meanings have concepts in common, which is why the same word gets used. However, often there are important differences as well.

In the planar world, when there is math involved, shear is easily explained. If you have an axis formed by perpendicular lines, you can induce shear by holding one axis fixed and tilting the other axis.

Original

Shear1.jpg (13.62 KiB) Viewed 2233 times

Sheared

Shear2.jpg (14.47 KiB) Viewed 2233 times

The greater the tilt, the more the shear. Shear like this is a relative thing: “After” is sheared compared to “before”. Something does not start out with shear. In this next example, it is not the first image that “has shear”. It is the second image.

Another original, unsheared

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Sheared

Shear4.jpg (13.97 KiB) Viewed 2233 times

The corollary is that shear is an action, not an adjective, when considered carefully. That action is a “mapping”. In this case, it is a plane-to-plane mapping.

Another thing to consider is that the shape that gets sheared is distinct from the coordinate axes. In the examples above, the original images are rectangles. Their edges are parallel to the usual left-to-right, top-to-bottom axes. However, recall that shear is not a property of an object; it is an action. The images labeled “sheared” are not inherently sheared. There is no reason you could not start from them, shear them, and return to the images labeled “original”. Doing this does •not• mean that we started with an axis tilted right and then straightened it out to perpendicular. It means we started with the usual left-right/top-bottom axis, and tilted the vertical axis left in order to shear the image so that it became rectangular.

What does shear mean in the context of globe-to-plane maps? The analogy with plane-to-plane mappings can only be taken so far. The mapping goes from globe to plane, not plane-to-plane, and so there is no pair of axes to tilt with respect to each other to go from the original to the map.

It might seem obvious to look at the first map below and say it has no shear because none of the vertical or horizontal lines have skewed away from vertical or horizontal. It might seem obvious to look at the second map and say that it has much shear because the meridians meet the parallels at heavily skewed angles.

Gall–Peters. Unsheared?

GallPeters.jpg (79.02 KiB) Viewed 2233 times

Sinusoidal. Sheared?

Sinusoidal.jpg (107.09 KiB) Viewed 2233 times

It is true that the sinusoidal shows strong shearing. When parallels meet meridians at angles other than perpendicular, that is evidence of shearing.

However, the converse is not true. When parallels do meet meridians at right angles, that does not thereby mean there is no shearing. It •might• mean that, but there is more to the story.

Let’s look at the Gall—Peters with just its graticule and some Tissot ellipses to show the distortion at the meridian/parallel intersections:

Gall–Peters, 15° graticule with Tissot indicatrix

EquatorialTissot.jpg (177.32 KiB) Viewed 2233 times

All the Tissot ellipses are oriented with the long edge either left-right or top-bottom. These ellipses show how a circle from the sphere gets distorted when mapped. Notice that all the ellipses at the same latitude are identical.

But we don’t have to project the earth as if it is upright. When we model the earth as a sphere, which we do here, we can slide the skin of the sphere around on the globe however we want, and then project. This does not change the behavior of the projection at all; it just changes which part of the surface gets projected where. Here is an example of the Gall—Peters when we rotate the globe to 163°E, then tilt 63° to the middle, and then rotate around the center by 43°:

Skewed Gall—Peters

GallPetersSkew.jpg (96.67 KiB) Viewed 2233 times

Remember, the projection itself has not changed at all. If we were to map the Tissot ellipses at the same locations •on the map• (not as they were on the sphere), then they would be identical to what we see above even though the graticule would be very different.

Let’s see the Tissot ellipses at positions adjusted to the new graticule:

Gall—Peters, rotated globe

GallPetersSkewTissot.jpg (199.15 KiB) Viewed 2233 times

To be clear, even though it is not necessarily obvious, all of the ellipses are oriented either left-right or top-bottom. Some of them look a little slanted because the graticular lines going through them trick the eye.

Look at the Tissot ellipse pointed to by the green arrow. The graticule in that ellipse looks pretty much like the meridian meets the parallel at right angles. But look at the one pointed to by the blue arrow. It’s heavily sheared!

Did something about the projection change? No. The distortion is the same at the location pointed to by the blue arrow as it is at the same place on the map of the unrotated globe. The amount of the distortion is the same, and the orientation of the distortion on the plane is the same. The only thing that has changed is which part of the skin of the globe now covers the same place. If the distortion is the same, is the shear the same?

The reason the graticule now looks skewed is because the meridian and the parallel enter the Tissot ellipse at different angles than they did on the map of the unrotated globe. If we rotated the meridian/parallel axis around the center of that ellipse, we would see it at right angles when the axis was north-south/east-west. We see it heavily skewed when we rotate it to the direction of the place pointed to by the blue arrow. The angle between the meridian and parallel expands or collapses as we rotate the meridian/parallel axis around the center of the ellipse. This is true for •any• Tissot ellipse, even the ones in the sinusoidal projection that has heavy shearing. Every one of those ellipses has some rotational angle for which the meridian and parallel would meet at right angles.

Is there something special about projections that have some rotational aspect in which •every• meridian/parallel intersection occurs at right angles? Well. No, not beyond that fact itself. Such maps are not thereby more or less distorted than any other map.

There is a class of projections in which every meridian/parallel intersection in •all• rotational aspects occurs at right angles. Those maps are conformal. They have no shear. They have other interesting and useful properties as well.

Map projection texts and research papers do not use “shear” to mean a deviation from right angle of the meridian/parallel intersection. That is because the intersection angle depends on how the globe was rotated before projecting. It’s arbitrary, and therefore not so useful. What’s more important is the underlying distortion characteristics as given by the Tissot indicatrix. The indicatrix does not vary by rotational aspect.

Here are final examples. The polar aspect of the Lambert azimuthal equal-area projection has meridians that meet parallels at right angles.

Lambert polar. No shear?

LambertPolar.jpg (198.3 KiB) Viewed 2233 times

Notice the upper left shoulder of this second map, where we have rotated the globe before projecting. Massive shearing—and yet the projection itself is identical:

Lambert oblique. Shear?

LambertOblique.jpg (198.49 KiB) Viewed 2233 times

In all of these examples, the presence of shear coincides with what is called “angular deformation”. The more angular deformation, the more shear. The less angular deformation, the less shear. Saying a part of the map has lots of shear is shorthand for saying that the angular deformation is high in that part of the map.

In summary, the analogy of shear in sphere-to-plane mappings to plane-to-plane mappings is only shallow. In the case of plane-to-plane mapping, shear is an operation on the entire plane. It does not matter if you rotate the coordinate plane and then apply shear; you will get the same magnitude of effect but just in a different direction. If we go with an idea that shear in sphere-to-plane mappings must refer to the angle between meridian and parallel, then we would be making shear a property of the local coordinate axis only, rather than an operation on the underlying space. If, instead, we consider shear to be the angular deformation at a point, then shear is an operation on the underlying spherical space being transferred to the plane, and is independent of the rotational aspect.

— daan

[Piotr]

Lambert Oblique doesn't look sheared. It's just squished in that direction.

[RogerOwens]

I don't know why it seemed so mysterious, what is meant by shear-distortion for maps. It just means that the mapped surface of the Earth is sheared with respect to how that surface was on the globe--by the ordinary, familiar and plain dictionary definition of shear:

Displacement of adjacent layers (in a 3-dimensional body) or strips (in a 2-dimensional surface) in a direction parallel to the border that separates them.

Alternatively it could referred to as displacement within a body or surface, by which two nearby points in the body or surface are displaced with respect to eachother in a direction perpendicular to the line that previously connected them them.

...it being a surface, in the case of map-shear, of course.

So, shear distortion in a map doesn't mean anything other than the familiar dictionary definition of shear, applied to the surface of the Earth in the map, compared to the surface of the Earth on the globe.

Michael Ossipoff

[Atarimaster]

So, Micheal and Piotr… just to get this straight:
If I get it right, by your definition, the word »shear« would not describe that kind of distortion that is applied by Photoshops’s Shear Distortion Filter in this little video:
https://youtu.be/USQzucw60Lc?t=78
(I've linked to position 1:18 of that video, the point where he starts using that filter.)

Do I get you right?

[RogerOwens]

So, Micheal and Piotr… just to get this straight:
If I get it right, by your definition, the word »shear« would not describe that kind of distortion that is applied by Photoshops’s Shear Distortion Filter in this little video:
https://youtu.be/USQzucw60Lc?t=78
(I've linked to position 1:18 of that video, the point where he starts using that filter.)

Do I get you right?

The image of the oblique Gall Orthographic map with obliquely-intersecting meridians and parallels makes the point well enough.

Let's just say that my statement that all cylindrical maps are un-sheared seemed, to me, well-supported when I said it. Contrary to what you might have heard, I do acknowledge mis-statements. Evidently, contrary to what I said, and contrary to what seemed so certain, most cylindrical maps have shear (as I defined it in my previous post to this thread).

It's completely plausible that conformal maps are un-sheared. Then Mercator, a conformal map, is un-sheared.

Starting with a Mercator in equatorial-aspect, deform it into (say) a Cylindrical-Equidistant. The motion is entirely north-south, and is uniform on each parallel. So it seems fair enough to say that there's no relative motion between adjacent strips of the surface, parallel to their border.

And, if the Mercator was un-sheared to begin with, and because no shearing was done when deforming it into Cylindrical-Equidistant, I concluded that Cylindrical-Equidistant must not be sheared either.

If that conclusion isn't true, then there must be some fallacy in the way that I reached that conclusion. It seemed so certain.

Michael Ossipoff

[RogerOwens]

Actually, after my posting of the statement, I was already having just a bit of doubt about all cylindricals being unsheared. There was the nagging question, "What about shear between strips oriented obliquely on that equatorial-aspect Cylindrical-Equidistant?" I intended to check-out that doubt.

But the argument in my post just before this one sounded so compelling that, even with the doubt, I didn't know for sure that that argument's conclusion wasn't right. The doubt was only a doubt, until I looked at the image of the obliquely-crossing grid-lines of oblique Gall Orthographic.

Michael Ossipoff

[Piotr]

The sheared one is actually elongating some edges, without shorting others or reducing area.

[Atarimaster]

Yeah, but daan’s posting was just about that »shear« has, in the domain of map projections (and obviously in Photoshop, too), a different connotation than in »everyday language«.
Which wouldn’t be uncommon.

I can’t say since I never stumbled across the english word »shear« before I read it here in the forum.
So, I’m out of this thread now and simply gonna say »angular distortion« when there is one.

[daan]

The sheared one is actually elongating some edges, without shorting others or reducing area.

Piotr,

This is a great illustration of why shear is an operation, rather than a property. It turns out, for plane-to-plane mappings, that shear, stretch, and rotation are not separate things. We can decompose them into separate things that are easy to conceptualize, but they are actually fundamentally interdependent.

You can think of your not-sheared-but-stretched-diagonally example just as you suggest. However, you can also think of it as a horizontal shear followed by a vertical shear. Or, you can think of it as a vertical shear followed by a horizontal shear. Or, you can think of it as a rotation, followed by a horizontal shear. It turns out you can think of it in an unlimited number of ways as long as you are willing to permit an unlimited number of operations.

This kind of transformation is called affine. It does not mean any one of shear, scale, or rotation. Rather, it means all of them together, regardless of how you choose to decompose them. A single affine transformation expresses an infinite diversity of combinations of shear, rotation, and scaling. Because the decomposition is a matter of choice, it should be clear that the concept of shear, stretch, and rotation as independent things is an illusion.

Now, it turns out that your illustration cannot be expressed as any one of shear, rotate, or scaling if your coordinate axis is horizontal and vertical. That is because the scaling is either horizontal or vertical, and not diagonal, and shear is either horizontal or vertical, and not both. However, that choice of axis is arbitrary. Merely by changing the orientation of your coordinate system, suddenly your operation is a simple shear.

Again, the analogy to sphere-to-plane mappings is hazardous, but these concepts are part of why a perpendicular parallel/meridian intersection does not indicate anything special. Our choice of coordinate axis was arbitrary. By changing it, the relationship of the coordinate axis to the operation appears to change, but in reality, nothing fundamental has changed.

— daan

[RogerOwens]

Yeah, but daan’s posting was just about that »shear« has, in the domain of map projections (and obviously in Photoshop, too), a different connotation than in »everyday language«.

Certainly there's a difference in the circumstances, the fact that the surface is initially on a globe--including the fact that the surface must be cut, in addition to being deformed.

I just meant that the shear in map shear-distortion nevertheless fits the ordinary dictionary definition of shear. That's what it is, even though it can occur in different situations and conditions and circumstances.

...simply gonna say »angular distortion« when there is one.

Yes, if shear is just non-conformality, then I don't perceive a need to call it by a different name.

Another thing: If most cylindricals have shear, then shear isn't what distinguishes Sinusoidal from cylindricals. Sinusoidal has something objectionable that cylindricals don't have, and I and others have been calling it extreme shear. But maybe it's not shear per se, but rather slant (nonperpendicularity of grid-lines) and non-verticality of meridians.

But I've been saying that those things are globe-natural and realistic, being present in Orthographic. Yes, but there are differences:

Equatorial-aspect Orthographic's slant only increases much at higher latitudes. Sinusoidal has nearly all of its slant already, at mid latitudes.
We aren't used to that and don't expect it. And we expect that slant to curve as it does in Orthographic, but with Sinusoidal it's much more constant, from mid latitudes all the way up.

Additionally, Orthographic, though it foreshortens peripheral regions radially, it doesn't expand them circumferentially.

So, when we notice North America, near the Sinusoidal's periphery, highly slanted even though it's only mid-latitude, and greatly north-south expanded, it makes people say, "Whoa!"

As I said, one gets the impression that everything is on its way to a vacuum-cleaner at the North Pole.

I like Sinusoidal, but it isn't going to win any beauty-contests or popularity-contests.

Anyway, my point was that maybe "shear" isn't the best word for the distortion that Sinusoidal has that cylindricals don't have any of.

Michael Ossipoff