On distortion and optimal projections

[daan]

it appears to me that your projection has such a strong resemblance to the Hammer that one could probably come up with an empirical modification of the Hammer that comes very close to it, even if it isn't an exact match. That would allow for a projection with a closed form description that at least closely approaches optimality.

Likely true!

— daan

[daan]

Hello, dummy_index, and welcome to the forums.

How about Hammer without stretching?
longitude halving -> Lambert azimuthal equal-area -> scale 1.4x in x and y (to restore area)
Maximum angular deformation is about 74°. (unconfident)

You are close. The highest angular deformation I found on your map is about 76.25°. Puzzlingly, this seems to be lower than my “optimal” projection—although I am not confident in my measurements on the “optimal” projection due to the approximate nature of my prototype. The average angular deformation of my “optimal” projection is much lower, but that alone would not suffice to meet the conditions of my conjecture for optimality.

Needless to say, Eisenlohr projection is a projection least area-error in the meaning of "infinity norm." In contrast, attempt to minimize "Entire strain energy" is "Euclidean norm" context and may lead differ answer (parhaps August projection, or maybe Lagrange projection.)

Could you elaborate on how a least area-error metric could (or should) depend on anything except total normalized area? Why would any norm be relevant? Thanks.

— daan

[Atarimaster]

How about Hammer without stretching?
longitude halving -> Lambert azimuthal equal-area -> scale 1.4x in x and y (to restore area)

I’m not sure if I’m getting this description right – does the result look like this?



If this is right…

The highest angular deformation I found on your map is about 76.25°.

Hmm, where is that point exactly?
The highest angular deformation I found on the map posted above is 73.74°, all around at the boundary.
I found it by moving the mouse pointer and looking at Geocart’s info panel, so it is possible that I simply didn’t move the pointer to the right places or that the map still was too small so the point with the highest deformation was not “available” because of interpolation.
… Although I already tried it on a huge map, with a nomical scale of 1:10,000,000; resulting in a map size of 2548.636 × 2548.636 mm, using a resolution of 300 pixel/inch and viewed with “Actual Pixels of Selected Object”.

Note: The map was created in Geocart using the generalized Wagner with following parameters:

Code: Select all

a = 1.41421, b = 1.41421, m = 1, m2= 1, n = 0.5

[daan]

I’m not sure if I’m getting this description right – does the result look like this?

Yes, that is how I interpreted it.

Hmm, where is that point exactly?

It’s along the periphery at 1:30 AM, but otherwise nowhere. You (and dummy_index) are correct. In fact, in closed form, the maximum angular deformation is 2 arcsin(3/5) ≈ 73.7398, constant along the periphery.

Now I must digest what this means vis-à-vis my conjecture.

Well done, dummy_index!

— daan

[daan]

Angular deformation along the central meridian is 2 arcsin(cos φ/(2 + cos φ)). This means that the map has no distortion at the poles, but does everywhere else.

Interesting.

— daan

[daan]

Distortion diagram:

Circular Hammer distortion

Circular_Hammer_distortion.jpg (56.59 KiB) Viewed 11093 times

[dummy_index]

Puzzlingly, this seems to be lower than my “optimal” projection—although I am not confident in my measurements on the “optimal” projection due to the approximate nature of my prototype. The average angular deformation of my “optimal” projection is much lower, but that alone would not suffice to meet the conditions of my conjecture for optimality.

Thinking only "angular deformation is constant on the periphery" and "i want to minimize this", it is isoperimetric problem. So circular.
(except

• When many interruption makes periphery longer than 4 π R, or
• When making circle makes strong twitch and is problematic. (Circular Hammer already has a weak twitch)

So, efforting much for each case may result three distinct theories...)

Needless to say, Eisenlohr projection is a projection least area-error in the meaning of "infinity norm." In contrast, attempt to minimize "Entire strain energy" is "Euclidean norm" context and may lead differ answer (parhaps August projection, or maybe Lagrange projection.)

Could you elaborate on how a least area-error metric could (or should) depend on anything except total normalized area? Why would any norm be relevant? Thanks.

— daan

[no conclusion] Normalized by ...

• Scale of the center point? (sometimes need to use designer's intention.)
• Minimum scale? (can't apply to Littrow.)
• Conversely, normalize by total area?

OK, now use the scale of the center point. Then, Lagrange projection is smaller than August epicycloidal projection.

Lagarnge_vs_August.png (182.53 KiB) Viewed 11088 times

Lagrange: 50.3 R² = 16 π R²
August: 67.0 R²
I don't know whether Lagrange is minimum total normalized area. But, in https://en.wikipedia.org/wiki/Fracture_mechanics, it is natural that singular point will appear at the end of interruption (ex. both pole with cutting 180° meridian.)
But Lagrange is not fit for min/max criterion... [/no conclusion]

[daan]

Thinking only "angular deformation is constant on the periphery" and "i want to minimize this", it is isoperimetric problem. So circular.
(except

• When many interruption makes periphery longer than 4 π R, or
• When making circle makes strong twitch and is problematic. (Circular Hammer already has a weak twitch)

So, efforting much for each case may result three distinct theories...)

Could you provide the Japanese word (漢字で) for “twitch”? There must be a better translation. I am fluent in Japanese, so that would be no burden for me. Is it “discontinuity”? (不連続)

The Lambert azimuthal, as a circular projection with minimum error for a projection interrupted at one point, is no surprise. My surprise is over your modification to Hammer, giving a projection that is interrupted along an entire meridian, has constant angular deformation around the periphery, and might have lower maximum angular deformation than that of the projection I conjectured.

I am also not (very) surprised at Lagrange’s smaller normalized area than August has. Neither August nor Lagrange has constant scale along its periphery, and both of them have regions with higher scale factors than Eisenlohr has. I had no reason to suppose that August fulfilled some other theory of optimality. However, this does make me wonder about the nuances of the August’s argument with Eisenlohr. If Lagrange results in a map with less areal inflation than August, then why would August be better than both Lagrange and Eisenlohr? Do we have some sort of “compromise projection” within the domain of conformal projections?

Could you elaborate on how a least area-error metric could (or should) depend on anything except total normalized area? Why would any norm be relevant? Thanks.

[no conclusion] Normalized by ...

• Scale of the center point? (sometimes need to use designer's intention.)

I definitely disagree that the center should be special.

• Minimum scale? (can't apply to Littrow.)

Yes. Littrow’s scale factors range from infinitesimal to infinite, so I don’t see any problem with using minimum scale and disqualifying Littrow or other such projections. In a sense, they disqualify themselves from any sort of theory of optimal.

• Conversely, normalize by total area?

That is surely a legitimate theory for minimum-error conformal projection, but of course is different than the Chebyshev criterion, not resulting in min/max optimization (as you note in your posting).

— daan

[dummy_index]

Thinking only "angular deformation is constant on the periphery" and "i want to minimize this", it is isoperimetric problem. So circular.
(except

• When many interruption makes periphery longer than 4 π R, or
• When making circle makes strong twitch and is problematic. (Circular Hammer already has a weak twitch)

So, efforting much for each case may result three distinct theories...)

Could you provide the Japanese word (漢字で) for “twitch”? There must be a better translation. I am fluent in Japanese, so that would be no burden for me. Is it “discontinuity”? (不連続)

It was translation of "引き攣れる". Hmm... すみません。日本語で書きます。例えば地表を almost double hemisphere に断裂させて（100 kmだけつながった状態だとして）できた periphery を直径4 Rの円の形にしようとすると、100 kmしか離れていなかった2地点を4 R = 25,484 kmまで離さないといけないので、地図の内部で angular deformation がとても大きくなります。元々の180° meridianのみの断裂でも両極間π Rの距離を4 Rに引き伸ばすことになります。

The Lambert azimuthal, as a circular projection with minimum error for a projection interrupted at one point, is no surprise. My surprise is over your modification to Hammer, giving a projection that is interrupted along an entire meridian, has constant angular deformation around the periphery, and might have lower maximum angular deformation than that of the projection I conjectured.

It seems to me that your algorithm は地図の中心でconformalになっていることを前提としていて、かつ地図の中心からどの方向に向かっても（azimuthal equal area や Hammer がそうであるように）径方向に圧縮されて周方向に引き伸ばされることだけを想定しているようです。こうなると必然的にcentral meridianは本来のnatural lengthより短く表現されることになり、circular Hammerとは違う形にならざるを得ません。その結果周長がcircular Hammerより長くなりひずみが大きくなります。要するに盛り込まれていた条件を私が無視しただけとも言えます。

However, this does make me wonder about the nuances of the August’s argument with Eisenlohr. If Lagrange results in a map with less areal inflation than August, then why would August be better than both Lagrange and Eisenlohr? Do we have some sort of “compromise projection” within the domain of conformal projections?

GS50 compromises among 50 states?
In fact (i think) it seems that many people avoid areal singular point. (But why many people use flat-polar?)

[daan]

Could you provide the Japanese word (漢字で) for “twitch”?

It was translation of "引き攣れる".

I still can’t find an equivalent term in mathematics. The third derivative of the position vector is often called “jerk”, and the fourth derivative is sometimes called “jounce”. Those have similar everyday meanings to “twitch”. Maybe there are engineering fields that use “twitch”, but I was not able to find any. What is the mathematical principle that 引き攣れ expresses?

例えば地表を almost double hemisphere に断裂させて（100 kmだけつながった状態だとして）できた periphery を直径4 Rの円の形にしようとすると、100 kmしか離れていなかった2地点を4 R = 25,484 kmまで離さないといけないので、地図の内部で angular deformation がとても大きくなります。元々の180° meridianのみの断裂でも両極間π Rの距離を4 Rに引き伸ばすことになります

If you represent the sphere on the plane as a nearly double hemisphere, with only 100km of connection between the two near-hemispheres, then you will end up with points that were separated by a mere 100km on the sphere being separated by 4R = 25,484km equivalent on the plane. This means there must be large angular deformation within the map. Even if you only interrupt along the 180° meridian, as per the projection under discussion, the original πR distance between the poles will get extended to 4R.

Is this an argument for how to minimize the angular deformation of the boundary, or how to minimize the average angular deformation across the map? It is not clear to me that the argument necessarily achieves either one. For example, while it is possible that the circular Hammer has lower angular deformation along its boundary than my modified Eisenlohr, it definitely has higher average angular deformation.

It seems to me that your algorithm は地図の中心でconformalになっていることを前提としていて

It seems to me that your algorithm presumes conformality at the map’s center;

It’s much more complicated than that. There is nothing special about the center in the general case. To transform a conformal map to an equal-area map by my algorithm, first identify the isocol that you wish to be conformal on the equal-area map. Often, such as on the modified Eisenlohr that I gave as an example, that isocol will be a single point. If I had chosen some other isocol, then, on my example map, the equal-area result would have opened a hole interior to the isocol. If the isocol you choose is closed, then the resulting equal-area map will have a hole. In order to avoid a hole, you must choose some local minimum as your conformal isocol. That local minimum may well be a path, such as the equator on a Mercator. However, for Eisenlohr, it is a single point.

If the map has several local minima that are not part of the same path, and the values of those local minima differ, then you must choose the one with the lowest value if you want to avoid all holes.

かつ地図の中心からどの方向に向かっても（azimuthal equal area や Hammer がそうであるように）径方向に圧縮されて周方向に引き伸ばされることだけを想定しているようです。

furthermore, no matter which direction from the center (such as azimuthal equal-area and Hammer are), it seems like you are just assuming a radial compression and a circumferential stretch.

That is true only locally to the conformal isocol. The compression is not radial finitely. Instead, compression follows the “antisocol”, which is the path that crosses each isocol perpendicularly from the boundary to the conformal isocol, and that contains the point you want to compute.

こうなると必然的にcentral meridianは本来のnatural lengthより短く表現されることになり、circular Hammerとは違う形にならざるを得ません。その結果周長がcircular Hammerより長くなりひずみが大きくなります。

Thus, the central meridian will inevitably be displayed shorter than its natural length and the shape will differ from the circular Hammer.

Yes. The circular Hammer has a central meridian that is too long. The modified EIsenlohr has a central meridian that is too short.

As a result, the boundary becomes longer than the circular Hammer, and therefore the distortion is higher.

Certainly the boundary on the modified Eisenlohr is longer because the circular Hammer’s boundary is a circle, and of all shapes with the same area, a circle yields the shortest perimeter. Your assertion that this implies lower distortion along the boundary is novel to me. I have to think about it. Superficially, it sounds correct.

Do we have some sort of “compromise projection” within the domain of conformal projections?

GS50 compromises among 50 states?

Actually, GS50 is (nearly) a min/max optimization, which would be antithetical to a compromise.

Cheers,
— daan