Eisenlohr’s optimal conformal map of the world

General discussion of map projections.
Milo
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Re: Eisenlohr’s optimal conformal map of the world

Post by Milo »

I think you accidentally hit quote instead of edit?

A complication with trying to compute the derivative of flation is that while for a conformal projection flation itself is always a scalar value (per definition), its derivative is not. The Mercator projection, obviously, has zero flation change in the longitude direction, but nonzero in the latitude direction. This can probably be generalized to all projections having isolines of constant flation in some direction, so what we're interested in is the rate of change perpendicular to those isolines (as you pointed out to me). We can't just keep using complex analysis, since flation is the absolute value of the complex derivative, and also since we need to adjust the value of flation for the sphere (complex analysis by itself only does conformal projections from a Euclidean plane to another Euclidean plane).

The Mercator projection, then, is pretty easy. Its flation is 1/cos(latitude)^2, so its rate of change in flation is |2*tan(latitude)/cos(latitude)^2|.

Alternatively, you could measure the derivative of linear scale instead of flation (which is the square of linear scale), in which case its rate of change in linear scale is |tan(latitude)/cos(latitude)|.

Or you could even measure the rate of change in logarithmic scale/flation, giving a plain |tan(latitude)|.
daan
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Re: Eisenlohr’s optimal conformal map of the world

Post by daan »

Milo wrote: Wed Sep 20, 2023 8:31 am I think you accidentally hit quote instead of edit?
Ugh. Fixed. Thanks.
A complication with trying to compute the derivative of flation is that while for a conformal projection flation itself is always a scalar value (per definition), its derivative is not. The Mercator projection, obviously, has zero flation change in the longitude direction, but nonzero in the latitude direction. This can probably be generalized to all projections having isolines of constant flation in some direction, so what we're interested in is the rate of change perpendicular to those isolines (as you pointed out to me).
Right. I should have been specific and stipulated the gradient, rather than merely rate of change. Thanks.

— daan
Milo
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Re: Eisenlohr’s optimal conformal map of the world

Post by Milo »

Thinking about it, taking the derivative of logarithmic scale makes the most sense, since that's the one that's independent of overall map scale (and absolute scale not mattering is the entire point here).
PeteD
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Re: Eisenlohr’s optimal conformal map of the world

Post by PeteD »

daan wrote: Wed Sep 20, 2023 7:47 am Since angular deformation α has range [0..180]°, maybe something as simple as tan α/2 could be usefully compared against rate of inflation s, with range [0..∞].
If we define i = a/b, where a and b are the semi-axes of the Tissot ellipse, then tan α/2 = 1/2 * abs(sqrt(i) - 1/sqrt(i)), which is more compact than I'd expected but still seems a bit convoluted compared to ln^2(i) or abs(ln(i)), which also have a range of [0, ∞]. Incidentally, using 2*tan α/2 instead approximates abs(ln(i)) very closely.
PeteD
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Re: Eisenlohr’s optimal conformal map of the world

Post by PeteD »

Milo wrote: Wed Sep 20, 2023 8:31 am Or you could even measure the rate of change in logarithmic scale/flation, giving a plain |tan(latitude)|.
The derivative of the logarithmic scale is equal to the derivative of the linear scale normalized to the linear scale, e.g. for the Mercator projection, d/dϕ(ln(a)) = (da/dϕ) / a, where a is the vertical semi-axis of the Tissot ellipse. This is equal to skewness as defined by Goldberg and Gott, except they averaged the skewness over all azimuths, whereas you're proposing to take the value perpendicular to the isoline.
PeteD
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Re: Eisenlohr’s optimal conformal map of the world

Post by PeteD »

Milo wrote: Wed Sep 20, 2023 5:21 pm Thinking about it, taking the derivative of logarithmic scale makes the most sense, since that's the one that's independent of overall map scale (and absolute scale not mattering is the entire point here).
I agree, except that if we still want this to be a measure of areal distortion, then taking the derivative of logarithmic flation would make more sense than taking the derivative of the logarithmic scale since the derivative of logarithmic flation is zero for equal-area projections, whereas the derivative of the logarithmic scale is essentially skewness, which isn't zero for any projection and isn't even correlated with low areal distortion but rather with a good balance between areal and angular distortion (to some extent).
PeteD
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Re: Eisenlohr’s optimal conformal map of the world

Post by PeteD »

Milo wrote: Wed Sep 20, 2023 6:57 am I guess that for conformal maps, what's important isn't so much absolute distortion (which is arbitrary and depends on normalization anyway) but rather relative distortion
Yes, and also for compromise projections, any relevant measure of areal distortion must be a measure of relative areal distortion.
Milo wrote: Wed Sep 20, 2023 6:57 am The exaggeration of Greenland compared to Africa isn't that important because, even if they're on the same meridian, they're not that close to each other. The exaggeration of Greenland compared to Newfoundland is a bigger deal because it affects local shapes.
There are two separate issues here: distortion of areas and distortion of finite shapes. Although the Mercator and Lagrange projections preserve infinitesimal shapes, as you say, the expansion of Greenland compared to Newfoundland distorts finite shapes in that region, so the expansion of Greenland compared to Newfoundland is a bigger deal than its expansion compared to Africa because it results in both distortion of areas and distortion of finite shapes. However, when we only consider distortion of areas, in my opinion, the expansion of Greenland compared to Africa is just as important.
Milo wrote: Wed Sep 20, 2023 6:57 am So I think some sort of metric based on the second derivative of a conformal projection, measuring the rate of change of flation over the map, might be called for.
This is a very interesting idea and definitely worth pursuing, but it's not the only way to measure relative areal distortion:
  1. Canters, Goldberg and Gott, Györffy and Kerkovits all scale the graticule in order to minimize the overall areal distortion.
  2. Instead of taking the average of the local areal distortion over the globe in order to obtain an overall value, you could take the standard deviation of the local areal distortion over the globe.
  3. Or you could take the mean absolute deviation from the median of the local areal distortion over the globe.
If we define areal distortion as ln^2(p), where p = a*b, where a and b are the semi-axes of the Tissot ellipse, then 1 and 2 are equivalent. On the other hand, if we define areal distortion as abs(ln(p)), then 1 and 3 are equivalent.
daan
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Re: Eisenlohr’s optimal conformal map of the world

Post by daan »

PeteD wrote: Thu Sep 21, 2023 7:28 am
This is a very interesting idea and definitely worth pursuing, but it's not the only way to measure relative areal distortion:
  1. Canters, Goldberg and Gott, Györffy and Kerkovits all scale the graticule in order to minimize the overall areal distortion.
  2. Instead of taking the average of the local areal distortion over the globe in order to obtain an overall value, you could take the standard deviation of the local areal distortion over the globe.
  3. Or you could take the mean absolute deviation from the median of the local areal distortion over the globe.
If we define areal distortion as ln^2(p), where p = a*b, where a and b are the semi-axes of the Tissot ellipse, then 1 and 2 are equivalent. On the other hand, if we define areal distortion as abs(ln(p)), then 1 and 3 are equivalent.
The reason I have been considering distortion gradient is to make comparisons between projections relying on as few arbitrary choices as feasible. Scale is arbitrary, and therefore absolute areal measure is arbitrary, no matter how it is defined.

I’m skeptical of measures that depend on arbitrary decisions, such as the frequent “Let’s ignore the upper 15° [or 10° or whatever arbitrary value] of latitudes even in normal projections for computing distortion. While it’s true that most useful map projections do not have bizarre behavior beyond interruptions and infinite scales, the proposition of normalizing area measure requires more special pleading: It requires the projection not to have points with both infinite and infinitesimal area measure. In the presence of both, any normalization by scaling becomes arbitrary. Most are arbitrary even in the presence of “just” infinite measure, which is a feature of many common maps. That is beyond the normalization method choice itself being arbitrary. Using the gradient of flation, on the other hand, eliminates arbitrary choices. It also permits natural choices for “the point(s) of no distortion”, which on most conformal maps, coincides with the flation gradient being zero. This yields a natural analog to the unambiguous distortionless points on equal-area maps.

Cheers,
— daan
Milo
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Re: Eisenlohr’s optimal conformal map of the world

Post by Milo »

PeteD wrote: Thu Sep 21, 2023 6:47 amThe derivative of the logarithmic scale is equal to the derivative of the linear scale normalized to the linear scale,
Indeed. A nice consequence of the chain rule :)
PeteD wrote: Thu Sep 21, 2023 6:54 amI agree, except that if we still want this to be a measure of areal distortion, then taking the derivative of logarithmic flation would make more sense than taking the derivative of the logarithmic scale
Logarithmic flation is always equal to exactly twice logarithmic linear scale (for conformal projections), and so so is its derivative. Multiplying by a constant doesn't meaningfully change the metric.

For non-conformal projections, linear scale is not a scalar function, and so you can't easily take its derivative anyway.
PeteD wrote: Thu Sep 21, 2023 7:28 amIf we define areal distortion as ln^2(p), where p = a*b, where a and b are the semi-axes of the Tissot ellipse, then 1 and 2 are equivalent. On the other hand, if we define areal distortion as abs(ln(p)), then 1 and 3 are equivalent.
Areal distortion is not either of those things. It's simply ln(a*b), and allowed to be negative. You want to take the absolute value (or square) after taking the derivative, not before. (Strictly speaking, what I'm proposing to measure is the norm of the gradient vector of ln(a*b), which has nonnegativity baked in as a consequence of being a norm.)

Indeed, if we interpret flation in the most literal sense, ln(a*b) will always be negative for any map that's less than life-size :)
PeteD
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Re: Eisenlohr’s optimal conformal map of the world

Post by PeteD »

Milo wrote: Thu Sep 21, 2023 9:54 am Areal distortion is not either of those things. It's simply ln(a*b), and allowed to be negative.
Sorry, my last point was hurriedly written and sloppily worded. I'll start again, trying to be more rigorous this time. I'll start by defining my terms to make sure we're not talking past each other. If any of my nomenclature is non-standard, please let me know and I'll update the terms and amend the rest of this post accordingly. If any of my reasoning is flawed, please let me know specifically what is flawed.

a and b are the semi-axes of the Tissot ellipse.

Flation p is given by p = a b.

p' = a b c2, where c is a rescaling factor.

Local areal distortion εp is given by εp = ln p = ln a b.

ε'p = ln p' = ln a b c2 = ln a b + ln c2 = εp + ln c2.

Note that this definition of local areal distortion is far from universal. The following alternative definitions have all been used at some point:
  • p - 1
  • max(p, 1/p) - 1
  • abs(p - 1) / (p + 1).
Given the local areal distortion εp, the goal is to calculate a meaningful value for the global areal distortion Ep. There are several different ways of doing this.

1. Standard deviation of εp

As you said, what's important is not so much the value of εp but how this value varies across the globe. If you want to measure how much a quantity varies, it's natural to take its variance or the square root thereof, otherwise known as the standard deviation:

Ep = sqrt(⟨(εp - ⟨εp⟩)2⟩),

where the angle brackets (apparently known as "chevrons" in North America), common in physics, denote an average:

⟨x⟩ = 1/(4 π) ∫∫ x cos ϕ dϕ dλ for any x.

This definition of Ep is scale-invariant.

2. Mean absolute deviation from the median of εp

The standard deviation isn't the only measure of how much a quantity varies. An alternative measure is the mean absolute deviation from the median:

Ep = ⟨abs(εp - median(εp))⟩.

This definition of Ep is also scale-invariant. We'll come back to why this alternative measure might be of interest.

3. Root mean square of εp

As you say, εp can be negative. If we simply took Ep = ⟨εp⟩, we could end up with Ep = 0 even if εp is non-zero almost everywhere. Kavrayskiy proposed taking the root mean square (also known as the quadratic mean) instead:

Ep2 = ⟨εp2⟩.

The problem with this is that it's not scale-invariant. Canters, Goldberg and Gott, Györffy and Kerkovits all rescale such that Ep is minimized:

E'p2 = ⟨ε'p2⟩ = ⟨(εp + ln c2)2⟩.

E'p is minimized when dE'p/dc = 0. This occurs for ln c2 = -⟨εp⟩, which gives

E'p2 = ⟨(εp - ⟨εp⟩)2

or alternatively,

E'p = sqrt(⟨(εp - ⟨εp⟩)2⟩).

Wait, haven't we seen this somewhere before?

Maybe it's just me, but I find it quite remarkable that we've obtained exactly the same Ep in two different ways:
  1. taking the standard deviation of εp; and
  2. taking the root mean square of εp and then rescaling such that Ep is minimized.
Note that 1. didn't explicitly involve any rescaling.

4. Absolute mean of εp

An alternative way of getting around the negative values of εp is by taking the absolute mean:

Ep = ⟨abs(εp)⟩.

Again, this isn't scale-invariant, so we can rescale such that Ep is minimized:

E'p = ⟨abs(ε'p)⟩ = ⟨abs(εp + ln c2)⟩.

E'p is again minimized when dE'p/dc = 0. This time, this occurs for ln c2 = -median(εp), which gives

E'p = ⟨abs(εp - median(εp))⟩,

which we've also seen before as the mean absolute deviation from the median of εp.

5. Derivative of εp

As has been mentioned, the derivative of εp is another scale-invariant measure of how εp varies, and it is equivalent to (derivative of p) / p. The latter definition may appeal to those who find logarithms an unnatural function to apply to flation. As you said, its value depends on the azimuth. Before we can start thinking about how to obtain a global value, we therefore first have to obtain a scalar local value. There are three obvious ways of doing this:
  1. averaging over all azimuths, as Goldberg and Gott do with their closely related skewness;
  2. selecting the value perpendicular to the isoline; or
  3. selecting the maximum value at that point.
Are 2. and 3. always the same?

Once you have your scalar local value, you then have to decide whether to take the root mean square or the absolute mean or some other average in order to obtain a global value. Goldberg and Gott use the absolute mean for flexion and skewness despite using the root mean square for areal and angular distortion.
Milo wrote: Thu Sep 21, 2023 9:54 am Logarithmic flation is always equal to exactly twice logarithmic linear scale (for conformal projections), and so is its derivative. Multiplying by a constant doesn't meaningfully change the metric.
I'm not interested in metrics that are only applicable to conformal projections. For equal-area projections, whether you use logarithmic flation or logarithmic linear scale will make a huge difference since the former will give zero everywhere while the latter will be essentially skewness and therefore non-zero almost everywhere.
Milo wrote: Thu Sep 21, 2023 9:54 am For non-conformal projections, linear scale is not a scalar function, and so you can't easily take its derivative anyway.
We've already established that the derivative depends on the azimuth anyway, so simply take the linear scale in the same direction as you're taking the derivative.
Milo wrote: Thu Sep 21, 2023 9:54 am You want to take the absolute value (or square) after taking the derivative, not before.
Where did you get the impression that I wanted to take the absolute value before taking the derivative?
Last edited by PeteD on Thu Sep 21, 2023 11:35 pm, edited 1 time in total.
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