Since I've been talking about resolution-efficiency, it may be worth comparing how this one stacks up the the conformal projection, given how similar they look.
I have a better understanding of elliptic functions than I used to, so the resolution-efficiency of the Peirce/Guyou/Adams projection is something I should actually be able to calculate now.
The conformal hemisphere-in-a-square projection is given by the inverse formula:
sl(z) = r
Where z is the hemisphere-in-a-square projection in quincuncial arrangement, r is the stereographic projection with scale 1/2 at the center, and sl is the lemniscate sine function.
The real and imaginary parts of z are allowed to range over the diamond marked by the vertices (±ϖ/2,0) and (0,±ϖ/2]), where ϖ is the lemniscate constant. You can see this from the fact that sl(ϖ/2) = 1, and more generally, that |sl(±ϖ/2 + (1±i)t)| = 1 for all real t (confirmed by plotting). Thus, this projection has a total area of ϖ2/2 (for one hemisphere).
As with most functions called "sine", the derivative of the lemniscate sine function at 0 equals 1. Therefore, the conformal hemisphere-in-a-square projection has the same scale at the center as the stereographic projection used to define it, or 1/2. It's fairly clear that this is where the projection has its minimum scale.
The final resolution-efficiency can then be computed as √π/ϖ ≈ 0.6760.
By contrast, my resolution-preserving projection has an area of π2 (for one hemisphere), and a minimum scale of 1. Therefore its resolution-efficiency can be computed as √(2/π) ≈ 0.7979.
0.6760 isn't a bad value (notably, it's better than an equal-area hemisphere-in-a-square projection could theoretically achieve, based on a perimeter argument), but 0.7979 is a clear improvement of 18% or so.
A dihedral projection
Re: A dihedral projection
I know this isn't the most active forum, but I'm still disappointed at how little response this got. I thought I had made a pretty big breakthrough 
I guess I'll just have to show everyone by coming up with an even better projection!
This post is not that projection.
What I'm doing here is a very simple modification of the projection at the start of the thread: instead of using the semiminor axes, I'm using the semimajor axes. I actually came up with this even before the semiminor version, but I quickly dismissed it as "not the projection I'm looking for, move along". However, it does have some properties of theoretical interest, despite my earlier neglect. At the very least, its formula is even simpler.
In a sense, this is the opposite of the previous projection. Where the previous projection maximizes minimum scale for a given area (or minimizes area for a given minimum scale), this one minimizes maximum scale for a given area (or maximizes area for a given maximum scale). In that sense, where the previous projection can be seen as an approximation to the azimuthal equidistant projection (at least in as far as being a compromise projection, it's not actually equidistant in any meaningful sense), this one can be seen as an approximation to the orthographic projection.
To put (0,0) at the center of the projection, with positive y direction up (mathematical convention):
x(P) = (d(P,TL) + d(P,BL) - π) / 2 = (d(P,TL) - d(P,TR)) / 2
y(P) = (π - d(P,TL) - d(P,TR)) / 2 = (d(P,BL) - d(P,TL)) / 2
To put (0,0) at the top-left of the projection, with positive y direction down (computer graphics convention):
x(P) = (d(P,TL) + d(P,BL) - d(TL,BL)) / 2 = (d(P,TL) - d(P,TR) + d(TL,TR)) / 2
y(P) = (d(P,TL) + d(P,TR) - d(TL,TR)) / 2 = (d(P,TL) - d(P,BL) + d(TL,BL)) / 2
Inverse formulae can be computed using the same technique as before.
This is not a good projection. Like the orthographic projection, distortion is fairly severe. Unlike the orthographic projection, this is not an actual perspective projection. I decided to properly formulate this projection not because I expect it to see much use, but because I hoped that studying its properties would help in the development of other hemisphere-in-a-square projections, including my eventual goal of a smooth equal-area projection.
In fact, this projection isn't even smooth! Although it satisfies the same "Tissot indicatrix axes are always parallel to the edges" property as the previous projection, which would normally be enough to ensure that curves perpendicular to the edges on the sphere are also perpendicular to the edges on the plane, that doesn't happen this time due to the projection having scale zero at the edges, causing a singularity.
Still, one can imagine some niche applications, if for some reason you just really need an approximation to the orthographic projection that can fit in a square. (Or a rectangle: like the previous projection, this one can be generalized to rectangular dihedra, and unlike the previous projection, doing so produces rectangles on the plane as well as on the sphere, rather than always mapping the rectangle to a square. This means that the limit case produces a degenerate projection, as each hemisphere ends up as a rectangle of zero width. Here, though, I'll only be showing the square case.) Though, even then, there's probably better projections possible (at the very least try to find a smooth one!), because again, while this projection satisfies a certain optimality criterion for psuedoorthographic projections, it is probably not the unique projection satisfying that criterion.
Here's what the projection looks like:
Then again, the most iconic application of the orthographic projection lies elsewhere. If you count looking up at the moon, then the orthographic projection (or at least a very good approximation to it) is the first map projection ever used by humanity, even before we actually knew what map projections were.
This shows only the hemisphere of the moon that is visible from Earth, and is rotated 90° from the usual convention so that east is up and north is left. For why I chose that rotation, see the filename.
Compare the actual orthographic projection (made using G.Projector), scaled to have the same scale at the center:
There's a few black spots due to gaps in the Clementine data that I used for this map. Sorry about that. (You'd think by now we'd have good pictures of the moon...)
The important math here is analyzing the Tissot indicatrices. Using the same technique as before, I arrive at (mathematical convention):
x scale = √((sin(ψ)2 - sin(x)2) / (1 - sin(x)2 - sin(y)2))
y scale = √((cos(ψ)2 - sin(y)2) / (1 - sin(x)2 - sin(y)2))
Observe that these are never greater than 1, because |x| ≤ ψ and |y| ≤ π/2 - ψ, and therefore cos(x) ≥ cos(ψ) and cos(y) ≥ sin(ψ).
In the square case, ψ = π/4, and this simplifies to:
x scale = √((1/2 - sin(x)2) / (1 - sin(x)2 - sin(y)2))
y scale = √((1/2 - sin(y)2) / (1 - sin(x)2 - sin(y)2))
Once again, when x = y, we obtain conformality, and this time also with constant scale, equal to √(1/2).
This allows the following gnuplot representation of the Tissot indicatrices:
In particular, scale is constant along the perimeter. This is important, because an optimal equal-area projection would also be expected to have this property. Therefore, an equal-area projection could presumably be devised by keeping the same boundary as the pseudoorthographic projection, but shifting around the insides.
Scale perpendicular to the perimeter, meanwhile, is also constant... and equal to zero. Obviously, an equal-area projection does not want this. It's also inconvenient for smoothness, as mentioned before.
Likewise, the code for the graticules is:
No screenshots this time, but I think most people here know how to run these on your own if you're interested.
I guess I'll just have to show everyone by coming up with an even better projection!
This post is not that projection.
What I'm doing here is a very simple modification of the projection at the start of the thread: instead of using the semiminor axes, I'm using the semimajor axes. I actually came up with this even before the semiminor version, but I quickly dismissed it as "not the projection I'm looking for, move along". However, it does have some properties of theoretical interest, despite my earlier neglect. At the very least, its formula is even simpler.
In a sense, this is the opposite of the previous projection. Where the previous projection maximizes minimum scale for a given area (or minimizes area for a given minimum scale), this one minimizes maximum scale for a given area (or maximizes area for a given maximum scale). In that sense, where the previous projection can be seen as an approximation to the azimuthal equidistant projection (at least in as far as being a compromise projection, it's not actually equidistant in any meaningful sense), this one can be seen as an approximation to the orthographic projection.
To put (0,0) at the center of the projection, with positive y direction up (mathematical convention):
x(P) = (d(P,TL) + d(P,BL) - π) / 2 = (d(P,TL) - d(P,TR)) / 2
y(P) = (π - d(P,TL) - d(P,TR)) / 2 = (d(P,BL) - d(P,TL)) / 2
To put (0,0) at the top-left of the projection, with positive y direction down (computer graphics convention):
x(P) = (d(P,TL) + d(P,BL) - d(TL,BL)) / 2 = (d(P,TL) - d(P,TR) + d(TL,TR)) / 2
y(P) = (d(P,TL) + d(P,TR) - d(TL,TR)) / 2 = (d(P,TL) - d(P,BL) + d(TL,BL)) / 2
Inverse formulae can be computed using the same technique as before.
This is not a good projection. Like the orthographic projection, distortion is fairly severe. Unlike the orthographic projection, this is not an actual perspective projection. I decided to properly formulate this projection not because I expect it to see much use, but because I hoped that studying its properties would help in the development of other hemisphere-in-a-square projections, including my eventual goal of a smooth equal-area projection.
In fact, this projection isn't even smooth! Although it satisfies the same "Tissot indicatrix axes are always parallel to the edges" property as the previous projection, which would normally be enough to ensure that curves perpendicular to the edges on the sphere are also perpendicular to the edges on the plane, that doesn't happen this time due to the projection having scale zero at the edges, causing a singularity.
Still, one can imagine some niche applications, if for some reason you just really need an approximation to the orthographic projection that can fit in a square. (Or a rectangle: like the previous projection, this one can be generalized to rectangular dihedra, and unlike the previous projection, doing so produces rectangles on the plane as well as on the sphere, rather than always mapping the rectangle to a square. This means that the limit case produces a degenerate projection, as each hemisphere ends up as a rectangle of zero width. Here, though, I'll only be showing the square case.) Though, even then, there's probably better projections possible (at the very least try to find a smooth one!), because again, while this projection satisfies a certain optimality criterion for psuedoorthographic projections, it is probably not the unique projection satisfying that criterion.
Here's what the projection looks like:
- ortho.png (252.13 KiB) Viewed 14991 times
- bunny.png (106.53 KiB) Viewed 14991 times
Compare the actual orthographic projection (made using G.Projector), scaled to have the same scale at the center:
- moon.png (69.98 KiB) Viewed 14991 times
The important math here is analyzing the Tissot indicatrices. Using the same technique as before, I arrive at (mathematical convention):
x scale = √((sin(ψ)2 - sin(x)2) / (1 - sin(x)2 - sin(y)2))
y scale = √((cos(ψ)2 - sin(y)2) / (1 - sin(x)2 - sin(y)2))
Observe that these are never greater than 1, because |x| ≤ ψ and |y| ≤ π/2 - ψ, and therefore cos(x) ≥ cos(ψ) and cos(y) ≥ sin(ψ).
In the square case, ψ = π/4, and this simplifies to:
x scale = √((1/2 - sin(x)2) / (1 - sin(x)2 - sin(y)2))
y scale = √((1/2 - sin(y)2) / (1 - sin(x)2 - sin(y)2))
Once again, when x = y, we obtain conformality, and this time also with constant scale, equal to √(1/2).
This allows the following gnuplot representation of the Tissot indicatrices:
Code: Select all
set size ratio -1
unset key
unset xtics
unset ytics
set samples 13
set isosamples 13,13
x_scale(x,y) = sqrt((sin(psi)**2 - sin(x)**2) / (1 - sin(x)**2 - sin(y)**2))
y_scale(x,y) = sqrt((cos(psi)**2 - sin(y)**2) / (1 - sin(x)**2 - sin(y)**2))
psi=pi/4; plot [-psi:+psi] [psi-pi/2:pi/2-psi] sample [u=-psi:+psi] [v=psi-pi/2:pi/2-psi] '++' using ($1):($2):(abs($1)==psi&&abs($2)==pi/2-psi?0: x_scale($1,$2)/15):(abs($1)==psi&&abs($2)==pi/2-psi?0: y_scale($1,$2)/15) with ellipses fill solid # I don't know why gnuplot throws up an error message if I name the variable anything other than u.
Scale perpendicular to the perimeter, meanwhile, is also constant... and equal to zero. Obviously, an equal-area projection does not want this. It's also inconvenient for smoothness, as mentioned before.
Likewise, the code for the graticules is:
Code: Select all
set parametric
set size ratio -1
unset xtics
unset ytics
unset xzeroaxis
unset yzeroaxis
unset key
set multiplot layout 1,2
x(longitude, latitude) = (acos((sin(latitude) - cos(latitude)*sin(longitude)) / sqrt(2)) - acos((sin(latitude) + cos(latitude)*sin(longitude)) / sqrt(2))) / 2
y(longitude, latitude) = (asin((sin(latitude) - cos(latitude)*sin(longitude)) / sqrt(2)) + asin((sin(latitude) + cos(latitude)*sin(longitude)) / sqrt(2))) / 2
set title 'Guyou'
plot [-pi/2:+pi/2] [-pi/4:+pi/4] [-pi/4:+pi/4] for[i=-6:+6] x(t, pi*i/12), y(t, pi*i/12) lc 1, for[i=-5:+5] x(pi*i/12, t), y(pi*i/12, t) lc 1
x(longitude, latitude) = (acos(cos(latitude)*sin(longitude)) - acos(cos(latitude)*cos(longitude))) / 2
y(longitude, latitude) = (asin(cos(latitude)*sin(longitude)) + asin(cos(latitude)*cos(longitude))) / 2
set title 'Peirce'
plot [-pi/2:+pi/2] [-pi/4:+pi/4] [-pi/4:+pi/4] for[i=1:6] x(t*2, pi*i/12), y(t*2, pi*i/12) lc 1, for[i=0:23] x(pi*i/12, t/2+pi/4), y(pi*i/12, t/2+pi/4) lc 1
Re: A dihedral projection
For my own part, I believe you have, and I am remiss not to have remarked. I apologize for that. There are several threads now that I have left languishing for lack of time to engage them properly. So. I intend to comment when I can do so coherently. I am definitely intrigued by your developments, so — while I lack enough understanding yet to say this with deep conviction — I congratulate you for what looks to me like very insightful work.
I am particularly hopeful for an equal-area solution eventually.
Cheers,
— daan
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mapnerd2022
- Posts: 165
- Joined: Tue Dec 28, 2021 9:33 pm
Re: A dihedral projection
Congrats, Mr.Milo, for such interesting work! Well done!
Re: A dihedral projection
I'm afraid that this is too complicated for me to understand.
I can only guess, based on the term "resolution efficiency", and your mention of the plate carrée, that your projection combines the property of being rectangular, and thus having a resemblance to the Guyou projection, with a property shared by the plate carrée and the Azimuthal Equidistant projection: the minimum scale through each point on the map is the same, and so the map takes up, in some sense, as little space as possible for a map with that minimum scale.
Since this is a property that ignores shear and distortion, which strongly affect how well one can make out places on a map, I'm not expecting this to take the world by storm, but that it is an interesting mathematical achievement is indeed without doubt.
I can only guess, based on the term "resolution efficiency", and your mention of the plate carrée, that your projection combines the property of being rectangular, and thus having a resemblance to the Guyou projection, with a property shared by the plate carrée and the Azimuthal Equidistant projection: the minimum scale through each point on the map is the same, and so the map takes up, in some sense, as little space as possible for a map with that minimum scale.
Since this is a property that ignores shear and distortion, which strongly affect how well one can make out places on a map, I'm not expecting this to take the world by storm, but that it is an interesting mathematical achievement is indeed without doubt.
Re: A dihedral projection
Ah, alright!daan wrote: ↑Sun Mar 27, 2022 9:49 pmFor my own part, I believe you have, and I am remiss not to have remarked. I apologize for that. There are several threads now that I have left languishing for lack of time to engage them properly. So. I intend to comment when I can do so coherently. I am definitely intrigued by your developments, so — while I lack enough understanding yet to say this with deep conviction — I congratulate you for what looks to me like very insightful work.
I confess that I can have a bad habit of reporting my results without detailing every step of the calculation I took to get there, so some of my posts can feel a bit dense. (Though I also don't like it when papers spend too much time showing their work and then forget to actually summarize their conclusions at the end.) I can try to offer clarification if there's anything specific you're confused about. Though really, it'd be useful to have someone else independently double-check my math (particularly on the distortion metrics), because if someone else can't arrive at the same results I did, I probably made a mistake.
I can be patient, though, so long as I know what I'm waiting for. Thanks!
Resolution-efficiency is explained in this thread.quadibloc wrote: ↑Wed Mar 30, 2022 8:43 amI can only guess, based on the term "resolution efficiency", and your mention of the plate carrée, that your projection combines the property of being rectangular, and thus having a resemblance to the Guyou projection, with a property shared by the plate carrée and the Azimuthal Equidistant projection: the minimum scale through each point on the map is the same, and so the map takes up, in some sense, as little space as possible for a map with that minimum scale.
It's not quite "the minimum scale through each point is the same" (I call that property "constant-resolution", and this projection does not satisfy it). However, "as little space as possible for a map with that minimum scale" is completely accurate: the map minimizes overall area while keeping overall minimum scale the same (or maximizes overall minimum scale while keeping overall area the same, or maximizes the ratio of minimum scale to square root of area, which is how I defined it in that thread).
I wouldn't say it ignores those things.
I'm not sure what you mean by "shear", or for that matter by "distortion" which is a generic term applying to all kinds of bad things happening on maps, but those things are not going to be unrelated to minimum scale.
For example, start with a circle of radius 1. Now apply the shear transformation (x, y) -> (x+y, y). It is now an ellipse with the same area, but a semimajor axis of (√5+1)/2 and a semiminor axis of (√5-1)/2. The latter is approximately 0.618, significantly less than 1.
In particular, note that area flation at a point equals minimum scale times maximum scale at that point, therefore if minimum scale is kept constant, and total area is minimized, then maximum scale can't be too large over most of the map, or it would result in inflating the map's area. Maximum scale can still escape to infinity somewhere, but it has to be confined to a very small region.
Compared to the plate carree projection, my map has dramatically less angle distortion: angle distortion on the plate carree projection approaches infinity near the poles, whereas on my projection, it stays strictly bounded everywhere, even near the singularities. Considering that angle distortion is the most reasonable definition I can think of for what you call "shear", that means my projection is actually significantly better than the plate carree projection at this. Shear is never worse than a √2:1 Tissot aspect ratio, or if you prefer, angles are never distorted by more than 9.88°.
Area distortion approaches infinity near the singularities on both the plate carree projection and my new one, but I think mine still exaggerates areas less badly (considering how much Antarctica is exaggerated in the plate carree), though it's hard to tell because all my example pictures placed the singularities in the ocean. No wait, the Adams aspect doesn't. To compare them, I used an image editing program's "select by color" tool to identify Antarctica in both that image, and an equal-sized plate caree projection. The plate caree projection measures 33392 pixels (10.435% of the map), while the Adams-aspect dihedral projection measures 19325 pixels (6.039% of the map). Antarctica is actually only 2.784% of Earth's surface, so both are significant exaggerations, but my projection is less so. Then again, Antarctica is still slightly more than half its plate carree area, and there are twice as many singularities that are exaggerated like that, even if the other three are in the ocean...
The basic principle makes sense: by having twice as many singularities (four instead of two), the amount of distortion at/near each individual singularity is reduced. So long as you can shuffle the singularities off to somewhere you don't care about too much, that's a win.
I believe it should be theoretically possible to create a resolution-efficient projection that also has both angle and area distortions bounded to a finite value over the whole map, but I haven't yet.
(And thanks for commenting! Even criticism gives me something to respond to.)
Re: A dihedral projection
Sorry for not responding earlier. Certain events around 1000 miles from me have been occupying my thoughts and reading time for the last six weeks, so I haven't been checking this forum much. I also admit that I don't completely understand everything that you've written, so I'll restrict myself to a comment on the projection's appearance.
The Guyou projection looks a lot like the two-hemisphere stereographic projection but with the oceans stretched to fill in the gaps and make it all rectangular. Personally, I much prefer the two-hemisphere stereographic projection and have long questioned the desirability of forcing a non-cylindrical projection into a rectangular shape. I feel the same way about the AuthaGraph projection.
Just to check whether or not I've understood this correctly, is your projection to the two-hemisphere azimuthal equidistant projection what the Guyou projection is to the two-hemisphere stereographic projection? If so, that obviously is a big achievement from an academic point of view.
In any case, the first two aspects of your projection that you posted look like "equidistant" versions of the Guyou and AuthaGraph projections. Since I'm not a big fan of the Guyou or AuthaGraph projections, I have to admit that these two aspects of your projection aren't really favourites of mine either.
I do like the quincuncial aspect, though. I can't decide whether or not I prefer it to Peirce's original. It represents the size of Greenland and Antarctica much better, but Africa and South America do look quite squashed in the north-south direction.
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Atarimaster
- Posts: 446
- Joined: Fri Nov 07, 2014 2:43 am
Re: A dihedral projection
Me, too.
So why did I say that I liked the Guyou before? It’s not its rectangular shape that I care about, but its “uninterruptedness” and its comparatively low areal inflations. Of course I’m aware
a) that actually, there is no uninterrupted world map (but I think you all know what I mean),
b) that the Guyou projection is only seemingly uninterrupted – no matter how you combine two hemispheric maps, in the end, they are still hemispheric maps.
But I do think that, for whatever reason, many people seem like rectangular maps a lot. And it’s always good to be able to give people what they want. Regarding that we’ve got conformal and equivalent hemispheres in a square, it’s nice to have a compromise hemisphere in a square projection now.
Kind regards,
Tobias
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mapnerd2022
- Posts: 165
- Joined: Tue Dec 28, 2021 9:33 pm
Re: A dihedral projection
Well, yes, even if a map projection isn´t unbounded (with at least one or two points at infinity, so therefore it is unbounded) it's always interrupted. The so called interrupted projections are only further interrupted( they just add more boundaries).
Re: A dihedral projection
The Guyou projection is interrupted along 270°, so it's certainly less interrupted than the two-hemisphere stereographic projection, which is interrupted along 360°. On the other hand, it's more interrupted than such conformal projections as Lagrange, Eisenlohr and August, which are interrupted along 180°. As a result, its areal inflation is low compared to Lagrange, Eisenlohr and August but high compared to the two-hemisphere stereographic projection.
Note that projections with intermediate "interruptedness" between 180° and 360° don't necessarily have to be rectangular -- these Spilhaus projections, for example, allow you to arbitrarily vary the length of the interruption, though I don't know of any comparable conformal projections.
Note that projections with intermediate "interruptedness" between 180° and 360° don't necessarily have to be rectangular -- these Spilhaus projections, for example, allow you to arbitrarily vary the length of the interruption, though I don't know of any comparable conformal projections.
Yes, Milo has rectified a longstanding deficiency in the catalogue of known map projections, so hats off for that.Atarimaster wrote: ↑Wed Apr 06, 2022 7:00 am Regarding that we’ve got conformal and equivalent hemispheres in a square, it’s nice to have a compromise hemisphere in a square projection now.