Which projections you favor?
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Re: Which projections you favor?
But I just read in progonos.com(in the wayback machine) that the Ortelius Oval isn't pseudocylindrical because it lacks constant scale along the meridians.
Re: Which projections you favor?
Pseudocylindrical projections have constant scale along the parallels.
Re: Which projections you favor?
Of course. I think I must have read too quickly and the word "pseudocylindrical" didn't register.
Did you manage to find an optimal pseudocylindrical equal-area projection?
Re: Which projections you favor?
The cited link says:
In any case, the more obvious reason that the Hammer projection isn't pseudocylindrical is that it doesn't have straight parallels. If the parallels aren't even straight, it's pointless to worry about if they have constant scale.
Minimizing distortion at the poles (which in most case will have the worst distortion) is a good first step, and I sorta figured out how to do that. In fact, it's pretty easy: just squash a sinusoidal projection so it has the right pole angle. However, this alone wouldn't be enough to define a unique projection: numerous other projections with the same pole angle would be tied with it. For true minimax optimality, you'd want a "greedy algorithm": first figure out the constraint that will minimize distortion at 90°, then given that constraint, figure out the constraint that will minimize distortion at 89°, then given that constraint, figure out the constraint that will minimize distortion at 88°, etc., except continuously rather than at discrete 1° increments. I never got that far. I'm not sure if a continuous greedy algorithm is even possible, although it seems like it should be.
This refers to the spacing between meridians, not along them.Neither it is pseudocylindrical because the meridian spacing is not uniform in the inner and outer hemispheres.
In any case, the more obvious reason that the Hammer projection isn't pseudocylindrical is that it doesn't have straight parallels. If the parallels aren't even straight, it's pointless to worry about if they have constant scale.
No, except maybe in a very loose and unsatisfying sense.
Minimizing distortion at the poles (which in most case will have the worst distortion) is a good first step, and I sorta figured out how to do that. In fact, it's pretty easy: just squash a sinusoidal projection so it has the right pole angle. However, this alone wouldn't be enough to define a unique projection: numerous other projections with the same pole angle would be tied with it. For true minimax optimality, you'd want a "greedy algorithm": first figure out the constraint that will minimize distortion at 90°, then given that constraint, figure out the constraint that will minimize distortion at 89°, then given that constraint, figure out the constraint that will minimize distortion at 88°, etc., except continuously rather than at discrete 1° increments. I never got that far. I'm not sure if a continuous greedy algorithm is even possible, although it seems like it should be.
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Re: Which projections you favor?
Ah, okay, thanks for the correction. My mind must be playing tricks on me.
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Re: Which projections you favor?
Which do you prefer? Mollweide or Hammer?Milo wrote: ↑Fri Feb 17, 2023 4:48 am The cited link says:This refers to the spacing between meridians, not along them.Neither it is pseudocylindrical because the meridian spacing is not uniform in the inner and outer hemispheres.
In any case, the more obvious reason that the Hammer projection isn't pseudocylindrical is that it doesn't have straight parallels. If the parallels aren't even straight, it's pointless to worry about if they have constant scale.
No, except maybe in a very loose and unsatisfying sense.
Minimizing distortion at the poles (which in most case will have the worst distortion) is a good first step, and I sorta figured out how to do that. In fact, it's pretty easy: just squash a sinusoidal projection so it has the right pole angle. However, this alone wouldn't be enough to define a unique projection: numerous other projections with the same pole angle would be tied with it. For true minimax optimality, you'd want a "greedy algorithm": first figure out the constraint that will minimize distortion at 90°, then given that constraint, figure out the constraint that will minimize distortion at 89°, then given that constraint, figure out the constraint that will minimize distortion at 88°, etc., except continuously rather than at discrete 1° increments. I never got that far. I'm not sure if a continuous greedy algorithm is even possible, although it seems like it should be.
Re: Which projections you favor?
I assume you mean the 2:1 aspect ratio versions now, not the circular Hammer?
Personally I prefer Mollweide.
https://map-projections.net/single-view/mollweide
https://map-projections.net/single-view/hammer
Just comparing them by eye rather than with any formal distortion metrics, Hammer looks better near the center of the map (Africa in the examples), but Mollweide looks better pretty much everywhere else, especially North America, and also pretty obvious over South America and Australia. Plus, I've mentioned before that I consider being pseudocylindrical to be a valuable property in itself, in large part because of how important latitude is to climate and to everything that depends on climate.
Oh, and even the stretching of Africa can be fixed if you use the Bromley scaling (a simple variant of Mollweide), although then it no longer has the same aspect ratio as the Hammer projection.
It's even more obvious if you look at Daan's Double Mollweide and Double Hammer projections. Neither of them are good projections that you'd use for anything other than a joke, but the Double Mollweide actually manages to still look almost decent despite its theoretical flaws, while the Double Hammer just looks ridiculous (except on Antarctica and Greenland, which it somehow handles surprisingly well).
Which is ironic, because, as I understand, Ernst Hammer was aware of the Mollweide projection, and deliberately designed his own projection in an attempt to improve on it (certainly, the Hammer projection was invented almost a century later, despite actually having easier math). I feel that he missed the mark.
Still, I'd advise anyone who's inclined to like either the Hammer or Mollweide projection to at least take a look at the other before deciding for sure.
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Re: Which projections you favor?
Yes, he wanted to reduce distortion in the outer meridians, which is extreme in the Mollweide.
Re: Which projections you favor?
Let's try using them in an aspect that puts more land near the outer meridians: Nope, I still prefer Mollweide. South America is badly bent in both projections, but less so in Mollweide. The Atlas region of northwestern Africa also suffers in the Hammer projection.mapnerd2022 wrote: ↑Fri Feb 17, 2023 7:22 amYes, he wanted to reduce distortion in the outer meridians, which is extreme in the Mollweide.
Last edited by Milo on Fri Feb 17, 2023 7:44 am, edited 1 time in total.
Re: Which projections you favor?
Unlike Milo, I prefer lenticular projections to pseudocylindricals and would normally only choose a projection with straight parallels if the graticule isn't going to be shown on the map. However, even I prefer the Mollweide to the Hammer, and analogously Apian II to the Aitoff.
Having said that, I don't really like any of those four projections very much. My favourite equal-area projection is Francula XIV, or if it needs to have a point pole, then danseiji I, whose deviation from equivalence is negligible. If it needs to have a point pole and you insist on it being mathematically perfectly equal-area, then I'd go for Canters W32 (lenticular) or W34 (pseudocylindrical).
Having said that, I don't really like any of those four projections very much. My favourite equal-area projection is Francula XIV, or if it needs to have a point pole, then danseiji I, whose deviation from equivalence is negligible. If it needs to have a point pole and you insist on it being mathematically perfectly equal-area, then I'd go for Canters W32 (lenticular) or W34 (pseudocylindrical).