Tobler’s hyperelliptical projection

General discussion of map projections.
justlikeoldtimes
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Joined: Mon Feb 27, 2023 8:13 pm

Re: Tobler’s hyperelliptical projection

Post by justlikeoldtimes »

It looks like it can do hemispheres on smaller resolutions, but not larger ones?

Tried existing saved files and changing it to Hyper, then Hemispheres, and the problem happens.

Started a new file (with the default settings), then changing it to Hyper, then Hemispheres, increased the resolution, and the same problem happens still.

The issue seems to be unique for hyperelliptical. Hemispheres work fine for Sinusoidal, Eckerts, Mollweide, Equal Earth, etc.
quadibloc
Posts: 292
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Re: Tobler’s hyperelliptical projection

Post by quadibloc »

There is a reason that the exponent of 2.5 was "Tobler's favorite"; that's because it's the exponent used by Piet Hein when he famously proposed the hyperellipse as a shape intermediate between the true ellipse and the rectangle.
justlikeoldtimes
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Re: Tobler’s hyperelliptical projection

Post by justlikeoldtimes »

daan wrote: Sat Mar 30, 2024 9:46 am Tobler’s paper has some typos and his implementation suffers from poor numerical accuracy (it was 1973, after all), which might account for some problems. This is what I get using his parameters, and it’s very close to a 2:1 astroid:
astroid.jpg
Is that not what you want?
It's been some time, I've accepted that the astroid is the more "organic" shape than the space between four touching ellipses (which Google does not provide a name for, if any shape name exists).

The parameters in the paper only approximate a 2:1 astroid; it does not give me the exact 2:1 shape. I'm a pixel counter. Is there an easy way to figure out what those exact parameters would be? Mathematical development hasn't been my strong suit.

I've been fascinated by the entire family of projections that could be lumped together into the hyperelliptical category. I'm trying to think what other shapes besides the rectangle, ellipse, diamond, sinusoid, and astroid are "natural" choices for similar equal areas projections. (And also their equidistant counterparts). But I think that's about it?
Milo
Posts: 271
Joined: Fri Jan 22, 2021 11:11 am

Re: Tobler’s hyperelliptical projection

Post by Milo »

justlikeoldtimes wrote: Mon Jun 03, 2024 10:23 pmIt's been some time, I've accepted that the astroid is the more "organic" shape than the space between four touching ellipses (which Google does not provide a name for, if any shape name exists).
Eh, I don't there there's anything particularly "natural" about an astroid, or hyperellipses in general. It's taking the formula for a circle/ellipse and going "hmm, what if we plug a different number instead of a 2 there?", but that's really an abuse of the formula that doesn't have any theoretical justification beyond "I can do it, so why not?". You don't encounter hyperellipses much in actual nature, unlike plain ellipses.
justlikeoldtimes wrote: Mon Jun 03, 2024 10:23 pmThe parameters in the paper only approximate a 2:1 astroid; it does not give me the exact 2:1 shape. I'm a pixel counter. Is there an easy way to figure out what those exact parameters would be? Mathematical development hasn't been my strong suit.
I don't remember exactly what parameters Tobler's definition of his projection uses, and I don't feel like looking it up. As I've discussed before, I think his parametrization is somewhat awkward and more complicated than it needs to be. It's not that hard to make a "Tobler-in-spirit" projection that meets your requirements without using his exact formulae, though it probably requires numerical integration.
justlikeoldtimes wrote: Mon Jun 03, 2024 10:23 pmI've been fascinated by the entire family of projections that could be lumped together into the hyperelliptical category. I'm trying to think what other shapes besides the rectangle, ellipse, diamond, sinusoid, and astroid are "natural" choices for similar equal areas projections. (And also their equidistant counterparts). But I think that's about it?
Are you just talking about pseudocylindrical projections? Because a sinusoid isn't a hyperellipse.

Pseudocylindrical equal-area projections can be made in the shape of anything. Take a look at this one. (Arguably, a pseudocylindrical projection isn't the best choice for that shape, since it lacks sixfold symmetry... but it's what I have!)
justlikeoldtimes
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Re: Tobler’s hyperelliptical projection

Post by justlikeoldtimes »

I wasn't sure about Sinusoidal. I didn't mean to imply that it definitely was a hyperelliptical. But I think, like Mollweide, Lambert EA (and Smyth ES), Collignon, there's an appealing simplicity to it even if it's not immediately aesthetically pleasing.

I'm all for alternative ways generating hyperelliptical projections (as well as other equal area maps); but I'm sticking with the tools I have for now, mainly GeoCart.
daan
Site Admin
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Re: Tobler’s hyperelliptical projection

Post by daan »

justlikeoldtimes wrote: Mon Jun 03, 2024 10:23 pm The parameters in the paper only approximate a 2:1 astroid; it does not give me the exact 2:1 shape. I'm a pixel counter. Is there an easy way to figure out what those exact parameters would be? Mathematical development hasn't been my strong suit.
Being equal-area, you can stretch it in one direction and compress it reciprocally in the other to retain overall area and differential area, as well as get your exact 2:1 ratio. Maybe that doesn’t fit your idea of an astroid…?

— daan
justlikeoldtimes
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Re: Tobler’s hyperelliptical projection

Post by justlikeoldtimes »

Yeah it does. Sorry.
Attachments
astroid equal-area.png
astroid equal-area.png (591.36 KiB) Viewed 12080 times
Milo
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Re: Tobler’s hyperelliptical projection

Post by Milo »

That looks like a hyperellipse with exponent ½, which would make sense to call an "astroid" in as far as that it looks starlike, but Wikipedia claims that the name "astroid" is officially reserved specifically for the hyperellipse with exponent ⅔, and not similar-looking shapes. The exponent ⅔ at least has the unique property that in addition to being a hyperellipse, it is also a hypocycloid, which hyperellipses in general are not.

(Neither exponent will produce an "inverted ellipse" shape, so don't worry about that!)
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