Isoperimetric Cordiform

[sulyok]

Hello, everyone. I decided to derive the equations of the Isoperimetric Cordiform projection for the challenge, but I stucked and it is very frustrating. Based on its only property "Correct along all parallels and meridians in the default aspect." I have written the next equations for the north hemisphere:
(∂x(δ,λ)/∂δ)^2 + (∂y(δ,λ)/∂δ)^2 = 1
(∂x(δ,λ)/∂λ)^2 + (∂y(δ,λ)/∂λ)^2 = sin^2⁡(δ)
and conditions for the north pole at (x,y) = (0,0):
lim_(λ→0)⁡ x(δ,λ) = 0
lim_(λ→0) ⁡y(δ,λ) = -δ
lim_(δ→0) ⁡x(δ,λ) = 0
lim_(δ→0)⁡ x(δ,λ) = 0
where the signs are:
λ - longitude
δ – colatitude (=π/2 - ϕ) where ϕ is the latitude
x,y - projected coordinates (The wanted equations are x(δ,λ) and y(δ,λ).)

I tried many methods in the past two days to solve analytically these equations without any success and I ran out of ideas. I can solve them numerically like an evolutionary partial differential equation starting from the north pole, but the program is very slow when calculates the coordinates of a point far from the pole with acceptable accuracy. I feel there is an analytical solution, but I can't find it, although it can be very obvious.
I hope someone can give me a clue, what direction should I go. Even Mr. Strebe, if you are willing to share any details of your development of this projection, I would be very happy. Any advice or recommended step (that is not the final solution and not ruin my game) would be very helpful.
Thank you.
Ábel

[daan]

Hello Ábel, and welcome to the forums!

Like you, I thought there should be a closed-form representation, but wasn’t able to find it. However, I did not study the solution exhaustively. The Geocart calculation is numerical and relies on caching and interpolation for speed. Otherwise, it would be quite slow.

Cheers,
— daan

[quadibloc]

Having visited the page on the Geocart map projection directory which describes this projection, I see its importance.
You and Leszek Pernarowski have apparently achieved what had formerly been thought impossible: a projection on which (in conventional aspect) scale is correct along the entire length of every single parallel and along the entire length of every single meridian.
After some thought, it became clear that if one tried to change the standard parallel of this projection from the pole to, say, the Equator, it would not be possible to map the whole world - only the point on the Equator at the central meridian would be close enough to the poles for a meridian to be drawn from it all the way to the pole. But being able to change the standard parallel might still be valuable, as a projection with this property that could be centered at any point would be very useful for large-scale topographical maps.
Adjacent sheets would not have the same curve at their boundary, but the scale would be the same.

[Milo]

You and Leszek Pernarowski have apparently achieved what had formerly been thought impossible: a projection on which (in conventional aspect) scale is correct along the entire length of every single parallel and along the entire length of every single meridian.

The cordiform is not the only projection with this property. The isoperimetric pseudoazimuthal also has this property and is simpler to calculate.

Now, try coming up with a projection which has this property and also has all parallels and meridians perpendicular to each other...

[quadibloc]

The cordiform is not the only projection with this property. The isoperimetric pseudoazimuthal also has this property and is simpler to calculate.

True, but that projection (the more famous one by Leszek Pernarowski, as Martin Gardner gave it a mention) is clearly useless as a general-purpose map projection.
So modifying the cordiform to allow changing the standard parallel would create something of interest for large-scale topographical maps.

[Milo]

True, but that projection (the more famous one by Leszek Pernarowski, as Martin Gardner gave it a mention) is clearly useless as a general-purpose map projection.

I think that preserving scale along both parallels and meridians is a fairly useless property in general, since after all you still won't be preserving scale in other compass directions.

Also, if you want to travel somewhere due east of your location, the smart path to take is not to follow the parallel, but rather cut across a great circle.

It's possible to see from Tissot indicatrices that, provided the semiminor axis is less than one and the semimajor axis is greater than one (which will be true for equal-area projections, where they are each other's reciprocal), there will always be exactly two axes along which scale is preserved.

[quadibloc]

I think that preserving scale along both parallels and meridians is a fairly useless property in general, since after all you still won't be preserving scale in other compass directions.

Oh, that's true enough, I can't deny it. However, such a projection as I propose, like an azimuthal equidistant or a simple conic, will at least have good shape and area properties near its center, and the ability to measure distances along parallels and meridians is just a bonus.

[sulyok]

Like you, I thought there should be a closed-form representation, but wasn’t able to find it. However, I did not study the solution exhaustively. The Geocart calculation is numerical and relies on caching and interpolation for speed. Otherwise, it would be quite slow.

Thank you for your quick answer. I disappeared to waste more time on the solution. After spending a few hours more I gave up and implemented it numerically. I think there is no better way as your method. Below my result. (I also implemented the inverse with a tricky Newton-Rhapson.)

[daan]

After spending a few hours more I gave up and implemented it numerically… Below my result. (I also implemented the inverse with a tricky Newton-Rhapson.)

You probably have the second implementation ever attempted. Well done!


— daan

[mapnerd2022]

I guess it would also be a good projection for map with an antique look since it has a «look and feel» similar to Oronce Finé's two hemisphere cordiform or cardioid projection.