Looks promising! Good job.
...Should I be worried that the minor axis graph of your C0 projection looks like a radioactive hazard symbol?
dummy_index wrote: ↑Mon Aug 15, 2022 5:01 amI found a projection for Octants (face of regular octahedron) with no kinks over the edges.
It is based on the Chamberlin trimetric, but uses
trilinear coordinates in xy-plane. three points are placed on the pole and two points 90° distanced on equatorial. (this method works well only for octant. equatrial is constant distance from pole. sum of distance from other two points is constant on the equatorial.)
Can you explain in more detail what you mean about using trilinear coordinates?
According to my own analysis, I get the following formulae for computing to and from trilinear coordinates:
spherical trilinear coordinates (relative to 0°E 0°N, 90°E 0°N, 90°N):
k*a = latitude
k*b = asin(cos(latitude)*sin(longitude))
k*c = asin(cos(latitude)*cos(longitude))
Euclidean trilinear coordinates (relative to (-sqrt(1/3),0), (+sqrt(1/3),0), (0,1)):
k*a = y
k*b = (1 + sqrt(3)*x - y) / 2
k*c = (1 - sqrt(3)*x - y) / 2
k*(a+b+c) = 1
reverse Euclidean trilinear coordinates:
x = sqrt(1/3)*(b-c)/(a+b+c)
y = a/(a+b+c)
When I have gnuplot compute the spherical-to-trilinear-to-Euclidean transformation, I get the following graticule:
- trilinear.png (6.96 KiB) Viewed 11737 times
This is different from yours, and is not perpendicular to the edges in general.
dummy_index wrote: ↑Mon Aug 15, 2022 5:01 amThen how to gain resolution-efficiency... Modulate the distance function with keep some_func(distance from B)+some_func(distance from C)=pi/2. Define the function piecewise to correct the scale near the boundary.
If it helps, a necessary and sufficient condition for a projection of a
tangential polygon to have optimal resolution-efficiency for that polygon is that (A) it has constant scale along the
apothems, and (B) at no other point/direction is the scale lower than it is on the apothems. This is how I computed the theoretical optima you reference, in fact. All regular polygons are, of course, tangential polygons.
So perhaps you could accomplish this by determining the behavior on the apothem of your base projection, and then compute the inverse function?
Doesn't really solve the issue that, as an odd-sided polygon, a triangle's apothems don't continue all the way through, though.
dummy_index wrote: ↑Tue Aug 16, 2022 5:59 amhemisphere -> (Lagrange transform) -> 90° gore -> (rotate) -> (Lagrange transform) -> octant
By Lagrange transform, I assume you mean the "halve scale of Mercator projection" process used in the Lagrange projection?
This is a conformal transformation, so it's generally not ideal for designing non-conformal projections. On the plus side, though, I think that preservation of threefold symmetry is guaranteed by the uniqueness theorem of conformal transformations. (It's intuitively obvious that some symmetry-preserving transformation between the two triangles must exist, and therefore if you can devise a projection that gets enough details right to satisfy the uniqueness theorem, it must be the right one.)
However, what's the "Möbius transformation" that you're using to map the 120°-120°-120° points onto the 90°-90°-180° points? From my own understanding, a Möbius transformation of a great circle onto that same great circle will always preserve all points of the great circle as well (or rotate them), even if it shifts things around on the inside. Am I missing something? However, according to my own analysis, you could achieve the intended result by a variant of the Lagrange process where you translate instead of scale the Mercator projection, shifting it northwards by asinh(1/sqrt(3)) (so the previous 30°S parallel is now the equator). (You can't combine the two steps, though, because they use the Mercator projection in different aspects! The translation uses the Mercator projection where the hemisphere of interest lies between the 90°W and 90°E meridians, while the caling uses the Mercator projection where the hemisphere of interest lies north of the equator. Heh, the Mercator projection is pretty useful.)
If nothing else, this could make for a useful way of computing the conformal octant projection. This still requires an elliptic function (for the computation of the conformal hemisphere-in-a-triangle projection, which is then transformed into an octant projection), but probably a simpler one than whatever messy hypergeometric functions are necessary to compute the confomal octant projection directly from the Schwarz triangle function.
