Hello!
I've been dabbling in map projections for a while. Not professionally or anything, just someone who thinks recreational mathematics is a fun way to pass the time
Recently I've come across a link to this thread. I've found it an interesting read, because I'd been thinking about similar things for a while. Maps that are provably optimal by some metric appeal more to my sense of mathematical elegance than just something that someone thought looked good. (Of course, there are a bunch of different metrics you could use, so that still allows a library of projections to choose from.) I like the Eisenlohr's design philosophy of "do one thing perfectly, and then optimize as well as you can in other respects", but I prefer equal-area maps over conformal ones, so I've been searching for a while for what could fit the bill.
I'd actually independently come up with some of the things that got discussed in that thread. It's kind of embarrassing how close I got without ever realizing that a simple rescaling of the Hammer projection meets the optimization criteria I was looking for. (And also looks pretty ugly, which just serves as an indication that I need to come up with better optimization criteria.)
I registered an account, though, in order to comment on this:
daan wrote: ↑Sun Jan 26, 2020 11:30 amThe reality is the opposite: a closed path having no distortion on an equal-area projection can’t exist because the bounded area will always be too small. I was thinking of a different situation here and misapplying it.
daan wrote: ↑Fri Feb 14, 2020 5:23 pm- Concerning a region of interest on the ellipsoid, the equal-area projection that gives the shortest possible closed path of constant angular deformation (isocol) bounding that region is the projection whose bounding isocol has the smallest possible measure for a bounding isocol.
- The shortest closed path is circular. Therefore the map with the smallest min/max angular deformation will be a circle with the same surface area as the region from the ellipsoid as long as some interior point has no distortion.
Odd observation: These observations hold regardless of the region’s shape on the globe.
This is incorrect.
For any
circular shape on the globe, its area will always be greater than any Euclidean shape of the same perimeter, and therefore, projecting it as a circle is the closest you can do.
However, shapes on the globe can come with a wide variety of area-to-perimeter relationships, and for some shapes (particularly those which are long and thin), those values will fall within the range of what is also permissible for Euclidean shapes.
For example, consider a digon: a region bounded by two meridians, at an angle of theta from each other. It is obvious that, on the sphere, this region has a perimeter of 2*pi and an area of 2*theta.
In Euclidean space, the largest shape with a perimeter of 2*pi is the unit circle, with an area of pi. Therefore, when theta > pi/2, the shape cannot be mapped in a way that preserves both area and perimeter, and you're stuck approximating it as a circle.
When theta = pi/2 exactly (corresponding to one quadrant of the globe), the area and perimeter of the digon on the sphere match those of the circle on the plane. As such they can be projected onto each other... in this case rather easily, through a variant of the Hammer concept that dummy_index pioneered. I have attached an example of this projection, which is indeed an equal-area map that is perfectly conformal along the boundary, with the exception of singularities at the poles. (Examples without singularities probably exist, but I don't feel up to trying to find them.) This also shows that the map which has lowest distortion at the boundary will not necessarily have the lowest distortion everywhere, since while this example map is distortion-free at the boundary, it quite clearly is not inside it!
When theta < pi/2, the desired area and perimeter will be reached by something other than a circle. Generally, there will be more than one such shape.
None of this is likely to be very relevant to mapping the whole world, unless you want to interrupt the projection so the world is cut into a long narrow strip that wraps around it like a mummy (a "world tour" projection displaying the world as it would be perceived by someone trying to visit all of it in the most efficient way possible?). Still, I wanted to point it out.