I have been revising my thinking and coding to have enough generality to make icosahedrons easy. But I also thought of a way to fix most of the distortion in a cube mapping. But first I'll just review why we want that uniformity.
Grid criteria
The main reason why uniformity is good is that the error in integrating is determined by the largest cells. So with size variation, you need more cells in total. This becomes more significant with using a grid for integration of scattered points, because we expect that there is an optimum size. Too big and you have to worry about sample distribution within a cell; too small and there are too many empty cells. Even though I'm not sure where the optimum is, it's clear that you need reasonable uniformity to implement such an optimum.I wrote a while ago about tesselation that created equal area cells, but did not have the grid aspect of each cell exactly adjoining four others. This is not so useful for my diffusion infill, where I need to recognise neighbors. That also creates sensitivity to uniformity, since stepping forward (in diffusion) should spread over equal distances.
Optimised grid
I'll jump ahead at this stage to show the new grid. I'll explain below the fold how it is derived and of course, there will be a WebGL display. Both grids are based on a similarly placed cube. The left is the direct projection; you can see better detail in the previous post. Top row is just the geometry (16x16), the bottom shows the effect of varying data (as before 24x24, April 2015 TempLS). I've kept the coloring convention of s different checkerboard on each face, with drab colors for empty cells, and white lines showing neighbor connections that re-weight for empty cells.The right is the same with the new mapping. You can see that near the cube corner, SW, in the left pic the cells get small, and a lot become empty. IOn the right, the corner cells actually have larger area than in the face centre, and there is a minimum size in between. Area is between +-15% of the center value. In the old grid, corner cells were about 20% area relative to central. So there are no longer a lot of empty cells near the corner. Instead, there are a few more in the interior (where cell size is minimum).
In that previous post, I showed a table of discrepancies in integrating a set of spherical harmonics over the irregularly distributed stations:
L | 1 | 2 | 3 | 4 | 5 |
Full grid | 0 | 0 | 0 | 0 | 1e-06 |
Infilled grid | 8.8e-05 | 0.00029 | 0.001045 | 0.002015 | 0.003635 |
No infill | 0.007632 | 0.027335 | 0.049327 | 0.064493 | 0.075291 |
In the new grid, the corresponding results are:
L | 1 | 2 | 3 | 4 | 5 |
Full grid | 0 | 0 | 0 | 0 | 1e-06 |
Infilled grid | 8e-06 | 0.00019 | 0.000645 | 0.001348 | 0.002529 |
No infill | 0.004758 | 0.02558 | 0.047794 | 0.0601 | 0.069348 |
Simply integrating the SH on the grid (top row) works very well in either. Just omitting the empty cells (bottom row), the new grid gives a modest improvement. But for the case of interest, with the infilling scheme, the result is considerably better than with the old grid.
Optimising
I think of two spaces - z, the actual sphere where the grid ends up, and u, the grid on each cube face. But in fact u can be thought of abstractly. We just need 6 rectangular grids for u, and a mapping from u to z.But the mapping need not be simple projection. You could, for example, re-map the u space and then project. If f is a one-one mapping of the square of u onto itself, then the change in area going to z is:
det(f'(u))/(1+f2(u))^(3/2)
where the second factor comes from the projection, and includes the inverse square magnification and a cos term for the different angles of the du element and the sphere. f' is a Jacobian derivative on the 2D space, and the determinant gives the area ratio of that u mapping.
I use the mapping f(u)=1/r tan(r*u) on each parameter separately. The actual form of f doesn't matter very much. I choose tan because if tan(1/r)=sqrt(2) it gives the tesselation of great circles separated by equal longitude angle. That is already much better than simple projection (r=0). I can check by calculating what would happen to a square in the centre, mid-side, and corner, for various r:
1/tan(1/r) | 0 | 0.7071 | 1 | 1.1312 | 1.2 |
central | 1 | 1 | 1 | 1 | 1 |
midside | 0.3536 | 0.5303 | 0.7071 | 0.8059 | 0.8627 |
corner | 0.1925 | 0.433 | 0.7698 | 1 | 1.1458 |
So here finally is the 24x24 WebGL plot equivalent to that from last time. Again it shows with drab colors empty cells, and with white lines the direction of reallocation of weights for infill. The resulting improved integration is discussed in the head plot.
Nick,
ReplyDeleteAll this geometry and spherical harmonics stuff is way above my paygrade, and it does appear that VV is right about icosahedrons and generalization to a geodesic grid ...
Geodesic grid
https://en.wikipedia.org/wiki/Geodesic_grid
However, the Montreal Biosphère does appear to be made of six-sided equilateral triangles (hexagon) in what appears to be same pattern throughout (it does appear to have nice latitudinal bands, at least in the low-to-middle latitudes). That structure has some depth elements to it throughout and something appears to be happening to the surface geometry at higher latitudes (it's really hard to tell given all the structural elements that are being seen from a 2D perspective).
https://en.wikipedia.org/wiki/Montreal_Biosph%C3%A8re
https://upload.wikimedia.org/wikipedia/commons/1/16/Biosph%C3%A8re_de_Montr%C3%A9al_en_juillet_2011.jpg
Everett,
DeleteYes, I think icosahedrons might give more uniformity. But this (+-15%) is pretty good. A downside of icos is that you end up with triangles. The idea of gridding here is that the cell area is well represented by the average of data within. Compactness helps there. Triangles are, for the same area, less compact than a rectangle. Points can be further apart.
Does it matter, that the ocean data are already on a grid? There may be systematic or correlated shifts, if this grid points are near the corners of your grid.
ReplyDeleteDifferent question: It is possible to generate zonal instead of global means? GISS has something like this. This could be used to compare the zonal anomalies of different data sets. I suppose, that most differences are at high latitudes. Would the differences between the data set reduce, if the mean temperature is calculated f.e. 64S and 64N or even narrower to exclude ice regions? Maybe even between the ground and satellite data sets?
"Does it matter, that the ocean data are already on a grid?"
DeleteA little. But the grids aren't aligned, particularly in latitude. You can see stripes of grey in the WebGL plot where the 24x24 grid doesn't quite span the SST grid. That just means that there is infill of a narrow strip of empty cells, with some SST cells being upweighted, Since SST is more uniform anyway, that has small effect.
Obviously, a lat/lon grid is good for looking at latitude zones. My earlier method preserved that merit, at the expense of neighbor cell structure. The method in this post is mainly useful for global.