Degenerate Regular Tessellations & Spherical Delusions

I think I need to get a lot better at Spherical Geometry…

This paper:

https://mathstat.slu.edu/escher/index.php/Spherical_Geometry

has caused me to question my intuition about the geometric ‘shape’ of 2D regular polygons when projected onto a sphere. In particular, I’ve come to understand that as the projection covers more of the sphere, the angles at the “corners” becomes increasingly large.

Eventually, when one “polygon” covers an entire hemisphere, the “angles” at each vertex become 180 degrees. Essentially, a straight line at the equator. Yet, conceptually, it is just the limit case of increasing your one polygon incrementally over the globe.

Degenerate Regular Tessellations

Tessellation by two 12-gons with 180° “corner” angles.

Two strange “degenerate” types of regular tessellations show up in spherical geometry. The first is by polygons with corner angles equal to 180°. A 180° corner doesn’t look like a corner at all, and a regular n-gon with 180° corner angles simply looks like a hemisphere with n evenly spaced dots on its edge for the “vertices”. Two of these fit together to cover the sphere. One can argue about whether this should be a polygon at all, but we’ll see that it fits very nicely in a larger picture of regular tessellations and is worth including.

One could do the same thing with a regular hexagon of 180 degree angles…

Now a hexagon, even a “degenerate regular tessellation” one, can be divided by triangles (even spherical ones of greater than 180 degrees interior angles sum). This can result in another hexagon formed in the interior:

Which could be further so divided…

(For a nice mind twist, try to spot the 7 hexagons inside that tri-tiled hexagon. If you include the outer, larger, hexagon, I get 8 total. Can anyone visualize more somewhere in that image?)

Now it’s a bit of a stretch, but what I suspect from this is that one could in fact tessellate the surface of a sphere with spherical hexagons via a process somewhat along those lines.

Or, note that each set of 2 triangles around the interior hexagon, make a regular rhombus. Why not tessellate the globe in ‘circles’ of regular spherical rhombus shapes in nested “spherical hexagon” rings?

At the pole, the ‘regular hexagon’ approximates the spherical hexagon. The next ring out would be a little more distorted (by sin(theta) I think…) in every concentric ring. At the Equator, a hex tiled cylinder approximates reality. Hexagons increasingly distorted as they are projected onto the sphere as you move away from the equator.

In between those two, a series of distorted spherical hexagons would be needed to bring the two into agreement at the merge. Essentially, at the pole, each added ring of hexagons is 2 x the next innermost ring. At the Equator, each ring is equal to the next equatorial ring. In between, you must grade from A=B to A=2B in proportion to Sin(theta)… This can be done via variation in the number of equal vs 2x rings produced at each increment. These “hexagons” would be distorted spherically by the extent to which the N vs S ‘sides’ are required to be different lengths by the sphere, and by the extent to which some rings would get 2 x members and others equal number of members.

In the end, you would have a fully hex tiled world, but not REGULAR hexagons and without special adjustments of sizes, not equal area.

So which is worse? Having regular area and shape, but irregular lat / long as your sphere has 12 pentagons and lines of straight runs of hexagons go in all sorts of directions? Or having hexagons where their sides align with Lat and Long lines, but spherical shape distorts them and their areas are somewhat variable.?

Which “irregularities” are easiest to deal with? Alignment or size & shape?

I’m hoping a bit more exposure to “spherical geometry” will improve my intuition about it… I’m also finding that spherical geometry is just weird in some ways.

https://en.wikipedia.org/wiki/Spherical_geometry

http://www.math.ubc.ca/~cass/courses/m308-02b/projects/franco/index.htm

Interestingly enough, it was also Ptolemy and not Christopher Columbus who discovered that the earth was spherical and not flat, and stated his rationale in the Almagest 1300 years before Columbus sailed around the world:

“If the earth were flat from east to west, the stars would rise as soon for westerners as for orientals, which is false. Also, if the earth were flat from north to south and vice versa, the stars which were always visible to anyone would continue to be so wherever he went, which is false. But it seems flat to human sight because it is so extensive.”

Like geometry and geography, the worlds of spherical geometry (used in geography) and planar geometry (commonly taught in most geometry courses) are closely related and yet extremely different.

Anybody who has completed high school level geometry (or to some extent, elementary geometry) knows that in Euclidean or planar geometry, two parallel lines never meet, the sum of the three angles of a triangle add up to 180°, and the shortest route to get from one point to another is a straight line. In the world of spherical geometry, two parallel lines on great circles intersect twice, the sum of the three angles of a triangle on the sphere’s surface exceed 180° due to positive curvature, and the shortest route to get from one point to another is not a straight line on a map but a line that follows the minor arc of a great circle. Maps provide a way of translating the spherical view of the world to a planar view, by projecting the Earth’s topologies and locations to a flattened surface using Hammer, Mercator or cylindrical methods. A consistent and standard representation that minimizes projective distortions is yet to be established.

The discovery of spherical geometry not only changed the history and the face of mathematics and Euclid’s geometry, but also changed the way humans viewed and charted the world. Using this new knowledge, explorers and astronomers used the circular path of stars to navigate the earth to discover new lands and reason about the cosmos.

I think I need to think more spherically about what I think about…

A technical managerial sort interested in things from Stonehenge to computer science. My present "hot buttons' are the mythology of Climate Change and ancient metrology; but things change...
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47 Responses to Degenerate Regular Tessellations & Spherical Delusions

1. The True Nolan says:

In fact, the degenerate great circle triangle with three “corners” of 180 degrees is not the limit. If you keep expanding the triangle outward, you eventually get a gigantic triangle covering 99.99999…% of the globe, with the three corner interior angles adding to 900 degrees.

2. V.P. Elect Smith says:

Wouldn’t the interior angle start reducing at the other pole, ending back at 60 so 3 being 180 again?

3. V.P. Elect Smith says:

Oh, wait, I’m thinking pointy ends but the degenerate is sides, so… I have no clue…

4. The True Nolan says:

The corners cease being “outies” and become “inies”. Suppose you start at the north pole and just expand and expand and expand until you converge on the south pole. What looks like a conventional small triangle now at the south pole is actually OUTSIDE your triangle. Your triangle is the 99% of the globe, and the leftover tiny three sided bit at the south pole is the part which ISN’T in your triangle. Each interior angles of your original triangle is now 300-300-300 degrees (for an equilateral triangle) or 270-270-360 degrees (for an isosceles with an acute angle) or 360-360-180 degrees (for an isosceles with an obtuse angle).

5. V.P. Elect Smith says:

OK, I get it, but Ow my brain hurts…

Just to return the favor:

Things you run into while clicking around on geometry stuff…
https://en.wikipedia.org/wiki/Pseudo_great_rhombicuboctahedron

Gee. the Pseudo Great Rombicuboctahedron… and I didn’t even know there was a regular Great Rombicuboctahedron…

In geometry, the pseudo great rhombicuboctahedron is one of the two pseudo uniform polyhedra, the other being the convex elongated square gyrobicupola or pseudo rhombicuboctahedron. It has the same vertex figure as the nonconvex great rhombicuboctahedron (a uniform polyhedron), but is not a uniform polyhedron (due to not being isogonal), and has a smaller symmetry group. It can be obtained from the great rhombicuboctahedron by taking a square face and the 8 faces with a common vertex to it (forming a crossed square cupola) and rotating them by an angle of ​π⁄4. It is related to the nonconvex great rhombicuboctahedron in the same way that the pseudo rhombicuboctahedron is related to the rhombicuboctahedron.

Well, at least I didn’t have to look at the Convex Elengated Square Gyrobicupola aka the not-so-great Pseudo Rombicuboctahedron…

I sure picked a lousy day to run out of Scotch…

6. cdquarles says:

Yes, I see 7 smaller hexagons within the outer one, along with many triangles. It took some time to see them.

This reminds me why I found the Rubik’s cube fascinating.

7. cdquarles says:

Then again, the whole mathematical field of topology is fascinating (think abstraction from physical topology).

8. Graeme No.3 says:

You can add a few years before Ptolemy, Eratosthenes (died 195 BC) measured the size of the Earth (fairly accurately as 2 errors cancelled out) as he thought it was a sphere. Aristarchus (died 230 BC) postulated the Earth and planets orbited the Sun, implying that the Earth was a sphere.
It appears that the ancient greeks didn’t think the Earth was flat, unlike certain Climate Scientists these days.

9. My favourite regular polygon is the icosahedron. It is composed of 20 faces of equilateral triangles organised as a set of pentagons that can fully enclose a sphere. The regular icosahedron can be formed from a flat surface that can be cut and folded into an enclosing shape without distortion. The nice thing about an icosahedron is that each triangular face can be subdivided indefinitely while maintaining its regular triangular form. So, in theory you can create a planar surface map that approximates to a sphere to the highest degree of accuracy that is required. It is only at the 12 vertices that the 5 core triangular faces meet, while at all other subdivisions 6 triangular faces meet.

The beauty of this shape as a planar model of a sphere is that the triangular sub-faces form a network with a common surface area and similar distortion. They meet at sub vertices that are the same distance from the centre of the enclosing form, and so unlike a Mercator projection there is no latitudinal directed distortion anywhere on the planar grid.

10. The True Nolan says:

@V.P. Elect Smith Your brain hurts? If it is true that misery loves company please take some comfort in the fact that after trying to figure out the Pseudo Great Rombicuboctahedron I am ready to join in the brain pain. I think that geometry is something like Italian cuisine — once you learn the terminology, the dishes start to make sense. Some decades back I got very interested in geodesic structures and spent way too much time studying various tessellations of geometric shapes, but (thankfully?) the great bulk of that knowledge has gone into deep storage. Actually, I think I may have shuffled it into the mental compost bin…

I must admit though, that I still find myself thinking about spherical trig for historical reasons. Graeme No.3 remarks on the history of the-world-is-round theories, and while he is very much correct, I suspect that the round world idea goes back a lot farther than the books tell us. WARNING! Rank speculation and opinion follows!

I think there are some good indications that at some distant and poorly documented point in the past (6,000 BP? 10,000??) there was a maritime trading culture with world wide reach. Think, something like the Phoenicians on steroids. So, we all know the story of the invention of accurate timepieces in the 18th century which enabled longitude determination. (Though I would add that a good chronometer is pointless without a good knowledge of astronomy to go with it. No aliens need apply.) Anyway, how do you map the world without a timepiece, telescopes, or spherical trig? You watch the night sky and build an analog calculator out of sticks and stones. You know… Stonehenge! More accurately, something similar to Stonehenge that allows you to calculate lunar and solar positions. Knowing that allows you to predict lunar eclipses. Knowing lunar eclipse dates let you tell your long range ships what day to look for an eclipse. If they find themselves in Ancient Quasi-land that day, they watch for the eclipse and when the umbra crosses the face of the moon they note where in the sky the moon was as the shadow first darkens the moon. The astronomers back home do the same. Latitudes are easy for everyone, but now the eclipse data can be used with spherical trig to calculate the longitudinal difference between the two observers. Easy deal if you know spherical trig… But if you don’t, you build an analog spherical trig computer, which is just a big stone ball. You use that to draw on, plotting your observed Moon bearings and altitude observations, and that gives you the longitude of the distant land. (If interested I can give some details on the techniques.) Do that for enough times and you have a map of the globe. How accurate? On a wild seat of the pants guess, I would think something better than 100 miles, maybe better than 50 miles. Not bad for a Neolithic culture.

And meanwhile, we have modern people living in a technological civilization who think the earth is flat.

11. V.P. Elect Smith says:

The problem with my “just keep dividing a degenerate spherical hexagon” is that eventually you have a very small hexagon on the ‘vertex’ of the degenerate one. At that vertex, the 2 ‘edges’ of the little one “meet” at a 180 degree angle… i.e. it looks to us in the regular world like a pentagon…

@Philip Mulholland:

I suspect that I’m just re-creating your icosahedron “by other means” on a sphere… or at least a similar construction where the pentagon has one ‘long side’ approx 2 x the other sides…

It might still be worth it as, I think all the pentagons would end up around the equator… Easier to deal with them all in a row, maybe… and everywhere else spherical hexagons of various angularities…

There’s lots of software already written for the icosahedron tiling of a globe. I just don’t like the 12 pentagons and the way lines of hexagons run in all sorts of directions.

Maybe I just need to think a bit about “just how hard is it, or not, to deal with various directions of hexagonal rows and a pentagon IF you are using dots and vectors?”.

Only 12 cells where you use 5 vectors instead of 6 and area may vary but only 2 types of area of known size. Everything still meets at faces instead of points. That’s starting to sound a whole lot easier than dealing with spherical trig for every dot …

12. We are approaching the same solution but from different directions. Let’s do some cut and paste origami. Take a plane piece of paper and draw on it a regular hexagon consisting of 6 equilateral triangles. Cut out the hexagon and then also cut out one of the equilateral triangles to leave a flat shape of five triangles with a missing pie segment. Next fold and join the paper to create a five-sided cap. This cap forms one five-sided part of a regular icosahedron. Note that the sum of the angles of the 5 triangles around the apex of the cap is 300 degrees. (5 times 60).
Next draw a regular pentagon on a flat piece of paper. Triangulate the pentagon and note that the sum of the 5 internal angles at the apex is 360 degrees, so each triangle subtends an angle of 72 degrees at the centre (obviously). Next, we want to flatten the pentagon cap made earlier to make it more like the surface of an enclosing sphere. We do this by growing the apex angle of the 5 equilateral triangles that make up the pentagon cap. The point is this, because we subdivide the face of the five-sided cap as we model the surface of our sphere, we will be increasing the retained angle from 300 degrees towards 360 degrees, and so decrease the angle of the cut pie segment towards zero.
As we flatten the five-sided cap the angle of the gap in the plane sheet of cut paper, which starts at 60 degrees will decrease as the cap flattens towards a plane. The angle of the gap is a measure of the amount of flattening we have applied as we approach a planar tessellation. Clearly however, we will never achieve a flat planar pentagonal shape, there will always be a small angular gap in our paper as we are modelling the curved surface of a sphere.
OK so far? The next stage is to build out from our flattened pentagon cap. For illustration suppose the angles of apex for the flattened pentagonal cap sum to 359 degrees, then on each of the five edges of the central cap we will construct a hexagon but also with a cut pie gap. The purpose of the gap is to maintain curvature of the sphere that we are modelling. The apex angles of these hexagonal caps will also sum to less than 360 degrees. (I haven’t done the math to calculate what this angle will be).
Why is this so? When you draw the tile in plan view it appears that the angles are 60 degrees and the six triangles sum to 360 degrees. Instead think of your hexagonal cap as being a tent with 6 poles. When lain flat the angles will indeed be 60 degrees, but as soon as you raise the apex of the tent off the ground the poles will draw in, and the angle between each pole will now be less than 60 degrees. Another analogue for this is the angles between the collapsed spokes of a folded umbrella.
Look at this process of construction in plan view as a series of tiles on a flat plane, but at each growth outwards there will always be a small gap in each of the added hexagons as the tessellation proceeds. Remember that the gap is not missing surface it is the space for the join that moulds the cap back onto the curved surface of the sphere that we are modelling.
Please note that the location of each of the 12 apical caps of pentagons is rigorously defined by the geometry of the icosahedron. You can choose one apex for the north and its opposite for the south pole, but the latitude of the other 10 vertices is fixed. (You can of course define one of these latitudinal points as longitude zero).

As an interesting aside, it ought to be possible to determine the curvature of the surface of the earth by showing that the internal angles of a regularly laid out hexagon of suitable size on a flat surface (such as a salt flat) sum to less than 360 degrees. The trick of course is to accurately define the distance from the centre of the earth for each of the datum elevation points on the 6 marker poles that form the corners of the regular hexagon. To do this we would require an accurate gravity meter to determine a common gravity field datum at each pole elevation marker as a proxy for the distance to the centre of the earth.

13. Simon Derricutt says:

EM – As far as I’ve seen with geodesic spheres, the pentagons are pretty evenly-arranged around the sphere. The pentagons are not the same area as the hexagons, either.

I’ve been pondering this problem of how to get equal-areas. May be possible using successive subdivisions from a start-point. If we start with four points equidistant on the sphere, so tetrahedral symmetry, and then produce new dots at the central point of 3 old dots, then we then have 8 points after 1 iteration. Next iteration doubles it again, so providing you’re happy with a power of two points, the dots will have equal spacing and we can assume an equal area when we put the soap-bubbles around them. Each dot will be specified as being an angle in 3 axes, as well as the radius of the Earth. The triangles won’t be equilateral, but isosceles. As the density increases, seems they’ll get toward equilateral.

Yep, this will be a lot of trig, but it’s only done once to define the globe so not a major problem. Though I haven’t used OO languages, it does look as if you should be able to get this in Julia fairly easily. Looks like the main problem will be making sure you’ve found the triangles exactly once and put a new dot in the middle of them just once for each iteration.

At a certain density, you’ll have 3 nearest neighbours and their distance will be a certain maximum. At a lower density, could be more equidistant neighbours which would make life more difficult. Not too much of a problem if you’re sending a mass of air in a certain direction and dividing it between those neighbours according to the direction of them relative to your source dot. Finding the nearest neighbours to a dot only requires looking within a certain angle offset from the angle of that dot.

One pole will have a central dot, the other will have 3 dots around it to start with, but the first iteration puts a dot on the other pole. Though you may worry about polar vortex, and thus needing a central dot only, once you have enough dots specified then the spacing will be smaller than the polar vortex anyway.

It may be better to put more dots on at each iteration, so instead of dividing the angle by 2 you divide by 3 or 4 instead. Might get closer to equidistance between the dots. Also of course requires fewer iterations to build the matrix. Needs playing with to see what the results are.

Not a perfect solution to the problem, but it looks like there isn’t one anyway so this might be good enough and easier to use. If you used the pentagon solution then you don’t get equal areas and you have 5 or 6 nearest neighbours, and if you use the division method you started with here then you have up to 8 nearest neighbours you need to consider with the angles being unique to that cell so all need to be initially calculated. There’s also the problem that (as you increase the number of dots/areas in total) that the sum of the widths won’t exactly equal the diameter of the circle you’re dividing up, and if you adjust that then the areas won’t be the same as other strips unless you also adjust the angle of latitude taken up by each band. Getting equal areas without a gap or overlap looks to be an iterative process, since adjusting the latitude width of one band will affect the latitude of the others.

I’m at the moment assuming that the inaccuracy involved by using the Earth as a perfect sphere rather than a bit oblate (actually more pear-shaped) will be less than the inaccuracies from other things. When it comes to the upper air layers, though, the centripetal force needed to keep the air attached to the Earth (that force supplied by gravity) does make a big difference to the shape of the envelope of air, and it will be different depending on the direction of the wind relative to the rotation of the Earth. I wonder if the other models take that into account?

The act of dividing the wind out of one dot between the nearest neighbour cells depending upon their angular offset from the source point seems to me that it will spread the wavefront of that wind, with the amount of spreading increasing with the number of cells that wind passes through. This will not be accurate as regards winds, I think, but then again if I do blow with my mouth then the wave-front of that wind does expand. I’m not sure about whether the spreading will be correct or not when it comes to the climate model.

Maybe we’re going at this wrongly. The Sun drives everything, so maybe we should be dividing the Earth into almost-longitudinal slices with the “poles” relative to the Sun. Subdivide those slices into the size cells we want to deal with, and get equal areas per cell (shape varies with latitude, let’s say “square” at 45° latitude), and commence the sweep of calculations at the local midnight when what’s driving the weather is local conditions only (land, sea temperatures, cloudiness, and momentum of wind), and progress around the globe for one time-tick. The angle of movement of the globe (as it rotates around its own axis) will now vary by time of day and year, but that’s not a hard calculation. The drag of the actual rotation (offset to the Sun’s angle) will have a calculable effect, and the location of the cells we work with will have a calculable relationship to fixed locations on the actual Earth that might be a little tricky, but looking at the system that way could give a more-accurate representation of the effects, simply because it’s the Sun that drives the weather. You get less latitudinal resolution at the poles, but then that’s probably OK anyway. Less energy goes in there anyway, and the land conditions are far less variable.

Could be I’m making things over-complex here, and that just using the slices on actual longitude and then following where the Sun is in relation to that may also be good enough. Easy to get the nearest neighbours and the angles they are at, and to divide the guzouta of air into the guzintas.

14. V.P. Elect Smith says:

@Philip Mulholland:

Icosahedral tiling is a common way to tessellate a globe. Possibly the most common after ‘squares’ of latitude and longitude. It also isn’t hard to visualize. What I don’t like about it is that a wind circling the equator, for example, hits these discontinuities of a pentagon where all the straight runs of hexagons suddenly split and shift direction. I suspect that tessellation artifact would have a real impact on the model results.

Basically my motive for looking elsewhere has been the desire to avoid that potential effect. Splitting mass flow gets done between all cells anyway, so not that much harder to do a 5 than a 6. But will that bias the results?…

The degenerate hexagon / divided will have latitudinal bands of hexagons so will be air mass division friendly and straight wind friendly. But at the point where the equatorial hexagon edge is subdivided, 6 of those hexagons will have 2 sides with a 180 angle between them. I.e. it will be a pentagon from our reality point of view. But an irregular one with potentially ‘nice’ characteristics in that it doesn’t disrupt the parallel bands of hexagons. So I’m going to explore that one a bit more. Variable area and all…

The present approach of equal area dots has issues where the angle to neighbors continues to shift. An essential artifact of different dot counts per latitude band, equally spaced in a band. Makes finding neighbors a bit of a bother and dividing air flow worse.

There is no good solution, so the question is which bad solution has the fewest effects on the model; both results and difficulty in coding it up.

@Simon:

Interesting idea. One of the regular tessellations of a sphere is the beach ball slice (or orange slice shape). Don’t know if it makes things harder or easier.

Part of the problem is dividing an air mass that is, say, going NE, so that some ends up in a cell N and some in a cell E (assuming the tessellation has no NE neighbor) while still preserving a NE momentum… BOTH mass and momentum must be preserved even while this example changed the actual directions the mass was displaced.

Maybe just using a square grid of dots and ignoring that the NE wind has to magically flow through a point contact is the least disruptive. You get 8 ‘neighbors’ so any angular error is reduced. Just ignore the corners of the N and E that it had to ‘jump over’…

Maybe I’ll just make it so that I can try different tessellations and see what changes…

FWIW, Julia lets you do OO or imperative (procedural) styles.

15. V.P. Elect Smith says:

There’s another alternative I’ve been reluctant to mention as it has a giant number of “cells” to process…

Ignore the tessellation issue entirely. Make a “cell” a “packet of air” (or water in the oceans). Just let it have any vector value it likes, and change coordinates (in whatever system you choose) as it likes. A packet picks up water at surfaces, becomes cloud as conditions change (as it rises), rains out, descends again dry and dense, etc.

Easier to assign physical state (RH, altitude, Lat & Lon, V, …) BUT, you end up needing a LOT of them to capture things like a hurricane or ridge lift snow… “Billions and billions”… you can’t really average a packet… I can figure the average temperature and humidity over a 100 mile square of the Sahara, but in a packet world, one of them can drift in, or leave, so stays discrete.

An interesting alternative, if you have big bucks…

16. Simon Derricutt says:

EM – by using points to emit the air, when the air is divided between the two or more receiving cells the location of that air is bounced a bit so instead of coming from a line (drawn parallel to the direction) it now comes from the two or more points you’ve divided it between. Direction and momentum is preserved, position is quantised, and this will result in a widening of the front of the wind as it passes through cell points unless the wind direction is directly from one point to the next so it doesn’t get shared between receiving points. A bit difficult to describe that point, but I hope it is clear enough. You may need to draw a grid and what happens to a wind that travels at various angles to the grid. Along a grid-line it’s passed from point to point and stays narrow, but once you get an angle to the grid it starts to spread. It spreads most on a hexagonal grid. Relative worst spreading is 1 per cell for equilateral triangle, 1.414 for square, 2 for hexagon. If you take advantage of the full 8 cells around a square, then splitting between, say, N and NE (and “jumping over a point” gives you a spreading of around 0.707 per cell. Most of the time, you’ll only be dividing the output air between two cells/points, unless there’s a downdraught there and air spreads out from that point over possibly all neighbour cells.

Putting the air in packets means they’ll need to jostle each other, and getting that to work may be difficult. Sure, if you fill a swimming-pool with balls instead of water and jump into it you get some analogue of a splash, but it’s not the same as water. Maybe the hard bit of this though is that the computation load would likely go up as some power of the number of cells, because each one affects its neighbours at the same time that the neighbours affect it. The same sort of problem that SPICE was set up to solve, but that uses minimisation of matrices to do the job. The bigger the matrix, the longer it takes to solve.

It’s worth spending some time kicking around the “how the f%#* do we do this” discussion, since there will be gains and losses from each design. Some approximations may be near enough, others may blow up and give the wrong answers.

In the end, though, this will be a weather model. I suspect it is not possible to predict long-term climate changes because we can’t predict what the Sun will do, and we also don’t know what the effects of different volumes of space will have (increased/decreased cosmic rays to seed clouds). We can predict where we’ll be relative to the Sun, and thus what solar radiation we’ll probably get if the Sun remains constant, but we can’t predict what the absolute solar radiation will be with any confidence. A better weather model is however worth a lot.

Looking at the wind charts, I’d say the East-West axis maybe isn’t so important, so tesselation is a matter of choice (apart from that spreading problem). The spreading problem implies that a square cell may be the least-worst from that point of view, and it’s relatively easy to do. The nearest neighbours and their angles will be the same along one latitude line. I’d suggest a dot (of equal area) at the poles, but they could be feeding air to half the next areas or up to all of them with polar downdraught.

One more ground-level parameter to add is drag, depending on whether it’s buildings, trees, water, etc.. I can’t remember whether I mentioned that before. Still, the air and the ground will be moving in different directions a lot of the time, so the wind direction (and momentum vector) will be modified by that drag. When the packet of air (wind) moves to the next cell, it will retain its linear momentum which will now be no longer parallel to the ground (because the world isn’t flat) and we’ll need to take into account the gravitational attraction and thus a ground-level pressure slightly reduced from what would be calculated on a non-spinning world. I don’t know just how large a difference that correction will make, since though the correction is likely small it might have a larger effect since it varies with latitude (and wind speed/direction).

I’m not sure whether all this is actually useful (as in getting incorporated), but maybe the discussion is useful anyway in deciding upon the least-worst method.

17. V.P. Elect Smith says:

What one is persuaded not to do is just as important as what one does…

Per wind vs rotation. Rotational speed at the pole is near zero, at the equator about 1100 MPH IIRC (approx). Wind rarely gets over 100 to 200 mph and that only in intense cyclones or a jet stream. For now, I’ll be ignoring rotation as an effect on wind as wind is substantially local . The big effect is Coriolis as air moves pole / equator / pole and that will be accounted for.

Variable thickness of atmosphere I intend to handle by layers. Initially just 2 in Stratos and Tropos, but eventually a tropopause (due to special things that happen there) and then likely a Meso and Thermo too. Exact altitude of each layer limit has to be a variable. For example, Tropopause is much higher in the tropics and Stratosphere can touch the ground in Antarctica on high plateau areas. Not sure how to do that yet… but density is likely the key variable. (Tropopause happens at the point where cooling by IR radiation becomes possible. We have a Troposphere precisely because the air is too dense to radiatively cool so it must convect.) So I think a “total energy to dump == quantity you can radiate away (fn of pressure broadening @ your pressure)” is likely what will determine tropopause altitude. Perhaps with a bit of “mass intrusion in strong convection cells” disruption, but I’ll see if that’s an emergent effect before forcing it in (i.e. adding a tropopause layer function)… Surface drag likely just a ‘plug number’ for each cell until / unless I can figure something better.

Part of the attraction of a Hex Grid is that you can handle straight E / W winds by latitude without mass diffusion off the line (very common) and can handle winds in a generally circular pattern between cells (major cyclones) which is another major pattern. That “dividing wind location” between cells is why I’m trying to make hex cells work and what I don’t like about the icosahedron (where you WILL hit a pentagonal divide and your mass must now split on separating lines of Hex Cells and then ‘wiggle along’ with the threat of constant diffusion trying to hold a straight equatorial line…)

Square grid with 8 neighbors can work, but then one needs 2 classes of neighbors. Near and far… And work out how to do the “action at a distance through a point contact” that can’t be just regular physics and emergent behaviour… (Unless you will have a very minute time step so passing through a 2% of area of a corner on a N/W/S/E/ cube can get to the diagonal cubes eventually using regular physics… but even that I would expect would have some diffusion unless you subdivide that cell and then you are really just making your grid very much smaller…)

And yes, noodling though this to excess NOW saves a lot of time later. Be it in rewrites, ripping out stuff that would never work, doing 10 minutes of calculations when 2 seconds could do it ‘the other way’ so the model just runs too damn long, etc. etc. Why I’m willing to make “slow progress” early… That, plus I’m using this exercise to get good at Julia coding, so it isn’t wasted whatever happens.

18. V.P. Elect Smith says:

I probably ought to again mention “Wedding Cake World”. IRREGULAR hexagon tessellation.

At the pole, cells spread out as on a flat surface. 1 then 6 then 12.
At the equator, cells spread out as on a cylinder. Equal # each layer up / down.

In between, you do some x 2 and some other = depending on Sin(theta) so as to stay close to the surface of the sphere.

Each hex cell is ‘somewhat spherical geometry’ as the edges are slightly bowed over the globe and the ‘far’ edge a slightly different length from the ‘near’ edge (to the starting line)

The whole globe ends up tessellated in just hexagons, but at the cost of variable area and not quite really hexagonal shape…

I’m pretty sure one can use “just flat equal numbers” for about 15 degrees N / S at the equator without too much distortion. Similarly for about 15 degrees at the poles. Essentially the whole Tropics and Arctic / Antarctic circles. In the tropics the sun just wobbles back and forth each side of straight overhead. At the arctic, it mostly goes into / out of sunlight ‘all at once’ a couple of times a year.

In between, the temperate zone is the hard bit. Need to stair step it, but with variable sized steps… So 2 x flat-doubling cells per 1 equal-cylinder like near the poles, and 2 x equal cells per 1 flat-doubling near the equator (and other ratios in between).

Then conform the N / S edge lengths to the actual latitude length at that latitude line. Adjust other edges, and you are done. (Or place a point on the sphere and let ‘bubble physics’ set the edges… largely by ignoring their existence and assuming they happen… Hummm… Wedding Cake Bubble World? Kinda like it ;-) Makes volumes, areas, mass distribution a bit harder, but directions easier and neighbors are all pretty regular. Sample Issue: A unit of 1/10 the air in your cell leaves N to the next cell. It is a different volume. How do you handle it getting a different percentage air in? More pressure? Block some? More out? Likely can ignore that kind of stuff when cell count is very very large (like 10k to 100k), but in test / debug runs at 100 cells it’s gonna hurt.

I’ve not given up on Wedding Cake world just yet.

19. gallopingcamel says:

After winning a scholarship to Pembroke college, Cambridge (UK) my intention was to be a mathematician. Sadly spherical trigonometry, matrix algebra and chi squared undid my ambitions even though my tutor was Dr. J.C.P. Miller who was director of the Mathematical Laboratory at that time.
https://www.cl.cam.ac.uk/events/EDSAC99/history.html

So I switched my major to physics with less challenging mathematics but tensors and vector calculus tripped me up. Those darned “Divs”, “Grads” and “Curls”!

So I switched my major to Electrical Engineering and loved every minute of it with Fourier analysis, Laplace transforms and network transformations.

It is simply a question of finding your level of incompetence.

20. “It is simply a question of finding your level of incompetence.”
Ah, the Peter Principle,
Rather you are – Floating at your level of competence.

21. M Simon says:

Philip Mulholland says:
26 December 2020 at 9:48 pm

I studied geodesic geometry extensively. For a few years I was on a dome building craze.

22. cdquarles says:

This discussion reminds me of the seminars we had at a local bar, after chemistry lab work. Good times!

23. Jim Masterson says:

@gallopingcamel:

>>
Those darned “Divs”, “Grads” and “Curls”!

So I switched my major to Electrical Engineering . . . .
<<

Seriously? I took EE, and if you studied electromagnetic waves, then Maxwell’s Equations have those darn Divs and Curls (not to mention a few Grads and Laplacians scattered around here and there). I switched my specialty from electronics to digital electronics. Treating a transistor as a switch was considerably easier than dealing with it as an amplifier. Although, we didn’t have SPICE to simplify network analysis in those days–slide rulers only.

Jim

24. Steven Fraser says:

@EM: Found the hexagons straightaway, 7 with 1 triangle side, and the larger one with 2 triangles per side.

I’ve nothing of substance on your topic to offer, other than a fun puzzle:

From how many starting points on the surface of the earth is it possible to ….

Walk South one mile,
Walk East 1 mile,
Walk North one mile,

And be back where you started?

25. President Elect H.R. says:

@Steven Fraser re less-than-random walks:

It’s pretty much an infinite number – 2, but why couldn’t you also walk West and back one mile so long as you’re out for a stroll? I don’t get that.

26. @Steven Fraser
East is not a great circle direction. The north pole is one point, the only thing that can change there is your starting direction for which the choice is infinite. For the southern hemisphere however there are an infinite number of starting points located on a line of latitude (small circle) that is 1 mile north of the inner latitude that has a circumference of exactly one mile round the south pole.

27. President Elect H.R. says:

@Steven F – Oh dear. By my question about walking West and Philip’s reply to you, I see that I misread your puzzle as “and back from each direction.” It’s step-by-step instructions.

I see that Philip read the puzzle correctly as step-by-step sequential.

It’s true my coffee was still brewing, but I’ll pass on the no-coffee excuse and go with carelessness for my misreading. I didn’t read the puzzle carefully until I saw Philip’s reply and went back to see what I’d missed.

28. @President Elect H.R.
This was an old favourite of my father, so I have prior!

29. President Elect H.R. says:

@Philip M – Ha! A legacy puzzle, eh?

An old favorite of your father? What, the no-coffee excuse or carelessness?
😜

😜😜

30. Jason Calley says:

@ Steve Fraser RE the walking puzzle.

But, but, but…. the North Pole answer is WRONG!! Everyone knows that global warming melted all the ice almost 20 years ago. There’s no place to walk! :)

I doubt this will help, but regarding tessellation of the sphere — When faceting a sphere, the highest number of EQUAL faces possible (regarding chirality) is 120, the result of dividing each face of an icosahedron into 6 equal 30-60-90 triangles. Projecting each of those triangles into onto a sphere and then filling in hexagons gives very slightly less distortion than starting directly with the icosahedron face and then projecting. But you still have the 12 pentagons. Of course you can jigger the divisions so that the pentagons have the same area as the hexagons.

All in all, I think your wedding cake idea might be best, and it is certainly superior for east to west symmetry.

31. The True Nolan says:

@ Philip Mulholland “For the southern hemisphere however there are an infinite number of starting points located on a line of latitude (small circle) that is 1 mile north of the inner latitude that has a circumference of exactly one mile round the south pole.”

Or one mile north of any line of latitude that has a circumference of 1/2 mile. Or 1/3 mile. Or 1/4 mile etc.

32. @ The True Nolan
Nice! Congratulations that really ups the game!

33. The True Nolan says:

Just for clarity, “Jason Calley” and “The True Nolan” are both me. I moved locations and computers and in the change over lost access to my old Jason Calley email account. I find it more convenient to use “The True Nolan”, but sometimes I forget… I now return you to your regularly scheduled program.

34. V.P. Elect Smith says:

It seems to me that you can not start from 1 mile north of the south pole, as when you are AT the pole, there is no East or West. All directions are North…

35. The True Nolan says:

@VPE Smith “you can not start from 1 mile north of the south pole”
Sure. You don’t start from one mile north of the south pole; you start from one mile north of a line of latitude whose circumference is one mile, or whose circumference is an integer fraction of a mile (1/2, 1/3, 1/4, etc.)

“you are AT the pole, there is no East or West”
True. The limiting case is when you start at one mile north plus an infinitismal bit more — at which point you would walk south one mile, walk a bazillion times around a tiny circle going around the south pole with almost zero circumference (as long as “bazillion times” comes out to an integer) and then head north a mile.

36. Steven Fraser says:

@The True Nolan: Well done!

The visualization I like to use is a ‘lollipop’, with a candy disk 1 mile in Circumference, and a stick 1 mile long. Set the lollipop with its center on the South Pole. From wherever the free end of the stick may be, walk along it until you get to the disk, then turn left (East) and walk around the candy disk until reaching the stick again. Turn left (North) and walk until reaching the free end of the stick.

As you say, the position of the lollipop can be rotated to give an infinite number of starting points, and, and the disk size can be infinitely divided to provide 1…infinite numbers of circuits around the pole from each circle of latitude that meets the general conditions of the puzzle.

Now, the original question was not to describe the solutions, but to identify ‘how many starting points on the surface of the earth is it possible to…’ Its a number, or at least a numeric expression.

@All: Anyone like to take a stab? An answer like ‘1 more than infinite’ may be a little insouciante, but I think there must be an expression which reasonably represents this mathematically.

37. President Elect H.R. says:

@Steven Fraser: I was wrong for different reasons, as I commented above, But I think it is infinity minus one because the North pole is out of play.

I’m tired. I’m just getting settled in. I may be wrong in my answer, but I agree that there should be some sort of of expression that describes the solution, and I think the argument is whether it’s infinity plus one or infinity minus one.

Somebody needs to chime in here because my patience is NOT infinite 😜 I wanna know what others think, too.

38. @ President Elect H.R. says:
“But I think it is infinity minus one because the North pole is out of play.”
The North pole is in play. Start at the North pole,
Go one mile due South along any line of longitude (which is a great circle).
After one mile turn 90 degrees due East and go one mile along a small circle line of latitude.
(When travelling due East to maintain your bearing you are always turning).
After one mile along the line of latitude turn 90 degrees due North.
Travel one mile along this great circle line of longitude and you must arrive back at the North pole.
In spherical geometry the three internal angles of a triangle sum to more than 180 decrees, in this case the sum is 270 degrees.

39. President Elect H.R. says:

@Philip – It’s the water thingy. You can’t always walk at the North pole, as someone brought up early on. But you can most of the time, so it’s one of those never ending arguments, like a bar bet on who has the best team evah!

Geographically, yeah; the N. Pole is included.

40. I choose winter to do the experiment. YMMV ;-)
The beauty of this discussion is the way that The True Nolan so magnificently raised the bar for the South pole.
A worthy winner!

41. President Elect H.R. says:

Agreed about The True Nolan, Philip M. I haven’t had time to think about the puzzle for about 3 days so I’m just (barely) getting my head back in the game. And it will be another day or so before I get the last bit of everything set up here.

There is nothing like being surrounded by palm trees when you know that if you had stayed home, you’d be shoveling a foot of snow 😎

42. p.g.sharrow says:

Damn, now you guys have me wasting my time in this pointless argument. I can’t see any way you can travel 1 mile south, turn 90 degrees east, travel 1 mile in a straight line and wind up 1 mile south of your starting point. You must travel a curve, great circle, line or start out a bit less then 90 degrees to walk a straight line to the 1 mile south point.
I have fallen asleep twice visualizing that trek, A total waste of time because, no one in their right mind would even do such a foolish experiment in such a inhospitable place, 600 miles north of the nearest Bar…pg 8-)

43. “travel 1 mile in a straight line”
On the surface a sphere due East is a bearing, it is not a direction per se it is a line of latitude and this is not a straight line, it’s a small circle.

44. President Elect H.R. says:

OK, p.g. You raised another bar bet type point. Since the surface of the Earth is curved, there is nowhere on Earth where you can walk a straight line.

But, without scrolling up, I recall that it was walk N, S, E, and straight wasn’t specified.

But also… the way I misread the puzzle as walk S, N, E, and back instead of “and wind up back at the start” is another puzzle.

45. ossqss says:

Alright, are we talking magnetic North Pole or the actual one? :-)

I wonder if the minute physics guy did a video on this>>> I shall find out>>>

46. ossqss says:

Well, this is the closest I could get. So are we talking about a 3D walk or 1D?

47. p.g.sharrow says:

@HR; Now you get me confused about me attitude of my latitude or was my altitude at my latitude ? No, I’m of certitude, my attitude about my lattitude has something to do with my altitude and the with the distance to the nearest Bar! MAN ! I need another drink…………pg

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