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> Sure it's helpful to cover the earth in 3m squares with unique names ... I'm actually not convinced. Squares don't tessellate a sphere; you have singulariti

> Sure it's helpful to cover the earth in 3m squares with unique names ... I'm actually not convinced. Squares don't tessellate a sphere; you have singularities at the poles. Sure, there aren't many people living there, but it's important to think about the limitations of this approach. If you think this is really just an edge case, remember that projecting a square onto a sphere necessitates some distortion. You can choose to distort angles or distances, but you have to choose one. For distances, well, a "3m x 3m" square becomes essentially meaningless. The problem gets worse when you increase latitude. Canada, northern Europe and Russia are especially prone to this problem. The same argument applies to locations in the Southern hemisphere as well.

logfromblammo 11y ago
This is how I would do it. Your base unit is a circle with radius 180cm, which is approximately 10m^2. Place one circle at the north pole. Your starting angle is 0 degrees longitude. Place a circle at the south pole, and slide it along that line of longitude until your sliding circle touches any other already-placed circle. Then "roll" the circle to the local minimum distance between its center and the north pole; whenever moving it an epsilon east or west would not allow the center to get closer, stop. Add 222.5 degrees (actually (sqrt(5)-1)*180), modulo 360. Slide in another circle. Add 222.5 degrees, modulo 360. Slide in another circle. Repeat until the circle placed at the south pole intersects an already-placed circle. Move each center point to the nearest grid-based point at a given precision, then name them. If you stand on the named point, you should be able to touch anything using that point as its address, without moving your feet. A 10m^2 unit cell is a very human-sized measurement, but you are correct that forcing them into an orthographic grid is pointless when the addresses are names and not numbers. There's no way any living human could memorize the name-location mapping for all addresses, or even just all the land-based addresses. People would need to look them up in an index from a numeric coordinate, and vice versa. As such, there is no need to orient the address points to any existing coordinate system.
saraid216 11y ago
That's true for large chunks, but I'd imagine 3 meters is small enough to remain useful well past the 85 degrees latitude mark. Straight Dope informs me that the circumference up there is approximately 24k miles, which becomes 38,624,256 meters which means approximately 1 million divisions. As long as the system works, what's the big deal?
- anamexis 11y ago
  24k miles is Earth's circumference at the equator, not 85 degrees latitude.
  - saraid216 11y ago
    Dammit. I thought I had been careful enough to get the correct number. Alright. Picking the simplest formula off Wikipedia, cosine of 85 degrees is ~0.0871557 times a (6378137m) times pi is ~1746383.9m for the circumference. That's still approximately 500k squares at the exact latitude. My point is one of the sheer magnitude still available. What I should do is calculate how many decimal points down from 90 degrees 3 meters is and get that circumference, since that's the low bound, but I'm not really feeling up to the number-plugging. I could be incorrect about what a is; I presumed it is the semi-major axis on WGS84, but I might have picked the wrong thing. ... Ooh. I forgot there's a Wolfram|Alpha. ... And 10 minutes later I haven't convinced the input parsing to give me what I want. Oh well.

logfromblammo 11y ago

This is how I would do it. Your base unit is a circle with radius 180cm, which is approximately 10m^2. Place one circle at the north pole. Your starting angle is 0 degrees longitude. Place a circle at the south pole, and slide it along that line of longitude until your sliding circle touches any other already-placed circle. Then "roll" the circle to the local minimum distance between its center and the north pole; whenever moving it an epsilon east or west would not allow the center to get closer, stop. Add 222.5 degrees (actually (sqrt(5)-1)*180), modulo 360. Slide in another circle. Add 222.5 degrees, modulo 360. Slide in another circle. Repeat until the circle placed at the south pole intersects an already-placed circle. Move each center point to the nearest grid-based point at a given precision, then name them. If you stand on the named point, you should be able to touch anything using that point as its address, without moving your feet. A 10m^2 unit cell is a very human-sized measurement, but you are correct that forcing them into an orthographic grid is pointless when the addresses are names and not numbers. There's no way any living human could memorize the name-location mapping for all addresses, or even just all the land-based addresses. People would need to look them up in an index from a numeric coordinate, and vice versa. As such, there is no need to orient the address points to any existing coordinate system.

saraid216 11y ago

That's true for large chunks, but I'd imagine 3 meters is small enough to remain useful well past the 85 degrees latitude mark. Straight Dope informs me that the circumference up there is approximately 24k miles, which becomes 38,624,256 meters which means approximately 1 million divisions. As long as the system works, what's the big deal?

anamexis 11y ago
24k miles is Earth's circumference at the equator, not 85 degrees latitude.
- saraid216 11y ago
  Dammit. I thought I had been careful enough to get the correct number. Alright. Picking the simplest formula off Wikipedia, cosine of 85 degrees is ~0.0871557 times a (6378137m) times pi is ~1746383.9m for the circumference. That's still approximately 500k squares at the exact latitude. My point is one of the sheer magnitude still available. What I should do is calculate how many decimal points down from 90 degrees 3 meters is and get that circumference, since that's the low bound, but I'm not really feeling up to the number-plugging. I could be incorrect about what a is; I presumed it is the semi-major axis on WGS84, but I might have picked the wrong thing. ... Ooh. I forgot there's a Wolfram|Alpha. ... And 10 minutes later I haven't convinced the input parsing to give me what I want. Oh well.