7 ms·

I quibble with the description of the shapes bounded by 10 degrees of longitude and latitude as "rectangles" - they're not even planar shapes, and their adjacen

by caf 6y ago

I quibble with the description of the shapes bounded by 10 degrees of longitude and latitude as "rectangles" - they're not even planar shapes, and their adjacent sides certainly aren't at right angles. Some of them don't even have four sides. This bears considering when looking at the map, because some of those regions are much, much smaller than others.

deleted 6y ago
[deleted]
NopeNotToday 6y ago
He's not making a mathematical definition of the areas. 'area' or 'box' is a simple description that everyone can understand. And the map projection used does form squares.
- chrisseaton 6y ago
  > And the map projection used does form squares. Some of those squares have zero-length edges.
  - jtbayly 6y ago
    Which is really cool. Also very confusing!
deleted 6y ago
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mjd 6y ago
Since you want to quibble, I'll just point out that the word “rectangles” does not appear in the article.
- areyousure 6y ago
  https://i.imgur.com/El0JhG0.png https://i.imgur.com/El0JhG0.png
  - mjd 6y ago
    I stand corrected.
    - Klinky 6y ago
      I mean, technically that was in the image, not the actual article texts. ;-)
wcarey 6y ago
A fun thought experiment is to noodle through whether, given the method of choosing cities described, the choice of map projection would influence the resulting list of cities.
- greggyb 6y ago
  Latitude and longitude exist as boundaries outside the domain of map projections. A different projection may draw these lines or curves differently, but they still delineate the same geographic boundaries. E.g. a different projection may not show rectangles, but instead curved regions, but the cities existing in those regions would be the same. If you are thinking of drawing arbitrary rectangles on arbitrary projections, then you'd have to come up with a rigorous way to define your rectangles. If you don't, then you're just drawing random shapes on other random shapes. Either way, yes, a different projection with the same rectangles drawn over it would yield different answers, and this should be immediately obvious.
- jobigoud 6y ago
  Even the exact same projection but a slightly shifted 0° meridien would yield a completely different list of cities.
beerandt 6y ago
Quadrangle is the proper term (at least for four-sided boundaries). Hence the common name for USGS topographic maps: quads.
deleted 6y ago
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phreeza 6y ago
Maybe the author should have gone with the faces of a Disdyakis triacontahedron
- jsjohnst 6y ago
  For anyone else not familiar with the exact shape: https://en.wikipedia.org/wiki/Disdyakis_triacontahedron https://en.wikipedia.org/wiki/Disdyakis_triacontahedron
- dllu 6y ago
  The dual of a geodesic polyhedron, e.g. an icosahedron whose faces have been subdivided is in fact a common and excellent choice for geospatial applications, with https://github.com/uber/h3 https://github.com/uber/h3 being a good implementation. You can also subdivide the faces of a cube into smaller squares with a quadtree, e.g. https://s2geometry.io/ https://s2geometry.io/ The Disdyakis triacontahedron's faces are probably too long and skinny for most geospatial applications though.
lambdasquirrel 6y ago
Well it lends to an aliasing effect, that in Europe is even more comical. You completely lose the major cities of most countries like Ireland, Sweden, Denmark, and Poland. Central Europe is blotted out by Rome. The Romans finally conquer (most of) Germany. ;)
- qw 6y ago
  It also managed to include 5 Norwegian cities, but missed the largest city because it was in the same square as Berlin.
  - runarberg 6y ago
    It’s fun to see that out of the Nordic capitals: Nuuk, Reykjavík, Tórhavn, and Helsinki made it; while Copenhagen, Oslo, and Stockholm didn’t.
    - qw 6y ago
      Yes, they are all in the same quadrant as Berlin :-)
    - Delk 6y ago
      I'm not even sure how accurate the map is, though. St. Petersburg is at the eastern end of the Gulf of Finland, and it would seem that it should be in the same square as Helsinki. St. Petersburg is definitely bigger. I guess it ends up in the same square with Moscow and thus doesn't show up but either the map doesn't show the shoreline of the sea very accurately at all or it's wrong about the location of St. Petersburg.
      - runarberg 6y ago
        According to Wikipedia: Helsinki has a population of 650,058 and is located at 60°10′15″N 24°56′15″E while; Saint Petersburg has a population of 5,351,935 and is located at 59°56′15″N 30°18′31″E Saint Petersburg is one quadrant south and one quadrant east of Helsinki, or the same quadrant as Moscow as you said. The bay gets a little thin before Saint Petersburg, and there is a parallel passing exactly through it. I’m guessing that’s whats making Saint Petersburg appear more inline then it is, or—more accurately—makes the shoreline appear to be more to the west then it actually is.
tantalor 6y ago
> their adjacent sides certainly aren't at right angles They certainly are. Parallels and meridians always meet at 90 degrees. https://www.geogebra.org/m/kNtNZzsS https://www.geogebra.org/m/kNtNZzsS
- doersino 6y ago
  Right angles with regard to the latitude-longitude coordinate system, but not with regard to the surface of the earth. Recall that all meridians intersect at the poles.
  - colanderman 6y ago
    @tantalor is technically correct; meridians and parallels intersect at right angles, even with respect to the (idealized) surface of the earth. Proof sketch: if the angles involved in a crossing were not all 90°, then either the crossing would not be symmetric about the meridian (it is), or one or both lines must have a sharp "kink" (i.e., have a non-smooth first derivative) at the intersection (neither does). However, parallels, while smooth, are not straight, they "curl" toward the nearest pole (i.e., non-zero second derivative, relative to the earth's surface). This accounts for the non-"rectangular" shape of quadrangles and can be replicated on a 2D plane.
  - tantalor 6y ago
    Yes with regard to the surface of the earth. Take the lines tangent to the meridian & parallel at the point of intersection. Observe these lines are perpendicular. Hence the meridian & parallel meet at a right angle.
    - simonh 6y ago
      What about the ‘boxes’ where the top edge is, er, the point at one of the poles?
      - tantalor 6y ago
        Yeah not the pole of course, but you can cut them off at at ±89.999 degrees and it still works.
paulirwin 6y ago
Mostly an aside to your comment, but it's somewhat related. This post's divisions of 10° latitude/longitude is pretty close to the Maidenhead grid system, used heavily by amateur radio operators like myself. The difference is that each top-level grid "square" (to your point, not actually square, but that's what they're called) is 20° longitude by 10° latitude represented by two letters. While the computation of a 4 or more character grid locator code is complex enough that most people can't quite do that in their head, because of them being treated as if they were coordinate "rectangles", it is simple enough to translate lat/lon coordinates to a grid code and vice versa with pen and paper if needed in the case of an emergency, if you know the algorithm or have a reference sheet. The 10° latitude size means that the parallels are the same in the Maidenhead system (for the first two letters of a code) as in this post. It also has the added benefit of knowing that DN is north of DM, which are both west of EM. I've gotten to where I can roughly place someone I hear on the radio based on a mental map of grid codes, and have memorized many grid codes of large population centers. The beauty of the grid code system is that you can further refine an area by adding on subsequent numbers and letters, much like degrees/minutes/seconds in coordinates, but requiring significantly less characters to read to others over the radio. And, you can use phonetics for the letters, i.e. "delta mike seven niner" (DM79) is roughly the entire Denver metro area. Fort Collins, CO on this parent post falls under DN, above that 10°-sized parallel. More info on the Maidenhead system: https://en.wikipedia.org/wiki/Maidenhead_Locator_System https://en.wikipedia.org/wiki/Maidenhead_Locator_System Map with two-letter grid codes: https://www.mapability.com/ei8ic/maps/gridworld.php https://www.mapability.com/ei8ic/maps/gridworld.php
hammock 6y ago
You make a good point. What is an equal-area, tessellated shape that could be used in its place? I guess a triangle. Or one of these: https://upload.wikimedia.org/wikipedia/commons/thumb/6/67/Spherical_square_prism2.png/100px-Spherical_square_prism2.png https://upload.wikimedia.org/wikipedia/commons/thumb/6/67/Sp... https://upload.wikimedia.org/wikipedia/commons/thumb/9/9b/Spherical_square_bipyramid2.svg/100px-Spherical_square_bipyramid2.svg.png https://upload.wikimedia.org/wikipedia/commons/thumb/9/9b/Sp...
- mjevans 6y ago
  https://en.wikipedia.org/wiki/Geodesic_polyhedron https://en.wikipedia.org/wiki/Geodesic_polyhedron For just shapes the above is a nice visual start. However a striking feature of the linked article is how frequently there's a major city right near a cell edge. A more useful list might be constructed by some process involving assigning a "city" an imprint size based on population, and a surrounding 'metro area' based on absorbing any weaker cities that overlap until the process either repeats or is matched by a neighboring city. A starting point for results that are already similar to that: https://en.wikipedia.org/wiki/Metropolitan_area https://en.wikipedia.org/wiki/Metropolitan_area The US specific list: https://en.wikipedia.org/wiki/List_of_metropolitan_statistical_areas https://en.wikipedia.org/wiki/List_of_metropolitan_statistic...
  - hammock 6y ago
    Yeah I was hoping there was a shape more interesting than a triangle (or a soccer ball)... Those rounded squares and rounded triangles were the best I could find. I like that idea of building the regions bottom-up.
oh_sigh 6y ago
But they are rectangles on the projection - they just aren't rectangles if you map the projection back onto the globe.
joshvm 6y ago
The South Pole is a particularly interesting case. Clearly the Pole itself is 90 South and the choice of longitude is arbitrary there. However, Amundsen-Scott station, which is a hundred odd metres away, happens to be at ~140 E (89.997553 S!) and due to the map projection it appears as if it's in a totally different place. It would be interesting to see an interactive version with a free choice of meridians.

by caf 6y ago

deleted 6y ago

[deleted]

NopeNotToday 6y ago

He's not making a mathematical definition of the areas. 'area' or 'box' is a simple description that everyone can understand. And the map projection used does form squares.

chrisseaton 6y ago
> And the map projection used does form squares. Some of those squares have zero-length edges.
- jtbayly 6y ago
  Which is really cool. Also very confusing!

deleted 6y ago

[deleted]

mjd 6y ago

Since you want to quibble, I'll just point out that the word “rectangles” does not appear in the article.

areyousure 6y ago
https://i.imgur.com/El0JhG0.png https://i.imgur.com/El0JhG0.png
- mjd 6y ago
  I stand corrected.
  - Klinky 6y ago
    I mean, technically that was in the image, not the actual article texts. ;-)

wcarey 6y ago

A fun thought experiment is to noodle through whether, given the method of choosing cities described, the choice of map projection would influence the resulting list of cities.

greggyb 6y ago
Latitude and longitude exist as boundaries outside the domain of map projections. A different projection may draw these lines or curves differently, but they still delineate the same geographic boundaries. E.g. a different projection may not show rectangles, but instead curved regions, but the cities existing in those regions would be the same. If you are thinking of drawing arbitrary rectangles on arbitrary projections, then you'd have to come up with a rigorous way to define your rectangles. If you don't, then you're just drawing random shapes on other random shapes. Either way, yes, a different projection with the same rectangles drawn over it would yield different answers, and this should be immediately obvious.
jobigoud 6y ago
Even the exact same projection but a slightly shifted 0° meridien would yield a completely different list of cities.

beerandt 6y ago

Quadrangle is the proper term (at least for four-sided boundaries). Hence the common name for USGS topographic maps: quads.

deleted 6y ago

[deleted]

phreeza 6y ago

Maybe the author should have gone with the faces of a Disdyakis triacontahedron

jsjohnst 6y ago
For anyone else not familiar with the exact shape: https://en.wikipedia.org/wiki/Disdyakis_triacontahedron https://en.wikipedia.org/wiki/Disdyakis_triacontahedron
dllu 6y ago
The dual of a geodesic polyhedron, e.g. an icosahedron whose faces have been subdivided is in fact a common and excellent choice for geospatial applications, with https://github.com/uber/h3 https://github.com/uber/h3 being a good implementation. You can also subdivide the faces of a cube into smaller squares with a quadtree, e.g. https://s2geometry.io/ https://s2geometry.io/ The Disdyakis triacontahedron's faces are probably too long and skinny for most geospatial applications though.

lambdasquirrel 6y ago

Well it lends to an aliasing effect, that in Europe is even more comical. You completely lose the major cities of most countries like Ireland, Sweden, Denmark, and Poland. Central Europe is blotted out by Rome. The Romans finally conquer (most of) Germany. ;)

qw 6y ago
It also managed to include 5 Norwegian cities, but missed the largest city because it was in the same square as Berlin.
- runarberg 6y ago
  It’s fun to see that out of the Nordic capitals: Nuuk, Reykjavík, Tórhavn, and Helsinki made it; while Copenhagen, Oslo, and Stockholm didn’t.
  - qw 6y ago
    Yes, they are all in the same quadrant as Berlin :-)
  - Delk 6y ago
    I'm not even sure how accurate the map is, though. St. Petersburg is at the eastern end of the Gulf of Finland, and it would seem that it should be in the same square as Helsinki. St. Petersburg is definitely bigger. I guess it ends up in the same square with Moscow and thus doesn't show up but either the map doesn't show the shoreline of the sea very accurately at all or it's wrong about the location of St. Petersburg.
    - runarberg 6y ago
      According to Wikipedia: Helsinki has a population of 650,058 and is located at 60°10′15″N 24°56′15″E while; Saint Petersburg has a population of 5,351,935 and is located at 59°56′15″N 30°18′31″E Saint Petersburg is one quadrant south and one quadrant east of Helsinki, or the same quadrant as Moscow as you said. The bay gets a little thin before Saint Petersburg, and there is a parallel passing exactly through it. I’m guessing that’s whats making Saint Petersburg appear more inline then it is, or—more accurately—makes the shoreline appear to be more to the west then it actually is.

tantalor 6y ago

> their adjacent sides certainly aren't at right angles They certainly are. Parallels and meridians always meet at 90 degrees. https://www.geogebra.org/m/kNtNZzsS https://www.geogebra.org/m/kNtNZzsS

doersino 6y ago
Right angles with regard to the latitude-longitude coordinate system, but not with regard to the surface of the earth. Recall that all meridians intersect at the poles.
- colanderman 6y ago
  @tantalor is technically correct; meridians and parallels intersect at right angles, even with respect to the (idealized) surface of the earth. Proof sketch: if the angles involved in a crossing were not all 90°, then either the crossing would not be symmetric about the meridian (it is), or one or both lines must have a sharp "kink" (i.e., have a non-smooth first derivative) at the intersection (neither does). However, parallels, while smooth, are not straight, they "curl" toward the nearest pole (i.e., non-zero second derivative, relative to the earth's surface). This accounts for the non-"rectangular" shape of quadrangles and can be replicated on a 2D plane.
- tantalor 6y ago
  Yes with regard to the surface of the earth. Take the lines tangent to the meridian & parallel at the point of intersection. Observe these lines are perpendicular. Hence the meridian & parallel meet at a right angle.
  - simonh 6y ago
    What about the ‘boxes’ where the top edge is, er, the point at one of the poles?
    - tantalor 6y ago
      Yeah not the pole of course, but you can cut them off at at ±89.999 degrees and it still works.

paulirwin 6y ago

Mostly an aside to your comment, but it's somewhat related. This post's divisions of 10° latitude/longitude is pretty close to the Maidenhead grid system, used heavily by amateur radio operators like myself. The difference is that each top-level grid "square" (to your point, not actually square, but that's what they're called) is 20° longitude by 10° latitude represented by two letters. While the computation of a 4 or more character grid locator code is complex enough that most people can't quite do that in their head, because of them being treated as if they were coordinate "rectangles", it is simple enough to translate lat/lon coordinates to a grid code and vice versa with pen and paper if needed in the case of an emergency, if you know the algorithm or have a reference sheet. The 10° latitude size means that the parallels are the same in the Maidenhead system (for the first two letters of a code) as in this post. It also has the added benefit of knowing that DN is north of DM, which are both west of EM. I've gotten to where I can roughly place someone I hear on the radio based on a mental map of grid codes, and have memorized many grid codes of large population centers. The beauty of the grid code system is that you can further refine an area by adding on subsequent numbers and letters, much like degrees/minutes/seconds in coordinates, but requiring significantly less characters to read to others over the radio. And, you can use phonetics for the letters, i.e. "delta mike seven niner" (DM79) is roughly the entire Denver metro area. Fort Collins, CO on this parent post falls under DN, above that 10°-sized parallel. More info on the Maidenhead system: https://en.wikipedia.org/wiki/Maidenhead_Locator_System https://en.wikipedia.org/wiki/Maidenhead_Locator_System Map with two-letter grid codes: https://www.mapability.com/ei8ic/maps/gridworld.php https://www.mapability.com/ei8ic/maps/gridworld.php

hammock 6y ago

You make a good point. What is an equal-area, tessellated shape that could be used in its place? I guess a triangle. Or one of these: https://upload.wikimedia.org/wikipedia/commons/thumb/6/67/Spherical_square_prism2.png/100px-Spherical_square_prism2.png https://upload.wikimedia.org/wikipedia/commons/thumb/6/67/Sp... https://upload.wikimedia.org/wikipedia/commons/thumb/9/9b/Spherical_square_bipyramid2.svg/100px-Spherical_square_bipyramid2.svg.png https://upload.wikimedia.org/wikipedia/commons/thumb/9/9b/Sp...

mjevans 6y ago
https://en.wikipedia.org/wiki/Geodesic_polyhedron https://en.wikipedia.org/wiki/Geodesic_polyhedron For just shapes the above is a nice visual start. However a striking feature of the linked article is how frequently there's a major city right near a cell edge. A more useful list might be constructed by some process involving assigning a "city" an imprint size based on population, and a surrounding 'metro area' based on absorbing any weaker cities that overlap until the process either repeats or is matched by a neighboring city. A starting point for results that are already similar to that: https://en.wikipedia.org/wiki/Metropolitan_area https://en.wikipedia.org/wiki/Metropolitan_area The US specific list: https://en.wikipedia.org/wiki/List_of_metropolitan_statistical_areas https://en.wikipedia.org/wiki/List_of_metropolitan_statistic...
- hammock 6y ago
  Yeah I was hoping there was a shape more interesting than a triangle (or a soccer ball)... Those rounded squares and rounded triangles were the best I could find. I like that idea of building the regions bottom-up.

oh_sigh 6y ago

But they are rectangles on the projection - they just aren't rectangles if you map the projection back onto the globe.

joshvm 6y ago

The South Pole is a particularly interesting case. Clearly the Pole itself is 90 South and the choice of longitude is arbitrary there. However, Amundsen-Scott station, which is a hundred odd metres away, happens to be at ~140 E (89.997553 S!) and due to the map projection it appears as if it's in a totally different place. It would be interesting to see an interactive version with a free choice of meridians.