What’s the world’s center of population?
The center of population for a region is, roughly, the center of mass of the inhabitants. The Census bureau defines the center of population of the US (currently in Missouri) as
the point at which an imaginary, flat, weightless, and rigid map of the United States would balance perfectly if weights of identical value were placed on it so that each weight represented the location of one person on the date of the census.
This definition breaks down for populations on curved surfaces. For the earth as a whole, the center of mass obviously falls deep inside the planet.
This problem is easy to fix. I figure a better definition would be the point at which the sum of straight-line surface distances to each person is minimized. This is equivalent to the standard definition for a flat region, but it has the advantage that you can use it to define the center of population for a sphere.
I’ve never seen anyone who’s calculated the earth’s center of population so defined, but it doesn’t seem like it would be hard. Does anyone have the answer?
Bonus: find the center of population for other groups. What is the center of population of native English speakers? internet users? … bloggers?
Edit: I was standing the shower just now when I realized that the generalization I was using had to be wrong. I got it from this page on Wolfram Mathworld,
The centroid of
point masses also gives the location at which a school should be built in order to minimize the distance travelled by children from
cities, located at the positions of the masses, and with
equal to the number of students from city
(Steinhaus 1999, pp. 113-116).
and did try to check out the citation while writing, but it was to a book and I was much too lazy for that. However, I think the Wolfram paraphrasing is wrong — it’s not the distances that are minimized; it must be some other quantity. You can see that this is wrong for center-of-mass of two people at A and one person at B. It’s probably sums of squares that are minimized (as suggested in a comment, and which works for the three-person example) but I don’t see an obvious proof of this.
And I think that’s the same thing as the 3D centroid projected to the surface.
34 N 68 E
Hope this wasn’t meant to be the same as my other answer. Will recheck tomorrow.
I think that bradluen is right. So is the AMS
Which means that really, the population center on the earth is the geometric center in 3D space projected on Earth’s surface.
This is also coincident with the minimum of the sum of distances (squared or absolute, that does not matter) on the surface.
This holds true as long as the Earth is viewed as a sphere, but might change if we regard it as an ellipsoid or a “geoid” (meaning “an earth-like body”), i.e. all bumby and not regular at all.
The point at which the distance to everyone else is minimised?
That sounds a bit like a redefinition of the travelling salesman problem to me.
I’ve done zero research into that proof however.
Good luck coming up with a solution short of brute forcing it for a few hundred million points.
The obvious proof is essentially the definition of torque.
Daniel
Happy 420, Randall!
There are a couple problems I see, but I think there’s a simple solution for the best result:
1) Where do we start looking? A traditional map with the Americas to the west and China in the far east? A China-centric map? From the north pole or south pole? An arbitrary starting point will give only one of many results…The ‘center’ could be on (multiple?) axes.
2) World maps cannot be accurate for both longitude and latitude. 3D object on 2D surface.
The simple solution:
Take a map accurate for latitude and calculate to find the axes…It’ll be in the northern hemisphere. Take a map accurate for longitude and do the same…It’ll be in the eastern hemisphere. This will give a general idea, but won’t be fully accurate, as people who are on the far side of the Earth are weighted too heavily for one direction. It’s as if we’re trying to average numbers…The largest and smallest get the most weight. But if our “number line” is instead a loop, that’s a problem for us. I don’t know the formula to adjust for this…Maybe someone’s posted it already…But it’s 3 AM and I can’t read too much longer.
1. Cut a globe in half.
2. Measure out blobs of silly putty of weight proportional to each country’s population.
3. Stick blobs to their respective countries from inside the globe.
4. Put globe back together.
5. Roll it on a flat hard surface a few times to see what point/points are consistently balanced upon.
Or, alternatively:
1. Get an inflatable globe
2. Measure out small blobs of putty of weight proportional to each country’s population.
3. Stick blobs to the exact geometric centres of respective countries (you’d have to make them as small and flat as possible in order to minimise the torque due to gravity)
4. Drop it in a tub of water, and see where it settles. The point at the bottom of the globe is the centre of population mass for the globe.
Perhaps an easier way to think of the point on the surface is to identify the surface point with the shortest distance to the center-of-spherical-mass. This then gives you an interesting hint about how you might want to display the value.
If you merely note the point with a dot, you have no information about the degree to which that point is central. The center-of-spherical-mass is on some line which spans (inclusively) from the surface to the center of the sphere (okay, oblate spheroid). This means that you can identify the proportion of the distribution of mass. If the point sits on the surface the whole population is at one point. If the point is in the center of the sphere, the whole population is evenly distributed.
So… draw the point representing the surface-center-of-mass (pick any point if this coincides with the center of the sphere or special case this in a different way) and, around that draw a circle with a radius which is the proportion of distribution. If the distribution of the population mass is uniform, the circle is the size of the sphere. If the population exists at one point, the circle is point sized, at that point. Any other distributions will then be represented by a circle which represents the diffusion from the point.
Aesthetically, I’d probably saturate the colors between the surface-center-of-mass point and the distribution circle and desaturate the colors outside of the distribution circle. The surface center of mass point would be a heavy black dot.
Why, after finding the center of population, would you want to project it to the surface? You would lose a large portion of information about the population’s distribution.
Aaron Brooks got to that point first, and his solution is actually a lot more visually appealing with less even distributions in my opinion, but it would still become misleading as the distribution evens out. It would still be trying to center on a specific surface point when the surface point is hardly relevant in that situation. Still, I would recommend it for less even distributions.
dude, i really must say this.
your comics are SO great!
some friends and i laughed for an entire hour watching most of them.
and your math humour is wickeeeed!
cheers from europe
Shouldn’t there be 2? think about it, we have a 3-d sphere, and even if we keep it on the surface… we are going to wrap back around. SO, we can go with a balance of all people in the middle of where everyone is, and where everyone isn’t. So if we come up with some equation, it should end up with 2 answers. Take a look at a simplified 2-D example. Imagine having 2 equal point charges placed (asymmetrically) on a hula hoop and finding a point of balance on it. There will be two points. One between the two points in the middle of the smaller arc length, and the other in the center of the longer arclength. Makes it a tricky subject… OR if we wanted to.. we could try and do it 3 dimensionally, and then just project the 3 dimensional result outward radially to the surface. Should give us the center point.
In 1929, Edwin B. Wilson offered some interesting insights on a related topic1. Mathematics can be correct or incorrect; it can also be appropriate or inappropriate. When you say you want to find the center of mass, what is it you want to know? Do you want a point where you could balance the map if people were standing on it, flattened? With which projection: Mercator, equal-area, orange-slice? Or do you want to find the point that minimizes the total distance travelled by people? Or the point that minimizes the fuel spent to move those people (which accounts for differing weights of people)? Or the point that minimizes airfares (which accounts for populare destinations)? Or the point that minimizes the number of people who get pissed off by having to travel a long way?
How about the point where you would put the wormhole that the alternate-universe raptors come through, such that they have to run the least distnace in order to kill everyone?
The shower thing kinda made me laugh…
Something about the shower just makes it a good place to think about tough problems. Maybe I should stick a white board in there…
Come on. Don’t you get it? Randall is setting up the next xkcd meeting/reunion, but this time he’s too lazy to pick the spot himself
>Come on. Don’t you get it? Randall is setting up the next xkcd meeting/reunion, but this time he’s too lazy to pick the spot himself
Yes, he wants us to pick the densest population center in the united states to gain more rebels to his world take over plot in order to have to pay less for it out of his own pocket. For the brotherhood indeed.
To offer an answer for an ongoing discussion for where they should hold their annual department meeting, I once calculated the ‘center of population’ for the Univeristy of Wisconsin Colleges’ Math departments using a half dozen different criteria. Including unweighted: (each school no matter the size of department counted as 1, weighting each of the schools by the number of lecturers (some had as many as 13, or as little as 2), how many vehicles they would need to travel by etc.
The results for all the methods were within 8 miles of each other.
For those of you that are curious, it was in a field between Oshkosh and Berlin Wisconsin.
QUOTE:
http://blag.xkcd.com/2008/05/01/center-of-population/#comment-17492
The center of population of internet users/bloggers could be between North America, Europe and Australia.
That would mean that the center of internet users would be somewhere around africa/middle east.
I would think that asia would have a larger part and suggest that center actually be closer to India — of course, depends on the 2D map we are using — what is the center of that 2D map of the world?
Wikipedia already cites a center of population for a few counties in Asia, Europe and North America (and a few other countries). If you could find a pre-calculated center of population for each country (yeah, not every country will have one) you could just weight these by population data, then find the center of these points, which would be a far simpler calculation than working it out from every citizen, and yet give the same result.
“Yes, he wants us to pick the densest population center in the united states to gain more rebels to his world take over plot in order to have to pay less for it out of his own pocket. For the brotherhood indeed.”
Except that we would probably have to meet in some field in the middle of nowhere, Missouri. And then have to shift around by a few meters constantly as people move around the country. OK, maybe it’s not that much of a difference, but still.
I also came up with 34N and 68E. There is actually two balance points but its just the opposite of this one, which should be 146S and 112W.
In response to other comments, it would not be a good spot for an xkcd convention because it is in the mountains of Afghanistan or in the middle of Antarctica.
When downloading Google Earth I saw this:
http://earth.google.com/ig/directory?synd=earth&pid=earth&num=6&hl=en&url=http://www.google.com/mapfiles/mapplets/earthgallery/pop.xml&output=html
It’s a Google Earth file with the world’s population density. Oh, and I didn’t create it, some Russian dude did.
My mother is fairly sure the entire world revolves around her. If that’s really the case, and she’s the center of the universe, then I’d throw my guess out to be wherever she’s standing is probably the center.
Atleast, that’s how it works in her mind. Other than that, I haven’t a flying fkuc about how to figure it out. Dont much care though. =P
For this case your best bet would be
Min Sum r*arcos(sin(px)sin(a) +cos(px)cos(a)cos(lx-b) for all x’s
where r=radius of earth p=latitude of person & l=longitude of person
(a then will equal the latitude of centroid and b will equal the longitude)
I like the “Minimise time traveling” rather than “distance traveled”. It means that the vast bulk of countries can be reduced to their main airport. Once the approximate answer has been found – then some countries with large numbers of airports (ie US) can then be transitioned to individual airports. Since it is a loosely connected graph, it becomes a travelling salesman problem (ish).
For ease of calculation, (and to avoid having individual travel speeds) the average travel speed should be approximated as avg land speed (to airport) and avg airspeed. Also a flat overhead for transitioning from land to air added (call it customs)
Initially avg land speed is 60kph and avg airspeed is 900kph, customs time is 1 hour
Since it is likely that a large proportion of the travelers are flying in (rather than arriving at the airport through other means), and there is a significant transition cost, the end result is likely to be at an airport. (Yes the Premise were picked with that in mind)
Because I cannot be bothered to disconnect the graph based on current airline legs, I will assume that all airports visit each other. By data was for straight line distances (rather than great circle) – but I believe that the differences are roughly equivalent
Roughing out an answer
so an initial list of airports with the attached populations – I only bothered working out the top 10 populated countries in the world (that covers 2/3 of the worlds population and would probably give a reasonable answer) – I origionally had much grander plans as I wanted to include Australia – but that is about 55 on the list and I’m lazy
Means that the “total travel time” would be
PRC 1.83*10^10
India 1.81*10^10
United States 5.02*10^10
Indonesia 2.72*10^10
Brazil 6.22*10^10
Pakistan 1.89*10^10
Bangladesh 1.86*10^10
Nigeria 4.26*10^10
Russia 2.14*10^10
Japan 2.24*10^10
I’m not going to look at “mean of squares” – cos I’m slack
But my answer for “Where is the best spot for the ships to come and pick us up” would be New Delhi
It’s interesting to see how it differs from the previous answers
Paul
Someone probably already mentioned this, but I don’t have time to check before class, and I’ll forget if I don’t do it now.
What a centroid does is create a point in which the torque (distance times mass) is equally balanced in all directions. I’m thinking that this would mean that, if we give all people an arbitrary mass of one, it would in fact minimize distance travelled. I’ll read through all the posts after class, and try to confirm this.
A couple of points that might serve interesting:
1) It seems to be semi-popular among cosmologists that when the ripples in spacetime are averaged out that we live in a roughly hyperbolic universe which would greatly change geodesics used in the calculation. This would change the notion of distance (although the Earth or a neighborhood might be small enough such that the effect is negligible).
2) If we wanted to minimize distances so that we find to “best” place to put a school/fire department/etc it would depend heavily on roads and traffic volumes. I think this has a great impact on the utility of such a centroid so we might consider asking more specific questions.
“I also came up with 34N and 68E. There is actually two balance points but its just the opposite of this one, which should be 146S and 112W.”
Except that 146S doesn’t actually exist. If you go 146 degrees south from 0S, 112W, you go around the South Pole and get to 34S, 68E, which is the opposite of what you want. 34S, 112W is in the middle of the South Pacific. Possibly R’lyeh.
I first want to give you props for using the wolfram website, because I’m an avid Mathematica user. Anywho, it should be able to be done if you do use Country data in Mathematica. After all, it is the only program that actually knows more than the internet.
Mathematica gave me many headaches back in the physics labs…
While in the shower a while back I was wondering which was bigger, 2 gigainches or 1 nanoparsec.
I was frantically trying to work it out (while rinsing) without having to look up any conversion factors. I failed, extra points if you can do it!
Considering that half the population is in India and China I highly doubt that it’s at the center os the Earth. Even if you look at it from eastern/western hemisphere it’s heavily biased towards the eastern hemisphere.
Google works it out for you:
2 billion inch /10^-9 in parsecs
http://www.google.com/search?q=2+billion+inch+%2F10^-9+in+parsecs
looks like 2 gigainches ~= 1.65 nanoparsecs
Satisfyingly close though…
“Except that 146S doesn’t actually exist. If you go 146 degrees south from 0S, 112W, you go around the South Pole and get to 34S, 68E, which is the opposite of what you want. 34S, 112W is in the middle of the South Pacific. Possibly R’lyeh.”
Thanks Cesium.I should spend more time in the Geography labs. Sorry about that.
Well anyway, in response to the clay in the globe example, Antartica would be the top or lightest part and Afghanistan would be the bottom or heaviest part.
I think we tried something similar on the now defunct The-Dot [R.I.P] : we tried to work out the centre of population mass for the globe including approximations of average weight for different nations. We lost approximately 1/2 our brain mass through haemorridging out our ears and had to recalculate :’-(
Have fun :-/
Ha ha i just noticed dude. i pressed the random comic button and realized…
THERE IS NO comic 404 Please exclude http://www.xkcd.com/404/ from the randomizer code
It’s a joke, Salman.
Nice!!!
I think the center of population for internet users is either (a) somewhere near where it was invented or (b) some imaginary place inside the internet that no one can see or actually be for that matter (possibly inside the computer?).
The philosophy geek in me compels me to point out that the idea of a “centre of population” was discussed briefly in 1991 by Dan Dennett in his paper “real patterns”
*The more you know
Sorry if I’m jumping over answers that were already given on this… got only halfway or 2/3rds through….
Gravitation between sun and earth can basically be thought of as a force between two points, center of mass of the sun and center of mass of the earth. So in that sense, jumping might move the earth, but the earth+people system won’t move.
Kepler’s laws don’t apply to mass (I’d have to check calculations, I think its just because the Sun has so much mass) but that was the idea that period is proportional to the semi-major axis of the orbit. There’s not a mass dependence, just distance, so the earth gaining mass isn’t going to effect how long a year is. Basically, we can’t mess with the earth’s orbit in that sense by jumping to change distance or time.
For the running around the earth idea, though, the key is just conservation of energy. I’d need to work the problem out more detailed, probably, but my initial thought is that the work it takes to start running and stop running should cancel out if acceleration rates are the same, just in opposite directions, but that whatever happens at constant running should stick, since you’d be I think just turning chemical energy in one’s body into angular momentum for the earth.
On further thought, I think air resistance is the solution on the second one. It will be a significant factor for people, but negligble for the earth, so people’s angular momentum would be transfered to the atmosphere by way of air resistance in large part, while the earth wouldn’t.
Without air resistance, I believe the two would balance out perfectly, and so when you stop running, there’s no net change, but the real situation energy loss is no longer negligible.
Hey Randall,
This is really random. But I accidentally found your Authors@Google video on youtube, and I must say – you are awesome, and your comics are genius.
Bye.
Hey Randall,
This is really random. But I’m a giant monkey. Whelp, time for me to go sling some more shit around.
Bye.
Hey Randall,
This is really random. But I’m actually a giant monkey. Whelp, time for me to go sling some more crap around.
Bye.
This is an interesting question, Randall. However, you posed two different questions by definition. The first, a query into the straight-line distance average (relativity to squares notwithstanding) involved the superposition of a three-dimensional surface (a sphere) onto a non-euclidean 2-D plane. This is a relatively easy problem to solve. You were right, in part, by recognizing the fact that unaltered straight-line distances were not pure variables in a function. However, the function relating the straight-line variables to a point in a spherical plane is one of dimensional conversion (a prime example of which being the all-too-familiar Rcos(?) ; Rsin(?) conversion between polar and Cartesian coordinate systems).
When you proposed that earthlings that do not dwell on Earth be included in the census average, however, you proposed an entirely different problem. Based upon interpretation, one may choose either to revert to a 3-D euclidean space (yielding a center of population somewhere near the center of the Earth); or, instead, render, for the purpose of calculation, a 3-D non-euclidean space, in which a surface may be rendered, such that all points of consideration (people on and off the planet), as well as the averaged center of population, lie upon it.
I apologize for the brevity and generality of this discourse.
I will post a full analysis when I can get to it.
Android clones,
That comment came from the heart. Honest! But you had to spoil it, didn’t you.
Bye. ;D
Randall, some days you’re one of my favorite people in the world.
Except in bed.
The planet has no edges so there cannot be a surface point that is defined as it’s “center”. Perhaps I’m missing something here, I skipped half the comments, but there is no way to calculate the same “middle point” as in the example for the US which uses a defined area. Some said a few things about balancing the globe on a pin – fine – but the example in the original post really just has nothing to do with the actual problem in that case.
The world’s center of population is IN MA BELLAY