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.
superkp: “I think that as long as it makes more than one circuit (girl spinning nonstop more than one rotation, all the people of the earth running around the globe more than once) then the energy that would be canceled would only be from that first circuit, wouldn’t it? she could certainly stop, thereby taking out exactly as much energy as one rotation. but if she had continued for more than one rotation, the energy from that extra rotation would still be present in the system.”
Well, first of all, any energy transfer takes place within the Earth-people system, so nothing outside that system – the orbit of the Earth, for example, is affected. As for internal things like rotation of the Earth, you may be right. Taking the running around the Earth example, going at constant speed around the circumference would have no effect whatsoever. But because of friction, you need to expend energy constantly, which goes into the ground and atmosphere and that goes around to other places… I dunno, it might make a difference.
Wendel: “Actually, they are also made of energy that comes from the sun; so the earth mass should be increasing, little by little – and the sun mass should be decreasing too.”
Not really. Energy and mass are only interchanged in nuclear reactions, and the sun’s energy is used merely to form chemical bonds in organic molecules by photosynthesis. Energy is also being released from the Earth, so there’s little net change. There may be a little conversion of energy to mass, but it’s not the bulk of what the light is used for.
superkp: “I know it is true that all the energy would stay in the system, and hardly cause any change, but there would be -some- change in the orbit, wouldn’t there? just the fact that all that energy is focused on one (theoretical) spot, and there is obviously a slight time delay between jumping and landing, making it that the force from the jump and the force from the land would be in slightly different spots in the earth’s orbit.”
If you removed enough mass and stayed away long enough, you might make the Earth fall into the sun, I guess.
I think that the center of population would be the point at which, setting each individual equal to a vector of magnitude equal to the surface distance from the point and angle equal to direction of said individual from the point, the planar sum of all vectors would equal zero. This compares to the center of mass, where the total torque adds up to zero.
cesium: that’s really all I wanted to hear. Just that with enough effort and organization, we can kill ourselves on a global scale, without a nuclear war.
but once again, I am an english/psych undergrad. don’t know nothing about no physics.
okay. someone asked about what body of water the human population is equal to. correct me if i’m wrong
~ 150 lbs / person
~6.5 billion people
150 * 6.5 billion = 975 billion, round up to 1 trillion
1 gallon = 8.33 pounds
1 trillion / 8.33 ~ 12 billion gallons.
the size of Lake Michigan is about 1.3 quadrillion gallons.
heh heh, there’s 12 billion gallons of people on the planet.
anyways, I agree, center of mass of population will be around India and China. There’s about 4 billion gallons of people living in that area.
Maybe someone’s already thought of this (or, more likely, someone defined it in much more concise math terms I don’t understand), but what if you just kept going with the flat approach? Assume that the earth is an infinite plane that extends in every direction, but only the closest instance of a particular point of mass would be detectable. The result is that every point of mass would be arranged in a roughly spherical (since you are limited to distances, and not coordinates, and the greatest distance would be one half of the earth’s circumference at it’s widest point) pattern around focal points that represented some longitude/lattitude. The search would be for the focal point that results in a perfectly balanced circle.
I believe that using this approximation, you’d could conceivably come up with multiple focal points that balanced perfectly. But I’m not too good at math, so I’m probably wrong.
You could always use an accurate earth map and thus have a flat surface for the original definition.
Ok, I have to put a stop to this usage of ‘miles’ ‘gallons’ and ‘pounds.’ Are we in the dark ages? Do the people here truly understand a system of measurement only used in 2 countries in the entire world? Two countries that can’t even agree on what the definition of a ‘gallon’ is?
Let’s keep this scientific: meters, liters and kilograms kids, that’s what the future looks like.
It would be fun to find the effect that Montana’s reenactment of a maximum state speed limit had on the states populous’ net mass.
It would also be fun to define net mass of humans in eV/c^2.
@Paul; 1trillion=1,000,000,000,000 :. 1tn/8.33lb=120bn gallons; which makes more sense I think that humans would be roughly 6 gallons, rather than two.
“Does anyone have the answer?”
Yes, I : 42 :D
EDIT: “…roughly 18 gallons…” Imperial measurements make my head hurt :(
The center of mass problem is a 1-D problem (on a Euclidean plane), essentially, because it’s the arithmetic weighted mean of the vector positions of a set of points. So you can treat it as points on a line. In that case, minimizing distance clearly doesn’t work, because this would make it an n-D problem instead of a 1-D problem. Consider (for simplicity) equal masses at (0,0), (1,0), and (a,b). If you have a point (x,y), the sum of distances is sqrt(x^2 + y^2) + sqrt((x-1)^2 + y^2) + sqrt((x-a)^2 + (y-b)^2), and if you try to minimize that, you get a complicated expression that isn’t (x,y) = ((a+1)/2,b/2), the center of mass.
The center of mass does, however, minimize the sum of squares, because it minimizes the moment of inertia, which is the weighted sum of squares. Suppose I have points {a_i} in R, i from 1 to N, all with equal weight (let’s assume for simplicity that weight is quantized and if a_r has weight 2, then that just means that there are two identical points a_r in the “set”), and we want to find point x that minimizes the sum of squares. That sum is S(x – a_i)^2, so the derivative is 2S(x – a_i) = 2(Nx – S(a_i)), which is zero when x = S(a_i)/N, the arithmetic mean. This generalizes trivially to n dimensions, since then the “set” would be {(a_i,b_i,…)}, the sum of squares would be S(x – a_i)^2 + S(y – b_i)^2 + …, the derivatives 2(Nx – S(a_i)), 2(Ny – S(b_i)), …, and the solution (x,y,…) = (S(a_i)/N,S(b_i)/N,…). (Oh, S stands for sum over i, but you knew that.)
So there’s your proof that the centroid minimizes square distances, weighted by mass. Another, easier proof that I just came up with is that the vector sum of (X – A_i), where A_i are the position vectors of the masses and X is the centroid, is the 0 vector, so we want NX – S(A_i) = 0, which minimizes whatever quantity has this expression as its gradient, so you integrate that to get the sum of squares.
Stephen Paul Weber: “You could always use an accurate earth map and thus have a flat surface for the original definition.”
But as many have said, that depends on the projection. Since you can’t preserve both distances and angles on a flat map of the Earth, each projection would differ in its centroid. Maybe you could use the minimum (square of) distance approach with a distance preserving projection, but then you might have issues with having to split the Earth down some meridian as well as the poles.
Did we ever discover the meaning of the word list? Is it barrelkid reporting from the metaverse?
Idealogical
————
Say we place the earth on a very large pin, pointing in the z-direction. We neglect the gravity due to earth, and assume a gravitational field in the negative z-direction. Assuming the pin points through the center of the earth and that we have a non-symmetric population arrangement, there are two points of contact where the earth will balance perfectly. One is the center of population, the other is 180 degrees away from the center.
Mathematical
—————
Basically, we need to find the point(s) at which the sum of the torques due to people is zero. So Sum(r_i x (mg)) = 0, where r_i is the distance from each person to the point on the pin, m is the mass of the person, and g is the gravitational field. Since we’re giving everyone equal weight (every person counts as 1 in the census), m is constant and can come outside the sum. The magnitude of g doesn’t matter, and it’s pointing in the negative z-direction, so we can define it to be z = 0i + 0j + -1k. The cross product for any given r will then simply be r x g = -(r_y)i + (r_x)j, where the quantities in parentheses indicate the components of the displacement vectors. This gives us a vector sum, but each set of components must separately equal zero. So we get the two sums: Sum(r_y_i) = 0 and Sum(r_x_i) = 0. Solve that to get your center. Note: I threw out the negative sign on r_y, since the sum has to equal zero it doesn’t really matter.
I know it’s a touch irrelevant here, but I’m going to college in the US population center. I find that both coincidental and awesome.
Summary of solutions we’ve seen so far:
Anup:
Latitude = 29.9168 Longitude = 78.0803
Mine:
Latitude = 30N Longitude = 76E
Jon:
Latitude = 20N Longitude = 80E
Seems consistent. Any answers sufficiently different are probably suspect.
“How about the center of population of xkcd readers? :) You could probably pull that one off using geographical positioning based on ip (however inaccurate). Multiple hits increase the weight.
I?d love to find the centre of population of dumbasses. That would be useful information.”
Same thing. ;-)
Kobayashi_Maru:
It seems to me your ideological solution, at least, merely finds the 3D centroid of the Earth’s population. If you’re translating in the xy plane, there is one point directly under the centroid that balances; if you allow all rigid transformations there is an infinite number of them for which the centroid is on the z-axis, which means it’s not saying much.
Michael Swart:
I got 25.354574 N, 35.496325 E from the method posted earlier, which as I said was in the Red Sea. It’s a simplistic method, so it’s probably very far off, but at least the latitude is consistent.
How about we minimize the sum of cubes or higher powers, so that the center is more likely to lie inside a population clump? As we take the limit, the population center should wind up being under whichever location has stacked the most people over one set of coordinates.
Also, it makes sense that the projection of the centroid to the surface of the sphere is a point which would balance the sphere if the sphere were suspended from that point, since the centroid would seek to hang as low as possible.
Hi all. first post. XKCD. what you are looking for is called ‘least squares’ I hope this isn’t obvious, but what it is is finding a solution that ‘minimises the sums of the squares of the (weighted) residuals’. A simple solution is the mean, but it is used in surveying and geodesy where accuracy and precision are paramount. The above solutions that I read, and the debian one, used a simple average, or mean, which is probably good, but a least squares solution would be better!
Now I’d love to sit down and work this out, probably using first the center of population of each country and apriori starting coordinates from Michael above. then working out the spherical trig for distances to the point (centre of pop) then using least squares to get the best fit solution. but I don’t have time.
I was thinking of a way to actually solve the problem and I remembered a map of what the Earth looks like at night time. If one had a similar map but of population density then you could assign each square kilometer a 3d vector length (proportional to the density) and position on the globe. From there, assuming a round Earth, having a computer calculate then add all the vectors together would create a center of mass for the people on Earth. For English speakers and internet users it would only be a matter of finding a new density map.
Cesium
—-
Upon reading more of the thread, I see that this solution is the same or very similar to things that Hugo, Simon VC, and John Armstrong have posted. I made the stipulation that the pin points through the center of the sphere. In this configuration, if we neglect simply spinning the earth about the z-axis, there are only two orientations that place the centroid (or any other point besides the center) on the z-axis. These correspond to the two solutions I noted. If it weren’t for that stipulation, there would be an infinite amount of solutions, as you pointed out.
Just in case this hasn’t been mentioned already: one “problem” with your definition is that it can, in theory, be multivalued. While in practice, with almost 7 billion people, it is probably quite unlikely that any two points on the Earth (even considered as a perfect sphere) have exactly the same sum of geodesic distances – or squares of distances – a more practical concern is that the centre of population as you’ve defined it for a sphere is unstable. That is, if you move the people around even slightly, the centre of population can shift to (for example) the other side of the planet. Such issues might not be of concern to you, but then again you might want to bear them in mind.
By traveling, you move the center of population a slightest bit. Contribute now!
You forgot the primary rule of tableware!
Beware of Spooning, because Spooning leads to Forking. Beware of Forking, because Forking leads to Knifing.
So few, though, so precious few know that Knifing leads to zombies…
TRH
What if the earth were a frictionless sphere, and the people were proportionately-sized frictionless spheres also. Would gravity eventually bring all the people together in a mass about the centre of population?
@Amarantha
Short answer is most likely no. There are many other effects that go into how a system like this works, but here are a few important ones.
As shown above, the mass of the people on the earth, and thus the gravitational attraction of the people is much less than even a modest mountain.
If we are assuming no imperfections to the sphere (no mountains/valleys, and not even an oblate spheroid) The effect of the gravity of the moon is the next most important. The sun is a close second to the moon.
The main problem is that if we assume frictionless, then there is nowhere for the energy to go, and we will never coalesce around a point. (This is why air molecules don’t end up all lying on the bottom of a bottle of air)
End of long answer: If there was some way to bleed off energy, perhaps we would all gravitate towards the lunar side of the planet, with some perturbations towards the sun. IIRC the effect of the sun is about half the effect of the moon wrt tides. Keep in mind, if you are thinking air resistance, and the air was moving with the earth, it would be going about 1000mph at the equator
Jim: the energy will convert to heat as we bump into each other at the meeting point.
My quick and dirty calculation (let’s pretend all the world’s population is concentrated at 18 points) gives:
20N 73E if you minimise the sum of distances (where you put the school, but no centre-of-mass interpretation);
18N 55E if you minimise the sum of squared distances (consistent with Euclidean centre-of-mass calculation but Gauss knows what it means on the sphere).
Looking at squared distances pulls you closer to the New World, as you’d expect.
We could just do it the easy way:
The center of the world is Jerusalem because God said so. QED.
While I think a mass-based average would be interesting (in comparison to a average in which each person is weighted the same), it may be difficult to get mass data. For the most part, people report their weight, which is different from mass and will, in fact, differ over the surface of the earth due to elevation and the centripital effect being greater at the equator than at the poles. If I remember correctly (because I’m too lazy to work it out) a person’s weight will vary by approximately 1% just by moving from the poles to the equator. I see this as causing a significant skew in the data, although the unreliability in the data may outwiegh (outweigh! hah!) these weight vs. mass differences.
I have a question. Why not just use exactly the same definition as the one from the US Census bureau? An imaginary, flat, rigid, weightless map of the Earth? Obviously, this has the problem of where to cut the spherical surface of the Earth to “fold” it into a flat map. But then again, one could argue that nobody *lives* below 60 degrees South, in Antarctica – you just count the guys there in their respective countries.
Would this work? I think it’d point to somewhere in Russia or Central Asia.
Oh, and I’m thinking of equal weights for every person, just like the Census Bureau.
PS: I forgot to say – I’d suggest the polar projection for my map, centered on North Pole, just like the United Nations flag.
I’ll admit I don’t know any of the math being discussed here (I’m rather young still)
but it seems to me that you are all making it much too complicated anyways. If I were looking for the center of population for the world I would look for a point on the surface of the Earth that would minimize the total distance that everyone would have to travel to get there. Just by thinking a little bit and looking at a globe it seems to me that it would be somewhere in Siberia. Of course no one really lives in Siberia so when you call it the center of population in the world that’s a little misleading. It seems like the simplest way though.
Gratz on PA linking you.
PirateNinja:
Good idea, but there are probably a lot of people on Earth that don’t have lights on at night.
Kobayashi_Maru:
OK, I see. I didn’t see the part about going through the center. But I think that would just be the same as the person who said the point on the surface closest to the centroid – Rob was the first, I think. And then the other would be the antipode of that.
I’ve done something like this with my friends (or rather while my friends look at me with that look that just screams “you’re a nerd”) But I call it the social equilibrium point, or the center where the distance to that point from every person is minimized. I forget what formula I used it was something like:
x=(x1+x2…+xn)/n
y=(y1+y2…+yn)/n
And that should give you the equilibrium point, you can also weight the graph by including people multiple times relative to how much they contribute to the social collective. Then theres the friend that when they show up they throw the equilibrium point to Madagascar.
You can also use the persons distance as a measure of they social…ness by using a simple mid-point formula. I wrote a little program in java that lets you plot where people are sitting and you can weight their socialosity its pretty cool.
I think we should avoid “Mass” altogether.
Mass gets too interesting from a dimensional, cosmological, and background (tides, moons, stars, mountains) point of view. Also complicates things on Sundays for Catholics…
A simple sphere, a massless point for every living human, and a distance to our “center of souls” on earth.
(buried dead bodies don’t count because they fail the “living” test)
I think an excellent way to at least generalize this is to do the original thing of finding the center of mass to each nation. Then, using that generalization of each nation, find the final point using the method above.
From what I recall, you can use group weights. So for the example, the US has a centroid of Missouri, then you can just use that point and have count as one point with a weight of ~300 Million.
All you need to do is get the data for every relevant country, or better yet, continent, and then you can proceed from there. It couldn’t be hard finding the centroid for 7 points :-)
it would be interesting to do this based on average body weight per nation… we could reconfigure the whole system so it favors fat people.
I wonder if this missouri model accounts for the mechanical advantage of inhabitants farther away from the center? Alaska and Hawaii, even with their small populations, would have an immense leverage advantage, and with 30+ million in California added to that, I doubt the center, from a physics standpoint, could be as far east as Missouri.
Also, did they account for the fact that everything (including lovehandles) are bigger in Texas?
if the entire population of the world were to vote on a position for some community center, where everyone is simply voting to try to get it as close to them is possible, would there exist a point on the globe that would win in a one on one election to any other point?
in other words, is there a point for which every other point is further away for a majority of people?
I was going to write that minimising the squares of the distances can’t be the solution, but then I tested it with a simple case (9 people in a group, 1 person at a distance) and it works! Center is 10% of the distance away from the group.
Well, usually the center of a mass is computed just like a mean value. That is for n equally weighted points and in 1d (to avoid typing vectors in an ascii post:
1/n * sum(coordinate(n))
This point then fulfills the “balancing” condition, which means that the sum of moments around it is zero:
sum{ coordinate(n) – centercoord } = 0
Now, in all of these equations, distance from center enters only linearly, so what’s with the sqares?
Minimising squares means in this case:
centercoord = min[ sum{ (coordinate(n) - x)^ 2 } ]
This being a function of first order means it has one extreme point Minimum in this case) where the derivative function becomes zero.
The derivative function is:
sum [ 2*x -2*coordinate(n)]
setting it to zero:
sum [ 2*centercoord -2*coordinate(n)] = 0
you can divide the whole thing by two, invert the sign and there you go, it’s the original equation for the center of gravity.
I assume that this will also work in 2D and 3D space (and so on…)
I’ll try and figure out what this means to the plain sum of walking distances (given more than 2 points) … in another post … right now it looks as if it might lead to the same result
If you want to compute the population center on the surface of a sphere, you’d have to “unravel” the straight surface connections from each point to the center (or point of interest) so they are in a plane, and then compute the sum of distance squares. Expensive computation, there should be a better way. Also, there might be more than one local minimum on a closed surface.
Zak
Can we not simply project from the Earth (treating it as a Riemann sphere with a north pole) to the plane? My instinct says that, since you’re mapping straight lines to straight lines, you can simply map the result back to the sphere afterwards.
I bear news both of failure and success. Or anyway, something nifty. I tried a divide-and-conquer approach to the calculation, and reached a pretty serious contradiction. Take any set of 2^N points. In flat space, it’s possible to find their average position, and therefore centroid and center of mass, by finding the centers of 2^(N-1) disjoint pairs of points, then the centers of 2^(N-2) pairs of centers, then the centers of 2^(N-3) pairs of centers of pairs of centers, etc. This is because the center of mass of the centers of mass of two equal clusters, is the center of mass of the entire arrangment.
Try this on a sphere, though. Take any four points fairly close together, in general position, A B C and D. Let (AB) be the center of mass of A and B on the sphere. This would just be halfway between the two points. Then, ((AB)(CD)), ((AC)(BD)), and ((AD)(BC)) should be the same, and would define the collective center of mass of the four points. They’re all different, though. So, center of mass on the surface of a sphere can’t be well-defined, at least not if you want to keep its traditional properties.
The other difficulty with this approach is that at each step two antipodal centers would be available, leading to some serious ambiguity, but that’s why I was focusing on a local definition.
What’s wrong with the standard formula for (2D) center of mass, replacing x and y with theta and phi?
CJ: “Can we not simply project from the Earth (treating it as a Riemann sphere with a north pole) to the plane? My instinct says that, since you’re mapping straight lines to straight lines, you can simply map the result back to the sphere afterwards.”
But you’re not preserving distances. So the point of minimum distance might be changed.
If it’s any consolation, the Census Bureau got it wrong as well (linked in name, requires AMS access).
I think the method proposed there (population exerts force on surface, surface exerts opposing force, CoP is point where force vector thru centre of earth hits surface) generalises to the whole world but it’s too late for me to think about it seriously.