Center of Population
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.
May 1st, 2008 at 7:38 am
I think the center of population would be somewhere in the northern hemisphere, because there is more land up here. I also think, it’s near China.
The center of population of native Enlish speakers might be near North America.
The center of population of internet users/bloggers could be between North America, Europe and Australia.
May 1st, 2008 at 7:38 am
If it’s a “balance”, it would be pretty hard if not impossible to balance the earth from the inside of itself…
May 1st, 2008 at 7:38 am
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.
Cheers!
May 1st, 2008 at 7:39 am
Center of the population for hipsters: Brooklyn. I did the calculations.
May 1st, 2008 at 7:40 am
Isn’t that np-complete? I think that’s like finding the smallest square (or cube for that matter) to fit a 2- or 3-dimensional dot cloud. A graphic result would be interesting though.
May 1st, 2008 at 7:44 am
I am guessing the answer is somewhere around the Xinjiang/Tibet/Hindu Kush region of China/India. But I could be way off.
It’s not an insanely hard problem given a population density database. At worst, You can just calculate transit distances to each point for each of a few hundred thousand spots. I don’t think you’re going to hit a huge number of local minima (as you would if the metric were squared distances). Edit: Er, oops.
May 1st, 2008 at 7:49 am
Personally, I think it’s worthless unless we actually take into account the specific weights (generally) of the people involved. Ohio has a higher than average number of morbidly obese people, and that will adjust the balance point. Scoring a child as the same weight as a sumo wrestler just seems as through their really missing a great opportunity. Even better, calculate both, and see how much it shifts.
May 1st, 2008 at 7:50 am
I disagree with your generalization, since it is “unstable” (a slight perturbation could cause the center of mass to shift across the globe, which feels contrary to the idea of “mass” somehow).
But I can’t think of a _good_ generalization (even with all my differential geometry skillz), and ya gotta post about something!
May 1st, 2008 at 7:52 am
I think you want to minimize the sum of the squares of the distances. That would be like a mean. Minimizing the sum of the distances is like a median.
May 1st, 2008 at 7:54 am
You just want the expected position of a person, averaged over the sphere. If you take populations to the continuum limit, then this averaging should be easy to do.
May 1st, 2008 at 8:05 am
Ramsey: Good call! I was actually going to include this link in the post:
http://www.cdc.gov/nccdphp/dnpa/obesity/trend/maps/
but left it out while editing for concision.
It’s a great map, though.
May 1st, 2008 at 8:16 am
What’s wrong with the conventional definition? If you want to map it to a point you can visit by bicycle or canoe, just project the point (on the interior of he earth) to the surface of the earth along the ray from the center of the earth which goes through the calculated center of mass.
This is both easy to calculate and unambiguous/stable, unless somehow the earth’s population is amazingly balanced (the center of mass nearly coincides with the geometric center of the sphere).
My conjecture is that the center of mass for the world population is going to be somewhere closer to Longitude 30-40; this is opposite the middle of the Pacific ocean.
May 1st, 2008 at 8:19 am
Debian’s Centre of Mass is just off the coast of Greenland:
http://people.debian.org/~edward/globe/average/
May 1st, 2008 at 8:26 am
What if you did it as moment problem? Pretend the earth is a sphere in the presence of a “gravitational” field that can pivot. I’m not positive this would work, but I’m imagining putting weights on a sphere and I think it would work for that… So you’d basically be looking for the configuration where the sum of the moments was equal to 0.
Caity
May 1st, 2008 at 8:28 am
What about taking into account human activity on space stations… another interesting factor.
May 1st, 2008 at 8:31 am
> If you want to map it to a point you can visit by bicycle or canoe, just project the point (on the interior of he earth) to the surface of the earth along the ray from the center of the earth which goes through the calculated center of mass.
Does this necessarily give the same result as a minimize-distances approach? It’s not obvious to me either way.
May 1st, 2008 at 8:48 am
Is there necessarily a unique solution (and this is a question whether it is a straight distance metric or a squared-distance metric)? Imagine, for example, that everyone on Earth is distributed along equidistant points along the equator (or any great circle). Depending on the number of people, the set of minimal-distance points might be the equator (or great circle), but it also may be the poles (or their equivalent antipodes for other great circles).If everyone on Earth is divided equally between the poles (or other antipodes), then the set of minimal-distance points are going to lie along the equator (or the great circle between other antipodes).
(It would probably be an interesting challenge to figure out if you could distribute the population so that the set of minimal-distance points has an interesting configuration…)
This is all hypothetical: My guess is that those who predict the surface center somewhere in the PRC (or maybe between the PRC and India) are probably correct.
May 1st, 2008 at 8:55 am
The proof that the center of mass minimizes the sum of squared distances is easy:
To minimize
Sum m_i (x_i-x_0)^2
where x_i, m_i are given and x_0 can be varied (the x’s can be vectors of any dimension), just set the gradient to zero, to get:
Sum m_i x_i = Sum m_i x_0 = x_0 Sum m_i
and thus
x_0 = Sum m_i x_i / Sum m_i
Which is the centroid, or center of mass.
May 1st, 2008 at 8:59 am
I say we just take the earth and try to balance it on a giant pin. We’re not using it for much right?
Why do you suppose the BMI is so high in the middle of the US? I larger amount of space between people perhaps?
May 1st, 2008 at 9:00 am
A similar, but not identical problem has already been solved: the point where one should start to be able to visit every household once in minimum time is somewhere in Kyrgyzstan.
http://www.sweco.se/templates/Page____19592.asp
http://news.bbc.co.uk/2/hi/asia-pacific/7129175.stm
http://www.news.com.au/story/0,23599,22867702-13762,00.html
May 1st, 2008 at 9:01 am
Okay, here’s another question this raises. We’re talking about population “mass”. Is there a corresponding population gravity? Population inertia? If we examine population distributions over time, would we see a gravitational effect drawing them together into cities?
Highways become anti-gravity beams- they created the suburbs by pulling mass from the cities. They’re a gradient pump.
May 1st, 2008 at 9:06 am
I’m just going on intuition here… but I would think that the center would be somewhere around the north pole (skewed towards asia). The Debian Centre was in Greenland for a similar reason, I believe.
I think its easy to forget about the pole in relation to distances on a sphere. I remember an editorial criticizing the President (Bush1 or Clinton) for visiting Alaska on his way to Japan. The author accused him of campaigning on the tax-payer dollar or something, but.. flying through Alaska is shorter than across the US and then Pacific.
May 1st, 2008 at 9:06 am
Also, here’s a proof that projecting the centroid to the surface of the earth would give you the sum-of-squared-distances minimizer:
We want to minimize
Sum m_i (x_i-x_0)^2
with the constraint that x_0 lies on the surface of the earth. Rewrite the above as (where xbar is the unrestricted center of mass),
Sum m_i (x_i-xbar + xbar-x_0)^2 =
Sum m_i (x_i-xbar)^2 + 2 (xbar - x_0) Sum m_i (x_i - xbar) + (xbar-x_0)^2 Sum m_i^2
The middle term goes to zero, because of the definition of xbar, the first term doesn’t vary with x_0, so it’s just a constant, and the third term is minimized by setting x_0 to the point on the surface of the earth that’s closest to xbar. QED.
My bet is on somewhere in Bangladesh as the point.
May 1st, 2008 at 9:44 am
Someone hit something important in an earlier comment, which is that a median might be preferable. If you were going to build a school that served everyone in a certain region, you would want the median along all relevant axes. For example, in a city with rectangular blocks, you want the median street in both directions. If you deviate from there, there are more people adding distance than there are getting closer, so sum of distances is minimized.
Extrapolating this to the whole world, it seems like you would first need to pick a line of latitude above which 1/2 the world’s population lives, which would give you a circular cross-section. Then you’d have to find two other cross-sections along some axis that behaved similarly. It might be possible to find a line of longitude but it might not. It seems very unlikely you would find two different ones. In any case, with two circles you’d have a line through the earth and a third circle would give you a nice warm point somewhere near the middle.
May 1st, 2008 at 9:56 am
What Rob Said. I didn’t check Y.K.’s math.
Intuitively: Rob suggests calculating the 3D center of mass, no matter where you grab the earth (in arbitrary 3D coordinate), letting it hang from that point would result in the center of mass being directly below it.
The line through that point and the center of the earth becomes the axis about which you can spin the population without the (zero weight) earth “wobbling”. Though, I’m assuming it is a sphere, which it isn’t. The two points of intersection of this line with the surface therefore gets my vote, with the nearest being the one I think we should be looking for.
May 1st, 2008 at 10:03 am
Really? I must be missing something. Think of two people… Or think of two people together at one point and one at another… The center of a population minimizes the sum of the square distance, does it not?
May 1st, 2008 at 10:39 am
It wouldn’t be hard to do using spherical co-ods (theta,phi,r) and keeping r constant (or if you are really a keener then you can take into account that the earth is not a perfect sphere) to generate a surface.
Personally I think using population density (people/ sq.km) rather than mass is a good idea, unless you really want to hate on the Americans. The summing the density of each sq.km times a surface distance with respect to some point and trying to minimize that value. I’m not sure if it would be possible to get an actual center of gravity (zero moment about that point) but I’m sure that a point can be found that is a relatively small minimum.
This isn’t a hard problem if you just want to make an approximation. Of course, making a database that had population density data as a function of latitude and longitude would be a little tedious, but once thats there it just comes down to a fairly simple program to iteratively come up with a solution.
May 1st, 2008 at 10:48 am
It’s easy to prove that the centroid minimizes the sum of the squared distances.
To wit:
S = sum of squared distances from x
= m_i * ( x_i - x )^2, summed over i
The derivative with respect to x gives
2 * m_i * ( x_i - x ) summed over i = 0
and so
x = sigma ( m_i * x_i ) / sigma ( m_i ), which is the centroid.
This is a standard result in statistical learning, when we’re best trying to fit a set of points to a Gaussian. Of course, we’re talking Euclidian distances here. I’m not quite sure the “Geodesic_centroid” minimizes the sum of the Geodesic_squared_distances. An interesting problem, at any rate.
May 1st, 2008 at 10:53 am
I would do it like this.
Take the Acutal 3D Centre of mass (which would be in the middle of the planet somewhere)
Then extend a line from the center of the planet, through the centre of mass to the surface. And where that line intersects the surface, call that the Spot.
This will proabaly result in a point on earth somewhere very near the north pole, but i noticed that a lot international flights go over the pole (for obvious reasons) so its probably very close to the Randell defined “Center” anyway.
May 1st, 2008 at 10:58 am
I don’t know the math (a terrible admission on this site), but I pulled this off of worldatlas.com
CONTINENTS (by population) 2005 est.
#1 Asia - (3,879,000,000)
#2 Africa - (877,500,000)
#3 Europe - (727,000,000)
#4 North America - (501,500,000)
#5 South America - (379,500,000)
#6 Australia/Oceania - (32,000,000)
#7 Antarctica - (0)
Asia’s looking pretty tough, huh? Though I’m not factoring in population masses, rather area masses for large swaths of land, would your theoretical center of the world population be in northern africa/southern mediteranean?
Actually, rethinking this, given that more than half of the worlds population is in asia, with africa and europe far outweighing the rest, wouldn’t the center of the worlds population be in the middle east?
Again, to get an exact area, or country, I would need to look into population distributions, but I’m going to arbitrarily point at afghanistan.
May 1st, 2008 at 10:59 am
As someone else pointed out, taking the standard definition and projecting the answer along a radial line to the surface is simpler to calculate and probably just as valid. It might even be the same as your suggestion … I haven’t checked on that yet.
Of course if the population is evenly distributed, this will fail (you will get the exact center, which has no unique projection). However, in such a case, I think the problem is naturally degenerate (when restricted to the surface).
May 1st, 2008 at 11:00 am
Steve is right: you’re thinking too hard.
Yes, on a flat surface you use the same formula to find the centroid of a physical density and the expectation value of a probability density. So when you’re trying to find the expected location you can think of it in terms of the centroid. On the sphere the physical definition of the centroid breaks down, and you’re all panicking. But the expectation value of a probability distribution on the sphere is still perfectly well-defined. Just use the counting measure divided by the current population.
And you can even find a physical analogue. Think of the Earth as a sphere with weights for every person on it. If you suspend it from various points, the vertical line through the suspension point may or may not be a diameter of the sphere, depending on how the weights are distributed. What you’re looking for is the pair of antipodal suspension points where the sphere hangs perfectly vertically.
May 1st, 2008 at 11:03 am
On further review, both Hugo and Simon VC also have the physical interpretation right in different language.
For thems that know, we should be integrating with respect to Haar measure on SO(3,R).
May 1st, 2008 at 11:07 am
Now I’m wondering about applying different weights to the calculation. If we actually use mass as the weight (i.e. center of human mass), I wonder how different the answer will be.
All of the subpopulations suggested (bloggers, english speakers, etc) boil down to a weight of 1 or 0 based on the criteria.
Other interesting weights: wealth? age? height?
May 1st, 2008 at 11:33 am
Well, it’s got to be somewhere in the US given all the obese people you guys have. :-P
Here’s another question, if we got everyone in the world to travel to the same place (say the island of Ireland where I live) and we somehow packed them all in as densely as we could. Then on the count of 3 we got everyone to jump up and down as hard as they could, what would happen the planet?
May 1st, 2008 at 11:50 am
I like that Idea, and I think if we managed to get them to do it once every five seconds during daylight hours, and on the equator, we might be able to finally rid ourselves from the gravitational tyranny of the sun!
now we will be able to pick what we want to orbit! down with Sol!
I’m on a campaign now. I’m not sure if I want Jupiter (not alot of energy guarantees, but it is in the neighborhood), or the Centauri system (TONS of energy, and we can choose which of the three we really want to be most associated with, or even make them overturn it once every four…well, the time period of “year” will have to be redefined.)
May 1st, 2008 at 11:58 am
Well, taking a wild average of 60kg weight per person, that comes to 360 million metric tons of weight (from 6bn people, another average). Interestingly, wiki says that’s about the weight in CO2 emissions annually from fires occuring in coal mines in China. So, we’re fairly insignificant then.
May 1st, 2008 at 12:10 pm
And if 6bn prople were allowed to stand in their very own square meter, they could all stand in 600 km^2, or a square of side 24.5km. The M50 motorway is 45km long…
May 1st, 2008 at 12:13 pm
Well I’m wrong there, but my point was that it’s still a very small surface area.
May 1st, 2008 at 12:46 pm
There are definitely people in Antarctica.
May 1st, 2008 at 12:57 pm
will the max and min be at opposite ends of the globe?
May 1st, 2008 at 1:08 pm
Re: Jeremy
A larger amount of space between people is exactly why I think there are more obese people in the middle/heartland/rural parts of America as opposed to urban poplations. More distance=more driving. More driving=less walking. Less walking=less exercise.
I’m speaking mostly from my own experience on this, so it’s hardly scientific. But when I’ve lived in pedestrian-friendly cities, I’ve lost weight without changing my diet (and actually, possibly eating more). When I’ve lived in more rural areas or places you wouldn’t want to walk around, I’ve gained weight or maintained it. I’d really love to one day see some kind of study on this. I think it would be interesting.
May 1st, 2008 at 1:47 pm
I wish that we weren’t constrained to the surface of the earth.
How cool would it be to have a deep subterranean school serving all children of the earth (somewhere under asia?)…
May 1st, 2008 at 2:28 pm
I did a quick&dirty LP using MS Excel’s solver. I got population data for countries & a center point latitude & longitude from “GRUMP”, an initiative of Center for International Earth Science Information Network
(CIESIN) of the Earth Institute at Columbia University. Of course this presents a problem analogous to calculating “an average of averages”. In order to get a better answer, we would need a finer population “unit” such as a city, or village. Ideally it would be a small grid “square” on the surface of the earth. But for now, using countries as units will have to do.
To calculate Haversine distance (curved paths on the surface of a sphere, not straight lines), I used a commonly used formula:
=6377*ACOS(COS(RADIANS(90-”latitude of source point”))*COS(RADIANS(90-”latitude of destinantion point”))+SIN(RADIANS(90-”latitude of source point”))*SIN(RADIANS(90-”latutude of destination point”))*COS(RADIANS(”logitude of source point”-”longitude of destination point”)))
…and of course each country’s geographic center point was weighted by its population.
Anyway…to minimize the sum of the distances of each country (weighted by its population) to a point, I got an answer :
Latitude = 29.9168
Longitude = 78.0803
Which is somewhere in northern India.
May 1st, 2008 at 2:32 pm
I have been searching but have not been able to find a Buckminster Fuller Dymaxion map with a current world population overlay, which would be helpful here. Projection distortion is minimized, and you could cut the thing out and fold it up into a fair approximation of a globe after you did all the math.
The map is one of the best land mass geography visualization tools ever, too. Note the passage about the WWII approach to Tokyo on page 270 of Your Private Sky, which is on Google Books.
After you locate the center of population, it will be interesting to use the “dig a hole through the center of the Earth” tool on Google Maps to see where the, uh, “center of no population” is.
(When you do this with Cambridge you end up about 650 miles SW of Perth.)
May 1st, 2008 at 2:42 pm
A friend of mine showed me once that if you have a distribution of points in more than one dimension, you can’t find a line segment such that the sum of the distances is constant along the segment. Clearly, the distribution of population isn’t one-dimensional, so I assume you could find a unique centroid by linear distance.
Lori: Except for Detroit… funny, since it has about zero public transportation.
May 1st, 2008 at 2:47 pm
Also, Apotheosis:
I’m not sure on five seconds of thought that the center of no population would be antipodal, but the center of population being in India would put it somewhere in the southern United States.
May 1st, 2008 at 3:06 pm
I suggest finding the population density per square mile. Then get avg. weight per person in geographic region (possibly country wide avg??), multiply by pop density. I figured that people *move* and would constantly shift center of mass and taking averages in a square mile would account for this. Plus, population density figures are easily available.
this way the earth is broken into neat sq. mile grids. If I had taken a math course in the last 2 1/2 years I would probably be able to take this further.
May 1st, 2008 at 3:11 pm
Not related to the blag post, but I used to have nightmares about gravity pulling me sideways.
May 1st, 2008 at 3:45 pm
The Debian map is great evidence for the idea that this centre will be way off where a Mercator map would lead you to think.
If we were doing this for a Mercator 2D world: somewhere very close to China and India is guaranteed … perhaps pulled to the west by Europe and of course the New World.
But if we’re doing it for a spherical / oblate 3D world: I reckon it’ll be pulled north by the ring shape of our population. The southern hemisphere is almost empty of humans compared to the north, and we have concentrations on very distant longitudes. India and China will still make sure it’s close to them, but it could end up in Siberia yet!
May 1st, 2008 at 5:00 pm
I think my posts are getting filtered… I’ll try one last time.
With the huge and terrible assumption that a countries population is centered on it’s capital. (The assumption makes the math easier).
I built a database with statistics from wikipedia and captial locations from some other web site. I cross referenced them and filled in the wholes.
Wrote a function based on straight-line-surface distance and then picked a point and measured to each capital (weighted by population).
First I took latitude/longitude rounded off by 10. This got me 30N 80E.
Then I looked closer and found that 33N, 76E is my best bet. This ends up being in the himalayas. Opposite the pacific ocean… It seems about right.
May 1st, 2008 at 5:33 pm
Using the ever popular Google Maps we could have all participating readers of Xkcd to mark their positions on the map anonymous-wise to collect the data needed to calculate this…maybe?
captcha challenge : what on earth is a faet?
May 1st, 2008 at 6:16 pm
“If you want to map it to a point you can visit by bicycle or canoe, just project the point (on the interior of he earth) to the surface of the earth along the ray from the center of the earth which goes through the calculated center of mass.
Does this necessarily give the same result as a minimize-distances approach? It?s not obvious to me either way.”
I thought about that, but think about this case: 3 points arranged on the equator in an isosceles triangle with one side only slightly shorter than the other two. The center of mass, projected to the surface will be in the middle of the short side. The distance on the surface to the far side is pi*r, the distance to the closer two points is just less than pi*r/3 for a total of just less than 5pi*r/3. Now walk from there to one of the closer two points. The sum of the distance of the closer two remains constant, while the distance to the far point decreases, until reaching the point, where the sum will be 4pi*r/3
Even so, this definition would be useful, it’s just different.
May 1st, 2008 at 6:38 pm
You need to take into account the activity as well, people are not static! we move around. Each person in the world must be acounted as a probability distribution of its weight stretched over the globe.
Of course it would be a pretty sparse distribution with a high pick at home / work but hey, scientific accuracy above all
May 1st, 2008 at 8:22 pm
The centre of awesome is definitely located in Australia.
May 1st, 2008 at 8:35 pm
If we assume the Earth is a sphere, can’t we just find the standard moment of the population (somewhere inside the planet) and then point a ray through that from the Earth’s geometric center? Where ever the ray intersects the planet’s surface, that’s the center of population.
I’d like to know if that differs from the minimized straight lines approach, and if so, how.
May 1st, 2008 at 10:59 pm
Roy has it right. I dunno how we’d calculate the probability distribution of international business tycoons, but I imagine they might be statistically inconsequential.
May 1st, 2008 at 11:57 pm
anup’s answer of (29.9168N, 78.0803E) and the method behind it look promising, if the suggested refinements could be implemented.
May 2nd, 2008 at 12:14 am
anup’s system is a decent first try, but it’s quite skewed by not all countries being the same size. European countries tend to be small and heavily populated, while for example russia is huge… In a situation like this, it can lead to gross miscalculations.
My personal guess would be the centre of mass is in northern russia, not far from the north pole, leaning towards china.
May 2nd, 2008 at 12:27 am
I’m far too confused by this to give an intelligible comment, but I do agree with Ramsey that it would be interesting to see how this changes with weight.
May 2nd, 2008 at 12:37 am
I think before anyone goes through the hard work to calculate the exact CoP (Center of Population), a good rough estimate is in order so that a misplaced minus sign will return a noticeable error. Even if the team (it would really have to be a team) working to find it was really careful, the amount of work involved is likely to be messed up somewhere. It think the majority of their time would be spent tracking down Census data and hoping that each country predicts it’s own CoP.
Okay, rough estimate time:
http://en.wikipedia.org/wiki/World_population
Starting with this possibly accurate Wikipedia page, I’m using the world’s 15 most populated countries, about halfway down the page. I think it’s a fair estimate that the amount of weight provided by countries not on this list will be fairly marginal and we can neglect them for the purposes of this calculation. The total weight of these 15 countries is 65.34% of the world’s total population.
I will also be using this tool:
http://www.zefrank.com/sandwich/tool.html
created or hosted by Ze Frank because it provides an accurate opposite to any location on the planet.
China, India, Vietnam, Pakistan, Bangladesh, Indonesia and the Philippines, being very close, are lumped together and the resulting mass holds 47.73% of the world’s population, which is nearly 3/4 of our 15 countries’ populations. If the 7 countries are the entire population, the CoP is in Burma(n?) territory. Those damn Burmanites.
Now I’m going to add in the rest of the nations in small chunks by starting off assuming those 7 countries are the total population and adding new populations, offsetting the CoP each time by the weight of the addition.
Adding Brazil makes the total weight 50.53 and moves the CoP a startling half a percent of the way across the Earth towards Brazil.
The US and Mexico are lumped together, making the total weight 56.68, and the resulting 6.15 pulls the CoP 1% of the way towards Texas, where supposedly everything is bigger. Now the CoP is somewhere under 1.5% of the diameter of the Earth away from South Burma.
Now all that is left are Ethiopia, Germany, Russia, Japan and Nigeria. If vectors are drawn from each one to through the center of the Earth, because their relative weights are so close, their weight displacements from the center of the Earth cancel each other out, leaving the 5 country CoP at the center of the Earth. If we apply this concept to the current 10 country CoP (under South Burma), it doesn’t move noticeably.
Therefore, if the diameter of the Earth is 12,756.32km, then a very rough 2/3 estimate of Center of Population for the World is 191km beneath South Burma. Time to get a Burmanese Visa and start digging.
May 2nd, 2008 at 12:39 am
Never being one to do work that someone else has done for me, I just did some googling and found this:
http://projecteuclid.org/DPubS?verb=Display&version=1.0&service=UI&handle=euclid.cmp/1104252873&page=record
It’s fairly high-level math (I’m too tired right now to really work through it, but I skimmed the important bits) that details the solution to centroids in curved spaces, including spheres.
May 2nd, 2008 at 12:42 am
Oh, one other nice thing of note is that the paper mentions that one of the nice properties of centroids (that is preserved by the curved definition) is that you can do them piecewise and then take the centroid of centroids, as long as you preserve the masses. I bet you could use this property to approximate the global centroid by finding “flat” centroids of individual countries/continents/regions and then using the spherical formula to find the centroid of centroids.
May 2nd, 2008 at 12:56 am
Can’t you do this using the same method you’d calculate the centroid of any shape? x[c]=Sigma(x[i]*m[i]) / Sigma(m[i]), and similarly for y[c] and z[c].
For your mass value, simply research the average population weight(Or mass if you’re a stickler for terminology) for an area. You could probably go city by city, but that’d be a lot of work. However, going country by country may be too generalized. Maybe a variety at the users discretion is best? Regardless, find an average weight value, and multiply that by the areas population.
Finding your position value may be a bit trickier. Based on latitude and longitude, and elevation above mean sea level, and using Earth’s centre as a your origin, you can calculate the Cartesian coordinates. You’d need to keep a steady reference orientation. For example:
0* latitude, 0* longitude => x=(MSL+Elevation), y=0, z=0
90* latitude, 0* longitude => x=0, y=0, z=(MSL + Elevation)
So knowing the total population weight of an area, based on the centre of that area, you should be able to figure out the centroid of the world’s weight.
Of course, I’m probably missing something and/or over simplifying. But at nearly 1am it makes sense to me.
May 2nd, 2008 at 1:54 am
Going on from the “everyone jump up and down” theory of knocking the earth off balance, if we all start running clockwise can we make the day/night cycle longer via the same principle as running on a log in water?
In my heart I know we’re just too insignificant to manage it, but damnit I can never adjust to 24 hour days. I feel like my circadian rhythm is set to 30 hours or something.
May 2nd, 2008 at 2:01 am
I think that this is an awesome way of defining the center of the noosphere. Every person should probably be considered ‘of equal weighting’ at least until we all agree on a universally applied measure of the ideas people have. Make it a surface calculation and then you can visit the centre of the noosphere and put a library there, or just wear a tinhat and indulge in complete paranoia, after all the other peoples brainwaves would be strongest at that point.
May 2nd, 2008 at 3:11 am
According to this data, which I just sed and C-bashed on the simplistic basis of taking the weighted average of latitude and longitude, the center of population of the world is somewhere in the Red Sea. It’s almost definitely biased, however.
Paul: “this way the earth is broken into neat sq. mile grids. If I had taken a math course in the last 2 1/2 years I would probably be able to take this further.”
Except you can’t really break the earth up into neat grids… you could force it, but that would be getting into the whole projection thing.
Roy: “You need to take into account the activity as well, people are not static! we move around. Each person in the world must be acounted as a probability distribution of its weight stretched over the globe.”
Excellent idea. Let’s decohere our quantum wavefunctions and become probability distributions!
May 2nd, 2008 at 3:20 am
“Going on from the “everyone jump up and down” theory of knocking the earth off balance, if we all start running clockwise can we make the day/night cycle longer via the same principle as running on a log in water?”
The way I see it, you’d have to keep running to maintain the change… as soon as you stopped, the reaction from your stopping would recreate the old conditions. Like when everyone’s in the air, the earth has moved slightly, but as they come back down it comes back. Sorta. Which means the girl spinning doesn’t really change the length of the night, either…
May 2nd, 2008 at 3:44 am
if we told everyone on the planet to jump at the same time, calculating center of mass for that instant when everyone is airborne would be relatively simple. anybody who doesn’t jump shall be considered terrorists and will be dealt with accordingly; thereby negating their affect on our calculations.
May 2nd, 2008 at 3:45 am
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.
For the jumping up and down, I don’t think it does the same thing, because they initially push -off- the earth, theoretically moving it slightly off kilter. When they get back down, that is 6.7billion people -colliding- with the earth, causing it to become slightly -more- off kilter, isn’t it? If we all did that on the bit of the earth that coincides with the plane defined by the orbit of the earth, and only on the day side, once every five seconds, for a year, I would be willing to bet that we could get this hunk of rock and gas to be our personal ship to the other stars. or crash it into the sun, of course, by causing a more quickly-decaying orbit.
I am an undergrad, and an english major, so I do not know much physics and math. I still think what I said makes sense, nonetheless.
by the way, Iron Man is worth the money.
May 2nd, 2008 at 3:50 am
The practical problem I can see is that a large portion of the world’s nations do not have a census, or at least an accurate one. Working with the data available, it probably will be between the major world powers, most likely closest to China. Over one billion serving, and all that.
May 2nd, 2008 at 4:12 am
first, not counting human/animals/organisms, where is the center of the earth? second, how much does the earth weaigh? third, i dont see the human population being enough (compared to earth’s mass) to really skew the center of gravity of earth. and if it did, it would be toward india/china since about a third of the world lives their.
this actually brings to mind an interesting question. how does the orbit of the earth change in relation to population growth? are years getting longer or shorter with the huge population expansion in the last century (something like 3 billion people or more)? or maybe it doesnt. does adding new humans increase mass of earth since they are just made from matter that was on earth in first place?
May 2nd, 2008 at 4:47 am
shouldn’t be the center within the sphere?
i dunno i’m no mathematician.. o.o
May 2nd, 2008 at 5:16 am
My calculations are based on data from GRUMP which gives number of people for each “square” of one degree of latitude and longitude. Then I’m finding the point that minimises the total geodesic distance every person in the world would have to travel to get there.
I end up with 20N 80E, more or less in the middle of India.
That’s not what I would have expected without doing the calculation, so I’ve put a map of population density at http://www.ibbly.com/pop.html as a visual check.
May 2nd, 2008 at 5:21 am
@Cesium, “Excellent idea. Let’s decohere our quantum wavefunctions and become probability distributions!”
Dude. The pickup line is “let’s decohere are wavefunctions and /get entangled with each other/.”
Now I should probably write something that attempts to duplicate “anup”’s solution.
May 2nd, 2008 at 5:22 am
sp:”are”:”our”. I need my morning coffee…
May 2nd, 2008 at 6:44 am
A few hasty calculations lead me to believe that Wolfram’s method will only work for a collection of non-colinear points, or points with different weights.
May 2nd, 2008 at 6:59 am
The definition as “the point who is closest to all humans” is wrong; more precisely, your assertion that it is the same as the isobarycentre in Flatland is already wrong for three point (the point closest to the three summits of a triangle is its Fermat point, not its center of mass).
The definition from Hugo is fine though (take the barycentre in 3D, change to spherical coordinates, change r so that you are on the surface of Earth). But I think this calculation has already been done by several people (at least I’ve seen several definitions of “center of the EU” for instance).
May 2nd, 2008 at 7:01 am
Yeah, you’re minimizing
(x-a)^2+(x-b)^2+(x-c)^2….(x-z)^2
The derivative has to be 0.
0=2(x-a)+2(x-b)+2(c-c)…2(x-z)^2
2a+2b+2c…2z=52x
(a+b+c….z)/26=x
This method generalizes to all numbers, or situations with weights (not that our population needs them).
May 2nd, 2008 at 7:06 am
superkp:
>For the jumping up and down, I don’t think it does the same thing, because they initially push -off- the earth, theoretically moving it slightly off kilter. When they get back down, that is 6.7billion people -colliding- with the earth, causing it to become slightly -more- off kilter, isn’t it?
You’re neglecting the effect the gravity of people has on the earth. If earth pushes you around, its momentum changes by the same amount. So the earth’s momentum is going to be let’s say +1, then fall linearly to -1, then up to 0. The position of the earth in time will be —\ /—
U <- parabola
May 2nd, 2008 at 7:24 am
I do not have anything to add in re maths, but I am filled with an urge to visit that spot in Missouri. Unfortunately, it is in Missouri.
I wonder what the center of population is for scientists. Or mathematicians.
(Not quite related: I discovered the other day that the word “blague” means “a joke or piece of nonsense,” according to the Oxford American dictionary.)
May 2nd, 2008 at 7:29 am
“For the jumping up and down, I don’t think it does the same thing, because they initially push -off- the earth, theoretically moving it slightly off kilter. When they get back down, that is 6.7billion people -colliding- with the earth, causing it to become slightly -more- off kilter, isn’t it? If we all did that on the bit of the earth that coincides with the plane defined by the orbit of the earth, and only on the day side, once every five seconds, for a year, I would be willing to bet that we could get this hunk of rock and gas to be our personal ship to the other stars. or crash it into the sun, of course, by causing a more quickly-decaying orbit.”
I’m a mere first year studying physics (in engineering), but I believe although we are pushing the earth away, we are also pulling the earth with us, due to our mass - gravity. Lets play with some figures.
Mass population (mp): Estimated at 6.7 billion (wiki) * 70kgs (to allow for America) = 4.69 × 10^11 kilograms
Mass earth (me): 5.9742 × 10^24 kilograms
Distance between (d) = radius of earth = 6 378.1 kilometers
Assume the two are points, with exact centres of masses d km apart.
We can use Newton’s Law of Gravitiation, and the gravitational constant G, which is 6.67 × 10^24 m^2/kg^2.
F = G m1 m2 / r^2
So force exerted on earth by the worlds population = 4.59 × 10^12 N.
No small force.
(I hope this is correct, anyone care to work out the acceleration of the earth caused by the population jumping (1) and returning to earth (2)?) thud.
I imagine this obays conservation of energy as we are putting energy into the system by jumping, and pulling the earth (and it pulling us), then pushing the earth (landing) & restoring?
All things
by immortal power
near or far
to each other
hiddenly linked are.
That thou cans’t not stir a flower
without troubling a star.
May 2nd, 2008 at 8:19 am
@Paul
“does adding new humans increase mass of earth since they are just made from matter that was on earth in first place?”
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.
May 2nd, 2008 at 8:47 am
@Wendel
Baryogenesis? Don’t think so.
The Sun loses a huge mass every second thanks to its fusion process (which easily overcomes the mass of objects falling into it), while the Earth gains a little through meteorites and meteors (if their burned up mass is absorbed in the atmosphere), but also loses mass from its upper atmosphere to its tail. Probably a small net loss at this stage of the solar system’s evolution. (Yes: we do have a tail. One of the reasons none of the Apollo missions visited the Moon while it was full.)
May 2nd, 2008 at 8:54 am
oh oh oh oh oh
This is a totally intuitive solution, but it may be the right one:
Find the centre of mass (in the Earth’s core) and the surface centre is the surface point closest to the centre. I estimate this will not be as interesting as it seems, probably on a broad line between India and China, pulled slightly towards Indonesia.
May 2nd, 2008 at 9:34 am
The two methods suggested, distance along the surface of a sphere, and Euclidean distance projected onto the surface of the sphere, do not yield the same results, despite guesses to this effect.
For a simple example, consider the unit circle as a lower-dimensional sphere. Place two points 45 degrees, and 1 at 135 degrees, because I can’t type a “pi” symbol.
By the through-the-sphere method: arccos(sqrt(2) * 6) = ~76.27 deg
By the around-the-sphere method: 75 degrees
May 2nd, 2008 at 9:39 am
I just love that Randall edits this from the shower.
May 2nd, 2008 at 10:51 am
this may be of possible help:
http://www.freebase.com/
although I couldn’t find any data under “population”. This seems like the kind of thing that belongs on freebase mind you.
Going completely off on a tangent, I heard about freebase through boingboing, where this was linked to: http://blog.kiwitobes.com/?p=51
Hooray for the web 2.0 mashup revolution! I for one welcome our new internet overlords.
May 2nd, 2008 at 11:04 am
@Nickel
I don’t think he edits it from the shower, but he certainly is getting epiphanies while he’s in there. I should take more showers.
@Alex
That is an awesome poem.
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.
Its not that I just really want to try this, but I find it hard to believe that that much mass moving in that amazing coordination would have no effect. It would be like a mountain range punching the earth.
…uh-oh…what geographical feature has a similar amount of mass to ~6.7 billion humans? that may give us a good estimation of scale.
May 2nd, 2008 at 11:24 am
I’m starting to suspect that xkcd’s posting these things to keep us occupied in between comics.
May 2nd, 2008 at 11:27 am
It was Archimedes that said (more or less): “Give me a place to stand, and I shall move the world.” Going to Ireland (or any other place) to jump up and down won’t make an difference; we can’t permanently change anything about the earth’s orbit or rotation without interacting with something outside the earth.
Running in circles to change the period of rotation is an interesting idea, but only works so long as everyone keeps running. Once we all stop, the conditions are exactly as they were before. The total angular momentum of the earth/human system remains unchanged.
Now, if we take everyone to Ireland, and then launch them off the earth (taking mass and possibly angular momentum with them) at just the right times, then we can effect the period of rotation, the length of the year, distance from the sun, and various other aspects of our celestial motion. (Though I suspect that the total mass of people + reasonably sized space ships would still be too small to make significant changes; we’d have to start launching spaceships filled with trillions of tons of rock.)
May 2nd, 2008 at 11:39 am
I believe the lines will also form a Delaunay triangulation of the surface. Which is just another way of drawing a voronai diagram.
Of course to use this to solve the traveling salesmen problem for all of earth’s inhabitants you’d have to use great arcs rather than lines. Cause the travel distance would be shorter.
May 2nd, 2008 at 11:55 am
To go back to the “everyone jump at once” topic, you’re all overlooking something. It might work if the ground on which you land was entirely solid. However, it’s not. The force/pressure you apply on the ground will be dissipate through soil particles, causing them to move into void spaces, and any air/water that was in the voids will be moved out. If Earth was solid and continous, the force applied would cause the planet to move. However, you’re just simply causing a much of tiny particles in the Earth to move into “empty” spaces in the same Earth.
May 2nd, 2008 at 12:36 pm
Good point tabernaque86, and it kinda lends a bit more sense to my CO2 comment,
May 2nd, 2008 at 1:07 pm
What an excellent chance for my two favorite blogs to coincide! I strongly recommend someone send this idea with a probably solution to strangemaps.wordpress.com
That place is just a cornucopia for the geography nut.
May 2nd, 2008 at 1:19 pm
You could figure this out by splitting the earth in to hemispheres. Take the hemisphere with the higher total population and note it. Split it at a different hemisphere. Do the same. Do it diagonally and at a bunch of different angels. Where it all overlaps that ought to be the centroid. I don’t have the proper programs to figure that out. So if someone wants to work that out. Find latitude and longitude of a point and a depth below that point
May 2nd, 2008 at 1:51 pm
Use personal travel times. Find the point on the earth that could be reached by every person on the earth in the shortest time. Include transport speeds and delays at terminals, and select appropriate transportation for individuals using their income levels. Calculate this for every year since Homer and watch the point move around the globe with the progression of time.
May 2nd, 2008 at 2:21 pm
I’m seeing some kind of knock-off of that p”eople holding hands around the world” tshirt.
May 2nd, 2008 at 2:48 pm
I hereby present an elegant proof that sum of squares is the correct approach by assuring you that I remember learning something like that in college.
May 2nd, 2008 at 2:51 pm
If everyone jumped up and down on the earth, you’d see the earth move slightly away and slightly towards again, ending where it was. You’d also deform the earth a little bit (make it less/more spherical), if it were indeed a solid continuous object like a rubber ball. The moon does something similar (water moves but so does ground, which is part of why the earth and moon are now locked facing a certain way relative to one another). The change in the earth’s position could theoretically affect the orbit, for instance by moving it closer to the sun for a little while and “trapping” it.
(I got any of this from reading “Bad Astronomy” a few years back and I don’t remember it too well.)
May 2nd, 2008 at 3:29 pm
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.
May 2nd, 2008 at 5:35 pm
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.
May 2nd, 2008 at 5:42 pm
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.
May 2nd, 2008 at 8:15 pm
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.
May 2nd, 2008 at 9:46 pm
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.
May 3rd, 2008 at 12:24 am
You could always use an accurate earth map and thus have a flat surface for the original definition.
May 3rd, 2008 at 1:52 am
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.
May 3rd, 2008 at 5:34 am
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.
May 3rd, 2008 at 7:26 am
@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.
May 3rd, 2008 at 7:34 am
“Does anyone have the answer?”
Yes, I : 42 :D
May 3rd, 2008 at 7:36 am
EDIT: “…roughly 18 gallons…” Imperial measurements make my head hurt :(
May 3rd, 2008 at 1:43 pm
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.
May 3rd, 2008 at 2:15 pm
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.
May 3rd, 2008 at 4:54 pm
Did we ever discover the meaning of the word list? Is it barrelkid reporting from the metaverse?
May 3rd, 2008 at 5:45 pm
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.
May 3rd, 2008 at 7:53 pm
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.
May 3rd, 2008 at 8:46 pm
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.
May 4th, 2008 at 1:09 am
“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. ;-)
May 4th, 2008 at 2:07 am
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.
May 4th, 2008 at 3:41 am
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.
May 4th, 2008 at 8:24 am
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.
May 4th, 2008 at 11:27 am
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.
May 4th, 2008 at 12:11 pm
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.
May 4th, 2008 at 2:29 pm
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.
May 4th, 2008 at 8:51 pm
By traveling, you move the center of population a slightest bit. Contribute now!
May 5th, 2008 at 12:23 am
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
May 5th, 2008 at 12:46 am
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?
May 5th, 2008 at 1:58 am
@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
May 5th, 2008 at 4:28 am
Jim: the energy will convert to heat as we bump into each other at the meeting point.
May 5th, 2008 at 5:35 am
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.
May 5th, 2008 at 1:29 pm
We could just do it the easy way:
The center of the world is Jerusalem because God said so. QED.
May 5th, 2008 at 2:42 pm
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.
May 5th, 2008 at 3:51 pm
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.
May 5th, 2008 at 3:52 pm
PS: I forgot to say - I’d suggest the polar projection for my map, centered on North Pole, just like the United Nations flag.
May 5th, 2008 at 7:18 pm
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.
May 5th, 2008 at 8:03 pm
Gratz on PA linking you.
May 5th, 2008 at 8:12 pm
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.
May 5th, 2008 at 10:53 pm
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.
May 5th, 2008 at 11:47 pm
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)
May 6th, 2008 at 1:05 am
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.
May 6th, 2008 at 12:32 pm
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 :-)
May 6th, 2008 at 1:11 pm
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.
May 6th, 2008 at 2:45 pm
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?
May 6th, 2008 at 8:45 pm
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?
May 7th, 2008 at 3:54 am
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
May 7th, 2008 at 4:12 am
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.
May 7th, 2008 at 4:37 am
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.
May 7th, 2008 at 6:51 am
What’s wrong with the standard formula for (2D) center of mass, replacing x and y with theta and phi?
May 7th, 2008 at 7:02 am
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.
May 7th, 2008 at 7:03 am
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.
May 7th, 2008 at 7:12 am
And I think that’s the same thing as the 3D centroid projected to the surface.
May 7th, 2008 at 7:35 am
34 N 68 E
Hope this wasn’t meant to be the same as my other answer. Will recheck tomorrow.
May 7th, 2008 at 9:32 am
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.
May 7th, 2008 at 3:16 pm
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.
May 7th, 2008 at 4:03 pm
The obvious proof is essentially the definition of torque.
Daniel
May 7th, 2008 at 5:24 pm
Happy 420, Randall!
May 8th, 2008 at 3:09 am
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.
May 8th, 2008 at 3:51 am
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.
May 8th, 2008 at 9:01 am
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.
May 8th, 2008 at 10:07 am
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.
May 8th, 2008 at 12:45 pm
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.
May 8th, 2008 at 1:13 pm
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 ;)
May 9th, 2008 at 12:45 am
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.
May 9th, 2008 at 2:52 am
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?
May 9th, 2008 at 4:32 am
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…
May 9th, 2008 at 4:51 am
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 :)
May 9th, 2008 at 4:11 pm
>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.
May 9th, 2008 at 4:48 pm
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.
May 10th, 2008 at 6:13 am
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.