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.
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.
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?
“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.
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
The centre of awesome is definitely located in Australia.
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.
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.
anup‘s answer of (29.9168N, 78.0803E) and the method behind it look promising, if the suggested refinements could be implemented.
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.
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.
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.
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.
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.
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.
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.
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.
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!
“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…
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.
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.
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.
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?
shouldn’t be the center within the sphere?
i dunno i’m no mathematician.. o.o
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.
@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.
sp:”are”:”our”. I need my morning coffee…
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.
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).
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).
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
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.)
“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.
@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.
@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.)
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.
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
I just love that Randall edits this from the shower.
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.
@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.
I’m starting to suspect that xkcd’s posting these things to keep us occupied in between comics.
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.)
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.
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.
Good point tabernaque86, and it kinda lends a bit more sense to my CO2 comment,
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.
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
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.
I’m seeing some kind of knock-off of that p”eople holding hands around the world” tshirt.
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.
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.)