Proof that the world is not flat

[Whitehole]

I'm reading the book by Zee, I came across a paragraph saying that the world is not flat.

"Given an airline table of distances, you can deduce that the world is curved without ever going outside. If I tell you the three distances between Paris, Berlin, and Barcelona, you can draw a triangle on a flat piece of paper with the three cities at the vertices. But now if I also give you the distances between Rome and each of these three cities, you would find that you can’t extend the triangle to a planar quadrangle (figure 1). So the distances between four points suffice to prove that the world is not flat."

What does he mean by "You would find that you can’t extend the triangle to a planar quadrangle. So the distances between four points suffice to prove that the world is not flat."? I can't depict what he wants to say here, can anyone help explain this?

 

[Orodruin]

You cannot find a point on the paper to represent Rome with all of the distances to scale.

 

[Ibix]

If I give you "crow's flight" distances between three points reasonably close together on the Earth's surface, you can always draw a triangle with sides of those lengths on a piece of paper. If I give you the distances from two of these points to a fourth one you can add it to the diagram (there's an ambiguity about which side of the line between the two points the new one lies, but that's all). However, the distance you measure on your piece of paper between the fourth point and the third will not match the distance on the Earth's surface. This is what the dotted lines in the diagram represent - one of their lengths does not match the "crow's flight"

I'm sure Google will furnish the distances beteen the named cities if you'd like to try it. You only need a ruler and a compass.

 

[Hornbein]

Alfred Wallace of Darwin-Wallace fame won a cash prize for an experiment that proved the Earth wasn't flat.

 

[Whitehole]

If I give you "crow's flight" distances between three points reasonably close together on the Earth's surface, you can always draw a triangle with sides of those lengths on a piece of paper. If I give you the distances from two of these points to a fourth one you can add it to the diagram (there's an ambiguity about which side of the line between the two points the new one lies, but that's all). However, the distance you measure on your piece of paper between the fourth point and the third will not match the distance on the Earth's surface. This is what the dotted lines in the diagram represent - one of their lengths does not match the "crow's flight"

I'm sure Google will furnish the distances beteen the named cities if you'd like to try it. You only need a ruler and a compass.

Oh, so you mean Zee is simply comparing the distance measurements from the paper to the actual distances of those COUNTRIES/points?

 

[Samy_A]

Oh, so you mean Zee is simply comparing the distance measurements from the paper to the actual distances of those COUNTRIES/points?

Assume that, after giving the three distances between Paris, Berlin, and Barcelona, he only gives the distance between Rome and Paris, and the distance between Rome and Barcelona. You can then draw two circles on your flat piece of paper (one around Paris, one around Barcelona) with the given distances as radius (scaled), and you then know that Rome must lie on one of the two intersection points of these two circles.
That will give you two possible distances between Rome and Berlin. None of these will match the figure for the actual distance between Rome and Berlin.

 

[Ibix]

He's comparing the distances drawn on a Euclidean plane to the ones along the Earth's surface and concluding that the Earth's surface is not a Euclidean plane, yes.

 

[A.T.]

Oh, so you mean Zee is simply comparing the distance measurements from the paper to the actual distances of those COUNTRIES/points?

No, just the ratios of the distances.

 

[Whitehole]

No, just the ratios of the distances.

Yeah I got it but what does he exactly mean by saying "you can't extend the triangle to a planar quadrangle"? What does it have to do with the measurement mismatch?

 

[A.T.]

what does he exactly mean by saying "you can't extend the triangle to a planar quadrangle"?

See post #2.

 

[Ibix]

Yeah I got it but what does he exactly mean by saying "you can't extend the triangle to a planar quadrangle"? What does it have to do with the measurement mismatch?

You extend the triangle into a quadrilateral by adding a fourth point. But there is no way to add a fourth point to a Euclidean plane such that the distances between all four points match the distances measured on the surface of the Earth. That's all Zee means. From that, you infer that the surface of the Earth cannot be a Euclidean plane.

 

[robphy]

Following Ibix's suggestion, here's an interesting calculation...

http://www.wolframalpha.com/input/?i=distance+paris+to+berlin (546mi)
http://www.wolframalpha.com/input/?i=distance+paris+to+barcelona (515.2mi)
http://www.wolframalpha.com/input/?i=distance+barcelona+to+berlin (931mi)
http://www.wolframalpha.com/input/?i=area+of+a+triangle+with+sides+546,+515.2,+931 (118474 sq mi)

http://www.wolframalpha.com/input/?i=triangle+with+sides+(rome+to+barcelona),+(rome+to+berlin),+(barcelona+to+berlin) ( 196679 sq mi)
Those triangle areas total to 315153 sq mi.

Using the other pair of triangles,
http://www.wolframalpha.com/input/?i=triangle+with+sides+(barcelona+to+paris),+(barcelona+to+rome),+(paris+to+rome) (136579 sq mi)
http://www.wolframalpha.com/input/?i=triangle+with+sides+(berlin+to+paris),+(berlin+to+rome),+(paris+to+rome) (178933 sq mi)
These triangle areas total to 315512 sq mi.

The disagreement maybe suggestive of not being coplanar, but maybe not conclusive.

http://www.wolframalpha.com/input/?i=distance+paris+to+berlin (546mi)
http://www.wolframalpha.com/input/?i=distance+barcelona+to+paris (515.2 mi)
[not http://www.wolframalpha.com/input/?i=distance+barcelona+to+berlin (931mi) ]
http://www.wolframalpha.com/input/?i=distance+berlin+to+rome (735.5mi)
http://www.wolframalpha.com/input/?i=distance+rome+to+barcelona (535.5mi)
http://www.wolframalpha.com/input/?i=quadrilateral+with+edges+546,+515.2,+735.5,+535.5
[not http://www.wolframalpha.com/input/?i=quadrilateral+with+edges+546,+931,+735.5,+535.5 ]
(gives the area as function of the two diagonals)

Interesting, but I don't have more time to play with this...

(EDIT: some corrections above... and now a little more...)
So the diagonals are
http://www.wolframalpha.com/input/?i=distance+barcelona+to+berlin (931mi)
http://www.wolframalpha.com/input/?i=distance+paris+to+rome (689 mi)

Thus,
http://www.wolframalpha.com/input/?i=quadrilateral+with+edges+546,+515.2,+735.5,+535.5+with+diagonals+931,+689
gives an area of
312608 sq mi [presumably assuming that this quadrilateral with these edges and diagonals lie on a plane?].

If these cities were coplanar, this quadrilateral area 312608 sq mi
should be equal to the sum of the areas of the component triangles,
which gave 315153 sq mi and 315512 sq mi when decomposed two ways.
So, assuming that WolframAlpha is correct with its as-the-crow-flies-distances and area-calculations,
we can conclude that these cities are not coplanar.
(I'm too lazy to find a different way to check these numbers and this calculation.)

 

[Dale]

Yeah I got it but what does he exactly mean by saying "you can't extend the triangle to a planar quadrangle"? What does it have to do with the measurement mismatch?

Another way to see this is to look at the angles. For any three cities you can take the distances and find a flat triangle that fits the distances. If you do that at Rome then you will find that the angle between Barcelona and Berlin is not equal to the sum of the angle between Barcelona and Paris and the angle between Paris and Berlin

 

[Whitehole]

Another way to see this is to look at the angles. For any three cities you can take the distances and find a flat triangle that fits the distances. If you do that at Rome then you will find that the angle between Barcelona and Berlin is not equal to the sum of the angle between Barcelona and Paris and the angle between Paris and Berlin

Are you talking about on the paper or the actual angle?

 

[martinbn]

In Weinberg's book there is a similar problem. But it is about the middle earth. Probably the only thing I like in that textbook.

 

[Ibix]

Are you talking about on the paper or the actual angle?

This one's easier to see with a bigger triangle. Use the north pole and two points on the equator that are 90° apart. The distances are all equal and all three angles are right angles on the Earth. But the angles of an equilateral triangle on a Euclidean plane are not right angles. So the angles in a flat paper map do not match those seen on the surface of the Earth even though the distances do. The mismatch is smaller for smaller triangles, which is why we can get away with Euclidean geometry for small areas - like city maps.

 

[Dr. Courtney]

https://en.wikipedia.org/wiki/Foucault_pendulum

To me, this is the most convincing proof not only that the Earth is not flat, but that it rotates as explained in the rotating spheroid model.

For me, it is more convincing, because it can be done in a single room and does not depend on outside sources of data that may be less reliable.
If that's not a slam dunk, the repeating the experiment at several latitudes should be.

 

[pbuk]

The Earth as a rotating disc also explains Foucault's Pendulum.

 

[Dr. Courtney]

The Earth as a rotating disc also explains Foucault's Pendulum.

How does the model of the flat Earth as a rotating disk explain the dependence of Foucault's Pendulum on latitude?

 

[Dale]

Are you talking about on the paper or the actual angle?

I am talking about the paper angle. There is no way to get the angles to match up on paper.

The actual angle will differ from the paper angle because the surface is not flat. So if you had the actual angle that would be another way to show that it is not flat. You could do that with the distances between three cities and anyone angle.

 

[pbuk]

How does the model of the flat Earth as a rotating disk explain the dependence of Foucault's Pendulum on latitude?

It doesn't, but then the argument that it is more convincing because it can be done in a single room no longer works.

(Edit: added from here)

And if you are going to travel a few hundred miles (North or South) you don't need anything as sophisticated as Foucault's Pendulum to see the Earth is not flat, just measure the shadow from a vertical stick at noon.

 

[robphy]

I corrected my WolframAlpha calculation and
concluded that the cities are not coplanar.

 

[Dr. Courtney]

And if you are going to travel a few hundred miles (North or South) you don't need anything as sophisticated as Foucault's Pendulum to see the Earth is not flat, just measure the shadow from a vertical stick at noon.

Would that really disprove the hypothesis of flatness? Could the Earth not still be flat with that experiment, but be angled with respect to the position of the sun at noon?

 

[pbuk]

Would that really disprove the hypothesis of flatness? Could the Earth not still be flat with that experiment, but be angled with respect to the position of the sun at noon?

How would that give rise to a different length of shadow/angle of incidence at different latitudes? This is how Erastothenes calculated the circumference of the Earth.

 

[pervect]

It may be helpful to consider a simpler example. Suppose you have a square, 4 points, with the property that AB = BC = CD = AD, and equal diagonals $AC = BD$.

Then if the four points are on a plane, we must have $AC = \sqrt{2} AB$.

Suppose instead the 4 points are on a sphere. Then we have the surface notion of the distance $AB_s$ measured along the surface of the sphere, i.e. along a geodesic path (which will be a great circle), and a separate notion of the distance measured along a straight line in 3 dimensional space (a cuve that does not stay on the surface of the sphere, but goes underground). We will call this underground distance $AB_u$, u for "underground". We can find the value of central angle $\theta$ subtended by the arc $AB$ and note that the ratio of the surface distance to the underground distance is depends on the size of the central angle subtended by the arc $AB$, because $AB_s / AB_u = r \theta / r \sin \theta = \theta / \sin \theta$.

We know that $AC_u = \sqrt{2} AB_u$, as the 4 points lie on a plane in 3d space. We also know that the ratio $AC_s / AC_u$ is not equal to $AB_s / AB_u$ because the ratio depends on the central angle, and the central angle is different (it's larger for$AC$ than for $AB$). Thus we can conclude that $AC_s$ cannot be $\sqrt{2} AB_s$.

 

[Ibix]

I corrected my WolframAlpha calculation and
concluded that the cities are not coplanar.

I used the distances you provided to construct triangles of cities and calculate the interior angles of the quadrilateral. Adding up the eight half-angles (e.g. Barcelona-Rome-Paris) gets me a total of 359.6°, which a crude sensitivity analysis suggests is different from 360° if the error on the distances is less than plus or minus a mile.

I also did the same calculation summing the four full angles (e.g. Barcelona-Rome-Berlin). This time I got 360.4°. The difference from the Euclidean expectation is exactly the same as above (to three or four decimal places at least) but in the opposite direction.

Not sure if that symmetry is to be expected. I'm a bit surprised - does it look right to others?

 

[robphy]

Unfortunately, I again don't have time right now... But let me pose a related problem... Given the inter-city distances among those 4 cities, determine the radius of the earth.

 

[Vanadium 50]

Given the inter-city distances among those 4 cities, determine the radius of the earth.

Is 4 enough? Or do you need 5?

 

[BillyT]

Only working vision will do that when there is an eclipse of the moon - the round Earth's shadow shape is very differnent from that a flat disk Earth would usually make and always the same no matter what time of day or night it is.

The ancient Greeks knew this. They also measured the diameter of the Earth by using the fact that twice each year at noon sunlight fell to the bottom of a deep well for a few seconds without hitting the sides. On that day at noon from a know distance due North, they measured the sun's angular difference from straight up. As I recall, they got the correct answer with less than 10% error. A feat not repeated for 1000 years! Their "clock" tell when noon was was the shortest length shaddow of a vertical stick.

 

[PAllen]

After a few takes on this, I have the answer to Vanadium 50's question. My first intuition (finally confirmed) is that distance pairs between 4 points are sufficient to determine radius on the assumption that the points are on a 2-sphere. My reasoning started from part of an argument Synge uses to describe a 5 point curvature detector for spacetime. Part of his argument uses the Cayley-Menger determinant, specifically that this determinant being zero for 5 points is required for flatness of a Euclidean 3-space; and the deviation of this determinant from 0 gives some information about curvature. Reducing dimensions, a non-zero Cayley-Menger determinant for 4 points embedded in 2-surface must give some information about curvature. But if you assume constant curvature, there is only one one number needed, so 4 points should then be sufficient to determine the radius of curvature.

This is born out in the following reference (which initially confused by jumbling two related problems in one paragraph). Equation (3) in said reference, for our question, gives a formula for radius of a 2-sphere embedding 4 points in terms of their pairwise distances:

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.104.4826&rep=rep1&type=pdf

 

[syhprum1]

I have evidence that the world is flat, I see innumerable comments in the scientific press that missiles move in a parabolic path that only occurs when the Earth is flat and of infinite extent.
If the Earth was spherical they would take an elliptical path !

 

[Nugatory]

I have evidence that the world is flat, I see innumerable comments in the scientific press that missiles move in a parabolic path that only occurs when the Earth is flat and of infinite extent.
If the Earth was spherical they would take an elliptical path !

They do take an elliptical path. The parabola is a very good approximation used when the angle subtended by the path of the missile is small.

 

[fresh_42]

The ancient Greeks knew this.

Eratosthenes 200 BC

 

[mfb]

https://en.wikipedia.org/wiki/Foucault_pendulum

To me, this is the most convincing proof not only that the Earth is not flat, but that it rotates as explained in the rotating spheroid model.

For me, it is more convincing, because it can be done in a single room and does not depend on outside sources of data that may be less reliable.
If that's not a slam dunk, the repeating the experiment at several latitudes should be.

The Sagnac effect is a modern version of the experiment. It allows to measure the rotation of Earth with a parts per billion precision, and the axis of rotation can be determined with a precision of millimeters. http://www.fs.wettzell.de/LKREISEL/G/LaserGyros.html

 

[Vanadium 50]

After a few takes on this, I have the answer to Vanadium 50's question. My first intuition (finally confirmed) is that distance pairs between 4 points are sufficient to determine radius on the assumption that the points are on a 2-sphere.

Thanks. I thought about this a bit after my question and came up with this. Three points is clearly too few. Four points takes 6 numbers to specify. You have 8 coordinates of the 4 cities, but one city can be placed at (0,0) and the whole setup can be rotated around the (0,0) point, making 5 - but you also need (as PAllen said) a single number R to specify the radius, so you have 6.

Now, how many numbers do you have? Between 4 cities you have (4 x 3)/2 = 6 distances. So in principle you have enough information. I think you also do in practice, provided the 4 points are convex.

How might this work? If A, B, C and D form a convex quadrilateral, you could find the area of ABC using Brahmaguptra's formula, and the sum of the areas of ABC and BCD using Heron's formula, and note that they don't match. The spherical versions of these are called (I looked this up) Lhulier's and Cagnoli's formulas, and they also include a parameter E from which R can be extracted. It's definitely quadratic, but you want the positive solution for R, so I think it's still OK.

 

[PAllen]

Now, how many numbers do you have? Between 4 cities you have (4 x 3)/2 = 6 distances. So in principle you have enough information. I think you also do in practice, provided the 4 points are convex.

I'm not seeing how 4 non-coplanar points in [Euclidean] 3-space can fail to be convex; or that there are any 4 non-coplanar points in 3-space that you can't embed in a 2-sphere of unique radius.

 

[Vanadium 50]

'm not seeing how 4 non-coplanar points in [Euclidean] 3-space can fail to be convex

Like a chevron.

 

[PAllen]

Like a chevron.

But if it is only 4 points, when you add all the sides it still makes a convex tetrahedron.

 

[PAllen]

There is one problem with my post #30. The logic and conclusions are all fine, but the formula referenced is not actually what we want. The distances used in the formula based on the Cayley-Menger determinant would be the 'through the earth' distances, which are not the ones we want to use. These can be converted in terms of the radius we seek, and you have a complex implicit formula for radius (it being on both sides of the equation). Most likely, it is not solvable in closed form, but could readily be solved numerically.

 

[mfig]

If the Earth is flat, then why has nobody ever taken a picture of the edge?
I don't get it. Are there really people out there who honestly believe the Earth is flat?

 

[WWGD]

If the Earth is flat, then why has nobody ever taken a picture of the edge?
I don't get it. Are there really people out there who honestly believe the Earth is flat?

I think it is more of an abstract exercise in proving it, sort of assuming you are arguing with someone who legitimately believes it.

 

[PAllen]

I think it is more of an abstract exercise in proving it, sort of assuming you are arguing with someone who legitimately believes it.

Also, in terms of motivation of the original source (differential geometry lite for physicists) discussing the minimum quasi-local geometric information needed to distinguish curvature, particularly the idea pure metric (distance) quantities are enough (if you allow angle measurements, you need fewer points).

The answer is you need pairwise distances between 4 points to detect curvature of a two manifold, and 5 points for a 3 manifold. Then we had a further discussion of whether pairwise distances between 4-points allows determination of the radius of curvature of a 2-manifold on the assumption that curvature is constant. I believe I have clearly demonstrated that this is so.

 

[PAllen]

One more note on the idea of using only distances rather than angles is that in GR, the notion of rigid grids whose angles you can measure is highly problematic; even angles between light rays is non-trivial in the presence curvature. This motivated Synge's purely metric approach; the real subtlety of his argument is going from observations about curvature of Riemannian 3-manifolds from 5 points to measuring curvature of a Lorentzian 4-manifold using only round trip times times between 5 world lines; this relies on a whole series of earlier results in the book on Fermi-Walker transport, and it is very surprising (to me) that it is sufficient.

 

[mfb]

Hmm... good gravimeters can measure g with a precision of 10^-9. Let them measure the horizontal component along the line of a laser, and you should get visible deviations within a meter. They can also measure how g decreases with height - again, a meter difference is sufficient. A measurement of the radius of Earth to ~1%, done in a lab, if you know the mass distribution of the lab itself with sufficient precision to account for its effect.

 

[PAllen]

Hmm... good gravimeters can measure g with a precision of 10^-9. Let them measure the horizontal component along the line of a laser, and you should get visible deviations within a meter. They can also measure how g decreases with height - again, a meter difference is sufficient. A measurement of the radius of Earth to ~1%, done in a lab, if you know the mass distribution of the lab itself with sufficient precision to account for its effect.

But a gravimeter is providing direct information about curvature by a different means. So, having determined this, you then measure that it has predictable effects on light paths. That is interesting, but not the same as, e.g. making a series of measurements of angles between light rays and trying to deduce geometry from that. E.g. the idea that you can use sum of angles of triangle, when your triangle consists of light rays following null geodesics of unknown geometry, is an extremely non-trivial undertaking.

 

[mfb]

But a gravimeter is providing direct information about curvature by a different means.

Well, there is no GR solution that would make sense with a flat Earth, but I think if you trust GR then you see enough proofs of a round Earth anyway.
The gravimeters are a GR-independent way to see that the Earth is not flat (at least not relative to a light path).

 

[PAllen]

Well, there is no GR solution that would make sense with a flat Earth, but I think if you trust GR then you see enough proofs of a round Earth anyway.
The gravimeters are a GR-independent way to see that the Earth is not flat (at least not relative to a light path).

On further thought, I think you are not measuring curvature at all this way. The aim is to detect curvature over the 'lab size'. Curvature is divergence from SR spacetime. Change in g over the lab size has only a second order difference from the Rindler flat spacetime prediction. My understanding is that this difference has never been measured on Earth (at lab scale). [edit: Tides, of course, measure curvature over Earth as a whole. ]

Validating GR and measuring a direct 4-geometric affect of curvature are different endeavors.

 

[mfb]

I never said you would measure curvature of spacetime. You measure an inhomogeneous gravitational field.

 

[PAllen]

I think further discussion of space-time curvature measurement should be in another thread, as it is physics of GR. This thread should stay on the original topic of geometric measurement of curvature within Riemannian manifold, specifically the case of a 2-surface.

 

[mgkii]

Yeah I got it but what does he exactly mean by saying "you can't extend the triangle to a planar quadrangle"? What does it have to do with the measurement mismatch?

Hi Whitehole

I don't think anyone actually got around to giving you the answer to your specific question. What he means is if you only use 3 cities and make triangles, then you'll never see a problem; you can take a hundred cities and draw a hundred different "planar triangles" (i.e. triangles on a flat piece of paper) and the Earth's curvature will not "show up"; the curvature will simply "get lost" as each triangle will have a small error in each of its angles that makes up for the "lost" curvature as you translate the triangles from the curved Earth onto the flat planar surface of your paper.

However, if you then go back and add a forth city onto each of your planar triangles - extending each triangle into a planar quadrangle, you'll see the curvature problem come into play - if your original triangle was Paris, Berlin, Barcelona (as in the example you used), then you'll not be able to find a single point for the fourth corner that is the correct distance away from all these three cities to Rome.