# Proof that the world is not flat

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1. Jan 28, 2016

### 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?

2. Jan 28, 2016

### Orodruin

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

3. Jan 28, 2016

### 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.

4. Jan 28, 2016

### Hornbein

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

5. Jan 28, 2016

### Whitehole

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

6. Jan 28, 2016

### Samy_A

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.

7. Jan 28, 2016

### 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.

8. Jan 28, 2016

### A.T.

No, just the ratios of the distances.

9. Jan 28, 2016

### Whitehole

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?

10. Jan 28, 2016

### A.T.

See post #2.

11. Jan 28, 2016

### Ibix

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.

12. Jan 28, 2016

### 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)
]
(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+paris+to+rome (689 mi)

Thus,
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.)

Last edited: Jan 28, 2016
13. Jan 28, 2016

### Staff: Mentor

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

14. Jan 28, 2016

### Whitehole

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

15. Jan 28, 2016

### 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.

16. Jan 28, 2016

### Ibix

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.

17. Jan 28, 2016

### 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.

18. Jan 28, 2016

### MrAnchovy

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

19. Jan 28, 2016

### Dr. Courtney

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

20. Jan 28, 2016

### Staff: Mentor

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 any one angle.

21. Jan 28, 2016

### MrAnchovy

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

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.

Last edited: Jan 28, 2016
22. Jan 28, 2016

### robphy

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

23. Jan 28, 2016

### Dr. Courtney

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?

24. Jan 28, 2016

### MrAnchovy

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.

25. Jan 28, 2016

### pervect

Staff Emeritus
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 $\sqrt{2} AB_s$.