Proof that the world is not flat

[fresh_42]

Isn't it a lot easier? How can we have day and night at the same time in different places if Earth was flat?

 

[Hornbein]

Isn't it a lot easier? How can we have day and night at the same time in different places if Earth was flat?

Big squares blocking the Sun? Gravity bending the Sun's rays so they hit they Earth before they get to the nighttime region?

 

[WWGD]

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.

It is interesting how we reach this "breakaway" value of 4 , the least number of points needed to determine whether there is non-trivial curvature. Is there a name for this least value needed?

 

[WWGD]

Big squares blocking the Sun? Gravity bending the Sun's rays so they hit they Earth before they get to the nighttime region?

But how do we then switch from night to day in the dark region?

 

[Hornbein]

But how do we then switch from night to day in the dark region?

The Sun gets teleported back to its starting point? (Once the Earth is flat, anything is fair game.)

 

[mgkii]

It is interesting how we reach this "breakaway" value of 4 , the least number of points needed to determine whether there is non-trivial curvature. Is there a name for this least value needed?

Just throwing a thought on this; I could well be wrong, but I think you might need to think of this from a different starting point. Moving from a 3 dimensional Earth to a 2 dimensional plane by ignoring angles of the triangle is a form of dimensional reduction. You may be able to maintained angles by choosing a different form of reduction, and find a different type of breakaway - i.e. the value of 4 may simply be a feature of the chosen reduction.

 

[PAllen]

It is interesting how we reach this "breakaway" value of 4 , the least number of points needed to determine whether there is non-trivial curvature. Is there a name for this least value needed?

I don't know if there is a name, but the theory can be seen as follows:

You hypothesize your manifold is flat, and is embedded in a flat manifold of one higher dimension. Then, ask what is the fewest points in the higher dimensional flat manifold to make a simplex (polytope) with volume. For 2 space it is 3, 3-space it is 4, for 4-space, it is 5, etc. Then, consider e.g. 4 points in the two manifold. They would form a degenerate zero volume simplex in the embedding 3-space. This means that some expression of pairwise lengths (e.g. the Cayley-Menger determinant) that is monotonically related to simplex volume, must be zero, and this gives a criterion for flatness of the two manifold. Summary: the number of points needed determine flatness (purely by distance measures) is the minimum number to form a simplex in one higher dimension.

 

[PAllen]

Just throwing a thought on this; I could well be wrong, but I think you might need to think of this from a different starting point. Moving from a 3 dimensional Earth to a 2 dimensional plane by ignoring angles of the triangle is a form of dimensional reduction. You may be able to maintained angles by choosing a different form of reduction, and find a different type of breakaway - i.e. the value of 4 may simply be a feature of the chosen reduction.

This is correct. It has already been pointed out earlier in the thread that you need fewer points if include angle measurements.

 

[Mark Harder]

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.

Nice. I think you've hit on the weakness in all the comments I've read so far. Namely, what do we mean by 'distance'? How would one know the 'distance'? By definition, the metric applicable to the Earth's surface is the contour length the great circle connecting any two cities. But this is a non-Euclidean geometry, a spherical geometry. (For one thing, the angles of a spherical triangle do not add up to 180 deg, which may play a role in our problem.) Therefore, the inconsistency between the distances defined by the spherical metric and the Euclidean distances is the result of the naive isomorphism between the 2 geometries. Perhaps there is a projection of the sphere onto the plane which preserves the distances between points on the two surfaces and the angles of triangles. I wouldn't know. But I do know what any 6th grader knows - that the Mercator projection is not the right one - all that business about Greenland being bigger than South America and so forth.

 

[WWGD]

Nice. I think you've hit on the weakness in all the comments I've read so far. Namely, what do we mean by 'distance'? How would one know the 'distance'? By definition, the metric applicable to the Earth's surface is the contour length the great circle connecting any two cities. But this is a non-Euclidean geometry, a spherical geometry. (For one thing, the angles of a spherical triangle do not add up to 180 deg, which may play a role in our problem.) Therefore, the inconsistency between the distances defined by the spherical metric and the Euclidean distances is the result of the naive isomorphism between the 2 geometries. Perhaps there is a projection of the sphere onto the plane which preserves the distances between points on the two surfaces and the angles of triangles. I wouldn't know. But I do know what any 6th grader knows - that the Mercator projection is not the right one - all that business about Greenland being bigger than South America and so forth.

Maybe one can somehow generalize the fact that any projection will produce distortions of some sort to make a general argument on the inequality/inequivalence between Euclidean and Spherical.

 

[PAllen]

Nice. I think you've hit on the weakness in all the comments I've read so far. Namely, what do we mean by 'distance'? How would one know the 'distance'? By definition, the metric applicable to the Earth's surface is the contour length the great circle connecting any two cities. But this is a non-Euclidean geometry, a spherical geometry. (For one thing, the angles of a spherical triangle do not add up to 180 deg, which may play a role in our problem.) Therefore, the inconsistency between the distances defined by the spherical metric and the Euclidean distances is the result of the naive isomorphism between the 2 geometries. Perhaps there is a projection of the sphere onto the plane which preserves the distances between points on the two surfaces and the angles of triangles. I wouldn't know. But I do know what any 6th grader knows - that the Mercator projection is not the right one - all that business about Greenland being bigger than South America and so forth.

There is a physical notion of distance which implements the mathematical notion, and abstracts from any knowledge of the geometry (metric). Then it is valid to ask whether how a set of such measurements can give you information about the geometry.

The physical notion is: take an ideal chain, fix it in e.g. Paris, extend it to Berlin, pull on it till it has no slack, then pull it to one place and measure it. The pulling till no slack implements the definition of geodesic as a distance minimizer.

 

[Mark Harder]

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?

Yes, there was a recent news item about a rapper who was a flat-earther. In my lifetime, there was a prime minister of South Africa who believed in a flat earth. "It certainly looks flat to me!" he is quoted as saying. I surmise he was a fundamentalist with little use for books other than the Bible...

 

[1oldman2]

This thread seems to be taking on a life of its own, or am I imagining that?. Its interesting to see serious people discussing such an off the wall fringe notion. I sense a little entertainment value here.

 

[Whitehole]

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.

Hi mgkii, thanks for your explanation. You've got a point in your first statement, haha. Their explanation kinda pointed out what you are saying and I know what they mean but I wanted to understand what Zee is saying. I know this topic can be explained in a lot of ways but your explanation exactly answered my question. Thanks for clearing it up!

 

[PAllen]

Closing the loop on the original geometric aspects of the problem, I decided to program the method in equation (3) of the reference I gave in my post #30, dealing numerically with the complication described in my post #38. Surprisingly little code was needed (about 45 lines at the level of C, with no use of libraries for determinants or numeric solvers - I programmed those out in direct arithmetic for the specific case at hand, since the matrices are special, with lots of simplifications and a complete factorization is given for one of them in the reference). The expression we sought a root of was so unwieldy (when the arc to chord adjustment is included) that I didn't trust any fancy methods and just coded brute force binary interpolation given root bracketing.

One comment on my discussion with Vanadium 50 on convexity: a tight argument that there is no such thing as non-convex tetrahedron in 3-space. Take any 3 non-colinear points. They form a triangle. Take any 4th point non-coplanar, and the result is a convex tetrahedron. Whether the points embedded in 2-sphere form a convex polygon or not is irrelevant. As concrete demonstration, one of the two solved examples below shows solution for the radius from pairwise distances between cities that form a large chevron on the Earth's surface.

1) Paris, Berlin, Barcelona, Rome using pairwise distances mentioned earlier (P-Be, P-Ba,P-R,Be-Ba,Be-R,Ba-R) = (546,515.2,689,931,735.535.5)
The first thing I was curious about was how much the last pairwise distance would have to change to be compatible with a plane. The code described above includes a solver for this (as well as the radius problem). The not very surprising answer (since the longest distance is < 4% of circumference, so we are still near flat) is if you change 535.5. to 537.74, the distances become compatible with flatness. Given this, and the low precision of the inputs, I was expecting the worst for solving for the radius. I was pleasantly surprised to get 3871 for the radius from these number - not very far off at all.

2) Paris, Berlin, Boston, Cuzco (Peru). These form a concave polygon (in any connection order) on the Earth's surface. With the same convention for giving distances for a list points as above, the distances are: (546,3445,6225,3788,6760,3843). These also span a much bigger area. Not surprisingly, the last distance would have to shift to be 4360 for these to be compatible with flatness. The radius computed from these inputs is 3994, not bad given the limited precision of the inputs - benefiting from the much greater significance of curvature. (correct figure is average Earth radius of 3957).

Thus you can unambiguously compute radius from pairwise distances of 4 points on the sphere, and there is no requirement for convexity of the on sphere polygon.

If anyone is curious about other cases, it now just takes typing in the distances to solve.

 

[Buzz Bloom]

I would like to suggest a thought that might well be considered to be only a nit.

The use the distances between six pairs of cities (using four cities) to PROVE that the Earth is not flat ignores that the practical measurement of these six distances seems to require prior knowledge (e.g., GPS technology) that the Earth is not flat.

Regards,
Buzz

 

[mfb]

You can use a (very long) ruler or comparable devices. That's how length measurements were done before GPS was set up. Triangulation helps as well, but there assuming a flat Earth will lead to trouble already (see the multiple angle measurements).

 

[PAllen]

I would like to suggest a thought that might well be considered to be only a nit.

The use the distances between six pairs of cities (using four cities) to PROVE that the Earth is not flat ignores that the practical measurement of these six distances seems to require prior knowledge (e.g., GPS technology) that the Earth is not flat.

Regards,
Buzz

No, following a straightest possible path locally, where difference from flatness cannot be discerned, is completely well defined and can, in principle, be carried out over any distance. All definitions of intrinsic curvature assume you can perform a variety of operations based on local asymptotic flatness without any apriori knowledge of curvature. For example, in principle, if radar distance between 5 objects in some some non-coplanar configuration were done on earth, and all the pairwise distnances were put in the order 4 Cayley-Menger determinant, the result would not vanish, indicating curvature of our universe. Of course, the precision required for this is unattainable in practice for a laboratory sized device (about one part in 10^30).

Anyway, the OP raised a purely geometric question - they wanted to know how distances between between 4 points could be inconsistent with flatness, as stated in Zee's text.

 

[Buzz Bloom]

No, following a straightest possible path locally, where difference from flatness cannot be discerned, is completely well defined and can, in principle, be carried out over any distance.

Hi Paul:

My admitted nit was intended to raise a question about practicality, rather than what might be doable in principle. If the four cities were too close together, then the distance measurements would perhaps be consistent with flatness. If the cities were far enough apart for the measurements to reveal the non-flatness, then would radar technology be a practical means of measurement?

I am curious whether any current technology not dependent on prior knowledge of non-flatness would be practical to make the six measurements to show non-flatness.

Regards,
Buzz

 

[PAllen]

Hi Paul:

My admitted nit was intended to raise a question about practicality, rather than what might be doable in principle. If the four cities were too close together, then the distance measurements would perhaps be consistent with flatness. If the cities were far enough apart for the measurements to reveal the non-flatness, then would radar technology be a practical means of measurement?

I am curious whether any current technology not dependent on prior knowledge of non-flatness would be practical to make the six measurements to show non-flatness.

Regards,
Buzz

Practical measurements, probably not. I would say the biggest practical difficulty on Earth (rather than a mathematical surface) is lack of local flatness. Even building roads, you always find the need to go around and up and down because of obstacles. But if you could build roads that are everywhere locally flat and straight, then measure them with an odometer, you could get enough precision. The thing is, we don't really live in a 2-manifold; the geometry problem is really answering questions about measurements within a 2-manifold.

 

[Hornbein]

This thread seems to be taking on a life of its own, or am I imagining that?. Its interesting to see serious people discussing such an off the wall fringe notion. I sense a little entertainment value here.

Well, it would be pretty sad if we couldn't prove the Earth was flat, wouldn't it? My fave is applying general relativity to the flat Earth hypothesis.

 

[fresh_42]

Well, it would be pretty sad if we couldn't prove the Earth was flat,

ehm, ehm, on a word ...

 

[Hornbein]

ehm, ehm, on a word ...

Oops. I'd also like to know who that South African flat Earth prime minister was. That was pre-Internet.

 

[1oldman2]

Well, it would be pretty sad if we couldn't prove the Earth was flat, wouldn't it? My fave is applying general relativity to the flat Earth hypothesis.

I think the main point was a slam dunk right from the go, but I have enjoyed this thread immensely. Thus the bonus entertainment value.