There are many ways to see that the Earth is round. Some of them require traveling to specific places.
Question: Where can I go, besides Warren Dunes (see (3) below), to see that the Earth is round in a fun, easy, and convincing way?
Some examples of ways to see the Earth is round:
One argument from antiquity that the Earth is round is to observe boats disappearing over the horizon (you see that the bottom of the boat disappears first, and the horizon gradually progresses up to the top. If you think about the geometry, this only makes sense if the Earth is round.) To see this, you need to travel to a body of water with ships. It takes some effort to do this properly, since you have to keep watching while a boat slowly travels a long distance, and it's only really convincing if you invest some time to watch it happen repeatedly, and convince yourself that the effect has nothing to do with waves or any number of other issues.
I think you can also get on an airplane and look out the window, and maybe convince yourself that the horizon looks curved rather than flat. But especially on a commercial airliner, the view is not really good enough to do this convincingly, and it's very easy to believe that you're just fooling yourself into seeing the effect. (The answer below appears to confirm that even looking out from an airplane is marginal at best for these purposes).
Key example which I'd like to replicate:
- But if you're willing to travel somewhere more specific than just any old airport or seaport, you can see a similar effect to the one with the boats more easily and the experience can be more fun. For instance, go to Warren Dunes State Park in Michigan on a clear day. From the beach, you can't see the Chicago skyline (80 miles drive away, 53 miles as the crow flies across Lake Michigan according to this article) (I'm going by memory -- it's possible you can see the tops of the tallest skyscrapers like the Willis Tower from the beach). But as you climb the large dune behind the beach (probably a few hundred feet high), the Chicago skyline comes into view. Moreover, as you climb, you first see the tops of the skyscrapers, and as you go higher you see lower and lower down the skyscrapers. This makes perfect sense for a round Earth, but zero sense for a flat Earth.
Discussion of the key Warren Dunes example:
There's a lot to consider about what makes this so convincing:
The conditions are pretty specific: the body of water allows you to see very far with an unobstructed view, and the dune enhances your perspective even more. The change in elevation is essential too.
Everything is very static. You can't really argue that the skyscrapers in Chicago are moving or anything.
You can walk back up and down the dune as many times as you like to repeat the experiment.
You can't argue that there's anything obstructing your line of sight. The only thing moving between you and Chicago are the waves on the lake, and they are obviously not tall enough to block out skyscrapers.
The fact that you see the skyscrapers from one spot, and then don't from a very nearby spot means you can't argue that it's haze in the atmosphere blocking your view or anything like that.
It's clear and easy to see when more or less of the skyscrapers are in view. This stands in contrast to the way mountain vistas change as you climb a mountain, because the surrounding mountains are not as vertical as a skyscraper, so the changes in the way they look are more subtle.
Here's a picture of the beach with the dune behind it. Unfortunately I didn't take pictures when I went of the skyline, but here's a news article indicating I'm not the first to consider the round-Earth implications of the view from Warren Dunes, but also indicating that things may be more complicated than I realized (I tentatively think that even if there's significant temperature inversion affecting the view, the fact that the view changes as you go up and down the hill is still pretty good evidence that the surface of the lake is curved.)
Parameters for replicating the Warren Dunes Example:
Tentatively, it seems that the ideal location will satisfy the following basic criteria:
There will be two components: an observing target (e.g. the Chicago skyline), and an observing platform (e.g. the big dune), separated by a distance most likely on the order of 10-100 miles, and both static, non-moving objects.
The observing platform will be something tall which you can go up and down (probably a hill or a building).
From the top of the observing platform, the observing target will be mostly or completely visible.
From the bottom of the observing platform, the observing target will be mostly or completely below the horizon.
The ideal horizon is the only thing substantially preventing you from seeing parts of the observing target from any part of the observing platform. In practice, I think this means there needs to be a large body of water or else a flat expanse of land between the platform and target, and the observing target needs to be much bigger than any small features like waves or trees breaking up the horizon.
(Mountain settings are again problematic for (5) -- the surrounding mountains modify where the horizon is so that it's not set by the curvature of the Earth. Similarly, (5) is an annoyance with the boat example because waves around the boat modify the horizon. In the Warren Dunes example, waves are still present, but they are way smaller than the skyscrapers we're using as an observing target, so they can be neglected more comfortably.)
In addition, there are a few more specific criteria:
You can go up and down the observing platform freely at your own will, ideally in a matter of minutes rather than hours.
You can observe the observing target continuously as you go up and down, or at least at regular intervals.
The observing target is such that it's easy to compare what it looks like when more or less of it is hidden by the horizon (e.g. a vertical skyscraper is better than a complex-shaped mountain EDIT: On second thought -- a complex-shaped mountain might actually be better for these purposes. There are other issues with mountain ranges, but if there's a lone mountain on the edge of an ocean or plain, that might be a very good observing target).
The observing target is an object that you can also easily travel to (In the interest of not making this question a moving target, I will not insist on this, but I think it's desirable.)
(It occurs to me that there are definitely a lot of places you can get a similar effect by changing your observing position horizontally rather than vertically. The problem with these is that you necessarily have to move horizontally a much bigger distance than is required in this "vertical observing platform" scenario. This makes it less convincing, even if you're in a car or something and so it doesn't take much time. It's less convincing because necessarily there's a lot of other things changing about your view at the same time which are distracting. You might also start worrying about atmospheric changes as you move, etc.)
For a start, I bet the Great Lakes afford some other similar opportunities. E.g. from the docks in Toronto, you can see a couple of American cities across Lake Ontario. Maybe going up and down the CN tower or something could allow you to see a similar effect, but I don't actually know (and as noted in the comments, it likely fails some of the "ideal criteria"). I think there are probably examples not requiring water or skyscrapers, but I haven't come up with any.
I asked an early form of this question on Skeptics.SE. It was closed there as not being in scope for the site, but Nate Eldredge suggested it might fit here.
Which distances / heights work for a Warren Dunes-like demonstration?
There are several variables at play in the above setup. Let's consider the most basic ones to get a feel for what's going on. That is, we'll see how it works when we ignore atmospheric effects, which I don't feel qualified to comment on.
Assume the ground is basically "flat" (i.e. does not deviate from a perfect sphere) apart from the observing platform and target -- as long as there are no other hills around, this makes sense. Then geometrically, if you look out from a certain distance above the ground (say ~ 2m if you're standing on the ground, or maybe ~100 m if you're standing on the platform), the horizon is the maximum distance away where you can still see the ground. The distance to the horizon is set by basic trigonometry, which we'll get to in a minute. But the phenomenon at play here hinges on observing not the ground, but the observing target, which pokes out above the ground, so that you can see it even when it's farther away than the horizon. What we really want to know is: If I am observing from a certain height h above the ground, and I'm looking at something of height H, how far away must H be before it's too far past the horizon and I can't see it anymore? The thing to notice is that the situation is symmetric: the observing target disappears underneath the horizon precisely when my line of sight to it just skims the horizon. So:
- The distance from which an observer of height h can see a target of height H is the sum of the horizon distances for an observer of height h and an observer of height H.
So it boils down to the distance to the horizon. After drawing a picture applying the Pythagorean theorem, and dropping some terms which are small because the Earth is big, we see that:
- The horizon distance for an observer of height h on an Earth of diameter D is √(hD) (the square root of the product of h and D -- a.k.a. their geometric mean.
(The diameter of the Earth is D ~ 12,742 km ~ 7,917.5 mi according to Google.)
Putting these together, if h is the elevation (above the ground) that you're observing from and H is the minimum elevation (above the ground) that you can see at the target, and d is the distance from observer to target, then
- d = √(hD) + √(HD)
- H = (d/√D - √h)^2 for h ≤ d^2/D
So as you vary h by going up and down the observing platform, the minimum height H above the ground that you can see at the observing target changes according to this formula. The inequality ensures that the observing target is actually beyond the horizon to begin with. This formula is not bad, but it is complicated enough, and the amount of change in H required to be obvious to the naked eye is dependent enough on further details, that I'm not sure what more to say other than "now just plug in numbers to see what kinds of setups will work".