So I stumbled across this flat-earth, crackpot post on another board ‘proving’ the earth was flat because water doesn’t curve.
Someone debunked it by posting a picture from space showing the oceans curving across the globe.
But all that silliness aside, it made me curious about how big of a surface area is required for the surface to follow the curvature of the planet.
Instinctually I would guess all surface water curves if you have sensitive enough instruments to measure it. But I’m guessing I’m wrong due to factors I’m unaware of like surface tension, effects of gravity, etc.
So what would be the smallest area of surface water that follows the earth’s curvature, if even by the tiniest amount? The Great Lakes? A local pond? Water in the pan on my stove?
Clearly there is a point where surface tension overwhelms the curvature due to the shape of the gravitational field. Where that lives would be quite interesting to calculate. Intuition suggests it is somewhere in the range of one to one hundred metres, but I’m more than prepared to be surprised.
Anomalies in the gravitational field certainly affect the shape of the surface. Gravity anomalies are mapped in oceans by satellite sea surface mappers. Quite literally it is possible to measure the difference in sea surface height from nominal from orbit and from this determine the local gravitational field, and thus the presence of local density variations. Whether there are any anomalies of sufficient local strength to create a locally flat surface is another matter.
Mount Schiehallion is at the end of Loch Rannoch. This mountain has a particularly special place in the annals of physics. It would be amusing to determine how much it perturbs the surface of the loch.
Normal ocean waves are up to 1/2 mile long. Anything shorter than that, the curve you’re seeing is a dynamic curve, not the effect of local gravity. The “Bedford Canal” experiment demonstrating a round earth was done on a flat stretch of water 6 miles long, using flags spaced at 1 mile.
You don’t need a lot of distance, a few miles of straight canal or a lake seems to do. There is a great YouTube that shows it on Lake Minnewanka near Calgary in Canada. “In Search of a Flat Earth”
One thing I’ve wondered about how flat-earthers like to show photos of flooded salt flats reflecting clouds like a gigantic mirror. I bet you could compare the clouds in the picture vs their reflection and show the lake is slightly convex.
The curvature of the drop is a combination of factors, including (1) surface tension, and (2) “down” on one side of the drop is not exactly the same direction as “down” on the other side of the drop. Right now, those “down” lines will meet at a point about 4000 miles below the drop. If the planet were larger or smaller, such that the distance would be 5000 miles or 3000 miles, then (all else being equal) the shape of that drop would be different. The difference would be too small to measure, but it would meet the OP’s criteria of “even by the tiniest amount”.
Yes, local variations and movement aside, every point on a body of water is “flat” in the sense that it lies on a tangent perpendicular to the line between that point and the center of the Earth (or its center of gravity), broadly speaking. But the next drop over will be on a very, very slightly different tangent line. It’s only when you can see a large enough expanse of water that you are able to see the grander curve with the naked eye.
And in cartography, as well - Charles Hutton, the surveyor with the 1774 expedition to which you refer, which measured Schiehallion to calculate its density, and derive the density and mass of the Earth, took bearings of over a thousand points on the mountain. Then he got the bright idea to draw lines linking points of the same altitude, thus inventing contour lines.
Perhaps (in re salt flats photos), but you have the confounding problem of lens distortion to account for. Yes, there are ways to measure and correct for it, but I feel that trying to account for such a subtle effect, it’s going to be difficult to tease out whether the curvature is partly due to optical distortion. (In a lot of those high altitude photos and videos, the curvature of the earth is quite exaggerated due to the optics. It’s there, just not to the degree it seems, depending on the lens being used. Some lenses have a more “objective” representation.)
And what counts as an “objective” lens representation is entirely a matter of taste, because you can’t accurately represent the field viewed by a lens in a flat picture. It’s the same problem as representing a curved Earth on a flat map.
Back of the envelope calculation, and using sin(x)~=x for very small angles. 1 degree at the equator is about 70 miles; 1 mile is about 1.2 minutes. That should be a measurable amount of curvature. (so 1000 feet would be show 0.24 minutes curvature - draw right triangles using the radius at the center versus tangents at both ends of this pie slice, it looks like the height difference is about 2xsin(0.12’)x500 feet - which, if I can do math I haven’t studied in decades, means about a 1 inch dip over 1000 feet.
Minor nitpick: That would be true if the earth were a sphere. If it were a genuine oblate spheroid of uniform density in spheroidal shells a plumb line in the northern hemisphere would point to a place on the axis south of the center and vice versa for the southern hemisphere. This is because level lines are always perpendicular to gradients (which is where plumb lines point), a standard result of advanced calculus.
I would think that since every drop of water is curved, and oceans are very large collections of droplets of water, it would be fairly straightforward to infer the curvature of water on the earth’s surface.
I don’t know why people try to use that as proof - flat-earthers reject all pictures like this as faked. They tend to rely on only things they can observe themselves. “People in Australia aren’t standing upside down, the earth is flat!”
One of the best arguments I’ve seen against this is observing the stars in different hemispheres. In the Northern Hemisphere, they circle the North Star in a counter-clockwise motion. In the Southern, all of a sudden they’re going clockwise?!
That’s not much of a confounding problem. Lens distortion is easy to counter, and most sophisticated photo editing software maintains a database of the corrections needed for hundreds of lenses.
This is for things like pincushion/barrel distortion, chromatic aberration, etc. Beyond that, there’s computational deconvolution and more. Adaptive optics is sufficiently advanced to be difficult to distinguish from magic.
Most of the lenses used to capture those images are extremely wide—like, the equivalent of 10-15mm for a 35mm camera—and many of them are fisheye lenses. It’s not a side-effect of “the optics,” but rather an intentional tradeoff.
Wide lenses (and especially fisheyes) intentionally distort in order to capture the biggest portion of the sky possible.
Per Chronos’ point, some degree of distortion is inevitable when you project a 3D world onto a planar sensor, but the extreme barrel distortion of a fisheye lens is a different kettle of fish. There are lots of ways to use a camera to measure the curvature of the earth, but barrel distortion isn’t a barrier to any that I’m aware of.
If you’re doing startrail pictures (i.e., photos of the night sky with such a long exposure time that the stars will come out as bright lines), you can even see that in one shot. Look at this photo, for instance (it’s from the website of a German hobby astrophotographer; the caption is in German, but that doesn’t matter). In the top right corner, you see the concentric circles of the stars in the Northern hemisphere, rotating around Polaris. But the celestial equator is already well up high in the sky from that latitude (about 50° N), so in the lower left corner you can see the trails of some of the stars of the Southern hemisphere circling the Southern celestial pole.
This I know, as I’m a photographer and use lens corrections all the time. The ones I’ve used in Lightroom and Photoshop are not 100% perfect and for the level of precision we would need to detect the small bit of curvature on a picture of something like the salt flats – I am skeptical that such level of precision is available – maybe if your specific lens, as in actual lens, were profiled, but I’d have to see proof. Remember, we are talking the very, very slighest of curves. (Also, I’m curious about the math as to what resolution this curvature would be theoretically detectable.)
I’m not only talking about the fisheyes and non-rectilinear wides, but, by “optics” I simply mean the lens and what it does to and how it bends the light. Whether it is intentional or not (and of course I know it is) is irrelevant to my point. The point is lots of people go “hey!!! flat Earthers! Look at this photo or video! See there’s proof the earth is curved!” while pointing to a photo or video taken with one of these lenses, not realizing that the lens itself is causing much of the apparent curvature. Yes, the earth is curved; yes you can see it in photos but in many of the photos you see, some of that curvature comes from the lens being used. It’s a point to be aware of if, for some crazy reason, you choose to get into an argument with flat Earthers (advice: do not. There is no point.)
