Earth curvature help please (My brother-in-law is a flat earther and sent me this video about earth curvature.)

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On the other hand, the fact that there is a horizon at all (and that ships do disappear over it) supports the roundness of the earth.

This is my point really.

It cannot be both so small that ships disappear over the horizon after 4.5km, yet also so big that you cannot see any curvature. It must be one or the other.

Whenever I read about "Flat-Earthers" I can't help but think about Eratosthenes who measured the Earth's circumference to 40,000 km which is pretty accurate already 2,250 years ago! And yet, here we are.

Here is how he did it:

Here is how he did it:

I appreciate you posting. I saw that video last year and thought "genius! how easy is that!". However if you ever bother to go down that "flat plane" rabbit hole, you'll see that it does nothing to disprove that theory.

I'm not here to argue for that theory, but to establish whether the math can support a sphere this size with the repeatable, observable evidence I can collect myself, at ground level.

Curvature should be a given - on a sphere this size. No other proofs for the globe can make up for their lack of proof for curvature.

Since we're sharing videos, here is an amateur high altitude weather balloon video up to an altitude of almost 110,000 feet. You'll notice the horizon will look convex when the camera tilts down, and concave when it tilts up - however when it's not bobbing the horizon is still flat all the way.

This one is uncut, all the way to the stratosphere

No matter the explanation, our world is beautiful! Have a great Easter, thanks for talking to me.

It cannot be both so small that ships disappear over the horizon after 4.5km, yet also so big that you cannot see any curvature. It must be one or the other.
What do you mean by that? You cannot see curvature - you can feel it intuitively. It is almost like fourth dimension - you can prove it exists. Radius can be measured in terms of length (kM) and easily envisioned. Informally speaking, the "measure" of curvature is the reciprocal of radius.

For building say one/two storied houses - earth's curvature does not affect the design. Earth is considered flat. For building Eiffel tower - you would consider earth's curvature - otherwise you might end-up with a Leaning tower. of course the "leaning of leaning tower of Pizza" is not due to curvature of earth.

You'll notice the horizon will look convex when the camera tilts down, and concave when it tilts up - however when it's not bobbing the horizon is still flat all the way.
Surely you recognize lens distortion when you see it changing before your eyes. How can you trust your impression that it looks flat, when it sometimes looks concave, which is obviously wrong?

Go high enough (on a rocket, not just within the narrow layer of atmosphere), and you can see the entire globe; I take it you don't believe satellites are real?

I appreciate you posting. I saw that video last year and thought "genius! how easy is that!". However if you ever bother to go down that "flat plane" rabbit hole, you'll see that it does nothing to disprove that theory.
Yes, I myself have talked about how Eratosthenes' experiment (by itself) doesn't prove a round earth, but is based on the (already accepted!) understanding that the earth is round. You can explain the one pair of observations in terms of the sun being above a flat earth.

But if you make three measurements, not just two, you will find that the data do not fit a flat earth with the sun some distance above it.

In particular, if you went to the pole, you'd find that the sun appears to be on the horizon, at a zero distance above the earth. If the calculation based on a flat earth gives different values for the distance to the sun, it must be a faulty model.

And if you went a great distance to the east or west, you'd discover time zones when you tried to coordinate observations, showing the earth is curved in that direction as well.

I'm not here to argue for that theory, but to establish whether the math can support a sphere this size ...
As has been explained, the math has to be supplemented with physics, because light doesn't always travel in mathematically straight lines. Combined, it all works fine.

... with the repeatable, observable evidence I can collect myself, at ground level.

Just walk east as far as you can, and you'll come back where you started (in principle). Try that on a flat earth.

Or, again, just watch a ship disappear over the horizon. They really do, you know!

Cogito ... full marks for being half-honest! Note that per the calculations, the commentator states that the radius has to be much larger than that officially stated i.e. the experiment with the camera (marks/points for that too) implies 2 things:
1. The earth is spherical BUT bigger than we think it is (high probability)
2. The earth is flat (we don't see beyond the reference rock. If the earth were truly flat, we should be able to have a panoramic 360 degree view of the entire coastline of all other countries that are at the opposite end)

Or, again, just watch a ship disappear over the horizon. They really do, you know!

Well I've zoomed ships back into frame that were "meant" to be over the horizon. They would, as you say, all need to be mirages to allow for a globe this size. Perspective on a flat plane would also make ships appear to disappear from the bottom up as they sail away from you. So it's not a good proof for either argument (as with the Eratosthenes' experiment).

Go high enough (on a rocket, not just within the narrow layer of atmosphere), and you can see the entire globe; I take it you don't believe satellites are real?

My main observation was that I should see curvature from left to right *at ground level* on a sphere this size. The fact that there is none does not allow for a sphere this size. This has not yet been explained. I'm really after experiments that I can repeatedly measure/observe myself (I don't have a rocket, or the ability to walk on water). The video was for interest, not proof, as I can't speak to how it was made.

Cogito ... full marks for being half-honest! Note that per the calculations, the commentator states that the radius has to be much larger than that officially stated i.e. the experiment with the camera (marks/points for that too) implies 2 things:
1. The earth is spherical BUT bigger than we think it is (high probability)
2. The earth is flat (we don't see beyond the reference rock. If the earth were truly flat, we should be able to have a panoramic 360 degree view of the entire coastline of all other countries that are at the opposite end)

With point 2. - the atmosphere becomes opaque over vast distances. But on very clear days you can see much farther than you're "meant" to. I'd be interested to compare over dry vs humid (ocean vs salt lake).

Curvature from left to right, however, should be very obvious especially when standing at the shoreline of an ocean.

Perspective on a flat plane would also make ships appear to disappear from the bottom up as they sail away from you.
Please explain how that would work. It wouldn't. You have the burden of proof here.

My main observation was that I should see curvature from left to right *at ground level* on a sphere this size.
Curvature from left to right, however, should be very obvious especially when standing at the shoreline of an ocean.
Please prove that. It just isn't true. The formulas for curvature are about curvature going away from you, not side to side.

Imagine standing in the middle of a hula hoop with a 3-mile radius (about the distance to the horizon). You won't see much curvature, if any at all; as you rotate, you'll always see it at the same (very small) angle below your eye level.

Sorry, but you're believing nonsense.

Please prove that. It just isn't true. The formulas for curvature are about curvature going away from you, not side to side.

Imagine standing in the middle of a hula hoop with a 3-mile radius (about the distance to the horizon). You won't see much curvature, if any at all; as you rotate, you'll always see it at the same (very small) angle below your eye level.

Sorry, but you're believing nonsense.

Imagine standing on a ball - the edge would look curved.

All I am believing, Sir - is that a sphere curves in every direction. According to the math for a sphere with this radius - there would be 360meters of curvature across a distance of 68km.

Imagine instead that I am standing directly between my 2 destinations. When I turn to face the city the curvature curves down and away. I turn 180 to face the island and I also looking at something curving down and away. Slice that sphere in half along that line, and you have the curvature between those 2 points, an arc with a curvature height of 350meters (half either side of me). Why would that arc ever become flat? A sphere curves across the whole surface in every direction at the same rate.

Imagine standing on a ball - the edge would look curved.
On a small ball, like that silly picture, sure! Or the (round) earth as seen from the moon. But not on a large ball, when you're a tiny percentage of its radius off the surface. Are you really doing your thinking based on such pictures?? You need to start imagining reality.
According to the math for a sphere with this radius - there would be 360meters of curvature across a distance of 68km.
Suppose you're right; 360 meters is only 0.5% of 68 km. (360/68000 = 0.0053) That's hardly perceptible. Have you seriously thought about the numbers?

Here, I've drawn a circle (red) with a radius of 6378 units, and put two points (green) on it 34 units to the left and right of the point (red) at the top of the circle, for a total distance of 68 units. The green broken line joins the two points. Can you tell the difference between the line and the circle?

That is perhaps what you are seeing (though, again, your calculation is not really appropriate).

We can see that the actual rise between the points is 6378 - 6377.909 = 0.091 km, or 91 meters, about 1/4 of what you say. Going to the curvature site you used, it appears that you entered 68 km and got 362.89 meters, which is the drop to a location 68 km away from you. To get the bulge across 68 km, you need to enter 34 km (from the middle to either end), and you'll get 90.72 meters, in agreement with my graph.

Do you still want to make your claims? The world is bigger than your imagination.

Perspective on a flat plane would also make ships appear to disappear from the bottom up as they sail away from you.
I still want an explanation of this claim. How does perspective make an object at a distance on a flat plane seem to go below the plane??

Did someone make this claim to you, and you believed them? Or did you just make this up?

I'm genuinely curious, not just trying to make fun of you.

On a small ball, like that silly picture, sure! Or the (round) earth as seen from the moon. But not on a large ball, when you're a tiny percentage of its radius off the surface. Are you really doing your thinking based on such pictures?? You need to start imagining reality.

Suppose you're right; 360 meters is only 0.5% of 68 km. (360/68000 = 0.0053) That's hardly perceptible. Have you seriously thought about the numbers?

Here, I've drawn a circle (red) with a radius of 6378 units, and put two points (green) on it 34 units to the left and right of the point (red) at the top of the circle, for a total distance of 68 units. The green broken line joins the two points. Can you tell the difference between the line and the circle?

View attachment 37583

That is perhaps what you are seeing (though, again, your calculation is not really appropriate).

We can see that the actual rise between the points is 6378 - 6377.909 = 0.091 km, or 91 meters, about 1/4 of what you say. Going to the curvature site you used, it appears that you entered 68 km and got 362.89 meters, which is the drop to a location 68 km away from you. To get the bulge across 68 km, you need to enter 34 km (from the middle to either end), and you'll get 90.72 meters, in agreement with my graph.

Do you still want to make your claims? The world is bigger than your imagination.

I am glad that you are still genuinely curious and asking. I too want to understand your perspective, because from where I am at the moment the horizon should be curved. And I can't "unsee" it. I would prefer to remain "sold" on the sphere model. I love sci-fi and the flat model does not make life easier, or better, it just raises very uncomfortable questions. But this curved horizon thing is what I can't get past.

I don't feel the picture with the dog was any more silly than imagining a hula hoop around my head. And I used it to help with visualising the curved horizon we should see.

Yes - I have thought about those numbers. When I'm standing on this beach, the island opposite me has 2 sand dunes facing me (I know them well). The taller sand dune (according to topographic maps) is 50 meters tall. I can very easily perceive its height. The tallest peak on the island is 280 meters, that height is also very obvious opposite me. The island is not some slim blur on the horizon. Why would 360 meters (or 150meters of drop from center to A and center to B) be imperceptible?

I still want an explanation of this claim. How does perspective make an object at a distance on a flat plane seem to go below the plane??
Did someone make this claim to you, and you believed them? Or did you just make this up?

I tried this experiment at home, but with a small ball. This shows "the how" and you can try it out yourself.

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I tried this experiment at home, but with a small ball. This shows "the how" and you can try it out yourself.
Ummm ... the camera is obviously below the edge of the table most of the time! Of course the object is hidden when it's farther away -- but hidden by the near edge, not by a non-existent horizon.

When you look at an island, are you below sea level?

Try actually thinking about how perspective works; draw a picture of where the light rays would be going, and try explaining your claim.

But this curved horizon thing is what I can't get past.
And yet you haven't answered what I said about it. Do you not see that the curve is too small to notice? (Seeing a single object at a distance is very different from seeing a broad curve.)

I don't feel the picture with the dog was any more silly than imagining a hula hoop around my head. And I used it to help with visualising the curved horizon we should see.
What's silly about the dog picture is that the scale is so different from what you were claiming that it's irrelevant. It's the specific numbers that matter, not just the idea of imagining.

Ummm ... the camera is obviously below the edge of the table most of the time! Of course the object is hidden when it's farther away -- but hidden by the near edge, not by a non-existent horizon.

Did you watch to where it zooms in, and the base of the piece is visible again? The table top itself is also marginally visible then, showing that the line of sight of the observer/camera is slightly above the table as well.

The formulas for curvature are about curvature going away from you, not side to side.

When you said this further up, it didn't make sense to me as I didn't understand why I couldn't use the formulas 'left to right' on a sphere. So the 91m drop you proposed with your graph - didn't make sense to me either.

However, I think I understand what you mean now! Because I Myself am standing between the two points (albeit West of them), the curvature will still be 'as if' I am looking from Me to A, or from Me to B? I can't experience the full curvature from "A to B" unless I am standing at one of those points? And the curvature calculator can be used, but I need to put myself in the middle when it's assessing curvature from left to right? (which halves the distance I can input). Is this correct?

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