Need help debunking FE video

[AmberRobot]

That's a minor problem. The atmosphere refracts vertically, and you can actually see stuff near the horizon disappear under the right conditions.

The major problem isn't the elevation, it's the azimuth. The sun does not set in the right direction. Like, not at all if you're in Australia, Africa, or South America in the summer (like December), and slightly off almost everywhere else. The light needs to bend sideways for Flat Earth, and it basically never does, because there's no reason for that. FEers can get their followers confused about what we should and shouldn't see vertically, but they stay away from adressing the sideways problem because they know the jig is up once anyone starts questioning that: the only out is to stop believing what you see, and that reeks of brainwashing.

I agree azimuth is also a problem. It's very easy to show that. But altitude is definitely a problem. Middle school geometry shows that. Many people don't appreciate how the sun moves in azimuth but a sunset directly disproves the flat earth model.

there's no "bendy" light theory that explains this photo. Why you can see the sun and the horizon at the same point.

 

[Mendel]

I'm not talking about geometry, I'm talking about Flat Earthers, and what I've seen them believe, via youtube videos and comments on them. You may overestimate how many of them even care about geometry, or understand refraction except in the vaguest sense.

And the fact is that atmospheric refraction defies middle school geometry. If you pronounce your trust in this, prepare to be proven wrong by the next Flat Earther, especially when photographs near large water surfaces are concerned.

But having the sun move in a circle above the equator or thereabouts on the "Flat Earth Map" precludes it being seen in the Southeast or Southwest in Adelaide in December; you need no geometry whatsoever to understand this, and it's a really large effect. You have to avoid thinking about it to remain a FEer.

 

[AmberRobot]

I'm not talking about geometry, I'm talking about Flat Earthers, and what I've seen them believe, via youtube videos and comments on them. You may overestimate how many of them even care about geometry, or understand refraction except in the vaguest sense.

But having the sun move in a circle above the equator or thereabouts on the "Flat Earth Map" precludes it being seen in the Southeast or Southwest in Adelaide in December; you need no geometry whatsoever to understand this. You have to avoid thinking about it to remain a FEer.

I'm not overestimating. I don't think flat earthers care about geometry at all. They don't even begin to demonstrate even the slightest comprehension of it. none of that I've ever seen at least.

And I completely agree with you about the azimuth. It's trivial to show that the sun moves in demonstrably contrary ways to reality just by drawing a circle over a plane (their model).

*We are in violent agreement here.*

A couple of years ago I started interacting with some flat earth proponents/space deniers on the NASA Facebook comments section and how I ended up finding metabunk. ultimately it became a mental health issue for me, a former professional astrophysicist, so I had to stop. I think I'll bow out here, having said my part

 

[Martin SC]

Hi all, a noob here. I hope I am in the right place, if not can an admin please move this post to where it should be.

I am in a series of discussions with FE supporters in the comments sections of a couple of YT videos and the subject of the infamous 8ins/mile*2 formula has arisen. I have found out by asking on Quora that a more correct formula for calculating the curvature of the Earth and what should and should not be visible beyond it is:

h = r – r cos(s/2r)

However my maths has always been terrible and I do not understand that second equation. Can someone here who does know it please explain it so I'm armed with the correct information in exchanges? Many thanks.

 

[Mendel]

h = r – r cos(s/2r)

However my maths has always been terrible and I do not understand that second equation. Can someone here who does know it please explain it so I'm armed with the correct

Visibility depends on the height of the observer and refraction. Consult https://www.metabunk.org/curve/ and scroll down for an interactive diagram.

Your formula is easy to derive:

In radians, α=d/r , and α=(d/r *360⁰/2π) in degrees..
cos(α)=(r-h)/r by definition of the cosine in a right triangle.
Rearrange this to get (r-h)=r×cos(α).
h=r-(r-h)
h=r-r×cos(α)
h=r-r×cos(d/r) in radians and h=r-r×cos(d/r *360⁰/2π) in degrees.

The 8" formula is quite accurate for several hundred miles of distance, but doesn't include refraction, either. You can account for standard refraction in the cosine formula by using a 7/6 (17% larger) radius.

 

[Rory]

I am in a series of discussions with FE supporters in the comments sections of a couple of YT videos and the subject of the infamous 8ins/mile*2 formula has arisen.

Also to remember: 8 inches per mile squared calculates "the drop", which is kind of useless in and of itself. In the early days of flat earthing they often tended to confuse it with the equation used to calculate "hidden amount" - which also includes viewer height - and understandably came up with some very different figures to what reality showed (Eric Dubay was especially guilty of this, and the source for a lot of the related misinformation around flat earth "should be hidden/visible" claims).

 

[Martin SC]

Thank you all, my posting continues. I've taken Mick's information into account as well about "how to debate" so I don't say the wrong things.