1 - Technically speaking buoyancy is a force.

2 - He has no idea what he is talking about.

He wants to say the 8 inches per mile squared must be wrong (with the clear implication that the FEers using it are doing so when a different formula would give significantly different results), and provides the "correct" formula which is h=r-r*cos(s/2r)

However, he didn't even bother to check what result that gives and compare it to the other formula.

His dismissal of the miles squared being out of place is entirely wrong as well, because he isn't actually representing it properly and has no idea where it comes from.

It isn't out of place at all.

And the way he dismisses it as just coming from the FE bible without people being able to derive it is pure BS.

It is quite simple, take the actual formula:

h=r(1-cos(s/r))

That is for the drop due to curvature.

The formula they used only works for an observer and thing to be observed at equal height.

In reality you need to take that formula and apply it twice.

But now with the actual correct formula, make an approximation, specifically the small x approximation for cos

cos(x) ~= 1-x^2/2

Subbing that in we get:

h=r*(1-(1-(s/r)^2/2))

h=r*(1-1+(s^2/r^2)/2))

h=s^2/2r

And notice a key thing here? It is distance SQUARED, i.e. that squared, in its place, just like it should be.

It seems out of place because we have height corresponding to distance squared. But that is because you need to divide by the radius of Earth.

But why stop there. Why not make it simple for those who hate the metric system and math.

We want to get a height in inches and a distance in miles.

So instead we will rewrite it as s = S miles. (So we put in a pure number S, as S=s/miles) So if we have a distance of 10 miles, we have S=10 miles/miles = 10, so we just put in 10 into the formula.

That gives us:

h=S^2 miles ^2 / (2 r)

Now sub in the radius of Earth (6963 miles) and 1 mile=63360 inch.

h=S^2 miles * 63360 inch / (2* 3963 miles)

h=S^2 * (63360/7926) inch

h = (s/miles)^2 * 7.99394...

Now that 7.99394 is awfully close to 8.

Almost as if that is where it came from.

If they actually understood what they were talking about, they would understand where this came from, or at the very least, try to verify how well it works.

And they would see it works just fine for a small enough angle. You need to get to hundreds of km to start seeing a significant difference.

He also dismisses it as a "parabolic approximation", yet doesn't seem to connect the dots.

He also places a parabola over Earth, but just a thin one, not even attempting other possible parabolas, or trying to see what the parabola in question would look like. If he did that he would get something like this:

The parabola is in green, the circle representing Earth is in red.

Notice how well they overlap at the top of the image?

So if you are dealing with short enough distances, 8 inches per miles squared is fine (except the use of archaic units).

3 - Yes, water curves, but this depends on if FEers are saying water is observed to be flat over large scales, or if they are just saying it must always be flat.

4 - His explanation for why makes pretty much no sense and is wrong (other than the very over simplified "because gravity exists").

It really does nothing to address the question and instead just restates it in another way. Why is air thinner the higher you go? Not because of kinetic energy; but because it has less air above pushing down on it. Gravity acts on the air as well, compressing it close to the surface.

5 - It doesn't make much sense to say you should measure it in terms of angular velocity and then say the ball is small so it doesn't stick to it.

If you keep the density constant, the same angular velocity should produce the same ratio of gravitational force to centrifugal force.

6 - Another example of not understanding what they are arguing. No, it isn't meant to be showing Earth is flat, it is trying to refute arguments in support of a RE, specifically the ones relating the moon reflecting the sun's light, such as those showing how the phases of the moon relates to its position, which would require the moon and sun to be opposite each other during a full moon, placing the sun below a flat Earth during a full moon. This is likewise done with the moon terminator illusion.

8 - This doesn't fully address it, as it just ignores it and acts like distance doesn't have any impact. It also ignores the fact that during summer the summer portion of Earth receives more daylight hours.

9 - Again, this doesn't actually address the FE arguments.

They are saying that lots of people just accept Earth is round because they have been told. It actually ties into the 10th point.

And there is some element of truth to this.

In fact, even relating to the point above, plenty of people think that it is hotter in summer because that side of Earth is closer to the sun, appealing to distance.

We have even had people come to this forum and claim that the phases of the moon prove Earth is round because the phases are caused by Earth's shadow on the moon.

And we have some people that will dismiss the FE, regardless of what argument is made because science shows Earth is round. They don't even bother trying to address the argument.

10 - Except plenty of FEers contradict each other. Partly because Earth isn't flat and they start with different things to address reaching a contradictory position.

Some say things fall because density. Others say Earth is accelerating upwards. Others have an infinite FE with gravity.

Some say light travels straight, so you can see things really far away, right down to the bottom. Others say light bends upwards to explain why the bottom of distant objects are hidden.