If you aren't willing to even put in the effort to list what the points are, why should anyone put in the effort to respond to it?

Did you even put in the effort to search for this video?

Literally copying and pasting the URL into the search will get you a few threads.

e.g. this one:

https://www.theflatearthsociety.org/forum/index.php?topic=82146Did you bother reading through that thread to see what people have said?

But this is just a crappy video to try to make fun of FEers; and has several incorrect statements.

e.g. point 2 is the most BS claim of his.

He claims that 8 inches per mile squared is wrong, and claims that it is a parabola which in no way represents Earth.

He suggest people learn basic math, while entirely failing to understand basic math himself (although I suppose the math he is objecting to is slightly more complex).

He also complains that people post it with no explanation of how it works or why.

Yet what does he do? He posts a formula with no explanation of how it works.

He accepts that formula should be h=r*(1-cos(s/2r)).

But this is wrong.

This is the formula for a bulge between 2 points, rather than the drop from one to another.

If you want the drop, that would be h=r*(1-cos(s/r)).

This is from simple trig.

Construct a right angle triangle as shown:

s is the arc along Earth.

h is the drop.

r is the radius

The angle at the centre (in radians) is s/r.

The vertical distance along the purple line before the grey bit which is h is r*cos(s/r).

i.e. h=r*(1-cos(s/r))

But this math is hard, using a trig function. You can't easily do it in your head.

Fortunately, there is a way to approximate it.

for a small value of x, cos(x) ~=1-x^2/2

So sticking that into the above we get:

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

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

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

h=s^2/2r

Notice how this is the parabola he objects to.

We can also put in some values to get a simple number we can use of the form:

h=(k*inches/mile^2)*s^2.

Where s is a distance in miles and h is a height in inches.

So the value we need to calculate is (miles^2/inch)/2r

Putting in Earths radius as 3963.19 miles, this gives us:

(mile/inch)/(2*3963.19).

Putting in 1 mile = 63360 inch, this gives us:

63360 / (2*3963.19).

That is 7.99356.....

or roughly 8.

This is how the 8 inches per mile squared comes to be.

And this IS valid, at least as a good approximation.

But in reality, considering you want to talk about the height of a distant object which might be obscured, you really want a different triangle, this one:

You still have the angle as s/r, but now you have:

cos(s/r) = r/(r+h)

(r+h)*cos(s/r) = r

r*cos(s/r)+h*cos(s/r) = r

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

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

You may notice that is the same as the above, but with an extra factor thrown in, 1/cos(s/r).

That means we can still apply the small x approximation, and we get this:

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

And then we can make another approximation. As s/r << 1, ((s/r)^2/2) <<1, so (1-(s/r)^2/2) ~= 1.

For example, if s was 100 miles, that would be 0.99968. That means the drop would be slightly larger, but by a negligible amount.

So not only does he use the wrong triangle, to get the wrong formula out; he also uses the wrong distance; and is entirely incorrect with his claim that the 8 inches per mile squared is wrong.

And for a valid comparison, he should have used this image:

The red line is a circle (with a radius of 1), the green line is a parabola, specifically 1-(x^2)/2.

But of course, if he did that you would see quite good overlap.

With them being so horribly wrong about that, and you not bothering to mention that, why should anyone care about the rest?