Oh my! More quackery from the round earth box.

You mean more deflection from you, still failing to address the numerous shortcomings of your experiment.

I find it very humorous that many here are claiming such an experiment is impossible

Not impossible, just so impractical given the string cannot be perfectly straight and over such a long distance, you have provided no way to ensure the 3 reference heights are level, and there is such a tiny drop, which you felt the need to exaggerate 16 fold.

Even if you did manage to set it up with the references all perfectly level as needed, with the correct height as predicted for the RE, and the twine kept perfectly straight, it would burn either way as the flame of the candle would be large enough to burn it for the RE or a hypothetical FE.

You will need something much smaller to demonstrate the difference.

And again, the string wont be perfectly straight.

The equation for such a string is given as y=sqrt((T

_{x}/(λ g))^2+l^2) where y is the height, T

_{x} is the tension in the horizontal direction, λ is the mass per unit length, g is the acceleration due to gravity and l is the length of the string on one side of the point.

Also note that T

_{y}=λ g l, and T=sqrt(T

_{y}+T

_{x}), which must be less than the tensile strength of the twine or it will break.

We can also simplify this a bit.

We get y=sqrt((T/(λ g))^2-l

_{t}^2 + l^2)

Where l

_{t} is the length from the middle to the end and l is the bit we are looking at.

So the middle has a height of y=sqrt((T/(λ g))^2-l

_{t}^2), while the end has a height of T/(λ g)

So that means the sag will be T/(λ g) - sqrt((T/(λ g))^2-l

_{t}^2).

If we assume that l is insignificant compared to T/(λ g) (which is needs to be to get a small sag), this can be simplified to

sag= λ g l^2/(2 T)

As g and l are fixed, this means the sag will be dictated by λ/T.

We want the smallest lambda with the largest T.

Also, conveniently, it scales just fine with area. So a 1 m diameter string will follow the same path as a 1 mm diameter string, given the same stress (as a force per unit area). This is because it all scales as λ/T, where T is given by the tensile strength times the area, and λ is given by the density times the area. So for simplicity we can just pretend the string has an area of 1 m^2.

But now we need to put in some properties, which are hard to find.

Lets use nylon as that is quite strong and more well defined than twine.

I find a tensile strength of 900 MPa and a density of 1130 kg/m^3.

That gives a difference in height for our string of ~0.25 m.

That would be the best you could do.

If you try to pull tighter to make the sag any less you would break the string.

So you are looking for a drop due to curvature of ~3.175 mm using a piece of string with a sag of 250 mm.

A small change in tension will have a much larger effect than the curvature of Earth.

Even if you went to the extreme of a carbon fibre, such as one with a tensile strength of 7000 MPa and a density of roughly 1.79 kg/m^3 you still get a sag of roughly 50 mm, more than the drop due to curvature.

You would need something like a carbon nano-tube to be able to see the drop. And in order to do so, that will require a very large amount of tension, far more than required to rip those sticks out of the ground.

So your experiment simply will not work.

And before you suggest to use a longer or shorter distance, note that the drop due to the curvature is roughly proportion to l^2, and the sag due to the weight of the string is also roughly proportional to l^2. So that wont help.

If I have made a mistake with the math, feel free to explain why it is wrong and I will fix it.

Did you take note of the shadows which showed no change in angle?

No change, or just a change too small to easily notice?

After all, you would only expect a roughly 13 arc-second change in angle.