How much of the turbine would you lose from a view out to 30 km by using the 8 inches per mile squared global set up?

I already told you that. And this was one of the questions I keep on asking you, and you keep on ignoring.

If you think there is a problem, or want to disagree, provide your own math.

The simple equation to use is h=d^2/(2*R).

But you need to note that the amount hidden is from the horizon, not from your position.

To find the distance to the horizon you also use that simple formula.

So first from the person to the horizon, assuming a 2 m, i.e. 0.002 km vantage point (i.e. eyes 2 m above sea level)

h=d^2/(2*R)

d^2=2*h*R = 2 * 0.002 km * 6371 km = 25.484 km^2

d=5.05 km (to 3 s.f., which is being generous).

Now, the distance from the horizon to the target is 30 km minus the distance to the horizon.

d = 30 km - 5.05 km = 24.95 km.

So the amount hidden is now given as:

h=d^2/(2*R) = (24.95 km)^2/(2*6371 km) = 0.0489 km = 48.9 m.

Of course, this doesn't account for refraction, which does change it slightly.

A simple correction for refraction is to pretend Earth has a radius 7/6 of what it actually is.

Doing the math with that gives 40.5 m hidden.

And like I have done before, to determine the tilt you instead have t=d * 360 degrees/(2*pi*R), which works out to be 0.27 degrees.

Now can you say anything to refute that?

Alternatively, how much should be hidden from view using your FE pure magic? Can you show any calculation to show that any should be hidden from view with water obstructing the view to the bottom, or can you just repeatedly claim that the FE magically hides it?