Perhaps I should explain that my "round earth" example would be at sea on a clear, calm day.
Or just looking out to sea if you were on the shore on a beach.
There is nothing to obstruct your view to the horizon, which is a definite line where the sea and sky appear to meet.
If you were a 6 feet tall person, standing up in a lifeboat, the horizon would appear to be about 3 miles from you. You would be in the middle of a circle with a radius of 3 miles.
If you were in the crow's nest on a ship, 100 feet above the sea, the horizon would appear to be about 12.2 miles from you. You would be in the middle of a circle with a radius of 12.2 miles.
These figures were taken from the Training Manual For Lookouts in the United States Navy.
In their training, they compare their results of estimating distances to ships, for example, with those from the ship's radar.
I would not necessarily like to debate the issue, but just compare round earth and flat earth ideas in regard to the horizon.
There is a simple "round earth" equation :
d (distance to the horizon in miles) = 1.22 (a constant) x square root of h (height of the observer in feet)
Since there would be no curvature on a flat earth, how would this be done, with some equations and figures and examples similar to those above ?
I don't intend this to be a debate.
I would just like to know how it's done on a flat earth, since you don't have to contend with the curvature of the earth on a flat earth.
Once I get an answer, I shall cease and desist on this subject.
However.........My examples are good enough for the USN.....Speaking from personal experience I know it worked for me, too !