Let us look at an object directly out to sea from a beach position.. There is a problem that most do not see. If the circumference of the earth is 24,901 miles, then a one degree deviation would be 69.17 miles, 360 degrees making up a whole circle, or globe.
Every experiment shown on youtube and elsewhere, that I have seen, states that there are numerous degrees of “drop” within a much smaller distance. The ratio is 1 degree to every 69.17 miles. For every 69.17 miles you travel there is only a one degree deviation, or drop.
Can you see a one degree deviation at almost 70 miles?
My observation is that, at 70 miles the drop would be one degree. One degree divided by 360 degrees produces a ratio of 0.00278 to one. So if a ship is 200 feet tall only 0.00278 of it should disappear: 0.556 feet, at 69.71 miles.
A building of 500 feet, at a full 69.71 miles out, such as one used in one of many observations, should drop out of sight by only 0.00278 X 500 or about 1.389 feet.
When many explain that such and such a thing on the horizon drops due to curvature, they use a ratio from nowhere.
In the following youtube video:
the producer claims much of a wind turbine 29 nautical miles from shore has been lost from sight, as he presents, almost half of its height. I am astonished at how anti-mathematical this producer is. He claims he is looking at a wind turbine that is 29 nautical miles from the beach on which he stands. 29 nautical miles converts to 33.35 miles. At 33.35 miles the curvature of the earth would be less than one half of one degree: 33.35/69.17, or .482 degrees.
Even if the windmill is as tall as he claims it could be, 180 meters, or 590.4 feet, at less than one half of one percent of drop, the windmill should only disappear 590 feet X 0.00278 feet per 69.17 mile distance, X .482 + .7864 feet. This is about 9 inches.
We see a lot more distortion than a mere drop, over 33.35 miles,
It is impossible that a distance of 33.15 miles produces such a drastic drop in elevation. Clearly there is a different variable involved.
If we were to triangulate and look at a ship moving away from a point would we se the tilt of the mast way from the beach location, for example, due to curvature? Incidentally, I have never seen a study or experiment that tests this.
If I were to stand 33.35 miles away from the point on the beach in a line perpendicular to the line of sight of the original position, would I be able to see the mast tilt away from the beach? The distance I would be making my observation would be much more, but only due to angle of observation. Would I now see less than .482 of a degree of tilt from well over 33.35 miles? The hypotenuse would be 1.414 times the distance from the original beach point or 47.16 miles. How could one tell with the tossing of waves and ship?
If I could see that far I would only be looking for a 33.35/69.71 ratio of one degree of dip, or tilt of the mast. It is imperceptible. Furthermore, the angle is less due to the angle away from direct observation, which would be directly to the side of the windmill.
In the producer’s example he shows the windmill clearly losing half its height. Impossible.