I was once a believer in flat earth, for about a day. I remember watching all of the videos i was told to watch and it started to make sense when it came to the rate of curvature and that we seem to see things that are too far away. You see, the rate of curvature according to NASA is 8 inches per mile squared. So at 1 mile you would see an 8 inch drop. At 2 miles its almost 3 feet, and it just compounds from there. You see, at first I believed that was an official formula like many flat earthers believe, the difference is I decided to stop right there and confirm it. And you know what…? I could not confirm it, instead I learned it is NOT the formula and if it was the earth would be shaped like a parabola. But I will let this author from Quora explain it as he does a great job here.
From Daryl Mark on Quora: https://www.quora.com/What-exactly-is-8-inches-per-mile-squared-and-why-do-flat-Earthers-say-it-so-much
8″ per mile^2 is a variation of y=x^2, which (every kid learns in middle school) is the equation for a parabola. If you know what a parabola is it is not a circle or a sphere.
Samuel Rowbotham used it in his 19th-century book “Earth Not a Globe” because even though he dropped out of school before his teens he was enough of a con man that he understood that using this equation calculates a ridiculously-large amount of Earth curvature – many times more than is actually the case.
Flat Earthers like to continue to use it today because it supports their “I see more of this thing across the lake than I should be able to, thus proving the Earth is flat” argument.
An excellent example: people often like to photograph the Chicago skyline from the dunes in Indiana. The distance is roughly 32 miles across Lake Michigan, but using the equation for a parabola the Earth curvature should be 29 miles – far too much to see even the tallest skyscrapers, which you do in reality see. Thus they get to proclaim “flat Earth proof!”
Use the real formula of h = r∗(1−cosθ) however, and the “drop” at that distance is more like 550 feet, which nicely matches nearly any and all photos or videos taken of Chicago from the Indiana dunes (roughly the bottom half of the Willits/Sears tower is hidden by the water).
Here’s a nice visual to demonstrate. Using 8″ per mile^2 the Earth would HAVE to look like the purple line. Even more absurd, that formula assumes the vertex (the point) is always underfoot, so you have the steepest drop looking in any direction. So not only would the Earth be pointy, the point would move as you do.
The nice green and blue Earth’s curvature on the other hand is nicely described by the equation h = r∗(1−cosθ). I hope this makes the stupidity of using the wrong equation all the more obvious.
UPDATE: There’s an FE-run “Earth drop calculator” web app that I’d bookmarked some time back and had used to output the “wrong” value shown in my answer. I was in a hurry when I wrote the original answer above (and maybe a little lazy) and never bothered to double check the numbers, especially since the flattie insisted that 8” per mile^2 was correct. A little while later it occurred to me that number looked a little severe (and a few Quora users pointed it out too) – even for a misapplied equation – and sure enough when I manually calculated it his app was exaggerating the numbers much worse than the parabolic equation does already.
When I went back to query the author what was going on, no response. I queried a few more times, providing a little math evidence, and suddenly the web page disappeared. I am going to keep the original number in my answer though because I think it nicely demonstrates how far the FE community will go to deceive and prop up their bogus assertions.
In the end I stand by everything else in my answer. 8” per mile^2 is totally wrong, and the error gets worse the farther you go out. It’s really not even that accurate if one’s goal is to make real-world short-distance approximations because it can only calculate the height of targets hidden if the perspective was at 0 cm from the Earth’s surface. At 0 cm a 96′ object would be hidden at a distance of 12 miles. On the other hand, at 5’7″ above the Earth’s surface (the eye height of a six-foot man), a 12-mile distance would only obscure a 55′ object. That’s a nearly 50% discrepancy! The error increases as you get nearer, and of course go the other direction and the parabolic error takes over, so it is erroneous both near and far.
I will concede, however, that with the distance given across Lake Michigan the 29 mile “drop” is a bit too much… even for a misapplied parabolic equation.