What is the formula for figuring out the alleged 'curvature' of the Earth? I would like to test this myself.
Question: What is the formula for figuring out the alleged 'curvature' of the Earth? I would like to test this myself.
Answer: If the Earth were a sphere 25,000 miles in diameter, the formula for determining the curvature at a given distance would be 8 inches per mile squared. Examples:
At 58 miles, an object with an elevation of 60 ft. should be hidden behind 2,182 ft. of curvature:
- Miles squared = 3,364 (58 x 58)
- 3,364 x 8 inches = 26,912 inches
- 26,912 inches / 12 = 2,242.67 ft.
- 2,242.67 ft. - 60 ft. (elevation of object) = 2,182 ft.
At 42 miles, an object with an elevation of 180 ft. should be hidden behind 996 ft. of curvature:
- Miles squared = 1,764 (42 x 42)
- 1,764 x 8 inches = 14,112 inches
- 14,112 inches / 12 = 1,176 ft.
- 1,176 ft. - 180 ft. (elevation of object) = 996 ft.
"The Statue of Liberty in New York stands 326 feet above sea level and on a clear day can be seen as far as 60 miles away. If the Earth were a globe, that would put Lady Liberty at an impossible 2,074 feet below the horizon." (200 Proofs Earth is Not a Spinning Ball, Eric Dubay)
- Miles squared = 3,600 (60 x 60)
- 3,600 x 8 inches = 28,800 inches
- 28,800 inches / 12 = 2,400 ft.
- 2,400 ft. - 326 ft. (elevation of object) = 2,074 ft.
Time and time again, it has been demonstrated that there is no measurable or observable curvature. It just isn't there.
Many make the mistake of only calculating 8 inches per mile: https://www.youtube.com/watch?v=6a9wxl6FQbY
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