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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:

  1. Miles squared = 3,364 (58 x 58)
  2. 3,364 x 8 inches = 26,912 inches
  3. 26,912 inches / 12 = 2,242.67 ft.
  4. 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:

  1. Miles squared = 1,764 (42 x 42)
  2. 1,764 x 8 inches = 14,112 inches
  3. 14,112 inches / 12 = 1,176 ft.
  4. 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)

  1. Miles squared = 3,600 (60 x 60)
  2. 3,600 x 8 inches = 28,800 inches
  3. 28,800 inches / 12 = 2,400 ft.
  4. 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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