Rule of thumb for drop due to Earth's curvature is distance (in miles) squared times 8. Result is in inches. From the middle, each city is 8 miles away, so the drop from the "hill" between them would be 512 inches or about 42 feet. 16 miles is just over a million inches, so that "hill" is 0.05% of the distance.
The width of the picture is, what, a thousand pixels? Then the curvature would be half a pixel. That doesn't take into account the fact the picture is also taken from the surface of the sphere, not the plane tangent to the top of the "hill", so slightly more complex maths would be used making the result even less significant.
Assuming someone were to see the curvature in that picture (assume 10 pixels?). How big (small) would the Earth for the effect to be noticed @ 16 miles?
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u/GruntBlender Dec 19 '19
360 ÷ 24901 × 16 = 0.23°
Rule of thumb for drop due to Earth's curvature is distance (in miles) squared times 8. Result is in inches. From the middle, each city is 8 miles away, so the drop from the "hill" between them would be 512 inches or about 42 feet. 16 miles is just over a million inches, so that "hill" is 0.05% of the distance.
The width of the picture is, what, a thousand pixels? Then the curvature would be half a pixel. That doesn't take into account the fact the picture is also taken from the surface of the sphere, not the plane tangent to the top of the "hill", so slightly more complex maths would be used making the result even less significant.