Flat vs. Round Earth Calculator
Pick one of three classic experiments, enter your measurements, and the calculator shows what a round Earth predicts versus what a flat Earth would predict. All three methods were used historically to prove the Earth is a sphere: the disappearing-ship experiment shows how objects sink below the horizon by a predictable amount, the stick-shadow method lets you calculate the Earth's circumference from two shadows just as Eratosthenes did around 240 BC, and the double-sunset experiment shows how standing up after lying down can let you watch the sun set a second time. Switch between metric and imperial, and the results update instantly.
The disappearing ship: curvature in everyday life
Stand at the shoreline and watch a ship sail away. On a flat Earth, it would simply get smaller and smaller until it was too tiny to see, always showing its full height. On a spherical Earth, the ship eventually disappears hull-first, with the mast remaining visible after the body has gone below the horizon. This is what you actually observe. The rate of disappearance is determined by the Earth's radius of about 6,371 km, your eye height, and the distance to the ship. The formula for curvature drop at distance d from an observer at sea level is approximately d squared divided by twice the radius, which equals about 8 inches per mile squared (or about 0.078 m per km squared). The horizon distance from observer height h is the square root of 2 times the radius times h. These formulas are derived from the geometry of a sphere and are confirmed by every observation ever made of distant objects at sea.
Eratosthenes and the stick shadow experiment
Around 240 BC, the Greek mathematician Eratosthenes noticed that a vertical stick in Alexandria cast a shadow at noon on the summer solstice, while a well in Syene (modern Aswan) had no shadow at all: the sun was directly overhead. He realized this angular difference was possible only on a curved surface. By measuring the angle (about 7.2 degrees, or one fiftieth of a full circle) and knowing the north-south distance between the two cities was about 800 km, he calculated the Earth's circumference as 50 times 800, or 40,000 km, within about 1% of the modern value of 40,075 km. On a flat Earth with a local sun, the same two shadow angles would imply a sun at a specific height directly above the surface, typically thousands of kilometres up. But that same local-sun height would then give wrong shadow angles for any third city, making the flat Earth model internally inconsistent.
The double sunset experiment
If you lie on a west-facing beach and watch the sun just set below the horizon, then quickly stand up, you can see the sun briefly re-appear before it sets again. This works because standing up raises your line of sight, pushing your horizon slightly further away and revealing a sliver more of the sun. The extra time you see the sun depends on how much height you gain: the formula is approximately 8 times the square root of the height gained in metres, giving seconds of extra visibility. On a flat Earth, the sun would be equally visible at any height above the surface, so no second sunset would be possible. The Burj Khalifa experiment, riding an elevator to the observation deck immediately after sunset at ground level, makes the effect dramatic: you can see over 80% of a second full sunset.
Why these experiments are definitive
Each of the three experiments independently requires the Earth to be a sphere. The disappearing ship matches the curvature formula at every observed distance. The stick shadow gives a consistent circumference from any two north-south locations. The double sunset appears whenever height is gained after the sun disappears and disappears when you descend. No flat Earth model can explain all three simultaneously without introducing a local sun, modified light behaviour, and special atmospheric properties that then fail to predict other observations. Atmospheric refraction can slightly change the apparent horizon distance (usually by about 7 percent), which is why this calculator uses the geometric formulas without a refraction correction by default: the underlying spherical geometry is exact, and refraction is a small, separately understood effect.
Curvature drop at common distances
| Distance | Curvature drop (metric) | Curvature drop (imperial) | Significance |
|---|---|---|---|
| 1 km / 0.6 mi | ~0.08 m | ~3 in | Invisible to the naked eye |
| 3 km / 1.9 mi | ~0.71 m | ~28 in | Just detectable with instruments |
| 10 km / 6.2 mi | ~7.8 m | ~25 ft | Ship hull noticeably cut off |
| 20 km / 12 mi | ~31 m | ~102 ft | Structures visibly truncated |
| 50 km / 31 mi | ~196 m | ~643 ft | Mountains hidden below horizon |
| 100 km / 62 mi | ~785 m | ~2,575 ft | Significant terrain obscured |
| 400 km / 249 mi | ~12,566 m | ~41,230 ft | ISS orbit altitude regime |
Approximate curvature drop (how far the surface curves below a horizontal straight line) for an observer at sea level. Refraction not included.
Frequently asked questions
What is the curvature drop formula?
The drop of the Earth's surface below a perfectly horizontal straight line at distance d from the observer is approximately d squared divided by twice the Earth's radius (d^2 / 2R). For small distances relative to the Earth's size, this is very accurate. At 10 km, the drop is about 7.8 m; at 1 km it is about 8 cm. This formula comes from the geometry of a circle and is exact for a smooth sphere.
How far away is the horizon?
The horizon distance for an observer at height h above a spherical Earth of radius R is the square root of (2 x R x h), which for small h simplifies nicely. At eye level on the beach (about 1.7 m), the horizon is roughly 4.6 km (2.9 mi) away. At 10 m height it is about 11.3 km (7 mi). The formula ignores atmospheric refraction, which typically extends the visible horizon by about 7%.
Can you see the curvature from a plane?
Not clearly from a typical cruising altitude. Studies suggest you need a clear horizon, at least 60 degrees of field of view, and an altitude of around 35,000 ft (10.5 km) or more to perceive the curvature. At standard airliner altitude the curvature is barely detectable. It becomes obvious above about 60,000 ft, as seen from high-altitude aircraft like the SR-71 Blackbird and from spacecraft.
Does atmospheric refraction change these calculations?
Yes, slightly. Atmospheric refraction bends light downward around the curve of the Earth, effectively making the visible horizon about 7% further away than the geometric formula predicts. This means you can sometimes see objects that are technically below the geometric horizon. The curvature drop formula in this calculator is the geometric (no-refraction) value. Add about 7% to horizon distances to account for average atmospheric refraction.
Why did flat Earth believers reject the disappearing ship?
Flat Earth proponents have argued that perspective causes distant objects to appear lower, eventually sinking below the vanishing point even on a flat surface. However, perspective alone cannot make an object disappear hull-first while the mast remains visible: perspective would shrink the whole ship proportionally. Zooming in with a telescope or drone always recovers a hull that was hidden by curvature, not by perspective. This is the key observable difference.
How accurate was Eratosthenes's measurement?
Remarkably accurate. He calculated about 39,375 to 46,620 km depending on which interpretation of the ancient unit "stade" you use. The true polar circumference is 40,008 km and the equatorial is 40,075 km. His best estimate was within 1-2% of the actual value, using only two sticks and the distance between two cities, over 2,200 years ago.
Can I really see the sun set twice by standing up?
Yes, and it has been filmed and replicated many times. You need a clear, flat western horizon (a beach or large lake works best), a west-facing position, and ideally a helper. Lie down and watch the sun fully disappear. Stand up as quickly as possible. If you do it within a second or two you will briefly see the top of the sun re-appear before it sets again. The effect is small because the height difference is only 1-2 metres, giving only a few seconds of extra visibility, but it is real and geometrically predictable.