Find how much the Earth’s surface drops below a straight line of sight over a given distance — the reason distant objects and horizons disappear below the visible line.
How it works
Using Earth’s mean radius, drop = distance² ÷ (2 × R). Over 10 km, the surface drops about 7.85 meters below a perfectly straight line.
What this does not include
This is a standard simplified approximation, accurate for everyday distances. It doesn’t account for atmospheric refraction, which bends light slightly and lets you see a little further than the geometric drop alone would suggest.
How to use this calculator
- Enter the distance in kilometers.
A worked example
Looking 10 km across the Earth’s surface: the curvature drop is approximately 7.8481 m.
At 50 km: the drop grows to 196.2015 m — the drop grows faster than distance, since curvature compounds with the square of distance.
What the variables mean
| Variable | Meaning |
|---|---|
| Distance | Straight-line distance across the Earth’s surface, in km |
| Drop | How far the Earth’s surface curves away below a flat line of sight, in meters |
Edge cases worth knowing
The drop scales roughly with the square of distance, not linearly. Going from 10 km to 50 km (5× the distance) increases the drop by roughly 25×, not 5×, matching the squared relationship.
This uses Earth’s mean radius as a fixed constant — the actual local curvature varies very slightly due to the planet’s not-quite-perfect sphere shape, though the difference is negligible for practical purposes.
Why does a ship disappear hull-first over the horizon?
The curvature drop increases with the square of distance, so as a ship sails away its lower hull drops below the visible line before its taller masts do.
Does atmospheric refraction change the answer much?
It extends the visible horizon by roughly 7-8%, a real but modest effect compared to the geometric curvature itself.
How far away is the horizon for someone standing at sea level?
Roughly 5 km for eyes at about 1.7 m height — the same curvature formula, solved for the distance at which the drop equals the observer’s height.
