Earth Curvature Calculator
See how much the Earth curves over a distance. Enter a distance (miles or km) and your eye height (feet or meters) to get the curvature drop, your distance to the horizon, and how much of a far-off object sits hidden below it.
Example: with Distance to target 10 · Distance unit miles · Eye height (observer) 6 · Height unit feet → Curvature drop over the distance: 66.69 ft (20.33 m).
- Distance to your horizon3.00 mi (4.83 km)
- Hidden below the horizon32.7 ft (10.0 m) hidden below the horizon
Computed by the calculator below using its default values. Change any input to see your own numbers.
Over a distance d the surface drops about d²/(2R) below the line of sight, with Earth's radius R ≈ 6,371 km. That matches the familiar rule of about 8 inches per mile squared.
Three different curvature numbers
People mean different things by how much the Earth curves. The simplest is the drop: over a straight distance, how far the surface falls below a level line of sight. That value grows with the square of distance and follows the well-known 8-inches-per-mile-squared rule, so 1 mile drops about 8 inches and 10 miles drops about 67 feet.
But the drop alone overstates what you cannot see, because your eyes sit above the ground. Raising your viewpoint pushes the horizon farther out, and only the part of a distant object beyond that horizon is truly hidden. This tool reports all three: the geometric drop, your horizon distance, and the hidden height.
Eye height and refraction
Your horizon distance scales with the square root of eye height. At 6 feet it is about 3 miles; from a 100-foot tower it stretches past 12 miles. That is why lighthouses and ship crows-nests are built tall — height buys visible range faster than it costs.
Real sight lines curve slightly because air bends light downward, an effect called atmospheric refraction. It typically lets you see a bit farther than pure geometry predicts, hiding roughly 14 percent less than the numbers here. This calculator uses the clean geometric model and notes that refraction reduces the hidden amount.
How it’s calculated
Curvature drop ≈ d²/(2R), with distance d and mean Earth radius R = 6,371,000 m; this reproduces the ~8 inches per mile squared rule. Horizon distance = √(2R·h + h²) for eye height h. Hidden height beyond the horizon = (d − horizon)²/(2R) when the target is farther than the horizon. Units convert at 1 mi = 1,609.344 m and 1 ft = 0.3048 m.
Ignores atmospheric refraction, which bends light and typically reduces the hidden height by around 14 percent. Uses a spherical Earth of mean radius; local terrain and tides are not modeled.
Curvature drop and hidden height (6 ft eye height)
| Distance | Curvature drop | Hidden below horizon |
|---|---|---|
| 1 mi | 0.67 ft | 0 (within horizon) |
| 3 mi | 6.0 ft | 0 (near the horizon) |
| 6 mi | 24.0 ft | 6.0 ft |
| 10 mi | 66.7 ft | 32.7 ft |
| 20 mi | 266.8 ft | 192.7 ft |
| 50 mi | 1,667 ft | 1,473 ft |
Computed with R = 6,371 km: drop ≈ d²/2R; hidden height uses a 6 ft (1.83 m) eye height and ignores refraction.
Common mistakes
- Treating the curvature drop as the hidden amount — your eye height and the horizon shrink what is actually blocked.
- Forgetting eye height entirely; from ground level the horizon is only a couple of miles away.
- Ignoring refraction, which lets you see slightly farther than the bare geometry suggests.
- Using the drop formula for very long distances where the small-angle approximation loses accuracy.
Frequently asked questions
How do you calculate the curvature of the Earth?
Over a distance d the surface drops about d²/(2R) below your line of sight, with R the Earth's radius of 6,371 km. That matches the rule of roughly 8 inches per mile squared.
What is the 8 inches per mile squared rule?
It estimates the curvature drop: multiply 8 inches by the distance in miles squared. One mile drops about 8 inches, 2 miles about 32 inches, and 10 miles about 67 feet. It is the drop, not the hidden height.
How far away is the horizon?
It depends on your eye height and equals about √(2R·h). At 6 feet it is roughly 3 miles; at 100 feet it is over 12 miles. Higher viewpoints see much farther.
Why can I still see distant objects that math says are hidden?
Atmospheric refraction bends light downward, letting you see a bit beyond the geometric horizon and hiding about 14 percent less than the pure calculation. Tall objects also poke above the curve.
Does the drop equal what is hidden from view?
No. The drop is measured from a level line at your eye, but your raised viewpoint moves the horizon outward. Only the portion of a target beyond that horizon is actually hidden, which is less than the drop.