Earth Curvature Calculator

Calculate Earth's curvature drop over a given distance, or horizon distance from a given observer height.

Science 8 in/mi rule Refraction toggle
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Earth curvature drop and horizon

NASA radius · optional refraction

Instructions — Earth Curvature Calculator

1

Pick a mode

"Drop over distance" computes how far an object dips below the horizon at a given distance. "Horizon distance" computes how far you can see from a given observer height. Both use the same Earth-radius math.

2

Enter the input

For drop mode, enter distance (miles or km). For horizon mode, enter observer height (feet or meters). Use the unit toggle to switch.

3

Toggle refraction if you want

Atmospheric refraction bends light slightly, extending the visible horizon by ~7%. Surveyors use a refraction correction factor of 7/6 on Earth's radius. Toggle it on for real-world estimates, off for pure geometric drop.

The "8 inches per mile-squared" rule: Drop in inches ≈ 8 × miles². Useful for mental math: at 10 miles, drop is 8 × 100 = 800 in = 66.7 ft. Matches the d²/(2R) formula within 1%.
Curvature accelerates: Drop scales with distance squared. Doubling distance quadruples the drop. At 50 miles, the drop is 1,667 ft — taller than the Empire State Building.

Formulas

Earth's mean radius is 3,959 miles (6,371 kilometers) per NASA Earth Fact Sheet. The two main formulas come from spherical geometry.

Hidden height (drop)
$$ h = \frac{d^2}{2R} $$
How much an object dips below the horizon at distance d. h and d in same units; result is the vertical drop.
Horizon distance
$$ d = \sqrt{2Rh + h^2} \approx \sqrt{2Rh} $$
Maximum distance to the geometric horizon for an observer at height h. The h² term is negligible for typical observer heights.
8-inches-per-mile-squared rule
$$ h_{inches} \approx 8 \times d_{miles}^2 $$
A useful mental shortcut. Matches d²/(2R) within about 1% — the exact factor is 7.98 inches per mile-squared.
Imperial horizon approximation
$$ d_{miles} \approx 1.225\sqrt{h_{feet}} $$
For observer height in feet, horizon distance in miles. A 6-ft-tall observer at sea level sees about 3 miles to the horizon.
Metric horizon approximation
$$ d_{km} \approx 3.57\sqrt{h_{meters}} $$
For observer height in meters, horizon distance in kilometers. A 1.8-m observer sees about 4.8 km.
Refraction correction
$$ R_{eff} = R \times \frac{7}{6} $$
Atmospheric refraction extends the visible horizon by ~7%. Surveyors multiply Earth's radius by 7/6 to get an "effective radius" that includes the bending of light.

Reference

Drop over distance (no refraction)
DistanceDrop (feet)8 in/mi² rule
1 mile0.67 ft8 in
5 miles16.7 ft200 in (16.7 ft)
10 miles66.7 ft800 in (66.7 ft)
20 miles267 ft3,200 in (267 ft)
50 miles1,667 ft20,000 in (1,667 ft)
100 miles6,667 ft80,000 in

Horizon distance by observer height

Ground level
HeightHorizon (mi)
5 ft2.7 mi
6 ft3.0 mi
10 ft3.9 mi
50 ft8.7 mi
100 ft12.3 mi
Aviation
AltitudeHorizon (mi)
500 ft27 mi
1,000 ft39 mi
10,000 ft122 mi
35,000 ft229 mi
100 km (Karman)713 mi

Source: NASA Earth Fact Sheet, NOAA geodesy resources, US National Geodetic Survey atmospheric refraction standards.

Article — Earth Curvature Calculator

Earth Curvature Calculator: drop and horizon distance

An Earth curvature calculator computes how much an object drops below the horizon at a given distance, using the formula d-squared divided by twice Earth's radius. The popular mental rule "8 inches per mile-squared" is a close approximation. At 10 miles, the curvature drop is 66.7 feet. Earth's mean radius is 3,959 miles (6,371 km) per NASA's Earth Fact Sheet.

The same geometry works in reverse: from a given observer height, you can compute the horizon distance using the square-root-of-2Rh formula. A 6-foot person at sea level sees about 3 miles to the horizon; from 35,000 feet, a passenger sees about 229 miles. Atmospheric refraction extends both numbers by roughly 7%.

What is Earth curvature

Earth curvature is the deviation of Earth's surface from a flat plane, caused by Earth being approximately spherical. Over short distances (under a mile), the deviation is negligible — the ground looks flat. Over miles, the curvature becomes measurable; over tens of miles, it dominates what you can see.

An Earth curvature calculator turns this geometry into numbers. Given a distance, it returns the vertical drop. Given an observer height, it returns the maximum line-of-sight horizon distance. Both calculations use Earth's mean radius from NASA: 3,959 miles or 6,371 kilometers.

Did you know

Earth is technically an oblate spheroid, not a perfect sphere. The equatorial radius is 3,963 miles; the polar radius is 3,950 miles. The 13-mile difference is 0.33% — small enough that most calculations use the mean radius of 3,959 miles without meaningful error.

Earth curvature formula

Two related formulas cover most real questions:

Earth curvature formulas
Drop h = d² ÷ (2R)
Horizon d = √(2Rh + h²) ≈ √(2Rh)
R = 3,959 mi (6,371 km)
With refraction Reff = R × 7/6

For the drop formula, h and d are in the same units. For the horizon formula, h is the observer height above the surface; for short heights the h-squared term is negligible. The 7/6 refraction multiplier accounts for the bending of light through the atmosphere.

Earth curvature per mile rule

The 8-inches-per-mile-squared rule is the most useful mental shortcut for Earth curvature. The exact factor from the formula d-squared divided by (2 × 3,959 mi) is 7.98 inches per mile-squared, so 8 is a clean approximation.

Quick mental calculation: at 5 miles, drop = 8 × 25 = 200 inches = 16.7 feet. At 10 miles, drop = 8 × 100 = 800 inches = 66.7 feet. The rule is reliable up to about 100 miles, where the formula's flat-plane assumption starts to introduce visible error.

Tip

The "8 inches per mile" rule is sometimes quoted without the "squared." That version is wrong — it would predict a linear 80-inch drop at 10 miles instead of the actual 800 inches. Always include the "squared." Drop scales with distance squared, not linearly.

Horizon distance by observer height

The horizon distance formula gives a clean answer for any eye height. The imperial form is especially useful:

  • Standing adult (6 ft): Horizon = 1.225 × √6 = 3.0 miles.
  • On a 30-foot mast: Horizon = 1.225 × √30 = 6.7 miles. Same formula, taller observer.
  • On a 100-foot hill: Horizon = 12.3 miles.
  • Commercial flight (35,000 ft): Horizon = 229 miles. About 245 miles with atmospheric refraction.
  • Karman line (100 km): Horizon = 713 miles. Clearly curved to the naked eye.

The reason taller observers see farther is geometric: the line tangent to the sphere from a higher point touches the surface at a greater arc distance. Doubling the observer height does not double the horizon distance — the square root in the formula means height has diminishing returns. Going from 6 ft to 24 ft (4x) increases horizon distance by only 2x.

Atmospheric refraction and curvature

Light bends as it travels through layers of air with different densities. Cool dense air near the ground refracts light slightly downward, which extends the visible horizon beyond the pure geometric prediction.

The standard correction used by surveyors and the National Geodetic Survey is to multiply Earth's radius by 7/6, producing an "effective radius" of 4,622 miles or 7,433 km. With this correction, the horizon distance grows by about 7 to 8%, and the curvature drop at a given distance shrinks by the same fraction.

Refraction varies

The 7/6 factor is the standard atmosphere value. Real refraction varies with temperature, pressure, and humidity. Warm air over cold water can produce mirages — objects appearing higher than they actually are. Cold air over warm surfaces can do the opposite. The standard formula gives a typical answer, not an exact one for any given moment.

Observing Earth curvature

Earth's curvature is subtle at ground level and obvious from altitude. From a typical commercial flight at 35,000 feet (10.7 km), the curvature is mildly visible if you look at the horizon from a window seat on a clear day. The curve is gentle — about 1.5 degrees of arc across the visible horizon — but unmistakably present.

From higher altitudes, the visibility increases dramatically. SR-71 pilots at 80,000 feet (24 km) reported clearly curved horizons. Astronauts above the Karman line (100 km / 62 miles) see the full curve of Earth against the blackness of space. The transition from "subtle curve" to "obvious sphere" happens between about 50 km and 100 km of altitude.

Eratosthenes and the first measurement

Around 240 BCE, the Greek mathematician and geographer Eratosthenes calculated Earth's circumference using only shadows. He observed that at noon on the summer solstice in Syene (modern Aswan, Egypt), the sun was directly overhead and shone straight down a deep well. On the same day in Alexandria, about 500 miles north, the sun cast a shadow at a 7.2 degree angle.

Eratosthenes reasoned that this 7.2 degree angle was the central arc angle between the two cities. He estimated the city-to-city distance from camel caravans, divided by 7.2, multiplied by 360, and got 252,000 stadia. Converting to modern units gives approximately 39,375 km. The actual circumference is 40,075 km. He was within 2% — remarkable accuracy for the third century BCE.

Common Earth curvature mistakes

The most common error is conflating "curvature drop" with "horizon distance." Drop is how far an object dips below the geometric horizon; horizon distance is how far you can see from a given height. Both come from the same Earth radius, but they answer different questions.

The second error is ignoring refraction in real-world observations. Pure geometric calculations under-predict what you can actually see by about 7%. If a YouTube video claims to disprove Earth's curvature because someone could see a distant building "that should be hidden," check whether they accounted for refraction and the building's height above sea level — both usually explain the apparent discrepancy.

The third error is using sea-level formulas over mountainous terrain. The drop formula assumes flat sea-level ground between you and the distant point. Real terrain — especially mountains and elevation changes — dominates over Earth curvature at moderate distances. A 1,000-foot hill 5 miles away matters more than the 16.7 feet of curvature drop.

FAQ

About 8 inches per mile at short distances. More precisely: at 1 mile, the drop is 0.67 ft (8 inches). At 2 miles, 2.67 ft. Curvature scales with distance squared, so the per-mile number is misleading for longer distances. At 10 miles, total drop is 66.7 ft, not 80 inches.
Depends on your eye height. At 6 ft tall on flat ground, the horizon is 3 miles away. Standing on a 50-ft hill, you can see 8.7 miles. From a commercial airplane at 35,000 ft, you can see 229 miles (without atmospheric refraction; about 245 miles with it).
Yes, within about 1%. The exact factor from d²/(2R) is 7.98 inches per mile-squared at Earth's mean radius. The round 8 is good enough for any practical mental estimate. For high-precision work (surveying, geodesy), use the full formula and account for refraction.
At commercial altitude (35,000 ft / 10.7 km), the curvature is subtly visible if you have a wide unobstructed view of the horizon. From much higher (50+ km, suborbital flights), the curvature is unmistakable. The Karman line (100 km) is where every astronaut has reported clearly curved horizons.
Light bends as it travels through layers of atmosphere with different densities. This extends the visible horizon by about 7 to 8% beyond the geometric prediction. Surveyors multiply Earth's radius by 7/6 to get an "effective radius" that accounts for refraction. Use the toggle on this calculator to compare.
Because the Earth curves. As a ship moves away, more of its hull is hidden below the horizon line each second. If Earth were flat, the ship would shrink uniformly and disappear due to perspective. Ships disappearing bottom-first is one of the oldest direct demonstrations of curvature, recorded by ancient Greek sailors.
Around 240 BCE, Eratosthenes observed that at noon on the summer solstice, the sun was directly overhead at Syene (modern Aswan) but at a 7.2° angle at Alexandria. The angular difference equals the arc angle at Earth's center. He estimated the city-to-city distance, divided by the angle, and computed 252,000 stadia — about 39,375 km, within 2% of the modern value of 40,075 km.
For most purposes, no. Earth is an oblate spheroid: equatorial radius 3,963 mi, polar radius 3,950 mi (a 0.33% difference). Using the mean radius of 3,959 mi gives errors under 1%. For high-precision surveying at extreme latitudes, the WGS84 ellipsoid model is used instead.