Earth Curve Calculator: See the Horizon Drop

An Earth curve calculator works out how much of a distant object is hidden by the curvature of the Earth, or how far the horizon is from a given viewing height. You enter your eye height and the distance to the object (or just your height for horizon distance), and it returns the hidden height in your chosen unit, plus the horizon distance. This is the same geometry used in surveying, long-range photography, and line-of-sight planning.

Formula

The calculator uses the standard geometric drop of a sphere. For a sphere of radius R (Earth’s mean radius ≈ 6,371 km), the distance to the horizon from height h is:

Horizon distance = √(2 × R × h)

For the hidden height of an object at distance d (with both heights measured from the same reference, usually sea level), the hidden portion is approximately:

Hidden height ≈ (d² × 1000²) / (2 × R)

Where d is in kilometers, R is in meters (6,371,000 m), and the result is in meters. This approximation ignores atmospheric refraction, which typically makes objects appear slightly less hidden than pure geometry suggests.

Worked Example

Using the default values: eye height = 1.7 m, distance to object = 10 km.

1. Horizon distance: √(2 × 6,371,000 × 1.7) = √(21,661,400) ≈ 4,654 m ≈ 4.65 km.
2. Since 10 km > 4.65 km, the object is beyond the horizon, so some of it is hidden.
3. Hidden height = (10 × 1000)² × 0.0785 / (2 × 6,371,000) = (10,000)² × 0.0785 / 12,742,000 = 100,000,000 × 0.0785 / 12,742,000 ≈ 7,850,000 / 12,742,000 ≈ 0.616 m.

The calculator shows a hidden height of 0.62 m and a horizon distance of 4.65 km. So if you are 1.7 m tall at sea level looking at a 10 km distant object, the bottom 0.62 m of it is behind the curve.

When to Use This Calculator

This tool is useful for anyone who needs to know line-of-sight over long distances:

• Photographers planning shots of mountains, city skylines, or ships from across a lake or sea.
• Amateur radio operators estimating whether two antennas have a clear path.
• Sailors and coastal navigators estimating when a lighthouse or landmass will appear.
• Drone pilots checking if a distant subject is visible from their altitude.
• Curious minds checking whether a photo “proves” the Earth is flat or round.

Remember that the calculation assumes a perfectly smooth sphere with no terrain, buildings, or waves. In reality, local elevation changes and obstacles matter more than the Earth’s curvature over distances under a few kilometers.

How to Read the Result

The hidden height tells you how much of the object’s base is blocked. If the hidden height is less than the object’s total height, the top part is visible. For example, with a 10 m tall lighthouse at 10 km and your eye at 1.7 m, the hidden height is 0.62 m, so you can see the top 9.38 m of the lighthouse.

The horizon distance is how far you can see before the ground itself blocks your view. At sea level with eyes 1.7 m up, the horizon is about 4.65 km away. From an airplane at 10,000 m, the horizon is about 357 km away.

If the object is closer than the horizon, the hidden height is zero—nothing is hidden by curvature. The calculator correctly returns 0 for such cases because the formula only applies when the distance exceeds the horizon.

Common Mistakes and Misconceptions

• Mixing units: The formula expects distance in kilometers and height in meters. If you enter feet or miles, the result will be wrong. Convert first: 1 mile = 1.609 km, 1 foot = 0.3048 m.
• Ignoring refraction: Atmospheric refraction bends light downward, making the horizon appear slightly farther and hidden heights smaller. A common rule of thumb is to use an effective Earth radius of 7,430 km (about 1.17 × actual) for refraction, but the exact amount varies with temperature and humidity.
• Forgetting your own height: The hidden height depends on your eye height, not just the object’s distance. Standing on a hill or a tall building changes the result significantly.
• Assuming the Earth is a perfect sphere: The Earth is an oblate spheroid, but the difference is negligible for distances up to a few hundred kilometers.

Limits of the Formula

This geometry assumes a clear line of sight with no obstacles and a smooth spherical surface. It does not account for:

• Terrain elevation (hills, mountains, valleys) between you and the object.
• Atmospheric refraction, which is stronger near the horizon and can make objects appear “lifted” or shimmering.
• Diffraction and mirages, which can sometimes show objects that are geometrically hidden.

For distances under about 10 km, the curvature effect is small—often less than a meter. For distances over 100 km, the hidden height grows quadratically, so even tall mountains can be completely hidden. Always use the calculator as a first approximation, and verify with local maps or GPS if the result matters.

Frequently Asked Questions

Why does the hidden height not match what I see in photos?

Photos often use telephoto lenses that compress distance and can show mirages or refraction effects. Also, the camera might be higher than eye level, or the object might be on elevated ground. The calculator gives the geometric ideal, not the photographic reality.

How far away is the horizon from a typical person?

For an eye height of 1.7 m (5 ft 7 in), the horizon is about 4.65 km (2.9 miles). From a 10 m cliff, it is about 11.3 km (7 miles). From a commercial airplane at 10 km altitude, it is about 357 km (222 miles).

Does the calculator work for the Moon or other planets?

Yes, if you change the radius. For the Moon (radius 1,737 km), the horizon is much closer. But the same formula works for any spherical body. For Earth, the mean radius of 6,371 km is used.

What is the “8 inches per mile” rule?

That is a common approximation for the Earth’s drop over a mile, but it is only accurate for the first few miles. The true drop is quadratic: drop = 8 inches × (distance in miles)². For 10 miles, that gives 800 inches (66.7 ft), but the exact value is about 66.2 ft. The calculator uses the exact quadratic formula.

Why does the hidden height sometimes appear negative?

If the object is closer than the horizon, the formula would give a negative number, which means nothing is hidden. The calculator clamps the result to zero in that case, so you only see a positive value when the object is actually beyond the horizon.

Can I use this to prove the Earth is round?

Yes, but with caution. If you observe a distant object that is partially hidden and the hidden amount matches the calculator’s prediction (with refraction accounted for), that is consistent with a spherical Earth. However, refraction can make the match less exact, and flat-Earth models have their own explanations. This calculator is a tool for geometric calculation, not a proof by itself.

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in h | Eye height | m | 1.7 | 0
in d | Distance to object | km | 10 | 0
out hidden | Hidden height | m | (d * 1000)^2 * 0.0785 / (2 * 6371000) | 2
out horizon | Horizon distance | km | sqrt(2 * 6371000 * h) / 1000 | 2