Distance to Horizon Calculator

Introduction: why horizon distance is a square-root problem

Stand on a beach and the horizon is about five kilometres away. Climb a hundred-metre cliff behind the beach — fifty times your standing height — and it is only about thirty-six kilometres away, seven times as far, not fifty. That is the whole story of the horizon in one comparison: distance grows with the square root of height, so the first few metres of elevation buy you far more than the next few hundred.

The reason is geometry. The Earth curves away beneath you at a rate set by its radius, and your line of sight leaves your eye and runs tangent to that curve. The tangent point is the horizon. Raise the eye and the tangent point slides further away, but the surface is falling away quadratically, so the gain in distance goes as the square root of the gain in height. If the thing you are looking at also has height — a mast, a lighthouse, an antenna — it has its own horizon, and the two of you can see each other when the sum of your horizon distances exceeds the distance between you.

Where most horizon calculators stop being useful is the atmosphere. Air is denser near the surface, so light bends gently downward as it travels, following the curve of the Earth a little way past the geometric tangent point. The effect is not a rounding error: for visible light in a standard atmosphere it pushes the horizon about 8% further out, and for VHF and UHF radio it is closer to 15%. This calculator lets you choose which of those you want, because the answer to "how far can I see" and the answer to "how far will my radio reach" are genuinely different numbers over the same ground.

It also answers the question that follows immediately after the first one. If a ship is beyond your horizon, it is not invisible — the top of it may still be showing. Given a distance, the calculator reports how much of a distant object is hidden below the curve, which is the number a sailor watching a hull disappear, or a photographer planning a shot across a bay, actually needs.

How to use the distance to horizon calculator

1. Enter your eye height, not your standing height. Eye height above the surface is what the geometry uses. Standing on a beach that is 1.7 m for most adults; on a boat it is the deck height plus your eye height; on a cliff it is the cliff plus the same again.
2. Add the target height if there is one. A mast, a lighthouse focal plane, a building, another antenna. Leaving it blank gives the plain horizon distance for your eye alone.
3. Choose the refraction model. Geometric is pure spherical geometry with no atmosphere and is the right answer only in a vacuum. Standard optical refraction uses an effective Earth radius seven-sixths of the true one and is the conventional figure for seeing. The four-thirds Earth model is the long-standing convention for VHF and UHF radio propagation. Custom lets you enter your own k factor for unusual conditions.
4. Optionally give the actual distance to the object. If you do, the calculator reports how much of it is hidden below the horizon and how much is still showing, which is more useful than a yes-or-no visibility verdict.
5. Read the diagram. The curvature sketch under the result is drawn to the geometry you entered, with the tangent point marked, so the relationship between the two horizon distances is visible rather than implied.

The formula, with and without an atmosphere

Draw a right triangle from the Earth's centre through the observer to the tangent point. The radius to the tangent point is perpendicular to the line of sight, so with the Earth's radius and the eye height, Pythagoras gives the exact distance along the line of sight:

Because is 6,371 km and is typically metres, the term is negligible for anything on the ground and the familiar approximation follows:

Atmospheric refraction is handled the way surveyors and radio engineers handle it, by replacing the true radius with an effective radius . Light bending downward around a sphere of radius is geometrically identical to light travelling straight around a larger sphere:

which collapses to a single coefficient per model, with in kilometres and in metres: 3.57√h for pure geometry, 3.86√h for standard optical refraction at k = 7/6, and 4.12√h for the four-thirds Earth radio model. Two observers or an observer and a target simply add their horizons:

How much of a distant object is hidden

Inverting the same relation answers the more practical question. Anything beyond your own horizon distance is hidden up to the height whose horizon distance equals the remaining gap, so at a true separation :

Plain-text formula: horizonMetres = sqrt(2 * k * 6371000 * heightMetres); totalRange = horizon(observer) + horizon(target); hiddenMetres = (distance - horizon(observer))^2 / (2 * k * 6371000), with k = 1 for geometry, 7/6 for standard optical refraction and 4/3 for the radio model.

Worked example: a sailboat mast from the beach

You are on a beach with an eye height of 1.8 m, watching a sailboat whose masthead is 15 m above the water.

• Pure geometry. Your horizon is √(2 × 6,371,000 × 1.8) = 4.79 km; the masthead's is 13.82 km; the boat's mast tip drops out of sight at 18.61 km.
• Standard optical refraction (k = 7/6). Your horizon becomes 5.17 km, the masthead's 14.93 km, and the masthead stays visible out to 20.11 km — 1.5 km, or 8%, further than geometry alone predicts.
• Four-thirds Earth (k = 4/3). If instead of watching the mast you were talking to a VHF antenna at the masthead, the radio horizon is 21.49 km, another 7% further again.

Now put the boat at a real distance. At 20 km under standard refraction, your own horizon reaches 5.17 km, leaving 14.83 km of curve between the tangent point and the boat. That curve hides everything below 14.79 m — so of a 15 m mast, the top 21 cm is still showing and the hull, deck and almost the entire mast are below the horizon. Move the boat to 25 km and the hidden height rises to 26.4 m, comfortably more than the mast is tall, and the boat has gone completely.

That last calculation is the arithmetic behind the oldest piece of evidence for a round Earth. A ship does not shrink to a dot as it leaves; it sinks, hull first, because the hidden height grows as the square of the distance beyond your horizon. Doubling the distance past the tangent point quadruples how much of the ship is swallowed.

Doubling the height never doubles the distance. Going from 50 m to 200 m — four times the height — exactly doubles the horizon, because the square root of four is two. That is why a modest coastal lookout captures most of the benefit available, and why the enormous expense of building higher gives diminishing returns for anything that depends on line of sight, from lighthouses to mobile phone masts.

Practical uses of a horizon distance

At sea. A lighthouse's charted range is normally its geographic range for a stated observer eye height, computed exactly this way. Knowing your own eye height lets you correct the charted figure for where you are actually standing, which on a small boat with a 1.5 m eye height rather than the charted 5 m is a difference of several kilometres.

On the radio. VHF marine and aviation communication, and most terrestrial microwave links, are line-of-sight. The four-thirds Earth model is the standard planning assumption, and the sum-of-two-horizons formula is exactly how link budgets decide whether two antennas can see each other before path loss is even considered.

Behind a camera. Whether a distant island, wind farm or skyline will appear above the water in a long-lens shot is a hidden-height calculation. So is the question of how much of a subject you will lose, which is what decides whether the shot is worth the drive.

Walking and flying. From a summit the horizon is often quoted as evidence of a view, but on land the geometric horizon is usually irrelevant because terrain intervenes long before it. The calculation is honest only over water or genuinely flat ground.

Limitations and refinements

Refraction is variable, not a constant. The k = 7/6 figure is a standard-atmosphere convention, not a measurement. Strong temperature inversions over cold water can bend light far more, producing looming and superior mirages that let observers see objects hundreds of kilometres away; a hot surface under cold air does the reverse and pulls the horizon in. Treat the refraction models as typical, not guaranteed.

The Earth is not a sphere. It is an oblate spheroid whose radius varies from about 6,357 km at the poles to 6,378 km at the equator. This page uses the mean radius of 6,371 km, which is within about 0.3% everywhere — smaller than the uncertainty in refraction on any given day.

Terrain and obstructions are not modelled. Hills, buildings, trees and wave height all block a view well before the geometric horizon. Over open water the assumptions hold well; over land they usually do not.

Sea state matters at low eye heights. From a kayak with a 1 m eye height, a 1.5 m swell obscures far more than the curvature does. The calculation assumes a smooth surface.

Visibility is not the same as detectability. Being geometrically above the horizon says nothing about whether an object is large enough, bright enough or contrasty enough to be seen through haze at that range. The calculator gives the geometric limit, which is a ceiling on visibility rather than a prediction of it.

Common questions about the horizon

Sources and assumptions. The Earth's mean radius is taken as 6,371 km, the IUGG mean radius; the actual radius varies from about 6,357 km at the poles to 6,378 km at the equator. Refraction is modelled by the standard effective-Earth-radius method, in which light curving around a sphere of radius R is treated as travelling straight around a sphere of radius kR. The value k = 7/6 for visible light and k = 4/3 for VHF and UHF radio are the long-established conventions in surveying and radio propagation respectively, corresponding to the familiar coefficients of 3.86 and 4.12 kilometres per square root metre against 3.57 for pure geometry. Real refraction depends on the vertical temperature and pressure gradient and varies substantially with conditions, so these are typical values rather than guarantees; strong inversions can produce far larger effects. The calculation assumes a smooth surface with no terrain, obstructions, waves or haze, and gives a geometric limit on visibility rather than a prediction that an object will actually be seen. Everything runs in your browser and nothing is stored or transmitted.