Distance to the Horizon: the Geometry, and the Refraction Correction

Pure Pythagoras on a sphere, and one of the most-asked questions in physics. The headline here is the GEOMETRIC horizon, which is what an exam wants; the refracted figure, about eight per cent further, is what you actually see, and both are shown. So is the question behind the question: how far away can that lighthouse be before its top dips out of sight?

Last updated 11 Nov 2026

Horizon distance, dip, and what a distant object hides behind

Eye height in metres, kilometres out

Height of your eye above the surfaceIn metres. Eye height, not your height — about 1.60 to 1.75 m for an adult standing on flat ground, and the figure that makes the answer about 4.7 km. Lying on a beach puts it at 0.2 m and the horizon at 1.6 km. A ten-storey window is roughly 30 m, a cruising airliner 11 000 m, the International Space Station 400 000 m. The field goes to 500 km, where the simple formula is still exact because this page keeps the h² term.

Height of a distant objectIn metres, measured from the same surface — a lighthouse, a ship’s mast, a hill, a wind turbine. This is what turns the page from a curiosity into something useful: an object taller than you can be seen from further than your own horizon, because its top peeps over the curve. The two horizons add. A 50 m lighthouse is a typical one; the Eiffel Tower is 330 m and a modern offshore turbine about 260 m to the blade tip.

How far away that object isIn kilometres. Used to work out how much of the object is hidden by the bulge of the Earth between you and it. Anything nearer than your own horizon is fully visible; beyond that, the hidden height grows as the SQUARE of the extra distance, so the last few kilometres hide far more than the first few. The default of 30 km against a 50 m lighthouse is deliberately right on the boundary.

Radius of the EarthIn kilometres. 6371 is the mean radius and the usual choice; the equatorial radius is 6378 and the polar 6357, a spread of about a third of a per cent which moves the horizon distance by about 0.17 per cent — a few metres at eye level. Published horizon coefficients differ in the third figure for exactly this reason, which is worth knowing before you decide that a source is wrong. Change it to 1737 for the Moon, where the same arithmetic gives a horizon at 2.4 km.

Your eye above the Earth's arc, with the sight-line tangent to the surface. The eye height is drawn on a heavily exaggerated logarithmic scale — at your real height the step up would be invisible — but the blue line is genuinely tangent to the circle shown, and both the height mark and the tangent point move with the number you enter.

4.6542kmExample

An adult standing on a beach, eye 1.70 m above the water, looking for a 50 m lighthouse 30 km away

One triangle, one square root, and one correction that is bigger than you expect

d = √(2Rh + h²) · d[km] ≈ 3.5696√h[m] · refracted: use R′ = 7R/6, giving 3.8556√h · combined = d(h) + d(h_obj) · hidden = (D − d)²/2R

d = √(2Rh + h²): the exact geometric distance along the straight line of sight from your eye to the point where it grazes the surface. It comes from the right angle between the sight-line and the radius at the grazing point. The curved distance along the ground is very slightly shorter; at eye level the two differ by under a millimetre and nobody distinguishes them
3.5696√h: the shortcut, for h in metres and d in kilometres, obtained by dropping h² and putting R = 6371 km. You will also see 3.57 and, with the equatorial radius, 3.572 — the spread is the choice of radius, not an error. Good to better than one part in 10⁵ below 100 m, and drifting slowly above that
the 7/6: the standard refraction correction. The air’s density falls with height, so a horizontal ray bends gently downwards — by roughly one seventh of the Earth’s own curvature. Working with an effective radius R′ = 7R/6 reproduces that, and turns the coefficient from 3.5696 into 3.8556: about eight per cent further. It is an average, reliable to a few per cent on an ordinary day and badly wrong over a temperature inversion
combined distance: two horizons add. You can see a 50 m lighthouse from your own horizon distance plus its own, because its top peeps over the bulge. That is why lighthouse ranges are published for a stated observer height, and it is the version of this question that is actually useful at sea
(D − d)²/2R: how much of a distant object the curve hides, for an object beyond your horizon at distance D. The square is the important part: an object 1 km past your horizon loses 8 cm, one 10 km past it loses 7.8 m, one 30 km past it loses 71 m. This is the formula behind every photograph of a distant shoreline with its beach missing and its buildings standing in the sea

Worked example

An adult standing on a beach, eye 1.70 m above the water, looking for a 50 m lighthouse 30 km away
YOUR OWN HORIZON. d = √(2Rh + h²) with R = 6 371 000 m and h = 1.70 m: 2Rh = 2 × 6 371 000 × 1.70 = 2.1661×10⁷ m², and h² = 2.89 m², which is utterly negligible beside it. √(2.1661×10⁷) = 4654.2 m = 4.6542 km. The shortcut 3.5696√1.70 = 4.6547 km, agreeing to four figures.
THE LIGHTHOUSE'S OWN HORIZON. Repeat with h = 50 m: √(2 × 6 371 000 × 50) = 25 240.9 m = 25.2409 km. Notice how much further: the horizon distance goes as √h, so thirty times the height buys only 5.4 times the distance.
ADD THEM. The top of the lighthouse first appears at 4.6542 + 25.2409 = 29.8951 km. Your lighthouse is at 30 km, which is just past that — by about 105 m. Geometrically it is hidden, and only barely.
HOW MUCH IS HIDDEN. The part of it below the line of sight is (D − d)²/2R = ((30 − 4.6542) × 1000)²/(2 × 6 371 000) = 25 345.8²/1.2742×10⁷ = 50.4 m. That is more than the lighthouse is tall, so nothing of it shows. A hundred metres closer and the lamp would just break the horizon.
NOW PUT THE AIR BACK IN. With R′ = 7R/6 = 7 432 833 m, your horizon becomes 5.0271 km and the lighthouse's 27.2633 km, for a combined 32.2904 km. The lighthouse at 30 km is now comfortably visible. Refraction changed the answer from no to yes, which is exactly why nautical tables are built on the refracted figure and not the geometric one.
AND THE DIP. The horizon is not at eye level; it is below it, by cos⁻¹(R/(R + h)) = 0.0419°, about 2.5 arcminutes. Small, but a sextant measures altitudes from the horizon, so every sight has to be corrected for it — and the correction is one of the oldest uses of this formula there is.

How far you can see, geometric and refracted, for R = 6371 km

The square root is what makes this unintuitive. Going from standing on the ground to standing on a 10 m cliff — six times the height — multiplies the horizon by only 2.4. Getting to Everest, five thousand times eye height, multiplies it by 72. That is why a small gain in height near sea level is worth so much and a large one higher up so little, and why coastal lookouts were built on modest headlands rather than mountains. The refraction column is a flat 7.99 per cent further in every row at these heights, because it is just √(7/6) applied to the same formula; the last row is marked because at 400 km the refraction model, built for a ray skimming the ground, no longer really applies. The dip column is small enough to ignore in daily life and far too large to ignore in navigation.
Eye height	Geometric	With refraction	Nautical miles (geometric)	Dip
0.20 m, lying on a beach	1.596 km	1.724 km	0.862 NM	0.86′
1.70 m, standing	4.654 km	5.027 km	2.513 NM	2.51′
10 m, a small cliff	11.288 km	12.192 km	6.095 NM	6.09′
30 m, a ten-storey window	19.551 km	21.118 km	10.556 NM	10.55′
100 m	35.696 km	38.556 km	19.274 NM	19.27′
1000 m, a mountain road	112.885 km	121.929 km	60.952 NM	60.93′
8848 m, the summit of Everest	335.886 km	362.781 km	181.364 NM	3.02°
400 km, the Space Station	2292.771 km	2471.086 km	1238.00 NM	20.80°

The triangle behind it, why your answer is always a bit short, and the lighthouse question

Stand up and look out to sea, and the surface curves away from you. The line of sight that just grazes the water is tangent to the sphere, and tangent means perpendicular to the radius at the point of contact. So you have a right-angled triangle: hypotenuse R + h from the centre of the Earth to your eye, one side R from the centre to the grazing point, and the third side d, which is what you want. Pythagoras gives (R + h)² = R² + d², and therefore d = √(2Rh + h²). That is the whole derivation, and it is worth being able to produce it from scratch, because the formula on its own looks arbitrary and the triangle does not.

Almost every textbook then throws away the h², because at eye level 2Rh is 21.7 million square metres and h² is under three. That gives d ≈ √(2Rh), and with R = 6371 km the convenient d[km] = 3.5696√h[m]. It is a very good approximation — at 100 m it is wrong by eight parts in a million — and it stops being one somewhere above the stratosphere. This page keeps the h², so you can put the Space Station in and get the right answer.

And then the atmosphere moves the answer. Air thins with height, so its refractive index falls with height, so a ray travelling horizontally bends gently downwards — it follows the Earth round, a little. On an average day the ray’s curvature is about one seventh of the Earth’s own, and the standard way to handle it is to pretend the Earth is bigger and the ray straight: use R′ = 7R/6. The coefficient becomes 3.8556 instead of 3.5696, and the horizon moves about eight per cent further out. That is not a rounding error; it is bigger than the difference between the equatorial and polar radii by a factor of twenty-five. The same bending is ordinary refraction, applied continuously through a medium whose index changes with height.

The question people actually want answered is the lighthouse one. Not ‘how far is my horizon’ but ‘how far away can I see that thing’. Two horizons add: yours, and the object’s own, because its top peeps over the bulge from the other side. A 50 m lighthouse has a horizon of 25.2 km all by itself, so from a beach you can see it from nearly 30 km — six times your own horizon distance. Published lighthouse ranges always state an assumed observer height for exactly this reason.

Beyond that combined distance the object does not vanish all at once; it sinks. The hidden height is (D − d)²/2R, and because it is a square, the sinking accelerates. One kilometre past your horizon hides 8 cm. Ten kilometres past it hides 7.8 m. Thirty kilometres past it hides 71 m. That progression is what you see in long-range photographs across water, where the bases of distant buildings are missing and the tops appear to stand in the sea.

Where this goes wrong. Mixing metres and kilometres is the main one — the shortcut 3.5696√h takes h in METRES and returns kilometres, which is a deliberately odd combination and the reason it is quoted so often with the wrong coefficient. After that: using your full height instead of your eye height; forgetting that the object has a horizon of its own; and quoting a geometric figure in a context where everyone else is quoting a refracted one, which makes two correct answers look like a disagreement. If a source gives 3.86 and you have 3.57, neither of you has made a mistake — you are answering different questions, and this page answers both.

Frequently asked questions

How far away is the horizon?

For an adult standing on flat ground with eyes about 1.70 m up, the geometric horizon is 4.65 km, or about 2.5 nautical miles. With ordinary atmospheric refraction included it is 5.03 km. The general formula is d = √(2Rh + h²), and the handy version is d in kilometres = 3.5696√h with h in metres — 3.8556√h if you want the refracted figure.

Why does the horizon distance go as the square root of height?

Because the formula is d = √(2Rh + h²) and the h² is negligible at ordinary heights, leaving d ∝ √h. The practical consequence is that height gives diminishing returns: doubling your height multiplies the distance by only 1.41. Climbing from 1.7 m to 10 m, nearly six times the height, buys you only 2.4 times the distance. It is why a short climb at sea level helps so much and a long one high up helps so little.

What does atmospheric refraction do to the answer?

It pushes the horizon further away, by about eight per cent on an average day. Air density falls with height, so a near-horizontal ray bends gently downwards, curving roughly one seventh as sharply as the Earth does. The standard trick is to use an effective radius of 7R/6 and keep the geometry straight, which turns the coefficient 3.5696 into 3.8556. It is an average: over a strong temperature inversion the bending can be much greater and objects far beyond the calculated horizon become visible, which is where superior mirages come from.

How far away can I see a lighthouse or a mountain?

Add two horizons: your own, and the one belonging to the object. Its top becomes visible when the distance between you equals d(your height) + d(its height). For a 50 m lighthouse seen from a beach that is 4.65 + 25.24 = 29.9 km geometrically, or 32.3 km with refraction. This is why published lighthouse ranges always state the observer height they assume — the number is meaningless without it.

How much of a distant object does the curve hide?

For an object at distance D beyond a horizon at d, the hidden height is approximately (D − d)²/2R. The square matters: 1 km past your horizon hides 8 cm, 10 km hides 7.8 m, 30 km hides 71 m. Note that this is the height hidden below your particular line of sight, so it depends on your eye height as well as the object’s distance — raise your eye and some of it comes back.

Which radius of the Earth should I use?

6371 km, the mean radius, unless you have a reason to do otherwise. The equatorial radius is 6378 km and the polar 6357 km, a spread of about a third of a per cent, which moves the horizon distance by about 0.17 per cent — roughly eight metres at eye level. That is why published coefficients differ in the third figure, with 3.57 and 3.572 both in circulation. It is a choice of radius, not a disagreement about physics.

Related calculators

References

A. T. Young, Distance to the Horizon, in the atmospheric refraction pages, San Diego State University (read 11 November 2026). Derives OG = √(2Rh + h²), notes that h² is dropped because h << R, and with R = 6378 km gives the coefficients as about 3.57 km per √(metre) and 1.23 miles per √(foot). For refraction it treats the ray as a circular arc of curvature about one seventh of the Earth’s, giving an effective radius R′ = 7R/6 ≈ 7440 km and coefficients of 3.86 km per √(metre) and 1.32 miles per √(foot). It states plainly that these are reliable to a few per cent most of the time and can be far off where there are inversions or ducts, which is the caveat this page repeats.
The International System of Units (SI), 9th edition, Bureau International des Poids et Mesures, 2019. The metre and the nautical mile: the nautical mile is not an SI unit, and the conversion used here, 1 NM = 1852 m exactly, is the international definition adopted in 1929 and used by every maritime authority since. It was originally one minute of arc along a meridian, which is why the dip and the distance tables in navigation fit together so neatly.
Values of Fundamental Physical Constants, NIST Physical Measurement Laboratory wall chart, 2022 printing (read 11 November 2026). Cited for completeness on units; no constant on this page comes from physics rather than geometry. The Earth’s radius is a measured property of one planet, not a fundamental constant, and the 6371 km mean value used here is the standard figure quoted in geodesy and astronomy. The equatorial 6378 km and polar 6357 km figures are the conventional ones; they were not checked against a primary geodetic reference such as WGS 84 for this page.
The 7/6 effective-radius rule is an engineering and surveying convention rather than a law, and it is quoted here from the San Diego State page above. Its physical basis — that the ray’s curvature is about one seventh of the Earth’s under an average lapse rate — is stated there. The underlying refraction coefficient varies with temperature gradient, and no primary meteorological source was read for this page, so the eight per cent correction should be taken as a typical value and not a guaranteed one.