How Far Is the Horizon? Formula, Refraction, Examples • The Science 360

Q: What is the formula for distance to the horizon?

Distance d = √(2Rh), where R is Earth's radius (6,371 km) and h your eye height. In practice d ≈ 3.57 √h with d in kilometres and h in metres, or d ≈ 1.22 √h with d in miles and h in feet. Including typical refraction, use 3.86 √h km or 1.32 √h miles.

Stand at the water's edge and look out to sea. With your eyes about 1.7 m above the water, the line where sea meets sky is 4.7 km away: a little under three miles, a brisk 50-minute walk if you could walk on water. Every ship you see beyond that line is partly sunk below it, and climbing to a viewpoint just 10 m up, a tall dune or a third-floor balcony, moves the horizon out to 11 km.

The top results for "how far is the horizon" give the $3.57\sqrt{h}$ rule and a few examples, and that is a good start. What they mostly skip is everything that makes the number messy in real life: refraction pushes the horizon out by about 7 to 10% and can, on odd days, push it out a lot more; seeing a distant tall object depends on two horizons, not one; and the popular "8 inches per mile squared" curvature rule answers a different question from the one people use it for. Here are the numbers for all of it.

Quick answerThe horizon is

\sqrt{2Rh}

away, where

R = 6{,}371\ \mathrm{km}

and

h

is your eye height. That is about

3.57\sqrt{h}

km (

h

in metres) or

1.22\sqrt{h}

miles (

h

in feet). At 1.7 m eye height it is 4.7 km (2.9 miles). Normal refraction stretches this by about 8%, to roughly 5 km.

Infographic: How far away is the horizon? A line of sight from a person grazes a curved sea at the horizon; a ship beyond it has its hull hidden. With eyes 1.7 m above the sea the horizon is 4.7 km away, from d = √(2Rh), and refraction adds about 8%. A line chart of horizon distance against eye height shows distance growing with the square root of height: 11.3 km from 10 m, 19.6 km from 30 m and 35.7 km from 100 m. An airliner at 11 km sees 375 km. Myth: 8 inches per mile squared is not the hidden height; from 6 ft at 20 miles about 193 ft is hidden, not 267 ft. — Tap the image to view it full size. View the accessible, editable SVG.

The horizon distance formula, derived in three lines

Draw a line from your eye that just grazes the sea: that grazing point is the horizon. The line meets the radius of the Earth at a right angle, so you, the horizon and the centre of the Earth form a right triangle. Its sides are R (centre to horizon), d (you to horizon) and R + h (centre to you).

Pythagoras on a sphere

\begin{aligned}(R + h)^2 &= R^2 + d^2 \\ d^2 &= 2Rh + h^2\end{aligned}

Since $h$ is tiny compared with $R$ :

\mathbf{d \approx \sqrt{2Rh}}

Numbers for a standing adult, h = 1.7 m

\begin{aligned}d &= \sqrt{2 \times 6{,}371{,}000\ \mathrm{m} \times 1.7\ \mathrm{m}} \\ &= \sqrt{2.166\times10^{7}\ \mathrm{m^2}} = 4{,}654\ \mathrm{m} \\ &\approx \mathbf{4.7\ km}\ (2.9\ \text{miles})\end{aligned}

The handy versions

\begin{aligned}\sqrt{2 \times 6{,}371\ \mathrm{km} \times 1\ \mathrm{m}} &= \sqrt{12.742\ \mathrm{km^2}} \\ &= 3.57\ \mathrm{km}\end{aligned}

So $d \approx \mathbf{3.57\sqrt{h}}$ (km, $h$ in metres). In miles and feet: $d \approx \mathbf{1.22\sqrt{h}}$ . In nautical miles and feet: $d \approx 1.06\sqrt{h}$ , usually rounded to $\sqrt{h}$ .

Because distance grows with the square root of height, height is a weak lever. Double your height and the horizon moves out only 41%. To see three times further you have to climb about ten times higher. Figure 1 shows the whole range, from a beach to the top of Everest.

Log-log plot of horizon distance in kilometres against eye height in metres from 1 to about 15,000. A straight blue line shows geometry only and a dashed orange line just above it includes refraction. Marked points: standing on a beach at 1.7 m sees 4.7 km; a 30 m clifftop sees 19.6 km; the top of the 828 m Burj Khalifa sees 103 km; the 8,849 m summit of Everest sees 336 km. — Figure 1. Distance to the sea horizon versus eye height, from d = √(2Rh + h²). On log axes the relation is a straight line of slope one half. The dashed line adds typical refraction using the 7/6 Earth-radius rule.

A few useful values, all geometric (refraction comes next): 2 m sees 5.0 km, 10 m sees 11.3 km, 30 m sees 19.6 km, 100 m sees 35.7 km, and an airliner at 11 km sees 375 km. The $h^2$ term we dropped only matters when the height is a decent fraction of Earth's radius: from the International Space Station at about 400 km, the exact formula gives a horizon 2,290 km away.

Does refraction make the horizon farther?

Yes. Air gets thinner with height, so its refractive index falls, and a nearly horizontal light ray bends slightly downward, following the curve of the Earth a little. You see over the edge. Surveyors capture this with the refraction coefficient k, the ratio of Earth's radius to the radius of curvature of the light ray. The effect is the same as living on a bigger planet with radius R / (1 − k).

Horizon with refraction

d = \sqrt{\frac{2Rh}{1 - k}}

\begin{aligned}k = 0.13:\ & 1/\sqrt{0.87} = 1.072,\ d \approx \mathbf{3.83\sqrt{h}} \\ k = 1/7:\ & \sqrt{7/6} = 1.080,\ d \approx \mathbf{3.86\sqrt{h}} \\ k = 0.17:\ & 1.098,\ d \approx 3.92\sqrt{h}\end{aligned}

0.13 is a common surveying value; $k = 1/7$ is the "7/6 Earth" rule, $R' = 7R/6$ ; 0.17 is the US Standard Atmosphere (see below). At $h = 1.7\ \mathrm{m}$ , 4.65 km becomes 4.99 to 5.11 km.

You will see " $k = 0.13$ gives the 7/6 rule" in places, and it is a near miss: 7/6 corresponds to $k = 1/7 = 0.143$ , which is also the value behind the standard leveller's correction of $0.0675\,D^2$ metres ( $D$ in km) for curvature and refraction combined. The difference between 3.83 and 3.86 is under 1%, much less than the day-to-day variation, so pick either.

Where does $k$ come from? It depends almost entirely on how fast temperature changes with height in the lowest layer of air. A standard formula, quoted in Wikipedia's article on atmospheric refraction, is $k = 503\,(P/T^2)\,(0.0343 + dT/dh)$ , with $P$ in millibars, $T$ in kelvins and $dT/dh$ in kelvins per metre.

k for different weather near the ground (P = 1013 mbar, T = 288 K)

503 \times 1013/288.15^2 = 6.14

Standard atmosphere, $dT/dh = -0.0065\ \mathrm{K/m}$ :

k = 6.14 \times (0.0343 - 0.0065) = \mathbf{0.171}

Our own ray trace through the US Standard Atmosphere gives the same: $k = 0.170$ . Air cooling at 3.43 °C per 100 m gives $k = 0$ , no bending at all. For $k = 1$ (light follows the Earth's curve exactly):

\begin{aligned}dT/dh &= 1/6.14 - 0.0343 \\ &= \mathbf{+0.129\ K/m}\end{aligned}

That is a 13 °C rise per 100 m.

Plot of horizon distance divided by geometric distance against refraction coefficient k from minus 0.4 to 0.9. The curve 1 over the square root of 1 minus k rises slowly through 1.07 to 1.10 for typical k of 0.13 to 0.17 and 1.15 for the radar 4/3 Earth model, then climbs steeply toward infinity as k approaches 1. A top axis shows the matching temperature change with height, from minus 3 to plus 9 degrees Celsius per 100 metres. — Figure 2. How much refraction stretches the horizon, as a function of k, with the near-ground temperature gradient that produces each k on the top axis. Ordinary days sit in a narrow band near the left. Cold water under warm air pushes k up steeply.

Those strong inversions are not exotic. Warm air flowing over cold water, on a spring afternoon over a lake or the sea, can give k near or above 1. Then distant coastlines and ships "loom" into view that are normally hidden, and mirages like the Fata Morgana appear. The opposite case, a hot road or desert, gives a negative k near the surface: rays bend upward, the horizon comes closer, and you see the inferior mirage that looks like water. The same near-surface layers are what turn the textbook green rim on the setting Sun into a visible flash, which we worked through in what causes the green flash.

Source note We could not open a primary geodesy source for k = 0.13 as the "standard" surveying value; it comes from a search summary describing k as typically 0.13 to 0.16 with 0.143 most common. The values computed from the temperature-gradient formula above do not depend on it.

How far can you see a tall building, ship or mountain?

This is the question people usually mean. A distant object has its own horizon: the point from which its top would just be visible. If you stand at your horizon and it stands at its horizon, and the two points coincide, the line of sight just grazes the sea. So the maximum distance is the sum of the two horizon distances.

Seeing the top of a 100 m tower from the beach

\begin{aligned}\text{you (1.7 m)} &= 3.57\sqrt{1.7} = 4.65\ \mathrm{km} \\ \text{tower (100 m)} &= 3.57\sqrt{100} = 35.7\ \mathrm{km} \\ \text{max} &= 4.65 + 35.7 = \mathbf{40.4\ km} \\ k = 1/7: &\ 5.03 + 38.6 = \mathbf{43.6\ km}\end{aligned}

A 3,000 m mountain from the same beach

\begin{aligned}&3.57\sqrt{3000} + 4.65 \\ &= 195.5 + 4.65 = \mathbf{200\ km}\end{aligned}

That is about 216 km with typical refraction.

Notice how little your own height matters in that sum. The tall object does most of the work, which is why sailors spot a lighthouse or a mountain peak long before the coastline itself. It also means a tiny bit of refraction makes a big difference at long range: 8% of 200 km is 16 km.

If the object is closer than that limit, its top shows and its bottom is hidden. Figure 3 shows the geometry. The hidden part depends on how far the object is beyond your horizon, not on its distance from you.

Diagram, with heights hugely exaggerated, of an observer's eye at height h above a curved Earth of radius 6,371 km. A dashed line of sight grazes the surface at the horizon, a distance d equals the square root of 2Rh away, and continues to a tall object further on. The object's lower portion, below the line of sight, is labelled hidden part and its upper portion visible part. The distance from the horizon to the object is labelled x equals D minus d. — Figure 3. Why the bottom of a distant object disappears first. Beyond your horizon the line of sight climbs away from the surface, and anything below it is hidden. The hidden height is very nearly x² / 2R, where x is the distance from your horizon to the object.

Hidden height of a ship 20 km away, observer 10 m above the sea

\begin{aligned}\text{horizon} &= 3.57\sqrt{10} = 11.3\ \mathrm{km} \\ x &= 20 - 11.3 = 8.7\ \mathrm{km} \\ \text{hidden} &= \frac{x^2}{2R} \\ &= \frac{(8{,}710\ \mathrm{m})^2}{2 \times 6{,}371{,}000\ \mathrm{m}} \\ &= \mathbf{6.0\ m}\end{aligned}

With $k = 1/7$ (use $7R/6$ ): about 4.1 m.

The misconception: “8 inches per mile squared” is not the hidden height

The rule appears in countless arguments about whether the Earth is curved. It says the surface drops 8 inches times the distance in miles squared. The rule is real, and it comes straight from the same geometry.

Where 8 inches per mile squared comes from

Drop below a level line after distance $D$ is $D^2/2R$ (for $D \ll R$ ), with $R = 6{,}371\ \mathrm{km} = 3{,}958.8$ miles:

\begin{aligned}\frac{1}{2 \times 3{,}958.8}\ \text{mile} &= 0.0001263\ \text{mile} \\ &= 0.0001263 \times 63{,}360\ \text{in} \\ &= \mathbf{8.00\ inches}\ \text{per mile}^2\end{aligned}

The catch is the starting point. That level line starts at the surface: it is the drop seen by an eye lying on the ground. Raise your eyes even a couple of metres and you can see 3 miles further before the drop even starts. Use the rule as if it were the hidden height and you overstate it, often badly. People on both sides of the flat-Earth argument make this mistake, which is one reason those arguments go nowhere.

A target 20 miles away across a lake, observer's eyes 6 ft above the water

The "8 inches × miles²" rule says this much "should be hidden":

8 \times 20^2 = 3{,}200\ \text{in} = \mathbf{267\ ft}

Your horizon: $1.22\sqrt{6} = 3.0$ miles, so the target is 17.0 miles beyond it. True hidden height:

8 \times 17.0^2\ \text{in} = 2{,}313\ \text{in} = \mathbf{193\ ft}

With standard refraction ( $k = 1/7$ ): about 161 ft. The naive rule is off by 66% compared with what you would actually observe.

Plot of height hidden below the horizon in feet against distance in miles from 0 to 40. A purple curve labelled 8 inches times miles squared rises to over 1,000 feet at 40 miles and marks 267 feet at 20 miles. Blue curves for an eye 6 feet above the water sit lower, at 193 feet at 20 miles for geometry only and lower still with refraction. Green curves for a 30 foot eye sit lower again. — Figure 4. Hidden height of a distant object versus distance, for eye heights of 6 ft and 30 ft, with and without typical refraction, compared with the "8 inches per mile squared" curve. The rule only applies to an eye at water level.

Two smaller caveats. The 8-inch rule is a parabola approximating a circle, so it slowly under-reads at very long range: at 1,000 miles it gives 126 miles of drop below the tangent line, against 130 miles for the exact geometry. And it measures drop vertically from a level line, which is not the same as the drop along your line of sight to the target. For anything you can see from a beach, these differences are negligible next to refraction.

Is the horizon at eye level?

Not quite. Because the line to the horizon slopes down to graze the curve, the horizon sits slightly below eye level, by an angle called the dip. For small heights the dip is $\sqrt{2h/R}$ radians.

Dip of the horizon

\begin{aligned}1.7\ \mathrm{m}:\ &\sqrt{2 \times 1.7/6{,}371{,}000} \\ &= 7.3\times10^{-4}\ \mathrm{rad} = \mathbf{2.5\ arcmin}\end{aligned}

100 m: 19 arcminutes, about two thirds of the Moon's width. 1,000 m: 61 arcminutes, about 1°. 11,000 m (airliner): 3.4°.

At the beach, 2.5 arcminutes is a twelfth of the Sun's diameter, far too small to notice without an instrument. Navigators correct sextant sightings for it. From an aircraft window, 3.4° is enough that the horizon visibly sits below the wingtips.

What actually limits how far you can see?

Over open sea in clean air, geometry and refraction set the limit, as above. Over land, the horizon is usually terrain: hills, trees and buildings much closer than the geometric horizon. And on most days, haze limits what you can make out long before geometry does. Air molecules alone scatter enough light to wash out distant mountains into the sky colour, the same scattering that makes the sky blue, which we computed in why the sky is blue and not violet. Distant hills look bluer and paler for the same reason.

How far is the horizon on the Moon or Mars?

The formula works anywhere: just change R. Smaller worlds curve away faster.

Horizon at 1.7 m eye height

Earth ( $R = 6{,}371\ \mathrm{km}$ ): 4.7 km

\begin{aligned}\text{Mars:}\ &\sqrt{2 \times 3{,}389{,}500 \times 1.7} \\ &= \mathbf{3.4\ km} \\ \text{Moon:}\ &\sqrt{2 \times 1{,}737{,}400 \times 1.7} \\ &= \mathbf{2.4\ km}\end{aligned}

On the Moon there is no air, so no refraction and no haze, and the horizon is a perfectly sharp edge only 2.4 km away. The same roundness that makes the horizon possible is the subject of why planets are round.

FAQ

How far is the horizon at sea level?

For an adult standing at the water's edge with eyes about 1.7 m above the sea, the geometric horizon is 4.7 km (2.9 miles) away. Normal atmospheric refraction bends light slightly around the curve and pushes it out to about 5.0 km (3.1 miles).

What is the formula for distance to the horizon?

Distance $d = \sqrt{2Rh}$ , where $R$ is Earth's radius (6,371 km) and $h$ your eye height. In practice $d \approx 3.57\sqrt{h}$ with $d$ in kilometres and $h$ in metres, or $d \approx 1.22\sqrt{h}$ with $d$ in miles and $h$ in feet. Including typical refraction, use $3.86\sqrt{h}$ km or $1.32\sqrt{h}$ miles.

How far can you see from an airplane?

From a cruising height of 11 km (36,000 ft), the geometric horizon is about 375 km (233 miles) away, and roughly 400 km with refraction. Haze and cloud usually limit what you can actually make out to much less.

How far away can you see a tall building or mountain?

Add the two horizon distances. A 1.7 m observer's horizon is 4.7 km and a 100 m tower's is 35.7 km, so the tower top can just be seen from 40.4 km, or about 43.6 km with normal refraction. For a 3,000 m peak seen from a beach it is about 200 km before refraction.

Is the 8 inches per mile squared rule correct?

It correctly gives how far Earth's surface drops below a level line drawn from a point on the ground. It does not give how much of a distant object is hidden, because your eyes are above the ground. For a 6 ft observer, 20 miles away, the rule says 267 ft but the true hidden height is 193 ft, or about 160 ft with normal refraction.

Sources

Wikipedia, "Horizon" (geometric formula, 3.57 and 1.22 constants, 7/6 and 4/3 Earth models, examples, combined distances): en.wikipedia.org/wiki/Horizon
Wikipedia, "Atmospheric refraction" (terrestrial refraction formula k = 503 P/T² (0.0343 + dT/dh)): en.wikipedia.org/wiki/Atmospheric_refraction
NPTEL (Indian Institutes of Technology), surveying course, "Error due to Earth's curvature and refraction" (combined correction 0.0675 D² m): archive.nptel.ac.in
Live Science, "How far away is the horizon?" (an example of the standard explanation): livescience.com
US Standard Atmosphere 1976 temperature profile and the Ciddor (1996) refractive index of air, used for our own ray trace of k: refractiveindex.info

Method note: distances use a spherical Earth of mean radius 6,371 km. Refraction is modelled as a constant coefficient k (an effective Earth radius R / (1 − k)), which is good for typical conditions and poor for strong inversions, where rays curve unevenly and mirages form. Hidden heights use the exact tangent-line geometry, not the parabolic approximation.

Header image: NASA (image iss007e10807), regraded for this article.