# Horizon

For other uses, see Horizon (disambiguation).
A water horizon, in northern Wisconsin, U.S.

The horizon (or skyline) is the apparent line that separates earth from sky, the line that divides all visible directions into two categories: those that intersect the Earth's surface, and those that do not. At many locations, the true horizon is obscured by trees, buildings, mountains, etc., and the resulting intersection of earth and sky is called the visible horizon. When looking at a sea from a shore, the part of the sea closest to the horizon is called the offing.[1] The word horizon derives from the Greek "ὁρίζων κύκλος" horizōn kyklos, "separating circle",[2] from the verb ὁρίζω horizō, "to divide", "to separate",[3] and that from "ὅρος" (oros), "boundary, landmark".[4]

## Appearance and usage

View of Earth's horizon as seen from Space Shuttle Endeavour, 2002

Historically, the distance to the visible horizon at sea has been extremely important as it represented the maximum range of communication and vision before the development of the radio and the telegraph. Even today, when flying an aircraft under Visual Flight Rules, a technique called attitude flying is used to control the aircraft, where the pilot uses the visual relationship between the aircraft's nose and the horizon to control the aircraft. A pilot can also retain his or her spatial orientation by referring to the horizon.

In many contexts, especially perspective drawing, the curvature of the Earth is disregarded and the horizon is considered the theoretical line to which points on any horizontal plane converge (when projected onto the picture plane) as their distance from the observer increases. For observers near sea level the difference between this geometrical horizon (which assumes a perfectly flat, infinite ground plane) and the true horizon (which assumes a spherical Earth surface) is imperceptible to the naked eye[dubious ] (but for someone on a 1000-meter hill looking out to sea the true horizon will be about a degree below a horizontal line).

In astronomy the horizon is the horizontal plane through (the eyes of) the observer. It is the fundamental plane of the horizontal coordinate system, the locus of points that have an altitude of zero degrees. While similar in ways to the geometrical horizon, in this context a horizon may be considered to be a plane in space, rather than a line on a picture plane.

## Distance to the horizon

Ignoring the effect of atmospheric refraction, distance to the horizon from an observer close to the Earth's surface is about[5]

$d \approx 3.57\sqrt{h} \,,$

where d is in kilometres and h is height above ground level in metres.

Examples:

• For an observer standing on the ground with h = 1.70 metres (5 ft 7 in) (average eye-level height), the horizon is at a distance of 4.7 kilometres (2.9 mi).
• For an observer standing on the ground with h = 2 metres (6 ft 7 in), the horizon is at a distance of 5 kilometres (3.1 mi).
• For an observer standing on a hill or tower of 100 metres (330 ft) in height, the horizon is at a distance of 36 kilometres (22 mi).
• For an observer standing at the top of the Burj Khalifa (828 metres (2,717 ft) in height), the horizon is at a distance of 103 kilometres (64 mi).
• For an observer atop Mount Everest (8,848 metres (29,029 ft) altitude), the horizon is at a distance of 336 kilometres (209 mi).

With d in miles[6] and h in feet,

$d \approx 1.22\sqrt{h} \,.$

Examples, assuming no refraction:

• For an observer on the ground with eye level at h = 5 ft 7 in (1.70 m), the horizon is at a distance of 2.9 miles (4.7 km).
• For an observer standing on a hill or tower 100 feet (30 m) in height, the horizon is at a distance of 12.2 miles (19.6 km).
• For an observer on the summit of Aconcagua (22,841 feet (6,962 m) in height), the sea-level horizon to the west is at a distance of 184 miles (296 km).
• For a U-2 pilot, whilst flying at its service ceiling 70,000 feet (21,000 m), the horizon is at a distance of 324 miles (521 km)

### Geometrical model

Geometrical basis for calculating the distance to the horizon, secant tangent theorem
Geometrical distance to the horizon, Pythagorean theorem
Three types of horizon

If the Earth is assumed to be a sphere with no atmosphere then the distance to the horizon can easily be calculated. (The Earth's radius of curvature actually varies by 1% between the Equator and the Poles, so this formula isn't absolutely exact even assuming no refraction.)

The secant-tangent theorem states that

$\mathrm{OC}^2 = \mathrm{OA} \times \mathrm{OB} \,.$

Make the following substitutions:

• d = OC = distance to the horizon
• D = AB = diameter of the Earth
• h = OB = height of the observer above sea level
• D+h = OA = diameter of the Earth plus height of the observer above sea level

The formula now becomes

$d^2 = h(D+h)\,\!$

or

$d = \sqrt{h(D+h)} =\sqrt{h(2R+h)}\,,$

where R is the radius of the Earth.

The equation can also be derived using the Pythagorean theorem. Since the line of sight is a tangent to the Earth, it is perpendicular to the radius at the horizon. This sets up a right triangle, with the sum of the radius and the height as the hypotenuse. With

• d = distance to the horizon
• h = height of the observer above sea level
• R = radius of the Earth

referring to the second figure at the right leads to the following:

$(R+h)^2 = R^2 + d^2 \,\!$
$R^2 + 2Rh + h^2 = R^2 + d^2 \,\!$
$d = \sqrt{h(2R + h)} \,.$

Another relationship involves the distance s along the curved surface of the Earth to the horizon; with γ in radians,

$s = R \gamma \,;$

then

$\cos \gamma = \cos\frac{s}{R}=\frac{R}{R+h}\,.$

Solving for s gives

$s=R\cos^{-1}\frac{R}{R+h} \,.$

The distance s can also be expressed in terms of the line-of-sight distance d; from the second figure at the right,

$\tan \gamma = \frac {d} {R} \,;$

substituting for γ and rearranging gives

$s=R\tan^{-1}\frac{d}{R} \,.$

The distances d and s are nearly the same when the height of the object is negligible compared to the radius (that is, h ≪ R).

### Approximate geometrical formulas

If the observer is close to the surface of the earth, then it is valid to disregard h in the term (2R + h), and the formula becomes

$d = \sqrt{2Rh} \,.$

Using metric units and taking the radius of the Earth as 6371 km, the distance to the horizon is

$d \approx \sqrt{12742000h} \approx 3570\sqrt{h} \,,$

where d is in kilometres, and h is the height of the eye of the observer above ground or sea level in metres.

Using imperial units, the distance to the horizon is

$d \approx \sqrt{1.50h} \approx 1.22 \sqrt{h}$

where d is in statute miles (as commonly used on land) and h is in feet.

If d is in nautical miles and h in feet, the constant factor is about 1.06. which is close enough to 1 that it is often ignored, giving:

$d \approx \sqrt {h}$

These formulas may be used when h is much smaller than the radius of the Earth (6371 km), including all views from any mountaintops, aeroplanes, or high-altitude balloons. With the constants as given, both the metric and imperial formulas are precise to within 1% (see the next section for how to obtain greater precision).

### Exact formula for a spherical Earth

If h is significant with respect to R, as with most satellites, then the approximation made previously is no longer valid, and the exact formula is required:

$d = \sqrt{2Rh + h^2} \,,$

where R is the radius of the Earth (R and h must be in the same units). For example, if a satellite is at a height of 2000 km, the distance to the horizon is 5,430 kilometres (3,370 mi); neglecting the second term in parentheses would give a distance of 5,048 kilometres (3,137 mi), a 7% error.

### Objects above the horizon

Geometrical horizon distance

To compute the greatest distance at which an observer can see the top of an object above the horizon, compute the distance to the horizon for a hypothetical observer on top of that object, and add it to the real observer's distance to the horizon. For example, for an observer with a height of 1.70 m standing on the ground, the horizon is 4.65 km away. For a tower with a height of 100 m, the horizon distance is 35.7 km. Thus an observer on a beach can see the top of the tower as long as it is not more than 40.35 km away. Conversely, if an observer on a boat (h = 1.7 m) can just see the tops of trees on a nearby shore (h = 10 m), the trees are probably about 16 km away.

Referring to the figure at the right, the top of the lighthouse will be visible to a lookout in a crow's nest at the top of a mast of the boat if

$D_\mathrm{BL} < 3.57\,(\sqrt{h_\mathrm{B}} + \sqrt{h_\mathrm{L}}) \,,$

where DBL is in kilometres and hB and hL are in metres.

As another example, suppose an observer, whose eyes are two metres above the level ground, uses binoculars to look at a distant building which he knows to consist of thirty storeys, each 3.5 metres high. He counts the storeys he can see, and finds there are only ten. So twenty storeys or 70 metres of the building are hidden from him by the curvature of the Earth. From this, he can calculate his distance from the building:

$D \approx 3.57(\sqrt{2}+\sqrt{70})$

which comes to about 35 kilometres.

It is similarly possible to calculate how much of a distant object is visible above the horizon. Suppose an observer's eye is 10 metres above sea level, and he is watching a ship that is 20 km away. His horizon is:

$3.57 \sqrt{10}$

kilometres from him, which comes to about 11.3 kilometres away. The ship is a further 8.7 km away. The height of a point on the ship that is just visible to the observer is given by:

$h\approx\left(\frac{8.7}{3.57}\right)^2$

which comes to almost exactly six metres. The observer can therefore see that part of the ship that is more than six metres above the level of the water. The part of the ship that is below this height is hidden from him by the curvature of the Earth. In this situation, the ship is said to be hull-down.

### Effect of atmospheric refraction

If the Earth were an airless world like the Moon, the above calculations would be accurate. However, the real Earth is surrounded by an atmosphere of air, the density and refractive index of which vary considerably depending on the temperature and pressure. This makes the air refract light to varying extents, affecting the appearance of the horizon. Usually, the density of the air just above the surface of the Earth is greater than its density at greater altitudes. This makes its refractive index greater near the surface than higher, which causes light that is travelling roughly horizontally to be refracted downward, so it goes, to some small degree, around the curvature of the Earth's surface. This makes the actual distance to the horizon greater than the distance calculated with geometrical formulas. With standard atmospheric conditions, the difference is about 8%. This changes the factor of 3.57, in the metric formulas used above, to about 3.86. This correction can be, and often is, applied as a fairly good approximation when conditions are close to standard. When conditions are unusual, however, it can be badly wrong. Refraction is strongly affected by temperature gradients, which can vary considerably from day to day, especially over water. In extreme cases, usually in springtime, when warm air overlies cold water, refraction can allow light to follow the Earth's surface for hundreds of kilometres. Opposite conditions occur, for example, in deserts, where the surface is very hot, so hot, low-density air is below cooler air. This causes light to be refracted upward, causing mirage effects that make the concept of the horizon somewhat meaningless. Calculated values for the effects of refraction under unusual conditions are therefore only approximate.[5] Nevertheless, attempts have been made to calculate them more accurately than the simple approximation described above.

Warm High Desert evening displays Mirage effect on Horizon

Outside the visual wavelength range the refraction will be different. For radar (e.g. for wavelengths 30 to 300 mm i.e. frequencies between 10 and 1 GHz) the radius of the Earth may be multiplied by 4/3 to obtain an effective radius giving a factor of 4.12 in the metric formula i.e. the radar horizon will be 15% beyond the geometrical horizon or 7% beyond the visual. The 4/3 factor is not exact, as in the visual case the refraction depends on atmospheric conditions.

Integration method—Sweer
If the density profile of the atmosphere is known, the distance d to the horizon is given by[7]

$d={{R}_{\text{E}}}\left( \psi +\delta \right) \,,$

where RE is the radius of the Earth, ψ is the dip of the horizon and δ is the refraction of the horizon. The dip is determined fairly simply from

$\cos \psi = \frac{{R}_{\text{E}}{\mu}_{0}}{\left( {{R}_{\text{E}}}+h \right)\mu } \,,$

where h is the observer's height above the Earth, μ is the index of refraction of air at the observer's height, and μ0 is the index of refraction of air at Earth's surface.

The refraction must be found by integration of

$\delta =-\int_{0}^{h}{\tan \phi \frac{\text{d}\mu }{\mu }} \,,$

where $\phi\,\!$ is the angle between the ray and a line through the center of the Earth. The angles ψ and $\phi\,\!$ are related by

$\phi =90{}^\circ -\psi \,.$

Simple method—Young
A much simpler approach, which produces essentially the same results as the first-order approximation described above, uses the geometrical model but uses a radius R′ = 7/6 RE. The distance to the horizon is then[5]

$d=\sqrt{2 R^\prime h} \,.$

Taking the radius of the Earth as 6371 km, with d in km and h in m,

$d \approx 3.86 \sqrt{h} \,;$

with d in mi and h in ft,

$d \approx 1.32 \sqrt{h} \,.$

Results from Young's method are quite close to those from Sweer's method, and are sufficiently accurate for many purposes.

## Curvature of the horizon

The curvature of the horizon is easily seen in this photograph, taken from a space shuttle at an altitude of 226 km in 2008.

From a point above the surface the horizon appears slightly bent (it is a circle). There is a basic geometrical relationship between this visual curvature $\kappa$, the altitude and the Earth's radius. It is

$\kappa=\sqrt{\left(1+\frac{h}{R}\right)^2-1}\ .$

The curvature is the reciprocal of the curvature angular radius in radians. A curvature of 1 appears as a circle of an angular radius of 45° corresponding to an altitude of approximately 2640 km above the Earth's surface. At an altitude of 10 km (33,000 ft, the typical cruising altitude of an airliner) the mathematical curvature of the horizon is about 0.056, the same curvature of the rim of circle with a radius of 10 m that is viewed from 56 cm. However, the apparent curvature is less than that due to refraction of light in the atmosphere and because the horizon is often masked by high cloud layers that reduce the altitude above the visual surface. The Horizon curves by: sqrt(radius^2 + distance^2)-radius, equivalent to distance^2/R*2. At 100 km, it descends 784m.

## Vanishing points

Two points on the horizon are at the intersections of the lines extending the segments representing the edges of the building in the foreground. The horizon line coincides here with the line at the top of the doors and windows.
Main article: Vanishing point

The horizon is a key feature of the picture plane in the science of graphical perspective. Assuming the picture plane stands vertical to ground, and P is the perpendicular projection of the eye point O on the picture plane, the horizon is defined as the horizontal line through P. The point P is the vanishing point of lines perpendicular to the picture. If S is another point on the horizon, then it is the vanishing point for all lines parallel to OS. But Brook Taylor (1719) indicated that the horizon plane determined by O and the horizon was like any other plane:

The term of Horizontal Line, for instance, is apt to confine the Notions of a Learner to the Plane of the Horizon, and to make him imagine, that that Plane enjoys some particular Privileges, which make the Figures in it more easy and more convenient to be described, by the means of that Horizontal Line, than the Figures in any other plane;…But in this Book I make no difference between the Plane of the Horizon, and any other Plane whatsoever...[8]

The peculiar geometry of perspective where parallel lines converge in the distance, stimulated the development of projective geometry which posits a point at infinity where parallel lines meet. In her 2007 book Geometry of an Art, Kirsti Andersen described the evolution of perspective drawing and science up to 1800, noting that vanishing points need not be on the horizon. In a chapter titled "Horizon", John Stillwell recounted how projective geometry has led to incidence geometry, the modern abstract study of line intersection. Stillwell also ventured into foundations of mathematics in a section titled "What are the Laws of Algebra ?" The "algebra of points", originally given by Karl von Staudt deriving the axioms of a field was deconstructed in the twentieth century, yielding a wide variety of mathematical possibilities. Stillwell states

This discovery from 100 years ago seems capable of turning mathematics upside down, though it has not yet been fully absorbed by the mathematical community. Not only does it defy the trend of turning geometry into algebra, it suggests that both geometry and algebra have a simpler foundation than previously thought.[9]