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| {{other uses|Horizon (disambiguation)}}
| | I'm Emory and I live with my husband and our three children in Rydal, in the NSW south area. My hobbies are Vehicle restoration, Speed skating and Bowling.<br><br>Feel free to visit my homepage - [http://www.xn--m8s483bxsi.com/plus/guestbook.php FIFA coin Generator] |
| [[File:Water horizon.jpg|thumb|right|A water horizon, in northern [[Wisconsin]], U.S.]]
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| 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''.<ref name="WebstersThird">
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| "offing", ''Webster's Third New International Dictionary, Unabridged''.
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| </ref>
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| The word ''horizon'' derives from the [[Greek language|Greek]] "ὁρίζων κύκλος" ''horizōn kyklos'', "separating circle",<ref>[http://www.perseus.tufts.edu/hopper/text?doc=Perseus%3Atext%3A1999.04.0057%3Aentry%3Do%28ri%2Fzwn "ὁρίζων"], Henry George Liddell and Robert Scott, ''[[A Greek-English Lexicon]]''. On [[Perseus Digital Library]]. Accessed 19 April 2011.</ref> from the verb [[:wikt:ὁρίζω|ὁρίζω]] ''horizō'', "to divide", "to separate",<ref>[http://www.perseus.tufts.edu/hopper/text?doc=Perseus%3Atext%3A1999.04.0057%3Aentry%3Do%28ri%2Fzw "ὁρίζω"], Liddell and Scott, ''A Greek-English Lexicon''.</ref> and that from "ὅρος" (''oros''), "boundary, landmark".<ref>[http://www.perseus.tufts.edu/hopper/text?doc=Perseus%3Atext%3A1999.04.0057%3Aentry%3Do%28%2Fros "ὅρος"], Liddell and Scott, ''A Greek-English Lexicon''.</ref>
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| == Appearance and usage ==
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| [[File:Earth's horizon as seen from Shuttle Endeavour.jpg|thumb|right|View of Earth's horizon as seen from [[Space Shuttle Endeavour|Space Shuttle ''Endeavour'']], 2002]]
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| 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 [[Telegraphy|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 disorientation|spatial orientation]] by referring to the horizon.
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| In many contexts, especially [[Perspective projection|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| reason = in the former case objects would be visible at any distance, which is very obviously not the true on the spherical earth|date=December 2011}} (but for someone on a 1000-meter hill looking out to sea the true horizon will be about a degree below a horizontal line).
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| 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 (astronomy)|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.
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| == Distance to the horizon ==
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| Ignoring the [[#Effect of atmospheric refraction|effect of atmospheric refraction]], distance to the horizon from an observer close to the Earth's surface is about<ref name="ATYoungDistToHoriz">
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| Andrew T. Young, [http://mintaka.sdsu.edu/GF/explain/atmos_refr/horizon.html "Distance to the Horizon"]. Accessed 16 April 2011.</ref>
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| :<math>d \approx 3.57\sqrt{h} \,,</math>
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| where ''d'' is in kilometres and ''h'' is height above ground level in metres.
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| Examples:
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| * For an observer standing on the ground with ''h'' = {{Convert|1.70|m}} (average eye-level height), the horizon is at a distance of {{Convert|4.7|km}}.
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| * For an observer standing on the ground with ''h'' = {{Convert|2|m}}, the horizon is at a distance of {{Convert|5|km}}.
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| * For an observer standing on a hill or tower of {{Convert|100|m}} in height, the horizon is at a distance of {{Convert|36|km}}.
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| * For an observer standing at the top of the [[Burj Khalifa]] ({{Convert|828|m}} in height), the horizon is at a distance of {{Convert|103|km}}.
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| With ''d'' in miles<ref>
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| In this article, ''mile'' refers to a "land" mile of {{Convert|5280|ft|m|3}}.
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| </ref>
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| and ''h'' in feet, | |
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| :<math>d \approx 1.22\sqrt{h} \,.</math>
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| Examples, assuming no refraction:
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| * For an observer on the ground with eye level at {{nowrap|1=''h'' = 5 ft 7 in}} (5.583 ft), the horizon is at a distance of {{Convert|2.9|mi}}.
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| * For an observer standing on a hill or tower {{Convert|100|ft}} in height, the horizon is at a distance of {{Convert|12.2|mi}}.
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| * For an observer on the summit of [[Aconcagua]] ({{Convert|22,841|ft}} in height), the sea-level horizon to the west is at a distance of {{Convert|184|mi}}.
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| * For a [[Lockheed_U-2|U-2]] pilot, whilst flying at its service ceiling {{Convert|70000|ft}}, the horizon is at a distance of {{Convert|324|mi}}
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| === Geometrical model ===
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| [[File:CircleChordTangent.png|thumb|right|300px|Geometrical basis for calculating the distance to the horizon, secant tangent theorem]]
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| [[File:GeometricDistanceToHorizon.png|thumb|right|300px|Geometrical distance to the horizon, Pythagorean theorem]]
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| [[File:Horizons.svg|thumb|right|300px|Three types of horizon]]
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| 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.)
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| The [[Circle#Theorems|secant-tangent theorem]] states that
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| :<math>\mathrm{OC}^2 = \mathrm{OA} \times \mathrm{OB} \,.</math>
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| Make the following substitutions:
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| *''d'' = OC = distance to the horizon
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| *''D'' = AB = diameter of the Earth
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| *''h'' = OB = height of the observer above sea level
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| *''D+h'' = OA = diameter of the Earth plus height of the observer above sea level
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| The formula now becomes
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| :<math>d^2 = h(D+h)\,\!</math>
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| or
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| :<math>d = \sqrt{h(D+h)} =\sqrt{h(2R+h)}\,,</math>
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| where ''R'' is the radius of the Earth.
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| The equation can also be derived using the [[Pythagorean theorem]].
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| 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
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| *''d'' = distance to the horizon
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| *''h'' = height of the observer above sea level
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| *''R'' = radius of the Earth
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| referring to the second figure at the right leads to the following:
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| :<math>(R+h)^2 = R^2 + d^2 \,\!</math>
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| :<math>R^2 + 2Rh + h^2 = R^2 + d^2 \,\!</math>
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| :<math>d = \sqrt{h(2R + h)} \,.</math>
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| Another relationship involves the distance ''s'' along the curved surface of the Earth to the horizon; with ''γ'' in [[radian]]s,
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| :<math>s = R \gamma \,;</math>
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| then
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| :<math>\cos \gamma = \cos\frac{s}{R}=\frac{R}{R+h}\,.</math>
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| Solving for ''s'' gives
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| :<math>s=R\cos^{-1}\frac{R}{R+h} \,.</math>
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| The distance ''s'' can also be expressed in terms of the line-of-sight distance ''d''; from the second figure at the right,
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| :<math>\tan \gamma = \frac {d} {R} \,;</math>
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| substituting for ''γ'' and rearranging gives
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| :<math>s=R\tan^{-1}\frac{d}{R} \,.</math>
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| 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'').
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| === Approximate geometrical formulas ===
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| [[File:How far away is the horizon.png|thumb|right|300px|]]
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| If the observer is close to the surface of the earth, then it is valid to disregard ''h'' in the term {{nowrap|(2''R'' + ''h'')}}, and the formula becomes
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| :<math>d = \sqrt{2Rh} \,.</math>
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| Using metric units and taking the radius of the Earth as 6371 km, the distance to the horizon is
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| :<math>d \approx \sqrt{12.74h} \approx 3.57\sqrt{h} \,,</math>
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| where ''d'' is in kilometres, and ''h'' is the height of the eye of the observer above ground or sea level in metres.
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| Using [[imperial units]], the distance to the horizon is
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| :<math>d \approx \sqrt{1.50h} \approx 1.22 \sqrt{h} </math>
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| where ''d'' is in [[statute mile]]s (as commonly used on land) and ''h'' is in feet.
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| If ''d'' is in [[nautical mile]]s and ''h'' in feet, the constant factor is about 1.06. which is close enough to 1 that it is often ignored, giving:
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| :<math>d \approx \sqrt {h} </math>
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| These formulas may be used when ''h'' is much smaller than the [[Earth radius|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).
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| === Exact formula for a spherical Earth ===
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| 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:
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| :<math>d = \sqrt{2Rh + h^2} \,,</math>
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| where ''R'' is the radius of the Earth (''R'' and ''h'' must be in the same units). For example,
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| if a satellite is at a height of 2000 km, the distance to the horizon is {{convert|5430|km}};
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| neglecting the second term in parentheses would give a distance of {{Convert|5048|km}}, a 7% error.
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| === Objects above the horizon ===
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| [[File:HorizonDistance.png|thumb|right|300px|Geometrical horizon distance]]
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| 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 ({{nowrap|1=''h'' = 1.7 m}}) can just see the tops of trees on a nearby shore ({{nowrap|1=''h'' = 10 m}}), the trees are probably about 16 km away.
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| 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
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| :<math>D_\mathrm{BL} < 3.57\,(\sqrt{h_\mathrm{B}} + \sqrt{h_\mathrm{L}}) \,,</math>
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| where ''D''<sub>BL</sub> is in kilometres and ''h''<sub>B</sub> and ''h''<sub>L</sub> are in metres.
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| 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 [[storey]]s, 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:
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| :<math>D \approx 3.57(\sqrt{2}+\sqrt{70})</math>
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| which comes to about 35 kilometres.
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| 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:
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| :<math> 3.57 \sqrt{10} </math>
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| 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:
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| :<math>h\approx\left(\frac{8.7}{3.57}\right)^2</math>
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| 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]].
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| === Effect of atmospheric refraction ===
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| <!-- this subsection is linked from the beginning of the parent section -->
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| 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.<ref name="ATYoungDistToHoriz" /> Nevertheless, attempts have been made to calculate them more accurately than the simple approximation described above.
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| [[File:Hell on Earth.jpg|thumb|right|Warm [[High Desert (California)|High Desert]] evening displays Mirage effect on Horizon]]
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| 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.
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| '''Integration method—Sweer'''<br/>
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| If the density profile of the atmosphere is known, the distance ''d'' to the horizon is given by<ref name="Sweer1938">John Sweer,
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| [http://adsabs.harvard.edu/abs/1938JOSA...28..327S "The Path of a Ray of Light Tangent to the Surface of the Earth"], ''Journal of the Optical Society of America'', 28 (September 1938):327–29. Available as paid download.
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| </ref>
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| :<math>d={{R}_{\text{E}}}\left( \psi +\delta \right) \,,</math>
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| where ''R''<sub>E</sub> 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
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| :<math>\cos \psi = \frac{{R}_{\text{E}}{\mu}_{0}}{\left( {{R}_{\text{E}}}+h \right)\mu } \,,</math>
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| where ''h'' is the observer's height above the Earth, ''μ'' is the index of refraction of air at the observer's height, and ''μ''<sub>0</sub> is the index of refraction of air at Earth's surface.
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| The refraction must be found by integration of
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| :<math>\delta =-\int_{0}^{h}{\tan \phi \frac{\text{d}\mu }{\mu }} \,,</math>
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| where <math>\phi\,\!</math> is the angle between the ray and a line through the center of the Earth. The angles ''ψ'' and <math>\phi\,\!</math> are related by
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| :<math>\phi =90{}^\circ -\psi \,.</math>
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| '''Simple method—Young'''<br/>
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| 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 {{nowrap|1=''R′'' = 7/6 ''R''<sub>E</sub>}}. The distance to the horizon is then<ref name="ATYoungDistToHoriz"/>
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| :<math>d=\sqrt{2 R^\prime h} \,.</math>
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| Taking the radius of the Earth as 6371 km, with ''d'' in km and ''h'' in m,
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| :<math>d \approx 3.86 \sqrt{h} \,;</math>
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| with ''d'' in mi and ''h'' in ft,
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| :<math>d \approx 1.32 \sqrt{h} \,.</math>
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| Results from Young's method are quite close to those from Sweer's method, and are sufficiently accurate for many purposes.
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| == Curvature of the horizon ==
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| {{Multiple issues|section=yes|
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| {{Unreferenced section|date=June 2013}}
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| {{Disputed|section|date=June 2013}}
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| {{Expand section|1=examples and additional citations|demospace=main|date=June 2013}}
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| }}
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| [[File:A colorful view of airglow layers at Earth's horizon.jpg|thumb|right|The curvature of the horizon is easily seen in this photograph, taken from a space shuttle at an altitude of 226 km in 2008.]]
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| 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 <math>\kappa</math>, the altitude and the Earth's radius. It is
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| :<math>\kappa=\sqrt{\left(1+\frac{h}{R}\right)^2-1}\ .</math>
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| The curvature is the reciprocal of the curvature angular radius in [[radian]]s. 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.
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| The Horizon curves by: sqrt(radius^2 + distance^2)-radius, equivalent to distance^2/R*2. At 100 km, it descends 784m.
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| ==Vanishing points==
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| [[File:TwoPointPerspective.png|thumb|right|210px|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.]]
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| {{main|vanishing point}}
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| 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 lines|parallel]] to ''OS''. But [[Brook Taylor]] (1719) indicated that the horizon plane determined by ''O'' and the horizon was like any other [[plane (geometry)|plane]]:
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| :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...<ref>[[Brook Taylor]] (1719) ''New Principles of Perspective'', p. v, as found in [[Kirsti Andersen]] (1991) Brook Taylor’s Work on Linear Perspective, p. 151, Springer, ISBN 0-387-97486-5</ref>
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| 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#Projective geometry|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 (mathematics)|field]] was deconstructed in the twentieth century, yielding a wide variety of mathematical possibilities. Stillwell states
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| :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.<ref>[[John Stillwell]] (2006) ''Yearning for the Impossible'', Horizon, pp 47 to 76, [[A K Peters, Ltd.]], ISBN 1-56881-254-X</ref>
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| == See also ==
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| * [[Aerial landscape art]]
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| * [[Atmospheric refraction]]
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| * [[Dawn]]
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| * [[Dusk]]
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| * [[Horizontal and vertical]]
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| * [[Landscape]]
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| * [[Landscape art]]
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| * [[Sextant]]
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| == Notes and references ==
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| {{reflist}}
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| == External links ==
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| * [http://mintaka.sdsu.edu/GF/explain/atmos_refr/dip.html Dip of the Horizon]. Andrew T. Young.
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| [[Category:Horizontal coordinate system]]
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| [[Category:Celestial coordinate system]]
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