Horizon

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Artistic representation of the horizon by computer.

Three types of horizons.

The horizon (from Old French orizon, and this, via Latin, from the Greek ὁρίζων (horízōn) and ὅρος ( hóros, “boundary”)) is the line that apparently separates heaven and earth. This line is actually a circle on the Earth's surface centered on the observer.

In other domains, the horizon is defined as a plane passing through the center of the Earth and perpendicular to the zenith-nadir line (a radius from the center of the earth to the surface) or the vertical . Such is the astronomical horizon or rational horizon. The terms of its definition consider that the celestial sphere is not centered in the observer but in the center of the Earth. Since the radius of the earth is negligible compared to the magnitude of the celestial sphere, this plane coincides with the plane perpendicular to the radius of the Earth that passes through the eyes of the observer.

Other types of horizons are defined according to the point of view of the observer:

Apparent horizon: ideal plane tangent to the surface of the Earth at the point of observation.
Sensitive horizon u Real horizon: depends on the local landscape (mountains, buildings, etc.)
Geometric horizon: conical surface with vertex in the observer and tangent to the terrestrial surface.
Physical horizon u optical horizon: determined by the atmospheric refraction, which allows to see below the real horizon.
Horizon line is the line that is the projection of the end of the floor plan or Geometral in the TableIn the Conical Perspective. The representation coincides approximately with the apparent horizon when we are at sea level. It is important to draw because it is the place where all the straights and horizontal planes leak.

Except for the astronomical horizon and the apparent horizon, all the others are optical horizons as they are affected by the phenomenon of refraction.

A view of the horizon from the Spanish coast. Earth's curvature hides the building's base away

A boat that moves away, beyond the horizon

The horizon is a fundamental plane for some celestial coordinates, so the precision of the measurements achieved depends on its correct establishment. Such is the case of geocentric horizontal coordinates, in which it is necessary to take heights above the horizon of a star or a planet. The measurements obtained in situ will, in principle, be referred to the apparent horizon, and they will have to be corrected for atmospheric refraction and geocentric parallax to obtain the height referred to the horizon astronomical.

The geocentric —or height— parallax decreases with height above the horizon, until it becomes zero at the zenith. Its correction, for precision measurements, requires considering the Earth as an ellipsoid and not as a sphere (it really is a geoid), taking the value of the terrestrial radius at the point of observation —not the mean radius—, in addition to the height on the floor. For very distant stars the height parallax may not be significant.

As for refraction, at 0º above the horizon it is worth about 34'. Since the angular diameter of the Sun is about 32', when the disk of the Sun touches the sea what we see is its refracted image, since the Sun is above our optical horizon but already below of our geometric horizon. The refraction decreases with the height above the horizon, as it happened with the height parallax, canceling out at the zenith.

Distance to the horizon

Geometric distance to the horizon, theorem of Pythagoras

Assuming the Earth as a perfect sphere (rather than an Oblate spheroid) without atmospheric refraction, the horizon from a height $h$ is (by the theorem of Pythagoras) at a distance $d$ in a straight line of the observer

${displaystyle d={sqrt {(R+h)^{2}-R^{2}}}}$ ,

where

R

is the radius of the Earth (6378.1 km). Post

{displaystyle h}

It's much lower than

R

, the above expression can be approached as follows:

${displaystyle d={sqrt {2Rh+h^{2}}}approx {sqrt {2Rh}}={sqrt {2R}}{sqrt {h}}approx 3,572{sqrt {h}}}$

where

h

is given in meters and the distance is obtained in kilometers.

The distance $s$ (orthodomic diplomacy) along the terrestrial surface to the horizon is, by trigonometry (with radian angles),

${displaystyle s=Rarccos {R over R+h}}$ .

Like $h$ It's much lower than $R$ The three distances are very similar. As an example, when $h$ = 8844 meters (the height of Mount Everest on the sea level), the three previous expressions respectively give: 335.997, 335.920 and 335.687 meters, so it is evident that in practice it is enough to use the second expression, ${displaystyle 3,572{sqrt {h}}}$ which is the easiest of all three.

Maximum distance of mutual visibility between two elevations

Maximum distance of visibility between two elevations

Two elevations separated by the horizon can be joined by a straight line passing above the Earth, so that they can be seen from each other up to a certain distance. This distance is none other than the sum of their distances to the horizon, as seen in the figure.

If the vigía of the ship of the figure is at a height ${displaystyle h_{B}}$ and the height of the lighthouse on the sea level is ${displaystyle h_{L}}$ , then the vigía will see the lighthouse as long as the distance between the lighthouse and the ship is less than

$<math alttext="{displaystyle D_{BL}DBL.3,572(hB+hL){displaystyle D_{BL} vis3,572,({sqrt {h_{mathrm {B}} }} }+{sqrt {h_{mathrm {L} }}}}}}<img alt="{displaystyle D_{BL}$

where

{displaystyle D_{BL}}

It's in miles and

{displaystyle h_{B}}

and

{displaystyle h_{L}}

in meters.

Example

The vigía wishes to verify its position and, as the only reference point in its navigation area, sees from the ship's command bridge the top of a lighthouse. In the nautical chart you can see both the geographical position and the height on the sea level of the lighthouse ${displaystyle h_{L}}$ in this example of 20 meters. To calculate the distance from the boat to the lighthouse, the sailor knows that the height from the sea level to the command bridge where it is located, ${displaystyle h_{B}}$ is 6 meters, from there, and given that you only see the upper part of the lighthouse can conclude that the lower part of the lighthouse cannot see it due to the curvature of the earth, and can then calculate the distance to the lighthouse ${displaystyle D_{BL}}$ as follows:

${displaystyle D_{BL}=3,572,({sqrt {6}}+{sqrt {20}})}$ or approximately 25 km.

Effect of atmospheric refraction

Main article: Atmospheric reaction

Typical horizon of a desert

If Earth were an airless world like the Moon, light would travel horizontally and the above calculations would be accurate. However, Earth has an atmosphere of air, the density and refractive index of which vary considerably with temperature and pressure. This causes the air to refract light to different degrees, affecting the appearance of the horizon. Usually, the density of the air just above the Earth's surface is greater than its density at higher altitudes. This causes its refractive index to be greater near the surface than higher up, causing light traveling roughly horizontally to be refracted downward. This makes the actual distance to the horizon greater than the distance calculated using geometric formulas.. With standard or normalized atmospheric conditions, the difference is approximately 8%. This changes the factor of 3.57, in the metric formulas used above, to approximately 3.86. This correction can be a fairly good approximation under standard atmospheric conditions.

When conditions are unusual, this approach fails. Refraction is strongly affected by temperature gradients, which can vary considerably from day to day, especially over water. In extreme cases, usually in spring when warm air overcomes cold water, refraction can allow light to follow the Earth's surface for hundreds of kilometers. The 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 upwards, causing mirage effects that make the concept of the horizon meaningless. Values calculated for the effects of refraction under unusual conditions are therefore approximate. However, attempts have been made to calculate them with greater precision than the simple approximation described above.

Outside the visual wavelength range, the refraction will be different. For radar (for example, for wavelengths from 300 to 3 mm, that is, frequencies between 1 and 100 GHz), the radius of the Earth can be multiplied by 4/3 to get an effective radius giving a factor of 4.12 in the metric formula, that is, 15% beyond the geometric horizon or 7% beyond the visual. The factor 4/3 is not exact, since in the visual case the refraction depends on the atmospheric conditions.

Sweer method of integration

If the profile density of the atmospheres is known, the distance d from the horizon is given by

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

where R_E is the radio of the Earth END is the immersion of the horizon δ is the refraction of the horizon. Immersion is determined in a simple way through

{displaystyle cos psi ={frac {{R}_{text{E}}{mu }_{0}}{left({{R}_{text{E}}}+hright)mu }},,}

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

The refraction must be found by integrating

${displaystyle delta =-int _{0}^{h}{tan phi {frac {{text{d}}mu }{mu }}},,}$

where

phi ,!

is the angle between lightning and a line through the center of the Earth. The angles END and

phi ,!

{displaystyle phi =90{}^{circ }-psi ,.}

Young's simple method

A much simpler approach, which provides essentially the same results as the first order approximation presented above, uses the geometric model but uses a radius of R ′ = 7/6 R_E. The distance to the horizon is then

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

Taking the radius of the Earth to be 6371 km, with d in kilometers and h in meters,

{displaystyle dapprox 3.86{sqrt {h}},;}

with d in miles and h in feet,

{displaystyle dapprox 1.32{sqrt {h}},.}

The results of Young's method are quite close to those of Sweer's method, and are accurate enough for most purposes.

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