Visibility

The title of this article is ambiguous. For visibility when designing roads, see Visibility (traffic planning) . For underwater visibility, see Underwater visibility .

As sight or vision in the narrower sense is defined as the greatest horizontal distance a dark object in in the off-road ground can be detected just before a light background. It is also referred to as the meteorological visibility , in contrast to other visibility:

• Visibility at night ( range , night vision, fire visibility ), at which a light source is just barely perceived by an observer, also limited meteorologically.
• Geometric or geographical visibility is limited by the curvature of the earth and is influenced by the height position of the observer and the target as well as geographical obstructions.
  • Taking into account the atmospheric refraction, this results in the geodetic visibility.

Atmospheric visibility

Left: visibility in clear weather; Right: Fog-reduced visibility

Decrease in contrast relative to the sky with increasing distance of the mountains

Three effects limit the atmospheric visibility:

• atmospheric: hydrometeors such as precipitation , snowfall or fog or lithometeors such as dust or smoke dampen the light or the contrast
• Air pollution: Aerosols cause additional light attenuation

Atmospheric scattering and absorption reduce the contrast of an object relative to its surroundings. This phenomenon is called light attenuation. The contrast depends exponentially on the distance and an absorption coefficient : ${\displaystyle K}$ ${\displaystyle s}$ ${\displaystyle \sigma }$

${\displaystyle K=K_{0}\cdot e^{-\sigma \cdot s}}$

A minimum contrast of

${\displaystyle K=0{,}02\;{\hat {=}}\;2\,\%}$

required. Assuming that the output contrast is approximately 1, can be out of sight directly to be closed: ${\displaystyle K_{0}}$ ${\displaystyle s}$ ${\displaystyle \sigma }$

${\displaystyle \sigma ={\frac {\ln(50)}{s}}\approx {\frac {3{,}91}{s}}}$

A visibility of 50 km corresponds to an absorption constant of . In good conditions the visibility is a few hundred kilometers, see table. ${\displaystyle 10^{-4}/\mathrm {m} }$

In the example image, the contrast between the mountains and the sky decreases with increasing distance. The mountain range in the right picture can no longer be seen in fog.

Weather dependence of visibility

Weather condition	Visibility in km
Exceptionally clear	280
Very clear	50
Clear	20th
Slightly hazy	10
Hazy	4th
Heavy haze, light fog	2
Moderate fog	1
Thick fog, heavy rain	0.1
Extreme fog, blowing snow	0.01

Visibility in the water

→ Main article : Underwater visibility

Depending on the wavelength, pure sea water has an extinction length 1 / σ of 2–100 m. When diving in natural waters , a visibility of 40 meters is considered extremely good. The view can be clouded by suspended particles ( plankton , pollen, desert sand), by alluvial particles in currents (river mouth) or by sewage and the discharge of chemical substances.

Geodetic visibility

calculation

The geodetic curvature of the earth limits the maximum visual range for objects from a point of view on the earth's surface or on spherically curved bodies. The range of vision from an elevated observation point or a high object (e.g. mountain peak) from a plane or from the surface of the sea can be calculated according to the Pythagorean theorem , since the line of sight and the earth's radius form the cathetus of a right triangle and the distance from the elevated point from the center of the earth its hypotenuse :

Right-angled triangle for calculating visibility

(1) ${\displaystyle s^{2}+R^{2}=(R+h)^{2}}$

(2) ${\displaystyle s={\sqrt {(R+h)^{2}-R^{2}}}}$

According to the first binomial formula, this results in:

(3) ${\displaystyle s={\sqrt {R^{2}+2Rh+h^{2}-R^{2}}}={\sqrt {2Rh+h^{2}}}}$

Because except in the space opposite is negligibly small, the formula can be simplified to ${\displaystyle h^{2}}$ ${\displaystyle 2Rh}$

(4) ${\displaystyle s\approx {\sqrt {2Rh}}={\sqrt {2R}}\cdot {\sqrt {h}}}$

The following formulas for practical use (and numbered with additional letters) give the visibility s in km , whereby the height h is to be inserted in meters . (In order to arrive at these practicable units of measurement, the mean earth radius of R = 6370 km compared to (4) or (6) with 6.37 megameters was taken into account.)

(5a) ${\displaystyle {\underline {s\approx 3{,}57\cdot {\sqrt {h}}}}}$

The refraction of the atmosphere bends the rays of light and makes the earth appear larger. The mean apparent radius of the earth is R _k ≈ 7680 km. The optical range of vision is usually increased by around 10% (in exceptional cases, however, considerably more or less):

(5b) ${\displaystyle {\underline {s_{\mathrm {opt} }\approx 3{,}9\cdot {\sqrt {h}}}}}$

With the range of electromagnetic waves of very short wavelengths ( ultra-short wave and shorter), the apparent earth radius for UHF should be used. It is at R _k ≈ 8470 km:

(5c) ${\displaystyle {\underline {s_{\mathrm {UHF} }\approx 4{,}1\cdot {\sqrt {h}}}}}$

Visibility between two elevated points. Seen from each of the two mountain peaks, the other only rises above the horizon.

If the eyes and object are raised above the reference plane , which is already given by the height of the eyes of the person standing in the plane, the distances between the two add up from the point where the tangent connecting them touches the surface of the earth.

(6a) ${\displaystyle s=s_{1}+s_{2}\approx {\sqrt {2R}}\cdot \left({\sqrt {h_{1}}}+{\sqrt {h_{2}}}\right)}$

respectively

(6b) ${\displaystyle {\underline {s\approx 3{,}9\cdot \left({\sqrt {h_{1}}}+{\sqrt {h_{2}}}\right)}}}$

Examples

A ship on the horizon: the hull is hidden by the curvature of the earth.

In the right picture you can see a ship on the horizon , of which the curvature of the earth hides part of the hull. The picture was taken at a viewing height of  m. If one assumes that the covered part of the hull has a height of approx.  M above the water level, the ship is approx. 14.2 km away (taking into account the atmospheric refraction according to formula 6b above). ${\displaystyle h_{1}=2}$ ${\displaystyle h_{2}=5}$

The table summarizes some values ​​for the maximum geometric visibility. This shows the importance of the height of the lookout of old warships. From a 15 m high mast, the observer can make out a ship 15 km away. Conversely, from a height of 0 m, the guard will only see the small mast on the horizon with a lot of luck.

Geometric visibility ( h ₂ = 0 m; without taking atmospheric refraction into account)

Eye level Visibility 01 m 03.6 km 02 m 05.0 km 03 m 06.1 km 04 m 07.1 km 05 m 08.0 km 06 m 08.7 km 07 m 09.4 km 08 m 010.0 km 09 m 010.7 km

Eye level Visibility 010 m 011 km 020 m 016 km 030 m 019 km 040 m 022 km 050 m 025 km 060 m 027 km 070 m 029 km 080 m 031 km 090 m 033 km

Eye level Visibility 0100 m 035 km 0200 m 050 km 0300 m 061 km 0400 m 071 km 0500 m 079 km 0600 m 087 km 0700 m 094 km 0800 m 0100 km 0900 m 0107 km

Eye level Visibility 01000 m 0112 km 02000 m 0159 km 03000 m 0195 km 04000 m 0225 km 05000 m 0252 km 06000 m 0276 km 07000 m 0298 km 08000 m 0319 km 09000 m 0338 km

Geographic latitude

Scheme drawing for determining the angular visibility range b on earth

In the case of high-flying objects such as airplanes - but above all satellites - one is less interested in visibility than distance information. Instead, you want to know which area of ​​the earth, expressed in degrees, is accessible for observation or can receive signals. In the schematic drawing, an observer sees an aircraft at angle a above the horizon. It flies at height h above the earth and at height h + R above the center of the earth. The aircraft can be seen on earth with an elevation ≥ a in the angular range of 2 b (angle in arcs ):

(1) ${\displaystyle b=b(a)={\frac {\pi }{2}}-\arcsin \left({\frac {R}{R+h}}\cos(a)\right)-a}$

With an elevation of a = 0, when the aircraft can just be seen on the horizon, (1) simplifies to:

(2) ${\displaystyle b(0)=\arccos \left({\frac {R}{R+h}}\right)}$

Relationship (2) also indicates how much the notch has shifted from an elevated observation position.

As an approximation:

(2 B) : ${\displaystyle \kappa _{\mathrm {geom} }=1{,}93\cdot {\sqrt {H}}}$

or.

(2c): ${\displaystyle \kappa _{\mathrm {opt} }=1{,}75\cdot {\sqrt {H}}}$

Examples:

• From an altitude of h = 10 km, a pilot sees an area on the earth of 2 · b = 6.4 °, corresponding to an area with a radius of approx. 356 km. He only glimpses the edge area. With an elevation angle of a = 10 °, the angular range is reduced to 2 · b = 0.99 °, corresponding to an area with a diameter of approx. 110 km.
• A satellite at an altitude of 36,000 km covers an area of ​​a maximum of 2 · 81.3 ° (see also footprint ).
• Measurements made with a sextant at eye level 4 m above the surface of the water normally need to be corrected by 3.5 'to 3.8', depending on the state of the atmosphere close to the ground.

See also

• Distance measurement
• Visual navigation
• Higher geodesy

Web links

Wiktionary: Visibility  - explanations of meanings, word origins, synonyms, translations

Individual evidence

1. ↑ ^{a b} "Seeing" and "recognizing" are meant here purely theoretically. A commercial aircraft with a fuselage diameter of 8 meters can no longer be distinguished from the sky with the naked eye from a distance of 40 kilometers. See resolving power .