As Visibility or View The distance at which objects can just be detected is defined as the distance between the two. Two effects limit visibility on Earth: atmospheric disturbances such as snowfall or fog cause light to dim, and the curvature of the Earth obscures distant objects.
Limitation by light attenuation
Atmospheric scattering and absorption reduce the contrast of an object relative to its surroundings. The contrast K depends exponentially on the distance s and a Absorption coefficients σ from:
For perception, a minimum contrast of
required. Assuming that the initial contrast K0 approximately 1, can be seen immediately s on σ can be concluded:
A visibility of 50 km corresponds to an absorption constant of 10 − 4 / m . Under good conditions the long-distance view a few hundred kilometers, see table.
In the example image, the contrast between the mountains and the sky decreases with increasing distance. The mountain range in the right image is no longer visible in the fog.
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Atmospheric visibility |
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Weather conditions |
Visibility / km |
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Exceptionally clear |
280 |
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Very clear |
50 |
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Clear |
20 |
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Slightly hazy |
10 |
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Hazy |
4 |
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Heavy haze, light fog |
2 |
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Moderate fog |
1 |
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Dense fog, heavy rain |
0.1 |
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Extreme fog, blowing snow |
0.01 |
Pure seawater has, depending on the wavelength, a Extinction length 1/σ From 2 to 100 meters. For diving, visibility of 40 meters is considered exceptionally good. In clean lakes, it is approximately 10 meters.
Limitation by the curvature of the earth
distance
The curvature of the Earth limits the maximum visibility for objects on Earth. In the schematic drawing, a person is looking to the left at a tower. Because of the curvature of the Earth, they can only see the top. Eye level h 1 , the height of the tower h2 and is R = 6300 km the radius of the earth, the maximum viewing range is s :
(1)
The formula results from the calculation of the partial lengths s1 and s2 after Pythagorean theorem and neglecting the small square elements. If you look from a tower to the horizon, i.e. h 2 =0 , the formula is simplified. A tower of height h in m allows the geometric visibility s in km:
(1b)
The Refraction of the atmosphere bends the light rays and makes the Earth appear larger. The middle apparent radius of the Earth is at R k ≈ 7680km. The optical range of (1b) increases by up to 10% to:
(1c)
Visibility also determines the range of electromagnetic waves of very short wavelength, that is Ultrashortwave and shorter. Here, too, an apparent Earth radius is introduced as a correction. For UHF it is at R k ≈ 8470km:
(1d)
Examples
In the right picture you can see a ship at Horizon , where the curvature of the Earth obscures part of the fuselage. The photo was taken at a viewing height of h1 = 2 m. Assuming that the covered hull has a height of approx. h2 = 5 m and the Earth's radius R = 6370 km, the ship is about 13 km away
The table summarizes some values for the maximum geometric visibility. This illustrates the importance of the height of the lookout old warships. From a 15 m high mast, an observer can spot a ship 15 km away. Conversely, from a height of 0 m, the watchman can only see the small mast on the horizon with great luck.
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Geometric visibility (h 2 = 0 meters) |
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Eye level |
Visibility |
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Eye level |
Visibility |
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2 m |
5 km |
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600 m |
87 km |
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5 m |
8 kilometers |
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800 m |
101 km |
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10 m |
11 km |
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1000 m |
113 km |
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15 m |
14 km |
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1500 m |
138 km |
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20 m |
16 km |
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2000 m |
159 km |
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50 m |
25 km |
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3000 m |
195 km |
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100 m |
36 km |
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4000 m |
225 km |
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200 m |
50 km |
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8000 m |
318 km |
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400 m |
71 km |
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9000 m |
338 km |
Geographical latitude
For high-flying objects such as aircraft or satellites, one is less interested in the visibility as a distance indication. Instead, one wants to know which area of the Earth is accessible for observation, expressed in angular degrees. In the schematic drawing, an observer sees an aircraft at an angle a above the horizon. It flies at an altitude h above the Earth and an altitude h+R above the Earth's center. The aircraft can be seen on Earth with an elevation >=a in the angular range of 2*b (angle in Bow ):
(2) b= π /2 - arcsin( R/(R+h) * cos(a) ) - a
At an elevation of a=0 , if the aircraft is just visible on the horizon, (2) simplifies to:
(3) b max = arccos( R/(R+h) )
The relationship (3) also indicates by how much the Kimm from an elevated observation position.
As an approximation:
(3b) :
or
(3c):
Examples:
- From an altitude of h=10km a pilot sees an area on Earth from 2*b = 3.2° , corresponding to an area with a radius of approximately 350 km. The edge area is only glimpsed. If the elevation angle should be at least a=10° the radius is reduced to about 50 km.
- A satellite at an altitude of 36,000 km covers a maximum area of 2*81.3° (see also Footprint ).
- Measurements taken with a sextant at a height of 4m above the water surface must be corrected by 3.5' to 3.8' depending on the weather conditions.