Visibility measurement

Visibility is the distance at which objects can just be seen. 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.

Atmospheric visibility
Weather conditions Visibility / km
Exceptionally clear 280
Very clear 50
Clear 20
Slightly hazy 10
Hazy 4
Heavy haze, light fog 2
Moderate fog 1
Dense fog, heavy rain 0.1
Extreme fog, blowing snow 0.01

Depending on the wavelength, pure seawater has an extinction length 1/σ of 2–100 m. 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 viewing distance 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. If the eye height is h 1 , the height of the tower is h 2 , and R = 6300 km is the radius of the Earth, the maximum viewing distance s is:
(1) s = \sqrt{2 R} \cdot \left( \sqrt{h_1} + \sqrt{h_2}\right)

The formula is derived from calculating the partial lengths s 1 and s 2 according to the Pythagorean theorem , neglecting the small quadratic terms. If one looks from a tower to the horizon, i.e., h 2 =0 , the formula is simplified. A tower of height h in m allows for the geometric visibility s in km:

(1b) s_{geom} = 3.56 * \sqrt{h}

Refraction of the atmosphere bends light rays, making the Earth appear larger. The mean apparent radius of the Earth is R k ≈ 7680 km. The optical range of (1b) increases by up to 10% to:

(1c) s_{opt} = 3.9 * \sqrt{h}

Visibility also determines the range of very short-wavelength electromagnetic waves, i.e., ultrashort waves and shorter. Here, too, an apparent Earth radius is introduced as a correction. For UHF , it is R k ≈ 8470 km:

(1d) s_{UHF} = 4.1 * \sqrt{h}

Examples

The right image shows a ship on the horizon , with part of its hull obscured by the curvature of the Earth. The image was taken at a viewing height of h 1 = 2 m. Assuming that the obscured hull has a height of approximately h 2 = 5 m and the Earth's radius R = 6370 km, the ship is approximately 13 km away.

The table summarizes some values ​​for maximum geometric visibility. This highlights the importance of the height of the lookout on old warships. From a 15-meter-high mast, an observer can spot a ship 15 km away. Conversely, from a height of 0 meters, the watchman will only see the small mast on the horizon with great luck.

Geometric visibility (h 2 = 0 meters)
Eye level Visibility Eye level Visibility
2 m 5 km 600 m 87 km
5 m 8 kilometers 800 m 101 km
10 m 11 km 1000 m 113 km
15 m 14 km 1500 m 138 km
20 m 16 km 2000 m 159 km
50 m 25 km 3000 m 195 km
100 m 36 km 4000 m 225 km
200 m 50 km 8000 m 318 km
400 m 71 km 9000 m 338 km

Geographical latitude

For high-flying objects such as aircraft or satellites, visibility is less important than distance. Instead, one wants to know which area of ​​the Earth is accessible for observation, expressed in 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 arcs ):

(2) b= π/2 - arcsin( R/(R+h) * cos(a) ) - a

At an elevation of a=0 , when the aircraft is just visible on the horizon, (2) simplifies to:

(3) b max = arccos( R/(R+h) )

The relationship (3) also indicates how much the horizon has shifted from an elevated observation position.

As an approximation:

(3b): \kappa _{geom}=1.93\cdot\sqrt{H}

or

(3c): \kappa _{opt}=1.75\cdot\sqrt{H}

Examples:

  • From an altitude of h=10 km, a pilot sees an area on Earth of 2*b = 3.2° , corresponding to an area with a radius of approximately 350 km. The peripheral area is only glimpsed. If the elevation angle is at least a=10° , the radius is reduced to approximately 50 km.
  • A satellite at an altitude of 36,000 km covers an area of ​​maximum 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.

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 an absorption coefficient σ :
K = K_0 \cdot e^{-\sigma \cdot s}

For perception, a minimum contrast of

K = 0.02 \;\hat{=}\; 2%

required. Assuming that the initial contrast K 0 is approximately 1, σ can be directly deduced from the viewing distance s :

\sigma = \frac{3.91}{s}

A visibility of 50 km corresponds to an absorption constant of 10 − 4 / m . Under good conditions, visibility extends to several 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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