Koschmieder Visual Range

Look at a dark hill on the horizon. It looks grey rather than black, and the further away it is the greyer it looks, until at some distance it is indistinguishable from the sky behind it. The reason is airlight: along the whole path between you and the hill, air molecules and aerosol scatter sunlight into your line of sight, progressively washing out the contrast. Harald Koschmieder worked the problem out in 1924 and the result is the standard relation between how strongly air extinguishes light and how far you can see through it.

The contrast of a black target against the horizon sky falls exponentially, $C = e^{-\beta x}$, where $\beta$ is the extinction coefficient — scattering plus absorption per unit path length. Set that equal to the smallest contrast an observer can detect and solve for $x$. With the classical 2 % threshold, $V = -\ln(0.02)/\beta = 3.912/\beta$.

The numerator is a choice, and choosing differently gives a different number for identical air. This is the single most common source of confusion in visibility work. Koschmieder's classical visual range uses a 2 % contrast threshold, giving 3.912. The World Meteorological Organization's Meteorological Optical Range — the definition behind aviation reports, METARs and the automated sensors at airports — uses 5 %, giving $-\ln(0.05) = 2.996$. The MOR figure is therefore about 23 % shorter than the Koschmieder figure for exactly the same atmosphere. If you compare this page against a METAR without converting, the difference will look like an error in one of them and it is not. The conversion factor is 2.996/3.912 = 0.766 in that direction.

Calibration worth memorising. Perfectly clean air still scatters — Rayleigh scattering by the molecules themselves is about 0.012 km⁻¹ at green wavelengths, which caps the visual range at roughly 300 km and is the reason distant mountains look blue. A clear day at 0.1 km⁻¹ gives 39 km. Haze at 1 km⁻¹ gives 3.9 km. Mist at 4 km⁻¹ gives about a kilometre. Dense fog reducing visibility to 100 m is running near 39 km⁻¹, three thousand times the clean-air value — which tells you that visibility is almost entirely a story about aerosol and droplets, not about air.

That is what makes the relation useful backwards. Visibility is the quantity that gets observed and reported everywhere in the world, hourly, for free; extinction is the quantity that atmospheric models and air-quality regulations are written in. Inverting Koschmieder turns a century of visibility observations into an extinction record, and subtracting the Rayleigh contribution turns that into an aerosol record. A great deal of what is known about long-term regional haze trends came in through this equation.

The assumptions, and where they break. The derivation needs a horizontally uniform atmosphere, a black target, a horizon-sky background, daylight, and a single wavelength. Real fog is patchy, and a patchy path has no single $\beta$. Real targets are grey, which reduces the range. The relation says nothing about a vertical path, where aerosol loading and density change with height — slant visibility from an approaching aircraft is a different and harder problem. And it does not apply at night at all: after dark, visibility is set by how far away a light of known intensity remains detectable, which is Allard's law, a different relation with a different form. An automated visibility sensor measures $\beta$ directly over a short path and reports a visibility computed from it, which is why its number can disagree with a human observer looking at a real horizon.