## Atmospheric Visual Range

On a clear day can we really see forever? If not, how far can we see? To answer this question requires qualifying it by restricting viewing to more or less horizontal paths during daylight. Stars at staggering distances can be seen at night, partly because there is little skylight to reduce contrast, partly because stars overhead are seen in directions for which attenuation by the atmosphere is least.

We are careful to distinguish between visibility, a quality, and visual range, a quantity. One can say that visibility is good or poor but not that it is 10 km or 100 km. W. E. Knowles Middleton, whose book Vision in the Atmosphere is the standard work on atmospheric visibility, inveighed against the careless confutation of these two terms. He didn't have much effect, and neither will we, so all we can do is express our scorn for folks, especially learned doctors of science, so devoid of linguistic sense that they can't distinguish a quality from a quantity.

The radiance in the direction of a black object is not zero because of airlight (Sec. 8.1). At sufficiently large distances, this airlight is indistinguishable from the horizon sky. An example is a phalanx of parallel dark ridges, each ridge brighter than those in front of it (Fig. 8.16). The farthest ridges blend into the horizon sky. Beyond some distance we cannot see ridges because of insufficient contrast. Figure 8.16: The brightness of each of these ridges, all covered with the same dark vegetation, increases with increasing distance, and hence their contrast with the horizon sky decreases.

Equation (8.3) gives the airlight radiance L, from which we obtain the airlight luminance B, which is what humans sense, by integrating over the visible spectrum:

where V is the luminous efficiency of the human eye and K is a constant of no concern here (see Sec. 4.1). The contrast C between any object, with luminance B, and the horizon sky is

where Bis the horizon luminance. For a uniformly illuminated line of sight of length d, uniform in its scattering properties, and over black ground, the contrast obtained from Eq. (8.3) is fGVL0exp(-/3d) dX

where we assume an infinite horizon optical thickness. This ratio of integrals defines an average (over the visible spectrum) optical thickness

and a corresponding scattering coefficient

Equation (8.32) for contrast reduction with optical thickness is formally, but not physically, identical to the expression for exponential attenuation at any wavelength of radiance [Eq. (8.20)], which perhaps is responsible for the misconception that atmospheric visibility is reduced because of attenuation. But if there is no light from a black object to be attenuated, its finite visual range cannot be a consequence of attenuation.

The distance beyond which a black object cannot be distinguished from the horizon sky depends on the contrast threshold, the smallest absolute value of contrast detectable by a human observer. Although this depends on the particular observer, the angular size of the object observed, the presence of nearby objects, and the absolute luminance, a contrast threshold of 0.02 is often taken as a typical value. To find the visual range for this threshold we have to evaluate the integrals in Eq. (8.31) numerically for various values of d to find the one for which the right side of this equation is -0.02. But if 3 is independent of wavelength the solution is simply

This equation is often called Koschmieder's law, although W. E. Knowles Middleton notes that "there can be no doubt that Bouguer was quite clear about the main factors determining the horizontal visual range, and that he effectively stated the law which has lately been called Koschmieder's law."

The scattering coefficient for molecules, however, is not independent of wavelength, although the geometrical factor G is. The function V is fairly sharply, and symmetrically, peaked around 550 nm. Because the molecular scattering coefficient decreases, and hence the exponential function in Eq. (8.31) increases with increasing wavelength, the average molecular scattering coefficient must correspond to wavelengths somewhat greater than 550 nm. We numerically solved Eq. (8.31) for a pure molecular atmosphere at standard pressure (1013 mb) and temperature (0 °C) to obtain d = 330 km, which corresponds to a wavelength of about 560 nm. Because the wavelength dependence of scattering is steepest for molecules and particles small compared with the wavelength, we usually can estimate the visual range from Eq. (8.34) using the scattering coefficient at or around 560 nm.

According to this analysis, therefore, "forever" is around 330 km, the maximum visual range at which a black object at sea level can be distinguished from the horizon sky for a purely molecular atmosphere, assuming a contrast threshold of 0.02 and also ignoring the curvature of the Earth. This maximum is more than twice the visual range considered exceptionally high. From this we conclude that almost always visual range is limited by particles.

We also observe contrast between elements of the same scene, a hillside mottled with stands of trees and forest clearings, for example. The extent to which we can discern details in such a scene depends on sun angle as well as distance. The airlight radiance for a black object is given by Eq. (8.3), whereas that for a reflecting object is given by Eq. (8.22). These two equations can be combined to give the contrast between adjacent reflecting and non-reflecting objects