Airmass calculation

[jahangir]

Hello,
Can some one help me in calculating the varying Airmass over time for a country near equator?
OR if some one can suggest what is the average Airmass, as weather is clear and days are almost of the same length?
Thanks in advance.

[J.P.M.]

If you are referring to the mass of air that beam radiation passes through as it traverses the atmosphere, a rough 1st approximation for the ratio of the mass of air passed through to the mass of a vertical column of air above the receiving surface is the inverse of the cosine (or the secant) of the zenith angle of the sun. The mass of the vertical column can be found from the local, absolute barometric pressure (I;e.,uncorrected to sea level barometric pressure).

[JTElectric]

The "airmass" as a general term, describes the ratio of the amount of an attenuating medium integrated along a slanted path through the atmosphere to the amount integrated along a vertical path. It enables the amount of the medium in a vertical column to be derived from measurements of attenuation along the slanted path. In the ozone measurement equations, and the following equations, three attenuating media are described, ozone, air molecules and aerosol, and each has a slightly different airmass, owing principally to their different height distribution and the sphericity of the atmosphere. However, all are approximately equal to secant Z, where Z is the solar zenith angle at the observing site.
Aerosols are concentrated near the ground and the appropriate airmass is taken to be sec Z, though generally the aerosol term and hence its airmass are neglected. The airmass for scattering by air molecules, m, can be calculated to account for the atmosphere's sphericity and for atmospheric refraction, and has been empirically expressed by Kasten (1966) as:

m = (cos Z + 0.1500 (93.885 - Z)^-1.253)^-1 where Z is in degrees. The airmass appropriate to the ozone layer, μ_h, is taken as the secant of the zenith angle at the mean height of the ozone layer, and the mean height has been traditionally assumed to be 22 km (Dobson, 1957a). illustrates the geometry of the situation, and from this it is not difficult to show that: