Relative Optical Air Mass (Kasten and Young)

Relative air mass says how many atmospheres of air a ray from the sun crosses, compared with the straight-up path. Straight up it is 1, and the lower the sun sits, the longer the slanted path through the air. A flat atmosphere would give $1/\sin h$, which runs off to infinity at the horizon. The real Earth is curved and bends light, so the true path stays finite. Kasten and Young (1989) fitted this expression to their revised air mass tables, which allow for both effects, and it stays sensible right down to the horizon.

The only input is $h$, the apparent elevation of the sun in degrees, the angle you would see or measure. In the denominator, $\sin h$ is the flat-atmosphere geometry. The second term, $0.50572\,(h+6.07995^\circ)^{-1.6364}$, is a small correction high in the sky and does the work near the horizon. At 0 degrees $\sin h$ is zero, and this term alone keeps the denominator positive. At 30 degrees it adds only 0.0014207 to 0.5, so the air mass is 1.99432 where the plain rule gives 2. Put $h$ in degrees inside the power.

Read the answer as a multiplier on the thickness of the atmosphere. The formula gives 0.9997 at 90 degrees, about 2 at 30 degrees, near 5.6 at 10 degrees (5.5860 by hand) and about 38 at the horizon, where it gives 37.92, which is 0.2 percent from the published figure of approximately 38. Going backwards, an air mass of 2.9031 corresponds to a sun 20 degrees high. This is the first step for clear-sky sunlight, ultraviolet index and solar panel output, because the sunlight that arrives falls roughly as an exponential in the air mass.

Two limits apply. The page works from the apparent elevation and goes down to 0 degrees and no lower. This is the air mass at sea level pressure: a station at altitude has less air above it in proportion to its pressure. The formula as printed in two open references agrees with the one used here.