Note: Some graphical work is required for this column.
I walk daily to my office and back during the week (except when the weather is atrocious). On clear sunny days, especially during the summer months, I’m exposed to a lot of sunlight, particularly on the left side of my face as I walk in a southerly direction in the mornings. Walking home, of course, still exposes the same side to more direct sunlight. Perhaps I should walk backward on the way home …
In what follows, the intensity of sunlight on the side of my head1 will be expressed in terms of the solar zenith angle θ for consistency with the clear-day model mentioned in question 2. Assume that the sky is cloudless.
Question 1: If I0 is the solar irradiance (power per unit area, W/m2) reaching my head, express the intensity on the side of my face (Is) in terms of θ. Assume for now that the irradiance is independent of path length through the atmosphere and that my face is normal to the direction θ = 90°.
Using the 1962 U.S. Standard Atmosphere,2 Hottel (1976)3 expressed the solar irradiance using the formula
I = I0(a0 + a1e−k sec θ), where A is the elevation in kilometers and
a0 = 0.4237 − 0.00821(6 − A)2; a1 = 0.5055 + 0.00595(6.5 − A)2 and
k = 0.2711 + 0.01858(2.5 − A)2.
As can be inferred, the exponential term is a measure of the irradiance reduction as a consequence of pathlength though the atmosphere.
Question 2: Taking account of the area factor in question 1, sketch I for (a) A = 0 and (b) A = 2.5. For what angle θ is this a maximum?
References
Fermi Questions are brief questions with answers and back-of-the-envelope estimation techniques. To submit ideas, please email John Adam ([email protected]).
