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I guess that these questions were being asked by many people on the Northern Hemisphere during this summer (and other summers) and someone may give a nice, coherent answer. The general question is how many times more slowly one is getting suntan or damages his skin in the evening, relatively to the noon?

The Sun altitude (solar elevation angle) $\alpha_s$ apparently makes the ozone layer etc. look $1/\sin\alpha_s$ times thicker than when the Sun is directly above our head. Clearly, this makes the solar UV radiation weaker if we're further from the noon. So

  • is that right that a particular spectral line gets weakened by the factor of $\exp(-C_\lambda/ \sin\alpha_s)$?
  • is that true that the changes to the UV-B radiation are the most important ones because UV-A is almost completely transmitted and UV-C is almost completely blocked?
  • because 98% of the UV radiation is said to be absorbed by the atmosphere, one would expect that the exponential reduction above will be dramatic. However, sources suggest that the "total amount" of UV radiation is only suppressed by a power law, probably by $1/\sin^2\alpha_s$ (10% thicker atmosphere implies 20% less radiation). Where does it come from? Is it from some integration of the exponentially suppressed function over frequencies? What is the approximate integrand and how does the decreasing exponential become a power law?
  • is it OK to assume that all transmitted UV rays cause suntan and potential diseases at the same rate, I mean that the ratio of "suntan vs harm" obtained by a photon is constant, or is it true that softer UV rays are giving us suntan with less harm to the skin? That would imply that it's healthier to get suntan in the evening
  • above, it was assumed that the solar photons travel straight from the Sun. But does Rayleigh scattering matter here? UV photons could go via the shortest path through the atmosphere (velocity orthogonal to the ozone layer) and then reflect to our skin via Rayleigh scattering – in this way, they would effectively see the minimally thin ozone layer. Rayleigh scattering is probably substantial for UV radiation, isn't it? In this way, one could explain why the Sun seems more powerful in the evening than the exponentially decreasing formula suggests. In the evening, one could still be getting suntanned from "all directions" of the sky (all places where it's "blue").

Sorry for this mixture of facts, questions, and half-baked hypotheses. Please fix the claims that are incorrect. There seem to be many related questions above but I would really like to get some usable "rate of getting suntan" as a function of the Sun altitude.

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Not sure it helps, but here you have the UV index as a function of time, and you can get Sun angle (what you call "altitude") as a a function of time. But this would be valid just for NYC. At least you'd get an idea of the shape. – anderstood Oct 29 '14 at 17:19
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@anderstood zardoz.nilu.no/~olaeng/fastrt/README.html This seems to have info on how that plot was generated. – Rick Aug 27 '15 at 18:30
    
Good for reminding me about this question, Rick, I missed anderstood's comment before - even though I was interested about this question every summer including this one. ;-) The drop in the morning and afternoon seems almost linear - I sort of expected it to be more brutal, exponential, like exp(-t) in the evening, because of the exponentially decreasing transmission through the absorbing layer, but it apparently ain't so. – Luboš Motl Aug 27 '15 at 19:15
    
@LubošMotl Unfortunately the curves ends a bit too early (4pm) but it could still be in exp(-t) between 4pm and sunset... – anderstood Aug 27 '15 at 20:50

The $1/sin\alpha$ is a useful approximation for sizing solar power systems, but it falls apart as the solar angle approaches zero. Essentially, it assumes the earth is an infinite plane, so when the sun reaches the horizon the amount of absorbing atmosphere between the observer and the sun becomes infinite. Since the earth is a sphere, rather than an infinite plane, this has obvious drawbacks. Also note that this model assumes that only absorption matters, and ignores Rayleigh scattering.

In terms of absorption, a much better model for atmospheric density is the Barometer Equation, which treats the atmosphere as having an inverse exponential density with a constant of about 8.47 km. This is not quite correct, and provides a total density error from sea level to space in the area of -1%, since it assumes the atmosphere is isothermal, and this is not true.

It is perfectly possible to compute the density of the atmosphere for a given solar angle and distance from the observer. Calling the distance D (in km) and the height at that distance in line with the sun h and the radius of the earth R, $$h = -R +\sqrt{D^2 +R^2 - 2RD\sin({\alpha +\frac{\pi}{2}})} $$ then $$\rho = \rho_0e^{-\frac{h}{8.47}} $$ Integrating this for D from zero to infinity will give the effective mass of the atmosphere along the path, normalized to a vertical path length of 1. This is not an integral which resolves nicely, and numerical integration is indicated.

With the effective mass of the path determined, absorption can be calculated based on the absorption coefficient.

Having dealt with absorption, Rayleigh scattering needs to be addressed, and I'm not sure quite how to do this. The problem is that the goal is not to address the apparent brightness of the sun, but rather the total UV produced by the $2\pi$ steradian expanse of scattered radiation provided by the total sky.

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Because of variable atmospheric haze from day to day, especially in the lowest 15 degrees from the horizon, the correct theory answer above is probably not applicable. You can try testing by presuming that a uv indicator card happens to change in response to the same uv ranges as your skin will (afro-caribbeans, that won't apply to everyone), and test with the card, and guess that the evening haze tomorrow might be the same as today. So, any formula which you get will not be sufficient to plan your day on the beach without looking up and looking at the sky.

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