Consider a fixed point in the Milankovich cycle and the solar cycle, a fixed Earth-Sun distance, and a fixed horizontal location on Earth, and assume that the Sun is at the zenith for that location. Wikipedia says that on a clear day, the global horizontal irradiance at the Earth's surface (i.e. including both direct normal irradiance directly from the Sun and diffuse horizontal irradiance from atmospheric scattering) is about 1120 W/m$^2$ at sea level, although of course the exact value will depend on all of the quantities assumed fixed above.

By about what fraction is the solar irradiance at the Earth's surface (more precisely, the global horizontal irradiance) reduced on a cloudy day? I assume that the answer will depend significantly on the types and thickness of the clouds, and (much less strongly) on all of the variables assumed fixed above, but what is a reasonable range of values? Do clouds reduce the solar irradiance at the Earth's surface by 50%? 90%? 99.999%?

I assume that clouds block the large majority of the irradiance - at least in the visible range of the EM spectrum - because clouds have an easily visible effect on the Sun's brightness, but our eyes perceive brightness on a logarithmic scale, so if the clouds didn't reduce the irradiance by a significant factor than we'd barely be able to see the difference. But clouds might be more (or less) transparent to other frequencies of sunlight (although most of the Sun's luminosity is in the visible range, so maybe that doesn't matter too much).


2 Answers 2


Back in the days of mechanical cameras, it was common to use "rules of thumb" for setting exposure. A common rule was that dark clouds reduced light by three "stops", a factor of 8.


The technical terms you are looking for are cloud optical depth and clearness index¹. The clearness index is defined as:

$$k_t = \frac{I}{I_0} $$

where $I$ is the incident irradiance at the Earth's surface and $I_0$ is the extraterrestrial irradiance.

The cloud optical depth is defined as:

$$\tau = \ln(\frac{\Phi_e^i}{\Phi_e^t})$$

where $\Phi_e^i$ is the radiant flux received and $\Phi_e^t% the radiant flux transmitted.

The cloud optical depth is routinely available from satellite data and tends to vary between ~0 (no cloud or sub-visible cloud) and ~50 for very thick clouds. In principle, this is wavelength-dependent, but the dependency is very weak in the visible (which is why clouds are white/grey). However, values vary strongly when you consider infrared radiation.

Plugging this into the definition of optical depth suggests a factor $e^{-50}=1.9 \cdot 10^{-22}$ gets transmitted. This is too low. The reason is that many photons get scattered but still get out "on the other side", so the effective cloud transmittance depends not only on the optical depth, but also on the exact scattering behaviour in the cloud. In other words: at very low values of optical depth, with very thin clouds, one can still see the disk of the Sun. When clouds are thick, one has no chance of seeing the disk of the Sun; but it's not pitch black dark outside. Calculating how much reaches the surface requires a radiative transfer model.

Serrano et al. (2015) use a radiative transfer model to estimate the relation between the so-called clearness index, which indicates the degree of availability of solar irradiance at the Earth's surface, and the cloud optical depth, using various models:

relation between cloud optical depth and clearness index (4 panels with plots)

Their results seem to indicate that at a cloud optical depth of 50, around 10% of solar irradiance reaches the surface.

D. Serrano, M.J. Marín, M. Núñez, S. Gandía, M.P. Utrillas, J.A. Martínez-Lozano. Relationship between the effective cloud optical depth and different atmospheric transmission factors, Atmospheric Research, Volume 160, 2015, Pages 50-58, ISSN 0169-8095, https://doi.org/10.1016/j.atmosres.2015.03.004. (https://www.sciencedirect.com/science/article/pii/S0169809515000861)

¹As of July 2022, the English language Wikipedia states that a clearness index of 0 corresponds to the sky being completely covered. This is wrong. A clearness index of 0 would be a pitch black sky, which would require an infinitely thick cloud, which is not possible with water or ice clouds (but it's possible with volcanic ash, which — unlike water — absorbs solar radiation in significant amounts).

  • $\begingroup$ It looks like the optical depth is the natural logarithm of the reciprocal transmittance, not the reciprocal transmittance itself as you suggest? $\endgroup$
    – tparker
    Jul 11, 2022 at 13:13
  • $\begingroup$ @tparker To be honest, I'm confused. I agree that it looks the way you write, but that would mean an optical depth of 50 means a factor $e^{-50}=1.9 \cdot 10^{-22}$ gets transmitted for very thick thunderstorm clouds, which can't be right. From people who have solar panels, their output may decrease by a factor 10 under a cloudy sky, but not by a factor $\frac{1}{e^{-10}}≅20000$. $\endgroup$
    – gerrit
    Jul 11, 2022 at 13:23
  • 1
    $\begingroup$ @tparker Thinking again, I think the factor $10^{-22}$ refers to the fraction of photons that pass through the cloud without undergoing any scattering. Of course, most photons that do pass through undergo scattering. This means my answer is wrong. Will edit… $\endgroup$
    – gerrit
    Jul 11, 2022 at 13:27
  • $\begingroup$ Your empirical solar panel factor of 10 matches the empirical photographic factor of 3 stops to the nearest stop. So, that would appear to be the right answer. $\endgroup$
    – John Doty
    Jul 11, 2022 at 13:42
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    $\begingroup$ I have edited the answer, now I'm more confident that it's correct. $\endgroup$
    – gerrit
    Jul 11, 2022 at 13:53

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