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Is there a clear formula or table relating the illuminance in (lux) to status of the sky in terms of cloud cover? For example, mostly cloudy would be 2000-3000 lux.

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    $\begingroup$ That's going to depend on latitude and time of day at the very least. Possibly other factors too. $\endgroup$
    – gerrit ♦
    Dec 1, 2014 at 21:38
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    $\begingroup$ To add to @gerrit's comment, another thing to consider is that terms such as 'mostly cloudy' can mean anything above 4 okta. What type of clouds - some can be translucent etc. $\endgroup$
    – user889
    Dec 2, 2014 at 8:26
  • $\begingroup$ You might want to see this link for info on getting the clear sky radiance by date/time/location, which could be multiplied with cloud fraction data to get something like you are looking for quora.com/… $\endgroup$
    – f.thorpe ♦
    Dec 2, 2014 at 19:21

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I don't believe such a formula or table exists (but admittedly, I have not attempted a lit review in this subject). Sky coverage (cloudy, mostly cloudy, etc) is a the fraction of the sky covered by clouds, not how much light penetrates to the surface. Even with a mostly cloudy day, all it takes is one break in the clouds between the sun and your instrument to give the same illuminance as a clear sky day. You might take a time series of illuminance obs and consider the fraction of time with clear sky readings, but this will be faulty unless you can take into account cloud motion.

From the AMS Glossary:

cloud cover

(Also called cloudiness, cloudage.) That portion of the sky cover that is attributed to clouds, usually measured in tenths or eighths of sky covered.

The term "partly cloudy", for example, varies in interpretation but is generally around 0.3 or 0.4 to 0.6 or 0.7 cloud cover (where 0.0 is clear sky and 1.0 is overcast).

The type of cloud will also vary the light transmitted. A high cirrus or altocirrus cloud layer will transmit light to the surface, whereas a low stratocumulus overcast will only transmit a small fraction of light (note how grey and dreary a northern winter day can be under those clouds). The fraction of light transmitted will depend on the depth of the cloud layer. The light transmitted will also be a function of time of day and day of the year, both of which determine the position of the sun.

For a look into how these issues play out in a real experiment, take a look at Pfister et. al. (2003) who look at how cloud fraction and irradiance are related over New Zealand. I think you'll find most of the literature deals with solar irradiance, so it may help you in searches to look for that term rather than illuminance.

Note that the linked Ametsoc article is free to read, as area all Ametsoc articles at least 2 years old.

References

G. Pfister, R. L. McKenzie, J. B. Liley, A. Thomas, B. W. Forgan, and C. N. Long, 2003: Cloud coverage based on all-sky imaging and its impact on surface solar irradiance. J. Appl. Meteor., 42, 1421-1434. doi: http://dx.doi.org/10.1175/1520-0450(2003)042<1421:CCBOAI>2.0.CO;2

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Operationally, cloud cover is determined subjectively by visual contact with the sky from the human observer (meteorologist on duty). No instruments or objective measures are used, so the two measures cannot be related in an obvious way. The unit used to describe cloud cover is called Okta, and ranges from 0 (sky completely clear) to 8 (full cover). Each value has a graphical representation which can be included on synoptic weather maps for each station.

Another important reason why cloud cover cannot be unambiguously related to illuminance is explained in detail in Casey's answer - different types of clouds and cloud depths may lead to different illuminance values for same values of cloud cover.

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    $\begingroup$ Note that ASOS/AWOS stations do use a Ceilometer to provide an automated sky cover observation, but this just based off the time series of detected cloud heights and isn't always reliable for 100% automated stations, particularly in low visibility conditions (based on my experience landing at airports served only by automated reports). This observation depends on laser ranging of clouds and not on solar irradiance. $\endgroup$
    – casey
    Dec 2, 2014 at 17:37
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    $\begingroup$ @casey Good to know! Is ceilometer ever used to record cloud cover, as in time series of True/False? $\endgroup$
    – milancurcic
    Dec 2, 2014 at 17:56
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    $\begingroup$ The normal timeseries output includes multiple detection heights (for detecting multiple cloud layers), but you could squash that into a detection at any height as T and otherwise F to get a T/F timeseres. vaisala.com/Vaisala%20Documents/Brochures%20and%20Datasheets/… is a brochure of a commercial product but it has some basic info on how it operates as well as a description of its output data. They even provide some basic stats on how their algorithm compares to a human observer (and some citations to AMS conference talks). $\endgroup$
    – casey
    Dec 2, 2014 at 18:14

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