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Does anyone know a site, software, public API (or formula) to get the solar light spectrum on earth given the time of the day, day of the year, and earth meassurement position (lattitude, longitude)?

Basically I need to know sunlight spectrum in realtime, including UltraViolet and InfraRed part of the spectrum.

The clouds and other particles don't have to be taken into account, so basically, is there any way for calculating the angular incidence of the sun and thus the received light spectrum after being filtered by the atmosphere with the light incidence at the calculated angle?

Thanks in advance!

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    $\begingroup$ This is fundamentally impossible as it requires knowing the cloud cover. Not sure if there are accurate historical records or how far, but there certainly isn't anything for the future. Unless you mean above the atmosphere, in which case the latitude and longitude don't matter, only time does. (and even then it's just blackbody radiation +/- some hydrogen and helium lines, times a slightly varying scalar quantity) $\endgroup$
    – John Dvorak
    Commented Sep 25, 2019 at 15:14
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    $\begingroup$ That's more interesting, but then I believe the only thing that matters is the sun's elevation (unless you want to model effects of light that gets reflected off the surface and then scattered). UV also gets influenced by the ozone layer, for which I also don't believe there's much record about. $\endgroup$
    – John Dvorak
    Commented Sep 25, 2019 at 15:23
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    $\begingroup$ I do agree that elevation -> spectrum is an interesting question. There's a lot of interest in accurate sky shaders in the CGI community, so it might be worth checking out if their approach can derive the wide-spectrum behavior rather than just a RGB color. $\endgroup$
    – John Dvorak
    Commented Sep 25, 2019 at 15:31
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    $\begingroup$ To be honest I wouldn't be surprised if the calculator you use had even forgotten that the ozone hole exists and simply calculated f(sun elevation) * time sunbathing $\endgroup$
    – John Dvorak
    Commented Sep 25, 2019 at 15:35
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    $\begingroup$ The output of the Sun is not constant. For example, ultraviolet irrandiance (at the top of the atmosphere) varies by about 1.5% during the solar cycle, and it also has a secular variation (increasing by about 3% during the last 300 years or so). $\endgroup$
    – AlexP
    Commented Sep 25, 2019 at 15:57

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There are many models of this. Research keywords would be "Mie scattering," which controls how aerosols in the air scatter light, and "Rayleigh scattering," which is why the sky is blue to begin with.

Essentially it boils down to a path integral. You integrate the depth of the atmosphere at any point. You can figure this out by calculating the solar angle from your position and taking the average thickness of a ray along that point to space. Once you know the depth of the atmosphere along the ray, you can calculate the amount of inscatter, absorption, and outscatter for a given frequency of light. Most graphical models do this for a given red, green, and blue wavelength (chosen for a particular color space), but the data is available for UV and infrared wavelengths as well.

Above the Earth's atmosphere, the distance to the Sun becomes essentially irrelevant, as space doesn't scatter, and the insolation of the Sun on the Earth is a simple inverse square law calculation, and since the Earth is very small compared to the distance to the Sun, you can take that insolation as being from the center of the Sun to the center of the Earth (i.e. the orbital distance at that time).

One of the best available papers on the subject is Eric Bruneton's Atmospheric Scattering model This model precomputes a set of scattering tables as a function of the sun's angle in the sky. It then becomes a trivial problem to determine that angle using standard astronomy calculations and the observer's position.

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  • $\begingroup$ Nice. My answer for tower clock and eyeball, yours for portable computer. $\endgroup$
    – Zeiss Ikon
    Commented Sep 25, 2019 at 17:20
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Disregarding the Sun's (well characterized) ordinary cyclic and long term variation, all that changes with solar radiation (that you can calculate, anyway) is distance from the Sun and the Sun's apparently angle in the sky (which depends on date and time of day -- i.e. how much air the light must pass through).

The lower the Sun's apparent sky position, the more filtering the air gives. Air preferentially filters blue light (by scattering), but doesn't change the fundamental spectrum (number, position, and intensity of emission or absorption lines, for instance). You should be able to find information (for photographers, if no other source) for light intensity and color changes throughout the day -- I recall the correction for exposure by rule ("Sunny f/16" rule) being plus 1 stop before 10 AM or after 2 PM (standard time, correct for DST), 2 stops before 8 and after 4, three stops if the sun is within five degrees of the horizon, and a particular filter (or color sensitivity correction for some panchromatic black and white films) within an hour of sunrise or sunset, the so-called "golden hour". Add one additional stop if more than 45 degrees from the equator, another if within one month of the winter solstice.

Each "stop" corrects for a halving of the total light flux, so if you're down three stops, you have 1/8 the light level.

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    $\begingroup$ Color temperature also changes rather dramatically between noon and sunset. $\endgroup$
    – AlexP
    Commented Sep 25, 2019 at 16:42
  • $\begingroup$ That's why you need a filter (for color) or sensivity correction (for B&W) close to sunrise or sunset. it's not technically the black body temperature of the light that changes, it's a blue-preferred filtration by the atmosphere. A suitable minus-blue will restore color balance for your slides, if you don't want that "golden" cast. $\endgroup$
    – Zeiss Ikon
    Commented Sep 25, 2019 at 16:47

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