I understand that before sunrise, twilight shows several colors, so I'm looking for the average color (wavelength) and brightness (lux) of light that is being projected from the sky before the sun visibly rises.

enter image description here

I've seen a similar question here, but the resulting formula is a) apologies, way over my head (I'm a physics fan but not a physicist) and b) doesn't appear to take into account the Rayleigh scattering of sunlight seen before sunrise or atmospheric refraction and how we see the sun before it's above the horizon line.

I've come to learn that there are a lot of variables that can affect this color/brightness including latitude, elevation (longitude), time of year, weather conditions, pollution, particles in suspension, etc. -- So for simplicity's sake, the viewer would be at sea level on the equator on the morning of the Summer Solstice, and the sky would be clear across the earth. (If I'm missing another key variable, please let me know)

Twilight spans over time in three light phases that are all measured by the degree of the sun below the horizon:

  • Astronomical Twilight (from -18° to -12°)
  • Nautical Twilight (from -12° to -6°)
  • Civil Twilight (from -6° to 0/-0.59°) (which contains Blue Hour)

It ends when Sunrise begins (which contains Golden Hour). Because of atmospheric refraction sunlight bends around the earth, so technically the sun is visible before it gets to 0° to the horizon (roughly 35.4 arcminutes higher = 0.59 degrees).

A representation of morning twilights, golden hour and blue hour.

My question: What is the average color (wavelength) and brightness (lux) of light for each degree of the sun below the horizon starting at Astronomical Twilight (-18°) and ending just before the moment of sunrise (~0.59°)?

This will allow me to create a proper gradient that will show this light change over time. In short, I'd like to fill out this table:

enter image description here https://codepen.io/Rogue75/pen/OERKop

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    $\begingroup$ Other variables would be atmospherics (clouds, aerosols). I suspect this would be much easier to measure (camera/analysis) than to compute. $\endgroup$ – BowlOfRed Jun 7 '18 at 22:25
  • $\begingroup$ @BowlOfRed Agree... Mathematical modeling... It is easier to find/buy a software, than to calculate it analytically. $\endgroup$ – MsTais Jun 8 '18 at 0:32

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