originalfoo
commented
7 years ago See Projection effect and Absorption effect
From the same page further down:
Average annual solar radiation arriving at the top of the Earth's atmosphere is roughly 1361 W/m2
The radiation is distributed across the electromagnetic spectrum. About half is infrared light.
The Sun's rays are attenuated as they pass through the atmosphere, leaving maximum normal surface irradiance at approximately 1000 W /m2 at sea level on a clear day. When 1361 W/m2 is arriving above the atmosphere (when the sun is at the zenith in a cloudless sky), direct sun is about 1050 W/m2, and global radiation on a horizontal surface at ground level is about 1120 W/m2.
The latter figure includes radiation scattered or reemitted by atmosphere and surroundings. The actual figure varies with the Sun's angle and atmospheric circumstances. Ignoring clouds, the daily average insolation for the Earth is approximately 6 kWh/m2 = 21.6 MJ/m2.
If the sun kicks out roughly the same amount of energy at all times (is there a way to find actual amount at any given time? eg. from satellite data?), and we know where it is in relation to earth (more specifically: in relation to sensor array) as part of heliotrope calculations, and we know how much should land on earth if the sky is clear... it should be possible to estimate aerosols (precipitation, dust, etc) in the atmosphere based on the difference between actual and expected pyrheliometer direct (collimated) readings. The oktameter readings could be used to get some idea of ratio between cloud cover vs. other types of aerosols.