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To: ableChair
"We're talking about how well the atmosphere can 'block' light, no matter how it does it, which is what is relevant for the laser discussion."

EXACTLY!

And that is precisely what those university gifs in #387 were telling you! That the atmosphere only absorbs 16% (in the first gif) and 19% (in the second gif) of the incoming solar radiation, NOT 95%!

Here is another...

University of Saint Louis says atmospheric absorption is 19%, not 95%...


This "19%" is the atmospheric absorption of the incoming solar radiation, and as you said, this "is what is relevant for the laser discussion".

--Boot Hill

413 posted on 09/29/2004 3:14:53 AM PDT by Boot Hill (Candy-gram for Osama bin Mongo, candy-gram for Osama bin Mongo!!!)
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To: Boot Hill
And that is precisely what those university gifs in #387 were telling you! That the atmosphere only absorbs 16% (in the first gif) and 19% (in the second gif) of the incoming solar radiation, NOT 95%!

Not only that, the graphics represent a whole earth energy balance, not the physics of light attenuation in air.

Light attenuation in air is variable as well. Clear vs. haze being an easily understood example. Further, attenuation will be wavelength specific.

When the question is "how much attenuation of beam energy is provided by transmission in air," a reference work that examones the solar energy balance of the planet is not directly useful.

415 posted on 09/29/2004 3:18:26 AM PDT by Cboldt
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To: Boot Hill
Hey, that's from the paper I posted a link to sometime ago...

That 19% is only an average for the whole earth including typical cloud cover, water vapor, etc. Light also reflects off the atmosphere and clouds sending it out to space.

On a clear day with no clouds and low water vapor the amount of radiant energy reaching the ground has to be much higher.

It is widely accepted that the sun provides about 1380 watts per square meter of radiation just above the atmosphere. It takes 1000 watts per square meter for a 13% efficient solar cell to produce 130 watts per square meter. This is what they do on a clear day not too far from the equator. Therefore about 72% of the suns radiation has to penetrate the atmosphere and land on the solar cell's surface to produce this much electricity. Significantly higher than the 49% average world wide
419 posted on 09/29/2004 3:30:42 AM PDT by DB (©)
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To: Boot Hill

So, umm, what's the difference between 'absorbed' and 'reach the surface'? I'll let you figure out your error. The difference between your 'sources' and the Cambridge Atlas is that your sources are less detailed. Cambridge is breaking the radiative and convection 'blocking' down into greater detail. But, wait, I already said all that at 3 a.m. didn't I?


479 posted on 09/29/2004 1:26:00 PM PDT by ableChair
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