[DB]

It is mostly my fault.

I was arguing that a large percentage of sun light reaches the earth through many miles of atmosphere and that laser light can also pass through many miles of atmosphere without being substantially attenuated. That is, a 50 watt laser could easily cause serious eye damage from miles away. If the sun can pass through it, so can a laser.

The "other guy" is determined to believe that 95% of the sun's radiant energy doesn't make it to the ground. And that a 50 watt laser going a few miles would be completely absorbed by the atmosphere to render it harmless regardless of the weather conditions.

[Cboldt]

It is mostly my fault.

I read most of the thread, and disagree. While it's not unusal for a few folks to talk past each other, it is unusual for it to occur to the degree presented in this thread!

Anyway, the general operative physics and engineering is found in the Beers-Lambert Law (didn't know of it until about 30 minutes ago, Google is a great tool). For a laser, a simple formula to find the ratio of power at distance "R" to power at distance "0" is e^-(alpha)R. alpha is an attenuation factor, and for air ranges from 0.1 (.43 dB/km) for clear air to 1.0 (4.3 dB/km) in hazy air.

http://www.freespaceoptic.com/WhitePapers/Comparison_Of_Beam_in_Fog.pdf

[ableChair]

The "other guy" is determined to believe that 95% of the sun's radiant energy doesn't make it to the ground

Ahhh, you're flailing again. Let's see if you see any difference in the meaning of these two sentences:

1.) "95% of the sun's radiant energy doesn't make it to the ground".

2.) 95% of the suns energy does not reach the Earth's surface in radiative form.

Like I said, you're flailing.

Do you see the difference? Statement 2 is the form we want for the laser discussion, not form 1. Think about it.