[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

[DB]

So a 50 watt laser (with whatever frequency the attenuation info is based on) would be attenuated about 4 dB over a distance of 5 miles (using 0.5 dB/km). Therefore the laser power level at 5 miles would still be 19.9 watts under clear sky conditions.

Enough to fry someone's eye's in milliseconds I'd guess.

The 50 watt laser value comes from what is reasonably available on eBay...

So this story falls under the easily doable as far as equipment goes. Aiming is another story...

[ableChair]

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.

Ding, ding, ding, ding. We have a winner! If that equation is even reasonably correct my points are mathematically manifest. Do you understand what that equation is saying? It means that as R grows the attenuation grows exponentially. D$%n that was easy. But, um, why would the attenuation factor be measured in decibels? Are you sure this isn't an equation for some kind of sound propagation? Guys, the more you try to make your 'point' the more you bury your own arguments.