[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...

[Dan Evans]

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.

Nice find. Good work. I'm going out for coffee, it's too late to sleep.

[Cboldt]

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.

I come up with a similar number. Keep in mind, it's not power, per se, that results in injury. It is power density. The cross sectional area of the beam is a vital matter.

As for aiming, aim at the landing light. At takeoff, once the plane has rotated, I think illuminating the pilot from the ground would not be practical.

[ableChair]

Dude, you are a RIOT! Your units are all screwed up. Ask anybody who's taken basic algebra. This is an exponential function with a range on the interval [0, 1]. Try the units again, you'll find the attenuation WAY more than what you're expecting here. LMAO!!!!