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Fox: Commercial Pilots 'attacked' with laser
Fox News | Greta Van Susteren

Posted on 09/28/2004 8:12:49 PM PDT by ableChair

Greta Van Susteren reported that a Delta pilot enroute to Salt Lake City was lazed in the cockpit this last Wednesday. Only country I know that has that hardware (for lazing bomber pilots) was the Soviet Union. Pilot reportedly required medical treatment and this was not a minor injury (weak laser) wound. More will come out to tomorrow as this story hits the print press.


TOPICS: Breaking News; US: Utah
KEYWORDS: airlinesecurity; dal; kapitanman; laser
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To: Cboldt
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.
421 posted on 09/29/2004 3:37:58 AM PDT by DB (©)
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To: ableChair; Boot Hill

I'm done.

I accept my failure to communicate.


422 posted on 09/29/2004 3:40:59 AM PDT by DB (©)
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To: DB
"It takes 1000 watts per square meter for a 13% efficient solar cell to produce 130 watts per square meter."

LOL, that 1,000 watts spec the manufacturers like to cite is the game we call "specs-manship"! There is no place on earth that gets that much solar radiation. If you read the fine print on their specs you'll find that even they will admit that a more realistic number is 800 W/m2. But my comments are a bit off-topic.

Maybe after a nights sleep ableChair will have had a chance to think over what so many posters have been trying to explain to him and hopefully it will sink in. But other than that, I don't know what else to offer. I feel like I've been beating my head against a brick wall!

--Boot Hill

423 posted on 09/29/2004 3:46:46 AM PDT by Boot Hill (Candy-gram for Osama bin Mongo, candy-gram for Osama bin Mongo!!!)
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To: ableChair; Boot Hill; DB; Cboldt
"Only 49 percent of the available solar radiation reaches the surface of our planet [all types - my comment], 5 percent through direct radiation, 22 percent through clouds and 22 percent by downwards scattering in the atmosphere".

5 percent gets through directly but 22 percent is scattered. But the index of scattering of light is proportional to the fourth power of the frequency so blue light is scattered 16 times as much as red light (that's why the sky is blue). So about 20 percent of the red light is not scattered and a total of about 25 percent of the red light gets through directly (when you consider clouds).

About half the time it's cloudy, but if there were no clouds, then the 22 percent that we get through clouds would get through directly and therefore about 50 percent of the red light from the sun would get through directly.
424 posted on 09/29/2004 3:58:42 AM PDT by Dan Evans
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To: DB; ableChair
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

425 posted on 09/29/2004 3:59:36 AM PDT by Cboldt
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To: Boot Hill
Ya, many of the "specs" are based on 1000 watts/m^2 of light... And then at 25C panel temperature... Right...

I previously found 800 watts/m^2 as being more typical, perfect conditions, and then forgot about it...

The Shell Solar Cell documentation claims:

1000 watts/m^2 in,
175 watts electricity out (with optimum load for maximum power transfer)
and 13.3% efficiency.

That so doesn't compute... The 13.3% and 175 watts don't jive.

Not my area at all. Just did a fast Google search on Shell solar cells...
426 posted on 09/29/2004 4:12:13 AM PDT by DB (©)
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To: abner

bttt


427 posted on 09/29/2004 4:17:23 AM PDT by Guenevere
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To: 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.

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...


428 posted on 09/29/2004 4:21:17 AM PDT by DB (©)
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To: 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.

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

429 posted on 09/29/2004 4:26:51 AM PDT by Dan Evans
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To: ableChair

How large would a device like this be? Especially if it had the ability to target a pilot in an aircraft?


430 posted on 09/29/2004 4:30:41 AM PDT by gitmo (Thanks, Mel. I needed that.)
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To: 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.

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.

431 posted on 09/29/2004 4:32:46 AM PDT by Cboldt
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To: finnman69
 Revelaion 13
 
 11.  Then I saw another beast, coming out of the earth. He had two horns like a lamb, but he spoke like a dragon.
 12.  He exercised all the authority of the first beast on his behalf, and made the earth and its inhabitants worship the first beast, whose fatal wound had been healed.
 13.  And he performed great and miraculous signs, even causing fire to come down from heaven to earth in full view of men.
 14.  Because of the signs he was given power to do on behalf of the first beast, he deceived the inhabitants of the earth. He ordered them to set up an image in honor of the beast who was wounded by the sword and yet lived.
 15.  He was given power to give breath to the image of the first beast, so that it could speak and cause all who refused to worship the image to be killed.
 16.  He also forced everyone, small and great, rich and poor, free and slave, to receive a mark on his right hand or on his forehead,
 17.  so that no one could buy or sell unless he had the mark, which is the name of the beast or the number of his name.
 18.  This calls for wisdom. If anyone has insight, let him calculate the number of the beast, for it is man's number. His number is 666.

432 posted on 09/29/2004 4:33:18 AM PDT by Elsie (Heck is where people, who don't believe in Gosh, think they are not going....)
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To: July 4th

Looks like Michelle Malkin is keeping up on this too...

http://michellemalkin.com/archives/000595.htm


433 posted on 09/29/2004 4:37:23 AM PDT by OXENinFLA
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To: finnman69

http://michellemalkin.com/archives/000595.htm

(hat tip: FReeper finnman69, post #22 on this thread):


434 posted on 09/29/2004 4:39:08 AM PDT by OXENinFLA
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To: DB
That so doesn't compute... The 13.3% and 175 watts don't jive.

The 13.3% probably takes into account efficiency over a range of input power (light) levels, not just the efficiency represented at the peak power input/output level.

I'd look to IEC 61215 and related industry standards to find the definition of mu (efficiency).

435 posted on 09/29/2004 4:48:22 AM PDT by Cboldt
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To: Cboldt
Ya, the power density (really energy density exposure) is what is critical to cause eye damage. This was discussed a few hundred posts back...

I have no idea what the beam spreading would be at that distance. With the 50 watt, 5 mile example I'd bet the power density (with 19.9 watts of total beam power) still available it could easily do major damage in a very short period of time even if the beam increased many multiples of the original beam width.
436 posted on 09/29/2004 4:52:10 AM PDT by DB (©)
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To: Dan Evans

Credit goes to Cboldt.


437 posted on 09/29/2004 4:53:04 AM PDT by DB (©)
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To: Cboldt
I found 14 mW/cm2 was the power density for eye damage. So if a 50 watt laser were focused to a one square meter area that would be (after atmospheric attenuation) 20 watts/10,000 cm2 = 2 mW per /cm2. Not enough for eye damage. It would take a 350 watt laser with optical tracking gear for a 5 mi range.




http://64.233.167.104/search?q=cache:iJqFiyB6bHMJ:www.ensc.sfu.ca/people/faculty/chapman/e894/e894l13g.pdf+laser+eye+damage+mw+&hl=en&ie=UTF-8
438 posted on 09/29/2004 4:58:09 AM PDT by Dan Evans
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To: Dan Evans

Not really......


439 posted on 09/29/2004 4:59:58 AM PDT by blackdog (I survived John Dupont's wrestling camp and all I got was a lousy tee shirt and a prolapse.)
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To: ableChair; Pukin Dog

Let's ask our knowledgeable pilot if he has any word on this. This laser incident supposedly happened to a Delta pilot flying into Salt Lake City. Got any info or insight you can add, Pukin Dog?


440 posted on 09/29/2004 5:02:06 AM PDT by arasina (So there.)
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