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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: ableChair
No, you don't lose a fixed 50 watts (or some other value) for fixed distance regardless of the the starting level. You lose a percentage of energy (attenuation) over X distance.

As Boot Hill stated, if you lose 10% over a mile, you lose another 10% over the next mile and so on. If you went 10 miles with 10% loss for each mile the total loss over 10 miles would be 65.1%. Now it isn't quite that straight forward but it is pretty close.

If you track the energy coming from the sun to the ground, 19% is absorbed in the atmosphere (heating it) with about 32% being reflected back out to space and the remaining 49% making it to the ground. That's through a lot of miles of atmosphere

The laser is basically no different. A percentage of the energy is lost over a given distance. If you double the light level you start with you still get double the light at the other end regardless of the absolute level. There are exceptions to this at extreme (large or small) energy densities but that isn't relevant here.
341 posted on 09/29/2004 1:18:13 AM PDT by DB (©)
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To: DB
Oh, BTW, the Earth's atmosphere absorbs 95% of incident radiative energy, which is what we're concerned about here, not 81% as you suggested. This is a common error as people confuse radiative and convective energy.
342 posted on 09/29/2004 1:20:49 AM PDT by ableChair
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To: Boot Hill
All I think I know is that light is quantized into integer counts. You can count individual photons (or whatever they are).
343 posted on 09/29/2004 1:22:00 AM PDT by DB (©)
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To: DB
No, you don't lose a fixed 50 watts (or some other value) for fixed distance regardless of the the starting level. You lose a percentage of energy (attenuation) over X distance.

I didn't say it was fixed, that was only an example. See my post to Boot. What he posted doesn't matter, I'm only required to show/suggest that the energy loss results in incident energy on the eyeball insufficient to burn it out. My suspicion is that it would take something FAR more energetic than what you could easily assemble at home.
344 posted on 09/29/2004 1:22:56 AM PDT by ableChair
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To: DB

Sometimes you can count them, but SOMETIMES you should treat light as a wave!


345 posted on 09/29/2004 1:23:46 AM PDT by ableChair
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To: DB

Oh wow, my mistake. You suggested the atmosphere absorbs only 19%. You're way off. It absorbs 95% of radiative energy.


346 posted on 09/29/2004 1:25:03 AM PDT by ableChair
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To: ableChair

Actually I got the numbers from here:
http://www.ldeo.columbia.edu/edu/dees/ees/climate/lectures/radiation/Kushnir_Lecture1.pdf

Of course it depends on where you are on the earth among other things...


347 posted on 09/29/2004 1:29:56 AM PDT by DB (©)
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To: ableChair
That can't be required just to burn metal; it must be either for penetrating a media or for the radius of the beam (as one poster pointed out).

Yes, and a lot of other things too. Maybe that figure was was the power input to the laser and the laser was very inefficient. Do you have a link to that source? Was it referring to an xray laser? If so, they weren't very collimated. The beam could have expanded a thousand miles wide. And also what does this have to do with atmospheric absorption? SDI was to shoot down missiles during the boost phase after they left the atmosphere.

348 posted on 09/29/2004 1:32:24 AM PDT by Dan Evans
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To: ableChair

If the the atmosphere absorbed 95% of the radiative energy the atmosphere would be seriously hot and I don't think we'd be here down below.


349 posted on 09/29/2004 1:32:36 AM PDT by DB (©)
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To: DB

How about another thought experiment? The premise of my objection is that it would take something much more energetic than what could be easily assembled at home to burn out an eye at significant distances. Recall the space heater. Suppose we were to measure the heat of the air around a laser beam and find that the air heated up by, say, 1 degree C over the firing time. Now, measure ALL points for the FULL length of the laser beam, say, 10 miles. The total temperature increase would almost certainly exceed the heat generated by the space heater. You can even set t1=t2 for comparisons sake, and set the power equal, at say, 1000 watts. Same output energy, but clearly different heat energies. This suggests that the energy at the target is infinitesimally approaching zero, not enough to burn out an eye. In reality, the heating of the atmosphere would taper off as you approached the target, but the idea is the same.


350 posted on 09/29/2004 1:34:00 AM PDT by ableChair
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To: DB

Dude, that the Earth's atmosphere absorbs 95% of all incident radiative energy is a well known, publicly accessible fact. See the Cambridge Atlas of Astronomy. Trust me.


351 posted on 09/29/2004 1:34:58 AM PDT by ableChair
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To: ableChair
...the Earth's atmosphere absorbs 95% of incident radiative energy...

I'm assuming you meant re-radiated energy, in which case the wavelengths involved are extremely long IR and totally beyond the wavelengths of any laser. On the other hand, if you truly meant incident radiant energy, as in the energy arriving from the sun, then in that case, the atmosphere only absorbs 27%.

In either case, I fail to see the relevance to the present discussion.

--Boot Hill

352 posted on 09/29/2004 1:35:48 AM PDT by Boot Hill (Candy-gram for Osama bin Mongo, candy-gram for Osama bin Mongo!!!)
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To: Dan Evans
SDI was to shoot down missiles during the boost phase after they left the atmosphere.

No, see my previous posts. These numbers were for a land based laser that fired at mirrors in orbit. It's a two-way attenuation by the atmosphere.
353 posted on 09/29/2004 1:36:30 AM PDT by ableChair
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To: DB

Trust me, that link is wrong, or you read it wrong. This is a common mistake people make. You're confusing radiative and convective energies.


354 posted on 09/29/2004 1:37:17 AM PDT by ableChair
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To: Boot Hill

I'll say it again, it is a publicly accessible fact. Look it up...see the Cambridge Atlas of Astronomy.


355 posted on 09/29/2004 1:38:22 AM PDT by ableChair
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To: DB
All I think I know is that light is quantized into integer counts.

LOL, then explain the operation of a Fabry-Perot Interferometer based upon photons!   J

--Boot Hill

356 posted on 09/29/2004 1:39:02 AM PDT by Boot Hill (Candy-gram for Osama bin Mongo, candy-gram for Osama bin Mongo!!!)
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To: Boot Hill

BTW, this point was brought up by a previous poster who was trying to argue that the atmosphere doesn't absorb much light; even though the light he is talking about is non-coherent and I don't even know if that's a valid comparison.


357 posted on 09/29/2004 1:41:29 AM PDT by ableChair
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To: Boot Hill

Hey, that's off topic :-) Light is schizophrenic but don't tell them that!


358 posted on 09/29/2004 1:42:29 AM PDT by ableChair
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To: ableChair

I think it was your point to another poster to site references, not to make others find your "facts" for you.

Your making a claim. Find a link to support it and I'll read it.

I sited my source. Your turn. I believe you are wrong by a large factor. Prove me wrong and I'll accept it.


359 posted on 09/29/2004 1:43:16 AM PDT by DB (©)
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To: DB

Well, my source is a hard-copy. How about a direct quotation from the Atlas? Otherwise I have to go hunt it down on the internet at 3 a.m.


360 posted on 09/29/2004 1:45:45 AM PDT by ableChair
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