Laser Pointers - The Maths
I tried to post some analytical numbers about this in the other thread, but I mixed in some of my conjectures and opinions and it seems they upset people, so my posts were deleted and I was locked out of that thread.
I think the analysis is still worth an airing, and I’ve had some PMs asking for the details again. So I thought I would have another go, but this time sticking solely to the analysis and leaving my opinions out of it in the hope that no one feels it necessary to censor it again. So for what it’s worth…
Laser pointers – the maths:
Assume airliner flying at 8,000 feet.
Assume that to be visible to the pilot the ground level source must be no more than 30 degrees below the horizon.
Therefore the distance between source and target is 8,000/Sin(30) feet = 16,000 feet: let’s call it 3 miles.
Laser dispersion of 2mrad (0.06 degrees) at 3 miles gives a beam width at the target of roughly 10 metres (generous number).
That same number tells us that to keep any part of the beam impinging on the eye of the pilot needs the laser to be within 0.06 degrees of the nominal target. For those with target shooting experience – this is the equivalent of a six inch grouping at 100 yards (not hard to achieve with a rifle while lying prone, but rather optimistic with a pistol held in one hand while standing up).
The target is moving, of course, but at that range it moves quite slowly.
Of course if we consider the low-level case on approach at 300 feet the range becomes 600 feet and the beam diameter becomes four and a half inches, but the targeting accuracy (where the “target” is a small point like the pilot’s eye) is actually the same, although the target is moving much faster across the sights.
According to the open sources [lots of them; the wiki page is the most easily found and digested] the power density required to cause retinal damage for lasers in the visible spectrum (say 400-800nm) is around 2mW/cm^2 for a 10 second exposure, 4mW/cm^2 for a 1 second exposure or 7.5mw/cm^2 for a 0.1 second exposure. So the question becomes what source power is required to achieve those power levels.
The beam diameter at 3 mile range was 10 metres. So the beam area is 785,500cm^2. If we assume even power density across the disk that means that to achieve a damaging power density for a 10 second exposure needs a 1500W laser, for a 1 second exposure this becomes a 3000W laser and for a 0.1 second exposure it would need a 5,800W laser. Note that all three of these are more than three orders of magnitude (ie a thousand times) more powerful than the worst of the “illegal” laser pointers being discussed, and six over orders of magnitude (ie over a million times) more powerful than the standard laser pointers people like me use in lecture theatres. KiloWatt lasers are not hand-held devices.
For the 300 foot case it’s a bit different. At this range the beam diameter is only 11.5cm, so the disk area is 104cm^2. The laser power to achieve the damaging exposures then becomes 0.21W, 0.42W and 0.78W (or 210mW, 420mW and 780mW if you prefer) for the 10 second, 1 second and 0.1 second exposures respectively. “Illegal” laser pointers with *claimed* powers up to around 1watt can be bought from overseas [from the UK], so if someone can target them accurately enough they certainly could represent a risk of eye damage to the pilots of aircraft in the final few hundred feet on approach, and less so on departure (simply due to the angles involved).
These rough calculations ignore the attenuation effects of the cockpit glazing and the atmosphere. And the latter should not be overlooked, because visible lasers are just light – haze, cloud, dust/smog will block a laser just as much as the reduce visibility. Visible light lasers don’t “burn through” haze and cloud until you get to power levels of tens or hundreds of kilowatts, that’s why jet fighters still have guns and missiles instead of laser weapons and even the anti-satellite laser concepts need lasers so large and powerful that they need an airliner to lift them and they need to be carried up to over 30,000 feet to get to air that is clear enough not to disperse the beam.
Finally, this analysis only considers the danger of eye-damage from lasers. The distracting flash will still be visible, assuming the target can be held accurately enough, at both the ranges considered.
PDR