Did a meteor bring down Air France 447? | Hacker News Reader

Did a meteor bring down Air France 447?

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31 pointsAnon8432 comments

The probability calculation is incorrect, you can't just multiply the numbers. This should be obvious from the fact that if we wait long enough the probability according to that expression will exceed 1.

If I have a probability of throwing a 6 on a dice of 1/6, throwing 6 dice doesn't give me a probability of 1 of getting (at least) a six.

The correct calculation is:

If the probability of one meteor hitting an airplane is 5.7e-13, then the probability of it missing is 1-5.7e-13. For no plane to be hit, all of the 750e6 hours * 125 meteors/hour must miss. That probability is (1-5.7e-13)^(750e6*125)=0.948, ie the probability of at least one hit is 5.2%.

The answer is close to the same, but that's just luck. If the probability was higher, it would be increasingly off.

Thanks for pointing out the error in his math. But I think you got something wrong too - thrown off big error on his part (~4000x). The correct calculation is:

  p(one or meteor strikes) = 1 - ( (1 - p(strike of any meteor))^ (total # meteors) )
  
  (assuming statistical independence, etc).
  p(strike of any meteor) = 5.82e-13
  (total # meteors) = 3000 * 365 * 20     (for testing over 20 years)

  p(one or meteor strikes) = 1.3e-5 = 0.0013%

The error of multiplying 720e6 hrs (the total # of flight hours) by 125 meteors/hr is that it does not give you the total number of meteors in the time period (which is what you need), it gives you a number ~4000 times larger (because at any one time, there are ~4000 planes in the sky...)

Except that if there are 4000 planes in the sky, that does make (assuming the planes are not overlapping, which seems largely safe) the probability 4000 times larger than for one plane.

As long as the planes don't overlap, the flight hours is the correct number to use, because more planes in the sky gives higher probability. It's not the number of meteors that matter, but the number of "potential meteor strikes".

Equivalently, you can say that the probability of a meteor hitting any one plane at a given time is the fractional area taken up by planes, which if there are 4000 planes in the air is 4000*5.82e-13. (Again assuming the planes don't overlap, which since 5.82e-13<<1 is a safe assumption.)

Ah, yes - I mis calculated p(strike of any meteor) by a factor of 3500 (the # of planes in the sky, according to the article). I think this should be applied to the probability, not the count of the occasions - but the numbers are small enough that it doesn't make much difference.

  p(one or meteor strikes) = 1 - ( (1 - p(strike of any meteor))^ (total # meteors) )
  
  (assuming statistical independence, etc).
  p(strike of any meteor) = 5.82e-13 * 3500 = 2.04E-9
  (total # meteors) = 3000 * 365 * 20     (for testing over 20 years)

  p(one or meteor strikes) = 0.0436 = 4.36%

Yeah, it doesn't make a difference until the planes cover a significant area of the Earth, in which case just multiplying by the numbers of planes will be wrong for the same case that the initial calculation was wrong. Then it would be 1-(1-5.82e-13)^3500 in this case, too.

Just to play this game a little further, the probability is probably (!) smaller, because those meteors are not distributed uniformly across the Earth. Due to the direction of motion of the Earth through space and its sense of rotation, most meteors hit the area that is facing forward in space. That's the area on Earth where it's between midnight and noon, with the maximally forward one at 6am, ie sunrise.

Since I suspect, due to the way civilization works, there are more planes flying between noon and midnight than between midnight and noon, and the ones that are flying during that time are mostly intercontinental flights that go preferentially across the pole, there are probably less planes than average where those meteors hit.

But now I'm just speculating. ;-)

It is not luck. The probability of the article is almost correct, because the multiplication of the numbers is a first order approximation to the exact result.

Using Taylor series, if A<1

  (1-A)^B = 1 – B A + B(B-1)/2 A^2 -  B(B-1)(B-2)/6 A^3 +  B(B-1)(B-2)/24 A^4 - ...

  1-(1-A)^B ~= B A - B(B-1)/2 A^2 + ...

And if A is small enough (A<<1)

  1-(1-A)^B ~= B A - B(B-1)/2 A^2

So, AB is a good approximation and if B is big (B>>1) the second order correction is

  -B(B-1)/2 A^2 ~= -(AB)^2 /2

Then, with A=5.7e-13 and B=750e6*125=9.4E10, then

  Exact:          1-(1-A)^B =  0.0520
  Approximation:         AB =  0.0534
  Second Order:  -(AB)^2 /2 = -0.0014

Yeah, I meant luck in the sense that, if the author consciously used the first-order expansion over the true expression, the article doesn't state that only going to first order was reasonable. Hence one has to assume no such check was made.

Just stating the calculation like he did might induce others to think that the expression applies regardless of the magnitude of the result. And then it will be pure luck whether it works or not.

Birthday paradox strikes again?

I saw a meteor on a flight once. Granted it was pretty far away, but I also only fly a few times a year. It's certainly not impossible.

Not that I want to give any credence to this theory, but actually the statistics are better (or worse) than he thinks, because meteors are believed more likely to strike at the equator than the pole.

[Horrid URL follows, sorry]

http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_quer...

Wow, how on Earth did you find that article? The poor dude's been cited 3 times in 45 years...

No secret, just google. I speculated the distribution may not be symmetrical, thinking there could be an excess of strikes along the ecliptic being fed by the asteroid belt and guessed what terms would crop up in a paper that discussed the topic (I find that is the best way to search for something - don't ask the question, but search for what you think the answer would look like).

Actually what I thought I would find is radar data showing actual counts of meteor strikes, but no joy - I suspect the military knows what they are but they are not saying.

Of course there could be more recent papers hidden behind a paywall, but ADS didn't bring up any obvious abstracts.

While I highly doubt this, I do wonder why we rely on black boxes for crash investigations. Airbus already received some data from the plane after problems developed. Why not develop this technology further to make black boxes redundant?

And if you really do want to keep black boxes, why not develop an ejection mechanism for them so that they would separate from the plane at the last second and float in the ocean?

That seems to be the direction...

http://en.wikipedia.org/wiki/Flight_data_recorder#Future_dev...

Because it's simply not possible to make black boxes redundant.

You can fly into a storm or into a canyon where radio/satellite uplinks won't work. Hence you need to store the data on sturdy local storage aka a black box.

As much as people love to think that technology can completely tame mother nature, it can't. It's still a brutal place, with nasty conditions. While the uplink data is great, you still need backups (Higher resolution is a nice bonus too)

A better way to explain it would be to point out that the remote telemetry and the black boxes are already examples of redundant design. The black box has fewer failure modes recording data, but requires recovery. The telemetry is more easily disrupted, but can get data out even in the face of a total loss of the aircraft.

Redundancy doesn't always mean the same thing as simple duplication.

I know nothing about flight recorders, but offhand it would make sense to me for them to have some sort of transponder. Maybe the requirement of having a power source is unrealistic.

They do - they've got a pinger that sends out a signal after a crash has been detected to aid in locating the box, similar to an avalanche beacon.

Problem is that it's not all that strong, and is likely under a few tons of airplane and a few kilometers of water.

To make things even worse, sea water strongly attenuates electromagnetic radiation at radio frequencies...

It uses sonar. Radio would be useless under water.

I suppose it emits both in radio and sonar, right? What if the airplane crashes over land?

It only activates the sonar if it is under water:

http://trueslant.com/milesobrien/

And it only pings for 30 days.

"Why don't they make the whole plane out of that black box stuff?"

-steven wright

This question was previously asked here:

http://news.ycombinator.com/item?id=640289

I have thought about an ejection mechanism too. Maybe a combined system of ejecting small beacons and the main one, that stays in the plane, would work better. The beacons could be located in multiple locations on the plane, e.g. nose, tail, wings, and when an inevitable crash is detected they could start firing off at regular time intervals, depending on the altitude. If they are light-weight enough maybe some of them could be combined with an expandable helium balloon for easier spotting. Of course this requires having weather data available for the crash area so that one could deduct their starting position. (Disclaimer: I'm not an engineer so I don't know if my idea is stupid.)

We know that the jet was in a severe thunderstorm when it crashed. Severe thunderstorms have brought down many airplanes over the years.

This particular instance involved a very modern plane piloted by a very experienced crew. Google the kind of stress tests they put plane wings and such under. A wide body fuselage should not come apart in a thunderstorm.

It depends on speed and attitude. It's entirely possible, if a plane is flying fast enough, or in regions with strong enough wind shear, to overstress the airframe. That's why there is this thing called "maneuvering speed", it's the maximum speed you can fly where the wing will stall rather than rip off. If you fly faster than that, or if due to wind shear your airspeed suddenly finds itself above that, the airframe can fail either due to severe turbulence or just due to maximum control deflection (though the latter may be electronically limited on an Airbus, don't know about that).

Thunderstorms can kill in a variety of ways -- the turbulence isn't your only problem, although I wouldn't bet that a storm as severe as this one (50kft tops) couldn't have done it.

The kind of bumps you get in something like that aren't what you think of as turbulence in a plane. Airlines have radar and always go around the big cells. This one would have been strong enough to fling people out of their seats and kill them from blunt force impact with the interior of the plane.

Second, in a thunderstorm, you have lots of rain, which if strong enough can interfere with the engines. There is also hail which can do significant damage to the skin of the plane and make it very difficult to handle. Third, there is lightning, and while airliners are supposed to be able to take a lighting strike, there are different types of lightning, and "positive lighting" has been known to poke holes in airplanes otherwise believed to be protected. Fourth, there can be lots and lots of ice. Airbus has already issued a letter warning of the danger of pitot icing as a result of this crash, and they know that at least one of the airspeed indicators wasn't working properly, probably because it was iced over.

Furthermore, with extreme turbulence, it can be difficult to even read the instruments. With lighting going off all around you, no visible horizon, and you being vibrated and thrown around in your seatbelts, it is very easy to become disoriented and lose control, possibly to the point of overstressing the airframe and ripping it apart.

A thunderstorm like this is about the most dangerous place for an airplane to be. I'm certain that one way or another, that is what did them in.

I think they haven't found any verified debris yet.

What's the probability that what those pilots saw was an up beam?

No.

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