4 ms·

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:

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...)

lutorm 17y ago
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.)
- tomsaffell 17y ago
  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%
  - lutorm 17y ago
    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.
lutorm 17y ago
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. ;-)

lutorm 17y ago

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.)

tomsaffell 17y ago
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%
- lutorm 17y ago
  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.

lutorm 17y ago

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. ;-)