Timeline for Why are commercial flights not equipped with parachutes for the passengers?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 18, 2014 at 3:21 | history | edited | Danny Beckett | CC BY-SA 3.0 |
proper times signs
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| Apr 17, 2014 at 2:09 | history | edited | Danny Beckett | CC BY-SA 3.0 |
Added MathJaX
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| Mar 16, 2014 at 20:16 | history | edited | egid | CC BY-SA 3.0 |
moving equation to its own line for readability
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| Mar 15, 2014 at 21:23 | comment | added | reirab | @QuestionOverflow, Yes, as I said before, it would depend on a lot of factors, especially related to the particular aircraft and which exit door is chosen. For a typical airliner, though, I wouldn't advise trying it. The temperatures aren't going to be pleasant, either. | |
| Mar 15, 2014 at 12:33 | comment | added | Question Overflow | @reirab, thanks for making the effort to do the maths. There are many factors at play. Whether a person would struck anything also depends on the configuration of the aircraft. Based on your calculations, we can't conclusively say that a person would meet certain death by bailing out from the aircraft. | |
| Mar 14, 2014 at 15:33 | comment | added | reirab | However, the drop required to really be safe would depend largely on which door you exited from, the type of aircraft, and the attitude of the aircraft relative to the airstream. If you jump in front of the wing, you'd need to fall further to be safe in a normal commercial aircraft, since the bottom of the wing is usually well below the cabin exit doors. If you exit behind the wing, the distance you need to fall will depend on the position of the horizontal stabilizers and the attitude of the aircraft relative to the airstream, both of which can vary significantly. | |
| Mar 14, 2014 at 15:28 | comment | added | reirab | @QuestionOverflow, Based on my computations above from the drag equation, the initial acceleration due to drag would have a magnitude about 7x the acceleration due to gravity. 1 g = about 32.17 ft/s^2, so 7 gs is about 225 ft/s^2. If we used the above equation for a constant acceleration, that would give us a backwards displacement of 39 ft in the 0.59s it takes for your head to drop below the level of the bottom of the exit door. The real displacement in the first 0.59s would be a bit less since the acceleration isn't constant (requires a diff eq to get the exact answer.) | |
| Mar 13, 2014 at 12:46 | comment | added | Question Overflow | Is there a good reason why you choose 0.4s as the time it takes to decelerate and not 1s or even 5s? | |
| Mar 13, 2014 at 12:11 | comment | added | reirab | It turns out to be a lot more complicated than this because both the direction and magnitude of drag depends enormously on how your body is positioned relative to the airstream and because the airstream itself might not be exactly parallel to the aircraft depending on your jump location (it would be moving upwards over the leading edge of the wing and downwards aft of the trailing edge,) but this is good as a first approximation. Thanks. And, just to state the obvious, if you're exiting from behind the wing, you are within 40 m horizontally of the horizontal stabilizer on any current jetliner. | |
| Mar 13, 2014 at 4:10 | comment | added | casey | The terminal velocity of a human oriented perpendicular to the relative wind and spread out is somewhere in the neighborhood of 120 mph, so you might figure your worst case deceleration to use this forward speed and assume it happens instantly. | |
| Mar 13, 2014 at 4:08 | history | edited | casey | CC BY-SA 3.0 |
grammar, understandability, spelling, units.
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| Mar 13, 2014 at 3:40 | review | First posts | |||
| Mar 13, 2014 at 4:11 | |||||
| Mar 13, 2014 at 3:24 | history | answered | kangacHASHam | CC BY-SA 3.0 |