PPRuNe Forums - View Single Post - Qantas A380 uncontained #2 engine failure
Old 17th Nov 2010, 03:28
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FluidFlow
 
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The probability of being hit by one of the 3 large pieces of this rotor is roughly as follows. Hull penetration angle spread of 15 degrees from center of rotor. Rotor broke into 3 pieces giving a probability of heading in the direction of a person in the hull of 15/360x3= 1 in 8. (perhaps up to 18 deg giving 1 in 6.7)
Wing penetration angle spread of 40 degrees from center of rotor giving a probability of heading in the direction of the wing of 40/360x3= 1 in 3.
So with this type of failure from an inboard engine, a wing penetration is twice the chance of a person being hit. Not a lot different from the probability of Russian Roulette. However the wing also provides some shielding effect to the hull.
FF

Last edited by FluidFlow; 17th Nov 2010 at 21:49. Reason: Better sketches obtained to determine angles
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