4 ms·

It should; all of the material quotes in this article came from a SpaceX public relations agent and not from independent space analysts. Not exactly objective.

by codex 14y ago

It should; all of the material quotes in this article came from a SpaceX public relations agent and not from independent space analysts. Not exactly objective. The only quote from Orbicomm is practically a tautology.

deleted 14y ago
[deleted]
- aeontech 14y ago
  You are mixing your terms I think. (2/4)^3 is the chances of the next launch also having a single engine failure. The chances of a single engine failing in a launch based on these four launches should be 2/36, two engines failed out of 36 fired total. So, assuming that a single engine failure does not increase failure chance of nearby engines (which is not true probably), and assuming past performance is indicator of future performance, and discounting all the test fires that they have done with the rocket, the formula should look more like the following: Chances of any given 3 engines failing is the inverse of the chances of at least two engines failing at the same time. Chances of at least two engines failing equals no engines failing + 1 engine failing + 2 engines failing Based on current engine failure rate, that's (17/18)^9 + (1/18)^8 + C(9,2) * (1/18)^2 * (17/18)^7 where C(9,2) = 9!/(2! 7!) = 36 So that looks more like 98.88% chance of two or fewer engines failing, the converse of which is about 0.02% chance of a triple or more failure. (math is not my strong suit, please correct me if I'm wrong, but #math on IRC thought this looked correct).
  - ordinary 14y ago
    > Based on current engine failure rate, that's (17/18)^9 + (1/18)^8 + C(9,2) * (1/18)^2 * (17/18)^7 I think the second part of that calculation is incorrect. I'm getting: P(0 engine fail) = (17/18)^9 ~= 0.5978 P(1 engine fail) = (17/18)^8 * (1/18) * C(9,1) ~= 0.3165 P(2 engine fail) = (17/18)^7 * (1/18)^2 * C(9,2) ~= 0.0745 Adding those together gives: P(<=2 engine fail) = (17/18)^9 + (17/18)^8 / 2 + (17/18)^7 / 9 ~= 0.9888 That's the same answer you came up with, so I guess you have the correct calculation written down somewhere. Moving to 3 or more engine fails then gives: P(>=3 engine fail) = 1 - P(<=2 engine fail) ~= 0.0112
    - aeontech 14y ago
      you're absolutely right, i was doing the final subtraction in my head and not paying attention...
  - dbaupp 14y ago
    Not commenting on the correctness or otherwise of the rest, but 98.88 + 0.02 != 100. (You probably want 1.12% chance of triple or more failure.)
    - aeontech 14y ago
      d'oh! thanks :)
  - codex 14y ago
    I believe this is correct. I had too few digits of precision in my original calculation; rounding is bad!
codex 14y ago
In looking at this a bit further, it seems that the reliability of the Falcon 9 is not looking so good. It has only flown four times and suffered single engine shutdowns on two of them (the other: http://www.spacenews.com/civil/spacex-acknowledges-falcon-engine-anomaly-during-latest-launch.html http://www.spacenews.com/civil/spacex-acknowledges-falcon-en...). See also: http://www.americaspace.org/?p=9044 http://www.americaspace.org/?p=9044. Simple probability then gives a failure rate of 3%. I have not attempted the Bayesian analysis.
- saraid216 14y ago
  FWIW, they're going to be using a different engine soon.

by codex 14y ago

deleted 14y ago

[deleted]

aeontech 14y ago
You are mixing your terms I think. (2/4)^3 is the chances of the next launch also having a single engine failure. The chances of a single engine failing in a launch based on these four launches should be 2/36, two engines failed out of 36 fired total. So, assuming that a single engine failure does not increase failure chance of nearby engines (which is not true probably), and assuming past performance is indicator of future performance, and discounting all the test fires that they have done with the rocket, the formula should look more like the following: Chances of any given 3 engines failing is the inverse of the chances of at least two engines failing at the same time. Chances of at least two engines failing equals no engines failing + 1 engine failing + 2 engines failing Based on current engine failure rate, that's (17/18)^9 + (1/18)^8 + C(9,2) * (1/18)^2 * (17/18)^7 where C(9,2) = 9!/(2! 7!) = 36 So that looks more like 98.88% chance of two or fewer engines failing, the converse of which is about 0.02% chance of a triple or more failure. (math is not my strong suit, please correct me if I'm wrong, but #math on IRC thought this looked correct).
- ordinary 14y ago
  > Based on current engine failure rate, that's (17/18)^9 + (1/18)^8 + C(9,2) * (1/18)^2 * (17/18)^7 I think the second part of that calculation is incorrect. I'm getting: P(0 engine fail) = (17/18)^9 ~= 0.5978 P(1 engine fail) = (17/18)^8 * (1/18) * C(9,1) ~= 0.3165 P(2 engine fail) = (17/18)^7 * (1/18)^2 * C(9,2) ~= 0.0745 Adding those together gives: P(<=2 engine fail) = (17/18)^9 + (17/18)^8 / 2 + (17/18)^7 / 9 ~= 0.9888 That's the same answer you came up with, so I guess you have the correct calculation written down somewhere. Moving to 3 or more engine fails then gives: P(>=3 engine fail) = 1 - P(<=2 engine fail) ~= 0.0112
  - aeontech 14y ago
    you're absolutely right, i was doing the final subtraction in my head and not paying attention...
- dbaupp 14y ago
  Not commenting on the correctness or otherwise of the rest, but 98.88 + 0.02 != 100. (You probably want 1.12% chance of triple or more failure.)
  - aeontech 14y ago
    d'oh! thanks :)
- codex 14y ago
  I believe this is correct. I had too few digits of precision in my original calculation; rounding is bad!

codex 14y ago

In looking at this a bit further, it seems that the reliability of the Falcon 9 is not looking so good. It has only flown four times and suffered single engine shutdowns on two of them (the other: http://www.spacenews.com/civil/spacex-acknowledges-falcon-engine-anomaly-during-latest-launch.html http://www.spacenews.com/civil/spacex-acknowledges-falcon-en...). See also: http://www.americaspace.org/?p=9044 http://www.americaspace.org/?p=9044. Simple probability then gives a failure rate of 3%. I have not attempted the Bayesian analysis.

saraid216 14y ago
FWIW, they're going to be using a different engine soon.