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Night flying in the mountians in a cessna 150,



 
 
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  #102  
Old February 26th 05, 10:09 PM
Howard Nelson
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"George Patterson" wrote in message
...


Jose wrote:

But you would never say "the Appalachian ranges." For the same reason,

you
should never say "the Sierras" when you're talking about the Sierra

Nevada.

But we say "the Appalachians".


And it would be correct to say "the Nevadas." But not "the Sierras."

George Patterson
I prefer Heaven for climate but Hell for company.



Correct or not. I live out here and if you said in conversation:

I went hiking last weekend in the Sierra.

or

I think I will go skiing next week in the Sierra

or

Take I80 over the sierra.

You would sound kind of goofy to a local.

Howard


  #103  
Old February 26th 05, 10:39 PM
Jose
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1. The probability of experiencing an engine failure (or any other
improbable event for that matter) AT SOME POINT IN YOUR FLYING CAREER
goes up the more you fly. It goes up monotonically but nonlinearly
according to the formula 1-(1-P)^N, which asymptotically approaches 1 as
N gets large.


I have a feeling everyone in this discussion is talking past each other.
However, I'll still pick a nit (since after all, this is usenet).
Probability deals =only= with events whose outcome is not known or not
taken into account. If I take any random 10,000 hours, the probability
of some occurance (like an engine failure) is the same. Let's say the
probability over 10,000 hours is 70%. If I have =already= flown 9,999
hours without a failure, it is =not= true that my chance of failure on
the last hour is 70%. Likewise, if I have already flown those 9,999
hours and already had three engine failures, the chance of having
another in that last hour is =not= zero nor is it negative ("to make up
for the extra failures"). It is the same as the probability of a
failure on the =first= hour.

HOWEVER... the chance that, OVER THOSE 9,999 HOURS flown =plus= the one
not flown, I would =either have an engine failure shortly, =or= look
over my logbook and find that I already had one, would be the original
70%. The key here is including those flown hours without regard to
whether or not there was a failure there - iow as if we did not know the
result.

If you eliminate the hours flown because their outcome is known, then
you can only (correctly) apply probability to the unflown hours.

This is (of course) a different question from the one that says "Here's
my logbook. It has 10,000 one-hour flights in it. I had one engine
failure." and then, as one thumbs through the book saying "not this
one... not this one..." the chance of coming across a flight with an
engine failure =does= increase - because in this case an engine failure
is =guaranteed=. (it already happened).

Running out of fuel is not my idea of "engine failure".


If the fuel pump breaks and thus all four engines quit, did you have an
engine failure?

Jose
r.a.owning and r.a.student trimmed. I don't follow them.
--
Nothing is more powerful than a commercial interest.
for Email, make the obvious change in the address.
  #104  
Old February 26th 05, 11:22 PM
Matt Whiting
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wrote:
On Sat, 26 Feb 2005 11:45:19 -0500, Matt Whiting
wrote:


wrote:


On Fri, 25 Feb 2005 17:16:35 -0800, Ron Garret
wrote:



That's true, but the longer you fly (or play the lottery) the closer
your probability of experiencing an engine failure (or a lottery win)
some time your career approaches 1.

Of course, you might have to fly/play for a *very* long time before that
probability actually gets close to 1, but sooner or later it will be 1
to any desired degree of accuracy. So the statement "fly long enough and
you will experience an engine failure" is pretty close to being true.
The question is how long is "long enough."

rg



This just ain't so.

Every time you play the lottery, it's like the first time you ever
played it.

It doesn't matter whether you won a jillion yesterday, or haven't won
in 50 years, or never played. The odds are exactly the same.


Yes, for every given play you are correct. However, Ron is correct that
in aggregate, someone who plays more often has a higher overall chance
of winning at some point than a person who only plays once in their
lifetime. At least I think that is the point he was making.


Matt




no, I think his point is that you are more likely to have an engine
failure tomorrow if you have flown 10,000 hours than if you have flown
10 hours.


I just re-read what he wrote above, and while it could have been
clearer, I still think he meant the probably in aggregate, not the
specific probability on the next flight. However, I'll let him weigh in
with what he meant. :-)

Matt
  #107  
Old February 27th 05, 12:02 AM
Ron McKinnon
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"Ron Garret" wrote in message
...
In article ,
wrote:

I think the implication, with all due respect, in the way you worded
your post, is that the probability is increasing as you flying time is
increasing.


It depends on what you mean by "the probability". There
are two different probabilities being discussed: there is the
probability of a failure on any particular flight, which doesn't
change, and there is the cumulative probability of experiencing
failure on some flight, which does change (it increases with
each flight). This is clearly not the case, as I think we all
now agree.


There is also the probability (that Peter (I think) proposed)
stated as a cumulative probability in terms of an arbitrary
large number of trials (flights, or hours, or whatever).

If you convert this to a probability of occcurence with
a lower number of trials (flights, or hours, or whatever)
that probability will be lower. Looked at it this way,
if the probability of an 'occurrence sometime in (the
remainder of )one's career is known, then as the career
progresses, the probability of 'an occurrence sometime
(in the remainder of) one's career diminishes from
that value.

This is a direct consequence of

1) the premises (accepted by all here,
apparently) that

- the probability for any given trial (hour, flight, or
whatever) is assumed to be independent of any other
given trial (hour, flight or whatever) and
- the probability is assumed to be the same for each
such trial, and

2) the assertion that the probability of an occurrence
over n trials is (1-(1-p)^n, where p is the probability
of occurence in a single such trial.

Its the same problem worked back to front (or
front to back, depending on your point of view):

i.e.: Let p2 be the probability of an occurence in
n2 trials, and let p1 be the probability of an occurence
in n1 trials, if n1 n2, then p1 p2.

If you *start* with p1, as you consider an increased number
of trials the probability will increase, if you *start* with p2
and consider a decreased number of trials, the probability
will decrease.

Your statement is ambiguous because you don't say
which probability you're referring to.


Yes. The logical conclusion is determined from
the premises used. You only get out of it what you
put in.

Every day is a new day, and N gets reset to zero.


Not quite. Every day is indeed a new day, but with
every flight N is incremented by one.


It depends on upon from which premise you started.
If you're considering your probability in terms of
occurences per N trials, you might change N if you
start out with it being 'the number of trials in my
entire career', but the probability of an occurence
'in the next N trials' otherwise doesn't need any
change in N from day to day.

But 'the number of trials in my career' is moot
in the first place, and I'd argue that arbitrarily
specifiying the number of trials that are 'going to
occur' in your career is equally problematic, as
is coming up with such a probability in the first
place. The best you can get out this argument, I
think, starting out with a guess for the cumulative
probability of the 'entire carreer', is a qualitative
'probability is decreasing' as the career progresses,
and you can't really ever quantitatively say how
much.


  #108  
Old February 27th 05, 12:36 AM
Ron Garret
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In article P68Ud.515993$8l.368458@pd7tw1no,
"Ron McKinnon" wrote:

as the career
progresses, the probability of 'an occurrence sometime
(in the remainder of) one's career diminishes from
that value.


Yes, but only because N is lower. Whatever N is, after every flight N
is 1 less than it was before.

But 'the number of trials in my career' is moot
in the first place,


That is arguable. As a precise number you're probably right. But in
broad brushstrokes you can decide, e.g. never to try something, to try
something once and then never again, to try something a dozen times in
your lifetime, to do something once a month, once a week, once a day, or
multiple times a day. Each of these choices entails a monotonically
increasing risk of encountering certain kinds of disasters over your
lifetime.

My personal risk tolerance works out something like this:

Things I'm not willing to try even once: heroin, motorcycle racing
Things I'm willing to try once in my lifetime and never again: going
into space (assuming I ever have the opportunity)
Things I'll do a dozen times: aerobatics
Once a month (on average): skiing
Once a week: Flying GA aircraft
Once a day: getting out of bed in the morning :-)
Multiple times a day: driving on the freeway, eating sushi :-)

rg
  #109  
Old February 27th 05, 02:03 AM
Peter Duniho
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wrote in message
...
[...] the chance is actually only 75%.


How so?

The probability of both engines failing is .25, I agree, but I'm
talking a failure of either engine.


I am too. What chance do YOU think you have of having a failure of either
engine, if not 75% (in this example)?

If the chance of an engine failure is 50% (0.5), then the chance of either
engine failing when you have two engines is 1-(0.5)*(0.5). 75%.

The probability of both engines failing is indeed only 25%. The probability
of EITHER engine failure is 75%.

You need to do the subtraction because the chance of an engine failure is
actually the opposite of the chance of completing a flight without an engine
failure. To make the flight successfully without either engine failing
requires BOTH engines to not fail, and the way to calculate that is to
multiply the chances of each engine failing (which in this case is just two
engines, with identical chances).

The chance of you completing the flight without a failure is 25% (50% *
50%), so the chance of an engine failure on the flight is 75%.

Pete


  #110  
Old February 27th 05, 02:06 AM
Peter Duniho
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"Ron Garret" wrote in message
...
[...] If you think about it, there is absolutely no
difference in the risk calculation between making one flight with four
engines and four flights with one engine


The difference is that when you make a flight with four engines, you know up
front that you're carrying four engines. The calculation based on making
four flights with one engine is only useful when you know in advance you're
making four flights.

I certainly hope to make at least four more flights during my flying career,
but it's not certain that I will.

Sorry you can't see the difference. It's a crucial element to the question
of whether it makes sense to worry about the cumulative odds of an engine
failure.

Pete


 




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