ATPL question
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ATPL question
64. A runway is 5,000 ft, having a stopway of 1,500 f t
and clearway of 2,000 ft. A departing aircraft must have
sufficient take-off distance to accelerate to V1, experience
a failure of the critical engine and continue to
climb to reach 35 f t at V2 within a maximum of
a) 5,000 f t .
b) 6,500 f t .
c) 7,000 f t .
d) 8,500 f t .
Would anyone help me with this one. I believe the answer that was proposed is wrong.
and clearway of 2,000 ft. A departing aircraft must have
sufficient take-off distance to accelerate to V1, experience
a failure of the critical engine and continue to
climb to reach 35 f t at V2 within a maximum of
a) 5,000 f t .
b) 6,500 f t .
c) 7,000 f t .
d) 8,500 f t .
Would anyone help me with this one. I believe the answer that was proposed is wrong.
We must of course get to screen height before the end of the TODA to avoid hitting anything nasty. But we must plan to do better than that.
The following definitions of the Take-Off Distance Required are taken from the CAP698 provided by the UK CAA for ATPL exam purposes
2.1.1. Field Length Requirements
a) When no stopway or clearway is available the take-off distance when multiplied by 1.25 must not exceed TORA.
b) When a stopway and/or clearway is available the take-off distance must:
i) not exceed TORA
ii) when multiplied by 1.3, not exceed ASDA
iii) when multiplied by 1.15, not exceed TODA
In this question we have the following distances available
TORA is 5000
ASDA is 5000 + 15000 = 65000
TODA = 5000 + 2000 = 7000
There is a stopway and a clearway so we apply para (b).
Applying requirement (b) (i) gives a distance required of 5000
Applying requirement (b) )ii) gives a distance required of 6500 / 1.3 = 5000
Applying requirement (b) (iii) gives a distance required of 7000 / 1.15 = 6086
The shortest of these distances required is 5000.
All of the above assumes a dry, flat paved surface of course.
The following definitions of the Take-Off Distance Required are taken from the CAP698 provided by the UK CAA for ATPL exam purposes
2.1.1. Field Length Requirements
a) When no stopway or clearway is available the take-off distance when multiplied by 1.25 must not exceed TORA.
b) When a stopway and/or clearway is available the take-off distance must:
i) not exceed TORA
ii) when multiplied by 1.3, not exceed ASDA
iii) when multiplied by 1.15, not exceed TODA
In this question we have the following distances available
TORA is 5000
ASDA is 5000 + 15000 = 65000
TODA = 5000 + 2000 = 7000
There is a stopway and a clearway so we apply para (b).
Applying requirement (b) (i) gives a distance required of 5000
Applying requirement (b) )ii) gives a distance required of 6500 / 1.3 = 5000
Applying requirement (b) (iii) gives a distance required of 7000 / 1.15 = 6086
The shortest of these distances required is 5000.
All of the above assumes a dry, flat paved surface of course.
Join Date: May 1999
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Good Morning, Keith. you are quoting the rules for a Class B aircraft, which has no V1 and does not consider an engine failure on the runway. I had assumed from the question (which refers to both V1 and a continued take-off after an engine failure) that it must be a Class A aircraft, and on a dry runway, which leads me to (c) 7000ft, like Beags. I'm looking for the catch but can't see it!
The pedant would say that the parameters of the question are not quite correct, that the engine failure should occur at VEF, that there is a minimum one second gap between VEF and V1, and that V1 is the speed at which, in the event of a rejected take-off, the first action should be taken to abort and, in the event of a continued take-off, is the minimum speed to continue to make 35ft and V2 at the screen height.
The pedant would say that the parameters of the question are not quite correct, that the engine failure should occur at VEF, that there is a minimum one second gap between VEF and V1, and that V1 is the speed at which, in the event of a rejected take-off, the first action should be taken to abort and, in the event of a continued take-off, is the minimum speed to continue to make 35ft and V2 at the screen height.
Yes Alex you are quite right. I was also looking for the catch and (mistakenly) thought that I had found it.
It would be interesting to know what "correct answer" the OP was given. This might help in deducing what the catch really is.
It would be interesting to know what "correct answer" the OP was given. This might help in deducing what the catch really is.
Join Date: Oct 2000
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Hey Alex,
You taught me this stuff 12 years ago and I chose answer (c) 7000' before reading further down the thread. Guess you taught me right!!
Maybe there's no catch; sometimes life really is as simple as it seems!!
Happy landings
3 Point
You taught me this stuff 12 years ago and I chose answer (c) 7000' before reading further down the thread. Guess you taught me right!!
Maybe there's no catch; sometimes life really is as simple as it seems!!
Happy landings
3 Point
