ATPL theory questions
Joined: Jul 2016
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From: lisboa
What happen with the left wing AoA when rolling to the right?
Sometimes I read that the AoA is higher because of the deflection of the aileron downwards and every now and then is lower attending to the rolling effect -lift vector tilts back due to the change of the relative wind-.
Can anyone help me to understand this possible contradiction?
Sometimes I read that the AoA is higher because of the deflection of the aileron downwards and every now and then is lower attending to the rolling effect -lift vector tilts back due to the change of the relative wind-.
Can anyone help me to understand this possible contradiction?
Last edited by smthngdffrnt; 12th September 2017 at 15:33.
Joined: Nov 2000
Posts: 4,330
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From: White Waltham, Prestwick & Calgary
Both I suppose - for example, in a helicopter, a rising rotor blade has a reduced angle of attack because of the change in relative airflow, a similar situation to a rising wing, but helicopter blades don't have ailerons, so the AoA will also increase when they are used.
Joined: Aug 2015
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From: eu
Let's have a go...
We know distance to PET = (DH)/(O+H), substituting..
(60/100)*1200 = (1200*H)/(O+H) therefore
(60/100) = H/(O+H) and, if we wanted to, we could find O in terms of H
also distance to PSR = EO'H/(O'+H), using O' for distance out because groundspeed out to PSR/PET may not be the same as groundspeed on from PET (engine fail case etc. - similar arguments can be presented for H not being the same as H')
substituting...
(84/100)*1200 = (8.4*O'*H)/(O'+H)
Now at this point, if O and O' were the same, we could substitute our values for O in terms of H from the first formula and find a numerical value for H. We don't know this is the case, though, and so here I stop. My comments would be (i) that this is not a fair test of knowledge of formula, rather it is a test of maths and (ii) the solution relies on an assumption that we cannot make. All comments and corrections welcome.
Could you tell us which exam this appeared in, please, and which examining State?
We know distance to PET = (DH)/(O+H), substituting..
(60/100)*1200 = (1200*H)/(O+H) therefore
(60/100) = H/(O+H) and, if we wanted to, we could find O in terms of H
also distance to PSR = EO'H/(O'+H), using O' for distance out because groundspeed out to PSR/PET may not be the same as groundspeed on from PET (engine fail case etc. - similar arguments can be presented for H not being the same as H')
substituting...
(84/100)*1200 = (8.4*O'*H)/(O'+H)
Now at this point, if O and O' were the same, we could substitute our values for O in terms of H from the first formula and find a numerical value for H. We don't know this is the case, though, and so here I stop. My comments would be (i) that this is not a fair test of knowledge of formula, rather it is a test of maths and (ii) the solution relies on an assumption that we cannot make. All comments and corrections welcome.
Could you tell us which exam this appeared in, please, and which examining State?
Hi and thanks for the answer.
I got it at the new IR Exam in South Africa. (This exam is way out of anything from the difficult level I wrote in all my CPL exams). I thought I understand FP got at my PPL and CPL exam 100%...but this at the IR Exam is way another level...I wrote it now 3 times and always got an PET/PSR question which was way out of that what you learn in the theory.
Problem is it is marked with 3 to 4 points in the exam, which is quiet heavy.
Since I am writing this exams (April 2017, no one from the chopper guys in whole South Africa passed this exam and the pass rate at the fixed wings is what I saw 0-2%).
Since the CAA changed the syllabus and the questions since mid 2016 the DFE's and instructors running massively against this new exam, because only 4 of 96 DFE's passed the exam itself. But the authority say, it's ok...
Joined: Nov 2000
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From: White Waltham, Prestwick & Calgary
after consultation, simultaneous equations...
(D x H)/(O + H) = PET for a tail wind home, with O=2/3xH.
Plugging that into PSR=(ExOxH)/(O+H) and rearranging comes up with H=300
You're right - a completely impractical question
(D x H)/(O + H) = PET for a tail wind home, with O=2/3xH.
Plugging that into PSR=(ExOxH)/(O+H) and rearranging comes up with H=300
You're right - a completely impractical question
Joined: Aug 2015
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From: eu
There were 5 answers, I dont know if 300 was one of them, but Ill check next time, when and if Ill get it again.
But one thing is not logical for me in this calculation. I know that assuming O and H the same the calc would be possible, but what makes a knot in my head is, that O and H cannot be the same, due to PET is after the midpoint, which means I have a headwind in going out and a tailwind in going back. The only thing what I assume now is, that I use now O or H and make out of the two an equation with only one variable and put this then in the big PSR formula....but then I get a few knots in my head...
Joined: May 1999
Posts: 1,846
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From: Bristol, England
I haven't explained myself properly. In the formula for PSR 'O' stands for groundspeed out to PSR. In the formula for PET 'O' stands for groundspeed onwards from PET. Assuming TAS and wind components are the same of course then these two speeds would be the same. But this is not necessarily the case, for instance we could be calculating an engine failure PET where the speed out to PET is based on an all engines TAS and the speed either onwards or back from PET is based on an engine out TAS, perhaps at a completely different height. Its a common mistake to assume that, because both formulae use the same letter 'O' that the two formulae can be related. The examiner has made this mistake.
I did not mean that O=H, I meant I assumed, as the examiner seems to have done, that the O in the PSR formula is exactly the same as the O in the PET formula. Likewise in order to get the answer I have assumed that H has the same value in both formulae, and it need not have.
I did not mean that O=H, I meant I assumed, as the examiner seems to have done, that the O in the PSR formula is exactly the same as the O in the PET formula. Likewise in order to get the answer I have assumed that H has the same value in both formulae, and it need not have.
Joined: Nov 2000
Posts: 4,330
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From: White Waltham, Prestwick & Calgary
I think the only way to do that question suitable for a pilot is to take the total PSR distance flown (2016), divide it by the endurance (8.4) to get a mean speed of 240 kts - closer to 200 than 300 because more time is spent at 200. Then look for an answer that is faster than that. The PET I think is a red herring.
Joined: Apr 2013
Posts: 10
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From: UK
ATPL New Exam Format (Quadrant)
Whilst watching the demo for the ATPL exam format (Quadrant), I have come across this question, which seems very ambiguous to me.
Does anybody have any idea on what they would write for this type of question?
Screenshot attached.
Does anybody have any idea on what they would write for this type of question?
Screenshot attached.
Joined: Oct 2017
Posts: 19
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From: Ireland
Hi folks,
Met question from one of the Q banks.
Carburettor icing may be expected to occur in clear air in the following conditions:
a. relative humidity 20%, OAT +30 C and climb power set.
b. relative humidity 40%, OAT +20 C and descent power set.
c. relative humidity 30%, OAT +30 C and descent power set.
d. relative humidity 20%, OAT +30 C and cruise power set.
C is the correct answer, but why not B?
PS. I can't attach the well known carburettor icing graph here since I have less then 10 posts.
Thank you all in advance!
Met question from one of the Q banks.
Carburettor icing may be expected to occur in clear air in the following conditions:
a. relative humidity 20%, OAT +30 C and climb power set.
b. relative humidity 40%, OAT +20 C and descent power set.
c. relative humidity 30%, OAT +30 C and descent power set.
d. relative humidity 20%, OAT +30 C and cruise power set.
C is the correct answer, but why not B?
PS. I can't attach the well known carburettor icing graph here since I have less then 10 posts.
Thank you all in advance!
Last edited by Kakaru; 16th October 2017 at 11:15. Reason: spell check
Joined: Oct 2017
Posts: 19
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From: Ireland
Thanks Alex. Looks like this needs to be appealed / reported. My logic was that the higher relative humidity would be a deciding factor here (even if you don't know the graph) and according to graph, B is closer to the "red" zone.
Last edited by Kakaru; 17th October 2017 at 01:22.
) from the LBA once told me that the answer will always be the one with the higher temp, since they wish to make the point that carb-icing can happen even at higher temperatures.
Joined: Apr 2016
Posts: 11
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From: Bratislava
I have been study for Performans and one question make me very nervious because there is two answer I have to make a sure which one is the correct. this question coming from new EASA data base ECQB4.
(For this question use annex ECQB-032-023-v2015-04 or CAP 698 Figure 2.3). Given the following information, what is the Rate of Climb? Pressure Altitude: 10000 ft ISA conditions Aircraft Mass: 3400 lb Full Throttle, 2700 RPM”
A-) 640 ft/min
B-) 500 ft/min
C-) 720 ft/min
D-) 1180 ft/min
I founded 640 ft/min however the question bank show me the corect answer is 720 ft/min. does anyone explain this question.
(For this question use annex ECQB-032-023-v2015-04 or CAP 698 Figure 2.3). Given the following information, what is the Rate of Climb? Pressure Altitude: 10000 ft ISA conditions Aircraft Mass: 3400 lb Full Throttle, 2700 RPM”
A-) 640 ft/min
B-) 500 ft/min
C-) 720 ft/min
D-) 1180 ft/min
I founded 640 ft/min however the question bank show me the corect answer is 720 ft/min. does anyone explain this question.
Joined: Apr 2004
Posts: 734
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From: London, GB
That question is so poorly worded that it ceases to be meaningful. It seems to ask for the final forward azimuth of a geodesic (great circle) between the given stand- and forepoints. On the WGS84 ellipsoid this angle is 249 deg (east of true north). See `fazi2` in Charles Karney's GeodSolve utility here (add 360 degrees to negative values). Chris Veness's implementation of the older Vincenty algorithm for the inverse solution can also be used, here. In practice you'd lay a line down on a Lambert conformal conic projection, e.g. UK CAA VFR Air Chart, and measure the local direction at the forepoint 53N 003E.
"Final true track" is a mongrel construct and should be avoided. "Track" is not explicitly defined as a loxodrome (rhumb line), or a geodesic (great circle), in any major text on navigation. Paths with constant true track values are always loxodromes so the qualifier "final" is redundant.
"Final true track" is a mongrel construct and should be avoided. "Track" is not explicitly defined as a loxodrome (rhumb line), or a geodesic (great circle), in any major text on navigation. Paths with constant true track values are always loxodromes so the qualifier "final" is redundant.
Joined: May 1999
Posts: 1,846
Likes: 4
From: Bristol, England
Feedback suggests there's a bit missing from the Q above, rhumb line track is given as 250 deg.
Convergency = change of long x sin mean lat
= 4 deg x sin 52.5 deg (roughly)
= 3.2 deg (roughly)
conversion angle is half this, 1.6 deg
subtract from 250 to get 248.4 deg (roughly)
Otherwise what selfin said ^^^^, but by the standards of EASA exams its an OK question.
Convergency = change of long x sin mean lat
= 4 deg x sin 52.5 deg (roughly)
= 3.2 deg (roughly)
conversion angle is half this, 1.6 deg
subtract from 250 to get 248.4 deg (roughly)
Otherwise what selfin said ^^^^, but by the standards of EASA exams its an OK question.
