ATPL theory questions
Join Date: Sep 2014
Location: oxford
Posts: 2
Likes: 0
Received 0 Likes
on
0 Posts
stuck stab and CG position
Found this question in a book and not very satisfied with the answer:
What is the best CofG position with a stuck stabilizer, and why ?
What are your thoughts ?
Thanks
What is the best CofG position with a stuck stabilizer, and why ?
What are your thoughts ?
Thanks
Looking at your location and your statement that
I suspect that you are undergoing flight training of some kind, possibly with OATS (or whatever they are called now). I could of course be wrong but that is the basis of my answer to your question.
The answer depends on what your flight condition was at the time the stabilizer became stuck.
In cruise flight for example, the aircraft would be trimmed for a low pitch attitude / angle of attack and gear and flaps up. In order to land you would need to extend the gear and flaps and assume a higher angle of attack / pitch as you decelerated the aircraft to landing speed. These actions would generate a nose down pitching moment, which would normally be countered by adjusting the stabilizer trim.
The stabilizer stuck in cruise position would not only prevent you from trimming out the pitch-down moments, it would actually increase them. A forward C of G position would make the problem even worse. So an aft C if G would be the most favourable in this scenario.
The ATPL CQB contains (or did for many years) a question concerning how best to achieve a landing with the stabilizer stuck in the cruise position. The best answer was a combination of aft CG, not much flap, and higher than usual landing speed.
"Found this question in a book "
What is the best CofG position with a stuck stabilizer, and why ?
In cruise flight for example, the aircraft would be trimmed for a low pitch attitude / angle of attack and gear and flaps up. In order to land you would need to extend the gear and flaps and assume a higher angle of attack / pitch as you decelerated the aircraft to landing speed. These actions would generate a nose down pitching moment, which would normally be countered by adjusting the stabilizer trim.
The stabilizer stuck in cruise position would not only prevent you from trimming out the pitch-down moments, it would actually increase them. A forward C of G position would make the problem even worse. So an aft C if G would be the most favourable in this scenario.
The ATPL CQB contains (or did for many years) a question concerning how best to achieve a landing with the stabilizer stuck in the cruise position. The best answer was a combination of aft CG, not much flap, and higher than usual landing speed.
If your question is for the purpose of ATPL theory study then you need to consider the gradient of the drag curve.
If the drag curve gradient is very steep then a small change in speed will result in a large change in drag. At speeds below Vmd this would make the aircraft more speed unstable. But at speeds above Vmd it would make the aircraft more speed stable.
Swept wing aircraft tend to low coefficients of drag, which gives them shallower drag curves compared to straight winged aircraft. This reduces the speed instability below Vmd. But it makes it harder to maintain a given speed at or above Vmd and this might be interpreted as an increased speed instability.
If the drag curve gradient is very steep then a small change in speed will result in a large change in drag. At speeds below Vmd this would make the aircraft more speed unstable. But at speeds above Vmd it would make the aircraft more speed stable.
Swept wing aircraft tend to low coefficients of drag, which gives them shallower drag curves compared to straight winged aircraft. This reduces the speed instability below Vmd. But it makes it harder to maintain a given speed at or above Vmd and this might be interpreted as an increased speed instability.
Join Date: Oct 2014
Location: Essex
Posts: 1
Likes: 0
Received 0 Likes
on
0 Posts
ATPL General Navigation Question
Hi would anyone be able to help me with this question that our whole class are confused about?
A ground feature was observed on a relative bearing of 325° and five minutes later on a relative bearing of 280°. The aircraft heading was 165°(M), variation 25°W, drift 10°Right and GS 360 kt. When the relative bearing was 280°, the distance and true bearing of the aircraft from the feature was:
A. 30 NM and 240°
B. 40 NM and 110°
C. 40 NM and 290°
D. 30 NM and 060°
The correct answer is A but were unsure how you get the answer.
A ground feature was observed on a relative bearing of 325° and five minutes later on a relative bearing of 280°. The aircraft heading was 165°(M), variation 25°W, drift 10°Right and GS 360 kt. When the relative bearing was 280°, the distance and true bearing of the aircraft from the feature was:
A. 30 NM and 240°
B. 40 NM and 110°
C. 40 NM and 290°
D. 30 NM and 060°
The correct answer is A but were unsure how you get the answer.
Join Date: Oct 2014
Location: Broadstairs, kent
Posts: 1
Likes: 0
Received 0 Likes
on
0 Posts
Can't believe the whole class are confused.
Classic RTFQ/RTFA question.
You can spend 15 to 20 minutes drawing it all out or
Hdg 165(M) - Var 25(W) = 140(T) + RB 280 = 420-360 = 060 from a/c to feature which means 240(T) from feature to a/c so the answer must be (A).
You do not need to calculate the distance as the bearings are all different.
Don't forget that in your exam NO question should take more than 3 minutes, so look for clues in the question/answers
Classic RTFQ/RTFA question.
You can spend 15 to 20 minutes drawing it all out or
Hdg 165(M) - Var 25(W) = 140(T) + RB 280 = 420-360 = 060 from a/c to feature which means 240(T) from feature to a/c so the answer must be (A).
You do not need to calculate the distance as the bearings are all different.
Don't forget that in your exam NO question should take more than 3 minutes, so look for clues in the question/answers
Join Date: Feb 2002
Location: Sunny Solihull
Age: 67
Posts: 265
Likes: 0
Received 0 Likes
on
0 Posts
Probably best to post this in the ATPL questions thread - however.
Strongly recommend drawing a diagram to aid your answer however the key to the question is the 45 degree change in RB as this must mean it forms an isosceles triangle (ist)
Dist along track = 360 * 5 mins = 30 nm and due to the (ist) is also the distance from the feature.
True bearing from feature - first get everything into true and it's HEADING that matters (unless you are drawing a diagram then track is relevant also) so Hdg(T) = 165 -25 = 140(T) + the final RB of 280 = 420 - 360 = 060 true bearing of feature from a/c + 180 = 240 true bearing from feature.
So in real plotting you would draw a line 240 from feature and arc 30 miles gives you your position.
NB lots of questions like this around just look for the 45 RB difference - simples.
Strongly recommend drawing a diagram to aid your answer however the key to the question is the 45 degree change in RB as this must mean it forms an isosceles triangle (ist)
Dist along track = 360 * 5 mins = 30 nm and due to the (ist) is also the distance from the feature.
True bearing from feature - first get everything into true and it's HEADING that matters (unless you are drawing a diagram then track is relevant also) so Hdg(T) = 165 -25 = 140(T) + the final RB of 280 = 420 - 360 = 060 true bearing of feature from a/c + 180 = 240 true bearing from feature.
So in real plotting you would draw a line 240 from feature and arc 30 miles gives you your position.
NB lots of questions like this around just look for the 45 RB difference - simples.
Join Date: Jan 2006
Location: Europe
Posts: 404
Likes: 0
Received 0 Likes
on
0 Posts
Flight Planning & Monitoring
Hi all,
I'm currently working on Flight Planning & Monitoring (Chapter: Practical Completion of Flight Plan; subpart: Completion of fuel plan) and came across a few sneaky question where there is no reference to the respective Figure (to be used) given. Here are a few examples.
1. (For this question use Fuel Planning MRJT1)
The airplane gross mass at top of climb (TOC) is 61.500 kg. Distance to be flow is 385 NM at FL 350 and OAT -54.3°C. The wind component is 40 kt tailwind. Use long range cruise procedure what fuel is required? 2.150 kg
2. (Information given: use Route Manual chart E(HI)4)
Flight from Paris Charles de Gaulle (N49 00.9 E002 36.9) to London Heathrow (N51 29.2 W 000 27.9) (twin jet). The alternate is Manchester (N53 21.4 W002 15.7).
Preplanning:
Wind from London to Manchester 250°/30 kt
Distance from London to Manchester 160 NM
Estimated landing mass at alternate about 50.000 kg
Find alternate fuel and time: 1.450 kg and 32 min
How do I know which chart/Figure(s) to use to come up with the answer? And how do I work it out?
Any advise greatly appreciated!
I'm currently working on Flight Planning & Monitoring (Chapter: Practical Completion of Flight Plan; subpart: Completion of fuel plan) and came across a few sneaky question where there is no reference to the respective Figure (to be used) given. Here are a few examples.
1. (For this question use Fuel Planning MRJT1)
The airplane gross mass at top of climb (TOC) is 61.500 kg. Distance to be flow is 385 NM at FL 350 and OAT -54.3°C. The wind component is 40 kt tailwind. Use long range cruise procedure what fuel is required? 2.150 kg
2. (Information given: use Route Manual chart E(HI)4)
Flight from Paris Charles de Gaulle (N49 00.9 E002 36.9) to London Heathrow (N51 29.2 W 000 27.9) (twin jet). The alternate is Manchester (N53 21.4 W002 15.7).
Preplanning:
Wind from London to Manchester 250°/30 kt
Distance from London to Manchester 160 NM
Estimated landing mass at alternate about 50.000 kg
Find alternate fuel and time: 1.450 kg and 32 min
How do I know which chart/Figure(s) to use to come up with the answer? And how do I work it out?
Any advise greatly appreciated!
Join Date: Feb 2002
Location: Sunny Solihull
Age: 67
Posts: 265
Likes: 0
Received 0 Likes
on
0 Posts
Advice is - perfectly okay questions and these days it is normal not to mention the figure directly BUT there is enough information in the question to lead you to the appropriate page in CAP697 or appendix.
Q1 says LRC procedures at FL350 gives you MRJT page 33. If you don't know the method this is explained on page 24. However briefly, ISA so TAS 429 kt. Then convert the 385 NGM into 352 NAM (as tables work in air miles). TOC mass of 61500 gives an air distance of 5313 nam - 352 = 4961 nam. Enter table looking for 4961 (lies about half way between 4954 & 4971 with equates to 59350 kg. 61500 - 59350 = 2150 kg (no further corrections needed).
Q2 mentions ALTERNATE there is only one Alternate Planning graph for MRJT on P16. So it's London to sunny Manchester all details given except track which if you measure it on EHI4 is 330(T). Now work the WC, quickest and easiest in this case use CRP5 square section and you will get a 5 kt headwind.
Now use Alternate Planning graph as shown in example with a 50,000 kg LM and you should get 1450 kg & 32 mins.
Hope that helps but should be explained in your FTOs course material.
Q1 says LRC procedures at FL350 gives you MRJT page 33. If you don't know the method this is explained on page 24. However briefly, ISA so TAS 429 kt. Then convert the 385 NGM into 352 NAM (as tables work in air miles). TOC mass of 61500 gives an air distance of 5313 nam - 352 = 4961 nam. Enter table looking for 4961 (lies about half way between 4954 & 4971 with equates to 59350 kg. 61500 - 59350 = 2150 kg (no further corrections needed).
Q2 mentions ALTERNATE there is only one Alternate Planning graph for MRJT on P16. So it's London to sunny Manchester all details given except track which if you measure it on EHI4 is 330(T). Now work the WC, quickest and easiest in this case use CRP5 square section and you will get a 5 kt headwind.
Now use Alternate Planning graph as shown in example with a 50,000 kg LM and you should get 1450 kg & 32 mins.
Hope that helps but should be explained in your FTOs course material.
Join Date: Jan 2006
Location: Europe
Posts: 404
Likes: 0
Received 0 Likes
on
0 Posts
Richard, many thanks for your respond so far! I will take a look at this tomorrow (I'll hit the sack now) and if I need any further explanation I'll let you know. Once again thanks a lot!
PS. Just went over the questions again and worked them out, it's all clear now, I came up with the correct results (thanks to your explanation)! I must confess, Flight Planning is the subject I neglected most, but since exam time is coming up, I better spent some time going over the charts now, instead of having to retake the subject (which hopefully won't be the case) and I must say with some practice it isn't that hard and you know what charts you have to look up, at least most of the time.
PS. Just went over the questions again and worked them out, it's all clear now, I came up with the correct results (thanks to your explanation)! I must confess, Flight Planning is the subject I neglected most, but since exam time is coming up, I better spent some time going over the charts now, instead of having to retake the subject (which hopefully won't be the case) and I must say with some practice it isn't that hard and you know what charts you have to look up, at least most of the time.
Last edited by Transsonic2000; 14th Dec 2014 at 13:47.
Join Date: Jan 2006
Location: Europe
Posts: 404
Likes: 0
Received 0 Likes
on
0 Posts
Hello,
would need some help with the following question:
Given: maximum allowable take-off mass 64.400 kg, maximum landing mass 56.200 kg, Maximum zero fuel mass 53.000 kg.
Dry operating mass 35.500 kg, Traffic load 14.500 kg, Trip fuel 4.900 kg, Minimum take-off fuel 7.400 kg.
Find maximum allowable take-off fuel:
a) 11.400 kg
b) 14.400 kg
c) 11.100 kg (this is the correct answer according to the QDB)
d) 8.600 kg
Here is how I calculated it:
since the question is asking for the max allowable T/O fuel I simply subtracted Max. ZFM 53.000 kg from Max. allowable TOM 64.400 kg and came up with 11.400 kg but apparently this is not correct. What's my mistake or what did I miss?
would need some help with the following question:
Given: maximum allowable take-off mass 64.400 kg, maximum landing mass 56.200 kg, Maximum zero fuel mass 53.000 kg.
Dry operating mass 35.500 kg, Traffic load 14.500 kg, Trip fuel 4.900 kg, Minimum take-off fuel 7.400 kg.
Find maximum allowable take-off fuel:
a) 11.400 kg
b) 14.400 kg
c) 11.100 kg (this is the correct answer according to the QDB)
d) 8.600 kg
Here is how I calculated it:
since the question is asking for the max allowable T/O fuel I simply subtracted Max. ZFM 53.000 kg from Max. allowable TOM 64.400 kg and came up with 11.400 kg but apparently this is not correct. What's my mistake or what did I miss?
Join Date: May 2001
Location: England
Posts: 1,904
Likes: 0
Received 0 Likes
on
0 Posts
You are using the Max ZFW instead of Actual ZFW.
Maximum fuel allowed is the lowest of:
A.) Max TOW - Actual ZFW
B.) Max LW - Actual ZFW + Trip fuel
C.) Max Fuel Capacity
Actual ZFW = DOW (or DOM) + Traffic Load
= 35.5 + 14.5
= 50
Putting it all back into the above formula you'll find B.) gives you the lowest answer. E.g. 56.2 - 50 + 4.9 = 11.1 tonnes or 11,100 kg.
Thanks for asking that, I had to open up the books
Maximum fuel allowed is the lowest of:
A.) Max TOW - Actual ZFW
B.) Max LW - Actual ZFW + Trip fuel
C.) Max Fuel Capacity
Actual ZFW = DOW (or DOM) + Traffic Load
= 35.5 + 14.5
= 50
Putting it all back into the above formula you'll find B.) gives you the lowest answer. E.g. 56.2 - 50 + 4.9 = 11.1 tonnes or 11,100 kg.
Thanks for asking that, I had to open up the books
Join Date: Jun 2012
Location: South somewhere
Posts: 56
Likes: 0
Received 0 Likes
on
0 Posts
When working out the 3 Regulated Take-Off Masses
MTOM = RTOM 1
MLM + Trip Fuel = RTOM 2
MZFM + TOF (Which you don't know the TOF in this problem). = RTOM 3
(Lowest of the 3)
Therefore RTOM 2
- Dom
- Traffic Load
= Max Allowable Fuel Load
56200+ 4900 = 61100
- 35500 = 25600
- 14500 = 11100 Max fuel allowable to be able to land at MLM
MTOM = RTOM 1
MLM + Trip Fuel = RTOM 2
MZFM + TOF (Which you don't know the TOF in this problem). = RTOM 3
(Lowest of the 3)
Therefore RTOM 2
- Dom
- Traffic Load
= Max Allowable Fuel Load
56200+ 4900 = 61100
- 35500 = 25600
- 14500 = 11100 Max fuel allowable to be able to land at MLM
Join Date: Jun 2014
Location: Istanbul
Age: 39
Posts: 38
Likes: 0
Received 0 Likes
on
0 Posts
i do have a little question about mechanics of atpl exams, if you guys do not mind. i got caps for performance and flight planning from a friend who entered atpl exams in irish authority. especially in performance the first 3-4 pages contain, conversions between units, definitions, some formulas and wet,dry,grass etc. runway coefficients. i'll have my exams in polish caa, are they going to give the complete cap 698 or just the graphs? i am wondering because in aviationexam performance sections, explanations seems like we have to know the wet,dry,grass etc. coefficients by memory. thanks.
For a definitive answer you must address your question to the Polish CAA.
The UK CAA have in the past permitted candidates to have a complete copy of the CAPs 696, 697 and 698 for the relevant exams. This reduced the amount of material that student were required to memorize. I believe that candidates in other countries have had only the relevant annexes (graphs, tables and diagrams).
In future UK CAA exam candidates will not be permitted to have the full CAPs. I believe that this change takes effect with the next set of exams (January 2015). I have heard that candidates booked for these exams have been given outdated instructions which state that full CAPs will be provided. The best advice would be for all candidates to memorize as much material from the CAPs as possible, until the picture becomes clearer after the next few sets of exams.
The UK CAA have in the past permitted candidates to have a complete copy of the CAPs 696, 697 and 698 for the relevant exams. This reduced the amount of material that student were required to memorize. I believe that candidates in other countries have had only the relevant annexes (graphs, tables and diagrams).
In future UK CAA exam candidates will not be permitted to have the full CAPs. I believe that this change takes effect with the next set of exams (January 2015). I have heard that candidates booked for these exams have been given outdated instructions which state that full CAPs will be provided. The best advice would be for all candidates to memorize as much material from the CAPs as possible, until the picture becomes clearer after the next few sets of exams.
Join Date: May 1999
Location: Bristol, England
Age: 65
Posts: 1,803
Received 0 Likes
on
0 Posts
For the UK the CAA have said "Students sitting exams that previously required CAPs will be given workbooks which contain straight copies of the graphs from the CAPs required to answer questions, if any additional information is needed, then it will be contained in the question. Conversion formulas and factorisation requirements for grass, paved, wet or dry surfaces and slope will need to be memorised." It is not immediately clear to me whether this is with effect from January or February, I will try to confirm the effective date as soon as I can, unless anyone else knows for sure.
Join Date: Jun 2014
Location: Istanbul
Age: 39
Posts: 38
Likes: 0
Received 0 Likes
on
0 Posts
thank you all for the information, i really appreciate it. if it is not too much to ask, is it same with the jeppesen manual especially in flight planning. can we keep the whole manual (because there are lots of information) or just the charts and plates? thanks.