All else being equal, almost certainly down to a shallower logitudinal static stability curve around the operating AoA range.

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 :)

Depends upon the control system of the aeroplane.

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.

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.

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.

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

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.

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!

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.

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.

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?

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 :O

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

Many thanks for your explanations (superpilot & Stn120) it's very much appreciated!

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.

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.

1 Jan I believe

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.

Actually, the phrase used was "until January 2015" for the use of the CAPs so to play safe, regard them as not being available for the January exams, as we don't know whether Jan is included.

The Jep is staying for the foreseeable future, in UK at least. I believe Spain has the Jep stuff in the exams questions, as required.

Guys, a quick question. The ones who are going to start with the ATPL theory in 2015, will the QB be outdated? Because I've read a lot of things that make me a little bit nervous.

Why do you want to use the QB? If for revision and practice, there should be no problem if you keep your mind flexible. If you want to learn the answers, definitely a problem. More questions being written right now.

i belive they add new questions every year and even every month. so the question banks are not %100 accurate.

but what i belive captain.weird asked, is something different with year 2015 rather than the usual updates on the questions. am i right?

ATPL met question
 

Hi all, im on my last set of ground school exams and having problems with the following questions:-

The QNH at an airfield located 200 metres above sea level is 1009 hPa. The air temperature is 10°C lower than a standard atmosphere. What is the QFF?

A) More than 1009

B) Less than 1009

C) 1009

D) It is not possible to give a definitive answer


The QNH at an airfield located 200 metres above sea level is 1022 hPa. The air temperature is not available. What is the QFF?



A )It is not possible to give a definitive answer


B) More than 1022 hPa


C) Less than 1022 hPa


D) 1022 hPa

The QNH at an airfield in California located 69 meters below sea level is 1018 hPa. The air temperature is 10°C higher than a standard atmosphere. What is the QFF?

A) 1018 hPa




B) Less than 1018 hPa C) More than 1018 hPa




D) It is not possible to give a definitive answer


The QFF at an airfield located 400 metres above sea level is 1016 hPa. The air temperature is 10°C higher than a standard atmosphere. What is the QNH?

A) It is not possible to give a definitive answer



B) Less than 1016 hPa




C) More than 1016 hPa




D) 1016 hPa

If someone could please shed some light on theses questions it would be much appreciated as I cant seem to get my head around them!

Answers courtesy of the great guys at BGS

If the temp = ISA conditions
QNH = QFF

If above MSL
If the temp > ISA then QNH > QFF
If the temp < ISA then QNH < QFF

If below MSL then this is reversed:
If the temp > ISA then QNH < QFF
If the temp < ISA then QNH > QFF


and
In ISA conditions QNH = QFF; the setting which gives airfield elevation at touchdown (QNH) = the atospheric pressure at sea level (QFF). If it is warmer than ISA the theoretical column of air between the airfield and sea level would expand and the QNH would occur below sea level which means that QNH would be greater than QFF. Similarly, if it is colder the ISA the column of air would contract and the QNH would be less than QFF.


and
QFF and QFE problems are all linked to the fact that in colder denser air pressure changes more rapidly as you go up or down.

Imagine you are on an airfield at 1000ft elevation and at 1000mb and you want to calculate the pressure at msl. In round figures for ISA conditions, at 30ft/mb you would make it 33.3mb higher at 1033.3mb. This is your QNH.

If the air is colder than ISA you get more rapid pressure changes, and we?ll use 20ft/mb for convenience. Now, using ambient temperature and this ?real? lapse rate we calculate the pressure at msl and it comes out at 50mb higher at 1050mb This is your QFF.

Now QFF is our best guess of the actual pressure at msl, but QNH is a lower pressure and zero on the altimeter with QNH set will leave you well above msl. So, in cold conditions QFF is a higher pressure setting than QNH but the QNH pressure level is at a higher altitude. As an aside, this means you can safely fly down to zero indicated altitude with QNH set and not hit the sea.

If you look at this upside down, from an airfield below msl it means that now QFF is at a lower pressure than QNH. QNH set will give you a negative height and as you climb to zero on QNH you will still be below msl

The General Rule is that in below ISA temperatures you indicated altitude on QNH will always be nearer to your station level than your true height

This works for obstacle clearance as well. If you are climbing over a mountain in cold conditions your indicated altitude will put you nearer to your station level, which is lower than your true height

This has been about cold air> If the conditions given are for temperatures above ISA then reverse the logic!

A quick google search is your friend :)

hello fellow aviators, i have finished 7 of the exams and all of the ones that left, i belive the most scary one is the gennav. is there a good source to study for gennav, maybe shortcuts, tips or summaries? thank you for your help.

Gen Nav is not scary - it just depends how you are taught, but you need to be very skilled on the flight computer.

Variation in groundspeed
 

Hi everyone, could you please help me out with this one? :confused:

At 07:33 you are 500 NM from Z.

ATC Request: do not cross Z before 08:57.

You are cruising at FL300, OAT -38°C, M.79, WC -70 KTS.

Calculate the latest time you can slow down to a GS of 320 KTS in order to comply with ATC.

Many thanks!

Exam Date
 

Guys any idea about the ATPL exam dates for Jan Session??

First work out your groundspeed. Lacking a nav computer I will do it by calculation:

Temperature in Kelvin = 273 – 38 = 235K
Local speed of sound = 38.94*sqrt 235 = 597KT
TAS = Mach number * LSS = 472KT
Groundspeed = 472KT – 70KT = 402KT

At 07:33 you were 500NM from Z. You are required to reduce speed to 320KT groundspeed to cross at 08:57.

Consider where you would be at 08:57 if you did not change speed. You would have travelled for 1 hr 24 minutes (1.4 hours) at 402KT, a distance of:

1.4 * 402 = 562NM, an ‘overshoot’ of 62NM.

You need to get rid of this overshoot by flying 402 – 320 = 82KT slower.

This will take 63 ÷ 82 = 0.77 hours = 46 minutes.

Therefore change speed 46 minutes before 08:57, or 08:11.
I have rounded numbers here but preserved decimal places on my calculator to actually get 45.8 minutes. How did I do?

Spot on!
 

Thanks a lot for taking the time to come up with a detailed explanation Alex, really appreciate your input! :ok:

LBA: 6 Down 8 To Go
 

I wrote VFR Comms, IFR Comms, OPs, Air Law, Weather, and GenNav yesterday in Braunschweig. The Comms, OPs, Air Law were 1:1 from Peter's Software databank. Weather for the most was from it as well except for three questions:

The temperature on the land is 10C/05C. What air temperature from an air mass originating over the water need to be to form advection fog (these are the answers as best as I can remember them...):

A) 10/05
B) 20/05
C) 25/10
D) 20/15

I chose D since the air temp in D is higher than the air over the land therefore holding more moisture and the spread is smaller than 25/10. The warmer air flows over the land and is then cooled. Dunno...

Also there were 2 weather charts questions. For one, you have to find the route where there is no icing at FL180. There is then a list of routes i.e. Frankfurt-Madrid, Zurich-Marsailles, etc. I cannot remember the other question, but it was fairly easy.

Anyhoo, crash and burn in GenNav. I would say that 1/2 of the questions where not in the Peter databank. All of them do-able, but I had mostly questions with headings and NONE of the heading questions were multiple choice: ALL FILL IN THE BLANKS. Bastards.... A really (for me) tricky polar stereographic question: North Pole: A) is at 75N 146E; B) 78N 168E; what is the TT at 155E. Almost all of the questions requiring a number for an answer were fill in the blank. :ugh:

My big problem was time management. Next time, I will use some of the leftover time in one of the other subject and write out the formulas and memory-helpers BEFORE I start the NAV section (i.e. C-D-M-V-T). Saves some time.

I stand to be corrected here but I believe the LBA add in non multi-choice questions to the standard CQB, presumably to make it more difficult. You could always go and take your exams in another State if you wish, you'll get a more standard set of exams. On first sight your polar stereo question can only be solved by plotting, ie a scale diagram, which is unusual, unless the answers are widely enough spaced to make 'about 090deg' correct.

Where can we check the amount of questions in each exam? does the CAA provide this information at all? I ll be doing the Flight Planning exam in 3 weeks and would like to know the number of questions :rolleyes:

Hi Alex,

Yes, they like to make it more difficult. Lord knows why. Even the Lutfwaffe guys were complaining. I tried plotting the diagram, but I wasn't happpy with the result. I get the results next week and we shall see.

EU Regulation 1178-2011 Part ARA - AMC, page 21. I'm afraid you will have to find it on EASA's website.

LNAV AND VNAV for sure.

The question about QNH and QFF relationship is very textbook question and the answer is really clear no ambiguity...

11 gone 3 to go :) gennav and both agks left.