Professional Pilot Training (includes ground studies) A forum for those on the steep path to that coveted professional licence. Whether studying for the written exams, training for the flight tests or building experience here's where you can hang out.

ATPL theory questions

Old 12th Apr 2012, 15:13
  #81 (permalink)  
 
Join Date: Jun 2010
Location: Stockholm
Posts: 25
Likes: 0
Received 0 Likes on 0 Posts
OAA Flight Performance & Planning. It's likely to be a bogus answer if you don't agree with it, it wouldn't be the first time...

On another note, I have stumbled upon a couple of questions involving the dumping of fuel to ensure level-off altitude isn't below obstacle clearance altitude during a drift down procedure. Is this common practice, as the only thing I have read about fuel dumping so far is regarding the procedure of doing so prior to landing to ensure not exceeding MSLM?

-Anders
Anders S is offline  
Old 12th Apr 2012, 16:11
  #82 (permalink)  
 
Join Date: Jan 2011
Location: England
Posts: 660
Received 18 Likes on 11 Posts
The OAA answer to question 3 is incorrect. Hopefully you should be able to see that is the case by applying common sense. If the aircraft gets heavier it won't climb so well. So if we are happy to be limited to a lower climb gradient we can be heavier.

I'm not an airline pilot, so I cannot comment on what is common practice. But if I knew that keeping any unnecessary fuel on board would cause me to hit the mountain tops, I would certanly dump any fuel that I didn't need.
keith williams is offline  
Old 12th Apr 2012, 17:35
  #83 (permalink)  
 
Join Date: Jun 2010
Location: Stockholm
Posts: 25
Likes: 0
Received 0 Likes on 0 Posts
I agree with you regarding question 3. It's just that as with the case of questions 1 & 2, it's easy to miss corrections, that's why it's seems like a better bet to run it by here, rather than just assume that I'm right and they're wrong.

The fuel dumping questions were:

A:

If the level-off altitude is below the obstacle clearance altitude during a drift down procedure?

a. Fuel jettisoning should be started at the beginning of drift down.
b. The recommended drift down speed should be disregarded and it should be flown at the stall speed plus 10 kt.
c. Fuel jettisoning should be started when the obstacle clearance altitude clearance altitude is reached.
d. The drift down should be flown with flaps in the approach configuration.

B:

During a drift down following engine failure, what would be the correct procedure to follow?

a. Begin fuel jettison immediately, commensurate with having required reserves at destination.
b. Do not commence fuel jettison until en-route obstacles have been cleared.
c. Descend in the approach configuration.
d. Disregard the flight manual and descend at Vs + 10 kts to the destination.

Correct answer for is a. for both questions. Looking at them in hindsight I realize that those are the only reasonable answers, it's just that when I came across them there had been no mention of fuel dumping in the previous chapters that I had read, other than for the purpose of ensuring MSLM, and I suppose that the intention after engine failure is to land as soon as possible, thus fuel dumping should be considered pertinent. However there is no mention in the question of what phase of flight is being referred to, and dumping of fuel doesn't seem to be advisory unless absolutely necessary. What had been brought to attention in the previous chapter however, was the use of drift down profile graphs, where one was to determine if the desired altitude, with regard to obstacle clearance, could be met with the current mass, and if not, (to my understanding) a second graph should be used to determine whether or not vertical clearance could be achieved using horisontal distance instead. Thus the reason for my inquery.

You are however absolutely right, Keith, it definately seems like a better idea, if need be, to dump some fuel rather than run in to a mountain top.

Cheers
Anders
Anders S is offline  
Old 13th Apr 2012, 16:46
  #84 (permalink)  
 
Join Date: Nov 2008
Location: Land of Hope
Posts: 47
Likes: 0
Received 0 Likes on 0 Posts
Hi I Am really confused and finding it difficult. Can anyone please help

Question is: indicated altitude =20,000 feet and the temperature is -35*C. What is the true altitude?
The equation given is -

true altitude= indicated altitude + (ISA_Deviation) x Indicated Altitude.
-------------------
(T)(k)






And if somebody can really break down each part of the equation and how it works, it would help immensely.


Question 2)
Indicated altitude 32,500 temperature is -32*. What is true altitude?

Last edited by clkorm3; 13th Apr 2012 at 16:59.
clkorm3 is offline  
Old 13th Apr 2012, 17:54
  #85 (permalink)  
 
Join Date: Jan 2011
Location: England
Posts: 660
Received 18 Likes on 11 Posts
If you go back to post number 33 in this thread you will find something very similar to your question.
keith williams is offline  
Old 20th Apr 2012, 18:49
  #86 (permalink)  
 
Join Date: Nov 2008
Location: Land of Hope
Posts: 47
Likes: 0
Received 0 Likes on 0 Posts
Could somebody be kind enough to tell me how many questions there are in the EASA ATPL EXAMS for
Principles of flight, meteorology, mass & balance, human performance and limitation, communication?

Has anyone sat the new exams and whAt is the differnce if there is any. More difficult I.e. complex questions?

Keith William thank you for your input on the previous post. It was very helpfull.
clkorm3 is offline  
Old 23rd Apr 2012, 18:08
  #87 (permalink)  
 
Join Date: Jun 2010
Location: Stockholm
Posts: 25
Likes: 0
Received 0 Likes on 0 Posts
Hey again...

On a ground pressurisation test, if the cabin suffers a rapid depressurisation:

A. The temperature will rise suddenly.
B. Water precipitation will occur.
C. Damage to the hull may occur.
D. Duct relief valve may jam open.

I had originally marked answer C, but according to the solution it should be B. Is this correct, and if so, would anyone care to elaborate?

-Anders
Anders S is offline  
Old 23rd Apr 2012, 22:24
  #88 (permalink)  
 
Join Date: Jan 2011
Location: England
Posts: 660
Received 18 Likes on 11 Posts
The sudden pressure drop will cause a sudden temperature drop.

If there is sufficient moisture in the air, and if the temperature falls below the dew point, it will cause water droplets to precipitate out of the air.

The unstated assumptions in this question are that:

a. There is sufficient moisture in the air.
b. The pressure and temperature drops are sufficiently large.

So the use of the words "will occur" in option C is probably a bit too strong. "May occur" would be more accurate.

But the other options are all wrong, so option C is the best answer.
keith williams is offline  
Old 24th Apr 2012, 08:14
  #89 (permalink)  
 
Join Date: Jun 2010
Location: Stockholm
Posts: 25
Likes: 0
Received 0 Likes on 0 Posts
Thanks Keith.

-A
Anders S is offline  
Old 3rd May 2012, 02:16
  #90 (permalink)  
 
Join Date: Aug 2011
Location: Neath
Age: 44
Posts: 2
Likes: 0
Received 0 Likes on 0 Posts
Thumbs up ATPL Theory study

Just embarked on the Bristol GS study, any tips on helping me ease my pain?

Have read many of the posts in this forum and see a lot of people are struggling with the Gen Nav side of things, I've just purchased a book from Baz at Bristol GS 'Mathematics for Aviation', I sure hope that helps me understand the subject a lot better when I receive it
CP Pilot is offline  
Old 8th May 2012, 17:33
  #91 (permalink)  
 
Join Date: Jun 2010
Location: Stockholm
Posts: 25
Likes: 0
Received 0 Likes on 0 Posts
Hey,

The tendency to call to mind common experiences or scenarios from the past and link them incorrectly to a perceived mental model is called:

a. Confirmation bias.
b. Action slip.
c. Environmental capture.
d. Frequency bias.

Supposedly the right answer should be d.
Is this correct? Couldn’t seem to find anything about frequency bias in reference to HP. I suppose that if one breaks it down, then it could be the right answer. However the definition of environmental capture seems quite fitting in the context…

-Anders
Anders S is offline  
Old 23rd May 2012, 00:17
  #92 (permalink)  
 
Join Date: Dec 2011
Location: Vietnam
Age: 33
Posts: 2
Likes: 0
Received 0 Likes on 0 Posts
Question about weather radar of airbus a321!!

Why on A321 the weather radar system has two function (auto and manual)? What difference between those function and which function more reliable?
Frank321 is offline  
Old 25th May 2012, 03:45
  #93 (permalink)  
 
Join Date: Oct 2011
Location: ireland
Age: 32
Posts: 9
Likes: 0
Received 0 Likes on 0 Posts
Are you serious?..stupid question....

Question: Which combination of answers of the following parameters give an increase or decrease of the take off ground run: 1 decreasing take off mass 2 increasing take off mass 3 increasing density 4 decreasing density 5 increasing flap setting 6 decreasing flap setting 7 increasing pressure altitude 8 decreasing pressure altitude

A.2, 3, 6 and 7

B.1, 3, 5 and 8

C.2, 4, 5 and 7

D.1, 4, 6 and 8
shoeless is offline  
Old 25th May 2012, 07:29
  #94 (permalink)  
 
Join Date: Mar 2012
Location: UK
Posts: 35
Likes: 0
Received 0 Likes on 0 Posts
I think it would have to be B because it's the only one which lists conditions which all affect ground run in the same direction, as in, they all cause it to decrease. All the rest contain some conditions which cause and increase and some which cause a decrease.

Pretty bad wording all the same.
XiRho is offline  
Old 25th May 2012, 14:25
  #95 (permalink)  
 
Join Date: Oct 2011
Location: ireland
Age: 32
Posts: 9
Likes: 0
Received 0 Likes on 0 Posts
Yea correct answer is B!!
shoeless is offline  
Old 1st Jun 2012, 09:18
  #96 (permalink)  
 
Join Date: May 2012
Location: sky
Age: 37
Posts: 1
Likes: 0
Received 0 Likes on 0 Posts
new formulae

hi dear just try this ZFW=TRAFFIC LOAD+DOW 112500=traffic load+80400 traffic load=11250-80400 =32100
mehrabi is offline  
Old 28th Jun 2012, 04:27
  #97 (permalink)  
 
Join Date: Nov 2009
Location: PK
Posts: 195
Likes: 0
Received 0 Likes on 0 Posts
W/V Calculation

Hi

For a given track the:

Wind component = +45 kt
Drift angle = 15 left
TAS = 240 kt

What is the wind component on the reverse track?

a) -55 kt
b) -65 kt <-- Marked Correct
c) -45 kt
d) -35 kt

With an E6B I get around -55. It probably uses the formula Wind Speed x Cos (Wind Direction - Course).

Is it normal/possible to have a 10kt difference if done with a CRP-5 or I am missing out something?

Thanks
Haroon is offline  
Old 28th Jun 2012, 04:39
  #98 (permalink)  
 
Join Date: Nov 2011
Location: victoria
Posts: 40
Likes: 0
Received 0 Likes on 0 Posts
Are you accounting for decrease in TAS to ETAS due to large drift angle at high TAS?
Advs is offline  
Old 29th Jun 2012, 03:08
  #99 (permalink)  
 
Join Date: Nov 2009
Location: PK
Posts: 195
Likes: 0
Received 0 Likes on 0 Posts
can you show please how you are solving it. thanks
Haroon is offline  
Old 29th Jun 2012, 14:24
  #100 (permalink)  
 
Join Date: Jan 2011
Location: England
Posts: 660
Received 18 Likes on 11 Posts
To solve this type of problem using the CRP5:

a. No heading is specified so select one at random, 000 degrees for example.

b. Set the high-speed slide with the centre dot at 240 kts.

c. The question specifies a +45 wind component. This means a 45 kt tailwind component. This means that the ground speed is 240 TAS + 45 kts = 285 kts.

d. The question also specifies 150 left drift. So draw a cross where the 15 degrees left drift line crosses the 285 kt arc.

e. With 15 degrees left drift the track is 345 degrees. So the reverse track is 165 degrees.

f. Rotate window to align 165 degrees with true heading index.

g. The cross now indicates 19 degrees right drift. Align the 165 degrees track with 19 degrees right drift. The cross now indicates 15 degrees right drift. Align 165 degrees with 15 degrees right drift. The cross now indicates 16 degrees right drift. Align 165 degrees with 16 degrees right drift. Both the cross now indicates Move and the outer scale now indicate 16 degrees right drift, so the drift is balanced.

h. The cross is now on an arc 65 kts below the centre dot. The wind component was a headwind during the outward leg so it must be a tailwind during the return leg This means that the wind component is -65 kts.


There are quicker ways of doing the job (I saw one once here in pprune but I can't remember what it was), but the above is the method described in the CRP 5 instruction sheet.

Last edited by keith williams; 29th Jun 2012 at 14:28.
keith williams is offline  

Thread Tools
Search this Thread

Contact Us - Archive - Advertising - Cookie Policy - Privacy Statement - Terms of Service

Copyright © 2024 MH Sub I, LLC dba Internet Brands. All rights reserved. Use of this site indicates your consent to the Terms of Use.