ATPL theory questions
Join Date: May 2008
Location: UK
Posts: 1,365
Likes: 0
Received 0 Likes
on
0 Posts
So simply put that the reason for finding the pressure height is to find the actual altitude of an aircraft?
We use pressure to measure altitude, as pressure drops as you climb, typically at 27' per millibar of pressure difference.
Air pressure at the surface varies, both higher and lower, so obviously you need a starting point to refer from. That's why they came up with ISA, and the standard pressure, which is 1013 millibars at the surface.
When they then test aircraft, they calculate and write down how the aircraft will perform on an ISA day, and that is what you use when you calculate how it will perform in the actual conditions.
On any given day, if you set 1013 on your altimeter, that is your pressure altitude. It has nothing to do with your actual altitude, that would be shown by setting the sea level pressure (QNH) on your sub-scale. Its is purely used in performance calculations, and when flying IFR we set 1013 and fly flight levels to aid aircraft separation as we're all flying on the same sub-scale.
Join Date: Nov 2008
Location: Land of Hope
Posts: 47
Likes: 0
Received 0 Likes
on
0 Posts
March 2014 exam question mass and balance
The mass and balance information gives : Basic mass : 1 200 kg ; Basic balance arm : 3.00 m Under these conditions the Basic centre of gravity is at 20% of the mean aerodynamic chord (MAC). The length of MAC is 2m. In the mass and balance section of the flight manual the following information is given : Position Arm front seats : 2.5 m rear seats : 3.5 m rear hold : 4.5 m fuel tanks : 3.0 m The pilot and one passenger embark; each weighs 80 kg. 95kg fuel. The rear seats are not occupied. The position of the centre of gravity at take-off (as% MAC) is :
1200 x 3m aft of the datum. 180 x 2.5. 95 x 3. 4355 / 1475= 2.95 m The BEM had a CG at 20% MAC on a 2m chord so LEMAC 2.4m from the datum. 2.95 - 2.4= 0.55m 55 / 2 x 100% = 27.5% but the answer closest to this was 29 and 21/22. Can't remember. I went for 29% is that right?
Mass of 450 is loaded at a station 8.0 behind centre of gravity and 6.8 behind the datum. (Assume g=10 m/sec squared). the moment for that mass used for loading manifest is?
3060kgm
30600kgm
What would be the correct answer for this? As practise exams use nm as units rather than kgm.
Thank you in advance.
1200 x 3m aft of the datum. 180 x 2.5. 95 x 3. 4355 / 1475= 2.95 m The BEM had a CG at 20% MAC on a 2m chord so LEMAC 2.4m from the datum. 2.95 - 2.4= 0.55m 55 / 2 x 100% = 27.5% but the answer closest to this was 29 and 21/22. Can't remember. I went for 29% is that right?
Mass of 450 is loaded at a station 8.0 behind centre of gravity and 6.8 behind the datum. (Assume g=10 m/sec squared). the moment for that mass used for loading manifest is?
3060kgm
30600kgm
What would be the correct answer for this? As practise exams use nm as units rather than kgm.
Thank you in advance.
Last edited by clkorm3; 3rd Mar 2014 at 23:40.
The question states that the basic arm of 3.00 m is at 20% MAC.
MAC length is 2 m, so 20% of this is 0.4 m.
Subtracting 0.4 m from 3.0 m give a MAC Leading Edge of 2.6 m.
Item Mass Arm Moment (mass x Arm)
BEM 1200 kg 3 m 3600 kgm
Pilot + Pax 160 kg 2.5 m 400 kgm
Fuel 95 kg 3 m 285 kgm
Total 1455 kg 4285 kgm
C of G position = Moment / Mass = 4285 / 1455 = 2.945 m
Subtracting MAC Leading Edge position gives C of G at 2.945 – 2.6 = 0.345 m aft of MAC Leading Edge.
To convert to % MAC use 0.345 m / 2 m x 100% = 17.25% MAC
The final question states that g = 10 m/sec.
This means that 1 kg = 10 Newtons.
So you can now convert the kgm into Nm by multiplying by 10. But you do not need to do this to answer the question.
Mass of 450 is loaded at a station 8.0 behind centre of gravity and 6.8 behind the datum. (Assume g=10 m/sec squared). the moment for that mass used for loading manifest is?
3060kgm
30600kgm
Moment = mass x arm from datum
Moment = 450 kg x 6.8 m = 3060 kgm
Note that the fact that the load is 8 m behind the CG is not relevant to this question.
MAC length is 2 m, so 20% of this is 0.4 m.
Subtracting 0.4 m from 3.0 m give a MAC Leading Edge of 2.6 m.
Item Mass Arm Moment (mass x Arm)
BEM 1200 kg 3 m 3600 kgm
Pilot + Pax 160 kg 2.5 m 400 kgm
Fuel 95 kg 3 m 285 kgm
Total 1455 kg 4285 kgm
C of G position = Moment / Mass = 4285 / 1455 = 2.945 m
Subtracting MAC Leading Edge position gives C of G at 2.945 – 2.6 = 0.345 m aft of MAC Leading Edge.
To convert to % MAC use 0.345 m / 2 m x 100% = 17.25% MAC
The final question states that g = 10 m/sec.
This means that 1 kg = 10 Newtons.
So you can now convert the kgm into Nm by multiplying by 10. But you do not need to do this to answer the question.
Mass of 450 is loaded at a station 8.0 behind centre of gravity and 6.8 behind the datum. (Assume g=10 m/sec squared). the moment for that mass used for loading manifest is?
3060kgm
30600kgm
Moment = mass x arm from datum
Moment = 450 kg x 6.8 m = 3060 kgm
Note that the fact that the load is 8 m behind the CG is not relevant to this question.
Last edited by keith williams; 4th Mar 2014 at 13:54. Reason: Misread question
Join Date: Dec 2013
Location: Baleares
Posts: 70
Likes: 0
Received 0 Likes
on
0 Posts
Anyone feel like helping with this exam q?
This has got me stumped. I know how to calculate LF in a turn (1/cos alpha) and that LF is total lift/weight. But don't know how to solve this. Would love some help!
"In a 30˚ banked turn an extra 15% of lift is required. Stalling speed will be increased by ?"
a) 7%
b)15%
c)22%
d)30%
"In a 30˚ banked turn an extra 15% of lift is required. Stalling speed will be increased by ?"
a) 7%
b)15%
c)22%
d)30%
Join Date: Sep 2012
Location: 738
Posts: 79
Likes: 0
Received 0 Likes
on
0 Posts
sqrt(1/cos30)=1.07 -->7%
or
VSold= i.e. 100kts
VSnew=VSold*sqrt(n); 100*sqrt(1.15)= 107kts --> increases 7%
The statement "an extra 15%" is not clear for me. During a 30º bank angle do you need an extra 15% of lift?, that would increase our load factor, or do you need an extra 15% of lift to execute that turn from straight and level flight?
If the answer is, you need an extra 15% whilst turning, then I think the answer is different.
The load factor would be 1.15+ 15% = 1.32; Vs=100*sqrt(1.32)= 114.89 --> 15%;
or
VSold= i.e. 100kts
VSnew=VSold*sqrt(n); 100*sqrt(1.15)= 107kts --> increases 7%
The statement "an extra 15%" is not clear for me. During a 30º bank angle do you need an extra 15% of lift?, that would increase our load factor, or do you need an extra 15% of lift to execute that turn from straight and level flight?
If the answer is, you need an extra 15% whilst turning, then I think the answer is different.
The load factor would be 1.15+ 15% = 1.32; Vs=100*sqrt(1.32)= 114.89 --> 15%;
Last edited by Live your dream; 20th Mar 2014 at 20:42.
Join Date: May 2013
Location: have I forgotten or am I lost?
Age: 70
Posts: 1,126
Likes: 0
Received 0 Likes
on
0 Posts
the guy is trying to understand the problem not the answer, so show your working dak.
according to kershner's 'advanced pilots flight manual' the stall speed increases as the square root of the load factor increases.
1/cos 30 = 1.1547 load factor
square root of 1.1547 = 1.07456
7.456% increase
(pilots don't bother with decimals so 7%)
according to kershner's 'advanced pilots flight manual' the stall speed increases as the square root of the load factor increases.
1/cos 30 = 1.1547 load factor
square root of 1.1547 = 1.07456
7.456% increase
(pilots don't bother with decimals so 7%)
Answering this type of question requires you to know the relationships between stall speed, weight and load factor.
These relationships include:
1. Stall speed in straight and level flight is proportional to the square root of the weight.
This gives us the equation Vs at new weight = Vs at old weight x square root of( new weight / old weight ).
2. Stall speed in any manoeuvre is proportional to the square root of the load factor.
This gives us the equation Vs in any manoeuvre = Vs is Straight and level flight x the square root of the load factor in the manoeuvre.
In this question we have load factor increasing from 1 in straight and level flight to 1.15 in the turn.
So the new stall speed is proportional to the square root of 1.15, which is approximately 1.07. This means an increase of 7%.
These relationships include:
1. Stall speed in straight and level flight is proportional to the square root of the weight.
This gives us the equation Vs at new weight = Vs at old weight x square root of( new weight / old weight ).
2. Stall speed in any manoeuvre is proportional to the square root of the load factor.
This gives us the equation Vs in any manoeuvre = Vs is Straight and level flight x the square root of the load factor in the manoeuvre.
In this question we have load factor increasing from 1 in straight and level flight to 1.15 in the turn.
So the new stall speed is proportional to the square root of 1.15, which is approximately 1.07. This means an increase of 7%.
Join Date: Sep 2012
Location: 738
Posts: 79
Likes: 0
Received 0 Likes
on
0 Posts
Exactly, that's the point. Then I should understand that this 15% extra lift is the amount I will achieve in a 30º bank angle. If the bank angle is i.e. 45º, would they say an extra 41% of lift is required?
Join Date: Apr 2009
Location: down south
Age: 76
Posts: 13,226
Likes: 0
Received 0 Likes
on
0 Posts
Look you people. Keith has given you the answer.
Do it you own way if you like, and get the wrong answer.
Ask your specialist instructor at your school !!!!!
If you get the wrong answer then GO SOMEWHERE ELSE.
Do it you own way if you like, and get the wrong answer.
Ask your specialist instructor at your school !!!!!
If you get the wrong answer then GO SOMEWHERE ELSE.
Join Date: Dec 2013
Location: Baleares
Posts: 70
Likes: 0
Received 0 Likes
on
0 Posts
The question is reproduced verbatim from the exam paper. God knows its provenance!
I took it to mean whilst banking at 30˚ you need an EXTRA 15 lift which is what stumped me and I still wouldn't know how to solve that.
Sq. root of 1/cos 30 makes the most sense thinking about it more, but you never know with some of these questions!
thanks for all the input
I took it to mean whilst banking at 30˚ you need an EXTRA 15 lift which is what stumped me and I still wouldn't know how to solve that.
Sq. root of 1/cos 30 makes the most sense thinking about it more, but you never know with some of these questions!
thanks for all the input
I took it to mean whilst banking at 30˚ you need an EXTRA 15 lift which is what stumped me and I still wouldn't know how to solve that.
If we now interpret the question as meaning that we need an additional 15% on top of this, we will have 1.15 x 1.15 = 1.3225 times the straight and level lift. Or 1.15 + 0.15 = 1.3 if you take the "additional 15%" as meaning 15% of the S+L lift.
If we now use the equation
Vs in a manoeuvre = Vs in straight and level x square root of load factor we have
Vs = 1 x square root of 1.3225 = 1.15. (Or 1 x square root of 1.3 = 1.14.)
This means that our stall speed has increased by 15%. (Or 14%, which is not an option)
No matter how much effort the examiners put into constructing their questions, it will always be possible for some readers to interpret them in unexpected ways. Personally I would interpret this question as being about a simple constant altitude 15 degree banked turn, in which case the answer is 7%.
It is worth noting that we can carry out a banked turn without any increase in lift, but this will cause the aircraft to sink. If this is a real JAR/EASA ATPL exam question, the use of the words "additional 15% lift" could be the examiner's way of indicating that he/she meant a constant altitude turn.
