So simply put that the reason for finding the pressure height is to find the actual altitude of an aircraft?

The one that it thinks it is at.

No, you've missed the point here. Lets start at the beginning.

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.

I am a noob and still dont fully understand the idea. Lets just say that it is for traffic separation and aircraft performance.

One more question. dead reckoning is only used for VFR only and IFR uses nav aids? Is there any part of training where we have to use dead reckoning when flying IFR?

Some instrument approach procedures can have dead reckoning segments. The details are laid down in 3.3.9 of PANS-OPS, Volume I, Section 4, Chapter 3.

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.

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.

Thanks Keith.

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%

a..................

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%;

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%)

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%.

But if in the turn you need an extra 15% of lift, I think the load factor is no longer 1.15 as the lift increases giving a higher load factor.

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?

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.

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

Well to maintain altitude during a 30 degree banked turn you need to increase lift by 15%. This gives 1.15 times the straight and level lift.

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.

Aspect ratio
 

What is the formula for working out wingspan if you have the aspect ratio and the wing area. Ratio 7.5 wing area 845m2 .

I will do this in stages to enable you to learn to do it for yourself.

Aspect ratio = Span / Chord

If we multiply top and bottom by span we get

Aspect ratio = (Span x Span) / ( Chord x Span)

But Chord x Span is area so we have

Aspect ratio = Span squared / Area

Multiplying both sides by area gives us

Aspect ratio x Area = Span squared

Square rooting both sides gives us

Square root of ( Aspect ratio x Area) = Span

Inserting your numbers gives

Square root of (7.5 x 845) = Span = 79.6

Affects on Mcrit
 

What's up guys. Came across this question and it stumped me.

Which of these increases Mcrit:

A) decrease wing area
B) decrease sweep back
C) increase load factor
D) aft movement of CG

Narrowed down to A and D but can't differentiate.
I know if you decrease airfoil thickness Mcrit increases but A doesn't really suggest thickness. Aft CG movement make sense as less work required by wing to compensate for downforce produced by the horizontal tail

Thanks in advance :)

Mcrit is the lowest free stream mach number at which we get sonic (Mach 1) airflow at any point on the surface of the aircraft. For the purposes of this type of question the examiners usually we assume that this will occur over the wings.

This means that Mcrit = Mach 1 minus the airflow acceleration over the wings.

So
Anything that increases this acceleration will decrease Mcrit
Anything that decreases this acceleration will increase Mcrit.

If we decrease wing area while keeping aircraft weight unchanged, then each square foot of wing will need to produce more lift. This will require a greater angle of attack, which will increase acceleration over the wing. This means that airspeed over the wing will be closer to Mach 1 so Mcrit will be reduced. So option A is incorrect.

The wings must produce sufficient lift to carry the weight of the aircraft plus any downward force from the tail plane. If we move the CG aft we need less tail plane down force, so we need less lift. This enables us to use a lower angle of attack, so Mcrit is increased. Option d is the correct answer.

What is the effect on the aeroplane's static longitudinal stability of a shift of the centre of gravity to a more aft location and on the required control deflection for a certain pitch up or down?

I put: The static longitudinal stability is smaller and the required control deflection is larger.

Correct Answer: The static longitudinal stability is smaller and the required control deflection is smaller.

My thinking was that as the CG is aft, the arm between elevator and CG is smaller therefore a greater deflection would be required for the same movement. Now I'm thinking that it depends if you want to pitch up or down. Pitching up with aft CG would be easier than pitching down???

Yes, but at the same time the distance from the CG to the neutral point decreases. Try to imagine a fat man and a child sitting on a balanced see-saw. If the man moves closer to the pivot point, the child must produce less downward force to keep the see-saw balanced.

Try to imagine a fat man and a child sitting on a balanced see-saw. If the man moves closer to the pivot point, the child must produce less downward force to keep the see-saw balanced.

I still can't get my head around this - I accept what you say about the see-saw but in that example surely the fat man is the elevator and would have to provide a bigger downforce? Your analogy makes the child the elevator, but if the CG is moving aft towards the elevator it's closer to the fat man.........???

Still struggling to visualise it.

The analogy with the see-saw was probably imperfect. I shouldn't have posted it. This explanation by Genghis is certainly much better:

Q In OPS
 

The application of a type II anti-icing fluid on an aircraft on the
ground will provide a:

1-protection time up to 24 hours
2-limited time of protection,dependent on the outside temp,precipitation and fluid
3-limited time of protection independent of the outside temp
4-protection against icing for the duration of the flight

You can discard a and d for a start :) because the fluid is only meant to get you off the ground - in fact Type II (being thicker) should blow off before you get airborne as it will screw with the lifting capabilities of the wing. Its concentration counts, so b is the least worst answer.

To expand on Paco's answer, de-icing fluids are only meant to be effective on the ground and are required to blow off the airframe usually by 100KT, airborne protection comes from the aircraft's own anti-icing and de-icing systems. There are published tables of holdover times which show how long the fluid can be expected to be effective while the aircraft is parked or taxying. If you look at them you will see that the paramaters that affect holdover time include OAT, precipitation type and intensity and fluid concentration.

Hey guys.
Probably a stupid question but how long are the ATPLs valid for? I finished mine in September 2013 and still debating on if I should complete my conversion.

Squall, unless it has changed, you hve 36 months to get an instrument rating or ATPL.

The effect of a high wing with zero dihedral is as follows:

Zero dihedral effect
Positive dihedral effect
Negative dihedral effect
Its only purpose is to ease aeroplane loading

I went with zero dihedral effect (despite it looking a bit obvious), however correct answer given is positive dihedral effect. I'm keen to understand the theory behind this.

And just for good measure, this one

When the lift coefficient Cl of a negatively cambered aerofoil section is zero, the pitching moment is:

maximum.
nose down (negative).
nose up (positive).
zero.

I went with zero, given answer is nose up. My thinking is that if Cl is 0, then following the lift formula IAS2*S*Cl, will give an answer of 0 for lift, so how can a moment be generated when there is 0 lift?:ugh::ugh::ugh:

High wing itself provides a positive effect due to a balancing moment. A negative camber at zero lift provides a nose up moment.

The term “dihedral effect” means lateral stability, which is the tendency to roll away from sideslip.

If a disturbance causes an aircraft to drop a wing (roll to one side) it will sideslip towards the dropped wing. This will cause the airflow to approach the fuselage from the dropped wing side. When the air meets the fuselage it will split into two parts with some flowing over the top of the fuselage and some flowing under the bottom. The two parcels of air will then flow back together after passing over/under the fuselage.

For a high wing aircraft (with shoulder-mounted wings) the air which flows over the top of the fuselage is moving upwards when it meets the dropped wing, then downwards when it meets the raised wing. This increases the angle of attack of the dropped wing and decreases the angle of attack of the raised wing. So the dropped wing produces more lift and the raised wing produces less lift. This causes the aircraft to roll away from the sideslip and back towards the wings level condition.

So the answer to this question is “Positive dihedral effect”.



When a cambered aerofoil is set at zero degrees angle of attack it will produce some lift. In order to get it to produce zero lift it must be set at a slightly nose down angle. In this condition the upward lift force produced by the upper surface is equal to the downward lift force produced by the lower surface. So the two cancel out to give zero lift. But the pressure distributions of the upper and lower surface are not identical. The Centre of Pressure of the upper surface is further aft than that of the lower surface. So the upward lift of the upper surface is further aft than that of the lower surface. So these two forces produce a nose down pitching moment.


A “negatively cambered aerofoil” is one in which the greatest curvature is on the bottom surface. (Just imagine a standard aerofoil placed upside down).

So to get zero lift we must set it at a slightly nose up angle. In this condition the upward lift force and the downward lift force are again equal, so there is no overall lift. But the upward lift on the upper surface is now ahead of that on the lower surface, so they produce a nose-down pitching moment.

All of this should be clearly illustrated in your course notes.

Me again. This is making me tear my non-existent hair out

During an ILS approach on RWY 33, a northwesterly wind is blowing parallel to the runway. Its speed is increasing rapidly with height while its change in direction is negligible. What has the pilot to be aware of with respect to wind shear and glide path (no autopilot engaged)?

1. Without the pilot's intervention, the aircraft is likely to fly above the designated glide path with decreasing deviation from it.
2. Without the pilot's intervention, the aircraft is likely to fly above the designated glide path with increasing deviation from it.
3. Without the pilot's intervention, the aircraft is likely to fly below the designated glide path with increasing deviation from it.
4 .A deviation from the glide path will not have to be considered since there is no significant wind shear to be expected.

I chose 2, correct is given as 3. My reason being with a constant power setting applied and no a/p, as the wind decreases, speed over the ground will increase, so you will travel further forward for the same drop in height meaning you will overshoot.

I also considered reduction in lift due to reduction in IAS, but surely with a constant power setting, this wouldn't be the case as the IAS would increase with less wind??

The various versions of the ATPL CQB have for a number of years included two questions of this type. One uses the term “wind speed increasing rapidly with height” and the other uses the term “wind speed decreasing rapidly with height”.

The answer we will come to for each of these questions depends upon how we interpret these statements.

If the term “wind speed increasing rapidly with height” means that the wind speed increases as the height above the ground increases, then the correct answer is that the descending aircraft will fly into a decreasing wind, which will cause its ground speed to increase. This increased ground speed (with a constant rate of descent) will cause it to fly above the glide slope with increasing deviation. But this is not the answer that has been selected as being correct by the examiners.

If the term “wind speed increasing rapidly with height” means that the wind speed increases as the height above the ground decreases, then the correct answer is that the descending aircraft will fly into an increasing wind, which will cause its ground speed to decrease. This decreased ground speed (with a constant rate of descent) will cause it to fly below the glide slope with increasing deviation. But again, this is not the answer that has been selected as being correct by the authors of the question.

In both questions, the examiners have selected answers that appear to be based on a different interpretation of the terms “wind speed increases rapidly with height” and “wind speed decreases rapidly with height”. To understand how the examiners came up with these answers we must set aside the obvious conclusion that they are just plain stupid, and try to deduce what interpretation they are using.

Their answers to these questions suggest to me that the terms used should have been something like “wind speed increases rapidly with decreasing height” and “wind speed decreases rapidly with decreasing height”. If the rate of change of wind speed is sufficiently great, this second condition (wind speed decreasing rapidly with decreasing height) would be a case of vertical wind shear. In the first case we would fly below the glide path and in the second case we would fly above it.

Comms question
 

I have found a question in the comms exam which i can't seem to find an answer for. If anyone could help or knows the answer.

The question as far as i remember it was: " When contacting a station and not getting a reply, how long do you have to wait until you could contact again?
a. 10 seconds
b. 30 seconds
c. 1minute
d. 2minutes
e. wait until they will reply "

at least 10 seconds