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ATPL theory questions

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ATPL theory questions

6th Apr 2014, 08:20

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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

Square root of (7.5 x 845) = Span = 79.6
7th Apr 2014, 11:52

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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
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

7th Apr 2014, 13:34

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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.
23rd May 2014, 11:15

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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???
23rd May 2014, 12:04

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Originally Posted by Straighten Up
My thinking was that as the CG is aft, the arm between elevator and CG is smaller
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.
23rd May 2014, 14:24

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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.
24th May 2014, 09:32

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

CG is quite close to, but in front of, the centre of lift. Moving CG aft moves it closer to the centre of lift, so that the tailplane has to do less work to match the pitching moment effects of CG and CofL not being in the same place.

So, it becomes more effective, since any elevator movement will be more powerful in terms of pitch response.
25th May 2014, 17:40

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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
26th May 2014, 07:07

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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.
26th May 2014, 08:38

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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.
28th May 2014, 18:23

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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.
29th May 2014, 16:14

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Squall, unless it has changed, you hve 36 months to get an instrument rating or ATPL.
2nd Jun 2014, 16:20

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

Zero dihedral effect
Positive dihedral effect
Negative dihedral effect

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.
2nd Jun 2014, 16:30

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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?
3rd Jun 2014, 11:09

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High wing itself provides a positive effect due to a balancing moment. A negative camber at zero lift provides a nose up moment.
3rd Jun 2014, 13:07

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

Zero dihedral effect
Positive dihedral effect
Negative dihedral effect

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

maximum.
nose down (negative).
nose up (positive).
zero.
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.
9th Jun 2014, 11:56

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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??
9th Jun 2014, 13:08

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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.

Last edited by keith williams; 9th Jun 2014 at 19:13.
4th Jul 2014, 01:30

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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 "
4th Jul 2014, 07:58

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at least 10 seconds