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ATPL theory questions

Old 14th Nov 2012, 11:50
  #161 (permalink)  
Join Date: Jan 2010
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In regards to the question about the CVOR and 030 - the difference between AM and FM is 30 degrees, but I think I'm right in saying that the phase difference to be measured is between FM and AM, not AM to FM.

AM to FM is 'backwards,' and given that the phase scanning is clockwise in a CVOR you'd have to go right round - i.e. 330 degrees.

In relation to the Gen Nav question, you are following great circle tracks. I assume you've learnt about earth convergence (sin mean lat) and conversion angle (1/2 sin mean lat).

You're in the northern hemisphere so your initial track to wpt 2 will be less than 90, and the final track more than 90. At wpt 2 your heading will decrease again to maintain another great circle. I imagine the answer will be D!
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Old 14th Nov 2012, 13:16
  #162 (permalink)  
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thanks Skittles This is what I was thinking but I couldn't figure it out at first. I came across a similar question Which had a better worded explanation.

Thanks mate
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Old 16th Nov 2012, 11:51
  #163 (permalink)  
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Hello everyone,

The efficiency of a gas turbine engine increases with:
A) an increase in ambient temperature
B) an increase in volumetric efficiency
C)a decrease in ambient air pressure
D) a decrease in ambient air temperature

While A &C are obviously wrong, having already gotten this question incorrect, could you please tell me why the answer more fitting is D and not B? I understand that a decrease in air temp increases air density, which has the effect of "naturally" increasing mass flow thorough the engine, without making the engine work more to artificially produce (faster compressor spin rate) the same amount of thrust for any give throttle position. Mass flow and acceleration imparted upon a given mass of air are after all the main factors which influence thrust production as I have been taught.

Now, isn't "an increase in volumetric efficiency" a better choice and an overall better answer in terms of terminology since it implies than an engine with better VE has better mass flow through the engine at all times, in any given ambient conditions compared with one of less VE? Thus, the answer can cover a broader range of all variables than just lower amb. temp, although this answer is technically correct as well.

Or is this term more generally reserved to performance discussion relating to piston engines and should not be used with gas turbines?

Thank you
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Old 16th Nov 2012, 19:18
  #164 (permalink)  
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Volume doesn't tell you the mass. For that you'd need to know more about its density. With very hot air you may have very little mass and vice versa with cold air. This is to do with the link between temp. and density. Therefore, a colder temperature will give you a higher density which is a greater mass per volume and therefore greater thrust.

A measurement of volume alone isn't enough information.

*I'm only studying myself so hopefully others will contribute and confirm my suspicions.
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Old 17th Nov 2012, 10:28
  #165 (permalink)  
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First, differentiate between efficiency and thrust. You can have a grossly inefficient engine producing lots of thrust. Second, efficiency is always stated as a comparison - SFC is a good example, fuel flow against thrust produced.

So the question starts of badly. What is "the efficency" of a jet engine? We could look at thermal efficiency, volumetric efficiency or propulsive efficiency, to name but a few. But in this context - ATPL questions - it probably means SFC

So how do the answers stack up? Decreased intake temp does two things. It increases density and therefore mass flow and thrust but also increases thermal efficiency. Themal efficiency is driven by the difference between the fixed EGT and the inlet temp. Volumetric efficiency is driven by engine design and is usually a maximum at about 90%N1. It is, therefore under your control to some degree but I don't think the examiner means that. At the back of the examiner's mind he is linking this with the conditions for max range, when you would use 90%N1 as a given. That leaves increased pressure, which would increse density and therefore mass flow and thrust but will have no direct efect on "efficiency"

A bit woolly, I'm afraid, but I hope it helps.
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Old 17th Nov 2012, 10:54
  #166 (permalink)  
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Better a wooly answer than a scantily clad one sometimes, in this case it does actually make sense. The trickiest thing I've found about some questions (also valid when examining graphs) is not necessarily the question itself that is being asked, but the associated assumptions/conditions the question is set against, that the question doesn't tell you about and can potentially affect a correct answer being given.

Thanks guys

Last edited by Skymaster15L; 17th Nov 2012 at 11:09.
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Old 17th Nov 2012, 17:18
  #167 (permalink)  
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Need help on this mass and balance question please

A small (1ft x 1ft6in x 2ft 3in) heavy crate, mass 500lb is to be transported in mep specimen aircraft zones 2 and 3. Based on the maximum floor loading limit of 120lb/ftsquared can a box be loaded directly into the aircraft?
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Old 17th Nov 2012, 17:47
  #168 (permalink)  
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max floor loading limit is 120 lbs/ft. Area of crate is 1ft x 1,5ft= 1,5ft.


120 lbs ...180 lb/ft
_____ = _______

1 ft.........1,5 ft

The 500lbs crate is heavier than the maximum for the 1,5ft area, so you cannot load it.

However, I heard a retired redcap say once that in real life they multiply the cargo masses with 3 to get the actual cargo weight at the max load factor of 2,5 for transport aircraft+ 0,5g extra margin.

Last edited by Da-20 monkey; 18th Nov 2012 at 13:23.
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Old 18th Nov 2012, 12:12
  #169 (permalink)  
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Thank you man
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Old 19th Nov 2012, 12:26
  #170 (permalink)  
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gen nav question HELP!

question 3286 (atpl online)


I am trying to figure out this question and I dont understand the explanation of the question. can someone help me please.

Rotate to put track 090T next to HEADING index; read drift (12 right (Stbd)) and rotate to put track (090) under drift (12Stbd).
Drift has now changed to 13Stbd so put track 090 under 13Stbd.

Why do you have to rotate it again( in blue) What does this do? When I do it I get 11 degrees.

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Old 19th Nov 2012, 12:55
  #171 (permalink)  
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You're doing a drift shuffle.

If you look at the tope of your CRP-5 you'll notice that it says 'true heading' not 'true track.' When you first put the 90 under the true heading that's wrong, you've lined up the track with the heading marker. This doesn't take into account any drift.

To resolve this you have to do a drift shuffle, which is the process described in your explanation. Check the drift on the centre bit and then rotate the wheel in the same direction until the true track is on the relevant number. In your example you had 12 starboard drift, so you have to put 90 under 12. Then you have to check the drift again. Has it changed? In your case it had - it had changed to 13. Do the same process, move the 90 under 13. You keep 'shuffling' like this until it all lines up (i.e. the 90 is under the 13, and the drift shows as 13), then the value under the 'true heading' is the heading you have to fly to make good that track.

Remarkably difficult to explain via text, but if you google the CRP-5 drift shuffle you'll get better explanations.
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Old 19th Nov 2012, 14:22
  #172 (permalink)  
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It would be useful to copy full question and answers into forum as not everybody is on ATPL online.

That said "Skittles" has explained it. However I should like to add a couple of pointers.

The correct term is "balancing drift" and it only needs to be done when working out a heading and/or ground speed when using the wind down method. Do NOT balance/shuffle for any other problem.

As the correct term suggests you have given a track/course you have to turn into wind to allow for it. It is this very turn that changes the original angle of the wind which may lead to having to "shuffle" a couple of times. You ALWAYS turn in the direction of the dots new position.

You are "balanced" when your dot on the drift lines = the number of degrees (heading to track) difference in the opposite direction.

There is another & simpler way if you care to PM me.
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Old 20th Nov 2012, 18:51
  #173 (permalink)  
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Principles of Flight

"A stagnation point of a body in a moving airstream is a point where: (answer) the velocity of the relative airflow is zero and the surface pressure is higher than the ambient atmospheric pressure"

Could you please tell me why surface pressure is higher and not equal to ambient atmospheric pressure? As the book that I read on this says, (perhaps also failing to expand more on this topic),that for incompressible flows, the total pressure of still air is composed of 100% static pressure with 0 % dynamic pressure if there is no movement/velocity involved. As soon as there is velocity given to a body of air, static pressure starts to drop, and total pressure is now also composed of dynamic pressure. The sum of these pressures, P total, is always a constant.

Also, since at the stagnation point where local airflow flow velocity is 0, I take it that static pressure is then highest, meaning no dynamic pressure is present, so shouldn't this pressure then be equal to ambient air pressure of the nearby parcel of air that is unaffected by the wing's passage through it?

I case that I am possibly:
1)using the terms static pressure and stagnation pressure interchangeably, when they actually mean different things, or
2)worrying about a question that was written wrongly to begin with, or
3)not understanding something else that I have failed to mention or learn

How can the surface pressure at the stagnation point be higher than ambient, when the sums of P Static and P Dynamic, in whatever proportions, are always a constant anyway. What is adding that extra pressure?

Appreciate your input
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Old 20th Nov 2012, 19:11
  #174 (permalink)  
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Anyone feel free to correct me if I'm wrong here anyone, I've just done the same lesson, and I understand the stagnation point to be the point at which the air is brought to rest on the aerofoil. Therefore it would be total pressure i.e. Static (or ambient) + Dynamic, the reason being the Kinetic energy has to be transferred from the airflow to the aerofoil in order to bring it to rest.
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Old 20th Nov 2012, 20:38
  #175 (permalink)  
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Total Pressure = Static Pressure + Dynamic Pressure.

We need to be careful in using the statement that "Total pressure is constant".

It is more correct to say that: "Total pressure is constant at all points in a stream tube provided no energy is added to or subtracted from the airstream."

To understand this let's look at an aircraft sitting on the runway in still air at ISA mean sea level. Static pressure is approximately 15 PSI and because the air is still, the dynamic pressure is zero. So the total pressure is approximately 15 PSI.

If we now accelerate the aircraft, the dynamic pressure will increase with the square of the TAS. Eventually a speed will be reached at which the dynamic pressure is 15 PSI. But this does not mean that the static pressure has fallen to zero. The increased dynamic pressure was caused by the extra energy that we provided by accelerating the aircraft. So the ambient static pressure will still be approximately 15 PSI and the total pressure will be approximately 30 PSI.

If we continue to accelerate the dynamic pressure will become greater than 15 PSI, so the total pressure must be greater than its initial value of approximately 15 PSI.

At the stagnation point the airflow is brought to rest and this converts all of its dynamic pressure into static pressure. The total static pressure (the stagnation pressure) at the stagnation point is then the sum of ambient static pressure plus the dynamic pressure that has been converted. In the case of our aircraft flying at a speed at which dynamic pressure is 15 PSI we would have 30 PSI of stagnation pressure, but only 15 PSI of ambient static pressure.

Later on in your studies you will look at the effects of shock waves. When air flows through a shock wave it is abruptly compressed. This converts some of the pressure energy into heat. This reduces the total pressure. So as air flows through a shock wave the total pressure of the air stream decreases.
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Old 20th Nov 2012, 21:53
  #176 (permalink)  
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the part that "provided no energy is added or subtracted from the airstream" does shed light on the light on the topic now, and I wish that was included in the book to begin with. Just a few more questions about this however:

Does this mean that the Total P=Static P+Dynamic P formula now mean that the T. pressure being a constant only really applies to an air mass with Static or Dynamic pressure fluctuations only caused by the wind/ breeze blowing? As soon as an airplane passes thru such an air mass, it blows everything out of the water in terms of validity/simplicity?

Also, what happens to the value of total pressure in other areas of the wing, for example the low pressure region responsible for lift? Does this value also exceed ambient pressure because we are adding energy to the airflow, or does airflow here behave more like in the above said equation since the airflow velocity does not reach zero like at the LE stagnation point?

My question being rephrased to ask that: does the Total Pr. of 30 PSI in our hypothetical example then, reach such a value higher than ambient, only in the region of the stagnation point because of the unique nature of its airflow, whereas at any other points along the wings, the variable constituents of static and dynamic pressure are always a constant (and equal to ambient pressure) but never higher?
Thanks for the clarification
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Old 20th Nov 2012, 22:52
  #177 (permalink)  
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This whole subject is related to the principle of conservation of energy. If we assume that no energy is added to or removed from our air stream, then the total energy must remain constant.

If we look at a moving mass of air we can see that it's total energy is the sum of the following parts.

1. Mechanical energy due to its static pressure. (this is the energy that permits compressed gasses to carry out work when they expand, as for example in a piston engine).

2. Kinetic energy due to its velocity (this is evident in the form of the dynamic pressure).

3. Potential energy by virtue of its height above some reference point.

4. Thermal energy (this is evident by virtue of its temperature).

If we simplify our experiment by assuming that the flow is horizontal, then the potential energy will be constant.

If we now introduce our air stream into a tube at some selected velocity, its total pressure energy will be the sum of the static pressure plus the dynamic pressure.

If the tube becomes narrower, the velocity must increase to permit the flow to continue at the same mass flow rate. This will increase the dynamic pressure.

But(if we ignore friction) the narrowing of the tube will not introduce or remove any energy, so the total pressure energy must remain constant. This means that the static pressure must decrease to offset the increasing dynamic pressure. The above scenario is an example of a situation in which the total pressure remains constant.

Now let's look at what happens when an aeroplane accelerates during the take-off run. Its engines provide a great deal of energy to accelerate it up to flight speed. This increases the dynamic pressure. But this increase in dynamic pressure is being funded by the engines, so there is no compensating reduction in static pressure. The overall result is that as the aircraft accelerates, the total pressure of the air flowing over its surfaces increases.

But if we look closely at its curved surfaces we can still see that localized trade-offs between static pressure and dynamic pressure are taking place. As the air flows over the nose for example, the curved surfaces cause the local airspeed to increase. This increases the local dynamic pressure and decreases the local static pressure. But these local changes are being applied to a total pressure which is already much higher than the ambient value.
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Old 21st Nov 2012, 08:48
  #178 (permalink)  
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Thanks for setting me straight on this K.W. It's nice to finally bridge a gap btwn what textbooks present to you in simplified format, and what is actually going on. I am revising and expanding on previous knowledge not only for the ATPL exams I am studying for, but also to slowly set aside a folder of detailed notes & explanations on the most important knowledge topics for when I start my FI rating.

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Old 5th Dec 2012, 15:20
  #179 (permalink)  
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Hey guys.

I dont get this one, and I have seen several explanations for it, among others right here, but in another thread, but I havent considered them to be satisfactory, so Ill give it another go.

Given: Waypoint 1 is 60S 030W, waypoint 2 is 60S 020W. What will be the approximate latitude on the display of an inertial navigation system at longitude 025W?

a. 6006S
b. 5949S
c. 6000S
d. 6011S

The correct answer is supposed to be, a. 6006S.

My solution was:

0.5 x 10 x sin(60) = 4.33, conversion angle.
5 x cos(60) x 60 =150 nm, departure.

Then do some trig-stuff, or 1-60 rule, whichever, to come up with roughly 11 nm, which would make d. an appropriate answer in my opinion.

Most of the explanations I have seen for this question seem intent on halving the conversion angle, or rather getting the conversion angle from the ch.long between 030W and 025W, to get 2.2-ish, and obviously that would present an answer more consistent with the correct one. But why is this done? I dont see how thats relevant as it would imply that we are flying from 60N 030W to 60N 025W and then on to 60N 020W, in which case I would fully agree with the answer, but that would mean that 025W was an additional waypoint Unless I'm missing something else.

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Old 5th Dec 2012, 15:56
  #180 (permalink)  
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As I am a teacher in the subject, it would be interesting to know which stance
the current QB has.

Our studyguide maintains that a blockage of the ram
air opening of a pitot tube will cause the ASI to freeze at current speed
whereas in real life, such a blocking would cause the ASI to read 0 because of
pressure escaping through the waterdrain.
If you teach the subject then why do you not know the answer!

The water drain is sealed, otherwise there would be a permanent loss of pressure.
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