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!

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

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

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.

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.

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

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?

max floor loading limit is 120 lbs/ft². Area of crate is 1ft x 1,5ft= 1,5ft².

So:

120 lbs ...180 lb/ft²
_____ = _______

1 ft².........1,5 ft²

The 500lbs crate is heavier than the maximum for the 1,5ft² 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.

Thank you man

gen nav question HELP!
 

question 3286 (atpl online)

Hi

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

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

Thanks

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.

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.

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

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.

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.

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

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.

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.

Cheers :ok:

Hey guys.

I don’t get this one, and I have seen several explanations for it, among others right here, but in another thread, but I haven’t considered them to be satisfactory, so I’ll give it another go.

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

a. 60°06’S
b. 59°49’S
c. 60°00’S
d. 60°11’S

The correct answer is supposed to be, a. 60°06’S.

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 030°W and 025°W, to get 2.2°-ish, and obviously that would present an answer more consistent with the correct one. But why is this done? I don’t see how that’s relevant as it would imply that we are flying from 60°N 030°W to 60°N 025°W and then on to 60°N 020°W, in which case I would fully agree with the answer, but that would mean that 025°W was an additional waypoint… Unless I'm missing something else.

-Anders

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.

anders:
 

You should use 5 degrees here instead of 10. It's obvious if you draw a picture of the situation with 2 parallel lines representing the mentioned lines of longitudes, like on a mercator chart, with straight rhumb line between the points and a downwards pointing great circle line. The point where the g.c. is most south is in the middle between 20W and 30W, so 5 degrees longitude.

Even then I am not sure how accurate this answer is in the real world..

Back in the days we didn''t have this q. in the database. (pipe-smoking smiley)

Nope, still don’t get it. The way I see it, that would give me the most southerly point between 30W and 25W, ie 27.5W.


Strike that, I'm moling it over and hopefully it will come to me.

-Anders

0.5 x 5° x sin(60) = 2,17°, conversion angle.

I mean 5 deg. for change of longitude.

Tangens = opposite/ adjecent

Tan 2,16= opposite/ 150NM

opposite= 5,67NM.

this is appox. 6 minutes of logitude, so 60,06

I’m sorry, I thought I had it but now I’m just confused again. The way I’ve learnt it is:

Conversion angle = 0.5 x ch.long x sin (lat)

But what you’re saying is basically:

Conversion angle = 0.25 x ch.long x sin (lat)

The change of longitude is 10°, not 5°, as I see it…

-Anders

anders, i agree with your method and answer. i guess the answer printed is wrong.
the formula for conversion angle is correct as 0.5*ch long*sin(lat)

i guess i've figured it out:) with some help from another forum.
lets name the starting position A and the intermediate position on 25 degree W as B.
when we join points A and B to form a triangle for calculation purpose, it divides conversion angle into two equal parts of 2.16 degress each.
now solving the sum in the same manner:
tan 2.16= X (unknown)/150
X=5.65nm
X/60 = 0 degrees 5 minutes
hence position B= 60 degrees + 5 minutes
=60 degrees 5 minutes
:ok:

like bayblade said

The midway point between 30W and 20W is 25W.

At 60S the departure between 30W and 25W is 150 nm.

We can now draw a triangle with a horizontal side 60S 30W to 60S 25W. The length of this line will be 150 nm.

The second side is a vertical line from 60S 25W down to the unknown position south of 25W.

The third line is the hypotenuse form 60S 30W to the unknown position south of 25W.

The internal angle between the hypotenuse and the horizontal side is the conversion angle based on going from 60S 30W to 25W.

This angle = 0.5 x 5 degrees x Sin 60 = 2.165 degrees.

The tangent of this angle is the vertical side divided by the horizontal side.

Tan 2.165 = Ch lat / 150 nm

Rearranging this gives Ch lat = Tan 2.165 x 150 = 5.67 degrees.

This means that the change of latitude between 60S 30W and the unknown position south of 25W is 5.67 degrees. This makes the new latitude 60 degrees 5.67 minutes south. The closest option to this is 60°06’S

Sorry Anders my previous post did not address this part of your question.


To understand why they do this we need to sketch the whole picture. Draw a horizontal straight line to represent the 300 nm rhumb line track from 60S 30W to 60S 20W.

Now draw a shallow arc looping down between the two ends of the rhumb line track. This represents the great circle track.

Midway between the two ends of the rhumb line draw a vertical line down to the great circle arc. This vertical line is 150 nm from each end of the rhumb line track. The length of this vertical line represents the maximum change of latitude between the two tracks.

Now draw a sloping straight line from each end of the rhumb line to the lower end of the vertical line.

Using the conversion angle equation we can now calculate the angle between each end of the great circle arc and the ends of the rhumb line.
This is 0.5 x 10 degrees x Sin 60S = 4.33 degrees. Write 4.33 in each of these angles.

Now let’s look at what happens when we fly from 60S 30W to the midway position south of 25W. We are initially tracking 094.33 degrees, but our track is continuously turning to the north, such that we are tracking 090 when we reach our most southerly point.

During this first half of the trip we have reduced our track direction by 4.33 degrees from 094.33 to 090. This means that our mean track was 092.165 degree. This mean track is represented by the straight line from our starting point to our most southerly point. So the internal angles in the triangles at each end of our track are 2.165 degrees. This is half of the conversion angle.

So in solving this type of problem it is easier to go straight for ½ the conversion angle and work out the solution from there.

I figured it was something like that, but I was having a hard time visualizing it. Thanks for clearing that up, and for everyone else efforts in trying to explain it to me.

Question: A turbojet aeroplane has a planned take-off mass of 190 000 kg; the cargo load is distributed as follows: cargo 1: 3 000 kg; (3.50 m from reference point) cargo 4: 7 000 kg. (20.39 m from reference point) Distance from reference point to leading edge: 14m Length of MAC = 4.6m. Once the cargo loading is completed, the crew is informed that the centre of gravity at take-off is located at 38 % MAC (Mean Aerodynamic Cord) which is beyond the limits. The captain decides then to redistribute part of the cargo load between cargo 1 and cargo 4 in order to obtain a new centre of gravity location at 31 % MAC. Following the transfer operation, the new load distribution is:
cargo 1: 6 000 kg; cargo 4: 4 000 kg

Can someone please tell me how this is worked out.

It is not clear what your question is asking.

Is it asking how much cargo must be moved to move the C of G from 38% MAC to 31% MAC?

Or is it asking where the C of G will be after when there is 6000 kg in Hold 1 and 4000 kg in Hold 4?

Initial condition.
MAC length = 4.6 meters, so 38% MAC is 4.6 x 0.38 = 1.748 meters aft of the MAC leading edge.

The MAC leading edge is 14 meters aft of the datum, so this is 14 + 1.748 = 15.748 meters aft of datum.

Total moment = total mass (190000 kg) x CofG position (15.748 m) = 2992120 kn m.

Final condition.
CofG is at 31% MAC which is 4.6 m x 0.31 = 1.426 meters aft of the MAC leading edge.

The MAC leading edge is 14 meters aft of the datum, so this is 14 + 1.426 = 15.426 meters aft of datum.

Total moment = total mass (190000 kg) x CofG position (15.426 m) = 2930940 kg m.


Calculation of cargo to be moved
Required moment change = new moment (2930940) – Initial moment (2992120) = -61180 kg m.

Moment change = cargo mass moved x distance moved.

Distance moved = new position – initial position

Distance moved = Hold 1 at (3.5 m) – (hold 4 at(20.39 m) = -16.89 meters.

Required moment change = -61180 kg m

Dividing required moment change (-61180 kg m) by distance cargo is moved (-16.89 m) = 3622 kg.

This means that 3622 kg of cargo must be moved from hold 4 to hold 1 to move the C of G from 38% MAC to 31% MAC.


But the final line of your statement of the questions states that “ Following the transfer operation, the new load distribution is: cargo 1: 6 000 kg; cargo 4: 4 000 kg”. If this is correct then only 3000 kg has been moved, so the new C of G will not be at 31% MAC.

Moving 3000 kg a distance of (-16.89 m) give a moment change of 3000 kg x (-16.89 m ) = -50670 kg m.

Adding this to the initial moment gives 2992120 kg m. – 50670 kg m = 2941450 kg m.

Dividing this by the total mass gives a new C of G position of 2941450 kg m / 190000 kg = 15.48 meters.

Subtracting the position of the MC leading edge gives 15.48 – 14 = 1.48 meters.

Dividing this by the MAC length then multiplying by 100% gives 1.48 m / 4.6 m = 0.322, which is 32.2% MAC.

So the new C of G position after moving 3000 kg of cargo is 32.2% MAC.

Thank you so much. For breaking it down for me. It's one of the questions on ATPL and one that doesn't also give any explanation too. But thank you.

Does anyone have a useful nemonic/memory technique for remembering the answers to questions such as:

"Turbine blade stages may be cassified as either impulse or reaction. In an impulse turbine stage:

A. The pressure rises across the stator bades and remains constant across the rotor blades;

B. The pressure remains constant across the stator blades and drops across the rotor blades;

C. The pressure drops across the stator blades and remains constant across the rotor blades;

D. The pressure remains constant across the stator blades and rises across the rotor baldes."

(answer in this example is C).

They are a real bugbear of mine...

Tks.

This question is very confusing and one that i can not work out.
Although it doesn't state it but I checked cap mrjt1 but even then I don't get the answer it is saying. Can anyone please help

The MRJT 1 CAP STATES:
MS Taxi Mass: 63060
MSTOM 62800
MSLM 54900
MZFM 51300


Question: Prior to departure the medium range twin jet aeroplane is loaded with maximum fuel of 20100 litres at a fuel density (specific gravity) of 0.78. Using the following data - Performance limited take-off mass 67200 kg Performance limited landing mass 54200 kg. Dry Operating Mass 34930 kg. Taxi fuel 250 kg. Trip fuel 9250 kg, Contingency and holding fuel 850 kg, Alternate fuel 700 kg. The maximum permissible traffic load is
Answer: 13090 kg.

I get 12442 kg max traffic load.

The proposed traffic load of 13090 kg would exceed the max structural take-off mass, so there seems to be an error in the question.

IF you didn't take into account the MRJT from CAP696/7 and just worked with the given figures then 13092 is the maximum traffic load, but not the clearest question ever produced. Would be better saying "an aircraft" with no reference to MRJT.

20100 * .78 = 15678 kg total fuel - 250 taxi = 15428 kg T/O - 9250 kg trip = 6178 kg on landing.

67200 - 34930 - 15428 = 16842 maximum T/L at take off.
54200 - 34930 - 6178 = 13092 maximum T/L for landing.

Unless the question says "refer to CAP696/7" then I suggest you go with the raw data provided however poor it might be.

Thank you for your assistance. Highly appreciated.

can someone please tell me how this is worked out.

Question: If an aeroplane performs a steady co-ordinated horizontal turn at a TAS of 200 kt and a turn radius of 2000 m, the load factor (n) will be approximately: 1.1.

Many thanks

You can use the following two formulae to calculate the bank angle, and then the load factor:

tan phi = v² / (g * R)

phi: bank angle [°]
v: speed [m/s]
g: gravitational acceleration [m/s²] (g = 9.80665 m/s²)
R: radius of turn [m]

n = 1 / cos phi

n: load fator