ATPL theory questions

[keith williams]

If a radar pulse contains 300 cycles of RF energy at a frequency of 600 MHz, the physical length of the pulse is:

1550 metres
150 metres
1.5 metres
0.15 metres

Wavelength = c / frequency

c = 300 000 000 m/sec. f = 600 MHz = 600 000 000 cycles/sec

Wavelength = 300 000 000 / 600 000 000 = 0.5 metres

Pulse length =cycles x wavelength

Pulse length = 300 cycels x 0.5 m/cycle = 150 metres

[Haroon]

Thanks Keith

I was getting uptil 0.5 meters but could'nt figure out that Pulse length = cycles x wavelength :ugh:

Can you please explain this one too:

What is the PRF given 50 micro second pulse width and a range of 30 nm:

1620 pps
810 pps
3240 pps
3086 pps

Thanks

[keith williams]

To produce unambiguous range information the radar system must remain silent between pulses for sufficient time for the pulse to travel out to the furthest target and return to the antenna.

The distance traveled by the signal = c x travel time

The travel time = the time between the transmission of successive pulses, which is equal to 1/PRF.

So distance traveled by the signal = c x 1/PRF which = c / PRF

But the signal must travel out to the furthest target and back to the antenna, so maximum range to the furthest target is half of the distance that the signal travels.

So maximum range = c / ( 2 x PRF)

Rearranging this equation gives

PRF = c / ( 2 x max range)

C is the speed of light, which is approximately 300 000 KM/sec or 162 000 NM / second.


For a range of 30 Nm the above equation gives

PRF = 162 000 NM/sec / ( 2 x 30 NM) = 2700 pulses per second

So your initial calculation was correct, but 2700 pps is not an option in this question.


BUT EASA CQB 15 does contain the following similar question.



The maximum pulse repetition frequency (PRF) that can be used by a primary radar facility in order to detect targets unambiguously at a range of 50 NM is?

A 713 pps
B 610 pps
C 1620 pps
D 3240 pps

The correct answer to this question is option C 1620 pps

This can be calculated at follows

PRF = 162 000 NM/sec / ( 2 x 50 NM) = 1620 pulses per second



The 50 micro second pulse width quoted in your question does not affect the maximum range, but will determine the minimum range.

[Haroon]

Thanks for the explanation Keith.

That's what I thought. However I got confused with the post:

81000/50 = 1620

81000 is one million divided by 12.36, the latter being the "radar mile"

Regards

[keith williams]

I have not used the Radar Mile method before but doing a GOOGLE search revealed this.

Radar timing is usually expressed in microseconds. To relate radar timing to distances traveled by radar energy, you should know that radiated energy from a radar set travels at approximately 984 feet per microsecond. With the knowledge that a nautical mile is approximately 6,080 feet, we can figure the approximate time required for radar energy to travel one nautical mile using the following calculation:

A pulse-type radar set transmits a short burst of electromagnetic energy. Target range is determined by measuring elapsed time while the pulse travels to and returns from the target. Because two-way travel is involved, a total time of 12.36 microseconds per nautical mile will elapse between the start of the pulse from the antenna and its return to the antenna from a target.
This 12.36 microsecond time interval is sometimes referred to as a RADAR MILE, RADAR NAUTICAL MILE, or NAUTICAL RADAR MILE.

1 Radar Kilometer = 2 · 1000 m = 6.66 µs (1)

3 · 108 m/s


1 Radar Mile = 2 · 1852 m = 12.35 µs (2)

3 · 108 m/s

The range in kilometers to an object can be found by measuring the elapsed time during a round trip of a radar pulse and dividing this quantity by 6.66.

The range in nautical miles to an object can be found by measuring the elapsed time during a round trip of a radar pulse and dividing this quantity by 12.36.


If we use PRT = the Pulse Repetition Time

PRT = 1 000 000 micro seconds / PRF


Range = PRT / 12.36 from GOOGLE extract

PRT / Range x 12.36


For 39 NM range

30 NM x 12.36 = 370.8

PRF = 1000000 microseconds / PRT = 1000000 / 370.8 = 2696

Which is approximately 2700.

For 50 NM range

50 NM x 12.36 = 618

PRF = 1000000 microseconds / PRT = 1000000 / 618 = 1618

Which is approximately 1620

[Haroon]

Thanks Keith

[pilot in command]

Hi everyone,

I would just like some peoples thoughts on the ATPL Met subject.

I have the Oxford OAA CDs for Module 1 of the CATS course which are excellent and I'm going to be properly starting Met tomorrow. Generally, Met seems to be one of the harder subjects of the ATPL course and I'm just wondering how many weeks or even months people are taking to study for Met?

I have been studying 5 - 6 hours a day and would appreciate someone giving me an indication of just how big/complex the subject is and their experiences.

Thanks in advance

[pipersam]

Can't really comment on amount of time to spend as I went integrated and completed the whole syllabus in 2 weeks :yuk: but what I will say is that it's (for me) probably the most interesting subject, try to enjoy learning it - you'll find it one of the most useful subject and most important to retain when you're flying.

Good luck!

[cefey]

Get access to question bank, so you can test your self and see what you might get on the exam. A lot of tricky questions!;)

[Haroon]

Hi

The range to a required waypoint presented by RNAV system is:

plan range or slant range depending on RNAV settings
plan range
slant range
neither plan range nor slant range

Isnt it always plan range that is presented?

thanks

[Transsonic2000]

Hi all,
would need some help with the following two questions:

1. By what percentage does lift increase in a steady level turn at 45° angle of bank, compared to straight and level flight?
Correct answer would be 19%

2. Twin engine airplane (59.000 Kg) climbing with all engines operating. Lift-Drag ratio is 12. Each engine producing 60.000 N thrust. The gradient of climb is (assume g = 10 m/s²)?
Correct answer is 12%

[keith williams]

I think that you will find that there is an error in your first question.


By what percentage does the lift increase in a steady level turn at 45° angle of bank, compared to straight and level flight?

a. 52%.
b. 41%.
c. 19%.
d. 31%.

In order to carry out a constant altitude turn the aircraft must be banked in the direction of the turn. This tilts the lift force towards the lower wing, so that part of it acts horizontally towards the centre of the turn. It is this horizontal component of lift that accelerates the aircraft in the direction of the intended turn.

But tilting the lift force reduces the vertical component of lift, such that it no longer equals the weight of the aircraft. This means that more lift is required in order to prevent the aircraft from sinking. This increase in lift constitutes an increase in load factor.

The load factor at any bank angle in a constant altitude turn is equal to 1 divided by the cosine of the bank angle. So in a 45 degree banked turn the load factor is 1/Cos 45 which is 1/0.7071, or approximately 1.414.

The load factor is equal to the lift divided by the weight. In straight and level flight the lift is equal to the weight of the aircraft and the load factor is 1. So an increase in load factor from 1 to 1.41 means that lift has increased to 141% of its initial value. This means that the lift has increased by 41% (option b).


But had the question been about the change in stalling speed we would have:

By what approximate percentage will the stall speed increase in a horizontal coordinated turn with a bank angle of 45°?

a. 52%.
b. 19%.
c. 31%.
d. 41%.


In order to carry out a constant altitude turn the aircraft must be banked in the direction of the turn. This tilts the lift force towards the lower wing, so that part of it acts horizontally towards the centre of the turn. It is this horizontal component of lift that accelerates the aircraft in the direction of the intended turn.

But tilting the lift force reduces the vertical component of lift, such that it no longer equals the weight of the aircraft. This means that more lift is required in order to prevent the aircraft from sinking. This increase in lift constitutes an increase in load factor.

The load factor at any bank angle in a constant altitude turn is equal to 1 divided by the cosine of the bank angle. So in a 45 degree banked turn the load factor is 1/Cos 45 which is 1/0.7071, or approximately 1.414.

The stall speed in any manoeuvre can be calculated by using the standard equation:

Stall speed at new load factor = 1g stall speed x Square root of (new load factor).

Inserting the data above gives:

Stall speed in 45 degree turn = 1g stall speed x Square root of (1.414) which = 1g stall speed x 1.189.

This means that the stall speed has increased by 18.9% or approximately
19% (option b).



For your second question

2. Twin engine airplane (59.000 Kg) climbing with all engines operating. Lift-Drag ratio is 12. Each engine producing 60.000 N thrust. The gradient of climb is (assume g = 10 m/s²)?
Correct answer is 12%

If we make the simplifying assumption that lift = weight in a steady climb, then we can calculate climb gradient using the standard equation below.

% Gradient = 100 x ( (thrust/weight) – (1/Lift: Drag ratio)).

This can also be expressed as:

% Gradient = 100 x ( (thrust/weight) – (Drag/Lift)).

It is essential that both all forces are expressed in the same units.
In this question weight is in kg and thrust is in Newton, so one must be of these must be converted to match the other.
To convert Newton to kg divide by g.
So thrust per engine = 60000 N /10 m/s/s = 6000 kg thrust.

The question asks for the all engines climb gradient so the total thrust is 2 x 6000 = 12000 kg.

Inserting the data into the equation gives:
% Gradient = 100 x ( (thrust/weight) – (Drag/Lift) ).
% Gradient = 100 x ( (12000 kg /59000 kg) – (1/12)).

% Gradient = 12%

[Haroon]

Hello Keith

Have you seen this one:

The range to a required waypoint presented by RNAV system is:

plan range or slant range depending on RNAV settings
plan range
slant range
neither plan range nor slant range

Isnt it always plan range that is presented?

Regards

[keith williams]

I have not seen the question before, so I cannot say with any certainty what the examiner's answer is.

If the RNAV system is using DME as an input then this will introduce slant range errors.

I believe that modern RNAV systems compensate for slant range errors, but earlier systems probably do not do so.

[Haroon]

Thanks Keith, the examiner answer is highlighted in bold.

wasnt too sure about which RNAV settings is he talking about.

anyway thanks a lot.

[Transsonic2000]

Keith, thanks a lot for taking the time to answer my questions in such detailed manner!!!! Very much appreciated!!!! Your explanation helped me a lot, it now appears to be so simple and crystal clear. People like you and your competent posts really contribute to enhance this forum!

[keith williams]

Happy to be of help Transsonic 2000.

Most of my explanations are taken from my range of books and CDs, so in most cases providing them here involves little extra work. It's mainly a copy-paste job.

[Transsonic2000]

Hi there,
having trouble figuring out the following QDB questions, would appreciate if someone could shed some light on it:

1. Whilst maintaining straight & level flight with a lift coefficient of CL=1, what will be the new value of CL after the speed is increased by 41%:

a) 0.25
b) 0.50 (correct answer)
c) 0.60
d) 0.30

2. Airplane flying at 100 kt in straight & level flight is subject to a disturbance that suddenly increases the speed by 20 kt. Assuming AoA remains constant, load factor will initially be:

a) 1.41
b) 1.44 (correct answer)
c) 1.30
d) 1.04

3. In straight & level flight at a speed of 1.3 VS, the lift coefficient, expressed as a percentage of of its maximum CLMAX, would be:

a) 169%
b) 130%
c) 59% (correct answer)
d) 77%

Help appreciated, thanks a lot!

[keith williams]

1. Whilst maintaining straight & level flight with a lift coefficient of CL=1, what will be the new value of CL after the speed is increased by 41%:

a) 0.25
b) 0.50 (correct answer)
c) 0.60
d) 0.30

Lift = Cl 1/2Rho V squared S

Where V is the TAS.

If speed increase by 41% it becomes 1.41 of its initial value.

So V squared becomes approximate 2 times its initial value.

The question states that SL flight is to be maintained, so the lift must not change.

So to maintain level flight the Cl must become ½ of its initial value.


2. Airplane flying at 100 kt in straight & level flight is subject to a disturbance that suddenly increases the speed by 20 kt. Assuming AoA remains constant, load factor will initially be:

a) 1.41
b) 1.44 (correct answer)
c) 1.30
d) 1.04

Lift = Cl 1/2Rho V squared S

Where V is the TAS.

Cl is proportional to AoA, so if AoA remains unchanged then Cl remains unchanged.

The 20 kt increase from 100 kts increases speed to 120 kts. This is 1.2 times its initial value.
So V squared becomes 1.2 squared = 1.44 times its initial value.

So the new lift = 1.44 times its initial value.

If the aircraft was initially in SL flight the lift must have been equal to the weight and the load factor was Lift / weight = 1

So if we use 1 to represent the initial lift we can use 1.44 to represent the lift in the gust.

So in the gust the load factor = 1.44 / 1 = 1.44


3. In straight & level flight at a speed of 1.3 VS, the lift coefficient, expressed as a percentage of of its maximum CLMAX, would be:

a) 169%
b) 130%
c) 59% (correct answer)
d) 77%

In SL flight at any speed above Vs
Lift = Cl 1/2Rho V squared S

At the stall in SL flight
Lift = ClMax 1/2Rho Vs squared S

Assuming weight is unchanged in the two equations above we can say that the first equation is equal to the second.

If we take out the common factors of Lift and 1/2Rho S we leave the following:

Cl V squared = ClMax Vs squared

Rearranging this gives us

Cl = ClMax Vs squared / V squared)

If we use 1 to represent Vs we can use 1.3 to represent 1.3Vs. Putting these values into the equation gives us
Cl = ClMax x (1 squared / 1.3 squared)

Cl = ClMax x (1 / 1.69)

Cl = 0.59 17 ClMax.

This is approximately 59% of Cl Max

For this type of question just remember the following

Cl at any speed = ClMax x ( Vs squared /Speed chosen squared )

[Transsonic2000]

Keith, once again I must say thanks a lot! Your explanations are really outstanding, it makes things appear easy and simple to understand. I see myself ending up buying one of your books :ok:

[avionimc]

This page is always good: http://aviation.weathersa.co.za/codesexpl.php
but only for locals ???

Please note: Access will be limited to users operating in/from South African Air Space, for aviation purposes only. Although not all the fields are mandatory we require RSA contact details and pilot licence number - add license number in the motivation! If you are a handler or operator for an airline please state this in the motivation

[olav737NG]

Hi,

Could anybody tell me what a ''jammed'' control column or elevator is?

Thanks.

[hvogt]

'Jammed' means stuck, unable to move, e.g. because someone in the cockpit has dropped something that blocks the control column or because of ice accretion on the elevator.

[Haroon]

A flux valve detects the horizontal of the earth’s magnetic field

1) the flux valve is made of a pair of soft iron bars
2) the information can be used by a “flux gate” compass or a directional gyro
3) the flux gate valve signal comes from the error detector
4) the accuracy on the value of the magnetic field indication is less than 0.5%

Which of the following combinations contains all of the correct statements?

a) 2, 4
b) 1, 2, 4
c) 1, 4
d) 1, 2

a) marked correct. Why not (b)?

can some also pls explain the meaning of "the accuracy on the value of the magnetic field indication is less than 0.5%"

this seems pretty inaccurate

thanks

[Lightning Mate]

'b' is incorrect because a flux valve assembly comprises three pairs of cores at 120 degrees to each other.

[Haroon]

thanks lightning mate. I had a feeling it might be that but never thought they'll dig in that deep.

Do you have any idea abut this one:

the accuracy on the value of the magnetic field indication is less than 0.5%


thanks a lot

[Lightning Mate]

I have no idea what that means.

The highest accuracy in any system can only be 100%, so less than 0.5% is very very poor.

I don't know where you found the question, but I taught the JAA syllabus for twenty years, including some time setting and verifying questions in the CQB at Gatwick.

I know of many flawed/invented questions in commercial so-called "databases".

Beware.

[Haroon]

Thanks LM

I know of many flawed/invented questions in commercial so-called "databases".

probably this one is a victim as well unless they mean "inaccuracy"

regards

[outbacktourer]

The question
During a climb at maximum power, the IAS is progressively reduced from the maximum rate of climb speed to the stalling speed.
The effect on the Angle of climb is that it will..

A)Angle of Climb will Increase Continuously
B)Angle of Climb will Decrease Continuously
C)Angle of Climb will Increase then Decrease
D)Angle of Climb will Decrease then Increase

Now what's peoples answer?




I thought considering the stalling IAS with power on would be lower then the actual published Stall speed of the Aircraft the answer would have to be that the Angle would increase continously.. but am I wrong in assuming the question would factor in that the stall speed would be at a lower IAS then published due to power?

[keith williams]

You are missing the point of the question.

Vx is greater than Vs.
Vx is the speed at which angle of climb is best.
Vy is the speed at which rate of climb is best.
Vy is greater than Vx.

So our sped range is

Low speed...Vs.............Vx.............Vy...........High speed

In this question we are decelerating from Vy (at which ROC is best), through Vx (at which angle of climb is best) to Vs (at which we stall).

This will cause the angle of climb to increase (until we reach Vx) then to decrease (until we stall).

[clearblueskyy]

I think the AoA will increase continuously (decreasing rate ) up to the critical angle ,after which it reduces. Hence your answer should be C.

[Rithalic]

I would have thought that C is correct

Angle will Increase and then Decrease.

As you depart from Vy best rate, your angle of climb will increase as you progress towards Vx best angle. Remember that best climb is achievable with maximum excess thrust available. As you increase your AoA beyond that towards the critical AoA your speed reduces towards the stall speed you are progressively increase profile and induced drag and your thrust available is being progressively used up to balance the drag consequently is no longer excess.

Or in more simple terms Vy........Vx.........VS

Best Angle will be achievable at VX, any movement away from it either increase or decrease in speed will reduce your angle of climb.

[outbacktourer]

Okay, so basically think simply and don't read to far into the question.

Cheers ;)

[keith williams]

Okay, so basically think simply and don't read to far into the question.

In many questions that is certainly true.

But you also need to consider whether taking a really deep look will produce anything useful.

For example your statement that:

the stalling IAS with power on would be lower then the actual published Stall speed of the Aircraft


Is true, but this does not mean that:

the answer would have to be that the Angle would increase continously.

The best climb angle would still occur at Vx and Vx would still be between Vy and Vs. So the correct answer would still be "increase then decrease"

The fact that the power on stall speed would be a bit lower than the power off stall speed, would not change the answer. So there is nothing to be gained by spending much time considering this factor.

[Phantomas21]

Hi guys! I need your help asap! I am sitting ATPL exams studying from Obristol + aviation exam + e-atpl + aviationtire database. Although I study extremely hard from all these databases I have failed 3 times in air law. The questions of the exams in the local caa don't belong in any of the databases I study. Here I post some questions I copied in order someone to help me if you have found them in any database:


-ICAO organisation
The body of ICAO that considers and recommends modification to the Annexed of the Convention to the ICAO Council is the:
(answers)


-Internation Private Law
The Convention defining rules of the parties relating to the carriage of passengers, baggage and cargo and the liability of the carriers and extent of compensation for damage is the:
(answers)



-Certificate of airworthiness
Validity and renewal of airworthiness certificates shall be subject of the:
(answers)



-Classification of Airspace
Name the Class of Airspace with the following conditions and services:
IFR and VFR permitted.
Air traffic control service: all flights,
Separation: IFR from IFR
Traffic information: IFR receive traffic information in respect of VFR flights,
VFR receive traffic information in respect of all other flights:
(answers)

-Classification of Airspace
Name the Class of Airspace with the following conditions and services:
IFR and VFR permitted,
Air traffic control service: all flights,
Separation: IFR from IFR, IFR from VFR, VFR from IFR,
Traffic information: in respect of other VFR flights
(answers)


-Classification of Airspace
For an aerodrome in Controlled Airspace Class C
(answers)

[Haroon]

What is the wavelength of an ILS signal?

Centimetric
Hectometric
Metric
Decimetric

they've marked metric as correct but what about the glide slope signal which is decimetric?

[Lightning Mate]

Yet another flawed and invented question.

[Haroon]

Thanks LM.

[Alex Whittingham]

The ILS question was removed from the Central Question Bank about five years ago for just that reason. Have you seen it in an exam?

[Haroon]

its in some databases that are not meant to be updated till an elephant passes through a key hole

[Da-20 monkey]

Phantomas21:

-Classification of Airspace
Name the Class of Airspace with the following conditions and services:
IFR and VFR permitted.
Air traffic control service: all flights,
Separation: IFR from IFR
Traffic information: IFR receive traffic information in respect of VFR flights,
VFR receive traffic information in respect of all other flights:
(answers)


-Classification of Airspace
Name the Class of Airspace with the following conditions and services:
IFR and VFR permitted,
Air traffic control service: all flights,
Separation: IFR from IFR, IFR from VFR, VFR from IFR,
Traffic information: in respect of other VFR flights
(answers)

Controlled airspace can be class A, B, C, D, or E.
The right answer can't be class A, because VFR is not allowed in class A. Class A can be high-altitude airways and some terminal areas around large airports, like Schiphol.

class B airspace is more rare in Europe; it's like class A, but VFR is allowed. So highest security, everything seperated.

Class C airspace is typically a CTR around an airport: The ATC controller must separate all traffic exept VFR from VFR.
VFR traffic is supposed to see and avoid each other: However, all traffic in class C must maintain radio contact and have a flight plan, so the controller can see all traffic on his screen; and can provide traffic information from all existing traffic in his CTR on request.

Class D airspace: CTR's in Germany, for example, are often class D. The difference with class C is that the controller doesn't separate IFR from VFR either, only IFR from IFR.
In this type of CTR the Boeing on the ILS has to keep a lookout for VFR traffic if they are clear of cloud (in VMC).
But the controller can still give traffic info from all existing traffic on request or whenever he wants, depending on his workload. But he is not required to do so.

In Class E airspace, VFR traffic is not required to make radio contact, so a controller can never be sure that he knows where all traffic in his area is. Technically class E is only Controlled A/S for IFR traffic.

I hope this makes it clear that q.1 is class D, and q.2 class C.

[selfin]

Phantomas21,

ICAO document 7300 (http://www.icao.int/publications/pages/doc7300.aspx), a consolidated edition of the Convention on International Civil Aviation available from ICAO, answers your first question. Articles 54 & 57.

The Warsaw system of agreements addresses passenger, baggage and cargo liabilities. The principal instrument is the Convention for the Unification of Certain Rules relating to International Carriage by Air (Warsaw 1929) which has a number of amendments. A modern, but separate, agreement is the Convention for the Unification of Certain Rules for International Carriage by Air (Montreal 1999). You'll find more information on the ICAO Legal Affairs and External Relations Bureau treaty collection page (http://www.icao.int/secretariat/legal/Pages/TreatyCollection.aspx).

ICAO Annex 8, Part II, 3.2 requires that "a Certificate of Airworthiness shall be renewed or shall remain valid, subject to the laws of the State of Registry [...]"

Da-20 Monkey has addressed two of your three airspace classification questions. The summary table for ATS airspace classes (services provided and flight requirements) is in Annex 11, Appendix 4. The text of the ICAO standards is provided in Chapter 2. Be aware that the EASA ATPL air law syllabus mainly deals with ICAO standards, recommended practices and procedures. Contracting States may have established differences in respect of the classification and flight requirements of, and the provision of services in, their airspaces.

The third airspace question is unclear. Perhaps you'd reword it?

Your EASA ATPL theory provider is required to maintain a copy of the ICAO Annexes should you wish to pursue the details further. Additionally, EUROCONTROL's SKYbrary has published an invaluable ICAO SARPS & PANS Search Centre (http://sissy.skybrary.aero/) where you can electronically search the Annexes and other ICAO documents.

[Haroon]

Hi

Q.1. The main advantage of a ratiometer type temperature indicator is that it:

carries out an independent measurement of the supply voltage <-- Marked Correct
can operate without an electrical power supply
is very accurate
is very simple

Q.2. A ratiometer:

Can measure independently supply voltage
Is very accurate <-- Marked Correct
Does not require a power supply
None of the above

Which one is correct?
Thanks

[N1 Limit]

I have this one if any can help:
A large cumulus cloud with a base of 2000 ft is reported at an airfield,if the surface temperature is 21 degrees celsius the height of the freezing level in the cloud is likely to be:

A) 10330 feet

B) 7000 feet

C) 10500 feet

D) 18000 feet

Any one please to help me resolve this one with explanations if possible

Best regards

[Lightning Mate]

Here are a few questions from oxford school regarding instrumenation which,
actually, are not that hard but i prefer to ask for advise before submitting my
assessment.

Don't ask here - submit you answers and the school will respond.

The whole point of such questions is to see if you have correctly assimilated the information in the relevant section(s) of the notes.

You will never pass the ATPL exams by using the "experts" on this site.

[clearblueskyy]

I have this one if any can help:
A large cumulus cloud with a base of 2000 ft is reported at an airfield,if the surface temperature is 21 degrees celsius the height of the freezing level in the cloud is likely to be:

A) 10330 feet

B) 7000 feet

C) 10500 feet

D) 18000 feet




The freezing level in a cloud is calculated by using the the standard lapse rate 2celc./1000'. In your case however, the freezing level (i.e temp reaches 0 from the surface temp of 21 celcius ) at an altitude of 10,500' or 10,600 to be more precise .

The answer in my opinion would be (c) .

[Da-20 monkey]

A large cumulus cloud with a base of 2000 ft is reported at an airfield,if the surface temperature is 21 degrees celsius the height of the freezing level in the cloud is likely to be:

A) 10330 feet

B) 7000 feet

C) 10500 feet

D) 18000 feet

The freezing level in a cloud is calculated by using the the standard lapse rate 2celc./1000'. In your case however, the freezing level (i.e temp reaches 0 from the surface temp of 21 celcius ) at an altitude of 10,500' or 10,600 to be more precise .

The answer in my opinion would be (c) .

If you use the DALR from the surface up to the cloud base (3 deg./ 1000ft)
and from there on 1.8 deg. / 1000ft as the SALR
(only valid in the cloud itself ) , you get 10,330 ft as the FZ LVL.

I believe this is the answer they are looking for, hence A.

Is this the answer they give?

[cyrilroy21]

Hi

I would appreciate some help in explaining the answer to the following question

Q. A pilot wishes to turn right on to a northerly heading with 20 degree bank at a latitude of 40 degree North
Using a Direct reading Compass in order to achieve this he must stop the turn onto an approximate heading of

1) 330
2) 350
3) 030
4) 010

The correct answer is 330

Could someone care to explain as to how this answer is calculated

Thanks in advance

[Lightning Mate]

Your quoted question is not as per the original.

It says:

Turning right through ninety degrees on to north at rate two.......

Mind you, it might now have been changed.

The seemples way to remember it:

Turning through the nearest pole, roll out thirty degrees early.

Turning through the furthest pole, roll out twenty degrees late.

This takes into consideration acceleration/turning error and liquid swirl error.

[cyrilroy21]

Thanks

Does your answer work for the following questions as well


A pilot wishes to turn right on to a southerly heading with 20° bank at a latitude of 20° North. Using a direct reading compass, in order to achieve this he must stop the turn on an approximate heading of :
a) 210°
b) 150°
c) 170°
d) 190°



A pilot wishes to turn left on to a southerly heading with 20° bank at a latitude of 20° North. Using a direct reading compass, in order to achieve this he must stop the turn on an approximate heading of :
a) 160°
b) 200°
c) 170°
d) 190°

[Lightning Mate]

The answer to your first question is incorrtect - the second correct.

My method answers any questions - trust me - I have been teaching this stuff for twenty years.

BEWARE commercial "real questions".

[cyrilroy21]

Thanks

I am not doubting your methods , infact you explained it perfectly :D and I understand that commercial question banks have quite few errors in them as well

But there some non EASA CAA's around the world that follow the EASA question bank word for word and naturally incorporate these errors in their question banks as well :rolleyes:

[cyrilroy21]

Help Please

Q. A turn indicator is built around gyroscope with ?

1) 1 Degree of Freedom
2) 3 Degrees of Freedom
3) 2 Degrees of Freedom
4) 0 Degree of Freedom


The correct answer should be 1 Degree of Freedom . This is also given and explained in the Keith Williams questions banks as well

However in two questions banks that I have seen Professional Volare and JAR Databank Presentation the correct answer is given 2 Degrees of Freedom

If you were to see this question for an airline written exam what would you choose ?

[paco]

First of all, Volare was discredited years ago. Having said that, there are many questions that are just plain wrong, even now. If you feel that they are looking for a wrong answer, or one different from what it should be, then query the question at exam time - only don't mention any databases! :)

Here is what it should be:

An older method of classifying gyroscopes uses the
numbers of axes not including the spin axis, so one with 3
planes of freedom has 2 degrees of freedom. A degree of
freedom is the ability to move around an axis - for
example, a fuselage can pitch, roll and yaw. A turn
indicator has 1 degree of freedom. An airborne
instrument, with a gyro that has 2 degrees of freedom and
a horizontal spin axis could be a DGI.

Nudge, nudge, wink, wink ;)

[cyrilroy21]

@paco

Thanks

I re read the question again and I think the question is asking about the gyroscope used rather than the turn indicator itself

Here is an answer I got courtesy of the theflyingengineer.com

Definition: "A turn indicator is based on the ability of a gyroscope with two degrees of freedom to superpose the angular velocity vector of the gyroscopic rotor’s own spin and the angular velocity vector of the rotation of the device’s case.". The rotation of the gyro about its own axis constitutes one degree of freedom. And the rotation about the longitudinal axis is the second.

And while we are at it , can someone please tell me what the Volare Question bank is about ?
Is it some official JAA question bank ? Does it use CQB 14 or 15 ?

[Lightning Mate]

Is it some official JAA question bank ?

There's only one of those, and it sits at Gatwick.

The computer is stand-alone, cannot be accessed from outside, and is multiple password protected. I know - I've worked on it!

edit to say paco is correct - the turn indicator gyro has one degree of freedom.

[keith williams]

A quick GOOGLE search revealed the following material on the subject:

electriciantraining.tpub.com/14187/css/14187_139.htm
A gyro can have different degrees of freedom, depending on the number of gimbals in which it is supported and the way the gimbals are arranged. Do not confuse the term "degrees of freedom" with an angular value such as degrees of a circle. The term, as it applies to gyros, is an indication of the number of axes about which the rotor is free to precess. A gyro mounted in two gimbals has two degrees of freedom.

When two gimbals are used, the gyro is said to be UNIVERSALLY MOUNTED. This arrangement provides two axes about which the gyro can precess. These two axes and the spin axis intersect at the center of gravity of the entire system (excluding the support). Because of this arrangement, the force of gravity does not exert a torque to cause precession. The rotor, inner gimbal, and outer gimbal are balanced about the three principal axes.

Elmer A. Sperry - Degrees of Freedom (http://www.fi.edu/learn/case-files/sperry-2524/freedom.html)
A gyro has two basic mountings: one with three degrees of freedom, and one with only two degrees of freedom. A gyro having three degrees of freedom is able to rotate about three axes including its own axis of rotation. Foucault used such a model to illustrate the rotation of the earth, crafting a model with a gyro mounted in two rings. One ring rotated on a horizontal axis within the other, which in turn rotated on a vertical axis.

www.navymars.org/national/training/nmo_courses/nmoc/module15/14187_ch3.pdf
DEGREES OF FREEDOM
A gyro can have different degrees of freedom, depending on the number of gimbals in which it is supported and the way the gimbals are arranged. Do not confuse the term "degrees of freedom" with an angular value such as degrees of a circle. The term, as it applies to gyros, is an indication of the number of axes about which the rotor is free to precess.

A gyro mounted in two gimbals has two degrees of freedom. When two gimbals are used, the gyro is said to be UNIVERSALLY MOUNTED. This arrangement provides two axes about which the gyro can precess. These two axes and the spin axis intersect at the center of gravity of the entire system (excluding the support). Because of this arrangement, the force of gravity does not exert a torque to cause precession. The rotor, inner gimbal, and outer gimbal are balanced about the three principal axes.

TWO DEGREES-OF-FREEDOM GYROS
The two-degrees-of-freedom (free) gyros can be divided into two groups. In the first group, the gyro's spin axis is perpendicular to the surface of the Earth. Thus the gyro's rotor will spin in a horizontal plane. These gyros are used to establish vertical and horizontal planes to be used where stabilized reference planes are needed.

In the second group, the gyro's spin axis is either parallel to the surface of the Earth or at some angle other than perpendicular. The spin axis of the gyro in the gyrocompass, for example, is maintained in a plane parallel to the surface of the Earth It is aligned in a plane of the north-south meridian. Once set, it will continue to point north as long as no disturbing force causes it to precess out of the plane of the meridian.


en.wikipedia.org/wiki/Gyroscope
Within mechanical systems or devices, a conventional gyroscope is a mechanism comprising a rotor journaled to spin about one axis, the journals of the rotor being mounted in an inner gimbal or ring; the inner gimbal is journaled for oscillation in an outer gimbal for a total of two gimbals.
The outer gimbal or ring, which is the gyroscope frame, is mounted so as to pivot about an axis in its own plane determined by the support. This outer gimbal possesses one degree of rotational freedom and its axis possesses none. The next inner gimbal is mounted in the gyroscope frame (outer gimbal) so as to pivot about an axis in its own plane that is always perpendicular to the pivotal axis of the gyroscope frame (outer gimbal). This inner gimbal has two degrees of rotational freedom.

The axle of the spinning wheel defines the spin axis. The rotor is journaled to spin about an axis, which is always perpendicular to the axis of the inner gimbal. So the rotor possesses three degrees of rotational freedom and its axis possesses two. The wheel responds to a force applied about the input axis by a reaction force about the output axis.

en.wikipedia.org/wiki/Turn_and_balance_indicator
Turn indicator
The turn indicator is a gyroscope instrument that works on the principle of precession. The gyro is mounted in a gimbal. The gyro's rotational axis is in-line with the lateral (pitch) axis of the aircraft, while the gimbal has limited freedom around the longitudinal (roll) axis of the aircraft.


MY COMMENTS
The existence of the spin axis itself does not provide a degree of freedom because the spin axis cannot precess. But each gimbal enables the spin axis to precess in one additional direction. So the number of degrees of freedom is equal to the number of gimbals.

[Transsonic2000]

Hi all,

would need some help with the following question:

(Flight Planning MJRT refer to Fig. 4.5.3.1)
Given: flight time from TOC to enrout point in FL280 is 48 min. Cruise procedure is long range (LRC).
Temp. ISA -5° C
TOM 56.000 kg
Climb fuel 1.100 kg
Find: distance in nautical air miles (NAM) for this leg and fuel consumption.

Answer would be: 345 NM / 2000 kgHow do I get this figured out and which charts, beside the one mentioned (I used the one on page 26, Section 4, MJRT - 28.000 ft PA) do I have to refer to get the answer? Thanks in advance!

[keith williams]

Subtracting 1100 kg climb fuel from the 56000 kg take-off mass gives a mass of 56000 kg – 1100 kg = 54900 kg at the start of the cruise.


In Figure 4.5.3.1 the second cells in the 55000 kg row indicates that the TAS = 437 kts.

Note 2D at the base of the page states “decrease TAS by 1 kt for each 1 deg C below ISA.

So for the existing ISA deviation of -5 deg C the required correction is -5 kts.

So the corrected TAS = 437 – 5 = 432 kts.

The question states that the leg time is 48 min.

In 48 minutes at 432 kts TAS the distance flown = 432 x 48 / 60 = 345.6 NAM.

In Figure 4.5.3.1 the cell at the intersection of the 54000 kg row and the 900 kg column (for the mass of 54900 kg) indicates an initial cruise range of 3736 NAM.

Subtracting the 345.6 NAM flown in this leg gives 3736 – 345.6 = 3390.4 NAM remaining.

The closest figure to this in the table is 3395, which appears at the intersection of the 52000 kg row and the 900 kg column. This indicates a final mass of 52900 kg.

Subtracting this final mass from the initial mass gives a mass change of (54900 – 52900) = 2000 kg caused by the burning of fuel.

Note 2B at the base of the page states “decrease fuel by 0.6% for each 10 deg C below ISA.

So for the existing -5 deg C ISA deviation, the fuel must be decreased by 0.6% x 5/10 = 0.3%.

0.3% of 2000 kg = 6 kg. So the corrected fuel = 2000 kg – 6 kg = 1994 kg.

So the fuel consumed in the cruise is 1994 kg.

So for the cruise leg of this flight the distance flown is approximately 345 Nm and the fuel burned is approximately 1994 kg.

[Transsonic2000]

Hi Keith,

once again, thanks a lot! As always, very helpful and detailed explanation, great job, it's very much appreciated :ok:

[cyrilroy21]

Hey guys help with this question please

The approximate range of a 10 KW NDB over the sea is

a) 500nm
b) 1000nm
c) 50nm
d) 100nm

Option d 100nm is supposed to be the correct answer

I used the formula 3 (over the sea ) X Sqrt ( 10,000 Watts ) = 300nm

Anybody know how 100nm is the supposed to be the correct answer ?

[Lightning Mate]

The range of an NDB depends to a large extent on the frequency used.

The ICAO band allocated is 190 kHz to 1750 kHz.

Without the frequency stated the question is impossible to answer and is yet another "invented" question I suspect.

The original CAA/JAA question was not the one as stated.

[Romeo Kilo]

Doing some Aircraft Performance revision I've come across a question in my school's 'exam preparation' questions (so not necesserily previous exam questions, just revision tools):

"What is the effect on V1 if mass is increased?"

The answer given is that it increases. Indeed, looking at the MRJT in CAP698 shows an increase in V1 with mass. This is a general theory question, however.

I don't understand how this can be a hard and fast rule - Will there not be situations where an increase in mass will slow the acceleration and deceleration such that ASDA will be insufficient should V1 be increased or even left unchanged after adding mass? I'm thinking, in this case, a lower V1 would be necessary in order to guarantee a stop within ASDA should it be required. Obviously, this requires better OEI performance, as we will have to accelerate further on only one engine, but if we know we can still reach screen height at V2, what's the issue?

I'm just thinking about taking off from a short runway, with a clearway, with a FLLTOM dictated by ASDA, and we increase the a/c mass up to the FLLTOM. Would V1 really go up?

Probably a question for my instructor, but as its a holiday weekend and this was on my mind, I thought I might see if some other ppruner was feeling simlilarly pensive...

[Romeo Kilo]

So hopefully getting towards answering my own question....

In the case above, before adding mass, we have excess performance in terms of reaching screen height OEI, right? So would we have set V1 at the minimum speed that we can be assured to reach screen height/V2 after engine failure, rather than the maximum speed we can be assured of stopping within ASDA?

If the former is the case, I can see how an increase in mass would require an increase in V1. So there will be some speed range beyond V1 where in theory the aircraft will come to a stop within the (factored) ASDA, but we ignore this, because as long as we hit screen height on speed, we're happy? I suppose this fits in with a 'go-minded' philosophy..

Am I on the right lines?

[keith williams]

You are correct in saying that if the aircraft is ASDA limited you cannot increase V1 to accommodate an increase in mass. This is because the increase in mass and the increase in V1 would both increase the ASD.

But if the aircraft is limited by its ability to get to V2 and screen height within the TODA, then increasing V1 will reduce the distance to reach screen height.

In using the CAP 698 you need to take care to read the small print. Para 2.1 in Section 4 MRJT1 page 7, states that "The field length used in the graph is based on minimum V1 being equal to VMCG". This is true throughout the CAP698, so the speeds in tables 4.8 and 4.9 are also based on minimum V1.

As a general rule for the exams, unless the question contains some additional restrictions such as "the aircraft is climb limited", then the effects listed in the CAP are the ones that they are looking for in the questions.

The thread below contains material on this subject.

http://www.pprune.org/professional-pilot-training-includes-ground-studies/487179-balanced-v1-lower-than-vmcg.html?highlight=increasing+V1

[Romeo Kilo]

Thanks Keith. I shall have a read of the link.

Would your assessment be a fair representation of real life? That is to say the aim will tend to be to use as low a V1 as possible whilst guaranteeing screen height and directional control, despite the fact that ASDA may permit an abort at higher speed? (Before we start worrying about/considering reduced/debated thrust for the sake of engine life).

Thanks again.

[ahmedsalman]

hi this question cropped up in last mass & balance exam Given a crate at 1 ft 9 inches X 1 ft 7 inches X 1 ft 2 inches, what's the maximum load in it if RL is ____and load intensity is _____, what the method to tackle this question thanks sorry no figures i copied it of atp forum the answer was 257kg

[keith williams]

You have not provided sufficient information to solve the problem. But the two examples below illustrate the general method for solving these types of problems.

Example 1.
An aircraft has a floor load limit of 4500 N/m squared. What is the maximum mass that can be loaded in this hold on a 1.4m x 0.4m pallet? (g = 9.81 m/s/s).

(a) 464 Kg.
(b) 122 Kg.
(c) 177 Kg.
(d) 256 Kg.


The floor loading limit is 4500 n/m squared and the floor area of the pallet is 1.4m x 0.4m = 0.56m squared.

Floor loading = load / area.

So load required to exert a given floor loading = floor loading x area.

This means that the pallet load required to exert the limiting floor loading of 4500 N/m squared, is equal to 4500 N/m squared x 0.56 m squared = 2520 N.

The force exerted by a mass = F = mass x g
Rearranging this equation gives mass = F/g

Using g = 9.81, the mass required on the pallet must be 2520 N/9.81 = 256.9 Kg.



Example 2.
What is the maximum mass that can be loaded onto a 0.8m square 25 Kg pallet, if the running load limit is 800 Kg/m?

(a) 644 Kg.
(b) 615 Kg.
(c) 475 Kg.
(d) 519 Kg.

The running load exerted by a load is equal to the total mass of the load divided by the length of its base.

In order to obtain the maximum allowable mass, it is necessary to orient the pallet such that its longest possible base length is used.

If the mass loaded on the pallet = M, then the total mass is the load (M Kg) plus the pallet mass (25 Kg) which gives a total mass of (M + 25) Kg.

Both of the edges of the pallet are 0.8 m long.
So the running load exerted by the loaded pallet is:

Running Load = Total mass/ base length = (M + 25) Kg / 0.8m

But the maximum allowable running load in this question is 800 Kg/m, so the maximum mass that can be loaded on the pallet is that which gives the condition:

(M + 25) Kg / 0.8 m = 800Kg / m.

This equation can be simplified by multiplying both sides by 0.8 m, to give:

(M + 25) Kg = (0.8 x 800) Kg Which is 640 Kg

Subtracting 25 Kg from each side gives M = 615 Kg
So the maximum mass that can be loaded on the pallet is 615 Kg.


General factors to remember:

1. The static load intensity for any given load will be a minimum when the load is sitting on its largest side. So to maximize the permitted load, it must be placed on its largest side.

2. The running load imposed by any given load will be minimum when the load is moving along its longest side. So to maximize the permitted load, it must be moved along its longest side.

[ahmedsalman]

hi thanks for the reply this is a valid question any help will be appreciated.
what is the maxium load that can be placed in a baggage compartment with a running load of 75 kg/inch and a load intensity of 300 kg/sq ft if a package dimensions of 18x24x15 inches.
A.350kg
B.900kg
C.1200kg
D.1800kg

[keith williams]

You will learning nothing from simply having me or any other member giving you the answer. You must learn to do the calculations for yourself.

The examples in my previous post illustrate how to answer this type of question.

You need to do one calculation for the static load and another for the running load.

For the maximum load that does not exceed the static load limit use:

Max load = Static load limit x floor area of load

For the maximum load that does not exceed the running load limit use:

Max load = Running load limit x longest side of of the load.

Note that the running load limit is in kg/inch and the static load limit is in kg/sq ft. So for the static load limit you need to convert inches into feet (there are 12 inches in a foot).

The maximum load is the smaller of the two figures that you will derive from these calculations.

If you then post your step-by-step solutions I or someone else will comment on them.

[squall1984]

Hey guys,

Ive been trying to find a place where I can see the number of questions / marks per exam for the ATPLs.

Anyone written them recently for Principles, Performance, Gen Nav and Flight Planning would be super helpful.

Thanks :)

p.s. see everyone in FL next week.

[Matt7504]

I recently finished my ATPLs and still have access to the BGS bank, for Performance it is out of 35 marks, Flight Planning is out of 43 marks, Gnav is 60 marks and POF is 44 marks. Bare in mind that is marks not questions, as each question on the bank counts for one mark I think. If I remember rightly there were a few questions in my Flight Planning and Gnav exams that counted for 2 marks, so this would mean you may have less questions to do than the amount of marks I have posted.

[squall1984]

Well I did the performance exam today and all I can say is WTF...I did Bristol and a few question banks and I would say at least 10 questions weren't in the banks. Anyone have the same experience or is it just me that's stupid? :ugh:

[keith williams]

The BGS website contains feedback from 5 students who took this exam. One thought that 90% was new, but the others thought it was not a problem exam. The author of the most detailed post stated that the new questions were simply slight modifications of old questions.

It is true that the examiners are currently introducing new questions at a greater rate than was previously the case. This obviously increases the element of risk for candidates who have simply memorized the answers from the various question banks. But there is nothing unfair or unreasonable about introducing new questions. From the examiners’ point of view the ideal exam paper is one in which all of the questions are new to the candidates.

To maximize their chances of passing the exams, candidates must study the course material then use the question banks to develop their ability to recognize and solve the various problem types. The alternative method of simply memorizing answers from question banks requires less effort and has often been successful in the past. But it greatly increases the risk of “getting unlucky” in the new exams.

[P40Warhawk]

Well, I am busy with preparing for my last two Exams.
Meteo and GNAV.

I had feedback from classmates who have done MET, and said there are not many surprises to expect in the exams of new questions.
But in GNAV there are.
Since there are many calculations in the exams that will take a lot of time, what are the focus point to work on? What to expect? I passed all my exams in first attempt till now , and I would like to keep it that way.
Of course I am working also with books, to have accurate knowledge, not only QB work.
Anyone hints?

[paco]

The questions have been changed and checked for accuracy in Performance, and there is a greater spread between the answers. As Keith says, slight modifications to previous ones. Not before time.

As for Nav tips, remember that it is possible to pass without the convergency stuff (at 1 mark per question) but you won't pass without whizzwheel skills.

[Luvmathur]

Hi

I know has been a while. Would you remember that when you mentioned "drift left - T alt increases and drift right - T alt decreases" was it applicable only to the NH or both hemispheres? Will appreciate you're help.

Thanks & Cheers

[P40Warhawk]

Thank you so much. Well my skills with the CRP5 are pretty good. No problem for me.

Really those Grid Navi questions are a problem in general. I have to revise Polar Stereographic Nav and convergence, but in general that wont be a problem.

Anything else I have to look at?

Thanks again Paco.

[paco]

There are only a few grid nav questions.

One tip - if you're really stuck and are given a selection of four diagrams, look for two that are similar and reject the one that is wildly off.

[keith williams]

Luvmathur

Would you remember that when you mentioned "drift left - T alt increases and drift right - T alt decreases" was it applicable only to the NH or both hemispheres

Go back to basics and work it out for yourself.

At the Equator the surface of the Earth moves to the east at about 1000 knots. In still air the air moves at the same speed as the surface.

If we move north or south of the Equator the circumference of the Earth decreases so the speed at which the surface moves east also decreases. But if air is moved north or south it tends to retain its eastward speed. This means that the air is now moving east faster than the surface. The overall effect is that the air tends to turn to the east as it moves north or south of the Equator.

If we have a low pressure area in the north the air is drawn towards it. The eastward movement of the air causes it to rotate anticlockwise around the low pressure area.

If we have a low pressure area in the south the air is drawn towards it. The eastward movement of the air causes it to rotate clockwise around the low pressure area.

So the effect of drift in the southern hemisphere is the opposite of that in the northern hemisphere.

[galaxy flyer]

And pressure pattern questions are relevant how?

[keith williams]

Galaxy Flyer

Much of the material in the EASA ATPL syllabus is irrelevant, but students will gain nothing from wasting their energies in getting angry about this fact. The syllabus is what it is.

Whatever the arguments regarding relevance, it is far better to try to understand the fundamentals of any subject under study, rather than simply memorizing the answers to questions.

[Runway37]

Hi All,

For some reason, I always find this particularly IMC style question relatively tricky. Even when I have been given the answer, I find it difficult to reverse engineer the question to try and find that particular answer.

Can someone please throw some light on this type of question?

Question)

During a descent, the static vent becomes blocked, how would this effect the Altimeter, VSI and ASI?

Altimeter | VSI | ASI
1. Read Normal | Read Normal | Under Read
2. Read Normal | Read Normal | Over Read
3. Under Read | Remain Static | Over Read
4. Over Read | Over Read | Under Read

The correct answer, apparantly is 1...

As to my understanding, Vertical Speed Indicators read differential pressures changes from the static system, so if this is blocked, wouldn't this remain static (i.e No Pressure Changes measured), And also wouldn't the Altimeter Over read, (i.e the instrument still reads the less dense atmosphere of when the instrument was blocked)?

Any advice on this question would be greatly appreciated!

[mad_jock]

Wel if the staic pressure stayed the same.

1. Alt would remian at the same level.

2. Rate of decent would go to zero.

And Airspeed indicator would go through the roof as the static stayed at a constant pressure while the dynamic increased as the air density increased.

So I would say they are all a load of :mad: ****e answers.

[B737900er]

I would appeal the question! The possible answers given are crap, as Mad Jock has said

[laz219]

I always find these questions frustrating just because of the way they are written.

To work it out though, I find it a lot easier to just think of it as where the air was trapped, and therefore it's pressure relative to the new height to work it out, rather than just directly about the instrument readings.

[paco]

For the ASI - just remember the Peru accident where the engineers taped over the static ports on a 737 to give the machine a wash and forgot to take it off again. The machine stalled out of the air because the asi readings kept increasing and they kept pulling the control column back.

[Runway37]

I'm glad I'm not the only one thinking it maybe wrong.

I cant how pressure instruments relying on the static source + pitot pressure can 'read normal' when the static pressure is blocked or incorrect.

I find it impossible for the ASI to read normal with a locked in higher altitude pressure.

For the question database, I am using PPLQuiz. I thought I was nearly ready to take the exam, but after a lower than pass score on pplquiz, it makes me think there are numerous errors in their IMC question database. (I have been getting top marks in Air Pilot's Manual (Volume 5)).

[Ralis]

The way I got taught and how I remember ASI errors in climb and descent regarding pitot and static blockages is the 'COPS DUO' way
..PS
COU Climb pitot blocked over read-- Climb Static blocked under read
DUO Descent pitot blocked under read-- Descent Static blocked over read

VSI should return to 0 FPM (Full blockage)
ALT should indicate the level that the blockage occurred

[RTN11]

Something very odd here, the answers given sound more like a problem with the Pitot supply.

[keith williams]

The analysis and conclusions drawn by Mad Jock are correct for a blocked static port.

As RTN11 has said, option 1 would be correct if the question had specified a blocked Pitot source.

Methods like "COPS DUO" are OK provided simply want to pass exams and have no wish to understand anything at all about how your aircraft and its systems function.

[Ralis]

Methods like "COPS DUO" are OK provided simply want to pass exams and have no wish to understand anything at all about how your aircraft and its systems function.
Its a good efin place to start! knowing what the errors are and and relating that to how the instrument and system work can make it easier to understand for some people.
Is that an OK response :mad:

[squall1984]

I wasn't expecting the performance paper to be the same as the question banks by usually you can apply to ethos from the question bank to get the answer, some of the questions I didn't see any method to, for Istanance figuring out a airspeed such. Maximum endurance from the maximum range which was given, that may not be the actual question but it was like that for two or three of them. I did the oxford course for performance and didn't see it in there, will see the result and so it again I suppose.

As for gen nav I didn't see any major surprises in May, was pretty much typical questions you'd find in question banks, I'd say most where word for word from question bank. Unite a few calculations in ground speed and drift...at least 5 or 6 of those.

[squall1984]

Either in gen nav or fling planning(think gen nav) there was a calculation question involve RAS which wasn't in the banka

[Runway37]

In regards to RBI questions,

Does anybody have any tips or tricks to work them out?

For the questions I have read, it's almost like you have to reverse engineer the RBI method. e.g,

Q. Your are flying inbound to a VOR on a magnetic heading of 156 degrees M, the aircraft maintains the VOR QDM 150, which is located to the left of the NDB. What will the RBI indicate when the aircraft is abeam the NDB?

I understand it has to be slightly more than 090M, due to the angular displacement when abeam. But how would one work this out, as they give 094 & 096 as potential answers.

Many Thanks,

[hvogt]

This is another one of the many questions which are phrased very poorly. From the information the VOR or the "VOR QDM 150" or whatever was located "to the left of the NDB", the NDB could be virtually anywhere. I think the key aspect of the question is to understand you are following radial 330 indound to the VOR, i.e. your course is 150 degrees magnetic. If your magnetic heading is 156 degrees, your nose is obviously pointing 6 degrees to the right of your course line. With a little interpretation, the question seems to imply (1.) you were abeam the NDB when passing the VOR and (2.) the NDB was on the right side of your course. If your nose was pointing in the direction of your course (150 degrees magnetic) when passing the station, your RB to the NDB would be 90 degrees. Since your nose is 6 degrees further to the right, your RB will be 90-6=84 degrees.

By the way, a relative bearing is never magnetic, true or otherwise. It's just relative, i.e. relative to the aircraft's longitudinal axis.

[Runway37]

Apparently, the correct answer is 94 degrees.

I interpret the 'abeam' point as when you are at a 90 degree angle from the station, rather than 90 degree angle from the aircraft to the station.

For them to get 094 degrees, abeam must mean when the aircraft is at 90 degrees to the station, i.e have gone past the NDB, when your nose is pointed slightly towards it, having to go further is attain a 90 degree angle.

I would imagine, if the nose is pointed slightly away from the NDB, the abeam point would be reached before you actually get to the 90 degree angle from the station.

Very confusing.

[Okavango]

I'm probably going to let my ATPL's lapse passing the 3yr limit for the MEIR as happily flying in GA for the time being and the situation for progressing doesn't look good. Has there been any major change of syllabus in the ATPL theoreticals over the last few years or is anything planned? Got 98% average last time so not too concerned about repeating with a bit of revision and question bank.

[paco]

How many ways can you describe an altimeter? :)

Apart from a review of the LOs later this year and possibly action taken in 2014, I don't see much on the horizon for at least the next 2-3 years.

[cumbrianboy]

They rewrote the syllabus about 2 years ago but I think it was moving various topics between the various subjects.

I finished mine last month and although I never relied that much on question banks, those that did/do will be in for a shock as they're writing a lot of new questions (which is good, I do think it's an outrage you can go into an exam and have already seen the exact question and been told the correct answer!)

[flyme19]

Hi everyone. Just looking for abit of information regarding the question banks. I have my 1st ATPL sittings not too far away and was wondering how alike the questions on there are compared to what I will be actually faced with?Any info would be great!


Thanks

[flyB_777]

can some one solve the following one with some explaination

As the INS position of the departure aerodrome, coordinates 35o32.7’N 139 o 46.3’W are input instead of 35 o 32.7’N 139 o 46.3’E. When the aircraft subsequently passes point 52 o N 180 o W, the longitude value shown on the INS will be?
a. 099 o 32.6’E.
b. 099 o 32.6’W.
c. 199 o 32.6’W.
d. 299 o 32.6’W.

[flyB_777]

can some solve this please

As the INS position of the departure aerodrome, coordinates 35o32.7’N 139 o 46.3’W are input instead of 35 o 32.7’N 139 o 46.3’E. When the aircraft subsequently passes point 52 o N 180 o W, the longitude value shown on the INS will be?
a. 099 o 32.6’E.
b. 099 o 32.6’W.
c. 199 o 32.6’W.
d. 299 o 32.6’W.

[paco]

Some subjects are still similar, but the UK CAA have changed 3500 questions already and are steaming ahead to change them all. Performance is definitely not to be relied on any more.

[slawek_s]

Latitude is unimportant in that type of questions.

To get from real position of 139o 46.3’E to the real position of 180o W (it is the same position as 180o E - around International Date Line) one must fly 40o 13.7' east or (which makes calculations more complicated but still possible and valid) 319o 46.3' west.

Let's assume one flies east. INS thinks that one starts at 139 o 46.3’W and flies 40o 13.7' east. Therefore INS calculates: 139o 46.3'W - 40o13.7' = 99o 32.6'W

If one flies west - INS thinks that plane starts at 139 o 46.3’W and flies 319o 46.3' west. Therefore INS calculates: 139o 46.3'W + 319o 46.3' = 459o 32.6'W = (-360oW around the globe) 99o 32.6'W.

[flyB_777]

slawek_s thanks a lot !!i may post some ins numericals in this post.hope you would help in solving them too

[P40Warhawk]

Well about this type of questions.
I use BGS online and atplonline.
In explanation in BGS online it says that I should appeal these questions, since it is actually part of Instruments exams. And not of GNAV.

So : Do these type of questions appear in the official CAA Exams? Because I will have my GNAV exam in 2,5 week and I dont want to waste time on things which will not be asked during exams.

Maybe you know it Paco? If you read this comment?

[P40Warhawk]

Well about this type of questions.
I use BGS online and atplonline.
In explanation in BGS online it says that I should appeal these questions, since it is actually part of Instruments exams. And not of GNAV.

So : Do these type of questions appear in the official CAA Exams? Because I will have my GNAV exam in 2,5 week and I dont want to waste time on things which will not be asked during exams.

Maybe you know it Paco? If you read this comment?

[flyB_777]

i dont have much idea about CAA exams i am actually preparing for Indian DGCA ATPL and these are included.

ok..all the best with your exams

[P40Warhawk]

Ah Thnx. Yeah I have only GNAV left. Passed the others all first attempt. But I feel pretty not happy about the fact that lot of questions are changed.
You know. Its the fact that you cannot prepare properly with QB. You should not only repeat QB and learing questions and answers by heart, but the thing is, that the questions are written in bad English. So its hard to understand what they try to ask.
How many exams did you have?

[flyB_777]

oh thats good and got one to go all the best...:ok:
I ve three written gnav,radios and met followed by two orals on gnav and radios.

[P40Warhawk]

GL Then ;) .

Oral on GNAV? Never heard of that.

[paco]

You might get INS questions in Nav - and remember that the Nav exam has been traditionally a dustbin for all sorts of stuff that doesn't fit into other subjects - we tell our guys to expect the unexpected! :) Sometimes you might get a question that simply asks you to move figures around a formula.

One tip if you get a mention of an INS in Nav is to realise they are really talking about great circles.

[P40Warhawk]

Thank you so much Paco . Very helpfull info :D .
I will work on this stuff.

[flyB_777]

great info paco thanks a lot:)

[Capt.Abhishek.S]

Can anyone please solve this question for me

Flying at FL270, an AWR weather return at range 40nm is identifiable when the centre of the beam is tilted between +1/-1deg & -6deg. What is the height of the base & top of the rain bearing cloud?

[keith williams]

Are you sure that you are quoting the correct data?

This type of question usually includes a tilt up angle, a tilt down angle and a beam width. You can then use the range and the 1 in 60 rule to work out the solution.

But your quoted figures of between +1/-1deg & -6deg do not make any sense.

[MatteoLIPH]

Hi, i'm a bit confused as on my CQB some questions of Air Law are out of date i suppose:

1) The VMC minima for an airspace classified as "B" above 10 000 feet MSL are: 8 KM CLEAR OF CLOUDS

Seems that now ICAO regulations are 8km 1.5 vis and 1000ft as according to jar ops..?

2)The protection areas associated with instrument approach procedures are determined with the assumption that turns are performed at a bank angle of:
SIGNED CORRECT:
25° or the bank angle giving a 3°/s turn rate, whichever is lower, for departure, approach or missed approach instrument procedures, as well as circling-to-land (with or without prescribed flight tracks).

What about this one??should be 20° for circling and 15° for missed app?

3)
According to JAR-FCL, Class 2 medical certificate for private pilots will be valid for SIGNED CORRECT:
60 months until age of 30, 24 months until age of 50, 12 months until age of 65 and 6 months thereafter

Shouldn't it be 60 until 40, 24 until 50 and 12 thereafter..?

4)
Runway edge lights excepted in the case of a displaced threshold shall be: SIGNED CORRECT:

Fixed variable white.

Isn't fixed variable white or yellow?

Thank you

[aditya104]

The Oxford ATPL book of Principles of Flight has references that I need some help with.

For e.g. On Page 174; Chapter-Stalling
It talks about EASA 25.103(b) on a few occasions.

Now, where can I look up this regulation?

[hvogt]

It talks about EASA 25.103(b) on a few occasions.

Now, where can I look up this regulation?They are referring to the Certification Specifications for Large Aeroplanes CS-25. It would have been more appropriate to write "CS 25.103(b)". A copy can be downloaded from EASA's website, under Aviation Professionals//Legislation/Cetification Specifications if I remember correctly.

[2close]

ICAO Annex III - Meteorological Service for International Air Navigation, Edition 15, Appendix V (Page 107)

TX25/13Z TN09/05Z
TX05/12Z TNM02/03Z

Maximum temperature 25°C at 1300Z Minimum temerature 09°C at 0500Z
Maximum temperature 05°C at 1200Z Minimum temerature -02°C at 0300Z

:)

[aditya104]

Thanks hvogt. Found it (http://easa.europa.eu/agency-measures/certification-specifications.php#CS-25).

[Low-bypass]

Hi guys
I'm currently doing self-studiesfor the ATP preparation and need some helps from anyone who can give me hints to solve out this type of question:

An aircraft leaves position S01 20 W012 15 on a track of 360 degrees true for 01:06 mins.The heading is then altered to make good track of 090 degrees true for 03:12 mins to eestination.If the TAS is 320 kts and the wind is calm throughout the flight,What is the lat & long of the of the destination?

I just need the methodology to lead on how to solve the question

Many thanks in advance

NEXT TIME please post in the correct thread

HWB

[paco]

320 kts for 1.1 hours is 352 nm. That's 352 minutes or 5 degrees 52 mins so you end up at the end of the Northern sector at N4 32, then turn right. The problem is now that 1 minute no longer represents 1 nm.......

3 hrs 12 mins at 320 kts is 1024 nm.

See if you can get the rest? Hint: you need to look at departure :)

[P40Warhawk]

From there its the Cos of Latitude time NM.

[2close]

I take it that you are not self-studying for an EASA licence........because you can't!

The studying must be done through an Approved Training Organisation (ATO).

However, I've written out a painstaking, line by line solution below, for you to check your answer against.

As previously stated, 5°52' due north from S01°20' W012°15' lands you at N04°32' W012°15'.

To calculate the change of longitude from this position you need to use the "departure" formula:

Departure (NM) = Change of Longitude (mins) x Cosine of Latitude

The Departure (or distance) = 320 kts x 3.2 hours = 1,024 NM

1,024 NM = Change of Longitude (mins) x Cosine 04°32'

1,024 NM = Change of Longitude (mins) x 0.997 (Transpose formula)

Change of Longitude (mins) = 1,024 / 0.997

Change of Longitude (mins) = 1,027'

Change of Longitude = 1,027'/60' = 17.12°

Change of Longitude = 17°07'

To find out the Easterly longitude, subtract 012°15' from 17°07'

17°07' - 012°15' = E004°52'

Final Position = N04°32' E004°52'

Hope that helps. :)

[Low-bypass]

Thanks very much guys i got the picture,it's so great

[squall1984]

Hey guys,

Can anyone send me some examples with the answers regarding Drift Down, Im doing Performance second attempt and there were 2 questions on drift down but I havent seen any in the banks (using easaatp.com bristol and oxford). The driftdown questions were calculations, not just memorization.

I'm also finding different answers for calculating speeds from holding speeds and stall speeds, for prop and jet, any ideas?

Thanks

[keith williams]

Given the following information determine the greatest distance from an obstacle, at the failure of one engine would cause an MRJT1 aircraft to clear the obstacle by the statutory minimum during the drift-down, the fuel used during the drift-down, and time taken to reach the obstacle.

Cruise Pressure Altitude 34,000 ft.
Gross Weight at Engine Failure 54,000 kg.
Ambient Temperature ISA +15°C.
Wind Component 40 kt tailwind.
Engine and Wing Anti-icing System on.
Air Conditioning System on throughout the drift-down.
Obstacle Pressure Altitude 14,000 ft.
One engine inoperative.

1. Select the appropriate graph for the given cruise altitude. For 34000 ft this is figure 4.25, which covers altitude between 33000 ft and 35000 ft.

2. Calculate the height required to clear the obstacle during the descent. The statutory minimum clearance during drift down is 2000 ft, so the required attitude at the obstacle is 14,000 ft + 2,000 ft = 16,000 ft.

3. Using the data in the table at the top left corner of the page adjust the gross weight at engine failure to reflect the conditions of the anti-icing and the air conditioning system. For the stated condition of engine and wing anti-icing systems on and air conditioning system on, the corrections are 54,000 kg + 5,650 kg (for anti-icing)= 59,650 kg. Note that an adjustment is required for air conditioning only when it is switched off, so no adjustment is required in this question.

4. Using the graph at the top right cornet of the page adjust the corrected gross weight at engine failure to reflect the temperature deviation. To do this enter the left edge of the sub-graph at 59,650 kg, then move parallel to the sloping lines to a point vertically above ISA +15°C. From this point move horizontally to the right edge of the sub-graph and read off equivalent gross weight at engine failure on the right vertical axis 61,500 kg.

5. Enter the right edge of the main graph and interpolate between 60000 kg and 65000 kg to locate the 61500 kg point. From this point draw a curve to the left interpolating between the printed curved lines on the graph. In doing this take care to take account of the fact that the curved lines converge as they move to the left.

6. Enter the left edge of the graph at 16,000 ft and draw a horizontal line to the left to intercept the curve drawn in stage 5 above.

7. At the intersection point interpolate between the dotted curved lines to estimate the fuel used during the drift-down. In this question the fuel used is approximately 1000 kg.

8. From the same intersection point draw a line vertically down to the bottom of the gridded area and read off the time required to drift-down. In this question it is approximately 29 minutes.

9. Extend the vertical line down to the reference line in the lower gridded area. From this point draw a line parallel to the sloping lines. From the existing 40 kt tailwind condition marked at the left edge of the graph, draw a horizontal line to the right to intercept the sloping line drawn previously. From the point of intersection of these lines drop vertically down to the bottom of the graph and read of the ground distance of approximately 175 nm.



If you are happy with the one above please try this one (something very like it appeared in a recent EASA ATPL exam)

Given the following information determine the greatest distance from an obstacle, at the failure of one engine would cause an MRJT1 aircraft to clear the obstacle by the statutory minimum during the drift-down, the fuel used during the drift-down, and time taken to reach the obstacle.

Cruise Pressure Altitude 37,000 ft.
Gross Weight at Engine Failure 44,000 kg.
Ambient Temperature ISA -15°C.
Wind Component 30 kt headwind.
Engine Anti-icing System on.
Wing anti-icing off.
Air Conditioning System on throughout the drift-down.
Obstacle Pressure Altitude 23,000 ft.
One engine inoperative.

Answers:

Ground distance 122 nm.
Time = 24 minutes.
Fuel used = 700 kg.

[alpha-b]

Here's a question few days ago i found out the answer but forgot the methodology that led me to this answer:

The scale at the equator on a Mercator chart is 1cm to 5nm.How many nm to the inch are there at S 44 00?

[keith williams]

Start by converting 5 nm/cm at the Equator into nm/inch

There are approximately 2.54 cm in an inch so multiplying 5 nm by 2.54 gives 12.7 nm/inch at the Equator.


Then use the following equation to calculate the distance per inch at 44 degree latitude.

Distance at B = ( Distance at A x Cos B) / Cos A

Where A and B are latitudes.

Using A = 0 latitude at the Equator and B = 44 degrees latitude we have

Distance at 44 degrees = (12.7 nm x Cos 44) / Cos 0

Distance at 44 degrees = 9.14 nm

So the scale at 44 degrees is 9.14 nm/inch

[Gulfstreamaviator]

I would never pass muster these days....

I seriously wonder just what practical benefit these questions actually give to the human race.

I am impressed that my replacement in a few years time will actually have passed these hurdles.

Good luck guys and gals.

[P40Warhawk]

Haha. Absolutely true. Well there are more questions that are totally useless.

I was actually wondering if there are any pilots who use Grid Navigation in there daily life as a pilot?

Sunrise / Sunset times. Stuff like that.

I am also very happy that I passed all exams first Attempt few weeks ago. Nav. With score 94%.

Had also some of these converting questions of scale. And actually it is not very hard, but you have to see how it works. Since I am from the Netherlands, I always converted it to Metric, and from there find out the scale. Scored 100% on that part.

[paco]

You'd be surprised. I'm using convergency/change of longitude all the time in my head.

Sunrise/sunset, too, although I don't use the almanac, just the GPS

[P40Warhawk]

Well Convergency Conversion I can imagine, but Grid?

Yeah I gues no pilot will take Almanac with him or her ;) .

[Haroon]

Hi

Q. For which of the following is a flight plan, in accordance with Annex 2, to be submitted?

a) Any flight crossing an IFR boundary
b) Any IFR flight in class F airspace
c) Any flight in controlled airspace
d) Any flight more than 40 km from the coast

b) marked correct but whats wrong with (a) or (c).

Thanks

[hvogt]

According to 3.3.1.2 c) of ICAO Annex 2, a flight plan is required for "any flight across international borders", however there is no such thing as an "IFR boundary"; hence, answer a) is wrong.

Answer c) is wrong because a flight plan must be filed for "any flight or portion thereof to be provided with air traffic control service" (3.3.1.2 a) of ICAO Annex 2), not for any flight within controlled airspace. Think of VFR flights in airspace class E - controlled airspace, but no clearance or flight plan required.

[Haroon]

Thanks hvogt

May I also ask you:

In an advisory airspace, IFR traffic using the advisory service will be separated from other participating IFR traffic using the advisory service.

Is this universal or specific to Europe?

All the references I've seen just say that separation provided is between IFR flights and don't mention whether they are participating or not.

Thanks

[hvogt]

Haroon

The term "separation" is somewhat relative when used with air traffic advisory service because this service will not issue clearances, it will only give advice and suggestions. Needless to say, separation based on such suggestions can only be achieved between aircraft which are known to the unit providing them. In other words, aircraft which are not "participating" cannot be separated. Keep in mind that ICAO permits IFR flights in airspace class F without clearance, so in airspace class F there might be a wild mix of IFR and VFR flights known or unknown to the air traffic services unit in charge.

As for your question concerning universal or European application of the respective rules, each state will have its own rules, some do not even have class F airspace (Denmark and Chile if I'm not mistaken), others might not provide air traffic advisory service at all but offer flight information service instead. Here in Germany, for example, IFR flights in airspace class F are subject to an ATC clearance and there will never be more than one IFR flight at the same time in the same airspace.

[Haroon]

Thanks hvogt, much appreciated

[Haroon]

If there is sufficient time for information to be disseminated by other means, a NOTAM is not issued. What is the time limit?

2 days
28 days
14 days
7 days Marked Correct

i cant find any reference for this, can someone help pls

thanks

[Haroon]

Hi standing by for the answer regarding the previous question (post #374).

Got some more:

1) What documentation is required by persons travelling by air, for entry into a state?

Passport and visa
Passport and confirmation of inclusion on the general declaration passenger manifest
The same as would be required if the person arrived by ship
Passport, visa and any necessary health documentation (vaccination certificates)

Why not the first option or the last option?


2) Who is responsible for setting into movement the Alert Phase?

ATC and the FIR
The State and ATC
The Area Control and the RCC
RCC and the FIR

Do they mean FIC? Why not the first option?


:arrow: Which of the following applies to general declaration?

Accepted orally for crew or passenger baggage
Accepted orally for crew baggage
Accepted orally for passenger baggage
Never accepted orally

Cant find a reference to this one!

thanks

[P40Warhawk]

That about visa Passport.
Its a useless question.Like most questions in AL and OPS. Just remember.

Have also question about, what is MIN age of a Cabin Crew? So as you can see. These kind of questions you should just remember. There is no logic why you should know that.
Questions about seperation , those are usefull , and signs, colours etc.

[Haroon]

need some logic to remember otherwise will forget when the time comes :)

[P40Warhawk]

No logic needed in AL and OPS.

Other subjects, yes then you should definetely understand that theory, So not only QB but also study your books. Otherwise its hard to get through the exams. But AL and OPS are very dry stuff.

[alpha-b]

Hi guys
I've got this one here:

Flight time 03:30
Reserve fuel shouldn't be less than 25% of the trip fuel
Block fuel:260 kgs
Taxi. :10 kgs

How many kgs of fuel should remain after 2 hours?
Any help is appreciated,i'm looking for an easy method to solve this kind

[hvogt]

Take-off fuel is block fuel (260 kg) minus taxi fuel (10 kg), i.e. 250 kg.

Reserve fuel must not be less than 25 % of the trip fuel, so reserve fuel is (250 kg / 125) · 25 = 50 kg.

Trip fuel is take-off fuel minus reserve fuel, i.e. 250 kg - 50 kg = 200 kg.

After 2 hours of a 3.5 hours’ flight, 2/3.5 of the trip fuel are burnt: 2/3.5 · 200 kg ≈ 114 kg. The remaining trip fuel is hence 200 kg - 114 kg = 86 kg. The 50 kg reserve fuel must not be forgotten, so remaining fuel in tanks is 86 kg + 50 kg = 136 kg.

[RichardH]

The simple way, least calculator hit (other methods are available) I teach my students is:-

Take-off fuel / Total flight time * Time remaining.

250 / 3.5 * 1.5 = 107 kg.

The question usually says 25% reserve of REMAINING trip fuel, which means the actual amount of reserve fuel required is decreasing as the flight progresses and would be zero on arrival.

As this question didn't ask for a split of trip & reserve the answer is 107 kg (assuming it meant REMAINING). However to split 107 / 1.25 (back work 25%) = 86 kg trip and reserve would be 21 kg which is 25% of 86 kg.

This is a proportion question and the old whiz wheel excels at this problem.

[Watt001]

The question usually says 25% reserve of REMAINING trip fuel, which means the actual amount of reserve fuel required is decreasing as the flight progresses and would be zero on arrival.


RichardH, I don`t absolutely agree with you. Trip fuel is fuel for all the trip, which is 3:30. If the final reserve should be 25% Trip, it is related to the Total Trip fuel, not just remaining time, as you say.
Your theory would say, that after landing, you can have zero final reserve, which you surely know is not true.
So after my calculations, the result is also 136.
Please, anyone correct me if I`m wrong.

[RichardH]

This type of question has always caused issues like this as it is a fuel related question but not fuel policy directly.

It's really a light aircraft fuel question where a reserve of REMAINING trip is perfectly okay, though strange to get your head around especially landing in THEORY with no reserve, but of course you started with that 25% and assuming nothing happened then that reserve would be intact on landing.

Questions similar to this have never mentioned FINAL RESERVE as I should agree with you if it did. Trust me with years of instructing experience and knowing what the examiners are after this is a proportion question not fuel policy knowledge, though perhaps not the best way to test it by ignoring the "rules". The examiners are basically saying with inaccurate fuel gauges the reserve is set at 25%, you don't know how that is split.

[Watt001]

I definitely don`t wanna argue, but if you say this is true, then it is just another junk question, which has nothing simmilar with the student`s knowledge. Just a stupit guessing, which way is the correct way. But we all know, that there are many questions like this in ATPL Theory, so I just have to believe your calculations.. Good luck to the students getting this kind of question at the exam..

[RichardH]

Yes some of the questions make you wonder and FP is much better than most.

As stated this question does cause issues and is one I always explain as discussed in previous posts. It must be right as some students are getting 100% with this type of problem included!

[squall1984]

Given that the characteristics of a three engine turbojet aeroplane are as follows: Thrust = 50 000 Newton / Engine, g = 10 m/s², Drag = 72 569 N, Minimum gross gradient (2nd segment) = 2.7% SIN(Angle of climb) = (Thrust- Drag) / Weight. The maximum take-off mass under 2nd segment conditions is:

The only way i get the answer that the data base gives is by assuming one engine is out...but at no point in the question does it say that it is.

[paco]

perhaps you only need the segments when the engine quits......

[MarkerInbound]

First, second and third segment weights are based on having one engine inop and still meeting the required gradient.

[Watt001]

Yes, I also noticed when learning for my ATPL, that this kind of questions is always based at one engine inop situation. Pay attention on this and good luck

[squall1984]

Two performance questions that I have found question banks with different answer.


On a twin engined piston aircraft with variable pitch propellers, for a given mass and altitude, the minimum drag speed is 125 kt, and the holding speed is 95 its. The best rate of climb speed will be obtained for a speed of:

One question bank says 95, the other 125.

What other speeds can you derive from the above question?

The other question I found is for a piston engine maximum endurance is achieved at:

One bank says maximum rate of climb speed, the other at maximum lift to drag ratio, one and the same?

[Haroon]

Hi

Q.1. During a flight to Europe, scheduled in MNPS (Minimum Navigation Performance Specification) airspace, you expect to cross the 30oW meridian at 2330 UTC; you will normally be:

a) in a day flight route system
b) in random airspace
c) in a night flight route system
d) out of the organised route system

why not (d)?

Q.2. During a flight to Europe, planning in MNPS (Minimum Navigation Performance Specification) airspace, you expect to cross the 30oW meridian at 00H30 UTC, you will then normally be:

a) within the organised night-time flight track system
b) with the organised daytime flight track system
c) out of the organised flight track system
d) in a random space

why not (d)? (unless you want to rule out "space" instead of "airspace")

thanks

[RichardH]

OTS times are:-

East or night time - 0100 to 0800 UTC at 30w
West or day time - 1130 to 1900 UTC at 30w

Flying an OTS route is optional and a flight not entirely following an OTS track would be called a random flight/route as are flights outside the time periods.

There is no such thing as 'random airspace'. Where did you get these questions? I thought they were corrected years ago.

Both flights are operating outside the standard OTS times and are random flights. Q1 = D, Q2 = C (as the better answers).

[Haroon]

Thanks Richard

They still stand uncorrected in some QBs among many others.

Regards

[squall1984]

Anyone seen this question before? It was in the Performance Exam today and though I cant recall it verbatim, it was regarding the benefits of reduced thrust.

Out of the 4 options only 2 were viable,

1) Reduces trip fuel
2) Reduces a limited VMCG.

I can see arguments for both answers, one side you do use less fuel when you use reduced thrust, VMCG could be lower since theres less thrust. Any suggestions?

[Contacttower]

Re yesterday's Performance exam I actually thought the answer to the above question was the option involving Field limiting take off weight being a lot more/less (or whatever...) than Climb limiting. I think it may have been answer C. Either that or reduce a limiting VMCG...

As far as I could see...

Reduced thrust CAN be used on a wet runway as long as it is not considered contaminated. So it couldn't be that...

Trip fuel option just sounds wrong because that is not a reason to use it...the saving would probably be small and I have never seen that given as a reason to reduce take off thrust...

It will reduce the VMCG but as far as I'm aware there is no data to support that in the CAP 698 beyond the VMCG table for max thrust. I have heard of it being used as a technique in the real world but I'm just not sure whether it would be considered the right answer for the purposes of the exam...

[squall1984]

In hind sight I would have gone with the VMC option only because you can't count on the use of reduced thrust in case of windshear etc. maybe someone appealed it.

What did you put for drag corffiecent question? I put lowest point which question bank people say is correct but app examiner is looking for tangent answer. We appealed that one.

[Contacttower]

Don't remember that one unfortunately, do you mean the one with the graph in the annex?

[squall1984]

Yes thats the one

[Haroon]

Use of reduced thrust for climb increases total trip fuel and should be evaluated by each operator.... Takes more time to reach the optimum level

[squall1984]

See we use reduced thrust is replaced with climb thrust at 1500 so saves fuel up to that point as well as other benefits engine wear etc

[Contacttower]

What did you put for drag corffiecent question? I put lowest point which question bank people say is correct but app examiner is looking for tangent answer. We appealed that one.

Can you remember more specifically what the question was asking? I can't remember what point it was asking to identify on the graph?

[Alew Crew]

Dear,

I know i'm not in the good thread but i need an answer :

Question 16- REF 330076

(Refer to route manual Chart Nap) date 1999
From Reykjavik ( 64°10 ' N - 020 °N 020° 00'W) to Amsterdam ( 52°32'N 004°50'E), what is the distance ?

I don't understand how they found 18° of Lat difference... Can someone explain me how ? (In the correction they found 18°-Bristol)

Thanks

[keith williams]

Alan,

Many readers do not have access the material to which youare referring. You might get a betterresponse if you were to post a copy of the question and the explanation to enableeveryone to examine it.

From what you have posted I can see that there is one errorin that the longitude of Reykjavik should be 022W and not 020W.

[RichardH]

Alew, I agree with Keith. Plus I know i'm not in the good thread but i need an answer: PLEASE I think you meant to say too!

However if the question is just asking for a distance then you should just draw a straight line on the chart and using dividers/compasses/rule and the fact 1 degree of latitude = 60 nm you should get around 1080 nm. I do using both the NAT charts both using dividers and the appropriate chart scale and an inch ruler.

So on information provided don't know where the 18 lat difference comes from, sounds rubbish to me.

[RussianPilot]

Hey guys im stuck here, could you please please help me out with the question,
if possible with explanation how to do it too.

You are flying along the 315* radial from a VOR/DME. On a heading of 310*M and flying a TAS of 450 kt, at 0650 UTC you are 93NM from the DME. At 0655 UTC you hav reach 135 NM from the DME. Variation off the map is 16*E. What is the most likely W/V?

answers
A 005/54
B 165/70
C 105/65
D 180/70 probably correct

[Lightning Mate]

This is what you have:

TAS 450 kts
Groundspeed 504 kts
Drift 5 degrees R

Pick up a flight computer and it will reveal the W/V.

I don't have mine any more (gave up teaching this stuff 18 months ago) but answer D indeed looks a likely candidate.

[Watt001]

Hey guys im stuck here, could you please please help me out with the question,
if possible with explanation how to do it too.

You are flying along the 315* radial from a VOR/DME. On a heading of 310*M and flying a TAS of 450 kt, at 0650 UTC you are 93NM from the DME. At 0655 UTC you hav reach 135 NM from the DME. Variation off the map is 16*E. What is the most likely W/V?

answers
A 005/54
B 165/70
C 105/65
D 180/70 probably correct

Hello. I have done some calculations. My Result is a little bit different from all answers.

In the first stage, you disregard variation, because VOR radials are magnetic bearings and your heading is also magnetic. So the angle of heading 5 degrees to the left from your track. Your GS can be obtained from time and DME: 5 minutes = 42 NM -> 1 minute = 42/5 = 8.4 -> 60 minutes = 504 KTS.

From here you work with a Cosine sentence:
c^2=a^2+b^2-2*a*b*cos(Gamma)

you fill in:
a:450
b:504
Gamma: 5 degrees

you get C = 68 kts - wind velocity.

if you have this, you use Sine sentence:
a/sine(Alpha) = b/ sine(Beta)

you fill in:
a: 68
b: 450
Alpha: 5 degrees

Then you get angle of 35 degrees.
This is the angle, which is added to your track, from where the wind is blowing. So, if you add 35 to 315, you get wind blowing to 350, from 170.
Wind is 170/68 Magnetic, if you add variation, you get final wind blowing 186/68.

Please, someone, correct me if I`m wrong.

[RichardH]

As Lightning Mate said pick up a flight computer and it's a lot easier than cosines and sines.

To work out W/V you need heading, track/course, TAS & GS - basic triangle of velocities stuff just asked in a more practical manner.

GS = 42 NM in 5 mins = 504 kts.

In this case as both heading and track were magnetic work in magnetic on the CRP-5 and get a magnetic wind of 170/70 then ADD the variation back in to get true W/V 186/70 so D.

If you have a mixture of true and magnetic I would always work in true.

Gross error check - wind blows from heading to track so W/V must have a southerly element in this case, rules out answer A to start with.

[Lightning Mate]

Why do some people make the solution to a simple problem so difficult ? :ugh::ugh:

[Pittslover]

Hi Guys,

can anyone recommend me a good free online source to study for the Principle of Flight exam?

Thanks :)

[Lightning Mate]

Try your approved school - that's what you are paying them for.

Some while ago I worked for the CAA on the PoF CQB.

I have a whole collection of questions.

[Pittslover]

I am doing the atplonline QB at the moment and my distance learning provider has not the best material online. I thought someone can point me in the right direction. Preferable key facts especially about high speed aerodynamics.

[selfin]

Pittslover,

While not directly satisfying your request I will nevertheless recommend HH Hurt's classic Aerodynamics for Naval Aviators. The FAA hosts a PDF copy, not under US domestic copyright protection, here: http://www.faa.gov/regulations_policies/handbooks_manuals/aviation/media/00-80T-80.pdf

Although reading this from cover to cover will be overkill for ATPL PoF exam purposes, it is sufficiently simple to supplement learning without requiring much prior knowledge.

[sweetnothing]

hi, i have a question from met, I want to know if the explanation is correct.
if an aircraft in northern hemisphere flies from high to low, it will experience _________ drift
ans. starboard

explanation: if winds are from left, then it will experience right drift, it is because aircraft is flying towards ( or in the) low pressure, so the winds will be from left. hence starboard drift.
rule applied is coriolis force.

[mad_jock]

Do a search for Ballots law.

[keith williams]

All fluids tend to flow from areas of high pressure to areas of lower pressure. But because of the rotation of the Earth, Coriolis effect causes the wind to circulate around areas of high and low pressure. The direction of this circulation is depends on the location (northern or southern hemisphere), and whether the area is one of high or low pressure. The outcome of these effects is predicted by Buys Ballot’s Law, which states that when standing with one’s back to the wind, the low pressure area is on your left.

If an aeroplane is subjected to a crosswind it will drift in the direction of that wind. In this question the aeroplane is drifting to the right, so the wind must be coming from its left side. Applying Buys Ballot’s law this means that the aeroplane in this question must be flying towards the area of low pressure.

If you need a more detailed explanation you should read up on coriolis effect, which is part of the ATPL Met syllabus.

[sweetnothing]

thank you!

[Richard Phillips]

If in doubt, draw it out! :)

[sweetnothing]

yes, thank you very much! :)

[Danny boy]

From memory Tx means braking action...

[Treadstone1]

HI

Can anyone please tell me if questions on the Polar Stereographic chart are still being asked in the ATPL Gen Nav exam?

Hearing that it has been taken out of the syllabus?

Cheers

[RichardH]

Yes they are still there in GN and questions are asked.

[Treadstone1]

Thanks for this Richard..

Cheers

[laurenbrooke]

Hey guys,

I'm doing my ATPL at the moment and I'm struggling to find the right explanation to the following questions: (if any of guys could help me I would really appreciate it)


Number 1:

A Northern Lamberts conformal conic chart is overprinted with a false grid.
The constant of the cone of the chart is 0,80.
At 60°W the grid track is 090° and the true track is 010°.
At which longitude is the false grid aligned?

a) 160° W b) 140°W c) 020° E d) 040°E

Number 2:

A Northern Lamberts conformal conic chart is overprinted with a false grid which is aligned with the 25°W longitude.
The constant of the cone of the chart is 0,80.
If the true track at 120°W is 090° will the grid track
be:
a) 014° b) 166° c) 185° d) 355°

Number 3:

A Lamberts conformal conic chart is overprinted with a false grid.
The constant of the cone is 0.60.
At 40°S 70°W the grid track is 197° and the true track
is 239°.
At which longitude is the false grid aligned?
a) 112° W b) 028°W c) 000° E/W d) 030°E

Number 4:

North polar stereographic chart is overprinted with a false grid.
At 77°N 37°W the grid track is 175° and the true track is 093°.
At which longitude is the false grid aligned?
a) 045° W b) 045°E c) 082° E d) 119°W

[RichardH]

Hi Lauren, I am not surprised you want an explanation!. I quote the JAA LOS

061 01 04 05 Gridlines, isogrives
LO Explain the purpose of a Grid north (GN) based on a suitable meridian on a polar stereographic chart. (reference or datum meridian).


Grid questions should only be asked in relation to polar charts not Lamberts. You are potentially going to waste hours of time trying to sort out questions 1 to 3.

Even Q4 is somewhat obtuse but unless I am having a bad day its 45E. To understand PS questions these are best done by drawing diagrams. However there is always a relationship between GRID, TRUE, LONGITUDE. 175-93=82-37=45.

These questions usually ask what is the grid or true depending on north or south polar with grid aligned on either prime or anti-meridian. Certainly nowhere near as tricky as these questions.

Your FTOs notes should cover all that's required for PS charts and grid - refer back to them. Also try looking the the Polar 5AT chart in the Jeppesen and drawing some tracks between places, measure true then grid. Bit hard to explain in a written reply.

However if you have no joy then PM me as I have a dummies guide to PS charts you can have.

[GoProPilot]

Hi All,

Not ATPl question but didnt want to start a new thread:

Somebody wrote this ages ago on the forum:

They then give you 2 small written tests. The first one is again basic mathematics but applied to flying. Questions like "you want to be at 1000' 5 miles from the beacon. You are currently at 8000' and 20 miles from the beacon and your airspeed is 150kts. What descent rate do you need?" So basically, a few distance/speed/time questions. Know the 1/60 rule also because I think there's a question on that.

I make the answer 7000/150 x 60 = 2400ft per minute decent.

Can anyone confirm?

Cheers.

[keith williams]

You are at 8000 ft when 20 miles from the beacon.

You want to be at 1000 ft when 5 miles from the beacon.

So you need to lose 7000 ft while flying a distance of 15 miles.

At 150 kts you will take 15/150 = 0.1 hours (which is 6 minutes) to fly the 15 miles.

So you need to lose 7000 ft in 6 minutes.

Required ROD = 7000 ft / 6 mins = 1167 ft/min.

And from your post

7000/150 x 60 = 2400ft should read 7000 / 150 x 60 = 2800ft

[GoProPilot]

Thanks very much,

I now get it..
Understand I need to find how long to fly the 15 miles

gives me time 6mins

Then 7000ft divide 6 mins give the 1167ft per minute, to achieve the 1000ft

Many thanks, I keep rushing my workings!

[samdol1978]

I am currently a SPL holder with 40hrs of flying at Malaysia.Currently at Navigation phase

I understand formula for lift is-->L=1/2 s*p*v*sq*Cl where Cl is coefficient lift,v is velocity of air,s is surface area of wing and p is density of air

I have to following question to ponder while revising my unusual attitude recovery

1)Why does increasing pitch decrease airspeed and why does decreasing pitch increases airspeed?

2)I understand when we are using aileron to turn,one wing lift is more than the other,hence we roll to either right/left

[mikios]

1)Why does increasing pitch decrease airspeed and why does decreasing pitch increases airspeed?

2)I understand when we are using aileron to turn,one wing lift is more than the other,hence we roll to either right/left

1) although you are correct, you also have to consider drag.
Changing pitch will affect your Cl, and your Cd (Coefficient of drag), the latter being:
D = Cd * 1/2 * ρ * V^2 * s. Notice the similariles with the lift formula.
Make sense now?
To understand more, refer to the Lift and Drag diagrams as well.

2) absolutely correct. Deflecting a aileron will change the chordline of (part of) the wing (chordline being from leading edge to trailing edge) thereby changing that parts AoA (angle of attack, alpha) and its Cl and Cd.

[Transsonic2000]

Hi all,

I'd need some help figuring out the following two questions, been working on this for too many hours now. Many thanks in advance!

Aircraft maintaining straight and level flight at speed of 1.9 VS. If a vertical gust causes a load factor of 2.9 the load factor n caused by the same gust at a speed of 2 VS would be:
a) n = 3.00 (correct answer)
b) n = 3.61
c) n = 2.85
d) irrelevant, since airplane would already be in a stalled condition at 1.9 VS


Aircraft maintaining straight and level flight at speed of 1.5 VS. If a vertical gust causes a load factor of 2.5 the load factor n caused by the same gust at a speed of 2 VS would be:
a) n = 3.00
b) n = 2.30
c) n = 2.25
d) irrelevant, since airplane would already be in a stalled condition at 1.5 VS (correct answer)

[keith williams]

The increase in load factor caused by flying at different speeds in the same gust can be calculated using the following equation:

N increase at new speed = N increase at old speed x (V new / V old)

For the first question comparing the conditions at 1.9 VS and at 2.0 VS gives the following:

N increase at old speed = 2.9 - 1 (straight and level) = 1.9
Old speed = 1.9 VS
New speed = 2.0 VS

N increase at new speed = 1.9 x (2 / 1.9) = 2.0

Adding this increase to the initial straight and level load factor of 1 give a new load factor of 3.0



Using the same equation in the second question gives us the following:

N increase at new speed = 1.5 x (2 / 1.5) = 2

Adding this to the initial straight and level load factor of 1 g give new load factor = 3.0.


But option d in each question poses the possibility that the aircraft may stall before it attains these increased load factors, so we need to test this possibility.

We can do this using the following equation:

VS at new N = VS at old N x (square root of new N / square root of old N)


Inserting the data for question 1 gives the following:

VS at 2.9 g = VS at 1 g x (square root of 2.9 / square root of 1)

VS at 2.9 g = 1.7 VS at 1 g

The aircraft was initially flying at 1.9 VS, so it would not stall before attaining 2.9 g. So option d is incorrect.



Inserting the data for question 2 gives the following:

VS at 2.5 g = VS at 1 g x (square root of 2.5 / square root of 1)

VS at 2.5 g = 1.58 VS at 1 g

This means that the stalling speed at 2.5 g would be 1.58VS. But the aircraft is flying at only 1.5 VS, so it would have stalled before attaining 2.5 g. This means that the higher load factor of 3.33 g could not be attained. So option d is correct.

[Transsonic2000]

Hi Keith,

once again, many thanks for your excellent explanation! It is greatly appreciated! it's of much help to me, without it I'd probably just went on hopping that this type of question won't be asked in the exams or simply just guessing, choosing answer "d" and keeping the fingers crossed :uhoh:

[Flying_hog]

Hi All,

Does anyone of you know, why are sometimes different DA(H) for different aircraft categories specified on the approach chart? For example:

CAT 1 ILS (missed approach gradient of MIM 2,5%)

DA(H):

A: 250' (213')
B: 260' (223')
C: 270' (233')
D: 280' (243')

Usually there is only one figure for ALL aircraft categories? :ugh:

[Lightheaded]

What's the difference between a semi-monocoque and a reinforced shell aircraft structures? All I see in the explanations is that the reinforced shell is a development on the semi-monocoque type and has a reinforced skin supported by other strucural members. But isn't that what differentiates monocoque from semi-monocoque - longitudinal stringers?

[Veins101]

a monocoque has no supporting structures... at all... Look at the diagrams again.

[Transsonic2000]

@ Flying hog:

see Jeppesen Route Manual, TOC (Approach Chart Legend), page 113 (landing minimums)

[MarkerInbound]

The one I like is the ILS 16L at SEA. The category A minimums are 60 feet higher than the B, C and D minimums. When the TERPS people did all the sums they figured a slow aircraft with a strong crosswind could be blown too close to the control tower so they set the category A mins at the height of the control tower. A faster aircraft would not be displaced as much by the crosswind and so can go lower.

Flying Hog, without knowing which airport it would be hard to say why the minimums are different.

[alphaLaura]

Hi all,

Studying mass and balance at the moment and working through the Oxford book. One thing is puzzling me about one of the exam questions. The question is based off CAP MRJT but in the worked answer they use a different MZFM. The MRJT MZFM is 51300 kg but they use 53000 kg in the answer. Is there a reason for this?

For anyone who has this book, the question is on page 3-14 (Q23) and the worked answer is on page 3-19.

[RichardH]

I suggest you contact your FTO (Oxford) directly as they will know why more than anybody on this forum. This goes for any material from any FTO best to contact the source first.

You are likely to get 1 of 3 responses:
1. Yes it's known a mistake in our notes - sorry
2. It's a mistake thanks for letting us know
3. You got it wrong because of x and y

[alphaLaura]

Hi RichardH, apologies for the slow reply - yes, I think best bet is to contact the people who wrote the book! Thanks :)

[Flyboy_Josh]

P0041
What is the maximum vertical speed of an aircraft with a mass of 76 000 kg?

Given: Thrust = 160 kN

Drag = 110 kN

TAS = 190 knts

g = 10 m/s²


I get an answer of 1250 ft/min. But in the potential answers the correct is 1267 ft/min. 1250 isn't even an option. Why?

[Flyboy_Josh]

Ahhhh explains it thank you :)

[clkorm3]

Hi,

I am trying to revise for flight planning and mass and balance. I seem to come across quite a number of LRJT questions on question bank. Although I have been told they are unlikely to appear in the exam. Is there any truth to that. Anyone recently sat the exam and had such questions asked.

Secondly although I am trying to read materials provided on running load and distribution load but I am still encountering problems grasping the principle and if someone can provide help with that. It will be much appreciated.

[RichardH]

This post might be better in ATPL theory questions forum.

LRJT certainly have/are been asked.

[ThomasNL]

Use 50 nm instead of 30 nm. 2700 is correct for 30 nm, but the question is confusing.

[bumfished]

hello guys.
would really really appreciate it if someone on pprune could help me with the solution to this problem on navigation:ugh:

**) aircraft takes off from 00 N 170 W at 2100 LMT on 21/12/99 at a ground speed of 300 knots. .Another aircraft takes off from 15 N 165 E on 22/12/99 at 2100 LMT to intercept the first aircraft.Find

a) time of interception
b) ground speed of second aircraft

please please help me with the solution to this problem. would really appreciate your time and help :):)

[pak0pak]

Which of the following is not and active force on the aircraft? lift,weight, thrust, drag, torque?

[clkorm3]

If CAS is 190 kts, Altitude 9000 ft. Temp. ISA - 10°C, True Course (TC) 350°, W/V 320/40, distance from departure to destination is 350 NM, endurance 3 hours and actual time of departure is 1105 UTC. The distance from departure to Point of Equal Time (PET)

9000ft/ISA -10⁰C (OAT= -13⁰C) TAS =212 kt D = 350 NM … O = 176 kt … H =246kt Distance to PET = D x H / (O + H)…………………. = 350 x 246 / (176 + 246) = 204 NM

Answer 203nm


I am unsure about the temprature corrrection as i get is -7 { 15 - (9 x 2) }= 3 and -10 =-7
can someone just explain this -10 or +10 isa and correction made accordingly please
thank you in advance.

[RichardH]

ISA at 9000ft is -3, however the question states ISA -10 which means the ambient OAT is 10 COLDER than -3 so should be using -13 to get your TAS of about 214 its.

If it had said ISA +10 then your +10-3 = +7.

Understanding these corrections is an important skill for a number of the subjects.

[clkorm3]

Thank you.

[Gulfstreamaviator]

I have no idea how I managed to pass my ATPL written papers.


All I can say is please do not ask me any of the questions as shown above, I have wasted 1/2 day trying to get even one right.


Perhaps 45 years ago I was just clever, but now stupid.

[clkorm3]

You have a crate of mass 174 kg with dimensions of 1ft 3in x 1ft 9in x 2ft 4 in. What are the distribution and linear loads?

79.81 kg/sq.ft and 8.28 kg/in

can someone please help me break this down as i get different answer when i work out longest side for running load answer

[newbiepilot84]

Hi,
I know how to find the pressure height but I dont fully understand the reason for finding it. I know that it has to do with the performance of the aircraft and terrain clearance, and also to know the take off distance. Can someone kindly explain to me the purpose of finding these two?
I am a noob
Cheers

[paco]

Say you have a mountain that is 6000 feet high and you want to land your helicopter on it. Assuming you know what ISA conditions are (as a noob :)), in ISA conditions, your machine would think it was landing at 6000 feet. But say the QNH was 995 instead of 1013. There is an 18 mb difference which translates to roughly 540 feet. Now your machine thinks it is at 6540 feet, and the engine and rotors (or wings) have to work just that bit harder (in these types of questions, draw a diagram and place the larger pressures at the bottom).

Now let's look at the temperature. The ISA temperature at that height should be around 3 degrees C - but say it's actually 15 degrees - it is warmer than it should be by 12 degrees and the air is thinner (less dense), so your machine has to work harder still because it thinks it is another 1400 or so feet higher - actually around 8000 feet.

To find out density altitude, take the ISA temperature difference, multiply it by 120 and add or subtract it to the PA.

[N5748E]

We use it to calculate performance related stuff such as take off distance, and landing distance.

At my organization, we use the pressure alt at our aerodrome to calculate the take off and landing distance for our aircraft, and add a 15 percent safety factor.

[gpiper]

N5748E

Unfortunately ISA figures are only theoretical, in practise you will hardly be flying at altitudes or places where ISA conditions are present, personally I have been doing a lot of flying from a coastal Aerodrome, never experienced ISA conditions.

[N5748E]

I know. It's never an ISA day...

[N5748E]

Just get the airport elevation, the forecasted QNH, for e.g. at my airport today, the QNH was 996hpa, and airport elevation was 55ft. Gives u a pressure alt of 563ft.

Go to your POH and use the charts in section 5, performance, remembering the temperature as well...

[alphaLaura]

clkorm3 - first work out the load intensity:

174 kg/(1.25 ft x 1.75ft) = 79.5 kg/sq. ft.

(Note that this answer is given in feet so you must convert from inches to feet.)

If you take the longest length according to the first part of the answer, you must calculate the running load with a length of 1 ft 9 in rather than the 2 ft 4 in. In this case, your answer has to be in inches, rather than feet.

Running load: 174 kg/1 ft 9 in= 174 kg/21 inches = 8.23 kg/in.

I'm not sure if you had other answers to select from as you can definitely choose a better configuration of running load and intensity by working with the 2 ft 4 in length. Maybe all the other answers had incorrect units or scaling?

[RTN11]

They just needed a datum to refer and compare to so they could write the performance charts. Without a starting reference, it would be a complete mess.

The fact that it's never an ISA day is irrelevant.

[clkorm3]

thank you that was very helpful.

[RTN11]

You also mention the reason finding Density Altitude (you said height, but meant Altitude)

This is used to correct the readings from an airspeed indicator to give the true airspeed of the aircraft.

The steps are:

Indicated Airspeed (subject to instrument and position error)

Calibrated Airspeed (subject to Compressibility error)

Equivalent airspeed (subject to density error)

True Airspeed (once you've factored IAS for all the above errors)

[newbiepilot84]

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

[paco]

The one that it thinks it is at.

[RTN11]

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

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.

[newbiepilot84]

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

[newbiepilot84]

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?

[hvogt]

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.

[clkorm3]

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.

[keith williams]

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.

[clkorm3]

Thanks Keith.

[baleares]

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%

[Dak Man]

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

[Live your dream]

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%;

[dubbleyew eight]

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%)

[keith williams]

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%.

[Live your dream]

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.

[Live your dream]

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?

[Lightning Mate]

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.

[baleares]

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

[keith williams]

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.

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.

[guyleedsutd]

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