Hi

Q.1. When switching on the weather radar, after start-up a single very bright line appears on the screen. This means that the:

scanner is not rotating
scanning of the cathode ray tube is faulty

Q.2. On switching on the AWR a single line appears on the display. This means that:

the CRT is not scanning
the antenna is not scanning

Whats the main difference between the two?

Sounds like you might be one of my ex-students.

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

The main difference appears to be the use of the words "very bright" in question one. If the electron gun in the CRT is repeatedly scanning either vertically or horizontally, instead of both vertically and horizontally, this will concentrate the energy in a single line. This concentration of energy is likely to make the line very bright. In this case the most accurate answer would be "scanning of the cathode ray tube is faulty".

For question 2, if the CRT is working correctly but the antenna is not scanning, then only the radar returns from the area in line with the scanner would be displayed. Whether or not these would produce a very bright line would depend upon what returns were being received from that area.

Hello

Lightning Mate: Can you pls explain a little bit how to solve that question?

Keith: Thankyou!

Battery life
 

Hey,

I recently heard a few questions asked at some interviews but cant find the answer anywhere.

The question was, by law how long must a battery last if power is lost. I have a feeling its 30 mins but I cant be sure. I have checked EU ops and cant find the answer anywhere. Anyone know what it is for a fact?

On the Boeing 737-800 it's 60 mins in a good world... And that's using both the main battery and alternate battery!

Thanks for the reply. Do you know if that is the legal limit of one hour (on a good day) or is that Boeing's own choice to make it last longer?

Ground school for CPL (H)
 

Hello Everyone.

I am looking for regular or part time ground school for CPL(H) Exam. Is there any in the UK and how much it cost?

Any suggestion is much appreciated.

Cheers

How does one do this one:

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

thanks

"Q.1. When switching on the weather radar, after start-up a single very bright line appears on the screen. This means that the:

scanner is not rotating
scanning of the cathode ray tube is faulty"

Interesting - as one who used to service monitors my first port of call would be the power supply :)

Deben - check out the big ad on the right

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

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

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.

Thanks for the explanation Keith.

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

Regards

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


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

Thanks Keith

ATPL Met exam
 

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

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!

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!;)

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

Principles of Flight - Flight Mechanics - Forces Acting on Airplane
 

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%

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%

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

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.

Thanks Keith, the examiner answer is highlighted in bold.

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

anyway thanks a lot.

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!

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.

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!

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 )

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:

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

Jammed
 

Hi,

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

Thanks.

'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.

flux valve
 

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

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

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

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.

Thanks LM

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

regards

Stalls exam question right or wrong?
 

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?

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).