Professional Pilot Training (includes ground studies) A forum for those on the steep path to that coveted professional licence. Whether studying for the written exams, training for the flight tests or building experience here's where you can hang out.

# ATPL theory questions

Join Date: Apr 2011
Location: NL
Age: 38
Posts: 91
Likes: 0
Received 0 Likes on 0 Posts
anders:

Code:
`0.5 x 10° x sin(60) = 4.33°, conversion angle.`
You should use 5 degrees here instead of 10. It's obvious if you draw a picture of the situation with 2 parallel lines representing the mentioned lines of longitudes, like on a mercator chart, with straight rhumb line between the points and a downwards pointing great circle line. The point where the g.c. is most south is in the middle between 20W and 30W, so 5 degrees longitude.

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

Back in the days we didn''t have this q. in the database. (pipe-smoking smiley) Join Date: Jun 2010
Location: Stockholm
Posts: 25
Likes: 0
Received 0 Likes on 0 Posts
Nope, still don’t get it. The way I see it, that would give me the most southerly point between 30W and 25W, ie 27.5W.

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

-Anders

Last edited by Anders S; 5th Dec 2012 at 18:23. Join Date: Apr 2011
Location: NL
Age: 38
Posts: 91
Likes: 0
Received 0 Likes on 0 Posts
0.5 x 5° x sin(60) = 2,17°, conversion angle.

I mean 5 deg. for change of longitude.

Tan 2,16= opposite/ 150NM

opposite= 5,67NM.

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

Last edited by Da-20 monkey; 5th Dec 2012 at 19:03. Join Date: Jun 2010
Location: Stockholm
Posts: 25
Likes: 0
Received 0 Likes on 0 Posts
I’m sorry, I thought I had it but now I’m just confused again. The way I’ve learnt it is:

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

But what you’re saying is basically:

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

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

-Anders Join Date: Jan 2011
Location: mumbai
Age: 35
Posts: 236
Likes: 0
Received 0 Likes on 0 Posts
the formula for conversion angle is correct as 0.5*ch long*sin(lat) Join Date: Jan 2011
Location: mumbai
Age: 35
Posts: 236
Likes: 0
Received 0 Likes on 0 Posts
i guess i've figured it out with some help from another forum.
lets name the starting position A and the intermediate position on 25 degree W as B.
when we join points A and B to form a triangle for calculation purpose, it divides conversion angle into two equal parts of 2.16 degress each.
now solving the sum in the same manner:
tan 2.16= X (unknown)/150
X=5.65nm
X/60 = 0 degrees 5 minutes
hence position B= 60 degrees + 5 minutes
=60 degrees 5 minutes  Join Date: Apr 2011
Location: NL
Age: 38
Posts: 91
Likes: 0
Received 0 Likes on 0 Posts Join Date: Jan 2011
Location: England
Posts: 652
Received 0 Likes on 0 Posts
The midway point between 30W and 20W is 25W.

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

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

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

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

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

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

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

Tan 2.165 = Ch lat / 150 nm

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

This means that the change of latitude between 60S 30W and the unknown position south of 25W is 5.67 degrees. This makes the new latitude 60 degrees 5.67 minutes south. The closest option to this is 60°06’S Join Date: Jan 2011
Location: England
Posts: 652
Received 0 Likes on 0 Posts
Sorry Anders my previous post did not address this part of your question.

Most of the explanations I have seen for this question seem intent on halving the conversion angle, or rather getting the conversion angle from the ch.long between 030°W and 025°W, to get 2.2°-ish, and obviously that would present an answer more consistent with the correct one. But why is this done? I don’t see how that’s relevant as it would imply that we are flying from 60°N 030°W to 60°N 025°W and then on to 60°N 020°W, in which case I would fully agree with the answer, but that would mean that 025°W was an additional waypoint… Unless I'm missing something else.

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

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

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

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

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

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

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

So in solving this type of problem it is easier to go straight for ˝ the conversion angle and work out the solution from there. Join Date: Jun 2010
Location: Stockholm
Posts: 25
Likes: 0
Received 0 Likes on 0 Posts
I figured it was something like that, but I was having a hard time visualizing it. Thanks for clearing that up, and for everyone else efforts in trying to explain it to me. Join Date: Nov 2008
Location: Land of Hope
Posts: 47
Likes: 0
Received 0 Likes on 0 Posts
Question: A turbojet aeroplane has a planned take-off mass of 190 000 kg; the cargo load is distributed as follows: cargo 1: 3 000 kg; (3.50 m from reference point) cargo 4: 7 000 kg. (20.39 m from reference point) Distance from reference point to leading edge: 14m Length of MAC = 4.6m. Once the cargo loading is completed, the crew is informed that the centre of gravity at take-off is located at 38 % MAC (Mean Aerodynamic Cord) which is beyond the limits. The captain decides then to redistribute part of the cargo load between cargo 1 and cargo 4 in order to obtain a new centre of gravity location at 31 % MAC. Following the transfer operation, the new load distribution is:
cargo 1: 6 000 kg; cargo 4: 4 000 kg

Can someone please tell me how this is worked out. Join Date: Jan 2011
Location: England
Posts: 652
Received 0 Likes on 0 Posts

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

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

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

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

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

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

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

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

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

Moment change = cargo mass moved x distance moved.

Distance moved = new position – initial position

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

Required moment change = -61180 kg m

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

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

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

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

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

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

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

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

So the new C of G position after moving 3000 kg of cargo is 32.2% MAC. Join Date: Nov 2008
Location: Land of Hope
Posts: 47
Likes: 0
Received 0 Likes on 0 Posts
Thank you so much. For breaking it down for me. It's one of the questions on ATPL and one that doesn't also give any explanation too. But thank you. Join Date: Mar 2012
Location: Central London
Age: 40
Posts: 308
Likes: 0
Received 0 Likes on 0 Posts
Does anyone have a useful nemonic/memory technique for remembering the answers to questions such as:

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

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

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

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

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

(answer in this example is C).

They are a real bugbear of mine...

Tks.

Last edited by taxistaxing; 15th Dec 2012 at 15:01. Join Date: Nov 2008
Location: Land of Hope
Posts: 47
Likes: 0
Received 0 Likes on 0 Posts
This question is very confusing and one that i can not work out.
Although it doesn't state it but I checked cap mrjt1 but even then I don't get the answer it is saying. Can anyone please help

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

Question: Prior to departure the medium range twin jet aeroplane is loaded with maximum fuel of 20100 litres at a fuel density (specific gravity) of 0.78. Using the following data - Performance limited take-off mass 67200 kg Performance limited landing mass 54200 kg. Dry Operating Mass 34930 kg. Taxi fuel 250 kg. Trip fuel 9250 kg, Contingency and holding fuel 850 kg, Alternate fuel 700 kg. The maximum permissible traffic load is Join Date: Dec 2006
Location: Hamburg
Age: 46
Posts: 432
Likes: 0
Received 0 Likes on 0 Posts
I get 12442 kg max traffic load.

The proposed traffic load of 13090 kg would exceed the max structural take-off mass, so there seems to be an error in the question. Join Date: Feb 2002
Location: Sunny Solihull
Age: 66
Posts: 246
Likes: 0
Received 0 Likes on 0 Posts
IF you didn't take into account the MRJT from CAP696/7 and just worked with the given figures then 13092 is the maximum traffic load, but not the clearest question ever produced. Would be better saying "an aircraft" with no reference to MRJT.

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

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

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

Last edited by RichardH; 16th Dec 2012 at 20:55. Join Date: Nov 2008
Location: Land of Hope
Posts: 47
Likes: 0
Received 0 Likes on 0 Posts
Thank you for your assistance. Highly appreciated. Join Date: Nov 2008
Location: Land of Hope
Posts: 47
Likes: 0
Received 0 Likes on 0 Posts
can someone please tell me how this is worked out.

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

Many thanks Join Date: Dec 2006
Location: Hamburg
Age: 46
Posts: 432
Likes: 0
Received 0 Likes on 0 Posts
You can use the following two formulae to calculate the bank angle, and then the load factor:

tan phi = v˛ / (g * R)

phi: bank angle [°]
v: speed [m/s]
g: gravitational acceleration [m/s˛] (g = 9.80665 m/s˛)

n = 1 / cos phi  Show Printable Version Email this Page