ATPL(A) 031 Mass & BalanceRevision notes and practice questions

031 Mass & BalanceFree sample, 10 questions

1. 1An aircraft is loaded according to the data below. If 250 kg of the cargo is moved to a station at an arm of 8.21 m, which statement about the new take-off CG position is correct?
   Take-off mass: 17 800 kg
   Cargo mass: 420 kg (moment arm: 18.57 m)
   Take-off CG position: 13.41 m
   Forward CG limit: 13.28 m
   Aft CG limit: 15.65 m

   1. The take-off CG will exceed the forward limit by 2.47 m.
   2. The take-off CG will move to a position 28 cm aft of the forward limit.
   3. The take-off CG will exceed the forward limit by 16 mm.
   4. The take-off CG will move to a position 15 mm aft of the forward limit.
   Show the answer

   Answer C. The 250 kg moves forward by 18.57 − 8.21 = 10.36 m, so the CG shifts forward by 250 × 10.36 / 17 800 = 0.1455 m. New CG = 13.41 − 0.1455 = 13.2645 m. The forward limit is 13.28 m, so the CG is now 13.28 − 13.2645 = 0.0155 m, about 16 mm, ahead of it. The total mass is unchanged because the cargo is only repositioned; applying the shift in the wrong direction is what leads to the 'aft of the forward limit' options.

2. 2What can happen if an aircraft is loaded incorrectly so that its Centre of Gravity lies ahead of the forward limit?

   1. There may be insufficient forward pitch control available to keep the aircraft level.
   2. The aircraft becomes excessively stable and the control force may be too high to control the aircraft.
   3. The aircraft may become overly manoeuvrable and hard to control.
   4. There may be a tendency to over-pitch the aircraft during the flare just before landing.
   Show the answer

   Answer B. With the CG ahead of the forward limit the aeroplane is nose-heavy and very stable in pitch. The tailplane needs a large download, so the pilot must use large elevator deflections and high stick forces to rotate or flare, and the forces may become too great to control the aircraft properly. The aircraft is sluggish rather than over-manoeuvrable, and over-pitching in the flare or a lack of forward control are symptoms of an aft CG.

3. 3Using the information below, what is the underload or overload?
   Allowed take-off mass: 4300 kg
   Operating mass: 3253 kg
   Total fuel mass: 650 kg
   Mass of the crew and their baggage: 185 kg
   Traffic load: 1015 kg

   1. An Underload of 32 kg.
   2. An Overload of 153 kg.
   3. An Underload of 397 kg.
   4. An Overload of 618 kg.
   Show the answer

   Answer A. The Operating Mass already includes the DOM (with the crew and their baggage) and the fuel. The actual take-off mass is therefore 3253 + 1015 = 4268 kg, which is 4300 − 4268 = 32 kg below the allowed take-off mass: an underload of 32 kg. Adding the crew again (4453 kg, an overload of 153 kg) or the fuel again (4918 kg, an overload of 618 kg) counts those items twice.

4. 4The mean aerodynamic chord given in the flight manual of an aeroplane starts at 520 mm and ends at 1700 mm from the datum line. The centre of gravity of the loaded aeroplane is at 30% MAC. What is the corresponding distance of the CG from the datum?

   1. 1 336 mm
   2. 1024 mm
   3. 874 mm
   4. 2 200 mm
   Show the answer

   Answer C. MAC length = 1700 − 520 = 1180 mm. 30% of it is 0.30 × 1180 = 354 mm aft of the LEMAC. CG = 520 + 354 = 874 mm from the datum. The percentage always applies to the MAC length, measured from its leading edge, never to the distance from the datum.

5. 5You are departing on a flight with a planned landing fuel of 3000 kg. The aircraft limitations require the centre tank fuel to be used before the wing tank fuel. Using the following information, how many metres aft of the datum will the landing CG be?
   Maximum Take-Off mass: 99 800 kg
   Actual Take-Off Mass: 59 500 kg
   Take-off CG location: 16.26 m aft of the datum
   Fuel in centre tank at take-off: 16 000 kg
   Fuel in wing tanks at take-off: 18 000 kg
   Centre fuel tank centroid: 15.38 m aft of the datum
   Wing fuel tank centroid: 17.79 m aft of the datum

   1. 16.08
   2. 15.74
   3. 15.95
   4. 15.88
   Show the answer

   Answer C. Fuel burnt: 16 000 kg from the centre tank and 18 000 − 3000 = 15 000 kg from the wings, so the landing mass = 59 500 − 31 000 = 28 500 kg. Take-off moment = 59 500 × 16.26 = 967 470 kg.m; remove 16 000 × 15.38 = 246 080 and 15 000 × 17.79 = 266 850, leaving 454 540 kg.m. Landing CG = 454 540 / 28 500 = 15.95 m. The maximum take-off mass plays no part in the calculation.

6. 6Regarding the securing of traffic load, which of the following statements is correct?

   1. Unit load devices must be restrained with straps to the designated tie-down points in the cargo floor.
   2. Traffic load must be restrained thoughout flight so that only the predictable movement of CG with fuel burn can occur.
   3. During rotation on take-off, an unsecured load may damage bulkheads and shift the CG forward outside the CG limits.
   4. During flight, containers and pallets are permitted to remain on rollers within tracks, provided they are secured by latches.
   Show the answer

   Answer D. In the holds, containers and pallets move on roller tracks and ball mats, and in flight they may stay on those rollers as long as the floor locks (latches) are engaged to stop any movement. ULDs are held by these locks rather than strapped to tie-down points. An unsecured load slides aft, not forward, when the nose rises at rotation, moving the CG towards the aft limit.

7. 7The crew of a turbojet aeroplane prepares a Commercial Air Transport flight with the data below, and the flight is operated as planned. Which statement about the Landing Mass (LM) is correct?
   Maximum Certificated Landing Mass: 53 900 kg
   Performance Limited Landing Mass: 51 950 kg
   Dry Operating Mass: 35 500 kg
   Block Fuel: 9720 kg
   Taxi Fuel: 120 kg
   Trip Fuel: 5080 kg
   Estimated Traffic Load: 11 800 kg
   Ignore any change in trip fuel caused by mass changes.

   1. LM = 52 380 and it exceeds the Regulated LM.
   2. LM = 51 820 and it does NOT exceed the Regulated LM.
   3. LM = 51 940 and it does NOT exceed the Regulated LM.
   4. LM = 47 300 and it exceeds the Regulated LM.
   Show the answer

   Answer B. ZFM = DOM + traffic load = 35 500 + 11 800 = 47 300 kg. Fuel at landing = block − taxi − trip = 9720 − 120 − 5080 = 4520 kg, so LM = 47 300 + 4520 = 51 820 kg. The regulated landing mass is the lower of the certificated and performance limits, 51 950 kg, so the LM stays 130 kg below it. Adding the trip fuel to the ZFM (52 380 kg) or forgetting to remove the taxi fuel (51 940 kg) are the classic slips.

8. 8What is the correct way to determine the underload?

   1. Allowable LM + Trip Fuel - DOM = Underload
   2. Actual TOM - DOM - Useful Load = Underload
   3. Allowable TOM - DOM - Useful Load = Underload
   4. Allowable TOM + Useful Load - DOM = Underload
   Show the answer

   Answer C. The underload is the mass still available before the limiting take-off mass is reached. Start from the allowable TOM (the lowest of the RTOM, MZFM + take-off fuel and RLM + trip fuel) and subtract the DOM and the useful load, i.e. the take-off fuel plus the traffic load already booked. Using the actual TOM always gives zero, and adding the useful load makes no sense.

9. 9An aircraft with tricycle landing gear is weighed on jacks. Determine the BEM and the CG position aft of the datum from the following data:
   Nose jack, located 161 in aft of datum, reads: 6488 kg
   Each main wheel jack, located 775 in aft of datum, reads: 17 783 kg

   1. 24 271 kg and 936 in
   2. 24 271 kg and 610.9 in
   3. 42 054 kg and 570.3 in
   4. 42 054 kg and 680.3 in
   Show the answer

   Answer D. There are two main jacks, so BEM = 6488 + 2 × 17 783 = 42 054 kg. Moments: 6488 × 161 = 1 044 568 kg·in and 35 566 × 775 = 27 563 650 kg·in, a total of 28 608 218 kg·in. CG = 28 608 218 ÷ 42 054 ≈ 680.3 in aft of datum. Counting only one main jack gives 24 271 kg, the classic mistake.

10. 10Using the information below, what minimum additional load, in whole kilograms, must be placed in compartment 1 or compartment 2 to bring the CG within limits for take-off?
    Aircraft take-off mass: 1785 kg
    Initial CG position: +2.08 m
    Forward CG limit: +2.14 m
    Aft CG limit: +2.41 m
    Compartment 1: +1.23 m
    Compartment 2: +4.76 m

    1. 108 kg into compartment 2.
    2. 41 kg into compartment 2.
    3. 40 kg into compartment 2.
    4. 118 kg into compartment 1.
    Show the answer

    Answer B. At +2.08 m the CG is forward of the forward limit, so it must move aft by 2.14 − 2.08 = 0.06 m, which means loading compartment 2. Distance from compartment 2 to the target CG = 4.76 − 2.14 = 2.62 m, so m = 1785 × 0.06 / 2.62 = 40.9 kg, rounded up to 41 kg; 40 kg would leave the CG just forward of the limit. Loading compartment 1 would move the CG further forward.