Question 1: Given: Dry Operating Mass= 29 800 kg, Maximum Take-Off Mass= 52 400 kg, Maximum Zero-Fuel Mass= 43 100 kg, Maximum Landing Mass= 46 700 kg, Trip fuel= 4 000 kg, Fuel quantity at brakes release= 8 000 kg. The maximum traffic load is: A B C D

12 900 kg 9 300 kg 14 600 kg 13 300 kg

Explanation: NO Question 2: The loaded centre of gravity (cg) of an aeroplane is 713 mm aft of datum. The mean aerodynamic chord lies between station 524 mm aft and 1706 mm aft. The cg expressed as % MAC (mean aerodynamic chord) is: A B C D

16 60 41 10

% % % %

Explanation: MAC length = 1706-524 = 1182mm Distance between L.E. and CG = 713-524 = 189mm So, CG is 189mm behind the L.E., which is (189/1182) of the MAC = 0.16 = 16% MAC Question 3: The take-off mass of an aeroplane is 117 000 kg, comprising a traffic load of 18 000 kg and fuel of 46 000 kg. What is the dry operating mass? A B C D

53 64 99 71

000 000 000 000

kg kg kg kg

Explanation: Take Off Mass = Dry Operation Mass + Fuel + Traffic Load Transpose: DOM = TOM - Fuel - TL DOM = 117000 - 46000 - 18000 DOM = 53000 kg Question 4: A location in the aircraft which is identified by a number designating its distance from the datum is known as: A Station. B Moment. C MAC.

D Index. Explanation: Station numbers designate the distance of station from the datum. They are part of the build process. Question 5: The mass of an aircraft is 1950 kg. If 450 kg is added to a cargo hold 1.75 metres from the loaded centre of gravity (cg). The loaded cg will move: A B C D

33 40 34 30

cm. cm. cm. cm.

Explanation: mass added ÷ old total mass = change of CG ÷ distance from hold to new CG or mass added ÷ new total mass = change of CG ÷ distance from hold to old CG ..and you use the version that fits the data. Here we have the distance from the hold to the old CG so we must use the new total mass: mass added ÷ new total mass = change of CG ÷ distance from hold to old CG change of CG = (mass added ÷ new total mass) x distance from hold to old CG = (450 ÷ 2400) x 1.75 = 0.328m Question 6: The Dry Operating Mass includes: A crew and crew baggage, catering, removable passenger service equipment, potable water and lavatory chemicals. B fuel and passengers baggage and cargo. C passengers baggage and cargo. D unusable fuel and reserve fuel. Explanation: Dry Operating Mass (DOM) is the total mass of the aeroplane (excluding usable fuel and traffic load) Items contained in the DOM include: - Crew and crew baggage. - Food and beverages. - Potable water and lavatory chemicals. - Catering and removable passenger service equipment. The operator determines the DOM. Once the aircraft is delivered at the BEM the operator adds the masses of the crew and operational items to establish the DOM. Question 7: If 390 Ibs of cargo are moved from compartment B (aft) to compartment A (forward), what is the station number of the new centre of gravity (cg).

Given : Gross mass 116.500 Ibs, Present cg station 435.0, Compartment A station 285.5, Compartment B station 792.5 A B C D

433.3 436.7 506.3 463.7

Explanation: Use the formula for moving mass: Mass moved รท total mass = change of CG รท distance moved therefore change of CG = (mass moved รท total mass) x distance moved change of CG = (390 รท 116500) x 507 change of CG = 1.7 we're moving the load forwards so the CG moves forward, 435.0 - 1.7 = 433.3 Question 8: The centre of gravity of a body is that point A through which the sum of the forces of all masses of the body is considered to act. B where the sum of the moments from the external forces acting on the body is equal to zero. C where the sum of the external forces is equal to zero. D which is always used as datum when computing moments. Explanation: NO Question 9: The stalling speed of an aeroplane will be highest when it is loaded with a: A B C D

high gross mass and forward centre of gravity. low gross mass and aft centre of gravity. high gross mass and aft centre of gravity. low gross mass and forward centre of gravity.

Explanation: NO Question 10: With the centre of gravity on the forward limit which of the following is to be expected? A B C D

A A A A

decrease decrease tendency decrease

in of to in

Explanation: NO

range. the stalling speed. yaw to the right on take-off. the landing speed.

Question 11: The maximum load per running metre of an aircraft is 350 kg/m. The width of the floor area is 2 metres. The floor strength limitation is 300 kg per square metre. Which one of the following crates (length x width x height) can be loaded directly on the floor? A B C D

A A A A

of of of of

400 400 700 500

kg kg kg kg

in in in in

a a a a

crate crate crate crate

with with with with

dimensions dimensions dimensions dimensions

1.2 1.4 1.8 1.5

m m m m

x x x x

1.2 0.8 1.4 1 m

m m m x

x x x 1

1.2 m. 0.8 m. 0.8 m. m.

Explanation: Divide the load by the longest side eg. 400kg / 1.2 = 333.33 so this is in limits as the question gives a maximum load per running metre of 350kg/m. Then to check the Floor strength limit which is 300kg per square metre: Length X Width so eg: 1.2 X 1.2 = 1.44 400 / 1.44 = 277.77 per square metre also within limits. If you repeat this for each option then you can discard all the others either for the linear limit or the floor strength. Take the 700 kg crate: 700 / 1.8 (as this is the longest side) = 388.88 can be discarded as over 350. etc.... Question 12: Given the following : - Maximum structural take-off mass 48 000 kg Maximum structural landing mass: 44 000 kg - Maximum zero fuel mass: 36 000 kg -Taxi fuel: 600 kg -Contingency fuel: 900 kg -Alternate fuel: 800 kg -Final reserve fuel: 1 100 kg -Trip fuel: 9 000 kg. The actual TOM can never be higher than: A B C D

47 53 48 48

800 000 000 400

kg kg kg kg

Explanation: You need 3 columns RTOM = 48000 RLM+TRIP FUEL = 44000+9000 = 53000 MZFM+TAKE OFF FUEL = 36000+11800 = 47800 Then take the lowest of these as the regulated take off mass = MZFM+T/OFF Fuel @ 47800 Question 13: The maximum zero-fuel mass: 1- is a regulatory limitation 2- is calculated for a maximum load factor of +3.5 g 3- is based on the maximum permissible bending moment at the wing root 4- is defined on the assumption that fuel is consumed from the outer wings tank first 5- is defined on the assumption that fuel is consumed from the centre wing tank first 6- can be increased by stiffening the wing The combination of correct statements is: A 1, 3, 5

B 2, 3, 6 C 2, 3, 4, 6 D 1, 2, 3 Explanation: The maximum zero-fuel mass is a fixed limit and is calculated on the maximum permissible bending moment at the wing root. Load factor is type specific and wing stiffening is a trick answer. This leaves fuel dumping/jettisoning: Fuel jettisoning is a procedure used to quickly reduce the mass of the aircraft. If time allows it then it would be safer to dump centre tank fuel first and the inner wing tanks next to maintain wing bending relief. Question 14: If nose wheel moves aft during gear retraction, how will this movement affect the location of the centre of gravity (cg) on the aircraft? A It will cause the cg to move aft. B It will cause the cg to move forward. C The cg location will change, but the direction cannot be told the information given. D It will not affect the cg location. Explanation: When a gear retracts either forward or backwards it is held in the retracted position by an uplock. The uplock hooks onto the leg somewhere near the wheel. So now we have the weight acting on the hinge line and the uplock. The CG will have moved marginally. Question 15: At the flight preparation stage, the following parameters in particular are available for determining the mass of the aircraft: 1- Dry operating mass 2- Operating mass Which statement is correct: A The Dry Operating Mass includes fixed equipment needed to carry out a specific flight. B The Operating Mass includes the traffic load. C The Dry Operating Mass includes take-off fuel. D The Operating Mass is the mass of the aircraft without take-off fuel. Explanation: Dry Operating Mass (DOM) is the total mass of the aeroplane (excluding usable fuel and traffic load) Items contained in the DOM include: - Crew and crew baggage. - Food and beverages. - Potable water and lavatory chemicals. - Catering and removable passenger service equipment. The operator determines the DOM. Once the aircraft is delivered at the BEM the operator adds the masses of the crew and operational items to establish the DOM. Question 16: Which of the following corresponds to zero fuel mass?

A B C D

The take-off mass of an aeroplane minus all usable fuel. Take-off mass minus fuel to destination and alternate. Operating mass plus passengers and cargo. Operating mass plus luggage of passengers and cargo.

Explanation: Zero Fuel Mass (ZFM) = DOM + traffic load - fuel Question 17: Given that: - Maximum structural take-off mass: 146 000 kg, - Maximum structural landing mass: 93 900 kg, - Maximum zero fuel mass: 86 300 kg, - Trip fuel: 27 000 kg, - Taxi fuel: 1 000 kg, - Contingency fuel: 1350 kg, - Alternate fuel: 2650 kg, - Final reserve fuel: 3000 kg, Determine the actual take-off mass: A B C D

120 120 146 121

300 900 000 300

kg. kg. kg. kg.

Explanation: The limiting traffic load could normally be found by subtracting DOM and fuel from each limit. In this case, however, there is no DOM given so traffic load calculations cannot have any bearing on the answer. It must be a fuel thing or a structural limit. The first contender is the structural TOM, 146000kg, too easy? A second is the MZFM plus fuel at take-off.... 86300 + 27000 + 1350 + 2650 + 3000 = 120300kg A third is the MLM plus the trip fuel... 93900 + 27000 = 120900Kg It looks like 120,300kg. Question 18: On an aeroplane without central fuel tank, the maximum Zero Fuel Mass is related to: A B C D

the bending moment at the wing root. wing loaded trip fuel. variable equipment for the flight. Maximum Structural Take-Off Mass.

Explanation: NO Question 19: Given are:- Maximum structural take-off mass: 72 000 kg, - Maximum structural landing mass: 56 000 kg, - Maximum zero fuel mass: 48 000 kg, - Taxi fuel: 800 kg, - Trip fuel: 18 000 kg, - Contingency fuel: 900 kg, - Alternate fuel: 700 kg, - Final reserve fuel: 2 000 kg. The actual take-off mass can never be higher than: A 69 600 kg B 70 400 kg

C 72 000 kg D 74 000 kg Explanation: It would have to mean Maximum Take-Off Mass rather than actual. From the data given you're a bit restricted. It's going to be the lowest of the given structural TOM, 72000kg, or the MLM plus the trip fuel, 56000 + 18000 = 74000, or the MZFM plus all the fuel except the taxy fuel, 48000 + 21600 = 69600. Question 20: Where is the centre of gravity of the aeroplane in the diagram? A B C D

26.57 32.29 26.57 32.29

cm cm cm cm

forward of datum. aft of datum. aft of datum. forward of datum.

Explanation: NO Question 21: The centre of gravity location of the aeroplane is normally computed along the: A B C D

longitudinal axis. horizontal axis. lateral axis. vertical axis.

Explanation: This is a fact you need to learn! Question 22: In mass and balance calculations which of the following describes the datum? A It is the point on the aircraft designated by the manufacturer from which all centre of gravity measurements and calculations are made. B It is the distance from the centre of gravity to the point through which the weight of the component acts. C It is the most forward position of the centre of gravity. D It is the most aft position of the centre of gravity. Explanation: The Datum is found along the longitudinal axis - it is the point from where all CofG calculations and measurements are made. Question 23: An aircraft has its centre of gravity located 7 metres from the datum line and it has a weight of 49000 N. The moment about the datum is: A 343 000 Nm.

B 7000 Nm. C 343 000 N/m. D 1.43 Nm. Explanation: 7m x 49000N = 343000 Nm Check your units! Question 24: Which one of the following is correct? A B C D

Arm = Moment / Force Arm = Force X Moment Arm = Force / Moment Moment = Force / Arm

Explanation: Transpose the formula: Moment = Force x arm Question 25: Given: Total mass 2900 kg Centre of gravity (cg) location station: 115 Aft cg limit station: 116 The maximum mass that can be added at station 130 is: A B C D

207 kg. 317 kg. 140 kg. 14 kg.

Use the formula for adding mass: old total mass = change of CG ÷ distance from hold to new CG (change of CG ÷ distance from hold to new CG) x old total (1.0 ÷ 14.0) x 2900 207kg

Question 26: Given: Total mass: 7500 kg, Centre of gravity (cg) location station: 80.5, Aft cg limit station: 79.5. How much cargo must be shifted from the aft cargo compartment at station 150 to the forward cargo compartment at station 30 in order to move the cg location to the aft limit? A B C D

62.5 65.8 73.5 68.9

kg. kg. kg. kg.

Explanation: Use the formula for moving mass, mass moved ÷ total mass = change of CG ÷ distance moved. Turn the formula around to read: mass moved = (change of CG ÷ distance moved) x total mass substitute the numbers given in the question: mass moved = (1 ÷ 120) x 7500 = 62.499 kg

Question 27: A jet aeroplane, with the geometrical characteristics shown in the appendix, has a take-off weight (W) of 460 000 N and a centre of gravity (point G on annex) located at 15.40 m from the zero reference point. At the last moment the station manager has 12 000 N of freight added in the forward compartment at 10 m from the zero reference point. The final location of the centre of gravity, calculated in percentage of mean aerodynamic chord AB (from point A), is equal to: 311580.gif A B C D

27.5 16.9 35.5 30.4

%. %. %. %.

Explanation: Mass x arm = moment Original mass 460000N x 15.4m = 7084000Nm Add mass 12000N x 10m = 120000Nm Total mass = 472000 N total moment = 7204000Nm New CG = moment / Mass = 7204000 / 472000 = 15.26m LEMac is at 14m therefore CG is 1.26m from LEMac As a %MAC then: 1.26/4.6 x 100 = 27.5%MAC. Question 28: Which is true of the aeroplane empty mass? A B C D

It It It It

is is is is

a component of dry operating mass. dry operating mass minus fuel load. dry operating mass minus traffic load. the actual take-off mass, less traffic load.

Explanation: NO Question 29: The total mass of an aeroplane is 9000 kg. The centre of gravity (cg) position is at 2.0 m from the datum line. The aft limit for cg is at 2.1 m from the datum line. What mass of cargo must be shifted from the front cargo hold (at 0.8 m from the datum) to the aft hold (at 3.8 m), to move the cg to the aft limit? A B C D

300 kg 30.0 kg 196 kg 900 kg

Explanation: Mass Change ÷ Total Mass = Change of C of G ÷ Total Distance Moved Mass Change ÷ 9000 = 0.1 ÷ 3 Mass change = (9000 x 0.1) ÷ 3 Mass change = 300 kg Question 30: Assume: Aircraft actual mass: 4750 kg Centre of gravity at station: 115.8 What will be the new position of the centre of gravity if 100 kg is moved from the station 30 to station 120? A B C D

Station Station Station Station

117.69 120.22 118.25 118.33

Explanation: We're told 100 kg must be moved from Station ("STN") 030 to STN 120, with a CG presently at STN 115.8. Aircraft mass is 4750 kg. What will the new CG be? Look at your M&B notes, "Body Stations" - we'll take the formula (memorise it!): Δ Mass...............Δ CG ---------- = ---------------Total mass........distance moved Viz, 100 kg ...............Δ CG ---------- = ---------------4750 kg..............(120-30) becomes, 0.021 * 90 = 1.89 Original CG was at STN 115.8, so the new CG is STN 117.69. A rearward movement of mass, so a rearward movement of the CG. Question 31: An aeroplane with a two wheel nose gear and four main wheels rests on the ground with a single nose wheel load of 500 kg and a single main wheel load of 6000 kg. The distance between the nose wheels and the main wheels is 10 meter. How far is the centre of gravity in front of the main wheels? A B C D

40 cm. 25 cm. 4 meter. 41.6 cm.

Explanation: The c/g is moment/weight and the total weight is 4*6,000Kg+2*500Kg or 25,000Kg. To get moment you need a datum, and since we're not given one we can choose one. Let's choose the main wheels as the datum, in which case their moment is zero (4*6,000Kg*0m), and the moment of the front wheels will therefore be -10,000Kgm (2*500Kg*-10m); minus because the front wheels are 10m forward

of the mains). The c/g is therefore -10,000Kgm/25,000Kg ie. -0.4m or 40cm in front of the mains. Just to prove that the datum doesn't matter let's pick the front wheels, in which case their moment is zero (2*500Kg*0m), and the moment of the mains is 240,000Kgm (4*6000Kg*10m; positive because the mains are 10m behind the nosewheel. Now the c/g is 240,000Kgm/25,000Kg ie. 9.6m behind the nosewheel or 40cm in front of the mains. Question 32: Considering only structural limitations, on long distance flights (at the aeroplane's maximum range), the traffic load is normally limited by: A B C D

The The The The

Maximum Maximum Maximum Maximum

Take-off Mass. Landing Mass. Zero Fuel Mass plus the Take-off Mass. Zero Fuel Mass.

Explanation: You are always restricted by MZFM whether the flight is long or not, so put a small amount of fuel in and you are going to be well short of the MTOM and hopefully the MLM. If however you put a large amount of fuel in, MTOM and MLM is now a factor. Remember that fuel does not count as part of the traffic load hence it does not affect MZFM. If traffic load remains the same and you add more fuel, you get closer to the MTOM. Therefore to fill the tanks you may need to offload traffic. Question 33: The Zero Fuel Mass of an aeroplane is always: A B C D

the the the the

Take-off Mass minus the mass of take-off fuel. Take-off Mass minus the fuselage fuel mass. Take-off Mass minus the wing fuel mass. Maximum Take-off Mass minus the take-off fuel mass.

Explanation: ZFM is the TOM minus the (usable) take-off fuel mass. There being no exactly correct answer, Question 34: Given: Maximum structural take-off mass= 146 900 kg, Maximum structural landing mass= 93 800 kg, Maximum zero fuel mass= 86 400 kg, Trip fuel= 27 500 kg, Block fuel= 35 500 kg Engine starting and taxi fuel = 1 000 kg. The maximum take-off mass is equal to: A B C D

120 113 120 121

900 900 300 300

kg kg kg kg

Explanation: From the data given, to find the take off mass you need only to add the fuel at take off to the MZFM given of 86400 kg.

If the aircraft has 35500 kg of fuel on the blocks and 1000 kgs of fuel is used for taxi. The mass at take off will be 86400 + 34500 = 120900 kgs. The MZFM is a structural limit that cannot be exceeded. It includes the DOM + Traffic Load. An increase in mass can therefore only be achieved by adding the fuel required ! Question 35: Given: Aeroplane mass = 36 000 kg Centre of gravity (cg) is located at station 17 m What is the effect on cg location if you move 20 passengers (total mass = 1 600 kg) from station 16 to station 23? A B C D

It It It It

moves moves moves moves

aft by 0.31 m. aft by 0.157 m. forward by 0.157 m. aft by 3.22 m.

Explanation: Mass moved รท therefore change of CG change of CG change of CG

Use the formula for moving mass: total mass = change of CG รท distance moved = (mass moved รท total mass) x distance moved = (1600 รท 36000) 7 = 0.31 aft, because they are moving aft.

Question 36: In mass and balance calculations the "index" is: A is a figure without unit of measurement which represents a moment. B an imaginary vertical plane or line from which all measurements are taken. C a location in the aeroplane identified by a number. D the range of moments the centre of gravity (cg) can have without making the aeroplane unsafe to fly. Explanation: Loading Index (LI) is a unspecified figure that represents a scaled down value of a moment. Its purpose is to simplify mass and balance calculations. Question 37: Loads must be adequately secured in order to: A B C D

avoid unplanned centre of gravity (cg) movement and aircraft damage. allow steep turns. avoid any centre of gravity (cg) movement during flight. prevent excessive 'g'-loading during the landing flare.

Explanation: NO Question 38: Traffic load is the:

A B C D

Zero Fuel Mass minus Dry Dry Operating Mass minus Take-off Mass minus Zero Dry Operating Mass minus

Explanation: The Traffic Load is the total mass of passengers, cargo and baggage including any non-revenue load. Question 39: Given the following information, calculate the loaded centre of gravity (cg). Details at reference. 311964.gif A B C D

56.53 56.35 53.35 60.16

cm cm cm cm

aft aft aft aft

datum. datum. datum. datum.

Explanation: Total Moment = 1369350 Total Mass = 24224 Centre of Gravity = Moment รท Mass Centre of Gravity = 1369350 รท 24224 Centre of Gravity = +56.528 cm (Aft of the datum) Question 40: Given are the following information at take-off. Details at reference Given that the flight time is 2 hours and the estimated fuel flow will be 1050 litres per hour and the average oil consumption will be 2.25 litres per hour. The specific density of fuel is 0.79 and the specific density of oil is 0.96. Calculate the landing centre of gravity 311964.gif A B C D

61.28 61.27 61.26 61.29

cm cm cm cm

aft aft aft aft

of of of of

datum. datum. datum. datum.

Explanation: Fuel: 6045 - ( 1050*2*0,79 ) = 4386 Oil: 124 - ( 2,25*2*0,96 ) = 119,68 TOT mass = 22560,68kg TOTAL Moment = 1'382'449,2 kg cm CG = 1'382'449,2 / 22560,68 = 61,2769 Question 41: Given that the total mass of an aeroplane is 112 000 kg with a centre of gravity position at 22.62m aft of the datum. The centre of gravity limits are between 18m and 22m. How much mass must be removed from the rear hold (30 m aft of the datum) to move the centre of gravity to the middle of the limits: A 29 344 kg

B 43 120 kg C 8 680 kg D 16 529 kg Explanation: Mass change/total mass = Change of CG/distance from mass to New CG Transposed with known values substituted we get: Mass change = 112000 x 2.62/10 = 29334 Kgs Note: it is the final effective distance between the extra mass and new CG position that is used. Question 42: What determines the longitudinal stability of an aeroplane ? A The location of the centre of gravity with respect to the neutral point. B The effectiveness of the horizontal stabilizer, rudder and rudder trim tab. C The dihedral, angle of sweepback and the keel effect. D The relationship of thrust and lift to weight and drag. Explanation: The point where the weight vector and lift vector act through the same point is known as the 'neutral point'. An aft limit for the CG (weight vector) is established so that it is in front of the lift vector in order to give positive longitudinal stability. The further forward the CG the greater the longitudinal stability. Question 43: Assuming gross mass, altitude and airspeed remain unchanged, movement of the centre of gravity from the forward to the aft limit will cause A B C D

increased cruise range. reduced maximum cruise range. lower optimum cruising speed. higher stall speed.

Explanation: The CG moving to the aft limit decreases thrust required, decreases fuel burn and reduces trim drag. Therefore increasing range and endurance. Question 44: While making mass and balance calculation for a particular aircraft, the term 'Basic Empty Mass' applies to the sum of airframe, engine(s), fixed ballast plus A unusable fuel and full operating fluids. B all the oil, fuel, and hydraulic fluid but not including crew and traffic load. C all the oil and fuel. D all the consumable fuel and oil, but not including any radio or navigation equipment installed by manufacturer.

Explanation: Basic Empty Mass (Basic Mass) is the mass of an aeroplane plus standard items such as: - emergency oxygen equipment - lubricating oil in engine and auxiliary units - pyrotechnics - unusable fuel and other unusable fluids - supplementary electronic equipment - fire extinguishers Question 45: The term 'Maximum Zero Fuel Mass' consist of: A The maximum permissible mass of an aeroplane with no usable fuel. B The maximum mass authorized for a certain aeroplane not including traffic load and fuel load. C The maximum mass authorized for a certain aeroplane not including the fuel load and operational items D The maximum mass for some aeroplanes including the fuel load and the traffic load Explanation: Maximum Zero Fuel Mass is the maximum permissible mass of an aeroplane with no usable fuel. Question 46: The actual 'Zero Fuel Mass' is equal to the: A B C D

Dry Operating Mass plus the traffic load. Basic Empty Mass plus the fuel loaded. Operating Mass plus all the traffic load. Actual Landing Mass plus trip fuel.

Explanation: Zero Fuel Mass is: DOM + traffic load - fuel. Question 47: The actual 'Take-off Mass' is equivalent to: A B C D

Dry Operating Mass plus take-off fuel and the traffic load Actual Landing Mass plus the take-off fuel Dry Operating Mass plus the take-off fuel Actual Zero Fuel Mass plus the traffic load

Explanation: Take-Off Mass (TOM) is the mass of the aeroplane (all cargo, passengers, etc. included) at the start of the take-off run. Question 48: Calculate the centre of gravity in % MAC (mean aerodynamic chord) with following data: Distance datum - centre of gravity: 12.53 m Distance datum - leading edge: 9.63 m Length of MAC: 8 m

A B C D

36.3 47.0 23.1 63.4

% % % %

MAC MAC MAC MAC

Explanation: ML = 8m CG â&#x20AC;&#x201C;Lemac = 12.53 - 9.63 = 2.9 2.9 / Ml = ((2.9 / 8) * 100) = 36.6% Question 49: Given an aeroplane with: Maximum Structural Landing Mass: 125000 kg Maximum Zero Fuel Mass: 108500 kg Maximum Structural Take-off Mass: 155000 kg Dry Operating Mass: 82000 kg Scheduled trip fuel is 17000 kg and the reserve fuel is 5000 kg. Assuming performance limitations are not restricting, the maximum permitted take-off mass and maximum traffic load are respectively: A B C D

130500 125500 125500 130500

kg kg kg kg

and and and and

26500 21500 26500 31500

kg kg kg kg

Explanation: The traffic load is limited by the MZFM so by definition the maximum mass has to be the MZFM mass allowed can only be that of 108,500 MZFM plus the total fuel. Otherwise the MTOM and MLM limits will both be broken. Question 50: For the purpose of completing the Mass and Balance documentation, the Traffic Load is considered to be equal to the Take-off Mass A B C D

less less plus plus

the the the the

Operating Trip Fuel Trip Fuel Operating

Mass. Mass. Mass. Mass.

Explanation: The DOM + T/O Fuel + Traffic load= TOM. The Operating Mass = DOM + T/O fuel Therefore the Traffic load is TOM - Operating Mass. Also see definitions in CAP 696. Question 51: For the purpose of completing the Mass and Balance documentation, the Operating Mass is considered to be Dry Operating Mass plus A B C D

Take-off Fuel Mass. Ramp Fuel Mass less the fuel for APU and run-up. Ramp Fuel Mass. Trip Fuel Mass.

Explanation: Operating Mass - The Operating Mass is the APS (aircraft prepared for service) mass of the aircraft and the fuel for the flight, Traffic Load is not counted. It can be thought of as the wet Operating Mass as opposed to the DOM defined earlier." If you learn diagram 4-3-1 in the electronic notes this goes a long way to solving all of the probable mass and balance 'definition' questions you're likely to get in the exam. Question 52: When establishing the mass breakdown of an aeroplane, the empty mass is defined as the sum of the: A standard empty mass plus specific equipment mass plus trapped fluids plus unusable fuel mass B basic mass plus special equipment mass C empty mass dry plus variable equipment mass D basic mass plus variable equipment mass Explanation: Basic Empty Mass (Basic Mass) is the mass of an aeroplane plus standard items such as: - emergency oxygen equipment - lubricating oil in engine and auxiliary units - pyrotechnics - unusable fuel and other unusable fluids - supplementary electronic equipment - fire extinguishers Question 53: For the purpose of completing the Mass and Balance documentation, the Dry Operating Mass is defined as: A The total mass of the aircraft ready for a excluding all usable fuel and traffic load. B The total mass of the aircraft ready for a excluding crew and crew baggage. C The total mass of the aircraft ready for a excluding all traffic load. D The total mass of the aircraft ready for a excluding all usable fuel.

specific type of operation specific type of operation specific type of operation specific type of operation

Explanation: Dry Operating Mass (DOM) is the total mass of the aeroplane (excluding usable fuel and traffic load) Items contained in the DOM include: - Crew and crew baggage. - Food and beverages. - Potable water and lavatory chemicals. - Catering and removable passenger service equipment. The operator determines the DOM. Once the aircraft is delivered at the BEM the operator adds the masses of the crew and operational items to establish the DOM.

Question 54: An aeroplane's weighing schedule indicates that the empty mass is 57320 kg. The nominal Dry Operating Mass is 60120 kg and the Maximum Zero Fuel Mass is given as 72100 kg. Which of the following is a correct statement in relation to this aeroplane? A operational items have a mass for this aeroplane is 11980 kg. B operational items have a mass 11980 kg. C operational items have a mass 14780 kg. D operational items have a mass for this aeroplane is 14780 kg.

of 2800 kg and the maximum traffic load of 2800 kg and the maximum useful load is of 2800 kg and the maximum useful load is of 2800 kg and the maximum traffic load

Explanation: Operational items are the difference between DOM & BEM and are therefore 2800kg. The 11980kg is the difference between MZFM & DOM i.e. the traffic load. Question 55: An aeroplane is to depart from an airfield where the performance limited take-off mass is 89200 kg. Certificated maximum masses are as follows: Ramp (taxi) mass- 89930 kg, Maximum Take-off mass- 89430 kg, MaximumLanding mass- 71520 kg, Actual Zero fuel mass- 62050 kg, Fuel on board at ramp: Taxi fuel- 600 kg, Trip fuel- 17830 kg, Contingency, final reserve and alternate -9030 kg. If the Dry Operating Mass is 40970 kg the traffic load that can be carried on this flight is A B C D

21080 21220 20870 21500

kg kg kg kg

Explanation: Create a table. Deduct DOM and fuel from mass to find traffic load MRM MTOM MZFM MLM limit 89930 89200 62050 71520 -DOM 40970 40970 40970 40970 -Fuel 27460 26860 nil 9030 Traffic load 21500 21370 21080 21520 Max traffic load = 21080 kg. Question 56: The empty mass of an aeroplane, as given in the weighing schedule, is 61300 kg. The operational items (including crew) is given as a mass of 2300 kg. If the take-off mass is 132000 kg (including a useable fuel quantity of 43800 kg) the useful load is A B C D

68400 26900 70700 29600

kg kg. kg kg

Explanation: Useful Load = TOM - DOM Dry Operating Mass = Basic Empty Mass + Operational Items 132000 - (61300 + 2300) = 68,400kg. Question 57: The following data applies to an aeroplane which is about to take off: Certified maximum take-off mass - 141500 kg Performance limited take-off mass - 137300 kg Dry Operating Mass - 58400 kg Crew and crew hand baggage mass - 640 kg Crew baggage in hold - 110 kg Fuel on board - 60700 kg From this data calculate the mass of the useful load. A B C D

78900 17450 18200 78150

kg kg kg kg

Explanation: Take Off Mass: DOM + Fuel + Traffic Load Useful Load: Take Off Mass - DOM Question 58: A revenue flight is to be made by a jet transport. The following are the aeroplane's structural limits: -Maximum Ramp Mass: 69 900 kg, -Maximum Take Off Mass: 69 300 kg, Maximum Landing Mass: 58 900 kg, Maximum Zero Fuel Mass: 52 740 kg. The performance limited take off mass is 67 450kg and the performance limited landing mass is 55 470 kg. Dry Operating Mass: 34 900 kg. Trip Fuel: 6 200 kg. Taxi Fuel: 250 kg. Contingency & final reserve fuel:1 300 kg. Alternate Fuel: 1 100 kg. The maximum traffic load that can be carried is: A B C D

17 19 19 19

840 500 200 100

kg kg kg kg

Explanation: Create a table. Check which of the masses gives you the RLTOM and RLM. MRM MTOM MZFM MLM limit 69900 67450 52740 55470 DOM 34900 34900 34900 34900 fuel 8850 8600 nil 2400 Traffic load 26150 23950 17840 18170 Question 59: A revenue flight is to be made by a jet transport. The following are the aeroplane's structural limits: -Maximum Ramp Mass: 69 900 kg, -Maximum Take Off Mass: 69 300 kg, Maximum Landing Mass: 58 900 kg, Maximum Zero Fuel Mass: 52 740 kg. Take Off and Landing mass are not performance limited. Dry Operating Mass: 34 930 kg Trip Fuel: 11 500 kg Taxi Fuel:

250 kg Contingency & final reserve fuel: 1 450 kg Alternate Fuel: 1 350 kg The maximum traffic load that can be carried is: A B C D

17 20 21 21

810 420 070 170

kg kg kg kg

Explanation: Refer to Diagram Remember to check MRM Maximum traffic load = 17810 kg. Question 60: A revenue flight is to be made by a jet transport. The following are the aeroplane's structural limits: -Maximum Ramp Mass: 69 900 kg, -Maximum Take Off Mass: 69 300 kg, Maximum Landing Mass: 58 900 kg, Maximum Zero Fuel Mass: 52 740 kg Take Off and Landing mass are not performance limited. Dry Operating Mass: 34 900 kg Trip Fuel: 11 800 kg Taxi Fuel: 500 kg Contingency & final reserve fuel: 1 600 kg Alternate Fuel:1 900 kg The maximum traffic load that can be carried is: A B C D

17 19 19 19

840 500 100 200

kg kg kg kg

Explanation: Firstly, create a table to find allowable load. The fuel is that in the tanks at the point under consideration. There is an overload at the landing mass of 1100 kg. Reduce the traffic load by 1100kg to 17500kg. Question 62: The following data is extracted from an aeroplane's loading manifest: Performance limited take-off mass 93500 kg Expected landing mass at destination 81700 kg Maximum certificated landing mass 86300 kg Fuel on board 16500 kg During the flight a diversion is made to an en-route alternate which is not 'performance limited' for landing. Fuel remaining at landing is 10300 kg. The landing mass A is 87300 kg and excess structural stress could result B must be reduced to 81700 kg in order to avoid a high speed approach. C is 87300 kg which is acceptable in this case because this is a diversion and not a normal scheduled landing. D is 83200 kg which is in excess of the regulated landing mass and could result in overrunning the runway Explanation: We know the TOM (93500) and the fuel at T/O (16500). We also know the landing fuel at the diversion (10300). Fuel used is 16500-10300 = 6200. Take this away from the TOM gives LM of 87300. As this is greater than the max certified LM, structural stress could result. Question 63: At maximum certificated take-off mass an aeroplane departs from an airfield which is not limiting for either take-off or landing masses. During initial climb the number one engine suffers a contained disintegration. An emergency is declared and the aeroplane returns to departure airfield for an immediate landing. The most likely result of this action will be A a high threshold speed and possible undercarriage or other structural failure. B a landing short resultant from the increased angle of approach due to the very high aeroplane mass. C a landing further along the runway than normal. D a high threshold speed and a shorter stop distance. Explanation: NO Question 64: If other factors are unchanged, the fuel mileage (nautical miles per kg) is A lower with a forward centre of gravity position. B lower with an aft centre of gravity position.

C higher with a forward centre of gravity position. D independent from the centre of gravity position. Explanation: With a forward CG the trim drag increases which in turn increases the fuel burn. Question 65: At reference or see Loading Manual MEP1 Figure 3.4. With respect to multi-engine piston powered aeroplane, determine the ramp mass (lbs) in the following conditions: Basic empty mass: 3 210 lbs, Basic arm: 88.5 Inches, One pilot: 160 lbs, Front seat passenger : 200 lbs, Centre seat passengers: 290 lbs, One passenger rear seat: 110 lbs, Baggage in zone 1: 100 lbs, Baggage in zone 4: 50 lbs, Block fuel: 100 US Gal. Trip fuel: 55 US Gal. Fuel for start up and taxi (included in block fuel): 3 US Gal. Fuel density: 6 lbs/US Gal. 314742.gif A B C D

4 4 4 4

720 390 120 372

Explanation: Basic Empty Mass 3210 Pilot/Front pass 360 Passenger/BZ2 290 Passenger/BZ3 110 Baggage Z1 100 Baggage Z4 50 ZFM therefore equals 4120Lbs Fuel 100 Us Gal at 6Lbs/Gal = 600Lbs ZFM+Fuel = 4720Lbs Question 66: At reference or see Loading Manual MEP1 Figure 3.4. With respect to multi-engine piston powered aeroplane, determine the block fuel moment (lbs.In.) in the following conditions: Basic empty mass: 3 210 lbs. One pilot: 160 lbs. Front seat passenger : 200 lbs. Centre seat passengers: 290 lbs. (total) One passenger rear seat: 110 lbs. Baggage in zone 1: 100 lbs. Baggage in zone 4: 50 lbs. Block fuel: 100 US Gal. Trip fuel: 55 US Gal. Fuel for start up and taxi (included in block fuel): 3 US Gal. Fuel density: 6 lbs./US Gal. Total moment at take-off: 432226 lbs.In 314742.gif A B C D

56 160 9 360 433 906 30 888

Explanation: Refer to Diagram Block fuel = 100USG 6 lbs per USG = 600lbs

600lbs x 93.6 in = 56160 in / lbs Question 67: See Loading Manual MEP1 Figure 3.4. With respect to a multi-engine piston powered aeroplane, determine the total moment (lbs.In) at landing in the following conditions: Basic empty mass: 3 210 lbs. One pilot: 160 lbs. Front seat passenger : 200 lbs. Centre seat passengers: 290 lbs. (total) One passenger rear seat: 110 lbs. Baggage in zone 1: 100 lbs. Baggage in zone 4: 50 lbs. Block fuel: 100 US Gal. Trip fuel: 55 US Gal. Fuel for start up and taxi (included in block fuel): 3 US Gal. Fuel density: 6 lbs./US Gal. Total moment at take-off: 432226 lbs.In 314742.gif A B C D

401 377 433 432

338 746 906 221

Explanation: The total moment at landing would be the total moment at take-off less the moment of fuel used. The fuel used depends on what is meant by trip fuel, is it T/O to touchdown or engine start to engine stop? If the former, the trip fuel is 52 USG or 52 x 6 = 312lb. The balance arm is 93.6 so the moment change is 312 x 93.6 = 29203, the new moment is 432226 - 29203 = 403022. If the latter, the trip fuel is 55 USG or 55 x 6 = 330lb, the moment change is 30888 and the new moment is 432226 - 30888 = 401338. This is the only answer that matches so this must be what they mean. Question 68: At reference or see Loading Manual MEP1 Figure 3.4. With respect to a multi-engine piston powered aeroplane, determine the CG location at take off in the following conditions: Basic empty mass: 3 210 lbs. One pilot: 160 lbs. Front seat passenger : 200 lbs. Centre seat passengers: 290 lbs. (total) One passenger rear seat: 110 lbs. Baggage in zone 1: 100 lbs. Baggage in zone 4: 50 lbs. Zero Fuel Mass: 4120 lbs. Moment at Zero Fuel Mass: 377751 lbs.In Block fuel: 100 US Gal. Trip fuel: 55 US Gal. Fuel for start up and taxi (included in block fuel): 3 US Gal. Fuel density: 6 lbs./US Gal. 314742.gif A B C D

91.92 91.84 93.60 91.69

inches inches inches inches

aft aft aft aft

of of of of

datum datum datum datum

Explanation: ZFM = 4120 lbs ZFM moment = 377751 in.lbs Add fuel : 100 USG = 600lbs. moment = 56160 in.lbs Ramp mass = 4720 lbs moment 433911 in.lbs Minus 18 lbs taxi fuel: moment - 1684.8 in. lbs T/O mass 4702 lbs moment 432226.2 in.lbs To CG = 432226.2/4702 = 91.92in aft of the datum.

Question 69: The crew of a transport aeroplane prepares a flight using the following data: - Dry operating mass: 90 000 kg - Block fuel: 30 000 kg - Taxi fuel: 800 kg - Maximum take-off mass: 145 000 kg The traffic load available for this flight is: A B C D

25 55 25 55

800 800 000 000

kg kg kg kg

Explanation: since 145 000 - 90 000 - 30 000 + 800 is in fact 25 8000 this should be right. You have 30 000 fuel onboard when going off-block, burn 800 when taxiing and can take a load of 25 800 not violating the MTOM. Just consider the MTOM being performance limited, e.g. by field-length. In this case a possible Max. Ramp Weight will be way higher than what's possible for you to take off-block and does not matter. Question 70: The planned take-off mass of a turbojet aeroplane is 190 000 kg, with its centre of gravity located at 29 % MAC (Mean Aerodynamic Cord) . Shortly prior to engine start, the local staff informs the flight crew that 4 000 kg must be loaded in cargo 4. After the handling operation, the new centre of gravity location in % MAC will be: 60-0232.gif A B C D

33 31 27 25

% % % %

Explanation: From the graphic loading an extra 4000 kg into Cargo 4 causes the aircraft CG to move to 33%. Question 71: At reference or see Loading Manual SEP1 Figure 2.4. With respect to a single-engine piston powered aeroplane, determine the zero fuel moment (lbs.In./100) in the following conditions: Basic Empty Mass: 2415 lbs. Arm at Basic Empty Mass: 77,9 In. Cargo Zone A: 350 lbs. Baggage Zone B: 35 lbs. Pilot and front seat passenger : 300 lbs (total) 316564.gif A B C D

2548,8 2496,3 6675 2311,8

Explanation: Basic moment 77.9 x 2415 = 188128.5/100 = 1881.3 (to one decimal place)

Pilots 79 x 300 = 23700 /100 = 237 Zone A 108 x 350 = 37800 /100 = 378 Zone B 150 x 35 = 5250 /100 = 52.5 Total = 2548.8 Question 72: Determine the Zero Fuel Mass for the following single engine aeroplane. Given : Basic Empty Mass: 1799 lbs Optional Equipment: 35 lbs Pilot Front seat passenger : 300 lbs Cargo Mass : 350 lbs Ramp Fuel = Block Fuel : 60 Gal. Trip Fuel : 35 Gal. Fuel density : 6 lbs/Gal. A B C D

2449 2659 2414 2589

lbs lbs lbs lbs

Explanation: The optional equipment of 35lbs is (as far as the examiner is concerned) part of the BEM this gives the ZFM as: BEM 1799lbs + Pilot and Pax 300lbs +cargo 350 lbs = 2449lbs Question 73: Determine the Landing Mass for the following single engine aeroplane. Given: Standard Empty Mass :1764 lbs Optional Equipment : 35 lbs Pilot Front seat passenger : 300 lbs Cargo Mass : 350 lbs Ramp Fuel = Block Fuel : 60 Gal. Trip Fuel : 35 Gal. Taxi Fuel 1.7 Gal. Fuel density: 6 lbs/Gal Determine the expected landing mass. A B C D

2589 2449 2799 2659

lbs lbs lbs lbs

Explanation: Remaining fuel = block fuel - taxi fuel -trip fuel = 60gal 1.7gal - 35gal = 23.3gal 23.3gal = 139.8lbs (6lb.gal) Landing mass = 1764lb + 35lb+300lb+350lb+139.8lb = 2588.9lb Question 74: To calculate the allowable take-off mass, the factors to be taken into account include: A B C D

the the the the

sum of the Maximum Landing Mass and the trip fuel. Maximum Take-off Mass minus the trip fuel. sum of the Maximum Zero Fuel Mass and the trip fuel. sum of the Maximum Landing Mass and the fuel on board at take-off.

Explanation: The sum of the Maximum Landing Mass and the trip fuel - Is CORRECT because you can take- off at the MLM plus trip fuel because you

would land at MLM, in the MLM would be the reserve, contingency and diversion fuel. The sum of the Maximum Zero Fuel Mass and the trip fuel - This cannot be right because the allowable TOM is not the MZFM plus trip fuel but the MZFM plus take-off fuel, assuming it doesn't exceed the MTOM The Maximum Take-off Mass minus the trip fuel - This is nonsense! The MTOM minus trip fuel will not give the allowable TOM! The sum of the Maximum Landing Mass and the fuel on board at take-off It can't be the MLM plus take-off fuel because this would mean you would land above the MLM as you would only plan to burn the trip fuel. Thus you would be landing at MLM plus reserve, contingency and diversion fuel. Question 75: Allowed traffic load is the difference between : A B C D

allowed take off mass and operating mass allowed take off mass and basic mass operating mass and basic mass allowed take off mass and basic mass plus trip fuel

Explanation: NO Question 76: Given: Dry operating mass = 38 000 kg maximum structural take-off mass = 72 000 kg maximum landing mass = 65 000 kg maximum zero fuel mass = 61 000 kg Fuel burn = 8 000 kg Take-off Fuel = 10 300 kg The maximum allowed take-off mass and payload are respectively : A B C D

71 73 71 73

300 000 300 000

kg kg kg kg

and and and and

23 24 25 27

000 700 300 000

kg kg kg kg

Explanation: Refer to Diagram Assume that the Max Structural TOM is also the RTOM. The fuel at landing = take-off fuel - trip fuel 10,300 - 8,000 = 2,300kg Max Traffic Load is the lowest shown in the table - 23,000kg. This is half of the answer. Now, find the TOM with this traffic load DOM + traffic load + take-off fuel = TOM 38,000 + 10,300 + 23,000 = 71,300kg. Question 77: Maximum allowed take-off mass limit: 37 200kg Dry Operating Mass: 21 600 kg Take-off fuel: 8 500 kg Passengers on board: male 33,female 32, children 5 Baggages: 880 kg The company uses the standard passenger mass systems (see annex) allowed by regulations. The flight is not a holiday charter. In these conditions, the maximum cargo that may be loaded is 31021.gif

A B C D

585 kg 1 098 kg 1 105 kg 901 kg

Explanation: NO Question 78: Length of the mean aerodynamic chord = 1 m Moment arm of the forward cargo: -0,50 m Moment arm of the aft cargo: + 2,50 m The aircraft mass is 2 200 kg and its centre of gravity is at 25% MAC To move the centre of gravity to 40%, which mass has to be transferred from the forward to the aft cargo hold? A B C D

110 183 104 165

kg kg kg kg

Explanation: Mac=1m moving from 25% to 40% = moving from 0.25 to 0.40 = 0.15m Length between holds = -50 to +2.50 = 3m Total mass = 2200kg Formula : d/D = m/M (d=distance cg moves , D= distance between holds m=mass to be moved , M=total mass) 0.15/3x2200=110 Question 79: Which statement regarding the relationship between traffic load and range is correct? A The B The C The Mass. D The

traffic load can be limited by the desired range. maximum traffic load is not limited by the reserve fuel quantity. Maximum Landing Mass is basically equal to the Maximum Zero Fuel maximum zero fuel mass limits the maximum quantity of fuel.

Explanation: Traffic load may have to be exchanged for fuel in order to extend range. Question 80: Which of the following statements is correct? A A tail heavy aeroplane is less stable and stalls at a lower speed than a nose heavy aeroplane B If the actual centre of gravity is located behind the aft limit the aeroplane longitudinal stability increases. C The station (STA) is always the location of the centre of gravity in relation to a reference point, normally the leading edge of the wing at MAC

D The centre of gravity is given in percent of MAC calculated from the leading edge of the wing, where MAC always = the wing chord halfway between the centre line of the fuselage and the wing tip Explanation: The question is talking about the C of G being towards the rear of the envelope compared to near the front - as opposed to out of the envelope on either side. Within the envelope, the further forward (nose heavy) the greater the stability (good thing) but the price is paid in greater wing loading induced by tail-down elevator to balance the forward C of G. This leads to higher stalling speed (bad thing). If tail heavy, by contrast, stability is reduced (bad thing) but stall speed is lower (good thing) as there is less need for the wing-loadinducing tail-down elevator. Question 81: Which of the following statements is correct? A A tail heavy aeroplane is less stable and stalls at a lower speed than a nose heavy aeroplane B If the actual centre of gravity is located behind the aft limit of centre of gravity it is possible that the aeroplane will be unstable, making it necessary to increase elevator forces C The lowest stalling speed is obtained if the actual centre of gravity is located in the middle between the aft and forward limit of centre of gravity D If the actual centre of gravity is close to the forward limit of the centre of gravity the aeroplane may be unstable, making it necessary to increase elevator forces Explanation: NO Question 82: Which of the following statements is correct? A The Maximum Landing Mass of an aeroplane is restricted by structural limitations, performance limitations and the strength of the runway. B The Maximum Zero Fuel Mass ensures that the centre of gravity remains within limits after the uplift of fuel. C The Maximum Take-off Mass is equal to the maximum mass when leaving the ramp. D The Basic Empty Mass is equal to the mass of the aeroplane excluding traffic load and useable fuel but including the crew. Explanation: NO Question 83: Given an aeroplane with: Maximum Structural Landing Mass: 68000 kg Maximum Zero Fuel Mass: 70200 kg Maximum Structural Take-off Mass: 78200 kg Dry Operating Mass : 48000 kg Scheduled trip fuel is 7000 kg and the reserve fuel is 2800 kg, Assuming performance limitations are not

restricting, the maximum permitted take-off mass and maximum traffic load are respectively: A B C D

75000 75000 77200 77200

kg kg kg kg

and and and and

17200 20000 19400 22200

kg kg kg kg

Explanation: While this question would be much better if the comment "no extra fuel" was added after the reserve fuel figure, we have to work with what we're given! First calculate the ATOM, which is the lower of: 70200kg (MZFM)+ 9800kg (T/O fuel) = 80 000kg, 78200kg (RTOM), and 68000kg (RLLM) + 7000kg (trip) = 75 000kg ANSWER. Step 2 subtract from the ATOM the DOM and T/O fuel 75 000kg - 48 000kg - 9800kg = 17 200kg. Note that the T/O fuel is the trip fuel plus reserve fuel (contingency, alternate and final reserve)... Question 84: Moment (balance) arms are measured from a specific point to the body station at which the mass is located. That point is known as A B C D

the the the the

datum. focal point. centre of gravity of the aircraft. axis.

Explanation: The datum is the point on the longitudinal axis from which the centre of gravity of all masses are referenced. Question 85: At reference or see Loading Manual MRJT 1 page 20. For the medium range twin jet the datum point is located 319583A.gif A B C D

540 inches forward of the front spar. at the leading edge of the Mean Aerodynamic Chord (MAC). 540 cm forward of the front spar. on the nose of the aeroplane.

Explanation: From the diagram: The front spar is 540 inches behind the datum. Question 86: The centre of gravity of an aircraft is that point through which the total mass of the aircraft is said to act. The weight acts in a direction A parallel to the gravity vector. B always parallel to the aircraft's vertical axis.

C governed by the distribution of the mass within the aircraft. D at right angles to the flight path. Explanation: Centre of Gravity is the point through which the force of gravity is said to act on a mass. Question 87: When an aircraft is stationary on the ground, its total weight will act vertically A B C D

through through through through

its centre of gravity. its centre of pressure. a point defined as the datum point. the main wheels of its undercarriage assembly.

Explanation: Centre of Gravity is the point through which the force of gravity is said to act on a mass. Question 88: The weight of an aircraft, which is in level non accelerated flight, is said to act A B C D

vertically through the centre of gravity. always along the vertical axis of the aircraft. vertically through the centre of pressure. vertically through the datum point.

Explanation: Centre of Gravity is the point through which the force of gravity is said to act on a mass. Question 89: In relation to an aeroplane, the term ' Basic Empty Mass' includes the mass of the aeroplane structure complete with its powerplants, systems, furnishings and other items of equipment considered to be an integral part of the particular aeroplane configuration. Its value is A found in the latest version of the weighing schedule as corrected to allow for modifications. B printed in the loading manual and includes unusable fuel. C inclusive of an allowance for crew, crew baggage and other operating items. It is entered in the loading manifest. D found in the flight manual and is inclusive of unusable fuel plus fluids contained in closed systems. Explanation: Basic Empty Mass (Basic Mass) is the mass of an aeroplane plus standard items such as: - emergency oxygen equipment - lubricating oil in engine and auxiliary units - pyrotechnics - unusable fuel and other unusable fluids - supplementary electronic equipment

- fire extinguishers Question 90: An aeroplane is weighed and the following recordings are made: nose wheel assembly scale 5330 kg left main wheel assembly scale 12370 kg right main wheel assembly scale 12480 kg If the 'operational items' amount to a mass of 1780 kg with a crew mass of 545 kg, the empty mass, as entered in the weight schedule, is A B C D

30180 31960 32505 28400

kg kg kg kg

Explanation: Empty mass does not include „operational items‟ or crew so total the weights on the wheels, 5330 + 12370 + 12480 = 30,180. Question 91: If individual masses are used, the mass of an aircraft must be determined prior to initial entry into service and thereafter A B C D

at intervals of 4 years if no modifications have taken place. only if major modifications have taken place. at intervals of 9 years. at regular annual intervals.

Explanation: This is a fact you have to learn! Question 92: The empty mass of an aeroplane is recorded in A the weighing schedule and is amended to take account of changes due to modifications of the aeroplane. B the loading manifest. It differs from the zero fuel mass by the value of the 'traffic load'. C the weighing schedule. If changes occur, due to modifications, the aeroplane must be re-weighed always. D the loading manifest. It differs from Dry Operating Mass by the value of the 'useful load'. Explanation: NO Question 93: Prior to departure an aircraft is loaded with 16500 litres of fuel at a fuel density of 780 kg/m³. This is entered into the load sheet as 16500 kg and calculations are carried out accordingly. As a result of this error, the aircraft is A lighter than anticipated and the calculated safety speeds will be too high

B heavier than anticipated and the calculated safety speeds will be too low. C lighter than anticipated and the calculated safety speeds will be too low D heavier than anticipated and the calculated safety speeds will be too high Explanation: 16 500 x 0.78 = 12 870 kg 16 500 litres at SG 0.78 = 12 870kgs. Therefore, the aircraft will be lighter. The V1, Vr and V2 speeds will be too high. On CRP-5: Align 16 500 kg on Inner Scale with KM-M-LTR Index on Outer Scale Look up 0.78 (780 kg/mÂł) on Specific Gravity Sp.G KG Scale Read 12 900 kg on Inner Scale. Question 94: An additional baggage container is loaded into the aft cargo compartment but is not entered into the load and trim sheet. The aeroplane will be heavier than expected and calculated take-off safety speeds A B C D

will give reduced safety margins. are unaffected but V1 will be increased. will not be achieved. will be greater than required.

Explanation: The V1,Vr and V2 speeds will be too low. The Take Off Distance will be greater than expected which will reduce the safety margins Question 95: Fuel loaded onto an aeroplane is 15400 kg but is erroneously entered into the load and trim sheet as 14500 kg. This error is not detected by the flight crew but they will notice that A B C D

speed at un-stick will be higher than expected the aeroplane will rotate much earlier than expected. V1 will be reached sooner than expected V1 will be increased.

Explanation: Vr and V2 will definitely have increased in value and probably V1 subject to ASDA. But we didn't know about the increase in weight. Once V1 has been fixed and you've started the take-off you can't change it. What we will notice is that the aircraft doesn't lift off at the speed we expected but at some point later down the runway. We can therefore say that the safety margins have been reduced. Question 96:

At reference or see Loading Manual MRJT 1 Figure 4.11. At the maximum landing mass the range of safe CG positions, as determined from the appropriate graph in the loading manual, is: 319596.gif A B C D

Forward Forward Forward Forward

limit limit limit limit

7.4% 8.0% 8.6% 8.0%

MAC MAC MAC MAC

aft aft aft aft

limit limit limit limit

27.0% 26.8% 27.0% 27.2%

MAC MAC MAC MAC

Explanation: The general gist with the envelope is to obviously ensure that the calculation point lies inside the 'shape', (ie to the right of the 'forward' limit, to the left of the 'aft' limit), otherwise the a/c is operating outside its CG limits. Question 97: At a given mass the CG position is at 15% MAC. If the leading edge of MAC is at a position 625.6 inches aft of the datum and the MAC is given as 134.5 inches determine the position of the CG in relation to the datum. A B C D

645.78 inches aft of datum 228.34 inches aft of datum 20.18 inches aft of datum 605.43 inches aft of datum

Explanation: The MAC is 134.5" long. The CG is 15% of this distance back from the leading edge. 15% of 134.5" is 20.17". The leading edge of the MAC is 625.6" aft of the datum, the CG is 625.6 + 20.17 = 645.77" aft. Question 98: At reference or see Loading Manual MRJT 1 Figure 4.11. The aeroplane has a Take Off Mass of 58 000 kg. At this mass the range of safe CG positions, as determined from the appropriate graph in the loading manual, is: 319596.gif A B C D

Forward Forward Forward Forward

limit limit limit limit

8.0% 8.5% 9.5% 8.2%

MAC MAC MAC MAC

aft aft aft aft

limit limit limit limit

26.5% 26.1% 26.1% 26.2%

MAC MAC MAC MAC

Explanation: Refer to Diagram: The general rule with the envelope is to ensure that the calculation point lies inside the 'shape' on the diagram (and therefore within the 'forward' and 'aft' limits) - otherwise the a/c is operating outside its CG limits. Question 99: When preparing to carry out the weighing procedure on an aircraft, which of the following is not required?

A B C D

drain all removable drain all drain all

engine tank oil. passenger services equipment to be off-loaded. chemical toilet fluid tanks. useable fuel.

Explanation: Basic Empty Mass (Basic Mass) is the mass of an aeroplane plus standard items such as: - emergency oxygen equipment - lubricating oil in engine and auxiliary units - pyrotechnics - unusable fuel and other unusable fluids - supplementary electronic equipment - fire extinguishers Question 100: An aircraft is weighed prior to entry into service. Who is responsible for deriving the Dry Operational Mass from the weighed mass by the addition of the 'operational items' ? A B C D

The The The The

Operator. commander of the aircraft. aircraft manufacturer or supplier. appropriate Aviation Authority.

Explanation: The operator is responsible for the addition of the operational items to establish the DOM. Question 101: An aircraft may be weighed A B C D

in in in at

an enclosed, non-air conditioned, hangar. a quiet parking area clear of the normal manoeuvring area. an area of the airfield set aside for maintenance. a specified 'weighing location' on the airfield.

Explanation: JAR OPS states that the aircraft must be weighed in an enclosed building- no mention of "non air conditioned". We believe that the UK exam answer just reflects the requirements of JAR OPS. Question 102: At reference or see Loading Manual MRJT 1 Figure 4.11. An aeroplane has a landing mass of 53 000kg. The range of safe CG positions, as determined from the appropriate graph in the loading manual, is : 319596.gif A B C D

Forward Forward Forward Forward

limit limit limit limit

7.8% 8.7% 8.2% 7.3%

MAC MAC MAC MAC

aft aft aft aft

limit limit limit limit

27.0% 26.8% 27.0% 26.8%

MAC MAC MAC MAC

Explanation: Refer to Diagram The general rule with the envelope is to ensure that the calculation point lies within the 'shape' (and therefore within the 'forward' and 'aft' limits), otherwise the a/c is operating outside its CG limits! Question 103: At reference or see Loading Manual MRJT 1 Figure 4.11. The aeroplane has a mass of 61 000 kg in the cruise. The range of safe CG positions, as determined from the appropriate graph in the loading manual, is: 319596.gif A B C D

forward forward forward forward

limit limit limit limit

8.3% 8.0% 7.7% 7.6%

aft aft aft aft

limit limit limit limit

26.3% 27.2% 25.2% 26.9%

MAC MAC. MAC MAC.

Explanation: Refer to Diagram The general rule with the envelope is to ensure that the calculation point lies whithin the 'shape' (and therfore within the 'forward' and 'aft' limits), otherwise the a/c is operating outside its CG limits. Question 104: At reference or see Loading Manual MRJT 1 Figure 4.9. For the transport aeroplane the moment (balance) arm (B.A.) for the forward hold centroid is: 319605.gif A B C D

367.9 inches. 314.5 inches. 421.5 inches. 257 inches.

Explanation: FWD-367.9 AFT- 884.5 Be careful as each cargo hold has three compartments with different individual centroids. There are questions that ask you to match centroids with compartment loads. ie. the aft cargo hold fwd compartment between BA 731 and 940 3062 kgs and 835.5 inches Question 105: Which of the following is unlikely to have any effect on the position of the centre of gravity on an aeroplane in flight ? A B C D

Changing the tailplane (horizontal stabiliser) incidence angle. Normal consumption of fuel for a swept wing aeroplane. Lowering the landing gear. Movement of cabin attendants going about their normal duties.

Explanation: Changing the tailplane (horizontal stabiliser) incidence angle. Moving the stabiliser re-trims the aircraft by moving the CP to counteract CG changes. All the others will have some minor effect on the CG. Question 106: At reference or see Loading Manual MRJT 1 Figure 4.9. Referring to the loading manual for the transport aeroplane, the maximum running load for the aft section of the forward lower deck cargo compartment is: 319605.gif A B C D

13.12 kg per inch. 14.65 kg per inch. 7.18 kg per inch. 13.15 kg per inch.

Explanation: See annex Question 107: At reference or see Loading Manual MRJT 1 Figure 4.9. Referring to the loading manual for the transport aeroplane, the maximum load intensity for the lower forward cargo compartment is: 319605.gif A B C D

68 kg per square foot. 150 kg per square foot. 3305 kg in forward compartment and 4187 kg in aft compartment. 7288 kg in forward compartment and 9232 kg in aft compartment.

Explanation: See annex Question 108: The maximum floor loading for a cargo compartment in an aeroplane is given as 750 kg per square metre. A package with a mass of 600 kg. is to be loaded. Assuming the pallet base is entirely in contact with the floor, which of the following is the minimum size pallet that can be used ? A B C D

40 30 40 30

cm cm cm cm

by by by by

200 200 300 300

cm cm cm cm

Explanation: If 750kg can be spread over a square metre, 650kg could only be distributed over a smaller area, and smaller in proportion. 600 / 750 = 0.8 so we're looking for an answer that gives an area of 0.8Â˛ metres, Question 109:

The maximum intensity floor loading for an aeroplane is given in the Flight Manual as 650 kg per square metre. What is the maximum mass of a package which can be safely supported on a pallet with dimensions of 80 cm by 80 cm? A B C D

416.0 kg 101.6 kg 41.6 kg 1015.6 kg

Explanation: The floor supports 650kg per m². The area of the pallet is 0.8 x 0.8 = 0.64 m². If 650kg can be distributed over a m² the smaller area can only take a load of 650 x 0.64 = 416kg. Question 110: The distance from the datum to the Centre of Gravity of a mass is known as A B C D

the the the the

moment arm or balance arm. moment. index. force.

Explanation: Balance Arm (BA) is the distance from the datum to the centre of gravity of a mass. Question 111: At reference or see Loading Manual MRJT 1 Figure 4.9. A pallet having a freight platform which measures 200 cm x 250 cm has a total mass of 300 kg. The pallet is carried on two ground supports each measuring 20 cm x 200 cm. Using the loading manual for the transport aeroplane, calculate how much mass may be added to, or must be off loaded from, the pallet in order for the load intensity to match the maximum permitted distribution load intensity for lower deck forward cargo compartment. 319605.gif A B C D

Explanation: The ground supports are significant as they form the contact between the pallet and the floor of the cargo hold. The contact area is 0.2m x 2m twice or 0.8 m². There are 3.28 x 3.28 = 10.76 ft² in a m² so 0.8 m² is 0.8 x 10.76 = 8.61 ft². The maximum load distribution is 68kg per ft² so the maximum load on the pallet is 68 x 8.61 = 585.3kg, 285.3kg more than the present load. Question 112:

During take-off you notice that, for a given elevator input, the aeroplane rotates much more rapidly than expected. This is an indication that: A B C D

the the the the

centre of centre of centre of aeroplane

gravity may be towards the aft limit. pressure is aft of the centre of gravity. gravity is too far forward. is overloaded.

Explanation: A CG at the aft limit will result in reduced longitudinal static stability, this means it is easy to displace in pitch which means that for a given control force it will pitch more rapidly. An overloaded aircraft would be closer to the stall, but could be stable or unstable, a forward CG would make the pitch forces greater and the CP is usually behind the CG. Question 113: Dry Operating Mass is the mass of the aircraft less A B C D

usable fuel and traffic load. usable fuel, potable water and lavatory chemicals. traffic load, potable water and lavatory chemicals. usable fuel.

Explanation: Dry Operating Mass (DOM) is the total mass of the aeroplane (excluding usable fuel and traffic load) Items contained in the DOM include: - Crew and crew baggage. - Food and beverages. - Potable water and lavatory chemicals. - Catering and removable passenger service equipment. The operator determines the DOM. Once the aircraft is delivered at the BEM the operator adds the masses of the crew and operational items to establish the DOM. Question 114: The total mass of the aeroplane including crew, crew baggage; plus catering and removable passenger equipment; plus potable water and lavatory chemicals but excluding usable fuel and traffic load, is referred to as: A B C D

Dry Operating Mass. Aeroplane Prepared for Service ( APS) Mass. Maximum Zero Fuel Mass Zero Fuel Mass.

Explanation: Dry Operating Mass (DOM) is the total mass of the aeroplane (excluding usable fuel and traffic load) Items contained in the DOM include: - Crew and crew baggage. - Food and beverages. - Potable water and lavatory chemicals.

- Catering and removable passenger service equipment. The operator determines the DOM. Once the aircraft is delivered at the BEM the operator adds the masses of the crew and operational items to establish the DOM. APS is the old terminology for DOM, so in some of the older ODM's that is what they'll use so that's why our notes describe both. The examiner wants DOM. If we could challenge the question we would, but the CAA just bury their heads in the sand! Question 115: The responsibility for determination of the mass of 'operating items' and 'crew members' included within the Dry Operating Mass lies with A B C D

the the the the

operator. authority of the state of registration. person compiling the weighing schedule. commander.

Explanation: Dry Operating Mass (DOM) is the total mass of the aeroplane (excluding usable fuel and traffic load) Items contained in the DOM include: - Crew and crew baggage. - Food and beverages. - Potable water and lavatory chemicals. - Catering and removable passenger service equipment. The operator determines the DOM. Once the aircraft is delivered at the BEM the operator adds the masses of the crew and operational items to establish the DOM. Question 116: If the centre of gravity is near the forward limit the aeroplane will: A require elevator trim which will result in an increase in fuel consumption. B benefit from reduced drag due to the decrease in angle of attack. C tend to over rotate during take-off. D require less power for a given airspeed. Explanation: A CG towards the forward limit will cause a pitch down moment which has to be corrected with elevator/stabiliser trim. This increases the drag, which in turn increases the thrust required. Fuel burn increases and therefore range and endurance decrease. Question 117: An aeroplane is said to be 'neutrally stable'. This is likely to: A be caused by a centre of gravity which is towards the rearward limit. B cause the centre of gravity to move forwards.

C be totally unrelated to the position of the centre of gravity. D be caused by a centre of gravity which is towards the forward limit. Explanation: An aircraft is neutrally stable when the lift/weight couple acts through the same point. It is close to the aft limit. Question 118: The Dry Operating Mass is the total mass of the aircraft ready for a specific type of operation but excluding A B C D

usable fuel and traffic load. usable fuel and crew. usable fuel, potable water and lavatory chemicals. potable water and lavatory chemicals.

Explanation: NO Question 119: The Take-off Mass of an aeroplane is 66700 kg which includes a traffic load of 14200 kg and a usable fuel load of 10500 kg. If the standard mass for the crew is 545 kg the Dry Operating Mass is A B C D

42000 41455 42545 56200

kg kg kg kg

Explanation: The DOM includes the crew. Check out definitions in CAP696 and your notes. AUM - TL - FUEL = DOM 66700kgs - 14200kgs - 10500kgs = 42000kgs. Question 120: When the centre of gravity is at the forward limit, an aeroplane will be: A extremely pitch. B extremely C extremely D extremely pitch.

stable and will require excessive elevator control to change stable and require small elevator control to change pitch. unstable and require small elevator control to change pitch. unstable and require excessive elevator control to change

Explanation: The aircraft will be very stable. Elevator control forces will be high but within limits. Question 121: If the centre of gravity of an aeroplane moves forward during flight the elevator control will :

A become pitch B become C become D become pitch.

heavier making the aeroplane more difficult to manouevre in lighter making the aeroplane more easy to manouevre in pitch. heavier making the aeroplane more easy to manouevre in pitch. lighter making the aeroplane more difficult to manouevre in

Explanation: NO Question 122: An aeroplane is loaded with its centre of gravity towards the rear limit. This will result in: A B C D

an increased risk of stalling due to a decrease in tailplane moment a reduction in power required for a given speed. an increase in longitudinal stability. a reduced fuel consumption as a result of reduced drag.

Explanation: Though both “an increased risk of stalling due to a decrease in tailplane moment” and “a reduced fuel consumption as a result of reduced drag” are both factually correct answers to the question; “an increased risk of stalling due to a decrease in tailplane moment” is the answer the examiner will be looking for and the answer to memorize in this case. Question 123: An aeroplane must be re-weighed at certain intervals. Where an operator uses 'fleet masses' and provided that changes have been correctly documented, this interval is A B C D

9 years for each aeroplane. whenever the Certificate of Airworthiness is renewed. whenever a major modification is carried out. 4 years for each aeroplane.

Explanation: 4 years for non fleet aircraft 9 years for fleet aircraft. Question 124: A flight benefits from a strong tail wind which was not forecast. On arrival at destination a straight in approach and immediate landing clearance is given. The landing mass will be higher than planned and A B C D

the the the the

landing distance required will be longer. approach path will be steeper. approach path will be steeper and threshold speed higher. landing distance will be unaffected.

Explanation: NO

Question 125: At reference or see Loading Manual MRJT 1 Figure 4.9. From the loading manual for the jet transport aeroplane, the maximum floor loading intensity for the aft cargo compartment is : 319605.gif A B C D

68 kg per square foot. 68 Lbs per square foot. 68 kg per square metre. 150 kg per square foot.

Explanation: See annex Question 126: At reference or see Loading Manual MRJT 1 Figure 4.9. From the loading manual for the transport aeroplane, the aft cargo compartment has a maximum total load of : 319605.gif A B C D

4187 1568 3062 9232

kg kg kg kg

Explanation: See annex Question 127: At reference or see Loading Manual MRJT 1 Figure 4.9. From the Loading Manual for the transport aeroplane, the maximum load that can be carried in that section of the aft cargo compartment which has a balance arm centroid at : 319605.gif A B C D

835.5 835.5 421.5 421.5

inches inches inches inches

is is is is

3062 6752 2059 4541

kg. kg. Lbs. kg.

Explanation: See annex Question 128: A mass of 500 kg is loaded at a station which is located 10 metres behind the present Centre of Gravity and 16 metres behind the datum. (Assume: g=10 m/sec squared). The moment for that mass used in the loading manifest is : A B C D

80000 Nm 30000 Nm 130000 Nm 50000 Nm

Explanation: The SI unit of Mass is the Kg. The mass of a body is constant. Force is calculated by Mass x Acceleration and the SI unit of force is the Newton (N). The force of gravity can change. The mass of an aircraft is acted on by the acceleration due to gravity. In this question you are told the value of acceleration due to gravity to use. (10 meters per second squared.) 500 kg x 10 = 5000 N. This force acts at a distance of 16 m. 5000 x 16 = 80000 Nm Make sure you use the correct units in the exam. Question 129: The maximum certificated taxi (or ramp) mass is that mass to which an aeroplane may be loaded prior to engine start. It is: A a fixed value which is listed in the Flight Manual. B a value which varies only with airfield altitude. Standard corrections are listed in the Flight Manual. C a value which is only affected by the outside air temperature. Corrections are calculated from data given in the Flight Manual. D a value which varies with airfield temperature and altitude. Corrections are listed in the Flight Manual. Explanation: Maximum Structural Taxi Mass is the structural limitation of the mass of the aeroplane at commencement of taxi. This may sometimes be referred to as Ramp Mass and is a fixed value which is listed in the Flight Manual. It can also be limited by runway or taxiway limitations. Question 130: The maximum mass to which an aeroplane may be loaded, prior to engine start, is: A B C D

maximum maximum maximum maximum

certificated taxi (ramp) mass. regulated take - off mass. regulated taxi (ramp) mass. certificated take - off mass.

Explanation: Maximum Structural Taxi Mass is the structural limitation of the mass of the aeroplane at commencement of taxi. This may sometimes be referred to as Ramp Mass and is a fixed value which is listed in the Flight Manual. It can also be limited by runway or taxiway limitations. Question 131: The maximum taxi (ramp) mass is governed by: A B C D

structural considerations. tyre speed and temperature limitations. bearing strength of the taxiway pavement. taxi distance to take - off point.

Explanation: Maximum Structural Taxi Mass is the structural limitation of the mass of the aeroplane at commencement of taxi. This may sometimes be referred to as Ramp Mass and is a fixed value which is listed in the Flight Manual. It can also be limited by runway or taxiway limitations. Question 132: The Zero Fuel Mass and the Dry Operating Mass A B C D

differ by the value of the traffic load mass. differ by the sum of the mass of usable fuel plus traffic load mass. differ by the mass of usable fuel. are the same value.

Explanation: This question is based on the basic mass definitions for aircraft. The basic empty mass + variable load = Dry operating mass + traffic load = Zero Fuel Mass. â&#x20AC;&#x153;differ by the value of the traffic load massâ&#x20AC;? makes a correct statement when added to the wording of the question. Zero fuel mass does not include any fuel. Question 133: Mass for individual passengers (to be carried on an aircraft) may be determined from a verbal statement by or on behalf of the passengers if the number of A B C D

passenger seats available is less than 10. passengers carried is less than 20. passenger seats available is less than 20. passengers carried is less than 6.

Explanation: Passenger mass may be determined by verbal statement where the number of passenger seats available is less than 10. Question 134: At reference or see Loading Manual MRJT 1 Figure 4.13. A revenue flight is planned for the transport aeroplane. Take-off mass is not airfield limited. The following data applies: Dry Operating Mass 34930 kg, Performance limited landing mass 55000 kg, Fuel on board at ramp -Taxi fuel 350 kg, Trip fuel 9730 kg, Contingency and final reserve fuel 1200 kg, Alternate fuel 1600 kg, Passengers on board 130, Standard mass for each passenger 84 kg, Baggage per passenger 14 kg. Traffic load - Maximum possible. Use the loading manual provided and the above data. Determine the maximum cargo load that may be carried without exceeding the limiting aeroplane landing mass. 3112266.gif A 4530 kg. B 3185 kg. C 6350 kg.

D 5400 kg Explanation: PLLM = 55000 Kg DOM = 34930 Kg Fuel (at landing) = 2800Kg ( 1200+1600) TL (minus the cargo load) = 130x84 + 130x14 = 12740 Kg So the amount of Cargo you can still carry is = 55000 (34930+2800+12740) = 55000-50470 = 4530 Kg Question 135: The empty mass of an aeroplane is given as 44800 kg. Operational items (including crew standard mass of 1060 kg) are 2300 kg. If the maximum zero fuel mass is given as 65500 kg, the maximum traffic load which could be carried is: A B C D

18400 23000 19460 20700

kg kg kg. kg

Explanation: Very little data is available so all that can be done is to subtract the BEM and variable load from the MZFM. 65,500 - 44,800 - 2,300 = 18,400. Question 136: At reference or see Loading Manual MRJT 1 Figure 4.14. The following data relates to a planned flight of an aeroplane - Dry Operational mass 60520 kg Performance limited take-off mass 92750 kg, Performance limited landing mass 72250 kg, Maximum Zero Fuel mass 67530 kg. Fuel on board at take-off - Trip fuel 12500 kg, Contingency and final reserve fuel 2300 kg, Alternate fuel 1700 kg. Using this data, as appropriate, calculate the maximum traffic load that can be carried. 3112266.gif A B C D

7010 kg 15730 kg 11730 kg 7730 kg

Explanation: Refer to Diagram: Create a table. Take the regulated take-off and landing mass limits which is the lower of the performance limits and the structural limits. Question 137: At reference or see Loading Manual MRJT 1 Figure 4.14. Aeroplane Dry Operating mass 85000 kg, Performance limited take-off mass 127000 kg, Performance limited landing mass 98500 kg, Maximum zero fuel mass 89800 kg, Fuel requirements for flight -Trip fuel 29300 kg, Contingency and

final reserve fuel 3600 kg, Alternate fuel 2800 kg. The maximum traffic load that can be carried on this flight is: 3112266.gif A B C D

4800 kg 6300 kg 12700 kg 7100 kg

Explanation: NO Question 138: The Maximum Zero Fuel Mass is a structural limiting mass. It is made up of the aeroplane Dry Operational mass plus A B C D

traffic load and unuseable fuel. traffic load, unuseable fuel and crew standard mass. traffic load and crew standard mass. unuseable and crew standard mass.

Explanation: We don't write the questions and this is in the QB so you could get it in the exams. If you do then appeal it. None of the answers are correct. Question 139: The take-off mass of an aeroplane is 141000 kg. Total fuel on board is 63000 kg including 14000 kg reserve fuel and 1000 kg of unusable fuel. The traffic load is 12800 kg. The zero fuel mass is: A B C D

79000 93000 65200 78000

kg kg kg. kg

Explanation: To find the zero fuel mass (which includes unuseable fuel), you need to deduct the fuel on board excluding unuseable fuel (i.e. 63000 - 1000 = 62000kg) from the take off mass of the aeroplane: 141000 - 62000 = 79000kg Question 140: 'Standard Mass' as used in the computation of passenger load establish the mass of a child as A 35 kg irrespective of age provided they occupy a seat. B 35 kg for children over 2 years occupying a seat and 10 kg for infants (less than 2 years) not occupying a seat. C 35 kg for children over 2 years occupying a seat and 10 kg for infants (less than 2 years) occupying a seat. D 35 kg only if they are over 2 years old and occupy a seat.

Explanation: NO Question 141: The maximum certificated take - off mass is: A a structural limit which may not be exceeded for any take - off. B limited by the runway take off distance available. It is tabulated in the Flight Manual. C a take - off limiting mass which is affected by the aerodrome altitude and temperature. D a take - off limiting mass which is governed by the gradient of climb after reaching V2 . Explanation: The maximum permissible total aeroplane mass at the start of the take-off run is called the Maximum Structural Take-Off Mass. Question 142: For a particular aeroplane, the structural maximum mass without any fuel on board, other than unusable quantities, is: A B C D

a a a a

fixed value which is stated in the Aeroplane Operating Manual. variable value which is governed by the PCN and ACN. fixed value which will limit the amount of fuel carried. variable value which is governed by the traffic load carried.

Explanation: The maximum permissible mass of an aeroplane with no usable fuel is known as Maximum Zero Fuel Mass or MZFM. Question 143: An aeroplane, which is scheduled to fly an oceanic sector, is due to depart from a high altitude airport in the tropics at 1400 local time. The airport has an exceptionally long runway. Which of the following is most likely to be the limiting factor(s) in determining the take - off mass ? A B C D

altitude and temperature of the departure airfield. Maximum Zero Fuel Mass. maximum certificated take - off mass. en route obstacle clearance requirements.

Explanation: CLTOM which is based upon weight, altitude and temperature. The data given in the question is suggesting the conditions are hot and the altitude high and could therefore be limited by the CLTOM. It could, however, also be limited by the Maximum certificated take-off mass. An must aircraft be able to take off safely from the airfield it is operating from and regulations state that it must also be able meet a minimum value of climb performance. As the conditions are hot, high and humid the aircraft might eventually take off from the long runway but its performance will be degraded (less

dense air means reduced thrust and higher true airspeeds) for the take off climb. Question 144: On an aeroplane with a seating capacity of more than 30, it is decided to use standard mass values for computing the total mass of passengers. If the flight is not a holiday charter, the mass value which may be used for an adult is A B C D

84 88 76 84

kg kg (male) 74 kg (female). kg kg (male) 76 kg (female).

Explanation: NO Question 145: The standard mass for a child is A B C D

35 35 30 38

kg kg kg kg

for for for for

all flights. holiday charters and 38 kg for all other flights. holiday charters and 35 kg for all other flights. all flights.

Explanation: NO Question 146: On an aeroplane with 20 or more seats engaged on an inter-continental flight, the 'standard mass' which may be used for passenger baggage is A B C D

15 11 14 13

kg kg kg kg

per per per per

passenger. passenger. passenger. passenger.

Explanation: NO Question 147: The following data applies to a planned flight. Dry Operating Mass 34900 kg, Performance limited Take-Off Mass 66300 kg, Performance limited Landing Mass 55200 kg, Maximum Zero Fuel Mass 53070 kg, Fuel required at ramp:- Taxy fuel 400 kg, trip fuel 8600 kg, contingency fuel 430 kg, alternate fuel 970 kg, holding fuel 900 kg, Traffic load 16600 kg. Fuel costs at the departure airfield are such that it is decided to load the maximum fuel quantity possible. The total fuel which may be safely loaded prior to departure is : A 12700 kg B 10730 kg C 13230 kg

D 15200 kg Explanation: First up you need to look at which is the limiting weight. PLTOM 66300 PLLM 55200 MZFM 53070 PLTOM = 66300 PLLM 55200 + trip fuel 8600 = 63800 MZFM 53070 + Take off fuel 10900 = 63970 From the above we can see that the PLTOM is the most limiting at 63800. Next we need to find out the landing fuel DOM 34900 + Traffic 16600 = 51500 To find out the landing fuel PLTOM 55200 - 51500 = 3700. So we land with 3700 on board. Next we add the trip fuel of 8600 this will give us 12300. NOW REMEMBER The question asks how much fuel can be loaded PRIOR to departure..i.e. on the RAMP..So we include the Taxi fuel of 400 12300 + 400 = 12700 Question 148: In determining the Dry Operating Mass of an aeroplane it is common practice to use 'standard mass' values for crew. These values are A flight crew 85 kg., cabin crew 75 kg. each. These are inclusive of a hand baggage allowance. B flight crew 85 kg., cabin crew 75 kg. each. These do not include a hand baggage allowance. C flight crew (male) 88 kg. (female) 75 kg., cabin crew 75 kg. each. These do not include an allowance for hand baggage. D flight crew (male) 88 kg. (female) 75 kg., cabin crew 75 kg. each. These include an allowance for hand baggage. Explanation: NO Question 149: 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 A B C D

13090 12840 18040 16470

kg. kg kg kg

Explanation: 20100 litres of fuel with a specific gravity of 0.78 weighs approximately 15,678kg (using either the rule of thumb of 1ltr of water with SG 1 = 1kg or using the CRP-5). Using the performance limited take-off mass figure, the max traffic load is 67200-34930(DOM)-(15678-250) (fuel loaded less taxi fuel) = 16842kg

Using the performance limited landing mass figure, the max traffic load is 54200-34930-(15678-250-9250) (fuel loaded less taxi fuel less trip fuel) = 13092kg Taking the lowest figure and matching to answers provided, the max permissable traffic load is 13090kg Question 150: At reference or see Loading Manual MRJT 1 Paragraph 3.1. The medium range jet transport aeroplane is to operate a flight carrying the maximum possible fuel load. Using the following data as appropriate, determine the mass of fuel on board at start of take off. Departure airfield performance limited take-off mass: 60 400 kg. Landing airfield -not performance limited. Dry Operating Mass: 34930 kg Fuel required for flight - Taxi fuel: 715 kg Trip fuel: 8600 kg Contingency and final reserve fuel: 1700 kg Alternate fuel 1500 kg Additional reserve 400 kg Traffic load for flight 11000 kg 319660A.gif A B C D

14 16 15 13

470 080 815 655

kg kg kg kg

Explanation: MTM MTOM MZFM MLM Limit 63060 60400 51300 54900 DOM 34930 34930 34930 34930 T/load 11000 11000 11000 11000 Fuel 12715 12200 nil 3600 U/load 4415 2270 5370 5370 The maximum extra fuel that can be on board at the take-off is 2270 kg. 12200+2270 = 14470kg Question 151: An aeroplane is to depart from an airfield at a take-off mass of 302550 kg. Fuel on board at take-off (including contingency and alternate of 19450 kg) is 121450 kg. The Dry Operating Mass is 161450 kg. The useful load will be A B C D

141100 kg 39105 kg 121450 kg 19650 kg

Explanation: Useful load is the traffic load + fuel, therefore, to find the useful load you need to deduct the DOM from the Take-off mass i.e. 302550-161450 = 141100kg Question 152:

When considering the effects of increased mass on an aeroplane, which of the following is true? A B C D

Stalling speeds will be higher. Flight endurance will be increased. Stalling speeds will be lower. Gradient of climb for a given power setting will be higher.

Explanation: Increasing mass leads to an increase in induced drag. Vs will therefore increase. Question 153: If an aeroplane is at a higher mass than anticipated, for a given airspeed the angle of attack will A B C D

be greater, drag will increase and endurance will decrease. be decreased, drag will decrease and endurance will increase. remain constant, drag will increase and endurance will increase. remain constant, drag will decrease and endurance will decrease.

Explanation: An increase in mass demands more lift. At a constant airspeed: the angle of attack and the drag will increase the range and endurance will decrease Question 154: Conversion of fuel volume to mass A may be done by using standard fuel density values as specified in the Operations Manual, if the actual fuel density is not known. B must be done by using actual measured fuel density values. C must be done using fuel density values of 0.79 for JP 1 and 0.76 for JP 4 as specified in JAR - OPS, IEM - OPS 1.605E. D may be done by using standard fuel density values as specified in JAR OPS 1. Explanation: NO Question 155: Standard masses may be used for the computation of mass values for baggage if the aeroplane A B C D

has 20 or more seats. has 6 or more seats. is carrying 30 or more passengers. has 30 or more seats.

Explanation: NO Question 156:

Which of the following is most likely to affect the range of centre of gravity positions on an aeroplane? A Elevator and tailplane (horizontal stabiliser) effectiveness in all flight conditions. B The need to maintain a low value of stalling speed. C The need to minimise drag forces and so improve efficiency. D Location of the undercarriage. Explanation: NO Question 157: A jet transport has the following structural limits: -Maximum Ramp Mass: 63 060 kg -Maximum Take Off Mass: 62 800 kg -Maximum Landing Mass: 54 900 kg -Maximum Zero Fuel Mass: 51 300 kg The aeroplane's fuel is loaded accordance with the following requirements: -Taxi fuel: 400 kg -Trip fuel: 8400 kg -Contingency & final reserve fuel: 1800 kg -Alternate fuel: 1100 kg If the Dry Operating Mass is 34930 kg, determine the maximum traffic load that can be carried on the flight if departure and landing airfields are not performance limited. A B C D

16 16 16 17

370 430 570 070

kg kg kg kg

Explanation: You need to use each of the limiting masses and deduct the DOM and fuel to find the allowable traffic load. You would normally create a table MRM MTOM MZFM MLM limit 63060 62800 51300 54900 - DOM 34930 34930 34930 34930 -fuel 11700 11300 nil 2900 Traffic load 16430 16570 16370 17070 Maximum Traffic load from MZFM = 16370 kg Question 158: A flight has been made from London to Valencia carrying minimum fuel and maximum traffic load. On the return flight the fuel tanks in the aeroplane are to be filled to capacity with a total fuel load of 20100 litres at a fuel density of 0.79 kg/l. The following are the aeroplane's structural limits: -Maximum Ramp Mass: 69 900 kg -Maximum Take Off Mass: 69 300 kg -Maximum Landing Mass: 58 900 kg -Maximum Zero Fuel Mass: 52 740 kg The performance limited take off mass at Valencia is 67 330 kg. The landing mass at London is not performance limited. Dry Operating Mass: 34 930 kg Trip Fuel (Valencia to London): 5 990 kg Taxi fuel: 250 kg The maximum traffic load that can be carried from Valencia will be: A B C D

14 331 kg 16 770 kg 9 830 kg 13 240 kg

Explanation: The correct equation should be: MLM - DOM - FUEL REMAINING Assuming traffic is limited by landing mass at London, the fuel figure; before taxi it is 20,100 * 0.79 = 15,879. To get fuel remaining of 9,639, you must subtract taxi and trip fuel (250 and 5,990). Using the formula, you get the answer given: 58,900 - 34,930 - 9,639 = 14,331 Question 159: The term 'useful load' as applied to a aircraft includes A B C D

traffic load plus useable fuel. the revenue-earning portion of traffic load plus useable fuel. the revenue-earning portion of traffic load only. traffic load only.

Explanation: Please be aware: 'Useful Load' is not officially recognised terminology by EASA and EU-OPS. But has historically been used for informal general use. The term 'useful load' as applied to a light aircraft includes: - Pilot(s) - Passengers - Useable fuel - Cargo - Operating items - Baggage Question 160: An aeroplane is performance limited to a landing mass of 54230 kg. The Dry Operating Mass is 35000 kg and the zero fuel mass is 52080 kg. If the take-off mass is 64280 kg the useful load is A B C D

29280 10080 17080 12200

kg. kg. kg kg.

Explanation: By definition, the TOM = DOM + USEFUL LOAD. The fact that the TOM is 64280 and DOM is 35000, USEFUL LOAD must be 29280. So the other figures are red herrings I think. If you work it this way ZFM = 52080; and ZFM = DOM + (USEFUL LOAD-FUEL) so you could say TOM - ZFM = FUEL, which works out as 12200kg. Rearranging: ZFM = DOM + (USEFUL LOAD-FUEL) ZFM-DOM+FUEL = useful load 52080-35000+12200=29280 So it does give the same value. The reason the values aren't limiting is that it is not MZFM and MTOM, they are just stated as facts that they are the actual values. Question 161:

See Loading Manual MRJT 1 Figure 4.5 For the medium range transport aeroplane, from the loading manual, determine the maximum total volume of fuel which can be loaded into the main wing tanks. (Fuel density value 0.78kg/l) 60-0203.gif A B C D

11350 litres 5674 litres 11646 litres 8850 litres

Explanation: From figure 4.5 the capacity of the wing tanks is 2998 USG. The conversion factor to litres is found in Section 1 general notes. 2998 x 3.785 = 11347 litres Question 162: Contrary to the forecast given in the LOAD and TRIM sheet, cargo compartment 1 is empty. The take-off centre of gravity in MAC % (Mean Aerodynamic Chord) will be located at: 60-0232.gif A B C D

32.5 % 31 % 36 % 25 %

Explanation: First you need to correc the TOm for empty cargo compartment 1 211 600 - 2 500 Which gives 209 100kgs. Now redraw in the CG envelope (pretty non-existant!) Now correct the cargo data - looks like 32.5% Question 163: Contrary to the loading sheet forecasts you have : Cargo compartment 1: empty. Cargo compartment 2: 1000Kg. Cargo compartment 3: 3000Kg. Cargo compartment 4: 2000Kg. Cargo compartment 5: 1000Kg. Passengers in compartment OA: 20. Passengers in compartment OB: 20. Passengers in compartment OC: 30. The take-off centre of gravity in MAC % (Mean Aerodynamic Chord), will be located at: 60-0232.gif A B C D

31.5 24.5 32.5 35.5

% % % %

Explanation: This is not a very well written question that appears to give the wrong answer. Remember - we don't write the questions we just try to reproduce them as accurately to the exam as possible. If this question comes up in an exam we would reccomend you appeal it.

Question 164: Just prior to departure, you accept 10 passengers additional on board who will be seated in "compartment OA" and you have 750 kg unloaded from cargo compartment 5. The take-off centre of gravity in MAC % (Mean Aerodynamic Chord) will be located at: 60-0232.gif A B C D

27.8 27.2 30.5 28.5

% % % %

Explanation: This is not a very well written question that appears to give the wrong answer. Remember - we don't write the questions we just try to reproduce them as accurately to the exam as possible. If this question comes up in an exam we would reccomend you appeal it. Question 165: The weight and balance sheet is available and contrary to the forecast, cargo compartment 1 is empty. The zero fuel weight centre of gravity in MAC % (Mean Aerodynamic Chord) is located at: 60-0232.gif A B C D

35.5 % 26 % 32 % 31.5 %

Explanation: This is not a very well written question that appears to give an incorrect/inaccurate answer. Remember - we don't write the questions we just try to reproduce them as accurately to the exam as possible. If this question comes up in an exam we would reccomend you appeal it. Question 166: A turbojet aeroplane is parked with the following data: Corrected dry operating mass: 110 100 kg, Basic corrected index: 118.6, Take-off mass: 200 000 kg; centre of gravity (C.G.) location: 32 %. Distance from reference point to leading edge: 14m Length of MAC = 4.6m. Initial cargo distribution: cargo 1: 4 000 kg (2.73m from reference point); cargo 2: 2 000 kg (8.55 m from reference point) ; cargo 3: 2 000 kg (16.49m from reference point) ; cargo 4 = empty (21.13 m from reference point) For perfomance reasons, the captain decides to redistribute part of the cargo loading between cargo compartments, in order to take off with a new C.G. location of 34 %. He asks for a transfer of: A B C D

1 000 kg from cargo 1 500 kg from cargo 500 kg from cargo 1 1 000 kg from cargo

1 to cargo 4 3 to cargo 1 to cargo 3 3 to cargo 1

Explanation: First draw a sketch of the aircraft with the data. After that we must work out the actual distance we need the CG to move. The length of the MAC is 4.6m and we need to move the CG rearwards by 2%. 2% x 4.6m = 0.092m The distance between hold 1 and 4 is 18.4m Now using the standard formula: Mass change/total mass = change in CG/total distance moved. Transpose and substitute known values to find mass change. 0.092 x 200000/18.4 = 1000kg If we do the same calculation for cargo hold 1 to 3 you will find that the mass change would have to be 1337.2kg. Question 167: The flight preparation of a turbojet aeroplane provides the following data: Take-off runway limitation: 185 000 kg Landing runway limitation: 180 000 kg Planned fuel consumption: 11 500 kg Fuel already loaded on board the aircraft: 20 000 kg Knowing that: Maximum take-off mass (MTOM): 212 000 kg Maximum landing mass (MLM): 174 000 kg Maximum zero fuel mass (MZFM): 164 000 kg Dry operating mass (DOM): 110 000 kg The maximum cargo load that the captain may decide to load on board is: A B C D

54 61 55 55

000 500 000 500

kg kg kg kg

Explanation: The useful load is traffic load and fuel. The question states that fuel is already on the plane. So we can calculate the cargo/traffic load. Question 168: The crew of a transport aeroplane prepares a flight using the following data: - Block fuel: 40 000 kg - Trip fuel: 29 000 kg - Taxi fuel: 800 kg - Maximum take-off mass: 170 000 kg - Maximum landing mass: 148 500 kg Maximum zero fuel mass: 112 500 kg - Dry operating mass: 80 400 kg The maximum traffic load for this flight is: A B C D

32 32 40 18

100 900 400 900

kg kg kg kg

Explanation: NO Question 169: An aeroplane, whose specific data is shown in the annex, has a planned take-off mass of 200 000 kg, with its centre of gravity (C.G.) located at 15.38 m rearward of the reference point, representing a C.G. location at 30 % MAC (Mean Aerodynamic Cord). The current cargo load distribution is: front cargo: 6 500 kg; rear cargo: 4 000 kg. For performance purposes,

the captain decides to reset the value of the centre of gravity location to 33 % MAC. The front and rear cargo compartments are located at a distance of 15 m and 25 m from the reference point respectively. After the transfer operation, the new cargo load distribution is: 3111227.gif A B C D

front front front front

cargo: cargo: cargo: cargo:

3 6 4 9

740 760 550 260

kg; kg; kg; kg;

rear rear rear rear

cargo: cargo: cargo: cargo:

6 3 5 1

760 740 950 240

kg kg kg kg

Explanation: Length of MAC = 18.6m â&#x20AC;&#x201C; 14m = 4.6m CG change required = 3% of 4.6m = 0.138m Use formula to find mass change between forward and aft cargo compartment: Mass change = change of CG Total mass distance moved Mass change = change of CG x Total mass distance moved = 0.138 x 200000 10 = 2760kg As CG has moved aft then new cargo load distribution is: FWD cargo: 6500 -2760 = 3740kg AFT cargo: 4000+2760 = 6760kg Question 170: The planned take-off mass of an aeroplane is 180 000 kg, with its centre of gravity located at 31 % MAC (Mean Aerodynamic Cord). Shortly prior to engine start, the local staff informs the flight crew that 4 000 kg must be loaded in cargo 1. After loading this cargo, the new centre of gravity location will be: 60-0232.gif A B C D

25 37 34 28

% % % %

Explanation: From graphic addition of 4 000kg in cargo 1 moves the CG position to 25%. Question 171: Prior to engine start a turbojet aeroplane has the following data: Corrected Dry Operating Mass: 110 100 kg. Basic corrected index: 118.6. Take off mass: 200 000Kg. C of G location: 32% MAC (Mean Aerodynamic Cord). Leading edge to reference point: 14m. Length MAC: 4.6m. Initial cargo distribution: cargo 1 = 4 000 kg at 2.73m from ref point. Cargo 2 = 2 000 kg at 8.55m from ref point. Cargo 3 = 2 000 kg at 16.49m from ref point. Cargo 4 = empty at 21.13m. To maximize performance, the captain decides to redistribute part of the cargo load between cargo holds. In

order to take off with a new centre of gravity location at 34 % MAC, load must be transferred as follows: A B C D

1 000 Kg from cargo hold 500 Kg from cargo hold 3 1 000 kg from cargo hold 1 500 kg from cargo hold

1 to cargo hold 4 to cargo hold 1 1 to cargo hold 3 1 to cargo hold 4

Explanation: NO Question 172: The planned take-off mass of a turbojet aeroplane is 180 000 kg, with its centre of gravity located at 26 % MAC (Mean Aerodynamic Cord). Shortly prior to engine start, the local staff informs the flight crew that 4 000 kg must be unloaded from cargo 4. After the handling operation, the new centre of gravity location in % MAC will be: 60-0232.gif A B C D

21.8 30.2 23.0 20.0

% % % %

Explanation: If you read the question closely it seems impossible to remove 4000 kg from cargo hold 4 as it only contains 2000kg! But as this is an exam question if you do the calculation anyway: New TOM = 180 000 kg â&#x20AC;&#x201C; 4000 kg = 176 000kg Working in reverse using the Load and Trim Sheet diagram a MAC of approximately 21.8% will be given at a TOM of 176 000kg

Question 173: The mass displacement caused by landing gear extension: A creates a longitudinal moment in the direction (pitch-up or pitch-down) determined by the type of landing gear B creates a pitch-up longitudinal moment C creates a pitch-down longitudinal moment D does not create a longitudinal moment Explanation: Creates a longitudinal moment in the direction (pitch-up or pitch-down) determined by the type of landing gear. A pitch-up could happen by moving CG rearwards if it retracts rearwards Question 174: Without the crew, the mass and longitudinal CG position of the aircraft are 6 000 kg and 4,70m. - the mass of the pilot is 90 kg - the mass of the copilot is 100 kg - the mass of the flight engineer is 80 kg With the crew, the mass and longitudinal CG position of the aircraft are :

31019.gif A B C D

6 6 6 6

270 270 270 270

kg kg kg kg

and and and and

4.594 m 5.012 m 4.61 m 4.796 m

Explanation: To solve this question you need to work out the total moment and total mass, after putting the crew on. The BEM moment = 6000 x 4.7 = 28200 kg.m The pilot weighing 90kg has moment from the table , column A, of 184kg.m The co-pilot weighing 100kg has a moment from the table, column A, of 204kg.m The flight engineer weighing 80kg has a moment from the table, column B, of 215kg.m The total moment = 28803kg.m and the total mass = 6270 kg. CG = moment/mass = 28803/6270 = 4.594m Answer: 6270 kg CG 4.594m Question 175: The centre of gravity of an aeroplane is at 25% of the Mean Aerodynamic Chord. This means that the centre of gravity of the aeroplane is situated at 25% of the length of: A B C D

the the the the

mean aerodynamic chord in relation to the leading edge mean aerodynamic chord in relation to the trailing edge aeroplane in relation to the leading edge mean aerodynamic chord in relation to the datum

Explanation: The % value of Mean Aerodynamic Chord (MAC) is always taken from the leading edge . Question 176: The operator of an aircraft equipped with 50 seats uses standard masses for passengers and baggage. During the preparation of a scheduled flight a group of passengers present themselves at the check-in desk, it is apparent that even the lightest of these exceeds the value of the declared standard mass. A the operator should use the individual masses of alter the standard masss B the operator may use the standard masses for the correct these for the load calculation C the operator is obliged to use the actual masses D the operator may use the standard masses for the calculation without correction Explanation: NO Question 177:

the passengers or balance but must of each passenger load and balance

The datum used for balance calculations is: A chosen on the longitudinal axis of the aeroplane, but not necessarily between the nose and the tail of the aircraft B chosen on the longitudinal axis of the aircraft, and always at the fire-wall level C chosen on the longitudinal axis of the aircraft and necessarily situated between the leading edge and trailing edge of the wing D chosen on the longitudinal axis of the aeroplane, and necessarily situated between the nose and the tail of the aircraft Explanation: Datum is the point on the longitudinal axis from which the centres of gravities of all masses are referenced. Question 178: The Dry Operating Mass of an aircraft is 2 000 kg. The Maximum Take-off Mass, Landing and Zero Fuel Mass are identical at 3500 kg. The block fuel mass is 550kg, and the taxi fuel mass is 50 kg. The available mass of traffic load is: A B C D

1 000 kg 1 500 kg 1 450 kg 950 kg

Explanation: MTOM-DOM-Take-off fuel = Traffic load. So in this case 3500-2000-500= 1000kg Question 179: Based on actual conditions, an aeroplane has the following performance take-off mass limitations: Flaps : 0° 10° 15° Runway: 4100 4400 4600 Climb: 4700 4500 4200 Masses are in kg Structural limits: takeoff/landing/zero fuel: 4 300 kg The maximum take-off mass is : A B C D

4 4 4 4

300 100 200 700

kg kg kg kg

Explanation: Flap 5 limits are: Runway 4100, Climb 4700, Structural 4300 so at Flap 5 we're restricted to 4100kg. It is always the lowest value. Flap 10 limits are: Runway 4400, Climb 4500, Structural 4300 so at Flap 10 we're restricted to 4300kg. Flap 15 limits are: Runway 4600, Climb 4200, Structural 4300 so at Flap 15 we're restricted to 4200kg. Now compare the limit of each flap setting to find the MTOM of 4300 at Flap 10. Question 180: The basic empty mass of an aircraft is 30 000 kg. The masses of the following items are : - catering: 300 kg - safety and rescue material:

nil - fly away kit: nil - crew (inclusive crew baggage): 365kg - fuel at take-off: 3 000 kg - unusable fuel: 120 kg - passengers, baggage, cargo: 8 000 kg The Dry Operating Mass is : A B C D

30 38 30 30

665 300 785 300

kg kg kg kg

Explanation: Unuseable fuel is part of the BEM DOM is BEM + catering equipment and crew and their hand baggage 30000 + 300 + 365 = 30665 kg Question 181: The floor limit of an aircraft cargo hold is 5 000 N/m2. It is planned to load-up a cubic container measuring 0,4 m of side. It's maximum gross mass must not exceed: (assume g=10m/s2) A B C D

80 kg 320 kg 32 kg 800 kg

Explanation: The contact area of the container is 0.4m x 0.4m = 0.16m². If 5000N can be spread over 1m² then a reduced force of 5000 x 0.16 = 800N would be permitted over 0.16m². Using the conversion factor of g=10m/s², which means the same as 10N = 1kg, 800N is 80kg, Question 182: To measure the mass and CG-position of an aircraft, it should be weighed with a minimum of: A B C D

3 4 2 1

points of support point of support points of support point of support

Explanation: NO Question 183: At reference. Without the crew, the weight and the CG-position of the aircraft are 7 000 kg and 4,70m. - the mass of the pilot is 90 kg - the mass of the copilot is 75 kg - the mass of the flight engineer is 90 kg With this crew on board, the CG-position of the aircraft will be: 31019.gif A B C D

4,615 0,217 4,783 4,455

m m m m

Explanation: If you take the moments individually: Pilot = 184 Co-pilot= 153 (close enough) Eng =242 BEM = 32900 Total moment = 33479 kg.m Total mass = 7255 kg CG = moment/mass = 33479/7255 = 4.615 m Question 184: During a violent avoidance manoeuvre, a light twin aircraft, certified to FAR 23 requirements was subjected to an instantaneous load factor of 4.2. The Flight Manual specifies that the aircraft is certified in the normal category for a load factor of -1.9 +3.8. Considering the certification requirements and taking into account that the manufacturer of the twin did not include, during its conception, a supplementary margin in the flight envelope, it might be possible to observe; A a permanent deformation of the structure B rupture of one or more structural components C a elastic deformation whilst the load was applied, but no permanent distortion D no distortion, permanent or temporary of the structure Explanation: Aircraft structures are designed to withstand a specified load with a very small safety margin. That safety margin is 1.5. The question states that the aircraft is stressed to -1.9 to + 3.8g. The turbulence has stressed the aircraft to +4.2g. This exceeds the limit of 3.8 but has not exceeded the ultimate limit (3.8 x 1.5 = 5.7g) The structure will therefore not fail but be permanently damaged. Question 185: The centre of gravity is the A point where all the aircraft mass is considered to be concentrated B centre of thrust along the longitudinal axis, in relation to a datum line C neutral point along the longitudinal axis, in relation to a datum line D focus along the longitudinal axis, in relation to a datum line Explanation: Definitions for Centre of Gravity include: "A point, near or within a body, through which its weight can be assumed to act when considering forces on the body and its motion under gravity." "The point within something at which gravity can be considered to act; in uniform gravity it is equal to the centre of mass." Question 186:

The determination of the centre of gravity in relation to the mean aerodynamic chord: A consists of defining the centre of gravity longitudinally in relation to the length of the mean aerodynamic chord and the leading edge B consists of defining the centre of gravity longitudinally in relation to the length of the mean aerodynamic chord and the trailing edge C consists of defining the centre of gravity longitudinally in relation to the position of the aerodynamic convergence point D consists of defining the centre of gravity longitudinally in relation to the position of the aerodynamic centre of pressure Explanation: NO Question 187: By adding to the basic empty mass the following fixed necessary equipment for a specific flight (catering, safety and rescue equipment, fly away kit, crew), we get: A B C D

Dry Operating Mass Zero Fuel Mass Take-off Mass Landing Mass

Explanation: Dry Operating Mass (DOM) is the total mass of the aeroplane (excluding usable fuel and traffic load) Items contained in the DOM include: - Crew and crew baggage. - Food and beverages. - Potable water and lavatory chemicals. - Catering and removable passenger service equipment. The operator determines the DOM. Once the aircraft is delivered at the BEM the operator adds the masses of the crew and operational items to establish the DOM. Question 188: The floor of the main cargo hold is limited to 4 000 N/m2. It is planned to load a cubic container each side of which measures 0.5m. Its maximum gross mass must not exceed: (assume g=10m/s2) A B C D

100 kg 5 000 kg 500 kg 1 000 kg

Explanation: If each side is 0.5m the footprint is 0.5 x 0.5 = 0.25m². The limit is 4000N/m² so the max load over 0.25m² is 4000 x 0.25 = 1000N. If g = 10m/s² there are 10N in a kg so the max force of 1000N equates to 100kg Question 189:

An aircraft basic empty mass is 3000 kg. The maximum take-off, landing, and zero-fuel mass are identical, at 5200 kg. Ramp fuel is 650 kg, the taxi fuel is 50 kg. The maximum traffic load is: A B C D

1 2 2 1

600 150 200 550

kg kg kg kg

Explanation: This question cannot be answered properly without the DOM. It transpires the JAA were thinking of DOM when they wrote BEM, itâ&#x20AC;&#x;s just a bad question. What was intended was available payload = MTOM - TO fuel - DOM and they meant the sum to read available payload = 5200 - 600 - 3000 = 1600 We have drawn this to the CAA's attention. They agreed that the wrong weight was given and that it should have read DOM. They have suggested to the JAA that this question should be amended. We would expect to see BEM already modified to DOM in UK exams but possibly not yet in other European countries. Question 190: For the purpose of aeroplane mass and balance calculations, the datum point is defined as: A a fixed point from which all balance arms are measured. It may be located anywhere on the aeroplane's longitudinal axis or on the extensions to that axis. B a point near the centre of the aeroplane. It moves longitudinally as masses are added forward and aft of its location. C a variable point, that is dependent on the load distribution for its location, from which all balance arms are measured D the point through which the sum of the mass values (of the aeroplane and its contents) is assumed to act vertically. Explanation: Datum is the point on the longitudinal axis from which the centres of gravities of all masses are referenced. Question 191: An aeroplane has a mean aerodynamic chord (MAC) of 134.5 inches. The leading edge of this chord is at a distance of 625.6 inches aft of the datum. Define the location of the centre of gravity of the aeroplane in terms of percentage MAC if the mass of the aeroplane is acting vertically through a balance arm located 650 inches aft of the datum. A B C D

18,1% 85,5% 75,6% 10,5%

Explanation: With this type of question draw yourself a little sketch so you can see the problem.

The LEMAC (leading edge MAC) is at 625.6" aft of the datum and the CG is at 650" aft of the datum therefore the CG is 650-625.6= 24.4" in from LEMAC. As a % of the MAC this 24.4/134.5 x 100% = 18.1% Question 192: At reference or see Loading Manual MRJT 1 Figure 4.13. Using the load and trim sheet for the MRJT1 aircraft which of the following is the correct value for the index at a Dry Operating Mass (DOM) of 35000 kg with a CG at 14% MAC ? 3112266.gif A B C D

40.0 33.0 35.5 41.5

Explanation: You can calculate it from the DOM and the %MAC using the bottom right hand bit of the load sheet. There's only one exam question we know of like this. Q. Given a DOM of 35,000kg and a CG of 14%MAC determine the Dry Operating Index. A. Enter from the left or right with 35,000kg, go straight across to the 14%MAC line, read down to a DOI of 40. Question 193: Using the data given in the Load & Trim sheet, determine which of the following gives the correct values for the Zero Fuel Mass and position of the centre of gravity (% MAC) at that mass. 3112268.gif A B C D

46130 51300 41300 46130

Kg Kg Kg Kg

and and and and

17.8% 20.8% 17.8% 20.8%

Explanation: The easiest thing to do is to identify the DOM and Traffic Load from the left hand side. DOM is 34 900kg, Traffic Load 11 230kg. Add them to find the ZFM, 46 130kg. Draw a horizontal line at this mass across the CG envelope at the bottom right of the Annex to represent the ZFM. Now you need to trace through the grids at the top right of the Annex starting from the DOM index of 33.5. Make no correction for fuel index and drop down to cross the line that represents the ZFM. We get 46 130kg and CG at about 17.6%MAC. From the options given, this makes the correct answer 46130 kg and 17.8% Question 194: Using the data given in the Load & Trim sheet, determine from the following the correct values for the take off mass and the position of

the centre of gravity at that mass if the fuel index correction to be applied is given as - 0.9 3112268.gif A B C D

17.5 22.6 20.3 20.1

% % % %

Explanation: From the left side find the TOM of 55630 kg. Enter this mass as a horizontal line in the CG envelope. Enter the trim sheet at a DOM index of 33.5 and follow the scale and direction arrows for the individual compartment masses. The fuel index for take-off fuel = - 0.09 (almost negligible). From the fuel index draw a line straight down to intersect the TOM of 55630kg to read a TOM CG of 17.5%. Question 195: Using the data given at the appendix to this question, if the fuel index corrections (from ZFM index) are as follows 9500 kg - 0.9 6500 kg - 6.1 3500 kg - 4.7 3000 kg - 4.3 Which of the following represent the correct values for landing mass of the aeroplane and the position of the centre of gravity for this condition ? 3112269.gif A B C D

49130 kg and 19 % 52900 kg and 21.6 % 49130 kg and 21.8 % 52900kg and 19 %

Explanation: Find the landing mass of 49130 kg (Take-off mass - trip fuel) on the left side of the trim sheet. Enter the trim sheet at a DOM index of 33.5 and follow the scale and direction arrows for the individual compartment masses. The fuel index to use is that for the fuel remaining at landing of 3000kg index -4.3. From the fuel index draw a line straight down to intersect the landing mass at 49130kg with a CG of 19% MAC. Question 196: Using the data given at the appendix, determine which of the following correctly gives the values of the Zero Fuel Mass (ZFM) of the aeroplane and the load index at ZFM 3112270.gif A B C D

48600 46300 35100 51300

kg kg kg kg

and and and and

57.0 20.5 20.5 57.0

Explanation: To find the ZFM you must add the DOM + Traffic load =48600.

This info can be found from the left hand side of the load and trim sheet. There is only one answer with this value of ZFM. Enter right side of loadsheet at the DOM index, 33.5, follow direction and scale arrows for the individual compartment loads. Finally in compartment Og draw a vertical line back up to the index grid to read 57.0 Question 197: From the data given at the appendix and assuming a fuel index shift of 5.7 from the ZFM loaded index, determine which of the following is the correct value (percentage MAC) for the position of the centre of gravity at Take Off Mass. 3112271.gif A B C D

18.5 % 14 % 15 % 19.5 %

Explanation: NO Question 198: For the purpose of calculating traffic loads, an operator's loading manual gives the following standard mass values for passengers. (These values include an allowance for hand baggage) Male 88 kg, Female 70 kg, Child 35 kg, Infant 6 kg.The standard mass value to be used for hold baggage is 14 kg per piece.The loading manifest shows the following details :Passengers loaded, Males 40, Females 65, Children 8, Infants 5. Baggage in hold number 4: 120 pieces. Using the standard mass values given and the data in the appendix, select from the following the correct value for the mass of freight (all loaded in hold No1) which constitutes the remainder of the traffic load 31025.gif A B C D

260 kg no cargo can be loaded in hold number 1 280 kg 210 kg

Explanation: Males: 40 x 88 = 3520 kg Females: 65 x 70 = 4550 kg Children: 8 x 35 = 280 kg Infants: 5 x 6 = 30 kg Baggage: 120 x 14 = 1680 kg Total = 10060 kg From loadsheet current load = 10320 kg Remainder of load = 10320 -10060 = 260 kg Question 199:

From the data contained in the attached appendix, the maximum allowable take - off mass and traffic load is respectively : 3112273.gif A B C D

61600 60425 68038 66770

kg kg kg kg

and and and and

12150 10975 18588 17320

kg kg kg kg

Explanation: From the left hand side of the load and trim sheet the take off mass is the lower of the boxes A,B,C. in this case 61600. The take off mass - operating mass (fuel+ DOM) 61600 - 49450 = 12150. This is the maximum TL that could be carried. The aircraft actually only has a TL of 11230. There is an underload of 920 kg. Question 200: An aeroplane is carrying a traffic load of 10320 kg. Complete the necessary sections of the attached appendix and determine which of the answers given below represents the maximum increase in the traffic load. 3112274.gif A B C D

1830 kg 655 kg 7000 kg 8268 kg

Explanation: The ZFM and TOM limits shown in the diagram are in fact incorrect, they should be MZFM 51300kg and MTOM 62800kg, the structural limit. So the examiner has got this bit wrong. However regardless of this error, the answer still comes from column (c). 54245 = 7355= 61600kg. Subtract from this the Operating mass (DOM +take-off fuel) 35720+13700 =49450kg. 61600-49450 = 12150kg Subtract from this the current traffic load to find the underload: 1215010320=1830kgs. Question 201: Due to a mistake in the load sheet the aeroplane is 100kg heavier than you believe it to be. As a consequence: A B C D

VMU will be higher. Vâ&#x201A; VMU VR will all occur earlier. Vâ&#x201A; will be higher. VR will be higher.

Explanation: NO Question 202: If an extra load is loaded into an aircraft the stall speed is likely to:

A Increase B Change depending on whether the load was placed FWD or AFT of the C of G. C Decrease. D Stay the same. Explanation: Any increase in mass increases the stall speed! Question 203: The Centre of Gravity of an aeroplane: A Can be allowed to move between defined limits. B May only be moved if permitted by the regulating authority and endorsed in the aeroplane's certificate of airworthiness. C Must be maintained in a fixed position by careful distribution of the load. D Is in a fixed position and is unaffected by aeroplane loading. Explanation: The manufacturer establishes the forward and aft CG limits at the certification stage. They are based upon the requirements for positive longitudinal static stability. The CG can move between these defined limits. Question 204: Who establishes the limits C of G? A B C D

The The The The

manufacturer CAA JAA insurers

Explanation: The forward and aft CG limits are established by the manufacturer and based upon the requirements for positive longitudinal static stability at the certification stage. Question 205: The effect of operating an aeroplane with a CG too far forward is to experience: A Inability or difficulty in flaring on touchdown, resulting in nosewheel landing first. B Lower stick forces per G loading. C Lower stalling speed. D Inability or difficulty in trimming when flaps are retracted. Explanation: If the CG is beyond the forward limit then there will not be sufficient elevator authority to flare. This will mean in the nosewheel touches down first - possibly resulting in damage.

Question 206: The handling and performance problems encountered with a CG too far aft include: A Degrade or loss of nose wheel steering. B Improvement in nose wheel steering. C Higher stick forces per G loading with no risk of over-stressing the airframe in manoeuvres. D No likelihood of a nose up overbalance on a nose wheel aircraft on the ground resulting in tail damage. Explanation: A CG beyond the aft limit will mean that the majority of the weight will be over the mainwheels meaning there will be very little nosewheel adhesion leading to a loss of nosewheel steering. Question 207: Just prior to take-off, a baggage handler put a large put a large extra bag into the forward hold without recording it in the LMC's. What are the effects of this action? 1. VMC will increase if the extra load is forward of the datum. 2. Stick forces in flight will decrease if the extra load is behind the datum. 3. Stick forces at VR will increase. 4.VMU will occur later. 5. The safe stopping distance will increase. A B C D

3, 4 and 5 only 2,3 and 4 only all of the above 1 and 5 only

Explanation: A forward CG increases stick forces resulting in a slightly later VMU. The inertia from the forward CG could also cause the stopping distance to increase. Question 208: Which of the following would not affect the CG? A B C D

Stabiliser trim setting. Cabin crewmembers performing their normal duties. Fuel usage. Mass added or removed at the neutral point.

Explanation: You trim the stabiliser as a result of a CG change. It does not move the CG. The other options do! Question 209: What effect does the CG on the aft limit have on the fuel flow of an aeroplane? A Decreases

B Marginal increase C No effect D Increases Explanation: An aft CG reduces the trim drag. Therefore less thrust is required leading to a decrease in fuel consumption. Question 210: The operating mass of an aircraft is: A B C D

The The The The

dry operating mass plus the take-off fuel mass. empty mass plus crew, crew baggage and catering. empty mass plus the take-off fuel mass. empty mass plus the trip fuel mass.

Explanation: NO Question 211: The Regulated Take-off Mass: A Is the lower of maximum structural take-off mass and the performance limited take-off mass. B Is the higher of the maximum structural zero fuel mass and the performance limited takeoff mass. C The maximum structural take-off mass subject to any last minute mass changes. D The maximum performance limited take-off mass subject to any last minute mass changes. Explanation: NO Question 212: The Basic Empty Mass is the: A B C D

Explanation: NO Question 213: The Operating Mass: A B C D

Is Is Is Is

the the the the

take-off mass minus the traffic load. maximum zero fuel mass minus the traffic load. take-off mass minus the basic empty mass and crew mass. landing mass minus the traffic load.

Explanation: The Operating Mass = DOM + fuel. The take-off mass = DOM + fuel + traffic load. Therefore, if we subtract the Traffic load from the take off mas we get the Operating Mass! Question 214: Mass refers to: A B C D

The quantity of matter in a body as measured by its inertia. Weight X gravity. The force exerted on a body by gravity. None of the above.

Explanation: In physics mass can be defined as a quantitative measure of an object's resistance to the change in speed. Question 215: Basic Empty Mass: A Is usually determined by the aircraft manufacturer during weighing at the manufacturing plant and does not include any additional equipment. B Is the mass of an aeroplane including operating fluids, toilet and galley water and fuel. C Is usually the mass of the basic aeroplane without any operating fluids. D Is the mass of an aeroplane as produced by the manufacturer including operating fluids, toilet and galley water plus all required equipment plus unusable fuel. Explanation: Basic Empty Mass (Basic Mass) is the mass of an aeroplane plus standard items such as: - emergency oxygen equipment - lubricating oil in engine and auxiliary units - pyrotechnics - unusable fuel and other unusable fluids - supplementary electronic equipment - fire extinguishers Question 216: Who determines what items to include in the dry operating weight? A B C D

The The The The

operator manufacturer JAA commander

Explanation: Dry Operating Mass (DOM) is the total mass of the aeroplane (excluding usable fuel and traffic load) Items contained in the DOM include: - Crew and crew baggage. - Food and beverages.

- Potable water and lavatory chemicals. - Catering and removable passenger service equipment. The operator determines the DOM. Once the aircraft is delivered at the BEM the operator adds the masses of the crew and operational items to establish the DOM. Question 217: The chemical fluids used to charge the aircraft's toilets are counted as? A B C D

Part Part Part Part

of of of of

the the the the

Explanation: The variable load contains items (including toilets) required by the operator to operate the service and form part of the DOM. Question 218: The Traffic Load of an aeroplane is: A B C D

TOM minus Operating Mass. LM plus Trip Fuel. Useful Load minus Operating Mass. Tom minus Useful Load.

Explanation: The Traffic Load is the total mass of passengers, cargo and baggage (including any non-revenue load) Question 219: Define the useful load. A B C D

Traffic load plus usable fuel mass. Dry operating mass plus usable fuel load. Traffic load plus dry operating. That part of the traffic load which generates revenue.

Explanation: NO Question 220: 2588 USG of fuel has been loaded into an aircraft, what is the volume in litres? A B C D

9796.65 979.664 9979.66 9785.58

L L L L

Explanation: 9796.65 L is given as the correct answer. However, if the 2588 USG is x 3.785 (litre to USG) the more accurate answer should be 9795.58 litres. Out of the given options, 9796.65 L is the closest

answer. The conversion rate of 3.785 litres is contained in the CAP 696 General Notes Page 4. Question 221: 5600 USG is equivalent to how many Imperial gallons? A B C D

4663 4366 4848 6338

imp imp imp imp

Explanation: The correct answer is given as 4663 imp. However, if the 5600 USG is รท 1.203 (USG to Imp Gall) the more accurate answer should be 4655 Imp Galls. 4663 is the closest possible answer out of the choices given. Question 222: 125 USG of Avgas in Litres is? A B C D

473 L 460 L 37 L 358 L

Explanation: 125 USG x 3.785 litres = 473 litres. The rate for a conversion of litres to USG is quoted in the CAP 696 General Notes Page 4 Question 223: How would you convert US Gallons to Litres (L)? A B C D

US US US US

Gall Gall Gall Gall

x x x x

3.785 4.546 1.205 0.264

Explanation: Conversion rate of litres to USG is 3.785. quoted in CAP 696, General Notes Page 4.

The rate is

Question 224: If 567 Kgs of fuel at SG 0.812 are on board an aircraft, the amount of fuel in US gallons is: A B C D

184 168 201 161

USG USG USG USG

Explanation: Firstly convert the Kg to Litres by taking 567kg รท 0.812 (SG) = 698.3 litres. Then divide 698.2 litres by 3.785 = 184.5 USG

Question 225: 125,988 kg is how many pounds? A B C D

277,756 lbs 271,525 lbs 58,458 lbs 314,970 lbs

Explanation: 125,988kg x 2.205lb = 277,803.5lb is a more correct answer. The conversion rate of lb to kg is given in CAP 696, General Notes Page 4. You must always choose the answer closest to the correct one out of the choices given. In this example – 277,756 lbs. Question 226: To convert 1 US gallon of AVGAS to lbs: A B C D

1 1 1 1

USG USG USG USG

= = = =

6 lbs 4 lbs 6.8 lbs 3.8 lbs

Explanation: The conversion rate of USG to lb is accepted as 6lb per USG. Question 227: The weight of 867 US Gallons of fuel (SG 0.78) is: A B C D

5643 5361 2560 8122

lbs lbs lbs lbs

Explanation: 867 USG ÷ 1.203 (USG to Imp Gall) = 720.7 Imp Gall x 10 (1 Imp Gall water to lb) x 0.78 (SG) = 5621.4lb is a more accurate answer. Out of the given options, „5643‟ is the closest and best possible answer. Question 228: The maximum zero-fuel mass: A B C D

Is calculated for a maximum load factor of +2.5 g. Can be increased by stiffening the wing. Is calculated for a maximum load factor of +3.5 g. Imposes fuel dumping from the outer wings tank first.

Explanation: It is required by CS25 that transport aircraft be designed to a limit load when manoeuvring of +2.5g at the MZFM. Question 229: The Maximum Zero Fuel Mass is the mass of the aeroplane with no usable fuel on board. It is a limitation which is:

A Listed in the Flight Manual as a fixed value. It is a structural limit. B Tabulated in the Flight Manual against arguments of airfield elevation and temperature. C Governed by the traffic load to be carried. It also provides protection from excessive 'wing bending'. D Governed by the requirements of the centre of gravity limits and the structural limits of the aeroplane. Explanation: The Maximum Zero Fuel Mass (MZFM) is a structural limit listed in the Flight Manual. Question 230: With regards to the Maximum Zero-Fuel Weight (MZFW). A It is the maximum weight that an aircraft can be loaded to without useable fuel. B It is important as exceeding the MZFW may mean that there is insufficient lift to get the aircraft airborne. C It is lower than the Maximum Take-Off Weight by the weight of a payload. D Is more relevant to aircraft with fuselage fuel tanks. Explanation: CAP 696 Definition: Maximum Zero Fuel Mass (MZFM) is the maximum permissible mass of an aeroplane with no usable fuel. Question 231: To calculate a usable take-off mass, the factors to be taken into account include: A B C D

Maximum Maximum Maximum Maximum

landing mass augmented by the fuel burn. take-off mass decreased by the fuel burn. landing mass augmented by fuel on board at take-off. zero fuel mass augmented by the fuel burn.

Explanation: CAP 696 Definition: Maximum Structural Landing Mass (MSLM) is the maximum permissible total aeroplane mass on landing in normal circumstances. This is the calculated from the Maximum Take-Off Mass less Fuel Burn. Question 232: If the maximum structural landing mass is exceeded: A The undercarriage could collapse on landing. B The aircraft will be unable to get airborne. C No damage will occur providing the aircraft is within the performance limited landing mass. D No damage will occur providing the aircraft is within the regulated landing mass.

Explanation: CAP 696 Definition: Maximum Structural Landing Mass (MSLM) is the maximum permissible total aeroplane mass on landing in normal circumstances. Exceeding this could cause the undercarriage to fail on landing. Question 233: If an aeroplane lands below its Max Structural Landing Mass but above the Performance Limited Landing Mass for the arrival airfield: 1. It will not suffer physical damage as a result of the extra mass. 2. Tyre temperature limits could be exceeded. 3. It might not have sufficient runway length in which to stop safely. 4. A go-around might not be achievable. 5. It will increase structural fatigue. A B C D

2,3,4 3 and 1,3,4 1 and

and 5 only 4 only and 5 only 5 only

Explanation: 1. It will not suffer physical damage as a result of the extra mass - FASLSE - it can suffer physical damage 2. Tyre temperature limits could be exceeded - TRUE - excess braking of the overweight aircraft could cause temperature limits to exceed 3. It might not have sufficient runway length in which to stop safely TRUE - excess inertia in the landing aircraft may cause the aircraft to over-run the available runway length 4. A go-around might not be achievable - TRUE - there may be insufficient power for a Go-Around given the ambient conditions at landing 5. It will increase structural fatigue - TRUE landing an overweight aircraft will increase structural fatigue Question 234: A twin-engine aeroplane is certified for a Max Structural TOM and a Max LM of 58,000 kg and 55,000 kg respectively. Given the information below, what is the limiting take-off mass for the aeroplane? Performance Limiting TOM 61000 kg Performance Limiting LM 54000 kg Operating mass 55000 kg Trip fuel 3000 kg Contingency fuel 5% of trip fuel Alternate fuel 500 kg Final reserve 500 kg Flight duration 3 hours Fuel consumption 500 kg per hour per engine A B C D

57000 58000 61000 56545

kg kg kg kg

Explanation: MLM = 54000 kg MTOM is MLM + trip fuel. 54000 + 3000 = 57000 kg Question 235: An aeroplane takes off as normal on a scheduled flight however, shortly after take-off the aeroplane is diverted to another airfield. Max

Structural TOM 14000 kg Performance Limited TOM 12690 kg Max Structural LM 9600 kg Trip Fuel to original destination 6000 kg Contingency fuel 200 kg Alternate fuel 200 kg Final reserve fuel 750 kg Expected landing mass at original destination 4600kg Actual flight duration 2 hours Fuel consumption 1500 kg per hour Performance Limited LM at diversion airfield 9000 kg A The aeroplane can land safely because it is below its PLLM. B The aeroplane can land safely as it is below its MSLM. C The aeroplane cannot land safely because it is above its MSLM. D The aeroplane cannot land safely because its mass is beyond the limit for the runway limitations. Explanation: Expected LM = 4600kg which includes 1150kg of fuel (cont/alt and final reserve) so the ZFM = 3450kg The TOM = ZFM + TOF = 3450+7150 = 10600kg Trip fuel to diversion = 3000kg LM = TOM - div fuel = 10600-3000= 7600kg which is below structural and PLLM so the A/C can land safely. It has to be Answer (c) because it is referring to the lower PLLM. Question 236: An aeroplane develops a serious technical problem shortly after take-off and has to return to its departure airfield. In order to land safely the aircraft must dump fuel. How much fuel must be dumped? A The pilot calculates the amount of fuel to jettison to reduce the mass to a safe level at, or below the Regulated Landing Mass. B As much as the pilot feels is just sufficient to land safely. C The fuel system automatically stops the jettison at the Regulate Landing Mass. D Sufficient to reduce the mass to the zero fuel mass. Explanation: You would only jettison sufficient fuel to bring the AUM to the MLM. Modern systems allow input of "fuel to remain" and then this is controlled automatically. Regarding the requirements for a fuel jettison system under CS25 any system will only allow you to jettison down to stack pipe level (a mechanical system) that will ensure sufficient fuel for a climb from sea level to 10 000 ft and cruise for 45 minutes at the aircraft's speed for maximum range. Question 237: "A scheduled flight of three hours estimated time, within Europe, is to be conducted. Using the data given calculate the maximum mass of freight that may be loaded in the following circumstances: Performance limited take-off mass: 67,900 kg Performance limited landing mass: 56,200 kg MZFM: 51,300 kg DOM: 34,960 kg Fuel on board at ramp: 15,800 kg Taxi fuel: 450 kg Trip fuel: 10,200 kg Passengers: 115 adults + 6 children

Flight crew (each 85 kg): 2 Cabin crew (each 75 kg): 5 Allow standard baggage for each passenger (13 kg)" A B C D

4647 1047 6147 4764

kg kg kg kg

Explanation: Explanation: Firstly, calculate the most limiting TOM PLTOM PLLM MZFM TF = Trip Fuel 67900 56200 51300 TOF = Take-Off Fuel TF 10200 TOF 15350 67900 66400 66750 66400kg is the limiting TOM. Now calculate the mass of the passengers and bags using the Standard Mass tables in CAP 696. 115 Adults x 84kg = 9660kg 6 Children x 35kg = 210kg 121 bags x 13kg = 1573kg Total = 11443kg Find the maximum payload available by taking the limiting TOM and deducting the DOM and Take-Off Fuel 66400kg – 34960kg – 15350kg = 16090kg Max Payload Max Freight available = 16090kg – 11443kg = 4647kg Question 238: Given: DOM = 3415 lbs Trip fuel = 400 lbs Passenger Mass = 600 lbs Freight/baggage = 1050 lbs Maximum structural TOM = 5850 lbs Performance TOM = 5200 lbs The traffic load is to be reduced to bring the TOM into the regulated TOM limits by: A B C D

265 400 600 200

lbs lbs lbs lbs

Explanation: The limiting TOM is based on the PLTOM = 5200b The addition of DOM +Traffic Load + Trip Fuel comes to 5465lb Therefore a reduction of 265lb in the Traffic Load needs to be done. Question 239: Given: DOM(APS) = 3500 lbs Trip fuel = 500 lbs Passenger Mass = 400 lbs Freight/baggage = 1250 lbs Maximum structural TOM = 5850 lbs Performance Regulated TOM = 5200 lbs The traffic load is to be reduced to bring the TOM into the regulated TOM limits by:

A B C D

450 650 250 200

lbs lbs lbs lbs

Explanation: The limiting TOM is based on the PLTOM = 5200lb The addition of DOM + Traffic Load + Trip Fuel comes to 5650lb Therefore a reduction of 450lb in the Traffic Load needs to be done. Question 240: What is the allowed traffic load for a medium range jet aircraft where the DOM is 35058 kg, where the limitations are MZFM 52790 kg, RTOM 61875 kg, RLM 53871 kg. The TOF is 13358 kg of which 8900 kg is trip fuel. A B C D

13,459 kg 10, 376 kg 14,355 kg 17,732 kg

Explanation: Firstly, work out what the most limiting TOM would be RTOM RLM MZFM TF = Trip Fuel 61875 53871 52790 TOF = Take-Off Fuel TF 8900 TOF 13358 61875 62771 66148 61875kg is the most limiting TOM. To calculate the Max Traffic Load subtract the DOM and the TOF from the RTOM 61875kg – 35058kg – 13358kg = 13459kg Question 241: Standard masses for baggage can only be used when the aircraft has: A B C D

20 seats or more 30 seats or more Less than 30 seats 9 seats or more

Explanation: NO Question 242: For the purposes of mass and balance JAR-OPS 1 defines a child as a person aged: A B C D

Of Of Of Of

2 3 2 3

years years years years

but but but but

not having reached their 12th birthday. not having reached their 12th birthday. having reached 15 years old. not having reached their 15th birthday.

Explanation: Children are defined as being between the ages of 2 and 12.

Question 243: The term 'BAGGAGE' means: A B C D

Personal belongings. Any non-human, non-animal cargo. Any freight cargo not carried on by the person. Excess freight.

Explanation: This is a fact you should know! Question 244: Choose the correct statement as related to infants travel in aircraft: A An B An C An kg. D An

infant is a person between the ages of 0 and 2 years. infant seated on an adults lap increases the pax mass by 35 kg. infant must always be seated in its own seat and accounted for as 35 infant is a person between the ages of 0 years to 3 years.

Explanation: The term 'infant' applies up to the age of 2. Past this they are considered as 'children' until the age of 12. Question 245: Standard passenger masses used for a holiday charter flight are: A B C D

Male 83 kg, Female 69 kg, or an average mass of Adult person 76 Kg. Child up to the age of 12 years 30 Kg. Average mass of adult person 84 Kg. Male 80 Kg, Female 70 Kg.

Explanation: NO Question 246: If standard mass tables are being used for checked baggage and a number of passengers check in baggage that is expected to exceed the standard baggage mass, the operator: A Must determine the actual mass of such baggage by weighing or adding an adequate mass increment. B Need make no alterations if the Take-off mass is not likely to exceeded. C Determine the actual masses of such baggage. D Must determine the actual mass of such baggage by weighing or by deducting an adequate mass increment. Explanation: EU-OPS Subpart J The operator must determine the actual mass of such baggage by weighing or adding an adequate mass increment.

Question 247: On any flight identified as carrying a significant number of passengers whose masses, including hand baggage, are expected to exceed the standard passenger mass the operator: A Must determine the actual mass of such passengers or add an adequate increment to each of such passengers. B Must add an adequate mass increment to each of such passengers. C Must determine the actual mass of such passengers. D Need only determine the actual masses or apply an increment if the Take-off mass is likely to be exceeded. Explanation: EU-OPS Subpart J The operator must determine the actual mass of such passengers or add an adequate increment to each of such passengers. Question 248: The weighing machine used for passenger weighing shall have a capacity of at least â&#x20AC;Ś and shall be displayed at minimum graduations of â&#x20AC;Ś A B C D

150 100 200 250

kg, kg, kg, kg,

500 g 1 kg 500 g 1 kg

Explanation: EU-OPS Subpart J Appendix 1 to OPS 1.620: The weighing machine to be used for passenger weighing shall have a capacity of at least 150kg. The mass shall be displayed to minimum graduations of 500g and accurate to within 0.5% or 200g whichever is greater. Question 249: For an aircraft having 16 passenger seats, if no hand luggage is carried, how much weight may be deducted from the standard passenger weights for passengers over 12 years old? A B C D

6 Kg 12 kg 0 kg 10 Kg

Explanation: EU-OPS Subpart J: On flights where no hand baggage is carried or where hand baggage is accounted for separately 6 kg may be deducted from the male and female masses. Question 250: Select the correct mass allowed for cabin crew in a 19 seat aircraft used on a holiday charter:

A B C D

75 84 76 85

kg kg kg kg

Explanation: Refer to tables in CAP 696 Question 251: The centre of gravity is that … on an aircraft through this the total … is considered to act vertically … A B C D

point; datum; point; datum;

mass; downwards moment; downwards moment; upwards mass; upwards

Explanation: The centre of gravity of a body is the point through which the sum of the forces of all masses of the body is considered to act vertically downwards. Question 252: If all the moments are positive when calculating mass (weight) and balance, the position of the datum would be at the: A B C D

Nose, or forward of the aircraft. Centre line of the nose or tail wheel depending on the aircraft type. Main wheels centreline. Trailing edge of the wing.

Explanation: NO Question 253: Balance Arm (B) is: A The distance B The point on turns. C The distance D The distance

from the datum to the centre of gravity of a mass. which a lever is supported, balanced, or about which it from the centre of gravity to the centre of a mass. from the centre of pressure to the centre of a mass.

Explanation: NO Question 254: The aircraft datum is a … reference point that is defined on or relative to the aircraft about which the … of any load locations are known. A B C D

fixed, arms forward, arms movable, moments variable, movements

Explanation: NO Question 255: Select the correct statement for the Datum: A Is a fixed vertical plane from which all the arm distances are measured. B Is a vertical plane through which all the forces of gravity are said to act. C Is the point through which all the forces of gravity are said to act. D Is a point from which all the arm distances are measured. Explanation: NO Question 256: The datum for determining the CG has to be along the longitudinal axis: A B C D

But does not have to be between the nose and the tail. Between the nose and the tail. At the fire wall. Between the leading and trailing edge of the MAC.

Explanation: NO Question 257: Determine the attached drawing and consider the correct mass & balance calculation (balanced condition). 60-0204.gif A B C D

Fb Fb Fb Fb

= = = =

(Fa x A) / B A / (Fa x B) (A x B) / Fa (Fa x B) / A

Explanation: From the diagram A x Fa = B x Fb. Therefore Fb = A x Fa B Question 258: Select the correct statement for the CG safe range: A The safe range both limits. B The safe range include them. C The safe range includes the fwd D The safe range includes the aft

falls between the front and rear CG limits and includes falls between the front and rear CG limits but does not falls between the front and rear CG limits but only limit. falls between the front and rear CG limits but only limit.

Explanation: NO Question 259: The C.G. limits are from 5 inches forward to 7 inches aft of the datum. If the MAC is 41 inches and its leading edge is 15 inches forward of the datum, what are the C.G. limits as % MAC? A B C D

Fwd Fwd Fwd Fwd

limit limit limit limit

24.4%, 36.6%, 29.3%, 12.2%,

Aft Aft Aft Aft

limit limit limit limit

53.7% 53.6% 36.6% 29.3%

Explanation: If the LEMAC is 15” in forward of the Datum and the CG Forward limit is 5” forward of the Datum the CG Forward limit is 10” into the MAC. The MAC is 41” so 10” as a % of 41” is 24.39%. The CG aft limit is 22” into the MAC so 22” as a % of 41” is 53.66% Question 260: Given the data below calculate the C of G at TOM as a percentage of the mean aerodynamic chord: Basic Empty Mass: 1095 kg C of G at BEM: 1.98m (25% MAC) Pilot and front seat occupant 80kg each. Front seat arm: 2.5m Rear seat arm: 3.2m Fuel load: 139 litres (SG 0.72) Fuel arm: 1.9m MAC 2m A B C D

28% 32% 30% 31%

Explanation: The CG for the BEM is 1.98m aft of the Datum and 25% of MAC. MAC is 2m so 25% is 0.5m. That means that the LEMAC is 1.48m aft of the Datum. Now calculate the CG for the loaded aircraft Mass Arm Moment 1095 1.98 2168.1 160 2.5 400 100 1.9 190.15 1355kg 2758.25kg.m New CG is 2758.25kg.in ÷ 1355kg = 2.04m With the LEMAC at 1.48m the new CG is 2.04 – 1.48 = 0.56m into the MAC 0.56m as a % of 2m = 28% Question 261: An aircraft that is not within a fleet mass evaluation program must be weighed: A B C D

Every Every Every Every

fourth year. third year. other year. year.

Explanation: NO Question 262: When weighing an aircraft certain precautions have to be made. Which of the following is not a requirement: A B C D

Ensure all fuel and oil tanks are full. Ensure the aeroplane is clean. Check for completeness of aeroplane and equipment. Ensure the weighing is accomplished in an enclosed building.

Explanation: NO Question 263: Aircraft must be weighed: A: On initial entry into service. B: If the mass and balance records have not been adjusted for alterations or modification. C: Every four years after initial weigh. D: Whenever the cumulative changes to the dry operating mass exceed plus or minus 0.5% of the maximum landing mass. E: If the cumulative change in CG position exceeds 0.5% of the mean aerodynamic chord. A B C D

A, B, C, D and E A and C only A, B and C only A, C and E only

Explanation: NO Question 264: An operator has 19 aircraft of the same type and wants to use fleet mass values. Select the number of aircraft that will have to be weighted on the initial weighing and the subsequent weighings: A B C D

Initially Initially Initially Initially

19 10 19 10

thereafter thereafter thereafter thereafter

7. 7. 14. 14.

Explanation: NO Question 265: The operator must establish the mass of the Traffic Load: A By actual weighing or determine the mass of the traffic load in accordance with standard masses as specified in JAR-OPS sub part J. B Prior to initial entry into service. C Prior to embarking on the aircraft. D By using an appropriate method of calculation as specified in the JAROPS subpart J.

Explanation: EU-OPS Sub part J: “An operator must establish the mass of the traffic load, including any ballast, by actual weighing or determine the mass of the traffic load in accordance with standard passenger and baggage masses as specified in OPS 1.620” Question 266: The mass and centre of gravity of an aircraft must be established by actual weighing: A B C D

By By By By

the the the the

operator prior to initial entry of aircraft into service. pilot on entry of aircraft into service. engineers before commencing service. owner operator before the first flight of the day.

Explanation: This is a bad question! Remember - we don't write the questions we only try and reproduce them as accurately to the exam as possible. The answer given cannot be said to be strictly true but it is the best of the four options. The aircraft is weighed at its BEM by the manufacturer prior to delivery to the operator. Only if the operator is satisfied that the weighing schedule has been adjusted for modifications or alterations then it can be put directly into service. Question 267: Given the following information, calculate the loaded centre of gravity (cg). STATION MASS(kg) / Balance Arm (cm) / MOMENT (kgcm) Basic Empty Condition: 12045 / +30 / +361350 Crew: 145 / -160 / -23200 Freight1: 5455 / +200 / +1091000 Freight2: 410 / -40 / -16400 Fuel: 6045 / -8 / -48360 Oil: 124 / +40 / +4960 A B C D

56.53cm 60.16cm 56.35cm 53.35cm

aft aft aft aft

datum. datum. datum. datum.

Explanation: Explanation: To calculate the final CG use the following table Mass Arm Moment 12045 +30 +361350 145 -160 -23200 5455 +200 +1091000 410 -40 -16400 6045 -8 -48360 124 +40 +4960 24224kg +1369350kg.cm Therefore +1369350kg.cm ÷ 24224kg = +56.53cm

Question 268: The position of the centre of gravity can always be determined by: A B C D

Dividing the total moment by the total mass. Dividing the total mass by the total moment. Subtracting the total mass from the total moment. Subtracting the total moment from the total mass.

Explanation: The arm to the CG is calculated by dividing the total moment by the total mass. Question 269: If an aircraft weight = 2000kg and 400kg of freight is added to a hold 2m aft of the present CG position, the movement of the CG is: A B C D

0.33 0.4m 0.4m 0.33

aft aft forward forward

Explanation: To calculate the CG movement when the 400kg is added assume the Datum is at the current CG position and then use the following table Mass Arm Moment 2000 0 0 400 +2 +800 2400Kg +800kg.m Therefore +800kg.m รท 2400kg = +0.33m Question 270: An aircraft has a mass of 7900kg and the CG is located at 81.2 in aft of the datum. If a package of mass 250 kg was loaded in a hold situated 32 in aft of the datum, what would the new CG position be? A B C D

79.7 89.3 74.4 88.1

inches inches inches inches

Explanation: Explanation: To calculate the new CG position when the 250kg is added use the following table Mass Arm Moment 7900 +81.2 +641480 250 +32 +8000 8150kg +649480kg.in Therefore + 649480kg.in รท 8150kg = 79.7in Question 271:

Given: C of G is located at STN 15 Aeroplane mass is 3650 lbs. What is the effect on the C of G if you move baggage (total mass 64 lbs) from STN 14 to STN 20? A B C D

It It It It

moves moves moves moves

AFT AFT AFT FWD

by by by by

0.1 units 0.3 units 0.31 units 0.13 units

Explanation: By moving the baggage further aft you would expect move aft. You can do this calculation in two ways. Firstly use the Mass x Arm = Moment table Mass Arm 3650 Stn 15 - 64 Stn 14 + 64 Stn 20 3650lb +55134lb.in Therefore new CG = +55134lb.in รท 3650lb = 15.1units. Therefore 0.1unit aft. The second method uses the formula d = D x m M Where d = Distance the CG moves, D = Distance between the stations, m = Mass to be moved and M = Mass of the aircraft d= 6 x 64 = 0.1unit 3650

the CG to Moment +54750 - 896 +1280 CG moves

Question 272: An aircraft has three holds situated 10 in 100 in and 250 in aft of the datum, identified as holds A,B and C respectively. The total aircraft mass is 3500 kg and the CG is 70 in aft of the datum. The CG limits are from 40 in to 70 in aft of the datum. How much load must be removed from hold C to ensure that the CG is positioned on the forward limit? A B C D

500kg 400kg 250kg 350kg

Explanation: Draw a sketch of the aeroplane and establish the distances: Small distance (CG change) 70 -40 = 30in Big distance (effective arm: from hold to new CG) = 250-40 = 210in Put the numbers in the formula: SM/BM=SD/BD Transposed: SM = SD x BM/BD = 30 x 3500/210 = 501.428kg. Rounded down to 500kg

Question 273: An aircraft has a mass of 5000 lbs and the CG is located at 80 in aft of the datum. The aft CG limit is at 80.5 in aft of the datum. What is the maximum mass that can be loaded into a hold situated 150 in aft of the datum without exceeding the limit? A B C D

35.97 23.15 39.50 58.15

lbs lbs lbs lbs

Explanation: For this calculation use the following formula m= d x M D Where d = Distance the CG can move, D = Distance between the New CG (aft limit) and the hold that the load is added to, m = Mass of the load that can be added and M = Original Mass of the aircraft m= 0.5 x 5000 = 35.97lb 69.5 Question 274: The CG limits of an aircraft are from 83 inches to 93 inches aft of the datum. The CG as loaded is bound to be at 81 inches aft of the datum. The loaded mass is 3240 lbs. How much mass must be moved from the forward hold, 25 inches aft of the datum, to the aft hold, 142 inches aft of the datum, to bring the CG onto the forward limit? A B C D

55.38 74.96 22.49 82.09

lbs lbs lbs lbs

Explanation: For this calculation use the following formula m= d x M D Where d = Distance the CG has to move to the forward limit, D is the distance between the holds when the load is transferred, m = Mass to be transferred and M = Loaded mass m= 2in x 3240lb = 55.38lb 117in Question 275: The CG limits of an aircraft are from 72 inches to 77 inches aft of the datum. If the mass is 3700 kg and the CG position is 76.5 inches aft of the datum, what will the change to the CG position be if 60 Kgs is removed from the fwd hold located at 147 in fwd of the datum? A 3.68 inches

B 2.12 inches C 4.66 inches D 3.31 inches Explanation: The basic formula for this type of question is: Change in Mass/Total mass = change in CG/total distance moved. If you know three of the variables the fourth can be found. It is recommended that you draw a sketch of the aircraft. In this particular question the datum is in the middle of the aeroplane. The change in mass = 60kgs. The total mass will be 3700-60 = 3640. The total distance moved (the effective arm) is 147째 + 76.5째 = 223.5째 Transpose the formula and substitute known values. Change in CG = 60 x 223.5/3640 = 3.68째 Question 276: If the CG position is 30m aft of the datum after 3,000 kg was added to a hold 50m aft of the datum and the original weight was 24,000 kg, the original CG arm was: A B C D

27.5m 34.2m 50.0m 41.4m

Explanation: Explanation: For this calculation use the following formula d= D x m M Where d = Distance the CG will move once the 3000kg is loaded, D = Distance between the New CG and the hold that the 300kg is put into, m = the 3000kg that is added and the M = the original mass of the aircraft d= 20 x 3000 = 2.5m 24000 If the new CG is 30m aft of the Datum then the original CG must have been at 27.5m Question 277: An aircraft has a loaded mass of the datum. A passenger, mass 150 distance of 70 inches. What will all dimensions aft of the datum) A B C D

23.9 22.9 21.1 26.3

5,500 lbs. The CG is 22 inches aft of lbs, moves aft from row 1 to row 3 a be the new position of the CG (assuming ?

inches inches inches inches

Explanation: For this calculation use the table Mass x Arm = Moment. Remember that the mass of the aircraft does not change in this calculation.

Mass Arm 5500 22 150 70 5500 131500 Therefore the new CG = 131500lb.in Ăˇ 5500lb = 23.9in

Moment 121000 10500

Question 278: If the C.G. position is 21% MAC, the MAC is 73 inches, and the C.G. datum is 26 inches aft of the leading edge of the MAC, what is the C.G. position relative to the datum? A B C D

10.67 10.67 41.33 41.33

inches inches inches inches

fwd aft aft fwd

of of of of

datum. datum. datum. datum.

Explanation: If the CG is 21% of MAC (73in) that means it is 15.33in into the MAC. The CG Datum is 26in aft of the LEMAC therefore the CG is 26in â&#x20AC;&#x201C; 15.33in = 10.67in Forward of the Datum. Question 279: Once the mass and balance documentation has been signed prior to flight: A B C D

Acceptable last minute changes to the load must be documented. Documented last minute changes to the load may be incorporated. No load alterations are allowed. The documentation is not signed prior to flight.

Explanation: NO Question 280: An aircraft of mass 17,400 kg, has its CG at station 122.2. The CG limits are 118 to 122. How much cargo must be moved from the rear hold at station 162 to the forward hold at station -100 (forward of the datum) to bring the CG to the mid position of its range? A B C D

146.1 kg 119.9 kg 99.9 kg 55.3 kg

Explanation: For this question use the following formula m= d x M D Where d = Distance the CG has to move to the mid limit, D = Distance between the two stations that the load is transferred, m = Mass that has to be moved and M = the mass of the aircraft m= 2.2 x 17400 = 146.1kg 262

Question 281: The loaded mass of the aircraft is found to be 1850 lbs and the CG moment 154,000 lbs in. How much mass must be moved from the forward hold 40 inches aft of the datum, to the rear hold, 158 inches aft of the datum, to bring the CG on the forward limit? (CG limits are 86 inches fwd and 90 inches aft) A B C D

43.2 23.1 66.0 50.9

lbs lbs lbs lbs

Explanation: Before making the calculation we need to find out where the current CG is positioned. Take the total moment, 154,000lb.in and divide that by the aircraft mass, 1850lb, and that gives a CG position of 83.24in. That CG has to be move to the Forward limit, 86in, which means a movement of 2.76in Now use the following formula m= d x M D Where d = Distance the CG moves, D = Distance between the holds that the load is being moved, m = Mass to be moved between the holds and M = the mass of the aircraft m= 2.76 x 1850 = 43.27lb 118 Question 282: The loaded mass of an aircraft is 12,400 kg. The aft CG limit is 102 inches aft of the datum. If the CG as loaded is 104.5 inches aft of the datum, how many rows forward must two passengers move from the rear seat row (224 inches aft) to bring the CG on to the aft limit, if the seat pitch is 33 inches? Assume a passenger mass of 75kg each. A B C D

7 5 6 8

rows rows rows rows

Explanation: To make this calculation you have to take the aircraft mass and the two CG positions to get the moments. Mass Arm Moment 12400 104.5 1295800 12400 102 1264800 The difference between the Moments will be generated by the movement of the two passengers = 31,000kg.in Divide that 31,000kg.in by the 150kg (passenger mass) = 206in Divide that 206in by 33in (seat pitch) = 6.27 rows which means a minimum of 7 rows.

Question 283: An aeroplane has a zero fuel mass of 47,800kg and a performance limited take-off mass of 62,600kg. The distances of the leading edge and trailing edge of the MAC from the datum are 16m and 19.5m respectively. What quantity of fuel must be taken up to move the CG from 30% MAC to 23% MAC if the tank arm is 16m aft of the datum and the fuel SG is 0.72. A B C D

4455 4213 3990 8850

IMP Gallons US Gallons Kg L

Explanation: This question is not worded particularly well. What its really saying is that at the ZFM the CG is at 30%MAC and how much (volume) of fuel must be added to bring the CG to 23% MAC? Draw a sketch of the aircraft. The length of the MAC is 19.5m 16m = 3.5m Current CG is at 30% MAC and the new CG is to be at 23% MAC so the physical distance to move the CG is 7% MAC = 3.5 x 7% = 0.245m The distance from the fuel tank to the new CG is 16.805m 16m = 0.805m The mass for the formula is the ZFM. Use the formula and substitute known values and transpose to find mass change in kgs: Mass change/Old total mass = Change of CG/Distance from mass to new CG Mass change = 0.245 x 47800/0.805 = 15547kgs Convert to litres using given SG: 14547/0.72= 20204 litres Convert litres to IMP gallons = 20204/4.546= 4444 IMP gallons Convert litres to USG = 20204/ 3.785= 5337.9 USG From the given answers it has to be 4455 IMP gallons. Question 284: Given: Maximum Floor Strength = 300 kg/m² Maximum Running Load = 350 kg/m Which of the following loads is legally permissible? A B C D

400 400 500 700

Kgs Kgs Kgs Kgs

-

Dimensions Dimensions Dimensions Dimensions

1.2m 1.4m 1.5m 1.8m

x x x x

1.2m 0.8m 1m x 1.4m

x 1.2m x 0.8m 1m x 0.8m

Explanation: You have to approach this question from two sides. Firstly, you have to see which of the crates meet the floor loading limit of 300kg/m² and then, if more than one crate complies, you must check the remaining ones against the Running Load limit of 350kg/m. To work out the max load for each crate against the floor loading limit multiply the floor limit by the area of each crate. Crate A. 300kg/m² x 1.12m² = 336kg (Outside limit with 400kg) Crate B. 300kg/m² x 1.44m² = 432kg (Inside limit with 400kg) Crate C. 300kg/m² x 2.52m² = 756kg (Inside limit with 700kg) Crate D. 300kg/m² x 1.5m² = 450kg (Outside limit with 500kg) Now take Crates B & C and check them against the Running Load limit (350kg/m) by dividing the Load by the Length of each crate.

Crate B.

400kg = 333kg/m. Crate C. 700kg = 388.9kg/m 1.2m 1.8m Therefore Crate B meets both the floor and running load limits. Question 285: In cruise flight, an aft centre of gravity location will: A B C D

Decrease longitudinal static stability. Does not influence longitudinal static stability. Increase longitudinal static stability. Not change the static curve of stability into longitudinal.

Explanation: NO Question 286: The Arm is the ________ (i) distance of a load as measured from the aircraft ________ (ii). A B C D

(i) (i) (i) (i)

horizontal (ii) datum horizontal (ii) forward limit lateral (ii) datum vertical (ii) aft limit

Explanation: NO Question 287: Using the image attached or CAP 696 Figure 4.13 Given the data in the annex. A delayed flight resulted in 16 passengers (standard mass 84 kgs each) and their baggage (standard mass 14 kgs each ) not arriving for the flight. They are removed from the manifest. The baggage was in Hold 4 and the passengers were in Compartment Oc. Which statement is correct after these changes have been accounted for? 60-0363.gif A B C D

TOM CG 20% MAC ZFM index 76 TOM index 70 LM CG 18% MAC

Explanation: NO Question 288: Given: Maximum structural take-off mass: 8350 kg Maximum structural landing mass: 8350 kg Zero Fuel Mass: 6210 kg Taxi Fuel: 10 kg Contingency Fuel:90 kg. Alternate Fuel: 300 kg. Final Reserve Fuel: 400 kg. Trip Fuel: 780 kg The expected Landing Mass at destination will be: A 7000 kg B 7790 kg C 8350 kg

D 7780 kg Explanation: Zero Fuel Mass + expected landing fuel = expected landing mass Question 289: In calculations with respect to the position of the centre of gravity a reference is made to a datum. The datum is A a reference plane which is chosen by the aircraft manufacturer. Its position is given in the aircraft Flight or Loading Manual. B an arbitrary reference chosen by the pilot which can be located anywhere on the aircraft. C calculated from the data derived from the weighing procedure carried out on the aircraft after any major modification. D calculated from the loading manifest. Explanation: The datum is the point on the longitudinal axis from which the centre of gravity of all masses are referenced. Question 290: Given: Zero Fuel Mass: 6660 kg Trip Fuel: 990 kg Block Fuel: 1540 kg Taxi Fuel: 25 kg The actual Take-Off Mass is equal to: A B C D

8175 8110 8200 7210

kg kg kg kg

Explanation: TOM = ZFM + block fuel - taxi fuel 6660 + 1540 - 25 = 8175 kg Question 291: Given: Zero Fuel Mass: 4920 kg Trip Fuel: 880 kg Block Fuel: 1330 kg Taxi Fuel: 25 kg The actual Take-Off Mass is equal to: A B C D

6225 6360 6250 6810

kg kg kg kg

Explanation: TOM = ZFM + block fuel - taxi fuel 4920 + 1330 - 25 = 6225kg Question 292: Given: Maximum structural take-off mass: 8600 kg Maximum structural landing mass: 8600 kg Zero Fuel Mass: 6500 kg Taxi Fuel: 15 kg Contingency fuel: 110 kg. Alternate fuel: 600 kg. Final Reserve Fuel: 130 kg. Trip Fuel: 970 kg. The expected Landing Mass at destination will be

A B C D

7340 8600 8325 8310

kg kg kg kg

Explanation: ZFM + expected landing fuel = Expected landing mass 6500 + 840 = 7340 kg Question 293: Given: Maximum structural take-off mass: 7400 kg Maximum structural landing mass: 7400 kg Zero Fuel Mass: 5990 kg Taxi Fuel: 15 kg Contingency Fuel:110 kg. Alternate Fuel: 275 kg. Final Reserve Fuel: 250 kg. Trip Fuel: 760 kg The expected Landing Mass at destination will be A B C D

6625 7135 7400 7385

kg kg kg kg

Explanation: ZFM + expected landing fuel = The expected landing mass 5990 + 635 = 6625 kg Question 294: Given: Dry Operating Mass: 4920 kg Zero Fuel Mass: 5740 kg Trip Fuel: 670 kg Take-Off Fuel: 1050 kg The Traffic Load is: A B C D

820 kg 2160 kg 2480 kg 1340 kg

Explanation: TL = ZFM - DOM 5740 - 4920 = 820 kg Question 295: Given: Dry Operating Mass: 5320 kg Zero Fuel Mass: 6790 kg Trip Fuel: 770 kg Take-Off Fuel: 1310 kg The Traffic Load is: A B C D

1470 2940 1610 3080

kg kg kg kg

Explanation: TL = ZFM - DOM 6790 - 5320 = 1470 kg Question 296: Given: Dry Operating Mass: 5210 kg Zero Fuel Mass: 6230 kg Trip Fuel: 990 kg Take-Off Fuel: 1590 kg The Traffic Load is

A B C D

1020 2370 3390 2980

kg kg kg kg

Explanation: TL = ZFM - DOM 6230 - 5210 = 1020kg Question 297: For the purpose of completing the Mass and Balance documentation, the Dry Operating Mass is defined as: A The total mass of the aircraft ready for a excluding all usable fuel and traffic load. B The total mass of the aircraft ready for a excluding all usable fuel. C The total mass of the aircraft ready for a excluding crew and crew baggage. D The total mass of the aircraft ready for a excluding all traffic load.

specific type of operation specific type of operation specific type of operation specific type of operation

Explanation: NO Question 298: Given: Zero Fuel Mass: 4770 kg Trip Fuel: 1040 kg Block Fuel: 1960 kg Taxi Fuel: 20 kg The actual Take-Off Mass is equal to: A B C D

6710 6730 4970 5890

kg kg kg kg

Explanation: TOM = ZFM + block fuel - taxi fuel 4770 + 1960 - 20 = 6710 kg Question 299: Considering only structural limitations, on very short legs with minimum take-off fuel, the traffic load is normally limited by: A B C D

Maximum Zero Fuel Mass. Maximum Take-off Mass. Actual landing mass. Maximum Landing Mass.

Explanation: With minimum fuel the traffic load will be limited by the MZFM! Question 300: For a given configuration, the stall speed of an aeroplane will be highest when loaded:

A B C D

to to to to

the maximum the maximum a low total a low total

allowable allowable mass with mass with

mass with the most forward CG. mass with the most aft CG. the most forward CG. the most aft CG.

Explanation: The further forward the CG the higher the stall speed as picking the nose up with elevator increases the trim drag. Question 301: Allowed traffic load is the difference between: A B C D

allowed take off mass and operating mass. allowed take off mass and basic empty mass plus trip fuel. operating mass and basic empty mass. allowed take off mass and basic empty mass.

Explanation: NO Question 302: For the following see-saw to be in balance 31035.gif A B C D

Fb Fb Fb Fb

= = = =

A B A A

x x x +

Fa / B Fa / A B / Fa Fa / B

Explanation: From the graphic A x Fa = B x Fb With some carefrul transposition this gives: Fb =(A x Fa)/ B

Question 303: The Empty Mass of an aircraft is recorded in A the weighing schedule and is amended to take account of changes due to modifications of the aircraft. B the loading manifest. It differs from the Zero Fuel Mass by the value of the 'traffic load'. C the weighing schedule. If changes occur, due to modifications, the aircraft must be re-weighed always. D the loading manifest. It differs from Dry Operating Mass by the value of the 'useful load'. Explanation: The weighing schedule is part of the aircraft technical log. Question 304:

The longitudinal centre of gravity datum: A B C D

Can be located anywhere Must be located in line with the main rotor mast Must be located forward of the main rotor mast Must be located aft of the main rotor mast

Explanation: The datum is that point on the longitudinal axis, from which the centre of gravity of all masses is referenced.

Question 305: For the following see-saw to be in balance 31036.gif A B C D

Fc Fc Fc Fc

= = = =

Fa / 3 3Fa 3 / Fa Fa / 3A

Explanation: From the graphic you can see that A x Fa = 3A x Fc Carefully transposed to : Fc =(A x Fa)/ 3A You can see that with A on both top and bottom of the same side of the equation, they can be deleted. Fc = Fa / 3

Question 306: For the following boom to be in balance 31037.gif A B C D

B B B B

= = = =

Fa x A / Fb Fb + A / Fa - (Fa x A / Fb) Fb x A / Fa

Explanation: From the graphic B x Fb = A x Fa With careful transposition you get B =(Fa x A)/ Fb

Question 307: The maximum load specified in the loading chart is exceeded by 10 percent. What action must be taken?

A Reduce useful load B Increase rotor speed by 10 percent C None, since a 10 percent greater load is still within the specified safety margin D Take off carefully Explanation: The maximum load is exactly what it says - the maximum load - it cannot be exceeded so the useful load must be reduced.

Question 308: The Maximum Structural Take-Off Mass is A a limit which may not be exceeded for any take-off B a take-off limiting mass which is governed by the gradient of climb after reaching Vy C limited by the take-off distance available. It is tabulated in the flight manual D a take-off limiting mass which is affected by the aerodrome altitude and temperature Explanation: NO Question 309: The maximum quantity of fuel that can be loaded into an aircraft's tanks is given as 2200 l. If the fuel density (specific gravity) is given as 0.79 the mass of fuel which may be loaded is: A B C D

1738 1798 2785 2098

kg kg kg kg

Explanation: 2200 x 0.79 = 1738kg This works on the CRP5 as well. Set 2200 on the inner scale against the "km-m-ltr" index on the outer scale and read in the SPG KG scale against 0.79 a mass of 1740kgs Question 310: For the following boom to be in balance: 31038.gif A B C D

A A A A

= = = =

B B B B

x ( x +

Fb Fa Fa Fb

/ + / /

Fa Fb) Fb Fa

Explanation: From the graphic you can see A x Fa = B x Fb By careful transposition you get A =(B x Fb)/ Fa

Question 311: The datum is a reference from which all moment (balance) arms are measured. Its precise position is given in the control and loading manual and it is located A B C D

at at at at

a convenient point which may not physically be on the aircraft. or near the natural balance point of the empty aircraft. or near the focal point of the aircraft axis system. or near the forward limit of the centre of gravity.

Explanation: The Datum is the point on the longitudinal axis from which the centres of gravities of all masses are referenced. Question 312: In cruise flight, a centre of gravity moving aft will: A B C D

decrease longitudinal static stability have no effect on longitudinal static stability not change the manoeuvrability increase longitudinal static stability

Explanation: NO Question 313: A load placed aft of the datum: A Has a positive B Has a negative C Has a positive negative mass D Has a negative positive mass

arm and therefore generates a positive moment arm and therefore generates a negative moment and mass arm and therefore generates a positive moment but arm and therefore generates a negative moment but a

Explanation: - A load forward of the datum generates a negative moment - A load aft of the datum generates a positive moment Question 314: Longitudinal CG location is normally expressed: A B C D

as a with as a with

percentage respect to percentage respect to

of the MAC from its leading edge. the centre of pressure. of the MAC from its trailing edge. the neutral point.

Explanation: On a swept wing aircraft the CG is usually expressed as a % of the MAC from the leading edge.

Question 315: Overloading has the following effects on performance: A Increased take off and landing distance, reduced rate of climb and increased fuel consumption B Reduced take off and landing distance, increased VNE and reduced rate of climb C Increased take off and landing distance, increased rate of climb and increased fuel consumption D Reduced take off and landing distance, increased VNE and increased fuel consumption Explanation: Any increase in mass increases TOD and landing distance as well as reducing ceiling capablility and ROC and increases fuel consumption which therefore decreases range and endurance. Question 316: A load placed forward of the datum A B C D

Has Has Has Has

a a a a

negative positive positive negative

arm arm arm arm

and and and and

therefore therefore therefore therefore

generates generates generates generates

a a a a

negative positive positive negative

moment moment mass and moment mass and moment

Explanation: Typically, a load forward of the datum creates a negative moment and a load aft of the datum creates a positive moment. Question 317: Which of the following statements is correct? A The Maximum Landing Mass of an aeroplane is restricted by structural limitations, performance limitations and the strength of the runway. B The Basic Empty Mass is equal to the mass of the aeroplane excluding traffic load and useable fuel but including the crew. C The Maximum Zero Fuel Mass ensures that the centre of gravity remains within limits after the uplift of fuel. D The Maximum Take-off Mass is equal to the maximum mass when leaving the ramp. Explanation: Maximum landing mass is a structural limit. The strength (or weakness) of the runway may further limit the landing mass based on the PCN/ CAN classification. Question 318: What mass has to be entered in the loading chart for aviation fuel F 34 if 170 l may be refuelled? (Fuel density = 0.78 kg/l) A 133 kg B 218 kg

C 170 kg D 133 daN Explanation: multiply 170 litres by 0.78 sg = 132.6 Question 319: The Traffic Load is defined as: A The total mass revenue load B The total mass fuel C The total mass D The total mass

of passengers, baggage and cargo, including any non of flight crew, passengers, baggage, cargo and usable of crew and passengers excluding any baggage or cargo of passengers, baggage, cargo and usable fuel

Explanation: The Traffic Load is the total mass of passengers, cargo and baggage (including any non-revenue load). Question 320: At maximum certificated take-off mass an aeroplane departs from an airfield which is not limiting for either take-off or landing masses. During initial climb the number one engine suffers a contained disintegration. An emergency is declared and the aeroplane returns to departure airfield for an immediate landing. The most likely result of this action will be A a high threshold speed and possible undercarriage or other structural failure. B a landing short resultant from the increased angle of approach due to the very high aeroplane mass. C a landing further along the runway than normal. D a high threshold speed and a shorter stop distance. Explanation: If you work this question through logically youâ&#x20AC;&#x;ll realise that, as the actual landing mass will be much higher than normal this means higher stalling speed. This will require a higher approach speed to compensate and this will then increase the weight and is so likely to increase chances of damage. Question 321: In relation to an aeroplane, the term '' Basic Empty Mass'' includes the mass of the aeroplane structure complete with its powerplants, systems, furnishings and other items of equipment considered to be an integral part of the particular aeroplane configuration. Its value is A found in the latest version of the weighing schedule as corrected to allow for modifications. B found in the flight manual and is inclusive of unusable fuel plus fluids contained in closed systems. C printed in the loading manual and includes unusable fuel.

D inclusive of an allowance for crew, crew baggage and other operating items. It is entered in the loading manifest. Explanation: NO Question 322: 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 A B C D

13090 12840 18040 16470

kg. kg kg kg

Explanation: It's an appallingly constructed question as it implies that you use the MRJT1 limitations in the CAP to establish the RLTOM/RLLM. If you do this then you get your answer. The only way to get the given answer is to ignore the CAP and use the given PLTOM/PLLM. Try it again. Question 323: 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 25% 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. Fuel tanks contain 140 litres of petrol with a density of 0.714. The rear seats are not occupied.Taxi fuel is negligable. The position of the centre of gravity at take-off (as % MAC) is : A B C D

22 34 17 29

% % % %

Explanation: Difficult to explain without a drawing but you can draw one out yourself. Work out the change of CG from BEM to TOM. BEM Mass 1200kg x arm 3.0m = 3600kgm Pilot and pax in front seats: mass 160kg x arm 2.5m = moment 400kgm Fuel 100kg x arm 3.0m = moment 300kgm Add masses and moments to find: TOM = 1460kg TOM moment = 4300kgm TOM CG = moment/mass = 4300/1460=2.95m The BEM CG was at 25% MAC with an arm from the datum to 25% MAC of 3m. The MAC is 2m which means the CG is 0.5m from LEMAC Now the TOM CG is 2.95m which means it is 0.45m from LEMAC

As a % MAC then : 0.45/2 x 100 = 22.5% Question 324: The maximum floor loading for a cargo compartment in an aircraft is given as 750 kg per square metre. A package with a mass of 600 kg. is to be loaded. Assuming the pallet base is entirely in contact with the floor, which of the following is the minimum size pallet that can be used ? A B C D

40 30 40 30

cm cm cm cm

by by by by

200 200 300 300

cm cm cm cm

Explanation: NO Question 325: The operator of an aircraft equipped with 50 seats uses standard masses for passengers and baggage. During the preparation of a scheduled flight a group of passengers present themselves at the check-in desk, it is apparent that even the lightest of these exceeds the value of the declared standard mass. A the operator should use the individual masses of alter the standard masss B the operator is obliged to use the actual masses C the operator may use the standard masses for the calculation without correction D the operator may use the standard masses for the correct these for the load calculation

the passengers or of each passenger load and balance balance but must

Explanation: In order to use a different standard mass it must first be approved by the Authority This is a JAROPS / EUOPS requirement. Question 326: Given: MTOM 37200kg DOM Males, 32 females and 5 standard passenger mass flight is not a holiday that may be loaded is A B C D

21600kg Fuel at T/O 8500kg Passenger load:33 children Baggage 880kg The company uses the systems (see annex) allowed by regulations. The charter. In these conditions, the maximum cargo

585 kgs 965 kgs 340 kgs 1105 kgs

Explanation: Solution: The standard all adult mass value is 84 kg for a non holiday charter. See CAP 696. Therefore passenger weights are: 65 x 84 = 5460 kg (adults)

5 x 35 = 175 kg (children) Total 5635 kg MTOM 37200 DOM 21600 Fuel 8500 Pax 5635 Baggage 880 = Cargo 585 kg Question 327: Given: - The take-off mass of an aircraft is 8470 kg. - Total fuel on board is 1600 kg including 450 kg reserve fuel and 29 kg of unusable fuel. - The traffic load is 770 kg. What is the Zero Fuel Mass? A B C D

6899 6420 6870 6129

kg kg kg kg

Explanation: ZFM = TOM - Useable fuel 8470 - 1571 = 6899 kgs Question 328: Using the reference provided, without the crew, the weight and the CG position of the aircraft are 7 000 kg and 4.70m. The mass of the pilot is 90 kg, the mass of the co-pilot is 75 kg and the mass of the flight engineer is 90 kg. With this crew on board, the CG position of the aircraft will be: A B C D

4.615m 0.217m 4.455m 4.783m

Explanation: NO Question 329: Which is true of the aircraft basic empty mass? A B C D

It It It It

is is is is

a component of dry operating mass. dry operating mass minus fuel load. dry operating mass minus traffic load. the actual take-off mass, less traffic load.

Explanation: NO

Question 330: The maximum quantity of fuel that can be loaded into an aircraft's tanks is given as 3800 US Gallons. If the fuel density (specific gravity) is given as 0.79 the mass of fuel which may be loaded is A B C D

11364 14383 13647 18206

kg. kg. kg. kg.

Explanation: Conversion Factor: 1 x USG = 3.785 litres 3,800 x 3.785 = 14,383 litres 14,383 x 0.79 = 11,363 kg On CRP-5: Align 3,800 on Inner Scale with US.Gal on Outer Scale. On Specific Gravity Sp.G KGS Index locate 0.79 on Outer Scale Question 331: Given the information at take-off shown at the reference. Given that the flight time is 2 hours and the estimated fuel flow will be 1050 litres per hour and the average oil consumption will be 2.25 litres per hour. The specific density of fuel is 0.79. The specific density of the oil is 0.96 .The "Freight 2" will be dropped during flight within the scope of a rescue action. Calculate the CG position at landing. 3110035.gif A B C D

24 25 22 27

cm cm cm cm

aft aft aft aft

of of of of

datum. datum. datum. datum.

Explanation: Calculate the TOM = 19339 kg . Subtract : Fuel used 2100 litres x 0.79=1659 kg. Oil used 4.5 litres = 0.96= 4.32 kg Freight dropped 410 kg Therefore Landing mass = 17265.68 kg Calculate the TO moment = 392350 kgcm Calculate the moment change for fuel, oil and freight: Fuel 1659 x -8 = +13272 kgcm Oil 4.32 x +40 = - 172.8 kgcm Freight 410 x -40 = + 16400 kgcm New total moment = 421849.2 kgcm Landing CG = Moment Mass = 421849.2 = +24.43 cm (aft) 17265.68 Question 332: The Maximum Zero Fuel Mass is a mass limitation for the: A strength of the wing root

B total load of the fuel imposed upon the wing C strength of the fuselage D allowable load exerted upon the wing considering a margin for fuel tanking Explanation: NO Question 333: An aeroplane with a two wheel nose gear and four main wheels rests on the ground with a single nose wheel load of 725 kg and a single main wheel load of 6000 kg. The distance between the nose wheels and the main wheels is 10 meters. How far is the centre of gravity in front of the main wheels? A B C D

57 cm. 40 cm. 400 cm. 25 cm.

Explanation: Moment effect of nose gear = 10 x 1450 = 14500 kg/m CG = Moment / Mass CG = 14500 / 25450 CG = 0.57m (or 57cm)

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