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Question 89-1 : When the centre of gravity is at the forward limit, an aeroplane will be ? [ Validation Marking ]

Extremely stable and will require excessive elevator control to change pitch.

When the cg is located on the forward limit, the pitch down moment is at its allowed maximum the distance between cg and cp is great..the elevator will have to develop a high downward force to counteract this pitch down moment, and it will seem heavier to the pilot..of course, the aircraft will be extremely stable, which is the good news about a forward cg.

Question 89-2 : 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 ?

80000 nm.

Arm = moment / force..thus..moment = arm x force..16 m x 5000 n = 80000 nm. exemple 193 80000 nm.

Question 89-3 : Without the crew, the mass and longitudinal cg position of the aircraft are 6000 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.. 220 ?

6270 kg and 4.594 m.

Moment = 6000 x 4.7 = 28200 kg.m.pilot 90 kg > moment is 184 kg.m.co pilot 100 kg > moment 204 kg.m.flight engineer 80 kg > moment column b is 215 kg.m..total moment = 28200 + 184 + 204 + 215 = 28803 kg.m..total mass = 6000 + 90 + 100 + 80 = 6270 kg...cg = moment/mass.cg = 28803/6270 = 4.594 m. exemple 197 6270 kg and 4.594 m.

Question 89-4 : 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.. 221 ?

40.0

Img169 exemple 201 40.0

Question 89-5 : Using the data given in the load and 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.. 222 ?

46130 kg and 17.8%.

Img161 exemple 205 46130 kg and 17.8%.

Question 89-6 : Using the data given in the load and 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.. 223 ?

17.5%.

Img162 exemple 209 17.5%.

Question 89-7 : 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 ?

49130 kg and 19%.

Img163.first complete the load calculation, then complete the graph. do not forget to apply 'fuel index correction of 4.3' 3000 kg of fuel remain in tank at landing at the end. exemple 213 49130 kg and 19%.

Question 89-8 : 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.. 225 ?

48600 kg and 57.0.

Zero fuel mass zfm = dry operating mass + traffic load = 37370 kg + 11230 kg = 48600 kg...we have only one answer with this value of zero fuel mass zfm , no need to go further in the graph. exemple 217 48600 kg and 57.0.

Question 89-9 : For this question use annex ecqb 031 mb 02 v2015 03..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.. 226 ?

18%.

Img164.. exemple 221 18%.

Question 89-10 : From the data contained in the attached appendix, the maximum allowable take off mass and traffic load is respectively.. 227 ?

61600 kg and 12150 kg.

Img165 exemple 225 61600 kg and 12150 kg.

Question 89-11 : An aeroplane is carrying a traffic load of 10320 kg..complete the necessary sections and determine which of the answers given below represents the maximum increase in the traffic load.. 228 ?

1830 kg.

Img166 exemple 229 1830 kg.

Question 89-12 : When has the centre of gravity to be computed ?

Prior to every flight.

exemple 233 Prior to every flight.

Question 89-13 : 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 ?

133 kg.

170 litres x 0.78 = 132.6 kg...for information, aviation fuel f 34 is a military kerosene type turbine fuel with fuel system icing inhibitor. exemple 237 133 kg.

Question 89-14 : 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 ?

40 cm.

Total airplane weight = 500 x 2 + 6000 x 4 = 25 000 kg...the center of gravity at this weight is located at 1000 kg x 10 m / 25 000 kg = 0.4 m in front of the main wheels.

Question 89-15 : 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.. ?

4.615 m.

Take the moments individually..pilot 184 kg.m.copilot 153 kg.m.flight engineer column b 242 kg.m..basic empty mass moment 7000 x 4.7 = 32900 kg.m..total moment 33479 kg.m..total mass 7000 + 90 + 75 + 90 = 7255 kg...cg = moment/mass = 33479/7255 = 4.615 m. exemple 245 4.615 m.

Question 89-16 : Given that the flight time is 2 hours and the estimated fuel flow will be 1050 l/h and the average oil consumption will be 2.25 l/h..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 ?

24 cm aft of datum.

Take off mass is 19339 kg..minus fuel = 2100 x 0.79 = 1659 kg..minus oil = 4.5 liter x 0.96 = 4.32 kg..minus 'freight 2' = 410 kg.our landing mass will be = 17265.68 kg...take off moment is 392350 kg.cm. fuel moment = 1659 x 8 cm = +13272 kg.cm.. oil moment = 4.32 x 40 cm = 172.8 kg.cm.. 'freight 2' moment = 410 x 40 cm = +16400 kg.cm..total moment at landing = 421849.2 kg.cm..the cg position at landing = 421769.2 kg.cm / 17265.68 kg = 24.42 cm

Question 89-17 : 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 ?

57 cm.

Total mass 725 kg x 2 + 6000 kg x 4 = 25450 kg...the distance between the nose wheels and the main wheels is 10 meters..the centre of gravity, in front of the main wheels, is at..centre of gravity = total moment / total mass...centre of gravity = 1450 kg x 10 m / 25450 kg = 0,5697 m. exemple 253 57 cm.

Question 89-18 : An aeroplane has a planned take off mass of 200 000 kg..its cg is located at 15.38 m of the reference point representing a cg location at 30% mac mean aerodynamic cord..moment arm of the forward cargo is 15 m.moment arm of the aft cargo is 25 m.for performance purposes, the value of the centre of ?

4600 kg.

Length of the mean aerodynamic chord = 14 + 4.6 14 = 4.6 m..change in cg = 30% vers 35% = 5%..5% of the mac lenght = 5% de 4.6 = 0.23 m...change in mass / total mass = change in cg / total distance moved..change in mass = change in cg x total mass / total distance moved..change in mass = 0.23 m x 200000 kg / 10 m..change in mass = 4600 kg. exemple 257 4600 kg.

Question 89-19 : The index method in mass and balance calculations is used for... ?

Reducing the magnitude of the moment.

In mass and balance calculations the 'index' is a figure without unit of measurement which represents a moment..the value of the index is the moment divided by a constant, usually 1000. it is used to simplify the calculations by decreasing the values. exemple 261 Reducing the magnitude of the moment.

Question 89-20 : What are the advantages of using the index method to determine moments.it... ?

Reduces the magnitude of the moments, making it less time consuming to compute.

exemple 265 Reduces the magnitude of the moments, making it less time consuming to compute.

Question 89-21 : What is the principle of the index method ?

To divide high magnitude moments by a constant and make result more easier to use.

Ecqb03 july 2016 exemple 269 To divide high magnitude moments by a constant and make result more easier to use.

Question 89-22 : Define the 'under load'. ?

Allowed tom dom useful load

exemple 273 Allowed tom - dom - useful load

Question 89-23 : For this question use annex ecqb 031 046 v2015 01.the aircraft is loaded as shown in the table..calculate the new total moment if the mass of the crew is increased.. 240 ?

300000 kg.mm

Ecqb03 july 2016...sorry, we haven't yet recovered the annex. we get only the question and the correct answer this is nice, isn't it exemple 277 300000 kg.mm

Question 89-24 : 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 ?

22.1 %.

Img /com en/com031 291.jpg.... exemple 281 22.1 %.

Question 89-25 : A turbojet aeroplane has a planned take off mass of 190 000 kg. following cargo loading, the crew is informed that the centre of gravity at take off is located at 38 % mac mean aerodynamic cord which is beyond limits. the captain decides then to redistribute part of the cargo load between cargo 1 ?

3000 kg from cargo 4 to cargo 1.

Img /com en/com031 292.jpg exemple 285 3000 kg from cargo 4 to cargo 1.

Question 89-26 : The planned take off mass of an 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 an additional load of 4 000 kg must be loaded in cargo 4. after loading this cargo, the new ?

32,2 %.

exemple 289 32,2 %.

Question 89-27 : 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 crew that an additional load of 4 000 kg must be loaded in cargo 1. after loading this cargo, the new centre of ?

25%.

exemple 293 25%.

Question 89-28 : What is dry operating index doi ?

The index for the position of the centre of gravity at dry operating mass.

Dry operating index doi is the index for the position of the centre of gravity at dry operating mass...dry operation mass dom is the total mass of the aeroplane ready for a specific type of operation excluding usable fuel and traffic load. the mass includes items such as.i crew and crew baggage..ii catering and removable passenger service equipment..iii potable water and lavatory chemicals..iv food and beverages...balance arm ba is the distance from datum to the centre of gravity of a mass..centre of gravity cg is that point through which the force of gravity is said to act on a mass..moment is the product of the mass and balance arm. exemple 297 The index for the position of the centre of gravity at dry operating mass.

Question 89-29 : A 3 m long plank is on a pivot halfway along its length. a 1 kg mass is suspended on the left end and a 2 kg mass from other end..how far and in which direction should the plank be moved in order for the plank to be in balance ?

0.5m to the left.

Moment = mass x balance arm..1 kg x 1.5 m = 1.5 kgm.2 kg x 1.5 m = 3 kgm..the 1 kg mass must have the same moment than the 2 kg mass for the plank to be in balance..1 kg x 2 m = 2 kg x 1 m.2 kg.m = 2 kg.m. exemple 301 0.5m to the left.

Question 89-30 : For the following see saw to be in balance, with a mass of 35 kg suspended on the left end 14m left of pivot and 75 kg suspended on the right end, the mass required at position 5m left of the pivot must be.. 247 ?

22 kg.

14m x 35 kg + 5m x = 8m x 75 kg.. = 8 x 75 14 x 35 / 5.. = 600 490 / 5.. = 22 kg. exemple 305 22 kg.

Question 89-31 : Refer to figure 031 13..for a medium range twin jet aircraft with a cg located at 18% mac at 62000 kg gross mass, determine the stabilizer trim units required for a take off flap setting of 15° ?

3.0

exemple 309 3.0

Question 89-32 : Consider a conventional aircraft with three wheels. the nose jack is located 161 inches aft of the datum. the main wheel jacks are located 775 inches aft of the datum. after weighing the following results are reported..nose jack 2300 lb.each main wheel jack 19300 lb..what is the basic empty mass of ?

Bem 40900 lb, cg 740.5 in.

exemple 313 Bem 40900 lb, cg 740.5 in.

Question 89-33 : Refer to figure 031 57..calculate the cg position and moment for the basic empty mass by using the data given from the attached table ?

115.7 in aft of datum and the moment is 365074 in.lb

exemple 317 115.7 in aft of datum and the moment is 365074 in.lb

Question 89-34 : The mass and the cg of an aircraft must be established, by actual weighing, by the ?

Operator prior to initial entry into service.

Easa air ops.regulation eu no 965/2012.cat.pol.mab.100 mass and balance, loading.. a during any phase of operation, the loading, mass and centre of gravity cg of the aircraft shall comply with the limitations specified in the afm, or the operations manual if more restrictive..... b the operator shall establish the mass and the cg of any aircraft by actual weighing prior to initial entry into service and thereafter at intervals of four years if individual aircraft masses are used, or nine years if fleet masses are used. the accumulated effects of modifications and repairs on the mass and balance shall be accounted for and properly documented. aircraft shall be reweighed if the effect of modifications on the mass and balance is not accurately known..... c the weighing shall be accomplished by the manufacturer of the aircraft or by an approved maintenance organisation..... d the operator shall determine the mass of all operating items and crew members included in the aircraft dry operating mass by weighing or by using standard masses. the influence of their position on the aircraft's cg shall be determined..... e the operator shall establish the mass of the traffic load, including any ballast, by actual weighing or by determining the mass of the traffic load in accordance with standard passenger and baggage masses..... f in addition to standard masses for passengers and checked baggage, the operator can use standard masses for other load items, if it demonstrates to the competent authority that these items have the same mass or that their masses are within specified tolerances..... g the operator shall determine the mass of the fuel load by using the actual density or, if not known, the density calculated in accordance with a method specified in the operations manual..... h the operator shall ensure that the loading of.. 1 its aircraft is performed under the supervision of qualified personnel and.. 2 traffic load is consistent with the data used for the calculation of the aircraft mass and balance..... i the operator shall comply with additional structural limits such as the floor strength limitations, the maximum load per running metre, the maximum mass per cargo compartment and the maximum seating limit. for helicopters, in addition, the operator shall take account of in flight changes in loading..... j the operator shall specify, in the operations manual, the principles and methods involved in the loading and in the mass and balance system that meet the requirements contained in a to i. this system shall cover all types of intended operations. exemple 321 Operator prior to initial entry into service.

Question 89-35 : Refer to figure 031 59..the see saw is in a state of equilibrium. calculate the distance between a and c ?

30,49 m.

Admin.left side must be equal to right side = moment on the left side must be equal to moment on the right side. 520kg x 8m = 48kg x 7m + 170 kg x x..x is the distance between the pivot to mass c..x = 520kg x 8m 48kg x 7m / 170.x = 4160 336 / 170 = 22,49 m...distance between a and c is 8 m + 22,49 m = 30,49 m. exemple 325 30,49 m.

Question 89-36 : Given.length of the mac 114 inches.forward gc limit 12% mac.aft cg limit 38% mac.calculate the distance of cg at 10m mac with reference to the forward limit ?

2.28 inches.

exemple 329 2.28 inches.

Question 89-37 : Consider a conventional aircraft with three wheels. the nose jack is located 161 inches aft of the datum. the main wheel jacks are located 775 inches aft of the datum. after weighing the following results are reported..nose jack 6488 lb.each main wheel jack 17783 lb..what is the basic empty mass of ?

Bem 42054 lb, cg 680.3 in.

exemple 333 Bem 42054 lb, cg 680.3 in.

Question 89-38 : Refer to figure 031 61..originally there has been 225 l of fuel in the tank of the aircraft but the pilot decides do add 200 kg of additional fuel. what will happen to the centre of gravity ?

The cg will move towards the nose of the aircraft.

exemple 337 The cg will move towards the nose of the aircraft.

Question 89-39 : Refer to figure 031 05..given.force fa 100 n.distance a 6 m.distance b 3 m.calculate force fb to obtain equilibrium ?

200 n.

exemple 341 200 n.

Question 89-40 : Refer to figure 031 34..given.bem 1200 kg.bem cg 3.00.pilot and front pax 200 kg.cargo on pax floor 50 kg.fuel 250 kg..find the loaded centre of gravity cg by using the attached graph ?

2.96 m.

exemple 345 2.96 m.


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