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Without the crew the mass and longitudinal cg position of the aircraft are 6000 ? [ Session helipad ]

Question 73-1 : 6270 kg and 4 594 m 6270 kg and 5 012 m 6270 kg and 4 61 m 6270 kg and 4 796 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 173 6270 kg and 4.594 m.6270 kg and 4.594 m.

Using the load and trim sheet for the mrjt1 aircraft which of the following is ?

Question 73-2 : 40 0 35 5 41 5 33 0

Img169 exemple 177 40.040.0

Using the data given in the load and trim sheet determine which of the ?

Question 73-3 : 46130 kg and 17 8% 46130 kg and 20 8% 51300 kg and 20 8% 41300 kg and 17 8%

Img161

Using the data given in the load and trim sheet determine from the following ?

Question 73-4 : 17 5% 20 3% 22 6% 20 1%

Img162 exemple 185 17.5%.17.5%.

Using the data given at the appendix to this question if the fuel index ?

Question 73-5 : 49130 kg and 19% 52900 kg and 21 6% 49130 kg and 21 8% 52900kg 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

Using the data given at the appendix determine which of the following correctly ?

Question 73-6 : 48600 kg and 57 0 51300 kg and 57 0 46300 kg and 20 5 35100 kg and 20 5

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 193 48600 kg and 57.0.48600 kg and 57.0.

For this question use annex ecqb 031 mb 02 v2015 03 .from the data given at the ?

Question 73-7 : 18% 19% 15% 14%

Img164. exemple 197 18%.18%.

From the data contained in the attached appendix the maximum allowable take off ?

Question 73-8 : 61600 kg and 12150 kg 68038 kg and 18588 kg 66770 kg and 17320 kg 60425 kg and 10975 kg

Img165 exemple 201 61600 kg and 12150 kg.61600 kg and 12150 kg.

An aeroplane is carrying a traffic load of 10320 kg .complete the necessary ?

Question 73-9 : 1830 kg 655 kg 7000 kg 8268 kg

Img166 exemple 205 1830 kg.1830 kg.

When has the centre of gravity to be computed ?

Question 73-10 : Prior to every flight after every 400 hours inspection at least every 4 years during every yearly inspection

exemple 209 Prior to every flight.Prior to every flight.

What mass has to be entered in the loading chart for aviation fuel f 34 if 170 ?

Question 73-11 : 133 kg 133 dan 170 kg 218 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 213 133 kg.133 kg.

An aeroplane with a two wheel nose gear and four main wheels rests on the ?

Question 73-12 : 40 cm 25 cm 4 m 41 6 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 exemple 217 40 cm.40 cm.

Using the reference provided without the crew the weight and the cg position of ?

Question 73-13 : 4 615 m 0 217 m 4 783 m 4 455 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 221 4.615 m.4.615 m.

Given that the flight time is 2 hours and the estimated fuel flow will be 1050 ?

Question 73-14 : 24 cm aft of datum 25 cm aft of datum 22 cm aft of datum 27 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 exemple 225 24 cm aft of datum.24 cm aft of datum.

An aeroplane with a two wheel nose gear and four main wheels rests on the ?

Question 73-15 : 57 cm 93 cm 108 cm 176 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 229 57 cm.57 cm.

An aeroplane has a planned take off mass of 200 000 kg .its cg is located at 15 ?

Question 73-16 : 4600 kg 5600 kg 3600 kg it is impossible to move the cg at the requested value

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 233 4600 kg.4600 kg.

The index method in mass and balance calculations is used for ?

Question 73-17 : Reducing the magnitude of the moment reducing the magnitude of the useful load increasing the magnitude of the useful load increasing 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 237 Reducing the magnitude of the moment.Reducing the magnitude of the moment.

What are the advantages of using the index method to determine moments .it ?

Question 73-18 : Reduces the magnitude of the moments making it less time consuming to compute is digitalised and can be automatically uploaded to the aircraft fmc to determine stabiliser trim setting directly produces the position of the cg and the stabiliser trim setting enabled the pilots to calculate the position of the cg by mental arithmetic and thereby expedites mass and balance calculations


What is the principle of the index method ?

Question 73-19 : To divide high magnitude moments by a constant and make result more easier to use to refer every item of the aircraft to a position called the ba balance arm to compute the cg position and stabiliser trim setting without completing a load sheet to round all values for easier and much faster calculation

Ecqb03 july 2016 exemple 245 To divide high magnitude moments by a constant and make result more easier to use.To divide high magnitude moments by a constant and make result more easier to use.

Define the 'under load' ?

Question 73-20 : Allowed tom dom useful load actual tom dom useful load allowed tl tom actual tl tom


For this question use annex ecqb 031 046 v2015 01 .the aircraft is loaded as ?

Question 73-21 : 300000 kg mm 30000 kg mm 34800 kg mm 348000 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 253 300000 kg.mm300000 kg.mm

The planned take off mass of a turbojet aeroplane is 180 000 kg with its centre ?

Question 73-22 : 22 1 % 30 0 % 21 1 % 24 2 %

Img /com en/com031 291 jpg.. exemple 257 22.1 %.22.1 %.

A turbojet aeroplane has a planned take off mass of 190 000 kg following cargo ?

Question 73-23 : 3000 kg from cargo 4 to cargo 1 2000 kg from cargo 4 to cargo 1 1000 kg from cargo 4 to cargo 1 it is not possible to obtain the required centre of gravity

Img /com en/com031 292 jpg exemple 261 3000 kg from cargo 4 to cargo 1.3000 kg from cargo 4 to cargo 1.

The planned take off mass of an aeroplane is 190 000 kg with its centre of ?

Question 73-24 : 32 2 % 31 % 27 % 25 %

exemple 265 32,2 %.32,2 %.

The planned take off mass of an aeroplane is 180 000 kg with its centre of ?

Question 73-25 : 25% 34% 38% 21%

exemple 269 25%.25%.

What is dry operating index doi ?

Question 73-26 : The index for the position of the centre of gravity at dry operating mass the distance from datum to the centre of gravity of a mass the point through which the force of gravity is said to act on a mass the product of the mass and balance arm

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

A 3 m long plank is on a pivot halfway along its length a 1 kg mass is ?

Question 73-27 : 0 5m to the left 0 5m to the right 1m to the left 1m to the right

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 277 0.5m to the left.0.5m to the left.

For the following see saw to be in balance with a mass of 35 kg suspended on ?

Question 73-28 : 22 kg 136 kg 57 kg 14 kg

14m x 35 kg + 5m x = 8m x 75 kg. = 8 x 75 14 x 35 / 5. = 600 490 / 5. = 22 kg exemple 281 22 kg.22 kg.

Refer to figure 031 13 .for a medium range twin jet aircraft with a cg located ?

Question 73-29 : 3 0 2 5 3 25 2 75

exemple 285 3.03.0

Consider a conventional aircraft with three wheels the nose jack is located 161 ?

Question 73-30 : Bem 40900 lb cg 740 5 in bem 21600 lb cg 722 3 in bem 40900 lb cg 722 3 in bem 21600 lb cg 709 6 in

exemple 289 Bem 40900 lb, cg 740.5 in.Bem 40900 lb, cg 740.5 in.

Refer to figure 031 57 .calculate the cg position and moment for the basic ?

Question 73-31 : 115 7 in aft of datum and the moment is 365074 in lb 111 4 in aft of datum and the moment is 489666 in lb 95 3 in aft of datum and the moment is 380373 in lb 113 6 in aft of datum and the moment is 430998 in lb

exemple 293 115.7 in aft of datum and the moment is 365074 in.lb115.7 in aft of datum and the moment is 365074 in.lb

The mass and the cg of an aircraft must be established by actual weighing by the ?

Question 73-32 : Operator prior to initial entry into service owner operator before the first flight of the day pilot on entry of aircraft into service engineers before commencing 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 297 Operator prior to initial entry into service.Operator prior to initial entry into service.

Refer to figure 031 59 .the see saw is in a state of equilibrium calculate the ?

Question 73-33 : 30 49 m 37 1 m 29 49 m 22 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 301 30,49 m.30,49 m.

Given .length of the mac 114 inches.forward gc limit 12% mac.aft cg limit 38% ?

Question 73-34 : 2 28 inches 8 77 inches 25 08 inches 10 53 inches

exemple 305 2.28 inches.2.28 inches.

Consider a conventional aircraft with three wheels the nose jack is located 161 ?

Question 73-35 : Bem 42054 lb cg 680 3 in bem 24271 lb cg 610 9 in bem 42054 lb cg 570 3 in bem 24271 lb cg 936 in

exemple 309 Bem 42054 lb, cg 680.3 in.Bem 42054 lb, cg 680.3 in.

Refer to figure 031 61 .originally there has been 225 l of fuel in the tank of ?

Question 73-36 : The cg will move towards the nose of the aircraft the cg will move towards the tail of the aircraft the cg will move towards the bottom of the aircraft the cg will remain the same

exemple 313 The cg will move towards the nose of the aircraft.The cg will move towards the nose of the aircraft.

Refer to figure 031 05 .given .force fa 100 n.distance a 6 m.distance b 3 ?

Question 73-37 : 200 n 300 n 50 n 100 n

exemple 317 200 n.200 n.

Refer to figure 031 34 .given.bem 1200 kg.bem cg 3 00.pilot and front pax ?

Question 73-38 : 2 96 m 2 99 m 3 03 m 3 00 m

exemple 321 2.96 m.2.96 m.

Refer to figure 031 04 .given .force fa 100 n.distance a 6 m.distance b 3 ?

Question 73-39 : 50 n 200 n 300 n 50 n

exemple 325 50 n.50 n.

Select the correct statement for the cg safe range ?

Question 73-40 : The safe range falls between the front and rear cg limits and includes both limits the safe range falls between the front and rear cg limits but does not include them the safe range falls between the front and rear cg limits but only includes the aft limit the safe range falls between the front and rear cg limits but only includes the forward limit

exemple 329 The safe range falls between the front and rear cg limits and includes both limits.The safe range falls between the front and rear cg limits and includes both limits.


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