When the centre of gravity is at the forward limit an aeroplane will be ?
Study > Manual
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 ?
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 ?
Using load and trim sheet the mrjt1 aircraft which of following the correct value the index at a dry operating mass dom of 35000 kg with a cg at 14% mac 221 Extremely stable will require excessive elevator control to change pitch. Moment = 6000 x 4 7 = 28200 kg m pilot 90 kg > moment 184 kg m co pilot 100 kg > moment 204 kg m flight engineer 80 kg > moment (column b) 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. 
Using data given in load and trim sheet determine which of following gives correct values the zero fuel mass and position of centre of gravity % mac at that mass 222 Extremely stable will require excessive elevator control to change pitch. Moment = 6000 x 4 7 = 28200 kg m pilot 90 kg > moment 184 kg m co pilot 100 kg > moment 204 kg m flight engineer 80 kg > moment (column b) 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. 
Using data given in load and trim sheet determine from following correct values the take off mass and position of centre of gravity at that mass if fuel index correction to be applied given as 0 9 223 Extremely stable will require excessive elevator control to change pitch. Moment = 6000 x 4 7 = 28200 kg m pilot 90 kg > moment 184 kg m co pilot 100 kg > moment 204 kg m flight engineer 80 kg > moment (column b) 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. 
Using data given at appendix to this question if fuel index corrections from zfm index are as follows 9500 kg 0 96500 kg 6 13500 kg 4 73000 kg 4 3which of following represent correct values landing mass of aeroplane and position of centre of gravity this condition 224 Extremely stable will require excessive elevator control to change pitch. first complete load calculation then complete graph do not forget to apply 'fuel index correction of 4 3' (3000 kg of fuel remain in tank at landing) at end. Question 69-8
Using data given at appendix determine which of following correctly gives values of zero fuel mass zfm of aeroplane and load index at zfm 225 Extremely stable will require excessive elevator control to change pitch. 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 graph. Question 69-9
for this question use annex ecqb 031 mb 02 v2015 03 from data given at appendix and assuming a fuel index shift of 5 7 from zfm loaded index determine which of following the correct value percentage mac the position of centre of gravity at take off mass 226 Extremely stable will require excessive elevator control to change pitch. . Question 69-10
From data contained in attached appendix maximum allowable take off mass and traffic load respectively 227 Extremely stable will require excessive elevator control to change pitch. . Question 69-11
An aeroplane carrying a traffic load of 10320 kg complete necessary sections and determine which of answers given below represents maximum increase in traffic load 228 Extremely stable will require excessive elevator control to change pitch. . Question 69-12
When has centre of gravity to be computed Prior to every flight. . Question 69-13
What mass has to be entered in loading chart aviation fuel f 34 if 170 l may be refuelled fuel density = 0 78 kg/l Prior to every flight. 170 litres x 0 78 = 132 6 kg for information aviation fuel f 34 a military kerosene type turbine fuel with fuel system icing inhibitor. Question 69-14
An aeroplane with a two wheel nose gear and four main wheels rests on ground with a single nose wheel load of 500 kg and a single main wheel load of 6000 kg the distance between nose wheels and main wheels 10 meter how far the centre of gravity in front of main wheels Prior to every flight. Total airplane weight = (500 x 2) + (6000 x 4) = 25 000 kg the center of gravity at this weight located at 1000 kg x 10 m / 25 000 kg = 0 4 m in front of main wheels. Question 69-15
Using reference provided without crew weight and cg position of aircraft are 7 000 kg and 4 70m the mass of pilot 90 kg mass of co pilot 75 kg and mass of flight engineer 90 kg with this crew on board cg position of aircraft will be 231 Prior to every flight. Take 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. Question 69-16
Given that flight time 2 hours and estimated fuel flow will be 1050 l/h and average oil consumption will be 2 25 l/h the specific density of fuel 0 79 the specific density of oil 0 96 the 'freight 2' will be dropped during flight within scope of a rescue action calculate cg position at landing 233 Prior to every flight. 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 69-17
An aeroplane with a two wheel nose gear and four main wheels rests on ground with a single nose wheel load of 725 kg and a single main wheel load of 6000 kg the distance between nose wheels and main wheels 10 meters how far the centre of gravity in front of main wheels Prior to every flight. Total mass (725 kg x 2) + (6000 kg x 4) = 25450 kg the distance between nose wheels the main wheels 10 meters the centre of gravity in front of main wheels at centre of gravity = total moment / total mass centre of gravity = (1450 kg x 10 m) / 25450 kg = 0 5697 m. Question 69-18
An aeroplane has a planned take off mass of 200 000 kg its cg located at 15 38 m of reference point representing a cg location at 30% mac mean aerodynamic cord moment arm of forward cargo 15 mmoment arm of aft cargo 25 mfor performance purposes value of centre of gravity location need to be move aft to 35% mac what mass of cargo must be shifted from front cargo hold to aft hold 234 Prior to every flight. Length of mean aerodynamic chord = (14 + 4 6) 14 = 4 6 m change in cg = 30% vers 35% = 5% 5% of mac lenght = 5% 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. Question 69-19
The index method in mass and balance calculations used Reducing magnitude of moment. In mass balance calculations 'index' a figure without unit of measurement which represents a moment the value of index the moment divided a constant usually 1000 it used to simplify calculations decreasing values. Question 69-20
What are advantages of using index method to determine moments it Reduces magnitude of moments making it less time consuming to compute. In mass balance calculations 'index' a figure without unit of measurement which represents a moment the value of index the moment divided a constant usually 1000 it used to simplify calculations decreasing values. Question 69-21
What the principle of index method To divide high magnitude moments a constant make result more easier to use. Ecqb03 july 2016. Question 69-22
Define 'under loa Allowed tom dom useful load. Ecqb03 july 2016. Question 69-23
for this question use annex ecqb 031 046 v2015 01 the aircraft loaded as shown in table calculate new total moment if mass of crew increased 240 Allowed tom dom useful load. Ecqb03 july 2016 sorry we haven't yet recovered annex we get only question the correct answer (this nice isn't it?) !!!. Question 69-24
The planned take off mass of a turbojet aeroplane 180 000 kg with its centre of gravity located at 26 % mac mean aerodynamic cord shortly prior to engine start local staff informs flight crew that 4 000 kg must be unloaded from cargo 4 after handling operation new centre of gravity location in % mac will be 244 Allowed tom dom useful load. Img /com_en/com031 291 jpg . Question 69-25
A turbojet aeroplane has a planned take off mass of 190 000 kg following cargo loading crew informed that centre of gravity at take off located at 38 % mac mean aerodynamic cord which beyond limits the captain decides then to redistribute part of cargo load between cargo 1 and cargo 4 in order to obtain a new centre of gravity location at 31 % mac he asks a transfer of 244 3 kg from cargo 4 to cargo. Img /com_en/com031 292 jpg. Question 69-26
The planned take off mass of an aeroplane 190 000 kg with its centre of gravity located at 29 % mac mean aerodynamic cord shortly prior to engine start local staff informs flight crew that an additional load of 4 000 kg must be loaded in cargo 4 after loading this cargo new centre of gravity location will be 245 3 kg from cargo 4 to cargo. Img /com_en/com031 292 jpg. Question 69-27
The planned take off mass of an aeroplane 180 000 kg with its centre of gravity located at 31 % mac mean aerodynamic cord shortly prior to engine start local staff informs crew that an additional load of 4 000 kg must be loaded in cargo 1 after loading this cargo new centre of gravity location will be 246 3 kg from cargo 4 to cargo.. Question 69-28
What dry operating index doi The index the position of centre of gravity at dry operating mass. Dry operating index (doi) the index the position of centre of gravity at dry operating mass dry operation mass (dom) the total mass of aeroplane ready a specific type of operation excluding usable fuel traffic load the mass includes items such as i) crew crew baggage ii) catering removable passenger service equipment iii) potable water lavatory chemicals iv) food beverages balance arm (ba) the distance from datum to centre of gravity of a mass centre of gravity (cg) that point through which force of gravity said to act on a mass moment the product of mass balance arm. Question 69-29
A 3 m long plank on a pivot halfway along its length a 1 kg mass suspended on left end and a 2 kg mass from other end how far and in which direction should plank be moved in order the plank to be in balance The index the position of centre of gravity at dry operating mass. 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 same moment than 2 kg mass the plank to be in balance 1 kg x 2 m = 2 kg x 1 m 2 kg m = 2 kg m. Question 69-30
For following see saw to be in balance with a mass of 35 kg suspended on left end 14m left of pivot and 75 kg suspended on right end mass required at position 5m left of pivot must be 247 The index the position of centre of gravity at dry operating mass. 14m x 35 kg + 5m x ? = 8m x 75 kg ? = (8 x 75 14 x 35) / 5 ? = (600 490) / 5 ? = 22 kg. Question 69-31
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 stabilizer trim units required a take off flap setting of 15° The index the position of centre of gravity at dry operating mass.. Question 69-32
Consider a conventional aircraft with three wheels the nose jack located 161 inches aft of datum the main wheel jacks are located 775 inches aft of datum after weighing following results are reported nose jack 2300 lbeach main wheel jack 19300 lbwhat the basic empty mass of aircraft and centre of gravity from datum Bem 4 9 lb cg 74 5 in.. Question 69-33
Refer to figure 031 57 calculate cg position and moment the basic empty mass using data given from attached table 5 7 in aft of datum the moment 365 74 in lb.. Question 69-34
The mass and cg of an aircraft must be established actual weighing the Operator prior to initial entry into service. Easa air ops regulation (eu) no 965/2012 cat pol mab 100 mass balance loading (a) during any phase of operation loading mass centre of gravity (cg) of aircraft shall comply with limitations specified in afm or operations manual if more restrictive (b) the operator shall establish mass the cg of any aircraft actual weighing prior to initial entry into service 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 repairs on mass balance shall be accounted and properly documented aircraft shall be reweighed if effect of modifications on mass balance not accurately known (c) the weighing shall be accomplished the manufacturer of aircraft or an approved maintenance organisation (d) the operator shall determine mass of all operating items crew members included in aircraft dry operating mass weighing or using standard masses the influence of their position on aircraft's cg shall be determined (e) the operator shall establish mass of traffic load including any ballast actual weighing or determining mass of traffic load in accordance with standard passenger baggage masses (f) in addition to standard masses passengers checked baggage operator can use standard masses other load items if it demonstrates to competent authority that these items have same mass or that their masses are within specified tolerances (g) the operator shall determine mass of fuel load using actual density or if not known density calculated in accordance with a method specified in operations manual (h) the operator shall ensure that loading of (1) its aircraft performed under supervision of qualified personnel and (2) traffic load consistent with data used the calculation of aircraft mass balance (i)the operator shall comply with additional structural limits such as floor strength limitations maximum load per running metre maximum mass per cargo compartment the maximum seating limit for helicopters in addition operator shall take account of in flight changes in loading (j) the operator shall specify in operations manual principles methods involved in loading in mass balance system that meet requirements contained in (a) to (i) this system shall cover all types of intended operations. Question 69-35
Refer to figure 031 59 the see saw in a state of equilibrium calculate distance between a and c Operator prior to initial entry into service. left side must be equal to right side = moment on left side must be equal to moment on right side (520kg x 8m) = ((48kg x 7m) + 170 kg x x) x the distance between pivot to mass c x = ((520kg x 8m) (48kg x 7m)) / 170 x = (4160 336) / 170 = 22 49 m distance between a c 8 m + 22 49 m = 30 49 m. Question 69-36
Given length of mac 114 inchesforward gc limit 12% macaft cg limit 38% maccalculate distance of cg at 10m mac with reference to forward limit Operator prior to initial entry into service. left side must be equal to right side = moment on left side must be equal to moment on right side (520kg x 8m) = ((48kg x 7m) + 170 kg x x) x the distance between pivot to mass c x = ((520kg x 8m) (48kg x 7m)) / 170 x = (4160 336) / 170 = 22 49 m distance between a c 8 m + 22 49 m = 30 49 m. Question 69-37
Consider a conventional aircraft with three wheels the nose jack located 161 inches aft of datum the main wheel jacks are located 775 inches aft of datum after weighing following results are reported nose jack 6488 lbeach main wheel jack 17783 lbwhat the basic empty mass of aircraft and centre of gravity from datum Bem 42 54 lb cg 68 3 in.. Question 69-38
Refer to figure 031 61 originally there has been 225 l of fuel in tank of aircraft but pilot decides do add 200 kg of additional fuel what will happen to centre of gravity The cg will move towards nose of aircraft.. Question 69-39
Refer to figure 031 05 given force fa 100 ndistance a 6 mdistance b 3 mcalculate force fb to obtain equilibrium The cg will move towards nose of aircraft.. Question 69-40
Refer to figure 031 34 givenbem 1200 kgbem cg 3 00pilot and front pax 200 kgcargo on pax floor 50 kgfuel 250 kgfind loaded centre of gravity cg using attached graph The cg will move towards nose of aircraft..
Exclusive rights reserved. Reproduction prohibited under penalty of prosecution.
2719 Free Training Exam