The reference about which centre of gravity moments are taken is the ?
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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 load placed aft of the datum ?
A load placed forward of datum Has a negative arm therefore generates a negative moment. The datum the point on aircraft designated the manufacturer from which all centre of gravity measurements calculations are made. 
For following see saw to be in balance 236 Has a negative arm therefore generates a negative moment. Fb x b = fa x a therefore fb = (fa x a) / b. 
For following see saw to be in balance 237 Has a negative arm therefore generates a negative moment. Fc x 3a = fa x a thus fc = (fa x a) / 3a simplification 'a' gives fc = fa / 3. 
For following boom to be in balance 238 Has a negative arm therefore generates a negative moment. Fb x b = fa x a therefore b = (fa x a) / fb. Question 70-8
For following boom to be in balance 239 Has a negative arm therefore generates a negative moment. Fa x a = fb x b therefore a = (fb x b)/fa. Question 70-9
A tri wheel aircraft has its datum 25 inches aft of nose wheel and 180 inches forward of main wheels if nose jack shows 125 lb and each main wheel jack shows 3400 lb determine bem and cg 6925 lbs 2 2 inches from nose wheel of aircraft. 25 x 125 = 3125 180 x (2x3400) = 1224000 total moment = 1224000+( 3125)= 1220875 basic empty mass bem) = 125 + (3400x2) = 6925 lbs the cg will be total moment/ total mass = 1220875/6925= 176 2 inches from datum the datum located 25 inches aft of nose wheel cg at 25 inches + 176 2 = 201 2 aft of nose wheel. Question 70-10
The mass of an aircraft 1950 kg if 450 kg added to a cargo hold 1 75 metres from loaded centre of gravity cg the loaded cg will move 6925 lbs 2 2 inches from nose wheel of aircraft. Change of cg = (mass added x distance from hold to cg) / new total mass change of cg = (450 x 1 75) / (1950 + 450) change of cg = 0 3281 m (33 cm). Question 70-11
If nose wheel moves aft during gear retraction how will this movement affect location of centre of gravity cg on aircraft It will cause cg to move aft. If nose wheel moves aft cg moves aft the cg that point through which force of gravity said to act on a mass always acts parallel to gravity vector. Question 70-12
Where the centre of gravity of aeroplane in diagram 185 26 57 cm forward of datum. Total moments 1 x 1750 + 2 5 x 8130 = 22075 n sum of total moments / sum of total weight = cg position 22075 n / 9880 kg = 2 2343 2 5 2 2343 = 0 2657 m (26 57 cm). Question 70-13
Given total mass 2900 kgcg location station 115aft cg limit station 116the maximum mass that can be added at station 130 26 57 cm forward of datum. Mass added / old total mass = change of cg / distance from hold to new cg mass added = (change of cg / distance from hold to new cg) x old total mass mass added = ((116 115) / (130 116)) x 2900 mass added = 207 kg. Question 70-14
Given total mass 7500 kgcentre of gravity cg location station 80 5aft cg limit station 79 5how much cargo must be shifted from aft cargo compartment at station 150 to forward cargo compartment at station 30 in order to move cg location to aft limit 26 57 cm forward of datum. Mass moved / total mass = change of cg / distance moved mass moved = (change of cg x total mass) / distance moved mass moved = (1 x 7500 kg) / (150 30) mass moved = 7500 / 120 mass moved = 62 5 kg you need to move cg from station 80 5 to station 79 5. Question 70-15
A jet aeroplane with geometrical characteristics shown in 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 zero reference point at last moment station manager has 12 000 n of freight added in forward compartment at 10 m from zero reference point the final location of centre of gravity calculated in percentage of mean aerodynamic chord ab from point a equal to 186 26 57 cm forward of datum. Mass x arm = moment 460000 n x 15 4 m = 7084000 nm 12000 n of last minute cargo added to a hold 10 m from datum 12000 n x 10 m = 120000 nm new mass is 472000 n new moment is 7204000 nm cg position now 7204000 / 472000 = 15 26 m from zero reference point in percentage of mean aerodynamic chord it lemac (leading edge mean aerodynamic chord) at 14 m therefore cg at 1 26m from lemac (15 26m 14m) 1 26 / 4 6 x 100 = 27 5% mac. Question 70-16
The total mass of an aeroplane 9000 kg the centre of gravity cg position at 2 0 m from datum line the aft limit cg at 2 1 m from datum line what mass of cargo must be shifted from front cargo hold at 0 8 m from datum to aft hold at 3 8 m to move cg to aft limit 26 57 cm forward of datum. (2 1 2) x 9000 = mass to moved x (3 8 0 8) 900 = mass to moved x 3 mass to moved = 900 / 3 = 300 kg. Question 70-17
Assume aircraft actual mass 4750 kgcentre of gravity at station 115 8what will be new position of centre of gravity if 100 kg moved from station 30 to station 120 26 57 cm forward of datum. Mass moved / total mass = change of cg / distance moved mass moved x distance moved / total mass = change of cg 100 x 90 /4750 = 1 89 new cg location 115 8 + 1 89 = 117 69. Question 70-18
Given aircraft mass 36000 kgcentre of gravity cg located at station 17 mwhat the effect on cg location if you move 20 passengers total mass 1600 kg from station 16 to station 23 26 57 cm forward of datum. We move passangers aft thus cg will move aft mass moved / total mass = change of cg / distance moved change of cg = (mass moved x distance moved) / total mass change of cg = 1600 x 7 / 36000 = 0 31 m. Question 70-19
Given following information calculate loaded centre of gravity cg 197 26 57 cm forward of datum. centre of gravity = 1 369 350 / 24224 = 56 528 aft of datum. Question 70-20
Given are following information at take off given that flight time 2 hours and estimated fuel flow will be 1050 litres per hour and average oil consumption will be 2 25 litres per hour the specific density of fuel 0 79 and specific density of oil 0 96 calculate landing centre of gravity 198 6 28 cm aft of datum. Fuel = 2100 l x 0 79 = 1659 kg huile = 4 5 l x 0 96 = 4 32 kg centre of gravity at landing = 1382449 2 / 22560 68 = 61 28 aft of datum. Question 70-21
Given that total mass of an aircraft 112000 kg with a centre of gravity position at 22 62 m aft of datum the centre of gravity limits are between 18 m and 22 m how much mass must be removed from rear hold 30 m aft of datum to move centre of gravity to middle of limits 6 28 cm aft of datum. Mass moved / total mass = change of cg / distance moved mass moved = (change of cg x total mass) / distance moved mass moved = ((22 62 20) x 112000 kg) / (30 20) mass moved = 293440 kg / 10 mass moved = 29344 kg. Question 70-22
The total mass of an aeroplane 145000 kg and centre of gravity limits are between 4 7 m and 6 9 m aft of datum the loaded centre of gravity position 4 4 m aft how much mass must be transferred from front to rear hold in order to bring out of limit centre of gravity position to foremost limit 199 6 28 cm aft of datum. Change in mass / total mass = change in cg / total distance moved change in cg = 0 3 (from 4 4 m to reach 4 7 m) total distance moved = distance between front hold rear hold = 8 7 m 2 9 m = 5 8 m change in mass = total mass x change in cg / total distance moved change in mass = 145000 x 0 3 /5 8 = 7500 kg. Question 70-23
With respect to multi engine piston powered aeroplane determine block fuel moment lbs in in 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 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 201 6 28 cm aft of datum. Block fuel 100 us gal x 6 lbs = 600 lbs 600 x 93 6 = 56160. Question 70-24
With respect to a multi engine piston powered aeroplane determine total moment lbs in at landing in following conditions basic empty mass 3210 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 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 202 6 28 cm aft of datum. Take off moment 432226 lbs in trip fuel moment 55 x 6 = 330 330 x 93 6 = 30888 lbs in ldg 432226 30888 = 401338 lbs in (it's given take off moment so 3 us gal start up taxi has already been burned therefore 55 not 58). Question 70-25
Determine cg location at take off in following conditions basic empty mass 3210 lb one pilot 160 lb front seat passenger 200 lb centre seat passengers 290 lb total one passenger rear seat 110 lb baggage in zone 1 100 lb baggage in zone 4 50 lb zero fuel mass 4120 lb moment at zero fuel mass 377751 lb inblock fuel 100 us gal trip fuel 55 us gal fuel start up and taxi included in block fuel 3 us gal fuel density 6 lb/us gal 203 9 92 inches aft of datum. centre of gravity at take off = 432226 2 / 4702 = 91 92 aft of datum. Question 70-26
With respect to a single engine piston powered aeroplane determine zero fuel moment lbs in /100 in 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 204 9 92 inches aft of datum. centre of gravity at take off = 432226 2 / 4702 = 91 92 aft of datum. Question 70-27
The aeroplane has a mass of 51 000 kg in cruise the range of safe cg positions as determined from appropriate graph in loading manual 1050 Forward limit 4% aft limit 29 7% mac. centre of gravity at take off = 432226 2 / 4702 = 91 92 aft of datum. Question 70-28
Considering annex attached to 50 000 kg mass determine displacement of front limit of center of gravity when moving from clean configuration to flaps and landing gear landing configuration The centre of gravity move back from 4% to 5% mac. centre of gravity at take off = 432226 2 / 4702 = 91 92 aft of datum. Question 70-29
Using weight and balance graph provided in appendix determine range of permissible center of gravity positions the maximum landing weight of 56245 kg From 5% to 29 2% mac. centre of gravity at take off = 432226 2 / 4702 = 91 92 aft of datum. Question 70-30
Prior to departure an aircraft loaded with 16500 litres of fuel at a fuel density of 0 78 this entered into load sheet as 16500 kg and calculations are carried out accordingly as a result of this error aircraft is Lighter than anticipated the calculated safety speeds will be too high. The aircraft lighter than anticipated (16500 kg entered into load sheet instead of 12870 kg) the calculated safety speeds (v1 vr v2) will be too high they have been computed an heavier aircraft you will get airborne earlier (you will have much greater margins). Question 70-31
An additional baggage container loaded into aft cargo compartment but not entered into load and trim sheet the aeroplane will be heavier than expected and calculated take off safety speeds Will give reduced safety margins. The aircraft lighter than anticipated (16500 kg entered into load sheet instead of 12870 kg) the calculated safety speeds (v1 vr v2) will be too high they have been computed an heavier aircraft you will get airborne earlier (you will have much greater margins). Question 70-32
Fuel loaded onto an aeroplane 15400 kg but erroneously entered into load and trim sheet as 14500 kg this error not detected the flight crew but they will notice that Speed at which airplane will leave ground will be higher than expected. The airplane heavier than pilots think it is v1 will be reached later aeroplane will rotate later than expected the crew did not detect error thus v1 will not change. Question 70-33
Which of following unlikely to have any effect on position of centre of gravity on an aeroplane in flight Changing tailplane (horizontal stabiliser) incidence angle. Trimming stabiliser a response to a cg change it does not move cg. Question 70-34
Which of following unlikely to have any effect on position of centre of gravity on an aeroplane in flight Changing tailplane (horizontal stabiliser) incidence angle. Trimming stabiliser a response to a cg change it does not move cg. Question 70-35
When centre of gravity at forward limit an aeroplane will be Extremely stable will require excessive elevator control to change pitch. When cg located on forward limit pitch down moment at its allowed maximum (the distance between cg cp great) the elevator will have to develop a high downward force to counteract this pitch down moment it will seem heavier to pilot of course aircraft will be extremely stable which the good news about a forward cg. Question 70-36
A mass of 500 kg loaded at a station which located 10 metres behind present centre of gravity and 16 metres behind datum assume g=10 m/sec squared the moment that mass used in loading manifest Extremely stable will require excessive elevator control to change pitch. Arm = moment / force thus moment = arm x force 16 m x 5000 n = 80000 nm. Question 70-37
Without crew mass and longitudinal cg position of aircraft are 6000 kg and 4 70m mass of pilot 90 kg mass of copilot 100 kg mass of flight engineer 80 kgwith crew mass and longitudinal cg position of aircraft are 220 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. Question 70-38
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. Question 70-39
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. Question 70-40
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.
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