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Exam > pilot : The reference about which centre of gravity moments are taken is the ?

Question 70-1 : Datum centre of pressure centre of mass chord line

The datum is the point on the aircraft designated by the manufacturer from which all centre of gravity measurements and calculations are made exemple 170 datum.datum.

The datum is a reference from which all moment balance arms are measured its ?

Question 70-2 : At a convenient point which may not physically be on the aircraft at or near the forward limit of the centre of gravity at or near the focal point of the aircraft axis system at or near the natural balance point of the empty aircraft

exemple 174 at a convenient point which may not physically be on the aircraft.at a convenient point which may not physically be on the aircraft.

A load placed aft of the datum ?

Question 70-3 : Has a positive arm and therefore generates a positive moment has a negative arm and therefore generates a negative moment and mass has a negative arm and therefore generates a negative moment has a positive arm and therefore generates a positive moment but negative mass

exemple 178 has a positive arm and therefore generates a positive moment.has a positive arm and therefore generates a positive moment.

A load placed forward of the datum ?

Question 70-4 : Has a negative arm and therefore generates a negative moment has a negative arm and therefore generates a negative mass and moment has a positive arm and therefore generates a positive mass and moment has a positive arm and therefore generates a positive moment


For the following see saw to be in balance. 236 ?

Question 70-5 : Fb = a x fa / b fb = a + fa / b fb = a x b / fa fb = b x fa / a

Fb x b = fa x a.therefore .fb = fa x a / b exemple 186 fb = a x fa / bfb = a x fa / b

For the following see saw to be in balance. 237 ?

Question 70-6 : Fc = fa / 3 fc = 3fa fc = fa / 3a fc = 3 / fa

Fc x 3a = fa x a..thus fc = fa x a / 3a.simplification by 'a' gives fc = fa / 3

For the following boom to be in balance. 238 ?

Question 70-7 : B = fa x a / fb b = fb x a / fa b = fa x a / fb b = fb + a / fa

Fb x b = fa x a.therefore .b = fa x a / fb exemple 194 b = fa x a / fbb = fa x a / fb

For the following boom to be in balance . 239 ?

Question 70-8 : A = b x fb / fa a = b + fb / fa a = b x fa / fb a = b fa + fb

Fa x a = fb x b.therefore .a = fb x b /fa exemple 198 a = b x fb / faa = b x fb / fa

A tri wheel aircraft has its datum 25 inches aft of the nose wheel and 180 ?

Question 70-9 : 6925 lbs and 201 2 inches from the nose wheel of the aircraft 6925 lbs and 176 3 inches from the nose wheel of the aircraft 3525 lbs and 172 7 inches from the datum 3525 lbs and 201 2 inches from the datum

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 the datum.the datum is located 25 inches aft of the nose wheel the cg is at 25 inches + 176 2 = 201 2 aft of the nose wheel exemple 202 6925 lbs and 201.2 inches from the nose wheel of the aircraft.6925 lbs and 201.2 inches from the nose wheel of the aircraft.

The mass of an aircraft is 1950 kg .if 450 kg is added to a cargo hold 1 75 ?

Question 70-10 : 33 cm 40 cm 30 cm 34 cm

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 exemple 206 33 cm.33 cm.

If nose wheel moves aft during gear retraction how will this movement affect ?

Question 70-11 : It will cause the cg to move aft it will not affect the cg location it will cause the cg to move forward the cg location will change but the direction cannot be told the information given

If the nose wheel moves aft the cg moves aft.the cg is that point through which the force of gravity is said to act on a mass and always acts parallel to the gravity vector exemple 210 it will cause the cg to move aft.it will cause the cg to move aft.

Where is the centre of gravity of the aeroplane in the diagram. 185 ?

Question 70-12 : 26 57 cm forward of datum 32 29 cm forward of datum 26 57 cm aft of datum 32 29 cm aft 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 exemple 214 26.57 cm forward of datum.26.57 cm forward of datum.

Given.total mass 2900 kg.cg location station 115.aft cg limit station 116..the ?

Question 70-13 : 207 kg 317 kg 140 kg 14 kg

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 exemple 218 207 kg.207 kg.

Given.total mass 7500 kg.centre of gravity cg location station 80 5.aft cg ?

Question 70-14 : 62 5 kg 73 5 kg 68 9 kg 65 8 kg

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 exemple 222 62.5 kg.62.5 kg.

A jet aeroplane with the geometrical characteristics shown in the appendix has ?

Question 70-15 : 27 5 % 35 5 % 30 4 % 16 9 %

Mass x arm = moment.460000 n x 15 4 m = 7084000 nm..12000 n of last minute cargo is added to a hold 10 m from the datum .12000 n x 10 m = 120000 nm..new mass is 472000 n.new moment is 7204000 nm..cg position is now 7204000 / 472000 = 15 26 m from the zero reference point.in percentage of mean aerodynamic chord it is .lemac leading edge mean aerodynamic chord is at 14 m therefore cg is at 1 26m from lemac 15 26m 14m .1 26 / 4 6 x 100 = 27 5% mac exemple 226 27.5 %.27.5 %.

The total mass of an aeroplane is 9000 kg the centre of gravity cg position is ?

Question 70-16 : 300 kg 30 kg 196 kg 900 kg

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 exemple 230 300 kg.300 kg.

Assume .aircraft actual mass 4750 kg.centre of gravity at station 115 8..what ?

Question 70-17 : Station 117 69 station 118 33 station 120 22 station 118 25

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 exemple 234 station 117.69station 117.69

Given .aircraft mass 36000 kg.centre of gravity cg is located at station 17 ?

Question 70-18 : It moves aft by 0 31 m it moves forward by 0 157 m it moves aft by 3 22 m it moves aft by 0 157 m

We move passangers aft thus the 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 exemple 238 it moves aft by 0.31 m.it moves aft by 0.31 m.

Given the following information calculate the loaded centre of gravity cg . 197 ?

Question 70-19 : 56 53 cm aft datum 56 35 cm aft datum 60 16 cm aft datum 53 35 cm aft datum

133.centre of gravity = 1 369 350 / 24224 = 56 528 aft of datum

Given are the following information at take off .given that the flight time is ?

Question 70-20 : 61 28 cm aft of datum 61 29 cm aft of datum 61 26 cm aft of datum 61 27 cm aft of datum

Fuel = 2100 l x 0 79 = 1659 kg..huile = 4 5 l x 0 96 = 4 32 kg.. 134.centre of gravity at landing = 1382449 2 / 22560 68 = 61 28 aft of datum exemple 246 61.28 cm aft of datum.61.28 cm aft of datum.

Given that the total mass of an aircraft is 112000 kg with a centre of gravity ?

Question 70-21 : 29344 kg 16529 kg 8680 kg 43120 kg

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 exemple 250 29344 kg.29344 kg.

The total mass of an aeroplane is 145000 kg and the centre of gravity limits ?

Question 70-22 : 7500 kg 35000 kg 62500 kg 3500 kg

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 and 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 exemple 254 7500 kg.7500 kg.

With respect to multi engine piston powered aeroplane determine the block fuel ?

Question 70-23 : 56160 9360 433906 30888

Block fuel 100 us gal x 6 lbs = 600 lbs.. 141.600 x 93 6 = 56160 exemple 258 56160.56160.

With respect to a multi engine piston powered aeroplane determine the total ?

Question 70-24 : 401338 lbs in 432221 lbs in 433906 lbs in 377746 lbs in

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 for start up and taxi has already been burned therefore 55 not 58 exemple 262 401338 lbs.in401338 lbs.in

Determine the cg location at take off in the following conditions .basic empty ?

Question 70-25 : 91 92 inches aft of datum 91 84 inches aft of datum 91 69 inches aft of datum 93 60 inches aft of datum

141..centre of gravity at take off = 432226 2 / 4702 = 91 92 aft of datum exemple 266 91.92 inches aft of datum.91.92 inches aft of datum.

With respect to a single engine piston powered aeroplane determine the zero ?

Question 70-26 : 2548 8 6675 2496 3 2311 8

142 exemple 270 2548,8.2548,8.

The aeroplane has a mass of 51 000 kg in the cruise the range of safe cg ?

Question 70-27 : Forward limit 4% aft limit 29 7% mac forward limit 5% aft limit 29 7% mac forward limit 4% aft limit 29 2% mac forward limit 5% aft limit 29 5% mac

exemple 274 forward limit 4% aft limit 29.7% mac.forward limit 4% aft limit 29.7% mac.

Considering the annex attached to the 50 000 kg mass determine the displacement ?

Question 70-28 : The centre of gravity move back from 4% to 5% mac the centre of gravity move forward from 4% to 5% mac the centre of gravity move back from 5% to 4% mac the centre of gravity move forward from 5% to 4% mac

exemple 278 the centre of gravity move back from 4% to 5% mac.the centre of gravity move back from 4% to 5% mac.

Using the weight and balance graph provided in the appendix determine the range ?

Question 70-29 : From 5% to 29 2% mac from 5% to 29 5% mac from 4% to 29 5% mac if flaps and landing gear are retracted from 4% to 29 2% mac if flaps and landing gear are retracted

exemple 282 from 5% to 29.2% mac.from 5% to 29.2% mac.

Prior to departure an aircraft is loaded with 16500 litres of fuel at a fuel ?

Question 70-30 : Lighter than anticipated and the calculated safety speeds will be too high lighter than anticipated and the calculated safety speeds will be too low heavier than anticipated and the calculated safety speeds will be too high heavier than anticipated and the calculated safety speeds will be too low

The aircraft is lighter than anticipated 16500 kg entered into the load sheet instead of 12870 kg.the calculated safety speeds v1 vr v2 will be too high they have been computed for an heavier aircraft you will get airborne earlier you will have much greater margins exemple 286 lighter than anticipated and the calculated safety speeds will be too high.lighter than anticipated and the calculated safety speeds will be too high.

An additional baggage container is loaded into the aft cargo compartment but is ?

Question 70-31 : Will give reduced safety margins will not be achieved will be greater than required are unaffected but v1 will be increased

exemple 290 will give reduced safety margins.will give reduced safety margins.

Fuel loaded onto an aeroplane is 15400 kg but is erroneously entered into the ?

Question 70-32 : Speed at which the airplane will leave the ground will be higher than expected v1 will be increased v1 will be reached sooner than expected the aeroplane will rotate much earlier than expected

The airplane is heavier than the pilots think it is v1 will be reached later the aeroplane will rotate later than expected .the crew did not detect the error thus v1 will not change exemple 294 speed at which the airplane will leave the ground will be higher than expected.speed at which the airplane will leave the ground will be higher than expected.

Which of the following is unlikely to have any effect on the position of the ?

Question 70-33 : Changing the tailplane horizontal stabiliser incidence angle lowering the landing gear movement of cabin attendants going about their normal duties normal consumption of fuel for a swept wing aeroplane

Trimming the stabiliser is a response to a cg change it does not move the cg exemple 298 changing the tailplane (horizontal stabiliser) incidence angle.changing the tailplane (horizontal stabiliser) incidence angle.

Which of the following is unlikely to have any effect on the position of the ?

Question 70-34 : Changing the tailplane horizontal stabiliser incidence angle lowering the landing gear movement of cabin attendants going about their normal duties normal consumption of fuel for a swept wing aeroplane

Trimming the stabiliser is a response to a cg change it does not move the cg exemple 302 changing the tailplane (horizontal stabiliser) incidence angle.changing the tailplane (horizontal stabiliser) incidence angle.

When the centre of gravity is at the forward limit an aeroplane will be ?

Question 70-35 : Extremely stable and will require excessive elevator control to change pitch extremely stable and require small elevator control to change pitch extremely unstable and require excessive elevator control to change pitch extremely unstable and require small 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 exemple 306 extremely stable and will require excessive elevator control to change pitch.extremely stable and will require excessive elevator control to change pitch.

A mass of 500 kg is loaded at a station which is located 10 metres behind the ?

Question 70-36 : 80000 nm 30000 nm 50000 nm 130000 nm

Arm = moment / force..thus.moment = arm x force..16 m x 5000 n = 80000 nm exemple 310 80000 nm.80000 nm.

Without the crew the mass and longitudinal cg position of the aircraft are 6000 ?

Question 70-37 : 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 314 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 70-38 : 40 0 35 5 41 5 33 0

169 exemple 318 40.040.0

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

Question 70-39 : 46130 kg and 17 8% 46130 kg and 20 8% 51300 kg and 20 8% 41300 kg and 17 8%

161 exemple 322 46130 kg and 17.8%.46130 kg and 17.8%.

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

Question 70-40 : 17 5% 20 3% 22 6% 20 1%

162 exemple 326 17.5%.17.5%.


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