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Question 84-1 : During take off you notice that for a given elevator input the aeroplane rotates much more rapidly than expected .this is an indication that ? [ Questionnaire LAPL ]

The centre of gravity may be towards the aft limit

If a very small up deflection of the horizontal stabilizer generates a downward force on the tail sufficiently strong to raise the airplane nose this is an indication that the centre of gravity may be towards the aft limit exemple 184 The centre of gravity may be towards the aft limit.

Question 84-2 : The mass displacement caused by landing gear extension ?

Creates a longitudinal moment in the direction pitch up or pitch down determined by the type of landing gear

The mass displacement caused by landing gear extension creates a longitudinal moment in the direction pitch up or pitch down determined by the type of landing gear

Question 84-3 : The datum used for balance calculations is chosen on the longitudinal axis of the aircraft ?

But not necessarily between the nose and the tail of the aircraft

exemple 192 But not necessarily between the nose and the tail of the aircraft.

Question 84-4 : The centre of gravity is the ?

Point where all the aircraft mass is considered to be concentrated

Centre of gravity 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 196 Point where all the aircraft mass is considered to be concentrated.

Question 84-5 : For the purpose of aeroplane mass and balance calculations the datum point is defined as ?

A fixed point from which all balance arms are measured it may be located anywhere on the aeroplane's longitudinal axis or on the extensions to that axis

The datum point is the point on the aircraft designated by the manufacturer from which all centre of gravity measurements and calculations are made in most cases the datum is located in the vicinity of the aircraft nose usually the firewall it is not necessarily a point 'on' the aircraft exemple 200 A fixed point from which all balance arms are measured. it may be located anywhere on the aeroplane's longitudinal axis or on the extensions to that axis.

Question 84-6 : In calculations with respect to the position of the centre of gravity a reference is made to a datum the datum is ?

A reference plane which is chosen by the aircraft manufacturer its position is given in the aircraft flight or loading manual

The datum is that point on the longitudinal axis or extension thereof from which the centre of gravity of all masses are referenced its position is determined by the manufacturer and found in the loading manual and operating data manual exemple 204 A reference plane which is chosen by the aircraft manufacturer. its position is given in the aircraft flight or loading manual.

Question 84-7 : The mass of an item multiplied by it's distance from the datum is it's ?

Moment

Arm moment arm is the horizontal distance in inches from the reference datum line to the center of gravity of an item .moment is the product of the weight of an item multiplied by its arm moments are expressed in pound inches lb in total moment is the weight of the airplane multiplied by the distance between the datum and the cg exemple 208 Moment.

Question 84-8 : The moment for an item is ?

The mass of the item multiplied by it's distance from the datum

The moment is the product of the mass and the balance arm exemple 212 The mass of the item multiplied by it's distance from the datum.

Question 84-9 : The reference about which centre of gravity moments are taken is the ?

Datum

The datum is the point on the aircraft designated by the manufacturer from which all centre of gravity measurements and calculations are made exemple 216 Datum.

Question 84-10 : 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 ?

At a convenient point which may not physically be on the aircraft

exemple 220 At a convenient point which may not physically be on the aircraft.

Question 84-11 : A load placed aft of the datum ?

Has a positive arm and therefore generates a positive moment

exemple 224 Has a positive arm and therefore generates a positive moment.

Question 84-12 : A load placed forward of the datum ?

Has a negative arm and therefore generates a negative moment

exemple 228 Has a negative arm and therefore generates a negative moment.

Question 84-13 : For the following see saw to be in balance . 236 ?

Fb = a x fa / b

Fb x b = fa x a.therefore .fb = fa x a / b exemple 232 Fb = a x fa / b

Question 84-14 : For the following see saw to be in balance . 237 ?

Fc = fa / 3

Fc x 3a = fa x a.thus fc = fa x a / 3a.simplification by 'a' gives fc = fa / 3 exemple 236 Fc = fa / 3

Question 84-15 : For the following boom to be in balance . 238 ?

B = fa x a / fb

Fb x b = fa x a.therefore .b = fa x a / fb exemple 240 B = fa x a / fb

Question 84-16 : For the following boom to be in balance . 239 ?

A = b x fb / fa

Fa x a = fb x b.therefore .a = fb x b /fa exemple 244 A = b x fb / fa

Question 84-17 : A tri wheel aircraft has its datum 25 inches aft of the nose wheel and 180 inches forward of the main wheels .if the nose jack shows 125 lb and each main wheel jack shows 3400 lb determine the bem and cg ?

6925 lbs and 201 2 inches from the nose wheel of the 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 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 248 6925 lbs and 201.2 inches from the nose wheel of the aircraft.

Question 84-18 : The mass of an aircraft is 1950 kg .if 450 kg is added to a cargo hold 1 75 metres from the loaded centre of gravity cg .the loaded cg will move ?

33 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 252 33 cm.

Question 84-19 : If nose wheel moves aft during gear retraction how will this movement affect the location of the centre of gravity cg on the aircraft ?

It will cause the cg to move aft

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 256 It will cause the cg to move aft.

Question 84-20 : Where is the centre of gravity of the aeroplane in the 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 exemple 260 26.57 cm forward of datum.

Question 84-21 : Given .total mass 2900 kg.cg location station 115.aft cg limit station 116.the maximum mass that can be added at station 130 is ?

207 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 264 207 kg.

Question 84-22 : Given .total mass 7500 kg.centre of gravity cg location station 80 5.aft cg limit station 79 5.how much cargo must be shifted from the aft cargo compartment at station 150 to the forward cargo compartment at station 30 in order to move the cg location to the aft limit ?

62 5 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 268 62.5 kg.

Question 84-23 : A jet aeroplane with the geometrical characteristics shown in the 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 the zero reference point .at the last moment the station manager has 12 000 n of freight added in the forward compartment ?

27 5 %

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 272 27.5 %.

Question 84-24 : The total mass of an aeroplane is 9000 kg the centre of gravity cg position is at 2 0 m from the datum line the aft limit for cg is at 2 1 m from the datum line .what mass of cargo must be shifted from the front cargo hold at 0 8 m from the datum to the aft hold at 3 8 m to move the cg to the aft ?

300 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 276 300 kg.

Question 84-25 : Assume .aircraft actual mass 4750 kg.centre of gravity at station 115 8.what will be the new position of the centre of gravity if 100 kg is moved from the station 30 to station 120 ?

Station 117 69

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 280 Station 117.69

Question 84-26 : Given .aircraft mass 36000 kg.centre of gravity cg is located at station 17 m.what is the effect on cg location if you move 20 passengers total mass 1600 kg from station 16 to station 23 ?

It moves aft by 0 31 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 284 It moves aft by 0.31 m.

Question 84-27 : Given the following information calculate the loaded centre of gravity cg . 197 ?

56 53 cm aft datum

Img133.centre of gravity = 1 369 350 / 24224 = 56 528 aft of datum exemple 288 56.53 cm aft datum.

Question 84-28 : Given are the following information at take off .given that the flight time is 2 hours and the estimated fuel flow will be 1050 litres per hour and the average oil consumption will be 2 25 litres per hour .the specific density of fuel is 0 79 and the specific density of oil is 0 96 .calculate the ?

61 28 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 292 61.28 cm aft of datum.

Question 84-29 : Given that the total mass of an aircraft is 112000 kg with a centre of gravity position at 22 62 m aft of the datum .the centre of gravity limits are between 18 m and 22 m .how much mass must be removed from the rear hold 30 m aft of the datum to move the centre of gravity to the middle of the ?

29344 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 .notice the middle of the limits is at 20m cg limits are between 18 m and 22 m exemple 296 29344 kg.

Question 84-30 : The total mass of an aeroplane is 145000 kg and the centre of gravity limits are between 4 7 m and 6 9 m aft of the datum the loaded centre of gravity position is 4 4 m aft .how much mass must be transferred from the front to the rear hold in order to bring the out of limit centre of gravity ?

7500 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 300 7500 kg.

Question 84-31 : With respect to multi engine piston powered aeroplane determine the block fuel moment lbs in in the 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 ?

56160

Block fuel 100 us gal x 6 lbs = 600 lbs. 141.600 x 93 6 = 56160 exemple 304 56160.

Question 84-32 : With respect to a multi engine piston powered aeroplane determine the total moment lbs in at landing in the 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 ?

401338 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 308 401338 lbs.in

Question 84-33 : Determine the cg location at take off in the 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 ?

91 92 inches aft of datum

Img141.centre of gravity at take off = 432226 2 / 4702 = 91 92 aft of datum exemple 312 91.92 inches aft of datum.

Question 84-34 : With respect to a single engine piston powered aeroplane determine the zero fuel moment lbs in /100 in the 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 ?

2548 8

Img142 exemple 316 2548,8.

Question 84-35 : The aeroplane has a mass of 51 000 kg in the cruise the range of safe cg positions as determined from the appropriate graph in the loading manual is . 1050 ?

Forward limit 4% aft limit 29 7% mac

exemple 320 Forward limit 4% aft limit 29.7% mac.

Question 84-36 : Considering the annex attached to the 50 000 kg mass determine the displacement of the front limit of the center of gravity when moving from the clean configuration to the flaps and landing gear landing configuration ?

The centre of gravity move back from 4% to 5% mac

exemple 324 The centre of gravity move back from 4% to 5% mac.

Question 84-37 : Using the weight and balance graph provided in the appendix determine the range of permissible center of gravity positions for the maximum landing weight of 56245 kg ?

From 5% to 29 2% mac

exemple 328 From 5% to 29.2% mac.

Question 84-38 : Prior to departure an aircraft is loaded with 16500 litres of fuel at a fuel density of 0 78 this is entered into the load sheet as 16500 kg and calculations are carried out accordingly .as a result of this error the aircraft is ?

Lighter than anticipated and the calculated safety speeds will be too high

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 332 Lighter than anticipated and the calculated safety speeds will be too high.

Question 84-39 : An additional baggage container is loaded into the aft cargo compartment but is not entered into the load and trim sheet .the aeroplane will be heavier than expected and calculated take off safety speeds ?

Will give reduced safety margins

exemple 336 Will give reduced safety margins.

Question 84-40 : Fuel loaded onto an aeroplane is 15400 kg but is erroneously entered into the load and trim sheet as 14500 kg .this error is not detected by the flight crew but they will notice that ?

Speed at which the airplane will leave the ground will be higher 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 340 Speed at which the airplane will leave the ground will be higher than expected.


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