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The centre of gravity of an aircraft is that point through which the total mass ? [ Questionnaire LAPL ]

Question 84-1 : Parallel to the gravity vector always parallel to the aircraft's vertical axis at right angles to the flight path governed by the distribution of the mass within the aircraft

exemple 184 Parallel to the gravity vector. — Img145.the sum of the aircraft weight acts through the center of gravity towards the earth's center Parallel to the gravity vector.

When an aircraft is stationary on the ground its total weight will act ?

Question 84-2 : Through its centre of gravity through its centre of pressure through the main wheels of its undercarriage assembly through a point defined as the datum point

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

The weight of an aircraft which is in level non accelerated flight is said to ?

Question 84-3 : Vertically through the centre of gravity vertically through the centre of pressure vertically through the datum point always along the vertical axis of the aircraft

exemple 192 Vertically through the centre of gravity. — 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 Vertically through the centre of gravity.

The distance from the datum to the centre of gravity of a mass is known as ?

Question 84-4 : The moment arm or balance arm the index the force the moment

exemple 196 The moment arm or balance arm. — Balance arm is the distance from the datum to the centre of gravity of a mass The moment arm or balance arm.

During take off you notice that for a given elevator input the aeroplane ?

Question 84-5 : The centre of gravity may be towards the aft limit the aeroplane is overloaded the centre of gravity is too far forward the centre of pressure is aft of the centre of gravity

exemple 200 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 The centre of gravity may be towards the aft limit.

The mass displacement caused by landing gear extension ?

Question 84-6 : Creates a longitudinal moment in the direction pitch up or pitch down determined by the type of landing gear creates a pitch up longitudinal moment does not create a longitudinal moment creates a pitch down longitudinal moment

exemple 204 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 Creates a longitudinal moment in the direction (pitch-up or pitch-down) determined by the type of landing gear.

The datum used for balance calculations is chosen on the longitudinal axis of ?

Question 84-7 : But not necessarily between the nose and the tail of the aircraft and necessarily situated between the nose and the tail of the aircraft between the leading edge and trailing edge of the wing or at rotor mast for an helicopter and always at the fire wall level

exemple 208 But not necessarily between the nose and the tail of the aircraft. — The datum is an imaginary vertical line specified by the designer from which all horizontal c of g measurements are made . /com en/com031 158 jpg.in most cases the datum is located in the vicinity of the aircraft nose usually the firewall But not necessarily between the nose and the tail of the aircraft.

The centre of gravity is the ?

Question 84-8 : Point where all the aircraft mass is considered to be concentrated centre of thrust along the longitudinal axis in relation to a datum line focus along the longitudinal axis in relation to a datum line neutral point along the longitudinal axis in relation to a datum line

exemple 212 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 Point where all the aircraft mass is considered to be concentrated.

For the purpose of aeroplane mass and balance calculations the datum point is ?

Question 84-9 : 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 point through which the sum of the mass values of the aeroplane and its contents is assumed to act vertically a point near the centre of the aeroplane it moves longitudinally as masses are added forward and aft of its location a variable point that is dependent on the load distribution for its location from which all balance arms are measured

exemple 216 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 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.

In calculations with respect to the position of the centre of gravity a ?

Question 84-10 : A reference plane which is chosen by the aircraft manufacturer its position is given in the aircraft flight or loading manual calculated from the loading manifest an arbitrary reference chosen by the pilot which can be located anywhere on the aircraft calculated from the data derived from the weighing procedure carried out on the aircraft after any major modification

exemple 220 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 A reference plane which is chosen by the aircraft manufacturer. its position is given in the aircraft flight or loading manual.

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

Question 84-11 : Moment centre of gravity moment arm force

exemple 224 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 Moment.

The moment for an item is ?

Question 84-12 : The mass of the item multiplied by it's distance from the datum the mass of the item divided by it's distance from the datum the distance the item is from the datum divided by it's mass the square of the distance the item is from the datum divided by it's mass

exemple 228 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 The mass of the item multiplied by it's distance from the datum.

The reference about which centre of gravity moments are taken is the ?

Question 84-13 : Datum centre of pressure centre of mass chord line

exemple 232 Datum. — The datum is the point on the aircraft designated by the manufacturer from which all centre of gravity measurements and calculations are made Datum.

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

Question 84-14 : 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 236 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 84-15 : 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 240 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 84-16 : 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

exemple 244 Has a negative arm and therefore generates a negative moment. — Has a negative arm and therefore generates a negative moment.

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

Question 84-17 : Fb = a x fa / b fb = a + fa / b fb = a x b / fa fb = b x fa / a

exemple 248 Fb = a x fa / b — Fb x b = fa x a.therefore .fb = fa x a / b Fb = a x fa / b

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

Question 84-18 : Fc = fa / 3 fc = 3fa fc = fa / 3a fc = 3 / fa

exemple 252 Fc = fa / 3 — Fc x 3a = fa x a.thus fc = fa x a / 3a.simplification by 'a' gives fc = fa / 3 Fc = fa / 3

For the following boom to be in balance . 238 ?

Question 84-19 : B = fa x a / fb b = fb x a / fa b = fa x a / fb b = fb + a / fa

exemple 256 B = fa x a / fb — Fb x b = fa x a.therefore .b = fa x a / fb B = fa x a / fb

For the following boom to be in balance . 239 ?

Question 84-20 : A = b x fb / fa a = b + fb / fa a = b x fa / fb a = b fa + fb

exemple 260 A = b x fb / fa — Fa x a = fb x b.therefore .a = fb x b /fa A = b x fb / fa

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

Question 84-21 : 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

exemple 264 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 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 84-22 : 33 cm 40 cm 30 cm 34 cm

exemple 268 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 33 cm.

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

Question 84-23 : 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

exemple 272 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 It will cause the cg to move aft.

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

Question 84-24 : 26 57 cm forward of datum 32 29 cm forward of datum 26 57 cm aft of datum 32 29 cm aft of datum

exemple 276 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 26.57 cm forward of datum.

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

Question 84-25 : 207 kg 317 kg 140 kg 14 kg

exemple 280 207 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 207 kg.

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

Question 84-26 : 62 5 kg 73 5 kg 68 9 kg 65 8 kg

exemple 284 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 62.5 kg.

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

Question 84-27 : 27 5 % 35 5 % 30 4 % 16 9 %

exemple 288 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 27.5 %.

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

Question 84-28 : 300 kg 30 kg 196 kg 900 kg

exemple 292 300 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 300 kg.

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

Question 84-29 : Station 117 69 station 118 33 station 120 22 station 118 25

exemple 296 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 Station 117.69

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

Question 84-30 : 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

exemple 300 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 It moves aft by 0.31 m.

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

Question 84-31 : 56 53 cm aft datum 56 35 cm aft datum 60 16 cm aft datum 53 35 cm aft datum

exemple 304 56.53 cm aft datum. — Img133.centre of gravity = 1 369 350 / 24224 = 56 528 aft of datum 56.53 cm aft datum.

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

Question 84-32 : 61 28 cm aft of datum 61 29 cm aft of datum 61 26 cm aft of datum 61 27 cm aft of datum

exemple 308 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 61.28 cm aft of datum.

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

Question 84-33 : 29344 kg 16529 kg 8680 kg 43120 kg

exemple 312 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 29344 kg.

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

Question 84-34 : 7500 kg 35000 kg 62500 kg 3500 kg

exemple 316 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 7500 kg.

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

Question 84-35 : 56160 9360 433906 30888

exemple 320 56160. — Block fuel 100 us gal x 6 lbs = 600 lbs. 141.600 x 93 6 = 56160 56160.

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

Question 84-36 : 401338 lbs in 432221 lbs in 433906 lbs in 377746 lbs in

exemple 324 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 401338 lbs.in

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

Question 84-37 : 91 92 inches aft of datum 91 84 inches aft of datum 91 69 inches aft of datum 93 60 inches aft of datum

exemple 328 91.92 inches aft of datum. — Img141.centre of gravity at take off = 432226 2 / 4702 = 91 92 aft of datum 91.92 inches aft of datum.

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

Question 84-38 : 2548 8 6675 2496 3 2311 8

exemple 332 2548,8. — Img142 2548,8.

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

Question 84-39 : 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 336 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 84-40 : 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 340 The centre of gravity move back from 4% to 5% mac. — The centre of gravity move back from 4% to 5% mac.

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