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Question 91-1 : During the pre flight, the pilot assumed that the only passenger weighing 170 lb would sit at the front row seat located 85 inches aft of datum. the calculations gave a take off mass of 4100 lb and a take off cg located at 94 inches aft of datum...when the passenger arrives, he/she sits at the rear ? [ Level reports ]
No, it is out of the rear limit.
The cg is currently in limits, but we need to find the effect of the passenger sitting in the rear seat. this distance from the front to rear seat is +85 in to +158 in = 73 in...step 1. establish the parameters..m = mass to be moved 170 lb..m = total mass 4100 lb..d = cg movement ..d = relevant distance, between the seats 73 in...step 2. time for the formula..m/m = d/d..170/4100 = /73..cg movement = +3.03 in...the cg will move aft by 3.03 in to a new position of 97.03 in and is thus out of limits.
Question 91-2 : Determine the mass that must be moved aft by a distance of 2.9 m, in order to establish a cg at 2.9 m aft of the datum. given current cg position 2.8 m aft of the datum. actual take off mass 6 603 kg ?
228 kg
The new centre of gravity position can be calculated using the following formula m x d = m x d. where m stands for the total mass of the aircraft, d is the distance between the previous and new cg, m is the moved mass and d stands for the distance between the old and new position of the moved mass. 6 603 kg x 0.1 m = m x 2.9m = 660.3 / 2.9m = 227.7 kg 228 kg
Question 91-3 : A turbojet aeroplane has a planned take off mass of 190 000 kg. the cargo load is distributed as follows cargo 1 3 000 kg 3.50 m from reference point. cargo 4 7 000 kg 20.98 m from reference point. distance from reference point to leading edge 14m. length of mac 4.6 monce the cargo loading is ?
Cargo 1 6 500 kg. cargo 4 3 500 kg.
The centre of gravity must be displaced by 38% 31% = 7% mac, which corresponds to 7% of 4.6 mac = 0.322 m...the following formula helps to find out how to redistribute part of the cargo load between cargo 1 and cargo 4 in order to obtain a new centre of gravity location mass change / total mass = change of cg / total distance moved...solving for mass change we get mass change = change of cg x total mass / total distance moved = 0.322 x 190000 kg / 20.98 m 3.5 m = 3500 kg...cg displacement from 38% mac to 31% mac means that it must be displaced forward, therefore the shifting load 3500 kg must be added to the cargo 1 forward cargo and at the same time the same amount must be removed from cargo 4 aft cargo...after the cargo load redistribution, the cargo 1 will weigh 3000 kg + 3500 kg = 6500 kg and the cargo 4 will weigh 7000 kg 3500 kg = 3500 kg.
Question 91-4 : A loaded aircraft has its centre of gravity cg 58 inches forward of the datum, and its mass is 2906 lb. if a last minute load of 182 lb is added 23 inches aft of the datum, what is the new position of the cg. ?
53.23 inches forward of the datum.
Moment is the turning force created by the mass over a distance or lever arm moment = mass x arm...remember that, by convention, the items aft of the datum have a positive arm, and consequently a positive moment, and those which are forward of the datum have a negative arm and moment...generally, the centre of gravity cg can be found by adding all the moments and then dividing by the total mass cg = total moment / total mass...initially, the aircraft's moment was 2906 lb x 58 in. = 168548 lb in.. after adding the last minute load, the aircraft's moment became 168548 lb in. + 182 lb x 23 in. = 164362 lb in...thus, the new position of the cg, after loading the last minute load will be cg = total moment / total mass = 164362 lb in. / 2906 lb + 182 lb = 53.23 in....the minus symbol implies that the new cg position is 53.23 in. forward of the datum...or..alternatively, you can answer to this question using the formula mass change / new total mass = change cg / distance from mass to old cg...solving for change cg, you get change cg = mass change x distance from mass to old cg / new total mass = 182 lb x 58 in. + 23 in. / 2906 lb + 182 lb = 4.77 in....since the initial cg position is forward of the datum and the last minute load is added aft of the datum, the new position of cg will be shifted closer to the datum and will be 58 in. 4.77 in. = 53.23 in. forward of the datum.
Question 91-5 : A turbojet aeroplane is designed with swept back wings. it is fitted with a central fuel tank and, on each wing, an inner and outer fuel tank. a commercial air transport flight is operated with full fuel in tanks. at landing, all the fuel in the central tank is consumed. the landing cg position ?
Aft.
. what happens to the aircraft's centre of gravity cg as the fuel is burned, depends on the location of the fuel tanks. typically they are located in the wings and very near the cg, so shifts are minimal.in the case that a swept back wing aircraft has a central fuselage tank, additionally to the wing fuel tanks, when the fuel consumption begins from the central tank, as it is forward of a large part of the wing planform, the cg will progressively move aft, as the fuel is initially consumed from the central tank.despite the fuel consumption management, it is important that this aft cg is always within the forward and aft limits of the aircraft's cg envelope, as an excessively aft cg may lead to reduced longitudinal and/or directional stability, which is not desired during the approach and landing phases of flight.
Question 91-6 : The cg position of an aeroplane is 683.4 inches aft of the datum. given the following characteristics of the aeroplane's wing, calculate the cg position as % mac.leading edge of the wing root 575 inches aft of the datum. leading edge of the mac 598 inches aft of the datum. leading edge of the wing ?
51.95% mac
The centre of gravity cg is usually expressed as a distance relative to a datum or reference point. for the swept wing aircraft it is convenient, for aerodynamic purposes, to relate cg position with the mean aerodynamic chord mac of a wing. the mac is the chord at the aerodynamic centre of the wing, not exactly halfway to the tip, but close to it...the cg may be also expressed as a percentage of its position along the mac %mac from the leading edge, also called lemac and is given by the formula % mac = cg position lemac x 100 / length of mac...thus, % mac = 683.4 in. 598 in. x 100 / 164.38 in. = 51.95% mac...note ignore the leading edge wing root and wingtip data given, as the mean aerodynamic chord is neither at the wing root nor the wingtip.
Question 91-7 : An aeroplane should have been loaded so as to give a tom of 57 600 kg and a cg located 646 inches aft of datum cg limits are from 640 inches to 661 inches aft of datum. however, the ground crew incorrectly loaded 900 kg of baggage into the rear cargo hold instead of the front...if the distance ?
Yes, it is within limits.
It is vital that any incorrect loading, such as loading in reverse order, is picked up by a robust visual checking process. if deviation from the loading instructions exists, then the resulting effects upon aircraft weight/balance and take off trim settings should be determined and verified before take off...the following formula helps to find out by how much the cg has been moved, after loading the baggage of 900 kg... mass change / total mass = change of cg / distance moved....solving for change of cg we get change of cg = distance moved x mass change / total mass = 517 in. x 900 kg / 57 600 kg = 8.07 in...since, the 900 kg baggage was loaded by mistake into the rear cargo hold, instead of the front, then the cg will move further aft and its new position will be 646 in. + 8.07 in. = 654.07 in. aft of the datum...the aircraft's cg limits are between 640 in. and 661 in. aft of datum. thus, the aircraft's new cg position, 654.07 in. is within limits.
Question 91-8 : You are operating a light aircraft for humanitarian operations. today, you will transport food parcels of 10 kg each, stored in the cargo bay with a cg of 7.8 m aft of the datum line. before loading the cargo, the total mass of the aircraft is 3 645 kg and its cg is 5.62 m aft of the datum line...to ?
13 parcels
For this question the following formula helps to find out how much mass must be added to bring the cg to the aft limit of 5.7 m mass change / old total mass = change of cg / distance from mass to new cg.solving for the mass change, we get mass change = old total mass x change of cg / distance from mass to new cg = 3 645 kg x 0.08 m / 7.8 m 5.7 m = 138.8 kg.the food will be divided into parcels of 10 kg each. thus, the maximum number of food parcels that can be loaded is 138.8 kg / 10 kg = 13.88. therefore, the food parcels cannot be more than 13.
Question 91-9 : During dispatch, you perform the mass and balance calculations. given the following information, calculate the take off mass cg in inches aft of the datum.basic empty mass 2 635 lb. basic empty mass moment 204 397.5 in.lb. front seat occupant's mass 93 lb. front seat occupant's arm 78 in. aft seat ?
80.65
Moment is the turning force created by the mass over a distance or lever arm moment = mass x arm.generally, the centre of gravity cg can be found by adding all the moments and then dividing by the total mass cg = total moment / total mass. 1 calculate total mass . total mass = bem + front seat occupant's mass + aft seat occupant's mass + take off fuel . total mass = 2 635 lb + 93 lb + 186 lb + 850 320 lb = 3 444 lb 2 calculate the moments bem = 204 397.5 in.lb front seat occupant's moment = 93 lb x 78 in = 7 254 in.lb aft seat occupant's moment = 186 lb x 119 in = 22 134 in.lb take off fuel moment = 850 320 lb x 83 in = 43 990 in.lb 3 calculate total moment . total moment = 204 397.5 + 7 254 + 22 134 + 43 990 = 277 775.5 in.lb 4 calculate cg position . centre of gravity = total moment / total mass . centre of gravity = 277 775.5 lb.in / 3 444 = 80.65 in. aft of datumnote question asks for the take off mass cg, so the taxi fuel will be consumed before take off, therefore it must be subtracted from the block fuel.
Question 91-10 : The load sheet of a light aircraft shows that the tom is 4400 lb and the cg is located 95 in. aft of the datum...calculate the mass of bags that needs to be moved from the rear cargo compartment located 178 in. aft of the datum , to the forward cargo compartment 22 in. aft of the datum , in order to ?
56 lb
The centre of gravity must be displaced by 95 in. 93 in. = 2 inches...the mass of bags needs to be moved by 178 in. 22 in. = 156 inches...for this question the following formula helps to find out the mass of bags that needs to be moved from the rear cargo compartment to the forward cargo compartment, in order to re position the take off cg mass change / total mass = change of cg / distance moved...solving for mass change we get mass change = total mass x change of cg / distance moved = 4400 lb x 2 in. / 156 in. = 56.4 lb.
Question 91-11 : You are departing for a flight with a planned landing fuel of 1800 kg. according to aircraft limitations, the centre tank fuel must be used prior to the wing tanks fuel.given the following information, the location of the cg on landing will be how many metres aft of the datum maximum take off mass ?
15.78 m
These are the formulas we will need to solve this exercise moment = mass x arm..cg = total moment / total mass.for this question, we are looking for the cg on landing. thus, the moment and mass of the burnt fuel must be subtracted from the take off moment and take off mass correspondingly, so as to find the landing moment and landing mass.the cg on landing is given by the formula landing cg = landing moment / landing mass. 1 determine the landing moment landing moment = take off moment centre tank fuel moment wings tanks fuel burnt momenttake off moment = actual take off mass x take off cg locationtake off moment = 59 500 kg x 16.26 m = 967 470 kgmcentre tank fuel moment = fuel in centre tank x centre fuel tank centroidcentre tank fuel moment = 10 000 kg x 15.38 m = 153 800 kgmwing tanks fuel burnt moment = wing tanks fuel burnt at landing x wing fuel tank centroidwing tanks fuel burnt moment = 18 000 1800 kg x 17.79 m = 288 198 kgm.landing moment = 967 470 kgm 153 800 kgm 288 198 kgm = 525 472 kgm 2 determine the landing mass landing mass = actual take off mass fuel centre tank burnt wing tanks fuel burntlanding mass = 59 500 kg 10 000 kg 18 000 1800 kg = 33 300 kg 3 determine the landing cg landing cg = landing moment / landing masslanding cg = 525 472 kgm / 33 300 kg.landing cg = 15.78 m
Question 91-12 : How can you calculate the dry operating index doi ?
Dom x cg arm / constant
Moment is the turning force created by the mass acting over a distance or lever arm. it is calculated by multiplying the arm by the mass moment = mass x cg arm...generally, loadind index is a non dimensional figure that is a scaled down value of a moment and the effect of reducing the magnitude of the moment to one that is much easier to use. for large aircraft, it is convenient to use an index to represent the large numbers involved and to simplify the calculations...a loading index li is simply a moment load mass x cg arm divided by a constant li = load moment / constant = load mass x cg arm / constant...thus, the dry operating index doi is a dom moment dom x cg arm divided by a constant doi = dom moment / constant = dom x cg arm / constant...in simple words, doi is the index of the position of the centre of gravity at the dry operating mass dom.
Question 91-13 : One of the main advantages associated with fuel tanks in the aeroplane's horizontal stabiliser is to... ?
Keep the cg at the rear during the flight.
The fuel tanks will generally be located in the wings outer and inner and in a central position in the fuselage. but, they may also be found in the fin and the horizontal stabiliser...the centre of gravity cg position does change as fuel is burnt off, but within defined limits. if it is burnt from a tank aft of the aircraft's cg, then the cg will move forward. conversely, if the fuel is burnt from a tank positioned forward of the cg, then it will move aft...over the years, fuel tanks in the tail fin and horizontal stabiliser have been employed by different manufacturers for two main reasons... to provide extra fuel capacity. to maintain the aircraft cg position at the optimum aft position for minimum drag, improved manoeuvrability and, as a consequence, for maximum fuel economy, as less downforce on the horizontal stabiliser is required.
Question 91-14 : Given the following weighing data...mass on nose wheel 285 kg..mass on left main wheels 1465 kg..mass on right main wheels 1395 kg..distance between the nose and main wheels 7 ft 8 in...calculate the aircraft’s cg position. ?
83.7 in aft of the nosewheel.
With this type of question, you will need to assume the position of the datum. the 4 answers reference both the nose wheel and also the main wheels, so for simplicity, it is suggested to use the main wheels, as this reduces the amount of calculations required...the calculations of moments and ultimately, the centre of gravity cg can be seen in the table below. the datum has been selected in line with the main wheels and all arms are using inches. the question states that the distance between the nose and main wheels is 7 ft 8 inches, but it is normally easier to convert to inches 7'8' = 92 inches as calculations in feet can prove difficult...remember 1 foot = 12 inches...... . . item. mass kg. arm in.. moment kg.in.. . . nose wheel. 285. 92. 26 220. . . left main wheel. 1 465. 0. 0. . . right main wheel. 1 395. 0. 0. . . totals. 3 145. cg = 8.3 in.. 26 220. . .....therefore, the aircraft's cg position is 8.3 in forward of the main wheels or 83.7 in. aft of the nose wheel 92 in. 8.3 in....further information on moments, arms and the centre of gravity..the moment is the turning force created around a datum by the mass over a distance or lever arm moment = mass x arm...moments and arms forward of the datum are considered to be negative and those aft of the datum are positive...the centre of gravity cg can be found by adding all the moments and then dividing by the total mass cg = total moment / total mass.
Question 91-15 : A tri wheel aircraft has a datum 25 cm aft of the nosewheel. the weight on the nosewheel is 125 lb. the main gear is 180 cm aft of the datum. the distance between the nose gear and main gear is 205 cm. the weight on each main gear is 3 000 lb. calculate the bem and cg. ?
6 125 lb, cg 175.8 cm aft of the datum.
The moment is the turning force created around a datum by the mass over a distance or lever arm moment = mass x arm...moments and arms forward of the datum are by convention negative and those aft of the datum are positive...the centre of gravity cg can be found by adding all the moments and then dividing by the total mass cg = total moment / total mass...for this question, the following table would help... . . item. mass lb. arm cm. moment lb.cm. . . nose wheel. 125. 25. 3 125. . . main gear. 6 000.. 2 x 3 000. +180. +1 080 000. . . totals. 6 125 bem. cg = +175.8. +1 076 875. . .....therefore, the aircraft's cg position is either 175.8 cm aft of the datum, or 200.8 cm 175.8 cm + 25 cm aft of the nosewheel.
Question 91-16 : The loaded mass of an aircraft is 12 400 kg. the aft cg limit is 102 inches aft of the datum. if the cg as loaded is 104.5 inches aft of the datum, how many rows forward must two passengers move from the rear seat row 224 inches aft to bring the cg on to the aft limit, if the seat pitch is 33 inches ?
7 rows
The centre of gravity must be displaced by 104.5 in. – 102 in. = 2.5 in...the two passengers that must be moved weigh 75 kg x 2 = 150 kg...the following formula helps to find out how to redistribute part of the passenger load mass change / old total mass = change of cg / distance from mass to new cg...solving for 'distance from mass to new cg' we get distance from mass to new cg = change of cg x old total mass / mass change = 2.5 in. x 12 400 kg / 150 kg = 206.6 in...since each seat pitch is 33 in, then the two passengers must move from the rear seat row to bring the cg on to the aft limit at least by 206.6 in. / 33 in. = 6.26 rows...therefore, the passengers must move 7 rows to bring the cg to the aft limit.
Question 91-17 : The captain has received a calculated load sheet with his pre flight package. the ground handler now hands over a final load sheet. in this scenario, which one does the captain have to use ?
The final load sheet provided by the ground handler.
The load sheet is the document which records the weight and the centre of gravity settings of a commercial flight...the person supervising the loading of the aircraft shall confirm by hand signature, or equivalent, that the load and its distribution are in accordance with the mass and balance documentation given to the commander. also, the commander shall indicate his/her acceptance by hand signature, or equivalent...a new load sheet is required for every take off and an updated one before departure including any last minute change lmc...therefore, for this question, the commander must take into consideration the final and most updated load sheet provided by the ground handler, which may include lmcs.
Question 91-18 : According to the afm, the centre tank fuel has to be consumed prior to the wing tanks fuel. according to the pre flight fuel planning, the remaining fuel upon landing will be 2 000 kg. given the following information below, calculate your cg position upon landing..zero fuel mass 40 000 kg..cg ?
15.41 m
..step 1. as we are given the tom cg in the question, calculate the actual take off mass... zfm + take off fuel = take off mass tom. zfm + centre tank fuel + wing tank fuel = 40 000 + 28 000 + 26 000 = 94 000 kg tom...step 2. confirm the landing fuel from the question..remaining fuel upon landing will be 2 000 kg... fuel in centre tank at take off 28 000 kg > 0 kg on landing. fuel in wing tanks at take off 26 000 kg > 2000 kg remaining on landing...step 3. work backwards from take off mass to establish the landing mass using... mass x arm = moments. total moments / total mass = cg position.... . . item. mass kg. arm m. moments kg.m. . . take off mass. 94 000. 15.45. 1 452 300. . . centre tank fuel burn. 28 000. 15.3. 428 400. . . wing tank fuel burn. 24 000. 15.7. 376 800. . . landing mass. 42 000. 15.407. 647 100. . .....for the landing mass from the table above... landing moments 647 100 / landing mass 42 000 = landing cg position 15.407 = 15.41 m
Question 91-19 : The take off mass of an aircraft is 12 500 kg and the take off centre of gravity is located 4.2 m aft of datum, which is out of the forward limit. calculate the mass that must be moved from the forward cargo hold to the rear cargo hold, in order to move the cg to the forward limit 4.5 m aft of ?
250 kg
The cg is currently in front of the fwd limit and needs to move aft onto the limit at +4.5 m. this distance is +4.2 m to +4.5 m = 0.3 min order to move the cg into limits, we will need to move cargo from the forward compartment to the aft compartment.step 1. establish the parameters m = mass to be moved m = total mass 12 500 kg d = cg movement required 0.3 d = relevant distance, between the compartments given in the question as 15 m step 2. time for the formula m/m = d/dm/12 500 = 0.3/15m = 250 kgby moving 250 kg into the rear compartment, the cg is moved back into limits.
Question 91-20 : An aeroplane, with its cg limits from 638 inches to 658 inches aft of the datum, should have been loaded so as to give a calculated take off mass of 64 200 kg and the cg located at 640 inches aft of the datum, with 750 kg loaded into the front cargo hold and 500 kg into the rear cargo hold. ?
No, it is out of the forward limit.
It is vital that any incorrect loading, such as loading in reverse order, is picked up by a robust visual checking process. if deviation from the loading instructions exists, then the resulting effects upon aircraft weight/balance and take off trim settings should be determined and verified before take off...the following formula helps to find out by how much the cg has been moved, after loading the rear cargo 500 kg into the forward cargo hold mass change / total mass = change of cg / distance moved...solving for change of cg we get change of cg = distance moved x mass change / total mass = 517 in. x 500 kg / 64 200 kg = 4 in...since, the 500 kg baggage was loaded by mistake into the front cargo hold, instead of the rear, then the cg will move further forward and its new position will be 640 in. 4 in. = 636 in. aft of the datum...the aircraft's cg limits are between 638 in. and 658 in. aft of datum. thus, the aircraft's new cg position, 636 in. is out of the forward limit.
Question 91-21 : After the weighing procedure of an aircraft with the results given below, what is the cg position of the bem with reference to the main landing gear..nose landing gear jack point 1 850 kg..each main landing gear jack point 8 140 kg..longitudinal distance between the nose landing gear and the main ?
1.02 m forward
..moment is the turning force created by the mass over a distance or lever arm moment = mass x arm...remember that, by convention, the items aft of the datum have a positive arm, and consequently a positive moment, and those which are forward of the datum have a negative arm and moment...thus, since the nose landing gear jack point is forward of the datum main landing gears jack points , both its arm and moment will be negative...generally, the centre of gravity cg can be found by adding all the moments and then dividing by the total mass cg = total moment / total mass...the total mass of the aircraft is 1 850 kg + 2 x 8 140 kg = 18 130 kg..and its total moment 1 850 kg x 10 m + 2 x 8 140 kg x 0 = 18 500 kg.m...therefore, the cg position of the bem will be 18 500 kg.m / 18 130 kg = 1.02 m...the minus symbol implies that the cg position is 1.02 m forward of the main landing gears.
Question 91-22 : Prior to every commercial air transport flight of a large aircraft, the operator uses an electronic or a paper format trim sheet to determine the centre of gravity of the actual zero fuel mass, take off mass and landing mass. what is the purpose of the trim sheet ?
This is to ensure that the cgs are within the certified envelope, and is achieved by using the index method.
. load/trim sheet is a document which enables the pilot in command to determine that both the aircraft's load and its distribution throughout the fuselage are such that the mass and balance limits of the aircraft are not exceeded.the load/trim sheet is in two parts part a to the left is a loading summary. moreover, part a is divided into three sections. section 1 top is used to establish the limiting take off mass and the maximum allowabletraffic load that can be carried, section 2 middle shows the traffic load distribution and section 3 bottom sums up the load and cross checks that any limit has not been exceeded. part b to the right is the trim portion, which helps to establish the cg %mac and ensure that it lies within the certified envelope. it is where we plot graphically the positions of the cg of operational masses, such as take off mass, zero fuel mass and landing mass. when completing load sheets, particularly for large aircraft, it is convenient to use an index method to represent the large numbers involved and to simplify the calculations.generally, loading index is a non dimensional figure i.e. a figure without unit of measurement that is a scaled down value of a moment and the effect of reducing the magnitude of the moment to one that is much easier to use.
Question 91-23 : The basic empty mass bem of a light aircraft used for aerial photography is 2 415 lb and its centre of gravity is located 77.7 inches aft of datum. what will be the new cg of the bem, if a camera which weighs 190 lb is installed 108 in aft of datum ?
79.9 in
The moment is the turning force created around a datum by the mass over a distance or lever arm moment = mass x arm...moments and arms forward of the datum are by convention negative and those aft of the datum are positive...generally, the centre of gravity cg can be found by adding all the moments and then dividing by the total mass cg = total moment / total mass...the new basic empty mass bem of the aircraft, after installing the camera is 2 415 lb + 190 lb = 2 605 lb...and its new bem moment 2 415 lb x 77.7 in + 190 lb x 108 in = 208 165.5 lb.in...therefore, the cg position of the new bem will be 208 165.5 lb.in / 2 605 lb = 79.9 in. aft of the datum
Question 91-24 : To calculate the centre of gravity of a large aircraft, the practical use of the index method is to... ?
Reduce the size of the moment values and to simplify calculations.
When completing load sheets, particularly for large aircraft, it is convenient to use an index to represent the large numbers involved and to simplify the calculations...generally, loading index is a non dimensional figure that is a scaled down value of a moment and the effect of reducing the size of the moment to one that is much easier to use. for large aircraft, where large, awkward numbers are produced, it is convenient to use an index to represent the large numbers involved and to simplify the calculations...a loading index li is simply a moment load mass x cg arm divided by a constant li = load mass x cg arm / constant = load moment / constant.
Question 91-25 : An aircraft carrying wild animals, has four small cargo bays available for an animal cage. in favor of a wildlife operation, a caged wild animal of 144.75 kg has to be transported. before loading the wild animal, the aircraft had initially a take off mass of 6224.25 kg and a cg positioned 9.8 m aft ?
14.2 m
The following formulae help to find out in which cargo bay the animal has to be loaded so as to displace the cg by 0.1 m 9.9 m 9.8 m more aft of the datum... . mass change / old total mass = change of cg / distance between new mass and new cg. ...solving for distance between new mass and new cg, we get distance between new mass and new cg = old total mass x change of cg / mass change = 6224.25 kg x 0.1 m / 144.75 kg = 4.3 m...thus, the animal must loaded at a cargo bay new cg + distance between new mass and new cg = 9.9 m + 4.3 m = 14.2 m aft of the datum...or... . mass change / new total mass = change of cg / distance between new mass and old cg. ...solving for distance between new mass and old cg, we get distance between new mass and old cg = new total mass x change of cg / mass change = 6224.25 kg + 144.75 kg x 0.1 m / 144.75 kg = 4.4 m...thus, the animal must loaded at a cargo bay old cg + distance between new mass and old cg = 9.8 m + 4.4 m = 14.2 m aft of the datum.
Question 91-26 : If the cg of an aircraft is close to its forward limit, what will be achieved if fuel is transferred from the wing tanks to the aeroplane’s fin or horizontal stabiliser tanks ?
Efficient fuel consumption, by repositioning the cg aft.
This question is asking about the effect on performance with a rearwards cg position. the aft cg is achieved by fuel being transferred from the wing tanks to the aeroplane’s fin or horizontal stabiliser tanks.with the decrease in trim drag and also less main wing lift being required, the effects include lower stalling speeddecreased fuel consumptionincreased range and enduranceincreased climb performancedecreased stabilitydecrease in take off and landing speeds
Question 91-27 : A loaded aircraft has a tom of 3800 lb. the take off cg is at 95 inches, which is out of the rear limit. how many pounds of ballast need to be placed in the forward cargo compartment, which is located at 22.5 inches, to move the cg to 93 inches ?
108 lb
Current situation the take off cg is at 95 inches out of the rear limits..we need to move the cg to 93 inches, 2 inches forward.how may we bring the cg forward by 2 inches to meet the cg aft limit without re arranging the passengers we will use an extra ballast mass forward of the aircraft forward cargo to bring the cg to the aft limit.how much extra ballast the take off cg is at 95 inches and the tom is 3800 lb.old total moment = 3800 x 95 lb.into bring the cg forward to within limits, a ballast 'm' must be added to the forward cargo. the forward cargo is located at 22.5 inches.ballast moment = m x 22.5 lb.inthe new total mass will be new total mass = 3800 lb + mthis will result in a new total moment new total moment = 3800 lb + m x 93 lb.innow apply the formula old total moment + moment of the ballast = new total moment3800 x 95 + m x 22.5 = 3800 + m x 93.3800 x 95 + m x 22.5 = 3800 x 93 + m x 93.361 000 + 22.5 m = 353 400 + 93 m.7600 = 70.5 m.m = 108 lbalternatively, simply apply the following formula new mass x new cg = old mass x old cg ± mass x arm 3800 + m x 93 = 3800 x 95 + m x 22.5.353 400 + 93 m = 361 000 + 22.5 m.70.5 m = 7600.m = 107.8 lb 108 lb
Question 91-28 : Calculate the centre of gravity location for the take off mass, given bem 2 635 lb. bem moment 204 793.5 in.lb. mass on front seat 90 lb. arm on front seat 78 in. mass on aft seat 186 lb. arm on aft seat 117 in. block fuel 850 lb. taxi fuel 230 lb. fuel arm 86 in ?
81.25 inches aft of datum
Formulas needed. center of gravity = total moment / total mass. moment = mass x arm 1 calculate total mass. total mass = bem + front seat + aft seat + take off fuel. total mass = 2 635 lb + 90 lb + 186 lb + 850 lb 230 lb = 3 531 lb 2 calculate the moments bem = 204 793.5 in.lb front seat = 90 lb x 78 in = 7 020 in.lb aft seat = 186 lb x 117 in = 21 762 in.lb take off fuel = 620 lb x 86 in = 53 320 in.lb 3 calculate total moment. total moment = 204 793.5 + 7 020 + 21 762 + 53 320 = 286 895.5 lb.in 4 calculate cg position. centre of gravity = total moment / total mass. centre of gravity = 286 895.5 / 3 531 = 81.25 in aft of datumnote question asks for centre of gravity at take off taxi fuel will be consumed before take off, therefore is must be subtracted from our calculations.
Question 91-29 : How can the dry operating index doi be calculated ?
Doi = sum of the moments of the dom / 12 000
The clues for the examiner's thinking come from cap 696 ‘dry operating index doi is the index for the position of the centre of gravity at dry operating mass.’‘loading index li is a non dimensional figure that is a scaled down value of a moment. it is used to simplify mass and balance calculations.’so putting it together the dry operating index doi is the dom moment dom x cg arm divided by a constant.in the answer 'doi = sum of the moments of the dom / 12 000' we have to assume that 12 000 is the constant.important note this option has been confirmed as correct by students who have taken official exams. if you encounter this question in your exam, please inform us.
Question 91-30 : Given the following information, determine the take off mass cg...basic empty mass 2 635 lb and its associated moment 204 379.5 in.lb. forward seat passenger 93 lb and the balance arm 79 in. aft seat passenger 180 lb and the balance arm 119 in. block fuel 830 lb. taxi fuel 230 lb. fuel tank ?
81.17 in
Formulas needed . center of gravity = total moment / total mass. moment = mass x arm 1 calculate total mass . total mass = bem + front seat + aft seat + take off fuel. total mass = 2 635 lb + 93 lb + 180 lb + 600 lb = 3 508 lb 2 calculate the moments bem = 204 379.5 in.lb front seat = 93 lb x 79 in = 7 347 in.lb aft seat = 180 lb x 119 in = 21 420 in.lb take off fuel = 600 lb x 86 in = 51 600 in.lb 3 calculate total moment . total moment = 204 379.5 + 7 347 + 21 420 + 51 600 = 284 746.5 lb.in 4 calculate cg position . centre of gravity = total moment / total mass. centre of gravity = 284 746.5 / 3 508 = 81.17 in aft of datumnote question asks for centre of gravity at take off taxi fuel will be consumed before take off, therefore is must be subtracted from our calculations.
Question 91-31 : When preparing for the flight, you note that the take off cg is 8 cm beyond the cg limit. the aircraft weight is 7 132 kg. a 320 kg container in the hold is moved to bring the cg within limits... what is the minimum value of this load shift to the aircraft ?
1.783 m
Mass change / old total mass = change of cg / distance from mass to new cg...solving for distance from mass to new cg we get... distance from mass to new cg = change of cg x old total mass / mass change.....distance from mass to new cg = 8 cm x 7 132 kg / 320 kg..= 178.3 cm 1.783 m
Question 91-32 : After the weighing procedure of an aircraft with the results given below, what is the cg position of the bem with reference to the main landing gear nose landing gear jack point 1 850 kg. each main landing gear jack point 8 140 kg. longitudinal distance between the nose landing gear and the main ?
1.24 m forward
Moment is the turning force created by the mass over a distance or lever arm moment = mass x arm...remember that, by convention, the items aft of the datum have a positive arm, and consequently a positive moment, and those which are forward of the datum have a negative arm and moment...thus, since the nose landing gear jack point is forward of the datum main landing gears jack points , both its arm and moment will be negative...generally, the centre of gravity cg can be found by adding all the moments and then dividing by the total mass cg = total moment / total mass...the total mass of the aircraft is 1 850 kg + 2 x 8 140 kg = 18 130 kg..and its total moment 1 850 kg x 12.2 m + 2 x 8 140 kg x 0 = 22 570 kg.m...therefore, the cg position of the bem will be 22 570 kg.m / 18 130 kg = 1.24 m...the minus symbol implies that the cg position is 1.24 m forward of the main landing gears.
Question 91-33 : An aircraft has its datum at the nose. the front seats are 65 inches aft of datum, the passenger seats are 105 inches aft and the separate baggage compartment is 145 inches aft. the aft limit is 157 inches aft. the current mass is 2570 lb...three passengers are in the rear seats, with the baggage ?
84 lb
The cg is currently aft of the cg limit and needs to move forward by a distance of 3 inches.in order to move the cg into limits, we will need to load ballast onto the front seats.the following is extra and useless information from the examiner the passenger seats at 105 inches aft andthe separate baggage compartment at 145 inches aftstep 1. establish the parameters m = mass to be added m = total mass 2570 lb d = cg movement required 3 inches d = relevant distance, between the target cg position and the front seats being loaded distance between +65 and +157 = 92 step 2. time for the formula m/m = d/dm/2570 = 3/92m = 83.8 lb > 84 lbby adding 84 lb onto the front seats, the cg is moved back into limits.
Question 91-34 : Determine the mass that must be moved aft by a distance of 2.9 m, in order to establish a cg at 2.9 m aft of the datum. given current cg position 2.8 m aft of the datum. actual take off mass 6 030 kg ?
208 kg
The new centre of gravity position can be calculated using the following formula m x d = m x d. where m stands for the total mass of the aircraft, d is the distance between the previous and new cg, m is the moved mass and d stands for the distance between the old and new position of the moved mass.6 030 kg x 0.1 = m x 2.9. m = 603 / 2.9. m = 208 kg
Question 91-35 : The crew of an aircraft calculates that the take off mass is 16 800 kg, and the cg is found at station 260, which is out of limits. what is the mass of baggage that needs to be moved from the rear cargo bay located at station 302 to the forward cargo bay station 40 , in order to bring the take off ?
128 kg
Calculate the cg movement distance required. cg movement distance required = 260 258 = 2.. distance between rear and forward cargo bay 302 40 = 262mass change / total mass = change of cg / distance moved. mass change / 16 800 kg = 2 / 262. mass change = 2 x 16 800 kg / 262 m. mass change = 128 kg approx.
Question 91-36 : An aircraft has a total mass of 7 220 kg with a cg of 4.16 m aft of the datum line. a few minutes before dispatch, an additional load of 170 kg is loaded onto the aircraft. with the aircraft's new cg at 4.31 m, where should the additional load be placed ?
10.68 m aft of the datum line.
Mass change / new total mass = change of cg / distance moved... mass change 170 kg. new total mass 7 220 kg + 170 kg = 7 390 kg. change of cg 4.31 – 4.16 = 0.15 m. distance moved ...170 kg / 7 390 kg = 0.15 m / distance moved..distance moved = 0.15 m x 7 390 kg / 170 kg..distance moved = 6.52 m..=> to move the cg aft, we must add the mass 170 kg further back from the original position. therefore, 4.16 m + 6.52 m = 10.68 m.
Question 91-37 : Determine the mass that must be moved aft by a distance of 2.8 m, in order to establish a cg at 2.8 m aft of the datum, given current cg position 2.7 m. aft of the datum actual take off mass 6 630 kg ?
237 kg
The new centre of gravity position can be calculated using the following formula m x d = m x d. where m stands for the total mass of the aircraft, d is the distance between the previous and new cg, m is the moved mass and d stands for the distance between the old and new position of the moved mass.in this case, the total mass of the aircraft 'm' is 6 630 kg. the distance between the previous and new cg ' d' equals 2.8 m 2.7 m = 0.1 m. and the mass must be moved aft d by a distance of 2.8 m. thus 6 630 kg x 0.1 = m x 2.8. m = 663 / 2.8. m = 237 kg
Question 91-38 : An aircraft has its datum at the nose. the front seats are 65 inches aft of datum, the passenger seats are 105 inches aft and the separate baggage compartment is 145 inches aft. the aft limit is 149 inches aft. the current mass is 2570 lb...three passengers are in the rear seats, with the baggage ?
92 lb
.current situation.the aft cg limit is 149 inches aft of the datum..however, due to the current arrangement of passengers, the cg exceeds the aft cg limits by 3 inches making the current cg at 152 inches aft of the datum.how may we bring the cg forward by 3 inches to meet the cg aft limit without re arranging the passengers we will use an extra ballast mass forward of the aircraft front seats to bring the cg to the aft limit.how much extra ballast.cg is 3'' behind the aft limit. therefore, the cg is 152'' aft of datum.old total moment = 2570 x 152 lb.into bring the cg forward to within limits, a ballast 'm' must be added to the front seat. the front seat is located at 65'' aft of datum.ballast moment = m x 65 lb.inthe new total mass will be new total mass = 2570 + mthis will result in a new total moment new total moment = 2570 + m x 149 lb.innow apply the formula old total moment + moment of the ballast = new total moment..2570 x 152 + m x 65 = 2570 + m x 149.2570 x 152 + m x 65 = 2570 x 149 + m x 149.m = 91.8 lb.alternatively, simply apply the following formula.new mass x new cg = old mass x old cg ± mass x arm. 2570 + m x 149 = 2570 x 152 + 65 x m.382 930 + 149 m = 390 640 + 65 m.84 m = 7710.m = 91.8 lb 92 lb
Question 91-39 : For mass and balance calculations, an index is given which is determined by... ?
Dividing the sum of the moments by a factor, to reduce its value.
Moment. the turning force created by the mass acting over a distance or lever arm and is calculated by multiplying the arm by the mass....where large, awkward numbers are produced, the moments are sometimes divided by a constant to produce a moment index. it is described as a non dimensional figure, that is a scaled down value of a moment and is used to simplify mass and balance calculations...a loading index li is simply a moment load mass x cg arm divided by a constant li = load mass x cg arm / constant = load moment / constant...for instance, a moment of 126 400 kg.in, may be referred to as an index of 1 264, divided by 100. therefore, index is obtained by dividing the sum of the moments by a factor, in order to reduce its value.
Question 91-40 : An aircraft has a total mass of 1 508 kg with a cg of 3.68 m aft of the datum line. before departure, the ground crew discovers that they overlooked loading some additional cargo onto the aircraft, at 4.82 m aft of datum. to ensure the aircraft stays within the aft cg limit of 3.79 m, what is the ?
161 kg
Mass change / new total mass = change of cg / distance from mass to old cg... mass change . new total mass 1 508 kg + mass change. change of cg 3.79 – 3.68 = 0.11 m. distance from mass to old cg 4.82 3.68 = 1.14 m...mass change / 1 508 kg + mass change = 0.11 m / 1.14 m = 0.096..mass change = 0.096 x 1 508 kg + mass change = 0.096 x 1 508 kg + 0.096 x mass change..mass change 0.096 x mass change = 145 kg..mass change x 1 0.096 = 145 kg..0.9 x mass change = 145 kg mass change = 145 kg / 0.9 = 161 kg..maximum mass change for cg aft limit = 161 kg.. to move the cg to its aft limit, we must add a mass of 161 kg at a station 4.82 m aft of datum....note to make these caclulations, familiarity with the following mathematic rule is needed.... a x b + c = a x b + a x c
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