Question 91-1 : An aircraft has 100 lb of ballast installed in the forward cargo hold located 225 inches aft of the datum the load and balance calculation gives a tom of 4 000 lb and the cg at 85 inches aft of the datumwhat should the pilot do to the ballast to move the cg to 86 inches aft of the datum ? [ Level reports ]

Question 91-2 : 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 datumwhen the passenger arrives heshe sits at the rear seat ?

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 instep 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 instep 2 time for the formula mm = dd1704100 = 73cg movement = +303 inthe cg will move aft by 303 in to a new position of 9703 in and is thus out of limits

Question 91-3 : Determine the mass that must be moved aft by a distance of 29 m in order to establish a cg at 29 m aft of the datum given current cg position 28 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 01 m = m x 29m = 6603 29m = 2277 kg 228 kg

Question 91-4 : 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 350 m from reference point cargo 4 7 000 kg 2098 m from reference point distance from reference point to leading edge 14m length of mac 46 monce the cargo loading is completed ?

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 46 mac = 0322 mthe 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 movedsolving for mass change we get mass change = change of cg x total mass total distance moved = 0322 x 190000 kg 2098 m 35 m = 3500 kgcg 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 91-5 : 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 ?

5323 inches forward of the datum.

Moment is the turning force created by the mass over a distance or lever arm moment = mass x armremember 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 momentgenerally 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 inthus 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 = 5323 inthe minus symbol implies that the new cg position is 5323 in forward of the datumoralternatively you can answer to this question using the formula mass change new total mass = change cg distance from mass to old cgsolving 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 = 477 insince 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 477 in = 5323 in forward of the datum

Question 91-6 : 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 compared to ?

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 minimalin 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 tankdespite 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 andor directional stability which is not desired during the approach and landing phases of flight

Question 91-7 : The cg position of an aeroplane is 6834 inches aft of the datum given the following characteristics of the aeroplane's wing calculate the cg position as % macleading 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 tip ?

5195% 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 itthe 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 macthus % mac = 6834 in 598 in x 100 16438 in = 5195% macnote 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 91-8 : 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 frontif the distance between ?

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 weightbalance and take off trim settings should be determined and verified before take offthe 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 movedsolving for change of cg we get change of cg = distance moved x mass change total mass = 517 in x 900 kg 57 600 kg = 807 insince 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 + 807 in = 65407 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 65407 in is within limits

Question 91-9 : 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 78 m aft of the datum line before loading the cargo the total mass of the aircraft is 3 645 kg and its cg is 562 m aft of the datum lineto be within ?

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 57 m mass change old total mass = change of cg distance from mass to new cgsolving 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 008 m 78 m 57 m = 1388 kgthe food will be divided into parcels of 10 kg each thus the maximum number of food parcels that can be loaded is 1388 kg 10 kg = 1388 therefore the food parcels cannot be more than 13

Question 91-10 : During dispatch you perform the mass and balance calculations given the following information calculate the take off mass cg in inches aft of the datumbasic empty mass 2 635 lb basic empty mass moment 204 3975 inlb front seat occupant's mass 93 lb front seat occupant's arm 78 in aft seat ?

8065.

Moment is the turning force created by the mass over a distance or lever arm moment = mass x armgenerally 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 3975 inlb front seat occupant's moment = 93 lb x 78 in = 7 254 inlb aft seat occupant's moment = 186 lb x 119 in = 22 134 inlb take off fuel moment = 850 320 lb x 83 in = 43 990 inlb 3 calculate total moment total moment = 204 3975 + 7 254 + 22 134 + 43 990 = 277 7755 inlb 4 calculate cg position centre of gravity = total moment total mass centre of gravity = 277 7755 lbin 3 444 = 8065 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 91-11 : The load sheet of a light aircraft shows that the tom is 4400 lb and the cg is located 95 in aft of the datumcalculate 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 re ?

56 lb.

The centre of gravity must be displaced by 95 in 93 in = 2 inchesthe mass of bags needs to be moved by 178 in 22 in = 156 inchesfor 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 movedsolving for mass change we get mass change = total mass x change of cg distance moved = 4400 lb x 2 in 156 in = 564 lb

Question 91-12 : 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 fuelgiven the following information the location of the cg on landing will be how many metres aft of the datum maximum take off mass 99 ?

1578 m.

These are the formulas we will need to solve this exercise moment = mass x arm cg = total moment total massfor 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 massthe 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 1626 m = 967 470 kgmcentre tank fuel moment = fuel in centre tank x centre fuel tank centroidcentre tank fuel moment = 10 000 kg x 1538 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 1779 m = 288 198 kgmlanding 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 = 1578 m

Question 91-13 : 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 armgenerally 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 calculationsa 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 constantthus 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 constantin simple words doi is the index of the position of the centre of gravity at the dry operating mass dom

Question 91-14 : One of the main advantages associated with fuel tanks in the aeroplane's horizontal stabiliser is to ?

Question 91-15 : Given the following weighing datamass 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 incalculate the aircraft’s cg position ?

837 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 requiredthe 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 difficultremember 1 foot = 12 inches item mass kg arm in moment kgin nose wheel 285 92 26 220 left main wheel 1 465 0 0 right main wheel 1 395 0 0 totals 3 145 cg = 83 in 26 220 therefore the aircraft's cg position is 83 in forward of the main wheels or 837 in aft of the nose wheel 92 in 83 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 armmoments and arms forward of the datum are considered to be negative and those aft of the datum are positivethe 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 91-16 : 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 1758 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 armmoments and arms forward of the datum are by convention negative and those aft of the datum are positivethe centre of gravity cg can be found by adding all the moments and then dividing by the total mass cg = total moment total massfor this question the following table would help item mass lb arm cm moment lbcm nose wheel 125 25 3 125 main gear 6 000 2 x 3 000 +180 +1 080 000 totals 6 125 bem cg = +1758 +1 076 875 therefore the aircraft's cg position is either 1758 cm aft of the datum or 2008 cm 1758 cm + 25 cm aft of the nosewheel

Question 91-17 : 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 1045 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 1045 in – 102 in = 25 inthe two passengers that must be moved weigh 75 kg x 2 = 150 kgthe 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 = 25 in x 12 400 kg 150 kg = 2066 insince 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 2066 in 33 in = 626 rowstherefore the passengers must move 7 rows to bring the cg to the aft limit

Question 91-18 : 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 flightthe 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 hisher acceptance by hand signature or equivalenta 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 91-19 : 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 position at ?

1541 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 tomstep 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 landingstep 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 kgm take off mass 94 000 1545 1 452 300 centre tank fuel burn 28 000 153 428 400 wing tank fuel burn 24 000 157 376 800 landing mass 42 000 15407 647 100 for the landing mass from the table above landing moments 647 100 landing mass 42 000 = landing cg position 15407 = 1541 m

Question 91-20 : The take off mass of an aircraft is 12 500 kg and the take off centre of gravity is located 42 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 45 m aft of datum the ?

250 kg.

The cg is currently in front of the fwd limit and needs to move aft onto the limit at +45 m this distance is +42 m to +45 m = 03 min order to move the cg into limits we will need to move cargo from the forward compartment to the aft compartmentstep 1 establish the parameters m = mass to be moved m = total mass 12 500 kg d = cg movement required 03 d = relevant distance between the compartments given in the question as 15 m step 2 time for the formula mm = ddm12 500 = 0315m = 250 kgby moving 250 kg into the rear compartment the cg is moved back into limits

Question 91-21 : 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 weightbalance and take off trim settings should be determined and verified before take offthe 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 movedsolving 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 insince 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 datumthe 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 91-22 : 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 ?

102 m forward.

moment is the turning force created by the mass over a distance or lever arm moment = mass x armremember 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 momentthus since the nose landing gear jack point is forward of the datum main landing gears jack points both its arm and moment will be negativegenerally the centre of gravity cg can be found by adding all the moments and then dividing by the total mass cg = total moment total massthe total mass of the aircraft is 1 850 kg + 2 x 8 140 kg = 18 130 kgand its total moment 1 850 kg x 10 m + 2 x 8 140 kg x 0 = 18 500 kgmtherefore the cg position of the bem will be 18 500 kgm 18 130 kg = 102 mthe minus symbol implies that the cg position is 102 m forward of the main landing gears

Question 91-23 : 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.

loadtrim 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 exceededthe loadtrim 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 calculationsgenerally loading index is a non dimensional figure ie 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 91-24 : The basic empty mass bem of a light aircraft used for aerial photography is 2 415 lb and its centre of gravity is located 777 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 ?

799 in.

The moment is the turning force created around a datum by the mass over a distance or lever arm moment = mass x armmoments and arms forward of the datum are by convention negative and those aft of the datum are positivegenerally the centre of gravity cg can be found by adding all the moments and then dividing by the total mass cg = total moment total massthe new basic empty mass bem of the aircraft after installing the camera is 2 415 lb + 190 lb = 2 605 lband its new bem moment 2 415 lb x 777 in + 190 lb x 108 in = 208 1655 lbintherefore the cg position of the new bem will be 208 1655 lbin 2 605 lb = 799 in aft of the datum

Question 91-25 : 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 calculationsgenerally 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 calculationsa 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 91-26 : 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 14475 kg has to be transported before loading the wild animal the aircraft had initially a take off mass of 622425 kg and a cg positioned 98 m aft of the ?

142 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 01 m 99 m 98 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 = 622425 kg x 01 m 14475 kg = 43 m thus the animal must loaded at a cargo bay new cg + distance between new mass and new cg = 99 m + 43 m = 142 m aft of the datumor 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 = 622425 kg + 14475 kg x 01 m 14475 kg = 44 mthus the animal must loaded at a cargo bay old cg + distance between new mass and old cg = 98 m + 44 m = 142 m aft of the datum

Question 91-27 : 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 tankswith 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 91-28 : 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 225 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 forwardhow 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 limithow much extra ballast the take off cg is at 95 inches and the tom is 3800 lbold total moment = 3800 x 95 lbinto bring the cg forward to within limits a ballast 'm' must be added to the forward cargo the forward cargo is located at 225 inchesballast moment = m x 225 lbinthe 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 lbinnow apply the formula old total moment + moment of the ballast = new total moment3800 x 95 + m x 225 = 3800 + m x 93 3800 x 95 + m x 225 = 3800 x 93 + m x 93 361 000 + 225 m = 353 400 + 93 m 7600 = 705 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 225 353 400 + 93 m = 361 000 + 225 m 705 m = 7600 m = 1078 lb 108 lb

Question 91-29 : Calculate the centre of gravity location for the take off mass given bem 2 635 lb bem moment 204 7935 inlb 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 ?

8125 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 7935 inlb front seat = 90 lb x 78 in = 7 020 inlb aft seat = 186 lb x 117 in = 21 762 inlb take off fuel = 620 lb x 86 in = 53 320 inlb 3 calculate total moment total moment = 204 7935 + 7 020 + 21 762 + 53 320 = 286 8955 lbin 4 calculate cg position centre of gravity = total moment total mass centre of gravity = 286 8955 3 531 = 8125 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 91-30 : 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 constantin the answer 'doi = sum of the moments of the dom 12 000' we have to assume that 12 000 is the constantimportant 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 91-31 : Given the following information determine the take off mass cgbasic empty mass 2 635 lb and its associated moment 204 3795 inlb 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 balance arm ?

8117 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 3795 inlb front seat = 93 lb x 79 in = 7 347 inlb aft seat = 180 lb x 119 in = 21 420 inlb take off fuel = 600 lb x 86 in = 51 600 inlb 3 calculate total moment total moment = 204 3795 + 7 347 + 21 420 + 51 600 = 284 7465 lbin 4 calculate cg position centre of gravity = total moment total mass centre of gravity = 284 7465 3 508 = 8117 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 91-32 : 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 ?

1783 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= 1783 cm 1783 m

Question 91-33 : 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 ?

124 m forward.

Moment is the turning force created by the mass over a distance or lever arm moment = mass x armremember 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 momentthus since the nose landing gear jack point is forward of the datum main landing gears jack points both its arm and moment will be negativegenerally the centre of gravity cg can be found by adding all the moments and then dividing by the total mass cg = total moment total massthe total mass of the aircraft is 1 850 kg + 2 x 8 140 kg = 18 130 kgand its total moment 1 850 kg x 122 m + 2 x 8 140 kg x 0 = 22 570 kgmtherefore the cg position of the bem will be 22 570 kgm 18 130 kg = 124 mthe minus symbol implies that the cg position is 124 m forward of the main landing gears

Question 91-34 : 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 lbthree 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 inchesin order to move the cg into limits we will need to load ballast onto the front seatsthe 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 mm = ddm2570 = 392m = 838 lb > 84 lbby adding 84 lb onto the front seats the cg is moved back into limits

Question 91-35 : Determine the mass that must be moved aft by a distance of 29 m in order to establish a cg at 29 m aft of the datum given current cg position 28 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 mass6 030 kg x 01 = m x 29 m = 603 29 m = 208 kg

Question 91-36 : 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 cg to ?

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 91-37 : An aircraft has a total mass of 7 220 kg with a cg of 416 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 431 m where should the additional load be placed ?

1068 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 431 – 416 = 015 m distance moved 170 kg 7 390 kg = 015 m distance moveddistance moved = 015 m x 7 390 kg 170 kgdistance moved = 652 m=> to move the cg aft we must add the mass 170 kg further back from the original position therefore 416 m + 652 m = 1068 m

Question 91-38 : Determine the mass that must be moved aft by a distance of 28 m in order to establish a cg at 28 m aft of the datum given current cg position 27 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 massin this case the total mass of the aircraft 'm' is 6 630 kg the distance between the previous and new cg ' d' equals 28 m 27 m = 01 m and the mass must be moved aft d by a distance of 28 m thus 6 630 kg x 01 = m x 28 m = 663 28 m = 237 kg

Question 91-39 : 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 lbthree passengers are in the rear seats with the baggage ?

92 lb.

Current situation the aft cg limit is 149 inches aft of the datumhowever 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 limithow much extra ballast cg is 3'' behind the aft limit therefore the cg is 152'' aft of datumold total moment = 2570 x 152 lbinto 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 datumballast moment = m x 65 lbinthe new total mass will be new total mass = 2570 + mthis will result in a new total moment new total moment = 2570 + m x 149 lbinnow apply the formula old total moment + moment of the ballast = new total moment2570 x 152 + m x 65 = 2570 + m x 1492570 x 152 + m x 65 = 2570 x 149 + m x 149m = 918 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 m84 m = 7710m = 918 lb 92 lb

Question 91-40 : 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 masswhere 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 calculationsa 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 constantfor instance a moment of 126 400 kgin 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