Question: Intentional re-positioning of CG

► Adding Mass ► Ch Mass / New Total Mass = Ch CG / Distance Mass & (old) CG
► Removing Mass ► Ch Mass / Old Total Mass = Ch CG / Distance Mass & (new) CG
► Moving Mass ► Ch Mass / Total Mass = Ch CG / Distance Moved

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► A mass of 500kg is loaded at a station which is located 10 metres behind the present Centre of Gravity and 16 metres behind the datum. (assume g=10 m/s^2) The moment for that mass used in the loading manifest is

A) 30 000 Nm
B) 50 000 Nm
C) 80 000 Nm
D) 130 000 Nm

500kg x 10 = 5000N

Moment = Arm x Mass
Moment = 16m x 5000N
Moment = 80 000 Nm

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► Given: total mass 2900 kg centre of gravity (cg) location station: 115.0, Aft cg limit station: 116.0 The maximum mass that can be added at station 130.0 is:

A) 140 kg
B) 317 kg
C) 207 kg
D) 14 kg

Adding Mass -> Ch Mass / New Total Mass = Ch CG / Distance Mass & CG

New T Mass = 2900kg
Ch CG = 1 (From station 115 to 116)
Distance mass and new CG = 14 (130 to 116)

Ch Mass / 2900kg = 1 / 14
Ch Mass = 207 kg

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► Given: Total mass: 7500kg centre of gravity (cg) location station: 80.5, Aft cg limit station 79.5 How much cargo must be shifted from the aft cargo compartment at station 150 to the forward cargo compartment at station 30 in order to move the cg location to the aft limit?

A) 65.8 kg
B) 62.5 kg
C) 68.9 kg
D) 73.5 kg

Moving Mass -> Ch Mass / Total Mass = Ch CG / Distance moved

Total mass = 7500kg
Ch CG = 1 (station 80.5 to 79.5)
Distance moved = 120 (from aft cargo 150 to forward cargo 30)

Ch Mass / 7500 = 1 / 120
Ch Mass = 62.5 kg

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► The total mass of an aeroplane is 9000 kg. The centre of gravity (cg) position is at 2.0 m from the datum line. The aft limit for cg is at 2.1 m from the datum line. What mass of cargo must be shifted from the front cargo hold (at 0.8m from the datum) to the aft hold (at 3.8m), to move the cg to the aft limit?

A) 30.0 kg
B) 900 kg
C) 300 kg
D) 196 kg

Moving Mass -> Ch Mass / Total Mass = Ch CG / Distance moved

Total mass = 9000kg
Ch CG = 0.1 (station 2.0 to 2.1)
Distance moved = 3 (from forward cargo 0.8 to aft cargo 3.8 )

Ch Mass / 9000 = 0.1 / 3
Ch Mass = 300 kg

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► Given: Aeroplane mass = 36 000 kg Centre of gravity (cg) is located at station 17 m. What is the effect on cg location if you move 20 passengers (total mass = 1 600) from station 16 to station 23?

A) it moves aft by 3.22 m
B) it moves forward by 0.157 m
C) it moves aft by 0.31 m
D) it moves aft by 0.157 m

Moving Mass -> Ch Mass / Total Mass = Ch CG / Distance moved

Ch Mass = 1 600kg
Total Mass = 36 000kg
Distance moved = 7 (station 16 to 23)

1 600 / 36 000 = Ch CG / 7
Ch CG = 0.31 (Positive so aft)

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► The mass of an aeroplane is 1950 kg. If 450 kg is added to a cargo hold 1.75 metres from the loaded centre of gravity (cg). The loaded cg will move:

A) 40 cm
B) 33 cm
C) 30 cm
D) 34 cm

Adding Mass -> Ch Mass / New Total Mass = Ch CG / Distance Mass & CG

Ch Mass = 450kg
New total mass = 2400
Distance mass and new CG = 1.75

450k / 2400 = Ch CG / 1.75
Ch Mass = 0.328m (= 33cm)

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► Length of the mean aerodynamic chord = 1 m. Moment arm of the forward cargo: -0.50m , moment arm of the aft cargo: +2.50m . The aircraft mass is 2 200kg and its centre of gravity is at 25% MAC. To move the centre of gravity to 40%, which mass has to be transferred from the forward to the aft cargo hold?

A) 110 kg
B) 183 kg
C) 165 kg
D) 104 kg

Moving Mass -> Ch Mass / Total Mass = Ch CG / Distance moved

Total mass = 2 200kg
Ch CG = 0.15 (15% shift from 25% to 40%)
Distance moved = 3 (2.5 aft to 0.5 forward)

Ch Mass / 2200 = 0.15 / 3
Ch Mass = 110 kg

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► Assume: Aeroplane gross mass: 4750 kg centre of gravity at station: 115.8 What will be the new Position of the centre of gravity if 100 kg is moved from the station 30 to station 120?

A) station 118.25
B) station 118.33
C) station 120.22
D) station 117.69

Moving Mass -> Ch Mass / Total Mass = Ch CG / Distance moved

Ch Mass = 100kg
Total mass = 4750kg
Distance moved = 90 (30 to 120)

100 / 4750 = Ch CG / 90
Ch CG = 1.89

We moved the CG aft so 115.8 + 1.89 = 117.69

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► Given: Actual mass: 116 500 lbs
Original CG station: 435.0
Compartment A station: 285.5
Compartment B station: 792.5
If 390 lbs of cargo is moved from compartment B (aft) to compartment A (forward), what is the station number of the new CG?

A) 463.7
B) 506.3
C) 436.7
D) 433.3

Moving Mass -> Ch Mass / Total Mass = Ch CG / Distance moved

Ch Mass = 390lbs
Total mass = 116 500lbs
Distance moved = 507 (792.5 - 285.5)

390 / 116500 = Ch CG / 507
Ch CG = 1.69

We moved the CG forward so 435.0 - 1.69 = 433.3

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► The total mass of an aeroplane is 145.000 kg and the centre of gravity limits are between 4.7m and 6.9m aft of the datum. The loaded centre of gravity position is 4,4m aft. How much mass must be transferred from the front to the rear hold in order to bring the out of limit centre of gravity position to the foremost limit?

A) 3500kg
B) 35000kg
C) 62500kg
D) 7500

Moving Mass -> Ch mass / Total mass = Ch CG / Distance moved

Total mass = 145 000 kg
Ch CG = 0.3 (4.4 to 4.7)
Distance moved = 5.8 (2.6 to 8.4 from figure)

Ch mass / 145 000 = 0.3 / 5.8
Ch mass = 7 500kg

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► A Jet aeroplane, with the geometrical characteristics shown in the appendix, has a takeoff weight (W) of 460.000 N and a centre of gravity (point G on annex) located at 15,40 m from the zero reference point. At the last moment the station manager has 12.000 N of freight added in the forward compartment at 10m from the zero reference point. The final location of the centre of gravity, calculated in percentage of mean aerodynamic chord AB (from point A) is equal to:

A) 30.5 %
B) 35.5 %
C) 27.5 %
D) 16.9 %

Moment = mass x arm

Original mass: 460.000N
Arm: 15,40m
Original moment = 7084000Nm (460000 x 15,40)

Mass added: 12.000N
Arm: 10m
Moment = 120.000Nm

Total Moment = 7204000Nm
Total Mass = 472000N

- The new CG (after adding the extra freight):

CG = total moment / total mass
CG = 7204000 / 472000
CG = 15.26m

- Now find the %MAC as usual:

MAC Length = 4,60m
LE = 14m

15.26 = CG aft of datum
1.26 (15.26-14) = CG aft of LE

( 1.26 / 4.60 ) x 100 = 27.4%

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► A turbojet aeroplane is parked with the following data:
Corrected dry operating mass: 110 100 kg
Basic corrected index: 118.6
Take-off mass: 200 000 kg
centre of gravity (C.G.) location: 32 %.
Distance from reference point to leading edge: 14m
Length of MAC = 4.6m.

Initial cargo distribution:
cargo 1: 4 000 kg (2.73m from reference point)
cargo 2: 2 000 kg (8.55 m from reference point)
cargo 3: 2 000 kg (16.49m from reference point)
cargo 4 = empty (21.13 m from reference point)

For perfomance reasons, the captain decides to redistribute part of the cargo loading between cargo compartments, in order to take off with a new C.G. location of 34 %. He asks for a transfer of:

A) 1000 kg from cargo 1 to cargo 4
B) 500 kg from cargo 1 to cargo 3
C) 1000 kg from cargo 3 to cargo 1
D) 1500 kg from cargo 3 to cargo 1

- What do we want to do?

We want to move the CG aft by 2% MAC (From 32% to 34%)
= 0.092 m (2/100 x 4.6m)

! We want to move the CG (and thus the mass) backwards, we can therefore already cancel out answers C and D which would bring the mass to the front.

- Formula for moving mass:

Ch mass / total mass = Ch CG / distance moved

Total mass = 200 000kg
Ch CG = 0.092
Distance moved = 18.4 (From cargo 1 to cargo 4)

Ch mass / 200 000 = 0.092 / 18.4
Ch mass = 1000kg

Answer A seems correct! To be sure, we can also do the same for answer B. The values in the formula stay the same but now distance moved is 13.76 (cargo 1 to 3), the result will be 1337kg which does not correspond to what answer B states (500kg).

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► A turbojet aeroplane is parked with the following data:
Corrected dry operating mass: 110 100 kg
Basic corrected index: 118.6
Take-off mass: 200 000 kg
centre of gravity location: 32 %.
Distance from reference point to leading edge: 14m
Length of MAC = 4.6m.

Initial cargo distribution:
cargo 1: 4 000 kg (2.73m from reference point)
cargo 2: 2 000 kg (8.55 m from reference point)
cargo 3: 2 000 kg (16.49m from reference point)
cargo 4 = empty (21.13 m from reference point)

To Maximise performance, the captain decides to redistribute part of the cargo load between cargo 1 and cargo 4, in order to takeoff with a new centre of gravity position at 35% MAC. After loading, the new load distribution between cargo 1 and cargo 4 is:

A) 2000kg in cargo 1; 2000kg in cargo 4
B) 1000kg in cargo 1; 3000kg in cargo 4
C) 2500kg in cargo 1; 1500kg in cargo 4
D) 3000kg in cargo 1; 1000kg in cargo 4

- What do we want to do?

We want to move the CG aft by 3% MAC (From 32% to 35%)
= 0.138 m (3/100 x 4.6m)

- Formula for moving mass:

Ch mass / total mass = Ch CG / distance moved

Total mass = 200 000kg
Ch CG = 0.138
Distance moved = 18.4 (From cargo 1 to cargo 4)

Ch mass / 200000 = 0.138 / 18.4
Ch mass = 1500 kg (from cargo 1 to cargo 4)

New cargo 1 will be 4000-1500 = 2500
New cargo 4 will be empty+1500 = 1500

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► The planned takeoff mass of a turbojet aeroplane is 190.000 kg, with its centre of gravity located at 29% MAC (mean aerodynamic chord). Shortly prior to engine start, the local staff informs the flight crew that 4000kg must be unloaded from cargo 4 (23.89m aft of the reference point). Given:
Distance from reference point to leading edge: 14m
Length of MAC: 4.6m

After the handling operating, the new Centre of gravity location in % MAC will be:

A) 31%
B) 25%
C) 33%
D) 27%

Mass = 190.000kg at 29%MAC
We off load 4000kg from 23.89m aft

29%MAC = 1.334 from LE (29/100 x 4.6m)

LE = 14m aft of datum so to get CG aft of datum = 1.334 + 14 = 15.334m

Distance between CG and mass = (23.89 - 15.334) = 8.556m

- Formula for removing mass:

Ch mass (4000) / Old Mass (186000) = Ch CG (?) / Distance (8.556)

Ch CG = 0.184m

- Convert to %MAC

( 0.184  / 4.6 ) x 100 = 4%

4% CG shift , shift is to the front (we removed mass from the back so CG shifts forward)
29% - 4% = 25%

► Given: CG limits from datum: 82,0 - 94,6 inches
Arm from datum to baggage zone 1: 22,5 inches
Arm from datum to baggage zone 4: 178,7 inches
Loaded CG: 96,6 inches
Total weight of loaded aircraft: 3,400 lbs
Freight equally distributed between baggage zones 1 and 4.
The weight of freight to be moved between baggage zones to bring the aircraft into balance is:

A) 44 lbs from zone 4 to zone 1
B) 83 lbs from zone 1 to zone 4
C) 44 lbs from zone 1 to zone 4
D) 83 lbs from zone 4 to zone 1

Ch Mass / Total Mass = Ch CG / Distance moved

Total Mass = 3,400 lbs
Ch Mass = ?

Loaded CG was 96,6 , to bring it into balance we must move it forward to at least 94,6 (aft limit):

Ch CG = 2

Distance moved is between zone 1 and 4. Zone 1 is 22,5 and zone 4 is 178,7 = 156,2

Fill in all the values:

Ch Mass / 3400 = 2 / 156.2
Ch Mass = 44 lbs

We want to move the CG forward so only answer A (from zone 4 to 1) is correct.

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► Given:
CG is located at STN 15.
Aeroplane mass is 3.650 lbs.
What is the effect on the CG if you move baggage (total mass 64 lbs) from STN14 to STN20?

A) it moves fwd by 0.13 units
B) it moves aft by 0.1 units
C) it moves aft by 0.3 units
D) it moves aft by 0.31 units

- Formula for moving mass:

Ch Mass / Total mass = Ch CG / Distance moved
64 / 3650 = Ch CG / 6

Ch CG = 0.1 (and moves aft since we moved mass from STN14 to STN20)

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► A loaded aircraft weighs 4200 lbs with a CG at 9 inches aft of the datum. An extra 200 lbs is loaded into the aircraft 40 inches fwd of the datum. The new CG position is:

A) 6,8 inches FWD
B) 3,0 inches AFT
C) 6,8 inches AFT
D) 3,0 inches FWD

- Formula for adding mass:

Ch Mass / New Total mass = Ch CG / Distance mass & Old CG
200 / 4400 = Ch CG / 49 (from 9inch aft to 40 inch fwd = 49)

Ch CG = 2,227 (CG moves 2,227 forward of 9)

9 - 2,227 = 6,8 inches AFT (although we moved forward, we are still aft of the CG)

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► Given the following information, how much mass (to the nearest lb) has to be transferred from the front hold, the centroid of which is 20ft forward of the datum, to the rear hold 40ft aft of the datum to bring the centre of gravity within limits?

Aircraft mass: 2500 lbs
Total moment: -8.000 lbs/ft
Centre of gravity range from 0,5 ft to 2,0 ft fwd of datum

A) 50 lbs
B) 112 lbs
C) 1.120 lbs
D) 500 lbs

- Find CG

This time we are not given the CG directly but with the total mass and moment we can easily calculate it.

CG = -8000 / 2500
CG = -3.2 (3.2 FWD)

- Where do we want the CG to go?

The FWD limit is -2.0 and our CG is currently outside limits at -3.2 so we want it to move 1.2

- What is the distance moved?

We want to move mass between the front hold at 20ft FWD and aft hold at 40ft AFT. The distance between them is 60ft

- Fill in the values into the formula (for moving mass):

Ch mass (?) / Total mass (2500) = Ch CG (1.2) / Distance moved (60)
Ch mass = 50 lbs

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► The CG of an aircraft is at 25% MAC and MAC is 1 m. The aircraft has 2 holds, hold 1 is 7m aft of the datum and hold 2 is 22m aft of the datum. If the aircraft mass is 38.000kg, what load must be transferred from hold 1 to hold 2 to move the CG to 40% MAC?

A) 1520 kg
B) 259 kg
C) 480 kg
D) 380 kg

We want to move the CG 15% aft (from 25% to 40%) = 0.15m
Distance moved = between hold 1 and 2 = 15m

Ch mass / 38.000 = 0.15 / 15
Ch mass = 380 kg

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► The taxi mass of the aircraft is 62.500 kg and the centre of gravity is calculated at 20% MAC. A load weighing 1.000kg is moved from the forward section of the aft cargo compartment to the aft section of the forward cargo compartment. The new centre of gravity position is:

A) 10% MAC
B) 12% MAC
C) 15% MAC
D) 25% MAC

Total Mass = 62.500
Ch Mass = 1.000

Ch CG = ?
Distance moved = 835.5 inches to 421.5 inches = 414 inches (Extracted from table)

- Fill in values into the formula:

1000 / 62500 = Ch CG / 414
Ch CG = 6,624

- Find new CG location as %MAC

The first location was 20%MAC = 26.9 aft of LE
(20/100 x 134.5 <- Value extracted from table)

We calculated that the CG moved 6.624 forward (from aft cargo to forward cargo comp.)

26.9 - 6.624 = 20.276

Now convert this new location back to %MAC:

20.276 / 134.5 x 100 = 15%