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► At a given mass the CG position is at 15% MAC. If the leading edge of MAC is at a position 625.6 inches aft of the datum and the MAC is given as 134.5 inches, determine the position of the CG in relation to the datum.
A) 645.78 inches aft of datum
B) 20.18 inches aft of datum
C) 605.43 inches aft of datum
D) 228.34 inches aft of datum
To get CG as % MAC to inches aft of leading edge:
(CG% / 100) x MAC length
(15% / 100) x 134.5 = 20.175
The CG is 20.175 inches aft of the leading edge. However, they ask the distance in relation to the datum so add the LE distance from datum:
20.175 + 625.6 = 645.78 inches
► The loaded centre of gravity (CG) of an aeroplane is 713 mm aft of datum. The mean aerodynamic chord lies between station 524 mm and 1706 mm. The CG expressed as % MAC (mean aerodynamic chord) is:
A) 16 %
B) 41 %
C) 60 %
D) 10 %
MAC lies between 524mm (leading edge) and 1706mm (trailing edge)
MAC = 1182 mm long
CG = 713mm aft of datum, so
CG = 189mm aft of LE (713-524)
To get CG (distance) as % MAC:
(Distance CG aft LE / MAC length ) x 100
( 189 / 1182 ) x 100 = 16%
► The mass and balance information gives:
Basic mass: 1200 kg
Basic balance arm: 3.00 m
under these conditions the basic centre of gravity is at 25% of the mean aerodynamic chord (MAC). The length of the MAC is 2m.
In the mass and balance section of the flight manual the following information is given:
Position Arm front seats: 2.5 m
Position Arm rear seats: 3.5 m
Position Arm rear hold: 4.5 m
Position Arm fuel tanks: 3.0 m
The pilot and one passenger embark, each weighs 80kg, fuel tank contains 140 litres with a density of 0.714. The rear seats are not occupied. Taxi fuel is negligible. The position of the centre of gravity as % MAC is:
A) 29 %
B) 22 %
C) 34 %
D) 17 %
CG position = Total Moment / Total Mass
Moment = Arm x Mass
1- Calculate all the moments:
(BEM)---------------1200kg -- 3.0m -- 3600kg/m
(PILOT/PAX)--------160kg -- 2.5m -- 400kg/m
(FUEL)--------------100kg -- 3.0m -- 300kg/m
(TOM) CG = total moment / total mass
(TOM) CG = 4300 / 1460
(TOM) CG = 2.95m
2- Now we need to convert this to % MAC
Given;
(BEM) CG was 25% MAC with an arm of 3.0m
MAC Length = 2m
Distance (BEM) CG aft the LE = (25% / 100) x 2m = 0.5m
With this we can now find the distance from Datum to LE = (3m - 0.5m) = 2.5m
The TOM CG was 2.95m aft of datum - 2.5m = 0.45m aft of LE
(0.45m / 2m) x 100 = 22.5%
A) 645.78 inches aft of datum
B) 20.18 inches aft of datum
C) 605.43 inches aft of datum
D) 228.34 inches aft of datum
To get CG as % MAC to inches aft of leading edge:
(CG% / 100) x MAC length
(15% / 100) x 134.5 = 20.175
The CG is 20.175 inches aft of the leading edge. However, they ask the distance in relation to the datum so add the LE distance from datum:
20.175 + 625.6 = 645.78 inches
► The loaded centre of gravity (CG) of an aeroplane is 713 mm aft of datum. The mean aerodynamic chord lies between station 524 mm and 1706 mm. The CG expressed as % MAC (mean aerodynamic chord) is:
A) 16 %
B) 41 %
C) 60 %
D) 10 %
MAC lies between 524mm (leading edge) and 1706mm (trailing edge)
MAC = 1182 mm long
CG = 713mm aft of datum, so
CG = 189mm aft of LE (713-524)
To get CG (distance) as % MAC:
(Distance CG aft LE / MAC length ) x 100
( 189 / 1182 ) x 100 = 16%
► The mass and balance information gives:
Basic mass: 1200 kg
Basic balance arm: 3.00 m
under these conditions the basic centre of gravity is at 25% of the mean aerodynamic chord (MAC). The length of the MAC is 2m.
In the mass and balance section of the flight manual the following information is given:
Position Arm front seats: 2.5 m
Position Arm rear seats: 3.5 m
Position Arm rear hold: 4.5 m
Position Arm fuel tanks: 3.0 m
The pilot and one passenger embark, each weighs 80kg, fuel tank contains 140 litres with a density of 0.714. The rear seats are not occupied. Taxi fuel is negligible. The position of the centre of gravity as % MAC is:
A) 29 %
B) 22 %
C) 34 %
D) 17 %
CG position = Total Moment / Total Mass
Moment = Arm x Mass
1- Calculate all the moments:
(BEM)---------------1200kg -- 3.0m -- 3600kg/m
(PILOT/PAX)--------160kg -- 2.5m -- 400kg/m
(FUEL)--------------100kg -- 3.0m -- 300kg/m
(TOM) CG = total moment / total mass
(TOM) CG = 4300 / 1460
(TOM) CG = 2.95m
2- Now we need to convert this to % MAC
Given;
(BEM) CG was 25% MAC with an arm of 3.0m
MAC Length = 2m
Distance (BEM) CG aft the LE = (25% / 100) x 2m = 0.5m
With this we can now find the distance from Datum to LE = (3m - 0.5m) = 2.5m
The TOM CG was 2.95m aft of datum - 2.5m = 0.45m aft of LE
(0.45m / 2m) x 100 = 22.5%
