Last-minute fuel increase creates a weight and CG problem
At the start of a scheduled sector, the load sheet for a twin turboprop shows: basic empty weight 9,600 kg at 11.00 m; crew 170 kg at 4.50 m; 50 passengers at 84 kg each at average arm 12.20 m; front hold 300 kg at 6.50 m; rear hold 800 kg at 17.50 m; fuel 2,000 kg at 12.00 m. MZFW is 15,200 kg; MTOW is 17,400 kg; MLW is 16,000 kg; CG limits are 10.60 m to 11.50 m. The captain then asks for an additional 400 kg of fuel for expected holding. The loadmaster says the rear hold is full and the only space is in the front hold. Take-off and landing performance are not limiting.
1. Calculate ZFW, TOW and the corresponding CG positions, and determine whether the load is within limits before the extra fuel is added.
Read the worked answer
Weights:
- BEW 9,600
- Crew 170
- Passengers 50 × 84 = 4,200
- Front hold 300
- Rear hold 800
ZFW = 9,600 + 170 + 4,200 + 300 + 800 = 15,070 kg. This is below MZFW 15,200 kg.
Moments:
- BEW: 9,600 × 11.00 = 105,600
- Crew: 170 × 4.50 = 765
- Passengers: 4,200 × 12.20 = 51,240
- Front hold: 300 × 6.50 = 1,950
- Rear hold: 800 × 17.50 = 14,000
- Total ZFW moment = 173,555 kg·m
ZFW CG = 173,555 ÷ 15,070 = 11.516 m, which is already 0.016 m aft of the 11.50 m aft limit.
With 2,000 kg fuel: TOW = 17,070 kg, below MTOW. Fuel moment = 2,000 × 12.00 = 24,000. Total moment = 197,555 kg·m. TOW CG = 197,555 ÷ 17,070 = 11.572 m, which is 0.072 m aft of the limit. The load is therefore not within the CG envelope even though the weights are below maximum.
2. How much cargo must be moved from the rear hold to the front hold to bring the aircraft within the CG limit?
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Moving 1 kg from the rear hold at 17.50 m to the front hold at 6.50 m changes the moment by:
17.50 − 6.50 = 11.00 kg·m per kg moved (moment reduced).
The TOW CG must come back from 11.572 m to 11.500 m, a reduction of 0.072 m at 17,070 kg:
Required moment reduction = 0.072 × 17,070 = 1,229 kg·m.
Cargo to move = 1,229 ÷ 11.00 = 111.7 kg.
Move at least 112 kg; use 120 kg for a small margin. After moving 120 kg, the moment falls by 1,320 kg·m and the CG becomes (197,555 − 1,320) ÷ 17,070 = 11.495 m, inside the aft limit. The ZFW CG also improves because the same moment change applies before fuel.
3. The captain still wants the extra 400 kg of fuel. Recalculate and state the options.
Read the worked answer
With 2,400 kg fuel:
- TOW = 15,070 + 2,400 = 17,470 kg, which exceeds MTOW 17,400 kg by 70 kg.
- Fuel moment = 2,400 × 12.00 = 28,800; total moment = 173,555 + 28,800 = 202,355 kg·m.
- CG = 202,355 ÷ 17,470 = 11.583 m, further aft of the limit.
So the extra fuel cannot simply be added. Options:
- Offload at least 70 kg of payload, preferably from the rear hold to help the CG, then move about 100 kg of remaining rear cargo to the front hold.
- Reduce the fuel increase and move at least 112 kg of cargo forward.
- Re-plan the load: transfer cargo forward first, then add only the fuel that keeps TOW at or below 17,400 kg and CG at or before 11.50 m.
The load sheet must be recalculated and rechecked before departure; the extra fuel is not available without a compensating weight and CG change.