Mass and Balance: Loading a Freighter Sector with a Last-Minute Pallet
A widebody freighter is loading at Muscat for a sector to Frankfurt. Basic operating mass 118 400 kg, index 42.0. Dry operating index is valid. Cargo is loaded in three holds: forward hold 12 000 kg at arm 8.5 m, centre hold 18 000 kg at arm 22.0 m, aft hold 9 000 kg at arm 34.0 m. Reference (datum) is at the nose; the mean aerodynamic chord runs from 14.0 m to 30.0 m. Maximum zero-fuel mass is 178 000 kg; maximum take-off mass 233 000 kg; maximum landing mass 186 000 kg. Trip fuel 42 000 kg. A late pallet of 3 500 kg must be added. The forward hold is full; the aft hold can take it but the aft hold is already at its structural limit of 9 000 kg. The centre hold can take 3 500 kg but the centre hold is limited to 20 000 kg.
1. Compute the zero-fuel mass and the CG position as a percentage of MAC with the original load, and state whether the aircraft is within limits.
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Mass check.
ZFM = 118 400 + 12 000 + 18 000 + 9 000 = 157 400 kg ≤ 178 000 kg ✔
Take-off mass = 157 400 + 42 000 = 199 400 kg ≤ 233 000 kg ✔ Landing mass = 199 400 − 42 000 = 157 400 kg ≤ 186 000 kg ✔
CG. Total moment = Σ(mass × arm):
- BEM: 118 400 × arm_BEM - the BEM arm is not given directly, so use the index. The problem is set up so the index method gives the answer; treat the given index 42.0 as the BEM index and add load indices.
- Forward: 12 000 × 8.5 = 102 000 kg·m
- Centre: 18 000 × 22.0 = 396 000 kg·m
- Aft: 9 000 × 34.0 = 306 000 kg·m
- Cargo moment total = 804 000 kg·m over 39 000 kg → cargo arm = 20.62 m.
Without the BEM arm the absolute CG cannot be closed numerically; the method is: CG = (BEM × arm_BEM + 804 000) / 157 400, then %MAC = (CG − 14.0) / (30.0 − 14.0) × 100.
If the BEM arm is 22.5 m (typical for this class), CG = (118 400 × 22.5 + 804 000)/157 400 = (2 664 000 + 804 000)/157 400 = 22.04 m → %MAC = (22.04 − 14.0)/16.0 × 100 = 50.2 % MAC. That is within a typical 15–65 % MAC envelope, so the aircraft is within limits as loaded.
2. The 3 500 kg pallet must be loaded. Where can it go, and what is the effect on CG and on the forward-hold limitation? Show your reasoning.
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Where it cannot go:
- Forward hold: full - no space.
- Aft hold: already at its 9 000 kg structural limit; adding 3 500 kg would exceed the hold floor limit. Not permitted even though the aircraft's overall CG might tolerate it.
- Centre hold: currently 18 000 kg, limit 20 000 kg, so only 2 000 kg can be added. The 3 500 kg pallet cannot fit entirely.
Options:
- Split the pallet if it can be broken down: 2 000 kg to the centre hold and 1 500 kg elsewhere - but the aft hold is structurally full and the forward hold is full, so this does not close either.
- Re-shuttle: move some existing centre-hold cargo forward to free centre capacity. Moving 1 500 kg from centre (arm 22.0) to forward (arm 8.5) frees 1 500 kg of centre capacity and moves CG forward by (1 500 × (22.0 − 8.5))/157 400 = 0.129 m. Then load the 3 500 kg into centre: net CG shift from the swap = +1 500 kg at 22.0 minus 1 500 kg at 8.5 plus 3 500 kg at 22.0... compute directly: new cargo moment = original 804 000 − (1 500 × 22.0) + (1 500 × 8.5) + (3 500 × 22.0) = 804 000 − 33 000 + 12 750 + 77 000 = 860 750 kg·m over 42 500 kg → arm 20.25 m. CG moves forward by about 0.37 m relative to the original cargo arm - check against the forward MAC limit.
- Offload 1 500 kg of lower-priority cargo from the centre hold to make room, accepting the commercial cost.
CG effect of the pallet itself: 3 500 kg at 22.0 m against a current CG near 22.0 m produces almost no CG shift - the pallet is essentially on the CG. The binding constraint is therefore structural (hold floor limits), not the CG envelope. The correct answer is to re-shuttle or offload; you cannot simply add the pallet to the aft hold because the aircraft CG would tolerate it.
3. After loading, the load sheet shows ZFM 160 900 kg and 48 % MAC. The refueller then reports that only 38 000 kg of fuel can be uploaded due to a bowser shortage. Recompute the take-off mass and state the operational consequence.
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Recompute.
ZFM = 157 400 + 3 500 = 160 900 kg (matches the load sheet) ≤ 178 000 kg ✔
Take-off mass with 38 000 kg fuel = 160 900 + 38 000 = 198 900 kg ≤ 233 000 kg ✔ Landing mass = 198 900 − trip fuel burned. If trip fuel is still planned at 42 000 kg, the aircraft cannot complete the sector: it would land with 198 900 − 42 000 = 156 900 kg, which is below the ZFM - impossible, because the fuel available at destination would be negative relative to the planned trip fuel.
Consequence: the sector is fuel-limited, not mass-limited. The aircraft has 4 000 kg less fuel than planned. Options:
- Tech stop en route to uplift fuel - the usual solution.
- Reduce payload - but payload is already loaded and the ZFM is fixed; offloading payload does not help unless it is re-loaded as fuel, which is a mass-for-mass trade only if the tankage allows. Here the constraint is bowser supply, not tank capacity, so offloading payload does not create fuel.
- Re-plan the sector with a reduced contingency and alternate closer to destination, if the operator's fuel policy permits and the destination weather allows. This must be a commander's decision with a documented fuel plan.
The key exam point: a fuel shortage is a flight-planning and fuel-policy problem, not a mass-and-balance problem. The mass and CG remain within limits; the sector is not dispatchable as planned.