Mass and Balance with a Last-Minute Load Change
A narrow-body twin-jet is being loaded for a 2-hour sector. The load sheet shows: basic operating mass 41,200 kg, index at that mass 42.0; 150 passengers at 84 kg standard mass = 12,600 kg in the forward cabin (arm 12.0 m from datum, index factor +0.5 per 1000 kg·m); 40 passengers aft at 84 kg = 3360 kg (arm 24.0 m, index factor −0.5 per 1000 kg·m); cargo 2000 kg in hold 1 (arm 10.0 m, +0.5) and 1500 kg in hold 4 (arm 26.0 m, −0.5). The maximum zero-fuel mass is 62,000 kg, maximum take-off mass 69,000 kg, and the CG envelope at take-off is 18%–32% MAC, where the datum-to-MAC relationship gives CG% = (moment/ mass − 15.0) × 4. Fuel load is 9000 kg at arm 17.0 m. The dispatcher calls: 6 extra passengers (84 kg each) must be added, and the only space is in the aft cabin.
1. Compute the zero-fuel mass and the CG position before and after adding the 6 aft passengers, and determine whether the aircraft remains inside the CG envelope.
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First compute the zero-fuel mass without fuel. Basic operating 41,200 + forward pax 12,600 + aft pax 3,360 + cargo 2,000 + 1,500 = 60,660 kg. That is below MZFM 62,000 kg by 1,340 kg.
Moments (kg·m): basic operating - the index is given, but for CG we need the moment; assume the basic operating arm corresponds to index 42.0. To keep it tractable, compute the incremental moments from the loaded items and use the index factors as a cross-check. Forward pax: 12,600 × 12.0 = 151,200. Aft pax: 3,360 × 24.0 = 80,640. Hold 1: 2,000 × 10.0 = 20,000. Hold 4: 1,500 × 26.0 = 39,000. Sum of item moments = 290,840 kg·m. Adding 6 aft pax: 504 kg × 24.0 = 12,096 kg·m, new ZFM = 61,164 kg, new item moment = 302,936 kg·m.
The basic operating moment must be derived: index 42.0 with factor +0.5 per 1000 kg·m means the index contribution is moment/1000 × 0.5, so 42.0 = (moment/1000) × 0.5 → moment = 84,000 kg·m. Total ZFM moment before = 84,000 + 290,840 = 374,840 kg·m; CG arm = 374,840/60,660 = 6.18 m. That is forward of the 15.0 m datum reference, which suggests the datum convention differs; treat the MAC formula as the authority: CG% = (6.18 − 15.0) × 4 = negative, which is clearly inconsistent, so the index factors must be interpreted as index units, not direct arms. The safe conclusion: the added 504 kg at the aft arm moves the CG aft by ΔCG = 504 × (24.0 − 6.18)/61,164 ≈ 0.147 m, i.e. about 0.6% MAC aft. If the pre-load CG was within the aft limit by more than 0.6% MAC, the load is acceptable; otherwise it is not. The ZFM of 61,164 kg remains below 62,000 kg. The correct operational answer: recompute the load sheet formally, do not rely on the estimate, and if the aft limit is approached, move cargo forward or offload.
2. If the extra passengers cannot be moved, what other actions can restore the CG, and what is the take-off mass limit check with 9000 kg fuel?
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Take-off mass = ZFM 61,164 + fuel 9,000 = 70,164 kg, which exceeds MTOM 69,000 kg by 1,164 kg. So even before the CG question, the aircraft is overweight for take-off. Options: reduce fuel by at least 1,164 kg (plus taxi allowance) if range permits, or offload 1,164 kg of payload (about 14 passengers), or move cargo from hold 4 to hold 1 to shift the CG forward.
Moving 1,500 kg from hold 4 (arm 26.0) to hold 1 (arm 10.0) shifts the moment by 1,500 × 16 = 24,000 kg·m forward. On a ZFM of 61,164 kg, that is a CG shift of 24,000/61,164 ≈ 0.39 m forward, roughly 1.6% MAC - enough to counter the 0.6% aft shift from the extra passengers with margin. However, hold 1 may have a floor-loading limit; check it. Also confirm the forward CG limit is not breached: a 1.6% forward shift plus the aft 0.6% gives a net 1.0% forward of the original CG, so if the original CG was near the forward limit, this could breach it. The disciplined answer: recalculate the load sheet with the actual distribution, check both MTOM and the CG envelope, and only then sign.