Aft Baggage Fix That Is Only Just Enough
A light twin is being loaded for a 2 h 15 min VFR sector. Basic empty mass is 1,450 kg with a moment of 2,900 kg·m, giving a CG of 2.00 m. The operator uses a constant envelope: maximum take-off mass 2,100 kg, forward CG limit 1.95 m and aft CG limit 2.12 m. The load is pilot 85 kg at 1.60 m, front passenger 80 kg at 1.60 m, two rear passengers 75 kg each at 2.45 m, baggage 40 kg at 3.05 m and fuel 300 kg at 2.20 m. Planned fuel burn is 120 kg. The runway is short and the forecast temperature is 25°C, so the crew wants no unnecessary performance penalty. The dispatcher proposes removing 10 kg from the baggage hold because it is behind the CG and 'will fix the overweight condition'.
1. Compute the take-off mass and CG. Is the aircraft within the mass and CG limits? Show the moment calculation and state the operational decision.
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Sum the moments about the datum:
- Empty: 1,450 × 2.00 = 2,900 kg·m
- Pilot: 85 × 1.60 = 136 kg·m
- Front passenger: 80 × 1.60 = 128 kg·m
- Rear passengers: 150 × 2.45 = 367.5 kg·m
- Baggage: 40 × 3.05 = 122 kg·m
- Fuel: 300 × 2.20 = 660 kg·m
Total mass = 1,450 + 85 + 80 + 150 + 40 + 300 = 2,105 kg. Total moment = 2,900 + 136 + 128 + 367.5 + 122 + 660 = 4,313.5 kg·m. CG = 4,313.5 ÷ 2,105 = 2.049 m.
The CG is inside 1.95 m to 2.12 m, but the mass is 5 kg above the 2,100 kg maximum. The aircraft is therefore not dispatchable as loaded. The exceedance is small, but it is still a limit exceedance; performance and structural margins are not guaranteed.
2. Evaluate the dispatcher's proposal to remove 10 kg of baggage. Recompute mass and CG, comment on the claim that removing aft mass fixes the problem, and state what you would actually do.
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Removing 10 kg at 3.05 m removes 30.5 kg·m:
- New mass = 2,105 − 10 = 2,095 kg
- New moment = 4,313.5 − 30.5 = 4,283.0 kg·m
- New CG = 4,283.0 ÷ 2,095 = 2.044 m
The aircraft would be 5 kg under MTOM and the CG remains inside limits. Removing mass from behind the CG moves the CG forward, not aft; the dispatcher's reasoning about direction is correct in principle, but the original problem was mass, not an aft CG exceedance. The solution is technically legal but leaves only 5 kg of margin, which is easily consumed by passenger weight variation, baggage estimate error or fuel quantity error. A better practical action is to remove 20 kg of baggage: mass 2,085 kg, moment 4,252.5 kg·m, CG 2.040 m. That gives 15 kg margin and still keeps the CG comfortably inside the envelope. Re-weigh or re-check the final load sheet before departure.
3. Using the loading after removing 20 kg of baggage, compute the landing mass and CG after 120 kg of fuel has been burned. Is the forward CG limit a concern?
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Take-off mass with 20 kg baggage removed is 2,085 kg and the moment is 4,252.5 kg·m. Burning 120 kg at 2.20 m removes 264 kg·m:
- Landing mass = 2,085 − 120 = 1,965 kg
- Landing moment = 4,252.5 − 264 = 3,988.5 kg·m
- Landing CG = 3,988.5 ÷ 1,965 = 2.030 m
The forward limit is 1.95 m, so the landing CG is about 0.08 m aft of the forward limit. Forward CG is not a concern for this sector. In fact, because fuel is behind the forward limit, burning fuel moves the CG slightly forward; the landing CG is lower than the take-off CG but still has margin. The crew should still confirm the actual fuel load and baggage weight, because the margin is small enough that a loading error could change the result.