Modification Weight and Balance Sign-Off
An avionics modification is completed on a twin-engine business aircraft. Before modification the empty weight is 1,250 kg at an arm of 2.10 m. The work removes 12 kg at 1.20 m and installs 35 kg at 3.80 m. The aircraft is then loaded for a company flight: 180 L of fuel at a density of 0.72 kg/L and arm 2.60 m; pilot 80 kg at 2.35 m; front passenger 75 kg at 2.40 m; baggage 25 kg at 3.40 m. The maximum take-off weight is 1,650 kg and the certified CG range is 2.00 m to 2.55 m aft of the datum. The operator wants to add extra baggage at the same 3.40 m station. The hangar supervisor asks you to verify the load sheet before signing it, and to state clearly whether the aircraft is inside both weight and CG limits and how much more baggage can be carried.
1. Calculate the modified empty weight and empty CG after the modification.
Read the worked answer
Use moment = weight × arm.
- Original moment: 1,250 × 2.10 = 2,625.0 kg·m
- Removed item: −12 × 1.20 = −14.4 kg·m
- Installed item: +35 × 3.80 = +133.0 kg·m
- New weight: 1,250 − 12 + 35 = 1,273 kg
- New moment: 2,625.0 − 14.4 + 133.0 = 2,743.6 kg·m
- New empty CG: 2,743.6 ÷ 1,273 = 2.155 m
The empty CG has moved aft from 2.10 m to about 2.155 m. This is still inside the certified range, but the result is only valid if the modification weights and arms are final and the empty-weight record is updated. If the approved modification instructions require a reweigh, the calculated value cannot replace the actual weighing.
2. Calculate the take-off weight and CG with the stated load, and assess against the limits.
Read the worked answer
Fuel mass = 180 × 0.72 = 129.6 kg.
- Empty: 1,273 kg at 2.155 m → moment 2,743.6 kg·m
- Fuel: 129.6 kg at 2.60 m → 336.96 kg·m
- Pilot: 80 kg at 2.35 m → 188.0 kg·m
- Front passenger: 75 kg at 2.40 m → 180.0 kg·m
- Baggage: 25 kg at 3.40 m → 85.0 kg·m
Total weight = 1,273 + 129.6 + 80 + 75 + 25 = 1,582.6 kg.
Total moment = 2,743.6 + 336.96 + 188.0 + 180.0 + 85.0 = 3,533.56 kg·m.
CG = 3,533.56 ÷ 1,582.6 = 2.233 m.
The take-off weight is 1,582.6 kg, which is 67.4 kg below the 1,650 kg limit. The CG is 2.233 m, inside the 2.00 m to 2.55 m range. The load sheet is acceptable as stated, but compartment volume and local baggage limits must also be checked.
3. Determine the maximum additional baggage at 3.40 m that can be added without exceeding MTOW or the aft CG limit.
Read the worked answer
Let x be the additional baggage in kg at 3.40 m.
Weight limit: 1,582.6 + x ≤ 1,650 → x ≤ 67.4 kg.
CG limit: (3,533.56 + 3.40x) ÷ (1,582.6 + x) ≤ 2.55.
3,533.56 + 3.40x ≤ 2.55(1,582.6 + x)
3,533.56 + 3.40x ≤ 4,035.63 + 2.55x
0.85x ≤ 502.07 → x ≤ 590.7 kg.
The weight limit is much more restrictive. Maximum additional baggage is therefore 67.4 kg, giving exactly 1,650 kg and a CG of (3,533.56 + 67.4 × 3.40) ÷ 1,650 = 2.280 m. That remains inside the aft limit, so weight governs. In practice, the baggage compartment structural and volume limits may reduce the figure further.