Aft CG with a Full Cabin
You are loading a four-seat piston single for a 2 h 15 min VFR flight. The aircraft has a basic empty mass of 720 kg with its CG at 1.00 m aft of datum. Max takeoff mass is 1,150 kg and the certified CG range is 0.98 m to 1.06 m aft of datum. Station arms are: pilot and front passenger 0.85 m; rear passengers 1.75 m; baggage 2.30 m (maximum 30 kg); usable fuel 1.20 m. Fuel load is 120 kg. On board are the pilot (80 kg), one front passenger (70 kg), one rear passenger (65 kg), and 20 kg of baggage. The day is hot and the runway is 1,100 m long at 1,200 m elevation. The operator wants all four occupants and the baggage to remain on board, and asks you to confirm the loading is within limits before engine start. No ballast is available, and the baggage compartment is the only compartment with a 2.30 m arm.
1. Calculate total mass and CG. Is the aircraft within mass and CG limits? Show the moment calculation.
Read the worked answer
Use mass × arm = moment.
- Basic empty: 720 × 1.00 = 720.00
- Pilot: 80 × 0.85 = 68.00
- Front passenger: 70 × 0.85 = 59.50
- Rear passenger: 65 × 1.75 = 113.75
- Baggage: 20 × 2.30 = 46.00
- Fuel: 120 × 1.20 = 144.00
Total mass = 1,075 kg; total moment = 1,151.25 kg·m; CG = 1,151.25 / 1,075 = 1.0709 m.
Mass is 75 kg below MTOW, but CG is 0.0109 m (1.09 cm) aft of the 1.06 m aft limit. The aircraft is not within CG limits. Do not depart until the loading is corrected.
2. The operator refuses to leave any occupant behind and wants to keep as much baggage as possible. With fuel still 120 kg, what is the maximum baggage that can be carried, and how much must be offloaded?
Read the worked answer
Let B be the baggage retained in kg. Occupants and fuel are fixed.
Total mass = 1,075 − (20 − B) = 1,055 + B. Total moment = 1,151.25 − 2.30(20 − B) = 1,105.25 + 2.30B.
Require CG ≤ 1.06 m: (1,105.25 + 2.30B) / (1,055 + B) ≤ 1.06 1,105.25 + 2.30B ≤ 1,118.30 + 1.06B 1.24B ≤ 13.05 B ≤ 10.52 kg.
In whole kilograms, carry 10 kg and offload 10 kg. Check B = 10: mass 1,065 kg, moment 1,128.25, CG 1.0594 m - inside. B = 11: mass 1,066 kg, moment 1,130.55, CG 1.0606 m - outside. The maximum baggage is therefore 10 kg.
3. Instead, the rear passenger is moved to the front seat, within the seat limit, and all 20 kg of baggage stays. Calculate the new CG and state whether any operational limitation remains.
Read the worked answer
Move the 65 kg rear passenger from 1.75 m to 0.85 m.
Total mass remains 1,075 kg. Moment change = 65 × (0.85 − 1.75) = −58.5 kg·m. New moment = 1,151.25 − 58.5 = 1,092.75 kg·m. New CG = 1,092.75 / 1,075 = 1.0165 m.
This is 0.0435 m forward of the aft limit and 0.0365 m aft of the forward limit, so it is inside the 0.98–1.06 m envelope. Mass remains below MTOW. The operational caveat is that the front seat must be approved for that occupant and the change must be recorded in the load sheet; if the front seat limit or occupant restraint is unsuitable, offload baggage instead.