Aft CG on a Loaded Twin Turboprop
You are preparing a load sheet for a twin turboprop departing on a 55-minute sector. Fuel is 550 kg at an arm of 4.30 m; it is already the minimum planned (trip, alternate and final reserve), so none can be removed. Basic empty mass is 3,400 kg at 4.20 m aft of datum. Crew 160 kg at 3.55 m. Six passengers totalling 480 kg are seated in the rear row at 5.40 m. Baggage 90 kg in the aft hold at 5.95 m. The CG envelope is 4.05 m to 4.32 m aft of datum. MTOW 5,000 kg, MZFW 4,700 kg, MLW 4,800 kg. The draft load sheet shows 4,680 kg and a CG of 4.346 m. The ramp agent says the aircraft "is still fine because it is 320 kg under MTOW". Two passengers of 80 kg each could be reseated in row 1 at 3.10 m if the cabin crew agrees.
1. Verify the draft load sheet CG and explain why the mass check alone does not clear the flight. What happens to the CG as fuel is burned?
Read the worked answer
Sum moments about the datum:
- BE 3,400 × 4.20 = 14,280
- Crew 160 × 3.55 = 568
- Row 6 480 × 5.40 = 2,592
- Baggage 90 × 5.95 = 535.5
- Fuel 550 × 4.30 = 2,365
Total mass 4,680 kg; total moment 20,340.5 kg·m; CG = 20,340.5 ÷ 4,680 = 4.346 m, which is 26 mm aft of the 4.32 m limit. Mass is legal (MTOW 5,000 kg; MZFW 4,680 − 550 = 4,130 kg), but mass and balance are separate limits - being under MTOW says nothing about CG.
The CG sits aft of the fuel arm (4.30 m), so burning fuel moves the CG further aft: after 300 kg, 19,050.5 ÷ 4,380 = 4.349 m. The aircraft would therefore depart outside the envelope and diverge further, with reduced static stability and a longer rotation/lower stick-force gradient on a short sector where the loading never gets a chance to improve.
2. Quantify the effect of reseating the two 80 kg passengers in row 1 and state whether the aircraft is then acceptable.
Read the worked answer
Change in moment = 160 × (3.10 − 5.40) = −368 kg·m. New moment 19,972.5 kg·m, mass unchanged at 4,680 kg, CG = 19,972.5 ÷ 4,680 = 4.268 m - about 52 mm inside the aft limit and 218 mm inside the forward limit of 4.05 m.
Because the CG is now forward of the fuel arm (4.30 m), fuel burn moves the CG forward, which is the stabilising trend you want: after 300 kg burned, 18,682.5 ÷ 4,380 = 4.265 m.
Note that the mass is identical before and after the fix: CG is controlled by distribution, not by total mass. The seat change must be recorded on the load sheet and reflected in the trim sheet before departure; a verbal agreement with the cabin crew is not load control. Also confirm both passengers can actually be seated in row 1 (seat allocation, overweight/exit-row restrictions) and that the forward limit is still respected at zero fuel.
3. With the corrected loading, how much additional aft-hold baggage (arm 5.95 m) could be accepted before the aft limit is reached, and what else must be checked?
Read the worked answer
Solve for added mass m with CG ≤ 4.32 m:
(19,972.5 + 5.95m) ÷ (4,680 + m) ≤ 4.32 19,972.5 + 5.95m ≤ 20,217.6 + 4.32m 1.63m ≤ 245.1 → m ≤ 150 kg
So roughly 150 kg of extra aft-hold load would reach the aft limit exactly. Before accepting it, check the other limits: MZFW becomes 4,130 + 150 = 4,280 kg (limit 4,700 kg, satisfied) and MTOW becomes 4,830 kg (limit 5,000 kg, satisfied).
The hold's structural and volume limits must also be confirmed, and the answer assumes fuel stays at 550 kg. If fuel were increased, the result changes: the fuel arm of 4.30 m is aft of the current CG of 4.268 m, so adding fuel itself pushes the CG aft. That is why fuel cannot be used as ballast to fix an aft CG problem - it sits almost on the CG and, depending on the starting point, can move it the wrong way.