Hot Day, Aft Load at Kotoka
Accra Kotoka (DGAA): elevation 203 ft, runway 03 dry asphalt 3,400 m, 14:20 local, OAT 33 °C, QNH 1012 hPa, wind calm. The Cessna 172S you are to fly has MTOM 1,157 kg and an empty mass of 750 kg at an arm of 2.65 m from datum. Load: pilot 80 kg at 2.30 m; two rear passengers 75 kg each at 3.20 m; 100 kg fuel at 2.70 m; 40 kg baggage in the aft hold at 3.80 m. At this mass the approved envelope runs from 2.40 m forward to 2.72 m aft. An empty forward hold is available at 1.95 m. The POH, for 1,120 kg and 2,450 ft density altitude, gives 520 m ground roll, 960 m over a 50 ft screen and 340 ft/min climb; the departure procedure needs at least 300 ft/min to clear rising ground. Your passenger has a connecting flight and is pressing you to go.
1. Compute the take-off mass and CG position, and state whether the aircraft is inside the envelope.
Read the worked answer
Sum the moments about the datum (mass × arm):
| Item | Mass kg | Arm m | Moment kg·m |
|---|---|---|---|
| Empty | 750 | 2.65 | 1,987.5 |
| Pilot | 80 | 2.30 | 184.0 |
| Rear pax ×2 | 150 | 3.20 | 480.0 |
| Fuel | 100 | 2.70 | 270.0 |
| Aft baggage | 40 | 3.80 | 152.0 |
| Total | 1,120 | 3,073.5 |
CG = 3,073.5 ÷ 1,120 = 2.744 m.
Mass: 1,120 kg is 37 kg below MTOM - legal. CG: 2.744 m is 0.024 m aft of the 2.72 m aft limit - not legal, and the aft limit is the one that matters most on a hot, heavy departure.
Note the trap: the aircraft is under maximum mass, which is exactly why a pilot who only checks mass would sign it off. Mass and balance are separate checks, and the load sheet must be recomputed after any change - not estimated.
2. Estimate the pressure and density altitude, and assess the take-off and climb margins.
Read the worked answer
Pressure altitude = elevation + 30 ft per hPa below 1013: 203 + (1013 − 1012) × 30 ≈ 233 ft.
ISA temperature at that level = 15 − 1.98 × (233/1000) ≈ 14.5 °C. ISA deviation = 33 − 14.5 = +18.5 °C.
Density altitude ≈ pressure altitude + 120 × ISA deviation = 233 + 2,220 ≈ 2,450 ft - which matches the POH column, so the figures are for the right conditions.
Margins:
- Take-off over 50 ft = 960 m against 3,400 m available: about 28 % of the runway. Comfortable today, but a 10 kt tailwind would push it to roughly 1,440 m, and a wet runway adds more again.
- Climb: 340 ft/min available against 300 ft/min required is only a 13 % margin. That margin is being bought at 1,120 kg - any extra fuel, a later departure at 36 °C, or the loss of one engine-driven item and the terrain clearance disappears.
Conclusion: performance is acceptable; the loading is not. Fix the CG before discussing the schedule.
3. What is the correct way to bring the aircraft inside limits, and what does an aft CG do to handling?
Read the worked answer
Move mass forward rather than simply removing it, since the passengers' baggage is the item generating the problem:
New moment = 3,073.5 − 40 × (3.80 − 1.95) = 3,073.5 − 74.0 = 2,999.5 kg·m. New CG = 2,999.5 ÷ 1,120 = 2.678 m - inside the envelope, 0.042 m forward of the aft limit and 0.278 m aft of the forward limit, with mass unchanged at 1,120 kg.
Alternatives if the forward hold is not usable: offload baggage (recompute), move a rear passenger forward, or reduce fuel - but fuel reduction affects range and reserves, so the load move is preferred.
Handling effects of an aft CG that the exam expects you to state:
- Reduced longitudinal static stability and lighter elevator forces, so the aircraft is easier to over-rotate and harder to recover from a stall or spin.
- Higher cruise TAS for the same power, which is precisely why pilots drift aft of the limit without noticing.
- In the flare, a tendency to balloon and to land long.
Brief the change, re-sign the load sheet, and do not let a connecting flight override an aft-of-limit condition.