Wet grass take-off and tailwind landing
A normally aspirated SEP with MTOW 1,150 kg is planned from a 1,050 m grass runway at elevation 2,500 ft. TOW is 1,090 kg. QNH 1003, OAT 26°C, wind 240/08, runway 27, upslope 1.5%, grass wet with about 5 mm water. The POH, for 1,100 kg, dry grass, ISA, sea level, gives 420 m ground roll and 690 m to 50 ft. Use these corrections: +10% per 1,000 ft pressure altitude; +10% per 10°C above ISA; +15% for wet grass; +5% per 1% upslope; -10% per 10 kt headwind. The aircraft's climb gradient in these conditions is 4.8% and the runway has no obstacles in the take-off path. The destination runway is 850 m, dry, with an 8 kt tailwind forecast and an OAT of 30°C at 1,500 ft; landing distance at MLW is 560 m before corrections. Apply +20% for the tailwind and a 1.3 safety factor for landing.
1. Calculate the take-off distance to 50 ft and decide whether the departure is legal.
Read the worked answer
Pressure altitude = 2,500 + 30 × (1013 - 1003) = 2,800 ft. ISA temperature at 2,800 ft ≈ 15 - 2 × 2.8 = 9.4°C; OAT 26°C is 16.6°C above ISA. Headwind component = 8 × cos 30° ≈ 6.9 kt.
Multiply the POH 50-ft distance: 1.28 (altitude) × 1.166 (temperature) × 1.15 (wet grass) × 1.075 (upslope) × 0.931 (headwind) ≈ 1.72. 690 m × 1.72 ≈ 1,187 m.
TODA is 1,050 m, so the take-off is not legal. The ground roll would be about 420 × 1.72 ≈ 722 m, but the limiting distance is the 50-ft point. The 1,090 kg weight is slightly below the 1,100 kg POH weight, so it does not rescue the calculation.
2. What change would make the departure legal, and what must not be assumed?
Read the worked answer
To make it legal, reduce the distance by about 12%. Practical options:
- Wait for a lower OAT. A 10°C reduction cuts about 10%, giving ~1,068 m - still marginal; about 12°C is needed.
- Reduce weight: offload fuel or payload. A rough 8-10% weight reduction is needed; check the POH rather than assuming linearity.
- Use a runway with less upslope or a more favourable wind. A 10 kt headwind instead of 6.9 kt saves about 3%, not enough alone.
- Wait for the grass to dry: removing the 15% wet-grass penalty alone gives ~1,032 m, which is just inside 1,050 m.
Do not plan to 'be lighter after fuel burn'; take-off performance must use the weight at brake release. Also do not use the 4.8% climb gradient as a substitute for the 50-ft distance calculation.
3. Can the destination landing be made on the 850 m runway with the forecast tailwind?
Read the worked answer
At the destination: 560 m × 1.20 (8 kt tailwind) × 1.30 (safety factor) ≈ 874 m, which exceeds the 850 m runway. The book figure is not achievable.
The correct decision is to carry enough fuel to divert to a runway where the landing can be made into wind, or to land elsewhere if the tailwind persists. If the destination is the only option, landing in the opposite direction may be preferable even if it means a circling manoeuvre - but only if the runway and obstacle clearance allow it. Do not use a tailwind landing on a short runway to 'make it work', and remember that the 30°C temperature at 1,500 ft will also increase the true landing distance.