Wet Runway Departure with Tailwind at Queenstown
A twin-engine turboprop with MTOW 7,500 kg is scheduled to depart Queenstown runway 05 (TORA 1,891 m, elevation 1,171 ft, 1% uphill slope). OAT is 24°C, QNH 1013 hPa, and the reported wind is 230°M/10 kt, giving an 8 kt tailwind component. The runway is wet, with no standing water. The AFM gives a sea-level ISA dry takeoff distance to 35 ft of 1,250 m at MTOW. The operator's performance manual applies these corrections: +10% per 1,000 ft elevation; +1% per 5°C above ISA; +5% for the uphill slope; +15% for a wet runway; +10% per 5 kt of tailwind. The runway has no clearway or stopway, and the crew has no improved-climb procedure available.
1. Calculate the estimated takeoff distance required at MTOW and determine whether the departure is legal. Show each correction and explain why the corrections are applied multiplicatively.
Read the worked answer
Start with 1,250 m. Elevation: 1,171 ft × 10%/1,000 ft = 11.7% → 1,250 × 1.117 = 1,396 m. Temperature: ISA at 1,171 ft ≈ 15 − 1.98 × 1.171 = 12.7°C; OAT 24°C is 11.3°C above ISA, so +11.3/5 × 1% = +2.26% → 1,396 × 1.0226 = 1,428 m. Slope: +5% → 1,499 m. Wet: +15% → 1,724 m. Tailwind: 8 kt = 1.6 × 5 kt, so +16% → 1,724 × 1.16 = 2,000 m. TODA is 1,891 m, so the required distance exceeds TODA by about 109 m. The departure at MTOW is not legal on this runway and in these conditions. The corrections are multiplicative because each percentage applies to the distance after the previous effect; adding 11.7 + 2.26 + 5 + 15 + 16 = 49.96% would give 1,874 m, which understates the true 60% total increase and would incorrectly appear legal.
2. Using the operator's rule that each 100 kg below MTOW reduces takeoff distance by about 1.5%, estimate the minimum weight reduction. Give two other options and their operational trade-offs.
Read the worked answer
Required reduction = 2,000 − 1,891 = 109 m, which is 109/2,000 = 5.45% of the current distance. At 1.5% per 100 kg, that is 5.45/1.5 = 3.6 increments, so reduce weight by about 360–370 kg. That might be achieved by offloading fuel or cargo, but fuel offload must preserve reserves and may require a fuel stop. Alternative 1: use runway 23, which gives a headwind component and removes the 16% tailwind penalty, but check obstacle/terrain, circuit direction and runway length; the headwind runway may also have a different slope. Alternative 2: wait for a lower OAT; the temperature correction is only about 1% per 5°C, so a 5°C reduction saves only about 20 m and would not solve the problem alone. A runway change or a combination of weight reduction and cooler temperature is more realistic.