Wet Runway Landing with Tailwind at Wellington
A twin-jet is planning to land at Wellington runway 16 (LDA 1,815 m, elevation 41 ft). Landing weight is 60,000 kg. OAT is 22°C, QNH 1005 hPa, and the wind is 340°M/12 kt, giving a 10 kt tailwind component on runway 16. The runway is wet, with a reported braking action of medium. The AFM gives a dry, sea-level, ISA landing distance of 1,400 m at 60,000 kg. The operator's manual applies: +5% per 1,000 ft pressure altitude; +1% per 5°C above ISA; +15% for a wet runway; and +10% per 5 kt tailwind. No reverse-thrust credit is included. The crew has enough fuel to hold or divert, but the destination is the only suitable airport within 30 minutes.
1. Calculate the estimated landing distance required. Determine whether the landing is legal on runway 16 and state the margin.
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Pressure altitude = 41 ft + 30 ft per hPa × (1013 − 1005) = 41 + 240 = 281 ft. Elevation/pressure correction = +5%/1,000 ft × 0.281 = +1.4%, factor 1.014. Temperature: ISA at 41 ft is about 15°C; OAT 22°C is +7°C, so +7/5 × 1% = +1.4%, factor 1.014. Wet runway: +15%, factor 1.15. Tailwind 10 kt: 10/5 × 10% = +20%, factor 1.20. Required distance = 1,400 × 1.014 × 1.014 × 1.15 × 1.20 = 1,987 m. LDA is 1,815 m, so the required distance exceeds LDA by about 172 m, or 9.5%. The landing is not legal as planned. The calculation assumes the operator's stated corrections and no reverse-thrust credit; medium braking action on a wet runway may be worse than the generic 15% wet correction, so the real margin is if anything smaller.
2. Explain the brake-energy and directional-control risks of the 10 kt tailwind on this wet runway. Recommend a practical alternative and a landing technique if the approach must be continued.
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A 10 kt tailwind raises the ground speed at touchdown by about 10 kt for the same IAS. Brake energy rises with the square of ground speed, so the brakes must absorb significantly more energy, while the wet surface gives less friction and may promote aquaplaning. The higher ground speed also means the aircraft covers more runway during the flare and initial roll, and reverse thrust and wheel brakes have less distance to work. Directional control is more demanding, especially with medium braking action. The practical alternative is to use runway 34, which converts the tailwind into a headwind, or to divert to a longer dry runway. If the landing must continue, fly a stabilised approach at the correct Vref, avoid a fast or floated touchdown, touch down firmly in the touchdown zone, select reverse thrust promptly, use autobrake/braking as recommended, and do not delay braking to achieve a smooth landing.
3. Using the operator's rough rule that each 1,000 kg below the planned landing weight reduces landing distance by 1.5%, estimate the weight reduction needed. Is burning fuel a practical solution?
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Required reduction = 1,987 − 1,815 = 172 m, or 172/1,987 = 8.7% of the required distance. At 1.5% per 1,000 kg, the reduction is 8.7/1.5 = 5.8 increments of 1,000 kg, so about 5,800 kg would have to be removed. That is a very large reduction and would not be achievable by simply burning fuel on a short diversion without compromising reserves; it is also uncertain because the rule is approximate. A runway change, a different airport, or a landing with a headwind is far more practical. If no alternative exists, the crew should hold to burn fuel only if the resulting weight and CG remain legal and the time/fuel reserves are preserved, but the better decision is not to accept the tailwind landing on the limiting wet runway.