Wet Runway Contamination and the Take-off Decision
You are the commander of a twin-engine turbofan aircraft at a sea-level airport on a day with OAT 28 degrees C, QNH 1013 hPa, and a reported surface wind of 240/08 kt. The runway in use is 27, 2,400 m long, dry asphalt. Just before taxi, heavy rain leaves standing water of about 4 mm depth over the first 900 m of runway 27, with the remainder damp. Your aircraft's AFM gives a dry-runway take-off distance of 1,780 m at these conditions, a wet-runway factor of 1.15, and a contaminated-runway penalty of 1.35 for standing water. There is no clearway, and the stopway is 60 m. The aircraft's maximum certified take-off mass is 68,000 kg; your current mass is 64,500 kg. The first segment climb gradient requirement is 0.5 percent in the take-off configuration, and your AFM shows a net gradient of 0.9 percent at this mass and temperature.
1. Calculate the effective take-off distance required and determine whether the runway length is adequate for a wet/contaminated surface.
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Start with the dry-runway take-off distance of 1,780 m. Apply the wet factor of 1.15: 1,780 x 1.15 = 2,047 m. Because the first 900 m is covered by 4 mm of standing water, the contaminated penalty of 1.35 applies to that portion, not the whole runway. A conservative method is to apply the contaminated factor to the full distance: 1,780 x 1.35 = 2,403 m, which already exceeds the 2,400 m runway. A more precise split gives the contaminated segment 1,780 x (900/2,400) x 1.35 = 901 m, and the damp segment 1,780 x (1,500/2,400) x 1.15 = 1,280 m, totalling 2,181 m. However, the AFM contaminated factor is defined for a uniformly contaminated surface; using the full-runway method is the safer dispatch choice. Add the 60 m stopway only if the AFM permits it in the contaminated case. The conclusion is that the take-off is not legally permissible without reducing mass or waiting for the water to drain.
2. The crew considers reducing mass to 62,000 kg. Estimate the effect on the required distance and state whether this resolves the problem.
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Take-off distance scales approximately with the square of the mass ratio for the ground roll and with a lower power for the airborne segment. Using the square approximation for the ground roll: (62,000/64,500)^2 = 0.924, so the dry distance falls to about 1,780 x 0.924 = 1,645 m. Applying the full contaminated factor of 1.35 gives 1,645 x 1.35 = 2,221 m, which is below the 2,400 m runway. The margin is about 179 m, or 7.5 percent. Because the square-law approximation is rough and the AFM factor is conservative, the crew should confirm with the actual AFM chart rather than rely on the estimate. If the chart confirms a value below 2,400 m, the reduced mass resolves the performance problem, but the crew must also recheck the climb gradient at the lower mass, which will improve, and the landing distance at destination.
3. Explain how the reported wind of 240/08 kt affects the take-off distance calculation for runway 27.
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Runway 27 has a magnetic heading of 270 degrees. The wind is from 240 degrees at 8 kt, so the angle between runway heading and wind direction is 30 degrees. The headwind component is 8 x cos 30 = 6.9 kt, and the crosswind component is 8 x sin 30 = 4.0 kt. A headwind reduces the ground speed at lift-off and therefore shortens the ground roll. A useful approximation is that each knot of headwind reduces the take-off distance by about 1.5 to 2 percent for a light twin, but the AFM chart should be used. For a 6.9 kt headwind, the reduction is roughly 10 to 14 percent, which would bring the 2,403 m contaminated distance down to about 2,070 to 2,160 m. The crosswind of 4 kt is well within limits. However, the wind is reported, not guaranteed, and a gust or shift to a tailwind would remove the benefit, so the calculation should be based on the still-air distance unless a measured, steady headwind is confirmed.