Density Altitude and Runway Performance on a Hot Gulf Afternoon
You plan to depart a 1,400 m asphalt runway at an airfield with elevation 1,200 ft. The afternoon temperature is 44 °C and the QNH is 1002 hPa. Your aircraft's manual gives, for a sea-level airfield, ISA conditions, maximum take-off mass and zero wind, a take-off ground roll of 480 m and a take-off distance to 50 ft of 870 m. The manual's correction notes state: increase distances by 10% per 1,000 ft of pressure altitude and by 10% per 10 °C above ISA temperature. The runway has a 0.7% upslope in the take-off direction and the wind is reported as 5 kt tailwind. Your planned take-off mass is 60 kg below maximum.
1. Estimate the pressure altitude and the ISA temperature deviation, and explain what each does to performance.
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Pressure altitude is calculated from elevation and QNH: PA = elevation + 30 × (1013 − QNH) in feet, with QNH in hPa. Here PA = 1,200 + 30 × (1013 − 1002) = 1,200 + 330 = 1,530 ft.
ISA temperature at 1,530 ft: sea-level ISA is 15 °C, and the lapse rate is about 1.98 °C per 1,000 ft. So ISA temp ≈ 15 − (1.53 × 1.98) ≈ 15 − 3.0 = 12.0 °C.
Deviation = 44 − 12 = +32 °C, i.e. ISA +32.
Both effects reduce performance. Pressure altitude of 1,530 ft means lower air density than sea level, so the engine produces less power and the propeller and wing produce less thrust and lift for a given IAS. The high temperature further reduces density, compounding both effects: take-off run lengthens, the climb gradient falls, and the aircraft accelerates more slowly. The combination is the classic hot-and-high problem, and it is why the manual gives corrections rather than a single figure.
2. Apply the manual's corrections to estimate take-off distance to 50 ft, and judge whether the departure is acceptable.
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Start from the sea-level ISA, maximum-mass, zero-wind figure of 870 m to 50 ft.
- Pressure altitude: 1,530 ft, so factor = 1 + (1.53 × 0.10) = 1.153. Distance = 870 × 1.153 ≈ 1,003 m.
- Temperature: +32 °C above ISA, so factor = 1 + (32/10 × 0.10) = 1 + 0.32 = 1.32. Distance = 1,003 × 1.32 ≈ 1,324 m.
- Upslope 0.7%: treat as an increase of roughly 1% per 0.1% slope in the conservative direction, i.e. about 7% here. Distance ≈ 1,324 × 1.07 ≈ 1,417 m.
- Tailwind 5 kt: approximately a 10% increase per 2 kt of tailwind is a common rule of thumb, so 5 kt is about 25%. Distance ≈ 1,417 × 1.25 ≈ 1,771 m.
- Mass 60 kg below maximum gives a small reduction, perhaps 1–2%, so ≈ 1,740 m.
Against a 1,400 m runway, the required distance to 50 ft exceeds the available runway by roughly 340 m. Even allowing for the roughness of these corrections, the margin is negative, not marginal. The departure is not acceptable as planned.
3. List the practical options the pilot has to make the departure safe, and rank them.
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Options, best first:
- Wait for cooler conditions. Temperature is the dominant multiplier here (1.32 alone). An early-morning departure at, say, 28 °C would give a deviation near +16 °C and cut a large part of the penalty, plus typically less wind and better density. This is the simplest and safest fix.
- Reduce mass. Offloading fuel or payload reduces both the take-off distance and, more importantly, improves the climb gradient after lift-off. Note the 60 kg already assumed is small; substantially more reduction would be needed to close a 340 m gap, so this alone is unlikely to be sufficient.
- Change runway or airfield. A longer runway, a downslope, or a field at lower elevation transforms the calculation. Check other runways for slope and wind alignment - a 5 kt headwind instead of a tailwind is worth far more than any other single change.
- Reject the departure if none of the above gives a comfortable margin. A take-off distance that only just fits leaves nothing for a slower-than-expected acceleration, a late rotation or an engine problem at the critical point.
Whichever is chosen, the pilot should recompute rather than assume, confirm the climb gradient clears obstacles in the departure path, and brief a rejected take-off decision point.