High-density-altitude short-field departure
You are planning a short-field takeoff from a 2,400 ft dry-grass runway at 2,800 ft MSL. The QNH is 30.12 inHg and the outside air temperature is 29 °C. The aircraft weighs 2,150 lb; MTOW is 2,550 lb. The POH sea-level ISA takeoff distance over 50 ft at MTOW is 1,180 ft. Use these corrections: 12% per 1,000 ft of density altitude, 10% per 1% of runway upslope, 15% for dry grass, and 1.5% per 100 lb below MTOW. The runway has a 1.5% upslope. There is no obstacle at the departure end, but the runway end is 300 ft from a public road. The wind is calm. Your passenger asks whether you can depart with full fuel, which is already included in the 2,150 lb.
1. Calculate the pressure altitude and density altitude for the departure.
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Pressure altitude = elevation + (29.92 − QNH) × 1,000 = 2,800 + (29.92 − 30.12) × 1,000 = 2,800 − 200 = 2,600 ft.
ISA temperature at 2,600 ft = 15 − (2 × 2.6) = 9.8 °C. The actual temperature is 29 °C, so the ISA deviation is 29 − 9.8 = 19.2 °C.
Using the common approximation of 120 ft per °C of ISA deviation: density altitude = 2,600 + (120 × 19.2) = 2,600 + 2,304 = about 4,900 ft.
The takeoff will therefore be performed at roughly 2,100 ft higher density altitude than the field elevation suggests.
2. Using the stated corrections, estimate the takeoff distance over 50 ft and compare it with the available runway.
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Apply the corrections multiplicatively to the 1,180 ft book figure:
- Density altitude: 4.9 × 12% = 58.8%, so factor 1.588 → 1,180 × 1.588 = 1,874 ft.
- Upslope: 1.5% × 10 = 15%, factor 1.15 → 1,874 × 1.15 = 2,155 ft.
- Dry grass: 15%, factor 1.15 → 2,155 × 1.15 = 2,478 ft.
- Weight: 2,150 lb is 400 lb below MTOW; 4 × 1.5% = 6% reduction, factor 0.94 → 2,478 × 0.94 = 2,329 ft.
The calculated distance is about 2,329 ft against a 2,400 ft runway, a margin of only 71 ft, or about 3%. That is less than the variability introduced by surface condition, technique, and the absence of any headwind. The road 300 ft beyond the end also removes the comfort of a stopway. Treat the departure as operationally unacceptable despite being just inside the book distance.
3. What is your decision, and what single change gives the greatest safety improvement without exceeding MTOW?
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I would not depart as loaded. The calculation is inside the runway length only on paper, with no allowance for a soft patch, late rotation, or engine deterioration.
Offloading weight helps, but modest amounts do not fix the problem: 100 lb off gives only 1.5% shorter takeoff, about 35 ft. A 10% margin would need roughly 240 ft less distance, which corresponds to about 10% of the current takeoff distance. At 1.5% per 100 lb, that would require removing roughly 650–700 lb - more than is available without major fuel or passenger changes.
A better single change is to wait for a lower temperature. The no-density-altitude distance after slope, grass, and weight corrections is about 1,467 ft. To bring the result to about 2,160 ft, the density-altitude factor must fall to about 1.47, i.e. density altitude about 3,900 ft. That is roughly 1,000 ft lower than the current 4,900 ft, requiring about 8 °C cooler air. Departing in the evening or early morning, or using a longer paved runway, is the correct decision.