Tailwind, Density Altitude and the Insurance Margin
A four-seat piston single at maximum take-off mass 1,150 kg is planned from a 1,000 m dry asphalt runway at elevation 1,200 ft. QNH is 1,003 hPa and OAT is 24°C. For 1,150 kg, sea level ISA, zero wind and dry paved surface, the AFM gives a ground roll of 300 m and a take-off distance to 50 ft of 520 m. The operator's approved corrections, applied multiplicatively, are: pressure altitude +10% per 1,000 ft; temperature +10% per 10°C above ISA; tailwind +10% per 2 kt; dry grass +15%; wet grass +40%. The forecast wind is 8 kt tailwind. The crew requires a 15% margin over the 50 ft distance for insurance. The runway has 1,000 m available and a 15 m obstacle at the end.
1. Compute pressure altitude, ISA deviation and take-off distance to 50 ft. Can the flight depart under the insurance rule?
Read the worked answer
Pressure altitude = elevation + 30 ft per hPa below 1,013 hPa: PA = 1,200 + 30 × (1,013 − 1,003) = 1,200 + 300 = 1,500 ft.
ISA temperature at 1,500 ft = 15 − 1.98 × 1.5 = about 12°C. OAT is 24°C, so the deviation is about +12°C.
Apply the factors multiplicatively:
- Pressure altitude: 1,500 ft gives +15%, factor 1.15
- Temperature: +12°C gives +12%, factor 1.12
- Tailwind: 8 kt gives +40%, factor 1.40
Distance to 50 ft = 520 × 1.15 × 1.12 × 1.40 = 937.7 m. Required distance with 15% margin = 937.7 × 1.15 = 1,078 m.
That exceeds the 1,000 m available, so the flight cannot depart under the insurance rule. The ground roll alone is 300 × 1.15 × 1.12 × 1.40 = 541 m, but the obstacle distance is the limiting case.
2. What maximum tailwind component would meet the 15% margin with all other corrections unchanged? Is 5 kt acceptable?
Read the worked answer
The maximum permitted 50 ft distance is 1,000 ÷ 1.15 = 869.6 m. With no tailwind, the corrected distance is 520 × 1.15 × 1.12 = 669.8 m. The permitted tailwind factor is therefore 869.6 ÷ 669.8 = 1.298.
Each 2 kt of tailwind adds 10%, so each 1 kt adds 5%. A factor of 1.298 corresponds to about 5.96 kt. In practice the maximum whole-knot tailwind is 5 kt, because 6 kt would give a factor of 1.30 and a distance of 520 × 1.15 × 1.12 × 1.30 = 870.7 m, which with the 15% margin becomes 1,001 m - just over the available runway.
At 5 kt the factor is 1.25, distance is 837 m and the required margin distance is 963 m, so 5 kt is acceptable. The forecast 8 kt is not.
3. The crew considers a 1,400 m wet grass strip with no tailwind. Using the same corrections and 15% margin, is it acceptable? What other factor is not included?
Read the worked answer
With no tailwind, apply pressure altitude 1.15, temperature 1.12 and wet grass 1.40: Distance to 50 ft = 520 × 1.15 × 1.12 × 1.40 = 937.7 m. Required with 15% margin = 1,078 m, which is less than 1,400 m. On the numbers, the wet grass strip is acceptable.
However, the correction table does not include standing water, soft or rutted ground, uphill slope, or the effect of a contaminated surface on braking and directional control. Wet grass can also reduce acceleration more than the nominal factor if the surface is soft. The crew should inspect the strip, use the actual wind, and retain a larger margin rather than treating 1,078 m as a precise result.