High-Elevation Take-off on a Hot Day
You are planning a VFR departure from a mountain airfield, elevation 4,200 ft AMSL, runway 24, 1,250 m long, dry asphalt. Current METAR: wind 060/05 kt, visibility 20 km, temperature 28°C, dew point 9°C, QNH 1015 hPa. The aircraft POH gives a sea-level ISA take-off ground roll of 480 m and a 50 ft obstacle-clearance distance of 750 m. For non-ISA conditions the POH says: increase the ground roll by 10% per 1,000 ft of pressure altitude and by 12% per 10°C that the OAT exceeds ISA at the airfield. Aircraft mass is 1,050 kg, 40 kg below MTOW. A 1.5% upslope on runway 24 and a 5 kt tailwind component are reported. You have two passengers and full tanks; the only alternate is a 35-minute flight away. The flight manual does not publish a separate tailwind or upslope correction.
1. Calculate the pressure altitude, ISA temperature at the field, ISA deviation, and approximate density altitude. Explain which of these numbers most directly affects take-off performance and why.
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Pressure altitude = 4,200 + (1013 - 1015) × 30 = 4,140 ft. ISA at 4,200 ft = 15 - 2 × 4.2 = 6.6°C. OAT 28°C gives a deviation of about 21.4°C above ISA. Density altitude ≈ 4,140 + 120 × 21.4 ≈ 6,700 ft. Density altitude is the performance-relevant number because engine power, propeller thrust and lift all depend on air density, not on pressure altitude alone. The high OAT expands the air; the aircraft behaves as if it were at about 6,700 ft on a standard day. Pressure altitude alone would understate the penalty by more than 2,500 ft. On a 1,250 m runway with a 5 kt tailwind and upslope, this is already a marginal situation before any POH corrections are applied.
2. Using the POH corrections, estimate the ground roll and the 50 ft obstacle-clearance distance. Compare them with the runway and identify the remaining margins or lack of margins.
Read the worked answer
Pressure-altitude correction: 4.14 × 10% = 41.4%. Temperature correction: 21.4/10 × 12% = 25.7%. Total factor = 1 + 0.414 + 0.257 = 1.671. Ground roll ≈ 480 × 1.671 ≈ 802 m. 50 ft distance ≈ 750 × 1.671 ≈ 1,253 m. The 50 ft distance already exceeds the 1,250 m runway, so there is no runway margin even before the 5 kt tailwind and 1.5% upslope are applied. The POH does not publish corrections for those two items, but both increase distance: a tailwind reduces initial lift and increases time on the ground, and upslope adds a gravitational component. With two passengers and full tanks, the aircraft is 40 kg below MTOW but still heavy. The correct conclusion is that the take-off is not assured; the calculation should be treated as a stop, not as a go with a little margin.
3. What is your operational decision, and what specific alternative would you choose? Include the altimeter setting you would use and why.
Read the worked answer
Decision: no-go on runway 24 at 28°C with full tanks. The 50 ft distance is at or beyond the runway length, and the uncorrected tailwind and upslope make it worse. Options: wait for a cooler period and recalculate; reduce fuel or passengers to lower mass; use a longer runway or an airfield with no upslope; or depart from a runway into wind if available. Do not accept the take-off on the basis of density altitude alone. Set QNH 1015 on the altimeter subscale so the instrument reads field elevation on the ground and gives terrain clearance after take-off. QFE would read zero at the field but is not useful for en-route terrain clearance. After departure, reset to the local QNH as soon as one is available.