Wanaka takeoff at max weight on a hot afternoon
Wanaka (elevation 1,142 ft) at 1430 local on a summer afternoon: OAT 24°C, QNH 1013 hPa, wind 250°M/8 kt. You plan to depart on runway 11 (sealed, 1,150 m TODA, 1.2% upslope) in a Cessna 172S at MTOW 1,157 kg. The POH gives, for 2,500 ft pressure altitude and 25°C, nil wind, sealed runway, max weight: ground roll 480 m and distance over 50 ft 880 m. Your operator's correction card says: add 20% to both distances for each 5 kt of tailwind; add 10% to both distances for each 1% of upslope; add 15% to both for dry grass. There is a 15 m fence 60 m beyond the departure end. The next runway (29) is downslope and would give a headwind, but has a 12 m tree line 200 m from the end. You have four passengers and baggage at max weight; the temperature is forecast to fall to 18°C by 1700.
1. Calculate pressure altitude, ISA temperature and density altitude for the departure.
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QNH equals 1013 hPa, so pressure altitude equals aerodrome elevation: PA = 1,142 ft.
ISA temperature at 1,142 ft = 15 − 1.98 × 1.142 = 12.7°C.
Density altitude = PA + 120 × (OAT − ISA) = 1,142 + 120 × (24 − 12.7) = 1,142 + 1,356 = 2,498 ft.
The POH row for 2,500 ft and 25°C is therefore almost exactly applicable (DA 2,498 ft, OAT 24°C). Note that the aerodrome elevation alone understates the performance penalty by about 1,350 ft.
2. Apply the wind and slope corrections and determine whether the takeoff on runway 11 is acceptable at MTOW.
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Tailwind component on runway 11 (110°): wind 250°/8 kt, angle 140°, so tailwind = 8 × cos 40° = 6.1 kt, say 6 kt.
Correction for 6 kt tailwind = 20% × 6/5 = 24%. Upslope correction = 1.2 × 10% = 12%. Combined factor = 1.24 × 1.12 = 1.389.
Ground roll = 480 × 1.389 = 667 m. Distance over 50 ft = 880 × 1.389 = 1,222 m.
TODA is 1,150 m, so the aircraft would not reach 50 ft within the runway, and the 15 m fence 60 m beyond the end makes it worse. The ground roll fits, but the takeoff distance does not. At MTOW the takeoff is not acceptable on runway 11.
3. Recommend a course of action and justify it, including the effect of waiting until 1700 and the suitability of runway 29.
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Wait until 1700 (OAT 18°C): ISA remains 12.7°C, so DA = 1,142 + 120 × 5.3 = 1,778 ft. Using roughly 10% per 1,000 ft of DA, distance falls by about 7%: 880 × 0.93 = 818 m; corrected 818 × 1.389 = 1,136 m. That is inside 1,150 m, but the margin is only about 14 m - not a sensible operational margin.
Runway 29: it gives a headwind of about 6 kt and a downslope, so the 50 ft distance would drop sharply (roughly 880 × 0.88 × 0.88 = 680 m). However, the 12 m trees only 200 m from the end are a close-in obstacle; the 50 ft screen does not protect against them, and the climb path at 200 m would be well below 12 m. Runway 29 is not automatically the answer.
Best decision: do not depart at MTOW on 11. Offload 80–100 kg, wait for cooler air, and use a personal safety factor; do not treat a 14 m margin as acceptable.