Cruise Level Choice with a Stronger Headwind Aloft
You are planning a 420 NM leg with a true track of 270°T. The forecast wind at 7,500 ft is 300°/15 kt; at 9,500 ft it is 300°/35 kt. TAS is 135 kt at 7,500 ft and 130 kt at 9,500 ft. Fuel burn is 38 L/h at 7,500 ft and 42 L/h at 9,500 ft. Usable fuel is 240 L, the fixed reserve is 45 min, and taxi allowance is 8 L. You have 3 POB and baggage; MTOW is not an issue. The aircraft has no oxygen, and you plan to remain below 10,000 ft. Cloud tops are 8,000 ft; turbulence is reported moderate at 9,500 ft and light at 7,500 ft. ATC can offer either level. You depart at 2300 UTC; last light at destination is 0600 UTC. The GNSS is serviceable.
1. Compute the groundspeed, flight time and fuel required at each level.
Read the worked answer
At 7,500 ft: wind 300/15, track 270, so the wind is 30° off the track from the right. Crosswind = 15 sin 30 = 7.5 kt; headwind = 15 cos 30 = 13.0 kt. Drift = arcsin(7.5/135) = 3.2°; GS ≈ 135 cos 3.2 - 13.0 = 121.8 kt. Time = 420/121.8 = 3.45 h = 3 h 27 min. Fuel = 3.45 × 38 = 131.1 L; reserve = 0.75 × 38 = 28.5 L; taxi 8 L; total 167.6 L. At 9,500 ft: crosswind = 35 sin 30 = 17.5 kt; headwind = 35 cos 30 = 30.3 kt; drift = arcsin(17.5/130) = 7.7°; GS ≈ 130 cos 7.7 - 30.3 = 98.5 kt. Time = 420/98.5 = 4.26 h = 4 h 16 min. Fuel = 4.26 × 42 = 178.9 L; reserve = 0.75 × 42 = 31.5 L; taxi 8 L; total 218.4 L. Both are within 240 L, but 9,500 ft leaves only 21.6 L above the planned reserve/taxi, and the wind is a headwind, not a tailwind.
2. Which level do you request and why?
Read the worked answer
Request 7,500 ft. It is about 49 min faster (3 h 27 min vs 4 h 16 min) and uses about 48 L less fuel. The forecast wind at 9,500 ft is stronger from the same direction, so the higher TAS does not compensate: GS is 23 kt lower. The turbulence at 9,500 ft is moderate, which increases fatigue and reduces precision, and the fuel margin is thinner. The higher level also puts you closer to the 10,000 ft oxygen threshold, although 9,500 ft is still below it. Unless the 7,500 ft level is unavailable or the turbulence is worse than reported, 7,500 ft is the better operational choice.
3. After 1 h at 7,500 ft, ATC offers 9,500 ft because the wind is 'better'. Actual winds are 300/30 at 7,500 and 300/45 at 9,500. Recalculate and decide.
Read the worked answer
At 7,500 with 300/30: headwind = 30 cos 30 = 26.0 kt; crosswind = 15 kt; drift = arcsin(15/135) = 6.4°; GS ≈ 135 cos 6.4 - 26.0 = 108.2 kt. At 9,500 with 300/45: headwind = 45 cos 30 = 39.0 kt; crosswind = 22.5 kt; drift = arcsin(22.5/130) = 10.0°; GS ≈ 130 cos 10 - 39.0 = 89.0 kt. Climbing would reduce GS by about 19 kt and increase fuel burn from 38 to 42 L/h. It would also put you in moderate turbulence. Decline the offer and stay at 7,500 ft. The word 'better' in an ATC offer is not a wind calculation; the stronger headwind aloft makes 9,500 ft worse for this track. If the offer were a tailwind, the answer could change, so always compute the along-track component.