Cruise Altitude Choice with a Deteriorating Headwind
You are planning a 640 nm leg in a normally aspirated single (best-range TAS 128 KT at 8,000 ft, 138 KT at 11,000 ft). Forecast winds at 8,000 ft are 350/25 and at 11,000 ft are 340/45. The aircraft has no oxygen for the pilot above 10,000 ft for more than 30 minutes, so 11,000 ft is only permissible if the leg above 10,000 ft is short. Fuel at start of cruise is 210 L; consumption is 34 L/h at 8,000 ft and 38 L/h at 11,000 ft. Reserve is 45 minutes at cruise consumption. En route there is a line of convective build-up forecast to top 14,000 ft, with a gap reported near the midpoint. The aircraft has a service ceiling of 14,000 ft and the outside air temperature at 11,000 ft is forecast ISA +15°C.
1. Compute the headwind components at both levels and decide which altitude gives the shorter still-air-equivalent time, showing the arithmetic.
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At 8,000 ft the wind is 350/25 and the track is assumed 360°, so the headwind component is 25 × cos(10°) ≈ 24.6 kt, giving a ground speed of 128 − 24.6 ≈ 103 kt. At 11,000 ft the wind is 340/45, so the headwind is 45 × cos(20°) ≈ 42.3 kt, giving 138 − 42.3 ≈ 95.7 kt. The 11,000 ft level loses more to wind than it gains in TAS. Time at 8,000 ft is 640 / 103 ≈ 6.21 h; time at 11,000 ft is 640 / 95.7 ≈ 6.69 h. The lower level is faster over the ground by about 0.48 h, or nearly 29 minutes. The crossover point can also be found by equating (TAS − HW)/distance: the higher level only wins if the headwind there is less than about 45 − (138 − 128) × (something), i.e. if the wind advantage exceeds the 10 kt TAS gain, which it does not. Conclusion: file 8,000 ft unless turbulence or cloud forces a change.
2. Check fuel against the chosen level and state whether the plan is legal and comfortable.
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At 8,000 ft the cruise burn is 34 L/h for 6.21 h, about 211 L, which already exceeds the 210 L on board before reserve is counted. That is a hard no-go. Adding the 45-minute reserve at 34 L/h (about 25.5 L) gives a required fuel of roughly 237 L, some 27 L short. Even allowing a 5 kt improvement in the headwind forecast, the time falls only to about 5.9 h and the burn to about 201 L, still leaving no reserve. The 11,000 ft option is worse on time and burns 38 L/h, so it cannot rescue the plan. The correct decision is to reduce payload or take a fuel stop, or to replan at a lower level where TAS drops but the headwind may also drop; a 6,000 ft check would be needed. The lesson is that the fastest level is not automatically the legal one, and the fuel check must be done after the wind, not before.
3. The convective line tops 14,000 ft with a reported gap. Discuss the decision to press through the gap versus diverting, using the aircraft's ceiling and the temperature.
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The service ceiling of 14,000 ft is a density-altitude-limited figure, and at ISA +15°C the density altitude at 14,000 ft is roughly 16,000 ft, so the aircraft cannot realistically climb over the tops. Penetrating a gap between cells is a judgement about the gap's width, the echo intensity on any available radar, and the aircraft's turbulence penetration speed. A single normally aspirated light aircraft without weather radar should treat an unseen gap as unreliable: cells move and the gap can close. The safer plan is to divert around the line, accept the extra time and fuel, and re-check the fuel against the diversion. If the gap is confirmed by ATC or another aircraft as wide and the cells are clearly separated, a pass at the highest safe level below the freezing level, with the aircraft slowed to turbulence penetration speed and the autopilot off in the worst of it, is defensible. The decision must be made before the fuel margin is spent, not at the gap.