Headwind, fuel reserve and the en-route replan
You are planning a 180 NM leg on track 090°T in an aircraft with TAS 120 kt. The forecast wind is 020°/30 kt. Fuel on board is 80 US gal, consumption is 32 US gal/hr, and the operator's fixed reserve is 30 minutes. After 45 minutes in the air, the observed ground speed has been only 95 kt because the headwind is stronger than forecast. You must decide whether to continue to destination, divert, or reduce power. The destination has no fuel, and the nearest alternate is 25 NM off track. You have no tankering option. You need to show the original heading and ground-speed calculation, the original fuel plan, and a revised fuel assessment using the observed 95 kt.
1. Calculate the heading required to maintain track 090°T and the resulting ground speed.
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Resolve the wind relative to the track. The wind is from 020°, which is 70° to the left of the 090° track. Headwind component = 30 cos 70° = 10.3 kt. Crosswind component = 30 sin 70° = 28.2 kt from the left. The wind correction angle satisfies sin WCA = crosswind/TAS = 28.2/120 = 0.235, so WCA = 13.6° into wind, to the left. Heading = 090° − 13.6° = 076.4°T. Ground speed = TAS cos WCA − headwind = 120 cos 13.6° − 10.3 = 116.6 − 10.3 = 106.3 kt. The headwind component is the along-track component, not the full wind speed; using 30 kt as a pure headwind would underestimate the ground speed by about 20 kt and overstate the fuel required.
2. Calculate the original fuel required and the reserve margin at destination.
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Time en route = 180/106.3 = 1.693 hr, or 101.6 min. Fuel burn = 32 × 1.693 = 54.2 US gal. The 30-minute reserve is 32 × 0.5 = 16.0 gal. Total fuel required = 54.2 + 16.0 = 70.2 gal. With 80 gal on board, the surplus over requirement is 9.8 gal, equivalent to about 18 minutes at 32 gal/hr. The plan is legal but not generous: any increase in headwind or route extension eats rapidly into the margin. The reserve must remain intact at destination, so the usable contingency is the 9.8 gal, not the full 25.8 gal above the trip fuel.
3. After 45 minutes at 95 kt ground speed, reassess the fuel situation. If the ground speed falls to 88 kt for the remainder, can you reach destination with the reserve intact?
Read the worked answer
At 95 kt for 0.75 hr, distance covered = 71.25 NM. Remaining distance = 180 − 71.25 = 108.75 NM. Fuel used = 32 × 0.75 = 24.0 gal, so fuel remaining = 80 − 24 = 56.0 gal. At 95 kt the remaining time would be 108.75/95 = 1.145 hr and fuel 36.6 gal; with the 16.0 gal reserve the total is 52.6 gal, leaving 3.4 gal spare. If the ground speed deteriorates to 88 kt, remaining time = 108.75/88 = 1.236 hr and fuel = 39.6 gal. Total required = 39.6 + 16.0 = 55.6 gal, leaving only 0.4 gal. That is effectively no margin. A further small headwind increase, or a 25 NM diversion, would breach the reserve. The correct judgement is to treat the destination as marginal and prepare to divert before the reserve is eroded.