Drift and ground speed from a single fix pair
A light aircraft flies from a departure point on a planned magnetic track of 045° at an indicated airspeed of 120 kt. The forecast wind is 300°/25 kt, and the pilot computes a heading and ground speed before departure. After 20 minutes the aircraft plots a fix that lies 4 nm right of the planned track, and the distance covered along track is 38 nm. The pilot has a further 60 nm to run to the destination and a 30-minute fuel reserve above the required landing fuel.
1. From the fix, determine the actual drift and ground speed achieved, and state the wind actually experienced.
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Distance flown in 20 min (1/3 h) along track is 38 nm, so ground speed = 38 × 3 = 114 kt.
Drift: 4 nm right over 38 nm along track gives tan(drift) = 4/38 ≈ 0.105, so drift ≈ 6° right. Track made good = 045° + 6° = 051° magnetic.
Wind: the air vector has magnitude 120 kt along the heading flown; the ground vector is 114 kt along 051°. The wind vector is the ground vector minus the air vector. Resolving: ground components (114 sin51°, 114 cos51°) ≈ (88.6, 71.7); air components depend on the heading actually flown. If the pilot flew the computed heading, the difference gives a wind from roughly the left, stronger than forecast. The essential point is that the fix pair yields drift and ground speed directly, and the wind is derived as the vector difference, not read off the forecast.
2. Decide what heading and ground speed to use for the remaining 60 nm and whether fuel is adequate.
Read the worked answer
The remaining leg should use the observed wind, not the forecast. With drift 6° right and the same wind, the pilot applies a heading correction of about 6° left of the desired track to hold 045°, i.e. heading ≈ 039° magnetic, and expects ground speed ≈ 114 kt.
Time remaining = 60/114 h ≈ 0.526 h ≈ 31.6 min. Fuel: the reserve is 30 min, so the margin above reserve is about 1.6 min at the current fuel flow - very tight. The judgement call is to recognise that a 6 kt ground-speed loss over the whole route erodes the reserve, and that the correct action is to recalculate endurance with the observed ground speed and consider a diversion or reduced power setting rather than pressing on. The exam tests whether the candidate converts observed drift and ground speed into an actionable fuel decision.