Oceanic Wind Error: Recalculating Heading and Fuel
You are on an oceanic leg from Auckland to Nadi, planned track 020°T, distance 1,200 nm. TAS is 450 kt and fuel flow is 2,400 kg/h. The forecast wind is 250°T/60 kt. After 30 minutes the GNSS shows actual track 028°T and groundspeed 470 kt while the aircraft is holding heading 015°T. This does not match the forecast. Fuel remaining is 7,000 kg. Company reserves are 30 minutes holding; no alternate is required. The aircraft is outside radar coverage and the next waypoint is 1,200 nm ahead. The crew must decide whether to continue with the current heading or recalculate the wind and revise the heading and fuel plan. The wind error is significant. You must determine the actual wind velocity from the observed heading, track and groundspeed, then calculate the heading and groundspeed required to maintain 020°T, and assess the fuel margin.
1. Determine the actual wind velocity from the observed heading, track and groundspeed.
Read the worked answer
Use the ground vector and the air vector; wind = ground vector − air vector.
Ground vector: track 028°T, GS 470 kt. North component = 470 cos 28° = 414.9 kt. East component = 470 sin 28° = 220.7 kt.
Air vector: heading 015°T, TAS 450 kt. North component = 450 cos 15° = 434.7 kt. East component = 450 sin 15° = 116.5 kt.
Wind components: North = 414.9 − 434.7 = −19.8 kt. East = 220.7 − 116.5 = +104.2 kt.
Wind speed = sqrt(19.8² + 104.2²) = 106.1 kt. Wind direction toward = atan2(104.2, −19.8) = 100.8°, so wind is from 280.8°, approximately 281°.
Actual wind is therefore about 281°T/106 kt, much stronger than the forecast 250°T/60 kt.
2. Calculate the revised heading and groundspeed needed to maintain 020°T.
Read the worked answer
Required track is 020°T. Wind from 281°T/106 kt gives an angle of 99° between track and wind direction.
Crosswind = 106 × sin 99° = 104.8 kt. The wind is from the left, so drift is right. Drift = asin(104.8 / 450) = 13.5° right.
True heading = 020° − 13.5° = 006.5°T.
Along-track component: wind toward 101°T is 81° from track 020°T, giving 106 × cos 81° = 16.6 kt tailwind.
Groundspeed = 450 cos 13.5° + 16.6 = 437.5 + 16.6 = 454.1 kt, approximately 454 kt.
3. Assess whether 7,000 kg of fuel is sufficient with the revised groundspeed and 30-minute reserve.
Read the worked answer
Revised time = 1,200 / 454.1 = 2.643 h.
Trip fuel = 2,400 × 2.643 = 6,343 kg.
Fuel remaining is 7,000 kg, giving a margin of 657 kg, or 657 / 2,400 × 60 = 16.4 min.
The required 30-minute reserve is 1,200 kg. Required fuel with reserve = 6,343 + 1,200 = 7,543 kg, which exceeds the 7,000 kg on board. The aircraft cannot continue to destination and still retain the full reserve. The crew should consider a diversion, a lower-power setting, or a revised destination; continuing unchanged is not acceptable.