Three-leg wind-triangle and fuel-feasibility check
A light twin is planned on a three-leg flight. Leg 1: distance 120 NM, track 090°, TAS 160 kt, wind 020°/20 kt. Leg 2: distance 180 NM, track 150°, TAS 160 kt, wind 200°/30 kt. Leg 3: distance 90 NM, track 240°, TAS 160 kt, wind 300°/25 kt. Cruise fuel flow is 90 kg/h. Taxi fuel is 10 kg, contingency is 5% of trip fuel, alternate fuel is 30 kg and final reserve is 45 min at 90 kg/h. The aircraft has 240 kg usable fuel and maximum usable tank capacity is 300 kg. No refuelling is available en route. The crew must decide whether the plan is fuel-feasible.
1. Calculate the heading and ground speed for each leg.
Read the worked answer
Leg 1: wind 020/20 relative to track 090° gives headwind 20 cos70° = 6.8 kt and crosswind 20 sin70° = 18.8 kt from the left. Drift = arcsin(18.8/160) = 6.8°. Heading = 090 − 6.8 = 083°. GS = 160 cos6.8° − 6.8 = 152 kt.
Leg 2: wind 200/30 relative to track 150° gives headwind 30 cos50° = 19.3 kt and crosswind 30 sin50° = 23.0 kt from the right. Drift = arcsin(23.0/160) = 8.3°. Heading = 150 + 8.3 = 158°. GS = 160 cos8.3° − 19.3 = 139 kt.
Leg 3: wind 300/25 relative to track 240° gives headwind 25 cos60° = 12.5 kt and crosswind 25 sin60° = 21.7 kt from the right. Drift = arcsin(21.7/160) = 7.8°. Heading = 240 + 7.8 = 248°. GS = 160 cos7.8° − 12.5 = 146 kt.
2. Calculate total elapsed time and trip fuel, then determine whether the 240 kg usable fuel is sufficient including reserves.
Read the worked answer
Leg times: 120/152 = 0.789 h = 47.3 min; 180/139 = 1.295 h = 77.7 min; 90/146 = 0.616 h = 37.0 min. Total = 161.9 min = 2 h 42 min. Trip fuel = 90 × 2.698 h = 242.8 kg.
Contingency = 5% × 242.8 = 12.1 kg. Final reserve = 45 min at 90 kg/h = 67.5 kg. Required fuel = taxi 10 + trip 242.8 + contingency 12.1 + alternate 30 + final reserve 67.5 = 362.4 kg. Available is only 240 kg, so the plan is not fuel-feasible; the shortfall is 122.4 kg.
3. What is the minimum extra fuel needed, and if the tank capacity is 300 kg, what must the crew change?
Read the worked answer
Minimum extra fuel = 362.4 − 240 = 122.4 kg, i.e. about 122 kg. With maximum usable capacity of 300 kg, the aircraft can carry at most 60 kg more than the current 240 kg, leaving a remaining shortfall of about 62 kg. Therefore fuel alone cannot solve the problem. The crew must change the plan: reduce payload or baggage, shorten a leg, choose a closer alternate, reduce contingency only if company policy permits a documented calculation, or plan a refuelling stop. The decision must be made before departure, not in flight, because no en-route fuel is available.