Ardmore to Tauranga: cruising level, wind triangle and fuel margin
You are planning a VFR flight in a C172 from Ardmore (NZAR) to Tauranga (NZTG), 78 NM, planned track 115°M. The aircraft has 110 L usable fuel, burns 32 L/h in cruise, and you allow 8 L for taxi and a 45-minute reserve. Forecast wind at 3,500 ft is 230/25. You initially write 4,500 ft on the flight plan because you want terrain clearance over the Kaimai Range, but your instructor questions the level. Departure is 1000 NZDT with 10 minutes taxi. Tauranga is controlled and you expect a clearance. There is no GNSS in the aircraft; you have a current Topo50 chart and a VNC. Your passenger weighs 80 kg and baggage is 15 kg, both within W&B. You need to confirm the cruising level, compute heading and groundspeed, and decide whether the fuel plan gives an acceptable margin. The weather is VFR with isolated showers, but the forecast wind is expected to remain steady.
1. Which VFR cruising altitude should you use for a track of 115°M above 3,000 ft, and why is 4,500 ft unsuitable? Show the rule and the available levels.
Read the worked answer
In New Zealand, VFR cruising levels above 3,000 ft AMSL are based on magnetic track: tracks 000–179° use odd thousands plus 500 ft; tracks 180–359° use even thousands plus 500 ft.
A track of 115°M falls in the 000–179° band, so the available levels are 3,500 ft, 5,500 ft, 7,500 ft and so on. A level of 4,500 ft is an even-thousand-plus-500 level intended for tracks 180–359°, so it is not a valid VFR cruising level for this track.
The practical choice is 3,500 ft if terrain, cloud and turbulence allow; otherwise 5,500 ft. Because the route passes near rising terrain toward the Kaimai Range, you must confirm minimum safe altitude and cloud clearance before selecting 3,500 ft. The flight plan should show 3,500 ft or 5,500 ft, not 4,500 ft.
2. Using TAS 100 kt and wind 230/25, calculate the magnetic heading and groundspeed for the 3,500 ft leg. State any assumptions.
Read the worked answer
Assume still air TAS 100 kt and a steady wind 230°M/25 kt.
Track = 115°M. The wind is from 230°M, so the relative angle is 230 − 115 = 115°. The wind is from the right and slightly behind the aircraft.
Crosswind component = 25 × sin 115° = 25 × 0.906 = 22.7 kt from the right. Headwind component = 25 × cos 115° = 25 × (−0.423) = −10.6 kt, i.e. a 10.6 kt tailwind.
Wind correction angle = arcsin(22.7/100) = 13.1°, applied to the right to counter the wind from the right. Magnetic heading = 115 + 13 = 128°M.
Groundspeed = TAS × cos(WCA) + tailwind = 100 × cos 13.1° + 10.6 = 97.4 + 10.6 = 108 kt.
So use about 128°M and expect about 108 kt groundspeed. These are approximate because the wind is forecast, not measured.
3. Calculate trip fuel, reserve, and total fuel required, and state whether 110 L is sufficient. If the ETA slips by 20 minutes due to holding, what is the fuel state?
Read the worked answer
At 108 kt, the 78 NM leg takes 78/108 = 0.722 h = 43.3 min.
Trip fuel = 32 L/h × 0.722 h = 23.1 L. Taxi fuel = 8 L. Reserve = 45 min at 32 L/h = 24 L. Total required = 23.1 + 8 + 24 = 55.1 L.
With 110 L usable, the margin is 110 − 55.1 = 54.9 L. The fuel plan is sufficient.
A 20-minute hold uses 32 × 20/60 = 10.7 L. The margin falls to about 44.2 L, which is still acceptable, but you should reassess if the hold extends or if the actual groundspeed is lower than 108 kt. The reserve is not available for normal planning; it is there for unforeseen delay.