Chart Scale, Fuel and Diversion on a Short VFR Leg
You plan a VFR flight using a 1:500,000 chart. The straight-line track from A to B measures 96 mm on the chart, and the true course is 090°T. Forecast wind is 300°/40 kt, TAS 110 kt, variation 4°E, and the compass deviation for the planned heading is 1°W. The aircraft burns 32 L/h and carries 40 L usable, including a 45-minute reserve and 3 L for taxi. At 12 NM after departure, the destination weather drops to 1,500 ft broken with 8 km visibility, while your alternate C is 30 NM from your present position on a track of 150°T. You are at 3,500 ft, TAS 110 kt, and the wind at that level is 310°/35 kt.
1. Calculate the chart distance in NM and the leg time and fuel from A to B, and decide whether the planned fuel is sufficient.
Read the worked answer
Chart distance: 96 mm × 500,000 = 48,000,000 mm = 48 km = 25.9 NM. Wind: 300°/40, TC 090°. Angle between course and wind source = 150°; crosswind = 40 sin 150° = 20 kt from the left, tailwind = 40 cos 30° = 34.6 kt. WCA = arcsin(20/110) = 10.5° left → TH 079.5°T, MH 075.5°M, CH 076.5°C. GS ≈ 110 cos 10.5° + 34.6 = 142.8 kt. Time = 25.9/142.8 = 0.181 h = 10.9 min. Fuel = 32 × 0.181 = 5.8 L + 3 L taxi + 24 L reserve = 32.8 L. With 40 L, margin 7.2 L (13.5 min). Sufficient.
2. At the diversion point, compute the heading and time to alternate C, and determine whether remaining fuel permits diversion plus reserve.
Read the worked answer
Fuel used to the diversion point: 12/142.8 h = 0.084 h; fuel = 2.7 L + 3 L taxi = 5.7 L; remaining = 40 − 5.7 = 34.3 L. Reserve 24 L leaves 10.3 L for diversion = 19.3 min. Diversion: TC 150°, wind 310°/35. Angle = 160°; crosswind = 35 sin 160° = 12 kt; WCA = arcsin(12/110) = 6.3° right → TH 156.3°T. Tailwind = 35 cos 20° = 32.9 kt; GS ≈ 110 cos 6.3° + 32.9 = 142.2 kt. Time = 30/142.2 = 0.211 h = 12.7 min; fuel = 6.7 L. Remaining after diversion = 34.3 − 6.7 = 27.6 L, which is 3.6 L (6.8 min) above reserve. Divert now; it is legal but tight, so do not delay and monitor fuel and headwind.
3. If the chart is later found to be printed at 1:475,000 rather than 1:500,000, explain the effect on the distance and fuel plan.
Read the worked answer
New scale 1:475,000: 96 mm × 475,000 = 45.6 km = 24.6 NM, 5.1% less than 25.9 NM. Trip time and fuel fall by about 5%: time 10.4 min, fuel 5.5 L, total requirement 32.5 L. The original plan is therefore conservative by about 0.3 L and 0.5 min. The serious case is the opposite error (e.g. 1:525,000), which would underestimate distance by 5% and could erode the reserve. Check the scale bar, not just the printed ratio, and cross-check the leg length against GPS or two known features before accepting the fuel plan.