One complete cross-country scenario using all six E6B functions in sequence.
Follow along using the calculator above.
Scenario: VFR cross-country flight
Route
EGBJ Gloucestershire → EGTK Oxford Kidlington
Distance
56 nm (measured from chart)
Cruise altitude
4,500 ft, QNH 1018 hPa
CAS
100 kt (from POH cruise table)
Forecast wind
270°/20 kt (from UAV/ATIS)
Magnetic variation
2°W (from chart)
Fuel available
34 US gal (usable, 100LL)
POH burn rate
9.5 US gal/hr at 65% power
Step 1 — True Airspeed (TAS tab)
PA = 4,500 + (1013.25 − 1018) × 30 = 4,500 − 142.5 = 4,358 ft ≈ 4,400 ft
T_actual = 5°C + 273.15 = 278.15 K
T_ISA = 288.15 − 0.0019812 × 4400 = 279.43 K
TAS = CAS × √(T_actual/T_ISA) = 100 × √(278.15/279.43) = 100 × 0.9977 = 99.8 kt ≈ 100 kt
At low altitude with near-ISA temperature, TAS ≈ CAS. Use 100 kt TAS.
Step 2 — Wind Triangle (Wind tab)
TC = 128° (measured from chart, Gloucester to Oxford)
TAS = 100 kt, Wind FROM 270° at 20 kt
sin(WCA) = (20/100) × sin(270 − 128) = 0.20 × sin(142°) = 0.20 × 0.6157 = 0.1231
WCA = arcsin(0.1231) = +7.1° (positive = crab right, wind from the right/south)
TH = TC + WCA = 128 + 7.1 = 135.1° True Heading
GS = 100 × cos(7.1°) − 20 × cos(270 − 135.1)
= 100 × 0.9923 − 20 × cos(134.9°)
= 99.2 − 20 × (−0.7059) = 99.2 + 14.1 = 113.3 kt Groundspeed
Step 3 — Magnetic Heading
Variation = 2°W → West is Best → add
MH = TH + variation = 135.1 + 2 = 137° Magnetic Heading
(Apply compass deviation from the aircraft deviation card for final Compass Heading)
Step 4 — Time en route (T·S·D tab)
Distance = 56 nm, Groundspeed = 113 kt
Time = 56 ÷ 113 = 0.4956 hr = 29.7 min ≈ 30 minutes
Departing 10:00Z → ETA Oxford: 10:30Z
Step 5 — Fuel required (Fuel tab)
Trip time = 0.50 hr (30 min), Burn rate = 9.5 US gal/hr
Trip fuel = 0.50 × 9.5 = 4.75 US gal
VFR reserve = 30 min = 0.50 hr × 9.5 = 4.75 US gal
Total req = 4.75 + 4.75 = 9.5 US gal
Fuel avail = 34 US gal → SUFFICIENT (24.5 US gal spare → 2h 35m additional endurance)
Step 6 — En-route off-course correction (60:1 tab)
After 30 nm, GPS shows 3 nm right of track, 26 nm remaining.
TEA = atan(3/30) = 5.7° (turn LEFT by 5.7° to fly parallel to track)
CA = TEA × (30/26) = 5.7 × 1.15 = 6.6°
Total correction = 5.7 + 6.6 = 12.3° → Turn LEFT 12.3° to fly direct to Oxford.