Most Air Navigation calculation questions need fewer than 20 formulas. Here they are, grouped by topic, each with a short worked example. Learn them, then practise until you can do each one in under two minutes.
Time, speed and distance
| Find | Formula | Example |
|---|---|---|
| Time (minutes) | Distance ÷ GS × 60 | 120 NM at 240 kt = 30 min |
| Distance | GS × time (min) ÷ 60 | 150 kt for 18 min = 45 NM |
| NM per minute | GS ÷ 60 | 300 kt = 5 NM a minute |
Habit that saves marks: work in minutes until the last step. Most wrong answers come from mixing hours and minutes.
The 1-in-60 rule
One nautical mile off track after 60 NM is an error of about 1°. It holds well for angles up to about 20°.
- Track error = distance off × 60 ÷ distance gone
- Closing angle = distance off × 60 ÷ distance to go
- To fly direct to the destination, turn by track error + closing angle.
- To regain track, turn by twice the track error, then back by the track error once you are on it.
Worked example. After 40 NM of a 120 NM leg you are 4 NM right of track. Track error = 4 × 60 ÷ 40 = 6°. Closing angle = 4 × 60 ÷ 80 = 3°. Turn left by 9° to fly direct to the destination.
Earth and charts
| Find | Formula | Example |
|---|---|---|
| Departure (NM) | Change of longitude (minutes) × cos latitude | 4° (240′) at 60° = 240 × 0.5 = 120 NM |
| Convergency | Change of longitude × sin mean latitude | 10° at 30° = 5° |
| Conversion angle | ½ × convergency | 5° convergency = 2.5° |
| Chart scale | Ground distance = chart distance × scale | 1 NM = 185,200 cm |
Glide path and descent
- A 3° path is about 300 ft per NM, so height ≈ 300 × distance. At 4 NM you should be about 1,200 ft above the threshold.
- Rate of descent on a 3° path ≈ 5 × groundspeed. At 140 kt, about 700 ft per minute.
- Rule of thumb for planning: about 3 NM for every 1,000 ft to lose.
Altimetry
| Find | Formula | Example |
|---|---|---|
| ISA temperature | 15 − 2 × (altitude in thousands of ft) | At 10,000 ft: −5 °C |
| Pressure altitude | Elevation + (1013 − QNH) × 30 | 800 ft, QNH 1003: 800 + 300 = 1,100 ft |
| Density altitude | Pressure altitude + 120 × ISA deviation | PA 2,000 ft at 31 °C (ISA +20): 4,400 ft |
| Temperature error | 4 ft × (altitude ÷ 1,000) × ISA deviation | 5,000 ft at ISA −10: about 200 ft lower than indicated |
The 30 ft per hPa figure is a rounded exam value; near sea level it is closer to 27 to 28 ft. Use the value your question or textbook gives.
Remember: "High to low, look out below." Flying from high to low pressure, or into colder air, puts you lower than the altimeter shows.
Speed and turns
- Local speed of sound (kt) = 38.95 × √(temperature in °C + 273). At −56.5 °C it is about 573 kt. Mach = TAS ÷ local speed of sound.
- Bank angle for a rate one turn ≈ TAS ÷ 10 + 7. At 120 kt, about 19°.
Radio
| Find | Formula | Example |
|---|---|---|
| Wavelength (m) | 300 ÷ frequency (MHz) | 110 MHz ≈ 2.73 m |
| VHF range (NM) | 1.23 × (√ aircraft height + √ station height), heights in ft | 10,000 ft and 100 ft: 1.23 × (100 + 10) ≈ 135 NM |
Critical point and point of no return
Let O be the groundspeed out, H the groundspeed home, D the total distance and E the safe endurance.
- Distance to the critical point = D × H ÷ (O + H)
- Time to the point of no return = E × H ÷ (O + H); distance = that time × O
Worked example. D = 600 NM, TAS 240 kt, 20 kt headwind on the way out, safe endurance 5 hours. O = 220 kt and H = 260 kt. Critical point = 600 × 260 ÷ 480 = 325 NM, reached in about 1 hour 29 minutes. Point of no return = 5 × 260 ÷ 480 = 2.71 hours, or about 596 NM.
Remember: with no wind the critical point is halfway. Wind moves it into wind, and any wind reduces the PNR distance.
Test yourself
Original practice questions with every answer explained. The first set is free.
Practise Air Navigation calculations