Time, Speed, Distance and Fuel Calculations

Every flight plan rests on one proportion: distance equals speed multiplied by time. From it come the time to fly each leg, the estimated time of arrival (ETA), the fuel each leg will burn, how long the fuel will last and where the descent must begin. The same few lines of arithmetic run through the EASA PPL, CPL and ATPL flight planning and navigation papers and the FAA knowledge tests, usually with a trap set in the units or the wind.

Fuel follows time and time follows groundspeed, so a wrong wind, a decimal hour read as minutes or a volume entered as a mass carries straight through to the fuel left on landing. The wind triangle that produces the groundspeed is covered in triangle of velocities, and the rules that set the minimum fuel in fuel planning and fuel reserves.

The basic relationship

Speed is distance covered per unit of time, so:

distance = speed × time, time = distance ÷ speed and speed = distance ÷ time

Distances are in nautical miles (NM), speeds in knots (kt, nautical miles per hour) and times in hours and minutes. Two habits prevent most errors.

• Use groundspeed, not airspeed. The time to cover a ground distance depends on the speed over the ground: the true airspeed (TAS) corrected for the wind component. TAS 105 kt into a 15 kt headwind gives a groundspeed of 90 kt, so a 45 NM leg takes 30 minutes; ignoring the wind gives about 26.
• Work in minutes, or convert decimal hours with care. 1 h 20 min is 80 minutes, or 1.33 h. At 24 litres per hour it needs 24 × 80 ÷ 60 = 32 litres. Reading 1:20 as 1.2 h gives about 29 litres and understates the fuel.

Converting the groundspeed to nautical miles per minute makes mental arithmetic easy: 60 kt is 1 NM a minute, 90 kt 1.5, 120 kt 2 and 150 kt 2.5. At 110 kt, 22 NM takes 22 ÷ 110 × 60 = 12 minutes.

Cruise tables and flight plans work in TAS. A rule of thumb for low levels is that TAS exceeds the calibrated airspeed (CAS) by about 2 % per 1,000 ft, so 100 kt CAS at 6,000 ft is roughly 112 kt TAS. The rule errs slightly high: the flight computer, set with the actual pressure altitude and temperature, gives the exact figure, about 109 kt in the standard atmosphere. The flight computer, the CRP-5 or E6B, solves the proportion on a circular slide rule: with 105 on the outer scale over the 60 index, 48 minutes reads on the inner scale under 84 NM. Its other side, the wind side, solves the triangle of velocities.

Time en route and ETA

The time en route for a leg is its distance divided by the planned groundspeed: 84 NM at 105 kt is 0.8 h, or 48 minutes. Adding the legs, with the climb and descent allowances, gives the total estimated elapsed time (EET). On the ICAO flight plan it goes in Item 16 with the destination and alternates; elapsed times to FIR boundaries go in Item 18 after EET/, so that EET/EBUR0029 means an estimated 29 minutes from take-off to the EBUR boundary.

The ETA is the time of departure, or of the last fix, plus the time en route. Flight plans, clearances and meteorological messages all use UTC, so local times are converted once, at the start. A take-off at 0930 local from an aerodrome on UTC+2 in summer is at 0730 UTC, and a flight of 1 h 45 min then has an ETA of 0915 UTC (see time, sunrise and twilight).

Groundspeed checks and revised ETAs

The planned groundspeed rests on a forecast wind, so the real one is measured early in flight. A groundspeed check times the aircraft between two fixes a known distance apart: 6 NM in 4 minutes is 1.5 NM a minute, or 90 kt. The ETA at the next point is then the time at the last fix plus the remaining distance at the new groundspeed.

An aircraft overhead A at 1012 UTC and overhead B, 78 NM further on, at 1042 has covered 78 NM in 30 minutes: 156 kt. Point C lies 104 NM beyond B, which takes 104 ÷ 156 h, or 40 minutes, so the ETA at C is 1122 UTC. A later ETA means more fuel, so a revised ETA is always followed by a fuel check (see dead reckoning and visual navigation).

A GNSS receiver shows groundspeed directly. DME groundspeed and time to station are valid only when the aircraft is tracking directly to or from the station, because they are derived from the rate of change of the range; on any other track they read low, and on a DME arc almost zero.

Fuel burn, endurance and range

Fuel burn for a leg is the time flown multiplied by the fuel flow. Because the time depends on groundspeed, a headwind costs fuel: a 120 NM leg at TAS 100 kt into a 20 kt headwind takes 1.5 h at 80 kt and, at 36 litres per hour, burns 54 litres. Light aeroplane consumption is usually given in litres or US gallons per hour, jet consumption in kilograms or pounds per hour: 2.5 h at 5,000 kg/h is 12,500 kg.

Endurance is the time the usable fuel will last: usable fuel ÷ fuel flow. 133 litres at 38 litres per hour gives 3.5 h. Endurance is a time, so the wind does not change it. What the wind changes is the range, the distance flown in that time: range = endurance × groundspeed. Only usable fuel counts; unusable fuel is part of the basic empty mass.

Planning uses the safe endurance, the endurance with the required reserve left untouched. An aeroplane with 136 litres usable burning 32 litres per hour has a total endurance of 4 h 15 min, but only 3 h 45 min to plan with when a 30-minute reserve must remain on landing. The endurance entered after E/ in Item 19 of the ICAO flight plan is different again: the total usable fuel on board expressed as time, for search and rescue, not the legal minimum.

The reserve itself is set by the rules. For private flights in non-complex aircraft, AMC1 NCO.OP.125(b) gives a final reserve of 30 minutes by day under VFR and 45 minutes by night under VFR or under IFR, at holding speed at 1,500 ft above the destination, or above the alternate when one is required; a local day VFR flight that stays in sight of the aerodrome needs only 10 minutes at maximum continuous cruise power. The FAA's 14 CFR 91.151 requires 30 minutes by day and 45 at night under VFR, at normal cruising speed. Commercial aeroplanes under Part-CAT carry 30 minutes for turbine engines and 45 for piston engines at holding speed at 1,500 ft above the aerodrome and, in the basic scheme, contingency fuel of 5 % of the trip fuel or five minutes' holding, whichever is greater (see fuel and energy planning for general aviation).

View full schematic

Specific range, distance per unit of fuel, compares power settings: 110 kt for 34 litres per hour gives about 3.2 NM per litre, 95 kt for 25 litres per hour about 3.8. In a wind, divide the groundspeed instead: TAS 150 kt into a 30 kt headwind at 60 litres per hour gives 2.0 NM per litre over the ground. Safe endurance is also the starting point of the point of no return calculation.

Climb and descent distance

The top of climb (TOC) follows from the time to climb: from an aerodrome at 2,000 ft to a cruising altitude of 8,000 ft at an average 500 ft/min the climb takes 12 minutes, which at a groundspeed of 90 kt covers 18 NM. Flight manual climb tables are cumulative from mean sea level, so the values for the aerodrome elevation are subtracted from those for the cruising level.

Climb and descent tables give nautical air miles (NAM), distances through the air, which must be converted to nautical ground miles (NGM) for the wind. Both are flown in the same time, so NAM ÷ NGM = TAS ÷ GS. Jet cruise data work the same way: integrated range tables give air distance, and the wind is applied separately.

The top of descent (TOD) is placed by working backwards from where the descent must end. Height to lose divided by rate of descent gives the time, and time multiplied by groundspeed gives the distance. From a cruising altitude of 6,500 ft to a circuit altitude of 1,500 ft at 500 ft/min takes 10 minutes, which at 120 kt is 20 NM. Pilots cross-check with rules of thumb (see climb and descent planning):

• Distance: a jet in an idle descent needs about 3 NM per 1,000 ft, so from FL350 to sea level about 105 NM, plus distance to slow down and an allowance for any tailwind.
• Rate of descent on a 3° path: a 3° path descends about 318 ft per NM, or 5.2 %, so the rate in ft/min is about five times the groundspeed. At 90 kt the rule gives 450 ft/min; the exact figure is 477.
• Required rate: to descend from 9,000 ft to 3,000 ft within 20 NM at 150 kt, the 8 minutes available need 6,000 ÷ 8 = 750 ft/min.
• Gradient to rate: a gradient in per cent times 60.76 gives feet per nautical mile, because a nautical mile is about 6,076 ft. A 6.5 % climb gradient is about 395 ft per NM, which at 150 kt (2.5 NM a minute) needs about 990 ft/min.

A tailwind in the descent carries the aircraft further for each 1,000 ft lost, so the TOD moves out; a headwind brings it in.

Unit conversions

Exam questions mix units on purpose. Fuel is the commonest source of error: it is bought by volume, but load sheets and jet fuel plans work in mass.

Fuel density links volume and mass. The specific gravity (SG) of a fuel is its density relative to water, so an SG of 0.72 means 0.72 kg per litre. Avgas is about 0.72 and Jet A-1 about 0.79 to 0.82, and the SG falls as the fuel warms. 140 litres of avgas at 0.72 is 140 × 0.72 = 101 kg; dividing instead of multiplying gives 194 kg. Under the EASA rules the operator uses the actual density, normally from the fuel supplier's delivery note, or a standard density from its Operations Manual. FAA weight and balance uses 6 lb per US gallon for avgas; jet fuel is heavier, at about 6.7 lb per gallon (see fuel mass and fuel loading).

Exam-style worked examples

A VFR leg (EASA). A 120 NM leg is planned at TAS 100 kt with a 20 kt headwind component. The aeroplane burns 36 litres per hour, has 140 litres usable, and the flight is day VFR under Part-NCO.

1. Groundspeed: 100 − 20 = 80 kt.
2. Time: 120 ÷ 80 = 1.5 h, or 1 h 30 min.
3. Trip fuel: 1.5 × 36 = 54 litres.
4. Final reserve: 30 minutes, worked at the cruise flow as exam questions usually do, 0.5 × 36 = 18 litres, so at least 72 litres before taxi fuel and any contingency the pilot chooses.
5. Mass of 140 litres at SG 0.72: about 101 kg.

Range with a reserve (FAA). An aeroplane has 38 US gallons usable, burns 8.5 gallons per hour and makes good 115 kt, and a 45-minute reserve is to be kept.

1. Reserve: 0.75 × 8.5 = 6.4 gallons.
2. Fuel for the trip: 38 − 6.4 = 31.6 gallons.
3. Safe endurance: 31.6 ÷ 8.5 = 3.72 h.
4. Range: 3.72 × 115 = about 428 NM. The 38 gallons weigh 38 × 6 = 228 lb.

A jet cruise segment (ATPL). A segment of 600 NGM is flown at TAS 450 kt with a 50 kt headwind component and a fuel flow of 2,400 kg/h.

1. Groundspeed: 450 − 50 = 400 kt.
2. Time: 600 ÷ 400 = 1.5 h.
3. Fuel: 1.5 × 2,400 = 3,600 kg.
4. Air distance: NAM = NGM × TAS ÷ GS = 600 × 450 ÷ 400 = 675 NAM, the figure to use when entering an integrated range table.

Library articles are written for study and exam preparation. They do not replace your aircraft's approved documentation, your operator's procedures or the regulations themselves.