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Runway length corrections

Intermediate Airport Engineering Reviewed 24 Oct 2026 2 min read

An aircraft needs more runway when the air is thinner and when the runway slopes uphill. Thin air reduces engine thrust and wing lift, so the aircraft must reach a higher ground speed before it can fly. Air density falls with altitude and with temperature, which is why high, hot airports need long runways.

Planners start from the basic runway length: the length the critical design aircraft needs at sea level, in standard atmosphere, on a level runway with no wind. This is then corrected. The ICAO Aerodrome Design Manual suggests, where aircraft performance data are not available, increasing the length by 7 percent for every 300 m of elevation, then by 1 percent for every degree Celsius by which the aerodrome reference temperature exceeds the standard-atmosphere temperature at that elevation. If the two corrections together exceed 35 percent, a specific study is needed rather than these rules. A further increase of 10 percent per 1 percent of effective runway gradient is applied for take-off.

Diagram: Indiacon original, not to scale.

The aerodrome reference temperature is the monthly mean of the daily maximum temperatures for the hottest month. The standard-atmosphere temperature falls by about 6.5 °C per 1 000 m from 15 °C at sea level.

These rule-of-thumb corrections are for early planning. Final runway length is set from the aircraft manufacturer's performance charts for the design aircraft and the routes to be flown, including take-off weight, which depends on stage length.

Tip

Confirm the design aircraft and stage length before fixing runway length.

Key formulas
  • Elevation: L₁ = L₀ (1 + 0.07 E / 300)
  • Temperature: L₂ = L₁ [1 + 0.01 (T_ref − T_std)], T_std = 15 − 0.0065 E
  • Gradient: L₃ = L₂ (1 + 0.10 G), G in percent
Common mistake

Applying the temperature correction to the basic length instead of the elevation-corrected length.

Solved example

Problem. The basic runway length is 2 100 m. The aerodrome is at 450 m elevation, the reference temperature is 33 °C and the effective gradient is 0.4 percent. Find the corrected length.

  1. Elevation: L₁ = 2 100 × (1 + 0.07 × 450 / 300) = 2 100 × 1.105 = 2 320.5 m.
  2. T_std = 15 − 0.0065 × 450 = 12.08 °C; excess = 33 − 12.08 = 20.93 °C; L₂ = 2 320.5 × 1.2093 = 2 806.1 m.
  3. Combined elevation and temperature increase = 2 806.1 / 2 100 − 1 = 33.6 percent < 35 percent, so the rules apply.
  4. Gradient: L₃ = 2 806.1 × (1 + 0.10 × 0.4) = 2 918.3 m.

Answer. About 2 920 m

Quick quiz

1Why do high, hot airports need longer runways?

Thinner air reduces thrust and lift, so a higher ground speed is needed.

2What is the aerodrome reference temperature?

The monthly mean of daily maximum temperatures for the hottest month.

3When should the rule-of-thumb corrections not be used?

When elevation and temperature corrections together exceed 35 percent.

Code note
  • ICAO Annex 14 (International): International standards for aerodrome design and operation.
  • FAA AC 150/5300-13 (USA): Airport design advisory circular.

Clauses and documents to read

  • ICAO Doc 9157 (Aerodrome Design Manual), Part 1 Runways: corrections for elevation, temperature and slope

Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • S. K. Khanna, M. G. Arora, S. S. Jain, Airport Planning and Design, Nem Chand and Bros, latest edition.
  • R. Horonjeff, F. X. McKelvey, W. J. Sproule, S. B. Young, Planning and Design of Airports, McGraw Hill, latest edition.

Free and official sources

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