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DGCA Aviation Meteorology · formula sheet

DGCA Aviation Meteorology Formulas and Key Concepts

17 formulas and the ideas behind them, grouped by topic. Understand each one, then practise it on the question bank.

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How is air density calculated?

From the equation of state for dry air.

ρ=pR TR=287 J kg−1K−1\rho=\frac{p}{R\,T}\qquad R=287\ \text{J\,kg}^{-1}\text{K}^{-1}
  • Density falls with height and with higher temperature
  • Humid air is less dense than dry air (water vapour is lighter than N₂ and O₂)

What are the ISA sea level values and lapse rate?

15 °C, 1013.25 hPa, 1.225 kg/m³. Temperature falls 1.98 °C / 1 000 ft (6.5 °C/km) up to 11 km (36 090 ft).

TISA=15−1.98×h1000T_{ISA}=15-1.98\times\dfrac{h}{1000}
  • Above 11 km the temperature is constant at −56.5 °C to 20 km
  • Tropopause height ≈ 8 km at poles, 16–18 km at the equator
  • ISA deviation = actual − ISA temperature

How do you find pressure altitude?

Correct the elevation for the difference between 1013 hPa and the QNH.

PA=elevation+(1013−QNH)×30 ftPA=\text{elevation}+(1013-QNH)\times 30\ \text{ft}
  • QNH 1003 hPa at elevation 500 ft gives PA = 500 + 300 = 800 ft

What temperature correction applies to an altimeter?

True altitude differs from indicated by about 4 ft per 1000 ft per °C of ISA deviation.

Δh≈4×h1000×ΔTISA\Delta h\approx 4\times\frac{h}{1000}\times\Delta T_{ISA}
  • Colder than ISA: true altitude is lower than indicated (dangerous over high ground)
  • Warmer than ISA: true altitude is higher

How do you convert QNH to QFE?

Subtract the elevation expressed in hPa.

QFE=QNH−elevation (ft)30QFE=QNH-\frac{\text{elevation (ft)}}{30}
  • QNH 1013 at elevation 900 ft gives QFE ≈ 983 hPa

How does pressure altitude relate to true altitude?

True altitude = pressure altitude corrected for the actual (non-ISA) temperature.

DA=PA+120 (TOAT−TISA)DA=PA+120\,(T_{OAT}-T_{ISA})
  • Cold air: true altitude below indicated
  • Correction ≈ 4 ft per 1 000 ft per °C of ISA deviation
  • Density altitude = pressure altitude + 120 × (OAT − ISA)

Temperature, Humidity & Cloud Base

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How do you estimate cloud base?

Use the dew-point spread. The temperature falls 3 °C per 1000 ft and the dew point 0.5 °C, so they close 2.5 °C per 1000 ft.

Cloud base (ft)≈400×(T−Td)\text{Cloud base (ft)}\approx 400\times(T-T_d)
or  122×(T−Td) m\text{or }\ 122\times(T-T_d)\ \text{m}
  • T = 24 °C, Td = 14 °C: spread 10, base ≈ 4000 ft above the surface
T 24 °CTd 14 °CT −3 °C / 1000 ftTd −0.5 °C/ 1000 ftcloud baseBase (ft) ≈ 400 × (T − Td)4000 ft

How do you find the cloud base from temperature and dew point?

Cloud base height (ft) ≈ 400 × (T − Td), or 125 m per °C of spread.

Base (ft)≈400 (T−Td)\text{Base (ft)}\approx 400\,(T-T_d)
  • T 25 °C, Td 13 °C: 12 × 400 = 4 800 ft above the station
  • Convection cloud forms at the condensation level
  • Dew point falls about 0.5 °C/1 000 ft in rising air

What is the geostrophic wind and when does it apply?

The wind balanced between the pressure gradient force and the Coriolis force, blowing parallel to straight isobars.

Vg=12Ωρsin⁡ϕ ΔpΔnV_g=\frac{1}{2\Omega\rho\sin\phi}\,\frac{\Delta p}{\Delta n}
  • Closer isobars: stronger wind
  • Speed is inversely proportional to sin latitude and density
  • Not valid near the equator (within about 15°)
LWindPGFCoriolisNorthern hemisphere: wind blows anticlockwise round a lowBuys Ballot: back to the wind,low pressure is on your left

What is the geostrophic wind?

The wind that blows parallel to the isobars when the pressure gradient force balances the Coriolis force.

Vg=12Ωρsin⁡ϕΔpΔnV_g=\dfrac{1}{2\Omega\rho\sin\phi}\dfrac{\Delta p}{\Delta n}
  • Geostrophic speed ∝ gradient / (ρ sin latitude)
  • No geostrophic wind at the equator
  • Thermal wind: wind shear due to horizontal temperature differences, blows with cold air to the left (N)

Meteorological Calculations

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How do you calculate density altitude?

Start from pressure altitude and add 120 ft for every °C above ISA.

DA=PA+120 (OAT−TISA)DA=PA+120\,(OAT-T_{ISA})
TISA≈15−2×PA1000T_{ISA}\approx 15-2\times\frac{PA}{1000}
  • PA 4000 ft, OAT 25 °C: ISA 7 °C, deviation +18, DA = 4000 + 2160 = 6160 ft

How do you find temperature at altitude in ISA?

Take off 2 °C per 1000 ft from the surface value.

Th=T0−2×h1000T_h=T_0-2\times\frac{h}{1000}
  • ISA deviation = actual temperature − ISA temperature

How do you estimate the freezing level?

Divide the surface temperature by the lapse rate.

h0∘≈T02×1000 fth_{0^\circ}\approx\frac{T_0}{2}\times 1000\ \text{ft}
  • Surface 12 °C: freezing level ≈ 6000 ft

True altitude with a temperature correction: how?

Work out the ISA deviation at the level, then multiply.

True alt≈Indicated+4×h1000×ΔTISA\text{True alt}\approx\text{Indicated}+4\times\frac{h}{1000}\times\Delta T_{ISA}
  • Indicated 10 000 ft, ISA −15 °C: error −600 ft, true altitude 9400 ft

How is the temperature at altitude estimated from the surface?

T = T₀ − 2 °C per 1 000 ft (ISA lapse rate 1.98).

Th≈T0−2 h1000T_h\approx T_0-2\,\dfrac{h}{1000}
  • If the surface is 25 °C, at 10 000 ft the temperature is ≈ 5 °C (ISA)
  • Freezing level: height where T = 0 °C
  • Height of 0 °C isotherm = surface temp (°C) / 2 × 1 000 ft

Make the Aviation Meteorology formulas stick

In class every formula is built from a diagram first, then practised until it is automatic.

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