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DGCA Aviation Meteorology · 174 questions

Meteorological Calculations: DGCA Aviation Meteorology Questions and Answers

174 practice questions on Meteorological Calculations with answers and explanations, plus the key concepts and formulas, from the TrueHeading question bank.

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About this topic

Use this page to revise Meteorological Calculations for the DGCA Aviation Meteorology paper. Below are the key ideas first, then 12 practice questions with answers and explanations, picked from the 174 questions in the TrueHeading bank for this topic.

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

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Key concepts: Meteorological Calculations

5 ideas to know cold.

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

Practice questions: Meteorological Calculations

Tap “Show answer” after you have tried each one.

Q1What is the ISA temperature at FL040?

  1. 15 °C
  2. 7 °C
  3. 1 °C
  4. 13 °C
Show answer

Answer: B. 7 °C

ISA: 15 °C − 1.98 °C per 1 000 ft × 4 = 7 °C (tropopause isothermal −56.5 °C above 36 090 ft).

Q2At FL300 the outside air temperature is -12 °C. The temperature deviation from ISA is:

  1. ISA +32 °C
  2. ISA +40 °C
  3. ISA +28 °C
  4. ISA +24 °C
Show answer

Answer: A. ISA +32 °C

ISA at FL300 = -44 °C, so deviation = -12 − (-44) = +32 °C.

Q3At FL380 the outside air temperature is 21 °C. The temperature deviation from ISA is:

  1. ISA +68 °C
  2. ISA +84 °C
  3. ISA +78 °C
  4. ISA +88 °C
Show answer

Answer: C. ISA +78 °C

ISA at FL380 = -56 °C, so deviation = 21 − (-56) = +78 °C.

Q4An airfield elevation is 500 ft and the QNH is 1027 hPa. Assuming 1 hPa = 30 ft, the pressure altitude of the airfield is:

  1. 80 ft
  2. 90 ft
  3. 60 ft
  4. 130 ft
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Answer: A. 80 ft

PA = elevation + (1013 − QNH) × 30 = 500 + (-14) × 30 = 80 ft.

Q5Airfield elevation 3000 ft, QNH 1026 hPa. Assuming 1 hPa = 30 ft, the QFE is approximately:

  1. 936 hPa
  2. 926 hPa
  3. 918 hPa
  4. 930 hPa
Show answer

Answer: B. 926 hPa

QFE = QNH − elevation/30 = 1026 − 100 = 926 hPa.

Q6Airfield elevation 3000 ft, QNH 1012 hPa. Assuming 1 hPa = 30 ft, the QFE is approximately:

  1. 912 hPa
  2. 920 hPa
  3. 908 hPa
  4. 915 hPa
Show answer

Answer: A. 912 hPa

QFE = QNH − elevation/30 = 1012 − 100 = 912 hPa.

Q7Pressure altitude 5000 ft and OAT 20 °C. Using 120 ft per °C of ISA deviation, the density altitude is approximately:

  1. 7500 ft
  2. 8500 ft
  3. 5800 ft
  4. 6800 ft
Show answer

Answer: D. 6800 ft

ISA temp 5.1 °C; deviation +14.9 °C; DA = 5000 + 120 × 14.9 ≈ 6788 ft.

Q8Surface temperature 12 °C and dew point 10 °C. Using the 400 ft per °C rule, the cloud base of convective cloud is approximately:

  1. 500 ft
  2. 700 ft
  3. 900 ft
  4. 800 ft
Show answer

Answer: D. 800 ft

Spread 2 °C × 400 ft = 800 ft above ground.

Q9The surface temperature is 34 °C. Assuming a standard lapse rate of about 2 °C per 1 000 ft, the freezing level is approximately at:

  1. 19600 ft
  2. 15300 ft
  3. 23800 ft
  4. 17000 ft
Show answer

Answer: D. 17000 ft

34 ÷ 2 × 1 000 = 17000 ft above the surface.

Q10A wind speed of 85 kt is approximately:

  1. 26 m/s
  2. 44 m/s
  3. 48 m/s
  4. 50 m/s
Show answer

Answer: B. 44 m/s

1 kt = 0.514 m/s; 85 × 0.514 = 43.7 m/s.

Q11-15 °C converts to:

  1. 9 °F
  2. 11 °F
  3. 5 °F
  4. 8 °F
Show answer

Answer: C. 5 °F

°F = °C × 9/5 + 32 = 5 °F.

Q12Air at sea level on the windward side has a temperature of 24 °C and a dew point of 14 °C. It is forced over a ridge 7000 ft high. Use DALR 3 °C/1 000 ft, SALR 1.5 °C/1 000 ft and condensation level = spread × 400 ft. The temperature at sea level on the lee side is approximately:

  1. 24.5 °C
  2. 28.5 °C
  3. 31.5 °C
  4. 34.5 °C
Show answer

Answer: B. 28.5 °C

Condensation level 4000 ft; temp there 12.0 °C; at ridge top 7.5 °C (SALR); descending dry to sea level adds 3 °C/1 000 ft: 28.5 °C.

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