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.
- T = 24 °C, Td = 14 °C: spread 10, base ≈ 4000 ft above the surface
94 practice questions on Temperature, Humidity & Cloud Base with answers and explanations, plus the key concepts and formulas, from the TrueHeading question bank.
Use this page to revise Temperature, Humidity & Cloud Base for the DGCA Aviation Meteorology paper. Below are the key ideas first, then 12 practice questions with answers and explanations, picked from the 94 questions in the TrueHeading bank for this topic.
In short: 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.
Open the full set in the student zone to answer every question, track your accuracy and take timed mock tests.
5 ideas to know cold.
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.
Dew point is the temperature at which the air becomes saturated. Relative humidity is the vapour present as a percentage of the most the air could hold at that temperature.
A layer where temperature increases with height.
Maximum about 1–2 hours after local noon. Minimum around sunrise.
Radiation, conduction, convection (vertical) and advection (horizontal), plus latent heat from phase changes.
Tip: The atmosphere is warmed mostly from below by the Earth’s surface, not directly by the Sun.
Tap “Show answer” after you have tried each one.
Answer: A. 1 200 ft
Spread = 14 − 11 = 3 °C. Base ≈ 400 ft × spread = 1 200 ft (DALR 3 °C/1 000 ft minus dew-point fall of 0.5 °C/1 000 ft leaves 2.5 °C of closure per 1 000 ft).
Answer: D. 4 800 ft
Spread = 20 − 8 = 12 °C. Base ≈ 400 ft × spread = 4 800 ft (DALR 3 °C/1 000 ft minus dew-point fall of 0.5 °C/1 000 ft leaves 2.5 °C of closure per 1 000 ft).
Answer: A. 6 400 ft
Spread = 15 − −1 = 16 °C. Base ≈ 400 ft × spread = 6 400 ft (DALR 3 °C/1 000 ft minus dew-point fall of 0.5 °C/1 000 ft leaves 2.5 °C of closure per 1 000 ft).
Answer: C. 17 °C
Spread = base ÷ 400 = 7 °C. Dew point = 24 − 7 = 17 °C.
Answer: C. 7 °C
Spread = base ÷ 400 = 7 °C. Dew point = 14 − 7 = 7 °C.
Answer: C. 17.4 °C
Condensation level = 8 × 400 = 3200 ft. Cooling = 3 × 3.2 = 9.6 °C. Temperature = 27 − 9.6 = 17.4 °C.
Answer: B. 11.6 °C
Condensation level = 7 × 400 = 2800 ft. Cooling = 3 × 2.8 = 8.4 °C. Temperature = 20 − 8.4 = 11.6 °C.
Answer: D. 12 000 ft
Height = temperature ÷ lapse rate = 24 ÷ 2 = 12.0 thousand ft = 12 000 ft.
Answer: C. 375 m
Base ≈ 3 × 125 m = 375 m (equivalent to 400 ft per °C).
Answer: B. increases
Colder air can hold less vapour, so the RH rises.
Answer: C. close to the ground
Terrestrial radiation cools the surface, giving a ground inversion.
Answer: D. is constant with height
Isothermal layers are stable.
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