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An aircraft is departing from an airport which has an elevation of 2000 ft and the QNH is 1003 hPa. The TAS is 100 kt, the head wind component is 20 kt and the rate of climb is 1000 ft/min. Top of climb is FL 050. At what distance from the airport will this be achieved?
  • A
    5.4 NM
  • B
    4.0 NM
  • C
    3.6 NM
  • D
    4.4 NM

(1) Convert True Altitude into Pressure Altitude:

1013 hPa - 1003 hPa = 10 hPa

Assuming 30 ft/hPa:

10 hPa x 30 ft/hPa = 300 ft
Standard pressure (ISA) is higher than local QNH => pressure altitude is higher than true altitude: 2 000 ft + 300 ft = 2 300 ft

(2) Determine change in Height:

Climbing from 2 300 ft to 5 000 ft equals a change in height of 2 700 ft.

(3) Calculate time to climb 2 700 ft at a rate of climb of 1 000 ft/min:

2 700 ft : 1 000 ft/min = 2.7 min

(4) Calculate Ground distance:
  • Ground speed = TAS - headwind
    Ground speed = 100 kt - 20 kt = 80 kt
  • Ground distance = (80 kt x 2.7 min) / 60 min
    Ground distance = 3.6 NM

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