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Refer to figure.
You are taking-off with a single engine aeroplane and field length limited take off mass from runway 27 in VMC, for a straight out departure. Given the following information, what will be the expected obstacle clearance over the water tank on the extended centerline?

Climb gradient: 8%
TAS: 111 kt
Wind component: 5 kt tailwind
1 NM = 6 080 ft
Height Difference = ((GD x TAS)/GS) x Gradient/100

  • A
    215 ft
  • B
    465 ft
  • C
    515 ft
  • D
    265 ft

Using the given formula and the figure's data, we get the height difference, when over the water tank at the extension of Rwy 27 (1 NM after the end of TODA): Height Difference = ((GD x TAS)/GS) x Gradient/100 = ((1 NM x 6 080 ft x 111 kt) / (111 kt + 5 kt)) x 8/100 = 465 ft AMSL.

For a single-engined aeroplane, we assume that the required climb gradient will be achieved at the end of TODA at the screen height of 50 ft. Thus, the height difference above the water tank is: 465 ft + 50 ft = 515 ft AMSL.

Since, the the water tank's elevation is 250 ft AMSL, the aircraft will clear the water tank on the extended centre line by: 515 ft - 250 ft = 265 ft.

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