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A helicopter is flying from A to B, a distance of 90 NM, with a heading of 090°. The TAS is 120 kt and 15 kt headwind is experienced from 090°. After arriving at point B, the helicopter is flying back to point A. The helicopter has an average fuel consumption of 4 kg/min, carrying an additional fuel of 92 kg for safety reasons. Calculate its maximum endurance and the flight time required.

  • A

    Flight time: 1 h 55 min; Endurance: 1 h 54 min

  • B

    Flight time: 1 h 31 min; Endurance 1 h 54 min

  • C

    Flight time: 1 h 45 min; Endurance 2 h 05 min

  • D

    Flight time: 2 h 05 min; Endurance 3 h 15 min

The headwind component of 15 kt from A to B switches into a tailwind component of 15 kt, because the heading from A-B is 090°, thus it is 180° from B-A​,​​​​​​ and the wind comes from 090°.

1. Groundspeed from A to B: TAS - Headwind = 120 kt - 15 kt = 105 kt.

2. Flight time from A to B: Distance / Groundspeed = 90 NM / 105 kt = 0.85 hr.

3. Grounspeed from B to A: TAS + Tailwind = 120 kt + 15 kt = 135 kt.

4. Flight time from B to A: Distance / Groundspeed = 90 NM / 135 kt = 0.66 hr.

5. Total Flight Time from A to B and back to A: 0.85 hr + 0.66 hr = 1.53 hr or 01 and 31 minutes

6. The additional fuel of 92 kg corresponds to: 92 kg / 4 kg/min = 23 minutes.

7. Thus, the total endurance is: 01:31 + 00:23 = 01 and 54 minutes.

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