Equal Time Point and Critical Fuel
On an oceanic sector, the equal time point (ETP) between the departure and destination is computed with one engine inoperative. True track 090, distance 1800 NM. With all engines, TAS 480 kt and headwind 30 kt. One-engine-inoperative TAS 400 kt, and the same wind applies. The aircraft must be able to reach a suitable alternate from the ETP with 30 min holding plus 15 % of the one-engine-inoperative fuel. The one-engine-inoperative fuel flow is 3200 kg/h. The alternate is 300 NM beyond the ETP along track.
1. Explain how the ETP is determined and why wind must be applied consistently on both legs.
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The ETP is the point where time to the departure equals time to the destination under the assumed condition. With one engine inoperative, ground speed outbound = 400 - 30 = 370 kt (headwind), and ground speed inbound (reciprocal track) = 400 + 30 = 430 kt. The ETP distance from departure is found by equating times: d/370 = (1800 - d)/430, so 430d = 370(1800 - d), giving 800d = 666 000, d = 832.5 NM. Applying wind to only one leg, or using all-engine speeds, would move the ETP and understate the fuel needed. The wind must be the forecast at the relevant level and applied to the actual ground speed on each leg, not to TAS.
2. Compute the one-engine-inoperative fuel required from the ETP to the alternate and assess whether a typical reserve is adequate.
Read the worked answer
Distance ETP to alternate = 300 NM. Inbound ground speed = 430 kt, so time = 300/430 = 0.698 h = 41.9 min. Fuel at 3200 kg/h = 3200 x 0.698 = 2233 kg. Add 30 min holding at 3200 kg/h = 1600 kg, and 15 % of the one-engine-inoperative fuel = 0.15 x 2233 = 335 kg. Total = 2233 + 1600 + 335 = 4168 kg. This is the critical fuel that must be on board at the ETP. If the planned fuel at the ETP is less, the ETP must be moved or the route changed. The judgement is to check the critical fuel against the actual fuel state at the ETP, not against a nominal reserve figure.