Ferry Flight Fuel Judgement with a Worsening Headwind
A ferry flight is being planned. Usable fuel is 180 L and cruise burn is 32 L/h at 120 KTAS. Taxi and run-up consume 6 L. A fixed 45-minute reserve at cruise burn must remain untouched, and the destination alternate requires 20 minutes at 30 L/h. The planned leg is 420 NM with a forecast headwind of 15 kt at cruise level. No fuel can be added en route. The commander's standing order reads: 'Launch only if the planned leg can be flown without eating into either the fixed reserve or the alternate allowance.' A later forecast warns that the headwind will strengthen to 25 kt during the afternoon.
1. Calculate the fuel available for the leg and the corresponding endurance. Then test the planned leg against the standing order at the forecast 15 kt headwind.
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Step 1 - fuel that may not be used for the leg.
- Reserve: 45 min at 32 L/h = 0.75 x 32 = 24 L
- Alternate: 20 min at 30 L/h = (1/3) x 30 = 10 L
- Taxi/run-up: 6 L
- Total unavailable: 24 + 10 + 6 = 40 L
Step 2 - fuel and time for the leg.
- Available: 180 - 40 = 140 L
- Endurance: 140 / 32 = 4.375 h = 4 h 22.5 min
Step 3 - the leg.
- Ground speed = 120 - 15 = 105 kt
- Time = 420 / 105 = 4.00 h
- Fuel used = 4.00 x 32 = 128 L
- Surplus = 140 - 128 = 12 L, i.e. 12/32 h = 22.5 min
Verdict: the leg is inside the standing order - 4 h 00 min against 4 h 22.5 min available - but the margin is one reserve-width. The reserve is untouched by 12 L, which is the correct way to state it: the flight does not 'use' the reserve, it simply has 22.5 minutes less to spare than the fuel load suggests.
2. Determine the maximum headwind at which the flight may still launch under the standing order, and state whether the 25 kt afternoon forecast crosses it.
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Method: the endurance sets the minimum acceptable ground speed.
- Required ground speed = 420 / 4.375 = 96 kt
- Maximum headwind = 120 - 96 = 24 kt
Check by trial at 25 kt:
- Ground speed = 120 - 25 = 95 kt
- Time = 420 / 95 = 4.421 h = 4 h 25.3 min
- Fuel = 4.421 x 32 = 141.5 L against 140 L available
- Shortfall = 1.5 L, about 2.8 min
Verdict: the limit is 24 kt; a 25 kt headwind breaches the standing order by one knot and roughly three minutes. A one-knot violation is still a violation, and the honest reading of the numbers is that the launch decision now rests on a forecast that is itself uncertain by several knots in either direction.
3. The headwind is confirmed at 25 kt. Compare two options - shortening the route by 10 NM, or landing to refuel at the 210 NM point - and recommend one, using numbers.
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Option A - shorten by 10 NM.
- Leg = 410 NM, ground speed 95 kt
- Time = 410 / 95 = 4.316 h
- Fuel = 4.316 x 32 = 138.1 L against 140 L
- Surplus = 1.9 L, about 3.6 min It fits, but the margin is smaller than the forecast error. A further 1 kt of headwind removes it.
Option B - refuel at 210 NM.
- First half: 210 / 95 = 2.211 h, fuel = 70.7 L
- Land with 180 - 6 - 70.7 = 103.3 L still aboard, which is well above reserve plus alternate (34 L)
- Second half repeats the same figures after refuelling
- Every segment now has over 60 L of surplus instead of 2 L
Recommendation: Option B. Option A satisfies the arithmetic by three minutes and therefore depends on the forecast being right; Option B removes the forecast from the critical path. The commander's order is a floor, not a target - clearing it by 1.9 L is not the same as being safe. Option A should only be considered if the forecast is re-verified in flight before the point of no return.