Round-Trip Fuel Planning with Wind and a Reserve Floor
You are planning a round-trip liaison flight in a single-engine airplane. Usable fuel is 300 gal, and fuel density is 6.7 lb/gal. Published cruise burn is 480 lb/hr at 180 kt true airspeed. The squadron order requires a reserve equal to the greater of 45 minutes at cruise burn or 20 percent of total usable fuel. The forecast wind is 25 kt directly on the nose outbound and 25 kt on the tail inbound. The detachment site is 280 nm from base, no refueling is available there, and the only alternate is 40 nm beyond it. Cruise altitude is planned at 8,500 ft; a lower altitude would raise burn to about 520 lb/hr.
1. Compute the maximum still-air radius at which the required reserve is preserved. Then state whether the 280 nm round trip clears that test, and by how much.
Read the worked answer
Fuel on board: 300 gal × 6.7 lb/gal = 2,010 lb
Reserve floor:
- 45 min at 480 lb/hr = 360 lb
- 20% of 2,010 lb = 402 lb
- Reserve required = 402 lb (the greater of the two)
Fuel available for transit: 2,010 − 402 = 1,608 lb
Endurance and range:
- 1,608 lb ÷ 480 lb/hr = 3.35 hr
- 3.35 hr × 180 kt = 603 nm still-air range → radius ≈ 301 nm
Verdict: 280 nm radius is inside 301 nm, so the still-air test passes - but only by about 21 nm (roughly 14 minutes of fuel). That margin is thin enough that it is not the right number to brief. The wind must be flown through before the trip is called good, which is exactly what the next task does.
2. Using the forecast wind, compute the actual fuel required for the 280 nm out-and-back and the fuel remaining on arrival. Does the flight meet the reserve requirement, and by how many minutes?
Read the worked answer
Outbound: groundspeed = 180 − 25 = 155 kt → 280 ÷ 155 = 1.806 hr
Inbound: groundspeed = 180 + 25 = 205 kt → 280 ÷ 205 = 1.366 hr
Total time: 3.172 hr
Fuel burned: 3.172 hr × 480 lb/hr = 1,523 lb
Fuel remaining on arrival: 2,010 − 1,523 = 487 lb
Required reserve: 402 lb → surplus = 487 − 402 = 85 lb, or 85 ÷ 480 × 60 ≈ 10.6 minutes
Verdict: legal, but the margin is ~11 minutes of fuel at cruise burn, not the ~14 naive minutes the still-air number implied. Note how the headwind hurts more than the tailwind helps: the outbound leg is flown 14% slower for the same distance, and that penalty is not symmetric with the inbound gain.
3. Weather forces the flight down to a lower altitude where burn is 520 lb/hr, with the same winds and distance. Go or no-go? Support it with numbers, and state what you would change rather than what you would shave.
Read the worked answer
Time is unchanged (same winds and distance): 3.172 hr.
Fuel burned: 3.172 hr × 520 lb/hr = 1,650 lb
Fuel remaining on arrival: 2,010 − 1,650 = 360 lb
Required reserve: 402 lb → short by 42 lb, which is about 42 ÷ 520 × 60 ≈ 4.8 minutes.
Verdict: no-go at 280 nm. The airplane would land with a legal-looking fuel quantity but a violated reserve.
What to change (not the reserve):
- Reduce radius. Solve for distance d: (d ÷ 155 + d ÷ 205) × 520 = 1,608 → d(0.006452 + 0.004878) = 3.092 → d ≈ 273 nm. Even that leaves zero margin, so a practical cut is closer to 250 nm.
- Or wait for a better wind forecast - a 15 kt reduction in headwind adds usable range at no fuel cost.
- Or add fuel at the site / move the detachment to a field with fuel.
- The reserve is a floor, not a planning variable. Shaving it converts a planning problem into an emergency.