In-Flight Monitoring: Late Arrival Against a Slot and a Thin Fuel Margin
You depart A at 09:10 for B, 850 NM away, flight-planned at 425 kt ground speed with an ETA of 11:10 and cruise fuel flow of 1,050 kg/h. B has a landing slot that expires at 11:15; ATC has warned that aircraft arriving after the slot should expect 20 minutes of holding. At 10:40, after 1 h 30 min airborne, you have covered 600 NM and have 1,950 kg of fuel remaining. Weather at B is fine, but the wind at cruise level is stronger than forecast. Your operator's planning minima require arrival at B with fuel for a 25-minute flight to the alternate plus 30 minutes final reserve; the flight plan included 5% contingency. The commander asks whether to press on, speed up, or plan a diversion, and wants the numbers before deciding.
1. Establish the actual ground speed, the revised ETA and whether the 11:15 slot can be met.
Read the worked answer
Ground speed achieved = 600 NM ÷ 1.5 h = 400 kt, i.e. 25 kt below the planned 425 kt.
Remaining distance = 850 − 600 = 250 NM. Time remaining = 250 ÷ 400 × 60 = 37.5 min → revised ETA 11:17:30.
The slot expires at 11:15, so it is missed by about 2.5 minutes and you must plan on the 20-minute hold, putting arrival at B at roughly 11:37.
The monitoring lesson is that a 25 kt ground-speed shortfall is not noise - it is a wind or temperature deviation that should have been detected on the first leg check and fed back into the flight plan. Do not assume the loss will be recovered; assume it continues unless a later fix shows otherwise.
2. Compute the fuel at B with and without the 20-minute hold, and compare it with the required reserve.
Read the worked answer
Fuel to fly the remaining 250 NM at 400 kt = 0.625 h × 1,050 kg/h = 656 kg.
Fuel at B, no hold = 1,950 − 656 = 1,294 kg.
Required at B = alternate 25 min (25/60 × 1,050 = 437 kg) + final reserve 30 min (525 kg) = 962 kg (plus the unused 5% contingency of about 33 kg for the remaining trip if you wish to keep it intact).
So the no-hold case has about 332 kg of surplus - comfortable.
With a 20-minute hold at 1,050 kg/h = 350 kg: fuel at B = 1,950 − 350 − 656 = 944 kg, which is 18 kg below the 962 kg requirement, and about 51 kg below the figure that still preserves the contingency.
The hold cannot be absorbed without eating into the alternate or final reserve. Landing with 944 kg at B is still above the 525 kg final reserve, but if the approach at B then fails there is not enough fuel to divert to the alternate and land with 30 minutes remaining.
3. Recommend a course of action, with numbers, and state what you would monitor.
Read the worked answer
Option 1 - regain the slot. To cover 250 NM in 35 minutes you need 250 ÷ (35/60) = 429 kt, about 29 kt more. If the higher speed raises flow to about 1,180 kg/h, fuel to B = 35/60 × 1,180 = 688 kg, leaving 1,950 − 688 = 1,262 kg at B against a 962 kg requirement - 300 kg of margin and the slot met. The extra fuel burned is only about 32 kg, far cheaper than a 350 kg hold.
Option 2 - accept the hold. Not recommended: it leaves no fuel to divert if the approach fails, which is exactly the exposure you are trying to avoid.
Option 3 - divert now. Fuel at the alternate = 1,950 − 437 = 1,513 kg; very safe, but commercially unnecessary if option 1 works.
Recommendation: request the higher speed and ATC priority for the slot; if it cannot be granted, plan the diversion rather than drift into a hold that erodes the alternate reserve. Monitor the next position fix: if the achieved ground speed falls below about 390 kt, even the speed-increase case no longer works and the diversion decision must be taken immediately.