I'm interested to know how much time a descent and ascent to cruise altitude takes as part of a longer journey.

A standard jet-aircraft (for example a A340 or 777) flies for eight hours from City A to City C. While doing so it flies over City B.

Now the same jet flies from City A to City C but as it approaches City B it descends and lands at B. The plane taxis to the gate. As soon as it's there it pushes back, taxis, takes off and returns to cruise flying to City C.

Approximately how much time is added to the journey by virtue of having descended, landed, taken-off and climbed back to cruise? Assume no ATC delays etc.


1 Answer 1


Let's take the best case scenario to start with, and assume that the cruise ground speed of the airliner is around 400kts.

For our best case, we also assume that the runway is lined up with the flight path and that the airspace is quiet so the plane is cleared "straight in" with no vectoring for separation or holding patterns.

  1. Descent:

A typical descent profile would start 100nm out, during which the plane would slow pretty uniformly from around 400kts to around 140kt. This gives an average groundspeed of 270kt. So this 100nm is covered in around 25 minutes rather the 15 minutes at full cruise.

The descent has cost us only 10 minutes.

  1. Taxi and turnaround:

At a smallish (and well designed) airport the rollout and taxiing to the stand may only take 5 minutes, plus another 5 to attach the air bridge/stairs and open the door. Passenger and baggage turnaround would typically take around 30 minutes, but you have specifically excluded this in your question so we can ignore it. Then getting departure clearances, startup and taxiing to the active: say another 15 minutes.

So this phase has cost more or less 25 minutes.

  1. Take off and climb out:

Again assuming we're cleared for immediate take off, it takes around 20 minutes to climb to cruise altitude, at an average ground speed of (say) 200 knots. This means we've covered about 70 miles, which the cruising aircraft would have covered in just over 10 minutes.

So this final phase has cost us another 10 minutes.

Total for the best case scenario is 45 minutes.

Of course the best case is not that likely. If the airspace is busy, holding for another 10 minutes before final approach, and 10 minutes further wait for departure and takeoff clearances, would not be surprising. And if the active runway is in the "wrong" direction or vectoring is needed for traffic or noise reasons, that can easily add another 10 minutes or so.

So in the non-best case scenario, a total loss of 1h15 to 1h30 would not be unreasonable.

  • 3
    $\begingroup$ Excellent answer, thank you. $\endgroup$
    – glaucon
    Commented Feb 26, 2016 at 10:56
  • $\begingroup$ Great answer, thanks. I'm wondering if the total added time makes sense, given that many short flights take 1h10 (e.g. Zurich-Nice) and still have to traverse 400km/215nm $\endgroup$
    – Chris
    Commented Jul 3, 2017 at 15:33
  • $\begingroup$ I've compared the scheduled flight times of URC-TSE-SVO (where the non-stop URC-SVO flight directly overflies TSE), and the extra time needed seems to be around 30 minutes. How can it be possible? $\endgroup$ Commented Sep 19, 2018 at 17:07
  • $\begingroup$ Either one or more of my estimates/assumptions was wrong, or the two flights have different flight profiles (speeds/altitudes). Eg without looking, it's possible that the direct flight was flown at a slightly slower speed to save on fuel. $\endgroup$
    – IanF1
    Commented Sep 19, 2018 at 17:12
  • 1
    $\begingroup$ I was curious if I could find other sets of cities exhibiting this behavior. SFO-DEN-PIT offers an example. Quoted times vary slightly across airlines: SFO-DEN (2hr 30 min - 2hr 40 min); DEN-PIT (2hr 50 min - 2hr 58min). Selecting UA, we have SFO-DEN at 2hr 37 min, DEN-PIT at 2hr 56 min, and SFO-PIT at 4hr 47 min, for a total loss of 46 min. $\endgroup$ Commented Feb 5, 2023 at 0:36

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