If an aeroplane makes a round trip and a wind is blowing, is the trip time shorter, longer or the same as without wind?

Round trip mean aeroplane go from point A to point B and then return to point A.

The velocity of the wind is constant.

  • $\begingroup$ The easiest way to deduce the answer is to look at the case where the wind speed is the same as the plane's speed. The plane will take half the time one way, but infinite time the other. $\endgroup$ – Justin Jan 28 '15 at 21:05
  • $\begingroup$ This one smells to me of a math problem rather than a math puzzle, but I'll hold off on closing. I could see the result being seen as a bit surprising, and Gerhard posted an intuitive solution, though the algebraic solution is totally standard and textbook. $\endgroup$ – xnor Jan 29 '15 at 3:30
  • $\begingroup$ Equivalent problem: you are at the departure gate for your flight and you realize you forgot to buy a present at the tax free shop. In between the gate and the tax free shop is a moving walkway. Given a choice of entirely avoiding this moving walkway, or using this moving walkway both ways, what do you do? $\endgroup$ – Johannes Jan 29 '15 at 12:59
  • $\begingroup$ @Johannes There are a number of similar problem formulations that avoid the loopholes that Nattgew and I used. A boat traveling up and down a river, for example. $\endgroup$ – KSmarts Jan 29 '15 at 14:28

It takes longer. Let the distance be $d$, the plane speed be $v$, and the wind speed be $w$.
With no wind, the round trip takes $\frac {2d}v$.
With wind the round trip takes $\frac d{v+w}+\frac d{v-w}=\frac{d(v-w)+d(v+w)}{v^2-w^2}=\frac {2dv}{v^2-w^2}=\frac {2d}{v-\frac {w^2}v}$.
Decreasing the denominator increases the time.
If $w \gt v$ he can never return.

  • $\begingroup$ +1 if w > v it is probably facing a tornado F5. $\endgroup$ – Juan Carlos Oropeza Jan 28 '15 at 17:15
  • $\begingroup$ @JuanCarlosOropeza: That depends on the airplane. The Gossamer Condor was quite slow, though I didn't find a speed. $\endgroup$ – Ross Millikan Jan 28 '15 at 17:18
  • $\begingroup$ I just add $T_0$ over $T_W \Rightarrow $ $\frac{T_0}{T_W} = 1 - (\frac {W}V)^2 \Rightarrow T_0 < T_W$ for all non zero values of W $\endgroup$ – Juan Carlos Oropeza Jan 28 '15 at 17:49

The overall trip becomes longer: one way the wind speed is essentially added to the aeroplane speed, and the other way the wind speed is essentially subtracted. The trip with addition (and faster speed) takes shorter time than the trip with subtraction (and slower speed).

In other words, the plane is sped up for a short time and slowed down for a longer time. This makes the overall trip to become longer.


It will usually take longer, as others have said. However, under the right conditions, the wind can make for a shorter trip.

It depends on where point A and point B are, and the direction and speed of the wind. If, for example, you are flying round-trip between Washington D.C. and Singapore, the distance from one to the other is approximately the same whether you go east or west. So, if the wind is blowing to the east or west, you can travel from A to B by flying with the wind, and then from B to A, while still flying with the wind. In this case, the wind will make you go faster. Even if the two points aren't on approximately opposite sides of the world, it could still be faster to go "the long way around", if the winds are strong enough.


There is not enough information to answer the question.

It is stated that wind speed is constant, but not direction. So there could be a tail wind in both directions, which would be faster than without wind.

Or, if wind direction is constant, the plane could circumnavigate to go from A to B. Then circumnavigate back from B to A, in the same direction. This way there could be a tail wind for both flights. Near the poles, this wouldn't have to be an overly long flight. (I guess this is similar to the half-way idea that KSmarts answered)

The E6B calculator is helpful.

  • $\begingroup$ You can assume Wind vector, not Wind scalar. Also if you go from A to B and then B to A for a different route you can't compare both times no matter the Wind. $\endgroup$ – Juan Carlos Oropeza Jan 28 '15 at 22:39
  • $\begingroup$ I thought that we are comparing round trip with and without wind, not A-B with B-A. $\endgroup$ – Nattgew Jan 28 '15 at 22:45
  • $\begingroup$ Yes, we are comparing times with and without wind, but as you mention shorter distance. The comparation won't make sense. I understand your idea of find an optimal way using other variables, but that wasn't the question. $\endgroup$ – Juan Carlos Oropeza Jan 28 '15 at 22:53
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    $\begingroup$ @JuanCarlosOropeza The route from A to B is a different route than the one from B to A by its very nature, unless A and B are the same point. $\endgroup$ – KSmarts Jan 28 '15 at 22:56
  • $\begingroup$ @JuanCarlosOropeza You asked about how long the trip takes. You said nothing about the distance, and taking a longer-distance path can often result in shorter times in real life. The shortest route from my house to my job is a straight line, but I'll get there faster if I stay on the roads. $\endgroup$ – KSmarts Jan 28 '15 at 22:59

Not a proof but a strong hint that the trip should be longer comes from the case where the wind matches the normal flight speed of the plane. It will not be able to complete any segment of the tour against the wind. The travel becomes infinite. For intermediate speeds, the time should somewhere in between, between unchanged and infinite.

This being said, if the plane circumnavigates the globe and is pushed by a constant westward wind like the jetstream, of course the trip becomes shorter.


It should take almost the same time. If the wind was to remain completely constant for the whole trip, at all altitudes (extremely unlikely), the cruise legs added together would be the same.

During the phases of take off, climb, approach and landing, there would be differences. Depending on the procedures in use (VFR, IFR, SIDs, STARs, etc), the time will be different. It might be necessary to fly the approach into a headwind/tailwind during a joining procedure or instrument approach, while the take off may be straight out into a headwind. The wind direction will require the aircraft to turn around after take off when the cruise has a tailwind.

This supposes that the aircraft follows the exact same route for the cruise in both directions (unlikely), and flies at the same altitude (extremely unlikely, unless the entire flight is conducted outside controlled airspace and there is no conflict risk).

All of these factors can influence the flight time.

Generally, in reality, no wind provides the shortest flight time. There are exceptions, but not with constant wind. I'm a pilot myself and can say for sure that from experience the shortest flights are usually the ones with no wind.

Another point to take into consideration is that an aircraft moves slower while it's climbing and faster while it's descending. If it has a tailwind on both legs, the climb would take the same time, but would cover more distance at the slower speed, the descent would usually also take the same time, but cover more distance again at the faster speed. The reverse is true for a headwind. So again these factors would affect the flight.

The take off and landing is always flown into wind and is at a slower speed, that would make the overall trip longer the higher the wind speed.


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