From 15 October 2016, AI 173 from Delhi to San Francisco started flying over Pacific. Jet streams over Pacific helped Air India in reducing the flight time by 2 hours, despite increasing the travel distance by 1100 km. This also made AI 173 the world's longest flight, with total distance of over 9300 miles (15,000 kilometres). It completes the journey in 14.5 hours, with average speed of 1040 km/h; 560 knots.

(Adapted from here.)

As many people are wondering, could winds of up to 150 km/h impact the structural loads on the B777-200LR? Is it ever likely to be dangerous?

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    $\begingroup$ 150 km/h (8,000 fpm) vertical wind may be a problem while 300 km/h horizontal tail wind jet streams are usually welcome, but not always. $\endgroup$
    – mins
    Commented Oct 19, 2016 at 10:34
  • $\begingroup$ Wow! I wonder if pilots can actually "feel" that they are in a jetstream without having to compare speed relative to the ground vs air speed. $\endgroup$
    – Nav
    Commented Oct 19, 2016 at 15:45
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    $\begingroup$ @Nav: You cannot sense speed (only acceleration), especially when above clouds at the same velocity. Note that Earth is spinning very fast, but we don't feel it. $\endgroup$
    – mins
    Commented Oct 19, 2016 at 15:55
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    $\begingroup$ High wind does not overload aircraft for the same reason that the rotation of the Earth does not overload road vehicles. $\endgroup$ Commented Oct 19, 2016 at 17:37
  • $\begingroup$ Note that SIN-EWR still holds the record for the world's longest scheduled flight (direct route would be 9,534 sm, actual would be more.) This might be the longest currently-active scheduled flight by actual flown distance, though. By direct distance, the DXB and DOH to AKL routes are longer, though. SIN-SFO is also longer than DEL-SFO by direct distance. $\endgroup$
    – reirab
    Commented Oct 19, 2016 at 19:19

3 Answers 3


Could winds of up to 150 km/h impact the structural loads on the B777-200LR?

Not at all.

The dynamic pressures on the plane depend on the plane's velocity with respect to the air, not the ground speed.

Flying in 150 km/h tailwind is the same as flying with no wind, the plane's indicated and true airspeeds won't be affected. True airspeed is the plane's velocity inside the fast moving wind tube.

Since the jet stream itself is moving, it just moves the plane with it. With the added bonus of extra ground speed, or if you're the pilot, a dreaded return flight—unless you can avoid the now headwind, usually they are avoided for the Pacific and Atlantic routes, not as often for elsewhere.

Here's a 1200 km/h subsonic flight: Jet Stream Blasts BA Plane Across Atlantic in Record Time

  • $\begingroup$ I think it is worth mentioning that to get into the jet stream, you should go through a gradient. I guess this is smooth enough in real life, but if you would enter a 150km/h wind shear, it would not be pretty. $\endgroup$ Commented Oct 19, 2016 at 18:41
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    $\begingroup$ @MartinArgerami It's pretty bumpy in real life. Nothing dangerous to the airplane, but enough to get the FAs to sit down in my experience of Asia->N.Amer trans-Pac crossings. It's amazing watching the IFE's indicators go from showing 15 mph headwind to 180 mph tailwind, but it can get pretty bumpy for a while. It's pretty interesting watching the groundspeed rise to 770-780 mph, too. If your airspeed were actually that fast, that would be around Mach 1.1 at that altitude. $\endgroup$
    – reirab
    Commented Oct 19, 2016 at 19:06
  • $\begingroup$ Another example of riding the jet stream: goo.gl/6CVtvD (but the BA plane wins). $\endgroup$
    – pr1268
    Commented Feb 22, 2018 at 16:11

No, there's no considerations on structural load because the airplane is moving in relation to the air mass. Think of the jet stream as a river, as air is just another fluid. A boat heading downstream on a fast moving river will move faster in relation to the land, but in relation to the water it's moving the same speed as if it were on a lake with no current.


ymb1's and GdD's answers are very good, I'd like to expand them a bit further.

The loads on the airplane structure are caused by forces between air and the plane.

If the stream can be described as laminar flow, there is no difference between flying in headwind, tailwind or calm air. Except for ground speed, obviously. When onboard and near the wing, you won't notice anything at all.

When the stream is (highly) turbulent, that means there are significant changes in the wind directions, pressure, temperature and density these inhomogenities creates additional loads to the structure. You will notice this; the fight will be like rollercoaster ride and you will see the wings flapping and wild corrections.

This characteristic can be estimated using Reynolds number, where stream velocity is used, but it is not the only variable to form the criterion.


It depends.

  • $\begingroup$ What do you want to compute by the Reynolds number here? It is in millions whatever wind you have. What conclusion would you like to take out of that? The turbulence in the air itself is governed by other processes and the Reynolds number scaled by the airplane's dimensions doesn't play any role at all. $\endgroup$ Commented Oct 19, 2016 at 14:17
  • $\begingroup$ @VladimirF I don't want to estimate Re near the airplane, it is obvious that it will be high around (behind) it. I want to estimate the jet stream flow behaviour without the plane (several meters in front of it). Turbulences are governed by the air flow, the "weather" - humidity, temperature, termics. I just wanted to show, that the air speed is not the lonely factor, and Re came to me as a first rough estimate. $\endgroup$
    – Crowley
    Commented Oct 19, 2016 at 15:20
  • $\begingroup$ I can't how that would work. Maybe the Richardson number, but I simply can't see a useful direct application of the Reynolds number there. $\endgroup$ Commented Oct 19, 2016 at 16:20
  • $\begingroup$ The Reynolds numer tells you if molecular viscosity is important or not. In the jetstream you can forget the viscosity. It is completely negligible and therefore so is the Reynolds number. You must use criteria that describe the relevant physics in your situation. $\endgroup$ Commented Oct 19, 2016 at 16:27

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