> A more serious question: I assume this kind of event doesn't actually save fuel because the airplane still has to maintain the same airspeed to avoid stalling, right?
If a plane maintains the same airspeed, but gets extra ground speed thanks to a tailwind, it'll complete its journey in less time and therefore should save fuel. (Unless it ends up having to circle the destination airport while it waits for its original landing slot!)
Of course, planes going the other direction will use extra, so overall we don't win.
> Of course, planes going the other direction will use extra, so overall we don't win
Aren't some(all?) long-haul routes chosen on a flight by flight basis to take advantage of favourable winds where possible?
Compare yesterday's eastbound BA11 (LHR-Singapore) vs the westbound BA12 (Singapore-LHR) flights.[0] Neither route looks like a great circle.
Eastbound routing: London - north of Berlin - Minsk - Voronezh - Volgograd - cross Caspian sea - Turkmenistan - Lahore - New Delhi - KL - Singapore.
Westbound: Singapore - KL - south of Jaipur - Iran - just touched Turkmenistan - south of Baku - Tbilsi - Turkey (just) - south of Prague - south of Dortmund - LHR
Typically yes. Right now there is a general avoidance of Ukrainian (Crimean and nearby) and Syrian airspace so the flight plans will take that into account, the winds aloft, and the airways available.
Yes, most of the time. There is actually Winds aloft data that is made available to pilots to select routes. And in lots of cases IFR routes are designed keeping these speeds in mind.
> Of course, planes going the other direction will use extra, so overall we don't win
It's even worse than not winning. You actually lose if you have to fly in the same wind both ways, once as a headwind and once as a tailwind. The extra time in the into the headwind direction is more than the time saved in the with a tailwind direction.
If the ground distance each way is D, airspeed is V, and wind speed relative to the ground is w, total time is D/(V-w) + D(V+w) = 2DV/(V^2-w^2) = 2D/(V-w^2/V).
For 0 < w < sqrt(V), this is more than the 0 wind case. (For w >= sqrt(V), the headwind is so high that you can't make any progress toward the destination).
Just a small fix for those that were checking as I was
If the ground distance each way is D, airspeed is V, and wind speed relative to the ground is w, total time is D/(V-w) + D/(V+w) = 2DV/(V^2-w^2) = 2D/((V^2-w^2)/V).
For 0 < w < V: this is more than the 0 wind case.
For w >= V: the headwind is so high that you can't make any progress toward the destination).
Or just fly with the jetstream in all cases, all the way around the earth if necessary. Some trips would be quite long, but think how much fuel they'd be saving!
If a plane maintains the same airspeed, but gets extra ground speed thanks to a tailwind, it'll complete its journey in less time and therefore should save fuel. (Unless it ends up having to circle the destination airport while it waits for its original landing slot!)
Of course, planes going the other direction will use extra, so overall we don't win.