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The turbulence and tip vortices behind an airliner are a familiar hazard for smaller aircraft when near an airport. But a lightweight glider seems to have not suffered from this, when its tail was actually struck by an Airbus A 320 at altitude. Page 8 of the accident report mentions only the choc brutal felt by the glider's occupants, and nothing about being blown around like a leaf.

At the time of collision, the airliner was descending through 8600 feet at a true airspeed of 280 knots, if my French can be trusted. The glider was flying at 53 knots, almost perpendicular to the airliner's path. So the glider would have encountered the airliner's wake for a good fraction of a second. Its long wings would be quite sensitive to turbulence.

Compared to the Airbus's high thrust at takeoff, or slow speed and high drag at landing, would its turbulence be less hazardous here, at low thrust and high speed?

https://www.skybrary.aero/index.php/Wake_Vortex_Turbulence#Accidents_and_Incidents mentions some "en-route" turbulence, but only from one aircraft trailing another. So maybe a cross-track incident is much milder.

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    $\begingroup$ I'm not sure an unprepared pilot can distinguish between severe wake turbulence and being struck by an actual airliner... $\endgroup$
    – Sanchises
    Sep 11 '20 at 16:27
  • $\begingroup$ The forces transmitted by mechanical impact surely differ from those transmitted by turbulence to long skinny wings? $\endgroup$ Sep 11 '20 at 16:41
  • $\begingroup$ @CamilleGoudeseune - BTW, wake turbulence or wing tip vortices are the most severe when the aircraft is at its heaviest and slowest while in a “clean” configuration. This is usually at takeoff. High thrust really would not be a factor unless you are directly behind the engine. Thrust forces would dissipate in a lot shorter time and distance. Whereas wake turbulence can stick around for minutes. And, although the wing tip vortex will descend and spread outward away from the wingtip, it would be almost non-existent directly below the wing tip. $\endgroup$
    – Dean F.
    Sep 11 '20 at 23:40
  • $\begingroup$ Interestingly enough, a Cessna 172 in a 30 degree bank can circle around onto its own flight path and hit its own wake turbulence in about a minute. It would not hit its own thrust forces. $\endgroup$
    – Dean F.
    Sep 11 '20 at 23:44
  • $\begingroup$ I used to commute to work in my plane and one morning I was approaching my destination airport (uncontrolled FSS only, but with heavy freighters coming and going in the early hours) and was about 3 miles south of the approach that had been used by a DC-10 freighter that had landed several minutes before, planning to get on a downwind to land the opposite way. I'm pretty sure I passed through remnants of his wake several miles south of the approach path because in the dead smooth air there was a sudden jolt like driving over a curb, then smooth again. That's the only thing that makes sense. $\endgroup$
    – John K
    Sep 12 '20 at 3:31
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To answer the question directly: Yes, in cruise turbulence is less severe, but it is lowest when flying low and fast.

We can actually calculate the downward component $v_z$ of the wake (which is a good proxy for turbulence intensity) in cruise and during approach when we know how much lift is being created: $$m\cdot g = \bar{m}\cdot v_z = \epsilon\cdot\frac{b^2}{4}\cdot\pi\cdot v_{\infty}\cdot\rho\cdot v_z$$ $$v_z = \frac{m\cdot g}{\epsilon\cdot\frac{b^2}{4}\cdot\pi\cdot v_{\infty}\cdot\rho}$$ The strength of the downwash is inversely proportional to the product of flight speed and density. This is not the dynamic pressure, which is proportional to flight speed squared times density, so at the same dynamic pressure the downwash gets stronger the higher you fly. This means a descending A320 at only 8000 ft should have a comparably weak wake.

Nomenclature:
$\kern4mm m\kern6mm$airplane mass
$\kern4mm g\kern7mm$gravitational acceleration
$\kern4mm \bar{m}\kern6mm$mass flow of air which is accelerated downwards
$\kern4mm v_{\infty}\kern4mm$flight speed
$\kern4mm b\kern7mm$wing span
$\kern4mm \rho\kern6mm$air density
$\kern4mm \epsilon\kern6mm$Oswald factor

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    $\begingroup$ Would I be correct (enough) in oversimplifying, that the amount of downwash, and tip vortex "strength" is always the same, but when going faster it just spreads out over larger volume of space, thus presenting itself to a bystander as lower in energy? (excluding flying close to ground, of course) $\endgroup$
    – Jpe61
    Sep 11 '20 at 21:58
  • $\begingroup$ Would $v_z\cdot\rho$ be a better proxy? Thinner air exerts less force. $\endgroup$ Sep 12 '20 at 2:26
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    $\begingroup$ @Jpe61 going faster means higher mass flow over (and under) the wing. Spreading the energy out is one thing, the other is that less energy is needed because the amount of wake deflection goes down with increasing speed. That is why induced drag is inversely proportional to the square of airspeed. $\endgroup$ Sep 12 '20 at 5:24
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    $\begingroup$ @CamilleGoudeseune: $v_z\cdot\rho$ is removing altitude dependency but adds one step to the explanation. So it is better indeed once the dependency has been explained. $\endgroup$ Sep 12 '20 at 5:27
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The glider in question was "in touch" with the tip vortex of the A320 only with it's tail, and very briefly: section 1.16.2 on page 24 of the accident investigation report Picture from section 1.16.2 on page 24 of the Accident investigation report.

The glider was flying at 53kts at almost a perpendicular angle to the path of the A320, and the tail of it struck the outer third of the leading edge of the Airbus. The G103 is about 8 meters long, a wing of an A320 is about 15 meters long, so the wings and most of the gliders fuselage were never superimposed by the wing of the A320:

Picture from section 2.4 on page 39 of the report Picture from section 2.4 on page 39 of the report.

As the glider was traveling across the path of the A320 at almost 27 meters per second, even its tail was out of the tip vortex in a blink of an eye, and in any case, only its tail ever was in the effect of the tip vortex. As the tip vortex forms at the tip of the wing, it's diameter is initially very small, growing larger as the distance to the plane grows.

Although the small tip vortex has high energy (high velocity), as the tail of the glider was very briefly traveling through the tip vortex, it was actually traveling in pretty much the same direction as the flow of the tip vortex was. So no great forces were exerted on the tail (except for the collision with the wing, of course...).

As such this incident can not be used as an example of the turbulence or tip vortex strength during cruise (or descent, to be precise), but the the said phenomena are less prominent at higher speed and lower angles of attack (normal cruise / cruise descent). They are at their strongest during slow flight when a high angle of attack is required to maintain a desired flight path (this would also apply for situations of high turn rate).

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  • $\begingroup$ Thanks @FreeMan , sloppy writing from me... $\endgroup$
    – Jpe61
    Sep 11 '20 at 19:59
  • $\begingroup$ Thanks, I'd overestimated how quickly turbulence would spread spanwise beyond the wings' span. Wind tunnel photos of smoke trails, or CFD renderings, are rarely shown from above. $\endgroup$ Sep 11 '20 at 21:27
  • $\begingroup$ Btw, there are so many lucky ppl in that incident... a very close shave! $\endgroup$
    – Jpe61
    Sep 11 '20 at 21:59
  • $\begingroup$ no worries, @Jpe61, that's what friends are for! ;) $\endgroup$
    – FreeMan
    Sep 12 '20 at 12:32

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