A plane exerts downward force on the air, so the ground must exert greater upward force on the air while a plane is in flight than when it is not (in a steady state where the air has no net acceleration). This means the pressure is higher at the ground when a plane is in flight.

If a passenger jet flies several miles over me, directly above me, I don't expect to notice any change in the air pressure, even though the plane is very heavy. This suggests that the area over which pressure increases under the plane is pretty large. How large? And does it stay directly under the plane (if flying at constant speed) or lag behind?

  • $\begingroup$ When you say "stay" directly under the plane, is that in the plane's frame of reference? Keep in mind that if you're directly under the plane, then a longitudinal lag in the plane's frame translates to a temporal lag in the on-the-ground frame. I.e. that last question could be sharpened somewhat. $\endgroup$ Commented Apr 25, 2018 at 17:18
  • $\begingroup$ I don't see how the frame of reference chosen affects whether or not the area is directly under the plane. I follow your second to last sentence, but I don't see any ambiguity that it highlights. $\endgroup$ Commented Apr 25, 2018 at 17:28
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    $\begingroup$ Any reason why you're posing the question in terms of a moving airplane instead of a stationary helicopter or drone hovering in mid-air? Or are you interested in the issue of temporal lag in the pressure distribution with a moving aircraft? $\endgroup$
    – user93237
    Commented Apr 25, 2018 at 17:31
  • $\begingroup$ No, I just happened to see an airplane today, and not a drone, blimp, or helicopter. I added the last part out of curiosity when I got to the end of the question. $\endgroup$ Commented Apr 25, 2018 at 17:43

2 Answers 2


Break it down, and simplify. A 3D wing is creating lift via propulsive mechanism, by vortex-shedding (which is very unlike the venturi-effect forces of an airfoil-section in the 2D world of textbook flow diagrams.)

To construct an intuitive 3D fluid model, imagine that the aircraft is assembling bags of air, then flinging them downwards. (Let the bags' own mass be insignificant, like giant dry-cleaning bags, or imagine that the craft is flinging giant chunks of aerogel!) The aircraft receives an upwards F=mA force as it accelerates and releases each air-bag. That's the first N3 force-pair.

The aircraft flies away, and doesn't interact any further with each descending bag. But the bags experience viscous drag. They will accelerate the surrounding air downwards, and themselves be slowed. They produce an enormously wide pattern of entrained descending air. If no ground was present, they'd entrain a wide portion of the atmosphere, which would move downwards. But if the ground prevents this, then the bags would come to a halt wrt the ground, and a very wide pressure-pattern would appear on the ground. That's the second, independent N3 force-pair.

In other words, what will happen on the ground if a down-moving sphere of air is decelerating because of viscous drag? Now just sum the effects of a long string of such descending spheres, and we have the ground-pressure distribution caused by aircraft vortex-wakes.

(Now add force-propagation acoustic limits ...and this simplified picture doesn't apply very well.)

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    $\begingroup$ So your answer is "very wide"? I was hoping we could be a bit more specific than that. $\endgroup$ Commented Apr 26, 2018 at 6:25
  • $\begingroup$ @MarkEichenlaub the conventional version, based on Bernoulli force from airfoil moving at infinite aircraft velocity (Prandtl's un-tilted horseshoe diagram) is in Prandtl 1934 pp185-, see books.google.com/…, with pressure pattern inversely proportional to square of altitude. $\endgroup$
    – wbeaty
    Commented Apr 30, 2018 at 6:58

the phenomenon you describe only occurs when the plane in question is less than approximately one wingspan off the ground, as for instance just before touchdown and just after takeoff. this is called "ground effect" and is completely absent during normal flight.

When in ground effect, the pressure bulb that builds under the plane acts over an area of the ground approximately the same as the area of the underside of the plane itself. you can think of this pressure bulb as forming when the downwash from the wing (which is responsible for the lift it creates) is transiently entrained between the underside of the wing and the ground. when flying in this mode, the wing's apparent lift-to-drag ratio increases significantly and the pilot of the plane can feel the plane "float" along on that bubble of entrained air.

This effect can be seen when watching a seaplane take off or land, when a "shadow" of the plane appears in the surface of the water under it. you can also feel the pressure pulse pass over you when lying directly under the path of an airplane just before it lands. As I have done.

when a wing is not in ground effect, it generates lift by tipping the momentum vector of the air through it moves, propelling that air downward during its passage through it and thereby generating an upward reaction force on the wing. this process can be modeled in a variety of equivalent ways (circulation, vortex sheet, pressure distribution) but none of them involve the proximity of the ground in order to generate lift.

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    $\begingroup$ I assure you it is not completely absent for normal flight, that's an obvious violation of newton's laws. $\endgroup$ Commented Apr 25, 2018 at 17:42
  • $\begingroup$ @markeichenlaub, i did not say that downwash is absent in normal flight, I said that ground effect is absent during normal flight. the wing of a 767 at 35,000 feet is not propagating a pressure distribution which relies upon the presence of the ground in order to generate lift $\endgroup$ Commented Apr 25, 2018 at 20:40
  • $\begingroup$ I find you quite unclear. For example, I don't know what it means to "propagate a pressure distribution". Suppose a plane is flying at 30,000 feet above the ground. You look at the air pressure at ground level underneath the plane. Would it be the same, more than, or less than its value if the plane were on the ground? $\endgroup$ Commented Apr 25, 2018 at 23:41
  • $\begingroup$ at 30,000 feet the plane is probably going 550 mph. if the plane is directly above you, the pressure you experience will be the same as if you were standing under the plane at rest on the ground. any pressure pulse created by the plane at 30,000 feet will take about 30 seconds to reach the ground by which time the plane will have moved 24,000 feet laterally. at that point you will hear the plane very faintly and perceive the sound source as being vertically above you even though the plane is 4.58 miles downrange. $\endgroup$ Commented Apr 25, 2018 at 23:50
  • $\begingroup$ So are you saying the pressure at ground level would be higher somewhere at ground level than it would be if the plane were not flying? $\endgroup$ Commented Apr 26, 2018 at 0:04

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