I need to find an equation for the following problem: Suppose a full-size glider passes 10 feet over my head at high speed. No doubt I will feel the downward air pressure (downwash) caused by the angle of attack of the wings. But if the same glider flies over me at a distance of one mile, I will not feel any effect at all.

So obviously there is a diminishing strength of the downward air pressure that eventually lessens to a point where I cannot physically detect it. Therefore, is there an equation that, with all relevant criteria being inputted, such as air temp, air density, wing speed, wing characteristics, wing angle of attack, etc., can allow me to at least approximate how far I must be below a passing wing to be unable to detect its downward pressure wave?

  • $\begingroup$ It seems that you are confusing ground effect and wake. An airplane wake can move a long distance and still be felt, whether near the ground or not. Ground effect relates to how the flow around the aircraft interacts with the ground. $\endgroup$
    – Adam
    May 25, 2018 at 13:31

2 Answers 2


What you probably are looking for is the Biot-Savart law.

This is originally describing the magnetic field generated by an electric current; however, it was soon found that it equally well describes the induced speeds caused by a vortex. The vortex is the bound vortex of the wing plus the free vortices trailing behind it. Integration over their length will give the induced speed at a given point. This speed is proportional to the inverse of the distance to the vortex squared. The vortex strength depends on air speed and span loading; the proportionality being with the inverse of flight speed, aircraft mass and the inverse of wingspan squared.

There is also a dependency of downwash speed with the inverse of air density at the altitude where the glider flies, but this will be canceled out when the effect at ground level is calculated. What is left is an inverse dependency on the local air density at ground level.

Note that you will only get the induced speed, not the pressure from the Biot-Savart law. In order to calculate the pressure increase, you need to add the equation for the pressure rise of the downwash interfering with the ground. This pressure is proportional to the downwash speed squared and air density, so now all density effects are canceled out and you will be left with those factors for the pressure rise from an aircraft flying overhead:

  • the square of the induced downwash speed

which depends in turn on:

  • the inverse of height squared
  • aircraft mass
  • the inverse of wingspan squared
  • the inverse of flight speed

Note also that this does not consider viscous effects.

In order to know the limit of detectability, you also need to know the sensitivity of your measuring equipment.


There definitely is a commonly-accepted rule of thumb:

When a wing is traveling at or less than half the wing's length above a runway or other flat srface, it will experience ground effects.

Feeling a downdraft is one thing. But it is another thing for that pressure wave to interact with the ground closely enough to affect the airplane.

Anyway, if it was a small plane you would be more worried about getting hit by the wheels if it were in ground effect..

  • 1
    $\begingroup$ I really think your answer is flawed in what ground-effect is. It is not a "cushion of air bouncing up" or collecting under the wing, it is a reduction in induced drag. $\endgroup$
    – Ron Beyer
    May 25, 2018 at 2:31
  • 3
    $\begingroup$ @ronbeyer, that reduction in induced drag occurs because the wing in ground effect flies at a reduced angle of attack compared to the same wing not in ground effect and at the same airspeed. this trigonometrically reduces the size of the induced drag vector. $\endgroup$ May 25, 2018 at 5:18
  • 1
    $\begingroup$ @SDsolar: Yes, downdraft is what I was looking for, not wake or ground effect, and I was informed by a physics user that a good equation to use is the inverse square law as applied to the downward air pressure, whose answer, although an approximate one, is close enough. $\endgroup$
    – user32016
    May 26, 2018 at 19:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .