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The rule I have been taught is that ground effect begins within 1 wingspan of the ground.

What is the derivation of this measurement? Is it actually true, or just an urban-legend/rule-of-thumb style teaching aid?

If it is a valid way to assess where ground effect begins for any given airplane, what limitations does it have? Do particularly heavy or airplanes with exceptionally long wingspans not fit this rule the same as a smaller more common build of aircraft?

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It's most definitely not urban legend. However, it also does not scale 1-1 with wingspan. Rather, it scales nonlinearly with above ground height to span ratio ($h/b$). The simplest analytical prediction for ground effect can be derived from potential theory and the circulation theory of lift.

We know that the wing can be modeled as lifting line consisting of a line of vortex bound to the wing and trailing vortices that look like horseshoes. The trailing vortices induce downwash on the wing and everywhere else. But what about at the ground directly underneath the wing? Due to the tangency boundary condition of non-viscous flow, induced downwash must necessarily be zero everywhere on the ground.

Within the context of circulation theory of lift, the only way we can achieve this is if we have a mirror wing underneath the ground with the ground plane being the plane of symmetry, like this:

Image Plane

Since the image wing is exactly inverted, it produces upwash that reduces the overall downwash on the real wing. In the super simplified case of a single horseshoe vortex (which is incorrect), after some maths, the induced downwash at the wing root is now:

$$w_i=-\frac{\Gamma}{\pi b}\frac{16(\frac{h}{b})^2}{1+16(\frac{h}{b})^2}=w_{i_\infty}\frac{16(\frac{h}{b})^2}{1+16(\frac{h}{b})^2}$$

where $w_{i_\infty}$ is the downwash induced without the ground image (i.e. free-air). Turns out the latter factor is an ok approximation of total reduction of induced drag in ground effect as well. As you can see, at half wing span away from the ground, we have 80% of the free-air induced drag; at one full wing span away, we are at 94%. The effect falls away quickly.

You can read this MIT lecture on ground effect (from which I've taken the above diagram) for the derivation of the simplified ground effect prediction. More precise ground effect accounting can be found in references such as Raymer and ESDU Item 72023.

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  • $\begingroup$ Even though the derivation is ok, the conclusion is misleading. Please see my answer. $\endgroup$ – Zeus Oct 18 '19 at 0:49
  • $\begingroup$ @Zeus you do know the h/b relation hold in Raymer, Asselin, Roskam, wind tunnel tests and flight tests right? For conventional configurations, I don't see why you would say it is misleading. $\endgroup$ – JZYL Oct 18 '19 at 1:19
  • $\begingroup$ For a given configuration, it holds for h/c as well, just with a different coefficient. It is misleading exactly the way the OP got confused: if the wingspan is the factor (as is often explained to pilots), then longer wingspans will experience ground effect at a higher altitude, right? Wrong. On the contrary, for an infinite wingspan there will be no meaningful ground effect at all. If you try to land two similar airplanes with substantially different aspect ratios (say, PA-28 vs DA-40 or SR-20), you'll notice that shorter wing experiences more change. $\endgroup$ – Zeus Oct 20 '19 at 23:42
  • $\begingroup$ ...So, instead of saying "ground effect begins within 1 wingspan of the ground", it is more meaningful to say something like "ground effect begins within 10 chords from the ground". $\endgroup$ – Zeus Oct 20 '19 at 23:44
  • $\begingroup$ @Zeus If you have a large span, then ground effect will start showing up at a higher altitude. If you have infinite span, it will start at infinity. It'll also have zero induced drag to begin with. Why is this contradictory? I haven't seen ground effect being linked to h/c for standard wing configurations. Do you have any reference? $\endgroup$ – JZYL Oct 21 '19 at 0:14
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This is nothing more than a rule of thumb, and a rather poor one.

As Jimmy's answer correctly states, the ground effect scales expotentially with the height-to-span ratio, and the effect is primarily on reduction of the induced drag.

However, this approximation apples only to a given fixed aircraft. It often leads to incorrect conclusion that the wingspan itself is involved here, and the original 'one wingspan' rule of thumb reinforces it. The uneasiness expressed in the last paragraph of your question shows that it is easy to feel that it must be wrong. Unfortunately, this is a very common misconception.

The truth is, ground effect is negatively related to the wingspan, and positively related to the wing chord. The wing chord, or more accurately, the height-to-chord ratio $h/c$, is a better factor to use for approximations.

Of course, for a given aircraft (wing) with a fixed ratio of $b/c$ (read aspect ratio), one can express the effect based on either wingspan or chord. But wingspan is misleading.

Indeed, if we double the wingspan, and accordingly halve the chord, keeping everything else (particularly lift and height) the same, what will happen to the ground effect? It will reduce, contrary to the 'wingspan' rule.

This does not directly contradict the derivation in the Jimmy's answer; rather, one should remember that the induced drag ($w_{i_\infty}$) will also reduce in this case. As we approach infinite aspect ratio, both the induced drag and the ground effect will tend to zero. The near-zero chord clearly and intuitively indicates that.

You may notice that aircraft which rely on ground effect always have stubby wide-chord wings. The reason is exactly that: the height at which the ground effect becomes noticeable depends on the chord rather than the wingspan, for a given lift. Interestingly, in the Russian aerospace school (and USSR/Russia is known for its ekranoplans), wingspan has never been used as a proxy for ground effect calculations. Only the wing chord.

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    $\begingroup$ "If we double the wing span and halve the chord, we reduce the ground effect." How did you reach this conclusion? Ground effect is expressed as a factor on the free-air downwash; a reduction in free-air quantities themselves due to aspect ratio change doesn't factor into this. $\endgroup$ – JZYL Oct 18 '19 at 1:34
  • $\begingroup$ This is a very narrow definition of ground effect, but even then, with the same mirror wing model, it is obvious that this ground effect will not increase with the wingspan (for the same lift and other conditions). Generally speaking though, ground effect is the overall change of all aerodynamic characteristics due to ground proximity. And in this sense (most importantly for pilots), reduction of total drag (and increase of L/D) near ground will inversely depend on wingspan (and positively on chord). $\endgroup$ – Zeus Oct 21 '19 at 0:25

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