# Has wall to wall airfoil induced drag?

Walls in wind tunnel stop air leakage at outboard parts of airfoil,pressure distribution at center section is same as at outboard sections(if we neglect interferenece with walls).

-Does that mean that induced drag is zero?

BUT

Brilinat words from memeber @JanHudec:

"The infinite span has nothing at all to do with having no ends. The ratio of lift to induced drag increases with span. If you fix the lift, and vary span, the induced drag will tend to zero as the span tends to infinity. That means in limit a wing of infinite span generating finite lift (and therefore flying at zero coefficient of lift!) will have zero induced drag."

That mean wall to wall airfoil is not theoretical 2d airfoil,so it has induced drag but smaller than 3d wing BUT REMBER FOR SAME LIFT.

• A wall-to-wall wing in a windtunnel should best be compared to one in ground effect and not as a segment of an infinite wing. Both are similar since the infinite wing will be infinitely close to the ground, relative to its span. Since the airfoil in the tunnel is still above the ground (= tunnel wall), induced drag is less than in free flight but some induced drag is still there. Only when the airfoil merges with the bottom of the tunnel will induced drag disappear. Sep 6 '20 at 13:32

A wall-to-wall wing in a windtunnel should best be compared to one in ground effect and not as a segment of an infinite wing. Both are similar since the infinite wing will be infinitely close to the ground, relative to its span.

The opinion that an infinite wing has no induced drag is misleading. When we plot the amount of induced drag over aspect ratio, induced drag will tend to zero for an infinite aspect ratio. But what does that mean in practical terms? That infinite wing also produces infinite lift by pushing an infinite amount of air downwards by a finite amount. Induced drag is still there but becomes insignificant relative to lift.

Better to look at what happens in reality. It is worth to show the picture of the linked answer again here:

Induced drag is the consequence of lift. Lift is the consequence of air flowing around the obstacle that the airfoil poses to the air. While less than in free flight, some downward acceleration of air still happens but is soon stopped by the ground. Therefore, while less than in free flight, some induced drag is still there and will only disappear when the airfoil merges with the bottom of the tunnel.

Now it is crucial to know what induced drag really is. Air is accelerated downwards while flowing over the wing and the reaction force to this acceleration is lift. Since that does not happen instantly but gradually, some of the lift is created in an already downward-bent flow, so the reaction force to this, being perpendicular to the local flow, is pointing slightly backwards. This backward component is induced drag.

The ideal of the infinite span wing produces lift instantly because its chord is vanishingly small relative to span. The same cannot be said of the real wing in ground effect.

There is no free lunch. As long as lift is produced, induced drag will be present.

• If we striktly follow drag definition: total drag=zero lift drag(skin friction+pressure drag) + induced drag,then pressure drag and skin friction drag are measured only at zero lift angle and induced drag is measured whenever wing produce lift..So if wall to wall wing at zero lift angle show total=500N and at 15° AoA show total= 700N,then induced drag is 200N(if we assume that skin friciton is not changing with small AoA).......
– user52248
Sep 6 '20 at 13:57
• .......Look at this way it turns out that wall to wall wing produce more induced drag than wing with open wingtips,BUT REMEBER AT SAME AoA. If we measured at same lift, then wing with open wingtips has higher induced drag,because it must be set at HIGHER AoA to produce same lift as wall to wall wing. Do you agree?
– user52248
Sep 6 '20 at 13:58
• @Сократ The wing with open tips will have a smaller aspect ratio and produce proportionally more induced drag at the same lift and will need a higher angle of attack because of its lower lift curve slope. So far I can agree. However, I do not know whether induced drag would be higher for the wall-to-wall wing at the same AoA – this would be speculation without concrete data. Sep 6 '20 at 15:33
• However, I do not know whether induced drag would be higher for the wall-to-wall wing at the same AoA – this would be speculation without concrete data Here we have this topic,but I think accepted answer is wrong...aviation.stackexchange.com/questions/77573/…
– user52248
Sep 6 '20 at 15:41
• will only disappear when the airfoil merges with the bottom of the tunnel.Airfoil will still producing some lift with upper side curvature, but why then induced drag will be zero?
– user52248
Sep 6 '20 at 18:41

Yes, the leakage at the wingtips is what produces induced drag and since as your write it is blocked in the windtunnel, the induced drag is zero. Or as Wikipedia puts it:

...a wing of infinite aspect ratio (wingspan/chord length) and constant airfoil section would produce no induced drag. The characteristics of such a wing can be measured on a section of wing spanning the width of a wind tunnel, since the walls block spanwise flow

• I would like to see some studies with measurements to go. I wouldn't take Wikipedia as any authoritative source.
– JZYL
Sep 5 '20 at 22:23
• Wikipedia's explanation cites the usual equations defining $D_i$ and $C_{D,i}$. There, as wingspan or aspect ratio goes to infinity, those values go to zero. Since a wing of infinite span has no spanwise flow, a walled segment thereof (also lacking spanwise flow) is a valid approximation, except for drag at the wind tunnel's own walls. Sep 5 '20 at 23:24
• @Camille Induced drag comes from the trailing vortices. I don't see why wall bounded wing would not shed. The linear theory from which induced drag is predicted actually does not predict any spanwise flow over the wing.
– JZYL
Sep 6 '20 at 1:25
• True, the "spanwise flow" is beyond the wingtip, but Сократ likes to play it by the book, so I'm doing that instead of finding an equation that models induced drag as a function of vortex strength. Sep 6 '20 at 1:34