I understand that interference drag is the difference between the terms: the total drag of a configuration, and the sum of the individual drags of the components in the configuration. (It exists due to flow field interactions)

So if we assume a non lift generating configuration and components at subsonic speeds (so no induced or wave drag), is it right to say that there will be no drag in an inviscid flow as both the terms would reduce to zero (no form or skin friction drag) resulting in zero interference drag?

  • $\begingroup$ It is not always easy to know that your lift-free objects (when separate) remain lift-free when brought into close proximity. The total lift can be zero, but if there are forces anywhere within system, you will have induced drag. $\endgroup$ Commented May 5 at 2:49

2 Answers 2


A non-lifting body in inviscid flow can have induced drag if there is lift (that cancels) on parts of the body.

Imagine two rectangular, untwisted wings with symmetrical airfoils -- arranged in a biplane configuration. They are connected together with end plates simply to make them a single body.

If the chord lines are parallel and the configuration is at zero alpha, it will have zero lift (and should have zero drag).

However, imagine the chord lines have +-5 deg incidence. Each wing will experience an equal and opposite lifting force -- each generating induced drag. However, the total lift will be zero and the total drag will be non-zero.

You can accomplish the same thing with a twisted wing -- twisted (even cambered) wings will have non-zero induced drag at the zero lift condition.

You might not consider this interference drag. I would probably agree with you.

To me, the distinction is that if you move the two wings arbitrarily far apart -- but still consider them a single body -- you get the same effect. This is simply two wings with equal and opposite lift that result in zero lift and finite induced drag.

However, let us go back to the zero-incidence biplane. If the two wings are 'close' together, then there will actually be a 'venturi' effect (due to the airfoil thickness) between the airfoils. This will cause the flow acceleration 'between' the wings to be greater than on the 'outside' of the wings. This will cause (cancelling) lift forces -- that will have associated induced drag.

Biplane wings exhibiting Venturi effect

In this analysis, two NACA 0012 wings were placed a half-chord apart. Zero incidence, zero angle of attack, etc.

You can see from the shape of the pressure coefficient contour lines that the flow on the top of the top wing is different from the flow on the top of the bottom wing. This is because of the super-acceleration in-between the wings. In this case, the bottom wing will have an 'up' lift force and the top wing will have a 'down' lift force.

Lift distribution on biplane wings

Here is the lift distribution on these wings -- the important part is that they are non-zero and that they are equal and opposite.

Because they are non-zero, they will each have induced drag. However, the total lift will be zero.

I would say this is properly an interference drag. If you move the wings far apart, this effect will go to zero. If you move them closer together, this effect will increase.

That said, this effect is really small. However, it is exactly the root cause of viscous interference drag. If you place an object in a flow with other bodies such that it experiences super-velocities compared to when it was in isolation, the viscous drag will increase just as if it was in a faster flow.

  • $\begingroup$ Great explanation! $\endgroup$ Commented May 13 at 22:51

If you assume the flow to be incompressible, inviscid and stationary then the Navier-Stokes equations reduce to the Laplace's equation and the relevant aerodynamic force posseses no drag if the body is either 2D or 3D but not generating lift anywhere - just like in your hypothesis.


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