From formula, induced drag is function of lift. When wing stalls, lift is reduced so induced drag must also be reduced?

Stall decrease static pressure at suction side of wing, so I know pressure drag must increase, how then induced drag decrease if this is also pressure drag or how can I distinguish these two drag?

$D_i = \frac{L^2}{\frac{1}{2}\rho V^2 \pi b^2 \epsilon}$

  • $\begingroup$ @sophit I added in text $\endgroup$
    – user207141
    Apr 23 at 12:10

3 Answers 3


When wing stalls, lift is reduced so induced drag must also be reduced?

Yes. But the form drag increases more so the overall drag increases. And it's not like you could actually distinguish them anyway.

Stall decrease static pressure at suction side of wing, so I know pressure drag must increase

Does it?

Well, “suction side” is a bit problematic here, because lift and pressure drag depend on the pressure distribution differently. Lift is produced by low pressure on the top, and that basically ceases in stall, but pressure drag is produced by imperfect pressure recovery around the trailing edge, and that gets worse in stall.

how then induced drag decrease if this is also pressure drag

Induced drag is pressure drag, but not all pressure drag is induced drag.

or how can I distingusih these two drag?

Well, you can't really.

The formula for induced drag, with $\epsilon = 1$, is the minimum drag a passive aerodynamic device must produce to generate lift. Producing more lift violates conservation of momentum and energy—producing upward force, lift, requires accelerating air downward by principle of action and reaction, but then it's kinetic energy increases and if the device does not have additional power input (passive), it comes at expense of the vehicle as drag.

In the non-stalled regime, this formula with added efficiency factor $\epsilon$ plus another term proportional simply to dynamic pressure works well to approximate the drag.

But in the stalled regime the drag is non-linear. You can't use simple formulas to approximate it. So nobody bothers, and therefore nobody cares about which portion of drag is induced and which is parasite—when you do full fluid dynamics calculation, there is simply a pressure field and integrating it results in a force that has a lift and drag component, but the drag is not split to induced and parasite; it just is.

  • $\begingroup$ If I integrate static pressure around the wing at 14° AoA, this reslut will be the same as I use simple formula in my post(for non-stalled case)? $\endgroup$
    – user207141
    Apr 23 at 20:00
  • $\begingroup$ @user207141: For non-stalled, subsonic case the simplified formula for total drag, that is $C_{d0}q + \frac{L^2}{q\pi b^2\epsilon}$ ($q$ = $½\rho v^2$ to keep things simpler), will be close enough to what you get by integrating the pressure around the wing for practical purposes. $\endgroup$
    – Jan Hudec
    Apr 23 at 20:46
  • $\begingroup$ first part is pressure drag for zero lift , second part is induced drag, where in second part is pressure drag, why there is no pressure drag for case where wing produce lift? Why formula show that pressure drag is not changing with AoA, and where is here friction contribution? $\endgroup$
    – user207141
    Apr 23 at 21:03
  • $\begingroup$ @user207141 there is pressure drag when the wings produce no lift, it's major component of the parasite drag term. But yes, the parasite drag term includes also friction drag; my understanding is that in the subsonic regime they are both proportional to the dynamic pressure, so they can be easily approximated by one common term. As you approach transsonic, things get more complicated—and I haven't actually seen much in a way of simple approximation formulas there. $\endgroup$
    – Jan Hudec
    Apr 23 at 21:07
  • $\begingroup$ Pressure drag at zero lift and at 14° AoA can not be the same, so I dont understand why they neglect pressure drag at non zero lift case... $\endgroup$
    – user207141
    Apr 23 at 21:12

Well, considering the equation you posted if some part of the wing stalls then:

  • lift $L$ diminishes;
  • the spanwise lift distribution gets worse in respect to the optimal distribution it had before stall and this reduces the Oswald efficiency factor $\epsilon$.

Whichever of these two effects is predominant determines if $D_i$ increases or not.


The formula seems to describe the relationship between induced drag and Lift at angles of attack lower than stall.

Rather than to just plug it into the stalled condition, perhaps a better way to understand it would be to study the Lift to Drag ratio post stall.

Clearly, to make the same amount of lift, there is much more drag post stall.

Lift coefficient at 45 degrees can be nearly equal to that just before stall at a lower AoA, but with massively more drag$^1$

To make matters worse, the fuselage is also at a very high AoA.

With adequate power (afterburners may be needed) a very high pitch maneuver can be flow. It is known as the "Cobra".

The amount of thrust required to do this should reveal the amount of drag the wing is creating.

Don't forget the thrust contribution to vertical lift.

$^1$Flaps also can increase lift coefficient, but again with greater drag


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