Why does lift over drag decrease with AoA past the optimum AoA? Is it because the backwards vector for a given lift force on the wing is now larger for a given air particle compared to lower AoA?

For a single air particle interacting with the wing at higher vs lower AoA does the higher AoA cause a larger change in the velocity of the air particle?

Are there other factors involved?

  • $\begingroup$ I guess you wanted to ask why lift goes up past the optimum AoA. Lift over drag will already be at its maximum at the optimum AoA. $\endgroup$ Jan 18, 2015 at 8:22

4 Answers 4


The fundamental reason is that at high enough AoA, the lift becomes zero -- certainly no later than an AoA of 90°! At that point L/D is zero (there'll be lots of drag). So somewhere between 0° and 90° there will be an AoA with the highest L/D, and that is by definition the optimum AoA. Slightly beyond that angle there's nothing for the L/D to do except decrease, because if it didn't, the angle we're beyond wouldn't be the optimum.

Now if you were asking why that particular angle is optimal for a given airfoil, we would have to go into aerodynamics and boundary layers and whatnot. It's not merely about what happens to the molecules that (try to) hit the underside of the wing, but at least as much about how well the molecules above the wing can be convinced not to hit it.

Note that above the optimal AoA, until the critical (stall) angle, the lift generated still increases with increasing AoA -- it's just that in this range drag increases faster than the lift does.

  • 1
    $\begingroup$ I like this explanation and haven't seen it before. Somewhere between 0 and 90 really gets the point across. $\endgroup$
    – Simon
    Jan 18, 2015 at 10:31
  • $\begingroup$ I don't think its true, but your explaination by itself doesn't show that there is only one point on the L/D curve where lift is maximum. Nor does it show that it is not linear. Either of these would mean that there is no single maximum. $\endgroup$
    – rbp
    Jan 18, 2015 at 14:52
  • $\begingroup$ @rbp: Yes, there could be more than one local optimum. But the curve can't be linear, because at (somewhere near) both 0° and 90° the lift will be zero with nonzero drag (and thus L/D=0), and if we call it a wing at all there must be an angle where it delivers positive lift (and thus positive L/D). In theory there could in an interval with exactly constant L/D, but such a degenerate case does not arise in practice. Change the Reynolds number by 0.01 and you'd miss it. $\endgroup$ Jan 18, 2015 at 15:12
  • $\begingroup$ As I said, i don't think its true, but i think you might want to say something about the kind and order of the function you're describing. $\endgroup$
    – rbp
    Jan 18, 2015 at 15:40

Local speed at low angle of attack

At low angle of attack, the oncoming air hits the airfoil right at the tip of the leading edge. There is not much of curvature to negotiate before the long stretch with little curvature behind the nose follows. In order to change its flow path quickly, the air molecule must flow in a high pressure gradient, and this gradient accelerates it not only sideways, but also in flow direction. Consequently, high curvature areas see high local flow speed. (I explained the physics in a very simple way; if you want to know more, please read this).

CFD plot of airfoil at moderate AoA

Local speed at high angle of attack

With increasing angle of attack, the suction area over the airfoil needs to become stronger to force the airflow to follow the contour. Consequently, more of the oncoming flow is sucked over the wing, and the stagnation point shifts downwards on the airfoil's nose. Now the particles which hit the nose just above the stagnation point have to negotiate most of the nose's curvature, which needs more acceleration. This creates a suction peak in the flow around the nose and high local flow speed on the top side.

The flow over the bottom of the airfoil, however, will be decelerated near the stagnation point and see only little acceleration while it flows towards the trailing edge. See below for a plot of the inviscid pressure distribution. The pressure coefficient $c_p$ correlates with speed, and $c_p = 0$ means that local speed equals the flight speed of the airfoil. Negative values denote suction, and local speeds in excess of flight speed, whereas positive values of $c_p$ indicate slower speeds. Note that the y-axis is plotted reversed, with decreasing values up.

Pressure distribution over chord of the NACA 2412 at several AOAs

A high local speed will also cause high local friction. Also, the long pressure rise on the upper side at high angle of attack needs more energy from the outer flow to keep flow attached, so the boundary layer grows here and slows down more of the air than at low angle of attack (again, extremely simplified). Both effects mean that drag goes up at higher angle of attack.

If you increase the angle of attack further, the flow is not able to follow the upper contour and will separate. This causes the pressure rise to stop at the separation point, so downstream the pressure decreases. Since this part of the airfoil points slightly backwards, this local suction increase will drive up drag. This starts before stall and on a well-behaved airfoil the separation point will first be near the trailing edge and move upstream with increasing angle of attack. This means then drag will rise progressively with higher angle of attack. Since most of the flow is still attached, lift will still grow until stall.

Optimum angle of attack

The airfoil will be at its optimum angle of attack $\alpha$ with the best ratio of lift over drag $E_{opt}$ when the rate of lift increase with angle of attack equals the rate of drag increase times the optimum ratio of lift to drag. Once the local drag increase with angle of attack exceeds this value, the ratio of lift to drag decreases. Maybe this is easier to explain with a formula: $$E = \frac{L}{D}$$ To find the optimum we look at the point where the derivative of $E$ over $\alpha$ is 0: $$\frac{\delta L}{\delta\alpha} = E_{opt} \cdot \frac{\delta D}{\delta\alpha}$$ The rate of lift increase over angle of attack is quite constant until stall. At low angle of attack drag decreases with increasing $\alpha$ until the minimum drag point is reached. Then drag goes up slowly, so the ratio of lift to drag still grows. At one point both grow such that their ratio will not change much, and once the rate of drag increase grows even higher, the ratio will drop again. This all happens before stall.


I think you were asking this question related to wing physics and not to airfoil physics. Here comes the physics of induced drag. I will try to simplify the explanation.

As you might know, the upper surface of the wing has lower pressure and the lower part of the wing has higher pressure. This is what is producing lift.

But... what about the tip of wing? Both sides are in contact so they should be at the same pressure?

So, we have a lower surface of the wing with higher pressure and the tip with "normal pressure", so, naturally in the lower surface the air will try to go from the root to the tip.

On the upper surface the situation is the opposity, the root will have lower pressure and the tip higher, so the air will try to go from the tip to the root.

As in the tip both sides are connected, the air will try to scape from the lower to the upper level (that's why there are wing tips). And that's why the tip vortexes are created with the typical pictures

Ok, now let's talk about drag. Drag is actually "any energy we are losing just because we fly" (ok, is a simplification but allows me to explain the next step). So, the airplane is losing energy and gives it to the air to create the tip vortex. The bigger the tip vortex the more energy is lost.

So, know coming to the question, why efficiency (L/D) decreases? So, as the angle of attack increases the lift increases linearly (proportional to the increase of the angle), however, the energy lost in the tip vortex (that is a component of drag) increases quadratically (proportional to the square of the angle of attack), that means that the drag is increasing more quickly (but starting at a lower level) that lift.

(lift grows, but drag grows proportionally faster)

So, we reach a point where the increase of drag makes the efficiency starting to decrease. That point can be calculated with the formula provided in the previous post.

Notice that the physics of the wing is more complicated as there are other drag components also interacting with the wing. It is also true that "infinite" wings have a maximum L/D, but there is a different mechanism in place.


At higher AOA, flow is separated, consequently Stall happens. Due to stall, Lift decreased and Drag increased, so L/D decreased.

  • $\begingroup$ For a typical wing, as the AoA increases, L/D has already been decreasing long before the stall angle is reached. Other answers assumed (correctly, I think) that the "optimum" angle in the question is the angle of maximum L/D, not the stall angle. $\endgroup$
    – David K
    Jul 15, 2016 at 15:50

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