What is the relation between sweep angle and lift-to-drag ratio?

And does it depend on the wing planform?

For subsonic speeds.

  • $\begingroup$ I thought it didn't change L/D, but instead shifted the same behavior to a higher speed speed, and of course shockwaves. $\endgroup$
    – DKNguyen
    Commented Nov 16, 2020 at 15:56
  • $\begingroup$ are you sure about that? $\endgroup$
    – user52248
    Commented Nov 16, 2020 at 17:05
  • $\begingroup$ No, if I was I would have put it as an answer. $\endgroup$
    – DKNguyen
    Commented Nov 17, 2020 at 1:00
  • $\begingroup$ no one answer ? $\endgroup$
    – user52248
    Commented Nov 19, 2020 at 6:26
  • $\begingroup$ Yeah, seems oddly quiet. $\endgroup$
    – DKNguyen
    Commented Nov 19, 2020 at 14:08

3 Answers 3


This is my take on this.

We assume that an infinite-span swept wing is under compressible flow, and that the flow acts on it as in Figure 1 below:

Swept wing flow visualization

With that, we can now utilize the equation of motion to find the relation between the sweep angle and the coefficient of lift, which in turn, will make the relation with the ratio L/D

From the equation of motion, we have:

$$(1-M^2_{\infty}\cos^2\Lambda)\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2}$$

Now, only the perpendicular component acts we have that:

$$(u'_c)_n = \frac{\partial \phi}{\partial x} = \frac{1}{\sqrt{1-M^2_{\infty}\cos^2\Lambda}} \frac{\partial \Phi}{\partial x} = \frac{(u'_i)_n}{\sqrt{1-M^2_{\infty}\cos^2\Lambda}}$$ $$C_{p_c} = -2\frac{\cos\Lambda(u'_c)_n}{U_\infty} = -2\frac{\cos\Lambda(u'_i)_n}{U_\infty\sqrt{1-M^2_{\infty}\cos^2\Lambda}}$$

Now, since:

$$C_L = \frac{1}{c}\oint C_p dx$$

We have that:

$$C_L = \frac{C_{Li} }{\sqrt{1-M^2_{\infty}\cos^2\Lambda}}$$

The above equation shows that the lift coefficient is sweep angle dependent, therefore, the ratio L/D is also sweep angle dependent.

This is a very complex side of aerodynamics and understanding it completely requires some study. I recommend studying Chapter 11 from "Fundamentals of Aerodynamics" by J. Anderson (a classic) and Chapter 8 from "Aerodynamics for Engineering Students" by E.L. Hughton.

  • $\begingroup$ Does sweep reduce L/D or not? $\endgroup$
    – user52248
    Commented Nov 20, 2020 at 16:22
  • 3
    $\begingroup$ @Сократ Second last paragraph, but assumes things about the Cd relative to C_L $\endgroup$
    – DKNguyen
    Commented Nov 20, 2020 at 16:49
  • $\begingroup$ @DKNuyen all gliders use zero sweep,maybe that show that sweep reduce L/D ratio?gliders most import things is to has L/D as much as posssible.. $\endgroup$
    – user52248
    Commented Nov 22, 2020 at 6:15
  • 1
    $\begingroup$ please consider typing the equations out with mathjax, rather than posting images. it is helpful both for people that use screen readers and for the search results. $\endgroup$
    – Federico
    Commented Nov 22, 2020 at 7:38
  • $\begingroup$ The above equation shows that critical Mach is delayed by a factor cos$\Lambda$, and nothing about the $C_D$. Complex aerodynamics indeed. Kudo's for @Federico's skills with MathJax. $\endgroup$
    – Koyovis
    Commented Dec 22, 2021 at 3:15

The answer depends on the Mach number and the angle of attack. At low speed significant sweep provokes earlier boundary layer transition which negatively affects L/D, but at transsonic speed it delays shocks and improves L/D. At high angle of attack a swept wing will stall earlier and reach lower maximum lift, except for delta wings.

Sweep will transform bending moment into torsion and requires a stiffer wing, and it lowers maximum lift and the lift curve slope, so for the same landing speed a swept wing must be larger. This drives up wing weight for the swept wing and, indirectly, drag. Therefore, an unswept wing in subsonic flow will produce the best ratio of useable lift over drag.

For completeness: At low supersonic speed, when leading edge sweep is larger than the Mach cone angle, flow around the wing is still similar to subsonic flow and the leading edge thrust resulting from this greatly improves L/D. Once the Mach cone angle is larger than the sweep angle, flow is fully supersonic and L/D is little affected by sweep.

Back to subsonic flow:

At low sweep angles (< 20°) the impact on the stability of the boundary layer is low and very little change in L/D can be noticed at low and moderate angles of attack. The effect depends also on the Reynolds number and is largest between Re = 500,000 and 5,000,000 when significant laminar runs are possible but the boundary layer can be tripped rather easily.

Explanation: The acceleration of the flow normal to the lines of equal chord and deceleration from viscosity produce a 3D speed profile in the boundary layer which is twisted. This favors earlier turbulent transition.

Below approximately Re = 500,000 the laminar boundary layer is more stable and can lead to large separation bubbles or outright laminar separation, so active tripping of the boundary layer improves L/D. Sweep will not affect this.

Above approximately Re = 5,000,000 the boundary layer trips by itself rather quickly and again sweep will not affect this significantly.

At high angles of attack sweep will reduce lift at the wing's center and lead to early flow separation at the wingtips from a tipwise flow of the boundary layer. Now sweep can be very detrimental to L/D but its effect on the pressure distribution can be somewhat mitigated with changes in the airfoil and incidence over span.

At very high sweep angles and angles of attack you get flow separation at the leading edge and vortex lift when a straight wing would simply show separated flow. However, due to the reduction in the lift curve slope with increasing sweep angle, and the loss of nose suction at high angle of attack, LD in this condition is rather low.


Wing sweep delays compressibility effects. When we assume an infinitely long wing as in the top image below (Torenbeek Fig 7-19), the pressure distribution would be determined by the velocity component over the leading edge only.

Torenbeek Fig 7-19

According to this simple wing sweep theory, the following effects occur:

  • Mach number over the swept back wing $M_n = M_∞/$cos$\Lambda$, so the critical Mach number of the wing is delayed by the same factor as well.
  • The wing lift coefficient $C_L = c_l \cdot$(cos$\Lambda)^2$, with $c_l$ being the lift coefficient of the section normal to the leading edge.

So according to simple sweep theory, the normal section shape must be designed at a higher $c_l$ than the wing lift coefficient. From Torenbeek page 247:

For example, a wing sweptback by 35 degrees, and operating at M = .85 and C$_L$ = .4, will have normal sections designed for operation at M=.7 and C$_L$ =.6.

There are quite a few factors in optimising a wing, particularly a swept wing:

  • wing area S;
  • aspect ratio A;
  • wing twist;
  • wing taper;
  • sweep angle;
  • wing dihedral;
  • interference with the fuselage;
  • etc.

Sweep of a rectangular wing

Therefore we'll leave Torenbeek's design reference book alone for now, and have a look at the effect of wing sweep only on an untapered, untwisted wing.

enter image description here

If a rectangular planform wing is swept by angle $\Lambda$, the wing span decreases by factor cos$\Lambda$. And then using the quadratic approximation equation for induced drag $C_D = C_{D0} + \frac{{C_L}^2}{\pi Ae}$, $C_D$ increases with a factor 1/cos$^2\Lambda$. Using some values for a 1970 subsonic jet:

enter image description here

Note that this is a simplified approximation:

  • The Oswald factor e would decrease if a wing is only swept, since the tip loading becomes higher.
  • All compressibility effects are disregarded.
  • Weight is assumed constant, while a swept back wing requires a stiffer torsional construction.

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