Swept wings are used in aircraft that fly in the transonic regime because they 1) delay the critical Mach number 2) reduce the drag increase that occurs (smaller drag divergence).

I am trying to understand physically why a lower drag is observed. A common explanation is that only a component of the free stream is perturbed by the wing (see: Why are swept wings better for breaking the sound barrier?), but this does not explain why there is a lower drag when shocks do occur over the wing.

More specifically, I understand why the curve below shifts to the right as sweep increases, but want to know why the drag is not as high.

enter image description here

Source: https://history.nasa.gov/SP-468/ch10-4.htm


2 Answers 2


Shock waves form where air can no longer "get out of the way" and begins to compress.

Sweeping the wing effectively increases the chord and reduces the rate of wing thickness increases as airflow moves across it, essentially making the body longer and more "streamlined". Instead of running into a steep hill and compressing, it can flow over a more gently sloping barrier with greater ease.

What is most interesting is the decreasing drag of the 47 degree sweep at high subsonic. As sweep increases, airflow over the wing may transition from the classical front to back flow seen in straight wings to the deflected "spanwise" airflow of swept wings to the rolling vortices seen in highly swept "slender delta" wings.

At some combination of speed and sweep, airflow will roll up into a vortex rather than washing off the back or end of the wing. It may be that, at higher subsonic speeds, with greater sweep, the vortices roll into the back of the wing and push the aircraft forwards.

It has been thought the same thing occurs underneath thin undercambered wings, which may help account for some of their "magical" slow flight properties. Definitely worth studying.

As far as 11, and 35 degrees of sweep, they follow predictable patterns based on compression effects at transsonic speeds, as does the Area Rule.

Are swept wings better for breaking the sound barrier? Not really, as the X-1 proved. But they do significantly delay transsonic drag at high subsonic speeds.

  • $\begingroup$ By effectively i creasing the chord, are you reffering to spanwise flow? $\endgroup$
    – Jpe61
    Apr 6, 2020 at 18:55
  • $\begingroup$ That as well, but if you start with a straight wing and sweep it, relative to the wind flow (even if it is not deflected) the chord (distance from front to back) is greater. $\endgroup$ Apr 6, 2020 at 19:21
  • $\begingroup$ Yes, but the profile would be different. Say you want to use naca xyz, because it is suitable for flight characteristics you want, you cannot use the profile perpendicular to the span. It would not behave as naca xyz because the airflow is coming at an angle to this profile. You cannot compare profiles like that. $\endgroup$
    – Jpe61
    Apr 6, 2020 at 22:04
  • $\begingroup$ It would be good to see how 2D polars changed adjusted for sweep, both computer and wind tunnel data. Spanwise flow will also cause "variation in profile" towards the wing tips. I'm really interested in the conditions where spanwise flow begins to roll into vorticies as sweep increases. $\endgroup$ Apr 6, 2020 at 22:09
  • $\begingroup$ Me too. I'm not an aerodynamicist (?), but I bet the sweep enhances spanwise flow, and this gets stronger as the plane goes faster, thus "slendering" the wing profile as seen by the airflow. $\endgroup$
    – Jpe61
    Apr 6, 2020 at 22:17

Let's consider an infinite swept wing with a sweep angle $\Lambda$. We can divide the freestream velocity ($V_\infty$) into two components: the component perpendicular to the sweep ($V_{\perp}$) and the component parallel to the sweep ($V_{||}$). Similarly, we will divide the exo-boundary layer velocity field around the wing ($V_e$) into the perpendicular component ($u_e$) and parallel component ($w_e$).

Swept wing coordinates

Graph cited from Drela, Flight Vehicle Aerodynamics.

Due to infinite span, the flow quantities along the swept coordinate ($z$) must be invariant. Therefore, the flow parallel to the wing sweep ($w_e$) must also be invariant and equal to the freestream component ($V_{||}$) everywhere.

Assuming that the flow outside of the boundary layer contains only weak shocks and is therefore isentropic (except at the shock), we can use the isentropic relationship to relate pressure field and velocity field. Here, everything with subscript $_e$ denotes flow quantities outside of the boundary layer:

$$p_e(x)=p_\infty \left(\frac{h_e}{h_\infty} \right)^{\gamma/(\gamma-1)}=p_\infty \left[ 1 + \frac{\gamma-1}{2}M_\perp^2 \left(1-\frac{u_e^2}{V_\perp^2} \right)\right]^{\gamma/(\gamma-1)}$$

where $h$ is enthalpy, $p$ is pressure, $M$ is Mach number, $\gamma$ is specific heats ratio.

Notice that the flow parallel to the wing sweep has no bearing on the pressure field; it's as if each wing section only sees the perpendicular flow. The same goes for effective 2D Mach ($M_\perp$), which is a net reduction from the freestream Mach ($M$) by a factor of $\cos\Lambda$. Since a reduction in Mach reduces the normal shock strength in the 2D section, or even removes the shock altogether, it leads to a reduction in wave drag compared to no sweep.

Below is the 2D drag curve vs. Mach for NACA 0012 from Euler solution. Local shock is obvious starting 2D free-stream Mach 0.65. This goes to show that decreasing effective 2D Mach lowers the overall wing drag via sweeping.

NACA0012 drag rise

Graph cited from http://aerodesign.stanford.edu/aircraftdesign/drag/dragrise.html.

In a finite span wing, there will be gradient in $w_e$ that will lead to flow variation along the span. However, the principle of wave drag reduction through wing sweep readily applies.

  • $\begingroup$ Could you please provide the reference to the formula of $p_e$? $\endgroup$
    – Hans
    Apr 7, 2020 at 0:01
  • $\begingroup$ @Hans See grc.nasa.gov/www/k-12/airplane/isentrop.html relation 4. Under calorically perfect gas, Cp is a constant, thereby giving the relationship in my post. $\endgroup$
    – JZYL
    Apr 7, 2020 at 0:13
  • $\begingroup$ Thank you. It would be better to add the reference to the main text of your answer below the equation. $\endgroup$
    – Hans
    Apr 7, 2020 at 0:53
  • $\begingroup$ @JZYL At some point, Mcos(Lambda) is going to be such that a shockwave occurs over the wing. Your explanation does explain why the critical Mach number increases, but can you clarify how the actual drag decreases when the shock DOES occur? In particular, the shock strength is reduced at a given free stream Mach number, but how does this explain why the curve in my original question has a smaller peak (albeit at a higher Mach number)? $\endgroup$
    – Nick Hill
    Apr 7, 2020 at 20:12
  • $\begingroup$ @Nick Are you asking why the maximum wave drag decreases as a function of sweep? 2D shock strength is a function of Mach number. $\endgroup$
    – JZYL
    Apr 7, 2020 at 21:19

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