According to NASA's claims, when the swept back wing is inside the shock cone and the shock is not attached to the leading edge of the wing(subsonic leading edge), the wave drag of the swept wing is less than that of a straight wing, but with increasing speed and decreasing the angle of the shock , the shock attached to the leading edge of the swept wing(supersonic leading edge), so in this situation the wave drag of straight wing is less than the wave drag of the swept back wing! Why ? i mean wave drag(not total drag )


1 Answer 1


First, you should always provide a citation when you say something like "According to NASA's claims". It will provide the context that we can make sure we are answering the right question.

I actually think there is some confusion here depending on how formal people are being. In any case, we can probably look at this by just considering a 2D airfoil.

The transonic drag on an airfoil will often be higher than the supersonic drag.

In the supersonic case, the shock is in front of the airfoil (like the wing behind the shock case you describe) and the airfoil exists entirely in the post-shock flow.

In the transonic case, the shocks will lie on the surface of the airfoil -- likely both the top and bottom surface. There, the shocks cause a strong adverse pressure gradient on the airfoil surface (pressure jump). This is very likely to cause the boundary layer to separate. This separation causes a large increment in drag.

This is the cause of additional transonic drag. It isn't really wave drag, but it is separation drag caused by the shockwave.

The same thing happens in 3D. When the shock is in front of the wing, the wing exists in relatively benign post-shock flow. When the shock is on the wing surface, the wing sees a strong adverse pressure gradient at the location of the shock -- which causes separation and lots of drag.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .