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Inspired by this question, why do shocks move further back on a wing as the Mach number increases?

Normal shocks on the surfaces of wings form when the air molecules of the pressure recovery area on the wing don’t have time to react to the incoming air. Not having time to react causes a rapid compression of the air, making a shockwave.

Now if the air on the wing was traveling faster, when it got to the pressure recovery, it would just collapse into a stronger shock but in the same position, right? Why does it not do that and instead move further back in the wing?

Some pictures to see what I’m taking about :

enter image description here

This is when the shock is half down the wing. (Ignore the separation part of the photo)

enter image description here Then at a higher speed, the shocks for some reason move backwards instead of just staying where they were and getting stronger. Why is that?

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  • $\begingroup$ @sophit good point. Separation will still cause drag, but not wave drag in this case. $\endgroup$
    – Wyatt
    Commented Apr 21 at 20:53
  • $\begingroup$ The boundary layer separation aft of the shock wave does not cause wave drag as suggested in the first picture. Wave drag is an intrinsic drag generated at supersonic speeds and has nothing to do with boundary layer. As already suggested by @RobMcDonald beware of random pictures from the internet. Btw, what's the source of that picture? $\endgroup$
    – sophit
    Commented Apr 21 at 20:53
  • $\begingroup$ @sophit The latest response ever, but the source is here $\endgroup$
    – Wyatt
    Commented Jul 8 at 3:00
  • $\begingroup$ Again... $\endgroup$
    – sophit
    Commented Jul 8 at 3:59

1 Answer 1

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Why would it stay in the same place? Is there some physical object to keep it there? Something like a ridge or a bump or a throat of a nozzle?

On a smooth surface, the location a shock 'chooses' is determined by a balance of effects. One of those is the upstream Mach number -- as the airfoil goes faster, the supersonic region around the airfoil gets larger and the Mach number in it gets larger. Another is the back pressure.

Without worrying about the details, when we're faced with a balance of factors like this, there is no reason to think they will be the same as you vary the important conditions. In fact, it makes a lot more sense for them to be different.

Edit

Adding an image to illustrate the behavior the OP was asking about

enter image description here

enter image description here

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  • $\begingroup$ I see, but I assumed the back pressure stays the same as the Mach number increases. Is this true? If so, it seems like it’d collapse in the same place? $\endgroup$
    – Wyatt
    Commented Apr 21 at 20:10
  • $\begingroup$ It would make sense if the back pressure decreased as you speed up, then the shock would travel backwards. Not sure why it would decrease or change though. $\endgroup$
    – Wyatt
    Commented Apr 21 at 20:35
  • $\begingroup$ It doesn't actually matter if the back pressure stays the same or not. When the Mach number is higher, the normal shock will be stronger. A stronger shock means a larger pressure jump at the shock. Check out the normal shock tables. en.wikipedia.org/wiki/Normal_shock_tables $\endgroup$
    – Rob McDonald
    Commented Apr 22 at 4:36
  • $\begingroup$ I came back to this answer because I am a little confused. Imagine you’re going Mach 0.82 like in the picture at the bottom of your answer. When the air accelerates to supersonic, it is forced to decelerate through a normal shock when it arrives at the pressure recovery area. When you increase speed, why wouldn’t it just force this normal shock to be stronger and stay in the same spot? $\endgroup$
    – Wyatt
    Commented Oct 28 at 3:16
  • $\begingroup$ @Wyatt Unfortunately, the picture I originally posted does not show the whole picture. The experimental image was too zoomed in. Check out the illustration I've now added. Notice that when you get to supersonic flow, a bow shock forms before the rounded airfoil leading edge. The bow shock is a normal shock (strong shock solution, subsonic flow downstream) that then bends around and becomes an oblique shock away from the body. This illustration gives a good indication of the regions of subsonic and supersonic flow. $\endgroup$
    – Rob McDonald
    Commented Oct 28 at 4:53

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