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A supersonic plane will produce shock waves off the nose cone, as seen below:

oblique shocks

These oblique shocks reduce the speed of the air that the wing experiences. If the plane is at a low enough Mach number, the shocks will reduce the speed of the air to below Mach 1. For example, I calculate (with the Taylor-Maccoll equations, I could be wrong) that a Mach 1.5 plane with a nose cone total angle of 25° will produce Mach 0.7 flow following that shock wave.

Assuming the shocks off the body don't significantly alter the Mach number, the wings will experience this lower Mach number if their span isn't too large. The wings are then experiencing a Mach number lower than most commercial jets.

Why, then, do almost all supersonic planes use either a highly swept leading edge (most jet fighters) or a sharp leading edge (like the F-104)?

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    $\begingroup$ My guess is that the shock cone appears only when the nose (and the aircraft) is going faster that Mach 1. For example, if the aircraft is flying at Mach 0.9 there will be no shock cone (or not big enough), and your wings will fly through a Mach 0.9 air flow. As consequence, I think that your wings must be effective at least for the whole subsonic and transsonic domains. $\endgroup$
    – Quentin
    Commented Jan 19, 2017 at 8:07
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    $\begingroup$ If you want to reach M 1.5, you have to go through M 0.1, M 0.2, M 0.3... The aircraft does not magically pop at M 1.5, so the whole flight domain must be considered, and this may be an important factor in the design of the wing of supersonic aircraft. (Note that this does not hold if the aircraft/drone/missile is released by a supersonic carrier vehicle). $\endgroup$
    – Quentin
    Commented Jan 19, 2017 at 9:37

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Normally, the flow speed at the wing is about the same as the speed ahead of the bow shock. Only if the fuselage is still conical and expanding at the wing station would the local flow speed be smaller. In order to be subsonic, you would need an absurdly blunt fuselage at Mach 1.5. But the wing is in a way still subsonic if its leading edge sweep is sufficient.

If you look closely enough, every shock has a subsonic region, but it is normally very small. The size of the subsonic region depends on the flow speed and the bluntness of the body causing the shock. See the diagram below for the parts of a classical oblique shock.

Diagram of an oblique shock and its three areas
Diagram of an oblique shock and its three areas (picture source)

Supersonic aircraft are pointed at the front end in order to minimize the region of the strong shock, resulting effectively in an attached shock. The steeper the shock angle for a given flow speed, the higher the losses are. Only in hypersonic vehicles are blunt noses used on purpose, but nor for braking but for limiting local heat loads.

Why does the subsonic region disappear so soon? Because the cross section of the body stops to grow in flow direction, and the pressure energy past the strong shock is converted into speed again in expansion waves.

This means that the wings still fly in supersonic flow. If the fuselage has a constant cross section along the wing's chord, the flow speed is even the same as the speed ahead of the shock. Otherwise, the tail shock would have a smaller cone angle, which it does not have; the flow accelerates again in expansion waves when the conical nose changes to the regular fuselage, and again when the fuselage contracts at its end.

Now it makes sense to sweep the wing such that it will fit inside the Mach cone resulting from the inboard wing section. This results in a subsonic orthogonal speed component at the leading edge. The important insight here is that the spanwise speed component of the flow is not affected by the wing, and only the orthogonal component counts. If that component is subsonic, the whole wing behaves like a straight wing in subsonic flow. Note that all supersonic aircraft with a subsonic leading edge use blunt-nosed airfoils, because those create a suction force (PDF!) which reduces drag.

At higher speed sweep needs to be so steep that the problems of swept wings cannot be tolerated any more. Then and only then it makes sense to have a supersonic leading edge which has less sweep than the Mach cone and needs to be sharp in order to minimize drag. Another way to combine high sweep and good subsonic characteristics is to use variable sweep. This was the preferred solution until it was realised that high supersonic speed is rarely used in combat.

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  • $\begingroup$ Thanks for the answer. That makes sense. So the speed of the air in the flight direction will be the same as that in the freestream if the fuselage has a relatively constant cross section, because the expansion waves will exactly subtract the velocity change from the cone. A swept wing must be tucked back such that the normal component is below the critical Mach number of the airfoil. What about the density, etc. of the air along the fuselage and at the wing? I'd imagine it must be higher since the fuselage must push air out of the way. $\endgroup$
    – Gus
    Commented Jan 20, 2017 at 0:43
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    $\begingroup$ @Gus, no, pushing the air out of the way does not have to mean compressing it. At subsonic speeds it instead accelerates and expands. At supersonic speeds the shock waves compress the flow and expansion waves, well, expand it. In both cases, density decreases with increasing speed. $\endgroup$
    – Jan Hudec
    Commented Jan 20, 2017 at 19:45
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    $\begingroup$ @Gus: You find a strong density variation between the upper and the lower surface of the wing, but in lengthwise direction density varies little. $\endgroup$
    – Peter Kämpf
    Commented Jan 20, 2017 at 20:04
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Though the oblique shocks reduce the downstream mach number, the flow will still be supersonic usually; for the flow to be subsonic, the shock has to be either normal or detached (bow shock).

For every body angle (i.e. corner angle)-mach number combination, there is a maximum corner angle $\theta_{max}$, beyond which the shock will be detached from the body.

Oblique Shock

Oblique shock angle, β, as a function of the corner angle, θ; By -- Myth (Talk) 05:29, 27 October 2007 (UTC) - Own work (Original text: self-made), Public Domain, Link

For cases where the corner angle is less than the maximum, there are two solutions possible- strong and weak. The weak solution leads to downstream supersonic flow (and smaller shock wave angle β), while the strong one leads to subsonic flow downstream (and larger shock wave angle β).

Which solution is 'preferred' depends on the ratio of upstream and downstream pressure; in case of external flows which concern us, downstream pressure is usually close to upstream pressure (both near $P_{atm}$) and as a result, the weak solution (and downstream supersonic flow) is 'selected'.

From NACA Report 1135: Equations, tables and Charts for Compressible Flow:

... two solutions exist for each cone and Mach number, but it is believed that only the weaker shock wave can occur on an isolated convex body.

So, the wings have to be designed for supersonic regime in case the nose produces oblique shock waves (detached shock waves are no good- they increase drag tremendously and that is the main reason they are used in re-entry vehicles). Even if the flow is not supersonic, you're still in transonic regime, where you need design the wings keeping the critical Mach number in mind.

Though the 3D flow over a cone is similar to the flow over the wedge (in that it has strong-weak cases and separation above a certain limit), the maximum angle $\theta_{max}$ is higher in case of 3D flow. Again, the attached shock is the 'weak' one.

Because of the 3D relieving effect (which causes a weaker shock), the pressure on the cone surface is less than the wedge surface pressure and the cone surface Mach number is greater than that of the wedge surface. As a result, the designer has to contend with supersonic flow still.

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    $\begingroup$ Thanks for the answer. A shock off a 3D cone is different than a shock off a 2D wedge, though. Moving from 2D to 3D changes the shock's strength, which is why you're seeing 12.5° as exceeding the limit for the 2D case. I don't think a shock that causes supersonic flow to go to subsonic flow necessarily requires the shock to be detached, though I could be wrong and that may be the issue. $\endgroup$
    – Gus
    Commented Jan 19, 2017 at 15:52
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    $\begingroup$ I think what you're saying is that if the solution we see for the 3D cone is a strong shock (ie the downstream flow is subsonic), then the nose is poorly designed to begin with. We should never see strong shocks off the nose, otherwise the drag is very high. I don't think the strong shock necessarily creates a detached bow shock, so I'm still having trouble thinking about why the drag will be so much higher in the case of the strong shock off the nose cone. $\endgroup$
    – Gus
    Commented Jan 19, 2017 at 16:01
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    $\begingroup$ @Gus I added a note in response to your comment. You're correct that the shock will still be attached in the 3D case; however, it will be comparatively weaker compared to the 2D case and as a result of the 'weak' solution, the downstream Mach number is still likely more than 1. $\endgroup$
    – aeroalias
    Commented Jan 19, 2017 at 16:51

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