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Swept back wings can delay the formation of shock wave and increase the aircraft critical mach number of the aircraft, right? But how

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The key is to minimize how fast the air is going over the wing.

Aerodynamics for Naval Aviators has information on this subject in and around page 226. Aerodynamics for Naval Aviators.

enter image description here

The ANA will tell you that there is a spanwise component and a chordwise component of velocity associated with the swept wing, and that the only one that matters for critical Mach number is the chordwise component of flow. That is because the wing has to accelerate air over the top of the wing faster than the plane is flying, and only air going "across" the wing gets sped up. The air going "down the length" of the wing does not.

I do not think this is very intuitive of an explanation, so I would describe it a little differently than the ANA. I would say that swept wings delay the shockwave by appearing longer and thinner than the same wing would be if it was unswept. Air passing down the longitudinal axis of a swept wing will see a wing that is "longer" (meaning longer chord length) and with a lower relative thickness than if it passed "sideways" across the wing in the shortest direction. Please see the colored lines that I added to the ANA figure. The red line is the path air will actually take across the swept wing. The blue line is the path the air would take if the wing was unswept. Notice that they are different length lines. The red line is longer, and therefore the wing appears thinner when taking this path. This will have the effect of reducing the apparent thickness of the wing, and accelerating each parcel of air a bit less than if the wing was straight.

Thin wings with a long chord length do not accelerate air as much as a thick wing with a short chord length. Think of a fighter jet vs a STOL bush plane.

In the business of private aviation we get to see a lot of variation of airplane designs, much more so than in the modern airlines. Let's compare two different ways of raising the critical Mach number. We will look at a swept wing Citation Longitude first.

enter image description here

The longitude has a maximum Mach number of 0.84. It achieves that with a swept wing design that makes the wing appear "thinner" to the air that flows over it. The wing is not all that thin, but the sweep makes it appear thinner to the air flowing over it. Again, thinner wings cause the air to accelerate less during its trip over the top of the wing.

Next up, lets look at a different way of delaying the shockwave. A lear 25.

enter image description here

The Lear 25 is the answer to the natural question of "if sweeping the wings makes them seem thinner, why not make them actually thinner?". The Lear 25 has a maximum Mach number of 0.81. It goes nearly as fast as the Longitude because it really does have very thin wings. Since the wings are actually thin, the air wont accelerate as much over the front 1/3rd of the wing (typical location of both maximum air speed and lift), and therefore will delay the shockwave and allow the plane to fly fast. No sweepback needed because the wings are already thin.

Now let's examine one last thing. If making wings thinner from the start is just as good as sweepback, why would you ever use sweepback to achieve the same thing? Structural reasons and storage reasons.

Structural: the main load-bearing member in the wing is called the "wing spar". It runs the length of the wing and is what carries the fight loads. The thicker it is the stronger it is, but also the heavier it will become. You want a light wing spar if possible. Making the spar taller makes it stronger, but also makes your wing thicker. If you can get away with a wing that is taller, you can make the spar taller, and therefore use less material in the creation of the wing spar, creating an overall stronger and lighter wing.

Storage: The Longitude has a range of 3,500 nautical miles. The Lear 25 has a range of only 1,767 nautical miles. The Lear also must make use of the visible wing tip tanks to store fuel. Having a wing that has good internal volume is advantageous because you can put more fuel in the wings and more fuel means more range.

So sweepback delays the onset of the shockwave by appearing to have a lower thickness ratio than an unwept wing. This causes the air to accelerate less over the top of the wing surface and delays the shockwave. This is often considered better than making a wing actually thinner because making the wing thinner would lead to structural issues fitting an acceptably light wing spar and would reduce the internal volume for fuel storage.

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  • $\begingroup$ Nice answer. Anyway the comparison Cessna vs. Learjet is definitely unfair since they are two very different aircrafts: mtow, wingspan, aspect ratio, engine's thrust, airfoils,... they are all different. Just the fact that the "old" Learjet uses a NACA 64 airfoil while the new Cessna uses a supercritical airfoil is enough to lower significantly the critical Mach number even without considering the swept wing. Thickness is also not that important to reduce the shock wave's strength rather the pressure distribution on the upper surface of the airfoil. $\endgroup$
    – sophit
    Aug 17, 2023 at 20:27
  • $\begingroup$ Also range depends on the engines - the much smaller Lear 25 has wings and engines from the late 1950s with an SFC of 126 g/Nh while the Longitude uses modern Honeywell HTF7700L with an SFC of only 43.2 g/Nh - one third of the gas-guzzling J85s of the Lear. $\endgroup$ Aug 18, 2023 at 19:20
  • $\begingroup$ Just to make my point about the unimportance of the thickness reduction: let's take for example the NACA 64 airfoil of the Learjet 25 and let's see how the critical Mach number changes according to the thickness. Let's consider a $C_l$ of 0.4, midway between mtow and empty weight. As visible, the critical Mach number actually decreases (worsens) for thinner airfoils, basically contradicting your answer. $\endgroup$
    – sophit
    Aug 18, 2023 at 21:45
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    $\begingroup$ @ROIMaison: $C_L=0.4$ is midway between mtow and empty weight, I don't know if 0.1 is ever reached actually $\endgroup$
    – sophit
    Aug 24, 2023 at 15:53
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    $\begingroup$ @ROIMaison: understood and agreed. I just wanted to point out that the equation "thinner = higher critical Mach number" is not always true 🤗. But in my opinion this kind of "popular explanation" is anyway not that consistent: if we know that air simply flows straight on the wing, why shall we design the wing like if it were perpendicular to the leading edge instead? We just choose the airfoil taking into account a straight streamline and that's all. $\endgroup$
    – sophit
    Aug 24, 2023 at 17:10
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It increases the effective chord, the effective chord line being parallel to the longitudinal axis, whereas the true chord is perpendicular to the spar axis.

It sort of tricks the air into thinking a wing with, say, a 12% thickness, is an 11% thickness, and the pressure gradients are more gentle and shockwave formation is delayed a bit.

In other words, a swept wing, compared to a straight one with the same area, acts like a wing that is thinner, as far as shockwave formation is concerned.

enter image description here

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  • $\begingroup$ Also for this answer this comment of mine applies. $\endgroup$
    – sophit
    Aug 19, 2023 at 8:29
  • $\begingroup$ My point is that the two airfoil sections have the same thickness, but it's that the thickness ratio and therefore pressure gradient is reduced along the longitudinal axis by sweeping, because the airfoil is effectively "stretched", while the actual thickness, and the overall wing area, is unchanged. Is that the wrong way to think about it? $\endgroup$
    – John K
    Aug 19, 2023 at 13:04
  • $\begingroup$ "My point is that the thickness ratio and therefore pressure gradient is reduced" understood but a smaller thickness ratio might actually worsen the critical Mach number just like the airfoils in my link where a thinner airfoil has a smaller (i.e. worse) critical Mach number $\endgroup$
    – sophit
    Aug 19, 2023 at 14:20
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If you are unsure why air will accelerate when flowing over a wing, read this answer first.

Now keep in mind that local curvature will determine the amount of acceleration. In a straight wing that acceleration is in the direction of the original flow speed because the wing surface curvature is highest along chord lines while zero in spanwise direction.

Now sweep the wing. This will also sweep the direction of highest curvature, and with it the direction into which the flow is accelerated. The speed component of the air in the direction of highest curvature is the flow speed times the cosine of the sweep angle. This is the chordwise speed component in EquipmentOperator's answer. Since the acceleration is also a function of flow speed, the acceleration will not only start from a lower level, but is also reduced by the cosine of the sweep angle, leaving more margin before the speed of sound is reached.

With 60° of sweep the cosine of the sweep angle will drop to 0.5, doubling the flight Mach number at which the same acceleration is reached in a straight wing at Mach 1.0, which nicely confirms the graphics in the first picture of EquipmentOperator's answer.

As much as I support his answer, once you know that curvature is key to the understanding of lift, the explanation with the chordwise speed component becomes very intuitive. The apparent reduction in thickness actually has very little to do with the way sweep works. Also, air passing over a swept wing will actually first move towards the wing tip in the region just ahead of the stagnation line, then curve inwards when acceleration kicks in over the forward chord and finally return into its original direction when pressure rises again over the rear wing.

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  • $\begingroup$ If I were you I'd have elaborated an answer based on this very nice answer of your 🙃 $\endgroup$
    – sophit
    Aug 18, 2023 at 19:05
  • $\begingroup$ @sophit I see we think along the same lines :-) However, reading this old answer again makes me think I should improve the drawing. It does not show correctly what happens ahead of the wing. $\endgroup$ Aug 18, 2023 at 19:06
  • $\begingroup$ 😉 What's actually the source of the first picture in that answer? I can't find anything similar $\endgroup$
    – sophit
    Aug 18, 2023 at 19:09
  • $\begingroup$ @sophit Did it myself, using Inkscape. $\endgroup$ Aug 18, 2023 at 19:21
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The benefits of a swept wing have already been known at least since 1935 when A. Büsemann presented a paper at the Fifth Volta Conference in Rome about the advantages of a swept wing at supersonic speed (Büsemann, A. - "Aerodynamicher Auftrieb bei Überschall Geschwindigkeiten" Luftfahrtforschung, Bd 12, Nr 6, October 3, 1935, pp. 210-220).

Apparently nobody among the attendees (of several nationalities) was able to appreciate that discovery, except the German government that classified that paper the next year. In the following years, further theoretical and practical research was carried out in Germany that verified the property of wing sweep in delaying the onset of the compressibility drag associated with the formation of shock waves.

The same research was independently conducted in the USA (although with some ten years of delay) and have resulted in the seminal paper "Wing plan forms for high-speed flight" by Robert T. Jones.



The answer to this question is anyway of disarming simplicity and it is based on exactly the same principle by which, from an aerodynamic point of view, a fixed airplane invested by moving air in a wind tunnel is perfectly equivalent to the same airplane flying in free air: Galilean transformation.

By virtue of this same principle, the airflow $V_0$ investing a swept wing can be decomposed into:

  1. a 2D airflow perpendicular to the leading edge with speed $V_0cos\Lambda$; and
  2. a spanwise airflow with speed $V_0sin\Lambda$.

enter image description here (source)

Flowing spanwise from the tip to the root of the wing, the airflow in 2. sees a boring, never-changing aerodynamic body that basically resembles a flat plate at zero AoA. Therefore, this airflow generates no aerodynamic forces (except in the boundary layer but we are so far only interested in what happens in the inviscid flow outside it).

On the contrary, the 2D airflow of 1. sees an airfoil travelling at the speed $V_0cos\Lambda$ which is obviously smaller than $V_0$ of a factor $cos\Lambda$.

So, sweeping the wing we simply cheat a bit making the aerodynamic part of our wing see a smaller speed than the flight one. That's all!



In his paper, Robert T. Jones considers a straight wing invested by a flow with a sideslip$\beta$. Albeit not directly considering a swept wing, this approach is anyway formally equivalent to a swept wing with angle $\beta$ invested by a straight airflow; but it simplifies the theoretical treatment.

enter image description here

In particular, it is proved that "the flow patterns are similar to those occurring in an incompressible fluid except for an increase of the pressures in the ratio

$\frac{1}{\sqrt{1-(\frac{Vcos\beta}{c})²}}$"

The net outcome of this result is that the only component of the speed contributing to the aerodynamic forces is the one perpendicular to the leading edge ($Vcos\beta$) while the component parallel to the wing gives a negligible contribution to them.

Later in the paper it is demonstrated that the same conclusions are valid also at supersonic speeds if $\beta$ is bigger than the Mach cone: in that case the aerodynamic forces depend again only on the (subsonic) speed perpendicular to the leading edge of the wing.

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