How swept wing actually works?

[Dreamlinerwannabe]

Hi,

Fellow aviations, it's me again.

I am puzzled with the question how swept wing works. I read a book called Ace the technical pilot interview and found a lot of things inside are wrong, and not clear enough. I also used wikipedia to aid my studies, but it's still not enough.

What wikipedia says:

"Suppose a cylindrical wing
(constant chord, incidence, etc.) is placed in an airstream at an angle
of yaw - ie., it is swept back. Now, even if the local speed of the air on the upper surface of the wing becomes supersonic, a shock wave cannot form there because it would have to be a sweptback shock - swept at the same angle as the wing - ie., it would be an oblique shock. Such an oblique shock cannot form until the
velocity component normal to it becomes supersonic."

Why shock wave cannot form there because it would have to be a sweptback shock?
What does it mean by "velocity component normal to it becomes supersonic"?
So sweeping a wing make the shock wave can't form, then it backs to the question, why and how can it do that?

I have been thinking about it for 2 weeks!

Thank you very much!

[Runaway Gun]

In simple terms (because that's how I think), sweeping back the wing effectively increases the chord, which actually reduces the thickness/chord ratio, making it seem like a thinner wing, thereby delaying shock wave formation.

Does that help?

[Nomde plume]

Normal meaning perpendicular.
Normal component of a straight wing will just be whatever the relative air flow is. As the wing sweeps back the air flow goes from running normal to the leading edge to running parallel (imagine the extreme: theoretical wing with 90 degree sweep back, no perpendicular airflow, all airflow runs parallel and can never sit in the shockwave.)
To calculate the basic normal component of air flow involves simple trigonometry.

[FE Hoppy]

Easiest way to picture it is to look outboard at 90° any point along the chord. the if the wing is thinner that's where the air would rather go. If the wing is thicker then it gets more difficult so the pressure can't escape and a shock wave could form.

It's not a perfect explanation but it helps build a mental picture of the pressure distribution.

[JammedStab]

It seems these definitions are difficult to understand. I can't understand what Wikipedia is talking about.

But I do believe that thinner wings delay shockwave formation. Why, because it is thinner and therefore has less camber for airflow acceleration toward the speed of sound. So in the old days of straight wings where they wanted to go faster but delay shockwave formation, wings were made thinner. Of course you can only make a wing so thin and then you have strength problems(plus nor much fuel carrying capability).

So they swept back the wings. This delayed the shockwave even more. The engineers told us that this effectively increases the chord, which actually reduces the thickness/chord ratio, making it seem like a thinner wing.

But what does that mean? Seems difficult to understand. But if you use some imagination for your favourite jet transport and imagine that it was a swing wing airplane it might help(or maybe an F-111).

First you have to accept as fact that a lower thickness to chord ratio means delayed shockwave.

Then imagine the wings can sweep back and forth between 0° sweep and 35° sweep. You have a big saw and are going to cut each wing off to look at its camber(profile view). With the wings fully forward at 0° sweepback you make a marking at each wing equidistant from the wing root at the fuselage. Then you saw off the left wing parallel to the fuselage straight from front to back. Then you sweep the remaining wing back to the 35° position and then saw off the other wing parallel to the fuselage which will be a different cut(longer actually).

Now compare the cambers of the wing and measure them as well from leading edge to trailing edge. This is the equivalent of the route that the airflow takes for 0° sweep and 35° sweep. Your measurements will show that the sweepback cut is longer, therefore there is a lower thickness to chord ratio.

What does that mean?

For a sweptback wing, the lower thickness to chord ratio=less acceleration of airflow=lower airflow speed=less drag at high mach numbers=less power required to reach a given higher mach number=less fuel burn=less money spent=more smiles).

[Skyblade]

Wing is sensitive to the airspeed vector which is normal to the leading edge.

For the sake of explanation airspeed over the surface of a swept back wing can be considered as a force vector which is made up of two components.
One component perpendicular to the leading edge and one across. By triangle/vector law of addition these two components/vectors can be joined by one single resultant, which is bigger than both the vectors.

Since the wing is swept back, the component of airspeed perpendicular to the leading edge becomes smaller in magnitude than the resultant (Resultant here is the relative airflow which passes parallel to the longitudinal axis).
Hence a swept back wing can go faster as compared to an equivalent straight wing without the hazard of formation of shock waves.