Increased drag swept wing high Mach nr.

[Spir4]

Hello,

I've been struggling with this for a while, and no one was able to give me a clear answer yet.
Please have a look at the following graph:

This graph shows the development of the coefficient of drag versus the Mach number, comparing a straight wing, a 30° swept wing and a 45° swept wing.
The bigger the angle of sweep, the less drag, because more of the wing is inside the Mach cone.
But why, after a high Mach number (M2.1 on this graph), is the drag MORE on a 45° swept wing than on a 30°, even a straight wing?

Anyone knows why this is?

[hawk37]

Perhaps it is because for a higher sweep, the distance the air flows over the wing will be greater, ie straight wing chord / cos (sweep angle). Hence more drag?

Does this question pertain to the case where the wingspan (distance from wingtip to wingtip) remains the same for all sweep angles (ie, the wing would have to be made longer)?

Or does it decrease as the wing gets swept back?

An interesting question....

[BOAC]

Once the aircraft is well through the critical drag rise and wholly supersonic, sweepback does little for drag and as hawk says, merely increases the 'wetted' area. Also, as the mach 'cone' begins to play on the leading edge of a swept wing at the higher Machs, the advantage of a subsonic wing flow are degraded. This is why a lot of high Mach aircraft have 'straight' wings, accepting the greater drag rise trans and low sonic for the benefits at higher Mach, remembering also that these a/c have significant thrust excesses anyway and are not designed to spend much time in the critical drag rise regime.

[Spir4]

Perhaps it is because for a higher sweep, the distance the air flows over the wing will be greater, ie straight wing chord / cos (sweep angle). Hence more drag?

Then, why is the drag less below M2.1? Why suddenly, above M2.1, the drag gets higher for a higher swept-back wing?

Does this question pertain to the case where the wingspan (distance from wingtip to wingtip) remains the same for all sweep angles (ie, the wing would have to be made longer)?

Or does it decrease as the wing gets swept back?

I assume every other aspect of the wing remains the same. According to the graph the aspect ration is 3, and the t/c ration 5%, for the three wings.

Still puzzeled...

[fizz57]

Perhaps that's when the Mach cone hits the wingtip?

[henra]

Originally Posted by fizz57

Perhaps that's when the Mach cone hits the wingtip?

This does not really explain whys the drag lines for 0°, 30° and 45° wing sweep all meet at M2,1.

The cone angle of the Mach cone is Phi=arcsin(1/Ma).
That gives a cone angle of 28,4°, equalling a sweep of 61,6°.

Edit2: Deleted after valid comment from @Hazelnuts.

[HazelNuts39]

If the tree wings have the same span and aspect ratio, they have the same wing area. Why would the wetted areas be different?

[henra]

Originally Posted by HazelNuts39

If the tree wings have the same span and aspect ratio, they have the same wing area. Why would the wetted areas be different?

Edit: removed for late night stupidity

You are oviously right about the wing area.
Reminder to self: Don't post past 9pm.

Edit3:
And if posting start thinking before.

Anyway, here comes an explanation that has at least some 'official' backing:

The relative thickness of the profile perpendicular to leading and trailing edge of a swept wing is higher than the same for the straight wing or the yawed wing.
http://naca.central.cranfield.ac.uk/...rc/rm/2818.pdf
Chapter 6 + figures 8,9.
Sheared vs yawed wing.

[Mac the Knife]

Bell X-1

[Spir4]

Originally Posted by henra

The relative thickness of the profile perpendicular to leading and trailing edge of a swept wing is higher than the same for the straight wing or the yawed wing.
http://naca.central.cranfield.ac.uk/...rc/rm/2818.pdf
Chapter 6 + figures 8,9.
Sheared vs yawed wing.

Why would that be? And doesn't an increase in thickness creates an increase in drag?

[henra]

Originally Posted by Spir4

Why would that be? And doesn't an increase in thickness creates an increase in drag?

If you shear a wing as in above's description, i.e. constant apsect ratio, the wing area remains the same. When lloking at the chord length perpendicular to the leading edge this will be reduced. As the absolute thickness of the profiles remains the same, the relative thickness will increase

Due to the pressure gradient across the z- axis (i.e. in the spanwise direction, perpendicular to Fuselage) the air will be pulled towards the perpendicular axis of the Leading edge close in front of the LE.

There are two effects whose significance varies with the Mach Number.
The Sweep of the leading edge helps from M1 to the point where the Mach cone angle is the same as the sweep. That is where a shock wave emanating from one point of the leading edge will hit a neighbouring section of the LE.
At the same time there is the relative thickness perpendicular to the LE. As this is thicker the Mach induced flow detachment on the upper side will have a steeper angle and occur earlier and thus increase drag.

[Spir4]

Originally Posted by henra

There are two effects whose significance varies with the Mach Number.
The Sweep of the leading edge helps from M1 to the point where the Mach cone angle is the same as the sweep. That is where a shock wave emanating from one point of the leading edge will hit a neighbouring section of the LE.
At the same time there is the relative thickness perpendicular to the LE. As this is thicker the Mach induced flow detachment on the upper side will have a steeper angle and occur earlier and thus increase drag.

Okay, but if you take the relative thickness into account, and the aircraft is below M2.1, a higher relative thickness (as is the case with increased wing sweep) should also produce larger drag figures.
But on the graph the drag is lower for a higher swept wing below M2.1.

[Spir4]

Okay here are two theories:

1)

• Why the drag decreases with wing sweep in subsonic speeds below M2.1:
  The part of the wing behind the mach cone 'knows' the air is coming, so the airflow can adjust itself to the wingshape. This means no shock drag, but induced drag.
  The smaller the sweep angle, the bigger part of the wing is outside the mach cone, which means shock drag on the part outside of the wing.
  So a low swept wing at subsonic speeds: a lot of induced drag + a lot of shock drag = more drag in total.
  
• Why the drag increases with wing sweep in subsonic speeds above M2.1:
  The faster the aircraft, the smaller the mach cone.
  With a lower swept wing, more of the wing is now outside the cone. Now the drag is mainly shock drag again, and less induced drag: total drag is less.

2)
The higher the speed, the higher the compressibility effects, the higher the temperature, the higher the local speed of sound, the lower the local mach number.
So:

• Why the drag decreases with wing sweep in subsonic speeds below M2.1:
  with a highly swept wing, less compressibility, local mach nr. doesn't decrease that much. With a straight wing: a lot of compressibility, local mach nr. decreases back to the transonic range, a lot of shock drag.
  
• Why the drag increases with wing sweep in subsonic speeds above M2.1:
  At speeds higher than M2.1, the compressibility effects become greater, even at the highly swept wing, reducing the local mach nr. to the transonic range, meaning lots of shock drag.
  The straight wing has even greater compressibility effects, reducing the local mach nr. to a number below the transonic range, meaning subsonic speed, no shock drag at all anymore.

Or maybe it's a combination of both theories.

Those theories are probably not correct, but I'd rather have a semi correct theory that at least explains it a bit, than no theory at all and just accept it.
But of course, counterarguments are very much welcome!