In a turn the relative forward airspeed over the wings varies along the length of the wings. The wing tip on the inside of the turn has the lowest airspeed, and the tip on the outside the highest. The angle of attack is the vector sum of the forward speed in the air and the vertical speed. The plane is solid lump, so the vertical speed is the same for all parts of the plane.
So, in a turn, along the wings, the vertical speed is the same, but the forward speed varies. The wingtip on the outside of the turn has the lowest angle of attack, and the angle varies continuously over to the tip on the inside of the turn, which has the highest angle. Thus it is possible to create a situation where part of the wing on the inside of a turn stalls.
Thanks for you answer. However, that was not what I am trying to discuss. I know that the wings are designed with an angle of incidence that decreases towards the wing tip. But thats not the point. What I mean is that both wings with the same design, during a turn, should have the same relative wind, therefore same AOA, doesnt matter which one is down or up. I know that this is not right, but I cannot get why. I cant visualize two different relative winds, therefore different AOAs, on each wing. Any help?
Thanks again.
Forgive my English, Im from Argentina.
Regards
Agus.
Francis is not talking about decreasing angle of incidence. He is talking about the wing tip on the inside of a turn having lower forward speed than the outside wingtip.
Yes, I am assuming that the wing section is identical along its length.
After I wrote what I did, I was thinking some more about how to explain it. It isn’t easy without a diagram or two. But starting again.
Assume a long solid wind with the same section end to end. Flying forward, and with some known angle of attack. That angle of attack is made up of two components - the forward motion relative to the wing if it were at zero AoA, and vertical motion in the body of air the plane is flying in. The two components are summed as a vector to yield the airflow angle at the edge of the wing. When you turn, the vertical component is the same for the entire wingspan, but the forward component changes along the length of the wingspan. An increase in forward airspeed will reduce the angle of attack, and an increase in forward airspeed will increase it. So the AoA changes along the wingspan, being higher on the inside of the turn.
Angle of attack is not the result of airspeed. It is the angle between the wing’s chord line and the air approaching it. In a coordinated turn, angle of attack is the same along the entire wing, but the airspeed is higher on the outer wingtip than the inner one.
That higher airspeed means higher induced drag on the outer wing, which will tend to pull the nose away from the turn. That is called “adverse yaw”. Ailerons are designed to produce higher drag when raised than when lowered, adding drag to the inner wing of a turn to counteract adverse yaw.
Is there a chance to get a diagram? I cannot get the relation between the airspeed and the AoA (determinated by the chords line and the relative wind).
I will be glad for any of your answers.
Thank you.
I agree with you in the AoA definition, but Im sure that the AOA during a coordinated turn is not the same in both wings, despite I dont know why.
Thanks you for your answer.
Regards,
Agus.
As an aside, that site also contains the best primer on relativity I’ve ever seen, despite relativity being basically irrelevant to aviation. If he did as good a job on the aspects of physics that are relevant, I’ll wholeheartedly endorse that site.
Adverse yaw results from deflection of the ailerons when a turn is initiated. The downgoing aileron on (what will soon be) the outer wing creates more lift (desirable, as it banks the plane) and more drag (undesirable, but an inherent consequence of the additional lift); the upgoing aileron on the inner wing yields less lift and less drag. This means that when ailerons are deflected to initiate a left turn, the nose of the plane yaws right - “adverse” to the desired turn direction. The pilot uses rudder to counteract this and produce a coordinated turn.
Adverse yaw is seen the moment ailerons are deflected in level flight - there is no need for the plane to be turning, with the outer wing seeing a (marginally) higher airspeed.
A point worth noting is that induced drag is lower at higher airspeeds (it’s proportional to the inverse square of IAS). And “higher airspeed means higher lift” is not a very useful statement: an aircraft in unaccelerated flight at 200 kts produces lift equal to its weight - just as it does at 60 kts.
But the more significant point is that you definitely don’t need one wing moving faster than the other to see adverse yaw - deflecting the ailerons in straight & level flight will do the job every time.
Right. And in a coordinated turn, the lift produced by each wing will be awfully close to equal - otherwise, the bank angle would not remain constant.
You’ve noted that the outer wing sees a higher airspeed; this means it will need a lower AoA to produce lift equal to that of the inner wing. It will have this if the pilot holds a small amount of aileron against the bank (e.g. left aileron in a steady right turn).
Note that this effect is quite small unless the turn has a small radius (which implies a reasonably low airspeed and substantial bank angle).
Define your speeds and radius. C-150 straight back with 150 HP fixed pitch prop is routinely reversed in course in a track about 200 feet wide starting at 120 MPH. At a constant altitude. Slow it to 70 & it can be done in a stupidly small distance. Won’t make a claim that only acrobatic and pipeline patrol pilots would understand or believe. It is just really small. It is called ‘swinging around the post.’
Ask airline pilots how much they use the ruder in hand flying in a cruise condition. Ask an Aerostar pilot or a Swift pilot or any fighter war bird pilot.
Fly a U control model, swing a rock on a string and the rock is going through the air faster than a point 12" inboard of the rock.
A perfect ( for sake of clarity ) level circle flown by an aircraft in still air as far as it is concerned ( don’t go there, we are looking at theory ) the outboard tip will be going faster than the inboard wing tip. Control positions do not make any difference, if the circle is good the the wings, as in two of them, or as measured span wise will have the outboard tip going faster. I think everyone agrees with this but then promptly forget it with all this flight control position stuff which is vastly different for each aircraft
The total lift produced by the airplane, not just the wings, plus any upward pull of the engine & prop combination can result in a level turn. The controls can be all over the place but if the altitude indicator or radar altimeter or the GPS stays at the same altitude and the plane is not accelerating in speed, ( either to the front or back ) and the ball is centered, will be called a level coordinated turn.
Very few planes ( small non jets ) which we seem to be talking about, can not maintain a level coordinated turn with just power, the control surfaces will be in some degree of deflection adding or subtracting lift, drag, and the rudder will never be totally still.
So a coordinated turn for 360° is very rare.
In knife edge flight, most of the aerodynamic lift is supplied by the fuselage and the rest is mostly power. The wings are not doing much to keeping the airplane up.
IIRC, the DC-10 flies nose up because the body is used as a lifting element at high altitudes.
So the argument about which wing is doing how much is kind of irrelevant because too many other things are also in the mix. It is all a joint effort and only considering the vector straight through the aircraft as if it was perfectly level do the wings have balanced or equal lift. That in no way indicates the controls are neutral.
If flying in a straight line with one wing low, held that way with aileron and top rudder is used to maintain a straight line through still air, the wings are not balanced & not totally supporting the aircraft. Need added power and the actual airflow across the wings will be a bit different due to interference with the flow pattern caused by the fuselage track through the air.
The lift is not balanced.
If you still don’t get it, go take some flying lessons.
You will quickly learn that perfect theory of flying does not always mean survivable flying.
Always make an aircraft do what it can.
Never ask it do do what it was not designed to do.
Unless you are a wise old pelican.
And can actually do what you think you can.
At 120 mph, a level turn at at 45-degree bank has a diameter of 1934 ft. At a 70-degree bank (probably more than a C-150 can do in level flight), it’s 705 ft. At 70 mph, these values are 658 and 240 ft.
Yes, but not by a lot.
The C-150 has a wingspan of 33 ft. At a 45-degree bank, the outer wingtip is 23.3 ft further than the inner tip from the center of a turn. So, at 120 mph, it sees an airspeed 2.4% larger; at 70 mph, 7.1%.
And what happens during climbing turn stall? I think that in this case, the outside wing should have a greater AoA than the inner one, cause that one is finally the one that stalls.
True, but not relevant here: no one has said that in a turn the two wings see the same airspeed. I have claimed - and then supplied evidence - that they don’t differ greatly.
I’ve flown one (with an O-200 engine - 100 hp), but never on pipeline patrol.
Then perhaps you can say whether it’s capable of a level 70-degree banked turn. (Note that this implies nearly 3 G.)
Theory of flight is sufficiently well developed to comfortably explain what happens at an airshow.
Right - because none of these* except environment (specifically, air density) meaningfully affect the radius of a banked turn, which is proportional to:
TrueAirspeed[sup]2[/sup] / tan(BankAngle)
*Excluding certain special cases such as a fighter with vectored thrust.
Note that (though you may have to dig a bit for it) all the above is clearly explained by the excellent website linked in post #107, by Francis Vaughan.
