meritflyer
Well-Known Member
Can you give me a reference for your thoughts on this subject?
Overbanking v. Adverse Yaw
- Thread starter meritflyer
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Can you give me a reference for your thoughts on this subject?
tgrayson
New Member
Absolutely. I'm pleased you asked. Which thoughts in particular would you like references for?
meritflyer
Well-Known Member
Any and/or all if possible.
tgrayson
New Member
Hmmm....we'll start with this. Let me know if there's something else you'd like references to.
From Sears and Zemansky’s “University Physics”, 10th edition, by Young and Freedman, p. 140.
Caution:>In doing problems involving uniform circular motion, you may be tempted to include an extra outward force of magnitude m(v^2/R) to “keep the body out there” or to “keep it in equilibrium”; this outward force is usually called “centrifugal force” (“centrifugal” means “fleeing from the center”). Resist this temptation, because this approach is simply wrong! First, the body does not “stay out there”; it is in constant motion around its circular path. Its velocity is constantly in direction so it accelerates and is NOT in equilibrium. Second, if there were an additional outward (“centrifugal”) force to balance the inward force, there would then be no net inward force to cause the circular motion, and the body would move in a straight line, not a circle. Third, the quantity m(v^2/R) is not a force; it corresponds to the “ma” side of Sum(F)=ma and does not appear in Sum(F). It’s certainly true that a passenger riding in a car going around a circular path on a level road tends to slide to the outside of the turn, as though responding to a “centrifugal force.” But such a passenger is in an accelerating, non-inertial frame of reference in which Newton’s first and second laws don’t apply. As we discussed in Section 4-3, what really happens is that the passenger tends to keep moving in a straight line, and the outer side of the car “runs into” the passenger as the car turns (Fig. 4-8c). In an inertial frame of reference there is no such thing as a “centrifugal force” acting on a body. We promise not to mention this term again, and we strongly advise you to avoid using it as well.
http://www.amazon.com/Sears-Zemansk...=pd_bbs_3/104-7420168-3171959?ie=UTF8&s=books
For something more accessible, here’s the Physics FAQ:
http://math.ucr.edu/home/baez/physics/General/Centrifugal/centri.html
MidlifeFlyer
Well-Known Member
In other words, merit, what we =call= "centrifugal force" is actually a shorthand description of a body's tendency to keep moving in the same direction while being pulled to the center.
As I recall (it's been a long time) the standard elementary school "experiment" to show centrifugal force is swinging a ball around on a string. The apparent movement of the ball to the outside of the circle is actually the result of the ball's straight line momentum and the centripetal force exerted by the string. If the string broke, the ball would travel in a straight line at a tangent to the circle, not outward. If centrifugal force really existed, the ball would travel outward.
There's a pretty good graphic at http://www.phy.ntnu.edu.tw/oldjava/circularMotion/circular3D_e.html. Note the string (centripetal force) and the straight ahead (not curved) velocity vector of the ball.
That's why "centrifugal force" is not a "force" that acts on something. Ball or airplane, it's just a convenient shorthand to describe the net result of the centripetal force and velocity vectors. Maybe "centrifugal phenomenon" would appease those who object to the technically inaccurate use of the word "force?"
tgrayson, how'd I do?
tgrayson
New Member
Absolutely beautiful.
I, too, will use centrifugal force when it's useful, such as when I explain why the ball does what it does, or drawing a propeller governor.
tgrayson
New Member
Ah, I was wondering if that were you. Thanks for the kind word.
TXaviator
Well-Known Member
my groundschool instructor just drew a force diagram and completely drew in "centrifugal force".
he asks "so why does the ball go to the outside of the turn?.... because CENTRIFUGAL FORCE PULLS IT OUT THERE"
noooooooooooooooooooooooooooooooooooo BAD PHYSICS!
further, he went on to show, "when the centrifugal force equals the centripetal force, you have a coordinated turn"
which is just totally and blatantly FALSE!! NOT ACCURATE!! UNTRUE!!
:argue::argue::argue::argue::argue::argue:
he asks "so why does the ball go to the outside of the turn?.... because CENTRIFUGAL FORCE PULLS IT OUT THERE"
noooooooooooooooooooooooooooooooooooo BAD PHYSICS!
further, he went on to show, "when the centrifugal force equals the centripetal force, you have a coordinated turn"
which is just totally and blatantly FALSE!! NOT ACCURATE!! UNTRUE!!
:argue::argue::argue::argue::argue::argue:
fish314
Well-Known Member
Sorry, man. Been a while since I got on JC, I've been busy. Besides, Tgrayson has more geek power than I got! (Don't take that the wrong way Tgrayson, that's a good thing.)
Looks like you guys covered centrifugal and centripetal forces pretty well without me, though....
Now you guys have actually raised a question for me, which is the original question: Why does overbanking tendency exist at high bank angles and not at low ones? My initial guess is that it has something to do with the direction of the weight vector and that results in greater amounts of sideslip at high bank than low bank, but I'm not really sure.
How about you Tgrayson? Did you pay more attention in stability classes than I did? I've been looking through Jan Roskam's Airplane Flight Dynamics and Automatic Flight Controls, but it's been about 10 years since I studied this stuff.
tgrayson
New Member
You're right, that's a terrible explanation, and I hated regurgitating that when I was studying for checkrides. No one could give me another one, though. It took me years to finally understand.
A better answer is that the bank angle is not appropriate for the turn rate. That's because you have either too much inside rudder or not enough, meaning that you have other forces increasing or decreasing the centripetal force provided by the wing. These forces are
1) fuselage side force (because you're yawed)
2) thrust component.
Probably a more technically correct answer is that the relative wind is parallel to the plane of symmetry of the aircraft. The ball just reflects the resulting lateral acceleration.
tgrayson
New Member
Hey, I just read the books, I never took the classes. ;-) I'm sure I have lots of gaps, so I will appreciate any insight you have. I learn more by being wrong than by being right.
I have several of Jan Roskam's books, but I don't care for his stability stuff. I've learned more from Perkins and Hage's "Airplane Performance Stability and Control."
As for the overbanking tendency, I believe that it always exists. At small bank angles, it's just very small and gets overshadowed by other forces, such as small variations in coordination. Even at small bank angles, however, you will notice a strong overbanking tendency as the airspeed gets lower (CL gets higher.) I see this when doing turning stalls. Takes a lot of aileron to keep the bank at only 30 degrees at the stall.
If you have the Perkins book, it gives a formula on page 452 for the required aileron deflection:
Aileron deflection = - Lift Coefficient * Bank Angle (in radians) * (Rolling Coefficient due to yawing velocity)/(4*mu*Rolling Coefficient due to aileron deflection).
BTW, I'm just going from memory here on what the stability coefficients mean; if I were being graded, I'd look them up. ;-)
The text also says "In the pefect turn, then, the aileron must always [emphasis mine] be held against the turn to balance the rolling moment due to the yawing velocity, while the rudder must always be held into the turn to balance out the damping in yaw."
I don't think this book mentions it in this context, but the application of rudder produces another rolling moment not related to the yawing velocity. The effective lift of the vertical stabilizer will be above the longitudinal axis and will produce a rolling moment away from the turn. Once you have to put in a slight amount of opposite aileron, the adverse yaw will require less rudder into the turn, resulting in less rolling moment opposing the overbanking tendency resulting in more opposite aileron. This might mean that the increase in aileron deflection due to bank angle is non-linear. Just speculating here...better check my reasoning. ;-)
seagull
Well-Known Member
Without getting into too much detail, the answer is pretty simple for most of the scenarios. The dihedral effect is greater at higher AoAs. You are at a higher AoA in a steep bank, all else equal, due to the higher load factor. That increases the dihedral effect, so increases overbanking tendency.
Hey, I just read the books, I never took the classes. ;-) I'm sure I have lots of gaps, so I will appreciate any insight you have. I learn more by being wrong than by being right.
I have several of Jan Roskam's books, but I don't care for his stability stuff. I've learned more from Perkins and Hage's "Airplane Performance Stability and Control."
As for the overbanking tendency, I believe that it always exists. At small bank angles, it's just very small and gets overshadowed by other forces, such as small variations in coordination. Even at small bank angles, however, you will notice a strong overbanking tendency as the airspeed gets lower (CL gets higher.) I see this when doing turning stalls. Takes a lot of aileron to keep the bank at only 30 degrees at the stall.
If you have the Perkins book, it gives a formula on page 452 for the required aileron deflection:
Aileron deflection = - Lift Coefficient * Bank Angle (in radians) * (Rolling Coefficient due to yawing velocity)/(4*mu*Rolling Coefficient due to aileron deflection).
BTW, I'm just going from memory here on what the stability coefficients mean; if I were being graded, I'd look them up. ;-)
The text also says "In the pefect turn, then, the aileron must always [emphasis mine] be held against the turn to balance the rolling moment due to the yawing velocity, while the rudder must always be held into the turn to balance out the damping in yaw."
I don't think this book mentions it in this context, but the application of rudder produces another rolling moment not related to the yawing velocity. The effective lift of the vertical stabilizer will be above the longitudinal axis and will produce a rolling moment away from the turn. Once you have to put in a slight amount of opposite aileron, the adverse yaw will require less rudder into the turn, resulting in less rolling moment opposing the overbanking tendency resulting in more opposite aileron. This might mean that the increase in aileron deflection due to bank angle is non-linear. Just speculating here...better check my reasoning. ;-)
tgrayson
New Member
The dihedral effect (due to wing sweep) is stronger at high AOA's for swept wing aircraft, but I'm not sure that's true for a straight wing (which depends on geometric dihedral, mostly). I'll have to check my references.
Anyway, the overbanking tendency is caused by the relative velocity difference between the inside and outside wings, and this appears to be unrelated to the dihedral effect.
The dihedral effect depends on the existence of sideslip before it kicks in, and, in theory, as long as no sideslip develops (ball centered), no dihedral effect will materialize. In theory. ;-)
seagull
Well-Known Member
.
Dihedral effect is there regardless of how you get there. Swept wing aircraft get an extra dose, which is usually compensated for by doing other things that reduce it (like less actual dihedral).
I am not sure the relative difference in airspeed over the wings is much of a player in reality. I can't imagine there is much differential considering the actual radius of the turn -- with the exception of sail planes.
The constant rate of change of the turn would have to lead to some small amount of sideslip occurring. The aircraft is turning due to the natural stability added from the vertical stab, which is seeing a sideslip and applying force to change the heading of the aircraft. Additional rudder may be required as well, but I would not agree that there is no sideslip occuring. Sure, the ball is centered, but I think you are reading too much into that!
tgrayson
New Member
The aerodynamics books have the math to back up the claim that it is the wing velocity differential, including the one that I quoted. Other books agree, so I think the explanation is orthodox. I can show some of them to you if interested. I'll check and see if Carpenter discusses it. It really doesn't take much to produce the effect, considering that the greatest speed differential happens on the wing tips....long lever arm. And lift is proportional to V^2. I'm sure you don't see it in the MD11, with your wimpy bank angles.
Well, several issues here. Remember that an aircraft rotating around its vertical axis has an equilibrium rotation rate, which means the net moment is zero; otherwise, the rotation rate would be increasing or decreasing. Therefore, the existence of a rotation does not demonstrate the existence of a force applied via the vertical stabilizer. Yes, one was required to initiate the rotation, but isn't required to maintain it. Think of Newton's First Law.
Even if this force were required, however, the sideslip would be coming from the inside of the turn, not the outside, and would tend to roll the airplane away from the turn. This is the opposite of the overbanking tendency.
In actuality, however, the rotation around the vertical axis provides a damping effect for the vertical stabilizer and if not aided by the pilot, the aircraft would tend to yaw a bit towards the outside of the turn (one source of adverse yaw), producing a sideslip away from the direction of the turn. This is the same effect as in the above paragraph, the aircraft would tend to roll away from the turn. This is the classic slipping turn.
The only time that dihedral produces an overbanking tendency is in a skidding turn...too much inside rudder.
I'm sure that the ball is not a faultless indicator of coordination, because it's only located in one place in the airplane, but the principle is sound. An uncoordinated aircraft generates fuselage forces that produce an acceleration along the lateral axis. The ball measures this acceleration. Certainly you can fool it. In a ME airplane, flying along in a zero sideslip condition with one engine out has the ball 1/2 out of center, but the aircraft is perfectly coordinated, as verified with a yaw string. But that's an unusual situation.
One of my aerodynamics books stated that the ball is a poor indicator of coordination with aircraft that do not develop much fuselage lift in a sideslip. I think it used the F117 as an example.
seagull
Well-Known Member
I should know better than argue with you! That's what I get for writing this when my mind is on other matters, but that was good. Of course, any dihedral effect would not be overbanking, should have realized that, even in my current mental state. Well, better to be distracted here than while flying a jet, I guess!
tgrayson
New Member
And what more important could you have to think about than aircraft aerodynamics?
Anyway, Carpenter discusses the wingtip speed differential on p. 229 of the Stabilty and Control Book.
Realms09
Well-Known Member
In this thread two key concepts have been presented:
1) Overbanking tendency is present in all turns due to the velocity differential along the lateral axis of the aircraft.
2) A perfectly coordinated airplane is in a stable condition (no side slip) and therefore will have no tendency to roll out of the turn.
These are both correct.
From this correctness one may be tempted to conclude that the standard appraisal of overbanking tendency is bunk (stability counteracts overbanking tendency at low angles of bank, exactly balances at medium angles of bank, and is overwhelmed at large angles of bank). However, this appraisal is still correct - I've experienced it first hand, and so have you. The key to bridging the gap between these two schools of thought is to look at what happens when the overbanking tendency begins to act on an aircraft in a perfectly coordinated turn. The overbank will, quite naturally, increase the bank angle of the aircraft, but the overbank DOES NOT provide the necessary additional yaw moment to keep the aircraft coordinated. Bank has increased without additional rudder action, so a sideslip is initiated. If the angle of bank is much smaller (i.e., the side slip is much smaller) relative to the aircraft's lateral stability, then the system response could be so large as to completely roll the aircraft out of the turn. If the angle of bank is about on par with the aircraft's lateral stability, then the system response will be "about right" and eliminate the sideslip without much change in bank. If the angle of bank is very large compared to the lateral stability, the system response will be inadequate and the rolling motion will continue.
With this knowledge in hand, let's look at each sort of turn from the pilot's perspective.
SMALL BANK:
If you are flying a coordinated turn using a small amount of bank, one equillibrium point is to hold aileron against the turn to exactly counteract the overbanking tendency. In this equilibrium, there is no overbank thanks to the counteracting moments of the ailerons, and there is no stability response because there is no sideslip. However, this is an UNSTABLE equilibrium and maintaining such a turn would be much like balancing an inverted pendulum. If at any point you do not hold EXACTLY the right aileron pressure, the equilibrium is destroyed. If too little aileron is used, overbank will initiate a sideslip and the strong stability response will roll the aircraft out of the turn. If too much aileron is used, you're rolling the aircraft out of the turn yourself. Therefore, except for that SINGLE point, the aircraft will tend to roll out of the turn, and we find ourselves needing to counteract that tendency.
MEDIUM BANK:
In this case we begin with the same equilibrium as in the small bank - opposite ailerons to balance overbanking in a coordinated turn. If too little aileron is used in this case the same sideslip will occur, but the side slip is now large enough that the stability is appropriately calibrated. The aircraft rolls out of the sideslip back to the original bank angle. A departure from equilibrium returns to equilibirum, so the flight control position is in stable equilibrium. We don't need to do much in a medium bank turn. If we goof, the aircraft picks up the slack.
LARGE BANK:
The flight control position is once again in unstable equilibrium, but in the opposite sense of the small bank. Too little aileron in that perfectly coordinated turn will cause a sideslip insufficient to produce the appropriate lateral stability response. Except for the SINGLE equilibrium point, the aircraft will tend to overbank, and we have to fight that tendency.
Maybe I've beaten a dead horse, maybe I've helped, but either way, there it is.
1) Overbanking tendency is present in all turns due to the velocity differential along the lateral axis of the aircraft.
2) A perfectly coordinated airplane is in a stable condition (no side slip) and therefore will have no tendency to roll out of the turn.
These are both correct.
From this correctness one may be tempted to conclude that the standard appraisal of overbanking tendency is bunk (stability counteracts overbanking tendency at low angles of bank, exactly balances at medium angles of bank, and is overwhelmed at large angles of bank). However, this appraisal is still correct - I've experienced it first hand, and so have you. The key to bridging the gap between these two schools of thought is to look at what happens when the overbanking tendency begins to act on an aircraft in a perfectly coordinated turn. The overbank will, quite naturally, increase the bank angle of the aircraft, but the overbank DOES NOT provide the necessary additional yaw moment to keep the aircraft coordinated. Bank has increased without additional rudder action, so a sideslip is initiated. If the angle of bank is much smaller (i.e., the side slip is much smaller) relative to the aircraft's lateral stability, then the system response could be so large as to completely roll the aircraft out of the turn. If the angle of bank is about on par with the aircraft's lateral stability, then the system response will be "about right" and eliminate the sideslip without much change in bank. If the angle of bank is very large compared to the lateral stability, the system response will be inadequate and the rolling motion will continue.
With this knowledge in hand, let's look at each sort of turn from the pilot's perspective.
SMALL BANK:
If you are flying a coordinated turn using a small amount of bank, one equillibrium point is to hold aileron against the turn to exactly counteract the overbanking tendency. In this equilibrium, there is no overbank thanks to the counteracting moments of the ailerons, and there is no stability response because there is no sideslip. However, this is an UNSTABLE equilibrium and maintaining such a turn would be much like balancing an inverted pendulum. If at any point you do not hold EXACTLY the right aileron pressure, the equilibrium is destroyed. If too little aileron is used, overbank will initiate a sideslip and the strong stability response will roll the aircraft out of the turn. If too much aileron is used, you're rolling the aircraft out of the turn yourself. Therefore, except for that SINGLE point, the aircraft will tend to roll out of the turn, and we find ourselves needing to counteract that tendency.
MEDIUM BANK:
In this case we begin with the same equilibrium as in the small bank - opposite ailerons to balance overbanking in a coordinated turn. If too little aileron is used in this case the same sideslip will occur, but the side slip is now large enough that the stability is appropriately calibrated. The aircraft rolls out of the sideslip back to the original bank angle. A departure from equilibrium returns to equilibirum, so the flight control position is in stable equilibrium. We don't need to do much in a medium bank turn. If we goof, the aircraft picks up the slack.
LARGE BANK:
The flight control position is once again in unstable equilibrium, but in the opposite sense of the small bank. Too little aileron in that perfectly coordinated turn will cause a sideslip insufficient to produce the appropriate lateral stability response. Except for the SINGLE equilibrium point, the aircraft will tend to overbank, and we have to fight that tendency.
Maybe I've beaten a dead horse, maybe I've helped, but either way, there it is
.