Steep Turns

Sidious

Sidious

Well-Known Member
Hey everyone,
Im working on a lesson plan for steep turns and in the Flight Instructor PTS as well as the AFH they mention the "Overbanking Tendency"....

Why does this happen?

The only thing that I can think of is the Horizontal Component of Lift becomes so great that it overcomes the lateral stability Basically it "wants" to continue banking and if we let it, it would "overbank"

Thoughts? Corrections?

Thanks :)
 
The outside wing in a turn is traveling faster than the inside wing. The faster speed generates more lift and causes the plane to "Overbank" in the direction of the turn.
 
:yeahthat:

Try your best not to make up explanations, the FAA examiner will see right through it.

Also, you're suppose to demonstrate instructional knowledge of an element.

Saying "Because it WANTS to. . ." does not demonstrate this at all. Why does it want to? (See DD's post)
 
Not only is the wing turning faster, but the outside wing and it's down deflected aileron changes the camber and increases the angle of attack of the wing, resulting in more lift.....right? maybe? no? yes? man, too much studying...brains about to pop!
 
MikeOH58 said:
Not only is the wing turning faster, but the outside wing and it's down deflected aileron changes the camber and increases the angle of attack of the wing, resulting in more lift.....right? maybe? no? yes? man, too much studying...brains about to pop!
Click to expand...

the outside wing with the down aileron does have increased camber, but it also has increased drag to go along with...so you get 'adverse yaw'. so, here's a good practice question for ya, mike...what is the effect of adverse yaw and how do you compensate for it? :)
 
I will take a stab at it...The effect of adverse yaw is the nose of an aircraft turning or yawing to be more precise opposite the direction of turn or outside of turn because higer induced drag and more lift on the raised or outside wing....

To compensate, use proper rudder control and coordination...Manufacturer can also compensate by building in differential aileron range of movement.
 
Sidious said:
the AFH they mention the "Overbanking Tendency"....
Click to expand...
Better stick with the AFH - it is explained where you say it is "mentioned" - on pg. 3-9, in the paragraph left of fig. 3-10.
"Because the outside wing is faster developing more lift".
Period. Don't add any more. That's what the book says, and you have to back up what you say. With FAA material.

The only thing that I can think of is the Horizontal Component of Lift becomes so great that it overcomes the lateral stability.
Click to expand...

I think I have read this as a cause also, and I think it used to be in an older version of the PHAK, but on pg. 3-4 of the current PHAK, under fig.3-17, the explanation of lateral stability does not support this, and I don't know where else to find it.

Remeber that the aileron is not deflected in a turn, once the turn (bank) is established. The aileron is deflected to start the roll, but once the desired bank is established, the aileron is neutalized. If anything, the aileron is up slightly to counter the overbanking caused by the higher speed outside wing. So don't go there.

Get the information from the AFH or the PHAK. That is what is referenced in the PTS.
 
I could be wrong, but when you say the horizontal component becomes so great it overcomes lateral stability, do you mean lateral stability by dehedral? ie- aircraft is banked, and to a certain amount of bank the lowered wing will be strictly be developing a vertical component of lift, and when that bank is increased, both wings are now developing stronger horizontal components of lift?
 
MikeOH58 said:
I could be wrong, but when you say the horizontal component becomes so great it overcomes lateral stability, do you mean lateral stability by dehedral? ie- aircraft is banked, and to a certain amount of bank the lowered wing will be strictly be developing a vertical component of lift, and when that bank is increased, both wings are now developing stronger horizontal components of lift?
Click to expand...
Right, actually the outside wing, with dihedral, is deveoping stronger hcl than the inside wing, and this would be at banks greater than 45 degrees, noticably more at 60 and beyond. Also at these greater banks, the relative difference in speed diminishes. A bank of 90 degrees would have both wings flying at the same speed again.

However, this is speculation on my part. Maybe it is memory, but I can't find it in any current FAA AFH or PHAK, so I cannot trust it as accurate.
 
MikeOH58 said:
horizontal component becomes so great it overcomes lateral stability
Click to expand...

The HCL does not impact lateral stability, other than through the amount of sideslip it generates. Since the dihedral only kicks in when there is some amount of sideslip, a perfectly coordinated turn would have no tendency to fix itself.

Don't be mislead by which wing contributes more to the HCL. This isn't relevant to the amount of rolling tendency that the bank produces. The lift produced by both wings are still equidistant from the roll axis and produce the same amount of rolling tendency...they cancel.

See this thread:

http://forums.jetcareers.com/technical-talk/32512-overbanking-v-adverse-yaw.html
 
tgrayson said:
The HCL does not impact lateral stability, other than through the amount of sideslip it generates. Since the dihedral only kicks in when there is some amount of sideslip, a perfectly coordinated turn would have no tendency to fix itself.
Click to expand...
I agree that seems to be the limit of the explanation of dihedral in the current PHAK, and is all I would say on an FAA checkride about it, but where did I get this other explanation? I think it used to be in the old FTH, and it agrees with the aerodynamic principle of HCL vs VCL:

When the airplane is straight and level, with, let's say, 5 degrees dihedral, each wing is generating 5 degrees worth of hcl (we'll say 5%) inward with 95% generating vertical lift. That's lift directly opposing gravity.

When the airplane banks 5 degrees, coordinated so there is no sideslip, the inside wing goes level, 100% vertical lift, and the outside wing now is at 10% hcl and 90% vcl, so the outside wing loses vertical opposition to gravity and descends, returning the balance of both wings to 95/5.

That seems reasonable to me, and I think I read it. I don't think I made it up. But as you said, a lot of the FAA material is off a bit. or a lot.

What's your take on that, tgray?
 
nosehair said:
When the airplane banks 5 degrees, coordinated so there is no sideslip, the inside wing goes level, 100% vertical lift, and the outside wing now is at 10% hcl and 90% vcl, so the outside wing loses vertical opposition to gravity and descends, returning the balance of both wings to 95/5.
Click to expand...

That would all make sense if it weren't for the "10%" horizontal component of lift on the outside wing. This will act to roll the airplane further INTO the turn, completely offsetting the tendency of the other wing to roll OUT of the turn. As tgrayson said, the lift each wing produces is equidistant from the longitudinal axis and no matter which way you point the weight vector (by banking), there is no overall rolling moment.

The effects of dihedral require some sideslip. Say you are flying S+L and for some reason, the left wing drops a bit. Assuming relatively weak directional stability, the horizontal component of lift will pull the airplane into a sideslip to the left. For simplicity, assume the left wing is now level with the horizon. The airflow over the right wing (the now raised wing), is now a combination of the airplane's forward speed AND this slight component from the left of the aircraft due to the sideslip. This component from the left acts to decrease the AOA on the right wing, decreasing its lift. Now, even though the lift vectors are equidistant from the longitudinal axis, the lift on the left is greater than that on the right, and a right-rolling moment now exists.

So dihedral requires a sideslip, which yields higher lift on the low wing, which contributes to a rolling moment, which levels the wings.
 
nosehair said:
That seems reasonable to me, and I think I read it. I don't think I made it up. But as you said, a lot of the FAA material is off a bit. or a lot.
Click to expand...

You did read it in the old Flight Training Handbook. I remember reading the same thing myself during PPL training and the explanation bothered me. I made a note to myself that I'd have to investigate this aerodynamics stuff later.

The explantion makes intuitive sense, but it doesn't hold up during examination, using some basic physical principles you already know.

For an airplane to roll level, you need a rolling moment created that's stronger in one direction than the other. Gravity doesn't produce this because it acts through the CG, which is the definition of the CG. All you have is lift, and lift acts perpendicular to each wing semi-span. The rolling moment created by each semi-span is the lift force on that wing multiplied times the arm. The orientation of this lift vector with respect to level flight is not important, only the orientation and magnitude of the vector with respect to the aircraft geometry, and that hasn't changed.

It's a bit like mounting a prop on a drive shaft; if it's free to rotate, then the prop is just as happy in a vertical position as it is in a horizontal one. There's no tendency to float to a horizontal position. If you bent the prop to resemble dihedral, you'd have to mount the drive shaft through the new CG to duplicate the effect, which would be difficult since the CG would be in the empty space between the blades, just like the CG of a donut is in the hole. ;)
 
Hmm...haven't quite got it yet.

tgrayson said:
For an airplane to roll level, you need a rolling moment created that's stronger in one direction than the other.
Click to expand...
OK, here we have the level wing producing 100% VCL, which is 10% more than the 90% being produced on the other (outside) wing. The other 10% being HCL. If the 10% HCL cancels the 100%VCL, then this doesn't work.
lift acts perpendicular to each wing semi-span. The rolling moment created by each semi-span is the lift force on that wing multiplied times the arm. The orientation of this lift vector with respect to level flight is not important, only the orientation and magnitude of the vector with respect to the aircraft geometry, and that hasn't changed.
Click to expand...
What's changed, and is the point of discussion, is the lift vector's orientation to gravity.

A 90 degree banked wing still has 100% lift, but it is all horizontal, and produces no vertical component of lift.
 
nosehair said:
Hmm...haven't quite got it yet.


OK, here we have the level wing producing 100% VCL, which is 10% more than the 90% being produced on the other (outside) wing. The other 10% being HCL. If the 10% HCL cancels the 100%VCL, then this doesn't work.
Click to expand...

So you're saying the level wing has 10% excess lift in the vertical direction. That is 10% of the lift acting to roll the airplane back to level. Note this wing has 0% HCL.

That means the other wing has 10% excess lift in the horizontal direction. That is 10% of the lift acting to roll the airplane back into the turn. They will cancel out.

The wing doesn't care where gravity is. Unless you have different lift on one semispan than the other, the lift vector will simply point perpendicular to the lateral axis, and will run through the CG (or more accurately, the longitudinal axis)



nosehair said:
What's changed, and is the point of discussion, is the lift vector's orientation to gravity.
Click to expand...

The lift vector's orientation with respect to gravity has nothing to do with rolling moments. As tgrayson pointed out, it is the lift vector's orientation (and position) with respect to the aircraft itself that leads to rolling moments.


(using the % lift stuff here just to put it in your terms)
 
nosehair said:
A 90 degree banked wing still has 100% lift, but it is all horizontal, and produces no vertical component of lift.
Click to expand...

Doesn't matter. Do this: draw a picture of the airplane and put the lift vectors on each semi-span, with the vector perpendicular to the semi-span. Pick some arbitrary quantities of lift and an arbitrary arm to the longitudinal axis and calculate the moments. You'll see even before you begin that they must be equal and opposite. If you want to argue for any other rolling moments, you'll have to show where the other force is coming from and where it acts. Gravity acts at the CG and thus has a arm of 0, with a resulting moment of zero.

I'll draw a picture when I get home this evening. All I have where I am is Microsoft Paint. :(
 
tgrayson said:
I'll draw a picture when I get home this evening. All I have where I am is Microsoft Paint.
Click to expand...

Here is a picture of an airplane in straight flight. The dihedral is exaggerated for clarity:

StraightDihedral.png



Here is the same airplane in a left bank:

BankedDihedral.png
 
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