Stalling CP
Thread Starter
Join Date: Mar 2003
Location: Seat 1A or 1B
Posts: 112
Likes: 0
Received 0 Likes
on
0 Posts
Stalling CP
I have two ATPL's etc, so I should know this; but can anyone please explain why the Centre of Pressure on a cambered aerofoil is said to move rearwards to infinity after the point of stall, when any increase in alpha below critical or stalling angle moves the CP forward ?
I was asked this over the weekend, and not wishing to highlight my ignorance too overtly, mumbled something about a stalled wing still creating some lift, and it being a mathematical expression of the physical deterioration of lift after boundary layer separation...but it didn't convince me entirely !
Thanks in advance.
Miles Offtarget
I was asked this over the weekend, and not wishing to highlight my ignorance too overtly, mumbled something about a stalled wing still creating some lift, and it being a mathematical expression of the physical deterioration of lift after boundary layer separation...but it didn't convince me entirely !
Thanks in advance.
Miles Offtarget
Last edited by miles offtarget; 30th Jan 2005 at 10:26.
Join Date: Jan 2003
Location: The Heart
Posts: 811
Likes: 0
Received 0 Likes
on
0 Posts
Sounds like you answered correctly.
There must be some upward force beyond the critical AoA, though, because of the relative wind hitting the underside of the wing. This cannot be described in quite the same way as pre-stall aerodynamics but you can see that there is still a force acting perpendicular to the underside of the wing, it's vector being very much aft and sufficiently so as to be mathematially infinite.
You will remember that up to a third of the lift created by the wing comes from the underside. This is the physical pushing up, the remaining two thirds being the difference between the pressures, top and bottom (bernoullie's principle).
What I'm trying to say is that when the C of P for the top half of the wing ceases, all that is left is the C of P on the underside which has a rearward component.
I'll have a lie down now!
There must be some upward force beyond the critical AoA, though, because of the relative wind hitting the underside of the wing. This cannot be described in quite the same way as pre-stall aerodynamics but you can see that there is still a force acting perpendicular to the underside of the wing, it's vector being very much aft and sufficiently so as to be mathematially infinite.
You will remember that up to a third of the lift created by the wing comes from the underside. This is the physical pushing up, the remaining two thirds being the difference between the pressures, top and bottom (bernoullie's principle).
What I'm trying to say is that when the C of P for the top half of the wing ceases, all that is left is the C of P on the underside which has a rearward component.
I'll have a lie down now!
Thread Starter
Join Date: Mar 2003
Location: Seat 1A or 1B
Posts: 112
Likes: 0
Received 0 Likes
on
0 Posts
Thanks very much.
Have had chance to look in the old Bristol Groundschool and PPSC notes and it doesn't explain any more than that.
Miles Offtarget
Have had chance to look in the old Bristol Groundschool and PPSC notes and it doesn't explain any more than that.
Miles Offtarget
Last edited by miles offtarget; 30th Jan 2005 at 10:27.
Join Date: Sep 2002
Location: La Belle Province
Posts: 2,179
Likes: 0
Received 0 Likes
on
0 Posts
It's because the centre of pressure is an abstraction which does not describe what's really happening on the wing.
The reason it "goes to infinity" is that it's an attempt to define all the forces acting upon the wing as a SINGLE point force, the position of which tells you something about the pitching moment being applied.
Now, if a wing is generating 1kN of lift at the leading edge, 1kN at 25% chord, 0.5KN at mid chord and 0.25kN at 75% chord and at the trailing edge, thats a total of 3kN of lift.
Taking moments about the leading edge, it's 1.0*0% + 1.0*25% + 0.5*50% +0.25*75% + 0.25*100% = 0.9375kN-chords of moment.
So, in this case the CP would be at 0.9375/3.0=31.25% chord.
That's all straightforward.
Now, my wing truly stalls and generates ZERO lift, but being cambered it is still generating some forces, just they sum to zero.
0kN at 0%, 50% and 100%, 250N UP at 25% and 250N down at 75%; total LIFT = ZERO.
There is, however, a moment about the leading edge still - it's 0.25kN*25%-0.25kN*75%=-0.125kN-chords.
To get the CP, simply divide the moment by the lift and, er, oops. Divide by zero error. Infinity.
Simply, the attempt to reproduce the combined moments and forces acting on an airfoil by a single point force breaks down as the nature of the forces more closely approaches a pure couple - so as the lift gets smaller, the 'meaning' of CP becomes less and less physically plausible.
The reason it "goes to infinity" is that it's an attempt to define all the forces acting upon the wing as a SINGLE point force, the position of which tells you something about the pitching moment being applied.
Now, if a wing is generating 1kN of lift at the leading edge, 1kN at 25% chord, 0.5KN at mid chord and 0.25kN at 75% chord and at the trailing edge, thats a total of 3kN of lift.
Taking moments about the leading edge, it's 1.0*0% + 1.0*25% + 0.5*50% +0.25*75% + 0.25*100% = 0.9375kN-chords of moment.
So, in this case the CP would be at 0.9375/3.0=31.25% chord.
That's all straightforward.
Now, my wing truly stalls and generates ZERO lift, but being cambered it is still generating some forces, just they sum to zero.
0kN at 0%, 50% and 100%, 250N UP at 25% and 250N down at 75%; total LIFT = ZERO.
There is, however, a moment about the leading edge still - it's 0.25kN*25%-0.25kN*75%=-0.125kN-chords.
To get the CP, simply divide the moment by the lift and, er, oops. Divide by zero error. Infinity.
Simply, the attempt to reproduce the combined moments and forces acting on an airfoil by a single point force breaks down as the nature of the forces more closely approaches a pure couple - so as the lift gets smaller, the 'meaning' of CP becomes less and less physically plausible.
Join Date: Nov 2002
Location: Heart of Europe
Posts: 198
Likes: 0
Received 0 Likes
on
0 Posts
Sorry for the late addition.
Miserlou,
Try the link at the bottom for a pretty good explanation of lift. (I doubt that Bernoulli is so important - he surely helps saving fuel but lift? - Consider flying upside down when the camber and Bernoulli now work towards the earth surface. But even with planes using normal cambered wings the AoA is not that much in inverted flight as it should be when assuming two thirds come from Bernoulli (which would imply that the AoA in inverted flight must be at least 2 times greater)...
... Yes, I am a skeptic about that Bernoulli thing.
Mad(Flt)Scientist,
------------------------------------------------------------
Now, my wing truly stalls and generates ZERO lift...
------------------------------------------------------------
I hope that no wing will generate ZERO lift at the max CL AoA when it stalls. I've learned from aerodynamics that at an AoA higher than max lift (stall AoA) there is a pronounced drop in lift but not to zero. I've been told that increasing the AoA further you may find that lift drops first, then rounds out and starts to increase again. (This being the reason for some fighter jets to be able to slowly decrease forward speed to nearly zero, at the same time increase AoA and then accelerate straight up - provided enough thrust is available - looks pretty cool)
If the wing would produce zero lift (force perpendicular to the wing chord) it would at the stall AoA simply drop the nose or get uncontrollable Divide by ZERO - ERROR - Infinity
This may help to get a better explanation, as the lift is really only ZERO either at the zero lift AoA (which usually would be something negative) or when climbing in a straight vector upwards, or diving in a straight vector downwards
So I suggest that the movement of the CP may be rearwards first, then slowly forward again, once passed the stall AoA.
How about the CP moving further than the CG. Sounds like much fun.
Do you know if the CP can move outside the surface of the airfoil producing the lift? Never got that one answered.
Some idea may be found here
http://www.allstar.fiu.edu/aero/FlightTheory.htm
Miserlou,
Try the link at the bottom for a pretty good explanation of lift. (I doubt that Bernoulli is so important - he surely helps saving fuel but lift? - Consider flying upside down when the camber and Bernoulli now work towards the earth surface. But even with planes using normal cambered wings the AoA is not that much in inverted flight as it should be when assuming two thirds come from Bernoulli (which would imply that the AoA in inverted flight must be at least 2 times greater)...
... Yes, I am a skeptic about that Bernoulli thing.
Mad(Flt)Scientist,
------------------------------------------------------------
Now, my wing truly stalls and generates ZERO lift...
------------------------------------------------------------
I hope that no wing will generate ZERO lift at the max CL AoA when it stalls. I've learned from aerodynamics that at an AoA higher than max lift (stall AoA) there is a pronounced drop in lift but not to zero. I've been told that increasing the AoA further you may find that lift drops first, then rounds out and starts to increase again. (This being the reason for some fighter jets to be able to slowly decrease forward speed to nearly zero, at the same time increase AoA and then accelerate straight up - provided enough thrust is available - looks pretty cool)
If the wing would produce zero lift (force perpendicular to the wing chord) it would at the stall AoA simply drop the nose or get uncontrollable Divide by ZERO - ERROR - Infinity
This may help to get a better explanation, as the lift is really only ZERO either at the zero lift AoA (which usually would be something negative) or when climbing in a straight vector upwards, or diving in a straight vector downwards
So I suggest that the movement of the CP may be rearwards first, then slowly forward again, once passed the stall AoA.
How about the CP moving further than the CG. Sounds like much fun.
Do you know if the CP can move outside the surface of the airfoil producing the lift? Never got that one answered.
Some idea may be found here
http://www.allstar.fiu.edu/aero/FlightTheory.htm
Join Date: Oct 2004
Location: Body
Posts: 130
Likes: 0
Received 0 Likes
on
0 Posts
error 401
Isn't the fact that fighter jets fly straight up only due to the thrust available?
Don't forget that AoA is to the Relative Wind. In the case of going straight up the relative wind is now flowing over the wing normally i.e. from the sky to the earth so to speak. The wings will now produce lift in the horizontal plane. I don't think it has anything to do with lift increasing again and therefore allowing the a/c to fly straight up. That would be a new aerodynamic law!
Isn't the fact that fighter jets fly straight up only due to the thrust available?
Don't forget that AoA is to the Relative Wind. In the case of going straight up the relative wind is now flowing over the wing normally i.e. from the sky to the earth so to speak. The wings will now produce lift in the horizontal plane. I don't think it has anything to do with lift increasing again and therefore allowing the a/c to fly straight up. That would be a new aerodynamic law!
Join Date: Nov 2002
Location: Heart of Europe
Posts: 198
Likes: 0
Received 0 Likes
on
0 Posts
blueplume,
Absolutely correct!
Sorry for not having been too clear here. As stated in my post: "provided enough thrust is available" - meaning enough to hold up the full AC mass AND accelerate it vertically.
The example was meant to illustrate the movement of the CP with increasing the AoA to more than the max CL AoA thus stall AoA as I think the CoP cannot move backwards indefinitely and that is exactly what would happen during this manouvre.
As for the vertical flight it is not contrary to what I said but you are right - in this case the AoA on the wing must be slightly negative as for the wing to produce ZERO lift. This is the reason why I stated "climbing in a straight VECTOR" in order to emphasize the flight path and not the AoA associated with that manouvre.
Seen it myself MIG 29 and F/A 18 D on airshow - makes you go
Absolutely correct!
Sorry for not having been too clear here. As stated in my post: "provided enough thrust is available" - meaning enough to hold up the full AC mass AND accelerate it vertically.
The example was meant to illustrate the movement of the CP with increasing the AoA to more than the max CL AoA thus stall AoA as I think the CoP cannot move backwards indefinitely and that is exactly what would happen during this manouvre.
As for the vertical flight it is not contrary to what I said but you are right - in this case the AoA on the wing must be slightly negative as for the wing to produce ZERO lift. This is the reason why I stated "climbing in a straight VECTOR" in order to emphasize the flight path and not the AoA associated with that manouvre.
Seen it myself MIG 29 and F/A 18 D on airshow - makes you go
Join Date: Sep 2002
Location: La Belle Province
Posts: 2,179
Likes: 0
Received 0 Likes
on
0 Posts
OK, I used "zero lift" to make it clear why the phrase "CP moves to infinity" is used. Yes, in practice you don't get actual zero lift at a stall.
To answer the question, though - the CP is an artifical construction and does NOT represent the actual forces on the wing; it's an attempt to simplify a very complex force/moment situation into a single point force. So yes, the CP, being simply a theoretical device, is NOT constrained to lie on the wing.
If you prefer, take a cambered airfoil at the zero lift (attached flow) AoA. It almost certainly is generating a pitching moment. It has no lift. Therefore, CP either off the front or rear end of the wing, at infinity.
The phenomenom of the lift increasing again after the lift-loss at the classical stall AoA is generally associated with vortical flow from either the nose (weak effect), leading edge strakes (potentially powerful) or a sharp and highly swept leading edge. A more "normal" wing planform/configuration will not recover significant lift post-stall, but simply add more and more drag as the AoA increases.
To answer the question, though - the CP is an artifical construction and does NOT represent the actual forces on the wing; it's an attempt to simplify a very complex force/moment situation into a single point force. So yes, the CP, being simply a theoretical device, is NOT constrained to lie on the wing.
If you prefer, take a cambered airfoil at the zero lift (attached flow) AoA. It almost certainly is generating a pitching moment. It has no lift. Therefore, CP either off the front or rear end of the wing, at infinity.
The phenomenom of the lift increasing again after the lift-loss at the classical stall AoA is generally associated with vortical flow from either the nose (weak effect), leading edge strakes (potentially powerful) or a sharp and highly swept leading edge. A more "normal" wing planform/configuration will not recover significant lift post-stall, but simply add more and more drag as the AoA increases.
Join Date: Jan 2003
Location: The Heart
Posts: 811
Likes: 0
Received 0 Likes
on
0 Posts
Error_401
Thank you but...one cannot compare the angle of attack and lift produced between the upright and inverted wing; they are two different airfoil sections in that sense.
Having spent quite alot of time the other way up in aeroplanes I can assure you that very generally speaking, for a non-symmetrical airfoil, you do want the nose about twice as far above the horizon when inverted as for erect flight (depending on your seating position, of course).
Of course if the camber and Bernouille effect were acting towards the Earth when inverted, you wouldn't be able to fly inverted, that's surely not what you're saying?
If one considers a flat plane, this will create it's max lift when it reaches 45 degrees to the relative airflow. This effect is not considered in the normal lift equation.
So the disappearance of the forward CP at the 'aerodynamic' stall plus the rearward component of the underside make the CP disappear to a point mathematically infinte aft.
Simple really!
Thank you but...one cannot compare the angle of attack and lift produced between the upright and inverted wing; they are two different airfoil sections in that sense.
Having spent quite alot of time the other way up in aeroplanes I can assure you that very generally speaking, for a non-symmetrical airfoil, you do want the nose about twice as far above the horizon when inverted as for erect flight (depending on your seating position, of course).
Of course if the camber and Bernouille effect were acting towards the Earth when inverted, you wouldn't be able to fly inverted, that's surely not what you're saying?
If one considers a flat plane, this will create it's max lift when it reaches 45 degrees to the relative airflow. This effect is not considered in the normal lift equation.
So the disappearance of the forward CP at the 'aerodynamic' stall plus the rearward component of the underside make the CP disappear to a point mathematically infinte aft.
Simple really!
