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danandrews
16th Oct 2020, 15:35
Can someone explain how the coefficient of lift can be greater than one. My understanding (clearly incorrectly) was that the coefficient of lift was the proportion of the dynamic pressure contributing to the lift force. How can you have more force than the dynamic pressure that is available.

If you can help me with the answer I’ll consider ending lockdown!

regards Dan

megan
16th Oct 2020, 20:33
The coefficient of lift, which is a dimensionless number, expresses the ratio of the lift force to the force produced by the dynamic pressure times the area.

Cl = L / (q * A)

Cl = coefficient of lift
L = Lift
q = Dynamic pressure
A = Wing area

Tinstaafl
20th Oct 2020, 08:06
Keep in mind that total energy available with the air mass isn't just dynamic pressure. Think of static pressure & temperature for example.

Dont Hang Up
23rd Oct 2020, 13:52
Lift will equal the pressure differential between the top and bottom of a wing only if the wing passes through the volume of air and leaves it undisturbed.

At a positive angle of attack the volume of air the wing passes through is forced downwards. The reactive force to this acceleration of air downwards is additional lift.

john_tullamarine
25th Oct 2020, 07:15
Can someone explain how the coefficient of lift can be greater than one.

Megan provided the equation. It's really just a matter of when the particular section plus high lift bits and pieces lets go of the air and it all turns to turbulence. The numbers come from the measurements. No reason I can see why it ought not to be greater than one if that's what the section does ...

The reactive force to this acceleration of air downwards is additional lift.

It really is a matter of how you want to go about explaining stuff.

There is a pressure difference above and below - easy to show experimentally.

There is upwash and downwash - again, easy to show.

The two are not different/unrelated animals, though. To get the flow changes we need the pressure difference and the associated pressure gradients from the near-surface flow out to the free stream - one needs to recall that the only way to exert a force on a fluid flow is by the influence of pressure gradients - otherwise, the flow sort of just slips left, right, up, or down, as the case may be and continues to do its own thing. One needs the other and I don't think it is valid to try to separate them ?

only if the wing passes through the volume of air and leaves it undisturbed.

Now, that's an interesting concept. You don't happen to have a convenient video showing this happening in a tunnel, do you ?

megan
26th Oct 2020, 04:39
The reactive force to this acceleration of air downwards is additional liftThere is no "additional lift". Best think of a propeller, it's just an airfoil that gets its airspeed by rotating, the pressure difference between the forward and rear faces causes the airflow to flow in a rearwards direction at considerable speed at the beginning of take off, so providing thrust (lift as far as the airfoil is concerned).

Force(thrust or lift) = mass X acceleration
= d(mV)/dt

where d/dt = derivative with respect to time eg rate of change with time, mV = momentum, product of mass and velocity

Lift is often explained as throwing air at the ground, stand beneath a hovering helicopter for an example, or behind a stationary prop driven fixed wing running at high power.

blackmail
7th Nov 2020, 01:03
Hello All,

Along time ago, the 4 turboprop engined Breguet 941 had a blown high wing with 4 immense propellers(4m diameter each) for a half wing span of 9,88m. The 4 propellers were driven by one shaft so that an engine fail only made the 4 propellers turning more slowly & did not produce a yaw.
on my ATPL exam in 1972, there was a question about finding the CL coefficient for this STOL (Short Take Off & Landing) aircraft, the answer : CL = 8 !!

Theholdingpoint
7th Nov 2020, 14:52
Can someone explain how the coefficient of lift can be greater than one. My understanding (clearly incorrectly) was that the coefficient of lift was the proportion of the dynamic pressure contributing to the lift force. How can you have more force than the dynamic pressure that is available.

If you can help me with the answer I’ll consider ending lockdown!

regards Dan

You can't get it, because the "usual" lift formula is just a highly semplified equation, which has nothing to do with the real one.

Just think about the rest of the expression:

v: velocity of the undisturbed air. Not really useful when you realize that the air hitting your plane is not "undisturbed"
1/2: coefficient coming from the previous integral of v
rho: mean density of still air. Again, not really a factor, especially if M>0.3
S: top projection of the wing area. Not counting wing tickness (big factor), real wing area (another big factor), wing profile (huge factor), curvature (...), fuselage area (...), tailplane (...), engine nacelles and so on...

Lift is just the normal component of the total areodynamic force acting on an airplane, which is given by the 3D integral of the stresses acting on the entire machine.
So that formula has the following meaning: airplane A performance (in terms of lift) is equal to the performance of a model airplane flying at v, in a pocket of air with constant density rho, having a wing area S, multiplied by a factor called CL.
Airplane B, flying in the same air, with the same wing area but with a different wing profile (just an example of an influencing factor) will have a different value of CL.
This is useful in aircraft design, as it's a quick way to compare the performance of different airplanes at the same environmental conditions.

We could then say that CL is just a number used to sum up a lot of different variables (AoA, wing profile, Reynolds number, downwash, compressibility, local windspeeds, surfaces, etc), nothing more than that.
Well done with your critical thinking, keep up the good work!

Tinstaafl static pressure doesn't generate lift. Temperature is an influencing factor, but it's not really a useful form of energy for lift generation (we could even say that it's a negative contribution, as higher T-->lower rho, less lift, less thrust).

megan
7th Nov 2020, 22:54
Airplane B, flying in the same air, with the same wing area but with a different wing profile (just an example of an influencing factor) will have a different value of CLBeg to differ. Given the attributes you describe, same wing area different airfoil, the Cl will be the same, as the formula shows.

L=0.5*A*ρ*CL*V^2

where:

L = Lift force
A = Surface Area
V = Velocity of air
ρ= Density of Air
CL = Coefficient of Lift

Given a different airfoil the necessary Cl will likely be achieved at a different angle of attack, dependent on the lift slope of the airfoil used.

Theholdingpoint
8th Nov 2020, 04:08
No. You're implying that A and B are generating the same lift, an assumption that I didn't do.

megan
9th Nov 2020, 04:25
flying in the same air, with the same wing area but with a different wing profile (just an example of an influencing factor) will have a different value of CLYou're explanation is lacking in why the CL would be different. If B has a different CL to A, but wing area and velocity are the same, that would mean that the one with the higher CL would be climbing relative to the other, or descending less slowly. If both A and B are in level flight the CL will be exactly the same, though the angle of attack will more than likely differ.

Theholdingpoint
9th Nov 2020, 05:28
You're explanation is lacking in why the CL would be different. If B has a different CL to A, but wing area and velocity are the same, that would mean that the one with the higher CL would be climbing relative to the other, or descending less slowly. If both A and B are in level flight the CL will be exactly the same, though the angle of attack will more than likely differ.

Because lift generating performance comparison is made at the same angle of attack. It makes literally no sense to set up a comparison looking for an AoA to get the same value of generated lift.

danandrews
8th Apr 2022, 21:47
You can't get it, because the "usual" lift formula is just a highly semplified equation, which has nothing to do with the real one.

Just think about the rest of the expression:

v: velocity of the undisturbed air. Not really useful when you realize that the air hitting your plane is not "undisturbed"
1/2: coefficient coming from the previous integral of v
rho: mean density of still air. Again, not really a factor, especially if M>0.3
S: top projection of the wing area. Not counting wing tickness (big factor), real wing area (another big factor), wing profile (huge factor), curvature (...), fuselage area (...), tailplane (...), engine nacelles and so on...

Lift is just the normal component of the total areodynamic force acting on an airplane, which is given by the 3D integral of the stresses acting on the entire machine.
So that formula has the following meaning: airplane A performance (in terms of lift) is equal to the performance of a model airplane flying at v, in a pocket of air with constant density rho, having a wing area S, multiplied by a factor called CL.
Airplane B, flying in the same air, with the same wing area but with a different wing profile (just an example of an influencing factor) will have a different value of CL.
This is useful in aircraft design, as it's a quick way to compare the performance of different airplanes at the same environmental conditions.

We could then say that CL is just a number used to sum up a lot of different variables (AoA, wing profile, Reynolds number, downwash, compressibility, local windspeeds, surfaces, etc), nothing more than that.
Well done with your critical thinking, keep up the good work!

Tinstaafl static pressure doesn't generate lift. Temperature is an influencing factor, but it's not really a useful form of energy for lift generation (we could even say that it's a negative contribution, as higher T-->lower rho, less lift, less thrust).

Theholdingpoint Thank you I appreciate this reply, took me a while to get it at first. My incorrect assumption I believe was thinking that the rest of the equation was representative of all of the energy that could be extracted from the air. And that the co-efficient was simply the proportion that was extracted.

I think I see your point. When the relationship between some factor (wing size, velocity etc) to lift created, is simply observed, or rather, easily derived, it makes sense to put that into the equation. Nice simple relationships. The multitude of other contributing factors to the final value of lift are too complicated to measure and/or express simply. Thus they take the form of a dimensionless co-efficient? Am I getting that right?

I must say this has troubled me since I started my flying training many moons ago. So thank you!

And apologies for the late reply to all who took the time to answer my question in this thread.

MechEngr
9th Apr 2022, 01:06
Theholdingpoint Thank you I appreciate this reply, took me a while to get it at first. My incorrect assumption I believe was thinking that the rest of the equation was representative of all of the energy that could be extracted from the air. And that the co-efficient was simply the proportion that was extracted.

I think I see your point. When the relationship between some factor (wing size, velocity etc) to lift created, is simply observed, or rather, easily derived, it makes sense to put that into the equation. Nice simple relationships. The multitude of other contributing factors to the final value of lift are too complicated to measure and/or express simply. Thus they take the form of a dimensionless co-efficient? Am I getting that right?

I must say this has troubled me since I started my flying training many moons ago. So thank you!

And apologies for the late reply to all who took the time to answer my question in this thread.

Holdingpoint made it complicated and only applied it for the same wing area, but an entirely different weight of the airplane and different airfoil, so an entirely different airplane.

The first use of Cl is to characterize an airfoil for its performance as if it has an infinite wing span. For most small aircraft this will not need a correction for compressibility or Mach number effects - the presumption is the plane isn't traveling fast enough to cause the density of the air to change by moving through it. For a range of angle of attack there will be a chart relating the angle of attack to Cl. Pretty much every useful airfoil will have a chart, so every airfoil over it's unstalled range can have some arbitrary Cl.

When a plane is designed the first factor, after expected airspeed for take-off fully fueled and cruise with some fuel burned off, is an estimate of weight. Next an estimate of wing area is made - and poof there is the design range for the Cl - a high number for takeoff and a lower one for cruise. Then one goes to the big book of airfoil data and chooses an airfoil that covers that range with an eye to generating a minimum of drag while also being thick enough for a first estimate of the bending loads.

In any case, once the weight, speed, and wing area are chosen, then all airfoils for level flight have to operate at exactly the same Cl. What shifts with airfoil shape change is the amount of drag and the extra protection for staying away from stall. If the wing area is distributed in a stubby planform then the induced drag is high. If the wing area is long span and narrow chord, then flex is a potential problem. And so the trade-offs go. It will take a number of cycles where the weight is closer, the wing strength is changed based on the weight. If the structure weight goes down, then maybe bigger fuel tanks are designed for increased range or maybe not - the lower weight can result in lower structure weight, which in turn reduces the structure weight a little more - to some minimum amount of material for a minimum amount of raw material cost.

Many aircraft factors are dimensionless to cancel out the various contributors. The first airfoil data that the Wright brothers depended on was only useful for the exact airplane weight and speed and could not be used anywhere else. They lost nearly a year of development because of this and had to obtain their own data. By balancing the dynamic pressure term against the area, leaving just force, and then dividing that by the force due to lift/weight, then any size airplane (within reasonable limits**) could be designed using that airfoil data. It's possible to predict how an airfoil will perform on Mars, with 0.1% of the air pressure as at sea level on Earth using the same data as was used for a Piper Cub, even accounting for the fact that the weight of a plane is far less on Mars. No special correction to the airfoil data is required.

The other factor that sneaks in is that when the Cl is chosen for a particular cruise speed for a particular airfoil it is choosing the angle of attack, which affects the required pitch of the airfoil relative to the fuselage, which sets whether the fuselage is level in flight - a nice thing in a super-stretch commercial airliner to minimize fuel burn from drag and the slog uphill and downhill for the drink cart and bathroom users. Consider that at low speed, high Cl during takeoff, how slanted the plane is. One could operate the plane that way the entire flight- but it would burn too much fuel and be hard to move around in and be slow.

For the original question - it's a matter of leverage. The dynamic pressure is a measure of the momentum of the air heading towards the airfoil. If that was the only factor it's true that the maximum would be 1. However, there are also usually engines shoving the airfoil through the air, increasing the energy available to cause the air to change direction. High Cl is also a high drag condition. The additional push is either from the engines or by putting the weight of the plane on a steep slope. Look for a chart called the "drag polar" for airfoils. en.wikipedia.org/wiki/Drag_curve

**Reasonable limits involves Reynold's number, the ratio of dynamic properties to viscous properties. For very low speeds and over very short distances the Reynolds number is very low - this is where flies and dust particles operate. Most small aircraft operate at moderate Reynolds numbers and for supersonic aircraft the viscous properties almost don't matter. However almost all airfoil data is suitable for almost all small aircraft - less than 300 mph,

john_tullamarine
9th Apr 2022, 10:11
I think folks are trying to read too much razzle dazzle into CL (or, for that matter, any non-dimensional quantity used in engineering/physics/etc).

Suggest that a brief read on the subject of 'Dimensional Analysis' might be useful. See, for example, DA_unified.pdf (mit.edu) (http://web.mit.edu/2.25/www/pdf/DA_unified.pdf?msclkid=73f79527b7ee11ec9a20e5f9efc419cc)

There is a number of important non-dimensional quantities in engineering (such as mach number, reynolds number, lift coefficient, drag coefficient, and so it goes on ..). In the real olden days, before this all came about, we had reams and reams and reams and yet more reams of experimental data from which to figure out stuff with our pencils and paper. Eventually, brighter minds prevailed and dimensionless quantities came to be. The main value to us is that, by using these animals, we can reduce the amount of experimentally reported data by orders of magnitude and that was a great development.

There is nothing terribly magical about CL; as Megan noted, it is just a ratio of measured lift (usually what we get from wind tunnel work) to some other useful quantities which make for a non-dimensional result. Of sideline note, the 'S' for area is just some representative area - it doesn't matter all that much what area is used as it only needs to be there to complete the non-dimensionalising process. Obviously, it makes logical sense to use some sort of rationally pertinent area - while we could easily use the area of the local football field, wing planform area makes for a sensible quantity. It follows that any sets of data for CL should specify what area is being used as it will change the quantum of the CL numerical values which result from the exercise. Similarly, we might use frontal area for CD numbers .... same, same .... but we really need to specify what area it is that we are using lest our paper and pencil work comes up with strange numbers.

As others have stated or inferred, the CL with which we normally work only applies to the typical low speed aircraft. If you read up on the background, CL also is tied up with Mach Number and Reynolds Number which need to be incorporated when relevant. A simple story is given at The Lift Coefficient (nasa.gov) (https://www.grc.nasa.gov/www/k-12/airplane/liftco.html?msclkid=f796bf0ab7ed11ec976a897f39bf58b2)