Single engine Cessna tailplane lift

[digits_]

I'm on the edge of my academic qualification in discussing C of L and C of P topics, but, I will assert that the C of G of the aircraft will not ever be aft of the center of total lift provided by the wing, including any pitching moment for the wing in that configuration. The drawn schematic does not include the pitching moment of the wing.

The wing and the aircraft are one structure. A moment is created if 2 (or more) forces act on the same object. On our airplane, we have 3: the lift on the wing, the center of gravity (which takes into account the weight of the wings and the weight of the fuel in the wings etc) and the lift (up or down) by the tailplane. There is no other "pitching moment".

The higher the G loading acting on the plane, the more load acts downward through the C of G. Therefore, to remain in equilibrium, the higher the lift must be from the wing, through its center of lift. Therefore the greater the downforce must be from the tail to balance the increased lift produced by the wing, from the combination of increased AoA and/or speed. This is how Va is determined, the speed at which the aircraft stalls before it is overstressed. The tail stalls, and releases the download, which reduces that angle of attack of the wing, thus preventing overstressing. Faster than Va, the tail is capable of exerting so much down force that, the wing can be overstressed.

No, the G load is created by the fact that the wing is generating more lift than your weight. Not the other way around. Up untill Va, the aircraft will indeed stall before it overstresses. However, the 'normal' way would be that the angle of attack of the wing increases over the critical angle of attack, thus the main wing will stall. The tail has nothing to do with the stalling itself. It helps you to increase the AoA, and afterwards to decrease it again to get out of the stall. Maybe you can stall the tailplane first in some exotic cases, but that shouldn't happen in 'normal' stalls.

It's not the force on the elevator control, but the change in force per change in G (with no change in trim) which matters. In the graph I presented, the equilibrium point is 70 knots, where I have trimmed out the pitch force to zero, from there, I did not retrim, but measured the force on the stick as I slowed, and changed power. When you see my results showing a push force as the plane slowed, and worse a pitch force which did not change as the speed further reduced, you are seeing a plane which is very unstable, and difficult to fly. I cannot tell you if the tail was actually lifting at that point, or just providing a very small down force relative to all other forces, but it was not approvable. If it actually was lifting, it would just be the same or worse than what I measured, and certainly not approvable.

That's exactly my point. You can not deduce that from your graph. This has nothing to with upward or downward acting forces specifically, but with the stability characteristics of that plane.

[photofly]

The drawn schematic does not include the pitching moment of the wing.

There's no pitching moment about the centre of pressure, by definition of the centre of pressure. Since I'm taking moments about the CP, there's no contribution to the total moment from the wing, only from the weight of the aircraft acting at the centre of gravity, the pitching moment of the fuselage, and the pitching moment generated by the horizontal stabilizer. I don't believe the fuselage is holding the tail up while the tail is pushing down.

For those people who don't think the tail is lifting at rearward cg, let me ask you, what else do you think is holding the tail up? It's not skyhooks, or leprechauns.

And don't worry PilotDAR, I'm not going to change the way I teach on a whim.

One thing is clear, that it doesn't matter a fig what you teach. There are enough skilled pilots who've demonstrated in this thread that they don't understand this to show that it's obviously not important for a pilot to understand.

This is a simple discussion of balance - i.e. total moment is zero. It's not the same as the criterion for pitch stability, which is more complicated and needs to consider change of pitching moments as the aircraft pitches up and down. The equations for pitch stability are very standard, easily looked up, and none of them demand the tail provides an upforce. If you don't trust me on that, look them up; I'm happy to provide plenty of online references, upon request.

[PilotDAR]

the stability characteristics of that plane.

And exactly my point too. Because that plane perhaps was lifting at the tail, it demonstrated stability characteristics which made it not certifiable. All the GA planes are certified because they do not do that!

There is no other "pitching moment"

It's a part of the total lift of the wing, and has the apparent affect of displacing the total center of lift of the wing further aft than the center of pressure. I admit to being at the edge of my knowledge discussing those factors. However retired deHavilland aerodynamics who explained it to me last night seems to understand it very well

Wikipedia says:

"Pitching moment is, by convention, considered to be positive when it acts to pitch the airfoil in the nose-up direction. Conventional cambered airfoils supported at the aerodynamic center pitch nose-down so the pitching moment coefficient of these airfoils is negative."

As for my statement with respect to G, there might be a nuance of terms there, but I stand by the theme of my words.

I have stalled the stabilitaors of both Piper Arrows, and Cessna Cardinals, without stalling the wing, and the nose of the plane will drop, because the tail can no longer provide the downforce to maintain the desired attitude (control) for the aircraft. This is one of the definitions of a stall of the aircraft, though the wing itself never stalls.

[PilotDAR]

what else do you think is holding the tail up?

Gravity holds the tail up at all times, by being reacted through the lift of the wing.

[PilotDAR]

For the first plane in the video, when the bomb bounces and takes off the tail, the plane rapidly pitches down, so badly that the wings bend down because of negative G. That's because the tail was holding the nose up, until it was broken off then the nose went down!

[digits_]

what else do you think is holding the tail up?

Gravity holds the tail up at all times, by being reacted through the lift of the wing.

That can only be the case if the CoL is located behind the Cog
So if you have:
(nose) CoG -- CoL -- Tailplane

Our discussion is about the cases where the CoG is behind the CoL

There is no doub that you will find lots of examples of airplanes having a CoG -- CoL -- tail order, but the other ones do exist as well.

[digits_]

For the first plane in the video, when the bomb bounces and takes off the tail, the plane rapidly pitches down, so badly that the wings bend down because of negative G. That's because the tail was holding the nose up, until it was broken off then the nose went down!

Maybe, but the fact that there was a fat ass explosion under the tail might have had something to do with that as well.

[crazy_aviator]

A symetrical wing ( on a competition plane) is designed to fly at all attitudes and develop the same lift/stick force etc. MOST wings are NOT symetrical therefore are designed to fly ( non-inverted) but CAN fly upside down. Same with the tailplane , most are symetrical, some are asymetrical and are designed PRIMARILY for downforce only. A symetrical tailplane should be the HINT to answer your questions on this forum

[crazy_aviator]

Another thing, as the Cog G goes aft, the tailplane provides much less downforce , therefore the aircraft becomes more efficient ( less wasteful downforce) One thing that must be remembered is that the downwash coming from the back of the wing is meeting the tailplane leading edge at a greater angle than the ambient air. It would take a LOT of C of G change aft and/or aerobatic maneuvering for this tailplane to work to provide an upforce.

[Steve Pomroy]

There's no pitching moment about the centre of pressure, by definition of the centre of pressure. Since I'm taking moments about the CP, there's no contribution to the total moment from the wing, only from the weight of the aircraft acting at the centre of gravity, the pitching moment of the fuselage, and the pitching moment generated by the horizontal stabilizer. I don't believe the fuselage is holding the tail up while the tail is pushing down.

Actually, your taking moments about the aerodynamic center. The center of pressure, although well defined, is almost never used in these types of calculations because it moves around with changes in AOA. The AC is in a fixed location and is close to the c/4 point, which you're using. Measured about the AC, the NACA 2412 airfoil has a pitching moment coefficient of about -0.045 over most of the AOA range(http://www.tricity.wsu.edu/htmls/mme/me ... ca2412.jpg). This results in wing pitching moments of -130.8 ft-lbs just above the stall speed, and -1,373.4 ft-lbs at Vne.

Having said all of this, you still get a tail-up force over the entire speed range with a fully aft CG. This ignores moments caused by the engine and fuselage...

Cheers,
Steve
http://www.flightwriter.com
http://www.skywriters.aero

Edited: to correct a typo (-0.45 corrected to -0.045).

[Strega]

Mr Pie.

All wings will create a pitching moment at a given AOA... look up some airfoil info and you will see...

Taking the wing off of a balsa wing glider and throwing doesn't really prove anything.. other than that it will fall to the earth, and spin in some manner. In order to keep an airfoil section flying against a relative airflow at an angle of attack, a moment is required (the pitching moment)

All of the moments sum around the c of G of the aircraft, one moment that is VERY important is the pitching moment of the wing ( the tale plane pitching moment cannot be ignored either).

One aircraft I fly actually has a limitation on speed, until the main wing flaps can be reflexed. With flaps 0, the pitching moment of the wing is so high above 160 kts, it puts unreasonable stresses on the tail, and rest of the fuselage structure

FYI in all canard aircraft the c of G is not anywhere close to the aerodynamic c of p of any lifting surface...think about it for a bit.

[Strega]

Having said all of this, you still get a tail-up force over the entire speed range with a fully aft CG. This ignores moments caused by the engine and fuselage...

nope..


If I get some time tonight,, I'll sit down and show everyone whats really going on... (mind you might be beyond the scope of this forum and its included "Engineers")

[Strega]

This results in wing pitching moments of -130.8 ft-lbs just above the stall speed, and -1,373.4 ft-lbs at Vne

fyi most airplanes I fly,, you have to pull back "hard" to make them stall....

[Colonel Sanders]

Not sure anyone cares, but I frequently have an
"upwards" net force produced by my tail:



I transition all the time, between the tail producing
up and down force (shrug).

Here. I'll rotate the picture 180 degrees for you
straight and level guys. Note the near-stalling
-ve AOA, more visibile now:

[photofly]

Actually, your taking moments about the aerodynamic center. The center of pressure, although well defined, is almost never used in these types of calculations because it moves around with changes in AOA. The AC is in a fixed location and is close to the c/4 point, which you're using.

Hi Steve.

No, I'm not using the AC. I'm categorically and definitely calculating moments about the CP. Yes, I know the difference. This is my calculation, and I'll take moments about wherever I please.

Why am I using the CP? Because there's no pitching moment about the CP. I know the CP moves and the AC doesn't, and I don't care. All I need to know is that the CP is at chord/4 for a single sensible AoA.

Measured about the AC, the NACA 2412 airfoil has a pitching moment coefficient of about -0.45 over most of the AOA range(http://www.tricity.wsu.edu/htmls/mme/me ... ca2412.jpg). This results in wing pitching moments of -130.8 ft-lbs just above the stall speed, and -1,373.4 ft-lbs at Vne.

Interesting, but not relevant, because I'm not taking moments about the AC, I'm taking moments about the CP. And there's no moment about the CP.

If this was a stability issue, then the AC would be a sensible position to consider: it stays fixed as the AoA varies. Then PilotDAR's comments about there being an extra moment to account for would be true, and you've provided data as to what value it takes. That extra moment comes from the distance between the AC and the CP. But this is not a stability issue, it's way way simpler. It's a simple equilibrium balance, and I don't need to care how the CP moves with AoA because I'm only doing the calculation at a single angle of attack, the one for which the CP is at chord/4. Having shown you that there's a lift on the tail for one angle of attack I'm going to suggest that it doesn't vary much as the CP wanders about over a sensible range of AoA.

Having said all of this, you still get a tail-up force over the entire speed range with a fully aft CG. This ignores moments caused by the engine and fuselage...

Yes.... and taking moments about the CP is the easiest way to see why.

[Strega]

Yes.... and taking moments about the CP is the easiest way to see why.

please show us your work... Im curious as to how this "Flying tailplane theory" works.....

[Strega]

In aerodynamics, the pitching moment on an airfoil is the moment (or torque) produced by the aerodynamic force on the airfoil if that aerodynamic force is considered to be applied, not at the center of pressure, but at the aerodynamic center of the airfoil. The pitching moment on the wing of an airplane is part of the total moment that must be balanced using the lift on the horizontal stabilizer.[1]
The lift on an airfoil is a distributed force that can be said to act at a point called the center of pressure. However, as angle of attack changes on a cambered airfoil, there is movement of the center of pressure forward and aft. This makes analysis difficult when attempting to use the concept of the center of pressure. One of the remarkable properties of a cambered airfoil is that, even though the center of pressure moves forward and aft, if the lift is imagined to act at a point called the aerodynamic center the moment of the lift force changes in proportion to the square of the airspeed. If the moment is divided by the dynamic pressure, the area and chord of the airfoil, to compute a pitching moment coefficient, this coefficient changes only a little over the operating range of angle of attack of the airfoil. The combination of the two concepts of aerodynamic center and pitching moment coefficient make it relatively simple to analyse some of the flight characteristics of an aircraft.[2]
Contents [hide]
1 Measurement
2 Coefficient
3 References
3.1 Notes
4 See also
[edit]Measurement

The aerodynamic center of an airfoil is usually close to 25% of the chord behind the leading edge of the airfoil. When making tests on a model airfoil, such as in a wind-tunnel, if the force sensor is not aligned with the quarter-chord of the airfoil, but offset by a distance x, the pitching moment about the quarter-chord point, is given by

where the indicated values of D and L are the drag and lift on the model, as measured by the force sensor..
[edit]Coefficient

The pitching moment coefficient is important in the study of the longitudinal static stability of aircraft and missiles.
The pitching moment coefficient is defined as follows[3]

where M is the pitching moment, q is the dynamic pressure, S is the planform area, and c is the length of the chord of the airfoil. is a dimensionless coefficient so consistent units must be used for M, q, S and c.
Pitching moment coefficient is fundamental to the definition of aerodynamic center of an airfoil. The aerodynamic center is defined to be the point on the chord line of the airfoil at which the pitching moment coefficient does not vary with angle of attack,[2] or at least does not vary significantly over the operating range of angle of attack of the airfoil.
In the case of a symmetric airfoil, the lift force acts through one point for all angles of attack, and the center of pressure does not move as it does in a cambered airfoil. Consequently the pitching moment coefficient for a symmetric airfoil is zero.
Pitching moment is, by convention, considered to be positive when it acts to pitch the airfoil in the nose-up direction. Conventional cambered airfoils supported at the aerodynamic center pitch nose-down so the pitching moment coefficient of these airfoils is negative.[4]

Mr Pie.. see in bold...

(fyi this was from Wikipedia)

[photofly]

Actually, you're taking moments about the aerodynamic center.

I take it back. You're quite right. Chord/4 is the AC. I want to take moments about the CP, but in order to do so I have to pick a representative angle of attack and work out where the CP is, or find a graph of CP vs. AoA.

Instead, let's say the pitching moment about chord/4 (which is the AC, not the CP) is in the middle of the range you suggested, at 500 ft.lbs nose down. At 2950lbs MTOW acting at the CG, that means the CG must move 500/2950 = 2.03 inches further rearward before the tail starts to lift.

So my range of CG's for which the tail lifts should read about 39" to 48.5" (depending on airspeed).

Do you agree with the calculation?

[dr.aero]

Colonel...



Assuming you're flying level, it appears you've drawn the lift force on the tail in the wrong direction.

It is hard to say though from a picture as there will be a downwash angle imparted to the air flow over the horizontal stabilizer.

Pitching moments change drastically when you're inverted vs right-side-up. Notice that your attitude that you cruise at while inverted is noticeably higher than right-side-up.

[dr.aero]

photofly...

Is this what you're looking for?



What about this?

[Colonel Sanders]

you've drawn the lift force on the tail in the wrong direction

Really? I thought the tail provided negative lift, like this:





Are both of those diagrams wrong? They are the same as mine.

[dr.aero]

Colonel...

Look at the angle of attack on the horizontal stab - it doesn't magically produce lift down when it's got a positive angle of attack! As I mentioned, there is a downwash component from the wings that will affect the angle of attack on the horizontal stab but you're unable to determine that angle from a photograph.

How did you figure out where the Cp and Cg was in your picture?

[photofly]

Yes, thanks. That first graph is what I was looking for, albeit for a different airfoil. Let's proceed assuming the broad shape applies to NACA 2412 also.

My initial calculation was the correct calculation, but I shouldn't have used the quarter chord point for the CP. I should have used something at about 35% -40% MAC, for a typical AoA. The 10-15% difference in chord (from 25%) corresponds to putting the CP 10 inches further back, which does in fact take it to a station of around 48", which is the rear limit of the C of G. So if that curve applies to NACA 2412 then the tail does no (or very little) lifting and I was basically wrong from the start.

So the question is, what does the CP curve look like for NACA 2412? If Steve Pomroy's data is correct then the CP is closer to chord/4 and the tail does lift, at rearward cg.

How did you figure out where the Cp and Cg was in your picture?

I was wondering that, too.

[dr.aero]

Colonel...

I know that there is a noticeable amount of "push" needed to fly inverted vs right-side-up. That would lead you to believe that there is a upward force, relative to the airplane, being produced so the nose doesn't drop too low while inverted. But the only thing it really tells you is that the trim for right-side-up is not the same trim required for inverted.

It may very well be that your force labels are correct in your picture - I'm just pointing out that your image isn't definitive proof that that's the way it is.

Your elevator appears to not be deflected at all relative to the horizontal stab in your picture. That would mean that the angle of attack should primarily determine the lift of that surface. If you were flying level when you took the picture (which it appears you were), the angle of attack shows that your lift arrow is pointing in the wrong direction. It is possible that the downwash angle would be so significant that it would change the lift from being upwards (away from the ground while inverted) to downwards - but that's purely speculation.

[Colonel Sanders]

the trim for right-side-up is not the same trim required for inverted

Actually, it is. I set it at the start of the sequence
and never mess with it. You might be thinking
flat-bottom wing which needs a LOT of forward
stick for inverted flight.