Tailplane "Lift"

[Brian Abraham]

Esoteric I guess, but does anyone have any idea of the lift (I know which way it generally points so use the term loosely) produced by the tailplane/elevator combination on say 737 or 747. Seem to recall a figure of 30,000 pounds for a 747. Googled but no illumination.

[wiggy]

I thought that unless the aircraft was loaded with the C of G well aft the tailplane or Horizontal Stab. generates downforce not lift (at Crusing Mach numbers/speeds) ?

[On-MarkBob]

The tailplane does not produce any lift. You could say it produces a 'Negative Lift'. The reason many early aviators were killed is because the tailplanes produced lift it order to help it fly and the aircraft would then nose up in a tailplane stall, a condition in which it could never recover. Most modern aircraft are designed so that when the airflow decreases, the effect/moment produced by the tail surface is lessened and so the aircraft pitches nose down and therefore safe. This is also the reason why the aircraft will perform more economically with the C of G at its back limit, in this position the load on the tail is the least, whereas at the forward limit the tailplane is producing a high downward load countering the good work done by the mainplanes. This is also the reason why the 'Canard' type of configuration is so good, because the 'canard' will produce lift as well as the mainplanes making it more efficient.
As for the exact pounds/kilos lever-arm produced, it would depend entirely on the aircraft type, the weight to which it was loaded and the C of G.

[Brian Abraham]

Oh dear, read the question fellas. An aerofoil, or airfoil, produces lift when ever it has an angle of attack to the relative airflow, the zero lift position is usually at a negative angle of attack (symetrical profiles excepted). It produces lift irrespective of its orientation. And yes the lift vector of a tailplane generally is towards mother earth (unless I'm inverted in a loop). I think the profiles both you guys have are a bit suss.
Help Old Smokey.

[Miserlou]

Although I don't have the answer to the question, I also have heard a number of commercial pilots proclaim that there is only down force from the tailplane.

My question to them is always how does a canard work? Kind of ultimate tailplane lift, isn't it.

[wiggy]

Bleedin cheek, 10000 plus on the 744, A2 in a previous life blah, blah, and my profile is a bit "sus"..anyhow Brian, I will refrain from insult and I take your point, RTFQ and all that.

In fairness don't forget a poor question will often lead to a poor answer, if you had used "Force" instead of "lift" your question would have been even more clearer.

Rgds and all that;honest

[Brian Abraham]

Wiggy, I thought my statement

I know which way it generally points so use the term loosely

should have been clear enough. Once again the imprecision of the written language? The only text book I have to hand is Ira H. Abbott and Albert E. Von Doenhoff and they define as follows,
Lift is defined as the component of force acting in the plane of symmetry in a direction perpendicular to the line of flight
When talking of the properties of aerofoils we talk of "lift-curve slopes, lift distributions, section lift coefficients, design lift coefficients, maximum lift". The only time "force" is mentioned is to acknowledge that lift is but one of the forces acting on an aerofoil. Am interested in your comments.

[Capt Fathom]

Brian,

There was something written about this in the Boeing Aero Magazine several years ago.

At the moment the Boeing website appears to be down ??

The gist of the article was in reference to the B767.

By loading the aircraft towards it's rear limit, and therefore reducing the downforce required, ie lift, the subsequent reduction in drag enabled the aircraft to depart three(?) tons heavier than would normally be possible. (Can't recall the exact figure, but it was several tons).

It was of particular interest to airlines that operated out of marginal runways!

Here's the shortcut to Aero Magazine. You may find something there.

[BackPacker]

Brian, back to the original question. You wanted to have a rough idea of the "negative" lift of the tailplane. It shouldn't be that hard to work out, since basically the main wing holds up what gravity and the tailplane are pushing down. And how much the tailplane is pushing down depends on the distance between the C of G and the main wing.

So if you know the mass of a given airplane and the C of G, and you know the exact location of the MAC (mean aerodynamic chord, more or less the center of lift) of the main wing and tailplane, it's a fairly simple calculation, which closely resembles a normal M&B calculation.

This will give you a rough number. To refine it, you need to factor in things like:
- Center of lift of a wing changing with its angle of attack
- Fuselage-generated lift
- Center of lift changing with flaps & slats extension
and so forth.

But for a rough number, if you can cough up the mass and C of G for a given airplane, and the MAC for main and tailwings, I can show you how the calculation is done.

[Double Zero]

Tailplane force

So what everyone is getting at is that tailplanes produce a downward force; which is one reason besides stalling characteristics why canards are attractive - they push up !

[Miserlou]

I'd like to hear from some-one who knows what they are talking about here rather than the 'Flat Earth', 'tailplanes only produce down force' brigade.

People seem to be worried about lifting tails at stall but fail to see the reason that the regs require stick pushers and shakers, "where aerodynamic warnings are not sufficient or not present" or something similar it says, 'artificial means' rings a bell.

[BackPacker]

Miserlou, I don't know what you mean or what your post is trying to achieve.

If you have a conventional wing setup, meaning a main wing roughly halfway the fuselage but a little bit *behind* the center of gravity, and a smaller, tail wing at the back, that tailplane at the back produces a down force. Period. Otherwise the aircraft would be aerodynamically unstable (ie. more airspeed would force it to go down, less airspeed means up, very dangerous as this would force the aircraft, if left alone, to either overspeed into the ground or stall into the ground). As far as I can interpret all the posts below, everybody seems to agree on that (tailplane generates a downward force unless inverted), although it took a few posts to work that out.

If you have a canard-type setup, which means that you have the main wing more or less at the back of the airplane, some way behind the center of gravity, and you have a smaller wing (the canard) at the front of the airplane, far in front of the center of gravity, then the dynamic balance is completely different. Both wings will generate an "up" force but the angle of incidence of the canard is set up so that it will stall before the main wing stalls. So at a too-high angle of attack, the canard loses lift (because of the stall) and the nose pitches down, breaking the stall. OTOH - I'm not an expert on canard setups and I don't know how they become dynamically stable. I guess that the lift of the canard (due to the higher angle of incidence) increases relatively faster than the increase in lift of the main wings as the plane start to fly faster. So the aircraft pitches up and this reduces the airspeed again.

Now there's also an aircraft which has a three-wing setup. I think it's called a Pilatus or something. Probably even more complicated.

The thread so far never has touched upon the subject of stall warners at all, but the reason for stick shakers (or not) lies in the general aerodynamic effects of a stall. If the airflow over the main wing, in a small aircraft with straight (not swept wings) gets close to the stall, it separates at some point and becomes turbulent instead of smooth. This turbulent air flows over the tailplane and can be felt directly through the controls. Pilots learn to recognise these jittery controls as one of the indicators of an approaching stall. In fact, a lot of light aircraft have a slightly higher angle of incidence in the inboard sections of the wing as compared to the outboard sections. This ensures that the aerodynamic stall warning, due to the inboard section of the wing stalling, is present while the outboard sections of the wing are not yet stalled, and thus your ailerons remain effective.

In a large aircraft, with hydraulically actuated elevators, or fly-by-wire, this direct aerodynamic feedback is either dampened or completely absent, so an artificial stall warner (stick shaker) is added to simulate this effect. This is all in addition to the normal stall warners (based on a suction pipe, vane and/or AoA meter) because sometimes a tactile (feel) warning is easier noticed than an audible warning.

Large airlines also typically have swept wings for better performance at high speed. For some aerodynamic reason that was explained to me once but which I have now forgotten, the stall in these aircraft starts at the wing tips instead of wing root. So again, the burbling of air would not be felt through the controls because the tailplane does not sit in this ourboard turbulence. But a stall with a swept-wing aircraft is even more disastrous because, as the outboard areas of the wings stall first, the center of lift of the wing moves *forward* (because the wing is swept backwards and the outboard, backwards sections don't generate lift anymore). This changes the dynamic balance further: if the center of lift moves forward the aircraft pitches up, increasing the AoA and thus the stall. Possibly to a point where recovery is not possible anymore. Another reason for having very pronounced stall warnings in these types of planes.

Is all this dangerous? Not at all, as long as you understand the dynamics involved and don't let the aircraft get into a stall uninvited.

And the earth is flat. No doubt about it. Look at the horizon. Can you see a curve there? So there you go!

[Rainboe]

Your almost there, but there are large gaps in your knowledge. The Earth IS flat, but I see a curve from 39,000'. The reason is because it's flat and round, like a plate, so you can see the edge from high up. When I fly from Bangkok to London, I don't pitch over as I travel, see. You just get nearer the edge of the plate

For some aerodynamic reason that was explained to me once but which I have now forgotten, the stall in these aircraft starts at the wing tips instead of wing root

Very important the root stalls first. When the first wingtip goes into stall, the aeroplane would roll upside down. So to make the root stall first, you will see wings have a far greater angle of attack at the root, and it washes out towards the tip. When the root stalls first, as what it should, there will be a pitch down effect helping to correct the situation, which is what you want. Stall the tips first, and you lose that part of the lift nearest the rear, and the effect will be to exacerbate the stall.

[Miserlou]

BackPacker,
Just a kind of devil's advocate came up in me when people start on about the main wing stalling before the tail ergo tail must provide downforce.

The canard is only tailplane lift taken to it's logical extreme. Please explain how these work and lifting tailplanes on 'conventional' aircraft don't.

Any discussion of stalling characteristics is irrelevant anyway because the certifying standards require either sufficient aerodynamic warning of the impending stall or artificial warning. That is to say, a stick shaker and pusher system.

I think there may be some confusion also in that the stall occurs outside of the operating envelope where tailplane lift is most beneficial. It is in the cruise that you really want the efficiency of a lifting tail.

I seem to remember the ATR-72 has a take-off trim scale from -1.5 to +5 where the -42 only has a positive scale. Positive as in negative, of course, + being nose up.

Sorry for the thread shift.

[ahramin]

The A380's tailplane produces lift upwards instead of downwards.

Once you go to fly by wire you can start designing for performance rather than idiot proofing .

Oh, and in regards to the original question, i don't know what the actual force is.

[TURIN]

Just did a quick search on the A380 and found this.
http://http://designnews.com/article....html?ref=nbra
"To meet the loading (high) and weight (low), they employed a new, high-strength steel and titanium shaft and a bi-metal, spherical plain bearing designed to support loads up to 3,800 kN. The bearing's bronze inner ring allows a double rotation path. The plane is expected to enter service in 2005."
Not definitive I know as the loads quoted above will also be to cope with gusts and peaks, but it does give some idea.
By the way, is that true that the A380 is naturally unstable, as implied by ahramin?

[rubik101]

In S&L flight, any change in nose up pitch would tend to increase the lift of the wing, hence further increasing the angle of attack. This pitch up tendency is countered by the similar pitch up affecting the tailpane, hence lifting the tail and restoring the S&L flight situation. If the tail produced lift in a downwards direction, such a pitch up of the wing would result in further pitch up and hence lead to an unrecoverable stall.

Near the cruise condition most of the lift is generated by the wings, with ideally only a small amount generated by the fuselage and tail. We may analyse the longitudinal static stability by considering the aircraft in equilibrium under wing and tail lift, and weight. The moment equilibrium condition is called trim, and we are generally interested in the longitudinal stability of the aircraft about this trim condition.

(insert diagram of a/c with large arrow pointing up above the wing and a small arrow pointing up above the tailplane.)


Equating forces in the vertical direction:

W = Lw + Lt
where W is the weight, Lw is the wing lift and Lt is the tail lift.

The nature of stability may be examined by considering the increment in pitching moment with change in angle of attack at the trim condition. If this is nose up, the aircraft is longitudinally unstable; if nose down it is stable. Differentiating the moment equation with respect to α:

It is convenient to treat total lift as acting at a distance h ahead of the centre of gravity.

The total lift L is the sum of Lw and Lt so the sum in the denominator can be simplified and written as the derivative of the total lift due to angle of attack.

I'm sure this was covered some time ago but has since disappeared.