Longitudinal Dihedral
 

Longitudinal Dihedral

Old definition: Angle between the angles of incidence of the wing and tailplane. Positive if the wing incidence is greater.

All-moving tails or tails with trim systems that move the stabiliser have variable angles of incidence. Therefore the longitudinal dihedral is variable in flight. I suppose you could define a basic longitudinal dihedral in these cases by measuring the tailplane incidence at the designed zero deflection position.

Why would you design longitudinal dihedral into an aircraft? I have found two internet entries. One says it is to increase longitudinal stability, but the explanation of the way it works is fundamentally flawed. The other is on a Cranfield site that says it does improve longitudinal stability , but does not say why or how.

I have my own ideas, but would like to hear from any or all experts out there. What is the purpose or function of longitudinal dihedral

Dick W

I think you are asking about the following which is a very simple case describing what I think you are talking about;

Bolt on the wings at an incidence of 4 degrees.

Bolt on the horizontal stabilizer at an angle of incidence of 2 degrees.

Aircraft flying along level (wing 4deg angle of attack and tail 2 deg angle of attack) and gets disturbed, nose up slightly.

Momentum carries the aircraft along original path thus increasing angle of attack on both the wing and the tail.

Say direction of the relative airflow changes by 2 degrees.
Thus angle of attack of the wing increases from 4 to 6 degrees - a 50% increase.
Angle of attack on tail increases from 2 to 4 degrees..........a doubling of the angle of attack.

That increase in angle of attack of the horizontal stabilizer increases lift and thus the tail rises to restore the original situation.

Is that what you are after?

Regards,

DFC

DFC

Since the zero-lift angle of attack of a cambered airfoil isn't zero, you can't equate a % change in local AoA into a % change in lift/force.

Plus one would typically have to account for the delta downwash at the tail, due to the presence of the wing, which is going to roughly halve the delta AoA seen by the tail anyway.

As to the original question - why would one design for a different angle of incidence for the mainplane and tailplane...

The main driver, I would suggest, for the wing incidence is to obtain the best compromise between the fuselage deck angle for minimum drag, and the airfoil AoA for best L/D. Constraints on this optimisation would include ensuring that any geometric limits for takeoff/rotation/landing/flare were not penalising. (Barring a Crusader-style wing!)

Having thus arrived at a ideal wing incidence (almost as a \'tail-off\' consideration) the tail incidence is constrained by the desire for a minimum drag configuration at high speed - where we may desire that our mid c.g. cruise weight configuration has faired elevators and the best efficiency for our tail in creating trim lift - and the desire for the most efficient tail for purposes of low speed trim to ensure our field performance is not unduly affected by Vmu or trim limits.

Chances are, we\'re going to end up with a small positive root incidence on the wing, and a near-zero tail incidence.

In considering the effects the \'longitudinal dihedral\' might have on longitudinal S&C, let\'s consider a simplified case : symmetric wing and tail airfoils, and a fixed H-stab. Let\'s assume the wing is set at 2 degrees, and the tail at zero.

In cruise flight, assuming the body AoA is zero, we\'ll have a wing AoA of 2 degrees, a downwash of about 1 degree, and hence a tail AoA of -1 degree, giving us a bit of download for trim. (Which is good, assuming we\'re a nice simple +ve static stability type).

If we introduce a disturbance x to the body AoA, the wing AoA will be disturbed by X also, while the tail AoA will change by X/2, approximately (downwash ratio of 0.5 assumed previously). If our aircraft is stable the pitch axis should restore to the undisturbed condition.

Now, let\'s magically redesign the aircraft, and increase the wing incidence to 4 degs. Since a wing AoA of 2 deg is what we need for trimmed flight, that means we\'ll be at a body AoA of -2 degs for cruise. The tail AoA will be -3 degs now (-2 body -1 downwash) which means that to be in trim we\'ll need some elevator. Lets assume the elevator does indeed get adjusted.

Now, disturb this aircraft by X degrees. Again, delta of X to the wing AoA, delta of X/2 to the tail AoA, which means, all other things being equal, the aerodynamic response will be identical.

The only thing we\'ve changed is that the tail is going to be at -3+X/2, not at -1+X/2. As long as the tail lift curve is linear over a sufficiently large range, and the disturbances are relatively small, there will be no mathematical differences. If the tail is significantly nonlinear, it may matter - but in this case I\'d expect the case where the tail incidence is minimised to be the best case - and that\'s not directly a function of increasing or decreasing long-l dihedral, but comes from a consideration of the trim requirements.

In short, I can see no stability arguments in favour of increased or decreased long-l dihedral; other factors matter more.

Doesn't this have more to do with contolability rather than stability?
I'm thinking about high alpha situations where the stab alpha is + and thus a large elevator input is required to provide the required down force. max deflection is aerodynamicaly limited. so the designer has a choice of a bigger elevator and hence stab or some "decalage"sp. and smaller stab/elevator.

I concur with DFC. It's simple and the standard CFS A2 answer.

It's quite noticeable on the aircraft I did my A2 on, the Chipmunk. Wings at 4 degrees, tailplane at zero. One thing we noticed was that the students often bounced on their first solo. This was due to the c of g being further forward (without the weight of their often quite portly instructors) a greater elevator input being required, therefore greater longitudianl dihedral, therefore more pitch stability being encountered and touching down mainwheels first - leading to a bounce.

I'm sorry, but an increase of body AoA of 2 degrees will produce an increase of 2 deg AoA on the wing, but will NOT produce a 2 degree AoA increase at the tail unless the downwash gradient is ZERO - which is only possible if the wing has fallen off.

A fresh viewpoint:

Q. Why is it that when the CG is moved aft, the airplane is more efficient (albeit less stable)?

A. Because the aft CG means the horizontal stab doesn't have to work so hard - lower CL & Cd.

How can this be true if the stab is at a positive effective AOA? I was always taught the stab is at a negative AOA (effectively, allowing for the wing downwash). This implies the CG is forward of the wing CP to balance the bird.

This creates speed stability. A positive speed disturbance means both the wing lift (up) and tail lift (down) are increased, causing a nose-up moment to restore the trimmed speed.

Or do I need a revised aero text?

The latter.

The wing lift is not directly relevant.

What matters is the aircraft pitching moment, which is comprised of the wing (or wing-body if you prefer) pitching moment, and the tail lift * tail arm. Those make up the aircraft tail-on pitching moment, and it's THAT which matters.

(The reason wing lift isn't the key is that a wing with a nose down pitching moment - i.e. any cambered airfoil) will have an increased nose-down PM with increased speed at fixed alpha - even if the lift reference is at the cg.

This maybe a vast over-simplification of your question, barit1 - and I apologise if it is, but the MAIN reason why this is more efficient is that the tailplane lift is contributing to the overall lift component whereas if the tail has to produce a nose up force, it is not. Therefore the wing works less 'hard' therefore less induced drag. As 'mad' says, the wing pitching moment is almost irrelevant in the balance equation as it is acting MORE OR LESS around the cg. Aft movement of cg REDUCES the pitching moment of the tail and therefore REDUCES pitch stability.

Having just typed out a complete explanation from my much-copied A2 notes, that BASTARD PPRuNe 'you are not logged in' registration bollocks sent it off to cyberspace.

Why, oh why, must we have this bloody nonsense? It is an utter pain in the arse....


I might type it out again. On the other hand.....it is sufficient to say that the 'CFS' explanation is utter rot.

Thank you all for your inputs.

I have to admit that it was the "CFS" explanation that I described as fundamentally flawed. As far as I can see the main reason for designing in a longitudinal dihedral would be to achieve a cruise set-up with the wing at alpha for best TAS/Drag and the tail at minimum drag alpha, to minimise trim drag. This would rely on the wing/body moment at cruise alpha being near zero, achieved by careful CG positioning. By its nature this implies a longitudinal dihedral of about 4deg.

How does this seem to you all?

Dick W

Sounds too large at 4 degs, if we're talking about high-subsonic cruise.

I'd expect a cruise alpha in the range of 0-2 deg body, with a wing setting angle (incidence) of no more than 2 deg more. That means the wing alpha is going to be between 2 and 4 deg. Assuming, as ever, a downwash ratio of 0.5, that implies that a tailplane AoA of zero will be achieved with a tail incidence of about 0-1 deg:

For a body AoA of A and a wing incidence of W, the tail AoA as a function of Tail incidence will be:

tail AoA=A+T-(A+W)*0.5 (1)

So for a minimum tail AoA=0, we get:

T=(A+W)*0.5 - A = (W-A)*0.5

Therefore for A=0, W=2: T=1
for A=2, W=2 : T=0)

If the expression (1) is rewritten in terms of longitudinal dihedral D, where D=W-T, then we get:

tail AoA=A+ (W-D)-(A+W)*0.5

or

tail AoA = (A+W)*0.5 - D

so for a min drag tail AoA of 0:

D=(A+W)*0.5

Which for typical values, again, of A and W in the 0-2 deg range each, gives a D also in the 0-2 deg range

best alpha?
 

While that may be the objective, it's often missed by a mile. Ever notice the deck angle at cruise?

The fuselage is a pretty inefficient lifting body, and yet at typical cruise Mach, it's at maybe 1 or 2 degrees noseup. Shouldn't it be level? I think it's because the aero designer optimizes the wing incidence for unrealistically high Mach, and then misses that calculation on the low side. So, in order to cruise at real world conditions (crude oil approaching $60/bbl), the galley carts roll aft.

The DC-10 could definitely have benefitted from another 2 degrees incidence.

Cruise body AoA isn't the only constraint on wing incidence, as I mentioned above. You also have to consider field performance issues, both drag/lift optimisation and the need to get a decent Vmu limit. Then there's the issue of avoiding an overly nose-low approach, with attendant nose gear landing concerns.

This time I’ll draft in Word first!

Draw a diagram with the wing centre of pressure at distance a ahead of the centre of mass and the tailplane centre of pressure at distance b behind the centre of mass. Fairly obviously, for things to be in balance, then the lift generated by the wing, Lw and the lift generated by the tail, Lt, must be such that:

Lw x a = Lt x b.

Or, in other words:

CLw x ½ rho V² Sw x a = CLt x ½ rho V² St x b

Which can be re-arranged to CLw/CLt = St x b/Sw x a

Now, imagine our original diagram is given a small disturbance. Fairly obviously, to stop the whole thing tipping on its arse, then the increase in lift from the tail, dLt at the tail centre of pressure must be greater than the increase in lift from the wing, dLw, at the wing centre of pressure. That is:

dLt x b > dLw x a Or, in other words:

dCLt x ½ rho V² St x b > dCLw x ½ rho V² Sw x a

Which simplifies to:

dCLt x St x b > dCLw x Sw x a

Now, for aerofoils in the cruise range concerned, the lift curve slope may be assumed to be linear and the same for both wing and tail aerofoils, hence dCLt = dCLw.

So that:

St x b > Sw x a

Going back a few steps, we already showed that CLw/CLt = St x b/Sw x a. So it is clear that, for stability, CLw > CLt and this is achieved by ensuring that the wing is rigged at a greater incidence than the tailplane.

OK – a bit simplistic but far more algebraically correct than the so-called ‘CFS’ explanation which is utter hogwash!

And a tip for you boffins - K.I.S.S

BEagle

Well, that's a blow! I am going to copy that to Word and work on it.

Dick W

BEagle - top tip for PPruning,

No need to type into Word, when you have finsished typing your post, just highlight the whole lot then 'right click' and select copy. If your post fails because of the dreaded not logged in, log in then right click where you would type and select paste. I just do it automatically for every post.

Of course you could always find your Pprune cookie and use an editing tool to change the expiry date to next century, but thats too hard for me.

Of course you could say that the job of CFS is to provide the finest pilots for the RAF not to design aircraft.

If you give the studes not quite correct info then only the best survive out in the real world and so the standards of the RAF become second to none. Clever eh?

Mind you dunno how it all is today.

C'mon fellas, I'm a retired CFS*, and here I am asking for help.

BEagle's equations are internally consistent, but I still have an emotional problem with the concusions. If you consider an aircraft with an all-moving trimming tail the the tail angle for trim will be widely diferent at low and high speeds in level flight. This is the same thing as having widely different longitudinal dihedral at each extreme. Is the aircraft then more stable at low speed than at high speed?

At low speed the tail angle for trim would surely be more nose down. If this increased longitudinal dihedral makes it more stable, could you get away with an aft CG by flying slower?

Dick W

Dick

It may help to remember that the fixed tailplane has an elevator which of course will need to move when speed is changed. Same like the whole caboodle moving these days.

JF