C of G and pitch stability

[shlittlenellie]

Apologies for the simplicity of the question:

Why is an aircraft more pitch stable with a forward C of G than with an aft C of G? Accordingly, why are required control forces higher for the forward C of G.

My incorrect model is that the elevator exerts a force over a distance to effect a torque with the CofG position being the pivot. With an aft CofG, this torque is less due to a shorter arm. Conversely, with a forward CofG, the elevator has a much longer arm and so greater torque is available for a given control movement? Reality is the complete opposite of this and any explanation would be much appreciated.

Many thanks.

[L Peacock]

Before someone gives you the correct answer with theory and hard sums, try drawing the extreme cases on paper as balance/see-saw type arrangments.
a) with CG (acting downwards obviously) furthest forward, then centre of lift at the wing (upwards), then rear most, downforce at the elevator.
b) wing lift furthest forward, followed by CG, then lift provided by the elevator.

Very simplistic and not the whole story by you might start to see the answer for yourself.

[Genghis the Engineer]

The term "stability" in this context is a bit misleading and tends to confuse.


If you consider the standard diagram of an aeroplane - CG at the front, main lifting surface a little behind that, and a tailplane rather further behind that providing a downforce - you have a system in balance (hopefully!).

If you move the CG back to co-incide with the centre of pressure (CP) then the tailplane has no work to do. So, any movement of the elevator will have a very strong effect. This is referred to as neutral stability and the aeroplane will be virtually impossible to maintain at the desired pitch attitude or airspeed.

As you move the CG forward, the tailplane starts having to do some work. This means that the elevator is just modifying the amount of downforce from the elevator, and so you have a sensible amount of control over the aeroplane. The aft CG limit is set at the point where the company test pilots decided that the aeroplane was safely controllable by the average pilot.

Keep moving the CG forward and the elevator is doing more work. This means that the effect of elevator movement reduces -it needs to deflect more for the same effect. Test Pilots will measure this as either stick displacement per airspeed change, or (much more importantly since we fly by feel, not stick position) stick force per airspeed change.

Eventually you'll reach a point where it doesn't matter how much you move the stick you don't have enough control over the aeroplane - the most common way this'll exhibit itself is by the aeroplane refusing to pitch up into the flare properly at the approved approach speed. Again, the company test pilots will set the forward CG limit at the point where they consider that the aeroplane is just safely controllable by an average pilot. (Although occasionally you hit a tailplane structural limit first - the tail is working very hard at forward CG).


But what's important to realise is that what you are told is "pitch stability", is what a flight tester will more formally call "apparent longitudinal static stability" and is really just the amount you have to push or pull the stick to get a given airspeed or attitude change.

Hope that helps a bit,

G

N.B. If you like hard sums, I can recommend a couple of good textbooks, but frankly they are best avoided until you've got the basic mental model fixed anyhow. If you have, then most people don't need the maths!

[guclu]

Let me try to explain,


First of all you must understand what a moment is.

Moment = F X a (Force times arm.)

For an aircraft to be steady in pitch, the total moments acting through CG must be zero in the neutral plane. That is to say the moment of Lift from wings and moment of Lift from the tailplane (elevator) must be zero.

For the wings, lift is much greater than the lift of the tailplane (which is actually a lift in the direction of the earth for the tailplane) but the arm (the distance to cg) is much smaller.

Now, try to think that cg is at the most limit forward position. The arm of the tailplane is getting so much that its affect is much greater than before. So that is to say, for example, when your pitch is moved only small amount upwards, the lift of the wing and tailplane will increase. But consider the arms. If it is say 1 for the wings it is 100 for the tailplane. So even a small increase in angle of attack will affect the moment very much for the tailplane. In the most forward limit, the tailplane is very much stabilising. (Stabilising = Wanting to return the airplane to its original, not disturbed position). That is any change in pitch will be strongly resisted by the tailplane, because of the long arm.

The other scenario is the most limited rearword position of CG where the arm of the tailplane is at lowest value. At this position, the tailplane doesn't have that much stabilising affect and every movement you make to the elevator will not be stabilised by the tailplane. (Neutral tendency). This is aircraft is very sensitive to elevator control.

I hope it helped.

Guclu

[ant1]

Imagine a model airplane, in a wind tunel (moderate wind), sitting on a fulcrum (The fulcrum in flight would be at the intersection between the longitudinal axis and the wing's center of lift)

The aircraft is sitting horizontally in equilibrium.

Now, hit the aircraft at the nose or the tail. It will start oscillating around its horizontal equilibrium position due to the weathervaning effect of the elevator surfaces.

Repeat the experiment but now stick some weight on the aicraft nose: you have moved the CG forward.

Now, to maintain horizontal equilibrium, you have to set the elevator surfaces in such a position that they exert a higher downward force.

If you now hit the airplane nose or tail with the same force you used before, the aircraft won't move as far from its equilibrium as it did previously and will return faster to it through a lower amplitude and more dampened oscillation because both moments on each side of the fulcrum are higher. That is why it is more stable.

If you prefer, imagine that in the two previous experiments, instead of weight and elevator downward force, this time, the fuselage is linked to two springs (one forward and one aft of the fulcrum) each one exerting a downward force.

In the second experiment the springs are more tense and exerting a stronger pull, hence a stronger moment, around the fulcrum (that's the case witth a forward CG).

Cheers