How can an up-lifting tail have normal force direction and gradient when generating UP force?

Question

This article includes this statement about tail forces to validate the concept that the horizontal tail in conventionally tailed aircraft can be generating force DOWN or UP at different times and still have normal stability and control characteristics:

I took a Cessna 172 Skyhawk and put a couple of large pilots in the front seats, with no luggage and no other passengers. That meant the center of mass was right at the front of the envelope, so the tail had to produce considerable negative lift in order to maintain equilibrium. There was lots and lots of angle of attack stability. I took the same Skyhawk and put a small pilot in the front seat, a moderately large mad scientist in the back seat, and 120 pounds of luggage in the rear cargo area. That put the center of mass right at the rear of the envelope, so the tail had to produce considerable positive lift in order to maintain equilibrium. The airplane still had plenty of stability. (As far as the pilot could tell, it was just as stable as it ever was.) The easiest way to determine whether the tail lift is positive or negative is to observe the direction of motion of the tip vortices, as discussed in section 3.14. To observe the vortices, I attached a streamer of yarn, about half a yard long, to each tip of the horizontal tail, at the trailing edge. The streamer gets caught in the vortex, so its unattached end flops around in a circle. When the tail is producing positive lift, the circular motion is in the direction shown by the green “circulation” arrows in figure 3.29, i.e. downward on the inboard side. When the tail is producing negative lift, the direction of motion is the other way, i.e. upward on the inboard side.

Numerous posts on ASE, and too many comments to count, support this concept. That's great. There is a nagging problem though that prevents me from accepting the "consensus" just yet.

• For the tail to generate downforce, local flow must be above the zero lift line of the tail airfoil. The hinge moment imparted by air loads on the elevator is TE down, control stick forward, and aft stick force is required to counter it. For trim, the tab must move TE down to apply a servo force UP to counteract air loads.
• For the tail to generate upforce (lifting), the local flow must be below the zero lift line of the tail airfoil. The hinge moment imparted by airloads on the elevator is now TE UP, control stick aft, and FORWARD stick force is required to counter it. For trim, the tab must move TE UP to apply a local servo force DOWN to counter act air loads.

This is what you would call "stick force reversal" and I can't find any resources that say this is a good thing; actually quite the opposite, and I certainly wouldn't want to experience it in my own airplane.

So how exactly does that work? How can a lifting tail still have normal control force direction and gradients, same as a downforce tail, and how could the trim tab work given that it now has to work in the opposite direction to provide the required servo forces for trim, on the same aircraft?

Answer 1

Your thinking assumes that a specific elevator position produces a specific stick force. If I understand you correctly, a TE up elevator must be unstable because the required stick force to bring it back to neutral is the opposite of what a TE down elevator requires.

That is not how it works. Neutral elevator is just one of many possible positions for trimmed flight*. Next, you need to think in stick position changes, not absolute stick positions. Mathematically speaking you need to look at the derivatives.

A stable stick force is one which brings the airplane back to its equilibrium after an upset. This might be a gust or a jolt on the stick; details don't matter. All that counts is if that airplane finds itself in a state different that what had been trimmed, the stick force must act in the direction which brings the plane back to the trimmed state.

Let's look at the TE up case first. The airplane suffers an upset which raises speed by something above trimmed speed. Wing AoA is reduced and so is tail AoA, so the downforce on the tail is increased over the already existing downforce. The aircraft raises its nose and slows down. Mission accomplished. (Yes, quietflyer and Guy, now a phygoid motions starts, but we are only looking at static stability here).

What happens to the stick force? The lowered AoA would drive the elevator from a TE up position towards TE down and the stick more backward, as it should for positive stability, until the old, higher AoA is restored.

Now the same with the TE down position: The change in AoA is the same, and so the elevator would be driven even more TE down. Again, the stick moves backward and shows stable behavior. There is no fundamental difference to the TE up case.

It does not matter where the elevator TE stands for trimmed flight. All that counts is the change as a consequence of a deviation from the trimmed state. The only difference between down- and upload on the stabilizer at a given speed is the amount of change. With a downforce the center of gravity is located more forward and the aircraft is more stable, so the change in stick force as a consequence of an upset is more pronounced. But the direction of the change does not change until the lift per area on the stabilizer is higher than that on the wing. But then the center of gravity will be behind the neutral point and the aircraft will become unstable.

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* A naturally stable airplane with a fixed stabilizer will see the TE of the elevator rise as it is trimmed for slower speeds. At the same time, tail downforce decreases and becomes tail lift at low speed even as the elevator trim deflection is negative (TE up). The elevator (and with it, tail camber) runs against the required tail lift. This means tail lift is most likely to occur with negative elevator deflections.

A trim tab which produces mostly up forces could be combined with a spring in the elevator linkage in order to increase speed stability (read here how this works) or it serves to trim the airplane with flaps down because this case needs more negative elevator deflection. More down travel of the trim tab (meaning more trim range with elevator TE up) is completely normal and does not indicate that the tail will only produce a downforce over the speed envelope.

Stick forces are the difference in hinge moment between the free-floating elevator angle and the angle needed to trim the aircraft at that speed, multiplied by the elevator linkage gearing ratio. If the aircraft flies faster than what it has been trimmed for and the free floating angle is more positive (= TE down) than the trimmed elevator angle, stick forces will be in "pull" direction (the pilot has to actively push the stick away from the free-floating position in order to fly faster). This happens regardless of tail lift as long as the center of gravity is ahead of the airplane's stick-free neutral point.

Answer 2

Let me break it down in sections. It's rather long I'm afraid.

1. Tail lift direction

Consider the following diagram of a wing-body and tail combination, with a non-pitching aircraft in steady-state. Let's assume that a true wingbody mean aerodynamic centre ($h_{n_{wb}}\overline{c}$) exists and therefore the pitching moment does not vary with AOA at that point.

_{Image ref: Etkins, Dynamics of Flight}

The pitching moment for the whole aircraft about every point must be zero. Let's pick the point of CG ($h\overline{c}$) to sum our moments, since it also corresponds to the point of rotation for a free-body (this equation can be found in Etkins, Dynamics of Flight, but its derivation is simple and you should be able to deduce it from first principle):

$$C_m = C_{m_{ac_{wb}}} + C_L(h-h_{n_{wb}}) - \frac{\overline{l}_t S_t}{\overline{c} S} C_{L_t} = 0$$

where $C_m$ is the total pitching moment, $C_{m_{ac_{wb}}}$ is the wingbody pitching moment coefficient about the MAC, $C_L$ is the total lift coefficient (wingbody + tail), $C_{L_t}$ is the tail lift coefficient normalized over the tail surface area ($S_t$), $\overline{l}_t$ is the distance from the tail MAC to the wing MAC, and $\overline{c}$ is the wing reference chord.

The sign of the tail lift, $C_{L_t}$, to make the above hold can be either positive or negative, depending on the relative contribution of each of preceding terms. Note two facts:

1. For a positively cambered wing, $C_{m_{ac_{wb}}}$ is typically negative.
2. $C_L$ is positive for 1G flight.

Consider some extremes:

1. If the CG is way ahead of the wing MAC ($h \ll h_{n_{WB}}$), then the second term will be very negative, and $C_{L_t}$ must be negative to compensate. That is, negative tail lift.

2. If the CG is behind the wing MAC ($h > h_{n_{WB}}$), and $C_L$ is very large, then $C_{L_t}$ must be positive to compensate. That is, positive tail lift.

   Furthermore, if the CG is behind the wing MAC, then there will always be an AOA above which the trimmed tail lift becomes positive, assuming stall does not occur.

2. Neutral point

Neutral point ($h_n\overline{c}$) is the longitudinal position at which the aircraft does not experience any pitching moment with a variation in AOA or $C_L$ (in another word, NP = MAC of the aircraft). If the CG is ahead of the NP, the aircraft will experience a pitch down moment with increasing lift (statically stable); and vice versa for unstable.

I will cite the following result, once again from Etkins (but simplified):

$$h_n = h_{n_{wb}} + \frac{a_1}{a} \frac{\overline{l}_t S_t}{\overline{c} S} \left ( 1 - \frac{\partial \epsilon}{\partial \alpha} \right )$$

where $a=\frac{\partial C_{L}}{\partial \alpha}$ is the total lift slope of the aircraft, $a_1=\frac{\partial C_{L_t}}{\partial \alpha_t}$ is the isolated lift slope of the tail, $\epsilon$ is downwash from the wing on the tail.

Note that the second term is always positive. Therefore, the CG can be behind the wing MAC, yet ahead of the NP.

3. Elevator hinge moment and stick force

For a reversible aircraft where the column is directly connected with an elevator, stick force gradient is directly conferred by the aerodynamic hinge moment on the elevator (let's assume there is no down-spring or bob-weight for simplicity).

For a non-cambered tail, there are two main contributions:

• Hinge moment due to flow incidence ($b_1$)
• Hinge moment due to elevator deflection ($b_2$)
• Hinge moment due to trim tab ($b_3$)

You are right to point out that if a tail is lifting, and its flow incidence increases with aircraft AOA, then there will be increasingly TE up HM on the elevator, which would imply instability. However, static stability ensures that there is increasing TE up elevator needed to trim for increasing AOA, which will add TE down HM on the elevator.

The details are more nuanced than the above, and it will involve airspeed since HM is squared to airspeed, but it can be shown that the stick force reversal only happens when the CG is aft of the stick-free neutral point ($h_n^{'} \overline{c}$), and not at neutral point:

$$h_n^{'} = h_n - \frac{b_1}{b_2}K$$

where $K$ is a positive constant that is a function of the aircraft geometry (I've simplified from eqn 2.6.13 from Etkins). As you can see, if $b_2$ (HM due to elevator deflection) overpowers $b_1$ (HM due to flow incidence on tail), then we pretty much recover the NP. Down-spring and anti-servo tabs are some methods to increase the $b_2$ artificially.

More specifically, please consult the following figure, adapted from Bromfield, Factors affecting the apparent longitudinal stick-free static stability of a typical high-wing light aeroplane and Etkins, Dynamics of Flight:

In the graph, $\delta_t$ is the trim tab deflection.

Observations:

1. The first two curves are with the tab always set to 0 incidence against the elevator. The movement of the CG purely shifts the stick force vs. V curve up and down. When the CG is aft of the stick-free neutral point ($h>h_n^{'}$), any steady-state speed would require a pull force (despite the stick may be forward of neutral if the CG is still ahead of the neutral point!)

2. The last two curves are with the tab set for a trim speed ($V_{trim}$). Now, when the CG is aft of the stick-free neutral point, the stick force gradient also goes unstable! What effectively happens when the CG coincides with the stick-free neutral point is that the hinge moment contribution from the elevator+tab cancels that from the tail AOA; therefore, zero steady-state force is needed to hold any particular elevator/stick position.

Either way, whether the tail is lifting doesn't dictate the stability of the stick force vs. airspeed.

Answer 3

Keep the wind flow the same. Draw the chord line from each trailing edge deflection.

Indeed, this is how we switch from upforce to down force when we use our elevator.

This one twisted my brain a little too. But we must consider the lifting torques around the center of gravity. If the AOA of the tail is lower than the wing (up or down), then the percentage change in lift created will be greater for an increase of AOA for the tail than the wing.

But even if all lift is borne by the main wing, an increase in lift (from increasing speed) will create a vertical departure from the line of flight, which creates downforce on the tail. A sink from decreasing speed has the opposite effect.

So one can design the CG anywhere they want.

... on the same aircraft?

The problem comes from abusing the CG range. With aft CG, the torque forces move towards instability, and the elevator and trim authority to counter act it are used to a greater degree as lifters rather than pitchers or trimmers. Additionally the AOA ratio of wing to tail falls, further reducing stabilizing advantage.

The flip side of this is that too far forward a CG is harder to pull out of a dive, as now more pitch up authority is depleted. A wing that has a center of lift that shifts with AOA adds to the design considerations.

So one can have a "staticly stable" lifting tail. The old fashioned draggy way is a low aspect delta with a lot of area, set at a lower AOA than the wing. It's torque will be more stabilizing than the wing torque is destabilizing when pitch changes.

Answer 4

The force is still a download on the horizontal stab, even with an aft center of gravity.

There's a reason that a CG envelope exists.

Stick trim force may be reduced to zero, but this does not imply loss of a tail down force.

In the case of the 172 previously mentioned, consider a simple wings-level unaccelerated straight-ahead power-off stall. The pitch-down moment that occurs with the stall break, when the downwash off the wing is altered by airflow separation and the download on the horizontal stab decreases, is evidence of the download in the first place; this should occur at all points in the CG envelope (not just at the front of the envelope).

Consider also the consequence of a tailplane stall, typically associated with tailplane icing, or icing conditions, and the resulting loss of pitch authority. Which way does the aircraft pitch, if loaded within the CG envelope?