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We can see that most conventional aircraft have their horizontal tail surfaces arranged to provide a down force. This question asks if that is required for static longitudinal stability, and the answer is: not necessarily.

But if static stability is not the reason, what is? Or is it just a meme indeed?

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Because the safe situation is: a gravitational moment nose down, trimmed with an aerodynamic moment nose up. In all flight conditions. The critical situation is not cruise, but the conditions at low speed.

  1. At TO and landing the Angle of Attack is high: a nose up attitude, commanded by a downward tail force. Velocity is low, so the large force is achieved by a large upward deflection of the elevator and/or a negative stabiliser incidence. The AoA range is from around zero in cruise, to 15-20 deg nose up: the required elevator range is from zero deflection to large up deflection (plus superimposed manoeuvre requirements down & up). The average required force is downwards.
  2. You don't ever want the horizontal control surface to stall, but particularly not at low airspeeds = during take-off and landing, the most critical phases of the flight. Placing the tail at a lower average incidence angle than the main wing means that there is Angle of Attack margin: when the main wing approaches stall AoA, the tail surface has some AoA left before stalling. This implies that in cruise, if the wing has zero pitch angle, the horizontal tail has negative pitch angle.

From my decades old lecture book, paper copy only

  1. But despite the lower incidence, the horizontal tail could stall or partly stall, for instance in a sideslip where a large portion of the horizontal tail is in the wake of the aft fuselage. If it does, we want the aircraft to have a nose down moment, not a nose up moment. If the wing is not stalled yet, the lift force is now not in n.p.$_{fixed}$ anymore, but in a.c.$_W$ And that implies that in the above situation, the nose moves up and the main wing will stall as well. Only when the c.g. is in front of the a.c.$_W$ and the tail force is downwards, will there always be a nose down moment in a stalled tail situation.
  2. And in line with 2. and 3.: it must be certain that there is a stabilising moment in all flight circumstances, at all airspeeds, all angles of attack, all angles of sideslip. That is hard to guarantee when the main wing is optimised for cleanest aerodynamic flow and the stabiliser is in the wake of the main wing: efficiency of the main wing will be higher in all circumstances. By choosing an equilibrium state where the horizontal tail produces a downforce, and therefore less lift with increasing AoA, this problem is solved and the guarantee can be issued, without any additional analysis required. Very good in the olden days of aviation, before CFD.

Note that all above considerations are not really a huge concern in cruise: can the horizontal tailplane ever encounter a vertical wind gust with a velocity that can stall it? The velocity triangle says no, this is not likely to occur in 10$^9$ flying hours. A large aircraft can apply Relaxed Static Stability in cruise: pump fuel to the tail tanks and reduce average down force plus associated trim drag to a minimum, because AoA at cruise is very low. Or trim such that the stabiliser has an upforce, like in the drawing where there is static stability. But at start and landing, the trim tanks must be empty!

The additional burden for certifying this situation is a more elaborate analysis considering gusts at cruise altitude and the influence on static stability: are there any vertical gusts at a higher frequency than 10$^{-9}$ flying hour that can destroy the static stability? Such an analysis was impossible in the starting days of aviation, it is possible now of course.

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    $\begingroup$ Anonymous down voter: what is your issue with this answer? $\endgroup$ – Koyovis Jan 10 '18 at 10:34
  • $\begingroup$ The funny fact is that the tail load is the least negative at low speed, just when the elevator deflection optimizes the tail airfoil for maximum downforce. At least with positively cambered wings and no flaps, that is. The highest downforce is encountered in a dive at high speed. $\endgroup$ – Peter Kämpf Jan 12 '18 at 18:04
  • $\begingroup$ Actually, although this is kinda deep into the physics, the weight of the aircraft is not an aerodynamic force and does not play any part in this analysis. We think about it only because we are performing the mathematical calculations in an accelerated frame of reference. The analysis would be identical to that done for an aircraft in a giant box in deep outer space, that had a rocket on it that was generating 32 ft/sec2 linear acceleration, if we performed the calculations in the accelerated frame of reference of the box. That aircraft has no weight acting on it, it is in free fall. $\endgroup$ – Charles Bretana Jan 12 '18 at 22:04
  • $\begingroup$ Just like an aircraft in flight in the earth's atmosphere. It, too is in free fall, cause the force acting on it's wheels when it's parked on the ramp is not there. $\endgroup$ – Charles Bretana Jan 12 '18 at 22:05
  • $\begingroup$ another quick point to make, that does not dispute anything in this answer, is that the distinction made between low speed and high speed is totally dependent on the assumption that the aircraft is in a static One "G" flight condition. Although all the principals and factors affecting stability are of course true when at one "G", they must also be identically true at any G-Loading. as well as at any aircraft (bank or pitch) attitude. $\endgroup$ – Charles Bretana Jan 12 '18 at 22:10
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Simple answer: To provide enough static stability and a wider center of gravity range.

The lift on the tail will become negative if more stability is desired than what is achievable with no load on the tail. Another reason is a forward center of gravity location due to loading and securing the cargo on board or an uneven emptying of fuel tanks.

For usability, the center of gravity of every aircraft is allowed to vary, within clearly specified bounds. The forward limit is usually determined by:

  • Maximum stick forces in a pull-up maneuver (that used to be specified in MIL-F-8785C §3.2.3.5, for example). During the certification flights of the Do-228 for the CAA, the pilot flew a dive with full forward cg location and let the trim run away completely, waiting too long to take corrective action. His last words were "help me on the stick!".
  • Maximum stick forces over load factor. If the pilot needs to pull positive gs, the aircraft should not fight back too much.
  • Sufficient control authority in ground effect. When flying close to the ground, the effectiveness of the elevator is reduced. In order to have enough pitch control authority to rotate and lift off, the cg must not be too far forward. Neglecting this lead to an kludgy fix during the development of the F-18, the rudder toe-in.
  • Speed stability: The aircraft should maintain the trimmed speed, even in gusty weather, without requiring constant pilot inputs. This is the main reason for negative tail loads: Comfort.
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  • $\begingroup$ I agree with your last point, about comfort (aka, sufficient positive static stability to allow the aircraft to be manageable when flown manually), being the main reason for this. You are clearly more technically conversant than I. Do you know whether it is even possible to design an aircraft with tail-mounted pitch control so that at cruise, there is positive lift on the tail surface, yet there is sufficient static margin to allow reasonable range of CG, and not create negative stability conditions at high (close to Clmax) AOA? Was any aircraft ever so designed? $\endgroup$ – Charles Bretana Jan 16 '18 at 19:34
  • $\begingroup$ @CharlesBretana: Static stability doesn't change with AoA, only tail load does in planes with wing camber. Modern Airbus planes should have neutral to slightly positive tail loads when the tail tank is used correctly (I have spoken with pilots who refused to use it). Stability is ensured by the autopilot which is even required for comfort with negative tail loads at high altitude due to the low pitch damping there. Note that the tail tank is only used to shift the cg back, so cargo can be placed in a wider cg range without affecting performance. $\endgroup$ – Peter Kämpf Jan 16 '18 at 20:06
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    $\begingroup$ I don't mean to start another extended discussion, but what I am talking about is the forward motion of the aerodynamic center of any airfoil that occurs as the boundary layer separates at the trailing edge as AoA increases near the maximum (stall) AoA. Does not this forward motion caused by trailing edge boundary layer separation cause a decrease in static margin? $\endgroup$ – Charles Bretana Jan 16 '18 at 20:18
  • $\begingroup$ @CharlesBretana: Once separation starts, we leave the linear range of aerodynamics and all simplifications are no longer valid. Now the pressure over the separated area is lower, such that the center of pressure moves back with the onset of trailing edge separation. This actually stabilizes the aircraft! With full separation pressure is roughly constant over chord and the center of pressure is nearer 50%, not 25% as in attached subsonic flow. $\endgroup$ – Peter Kämpf Jan 16 '18 at 23:15
  • $\begingroup$ From Wikipedia article on Center of Pressure (en.wikipedia.org/wiki/… ), "For a conventionally cambered airfoil, the center of pressure lies a little behind the quarter-chord point at maximum lift coefficient (large angle of attack), but as lift coefficient reduces (angle of attack reduces) the center of pressure moves toward the rear". So as AOA increases, it moves forwards. It moves forwards because the boundary layer flow separates from the airfoil at the rear trailing edge and this separation propagates forwards as AOA increases. $\endgroup$ – Charles Bretana Jan 17 '18 at 1:41
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Although the answer is not absolutely yes, in almost all conventionally designed (where elevator control is behind the main wing, not a canard) aircraft, the answer is YES, that is the reason.

If the elevator control is behind the main wing, and there is any significant size difference between the main wing and the tail surface, the overall Aerodynamic center will be behind the CG, and the aerodynamic center (the center of pressure) of the main wing (which, subsonicly, is at the 25% of MAC point), will also be behind the CG, not in front of it. Therefore, lift from the main wing will produce a nose down pitching moment, and therefore the tail must be cambered and installed with negative incidence to produce negative (downwards) lift and nose up pitch moment to counteract it.

In order to position the aerodynamic center of the main wing in front of the CG in an aircraft designed with the elevator control surface at the tail, and still have the overall aerodynamic center be behind the CG (necessary for positive static stability), without considering the relative angles of incidence, the relative ratios of the Main wing moment arm to the tail surface moment arm, and the ratio of the lift produced by he main wing to the lift produced by the tail, as depicted in the diagram in @Koyovis' answer would have to satisfy the following inequality.

  1. W = Lift from main (forward) wing
  2. T = Lift from elevator control (Tail)
  3. w = length of Moment arm from main wing AC to CG
  4. t = length of Moment arm from elevator control AC to CG

then, in order for the overall AC to be behind the CG in an aircraft designed with Main wing AC in front of CG, the following would have to be true.

     t/w > W/T

i.e, the ratio of the tail control moment arm to the main wing moment arm must be greater than the ratio of Main wing Lift to Tail Lift. And it must be significantly greater, because the closer to equality these ratios are, the closer to neutral stability the aircraft is. So the inequality must be sufficiently large enough (I confess I don't know by how much that is), but it must be large enough to produce positive stability. To make it larger, you must make w smaller (move the wing AC closer to the CG), or t bigger (move the tail further back), or make the relative sizes of the wing and tail surface more equal (make the tail bigger and the wing smaller). These configurations are possible, and aircraft have been designed like this, but it is the exception, not the most common aircraft design.

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  • $\begingroup$ Yes the CoG must be in front of the lift neutral point, in all flight circumstances, at side slip angles with half the horizontal tail ineffective etc. What you are describing is that the wing and fuselage have a de-stabilising effect, the tail has a stabilising effect, and the tail stabilising effect must be higher in order to generate a stabilising total aeroplane configuration. That is certainly true, but does not explain why it is more beneficial to use tail downforce than upforce although both situations can have static stability in cruise. $\endgroup$ – Koyovis Jan 11 '18 at 3:11
  • $\begingroup$ Here I am not sure (perhaps I should post a question to see if someone can address this) but is it possible that this scenario is never used because the band in which the forward AC would have to be is so narrow that forward movement of the AC as AOA increases towards CLmax (Stall AOA) due to wing trailing edge boundary layer separation could too easily move the overall AC forward of the CG and change positive stability to unstable at the critical point of approach to stall? reddit.com/r/aviation/comments/2kibfo/… $\endgroup$ – Charles Bretana Jan 11 '18 at 14:45
  • $\begingroup$ If this analysis is correct, then it would mean that designing an aircraft this way (where wing AC is forward of CG and both Wing Lift and Tail Lift are positive [upwards] in normal cruise), would be inherently dangerous, as the forward movement of the Center of pressure as AOA increases towards stall would be reducing stability and potentially crossing the threshold and producing an unstable aircraft at approach to stall AOA (like during Landing phase) $\endgroup$ – Charles Bretana Jan 11 '18 at 14:51

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