Hinge moments and stick-free stability
Hinge-moment coefficients, the floating elevator, the free-elevator factor and stick-free neutral point, stick force, aerodynamic balance and tabs.
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Why it matters
The pilot does not feel angle of attack or elevator angle — he feels stick force. The aerodynamic moment about the elevator hinge line sets that force, and if the pilot lets go of the stick the elevator floats to wherever the hinge moment is zero. A floating elevator weakens the tail, so the stick-free neutral point lies ahead of the stick-fixed one and is often the real aft CG limit for manually controlled aircraft.
Key ideas
Hinge moment. The pressure distribution on a control surface produces a moment H about its hinge line. In coefficient form it is based on the control-surface area S_e and chord c_e (aft of the hinge), not on wing area. Sign: H is positive when it tends to deflect the surface trailing edge down (positive δe).
Linear model. For the elevator, Ch = Ch0 + Chα·α_t + Chδ·δe (+ Chδt·δt), where α_t is the tail angle of attack and δt a tab angle.
Chα(usually negative for a plain flap): a higher tail angle of attack pushes the trailing edge up.Chδ(negative): deflecting the surface down produces a moment pushing it back up — a restoring hinge moment. Both come from data sheets or tunnel tests and depend strongly on the nose shape, gap, and the amount of aerodynamic balance.
Free (floating) elevator. With the stick released and friction neglected, the elevator rests where Ch = 0: δe,free = −(Ch0 + Chα·α_t)/Chδ. When α_t increases, a negative Chα/Chδ ratio makes the elevator float trailing-edge up, which reduces the tail lift increment. The tail lift slope with free elevator becomes a_t·(1 − τ·Chα/Chδ).
Free-elevator factor. F = 1 − τ·Chα/Chδ, typically 0.7–0.9. It multiplies the tail's contribution to stability, so the stick-free moment slope is
Cm′α = a_w·(h − h_ac) − η·V_H·a_t·(1 − dε/dα)·F
and the stick-free neutral point is h_n′ = h_ac + η·V_H·(a_t/a)·(1 − dε/dα)·F. Since F < 1, h_n′ < h_n. The stick-free static margin is K_n′ = h_n′ − h.
Stick force. With a gearing G (radians of elevator per metre of stick travel), the stick force is F_s = G·η·q·S_e·c_e·Ch. Because Ch at trim depends on the stick-free margin, the gradient of stick force with speed about the trim speed is proportional to K_n′: a positive stick-free margin means the pilot must push to fly faster and pull to fly slower — good speed stability. The stick force is also proportional to the size of the surface (S_e·c_e ~ length³), so large aircraft need balance, tabs or powered controls.
Aerodynamic balance. Set-back hinges, overhang (nose) balance, horn balance and balance tabs all move part of the surface area ahead of the hinge or add a tab moment, reducing |Chδ| and |Chα|. Over-balancing makes Chδ positive — the surface would run away — so balance is limited. Reducing Chα toward zero makes F → 1 (stick-free ≈ stick-fixed). Mass balance (weight ahead of the hinge) is a different thing: it prevents flutter and does not change aerodynamic hinge moments.
Trim tabs. A tab deflected opposite to the required elevator produces a hinge moment that cancels Ch at trim, so stick force is zero. Tab setting changes the trim speed but not the stick-free neutral point.
Limits. Linear model, friction ignored; irreversible powered controls make stick-free stability a property of the feel system instead.
Formulas
Ch = H/(q·S_e·c_e)
- H: hinge moment (N·m); q: dynamic pressure at the surface (Pa); S_e: control-surface area aft of the hinge (m²); c_e: control-surface mean chord (m).
Ch = Ch0 + Chα·α_t + Chδ·δe + Chδt·δt
- Chα, Chδ, Chδt: hinge-moment derivatives (per rad or per degree, consistent with angles).
δe,free = −(Ch0 + Chα·α_t)/Chδ
F = 1 − τ·Chα/Chδ
- τ: elevator effectiveness dα_t/dδe.
h_n′ = h_ac + η·V_H·(a_t/a)·(1 − dε/dα)·F and h_n − h_n′ = η·V_H·(a_t/a)·(1 − dε/dα)·τ·Chα/Chδ
F_s = G·η·q·S_e·c_e·Ch
- G: stick gearing (rad/m); F_s: stick force (N).
Worked examples
Example 1 (standard). An elevator has S_e = 1.2 m², c_e = 0.3 m, Ch0 = 0, Chα = −0.003 per degree and Chδ = −0.006 per degree. At q = 2000 Pa (take η = 1) the tail angle of attack is 2° and the elevator is held at −3°. Find Ch, the hinge moment, the stick force for G = 1.5 rad/m, and the floating angle if the stick is released.
Ch = Chα·α_t + Chδ·δe= −0.003 × 2 + (−0.006)(−3) = −0.006 + 0.018 = +0.012.H = Ch·q·S_e·c_e= 0.012 × 2000 × 1.2 × 0.3 = 8.64 N·m (tending to push the trailing edge down).F_s = G·H= 1.5 × 8.64 = 12.96 N, the force the pilot must apply to hold the elevator up.δe,free = −Chα·α_t/Chδ= −(−0.006)/(−0.006) = −1.0°.
Answer: Ch = 0.012, H = 8.64 N·m, F_s ≈ 13.0 N, floating angle = −1.0° (trailing edge up).
Example 2 (GATE level). For an aircraft with h_ac = 0.25, η = 0.9, V_H = 0.5, a_t/a = 0.8, dε/dα = 0.4, τ = 0.5 and the hinge derivatives of Example 1, find the stick-fixed and stick-free neutral points and both static margins with the CG at 0.36c̄.
- Tail term: η·V_H·(a_t/a)·(1 − dε/dα) = 0.9 × 0.5 × 0.8 × 0.6 = 0.216.
h_n= 0.25 + 0.216 = 0.466.F = 1 − τ·Chα/Chδ= 1 − 0.5 × (−0.003/−0.006) = 1 − 0.25 = 0.75.h_n′= 0.25 + 0.216 × 0.75 = 0.25 + 0.162 = 0.412.- Margins: K_n = 0.466 − 0.36 = 0.106; K_n′ = 0.412 − 0.36 = 0.052.
Answer: h_n = 0.466c̄, h_n′ = 0.412c̄; K_n = 10.6 %, K_n′ = 5.2 %. Freeing the elevator moved the neutral point forward by 0.054c̄ and halved the margin.
Common mistakes
- Using wing area and wing chord in Ch; hinge moments use the control surface's own S_e and c_e.
- Assuming the stick-free neutral point is aft of the stick-fixed one. With the usual negative Chα/Chδ ratio it is forward.
- Confusing mass balance (anti-flutter) with aerodynamic balance (reduces hinge moment).
- Over-balancing: making Chδ positive gives control reversal of feel and an unstable surface.
- Thinking a trim tab moves the neutral point — it only changes the trim speed and zeroes the stick force.
- Dropping τ from the free-elevator factor.
For GATE AE
Expect a hinge-moment or stick-force numerical, the floating angle from Ch = 0, the free-elevator factor and stick-free neutral point, and conceptual questions on balance, tabs and why h_n′ < h_n. Practise sign bookkeeping for Chα, Chδ and δe.
Quick check
- On which area and chord is the hinge-moment coefficient based?
- Chα = −0.004, Chδ = −0.008 (per degree), α_t = 3°, Ch0 = 0. What is the floating angle?
- τ = 0.6, Chα/Chδ = 0.5. What is the free-elevator factor?
- Does a trim tab change the stick-free neutral point?
Answers: 1. The control surface's area and chord aft of the hinge. 2. −1.5° (trailing edge up). 3. 0.70. 4. No — it only changes the trim speed at which stick force is zero.
Interview questions
All Aircraft Stability and Control interview questionsTry answering each one aloud before you open it.
1.What is a hinge moment in the context of aircraft control surfaces?Concept
It is the aerodynamic moment of the pressure distribution on a control surface about its hinge line. It is written as Ch = H/(q·S_e·c_e) using the surface's own area and chord, and modelled as Ch = Ch0 + Chα·α + Chδ·δ. With reversible controls the pilot must balance it through the stick gearing, so it sets the stick force, and it decides where the surface floats when released.
2.Explain the concept of stick-free stability in aircraft.Concept
Stick-free stability is the static stability when the pilot releases the controls and the elevator floats to the angle where its hinge moment is zero. As angle of attack increases, a typical elevator floats trailing-edge up, which reduces the tail's lift increment by the free-elevator factor F = 1 − τ·Chα/Chδ. The stick-free neutral point therefore lies ahead of the stick-fixed one, and the stick-free margin governs the stick-force gradient with speed.
3.How does the hinge moment affect the pilot's control forces?Application
With reversible controls the stick force is F_s = G·η·q·S_e·c_e·Ch, where G is the gearing. Force grows with dynamic pressure and with the cube of surface size, so large or fast aircraft need balance, tabs or powered controls. Designers do not aim for zero force: they want forces and gradients that are light enough to fly comfortably but heavy enough to prevent over-control, within handling-quality limits.
4.Why must both stick-fixed and stick-free stability be checked?Application
Stick-fixed stability (elevator held) governs the elevator-angle-to-trim gradient, while stick-free stability (elevator floating) governs the stick-force gradient the pilot feels and the behaviour when controls are released. Because the floating elevator reduces tail effectiveness, the stick-free neutral point is further forward and is often the critical aft CG limit for aircraft with reversible controls. A design needs positive margins in both.
5.What happens to the hinge moment if the centre of pressure on a control surface moves aft?Application
The hinge moment is the surface's normal force times the distance from the hinge line to its centre of pressure, so moving the centre of pressure aft increases the moment arm and the hinge moment. This makes the surface heavier to move. Aerodynamic balance works the other way, putting some area ahead of the hinge so the centre of pressure moves closer to the hinge line.
6.Describe how aerodynamic balancing reduces hinge moments.Application
Aerodynamic balance places part of the surface's lift ahead of the hinge line, or adds an opposing tab force, so the net hinge moment falls. Methods include set-back hinges with overhang (nose) balance, horn balance at the tip, internal sealed balance and geared or balance tabs that deflect opposite to the surface. These reduce |Chδ| and |Chα|; too much balance makes Chδ positive and the surface would run away. Mass balance is different — it is used against flutter, not to reduce aerodynamic hinge moments.
7.Why might a designer use a horn balance on a control surface?Application
A horn balance is a portion of the surface ahead of the hinge line, usually at the tip, outside the fixed surface. Its lift acts ahead of the hinge and opposes the moment of the main surface, reducing stick force without changing the hinge line along the span. It is simple and effective on light aircraft, but the exposed horn can ice up and can cause buffet, so its size is limited.
8.The aerodynamic normal force on a control surface is 500 N, acting 0.3 m behind the hinge line. What is the hinge moment?Numerical
H = F × d = 500 × 0.3 = 150 N·m. Because the force acts behind the hinge, the moment tends to rotate the surface in the direction of the force, which the pilot (through the gearing) or the actuator must resist. Moving the hinge line aft (set-back hinge) would reduce this moment arm.
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