Elevator angle per g and manoeuvre point

Elevator angle per g in a steady pull-up, the pitch-damping contribution, the stick-fixed manoeuvre point h_m = h_n − Cmq/(2μ) and manoeuvre margin, with speed and altitude effects.

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Why it matters

Trimmed level flight is only half of longitudinal control; the pilot also pulls g — in a pull-up, a turn or a flare. The elevator deflection (and stick force) needed per g of extra load factor tells you how "heavy" the aircraft feels in manoeuvres. If it falls too low the pilot can over-stress the airframe with a small input; if it reaches zero the aircraft is neutrally stable in manoeuvres. The CG at which that happens, the manoeuvre point, is another aft CG limit.

Key ideas

Steady pull-up. Consider a symmetric pull-up through the bottom of a loop at speed V with load factor n = L/W. The lift must rise from W to nW, so ΔCL = (n − 1)·C_W, where C_W = W/(q·S) is the weight coefficient (equal to the trim CL in level flight). The flight path curves upward, so the aircraft pitches at a steady rate q = g·(n − 1)/V.

Pitch-damping contribution. Because the aircraft is rotating, the tail sees an extra upward relative wind of q·l_t, which increases its angle of attack by q·l_t/V and produces an extra nose-down moment. This is the pitch-damping derivative Cmq (negative), written with the non-dimensional rate q̂ = q·c̄/(2V). The elevator must overcome both the static stability (extra α) and this damping moment, so more elevator is needed in a pull-up than the static margin alone would suggest.

Elevator angle per g. Solving the lift and moment equations for the increments (neglecting CLq and CLδe): Δα = (n − 1)·C_W/CLα and Cmα·Δα + Cmq·q̂ + Cmδe·Δδe = 0. With Cmα/CLα = h − h_n and q̂ = (n − 1)·C_W/(2μ), where μ = 2m/(ρ·S·c̄) is the longitudinal relative density, this gives Δδe/Δn = −(C_W/Cmδe)·(h − h_m).

Stick-fixed manoeuvre point. h_m = h_n − Cmq/(2μ). Since Cmq < 0, h_m lies aft of the neutral point. The manoeuvre margin is h_m − h = K_n − Cmq/(2μ). Elevator angle per g is proportional to it and becomes zero when the CG reaches h_m. Because μ grows with altitude (ρ falls), the damping term shrinks and the manoeuvre point moves toward the neutral point at high altitude.

Stick-free. With the elevator floating the same argument with the free-elevator factor gives a stick-free manoeuvre point h_m′ < h_m and the stick force per g, dF_s/dn, proportional to h_m′ − h. Handling-qualities rules set minimum and maximum stick force per g; the minimum usually fixes the aft CG limit of fighters and aerobatic aircraft.

Dependence on speed and weight. Δδe/Δn is proportional to C_W = W/(qS), so it falls as 1/V²: at high speed a small elevator movement gives a large g. Stick force per g, by contrast, is roughly independent of speed in the simple theory, which is why it is the preferred handling parameter.

Assumptions. Steady symmetric pull-up, linear aerodynamics, CLq and CLδe neglected, rigid aircraft. A level turn gives a slightly different pitch rate, q = (g/V)(n − 1/n), so its numbers differ a little.

Formulas

C_W = W/(q·S), q_rate = g·(n − 1)/V, q̂ = q_rate·c̄/(2V)

  • W: weight (N); q = ½ρV² (Pa); S (m²); n: load factor; q_rate: pitch rate (rad/s); c̄ (m).

μ = 2m/(ρ·S·c̄)

  • m: mass (kg); ρ: air density (kg/m³). Dimensionless.

h_m = h_n − Cmq/(2μ)

  • Cmq: pitch-damping derivative (per rad, based on q̂); h_m, h_n as fractions of c̄.

Δδe/Δn = −(C_W/Cmδe)·(h − h_m)

  • Δδe/Δn: elevator angle per g (rad per g); negative = trailing-edge-up per g.

Manoeuvre margin = h_m − h = K_n − Cmq/(2μ)

Worked examples

Example 1 (standard). m = 5000 kg, S = 25 m², c̄ = 1.8 m, ρ = 1.0 kg/m³, V = 100 m/s, h_n = 0.40, h = 0.30, Cmq = −15 per rad, Cmδe = −1.0 per rad, g = 9.81 m/s². Find μ, h_m, C_W and the elevator angle per g.

  1. μ = 2m/(ρ·S·c̄) = 2 × 5000/(1.0 × 25 × 1.8) = 10 000/45 = 222.2.
  2. h_m = h_n − Cmq/(2μ) = 0.40 − (−15)/(444.4) = 0.40 + 0.03375 = 0.4338.
  3. C_W = W/(q·S) = 49 050/(0.5 × 1.0 × 100² × 25) = 49 050/125 000 = 0.3924.
  4. Δδe/Δn = −(C_W/Cmδe)·(h − h_m) = −(0.3924/−1.0) × (0.30 − 0.4338) = 0.3924 × (−0.1338) = −0.0525 rad per g.
  5. In degrees: −0.0525 × 57.30 = −3.01° per g.

Answer: μ = 222, h_m = 0.434c̄, elevator angle per g ≈ −3.0° per g (trailing edge up).

Example 2 (GATE level). The same aircraft flies at 150 m/s. (a) Find the elevator angle per g and the extra elevator for a 4 g pull-up from level flight with the CG at 0.30c̄. (b) A design rule requires at least 1.0° of elevator per g at this speed. Find the aft CG limit.

(a)

  1. C_W = 49 050/(0.5 × 1.0 × 150² × 25) = 49 050/281 250 = 0.1744.
  2. Δδe/Δn = −(0.1744/−1.0) × (0.30 − 0.4338) = −0.02333 rad per g = −1.34° per g.
  3. For n = 4, Δn = 3: Δδe = 3 × (−1.34) = −4.01°.
  4. Pitch rate during the pull-up: q = 9.81 × 3/150 = 0.196 rad/s.

(b)

  1. Require |Δδe/Δn| = 1.0° = 0.01745 rad per g, so (h_m − h) = 0.01745 × |Cmδe|/C_W = 0.01745/0.1744 = 0.1001.
  2. h = 0.4338 − 0.1001 = 0.3337.

Answer: (a) −1.34° per g; about −4.0° extra elevator for 4 g; (b) aft CG limit ≈ 0.334c̄. Note that the elevator per g at 150 m/s is (100/150)² = 0.444 times its value at 100 m/s.

Common mistakes

  • Using n instead of n − 1: from level flight the increment in lift is (n − 1)W.
  • Leaving out the pitch-damping term — then the manoeuvre point is wrongly taken equal to the neutral point.
  • Putting h_m ahead of h_n. With negative Cmq it is always aft.
  • Forgetting that μ uses mass (kg), not weight, and depends on density (altitude).
  • Expecting elevator angle per g to be constant with speed; it falls as 1/V².
  • Mixing the pull-up pitch rate g(n − 1)/V with the level-turn rate (g/V)(n − 1/n).

For GATE AE

Questions ask for the manoeuvre point from h_n, Cmq and μ, the elevator angle per g, its speed dependence, the pitch rate in a pull-up, and conceptual comparison of the neutral point and manoeuvre point (stick fixed and free). Practise computing μ and C_W with consistent SI units.

Quick check

  1. Is the stick-fixed manoeuvre point ahead of or behind the stick-fixed neutral point?
  2. μ = 200 and Cmq = −20 per rad. By how much does h_m exceed h_n?
  3. What is the pitch rate in a 3 g pull-up at 120 m/s?
  4. If speed doubles, how does the elevator angle per g change?

Answers: 1. Behind (aft). 2. 20/400 = 0.05c̄. 3. 9.81 × 2/120 = 0.1635 rad/s. 4. It falls to one quarter.

Interview questions

All Aircraft Stability and Control interview questions

Try answering each one aloud before you open it.

  1. 1.What is the elevator angle per g?Concept

    It is the additional elevator deflection needed per unit increase in load factor, dδe/dn, in a steady manoeuvre such as a pull-up. In simple theory Δδe/Δn = −(C_W/Cmδe)(h − h_m), so it is proportional to the manoeuvre margin and to the weight coefficient W/(qS). It is negative (trailing-edge up) for a stable aircraft and indicates how sensitive the aircraft is in manoeuvres.

  2. 2.Explain the concept of the manoeuvre point.Concept

    The stick-fixed manoeuvre point h_m is the CG position at which the elevator angle per g becomes zero, i.e. the aircraft is neutrally stable in manoeuvring flight. It equals h_n − Cmq/(2μ); since pitch damping Cmq is negative, it lies aft of the neutral point. The distance h_m − h is the manoeuvre margin, and a stick-free manoeuvre point is defined similarly from stick force per g.

  3. 3.How does the elevator angle (or stick force) per g affect handling qualities?Application

    It sets how much control input produces a given load factor. Too small a value means a small stick movement gives a large g, so the pilot can easily over-stress the airframe or induce pilot-induced oscillations; too large a value makes the aircraft heavy and tiring to manoeuvre. Handling requirements therefore specify upper and lower limits, usually on stick force per g, which in turn fix CG limits.

  4. 4.Why is it important to know the manoeuvre point of an aircraft?Application

    It defines an aft CG limit for manoeuvring flight: as the CG approaches h_m the elevator and stick force per g fall to zero and the aircraft becomes very sensitive in pitch. For fighters and aerobatic aircraft, the minimum stick-force-per-g requirement, linked to the stick-free manoeuvre point, is often more restrictive than the static neutral point. It is also used in flight test as an extrapolation target from measured elevator-per-g data.

  5. 5.What happens if the centre of gravity is behind the manoeuvre point?Application

    The manoeuvre margin is negative, so the elevator angle per g reverses sign: once a pull-up starts, the aircraft tends to tighten it, and the pilot must push to stop the load factor increasing. This manoeuvre instability makes over-stressing very easy and is unacceptable without a stability-augmentation system.

  6. 6.How is the elevator angle per g measured in flight test?Application

    The aircraft is flown in steady wind-up turns or pull-ups at constant speed and altitude, recording elevator angle (and stick force) against load factor from an accelerometer. The slope of the line gives dδe/dn. Repeating at two or more CG positions and extrapolating the slope to zero gives the manoeuvre point; stick-force data give the stick-free manoeuvre point.

  7. 7.An aircraft trimmed in level flight needs 5° of additional up-elevator to pull 2 g. What is its elevator angle per g?Numerical

    From level flight at 1 g, pulling 2 g is an increase of Δn = 1, so the elevator angle per g is 5°/1 = 5° per g (trailing-edge up). A common mistake is to divide by n = 2 instead of n − 1. The value would fall to a quarter if the same manoeuvre were flown at twice the speed, since it scales with W/(qS).

  8. 8.The manoeuvre point lies some distance behind the CG. What does this imply?Application

    The manoeuvre margin h_m − h is positive, so the aircraft is stable in manoeuvres: more load factor needs more up-elevator and pull force. The size of the margin (as a fraction of c̄) sets the elevator and stick force per g. Because pitch damping adds to static margin, the manoeuvre margin is always larger than the static margin for the same CG.

  9. 9.What factors influence the elevator angle per g?Application

    From Δδe/Δn = −(C_W/Cmδe)(h − h_m): CG position relative to the manoeuvre point, elevator control power Cmδe, and the weight coefficient C_W = W/(qS), so it falls with the square of speed and rises with wing loading. The manoeuvre point itself depends on the static neutral point and on pitch damping Cmq through μ = 2m/(ρSc̄), so it moves toward h_n at high altitude.

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