Longitudinal modes: short period and phugoid

Why it matters

Disturb a conventional aircraft in pitch and it responds with two superimposed oscillations: a quick, usually well-damped pitching motion and a slow, lightly damped exchange of speed and height. The first — the short period — decides how crisply the aircraft answers the stick; the second — the phugoid — decides how hard it is to hold speed and altitude by hand. Both are read straight from the longitudinal equations, and simple approximations give their frequency and damping in a few lines.

Key ideas

The characteristic equation. The linear longitudinal equations in Δu, Δw, Δq, Δθ give a quartic characteristic equation. For a conventional aircraft it factors into two pairs of complex roots, (s² + 2ζ_sp·ω_sp·s + ω_sp²)(s² + 2ζ_p·ω_p·s + ω_p²) = 0, the short-period and phugoid modes. Each mode is stable if its damping ratio is positive.

Short-period mode. A fast oscillation (period of a few seconds or less) mainly in angle of attack and pitch rate, at almost constant speed. The restoring "spring" is static stability (Mα); the damping comes from pitch damping Mq, downwash lag Mα̇ and the lift-curve slope (Zα, which makes the flight path follow the nose). It is usually well damped (ζ ≈ 0.3–0.8). Approximation: hold Δu = 0 and keep only the w (α) and q equations.

Phugoid mode. A slow oscillation (tens of seconds) in speed and height at nearly constant angle of attack. The aircraft trades kinetic for potential energy: as it climbs it slows, lift falls, it descends and speeds up, lift rises and it climbs again. Lanchester's approximation (constant α, no drag, thrust balancing drag) gives a frequency set only by speed, ω_p ≈ √2·g/u₀, and including drag gives damping ζ_p ≈ 1/(√2·L/D). Clean, efficient aircraft therefore have very lightly damped phugoids — easy for the pilot to correct because it is slow, but tiring on long flights without an autopilot.

Influence of design. Moving the CG aft reduces Mα and with it ω_sp; at the neutral point the short-period roots split into a real pair and one becomes unstable. A larger tail volume increases both Mα and Mq. Altitude reduces aerodynamic damping relative to inertia, so ζ_sp falls at high altitude — the reason many jets have pitch dampers. The phugoid period grows linearly with speed.

Why the approximations work. The two modes are well separated in frequency (often by a factor of 20–50), so during the short period the speed hardly changes, and during the phugoid the short-period dynamics have already settled (α constant).

Useful mode measures. Damped frequency ω_d = ω_n·√(1 − ζ²), period T = 2π/ω_d, time to half amplitude t_½ = 0.693/(ζ·ω_n), number of cycles to half amplitude N_½ = t_½/T.

Formulas

ω_sp ≈ √(Zα·Mq/u₀ − Mα) 2·ζ_sp·ω_sp ≈ −(Mq + Mα̇ + Zα/u₀)

• Zα (m/s²), Mα (s⁻²), Mq and Mα̇ (s⁻¹): dimensional derivatives (force or moment per unit mass or inertia); u₀: trim speed (m/s). Valid for conventional aircraft with well-separated modes.

ω_p ≈ √2·g/u₀, T_p ≈ π·√2·u₀/g, ζ_p ≈ 1/(√2·(L/D))

• g = 9.81 m/s²; L/D: lift-to-drag ratio. Lanchester approximation, low Mach number.

ω_d = ω_n·√(1 − ζ²), T = 2π/ω_d, t_½ = 0.693/(ζ·ω_n), N_½ = t_½/T

Worked examples

Example 1 (standard). An aircraft flies at u₀ = 100 m/s with L/D = 12. Estimate the phugoid natural frequency, period, damping ratio and time to half amplitude.

1. ω_p ≈ √2·g/u₀ = 1.4142 × 9.81/100 = 0.1387 rad/s.
2. T_p ≈ 2π/ω_p = 6.2832/0.1387 = 45.3 s (same as π√2·u₀/g).
3. ζ_p ≈ 1/(√2·L/D) = 1/(1.4142 × 12) = 0.0589.
4. t_½ = 0.693/(ζ·ω_n) = 0.693/(0.0589 × 0.1387) = 84.8 s.

Answer: ω_p ≈ 0.139 rad/s, T_p ≈ 45 s, ζ_p ≈ 0.059, t_½ ≈ 85 s — almost two full cycles before the amplitude halves.

Example 2 (GATE level). At u₀ = 100 m/s an aircraft has Zα = −250 m/s², Mα = −12 s⁻², Mα̇ = −0.8 s⁻¹ and Mq = −2.5 s⁻¹. Using the short-period approximation, find ω_sp, ζ_sp, the damped period and the time to half amplitude.

1. ω_sp² = Zα·Mq/u₀ − Mα = (−250)(−2.5)/100 − (−12) = 6.25 + 12 = 18.25 s⁻², so ω_sp = 4.272 rad/s.
2. 2·ζ·ω = −(Mq + Mα̇ + Zα/u₀) = −(−2.5 − 0.8 − 2.5) = 5.8 s⁻¹.
3. ζ_sp = 5.8/(2 × 4.272) = 0.679.
4. ω_d = 4.272 × √(1 − 0.679²) = 4.272 × 0.7342 = 3.137 rad/s; T = 2π/3.137 = 2.00 s.
5. t_½ = 0.693/(ζ·ω_n) = 0.693/2.9 = 0.239 s.

Answer: ω_sp = 4.27 rad/s, ζ_sp = 0.68, period ≈ 2.0 s, t_½ ≈ 0.24 s. The oscillation is essentially dead within one cycle.

Common mistakes

• Swapping the modes' characters: the short period is fast and well damped, the phugoid slow and lightly damped.
• Thinking the phugoid involves large angle-of-attack changes — α stays almost constant; speed and height change.
• Mixing non-dimensional (Cmα, Cmq) and dimensional (Mα, Mq) derivatives in the approximations.
• Forgetting the Zα·Mq/u₀ term in ω_sp² — it can be a third or more of the total.
• Using T = 2π/ω_n when damping is large; use the damped frequency.
• Assuming the phugoid frequency depends on aircraft size — in Lanchester's model it depends only on speed.

For GATE AE

Expect conceptual MCQs contrasting the two modes (variables involved, period, damping), Lanchester phugoid period or damping numericals, short-period frequency and damping from given dimensional derivatives, and the effect of CG, altitude or L/D on the modes. Practise ζ, ω_n, ω_d, period and t_½ conversions.

Quick check

1. Which state variables dominate the phugoid?
2. Estimate the phugoid period at 200 m/s.
3. L/D = 15. Estimate ζ_p.
4. What happens to the short period as the CG approaches the neutral point?

Answers: 1. Speed and height (pitch attitude), with α nearly constant. 2. π√2 × 200/9.81 ≈ 90.6 s. 3. 1/(1.414 × 15) ≈ 0.047. 4. ω_sp falls toward zero; aft of the neutral point one root becomes unstable.