Static lateral stability and dihedral effect

Why it matters

When a wing drops in a gust, the aircraft starts to slip sideways toward the low wing. Whether it then rolls back toward wings-level depends on the rolling moment due to sideslip — the dihedral effect, Clβ. It is set by geometric dihedral, sweep, wing position on the fuselage and the fin, and it has to be balanced against directional stability: too little and the spiral mode diverges quickly, too much and the Dutch roll becomes unpleasant and roll response to rudder excessive.

Key ideas

Conventions. Rolling moment L is positive right wing down, with coefficient Cl = L/(q·S·b). Sideslip β is positive with the wind from the right (v > 0).

Why sideslip is the trigger. A bank angle by itself produces no aerodynamic rolling moment. But when the aircraft is banked right, the lift vector tilts and gravity makes it slip to the right, so β > 0. A stable aircraft responds to that sideslip with a rolling moment to the left, raising the low wing. Hence the criterion for static lateral stability (positive dihedral effect) is Clβ < 0.

Geometric dihedral Γ. With the wing tips raised by Γ, a sideslip β gives the leading (into-wind) wing an upward component of relative velocity V·β·Γ, raising its angle of attack by about β·Γ, while the trailing wing loses the same amount. The leading wing lifts more and the aircraft rolls away from the sideslip. Strip theory gives, for a straight tapered wing, Clβ,Γ = −(a·Γ/6)·(1 + 2λ)/(1 + λ), which reduces to −a·Γ/4 for a rectangular wing.

Sweepback. In sideslip the leading swept wing has a smaller effective sweep and its normal velocity component rises, so it carries more lift; the trailing wing carries less. The effect is stabilising (negative Clβ) and proportional to CL — strongest at low speed. Swept transports often need little or even negative geometric dihedral (anhedral) for this reason.

Wing position. Cross-flow around the fuselage in sideslip raises the local angle of attack near the root of a high wing on the into-wind side and lowers it for a low wing. A high wing is therefore stabilising (worth roughly a few degrees of extra dihedral) and a low wing destabilising, which is why low-wing aircraft have visible dihedral and high-wing ones have little. Values come from data charts. (The "pendulum effect" of a low CG under a high wing is not the cause — the CG cannot produce a moment about itself.)

Vertical tail. The fin's side force in sideslip acts a height z_v above the roll axis, so a tall fin adds a negative (stabilising) Clβ. Ventral fins below the axis do the opposite.

Balance with directional stability. The lateral–directional modes depend on the ratio of Clβ to Cnβ. Large dihedral effect with weak Cnβ gives a lightly damped Dutch roll; strong Cnβ with weak Clβ gives spiral divergence. Designers aim for a compromise, typically with a mildly unstable or neutral spiral.

Assumptions. Linear, small β; the strip-theory dihedral formula ignores tip losses and fuselage interference.

Formulas

Cl = L/(q·S·b)

• L: rolling moment (N·m); q (Pa); S (m²); b: span (m).

Clβ < 0 for positive dihedral effect (static lateral stability).

Clβ,Γ = −(a·Γ/6)·(1 + 2λ)/(1 + λ); rectangular wing: Clβ,Γ = −a·Γ/4

• a: wing section/wing lift-curve slope (per rad); Γ: dihedral angle (rad); λ: taper ratio (tip chord/root chord).

Clβ,v = −η_v·(1 + dσ/dβ)·a_v·S_v·z_v/(S·b)

• a_v: fin lift-curve slope (per rad); S_v: fin area (m²); z_v: height of fin aerodynamic centre above the roll axis (m).

ΔCl = Clβ·β

Worked examples

Example 1 (standard). A rectangular wing has a = 5.0 per rad and Γ = 4°. The aircraft sideslips at β = 5° with q = 2000 Pa, S = 20 m², b = 10 m. Find Clβ from dihedral, the rolling-moment coefficient and the rolling moment.

1. Γ = 4° = 0.06981 rad, β = 5° = 0.08727 rad.
2. Clβ = −a·Γ/4 = −5.0 × 0.06981/4 = −0.0873 per rad.
3. ΔCl = Clβ·β = −0.0873 × 0.08727 = −0.00762.
4. L = Cl·q·S·b = −0.00762 × 2000 × 20 × 10 = −3046 N·m.

Answer: Clβ = −0.0873 per rad; Cl = −0.0076; L ≈ −3.05 kN·m (left wing down, rolling away from the sideslip).

Example 2 (GATE level). A low-wing aircraft has a tapered wing (λ = 0.5, a = 4.8 per rad, Γ = 3°), S = 25 m², b = 12 m, and a fin with S_v = 3 m², a_v = 3.0 per rad, η_v·(1 + dσ/dβ) = 0.9, whose aerodynamic centre is 1.2 m above the roll axis. A data chart gives the low-wing fuselage interference as ΔClβ = +0.020 per rad. Find the total Clβ.

1. Γ = 3° = 0.05236 rad.
2. Wing: Clβ,Γ = −(a·Γ/6)·(1 + 2λ)/(1 + λ) = −(4.8 × 0.05236/6) × (2.0/1.5) = −0.04189 × 1.3333 = −0.0559 per rad.
3. Fin: Clβ,v = −η_v(1 + dσ/dβ)·a_v·S_v·z_v/(S·b) = −0.9 × 3.0 × 3 × 1.2/(25 × 12) = −9.72/300 = −0.0324 per rad.
4. Total: −0.0559 − 0.0324 + 0.0200 = −0.0683 per rad.

Answer: Clβ ≈ −0.068 per rad — stable; the low-wing position cancels about a quarter of the wing-plus-fin dihedral effect.

Common mistakes

• Sign: positive dihedral effect means Clβ NEGATIVE.
• Thinking bank angle itself creates a restoring roll. It is the sideslip that follows the bank.
• Using degrees for Γ or β in the strip-theory formula.
• Basing Cl on c̄ instead of span b.
• Crediting a high wing to the "pendulum effect" rather than fuselage cross-flow.
• Assuming more dihedral is always better — it worsens Dutch roll and gust response.

For GATE AE

Expect sign questions on Clβ, qualitative effects of dihedral, sweep, wing position and fin height, and short numericals on rolling-moment coefficient from Clβ and β or Clβ from dihedral angle by strip theory. Remember the trade-off with Cnβ for the spiral and Dutch-roll modes.

Quick check

1. What sign of Clβ gives static lateral stability?
2. Rectangular wing, a = 6 per rad, Γ = 2°. Find Clβ.
3. Is a high wing stabilising or destabilising in roll?
4. Does sweepback's dihedral effect grow or fall with CL?

Answers: 1. Negative. 2. −6 × 0.0349/4 = −0.0524 per rad. 3. Stabilising. 4. It grows with CL.