Tools › Aerodynamics › Stability and Control › Longitudinal Static Stability › Example

Light airplane: stick-fixed neutral point, static margin, trim and the NACA Report 711 elevator-gradient criterion

Tool	Longitudinal Static Stability
Source	Gilruth, R. R., and White, M. D. (1941). Analysis and Prediction of Longitudinal Stability of Airplanes. NACA Report 711, National Advisory Committee for Aeronautics, Washington, DC. Equation (1) and the conventions of its Figure 2; Nelson, R. C. (1998). Flight Stability and Automatic Control, 2nd ed. McGraw-Hill, Boston. Chapter 2 (the component build-up and the approximate neutral-point formula, eq. 2.37).
Level	Undergraduate

Problem

A light airplane has a wing of 160 ft² and 4.5 ft mean aerodynamic chord (lift-curve slope 4.5 per radian, zero-lift angle −2°, incidence 1°, C_m,ac −0.05, aerodynamic centre at 25 % of the chord), a fuselage whose pitching-moment slope is 0.12 per radian, and a horizontal tail of 32 ft² whose aerodynamic centre is 15 ft behind the wing's (slope 3.5 per radian, efficiency 0.9, incidence −2°, elevator effectiveness 0.5, downwash gradient 0.35). With the centre of gravity at 25 % of the chord, find the lift-curve slope, C_m0 and C_mα, the stick-fixed neutral point and static margin, the angle of attack and elevator angle to trim 2,400 lb at 120 knots at sea level, the tail load, the elevator gradient dδ_e/dα that Report 711 uses as its stability criterion, and the c.g. range for a 5 % static margin and 25° of up elevator at C_Lmax = 1.6.

Inputs to Enter

Wing: S 160 ft², c̄ 4.5 ft, AR 7.4, a_w 4.5/rad, α_L0 −2°, i_w 1°, C_m,ac −0.05, h_ac 0.25; fuselage entered C_mα,f 0.12/rad, C_m0,f 0
Tail: S_t 32 ft², l_t 15 ft (a.c. to a.c.), a_t 3.5/rad, η_t 0.9, i_t −2°, τ 0.5, dε/dα 0.35
c.g. 0.25, W 2,400 lb, 120 KTAS, sea level, C_Lmax 1.6, up-elevator limit 25°, minimum static margin 0.05

Open the Longitudinal Static Stability tool (loaded with this example)

Answers

Tail volume V_H = 32 × 15/(160 × 4.5) = 0.6667; a = 4.5 + 0.9 × 0.2 × 3.5 × 0.65 = 4.9095/rad; C_L0 = 0.2021; C_Lδ = 0.315/rad
C_mα = 0 + 0.12 − 0.9 × 0.6667 × 3.5 × 0.65 = −1.245/rad; C_m0 = −0.05 + 0 + 0.1118 = 0.0618; C_mδ = −1.05/rad
Neutral point exactly h_np = 0.25 + (1.365 − 0.12)/4.9095 = 0.5036 (Nelson's approximation 0.5267); static margin 25.4 % of the chord; dC_m/dC_L = −0.2536
q = 48.75 lb/ft², C_L,trim = 0.3077; trim α = 1.100°, δ_e = +2.068° (trailing edge down); tail C_Lt = −0.0795, tail load −111.6 lb (download), wing lift 2,512 lb
dδ_e/dα = −C_mα/C_mδ = 1.186 with the elevator positive up (Report 711 eq. 1; the report suggests 0.5 for design), dδ_e/dC_L = −0.2614
Aft c.g. limit 0.4536 (SM 5 %), forward limit 0.1684 (elevator −25° at C_Lmax); range 28.5 % of the chord

Notes

The tool's default inputs are this airplane; press Submit. The case is the brief's light airplane with an illustrative fuselage slope, checked against an independent numpy solution of the same moment balance (References/Static-Stability/checks/stab_check.py) and against the closed forms in test_static_stability.php; Report 711 prints no neutral-point example of its own, its equation (1) is reproduced in the report's symbols and shown to equal −C_mα/C_mδ.