Equilibrium of the Glider

The AMAT12 has a rectangular main wing with span bm — 2.1m and constant chord cxm — 0.3 m. The tail is also rectangular with span bt — 1.0 m and chord cxt — 0.338 m. The equilibrium code provides the aircraft aerodynamic characteristics and a maximum take-off mass M — 18 kg. The reference area is Aref — Am + At. The reference length is lref — 2.12 m. The engine is turned off. The global aerodynamic
lift and nose pitching moment coefficients are found to be

CL(a, tt) = 3.853a + 0.902tt + 0.716
См, о(а, tt) = -1.336a – 0.894tt – 0.06

where a represents the geometric incidence and tt the tail setting angle in rd.

15.10.3.1 Equilibrium About the Center of Gravity

The pitching moment about the center of gravity CM, c.g.(a, tt) is given by

Xc

CM, c.g.(a, tt) = См, о (a, tt) + Cl (a, tt)

l-ref

Cm, c.g.(a, tt) = -1.336a – 0.894tt – 0.06 + 0.2868 (3.853a + 0.902tt + 0.716)

Cm, c.g.(a, tt) = -0.2367a – 0.6367tt + 0.1442

The airplane is statically stable since the slope dCm, c.g./da of the moment curve Cm, c.g.(a, tt) is negative.

The aerodynamic center of the glider is at xa. c. = 0.735 m. The static margin SM in %

SM = 100 ( — – —) = 100 (0.3467 – 0.2868) = 6%

lref I ref )

15.10.3.2 Equilibrium Incidence

The equilibrium condition is given by CM, c.g.(aeq, tt) = 0 and can be solved for

aeq(tt)

aeq(tt) = —2.69tt + 0.609

15.10.3.3 Trimming for Maximum Distance

The equilibrium code calculates the tail setting angle for the maximum CL /CD = 9.3 to be tt = 12° = 0.2094 rd.

Substitution of the value of tt in the equilibrium equation gives aeq = -2.69 0.2094 + 0.609 = 0.0457 rd = 2.6°.

The corresponding values of CL and CD, as well as the descent angle в are: Cl = 1.08, Cd = 0.116 and в = – Cd/Cl = -0.107rd = -6.15°.

The lift forces (in N) on the main wing and the tail, given that the velocity is U = 17.3m/s and the air density p = 1.1214 kg/m3 are given by

1 2 1 2
Lm — 2pU AmCLm7 Lt — ^pU AtCLt

The wing and tail lift curves are given respectively by

CLm(a) — 4.664a + 1.312 — 1.525
Cu (a, tt) — 2.342a + 2.583tt – 0.396 — 0.25

The results are: Lm — 161.2 N, Lt — 14.18 N.

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