# 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.

14.10.3.1 Equilibrium About the Center of Gravity

Calculate the pitching moment about the center of gravity CM, c.g. (a), given that the center of gravity is located at xc. g./lref = 0.2868. (Hint: use the change of moment formula). Verify your result as the rest depends on it.

Is the airplane statically stable?

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

14.10.3.2 Equilibrium Incidence

Write the equilibrium condition and solve the equation for aeq (tt). Check your algebra by substituting back into the equation.

14.10.3.3 Trimming for Maximum Distance

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

Find the corresponding values of CL and CD, as well as the descent angle в – Calculate 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 = 1.1214kg/m3. The wing and tail lift curves are given respectively by

CLm(a) = 4.664a + 1.312

CLt(a, tt) = 2.342a + 2.583tt – 0.396

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