Equilibrium of the AMAT11

The AMAT11 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 = 0.3 m. The equilibrium code provides the aircraft aerodynamic characteristics and a maximum take-off mass M = 19 kg. The reference area is Aref = Am + At.

14.9.3.1

Equilibrium of the AMAT11

Aspect Ratios of Lifting Elements—Global Lift Slope

give the global lift slope dCL /da for the wing+tail configuration.

14.9.3.2 Airplane Center of Gravity

The aerodynamic center is located at xa. c./lref = 0.3.

Find the center of gravity xc. g./ lref, given a 6 % static margin (SM).

14.9.3.3 Equilibrium Condition and Static Stability

The equilibrium code calculates the linear model for lift and moment coefficients for the complete configuration, at low incidences, to be:

CL(a, tf) = 3.955a + 0.984tf + 0.712
CM, o(a, tf) = -1.188a – 0.907tf – 0.008

Derive the moment coefficient at the center of gravity, CM, c.g.(a, tf) and write the condition for equilibrium. Verify your result as the rest depends on it. Is the equilibrium stable?

Solve for aeq (tf).

14.9.3.4 Take-Off Conditions

The take-off speed of U = 13.11 m/s is obtained for tf = 9.2°. Find aeq at take-off.

Find the lift coefficient of the tail at take-off : the tail aerodynamic lift curve is given by

CLt = 2.696a + 2.705tf – 0.347

Calculate the force on the tail in (N), given that p = 1.2 kg/m3. Is the force up or down?

14.10 Problem 10

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