Equilibrium of the AMAT10

The AMAT10 has a rectangular main wing with span bm = 3.6 m and constant chord cxm = 0.35 m. The tail is also defined with a 33% moving flap. The equilibrium code provides the aircraft characteristics and a maximum take-off mass M = 24 kg.

14.8.3.1 Airplane Aerodynamic Center and Static Margin

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

CL(a, tf) = 4.479a + 0.808tf + 0.9314
Cm, o (a, tf) = -1.469a – 0.7479tf – 0.1565

where a is the geometric incidence (in radians, measured from the fuselage axis) and tf is the tail flap setting angle (in radians). Cm, o is the aerodynamic moment about the origin of the coordinate system (located at the nose O). We will use this linear model for simplicity.

The center of gravity is located at xcg/lref = 0.268.

Find the aerodynamic center and the static margin SM in % of lref.

14.8.3.2 Equilibrium Condition and Static Stability

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 a(tf).

14.8.3.3 Take-Off Conditions

The take-off speed of U = 11.49 m/s is obtained for tf = 7.13° .Find aeq at take-off.

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

CLt = 3.032a + 2.886tf – 0.3259

Calculate the force on the tail in (N), given that the tail reference area is At = 0.49 m2, p = 1.2kg/m3.

Is the force up or down?

14.9 Problem 9

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