Equilibrium of the AMAT09

The best design for the AMAT09 has a rectangular main wing with span bm = 3.1m and constant chord cxm = 0.55 m. The lifting tail is also defined with a 33% moving flap. The equilibrium code provides the aircraft characteristics and a maximum take­off mass M = 30 kg.

14.7.3.1 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, tt) = 3.912a + 0.626tt + 0.844
CM, o(a, tt) = -1.124a – 0.625tt – 0.138

where a is the geometric incidence (in radians, measured from the fuselage axis) and tt is the tail 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, even for take-off conditions.

The center of gravity is located at xcg / lref = 0.227. Find the aerodynamic center and the static margin SM in % of lref.

14.7.3.2 Equilibrium Condition and Static Stability

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

Solve for a(tt).

14.7.3.3 Top Speed

The top speed is obtained for tt = 7.6°. Find aeq at top speed.

Find the top speed, given that Aref = 2.225 m2, p = 1.2kg/m3 (Hint: use the equilibrium equation for horizontal flight).

14.7 Problem 8

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