Equilibrium Condition and Static Stability

The moment coefficient at the center of gravity, CM, c.g.(a, tf) is given by the equation CM, c.g.(a) = Cm, o (a) + Cl (a)

lref

Substitution of the linear model yields

CM, c.g.(a) = -1.469a -0.7479tf -0.1565 + 0.268 (4.479a + 0.808tf + 0.9314)

CM, c.g.(a) = -0.2686a – 0.53141/ + 0.0931

The equilibrium is stable since dCM, c.g./da < 0.

The condition for equilibrium is CM, c.g.(aeq) = 0. Solving for aeq gives

aeq (tf) = -1.97841/ + 0.3466

15.8.3.2 Take-Off Conditions

The take-off speed of U = 11.49m/s is obtained for t/ = 7.13° = 0.1244rd. Hence, the incidence at take-off is aeq = 0.1rd = 5.75°.

The tail aerodynamic lift curve is given by

Cli = 3.032a + 2.8861/ – 0.3259

The lift coefficient of the tail at take-off is CLt = 0.336. The force on the tail, given that the tail reference area is At = 0.49 m2, and p = 1.2 kg/m3 will be

1 2

Lt = 2 pU2 AtCLt = 13.0 N The force is up. The tail is lifting.

15.9 Solution to Problem 9

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