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
The moment coefficient at the center of gravity, CM, c.g.(a, tf) is CMcg (a, tf) = CMo (a, tf) + XcgCL(a, tf) = -0.239a – 0.671tf + 0.163
The condition for equilibrium is that the moment at the center of gravity vanishes
CM, c.g.(aeqj tf ) = 0
The equilibrium is stable because dCM, c.g./da < 0.
Solving for the equilibrium incidence gives
aeq(tf) = —2.808tf + 0.682
22.214.171.124 Take-Off Conditions
The take-off speed of U = 13.11 m/s is obtained for tf = 9.2°. aeq at take-off is aeq = 0.231 rd = 13.2°.
The lift coefficient of the tail at take-off, given that
Cl = 2.696a + 2.705tf – 0.347
is Cu(aeq) = 0.711.
The force on the tail in (N) is Lt = 0.5 pU2AtCLt = 22 N.
The force is up.
15.10 Solution to Problem 10