Equilibrium of the AMAT09

15.7.3.1 Airplane Aerodynamic Center and Static Margin

The aerodynamic center is given by

1.124

= = 0.287 = 28.7/100

C 3.912 ‘

da

The static margin SM is

SM = Xac – Xcg = 0.287 – 0.227 = 0.06 = 6/100

lref lref

15.7.3.2 Equilibrium Condition and Static Stability

The moment coefficient at the center of gravity, CM, c.g.(a, tt) is given by

Xc

См,с. д.(а, tt) = Cm, o (a, tt) + Cl (a, tt)

lref

CM, c.g.(a, tt) = -1.124a – 0.625tt – 0.138 + 0.227(3.912a + 0.626tt + 0.844) One finds

CM, c.g.(a, tt) = -0.236a – 0.483tt + 0.054

The slope of the CM, c.g. is negative. The equilibrium is stable.

At equilibrium CM, c.g. = 0, hence

a. eq (tt) = —2.064tt + 0.229

15.7.3.3 Top Speed

The top speed is obtained for tt = 7.6° = 0.1326 rd. The corresponding value of

aeq is

aeq = -2.060.1326 + 0.229 = -0.0448rd = -2.6°

The top speed, given that Aref = 2.225 m2, p = 1.2kg/m3 is such that the lift balances the weight

12

2 pUeqArefC L, eq — Mg

But CLeq = 3.912(-0.045) + 0.626(0.1326) + 0.844 = 0.751 One finds

15.8 Solution to Problem 8