Airplane Longitudinal Equilibrium Global Coefficients

Aerobrick 2003 has the following lift and moment coefficients in terms of the geo­metric angle of attack a (rd) and tail setting angle tt (rd):

CL(a, tt) = 3.88a + 0.5 + 0.481tt
CM, o(a, tt) = -1.31a – 0.124 – 0.452tt

dCM, o

The aerodynamic center is given by ^ Щ = 0.338.

re ~da ‘

To satisfy the 4 % static margin, the center of gravity must be located in front of the aerodynamic center with

Tcg = ^ – 0,04 = 0,298

lref lref

Definition: the aerodynamic center is the point about which the moment is inde­pendent of the incidence angle.

It is clear from the above calculation of that CM o (a) + racCL (a) is indepen-

lref ’ lref

dent of a and reads

Cm, ac(tt) = 0.045 – 0.289ft Take-Off Conditions

The moment at the center of gravity is given by

Cm, cg(a, tt) = См, о(а, tt) + Cl (a, tt) = —0.154a + 0.025 — 0.309tt


At equilibrium the moment is zero, hence, solving for the equilibrium incidence:

aeq (tt) — 0.162 2.0tt

The equilibrium lift is then

CLeq(tt) = 1.129 — 7.28tt

At take-off, the airplane lift coefficient is CL = 1.44 = 1.129 — 7.28(tt)t—o, which can be solved to give

(tt )t—o = -0.0427 = -2.45°

At take-off the angle of attack will be

(aeq)t—o = 0.2474 = 14.17°

15.3 Solution to Problem 3

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