Equilibrium of the Aggie Micro Flyer

15.6.3.1 Airplane Lift and Moment Curves

The equilibrium code calculates the lift and moment coefficients for the complete configuration at low incidences to be:

CL (a, tt) = 4.391a + 0.757ft + 0.928
См, о(а, tt) = -1.733a – 0.686ft – 0.267

We will use this linear model, even for take-off conditions.

Definition of the aerodynamic center:

The aerodynamic center is the point about which the moment is independent of incidence.

Подпись: xac Iref Подпись: dCM,o d a dCi d a Подпись: 1.733 4.391 Подпись: 0.395

As seen in class, the location of the aerodynamic center is given by

That is xac = 0.632 m.

Given that xcg = 0.503 m, the static margin is

SM = Xac – Xcg = 0.08

Iref

a static margin of 8 %.

15.6.3.2 Take-Off Conditions

The longitudinal equilibrium equation for the moment reads

CM, cg(a) = См, о (a) + Cl (a) = 0

Iref

This equation represents the transfer of moment from the nose O of the aircraft to the center of gravity and states that the moment of the aerodynamic forces at the center of gravity must be zero.

The main wing lift curve is given by

CLm = 4.927a + 1.388

Since the take-off lift coefficient for the main wing is (CLm)t-o = 2.7, the inci­dence can be found to be (a)t-o = 0.266rd = 15.3°.

Find the tail setting angle at take-off is then given by the equilibrium equation as -0.686tt – 0.728 + 0.314(0.757tt + 2.096) = 0

which gives tt = -0.156rd = -8.9°.

Equilibrium of the Aggie Micro Flyer Equilibrium of the Aggie Micro Flyer

The location of the center of pressure is given by

Hence, xcp = 0.502 m.

The global lift coefficient can now be found to be

CL = 4.391 0.266 – 0.757 0.156 + 0.928 = 1.978 By definition we have

(am + at )CL = am CLm + atCLt

Here am = 0.709 m2 and at = 0.254 m2. Solving for CLt gives

CLt = -0.04

The tail lift coefficient at take-off is close to zero (within our linear model accuracy).

15.7 Solution to Problem 7

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