Equilibrium of the Aggie Micro Flyer
18.104.22.168 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.
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
a static margin of 8 %.
22.214.171.124 Take-Off Conditions
The longitudinal equilibrium equation for the moment reads
CM, cg(a) = См, о (a) + Cl (a) = 0
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°.
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