Equilibrium of the Aggie Micro Flyer
The AMF has a span bm = 1m and a constant chord cxm = 0.192 m.
188.8.131.52 Main Wing
The wing aspect ratio is ARm = bm/S = bm/cxm = 5.2.
The theoretical formula gives dCLm/da = 2n/ (1 + 2/ARm) = 4.54 < 4.8701. This discrepancy is probably due to viscous effects seen in the 2-D viscous polar, in particular a laminar separation bubble near the leading edge on the lower surface that affects the leading edge geometry making the airfoil appear blunter when a < 0. The 2-D lift slope is found to be (dCi/da)a=0 = 7.5 > 2n.
184.108.40.206 Lift Curves
The general formula for the lift CL (a, tt) of the combination of the wing and tail of areas Am and At, and reference area Aref = Am + At is:
AmCLm + AtCLt
Am + At
Here we get:
CL (a, tt) = 0.5573CLm + 0.4427CLt = 3.4374a + 1.5286tt + 0.2742
220.127.116.11 Moment Curve
The moment coefficient for the complete airplane is given by: CM, o(a, tt) = -1.378a – 1.268tt – 0.03492.
The equilibrium equation for the aerodynamic moment is given by:
CM, cg(aeq. tt) = CM, o(aeq. tt) + 7 CL (aeq. tt) = 0
This gives CM, cg(aeq, tt) = -0.3248aeq – 0.7996tt + 0.0491 = 0.
Solving for aeq(tt) = —2.4618tt + 0.1512.
Application: horizontal flight at tt = 1.63° = 0.02845 rd ^ aeq = 0.0812 rd = 4.7°.
18.104.22.168 Take-Off Conditions
At CLm = 1.9, using the wing lift curve one finds (aeq)t_o = 0.2464rd = 14.1°.
The corresponding tail setting angle is given by the equilibrium equation to be (tt)t-o = -0.0387 rd = -2.216°. The tail lift curve gives a small positive lift Cu = 0.0076.
22.214.171.124 Extra Credit
The AMF aerodynamic center is located at xac/lref = xcg/lref + SM = 0.3064 + 0.08 = 0.3864. The center of gravity is such that xacm/lref = 0.27 < xcg/lref = 0.3064 < xac/lref = 0.3864. At high take-off incidence, the center of pressure of the main wing is close to the aerodynamic center of the main wing, i. e. xcp/ lref = xacm/lref + є ~ 0.27 < xcg/lref = 0.3064. The main wing lift and the weight produce a nose up moment. Hence the tail must have a positive lift to counter that moment. See Fig. 15.13.
15.5 Solution to Problem 5