# Equilibrium of the Aggie Micro Flyer

The best design for the AMF IV has a span bm = 2.54 m and a constant chord cxm = 0.279 m. The corresponding tail is defined as bt = 1.0 m with constant chord cxt = 0.254 m. The equilibrium code provides the take-off velocity Vto = 12.5 m/s and a maximum take-off mass M = 19 kg.

14.6.3.1 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 + 0.757tt + 0.928

См, о(а, tt) = -1.733a – 0.686tt – 0.267

where a is the geometric incidence (in radians, measured from the fuselage axis) and tt is the tail setting angle (in radians). См, о is the aerodynamic moment about the origin of the coordinate system (located at the nose O). We will use this linear model, even for take-off conditions.

Give the definition of the aerodynamic center.

Find the location of the aerodynamic center xac in m given that lref = 1.6 m. The center of gravity is located at xcg = 0.503 m. Find the static margin SM in

% of lref.

14.6.3.2 Take-Off Conditions

Write the longitudinal equilibrium equation for the moment. How do you interpret this equation?

The main wing lift coefficient is given by

CLm = 4.927a + 1.388

At take-off the lift coefficient for the main wing is (CLm)t-o = 2.1. Find the take-off incidence (a)t-o.

Find the tail setting angle at take-off.

Find the location of the center of pressure in m at take-off.

Is the lift force on the tail up or down at take-off.

14.6 Problem 7

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