A1 Helicopter in flight with unbalanced stores
A helicopter in forward flight is shown in Figure 1. The helicopter in forward flight experiences a horizontal drag force DF . The drag force DF is assumed to pass through O. The helicopter is held up in
the air by its rotor which produces a lift force LF . The pilot modifies the lift force LF by changing the
pitch of the blades: the lift magnitude LF is modified by the collective pitch; the inclination of the lift force LF is modified by the cyclic pitch. Thus, the lift force vector LF can be increased or decreased and
can be also inclined in any direction; this gives the helicopter a 360 deg maneuverability. The helicopter rotor requires a torque moment M to turn against the air resistance. To balance the torque moment M
and prevent the helicopter from spinning, a tail-rotor force TF is applied laterally at point A( ,0, )
Figure 1: Forces and moments acting on the helicopter
− . ab
Because the point A is offset vertically from the Ox axis, the lateral force TF makes the helicopter roll
slightly and hence the helicopter center of gravity C is displaced distance d in the negative y direction. Another unbalancing effect on the helicopter is the weight SF of the left-hand stores (the right-hand
stores have been used but the stores on the left hand are still in place). The force SF acts at point
− . Because the stores location is slightly forward, the helicopter pitches forward about the Oy S( , , )
fg h
axis and its center of gravity C is displaced slightly backward by a distance e. Thus, the helicopter center of gravity finds itself at location C( , , )
−−−. The helicopter weight is W .
edc
If the weight W , drag DF , and geometry are known, then find vector components and magnitudes of lift LF , tail rotor thrust TF , store weight SF , torque moment M .
Numerical values: 10,250 lb 0.6 in
W = , 500 lb
e = , 0.51 ft
f = , 1.96 ft
DF = , 13.7 ft
g = , 1.38 ft
h =
a = , 1.4 ft
b = , 2.35 ft
c = , 0.57 ft
d = ,