Cross-Coupling

Knowing the ratio of rotational frequency to the undamped natural frequency and the damping ratio for the rotor with offset flapping hinges, the lag between the maximum aerodynamic input and the maximum flapping amplitude can be found from Figure 7.3. For the example helicopter, the Lock number, y, is 8.1; the hinge offset is 0.05; and the frequency ratio is 1.04. Thus the damping ratio, c/ccrj„ is

0. 42. From Figure 7.3, the phase angle, ф, is 84.8°. The difference between this angle and 90° represents one source of flapping cross-coupling that exists on a rotor with hinge offset but not on a rotor without offset. The magnitude of the coupling is:

where al{ and btj are shown in Figure 7.9.

The equation for the coupling can also be written directly. From the analysis of vibrating linear systems, the equation for the phase angle is:

C 0)

2 –

eft

Substituting in the expressions for the frequency and damping ratios gives the equation for coupling.

12 e

Y R 1 e ~ e ~ 1 +—— 1 -— 3 R _ —i

For the example helicopter, this means that a 1° change in longitudinal flapping will be accompanied by a lateral flapping of —.07°. If the pilot wishes to use cyclic pitch to tilt the tip path plane nose up in a purely longitudinal direction, he will have to move the stick slightly to the right as he pulls it aft in order to cancel out the left roll that would otherwise be generated. This type of cross-coupling is sometimes called acceleration cross-coupling because it is associated with the rotor moments, which provide the initial acceleration during a maneuver. Once steady rates are established in the maneuver, the cross-coupling changes to rate cross­

coupling, as will be shown later. Because these two types of cross-coupling have different values, it is not possible to compensate exactly for both by a simple mechanical rotation of control inputs between the stick and the swashplate, although some compromise may be used in an attempt to minimize both.

It is sometimes of interest to calculate the increment in cyclic angle of attack required to produce 1° of flapping. The change in angle of attack experienced by the blade is:

fir

CL(r’ + e

or at |/ = 180c

Да =

but, as was just shown:

12 _1_

, r R bl‘= ЇТ ‘

1 + 3 R

thus:

Да =

For the blade of the example helicopter at the 75% radius station, this

gives:

0.07

which says that 1° of cyclic flapping is maintained with only 0.07° change in angle of attack. If the hinge offset had been zero, the flapping could have been sustained with no cyclic change in angle of attack.