FLAPPING DUE TO PITCH AND ROLL VELOCITIES
The flapping associated with pitch and roll velocities was discussed earlier without deriving the equations. The derivations can now be made by using the same techniques as were used to analyze flapping in steady flight. For this derivation, the moment at the hinge must still be zero, but an extra contribution, due to gyroscopic effects, must be considered:
•d’f c. f. + №a + Mw + 2Vfgyro = 0
The equation previously derived for the moment due to centrifugal forces is valid for this case. Since we are interested in only the sine and cosine components, it can be written:
The increment of hinge moment due to the aerodynamics is:
AMA = r’ ^ UiaacAr’
For this increment we need only consider the change in local angle of attack caused by the angular rates and the associated flapping. The result can then be superimposed on the flapping due to the other independent variables such as collective pitch and shaft angle of attack. The change in the local angle of attack is caused by the components of vertical velocity due to the pitch rate, q the roll rate, p and the flapping angles and flapping velocities associated with these rates, shown in Figure 7.11.
The local angle of attack to be used is:
where:
Up = (r’ + e)(q cos \f + psin |/) — г’Сі(а^ sin |f — bx. cos |/) + V(aXj cos |/ + bx sin |/) cos |f
and:
UT= Cl(r’ + e) + V sin |i
Substituting these equations into the moment equation, integrating out the blade, and discarding all but the first harmonic sine and cosine terms, as was done in the analysis of flapping in steady flight, gives:
‘-L + { |
bx |
|
Cl |
V 2 ) |
+ |
cos Jf |
uisc Analogy
The hinge moment due to ^ cafl be found by examining the
upward acceleration of a blade e, 0SC0P1C orce jjCUlar to the tip path plane, assuming that the tip path plane and rolled as the solid disc of
Figure 7.1L ein& Pl C
The vertical velocity at the Ы н I is tf^de UP °f f°ur components: two due to the pitch and roll rat^ a e e ement two due to rotation on a
constantly pitching and tolling 4>lthout rotat‘°
Vm = – qr cos V – ^ ^ яп y qit – П’ “З у pit
The corresponding acceleration is:
^gyro= sin Ж — prto cos vj/ + sin у — pdr cos vj/
or:
а&[0 = 2^rfl sin VJ/ — 2/>rft cos V|/
The increment of hinge moment due to this acceleration is: Ao = A-r-a^dr’
and the total moment is:
f*-‘
^gyro = – I (?’ + + 0^ sin V — ?^(г’ + 0^ C0S
•’о
The sine and cosine components are:
^«yrosine =
and
^W«W =
The total sine and cosine components that must be zero are:
P
-2 qVtIb = 0
Q I £
Mcosinc = We — ah + – ас(Щ2 1——————————————–
і 8 V X
+ 2pClIb = 0
These two equations can be solved simultaneously, using the same techniques as were used for steady flapping, to give equations for longitudinal and lateral flapping, al t and bx, as a function of the pitch and roll rates, q and p.
These terms can be added to the previously derived equations for flapping in steady flight. In each of these equations, the first term may be thought of as the basic flapping due to pitch and roll velocities
That is, for a nose-up pitch rate, the tip path plane will lag behind longitudinally and tilt down to the left laterally. For a more or less conventional Lock number of 8, the lateral tilt will be one-half the longitudinal lag angle. This is rate cross-coupling as contrasted to acceleration cross-coupling, discussed earlier.
If the pitch rate is being produced by the pilot iq a deliberate maneuver using cyclic pitch, both the longitudinal and lateral flapping will be essentially zero and the trim value of lateral cyclic pitch will be approximately half of the longitudinal cyclic pitch. In this case, the longitudinal cyclic pitch, Bv will be
negative, and the lateral cyclic pitch, Aly will be positive—as required to prevent a left roll.
Two other sources of cross-coupling may be present during pitching maneuvers in forward flight. The first is due to the change in coning as rotor thrust changes during the maneuver. For a nose-up pitching maneuver—or pull- up—the increase in coning causes the rotor to tilt down to the right, which is the opposite to the left tilt caused by nose-up pitch rate just discussed. A second possible source is associated with the sideslip that will occur if the pilot holds his pedals fixed during a pull-up. The sequence of events is as follows: The helicopter slows down; the tail rotor thrust decreases, allowing the helicopter to sideslip such that the flight path is to the left of the nose (for counterclockwise main rotor rotation); the rotor longitudinal flapping aligns itself with the new flight path and, in so doing, produces a tilt of the tip path plane down to the right. This is the source of the rotor dihedral effect since it generates a rolling moment in the same sense that wing dihedral generates a rolling moment on an airplane. Because the various coupling effects have different characteristics, the magnitude and even the direction of roll during pitching maneuvers—or the lateral cyclic pitch required to suppress roll—will depend on the physical parameters of the helicopter, on the flight conditions throughout the maneuver, and on the pilot’s actions on the rudder pedals. To gain some insight into the coupling, assume that a helicopter is in a steady turn with no sideslip. If the lateral flapping is to be eliminated, the change in lateral cyclic pitch must be that due to both the pitch rate and that due to coning. The pitch rate effect is:
A Ax
From Chapter 5:
gO»2-[11])
V n
The coning effect can be found from the analysis made in Chapter 3, where it was shown that:
Лі ~K=~
4 v j
з^0+Ш
For illustration, assume the example helicopter in a steady zero sideslip turn at 115 knots with a load factor of 1.5.
For this maneuver:
Au tn =-2.2°
1 level flight
ft = 21.67 rad/sec
Thus
In other words, the pilot will have to hold left stick to prevent roll to the right because the effect of coning is more than the effect of pitch rate.
The pitching rate in this maneuver would be expected to produce a change in longitudinal flapping, but this flapping must be suppressed with longitudinal cyclic pitch in order to maintain zero pitching moments about the center of gravity (ignoring any damping effects of the fuselage and horizontal stabilizer until Chapter 9). The cyclic pitch required to do this is:
A. 1 16 і
‘ da. JdB, у a
From the trim equation of longitudinal flapping:
da
dB
Thus
For the example maneuver:
0.32 |
This means that the pilot will experience an almost one-to-one coupling when performing this maneuver.