Category Helicopter Performance, Stability, and Control

The lift and drag of the horizontal stabilizer play important roles in the longitudinal trim conditions of the helicopter. Figure 8.5 shows the conditions at the stabilizer, especially the relationships that define its angle of attack.

The lift and drag of the horizontal stabilizer enter into the longitudinal equilibrium equations as:

X„ = LH sin[0 – (e,„H + e, H + Y.-)] – D„ cos[0 – (e,„H + tfH + y,)]

Z„ = – L„ cos[0 – (ЄМн + Єрн + Y.)] – DH sin[0 – (Єд, н + s, H + Y.-)]

The basic equations for lift and drag are:

Lh =

LH = ^ ^qAH[a((l — CIlo)]h

and

These equations contain several types of parameters that must be evaluated. The first are the geometric terms, which are defined by the configuration description: stabilizer area, AH; stabilizer aspect ratio, ARH; incidence of the main rotor shaft with respect to the body axis, iM; incidence of the chord line of the horizontal stabilizer, iH and angle of zero lift of the airfoil used on the stabilizer, aLG. Three more parameters that may be found indirectly from the configuration description are the slope of the lift curve, aH, which is a function of aspect ratio and sweep of the stabilizer; the span-efficiency factor, 5,H; and the profile drag coefficient, CDq. Some methods for evaluating these will be given. The second set of parameters consists of the flight conditions: dynamic pressure, q, and the angle of climb, yf, which for most analyses are known beforehand. The third type are the environmental conditions in which the stabilizer operates: the dynamic pressure ratio qjq, and the downwash angles induced by the main rotor and fuselage, and eF{J. These terms must be evaluated by some more-or-less empirical methods based either on studies of previous designs or on wind tunnel tests of the configuration analyzed. Both types of methods will be discussed. The last set of parameters are the trim conditions, 0, arpp, and ax, which are to be solved for in the procedure to be oudined later.

The tail rotor contributes primarily to the side force and yawing moment equations of equilibrium. Nevertheless, a complete analysis will also account for small contributions to the other four equations of equilibrium. Figure 8.4 shows the geometric relationships that govern the individual tail rotor forces and moments.

XT = – HT

ZT = bls TT (for tail rotor blade closest to main rotor going up)

ZT = —KSr Tt (f°r tail blade closest to main rotor going down)
MT = —QT (for tail rotor blade closest to main rotor going up)

AfT = Qr (for tail rotor blade closest to main rotor going down)

Once the tail rotor thrust to provide the antitorque forces that the vertical stabilizer does not provide is known, the other tail rotor parameters, bl$, HT, and Qj> can be calculated by the methods of Chapter 3.

Main Rotor

The main rotor as the primary source of forces and moments on the helicopter is shown in Figure 8.3 and is described by the following equations for a rotor whose advancing blade is on the right side:

The equations contain several types of terms, as listed in Table 8.1.

EQUATIONS OF EQUILIBRIUM

Like any aircraft in steady flight, the helicopter must be in equilibrium with respect to three forces and three moments acting along and around three orthogonal axes through its center of gravity. The analysis an be based on one of three possible systems of axes: wind axes, stability axes, or body axes. These systems are distinguished from each other by relating them to the three types of balance systems that are used in wind tunnels as shown in Figure 8.1. If the balance system is always aligned with the tunnel center line, the measurements will be in the wind axes system in which the X axis points along the line of flight in both the side and top views. If the balance system is mounted on a yaw table that rotates to produce sideslip conditions, the measurements will be in the stability axes system. For this case, the X axis will be aligned with the flight path in the side view but with the body in the top view. If the balance system is contained in the body of the model, the forces and moments will be measured in the body axes system and the X axis will line up with the body in both the side and top views. Although each system is valid, there are two reasons for using the body axes system in helicopter analysis. First, the other systems lose their significance in hover. Second, many helicopters are equipped with stability augmentation systems using gyros or accelerometers
whose displacements are measured with respect to the airframe, and the analysis of the effects of these devices is easiest in the body axes system.

(One of the minor annoyances that may have to be faced with the body axis system is the reluctance of the design department to define waterlines as being perpendicular to the rotor shaft. It is more likely that the designers will lay out the fuselage using some such arbitrary line as a cabin floor for a waterline and then tilt the rotor shaft forward to obtain a level floor and a streamline fuselage attitude at

Axes System

FIGURE 8.1 (cont.)

the cruise condition. This produces unnecessary bookkeeping complications for the aerodynamicist in keeping track of center-of-gravity positions, moments of inertia, and horizontal stabilizer angle of attack. It would be helpful if the designers would use the shaft as their primary reference and then tilt the cabin floor and draw the fuselage contours for minimum drag. The finished helicopter would look the same but would be easier to analyze. This is the scheme that has been used for the example helicopter, but the analytical procedures that follow allow for those unfortunate cases where the shaft is tilted with respect to the body axis defined as a designer’s waterline.)

The aerodynamic moments and forces in the body axis system that are acting on the helicopter are shown in Figure 8.2. They are due to the main rotor, the tail rotor, the horizontal stabilizer, the vertical stabilizer, and the fuselage. (For some helicopters, add a wing and/or a propeller.)

The six equations of equilibrium are:

XM + XT + XH + Xv + XF = G. W. sin 0 YM + YT +Yv + Yf= – G. W. sin Ф Zft + Zj + Zfj + Zy + Zp = —G.^F. cos 0

Rm + YMhM + ZMyM + YThT + Yyhy + YFhF + RF = 0

+ ZMlM + MT XThT + ZTlT XHhH + ZHlH — Xvhv + Mp + Zp Ip — Xphp — 0

NM — YMlM — YTlT — Yyly + Nf — Ypip = 0

When the equilibrium equations were developed for airplanes many years ago, all engineering analysis and reports were hand-lettered, and the three moments were written in script: і? Л, and Ж Nowadays, typewriters and word processors don’t make script characters. This makes a special problem with rolling moment, since L is used for total lift and, if also used for total rolling moment, causes confusion, especially in distinguishing Ci, meaning lift coefficient, from CL meaning roll moment coefficient. For that reason, R is used in this book for rolling moment.

The moments about the aircraft’s center of gravity due to rotor flapping are produced by the couple at the hub as a result of the rotor stiffness, the tilt of the thrust vector perpendicular to the tip path plane, and the rotor’s inplane force. From Figure 7.12:

dMu

Atco. = (4 + HV°>^ ~ T/*

The rotor stiffness is produced by the vertical component of blade centrifugal force acting at the hinge offset; from Figure 7.10:

(P – «Ок

(r’ + ‘) C. F.„„ = ft!(3 – a„) MJg For pure longitudinal flapping:

P – a0 = – a. cos у

and the total rotor moment in the pitch direction for b blades is:

1 {2"

= ebCl2als Mk/g— cos2 J/ d\f

J0

where Mb = First static moment or

MM = – ebCt2ay Mjg
2 s

For uniform blade mass distribution, m, in slugs per foot: or

dMM з / e AhpR(CtR)2a 4 R) у

(Note: This equation could have also been derived by using the vertical shear at the hinge location associated with the incremental inertia forces acting normal to the blade elements due to the flapping motion.) For the example helicopter at sea level:

For hingeless rotors whose blades are cantilevered from the shaft, the rotor stiffness can be visualized as being produced by blade bending moments as pairs of blades are bent into an S shape. The stiffness of a hingeless rotor is usually computed from mode shape considerations at the same time the flapping natural frequencies are computed. Once the stiffness is known, an effective hinge offset can be determined:

1

3 bCt2It

4 dMJdals

For the hingeless rotor used on the Lockheed AH-56A, the effective hinge offset was approximately 13% of the radius.

For simple analyses, it is often satisfactory to assume that the total rotor vector is perpendicular to the tip path plane. It is more correct, however, to account for an effect of inflow that modifies this assumption. The geometry of this phenomenon is illustrated by a simple example in Figure 7.13. Here a helicopter is hovering with some nose-up rotor flapping, due in this case to the center of gravity being ahead of the shaft. The thrust vectors on two typical blade elements on the right and left blades are equal and tilted symmetrically with respect to the tip path plane but not to the shaft. Their contributions to the H-force, which is defined as being perpendicular to the shaft, are:

A#right = sin (ah + ф) and

AHkft= ATLsin(d, x-0)

or

A H = A Ttotal sin ax cos ф

Thus, when the tip path plane is tilted with respect to the shaft, the H-force is decreased by the cosine of the inflow angle in this simple example.

The effect can be derived for the entire rotor from the first equation for Сд/о derived in Chapter 3. Differentiating with respect to aXj gives:

but the second term, for ail practical purposes, is the equation for CT/<r, so that:

= CT/a + – X’

‘ 8

Thus, as the helicopter goes faster and А/ becomes more and more negative, the effect of the tilt of the thrust vector is reduced and in some extreme conditions might even change sign. This has the effect of reducing the damping that one might expect from rotor flapping.

This effect was first pointed out in reference 7.2. The equation can be written in the format of that report by assuming that the thrust coefficient is a function of the pitch at the three-quarter radius position:

Ст/°= 4 (з Є” + /

Solving for X’ and substituting in the equation for (dCH/o)/dat gives: dCH/<3 3 / ( a 0 75

————- = — Cj/о 1—————————–

dah 2 18 CT/o

This is a convenient form to use in hovering, but the one with А/ is recommended for forward flight. The reduction in the effect of the H-force due to flapping is proportional to the inflow which has to be compensated for by collective pitch.

For the example helicopter, the moment in foot-pounds produced about the center of gravity per radian of flapping is:

A derivation of the rotor Y-force equation shows that the effect is the same as with the H-force; that is:

EXAMPLE HELICOPTER CALCULATIONS

Frequency ratio 457

Phase angle 459

Damping ratio 459

Acceleration coupling 460

Ratio of cyclic angle of attack to flapping 461

Azimuth constant 462

Cyclic pitch required in turn 475

Rotor stiffness 476

Moments produced by flapping 479

HOW TO’S

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:

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:

This means that the pilot will experience an almost one-to-one coupling when performing this maneuver.

The complete equations for rotor flapping in forward flight can be derived by equating the increments of hinge moment due to aerodynamic, centrifugal, inertial, and weight forces to zero. The resultant equations will be similar to those derived for the trim values of cyclic pitch in Chapter 3 and are based on similar assumptions:

• Aerodynamic forces are considered to act from the hinge to the tip.

• The reverse flow region is ignored.

• The airfoil lift characteristics are linear and free of stall and compressi­bility effects.

• The blade motion consists of only coning and first harmonic flapping.

• Small-angle assumptions are valid.

The use of these assumptions has been found to be justified for conventional helicopters flying within conventional flight envelopes. For special applications, these assumptions can be relaxed at the expense of simplicity.

Since the blade motion consists only of coning and first harmonic flapping, the centrifugal forces may be assumed to he in the plane of rotation of the blade element, as shown in Figure 7.10. This assumption also eliminates all inertial moments due to flapping from the analysis. The increment of moment about the flapping hinge due to centrifugal force is:

ДЛГСР. = – AC. F. h

where

AC. F. = mAr(r’ + e)Cl2

and

h = r’a0 +

(f’P-гЧ)

For most rotors, the hinge offset ratio, e/R, will be small enough that it can reasonably be eliminated from those terms inside the square brackets.

The final contribution to the hinge moment that must be considered is that due to the blade weight, Mv. But:

R-e

mgr’dr’

I

The summation of the constant portions of the various contributions to the hinge moment can be used to give the equation for coning at the flapping hinge as it did in Chapter 3 for the rotor without hinge offset:

+ Мл + M-iw — О

For most applications, the second term in each denominator is’ small with respect to the first term. Taking it as negligible, the equation becomes:

The first term may be thought of as the basic flapping that is independent of hinge offset. The second term represents cross-coupling due to hinge offset and is significant when the determination of cross-coupling is a primary objective of the analysis. The similar equation for the lateral flapping, bxr, is:

– 0ojli + 20xp – Bx f 1 + – p2 ] + 2p ( pa, – Cr/a — 3 2 2py

The primary use of these equations in the analysis of stability and control is to yield flapping derivatives as a function of changes in flight conditions and control inputs. These flapping derivatives can then be converted into pitch and roll moment derivatives using the relationships between moments and flapping. The evaluation of the derivatives will be discussed in detail in Chapter 9.

The expression for blade damping derived several paragraphs ago allows the blade time constant to be determined. The time constant is the time required by a system

to achieve 63% of its final amplitude following the application of a step forcing function. For the blade subjected to a step moment, Msn the non-oscillatory part of the displacement, is:

-c/2

where MJk is the final value of p when time is very large. If this is evaluated at the time at which

c/2

then

The time that makes this true is called the time constant of the system:

~ c/2

But from one of the previous equations:

The amount of azimuth required for the blade to achieve 63% of its response is thus:

For the example helicopter, this equation gives an azimuth constant of 130°. This relatively fast response, compared to the response of the entire helicopter, will later be used to justify a simplifying assumption when writing the helicopter equations of motion.

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

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.

Since the ratio of rotational frequency to natural frequency for a rotor with hinge offset is lower than unity, the phase angle will be less than 90°, as can be seen from Figure 7.3. How much less depends on the damping ratio, c/cak, where ccrit is critical damping. The concept of critical damping comes from the study of a single pendulum-spring-damper system that has the differential equation:

/p + cp + £p = 0

The general solution to this equation is:

If the term under the radical is negative, the motion will be oscillatory. If it is positive, the motion will have a pure convergence with no oscillations. Critical damping is the value of c that makes the radical term zero—that is, that lies on the boundary between oscillatory and nonoscillatory motion:

(caJ22 = k

or

cr’c ^undamped

The damping of the blade, c, may be evaluated from the aerodynamic hinge moment due to flapping velocity:

MA = j r’ — acr’fi (r’ + e)Cldr

(Reader, please note that italic c stands for chord and roman c for damping; and that italic e stands for hinge offset and roman e for the base of natural logarithms.)

1 e

c =

e

Y _

A useful nondimensional parameter relating the inertia and aerodynamic characteristics of a blade is the Lock number, y, which was already introduced in Chapter 1.

cpaR4

For most blades, у is between 6 and 10. Adding a large tip weight such as a jet engine may lower

The factor, y/16, is a universal parameter that will arise again when rotor damping as a whole is discussed.