Category Helicopter Performance, Stability, and Control

The isolated rotor charts of Chapter 3 can be used with the method outlined there to calculate the steady rate of descent in autorotation. The results for the example helicopter are plotted in Figure 5.5 as a function of forward speed (the value at zero forward speed is from the example calculation in Chapter 2). The speed for minimum rate of descent is at the bottom of the curve, but the speed for minimum angle of descent—or maximum glide distance—is at a higher speed where a ray from the origin is tangent to the curve.

The figure shows that the example helicopter can make approaches down paths as steep as 9° at any speed; but if it is required to come down a steeper path, the allowable speed range is limited by the inability to dissipate fast enough the potential energy above that required just to maintain autorotation. The figure also shows that very steep approaches at intermediate rates of descent would be complicated by the possibility of power settling in the vortex/ring state, as discussed in Chapter 2.

As a practical matter, there are two requirements for safely doing steep approaches, especially with limited visibility: an airspeed (or possibly ground speed) indicator, which is accurate at low speeds, and an adequate field of view from the cockpit to the intended touchdown point.

In airplane studies, the ratio of forward speed to ntyof descent in a glide is used to define the lift-to-drag ratio, since:

L _ VM.

^ ^descent

For the example helicopter the maximum lift-to-drag ratio based on this definition is 6.3. This is lower than would be expected for a comparable airplane. The discrepancy can be charged to the drag of the rotor hubs and to the fact that, over a large portion of the rotor disc, the blade elements are not operating at their angles of attack for maximum airfoil lift-to-drag ratio, whereas on an airplane each wing element can be operating near its optimum angle of attack.

The entry into autorotation is usually an emergency—or at least a simulated emergency—maneuver, so there is a finite time delay between the power loss and the pilot’s corrective action. The decay of rotor speed during this period is a function of the power required and of the rotor’s level of kinetic energy. If the rotor were on a whirl tower, the decelerating torque could be assumed to be proportional to the square of the instantaneous rotor speed. Measurements of rotor speed decay during simulated power failures in flight such as those in references 5.11, 5.12, and 5.13, however, indicate that the actual rate of decay is somewhat less than would be predicted by this assumption. The reasons are several:

1. During a simulated failure using a throttle chop, the engine torque decays at a finite rate rather than instantaneously.

2. In some helicopters, the engine supplies a small amount of torque even after a throttle chop.

3. At low speeds, the helicopter immediately begins to descend, thus decreasing the power required.

4. At high speeds, the loss of rotor speed increases the tip speed ratio, which makes the rotor flap back, thus decreasing the power required.

The use of the assumption that the decelerating torque is proportional to the Square of the rotor speed is therefore somewhat conservative, but the degree will depend on the conditions existing at the time of the engine failure. Using the assumption anyway, the equation for the rotor speed decay is:

wherb the subscript 0 refers to conditions at the time of the power failure, and J is thfe total effective polar moment of inertia of the drive system, including the main rotor, the tail rotor, and the transmission referred to the main rotor speed; that is,

For the example helicopter, this gives:

An equation for the rotor speed time history an be derived by integrating the equation for the rate of deay:

f" — <m = –

ci0 Cl2 Л Jdl

Carrying out the integration gives:

ft 1

It is convenient to rewrite this as a function of the time during which all the rotor’s kinetic energy would be dissipated at the initial horsepower and rotor speed:

550 h. p.0

Using this definition, the rotor decay equation becomes:

ft 1

u 1 +—————

2^k. e.

The equation has been evaluated for several values of t^ and the results are plotted in Figure 5.4. The figure applies to any flight condition from hover to maximum speed. The example helicopter has a value of /KE of 1.2 seconds with full engine power. Thus for flight in this condition, the rotor speed would deay 30%

in the first second following a sudden failure of both engines. Failure of one engine from a full power condition would result in a decay of 17% in the first second if the other engine did not increase its power.

The pilot who has been asked to demonstrate the maximum load factor capability of a helicopter will build up the maneuver until he reaches one of the following limits:

• Maximum engine power

• Maximum stick displacement

• Unacceptable level of vibration

• High nose-up attitude or pitch rate from which recovery is uncertain

• Indication of abnormally high loads in the rotor or the control system

• Aircraft instability

• Ominous change in the rotor noise level

• Sudden rotor out-of-track condition

Some of these limits—such as reaching maximum engine power—are straight­forward and an be predicted by methods already developed in previous chapters. Others, however, are a function of the structural and dynamic characteristics of the rotor, the control system, and the remainder of the helicopter and of the pilot’s willingness to subject himself to uncomfortable or potentially dangerous flight con­ditions. There is as yet no analytical method for predicting the maximum attainable load factor when these latter considerations are involved, but there are enough experimental data to provide some insight into the problem. A convenient nondi – mensional representation of the maximum thrust capability is a plot of CT/cr versus |X such as Figure 5.2. (Note: if the blade is not of constant chord, the definition of thrust-weighted solidity given in Chapter 1 should be used in this analysis.) The plot has three boundaries depending on the flight condition.

The transient boundary an be achieved momentarily in flight or con­tinuously in a wind tunnel at high rotor angles of attack and corresponds to every blade element operating at its maximum lift coefficient. Test results indicate that this boundary is in the neighborhood of CT/o = 0.17, which is equivalent to an average lift coefficient of over 1. Values of this magnitude or higher have been

reported: in flight maneuvers by references 5.1 and 5.2; in autorotative flares by reference 5.3; and in wind tunnel tests by references 5.4 and 5.5. The theoretical limit of the transient capability is a value of CT/o of about 0.3, which could be achieved on a heavily stalled rotor at a very high angle of attack, where the contribution of the drag acting on each blade element parallel to the shaft is producing most of the rotor force. Other boundaries in Figure 5.2 represent level flight and steady turns and $re primarily based on the flight experience reported in references 5.6 and 5.7. For this reason, these boundaries probably reflect some of the nonaerodynamic limits listed in the beginning of this section. These limits vary from helicopter to helicopter and determine how heavily loaded a rotor may be before it gets into stall-related troubles. For example, it is common to generate high oscillating loads in the control system when the blade goes into stall on the retreating side. The resultant change in aerodynamic pitching moment can twist the blade nose down to an unstalled condition, only to have it spring back into stall. Depending on the torsional natural frequency of the blade and control system, each blade may go through several cycles of this oscillation—sometimes called stall flutter— while passing through the retreating side. A theoretical and experimental investigation of this problem is reported in reference 5.8 in which it is shown that control loads are higher when the torsional natural frequeqfy ratios are between 5 and 12 per revolution than for frequencies on either side of this range.

Another possible source of high vibration and high loads is the excitation of those blade-bending modes for which the forcing function Increases with tip speed

ratio and rotor thrust. One example of such a mode is the second blade flapping mode, whose natural frequency is near 3/rev (it would be exactly V27T or 2.51/rev if the blade acted like a chain stiffened only be centrifugal forces). As the tip speed ratio is increased, the 3/rev content of the aerodynamic forcing function—which, incidentally, is made larger by blade twist—increases, thus exciting the second blade flapping mode to a high amplitude with correspondingly high vertical shear at the flapping hinges. This particular vibration can be especially severe for a three – bladed rotor since all three blades will respond in unison to the 3/rev aerodynamics. On the two-, four-, or five-bladed rotor, the blades will be out of phase with each other and thus will produce less vibration in the helicopter as a whole.

Other factors that affect the apparent ability of the rotor to develop high load factors as limited by vibration are:

• The vibration isolation of the rotor and transmission from the rest of the helicopter;

• The natural frequency of the fuselage;

• The location of the cockpit with respect to fuselage nodal points (points that appear to stand still)

Thus it may be seen that there can be no single simple method for predicting the maximum load factor that will subject the pilot to unacceptable levels of vibration.

An exploration of the aerodynamic limits with a wind tunnel model is reported in reference 5.9. Figure 5.3 shows the limit of test points achieved at a tip speed ratio of 0.6. These limits are well beyond those normally accepted and were not limited by aerodynamic or structural phenomena but by straightforward mechanical interferences on flapping and cyclic pitch control on that particular model. Reference 5.9 uses these test results to predict high rotor performance for future rotors. A continuation of this wind tunnel study is reported in reference

5.10, which concludes that a 225-knot helicopter is feasible. The bottom portion of Figure 5.3 summarizes the maximum rotor thrust capability as measured on the wind tunnel model during that project. [7] [8]

Advance Ratio-fi

Source: McHugh & Harris, “Have We Overlooked the Full Potential for the Conventional Rotor?" JAMS 21 -3,1976; McHugh, “What are the Lift and Propulsive Force Limits at High Speed for the Con­ventional Rotor?” AHS 34th Forum, 1978

• What are the boundaries of the "Deadman’s Curve”?

• What is the minimum touchdown speed?

• How does this helicopter compare with others?

These questions can be answered using the following analytical methods.

INTRODUCTION

The previous chapters described methods for estimating the performance of the helicopter as an aircraft that hovers or simply flies from one place to another. Frequently, however, it is necessary to analyze the performance from other aspects. The aspects that will be discussed in this chapter are:

Turns and pullups Autorotation Maximum accelerations Maximum decelerations

Optimum takeoff procedure at high gross weights

Return-to-target maneuver

Towing

Aerobatic maneuvers

TURNS AND PULLUPS

Load Factor Relationships

Conditions in a steady turn or in a symmetrical pullup are similar to conditions in level flight except that the rotor thrust is significantly higher than the gross weight and the rotor has a pitching velocity that relieves the retreating tip angle of attack. The rotor thrust must overcome the vector sum of weight and centrifugal force, as shown in Figure 5.1

or, in terms of the bank angle:

»Шгп = V1 + tan2 Ф = —

The centrifugal force is:

where (0 is the rate of turn and jRturn is the radius of the turn. But

Kmrn © = V

and

r

Symmetrical

Pull-up

Yet another form of the equation can be written in terms of the rate of pitch, 0, by noting that

0 = 0) sin Ф

(in a vertical bank the rate of pitch would be equal to the rate of turn), but

J n2 — 1 radian/sec

The rate of pitch, 0, which is inherent in all these maneuvers, has a relieving effect on the retreating tip angle of attack. The rotor, which is producing the pitching motion of the helicopter, must precess itself nose up as a gyroscope; this requires increasing the angle of attack of the advancing blade with cyclic pitch while decreasing the angle of attack of the retreating blade. The change in cyclic pitch required to precess the rotor can be determined by setting the change in aerodynamic moment from trim equal to the gyroscopic moment. If the pitching velocity is constant, with no pitching acceleration, then no additional hub moment is required except for a small one due to damping moments from the fuselage and horizontal stabilizer. The analysis of the rotor with pitching velocity will be found in Chapter 7, "Rotor Flapping Characteristics.” From that analysis:

. 16 ‘

ДВ, ———– — 0 radians

yfl

For the example helicopter in a turn at 115 knots with a load factor of 1.2, the pitch rate is 0.08 rad/sec and the decrease of cyclic pitch required due to this rate is 0.2 degrees, which decreases the retreating angle of attack by this amount and thus allows the rotor to develop somewhat more thrust before stalling than it could in static conditions such as in a wind tunnel test.

The power required during a steady turn may be estimated from the level flight characteristics by using an effective gross weight equal to the actual gross weight multiplied by the load factor. Using this method and the information in Figure 4.38 of Chapter 4, the example helicopter requires a total power of 3,170 h. p. in a 1.2-g turn at 115 knots, compared to 1,470 h. p. for level flight at the same speed. Sometimes for demonstration purposes, it is permissible to lose some speed and some altitude during a "steady” turn. In this way, a significant amount of power can be extracted from the changes in kinetic and potential energy in order to demonstrate higher load factors within the installed power limitations. Assuming that a speed, AV, is lost during a 180° turn, the change in kinetic energy is:

Ш=^у Ду

8 avg

and, combining equations derived earlier, the time to make the 180° turn is:

gV»2-1

Thus the gain in power is:

/g. w.

Ah. p.„ = 2 —————— *———-

550 л

Similarly, the power associated with a loss of altitude during the same 180£ turn is:

For the example helicopter in a 1.5-g, 180° turn starting at 115 knots, ending at 100 knots, and losing 50 ft of altitude, the increment of available power due to the change of speed is 656 h. p. and due to the change in altitude, 230 h. p.

For analysis of nap-of-the-earth flight, it is necessary to consider load factors less than 1 developed during a pushover as would be used to fly over the top of a hill and into the valley beyond. For this case:

C. F.

G. W.

or

(h) 2 J?

^ ^ ^ -*v pushover

This can be used to calculate the load factor associated with flying over a hill with a given radius of curvature.

The altitude at which the rate of climb is zero is the absolute ceiling, and the altitude at which the rate of climb is 100 ft/min is the service ceiling. Both these ceilings can be found from Figure 4.49 for a gross weight of 20,000 lb. The same analysis at other gross, weights can be used to obtain the plot of ceiling as a function of gross weight, as in Figure 4.51.

A military mission may be considerably more complicated than simply flying from one place to another. A typical mission might include the following segments:

Since the fuel required for the mission is not known, the takeoff gross weight cannot be initially determined. The landing weight, however, can be closely estimated since it is simply the minimum operating weight plus a small fuel reserve. For this reason, it is convenient to analyse the mission in reverse as on Table 4.4.

A figure like the bottom part of Figure 4.48 can also be used to determine the maximum angle of climb, which might be of interest in clearing an obstacle in the flight path. In this case the maximum angle is 90° when both engines are running, since the helicopter an climb vertically. With one engine, however, it annot even hover, but it an climb at 1,100 ft/min at 48 knots, which gives a climb angle of

12.7 degrees. A similar analysis can be made when the gross weight or the altitude is too high for hover out of ground effect with two engines.

The minimum time to climb to a given altitude an be calculated using the maximum rates of climb on Figure 4.49 to determine the time required to climb in 1,000-ft increments. The summation of this process is shown on Figure 4.50. Knowing the speed corresponding to the climb rates makes it possible to find the distance required to climb. These values will later be useful in doing mission analyses.

Two world records for time-to-climb are spotted in Figure 4.50. These were set in 1972 with a Sikorsky CH-54B and were still valid in 1985.

The power required to climb at a given flight path angle, y, can be determined by using the charts of Chapter 3 just as for level flight but modifying the flat plate

area to account for the rearward component of gross weight along the flight path:

r r G. W. sin у

/climb — /

Figure 4.48 shows the results of the calculations for the example helicopter at rates of climb from zero to 3,000 ft/min at sea level conditions. A cross-plot at the various ratings gives corresponding rates of climb as a function of forward speed in the lower part of the figure. Carrying out the same analysis at several altitudes allows us to find the maximum rates of climb as a function of altitude, as in Figure 4.49.

Many military missions require the helicopter to loiter while observing or waiting for some specific action. For these mission legs, the optimum speed is at the bottom of the fuel flow curve of Figure 4.42, and the endurance at any speed is a function of the specific endurance, S. E., whose units are hr/lb:

1 1 hr

Fuel flow ’ lb/hr ’ lb

Figure 4.47 shows the calculated specific endurance curves for the example helicopter. Here again, the advantage of flying on one engine is evident. This might become important to a pilot who is forced to wait for a change in the weather or for a landing site to be prepared. The time spent in loiter flight while burning a given amount of fuel can be found by integrating under the specific endurance curves of Figure 4.47, since:

This equation can be used to generate payload-endurance curves similar to payload-range curves. /

Gross Weight, lb

FIGURE 4.47 Maximum Endurance Conditions, Example Helicopter

The calculation of the maximum ferry range uses the same general procedures as for the payload-range curves. Typical special considerations that might be used, however, are:

Conditions

No wind

Sea-level, standard day

Cruise at speed for 99% max. range

2-rnin warmup and takeoff (WUTO)

at max. cont. power

30-min reserve for cruise speed

at landing weight

Min. op. wt. = 10,430 lb

Std. fuel tank capacity = 3,000 lb

Auxiliary fuel tank

weight = 0.1 x fuel wt.

400 800 1,200 1,600

Range, N. M.

FIGURE 4.45 Payload-Range. Example Helicopter

• 20-knot headwind.

• Cruise at speed and altitude for best range.

• Warm up and takeoff: 2 min at maximum continuous power.

• Reserve no. 1: 45 minutes at speed and altitude for best range at minimum operating weight.

• Reserve no. 2: 10% of fuel used for cruise over 3 hours.

• External tanks may be dropped when empty.

• Noncritical items of equipment may be left behind (weapons, passenger seats, etc.).

• Special items of equipment may be required (life rafts, oxygen, etc.).

For illustration, the maximum ferry range of the example helicopter will be determined using these conditions. The assumed takeoff gross weight is 28,000 lb, which is almost the maximum that an be flown in level flight on maximum continuous power.

The Group Weight Statement for the ferry mission in Appendix A can be used to find the minimum operating weight as:

Empty weight 10,000 lb

Crew 360

Unusable fuel 30

Engine oil 40

Auxiliary cabin tank 481

Survival kits 100

Lift rafts 50

Oxygen equipment 200

Minimum operating weight 11,261 lb

The total usuable fuel for the ferry mission is contained in five tanks:

Forward internal 1,485 lb

Aft internal 1,485

Cabin auxiliary 4,809

First external 4,091

Second external 4,091

Total 15,961 lb

Each external tank weighs 409 pounds and is assumed to have an equivalent flat plate area of 2 square feet.

Figure 4.46 shows the specific range plot based on these conditions. The maximum range is calculated by integrating under the curve from the mission start weight, which is the takeoff gross weight minus fuel for warmup and takeoff, and the landing weight, which is the minimum operating weight plus reserves. The reserves consist of two amounts:

Reserve no. 1 = Fuel for 45 minutes at minimum operating weight.

Reserve no. 2 = .1/1.1 [Total fuel — (Fuel for first 3 hr + WUTO + reserve no. 1)]

For this example, reserve no. 1 is 420 lb when calculated at 11,261 lb, 20,000 ft altitude, and 120 knots air speed. The fuel required for the first 3 hours was found by calculating the distance and time represented by burning 1,000-pound increments of fuel from takeoff. In the first 3 hours, 4,400 lb of fuel will be used. Using a warm-up and takeoff (WUTO) fuel allowance of 56 lb, reserve no. 2 becomes 1,008 lb. Thus the landing weight is 12,689 lb, and the takeoff weight is 27,944 lb. Integrating the envelope on Figure 4.46 between these two weights and accounting for dropping the external tanks when empty gives a maximum ferry range of 1,289 nautical miles.

In this example, the optimum altitude is sea level until more than half the fuel is burned off. At that time a slow climb is initiated, which reaches 15,000 feet at the end of the mission. The extra fuel required for this climb can be calculated and subtracted from the fuel available for cruise; but unless it is a significant quantity it can be ignored by assuming that the same amount of fuel is saved during the descent at the end of the flight.

Gross Weight, lb

FIGURE 4.46 Ferry Mission Conditions, Example Helicopter

A payload-range analysis is done to provide an indication of the usefulness of the helicopter as a load-carrying vehicle. Figure 4.45 shows the payload-range curve for the example helicopter at two takeoff gross weights. The conditions are listed on the Figure and are typical for studies of this type. The curves are based on the evaluation of two fundamental equations:

f g. w.t. o.

Range = (S. R.yG. W., N. M.

*’G. W.ldng.

where G. W.lndg = G. W.T0 — (Expended fuel + WUTO fuel) lb

and Payload = (G. W.T0 — G. W.min0P.) — (Expended fuel + WUTO fuel + Reserves + Aux. fuel tank wt.) lb

(WUTO stands for warmup and takeoff.)

The Figure shows that as a transport aircraft the example helicopter, taking off at 20,000 pounds, can carry its design payload of 6,600 pounds—consisting of 30 passengers with baggage averaging 220 pounds apiece—for 330 nautical miles at

sea level with adequate reserves. Offloading payload and replacing with fuel in auxiliary tanks can extend the range significantly.