most airplanes, where discharging a passenger isunlikely to adversely affect the CG, off-loading a passenger from a helicopter could make the aircraft unsafeto fly. Another difference between helicopter and airplane loading is that most small airplanes carry fuel inthe wings very near the center of gravity. Burning offfuel has little effect on the loaded CG. However, helicopter fuel tanks are often significantly behind the centerof gravity. Consuming fuel from a tank aft of the rotormast causes the loaded helicopter CG to move forward.As standard practice, you should compute the weightand balance with zero fuel to verify that your helicopterremains within the acceptable limits as fuel is used.A BCDF1,6001,5001,4001,3001,2001,100104 105 106 107 108 109Baggage CompartmentLoading LinesFuel LoadingLinesEFigure 7-6. Loading chart illustrating the solution to sampleproblems 1 and 2.7-6SAMPLE PROBLEM 3The loading chart used in the sample problems 1 and 2is designed to graphically calculate the loaded center ofgravity and show whether it is within limits, all on asingle chart. Another type of loading chart calculatesmoments for each station. You must then add up thesemoments and consult another graph to determinewhether the total is within limits. Although this methodhas more steps, the charts are sometimes easier to use.To begin, record the basic empty weight of the helicopter, along with its total moment. Remember to use theactual weight and moment of the helicopter you are flying. Next, record the weights of the pilot, passengers,fuel, and baggage on a weight and balance worksheet.Then, determine the total weight of the helicopter.Once you have determined the weight to be within prescribed limits, compute the moment for each weightand for the loaded helicopter. Do this with a loadinggraph provided by the manufacturer. Use figure 7-7 todetermine the moments for a pilot and passengerweighing 340 pounds and for 211 pounds of fuel.Start at the bottom scale labeled LOAD WEIGHT.Draw a line from 211 pounds up to the line labeled“FUEL @ STA108.5.” Draw your line to the left tointersect the MOMENT scale and read the fuel moment(22.9 thousand lb.-inches). Do the same for the pilot/passenger moment. Draw a line from a weight of
340pounds up to the line labeled “PILOT & PASSENGER@STA. 83.2.” Go left and read the pilot/passengermoment (28.3 thousand lb.-inches).Reduction factors are often used to reduce the size oflarge numbers to manageable levels. In figure 7-7, thescale on the loading graph gives you moments in thousands of pound-inches. In most cases, when using thistype of chart, you need not be concerned with reduction factors because the CG/moment envelope chartnormally uses the same reduction factor. After recording the basic empty weight and moment ofthe helicopter, and the weight and moment for eachitem, total and record all weights and moments. Next,plot the calculated takeoff weight and moment on thesample moment envelope graph. Based on a weight of1,653 pounds and a moment/1,000 of 162 pound-inches,the helicopter is within the prescribed CG limits.COMBINATION METHODThe combination method usually uses the computation method to determine the moments and center ofgravity. Then, these figures are plotted on a graph todetermine if they intersect within the acceptable envelope. Figure 7-9 illustrates that with a total weight of2,399 pounds and a total moment of 225,022 pound-FUEL@ STA. 108.5PILOT& PASSENGER@ STA. 83.20100 200 300 400 5004812162024283236MOMENT (THOUSANDS OF LBS.-IN.)LOAD WEIGHT (LBS)Figure 7-7. Moments for fuel, pilot, and passenger.1901801701601501401301201101001,100 1,200 1,300 1,400 1,500 1,600 1,700LOADED WEIGHT (POUND)LOAD MOMENT/1000(POUNDS - INCHES)1. Basic Empty Weight..................2. Pilot and Front Passenger........3. Fuel...........................................5. Baggage...................................TOTALSWeight
(lbs.)Moment(lb.-ins./1,000)1,102 110.828.3 34022.9 211162.0 1,653Aft CG LimitStation 101.0Forward CG LimitStation 95.0Figure 7-8. CG/Moment Chart.7-7inches, the CG is 93.8. Plotting this CG against theweight indicates that the helicopter is loaded withinthe longitudinal limits (point A).CALCULATING LATERAL CGSome helicopter manufacturers require that you alsodetermine the lateral CG limits. These calculations aresimilar to longitudinal calculations. However, since thelateral CG datum line is almost always defined as thecenter of the helicopter, you are likely to encounternegative CGs and moments in your calculations.Negative values are located on the left side while positive stations are located on the right.Refer to figure 7-10. When computing moment for thepilot, 170 pounds is multiplied by the arm of 12.2 inchesresulting in a moment of 2,074 pound-inches. As withany weight placed right of the aircraft centerline, themoment is expressed as a positive value. The forwardpassenger sits left of the aircraft centerline. To computethis moment, multiply 250 pounds by –10.4 inches. Theresult is in a moment of –2,600 pound-inches. Once theaircraft is completely loaded, the weights and momentsare totaled and the CG is computed. Since more weightis located left of the aircraft centerline, the resultingtotal moment is –3,837 pound-inches. To calculate CG,divide –3,837 pound-inches by the total weight of 2,399pounds. The result is –1.6 inches, or a CG that is 1.6inches left of the aircraft centerline.Weight Arm Moment(pounds) (inches) (lb/inches)Basic Empty WeightPilotFwd PassengerRight Fwd BaggageLeft Fwd BaggageRight Aft PassengerLeft Aft PassengerRight Aft BaggageLeft Aft BaggageTotals with Zero FuelMain Fuel TankAux Fuel TankTotals with FuelCG1,40017025018550
502,1051841102,399107.7549.549.5444479.579.579.579.510610293.8150,8508,41512,37500014,7083,9753,975194,29819,50411,220225,022Longitudinal2,5002,3002,1001,9001,7001,500CL91 93 95 97 99 101 1032602562522482442402362321,1001,0501,000950900850800750700Fuselage Station (CM from Datum)Gross Weight - lb.Gross Weight - KGFuselage Station (in. from Datum)MainRotorMost FwdCG withFull FuelLongitudinal(Point A)Figure 7-9. Use the longitudinal CG envelope along with the computed CGs to determine if the helicopter is loaded properly.Figure 7-10. Computed Lateral CG.Weight Arm Moment(pounds) (inches) (lb/inches)Basic Empty WeightPilotFwd PassengerRight Fwd BaggageLeft Fwd BaggageRight Aft PassengerLeft Aft PassengerRight Aft BaggageLeft Aft BaggageTotals with Zero FuelMain Fuel TankAux Fuel TankTotals with FuelCG1,40017025018550502,1051841102,399012.2–10.411.5–11.512.2–12.212.2–12.2–13.513–1.602,074–2,600000–2,257610–610–2,783–2,4841,430–3,837Lateral7-8Lateral CG is often plotted against the longitudinal CG. In this case, –1.6 is plotted against 93.8,which was the longitudinal CG determined in the previous problem. The intersection of the two lines falls wellwithin the lateral CG envelope.CL260256252248244240236
2328R6R4R2R02L4L6L8LFuselage Station (CM from Datum)Lateral - in.Lateral CG - CMFuselage Station (in. from Datum)Lateral3R1R1L3LCLMainRotor(Point A)91 93 95 97 99 101 103Figure 7-11. Use the lateral CG envelope to determine if thehelicopter is properly loaded.8-1Your ability to predict the performance of a helicopteris extremely important. It allows you to determinehow much weight the helicopter can carry beforetakeoff, if your helicopter can safely hover at a specific altitude and temperature, how far it will take toclimb above obstacles, and what your maximumclimb rate will be.FACTORS AFFECTING PERFORMANCEA helicopter’s performance is dependent on the poweroutput of the engine and the lift production of therotors, whether it is the main rotor(s) or tail rotor. Anyfactor that affects engine and rotor efficiency affectsperformance. The three major factors that affect performance are density altitude, weight, and wind.DENSITY ALTITUDEThe density of the air directly affects the performanceof the helicopter. As the density of the air increases,engine power output, rotor efficiency, and aerodynamiclift all increase. Density altitude is the altitude abovemean sea level at which a given atmospheric densityoccurs in the standard atmosphere. It can also beinterpreted as pressure altitude corrected for nonstandard temperature differences.Pressure altitude is displayed as the height above astandard datum plane, which, in this case, is a theoretical plane where air pressure is equal to 29.92 in. Hg.Pressure altitude is the indicated height value on thealtimeter when the altimeter setting is adjusted to29.92 in. Hg. Pressure altitude, as opposed to true altitude, is an important value for calculating performance as it more accurately represents the air content ata particular level. The difference between true altitudeand pressure altitude must be clearly understood. Truealtitude means the vertical height above mean sea leveland is displayed on the altimeter when the altimeter iscorrectly adjusted to the local setting.For example, if the local altimeter setting is 30.12 in.Hg., and the altimeter is adjusted to this value, thealtimeter indicates exact height above sea level.However, this does not reflect conditions found at thisheight under standard conditions. Since the altimetersetting is more than 29.92 in. Hg., the air in this example has a higher pressure, and is more compressed,indicative of the air found at a lower altitude.Therefore, the pressure altitude is lower than the actualheight above mean sea level.To calculate pressure altitude without the use of analtimeter, remember that the pressure decreasesapproximately 1 inch of mercury for every 1,000-footincrease in altitude. For example, if the current localaltimeter setting at a 4,000-foot elevation is 30.42, thepressure altitude would be 3,500 feet. (30.42 – 29.92 =.50 in. Hg. 31,000 feet = 500 feet. Subtracting 500 feetfrom 4,000 equals 3,500 feet).The four factors that most affect density altitude are:atmospheric pressure, altitude, temperature, and themoisture content of the air.ATMOSPHERIC PRESSUREDue to changing weather conditions, atmospheric pressure at a given location changes from day to day. If thepressure is lower, the air is less dense. This means ahigher density altitude and less helicopter performance.Density Altitude—Pressure altitude corrected for nonstandard temperature variations. Performance charts for many older aircraft are basedon this value.Standard Atmosphere—At sea level, the standard atmosphere consistsof a barometric pressure of 29.92 inches of mercury (in. Hg.) or 1013.2millibars, and a temperature of 15°C (59°F). Pressure and temperaturenormally decrease as altitude increases. The standard lapse rate in thelower atmosphere for each 1,000 feet of altitude is approximately 1 in.Hg. and 2°C (3.5°F). For example, the standard pressure and temperature at 3,000 feet mean sea level (MSL) is 26.92 in. Hg. (29.92 – 3) and9°C (15°C – 6°C).Pressure Altitude—The height above the standard pressure level of29.92 in. Hg. It is obtained by setting 29.92 in the barometric pressurewindow and reading the altimeter.True Altitude—The actual height of an object above mean sea level.8-2ALTITUDEAs altitude increases, the air becomes thinner or lessdense. This is because the atmospheric pressure actingon a given volume of air is less, allowing the air molecules to move further apart. Dense air contains more airmolecules spaced closely together, while thin air contains less air molecules because they are spaced furtherapart. As altitude increases, density altitude increases.TEMPERATURETemperature changes have a large affect on density altitude. As warm air expands, the air molecules move further apart, creating less dense air. Since cool aircontracts, the air molecules move closer together, creating denser air. High temperatures cause even low elevations to have high density altitudes.MOISTURE (HUMIDITY)
The water content of the air also changes air densitybecause water vapor weighs less than dry air.Therefore, as the water content of the air increases, theair becomes less dense, increasing density altitude anddecreasing performance.Humidity, also called “relative humidity,” refers to theamount of water vapor contained in the atmosphere,and is expressed as a percentage of the maximumamount of water vapor the air can hold. This amountvaries with temperature; warm air can hold more watervapor, while colder air can hold less. Perfectly dry airthat contains no water vapor has a relative humidity of0 percent, while saturated air that cannot hold any morewater vapor, has a relative humidity of 100 percent.Humidity alone is usually not considered an importantfactor in calculating density altitude and helicopter performance; however, it does contribute. There are norules-of-thumb or charts used to compute the effects ofhumidity on density altitude, so you need to take thisinto consideration by expecting a decrease in hoveringand takeoff performance in high humidity conditions.HIGH AND LOWDENSITY ALTITUDE CONDITIONSYou need to thoroughly understand the terms “highdensity altitude” and “low density altitude.” In general,high density altitude refers to thin air, while low density altitude refers to dense air. Those conditions thatresult in a high density altitude (thin air) are high elevations, low atmospheric pressure, high temperatures,high humidity, or some combination thereof. Lowerelevations, high atmospheric pressure, low temperatures, and low humidity are more indicative of lowdensity altitude (dense air). However, high densityaltitudes may be present at lower elevations on hotdays, so it is important to calculate the density altitudeand determine performance before a flight.One of the ways you can determine density altitude isthrough the use of charts designed for that purpose.. For example, assume you are planning todepart an airport where the field elevation is 1,165 feetMSL, the altimeter setting is 30.10, and the temperature is 70°F. What is the density altitude? First, correctfor nonstandard pressure (30.10) by referring to theright side of the chart, and subtracting 165 feet fromthe field elevation. The result is a pressure altitude of1,000 feet. Then, enter the chart at the bottom, justabove the temperature of 70°F (21°C). Proceed up thechart vertically until you intercept the diagonal 1,000-foot pressure altitude line, then move horizontally tothe left and read the density altitude of approximately2,000 feet. This means your helicopter will perform asif it were at 2,000 feet MSL on a standard day.Most performance charts do not require you to compute density altitude. Instead, the computation is builtinto the performance chart itself. All you have to do isenter the chart with the correct pressure altitude and thetemperature.WEIGHTLift is the force that opposes weight. As weightincreases, the power required to produce the lift neededto compensate for the added weight must also increase.Most performance charts include weight as one of thevariables. By reducing the weight of the helicopter, youmay find that you are able to safely take off or land at alocation that otherwise would be impossible. However,if you are ever in doubt about whether you can safelyperform a takeoff or landing, you should delay yourtakeoff until more favorable density altitude conditionsexist. If airborne, try to land at a location that has morefavorable conditions, or one where you can make alanding that does not require a hover.In addition, at higher gross weights, the increasedpower required to hover produces more torque, whichmeans more antitorque thrust is required. In some helicopters, during high altitude operations, the maximumantitorque produced by the tail rotor during a hovermay not be sufficient to overcome torque even if thegross weight is within limits.WINDSWind direction and velocity also affect hovering, takeoff, and climb performance. Translational lift occursanytime there is relative airflow over the rotor disc.This occurs whether the relative airflow is caused byhelicopter movement or by the wind. As wind speedincreases, translational lift increases, resulting in lesspower required to hover.The wind direction is also an important consideration.Headwinds are the most desirable as they contribute tothe most increase in performance. Strong crosswinds8-3and tailwinds may require the use of more tail rotorthrust to maintain directional control. This increasedtail rotor thrust absorbs power from the engine, whichmeans there is less power available to the main rotorfor the production of lift. Some helicopters even have acritical wind azimuth or maximum safe relative windchart. Operating the helicopter beyond these limitscould cause loss of tail rotor effectiveness.Takeoff and climb performance is greatly affected bywind. When taking off into a headwind, effective translational lift is achieved earlier, resulting in more lift anda steeper climb angle. When taking off with a tailwind,more distance is required to accelerate through translation lift.PERFORMANCE CHARTSIn developing performance charts, aircraft manufacturers make certain assumptions about the condition of thehelicopter and the ability of the pilot. It is assumed that
the helicopter is in good operating condition and theengine is developing its rated power. The pilot isassumed to be following normal operating proceduresand to have average flying abilities. Average means apilot capable of doing each of the required tasks correctly and at the appropriate times.Using these assumptions, the manufacturer develops performance data for the helicopter based onactual flight tests. However, they do not test the helicopter under each and every condition shown on aperformance chart. Instead, they evaluate specificdata and mathematically derive the remaining data.HOVERING PERFORMANCEHelicopter performance revolves around whether ornot the helicopter can be hovered. More power isrequired during the hover than in any other flightregime. Obstructions aside, if a hover can be maintained,a takeoff can be made, especially with the additionalbenefit of translational lift. Hover charts are provided forin ground effect (IGE) hover and out of ground effect(OGE) hoverunder various conditions of gross weight,altitude, temperature, and power. The “in ground effect”hover ceiling is usually higher than the “out of groundeffect” hover ceiling because of the added lift benefitproduced by ground effect.28.028.128.228.328.428.528.628.728.828.929.029.129.229.329.429.529.629.729.829.929.9230.030.130.230.330.430.530.630.730.830.931.01,8241,7271,6301,5331,4361,3401,2441,1481,053957863768673579485392298205112200-73-165-257-348-440-531-622-712-803-893-983AltimeterSettingPressureAltitudeConversionFactor0-18°F°C -1210-720-13044010501660217027803290Outside Air TemperatureApproximate Density Altitude – Thousands of Feet13121110987654321SL12,00011,00010,0009,0008,0007,0006,0005,0004,0003,0002,0001,000-1,000PressureAltitude– FeetStandardTemperatureSeaLevelFigure 8-1. Density Altitude Chart.In Ground Effect (IGE) Hover—Hovering close to the surface (usuallyless than one rotor diameter above the surface) under the influence ofground effect.Out of Ground Effect (OGE) Hover—Hovering greater than one rotordiameter distance above the surface. Because induced drag is greaterwhile hovering out of ground effect, it takes more power to achieve ahover. See Chapter 3—Aerodynamics of Flight for more details on IGEand OGE hover.8-4Since the gross weight of your helicopter is less thanthis, you can safely hover with these conditions.SAMPLE PROBLEM 2Once you reach the remote location in the previousproblem, you will need to hover out of ground effectfor some of the pictures. The pressure altitude at theremote site is 9,000 feet, and you will use 50 poundsof fuel getting there. (The new gross weight is now1,200 pounds.) The temperature will remain at +15°C.Using figure 8-3, can you accomplish the mission?Enter the chart at 9,000 feet (point A) and proceed topoint B (+15°C). From there determine that the maximum gross weight to hover out of ground effect isapproximately 1,130 pounds (point C). Since yourgross weight is higher than this value, you will not beable to hover with these conditions. To accomplish themission, you will have to remove approximately 70pounds before you begin the flight.These two sample problems emphasize the importance ofdetermining the gross weight and hover ceiling throughoutDENSITY ALTITUDE12,600 FTSTANDARD DAY(Point A)900 1,000 1,100 1,200 1,300 1,400GROSS WEIGHT - LBS.425 450 475 500 525 550 57514131211109876543210PRESSURE ALTITUDE - HpX 1,000 FT.OAT°C °F– 20 – 4+ 14+ 32+ 50+ 68+ 86+ 104– 10+ 10+ 20+ 30+ 400OUT OF GROUND EFFECTFULL THROTTLE ( OR LIMIT MANIFOLDPRESSURE) AND 104% RPM
GROSS WEIGHT - KGS.MAX CONT. OR FULL THROTTLEOGE HOVER CEILING VS. GROSS WEIGHT600 625–20–10+10+20+30+400OAT°C(Point B)(Point C)Figure 8-3. Out of Ground Effect Hover Ceiling versus GrossWeight Chart.As density altitude increases, more power is required tohover. At some point, the power required is equal to thepower available. This establishes the hovering ceilingunder the existing conditions. Any adjustment to thegross weight by varying fuel, payload, or both, affectsthe hovering ceiling. The heavier the gross weight, thelower the hovering ceiling. As gross weight isdecreased, the hover ceiling increases.SAMPLE PROBLEM 1You are to fly a photographer to a remote location totake pictures of the local wildlife. Using figure 8-2, canyou safely hover in ground effect at your departurepoint with the following conditions?Pressure Altitude..................................8,000 feetTemperature...............................................+15°CTakeoff Gross Weight.....................1,250 poundsR.P.M..........................................................104%First enter the chart at 8,000 feet pressure altitude(point A), then move right until reaching a point midway between the +10°C and +20°C lines (point B).From that point, proceed down to find the maximumgross weight where a 2 foot hover can be achieved. Inthis case, it is approximately 1,280 pounds (point C).DENSITY ALTITUDE12,600 FTSTANDARD DAY1,370(Point A)(Point B)900 1,000 1,100 1,200 1,300 1,400GROSS WEIGHT - LBS.425 450 475 500 525 550 57514131211109876543210OAT°CPRESSURE ALTITUDE - HpX 1,000 FT.OAT°C °F– 20 – 4+ 14+ 32+ 50+ 68+ 86+ 104– 10+ 10+ 20+ 30+ 400IN GROUND EFFECT AT 2 FOOT SKID CLEARANCEFULL THROTTLE AND 104% RPMGROSS WEIGHT - KGS.–20–10+10+20+30+400IGE HOVER CEILING VS. GROSS WEIGHT(Point C)Figure 8-2. In Ground Effect Hover Ceiling versus GrossWeight Chart.8-5the entire flight operation. Being able to hover at the takeoff location with a certain gross weight does not ensure thesame performance at the landing point. If the destinationpoint is at a higher density altitude because of higher elevation, temperature, and/or relative humidity, more poweris required to hover. You should be able to predict whetherhovering power will be available at the destination byknowing the temperature and wind conditions, using theperformance charts in the helicopter flight manual, andmaking certain power checks during hover and in flightprior to commencing the approach and landing.TAKEOFF PERFORMANCEIf takeoff charts are included in the rotorcraft flight manual, they usually indicate the distance it takes to clear a 50-foot obstacle based on various conditions of weight,pressure altitude, and temperature. In addition, the valuescomputed in the takeoff charts usually assume that theflight profile is per the applicable height-velocity diagram.SAMPLE PROBLEM 3In this example, determine the distance to clear a 50-foot obstacle with the following conditions:Pressure Altitude..................................5,000 feetTakeoff Gross Weight.....................2,850 poundsTemperature .................................................95°FUsing figure 8-4, locate 2,850 pounds in the first column. Since the pressure altitude of 5,000 feet is not oneof the choices in column two, you have to interpolatebetween the values from the 4,000- and 6,000-footlines. Follow each of these rows out to the columnheaded by 95°F. The values are 1,102 feet and 1,538feet. Since 5,000 is halfway between 4,000 and 6,000,the interpolated value should be halfway between thesetwo values or 1,320 feet ( 42 = 1,320).CLIMB PERFORMANCEMost of the factors affecting hover and takeoff performance also affect climb performance. In addition,turbulent air, pilot techniques, and overall condition ofthe helicopter can cause climb performance to vary.A helicopter flown at the “best rate-of-climb” speedwill obtain the greatest gain in altitude over a given
period of time. This speed is normally used during theclimb after all obstacles have been cleared and is usually maintained until reaching cruise altitude. Rate ofclimb must not be confused with angle of climb.Angle of climb is a function of altitude gained over agiven distance. The best rate-of-climb speed results inthe highest climb rate, but not the steepest climb angleand may not be sufficient to clear obstructions. The“best angle-of-climb” speed depends upon the poweravailable. If there is a surplus of power available, thehelicopter can climb vertically, so the best angle-ofclimb speed is zero.Wind direction and speed have an effect on climb performance, but it is often misunderstood. Airspeed isthe speed at which the helicopter is moving throughthe atmosphere and is unaffected by wind.Atmospheric wind affects only the groundspeed, orspeed at which the helicopter is moving over theearth’s surface. Thus, the only climb performanceGrossWeightPoundsPressureAltitudeFeetAt–13°F–25°CAt23°F–5°CAt59°F15°CAt95°F35°CTAKE-OFF DISTANCE (FEET TO CLEAR 50 FOOT OBSTACLE)3734004284615675315686116548117437708619391,2014014344625106745696146607279758068769401,0641,5274304614945857796136607098481,1448649291,0171,255– –2,1502,5002,8504584915276778966527017599861,3559291,0111,1021,5381,320SL2,0004,0006,0008,000SL2,0004,0006,0008,000SL2,0004,0006,0008,000Figure 8-4. Takeoff Distance Chart.8-6affected by atmospheric wind is the angle of climb andnot the rate of climb.SAMPLE PROBLEM 4Determine the best rate of climb using figure 8-5. Usethe following conditions:Pressure Altitude................................12,000 feetOutside Air Temperature ...........................+10°CGross Weight..................................3,000 poundsPower ...........................................Takeoff PowerAnti-ice ..........................................................ONIndicated Airspeed .................................52 knotsWith this chart, first locate the temperature of +10°C
(point A). Then proceed up the chart to the 12,000-footpressure altitude line (point B). From there, move horizontally to the right until you intersect the 3,000-footline (point C). With this performance chart, you mustnow determine the rate of climb with anti-ice off andthen subtract the rate of climb change with it on. Frompoint C, go to the bottom of the chart and find that themaximum rate of climb with anti-ice off is approximately 890 feet per minute. Then, go back to point Cand up to the anti-ice-on line (point D). Proceed horizontally to the right and read approximately 240 feetper minute change (point E). Now subtract 240 from890 to get a maximum rate of climb, with anti-ice on,of 650 feet per minute.Other rate-of-climb charts use density altitude as astarting point. While it cleans up the chartsomewhat, you must first determine density altitude.Notice also that this chart requires a change in the indicated airspeed with a change in altitude.RATE OF CLIMB — MAXIMUMTAKEOFF POWER–40–20 0 20 40 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32OAT — °
C ANTI-ICE OFF RATE OF CLIMB — FT./MIN. (X 100)0100200300400500R/Correction — FT./MIN.This Chart is Based on:Indicated Airspeed 60 MPH 52 KNOTSN2 ENGINE RPM 100%PRESSUREALTITUDE— FT.20,00018,00016,00014,00012,00010,0006,0004,0002,000S.L.GROSSWEIGHTPOUNDS2,0002,2002,4002,6002,8003,0003,2008,000(Point A)(Point B) (Point C)(Point D)(Point E)890 ft/min.– 240 ft/min.650 ft/min.HOTDAYANTI-ICE ONFigure 8-5. Maximum Rate-of-Climb Chart.400 600 800 1,000 1,200 1,40012,00010,0008,0006,0004,0002,0000Rate of Climb, Feet Per MinuteDensity Altitude — FeetRATE OF CLIMB/DENSITY ALTITUDE2,350 LBS. GROSS WEIGHTBEST RATE OF CLIMB SPEED VARIES WITHALTITUDE; 57 MPH AT S.L. DECREASING TO 49MPH, IAS AT 12,000 FT.Figure 8-6. This chart uses density altitude in determiningmaximum rate of climb.9-1From the previous chapters, it should be apparent thatno two helicopters perform the same way. Even whenflying the same model of helicopter, wind, temperature,humidity, weight, and equipment make it difficult topredict just how the helicopter will perform. Therefore,this chapter presents the basic flight maneuvers in away that would apply to a majority of the helicopters.In most cases, the techniques described apply to smalltraining helicopters with:• A single, main rotor rotating in a counterclockwise direction (looking downward on the rotor).• An antitorque system.Where a technique differs, it will be noted. For example,a power increase on a helicopter with a clockwise rotorsystem requires right antitorque pedal pressure insteadof left pedal pressure. In many cases, the terminology“apply proper pedal pressure” is used to indicate bothtypes of rotor systems. However, when discussing throttle coordination to maintain proper r.p.m., there will beno differentiation between those helicopters with a governor and those without. In a sense, the governor is doingthe work for you. In addition, instead of using the termscollective pitch control and the cyclic pitch controlthroughout the chapter, these controls are referred to asjust collective and cyclic.Because helicopter performance varies with differentweather conditions and aircraft loading, specific noseattitudes and power settings will not be discussed. Inaddition, this chapter does not detail each and everyattitude of a helicopter in the various flight maneuvers,nor each and every move you must make in order toperform a given maneuver.When a maneuver is presented, there will be a briefdescription, followed by the technique to accomplishthe maneuver. In most cases, there is a list of commonerrors at the end of the discussion.PREFLIGHTBefore any flight, you must ensure the helicopter isairworthy by inspecting it according to the rotorcraftflight manual, pilot’s operating handbook, or otherinformation supplied either by the operator or the manufacturer. Remember that as pilot in command, it is
