讲义学习四性能airbus upfo

1、APPROACH AND LANDINGDescent preparation:determination of actual landing distance,determination of Vapp.ObjectiveLanding Distances on dry runwayStopActual Landing DistanceLanding distance available50 ftRequired Landing DistanceRequired Landing Distance (Dry) = Actual Landing Distance (Dry) 0.6Landing

2、 Distances on wet runwayStopRequired Landing Distance RLD (Dry)Landing distance available50 ftRequired Landing Distance RLD (Wet)R L D (Wet) = R L D (Dry) x 115%Actual Landing DistanceExample- Landing GW: 62 000 kg Wind calm Runway dry Landing elev 1000 ftDry runway, elevation = 0 ftActual landing d

3、istance = 840 m62840GW: 62 000 kg 2.03.10 P3+3 %CORRECTION IS 3 % PER 1000 ft ABOVE SEA LEVEL.840 x1.03=865 m 2.03.10 P3Actual Landing DistanceQRH 2.25Vapp5VappVapp = 127 + 5 = 132 ktsExample- Landing GW: 62 000 kg Wind calmQRH 2.25Abnormal ProcedureDetermination of Vapp and landing distance increme

4、nts.6 3 30 2.3 Example- Landing GW: 62 000 kg Wind calm G+Y HYD outQRH 2.26Vapp = 127+ 30= 157Example- Landing GW: 62 000 kg Wind calm G+Y HYD outQRH 2.25 3 30 2.3 Landing distance = 865 mCorrection x 2.3Landing distance = 1990 mVapp = 127+ 30= 157Example- Landing GW: 62 000 kg Wind calm G+Y HYD out

5、 3 30 2.3 Answer:Vapp = 149 ktALD = 967 mConf 3Exercise- Landing GW: 64 000 kg Airport elevation 0 ft Runway dry 30 kt headwind Initial Config Clean 1 FLAPS 2, Slats 1 3 15 1.3 Exercise- Landing GW: 64 000 kg Airport elevation 0 ft Runway dry 30 kt headwind Initial Config Clean 1 FLAPS 2, Slats 1Vap

6、p = 129 + (15 + Wind correction)= 129 + (25) Limited to 20= 149 kt 3 15 1.3 Exercise- Landing GW: 64 000 kg Airport elevation 0 ft Runway dry 30 kt headwind Initial Config Clean 1 FLAPS 2, Slats 1Exercise- Landing GW: 64 000 kg Airport elevation 0 ft Runway dry 30 kt headwind Initial Config Clean 1 FLAPS 2, Slats 1Landing Dist= (875 - 15%) x 1.3 = 967 mVapp = 129 + (15 + Wind correction)= 129 + (25) Limited to 20