At a given mass the reference stall speed of a twin engine turboprop aircraft is 100 kt in the landing configuration The minimum speed a pilot must ?
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Given OAT 25°C Pressure Altitude 3000 ft RWY 24L Wind 310°/20kt Take off Mass 4400 lbs Heavy Duty Brakes installed Other conditions as associated in the header of the graph What is the Accelerate ?
Use Performance Manual MEP1 Figure 3 2With regard to the graph for the light twin aeroplane if the brakes are released before take off power is achieved the accelerate/stop distance will be ERR ?
for this question use annex ecqb 032 016 v2015 10 or performance manual mep 1 figure 3 7 given following conditions what the one engine inoperative rate of climb oat 20°cpressure altitude 14000 ftgross mass 4000 lbother conditions as associated in header of graph err _a_032 387 Longer than graphical distance. Img /com_en/com032 387 jpg . 
Use performance manual mep 1 figure 3 1given oat 15°cpressure altitude 4000 ftrwy 12rwind 080°/12 ktstake off mass 4000 lbsother conditions as associated in header of graph what the take off distance under conditions given err _a_032 388 Longer than graphical distance. Runway 12 (120°) wind from 080° it a headwind headwind component 12 kt x cos (angle between wind the runway) 12 x cos 40° = 9 2 kt img /com_en/com032 388 png we are looking the total take off distance over 50ft obstacle not the ground roll distance. 
Use performance manual mep 1 figure 3 7given oat 20°cpressure altitude 18000 ftgross mass 4000 lbsmixture leaned to 25°f rich of peak egtother conditions as associated in header of graph what the two engine rate of climb the conditions given err _a_032 389 Longer than graphical distance. Img /com_en/com032 389 jpg . 
Other conditions as associated in header of graph oat 10°cpressure altitude 2000 ftgross mass 3750 lbsmixture full richgiven following conditions what the two engine rate of climb 2124 Longer than graphical distance. you must use longer lines the full rich mixture setting. Question 80-8
Use performance manual mep 1 figure 3 1given oat 15°cpressure altitude 4000 ftrwy 12rwind 080°/12 kttake off mass 4000 lbsother conditions as associated in header of graph what the ground roll distance under conditions given err _a_032 391 Longer than graphical distance. Runway 12 (120°) wind from 080° it a headwind headwind component 12 kt x cos (angle between wind the runway) 12 x cos 40° = 9 2 kt img /com_en/com032 391 png we are only looking the ground roll distance not the total take off distance over 50ft obstacle. Question 80-9
Given oat 24°cpressure altitude 3000 ftrwy 12lwind 080°/12 kttake off mass 3800 lbsother conditions as associated in header of graph what the take off distance under conditions given 2122 Longer than graphical distance. runway 12 (120°) wind from 080° it a headwind headwind component 12 kt x cos (angle between wind the runway) 12 x cos 40° = 9 2 kt we are looking the total take off distance over 50ft obstacle not the ground roll distance. Question 80-10
Given oat 24°cpressure altitude 3000 ftrwy 30rwind 060°/4 kttake off mass 3800 lbsother conditions as associated in header of graph what the take off distance under conditions given 2120 Longer than graphical distance. babar350 pay attention that 30r does not mean 030°m but 300°m absolutely runway 30r (300°) wind from 060° it's a tailwind tailwind component 4 kt x cos (angle between wind the runway) = 4 x cos 60° = 2 kt. Question 80-11
Given oat 20°cpressure altitude 14000 ftgross mass 4000 lbsmixture full richother conditions as associated in header of graph what the two engine rate of climb the conditions given 2119 Longer than graphical distance. you must use longer lines the full rich mixture setting. Question 80-12
Use performance manual mep 1 figure 3 1given oat 24°cpressure altitude 3000 ftrwy 12lwind 080°/12 kttake off mass 3800 lbsother conditions as associated in header of graph what the ground roll distance under conditions given err _a_032 395 Longer than graphical distance. Runway 12 (120°) wind from 080° it a headwind headwind component 12 kt x cos (angle between wind the runway) 12 x cos 40° = 9 2 kt img /com_en/com032 395 png we are only looking the ground roll distance not the total take off distance over 50ft obstacle. Question 80-13
Given oat 24°cpressure altitude 3000 ftrwy 30rwind 060°/4 kttake off mass 3800 lbsother conditions as associated in header of graph what the ground roll distance under conditions given 2116 Longer than graphical distance. runway 30r (300°) wind from 060° it a tailwind tailwind component 4 kt x cos (angle between wind the runway) = 4 x cos 60° = 2 kt we are only looking the ground roll distance not the total take off distance over 50ft obstacle. Question 80-14
Given oat 20°cpressure altitude 2000 ftrwy 07rwind 120°/ 15 kttake off mass 4500 lbsheavy duty brakes installed other conditions as associated in header of graph what the accelerate and stop distance under conditions given 2117 Longer than graphical distance. runway 07r (070°) wind from 120° it a headwind headwind component 12 kt x cos (angle between wind the runway) 12 x cos 50° = 9 7 kt accelerate stop distance = 3700 ft but heavy duty brakes are installed we must reduce distance 7% 3700 x 0 93 = around 3450 ft. Question 80-15
What the accelerate and stop distance under conditions given oat 25°cpressure altitude 3000 ftrwy 26lwind 310°/20kttake off mass 4400 lbheavy duty brakes installedother conditions as associated in header of of graph 2114 Longer than graphical distance. runway 26l (260°) wind from 310° it a headwind headwind component 20 kt x cos (angle between wind the runway) 20 x cos 50° = 13 kt accelerate stop distance = 3700 ft but heavy duty brakes are installed we must reduce distance 7% 3700 x 0 93 = around 3500 ft. Question 80-16
What the accelerate and stop distance under conditions given oat 10°cpressure altitude 4000 ftrwy 12rwind 180°/10 kttake off mass 4600 lbheavy duty brakes installed other conditions as associated in header of graph 2115 Longer than graphical distance. head wind component 10 kt x cos (angle between wind the runway) 10 x cos 60° = 5 kt accelerate stop distance = 3800 ft but heavy duty brakes are installed we must reduce distance 7% 3800 x 0 93 = 3534 ft. Question 80-17
Use performance manual mep 1 figure 3 2given oat 25°cpressure altitude 3000 ftrwy 24lwind 310°/20 kttake off mass 4400 lbsheavy duty brakes installedother conditions as associated in header of graph what the accelerate and stop distance under conditions given err _a_032 400 Longer than graphical distance. Head wind component 20 kt x cos (angle between wind the runway) 20 x cos 70° = 7 kt img /com_en/com032 400 jpg accelerate stop distance = 4000 ft but heavy duty brakes are installed we must reduce distance 7% 4000 x 0 93 = 3720 ft. Question 80-18
What the accelerate and stop distance under conditions given given oat 20°cpressure altitude 2000 ftrwy 24lwind 120°/ 8 kttake off mass 4500 lbheavy duty brakes installed other conditions as associated in header of graph 2112 Longer than graphical distance. runway 24l (240°) wind from 120° it a tailwind tailwind component 8 kt x cos (angle between wind the runway) 8 x cos 60° = 4 kt accelerate stop distance = 4500 ft but heavy duty brakes are installed we must reduce distance 7% 4500 x 0 93 = around 4200 ft. Question 80-19
What the accelerate and stop distance under conditions given oat 10°cpressure altitude 4000 ftrwy 30lwind 180°/10 kttake off mass 4600 lbheavy duty brakes installed other conditions as associated in header of graph 2111 Longer than graphical distance. runway 30l (300°) wind from 180° it a tailwind tailwind component 10 kt x cos (angle between wind the runway) 10 x cos 60° = 5 kt accelerate stop distance = 4580 ft but heavy duty brakes are installed we must reduce distance 7% 4580 x 0 93 = around 4250 ft. Question 80-20
Given oat 10°cpressure altitude 2000 ftgross mass 3750 lbsother conditions as associated in header of graph what the one engine inoperative rate of climb the conditions given 2109 Longer than graphical distance. . Question 80-21
Given oat 0°cpressure altitude 18000 ftgross mass 3750 lbsmixture leaned to 25°f rich of peak egtother conditions as associated in header of graph what the two engine rate of climb the condions given 2110 Longer than graphical distance. you must use shorter lines the leaned mixture setting. Question 80-22
Which engine considered critical in event of an engine failure during take off an aeroplane propelled two clockwise rotating piston engines Longer than graphical distance. Ecqb03 july 2016 on a jet engines aeroplane you don't have additional 'critical engine' notion added to your engine failure explanation clockwise rotation as viewed from pilot's seat critical engine will be left engine img /com_en/com080 341 jpg multi engine aeroplanes are subject to p factor just as single engine aeroplanes are the descending propeller blade of each engine will produce greater thrust than ascending blade when aeroplane operated under power at positive angles of attack the descending propeller blade of right engine also a greater distance from center of gravity therefore has a longer moment arm than descending propeller blade of left engine as a result failure of left engine will result in most asymmetrical thrust (adverse yaw) as right engine will be providing remaining thrust. Question 80-23
Unless otherwise specified in afm a performance class b aeroplane landing on a short grass runway what factor must be applied to landing distance Longer than graphical distance. Ecqb03 july 2016 on a jet engines aeroplane you don't have additional 'critical engine' notion added to your engine failure explanation clockwise rotation as viewed from pilot's seat critical engine will be left engine img /com_en/com080 341 jpg multi engine aeroplanes are subject to p factor just as single engine aeroplanes are the descending propeller blade of each engine will produce greater thrust than ascending blade when aeroplane operated under power at positive angles of attack the descending propeller blade of right engine also a greater distance from center of gravity therefore has a longer moment arm than descending propeller blade of left engine as a result failure of left engine will result in most asymmetrical thrust (adverse yaw) as right engine will be providing remaining thrust. Question 80-24
A pilot flying a twin engine piston aeroplane with all engines operating given following information what will be vertical clearance over obstacle listed below mtom 4750 kgairport elevation 500 ftobstacle elevation 644 ftdistance of obstacle from end of tod 2000 mtemperature isaqnh 1018 hpasid straight departure no specific climb requirements published Longer than graphical distance. Ecqb03 july 2016 multi engine class b multi engine class b aircraft have a requirement to clear obstacles 50 ft from end of tod up to 1500ft (using net performance) after which aircraft considered to be en route a) an operator shall ensure that take off flight path of aeroplanes with two or more engines determined in accordance with this sub paragraph clears all obstacles a vertical margin of at least 50 ft but performance class b (multi engined aircraft piston) required a minimum of 4% climb gradient at take off (all engine operating) at 2000 m (or 6560 ft) we are already at a height of 35 ft 4% of 6560 ft = 262 ft 35 + 262 = 297 ft will be our minimum height 2000m after tod obstacle height = 644 ft 500 ft = 144 ft 297 ft 144 ft = 153 ft. Question 80-25
What the minimum obstacle clearance above obstacle given perf class bcloud base above reference 0 300 ftwind calmobstacle distance since end of todr 15000 ftobstacle height above reference 0 600 ftrate of climb all engines 1830 ft/minrate of climb single engine 400 ft/mintas 101 kt Longer than graphical distance. perf class b failure of critical engine assumed to occur where visual reference lost the gradient to engine failure height the all engine gradient x 0 77 giving net gradient 15000 ft x 0 3048 = 4572 m 101 kt x 1 852 = 187 km/h to climb 250 ft at 1830ft/min x 0 77 it takes 250 / (1830x0 77) 0 177 minute at 187 km/h distance to climb 250 ft during 0 177 min will be 187 / 60 minutes = 3 12 km/min 3 12 x 0 177 = 0 552 km or 552 m it remains 4572 552 m before obstacle 4020 m the time taken a distance of 4008 m is 4 020 km / 3 12 km/min = 1 29 min during this time you will climb of 1 29 min x 400 ft/min = 516 ft 300 + 516 = 816 ft 816 600 = 216 ft. Question 80-26
During take off third segment begins When acceleration to flap retraction speed started. The first segment starts at 'reference zero' ends when gear comes up the second segment lasts until levelling off flap retraction the third segment ends when ready the enroute climb it usually a level burst at 400 ft during which acceleration made to climb speed flaps are retracted power reduced to max continuous. Question 80-27
What the maximum vertical speed of a three engine turbojet aeroplane with one engine inoperative n 1 and a mass of 75 000 kg using following tas 202 ktdrag 553000nthrust per engine 300000ng = 10 m/s²1 kt = 100 ft/minsin angle of climb = thrust drag / weight When acceleration to flap retraction speed started. Calculation the climb gradient aircraft weiht in newton 750000 n thrust (2 engines) = 300000 x 2 = 600000 n sin( angle of climb) = (thrust drag)/ weight sin( angle of climb) = (600000 553000) / 750000 = 0 0626 in percent it 6 26% rate of climb = climb gradient x tas rate of climb = 6 26 x 202 = 1265 ft/min. Question 80-28
During certification flight testing of a twin engine turbojet aeroplane real take off distances are equal to 1547 m with all engines running 1720 m with failure of critical engine at v1 with all other things remaining unchanged the take off distance adopted the certification file When acceleration to flap retraction speed started. Cs 25 113 take off distance takeoff run (2) 115% of horizontal distance along take off path with all engines operating from start of take off to point at which aeroplane 11 m (35 ft) above take off surface all engine take off distance 1547 x 1 15 = 1779 m one engine take off distance is 1720 m 1779 m the greatest distance. Question 80-29
For a turboprop powered aeroplane a 2200 m long runway at destination aerodrome expected to be 'wet' the 'dry runway' landing distance should not exceed When acceleration to flap retraction speed started. (2200/1 15) x 0 7 = 1339 m notice 0 7 turboprop 0 6 turbojet factor 1 15 a wet runway brudef isn't it asked 'dry runway' landing distance? in that case 1 15 shouldn't be taken into account the runway at destination expected to be 'wet' you must be able to stop your aeroplane in wet required landing distance (even if at time of arrival runway dry). Question 80-30
Characteristics of a three engine turbojet aeroplane are as follows thrust = 50 000 newton / engineg = 10 m/s²drag = 72 569 nminimum gross gradient 2nd segment = 2 7%sin angle of climb = thrust drag / weight the maximum take off mass under 2nd segment conditions When acceleration to flap retraction speed started. Sin(angle of climb) = (thrust drag) / weight or weight = (thrust drag) / sin(angle of climb) weight = (50000x2 72569) / 0 027 weight = 1015960 n divide newtons g = 101596 kg as we are calculating maximum take off mass a path (to avoid obstacles) we must assume one engine out unless question states otherwise our available thrust only 2x50000 n amassa i'm missing something with formula explanation it says weight = (thrust drag) / sin(angle of climb) weight = (50000 x2 72569) / 0 027 but sin(angle of climb) 0 047 0 027 angle of climb / 100 2 7% express as a decimal 0 027 angle = arctan(0 027) = 1 55° sin 1 55° = 0 027. Question 80-31
Minimum control speed on ground 'vmcg' based on directional control being maintained Primary aerodynamic control only. Sin(angle of climb) = (thrust drag) / weight or weight = (thrust drag) / sin(angle of climb) weight = (50000x2 72569) / 0 027 weight = 1015960 n divide newtons g = 101596 kg as we are calculating maximum take off mass a path (to avoid obstacles) we must assume one engine out unless question states otherwise our available thrust only 2x50000 n amassa i'm missing something with formula explanation it says weight = (thrust drag) / sin(angle of climb) weight = (50000 x2 72569) / 0 027 but sin(angle of climb) 0 047 0 027 angle of climb / 100 2 7% express as a decimal 0 027 angle = arctan(0 027) = 1 55° sin 1 55° = 0 027. Question 80-32
Which of following represents maximum value v1 assuming max tyre speed and max brake energy speed are not limiting Primary aerodynamic control only. note vmca (minimum control speed in air) located between v1 vr but vmca not maximum value v1 vr the speed at which rotation of airplane initiated to takeoff attitude. Question 80-33
During certification flight testing on a four engine turbojet aeroplane actual take off distances measured are 3050 m with failure of critical engine recognised at v1 2555 m with all engines operating and all other things being equalthe take off distance adopted the certification file Primary aerodynamic control only. you multiply all engine performance 1 15 then compare to one engine inoperative distance you take higher figure all engines operating distance = 2555 x 1 15 = 2938 m 3050 m (distance with failure of critical engine recognised at v1) the highest distance adopted the certification file. Question 80-34
In which of following distances can length of a stopway be included In accelerate stop distance available. you multiply all engine performance 1 15 then compare to one engine inoperative distance you take higher figure all engines operating distance = 2555 x 1 15 = 2938 m 3050 m (distance with failure of critical engine recognised at v1) the highest distance adopted the certification file. Question 80-35
At which minimum height will second climb segment end 4 ft above field elevation. the first segment starts at 'reference zero' ends when gear comes up the second segment lasts until levelling off flap retraction the third segment ends when ready the enroute climb it usually a level burst at 400 ft during which acceleration made to climb speed flaps are retracted power reduced to max continuous. Question 80-36
How does tas vary in a constant mach climb in troposphere under isa conditions 4 ft above field elevation. for those questions use very simple 'ertm' diagram the mach line vertical because question states in a constant climb mach ertm e as/r as(rectified air speed or cas)/t as/m ach mach number = tas / local sound speed the velocity of sound decreasing as temperature decreases to maintain constant mach as velocity of sound reduces tas has to reduce. Question 80-37
The optimum long range cruise altitude a turbojet aeroplane Increases when aeroplane mass decreases. for those questions use very simple 'ertm' diagram the mach line vertical because question states in a constant climb mach ertm e as/r as(rectified air speed or cas)/t as/m ach mach number = tas / local sound speed the velocity of sound decreasing as temperature decreases to maintain constant mach as velocity of sound reduces tas has to reduce. Question 80-38
How does specific range change when altitude increases jet aeroplane flying with speed maximum range First increases then decreases. specific air range = tas / fuel flow as altitude increases tas increases therefore specific air range increases but as jet aircraft maximum range in still air achieved at maximum tas/drag ratio approximately 95% rpm (engine efficiency) speed maximum still air range occurs at 1 32 times speed of minimum drag as you climb past a certain height engines go past their best efficiency so initially you increase specific fuel consumption then it decreases. Question 80-39
At reference assuming constant l/d ratio which of diagrams provided correctly shows movement of 'thrust required curve' mass m1 higher than mass m2 err _a_032 55 First increases then decreases. The curve the total drag (on 'thrust required curve' or 'drag or thrust required against airspee) with less mass you need less lift ==> less lift = less induced drag induced drag will decrease displacing total drag curve downwards to left the lowest point on curve vmd eas minimum drag (vmd velocity minimum drag) decreases. Question 80-40
Long range cruise a flight procedure which gives A specific range which approximately 99% of maximum specific range a higher cruise speed. The curve the total drag (on 'thrust required curve' or 'drag or thrust required against airspee) with less mass you need less lift ==> less lift = less induced drag induced drag will decrease displacing total drag curve downwards to left the lowest point on curve vmd eas minimum drag (vmd velocity minimum drag) decreases.
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