Use performance manual mep 1 figure 3 1.given .oat 15°c.pressure altitude ? [ Question security ]

Question 81-1 : 1550 ft 1270 ft 1220 ft 1830 ft

Runway 12 120° wind from 080° it is a headwind .headwind component .12 kt x cos angle between the wind and the runway 12 x cos 40° = 9 2 kt . /com en/com032 388 png.we are looking for the total take off distance over 50ft obstacle not for the ground roll distance

Use performance manual mep 1 figure 3 7.given .oat 20°c.pressure altitude ?

Question 81-2 : 1050 ft/min 1370 ft/min 870 ft/min 550 ft/min

exemple 185 1050 ft/min. — Img /com en/com032 389 jpg. 1050 ft/min.

Other conditions as associated in the header of the graph .oat 10°c.pressure ?

Question 81-3 : 1770 ft/min 1570 ft/min 1970 ft/min 430 ft/min

Admin . 1110.you must use the longer lines for the full rich mixture setting

Use performance manual mep 1 figure 3 1.given .oat 15°c.pressure altitude ?

Question 81-4 : 1270 ft 1600 ft 1500 ft 1830 ft

exemple 193 1270 ft. — Runway 12 120° wind from 080° it is a headwind .headwind component .12 kt x cos angle between the wind and the runway 12 x cos 40° = 9 2 kt . /com en/com032 391 png.we are only looking for the ground roll distance not for the total take off distance over 50ft obstacle 1270 ft.

Given .oat 24°c.pressure altitude 3000 ft.rwy 12l.wind 080°/12 kt.take off ?

Question 81-5 : 1700 ft 1600 ft 1420 ft 1950 ft

exemple 197 1700 ft. — Admin .runway 12 120° wind from 080° it is a headwind .headwind component .12 kt x cos angle between the wind and the runway 12 x cos 40° = 9 2 kt . 1112.we are looking for the total take off distance over 50ft obstacle not for the ground roll distance 1700 ft.

Given .oat 24°c.pressure altitude 3000 ft.rwy 30r.wind 060°/4 kt.take off ?

Question 81-6 : 2000 ft 1670 ft 1550 ft 2150 ft

exemple 201 2000 ft. — Admin . 1132.babar350 .pay attention that 30r does not mean 030°m but 300°m . .absolutely .runway 30r 300° wind from 060° it's a tailwind . 1115.tailwind component .4 kt x cos angle between the wind and the runway = 4 x cos 60° = 2 kt 2000 ft.

Given .oat 20°c.pressure altitude 14000 ft.gross mass 4000 lbs.mixture full ?

Question 81-7 : 1300 ft/min 170 ft/min 970 ft/min 1550 ft/min

exemple 205 1300 ft/min. — Admin . 1113.you must use the longer lines for the full rich mixture setting 1300 ft/min.

Use performance manual mep 1 figure 3 1.given .oat 24°c.pressure altitude 3000 ?

Question 81-8 : 1350 ft 1750 ft 1050 ft 1150 ft

exemple 209 1350 ft. — Runway 12 120° wind from 080° it is a headwind .headwind component .12 kt x cos angle between the wind and the runway 12 x cos 40° = 9 2 kt . /com en/com032 395 png.we are only looking for the ground roll distance not for the total take off distance over 50ft obstacle 1350 ft.

Given .oat 24°c.pressure altitude 3000 ft.rwy 30r.wind 060°/4 kt.take off ?

Question 81-9 : 1670 ft 2000 ft 1780 ft 2150 ft

exemple 213 1670 ft. — Admin .runway 30r 300° wind from 060° it is a tailwind . 1115.tailwind component .4 kt x cos angle between the wind and the runway = 4 x cos 60° = 2 kt . 1133.we are only looking for the ground roll distance not for the total take off distance over 50ft obstacle 1670 ft.

Given .oat 20°c.pressure altitude 2000 ft.rwy 07r.wind 120°/ 15 kt.take off ?

Question 81-10 : 3450 ft 3650 ft 3250 ft 3800 ft

exemple 217 3450 ft. — Admin .runway 07r 070° wind from 120° it is a headwind .headwind component .12 kt x cos angle between the wind and the runway .12 x cos 50° = 9 7 kt . 1116.accelerate and stop distance = 3700 ft .but heavy duty brakes are installed we must reduce distance by 7% .3700 x 0 93 = around 3450 ft 3450 ft.

What is the accelerate and stop distance under the conditions given .oat ?

Question 81-11 : 3500 ft 3800 ft 3350 ft 4300 ft

exemple 221 3500 ft. — Admin .runway 26l 260° wind from 310° it is a headwind .headwind component .20 kt x cos angle between the wind and the runway .20 x cos 50° = 13 kt . 1117.accelerate and stop distance = 3700 ft .but heavy duty brakes are installed we must reduce distance by 7% .3700 x 0 93 = around 3500 ft 3500 ft.

What is the accelerate and stop distance under the conditions given .oat ?

Question 81-12 : 3550 ft 3800 ft 4600 ft 4300 ft

exemple 225 3550 ft. — Admin .head wind component .10 kt x cos angle between the wind and the runway .10 x cos 60° = 5 kt . 1118.accelerate and stop distance = 3800 ft .but heavy duty brakes are installed we must reduce distance by 7% .3800 x 0 93 = 3534 ft 3550 ft.

Use performance manual mep 1 figure 3 2.given .oat 25°c.pressure altitude 3000 ?

Question 81-13 : 3750 ft 4000 ft 3350 ft 4300 ft

exemple 229 3750 ft. — Head wind component .20 kt x cos angle between the wind and the runway .20 x cos 70° = 7 kt . /com en/com032 400 jpg.accelerate and stop distance = 4000 ft .but heavy duty brakes are installed we must reduce distance by 7% .4000 x 0 93 = 3720 ft 3750 ft.

What is the accelerate and stop distance under the conditions given .given .oat ?

Question 81-14 : 4200 ft 4500 ft 3600 ft 3400 ft

exemple 233 4200 ft. — Admin .runway 24l 240° wind from 120° it is a tailwind .tailwind component .8 kt x cos angle between the wind and the runway .8 x cos 60° = 4 kt . 1120.accelerate and stop distance = 4500 ft .but heavy duty brakes are installed we must reduce distance by 7% .4500 x 0 93 = around 4200 ft 4200 ft.

What is the accelerate and stop distance under the conditions given .oat ?

Question 81-15 : 4250 ft 4600 ft 3550 ft 3800 ft

exemple 237 4250 ft. — Admin .runway 30l 300° wind from 180° it is a tailwind .tailwind component .10 kt x cos angle between the wind and the runway .10 x cos 60° = 5 kt . 1121.accelerate and stop distance = 4580 ft .but heavy duty brakes are installed we must reduce distance by 7% .4580 x 0 93 = around 4250 ft 4250 ft.

Given .oat 10°c.pressure altitude 2000 ft.gross mass 3750 lbs.other conditions ?

Question 81-16 : 430 ft/min 500 ft/min 890 ft/min 200 ft/min

Admin . 1134

Given .oat 0°c.pressure altitude 18000 ft.gross mass 3750 lbs.mixture leaned ?

Question 81-17 : 1050 ft/min 870 ft/min 1200 ft/min 500 ft/min

exemple 245 1050 ft/min. — Admin . 1122.you must use the shorter lines for the leaned mixture setting 1050 ft/min.

Which engine is considered critical in the event of an engine failure during ?

Question 81-18 : The left engine the right engine the left engine during ground run afterward the right engine both engines are equally critical

Ecqb03 july 2016 ..on a jet engines aeroplane you don't have the additional 'critical engine' notion added to your engine failure .explanation .clockwise rotation as viewed from the pilot's seat the critical engine will be the left engine . /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 the ascending blade when the aeroplane is operated under power and at positive angles of attack the descending propeller blade of the right engine is also a greater distance from the center of gravity and therefore has a longer moment arm than the descending propeller blade of the left engine as a result failure of the left engine will result in the most asymmetrical thrust adverse yaw as the right engine will be providing the remaining thrust

Unless otherwise specified in the afm for a performance class b aeroplane ?

Question 81-19 : 1 15 1 45 0 65 1 05

exemple 253 1.15 — Ecqb03 july 2016 1.15

A pilot is flying a twin engine piston aeroplane with all engines operating ?

Question 81-20 : 150 ft 50 ft 115 ft 70 ft

exemple 257 150 ft. — Ecqb03 july 2016..multi engine class b .multi engine class b aircraft have a requirement to clear obstacles by 50 ft from the end of the tod up to 1500ft using net performance after which the aircraft is considered to be en route .a an operator shall ensure that the take off flight path of aeroplanes with two or more engines determined in accordance with this sub paragraph clears all obstacles by 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 150 ft.

What is the minimum obstacle clearance above obstacle .given perf class b.cloud ?

Question 81-21 : 215 ft 815 ft 50 ft 235 ft

exemple 261 215 ft. — Admin .perf class b .failure of the critical engine is assumed to occur where visual reference is lost .the gradient to engine failure height is the all engine gradient x 0 77 giving net gradient . 1124.15000 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 the 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 the obstacle .4020 m .the time taken for 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 215 ft.

During take off the third segment begins ?

Question 81-22 : When acceleration to flap retraction speed is started when landing gear is fully retracted when acceleration starts from vlof to v2 when flap retraction is completed

exemple 265 When acceleration to flap retraction speed is started. — The first segment starts at 'reference zero' and ends when the gear comes up .the second segment lasts until levelling off for flap retraction .the third segment ends when ready for the enroute climb .it is usually a level burst at 400 ft during which acceleration is made to climb speed flaps are retracted and power is reduced to max continuous When acceleration to flap retraction speed is started.

What is the maximum vertical speed of a three engine turbojet aeroplane with ?

Question 81-23 : +1267 ft/min 1267 ft/min 0 ft/min +3293 ft/min

exemple 269 +1267 ft/min. — Calculation for 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 is 6 26%.rate of climb = climb gradient x tas.rate of climb = 6 26 x 202 = 1265 ft/min +1267 ft/min.

During the certification flight testing of a twin engine turbojet aeroplane the ?

Question 81-24 : 1779 m 1978 m 1547 m 1720 m

exemple 273 1779 m. — Cs 25 113 take off distance and takeoff run . . 2 115% of the horizontal distance along the take off path with all engines operating from the start of the take off to the point at which the aeroplane is 11 m 35 ft above the take off surface .all engine take off distance is 1547 x 1 15 = 1779 m .one engine take off distance is 1720 m.1779 m is the greatest distance 1779 m.

For a turboprop powered aeroplane a 2200 m long runway at the destination ?

Question 81-25 : 1339 m 1771 m 1540 m 1147 m

exemple 277 1339 m. — 2200/1 15 x 0 7 = 1339 m..notice .0 7 turboprop .0 6 turbojet ..factor 1 15 for a wet runway . brudef .isn't it asked the 'dry runway' landing distance in that case 1 15 shouldn't be taken into account ..the runway at destination is expected to be 'wet' you must be able to stop your aeroplane in the wet required landing distance even if at the time of arrival the runway is dry 1339 m.

Characteristics of a three engine turbojet aeroplane are as follows .thrust = ?

Question 81-26 : 101 596 kg 74 064 kg 209 064 kg 286 781 kg

exemple 281 101 596 kg. — 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 by g = 101596 kg .as we are calculating maximum take off mass for a path to avoid obstacles we must assume one engine out unless the question states otherwise our available thrust is only 2x50000 n . amassa .i'm missing something with the formula explanation .it says .weight = thrust drag / sin angle of climb .weight = 50000 x2 72569 / 0 027.but sin angle of climb is 0 047 .0 027 is angle of climb / 100..2 7% express as a decimal is 0 027 .angle = arctan 0 027 = 1 55°.sin 1 55° = 0 027 101 596 kg.

Minimum control speed on the ground 'vmcg' is based on directional control ?

Question 81-27 : Primary aerodynamic control only primary aerodynamic control and nose wheel steering primary aerodynamic control nose wheel steering and differential braking nose wheel steering only

exemple 285 Primary aerodynamic control only. — Primary aerodynamic control only.

Which of the following represents the maximum value for v1 assuming max tyre ?

Question 81-28 : Vr vmca v2 vref

exemple 289 Vr. — Admin . 1459.note vmca minimum control speed in the air is located between v1 and vr but vmca is not the maximum value for v1 .vr is the speed at which the rotation of the airplane is initiated to takeoff attitude Vr.

During certification flight testing on a four engine turbojet aeroplane the ?

Question 81-29 : 3050 m 2938 m 3513 m 2555 m

exemple 293 3050 m. — .you multiply the all engine performance by 1 15 and then compare to the one engine inoperative distance you take the higher figure .all engines operating distance = 2555 x 1 15 = 2938 m .3050 m distance with failure of the critical engine recognised at v1 is the highest distance adopted for the certification file 3050 m.

In which of the following distances can the length of a stopway be included ?

Question 81-30 : In the accelerate stop distance available in the one engine failure case take off distance in the all engine take off distance in the take off run available

exemple 297 In the accelerate stop distance available. — In the accelerate stop distance available.

At which minimum height will the second climb segment end ?

Question 81-31 : 400 ft above field elevation 35 ft above ground when gear retraction is completed 1500 ft above field elevation

exemple 301 400 ft above field elevation. — .the first segment starts at 'reference zero' and ends when the gear comes up .the second segment lasts until levelling off for flap retraction .the third segment ends when ready for the enroute climb it is usually a level burst at 400 ft during which acceleration is made to climb speed flaps are retracted and power is reduced to max continuous 400 ft above field elevation.

How does tas vary in a constant mach climb in the troposphere under isa ?

Question 81-32 : Tas decreases tas increases tas is constant tas is not related to mach number

exemple 305 Tas decreases. — Admin .for those questions use the very simple 'ertm' diagram . 1039.the mach line is vertical because the question states in a constant climb mach . ertm for e as/ r as rectified air speed or cas / t as/ m ach .mach number = tas / local sound speed .the velocity of sound is decreasing as temperature decreases to maintain constant mach as the velocity of sound reduces tas has to reduce Tas decreases.

The optimum long range cruise altitude for a turbojet aeroplane ?

Question 81-33 : Increases when the aeroplane mass decreases is always equal to the powerplant ceiling is independent of the aeroplane mass is only dependent on the outside air temperature

exemple 309 Increases when the aeroplane mass decreases. — Increases when the aeroplane mass decreases.

How does the specific range change when the altitude increases for jet ?

Question 81-34 : First increases then decreases decreases does not change increases only if there is no wind

exemple 313 First increases then decreases. — .specific air range = tas / fuel flow.as altitude increases tas increases therefore specific air range increases .but as for jet aircraft maximum range in still air is achieved at maximum tas/drag ratio and approximately 95% rpm engine efficiency the speed for maximum still air range occurs at 1 32 times the speed of minimum drag as you climb past a certain height the engines go past their best efficiency so initially you increase specific fuel consumption then it decreases First increases then decreases.

At reference .assuming constant l/d ratio which of the diagrams provided ?

Question 81-35 : C a b d

exemple 317 C — The curve is the total drag on the 'thrust required curve' or 'drag or thrust required against airspeed' .with less mass you need less lift ==> less lift = less induced drag .induced drag will decrease displacing the total drag curve downwards and to the left .the lowest point on the curve is vmd eas for minimum drag vmd velocity minimum drag decreases C

Long range cruise is a flight procedure which gives ?

Question 81-36 : A specific range which is approximately 99% of maximum specific range and a higher cruise speed a 1% higher tas for maximum specific range an ias which is 1% higher than the ias for maximum specific range a specific range which is approximately 99% of maximum specific range and a lower cruise speed

exemple 321 A specific range which is approximately 99% of maximum specific range and a higher cruise speed. — A specific range which is approximately 99% of maximum specific range and a higher cruise speed.

A commercial flight is planned with a turbojet aeroplane to an aerodrome with a ?

Question 81-37 : 1 440 m 1 250 m 1 090 m 1 655 m

exemple 325 1 440 m. — .eu ops 1 515 landing dry runways . a an operator shall ensure that the landing mass of the aeroplane determined in accordance with eu ops 1 475 a for the estimated time of landing at the destination aerodrome and at any alternate aerodrome allows a full stop landing from 50 ft above the threshold . 1 for turbo jet powered aeroplanes within 60% of the landing distance available or. 2 for turbo propeller powered aeroplanes within 70% of the landing distance available . 2400 x 0 6 = 1440 m 1 440 m.

At the destination aerodrome the landing distance available is 3000m .the ?

Question 81-38 : 1565 m 2070 m 1800 m 2609 m

exemple 329 1565 m. — .eu ops 1 515 landing dry runways allows a full stop landing from 50 ft above the threshold . 1 for turbo jet powered aeroplanes within 60% of the landing distance available .landing wet and contaminated runways a the landing distance available is at least 115% of the required landing distance determined in accordance with eu ops 1 515 .3000 m x 0 6 = 1800 m.1800 m / 1 15 = 1565 m 1565 m.

With zero wind the angle of attack for maximum range for an aeroplane with ?

Question 81-39 : Lower than the angle of attack corresponding to maximum endurance equal to the angle of attack corresponding to maximum endurance equal to the angle of attack corresponding to zero induced drag equal to the angle of attack corresponding to maximum lift to drag ratio

exemple 333 Lower than the angle of attack corresponding to maximum endurance. — Admin . 1100.for a jet aeroplane maximum endurance is achieved at a speed corresponding to the maximum l/d ratio which is vmd where the gap between power required and power available is greatest at this speed for a conventional aerofoil the angle of attack is about 4° .the speed for maximum range occurs at 1 32 times the speed of minimum drag vmd in level flight as speed goes up the angle of attack goes down angle of attack is less than 4° typically at around 2 5° .example .at 10° you are at the maximum range aoa .at 16° you are at the maximum endurance aoa . 2416 Lower than the angle of attack corresponding to maximum endurance.

Two identical turbojet aeroplane whose specific fuel consumptions are ?

Question 81-40 : 3804 kg/h 3578 kg/h 3365 kg/h 4044 kg/h

exemple 337 3804 kg/h. — 115/130 x4300 = 3803 kg/h 3804 kg/h.

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