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Exam > pilot : At a given mass the reference stall speed of a twin engine turboprop aircraft ?

Question 80-1 : 123 kt 115 kt 125 kt 120 kt

Given .oat 25°c .pressure altitude 3000 ft .rwy 24l .wind 310°/20kt .take off ?

Question 80-2 : 3750 ft 4300 ft 4000 ft 3350 ft

.runway 24l 240° wind from 310° it is a headwind .headwind component.20 kt x cos angle between the wind and the runway .20 x cos 70° = 7 kt . 1106.accelerate and stop distance = 4000 ft.but heavy duty brakes are installed we must reduce distance by 7%.4000 x 0 93 = around 3750 ft exemple 184 3750 ft.3750 ft.

Use performance manual mep1 figure 3 2.with regard to the graph for the light ?

Question 80-3 : Longer than the graphical distance shorter than the graphical distance the same as the graphical distance because both techniques are accounted unaffected because all take off techniques are accounted

On the associated condtions you can read 'full power before brake release' .if you release the brake before the engine is in full power your acceleration will be slower a longer runway distance is required to accelerate until v1

For this question use annex ecqb 032 016 v2015 10 or performance manual mep 1 ?

Question 80-4 : 175 ft/min 1250 ft/min 375 ft/min 625 ft/min

/com en/com032 387 jpg. exemple 192 175 ft/min.175 ft/min.

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

Question 80-5 : 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 exemple 196 1550 ft.1550 ft.

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

Question 80-6 : 1050 ft/min 1370 ft/min 870 ft/min 550 ft/min

/com en/com032 389 jpg. exemple 200 1050 ft/min.1050 ft/min.

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

Question 80-7 : 1770 ft/min 1570 ft/min 1970 ft/min 430 ft/min

. 1110.you must use the longer lines for the full rich mixture setting exemple 204 1770 ft/min.1770 ft/min.

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

Question 80-8 : 1270 ft 1600 ft 1500 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 391 png..we are only looking for the ground roll distance not for the total take off distance over 50ft obstacle exemple 208 1270 ft.1270 ft.

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

Question 80-9 : 1700 ft 1600 ft 1420 ft 1950 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 . 1112.we are looking for the total take off distance over 50ft obstacle not for the ground roll distance exemple 212 1700 ft.1700 ft.

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

Question 80-10 : 2000 ft 1670 ft 1550 ft 2150 ft

. 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 exemple 216 2000 ft.2000 ft.

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

Question 80-11 : 1300 ft/min 170 ft/min 970 ft/min 1550 ft/min

. 1113.you must use the longer lines for the full rich mixture setting exemple 220 1300 ft/min.1300 ft/min.

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

Question 80-12 : 1350 ft 1750 ft 1050 ft 1150 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 exemple 224 1350 ft.1350 ft.

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

Question 80-13 : 1670 ft 2000 ft 1780 ft 2150 ft

.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 exemple 228 1670 ft.1670 ft.

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

Question 80-14 : 3450 ft 3650 ft 3250 ft 3800 ft

.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 exemple 232 3450 ft.3450 ft.

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

Question 80-15 : 3500 ft 3800 ft 3350 ft 4300 ft

.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 exemple 236 3500 ft.3500 ft.

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

Question 80-16 : 3550 ft 3800 ft 4600 ft 4300 ft

.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 exemple 240 3550 ft.3550 ft.

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

Question 80-17 : 3750 ft 4000 ft 3350 ft 4300 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 exemple 244 3750 ft.3750 ft.

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

Question 80-18 : 4200 ft 4500 ft 3600 ft 3400 ft

.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 exemple 248 4200 ft.4200 ft.

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

Question 80-19 : 4250 ft 4600 ft 3550 ft 3800 ft

.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 exemple 252 4250 ft.4250 ft.

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

Question 80-20 : 430 ft/min 500 ft/min 890 ft/min 200 ft/min

. 1134 exemple 256 430 ft/min.430 ft/min.

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

Question 80-21 : 1050 ft/min 870 ft/min 1200 ft/min 500 ft/min

. 1122.you must use the shorter lines for the leaned mixture setting exemple 260 1050 ft/min.1050 ft/min.

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

Question 80-22 : 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 exemple 264 the left engine.the left engine.

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

Question 80-23 : 1 15 1 45 0 65 1 05

Ecqb03 july 2016 exemple 268 1.151.15

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

Question 80-24 : 150 ft 50 ft 115 ft 70 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 exemple 272 150 ft.150 ft.

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

Question 80-25 : 215 ft 815 ft 50 ft 235 ft

.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 exemple 276 215 ft.215 ft.

During take off the third segment begins ?

Question 80-26 : 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

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 exemple 280 when acceleration to flap retraction speed is started.when acceleration to flap retraction speed is started.

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

Question 80-27 : +1267 ft/min 1267 ft/min 0 ft/min +3293 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 exemple 284 +1267 ft/min.+1267 ft/min.

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

Question 80-28 : 1779 m 1978 m 1547 m 1720 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 exemple 288 1779 m.1779 m.

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

Question 80-29 : 1339 m 1771 m 1540 m 1147 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 exemple 292 1339 m.1339 m.

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

Question 80-30 : 101 596 kg 74 064 kg 209 064 kg 286 781 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 exemple 296 101 596 kg.101 596 kg.

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

Question 80-31 : 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 300 primary aerodynamic control only.primary aerodynamic control only.

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

Question 80-32 : Vr vmca v2 vref

. 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 exemple 304 vr.vr.

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

Question 80-33 : 3050 m 2938 m 3513 m 2555 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 exemple 308 3050 m.3050 m.

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

Question 80-34 : 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 312 in the accelerate stop distance available.in the accelerate stop distance available.

At which minimum height will the second climb segment end ?

Question 80-35 : 400 ft above field elevation 35 ft above ground when gear retraction is completed 1500 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 exemple 316 400 ft above field elevation.400 ft above field elevation.

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

Question 80-36 : Tas decreases tas increases tas is constant tas is not related to mach number

.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 exemple 320 tas decreases.tas decreases.

The optimum long range cruise altitude for a turbojet aeroplane ?

Question 80-37 : 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 324 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 80-38 : First increases then decreases decreases does not change increases only if there is no wind

.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 exemple 328 first increases then decreases.first increases then decreases.

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

Question 80-39 : C a b d

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 exemple 332 cc

Long range cruise is a flight procedure which gives ?

Question 80-40 : 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 336 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.

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