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Question 94-1 : Which margin above the stall speed is provided for the landing speed reference vref ? [ Experience landing ]

1 30 vso

exemple 194 1.30 vso

Question 94-2 : With regard to the graph for the light twin aeroplane will the accelerate and stop distance be achieved in a take off where the brakes are released before take off power is set . 2147 ?

No the performance will be worse than in the chart

Admin .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 exemple 198 No, the performance will be worse than in the chart.

Question 94-3 : The critical engine inoperative ?

Increases the power required and the total drag due to the additional drag of the windmilling engine and the compensation of the yaw moment

When an engine failure occurs in a multi engine aircraft asymmetric thrust and drag produce the following effects on the aircraft's axes of rotation . pitch down along the lateral axis loss of accelerated slipstream over the horizontal stabilizer produces less negative lift . roll down toward the inop engine along the longitudinal axis wing produces less lift on side of failed engine due to loss of accelerated slipstream . yaw toward the inop engine along the vertical axis loss of thrust and increased drag from the windmilling propeller .to compensate for these effects a pilot must add additional back pressure deflect the ailerons into the operating engine and apply rudder pressure on the side of the operating engine .a loss of one engine on a multi engine results in a loss of 50% of all available power and up to 80% of the aircraft's excess power and climb performance due to increased drag from the inoperative engine exemple 202 Increases the power required and the total drag due to the additional drag of the windmilling engine and the compensation of the yaw moment.

Question 94-4 : Following a take off determined by the 50ft 15m screen height a light twin climbs on a 10% over the ground climb gradient .it will clear a 900 m high obstacle in relation to the runway horizontally situated at 10 000 m from the 50 ft clearing point with an obstacle clearance of ?

115 m

Climb gradient = change in height / distance travelled.change in height = gradient x distance travelled.change in height = 10 x 10000 = 100000.divide 100000 by 100 to express climb gradient as percentage .100000 / 100 = 1000 m.add 15m screen height = 1015 m..the aircraft will clear a 900 m high obstacle with an obstacle clearance of 115 m exemple 206 115 m.

Question 94-5 : A runway is contaminated with 0 5 cm of wet snow .the flight manual of a light twin nevertheless authorises a landing in these conditions .the landing distance will be in relation to that for a dry runway ?

Increased

exemple 210 Increased.

Question 94-6 : Following a take off limited by the 50 ft screen height a light twin climbs on a gradient of 5% .it will clear a 160 m obstacle in relation to the runway horizontally situated at 5 000 m from the 50 ft point with an obstacle clearance margin of ?

105 m

A 5% climb on 5000 m means a 250 m gain in height 5% > for 100 m horizontal distance you gain 5 m in height + 50 ft 15 m it gives 265 m .265 160 = 105 m exemple 214 105 m.

Question 94-7 : The pilot of an aircraft has calculated a 4000 m service ceiling based on the forecast general conditions for the flight and a take off mass of 3250 kg.if the take off mass is 3000 kg the service ceiling will be ?

Higher than 4000 m

exemple 218 Higher than 4000 m.

Question 94-8 : The flight manual of a light twin engine recommends two cruise power settings 65 and 75 % .the 75% power setting in relation to the 65 % results in ?

An increase in speed fuel consumption and fuel burn/distance

exemple 222 An increase in speed, fuel consumption and fuel-burn/distance.

Question 94-9 : 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 maintain in short final is ?

123 kt

exemple 226 123 kt.

Question 94-10 : 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 and stop distance under the conditions given . 2129 ?

3750 ft

Admin .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 230 3750 ft.

Question 94-11 : Use performance manual mep1 figure 3 2.with 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 a 032 378 ?

Longer than the graphical distance

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 exemple 234 Longer than the graphical distance.

Question 94-12 : For this question use annex ecqb 032 016 v2015 10 or performance manual mep 1 figure 3 7 .given the following conditions what is the one engine inoperative rate of climb .oat 20°c.pressure altitude 14000 ft.gross mass 4000 lb.other conditions as associated in the header of the graph . err a 032 ?

175 ft/min

Img /com en/com032 387 jpg. exemple 238 175 ft/min.

Question 94-13 : Use performance manual mep 1 figure 3 1.given .oat 15°c.pressure altitude 4000 ft.rwy 12r.wind 080°/12 kts.take off mass 4000 lbs.other conditions as associated in the header of the graph .what is the take off distance under the conditions given . err a 032 388 ?

1550 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

Question 94-14 : Use performance manual mep 1 figure 3 7.given .oat 20°c.pressure altitude 18000 ft.gross mass 4000 lbs.mixture leaned to 25°f rich of peak egt.other conditions as associated in the header of the graph .what is the two engine rate of climb for the conditions given . err a 032 389 ?

1050 ft/min

Img /com en/com032 389 jpg. exemple 246 1050 ft/min.

Question 94-15 : Other conditions as associated in the header of the graph .oat 10°c.pressure altitude 2000 ft.gross mass 3750 lbs.mixture full rich.given the following conditions what is the two engine rate of climb . 2124 ?

1770 ft/min

Admin . 1110.you must use the longer lines for the full rich mixture setting exemple 250 1770 ft/min.

Question 94-16 : Use performance manual mep 1 figure 3 1.given .oat 15°c.pressure altitude 4000 ft.rwy 12r.wind 080°/12 kt.take off mass 4000 lbs.other conditions as associated in the header of the graph .what is the ground roll distance under the conditions given . err a 032 391 ?

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 exemple 254 1270 ft.

Question 94-17 : Given .oat 24°c.pressure altitude 3000 ft.rwy 12l.wind 080°/12 kt.take off mass 3800 lbs.other conditions as associated in the header of the graph .what is the take off distance under the conditions given . 2122 ?

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 exemple 258 1700 ft.

Question 94-18 : Given .oat 24°c.pressure altitude 3000 ft.rwy 30r.wind 060°/4 kt.take off mass 3800 lbs.other conditions as associated in the header of the graph .what is the take off distance under the conditions given . 2120 ?

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 exemple 262 2000 ft.

Question 94-19 : Given .oat 20°c.pressure altitude 14000 ft.gross mass 4000 lbs.mixture full rich.other conditions as associated in the header of the graph .what is the two engine rate of climb for the conditions given . 2119 ?

1300 ft/min

Admin . 1113.you must use the longer lines for the full rich mixture setting exemple 266 1300 ft/min.

Question 94-20 : Use performance manual mep 1 figure 3 1.given .oat 24°c.pressure altitude 3000 ft.rwy 12l.wind 080°/12 kt.take off mass 3800 lbs.other conditions as associated in the header of the graph .what is the ground roll distance under the conditions given . err a 032 395 ?

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 exemple 270 1350 ft.

Question 94-21 : Given .oat 24°c.pressure altitude 3000 ft.rwy 30r.wind 060°/4 kt.take off mass 3800 lbs.other conditions as associated in the header of the graph .what is the ground roll distance under the conditions given . 2116 ?

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 exemple 274 1670 ft.

Question 94-22 : Given .oat 20°c.pressure altitude 2000 ft.rwy 07r.wind 120°/ 15 kt.take off mass 4500 lbs.heavy duty brakes installed .other conditions as associated in the header of the graph .what is the accelerate and stop distance under the conditions given . 2117 ?

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 exemple 278 3450 ft.

Question 94-23 : What is the accelerate and stop distance under the conditions given .oat 25°c.pressure altitude 3000 ft.rwy 26l.wind 310°/20kt.take off mass 4400 lb.heavy duty brakes installed.other conditions as associated in the header of the of the graph . 2114 ?

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 exemple 282 3500 ft.

Question 94-24 : What is the accelerate and stop distance under the conditions given .oat 10°c.pressure altitude 4000 ft.rwy 12r.wind 180°/10 kt.take off mass 4600 lb.heavy duty brakes installed .other conditions as associated in the header of the graph . 2115 ?

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 exemple 286 3550 ft.

Question 94-25 : Use performance manual mep 1 figure 3 2.given .oat 25°c.pressure altitude 3000 ft.rwy 24l.wind 310°/20 kt.take off mass 4400 lbs.heavy duty brakes installed.other conditions as associated in the header of the graph .what is the accelerate and stop distance under the conditions given . err a 032 ?

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 exemple 290 3750 ft.

Question 94-26 : What is the accelerate and stop distance under the conditions given .given .oat 20°c.pressure altitude 2000 ft.rwy 24l.wind 120°/ 8 kt.take off mass 4500 lb.heavy duty brakes installed .other conditions as associated in the header of the graph . 2112 ?

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 exemple 294 4200 ft.

Question 94-27 : What is the accelerate and stop distance under the conditions given .oat 10°c.pressure altitude 4000 ft.rwy 30l.wind 180°/10 kt.take off mass 4600 lb.heavy duty brakes installed .other conditions as associated in the header of the graph . 2111 ?

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 exemple 298 4250 ft.

Question 94-28 : Given .oat 10°c.pressure altitude 2000 ft.gross mass 3750 lbs.other conditions as associated in the header of the graph .what is the one engine inoperative rate of climb for the conditions given . 2109 ?

430 ft/min

Admin . 1134 exemple 302 430 ft/min.

Question 94-29 : Given .oat 0°c.pressure altitude 18000 ft.gross mass 3750 lbs.mixture leaned to 25°f rich of peak egt.other conditions as associated in the header of the graph .what is the two engine rate of climb for the condions given . 2110 ?

1050 ft/min

. 1122.you must use the shorter lines for the leaned mixture setting exemple 306 1050 ft/min.

Question 94-30 : Which engine is considered critical in the event of an engine failure during take off for an aeroplane propelled by two clockwise rotating piston engines ?

The left engine

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 310 The left engine.

Question 94-31 : Unless otherwise specified in the afm for a performance class b aeroplane landing on a short grass runway what factor must be applied to the landing distance ?

1 15

Ecqb03 july 2016 exemple 314 1.15

Question 94-32 : A pilot is flying a twin engine piston aeroplane with all engines operating .given the following information what will be the vertical clearance over the obstacle listed below .mtom 4750 kg.airport elevation 500 ft.obstacle elevation 644 ft.distance of the obstacle from the end of the tod 2000 ?

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 exemple 318 150 ft.

Question 94-33 : What is the minimum obstacle clearance above obstacle .given perf class b.cloud base above reference 0 300 ft.wind calm.obstacle distance since end of todr 15000 ft.obstacle height above reference 0 600 ft.rate of climb all engines 1830 ft/min.rate of climb single engine 400 ft/min.tas 101 kt ?

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 exemple 322 215 ft.

Question 94-34 : During take off the third segment begins ?

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 exemple 326 When acceleration to flap retraction speed is started.

Question 94-35 : What is the maximum vertical speed of a three engine turbojet aeroplane with one engine inoperative n 1 and a mass of 75 000 kg .using the following .tas 202 kt.drag 553000n.thrust per engine 300000n.g = 10 m/s².1 kt = 100 ft/min.sin angle of climb = thrust drag / weight ?

+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 exemple 330 +1267 ft/min.

Question 94-36 : During the certification flight testing of a twin engine turbojet aeroplane the 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 for the certification file is ?

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 exemple 334 1779 m.

Question 94-37 : For a turboprop powered aeroplane a 2200 m long runway at the destination aerodrome is expected to be 'wet' the 'dry runway' landing distance should not exceed ?

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 exemple 338 1339 m.

Question 94-38 : Characteristics of a three engine turbojet aeroplane are as follows .thrust = 50 000 newton / engine.g = 10 m/s².drag = 72 569 n.minimum gross gradient 2nd segment = 2 7%.sin angle of climb = thrust drag / weight .the maximum take off mass under 2nd segment conditions is ?

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 exemple 342 101 596 kg.

Question 94-39 : Minimum control speed on the ground 'vmcg' is based on directional control being maintained by ?

Primary aerodynamic control only

exemple 346 Primary aerodynamic control only.

Question 94-40 : Which of the following represents the maximum value for v1 assuming max tyre speed and max brake energy speed are not limiting ?

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 exemple 350 Vr.


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