Un compte Premium offert sur le site FCL.055 ! Rendez-vous sur www.fcl055-exam.fr puis créez un compte avec le même email que celui...   [Lire la suite]


Given .distance from departure to destination 250 nm.gs out 130 kt.gs home 100 ? [ Diploma registration ]

Question 92-1 : 109 nm 141 nm 125 nm 192 nm

Ground speed out gso = 130 kt.ground speed home gsh = 100 kt.distance to pet = distance x gsh / gso + gsh .distance to pet = 250 x 100 / 130 + 100 .distance to pet = 25000 / 230 = 109 nm exemple 192 109 nm.109 nm.

Given .distance from departure to destination 550 nm.endurance 3 6 h.true track ?

Question 92-2 : 231 nm 305 nm 116 nm 319 nm

..under index set true track 200° centre dot on tas 130 kt with the rotative scale set wind 220°/15 kt you find a left drift of 3° .now drift is always measured from heading to track .turn to set true heading 203° 200° + 3° left drift under index you now read your ground speed out of 115 kt.proceed in the same way to find the ground speed home of 142 kt . true track of 020° right drift of 2° true heading of 018° .ground speed out gso = 115 kt.ground speed home gsh = 142 kt.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .point of safe return psr = 3 6 x 142 / 115 + 142 .point of safe return psr = 511 2 / 257.point of safe return psr = 1 99 h.distance of the psr from the departure point at a speed of 115 kt .1 99 x 115 = 229 nm closest answer is 231 nm exemple 196 231 nm.231 nm.

Given .distance from departure to destination 180 nm.endurance 2 h.tas 120 ?

Question 92-3 : Distance 118 nm time 53 min distance 79 nm time 45 min distance 59 nm time 30 min distance 62 nm time 28 min

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .ground speed out = 135 kt.ground speed home = 105 kt.point of safe return psr = 2 x 105 / 135 + 105 .point of safe return psr = 210 / 240.point of safe return psr = 0 875 h.0 875 x 60 = 52 5 minutes.distance of the psr from the departure point at a speed of 135 kt .52 5 min x 135/60 = 118 125 nm exemple 200 Distance: 118 nm, time: 53 min.Distance: 118 nm, time: 53 min.

Given .distance from departure to destination 2500 nm.gs out 540 kt.gs home 470 ?

Question 92-4 : 129 min 171 min 28 min 149 min

.pet = d x vsr / vsa + vsr.pet = 2500 x 470 / 540 + 470.pet = 1163 nm.1163 / 540 = 2 15 h. 2 x 60 + 0 15 x 60 = 129 minutes exemple 204 129 min129 min

Given .distance from departure to destination 875 nm.true track 240 .wind ?

Question 92-5 : Distance 394 nm time 43 min distance 481 nm time 64 min distance 716 nm time 78 min distance 438 nm time 53 min

.true track 240°.wind 060/50 kt.wind is parallel to our course thus .ground speed out 500 + 50 = 550 kt.ground speed home 500 50 = 450 kt.pet = d x gsh / gso + gsh .pet = 875 x 450 / 550 + 450 = 393 75 nm .393 75 / 550 = 0 716 minutes .60 x 0 716 = 43 minutes exemple 208 Distance: 394 nm, time: 43 min.Distance: 394 nm, time: 43 min.

The forecast period covered by the paris/charles de gaulle tafs totals hours . ?

Question 92-6 : 27h 9h 18h 20h

. /com en/com033 122 jpg.total tafs duration 27h exemple 212 27h.27h.

If cas is 190 kts.altitude 9000 ft.temp isa 10°c.true course tc 350°.w/v ?

Question 92-7 : 1213 utc 1221 utc 1233 utc 1203 utc

Convert cas to tas on your computer . /com en/com033 1156 jpg.calculate the outbound and inbound groud speed start first with the wind .40 x cos30 = 34kt .outbound ground speed 215 34 = 181 kt.inbound ground speed 215+34 = 249 kt..point of equal time = 350x249/ 181+249 = 202 67 nm .202 67/181 = 1 12 h .1 12x0 6 = 67 minutes or 1h07 ..11h05+01h07 = 12h33 exemple 216 1213 utc1213 utc

At reference or see flight planning manual mrjt 1 figure 4 4.planning a flight ?

Question 92-8 : 48 125 kg 2 250 kg 48 675 kg 49 250 kg

Landing mass at alternate = dry operating mass + traffic load + final reserve fuel.notice you must land at destination or alternate when pre planning with final reserve fuel in your tanks . /com en/com033 129 jpg.34000+13000 = 47000 kg.interpolate from the table 2280 + 2220 /2 = 2250 kg/h.for 30 minutes = 1125 kg .47000 + 1125 = 48125 kg exemple 220 48 125 kg.48 125 kg.

Given .distance from departure to destination 1385 nm .gs out 480 kt .gs home ?

Question 92-9 : 74 min 128 min 96 min 50 min

.ground speed out = 480 kt.ground speed home = 360 kt..pet = distance x gsh / gso + gsh .pet = 1385 x 360 / 480 + 360 .pet = 498600 / 840 = 593 nm..time of the pet from the departure point .593 / 480 = 1 23 h.1 23 x 60 = 74 minutes exemple 224 74 min.74 min.

Given .distance from departure to destination 256 nm .gs out 160 kt .gs home ?

Question 92-10 : 104 nm 152 nm 128 nm 176 nm

.ground speed out gso = 160 kt.ground speed home gsh = 110 kt.distance to pet = distance x gsh / gso + gsh .distance to pet = 256 x 110 / 160 + 110 .distance to pet = 28160 / 270 = 104 nm exemple 228 104 nm.104 nm.

Given .distance from departure to destination 480 nm.safe endurance 5 h.true ?

Question 92-11 : 280 nm 205 nm 141 nm 199 nm

..start by searching outbound ground speed on nav computer .set 115 kt under center dot true track 315° to true index put wind direction 100° under the red compass rose under 20 kt you read a left drift of 4° now drift is always measured from heading to track turn to set true heading 319° 315° + 4° left drift under index you now read your ground speed out of 130 kt .repeat the operation to find homeward ground speed .outbound gs 130 kt.homeward gs 99 kt.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .point of safe return psr = 5 x 99 / 130 + 99 .point of safe return psr = 495 / 229.point of safe return psr = 2 16 h.0 16 x 60 = 10 minutes.10 + 120 minutes = 130 min.distance of the psr from the departure point at a speed of 132 kt .130 min x 130/60 = 281 6 nm exemple 232 280 nm.280 nm.

Given .distance from departure to destination 150 nm .true track 142° .wind ?

Question 92-12 : 79 nm 71 nm 75 nm 134 nm

..under index set true track 142° centre dot on tas 132 kt with the rotative scale set wind 200°/15 kt you find a left drift of 5° .now drift is always measured from heading to track .turn to set true heading 147° 142° + 5° left drift under index you now read your ground speed out of 124 kt. /com en/com033 142 jpg.proceed in the same way to find the ground speed home of 139 kt . right drift of 5° true heading of 317° .ground speed out gso = 124 kt.ground speed home gsh = 139 kt.distance to pet = distance x gsh / gso + gsh .distance to pet = 150 x 139 / 124 + 139 .distance to pet = 20850 / 263 = 79 2 nm exemple 236 79 nm79 nm

A metar reads .1430z 35002kt 7000 skc 21/03 q1024 =.which of the following ?

Question 92-13 : Temperature/dewpoint runway in use day/month period of validity

1430 = time 1430 utc .35002kt = wind 350°/02 kt.7000 = visibility 7000 meters.skc = sky clear. 21/03 = temperature/dewpoint .q1024 = qnh 1024 hpa exemple 240 Temperature/dewpoint.Temperature/dewpoint.

The wind °/kt at 40°n 020°w is . err a 033 157 ?

Question 92-14 : 310/40 334/40 135/40 155/40

. /com en/com033 157 jpg.10 + 10 + 10 + 10 = 40 kt exemple 244 310/40.310/40.

Given .maximum allowable take off mass 64 400 kg.maximum landing mass 56200 ?

Question 92-15 : 3 000 kg 7 000 kg 5 600 kg 4 000 kg

. /com en/com033 162 jpg.we are able to add 3000 kg beore reaching our first limitation which comes from the maximum zero fuel mass 17500 14500 = 3000 kg exemple 248 3 000 kg.3 000 kg.

What mean temperature °c is likely on a course of 360° t from 40°n to 50°n ?

Question 92-16 : Mean temperature 47°c mean temperature 46°c mean temperature 49°c mean temperature 50°c

. /com en/com033 167 jpg.temperatures are negative unless prefixed by ps . 46 + 47 + 47 + 48 + 49 /5 = 47 4°c exemple 252 Mean temperature : -47°c.Mean temperature : -47°c.

For flight planning purposes the landing mass at alternate is taken as ?

Question 92-17 : Zero fuel mass plus final reserve fuel and contingency fuel landing mass at destination plus alternate fuel zero fuel mass plus final reserve fuel and alternate fuel zero fuel mass plus final reserve fuel

.planned landing mass at alternate = dom + traffic load + final reserve fuel + contingency.if everything goes to plan on the sector you won't use contingency fuel so landing mass at alternate will include final reserve fuel + contingency fuel exemple 256 Zero fuel mass plus final reserve fuel and contingency fuel.Zero fuel mass plus final reserve fuel and contingency fuel.

Given .distance from departure to destination 220 nm.true track 175°.wind ?

Question 92-18 : 116 nm 103 nm 110 nm 136 nm

..under index set true track 175° centre dot on tas 135 kt with the rotative scale set wind 220°/10 kt you find a left drift of 7° .now drift is always measured from heading to track .turn to set true heading 182° 175° + 7° left drift under index you now read your ground speed out of 119 kt. /com en/com033 169 jpg.proceed in the same way to find the ground speed home of 148 kt . right drift of 5° true heading of 350° .ground speed out gso = 119 kt.ground speed home gsh = 148 kt.distance to pet = distance x gsh / gso + gsh .distance to pet = 220 x 148 / 120 + 148 .distance to pet = 32780 / 268 = 121 nm closest answer is 116 nm exemple 260 116 nm116 nm

At references or see flight planning manual mrjt 1 figure 4 2 and figure 4 5 3 ?

Question 92-19 : 25000 ft 435 kt 33500 ft 430 kt 33900 ft 420 kt 24000 ft 445 kt

. /com en/com033 178 jpg.temperature isa 10°c .445 10 = 435 kt exemple 264 25000 ft, 435 kt25000 ft, 435 kt

Given .distance from departure to destination 950 nm.gs out 275 kt .gs home 225 ?

Question 92-20 : 93 min 139 min 114 min 39 min

.distance to pet = distance x gsh / gso + gsh .distance to pet = 950 x 225 / 275 + 225 .distance to pet = 213750 / 500 = 427 5 nm .time of the pet from the departure point .427 5 nm / 275 = 1 55 h.1 55 x 60 min = 93 minutes exemple 268 93 min.93 min.

Given .distance from departure to destination 950 nm .safe endurance 3 5 h .tas ?

Question 92-21 : Distance 622 nm time 117 min distance 528 nm time 79 min distance 311 nm time 52 min distance 328 nm time 62 min

exemple 272 Distance: 622 nm time: 117 minDistance: 622 nm time: 117 min

Given .distance from departure to destination 1000 nm .safe endurance 4 h .tas ?

Question 92-22 : 990 nm 450 nm 495 nm 10 nm

.ground speed out 550 kt.ground speed home 450 kt..point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .point of safe return psr = 4 x 450 / 550 + 450 .point of safe return psr = 1800 / 1000.point of safe return psr = 1 8 h..0 8 x 60 = 48 minutes.48 + 60 minutes = 108 min..distance of the psr from the departure point at a speed of 550 kt .108 min x 550/60 = 990 nm exemple 276 990 nm.990 nm.

Given .distance from departure to destination 2200 nm.true track 150° .wind ?

Question 92-23 : Distance 980 nm time 115 min distance 1120 nm time 179 min distance 1100 nm time 179 min distance 980 nm time 144 min

.track 150° wind from 330° it's a tailwind of 50 kt .ground speed out gso = 460 + 50 = 510 kt..return track 330° wind from 330° it's a headwind of 50 kt .ground speed home gsh = 460 50 = 410 kt..distance to pet = distance x gsh / gso + gsh .distance to pet = 2200 x 410 / 510 + 410 .distance to pet = 902000 / 920 = 980 nm ..time of the pet from the departure point .980 nm / 510 = 1 92 h.1 92 x 60 min = 115 minutes exemple 280 Distance: 980 nm time: 115 minDistance: 980 nm time: 115 min

Route manual chart nap.the initial true course from a 64°n006°e to c ?

Question 92-24 : 271° 259° 247° 279°

. /com en/com033 202 jpg.put your protractor on a align it with the true north you find an initial true course of 271° .departure is almost superposed on the 64° parallel . maercin .i couldn't found those points on the map because there is written that a is on 006e so its impossible to find in this map .nevertheless you may also substract 64 62 = 2 degresses so if we multiply it by 60nm there is 120nm in vertical range if we use formula for distance = 60nm x cos mean longitude x g difference in latitude we have 708nm so now we take usual calculator input arc tg from 120/708 = 9 deg and we know that track from a to b is 270 9 = 261 .after that we count convertion angle = 1/2 x sin mean latitude x difference in latitude and its equal to 10 .so we add 10 to 261 and here it is 271 .i know that this is a little bit more difficult than reading from the map but on the other hand if you wasn't passed gen nav and flight planning its possible to resolve all maps questions without even look on them exemple 284 271°.271°.

Which best describes the maximum intensity of icing if any at fl160 in the ?

Question 92-25 : Moderate severe light nil

. 629. exemple 288 Moderate.Moderate.

At reference or see flight planning manual sep 1 figure 2 1. given . fl 75.oat ?

Question 92-26 : 18 nam 15 nm 14 nam 18 nm 16 nam 18 nm 18 nam 13 nm

. /com en/com033 212 jpg.time to climb 9 minutes .distance to climb 18 nam .with a headwind ground distance to climb will be lower than air distance to climb to fl75 .during 9 minutes 20 kt of wind will reduce our ground distance by 20 kt x 9/60 = 3 nm .18 nam 3 nm = 15 nm exemple 292 18 nam. 15 nm.18 nam. 15 nm.

Given .distance from departure to destination 285 nm .true track 348°.wind ?

Question 92-27 : 154 nm 131 nm 143 nm 123 nm

Img /com en/com033 213a jpg. /com en/com033 213b jpg.ground speed out = 117 kt.proceed the same way to find ground speed home 136 kt .distance to pet = d x gsh / gso + gsh .distance to pet = 285 x 136 / 117 + 136 .distance to pet = 38760 / 253 = 153 2 nm exemple 296 154 nm.154 nm.

Given .distance from departure to destination 435 nm .gs out 110 kt .gs home ?

Question 92-28 : 236 nm 199 nm 218 nm 368 nm

exemple 300 236 nm236 nm

The approximate mean wind component kt along true course 180° from 50°n to ?

Question 92-29 : Tail wind 55 kt tail wind 40 kt tail wind 70 kt headwind 55 kt

. /com en/com033 225 jpg.wind is coming from 320° with an average speed of 70 kt .tailwind component = 70 kt x cos angle between the wind and the course .tailwind component = 70 kt x cos 40°.tailwind = 54 kt exemple 304 Tail wind 55 kt.Tail wind 55 kt.

The wind direction and velocity °/kt at 40°n 040°e is . err a 033 233 ?

Question 92-30 : 330/75 330/85 150/75 300/75

Img /com en/com033 233 jpg.wind is coming from 330° with a speed of 75 kt exemple 308 330/75.330/75.

In the vicinity of shannon 52° n009°w the tropopause is at about . err a 033 ?

Question 92-31 : Fl 360 fl 350 fl 300 fl 270

. /com en/com033 236 jpg. exemple 312 Fl 360.Fl 360.

At reference or see flight planning manual mrjt 1 figure 4 5 1.given .brake ?

Question 92-32 : 62 nam 59 nm 59 nam 62 nm 67 nam 71 nm 71 nam 67 nm

. /com en/com033 237 jpg.we are just below 58000 kg thus 62 nam is a good choice .with a headwind component our ground distance will be less than our air distance 59 nm sounds good exemple 316 62 nam, 59 nm.62 nam, 59 nm.

Route manual chart nap.the average magnetic course from c 62°n020°w to b ?

Question 92-33 : 119° 109° 099° 118°

.to find the average magnetic course from c to b with a protractor between the two points you will find an average true track of 109° then you must add the 10°w magnetic variation we have right in the middle of those two points .109° + 10° = 119° . gomis01 .in a past question asks the true course from c to b and the answer was 098º if right now asks about the magnetic course i am agree about the variation but i think that it will be 098+10 variation = 108º .the answer more near is 109º please tell me if i wrong thank you ..the other question asks for the initial true course from c to b not the average true course exemple 320 119°.119°.

Given .distance from departure to destination 360 nm .safe endurance 4 5 h ?

Question 92-34 : 308 nm 185 nm 154 nm 52 nm

..under index set true track 345° centre dot on tas 140 kt with the rotative scale set wind 260°/30 kt you find a right drift of 12° .now drift is always measured from heading to track .turn to set true heading 333° 345° 12° right drift under index you now read your ground speed out of 135 kt.proceed in the same way to find the ground speed home of 141 kt . left drift of 9° true heading of 174° .ground speed out gso = 135 kt.ground speed home gsh = 141 kt.apply the psr formula .psr = time x gs out x gs home / gs out + gs home .psr = 4 5 x 135 x 141 / 135 + 141 .psr = 310 nm .this is a 4 points question at the exam .mathematical calculation on this kind of exercise is valid only for one right angled triangle which is not the case here only the computer enables you to find the good answer exemple 324 308 nm308 nm

What mean temperature °c is likely on a true course of 270° from 025°e to ?

Question 92-35 : Mean temperature is 50°c mean temperature is 48°c mean temperature is 52°c mean temperature is 54°c

. /com en/com033 246 jpg.temperatures are negative unless prefixed by ps . 53 + 51 + 50 + 47 /4 = 50 25°c exemple 328 Mean temperature is -50°c.Mean temperature is -50°c.

At reference or see flight planning manual mrjt 1 figure 4 3 3c. given .ground ?

Question 92-36 : 12 400 kg 03h 55 min 12 400 kg 04h 12 min 11 400 kg 03h 55 min 11 400 kg 04h 12 min

Explanation . /com en/com033 1017 jpg. exemple 332 12 400 kg, 03h 55 min.12 400 kg, 03h 55 min.

Given .distance from departure to destination 260 nm .safe endurance 4 1 h ?

Question 92-37 : 213 nm 154 nm 107 nm 47 nm

..under index set true track 150° centre dot on tas 110 kt with the rotative scale set wind 100°/30 kt you find a right drift of 9° .now drift is always measured from heading to track .turn to set true heading 141° 150° 9° right drift under index you now read your ground speed out of 90 kt.proceed in the same way to find the ground speed home of 127 kt . left drift of 10° true heading of 340° .ground speed out gso = 90 kt.ground speed home gsh = 127 kt.apply the psr formula .psr = time x gs out x gs home / gs out + gs home .psr = 4 1 x 90 x 127 / 90 + 127 .psr = 215 nm .this is a 4 points question at the exam .mathematical calculation on this kind of exercise is valid only for one right angled triangle which is not the case here only the computer enables you to find the good answer exemple 336 213 nm.213 nm.

Route manual chart nap. the average true course from c 62°n020°w to b ?

Question 92-38 : 109° 119° 099° 120°

. /com en/com033 262 jpg.we are looking for average true course .with you protractor aligned on true north between c and b you will find a true course of 109° exemple 340 109°.109°.

Given .distance from departure to destination 2450 nm .safe endurance 7 5 h ?

Question 92-39 : 252 min 198 min 111 min 156 min

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .ground speed out = 360 kt.ground speed home = 460 kt.point of safe return psr = 7 5 x 460 / 360 + 460 .point of safe return psr = 3450 / 820.point of safe return psr = 4 20 h.4 20 x 60 = 252 minutes exemple 344 252 min252 min

Which describes the worst hazard if any that could be associated with the type ?

Question 92-40 : Engine flame out and windscreen damage severe attenuation in the hf r/t band reduced visibility there is no hazard

At 38°n 015°e we have the etna volcano . 1242. read the story of british airways flight nine you will find and understand the answer .on 24 june 1982 a 747 flew into a cloud of volcanic ash thrown up by the eruption of mount galunggung south east of jakarta indonesia resulting in the failure of all four engines .at approximately 13 42 utc 20 42 jakarta time engine number four began surging and soon flamed out the flight crew immediately performed the engine shutdown drill quickly cutting off fuel supply and arming the fire extinguishers less than a minute later at 13 43 utc 20 43 jakarta time engine two surged and flamed out within seconds and almost simultaneously engines one and three flamed out prompting the flight engineer to exclaim 'i don't believe it all four engines have failed ' .the flight crew quickly determined that the aircraft was capable of gliding for 23 minutes and covering 91 nautical miles 169 km from its flight level of 37 000 feet .at 13 44 utc 20 44 jakarta time the senior first officer declared an emergency to the local air traffic control authority stating that all four engines had failed however jakarta area control misunderstood the message interpreting the call as meaning that only engine number four had shut down it was only after a nearby garuda indonesia flight relayed the message to air traffic control that it was understood despite the crew 'squawking' the emergency transponder setting of 7700 the aeroplane could not be located by air traffic control on their radar screens .due to the high indonesian mountains on the south coast of the island of java an altitude of at least 11 500 feet was required to cross the coast safely the crew decided that if the aircraft was unable to maintain altitude by the time they reached 12 000 feet they would turn back out to sea and attempt to ditch into the indian ocean the crew began the engine restart drills despite being well above the recommended maximum engine in flight start envelope altitude of 28 000 feet the attempts failed .at 13 500 feet they were approaching the altitude at which they would have to turn over the ocean and attempt a risky ditching although there were guidelines for the procedure no one had ever tried it in a boeing 747 nor has anyone since as they performed the engine restart procedure engine number four started and at 13 56 utc 20 56 jakarta time the captain used its power to reduce the rate of descent shortly thereafter engine three restarted allowing him to climb slowly shortly after that engines one and two successfully restarted as well .the crew subsequently requested and expedited an increase in altitude to 11 500 feet in order to clear the high mountains of indonesia . as flight 9 approached jakarta the crew found it difficult to see anything through the windscreen and had to make the approach almost entirely on instruments despite reports of good visibility .although the runway lights could be made out through a small strip of the windscreen the landing lights on the aircraft seemed to be inoperable after landing the flight crew found it impossible to taxi due to glare from apron floodlights which made the already sandblasted windscreen opaque .aftermath it was found that the b747's problems had been caused by flying through a cloud of volcanic ash from the eruption of mount galunggung because the ash cloud was dry it did not show up on the weather radar which is designed to detect the moisture in clouds the cloud sandblasted the windscreen and landing light covers and clogged the engines as the ash entered the engines it melted in the combustion chambers and adhered to the inside of the power plant as the engine cooled from not running and as the aircraft descended out of the ash cloud the molten ash solidified and enough broke off to allow air to flow smoothly through the engine allowing a successful restart the engines had enough electrical power to restart because one generator and the onboard batteries were still operating exemple 348 Engine flame out and windscreen damage.Engine flame out and windscreen damage.

~

Exclusive rights reserved. Reproduction prohibited under penalty of prosecution.

3639 Free Training Exam