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Question 106-1 : The approximate mean wind component kt along true course 180° from 50°n to 40°n at 005° w is.. err a 033 225 ? [ Exam pilot ]
Tail wind 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.
Question 106-2 : The wind direction and velocity °/kt at 40°n 040°e is .. err a 033 233 ?
330/75.
Img /com en/com033 233.jpg..wind is coming from 330°, with a speed of 75 kt.
Question 106-3 : In the vicinity of shannon 52° n009°w the tropopause is at about.. err a 033 236 ?
Fl 360.
. /com en/com033 236.jpg..
Question 106-4 : At reference or see flight planning manual mrjt 1 figure 4.5.1..given.brake release mass 57 500 kg, temperature isa 10°c.head wind component 16 kt.initial fl 280.find still air distance nam and ground distance nm for the climb.. err a 033 237 ?
62 nam, 59 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.
Question 106-5 : Route manual chart nap.the average magnetic course from c 62°n020°w to b 58°n004°e is .. err a 033 240 ?
119°.
.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.
Question 106-6 : Given.distance from departure to destination 360 nm.safe endurance 4.5 h.true track 345°.w/v 260/30.tas 140 kt.what is the distance of the psr from the departure point ?
308 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.
Question 106-7 : What mean temperature °c is likely on a true course of 270° from 025°e to 010°e at 45°n .. err a 033 246 ?
Mean temperature is 50°c.
. /com en/com033 246.jpg..temperatures are negative unless prefixed by ps. 53 + 51 + 50 + 47 /4 = 50.25°c.
Question 106-8 : At reference or see flight planning manual mrjt 1 figure 4.3.3c. given.ground distance to destination aerodrome 1600 nm. headwind component 50 kt.fl 330.cruise speed 0.78 mach. isa + 20°c. estimated landing weight 55000 kg. ?
12 400 kg, 03h 55 min.
Explanation.. /com en/com033 1017.jpg..
Question 106-9 : Given.distance from departure to destination 260 nm.safe endurance 4.1 h.true track 150°.w/v 100/30.tas 110 kt.what is the distance of the psr from the departure point ?
213 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.
Question 106-10 : Route manual chart nap. the average true course from c 62°n020°w to b 58°n004°e is.. err a 033 262 ?
109°.
. /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°.
Question 106-11 : Given.distance from departure to destination 2450 nm.safe endurance 7.5 h.tas 410 kt.ground speed out 360 kt.ground speed home 460 kt.what is the time of the psr from the departure point ?
252 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.
Question 106-12 : Which describes the worst hazard, if any, that could be associated with the type of feature at 38°n 015°e . 2240 ?
Engine flame out and windscreen damage.
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.
Question 106-13 : Route manual chart nap.the distance nm from c 62°n020°w to b 58°n004°e is .. err a 033 266 ?
760 nm.
.report the track distance along latitude 60°n average latitude between a and c...you will count a little bit more of 25° separation between a and c..25° x 60 nm x cos 60° = 750 nm... /com en/com033 266.jpg..you can also use meridian 1° = 60 nm....take care if you simply calculate latitude distance between 020°w and 004°e, you will have 24° of latitude... but points are not on the same longitude, this is the reason that we need to report the track distance along latitude 60°n, you will now count 25° of latitude on the mid latitude.
Question 106-14 : The planned flight is over a distance of 440 nm.based on the wind charts at altitude the following components are found.fl50 30kt.fl100 50kt.fl180 70kt.the operations manual in appendix details the aircraft's performances.which of the following flight levels fl gives the best range performance.. ?
Fl 180.
.you have to extrapolate data....fl50 tas= 194kt ho y fuel flow= 206 l/hr gs= 164kt flight time= 2.6h fuel burn= 552 l.....fl100 tas= 200kt ho y fuel flow= 192 l/hr gs= 150kt flight time= 2.9h fuel burn= 563.2 l.....fl180 tas= 216kt ho y fuel flow= 163 l/hr gs= 146kt flight time= 3.01h fuel burn= 491 l.....flight level 180 gives the best range performance lowest comsumption.
Question 106-15 : At reference or see flight planning manual mrjt 1 figure 4.3.5.for a flight of 2800 ground nautical miles the following apply. head wind component 20 kt.temperature isa +15°c.brake release mass 64700 kg.the a trip fuel, and b trip time respectively are .. err a 033 268 ?
A 17000 kg b 6h 45 min
. /com en/com033 268.jpg..you always need to go first to the ref line, and then, apply the condition mass, temperature, wind.
Question 106-16 : Given.distance from departure to destination 1950 nm.gs out 400 kt.gs home 300 kt.what is the time of the pet from the departure point ?
Question 106-17 : Given.distance from departure to destination 95 nm.true track 105.wind 060/15.tas 140 kt.what is the distance of the pet from the departure point ?
51 nm.
Question 106-18 : Given.distance from departure to destination 150 nm..true track 020°..wind 180/30.tas 130 kt.what is the distance of the pet from the departure point ?
59 nm.
.find the wind correction angle and the ground speed on the computer.. computer solution.a set true track to true index..b turn the indicator to the wind direction, in this case using the black azeimuth graduation the angle being upwind counting anti clockwise..c shift the speed arc corresponding to the true air speed so as to coincide with the wind speed on the indicator..d read the wind correction at the same place. read the ground speed under the center bore from the scale on the axis slide...gs out = 157 kt..gs home = 102 kt...distance to pet = d x h / o + h.distance to pet = 150 x 102 / 157 + 102 = 59 nm.
Question 106-19 : The surface wind velocity °/kt at paris/charles de gaulle at 1330 utc was.. err a 033 281 ?
270/04.
Question 106-20 : From which of the following would you expect to find the dates and times when temporary danger areas are active ?
Notam and aip
Question 106-21 : What is the earliest time utc , if any, that thunderstorms are forecast for tunis/carthage .. err a 033 284 ?
1800 utc.
If any, the earliest time is 1800 utc..long taf terminal area forecast released at 1020 utc, specify a tempo from 1800 to 0200 next day, with shra or tsshra... /com en/com033 284.jpg..
Question 106-22 : A sector distance is 450 nm long..the tas is 460 kt..the wind component is 50 kt tailwind..what is the still air distance ?
406 nautical air miles nam
.with a tailwind of 50 kt, your ground speed will be 460 kt + 50 kt = 510 kt...450 nm at 510 kt = 450/510 = 0.882 h...nautical air miles = 460 x 0.882 = 406 nam...you can use the following formula.nam = ngm x tas/gs.nam = 450 x 460/510 = 406 nam.
Question 106-23 : Find the distance to the point of safe return psr..given.maximum useable fuel 15000 kg.minimum reserve fuel 3500 kg.outbound tas 425 kt.head wind component 30 kt.fuel flow 2150 kg/h.return tas 430 kt.tailwind component 20 kt.fuel flow 2150 kg/h ?
1125 nm.
.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..outbound gs = 425 30 = 395 kt.homeward gs = 430 + 20 = 450 kt.endurance = 15000 3500 / 2150 = 5.3488 h..point of safe return psr = 5.3488 x 450 / 395 + 450.point of safe return psr = 5.3488 x 450 / 845.point of safe return psr = 2.8484 h..2.8484 h x 395 kt = 1125 nm..we are talking about a point of safe return, a decision in case of our destination is finally unreachable weather issue for example and legally we need to return to our departure airport with our minimum reserve fuel.
Question 106-24 : What is the temperature deviation °c from isa over 50° n 010°e .. err a 033 295 ?
Deviation is 10°.
.at fl300, outside air temperature is 55°c... /com en/com033 295.jpg..at fl300, standard temperature is.15°c 30 x 2°c = 45°c..deviation is 10°c isa 10°c.
Question 106-25 : Given.distance from departure to destination 2000 nm.safe endurance 5 h.tas 500 kt.ground speed out 480 kt.ground speed home 520 kt.what is the distance of the psr from the departure point ?
1248 nm
Question 106-26 : At reference or see flight planning manual mrjt 1 figure 4.3.1c..for a flight of 1900 ground nautical miles the following apply. head wind component 10 kt.temperature isa 5°c.trip fuel available 15000 kg.landing mass 50000kg.what is the minimum cruise level pressure altitude which may be planned .. ?
17000 ft.
. /com en/com033 308.jpg.above 'landing weight' on the right part of the graph , you have to follow an imaginary line between the the dashed line and the continuous line continuous line is for a pressure altitude flight of 37000 ft, dashed line is for pressure altitude flight of 10000 ft.
Question 106-27 : At reference or see flight planning manual mrjt 1 figure 4.4.holding planning...the fuel required for 30 minutes holding, in a racetrack pattern, at pressure altitude 1500 ft, mean gross mass 45 000 kg, is .. err a 033 309 ?
1090 kg.
.you have to interpolate. /com en/com033 309.jpg.. 2220 + 2140 /2 = 2180 kg for one hour..2180/2 = 1090 kg for 30 minutes.
Question 106-28 : The wind °/kt at 60° n015° w is.. err a 033 315 ?
300/60.
. /com en/com033 315.jpg..50 + 10 = 60 kt.
Question 106-29 : Route manual chart nap..the initial true course from c 62°n020°w to b 58°n004°e is.. err a 033 319 ?
098°.
Img /com en/com033 319.jpg..true course 098°.
Question 106-30 : Find the distance from waypoint 3 wpt 3 to the critical point..given.distance from wpt 3 to wpt 4 = 750 nm.tas out 430 kt.tas return 425 kt.tailwind component out 30 kt.head wind component return 40 kt ?
342 nm.
.ground speed out = 430 + 30 = 460 kt..ground speed home = 425 40 = 385 kt....distance to pet = d x gsh / gso + gsh..distance to pet = 750 x 385 / 460 + 385..distance to pet = 288750 / 845 = 341.7 nm.
Question 106-31 : At reference or see flight planning manual mrjt 1 figure 4.5.1.given.brake release mass 57500 kg.initial fl 280.average temperature during climb isa 10°c.aaverage head wind component 18 kt.find climb time for enroute climb 280/.74.. err a 033 325 ?
13 minutes.
. /com en/com033 356.jpg..at 57500 kg we are closer to 58000 kg than 56000 kg, 13 minutes is our answer, no need to interpolate.
Question 106-32 : Given.distance from departure to destination 180 nm.true track 310.wind 010°/20 kt.tas 115 kt.what is the distance of the pet from the departure point ?
98 nm.
Ducksherminator.with my calculations, using crp 5w, i find..gs o = 110kts..gs h = 120kts....pet=dxh/ o+h =93,91nm, which is much closer to 92nm than 98nm......you need to be a little more accurate with the wiz wheel..you will get 104 kt for the outbound leg and 124 kt for the home leg...ground speed out gso = 104 kt..ground speed home gsh = 124 kt...distance to pet = distance x gsh / gso + gsh..distance to pet = 180 x 124 / 104 + 124..distance to pet = 22320 / 228 = 97.89 nm.
Question 106-33 : The surface system over vienna 48°n016°e is a.. err a 033 330 ?
Cold front moving east.
. /com en/com033 330.jpg.cold front moving east at 10 km/h.
Question 106-34 : Given.dry operating mass = 33510 kg.traffic load= 7600 kg.trip fuel = 2040 kg.final reserve fuel= 983 kg.alternate fuel= 1100 kg.contingency fuel= 5% of trip fuel.which of the listed estimated masses is correct ?
Estimated landing mass at destination= 43295 kg.
Question 106-35 : The lowest cloud conditions oktas/ft at bordeaux/merignac at 1330 utc were.. err a 033 340 ?
1 to 2 at 3000 ft.
. /com en/com033 340.jpg..the cloud cover classification is.few 1/8 to 2/8 cloud coverage..sct scattered 3/8 to 4/8 cloud coverage..bkn broken 5/8 to 7/8 cloud coverage..ovc overcast 8/8...030 means 3000 ft... dizertie.tempo indicate a could layer of scatered at 500ft...do you see an answer 3 to 4 at 500 ft .no, and there is a reason tempo doesn't refer to the actual conditions. the question asks for the actual conditions, not the forecasted conditions.
Question 106-36 : At reference or see flight planning manual mrjt 1 figure 4.2. find the short distance cruise altitude for the twin jet aeroplane. given.brake release mass=45000 kg.temperature=isa + 20°c.trip distance=50 nautical air miles nam.. err a 033 343 ?
10000 ft
Question 106-37 : The maximum wind velocity °/kt shown in the vicinity of munich 48°n 012°e is .. err a 033 346 ?
300/140
. arrows, feathers and pennants.arrows indicate direction. number or pennants and/or feathers correspond to speed...example with a 270°/115 kt wind.. /com en/com033 346a.jpg..pennants correspond to 50 kt..feathers correspond to 10 kt..half feathers correspond to 5 kt... /com en/com033 346b.jpg..munich is below the jet axis. direction of the jet is 300°, there is two pennants and four feathers, so 140 kt..there is a decrease of speed after the two oblic lines highlighted in red.
Question 106-38 : At references or see flight planning manual mrjt 1 paragraph 5.2 and figure 4.5.1. planning an ifr flight from paris to london for a twin jet aeroplane..given.estimated take off mass tom 52000 kg.airport elevation 387 ft.fl 280.w/v 280°/40 kt.isa deviation 10°c.average true course 340°. ?
50 nm.
. /com en/com033 347.jpg..tas 353 kt and air distance 53 nm..on nav computer, set under true index the true course 340°, under the center dot, tas 353 kt. with the rotating scale, set wind 280° and under the wind speed 40 kt, you read a right drift of 6°..it means that we need to fly on a true heading of 340° 6° = 334° to stay on the course..set 334° under true index, you read a ground speed of 332 kt.....to determine the ground distance travelled in the climb, multiply the air distance by the groundspeed and divide by the tas..53 x 332 / 353 = 49.84 nm.
Question 106-39 : Flight planning manual mrjt 1 figure 4.5.1. planning an ifr flight from paris charles de gaulle to london heathrow for the twin jet aeroplane..given estimated take off mass tom 52000 kg.airport elevation 387 ft.fl 280.w/v 280°/40 kt.isa deviation 10°c.average true course 340°. ?
11 min
Question 106-40 : Which describes the maximum intensity of turbulence, if any, forecast for fl260 over toulouse 44°n001°e .. err a 033 350 ?
Severe.
.this chart goes from fl100 up to fl450, embedded cumulonimbus clouds with base below fl100 and top up to fl270 are located in the area enclosed by scalloped lines, over toulouse. /com en/com033 350.jpg..even though we are in cat n°1 aera, showing moderate turbulence symbol, isol emb cb isolated embedded cb always means moderate to severe turbulence. thus, the maximum intensity can be severe.
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