In the vicinity of SHANNON 52° N009°W the tropopause is at about ERR _a_033 236 ?
Exam > pilot
At reference or see Flight Planning Manual MRJT 1 Figure 4 5 1Given brake release mass 57 500 kg temperature ISA 10°CHead wind component 16 ktInitial FL 280Find still air distance NAM and ground ?
Route Manual chart NAPThe average magnetic course from C 62°N020°W to B 58°N004°E is ERR _a_033 240 ?
Given distance from departure to destination 360 nm safe endurance 4 5 h true track 345°w/v 260/30 tas 140 ktwhat the distance of psr from departure point under index set true track 345° centre dot on tas 140 kt with rotative scale set wind 260°/30 kt you find a right drift of 12° now drift 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 same way to find 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 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 a 4 points question at exam mathematical calculation on this kind of exercise valid only one right angled triangle which not case here only computer enables you to find good answer. 
What mean temperature °c likely on a true course of 270° from 025°e to 010°e at 45°n err _a_033 246 Mean temperature 5 °c. img /com_en/com033 246 jpg temperatures are negative unless prefixed ps (53 + 51 + 50 + 47)/4 = 50 25°c. 
At reference or see flight planning manual mrjt 1 figure 4 3 3c given ground distance to destination aerodrome 1600 nm headwind component 50 ktfl 330cruise speed 0 78 mach isa + 20°c estimated landing weight 55000 kg find simplified flight planning to determine estimated trip fuel and trip time err _a_033 252 2 4 kg 3h 55 min. Explanation img /com_en/com033 1017 jpg . 
Given distance from departure to destination 260 nm safe endurance 4 1 h true track 150°w/v 100/30 tas 110 ktwhat the distance of psr from departure point 2 4 kg 3h 55 min. under index set true track 150° centre dot on tas 110 kt with rotative scale set wind 100°/30 kt you find a right drift of 9° now drift 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 same way to find 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 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 a 4 points question at exam mathematical calculation on this kind of exercise valid only one right angled triangle which not case here only computer enables you to find good answer. Question 90-8
Route manual chart nap the average true course from c 62°n020°w to b 58°n004°e err _a_033 262 2 4 kg 3h 55 min. img /com_en/com033 262 jpg we are looking average true course with you protractor aligned on true north between c b you will find a true course of 109°. Question 90-9
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 the time of psr from departure point 2 4 kg 3h 55 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 90-10
Which describes worst hazard if any that could be associated with type of feature at 38°n 015°e 2240 Engine flame out windscreen damage. At 38°n 015°e we have etna volcano read story of british airways flight nine you will find understand answer on 24 june 1982 a 747 flew into a cloud of volcanic ash thrown up the eruption of mount galunggung (south east of jakarta indonesia) resulting in failure of all four engines at approximately 13 42 utc (20 42 jakarta time) engine number four began surging soon flamed out the flight crew immediately performed engine shutdown drill quickly cutting off fuel supply arming fire extinguishers less than a minute later at 13 43 utc (20 43 jakarta time) engine two surged flamed out within seconds almost simultaneously engines one three flamed out prompting flight engineer to exclaim 'i don't believe it all four engines have failed!' the flight crew quickly determined that aircraft was capable of gliding 23 minutes covering 91 nautical miles (169 km) from its flight level of 37 000 feet at 13 44 utc (20 44 jakarta time) senior first officer declared an emergency to local air traffic control authority stating that all four engines had failed however jakarta area control misunderstood message interpreting call as meaning that only engine number four had shut down it was only after a nearby garuda indonesia flight relayed message to air traffic control that it was understood despite crew 'squawking' emergency transponder setting of 7700 aeroplane could not be located air traffic control on their radar screens due to high indonesian mountains on south coast of island of java an altitude of at least 11 500 feet was required to cross coast safely the crew decided that if aircraft was unable to maintain altitude the time they reached 12 000 feet they would turn back out to sea attempt to ditch into indian ocean the crew began engine restart drills despite being well above recommended maximum engine in flight start envelope altitude of 28 000 feet the attempts failed at 13 500 feet they were approaching altitude at which they would have to turn over ocean attempt a risky ditching although there were guidelines the procedure no one had ever tried it in a boeing 747 nor has anyone since as they performed engine restart procedure engine number four started at 13 56 utc (20 56 jakarta time) captain used its power to reduce rate of descent shortly thereafter engine three restarted allowing him to climb slowly shortly after that engines one two successfully restarted as well the crew subsequently requested expedited an increase in altitude to 11 500 feet in order to clear high mountains of indonesia as flight 9 approached jakarta crew found it difficult to see anything through windscreen had to make approach almost entirely on instruments despite reports of good visibility although runway lights could be made out through a small strip of windscreen landing lights on aircraft seemed to be inoperable after landing flight crew found it impossible to taxi due to glare from apron floodlights which made already sandblasted windscreen opaque aftermath it was found that b747's problems had been caused flying through a cloud of volcanic ash from eruption of mount galunggung because ash cloud was dry it did not show up on weather radar which designed to detect moisture in clouds the cloud sandblasted windscreen landing light covers clogged engines as ash entered engines it melted in combustion chambers adhered to inside of power plant as engine cooled from not running as aircraft descended out of ash cloud molten ash solidified enough broke off to allow air to flow smoothly through engine allowing a successful restart the engines had enough electrical power to restart because one generator the onboard batteries were still operating. Question 90-11
Route manual chart napthe distance nm from c 62°n020°w to b 58°n004°e err _a_033 266 Engine flame out windscreen damage. report track distance along latitude 60°n (average latitude between a c) you will count a little bit more of 25° separation between a c 25° x 60 nm x cos 60° = 750 nm img /com_en/com033 266 jpg you can also use meridian 1° = 60 nm take care if you simply calculate latitude distance between 020°w 004°e you will have 24° of latitude but points are not on same longitude this the reason that we need to report track distance along latitude 60°n you will now count 25° of latitude on mid latitude. Question 90-12
The planned flight over a distance of 440 nmbased on wind charts at altitude following components are foundfl50 30ktfl100 50ktfl180 70ktthe operations manual in appendix details aircraft's performanceswhich of following flight levels fl gives best range performance err _a_033 267 Engine flame out windscreen damage. you have to extrapolate data fl50 tas= 194kt hourly fuel flow= 206 l/hr gs= 164kt flight time= 2 6h fuel burn= 552 l fl100 tas= 200kt hourly fuel flow= 192 l/hr gs= 150kt flight time= 2 9h fuel burn= 563 2 l fl180 tas= 216kt hourly fuel flow= 163 l/hr gs= 146kt flight time= 3 01h fuel burn= 491 l flight level 180 gives best range performance (lowest comsumption). Question 90-13
At reference or see flight planning manual mrjt 1 figure 4 3 5for a flight of 2800 ground nautical miles following apply head wind component 20 kttemperature isa +15°cbrake release mass 64700 kgthe a trip fuel and b trip time respectively are err _a_033 268 (a) 7 kg (b) 6h 45 min. img /com_en/com033 268 jpg you always need to go first to ref line then apply condition (mass temperature wind). Question 90-14
Given distance from departure to destination 1950 nm gs out 400 kt gs home 300 kt what the time of pet from departure point (a) 7 kg (b) 6h 45 min. pet = distance x gsh / (gso+ gsh) pet = 1950 x 300 / (400 + 300) pet = 585000 / 700 = 835 nm 835 nm / 400 = 2 09 h (2 hour 05 minutes = 125 minutes). Question 90-15
Given distance from departure to destination 95 nm true track 105 wind 060/15 tas 140 kt what the distance of pet from departure point (a) 7 kg (b) 6h 45 min. pet = distance x gsh / (gso+ gsh) pet = 1950 x 300 / (400 + 300) pet = 585000 / 700 = 835 nm 835 nm / 400 = 2 09 h (2 hour 05 minutes = 125 minutes). Question 90-16
Given distance from departure to destination 150 nm true track 020° wind 180/30tas 130 ktwhat the distance of pet from departure point (a) 7 kg (b) 6h 45 min. find wind correction angle the ground speed on computer computer solution a) set true track to true index b) turn indicator to wind direction in this case using black azeimuth graduation (the angle being upwind counting anti clockwise) c) shift speed arc corresponding to true air speed so as to coincide with wind speed on indicator d) read wind correction at same place read ground speed under center bore from scale on 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 90-17
The surface wind velocity °/kt at paris/charles de gaulle at 1330 utc was err _a_033 281 (a) 7 kg (b) 6h 45 min. find wind correction angle the ground speed on computer computer solution a) set true track to true index b) turn indicator to wind direction in this case using black azeimuth graduation (the angle being upwind counting anti clockwise) c) shift speed arc corresponding to true air speed so as to coincide with wind speed on indicator d) read wind correction at same place read ground speed under center bore from scale on 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 90-18
From which of following would you expect to find dates and times when temporary danger areas are active (a) 7 kg (b) 6h 45 min. find wind correction angle the ground speed on computer computer solution a) set true track to true index b) turn indicator to wind direction in this case using black azeimuth graduation (the angle being upwind counting anti clockwise) c) shift speed arc corresponding to true air speed so as to coincide with wind speed on indicator d) read wind correction at same place read ground speed under center bore from scale on 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 90-19
What the earliest time utc if any that thunderstorms are forecast tunis/carthage err _a_033 284 (a) 7 kg (b) 6h 45 min. If any earliest time 1800 utc long taf (terminal area forecast ) released at 1020 utc specify a tempo from 1800 to 0200 next day with shra or tsshra img /com_en/com033 284 jpg . Question 90-20
A sector distance 450 nm long the tas 460 kt the wind component 50 kt tailwind what the still air distance 4 6 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 following formula nam = ngm x (tas/gs) nam = 450 x (460/510) = 406 nam. Question 90-21
Find distance to point of safe return psr given maximum useable fuel 15000 kgminimum reserve fuel 3500 kgoutbound tas 425 kthead wind component 30 ktfuel flow 2150 kg/hreturn tas 430 kttailwind component 20 ktfuel flow 2150 kg/h 4 6 nautical air miles (nam). 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 finally unreachable (weather issue example) legally we need to return to our departure airport with our minimum reserve fuel. Question 90-22
What the temperature deviation °c from isa over 50° n 010°e err _a_033 295 4 6 nautical air miles (nam). at fl300 outside air temperature 55°c img /com_en/com033 295 jpg at fl300 standard temperature is 15°c (30 x 2°c) = 45°c deviation 10°c (isa 10°c). Question 90-23
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 the distance of psr from departure point 4 6 nautical air miles (nam). at fl300 outside air temperature 55°c img /com_en/com033 295 jpg at fl300 standard temperature is 15°c (30 x 2°c) = 45°c deviation 10°c (isa 10°c). Question 90-24
At reference or see flight planning manual mrjt 1 figure 4 3 1cfor a flight of 1900 ground nautical miles following apply head wind component 10 kttemperature isa 5°ctrip fuel available 15000 kglanding mass 50000kgwhat the minimum cruise level pressure altitude which may be planned err _a_033 308 4 6 nautical air miles (nam). img /com_en/com033 308 jpg above 'landing weight' (on right part of graph) you have to follow an imaginary line between the dashed line the continuous line (continuous line for a pressure altitude flight of 37000 ft dashed line for pressure altitude flight of 10000 ft). Question 90-25
At reference or see flight planning manual mrjt 1 figure 4 4holding planning the fuel required 30 minutes holding in a racetrack pattern at pressure altitude 1500 ft mean gross mass 45 000 kg err _a_033 309 4 6 nautical air miles (nam). you have to interpolate img /com_en/com033 309 jpg (2220 + 2140)/2 = 2180 kg one hour 2180/2 = 1090 kg 30 minutes. Question 90-26
The wind °/kt at 60° n015° w err _a_033 315 4 6 nautical air miles (nam). img /com_en/com033 315 jpg 50 + 10 = 60 kt. Question 90-27
Route manual chart napthe initial true course from c 62°n020°w to b 58°n004°e err _a_033 319 4 6 nautical air miles (nam). Img /com_en/com033 319 jpg true course 098°. Question 90-28
Find distance from waypoint 3 wpt 3 to critical point given distance from wpt 3 to wpt 4 = 750 nmtas out 430 kttas return 425 kttailwind component out 30 kthead wind component return 40 kt 4 6 nautical air miles (nam). 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 90-29
At reference or see flight planning manual mrjt 1 figure 4 5 1given brake release mass 57500 kginitial fl 280average temperature during climb isa 10°caaverage head wind component 18 ktfind climb time enroute climb 280/ 74 err _a_033 325 4 6 nautical air miles (nam). img /com_en/com033 356 jpg at 57500 kg we are closer to 58000 kg than 56000 kg 13 minutes our answer no need to interpolate. Question 90-30
Given distance from departure to destination 180 nm true track 310wind 010°/20 kttas 115 kt what the distance of pet from departure point 4 6 nautical air miles (nam). ducksherminator with my calculations using crp 5w i find gs o = 110kts gs h = 120kts pet=dxh/(o+h)=93 91nm which much closer to 92nm than 98nm you need to be a little more accurate with wiz wheel you will get 104 kt the outbound leg 124 kt 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 90-31
The surface system over vienna 48°n016°e a err _a_033 330 Cold front moving east. img /com_en/com033 330 jpg cold front moving east at 10 km/h. Question 90-32
Given dry operating mass = 33510 kg traffic load= 7600 kgtrip fuel = 2040 kg final reserve fuel= 983 kg alternate fuel= 1100 kg contingency fuel= 5% of trip fuelwhich of listed estimated masses correct Estimated landing mass at destination= 43295 kg. img /com_en/com033 330 jpg cold front moving east at 10 km/h. Question 90-33
The lowest cloud conditions oktas/ft at bordeaux/merignac at 1330 utc were err _a_033 340 Estimated landing mass at destination= 43295 kg. img /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 there a reason tempo doesn't refer to actual conditions the question asks the actual conditions not forecasted conditions. Question 90-34
At reference or see flight planning manual mrjt 1 figure 4 2 find short distance cruise altitude the twin jet aeroplane given brake release mass=45000 kgtemperature=isa + 20°ctrip distance=50 nautical air miles nam err _a_033 343 Estimated landing mass at destination= 43295 kg. img /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 there a reason tempo doesn't refer to actual conditions the question asks the actual conditions not forecasted conditions. Question 90-35
The maximum wind velocity °/kt shown in vicinity of munich 48°n 012°e err _a_033 346 Estimated landing mass at destination= 43295 kg. arrows feathers pennants arrows indicate direction number or pennants and/or feathers correspond to speed example with a 270°/115 kt wind img /com_en/com033 346a jpg pennants correspond to 50 kt feathers correspond to 10 kt half feathers correspond to 5 kt img /com_en/com033 346b jpg munich below jet axis direction of jet 300° there two pennants four feathers so 140 kt there a decrease of speed after two oblic lines (highlighted in red). Question 90-36
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 a twin jet aeroplane given estimated take off mass tom 52000 kgairport elevation 387 ftfl 280w/v 280°/40 ktisa deviation 10°caverage true course 340° find ground distance to top of climb toc err _a_033 347 Estimated landing mass at destination= 43295 kg. img /com_en/com033 347 jpg tas 353 kt air distance 53 nm on nav computer set under true index true course 340° under center dot tas 353 kt with rotating scale set wind 280° under 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 course set 334° under true index you read a ground speed of 332 kt to determine ground distance travelled in climb multiply air distance the groundspeed divide the tas 53 x 332 / 353 = 49 84 nm. Question 90-37
Flight planning manual mrjt 1 figure 4 5 1 planning an ifr flight from paris charles gaulle to london heathrow the twin jet aeroplane given estimated take off mass tom 52000 kgairport elevation 387 ftfl 280w/v 280°/40 ktisa deviation 10°caverage true course 340°find time to top of climb toc err _a_033 348 Estimated landing mass at destination= 43295 kg. img /com_en/com033 347 jpg tas 353 kt air distance 53 nm on nav computer set under true index true course 340° under center dot tas 353 kt with rotating scale set wind 280° under 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 course set 334° under true index you read a ground speed of 332 kt to determine ground distance travelled in climb multiply air distance the groundspeed divide the tas 53 x 332 / 353 = 49 84 nm. Question 90-38
Which describes maximum intensity of turbulence if any forecast fl260 over toulouse 44°n001°e err _a_033 350 Estimated landing mass at destination= 43295 kg. this chart goes from fl100 up to fl450 embedded cumulonimbus clouds with base below fl100 top up to fl270 are located in area enclosed scalloped lines over toulouse img /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 maximum intensity can be severe. Question 90-39
At reference or see flight planning manual sep 1 figure 2 4given aeroplane mass at start up 3663 lbsaviation gasoline density 6 lbs/gal fuel load 74 galtake off altitude sea levelheadwind 40 ktcruising altitude 8000 ftpower setting full throttle 2300 rpm 20°c lean of peak egt calculate ground range err _a_033 351 Estimated landing mass at destination= 43295 kg. we are looking ground range as answers are in nm a wind component given img /com_en/com033 351 jpg on reference we find 844 nam range in nm = nam x (gs/tas) tas 160 kt ground speed tas wind = 160 40 = 120 kt range in nm = 844 x (120/160) = 633 nm milinoo how we know that true air speed 160 kt please? it's written in center of graph 'true airspeed knot. Question 90-40
In vicinity of paris 49°n 003°e tropopause at about err _a_033 352 Estimated landing mass at destination= 43295 kg. 200 km west of paris we have a box showing 400 (this the tropopause height) north of paris (near amsterdam) tropause height 350 let's have a look to a jet stream cross section img /com_en/com033 352 jpg before approaching jet tropopause subsides fl380 the correct answer at exam.
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