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Diploma > registration : Given .distance from departure to destination 260 nm .safe endurance 4 1 h ?

Question 92-1 : 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 192 213 nm.213 nm.

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

Question 92-2 : 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 196 109°.109°.

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

Question 92-3 : 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 200 252 min252 min

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

Question 92-4 : 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 204 engine flame out and windscreen damage.engine flame out and windscreen damage.

Route manual chart nap.the distance nm from c 62°n020°w to b 58°n004°e is . ?

Question 92-5 : 760 nm 725 nm 700 nm 775 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 exemple 208 760 nm.760 nm.

The planned flight is over a distance of 440 nm.based on the wind charts at ?

Question 92-6 : Fl 180 fl 100 fl 050 either fl 050 or fl 100

.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 exemple 212 fl 180.fl 180.

At reference or see flight planning manual mrjt 1 figure 4 3 5.for a flight of ?

Question 92-7 : A 17000 kg b 6h 45 min a 15800 kg b 6h 15 min a 16200 kg b 6h 20 min a 18400 kg b 7h 00 min

. /com en/com033 268 jpg..you always need to go first to the ref line and then apply the condition mass temperature wind exemple 216 (a) 17000 kg (b) 6h 45 min(a) 17000 kg (b) 6h 45 min

Given .distance from departure to destination 1950 nm .gs out 400 kt .gs home ?

Question 92-8 : 125 min 223 min 167 min 29 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 exemple 220 125 min125 min

Given .distance from departure to destination 95 nm .true track 105 .wind ?

Question 92-9 : 51 nm 47 5 nm 82 nm 44 nm

exemple 224 51 nm.51 nm.

Given .distance from departure to destination 150 nm .true track 020° .wind ?

Question 92-10 : 59 nm 75 nm 91 nm 65 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 exemple 228 59 nm.59 nm.

The surface wind velocity °/kt at paris/charles de gaulle at 1330 utc was. err ?

Question 92-11 : 270/04 300/05 270/08 180/12

exemple 232 270/04.270/04.

From which of the following would you expect to find the dates and times when ?

Question 92-12 : Notam and aip only aip sigmet rad/nav charts

exemple 236 notam and aipnotam and aip

What is the earliest time utc if any that thunderstorms are forecast for ?

Question 92-13 : 1800 utc 1300 utc 0800 utc nil

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. exemple 240 1800 utc.1800 utc.

A sector distance is 450 nm long .the tas is 460 kt .the wind component is 50 ?

Question 92-14 : 406 nautical air miles nam 499 nautical air miles nam 414 nautical air miles nam 511 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 exemple 244 406 nautical air miles (nam)406 nautical air miles (nam)

Find the distance to the point of safe return psr .given .maximum useable fuel ?

Question 92-15 : 1125 nm 1143 nm 1463 nm 1491 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 exemple 248 1125 nm.1125 nm.

What is the temperature deviation °c from isa over 50° n 010°e . err a 033 ?

Question 92-16 : Deviation is 10° deviation is 55° deviation is 2° 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 exemple 252 deviation is -10°.deviation is -10°.

Given .distance from departure to destination 2000 nm .safe endurance 5 h .tas ?

Question 92-17 : 1248 nm 1040 nm 624 nm 752 nm

exemple 256 1248 nm1248 nm

At reference or see flight planning manual mrjt 1 figure 4 3 1c..for a flight ?

Question 92-18 : 17000 ft 22000 ft 14000 ft 10000 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 exemple 260 17000 ft.17000 ft.

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

Question 92-19 : 1090 kg 1010 kg 1310 kg 2180 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 exemple 264 1090 kg.1090 kg.

The wind °/kt at 60° n015° w is. err a 033 315 ?

Question 92-20 : 300/60 300/70 320/60 115/60

. /com en/com033 315 jpg..50 + 10 = 60 kt exemple 268 300/60.300/60.

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

Question 92-21 : 098° 116° 080° 278°

/com en/com033 319 jpg..true course 098° exemple 272 098°.098°.

Find the distance from waypoint 3 wpt 3 to the critical point .given .distance ?

Question 92-22 : 342 nm 375 nm 408 nm 403 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 exemple 276 342 nm.342 nm.

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

Question 92-23 : 13 minutes 11 minutes 15 minutes 14 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 exemple 280 13 minutes.13 minutes.

Given .distance from departure to destination 180 nm .true track 310.wind ?

Question 92-24 : 98 nm 82 nm 90 nm 92 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 exemple 284 98 nm.98 nm.

The surface system over vienna 48°n016°e is a. err a 033 330 ?

Question 92-25 : Cold front moving east warm front moving north stationary occluded front cold front moving west

. /com en/com033 330 jpg.cold front moving east at 10 km/h exemple 288 cold front moving east.cold front moving east.

Given .dry operating mass = 33510 kg .traffic load= 7600 kg.trip fuel = 2040 kg ?

Question 92-26 : Estimated landing mass at destination= 43295 kg estimated take off mass= 45233 kg estimated landing mass at destination= 43193 kg estimated take off mass= 43295 kg

exemple 292 estimated landing mass at destination= 43295 kg.estimated landing mass at destination= 43295 kg.

The lowest cloud conditions oktas/ft at bordeaux/merignac at 1330 utc were. err ?

Question 92-27 : 1 to 2 at 3000 ft 1 to 4 at 3000 ft 3 to 4 at 2000 ft 3 to 4 at 800 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 exemple 296 1 to 2 at 3000 ft.1 to 2 at 3000 ft.

At reference or see flight planning manual mrjt 1 figure 4 2. find the short ?

Question 92-28 : 10000 ft 11000 ft 7500 ft 12500 ft

exemple 300 10000 ft10000 ft

The maximum wind velocity °/kt shown in the vicinity of munich 48°n 012°e is ?

Question 92-29 : 300/140 300/100 300/160 290/110

. 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 exemple 304 300/140300/140

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

Question 92-30 : 50 nm 47 nm 53 nm 56 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 exemple 308 50 nm.50 nm.

Flight planning manual mrjt 1 figure 4 5 1. planning an ifr flight from paris ?

Question 92-31 : 11 min 15 min 3 min 12 min

exemple 312 11 min11 min

Which describes the maximum intensity of turbulence if any forecast for fl260 ?

Question 92-32 : Severe moderate light nil

.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 exemple 316 severe.severe.

At reference or see flight planning manual sep 1 figure 2 4.given .aeroplane ?

Question 92-33 : 633 nm 547 5 nm 844 nm 730 nm

.we are looking for ground range as the answers are in nm and a wind component is given. /com en/com033 351 jpg..on the reference we find 844 nam .range in nm = nam x gs/tas.tas is 160 kt ground speed is tas wind = 160 40 = 120 kt..range in nm = 844 x 120/160 = 633 nm. milinoo .how we know that true air speed is 160 kt please..it's written in the center of the graph 'true airspeed knots' exemple 320 633 nm.633 nm.

In the vicinity of paris 49°n 003°e the tropopause is at about . err a 033 352 ?

Question 92-34 : Fl380 fl340 fl350 fl400

.200 km west of paris we have a box showing 400 this is the tropopause height north of paris near amsterdam tropause height is 350 .let's have a look to a jet stream cross section. /com en/com033 352 jpg..before approaching the jet tropopause subsides .fl380 is the correct answer at the exam exemple 324 fl380.fl380.

Given .distance from departure to destination 1860 nm .gs out 360 kt.gs home ?

Question 92-35 : 163 min 132 min 147 min 22 min

exemple 328 163 min163 min

Given .distance from departure to destination 3000 nm .safe endurance 8 h .tas ?

Question 92-36 : 203 min 173 min 277 min 117 min

exemple 332 203 min203 min

At reference or see flight planning manual mrjt 1 figure 4 3 1c.for a flight of ?

Question 92-37 : A 17100kg b 6h 07 min a 16000kg b 6h 25 min a 18000kg b 5h 50 min a 20000kg b 6h 40 min

. /com en/com033 371 jpg..6 13 h = 6h 08 minutes close to the answer exemple 336 (a) 17100kg (b) 6h 07 min(a) 17100kg (b) 6h 07 min

Given .distance from departure to destination 150 nm.safe endurance 2 4 h.true ?

Question 92-38 : 142 nm 83 nm 71 nm 98 nm

.....start by searching outbound ground speed on nav computer .set 120 kt under center dot true track 250° to true index put wind direction 280° under the red compass rose under 15 kt you read a groudspeed of 107 kt...repeat the opeartion to find homeward ground speed...outbound gs 107 kt..homeward gs 134 kt....point of safe return psr = endurance x homeward gs / outbound gs + homeward gs.point of safe return psr = 2 4 x 134 / 107 + 134.point of safe return psr = 321 6 / 241..point of safe return psr = 1 33 h....0 33 x 60 = 20 minutes..20 + 60 minutes = 80 min....distance of the psr from the departure point at a speed of 107 kt.80 min x 107/60 = 142 6 nm exemple 340 142 nm.142 nm.

At reference or see flight planning manual mrjt 1 figure 4 3 1c .for a flight ?

Question 92-39 : A 14200 kg b 5 h 30 min a 16200 kg b 5 h 45 min a 13600 kg b 6 h 30 min a 12000 kg b 5 h 15 min

. /com en/com033 380 gif. exemple 344 (a) 14200 kg (b) 5 h 30 min.(a) 14200 kg (b) 5 h 30 min.

Which best describes the maximum intensity of cat if any forecast for fl330 ?

Question 92-40 : Nil light moderate severe

. /com en/com033 382 jpg..cat area n°1 extends fl350 to fl450 so no clear air turbulence at fl330 exemple 348 nil.nil.

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