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Given .distance x to y 2700 nm .mach number 0 75.temperature 45°c.mean wind ? [ Experience landing ]

Question 94-1 : 1386 nm 1350 nm 1313 nm 1425 nm

..set corresponding mark m kt against outside temperature at flight altitude read in front of mach number on the outer scale the true air speed .example . 45° and mach 0 75 => tas is 437 kt .. /com en/com033 48 jpg..ground speed out gso = 437 + 10 kt = 447 kt.ground speed home gsh = 437 + 35 kt = 472 kt..distance to pet = distance x gsh / gso + gsh .distance to pet = 2700 x 472 / 447 + 472 .distance to pet = 1274400 / 919 = 1386 nm exemple 194 1386 nm.1386 nm.

At reference or see flight planning manual mep1 figure 3 6.a flight is to be ?

Question 94-2 : 20 nm 29 nm 36 nm 25 nm

. /com en/com033 398 jpg.distance to descend = 29 8 = 21 nm close to the answer exemple 198 20 nm.20 nm.

Given .distance from departure to destination 2800 nm .true track 140 .w/v ?

Question 94-3 : Distance 1680 nm time 252 min distance 1120 nm time 112 min distance 1400 nm time 168 min distance 1120 nm time 134 min

We start on a true track of 140° we have a 100 kt headwind thus ..ground speed out = 500 kt 100 kt = 400 kt.ground speed home = 500 kt + 100 kt = 600 kt..pet = distance x gsh / gso+ gsh .pet = 2800 x 600 / 400 + 600 .pet = 1680000 / 1000 = 1680 nm ..time of the pet from the departure point .1680 / 400 = 4 2 h.4 2 x 60 = 252 minutes exemple 202 Distance: 1680 nm time: 252 minDistance: 1680 nm time: 252 min

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

Question 94-4 : 116° 080° 098° 113°

. /com en/com033 410 jpg.true course is 098° + 18°w magnetic variation = 116° exemple 206 116°.116°.

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

Question 94-5 : 2 h 51 min 3 h 43 min 2 h 59 min 2 h 43 min

.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 34 h.point of safe return psr = 5 34 x 450 / 395 + 450 .point of safe return psr = 5 34 x 450 / 845 .point of safe return psr = 2 84 h.2 84 h = 2 h 51 min.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 210 2 h 51 min2 h 51 min

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

Question 94-6 : 34500 ft 33800 ft 35300 ft maximum operating altitude

Img /com en/com033 416 jpg. exemple 214 34500 ft34500 ft

Given .distance from departure to destination 3750 nm .safe endurance 9 5 h ?

Question 94-7 : 2255 nm 2070 nm 1128 nm 1495 nm

exemple 218 2255 nm2255 nm

Which best describes be maximum intensity of icing if any at fl150 in the ?

Question 94-8 : Moderate light severe nil

. 629. /com en/com033 430 jpg.from below the chart base to fl 200 icing is moderate over bucarest exemple 222 Moderate.Moderate.

Given .distance from departure to destination 210 nm.safe endurance 3 5 h .true ?

Question 94-9 : 200 nm 125 nm 100 nm 10 nm

exemple 226 200 nm200 nm

What maximum surface windspeed kt is forecast for bordeaux/merignac at 1600 utc ?

Question 94-10 : 30 kt 25 kt 10 kt 5 kt

. fc1100r 121100z 121221 amended r short taf prepared on the twelfth day of the month at 1100z valid from 12h to 21h utc .there is a tempo from 12h to 18h indicating wind from 280° for 20 kt with gust up to 30 kt tempo 1218 28020g30kt exemple 230 30 kt.30 kt.

Which best describes the significant cloud forecast over toulouse 44°n001°e ?

Question 94-11 : Broken ac/cu base below fl100 tops fl150 embedded isolated cb base below fl100 tops fl270 well separated cb base fl100 tops to fl 270 isolated cb embedded in layer cloud surface to fl270 5 to 7 oktas cu and ac base below fl100 tops to fl270

. /com en/com033 435 png. exemple 234 Broken ac/cu base below fl100 tops fl150, embedded isolated cb base below fl100 tops fl270Broken ac/cu base below fl100 tops fl150, embedded isolated cb base below fl100 tops fl270

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

Question 94-12 : 1635 kg 1090 kg 1690 kg 1125 kg

.fuel flow for 47000kg at 5000 ft is 2180 kg/h interpolating between 2220 and 2140 kg/h .the fuel required for 45 minutes holding is . 2180 x 45/60 = 1635 kg exemple 238 1635 kg.1635 kg.

The wind °/kt at 50°n 015°w is . err a 033 438 ?

Question 94-13 : 290/75 310/85 310/75 100/75

. /com en/com033 438 png.wind is coming from 290° .number or pennants and/or feathers correspond to speed .pennants correspond to 50 kt .feathers correspond to 10 kt .half feathers correspond to 5 kt .50+10+10+5 = 75 kt

At reference or see flight planning manual mrjt 1 figure 4 3 6. in order to get ?

Question 94-14 : Distance nm wind component landing mass at alternate still air distance wind component zero fuel mass flight time wind component landing mass at alternate distance nm wind component zero fuel mass

exemple 246 Distance (nm), wind component, landing mass at alternate.Distance (nm), wind component, landing mass at alternate.

Given .distance from departure to destination 320 nm.safe endurance 4 3 h.true ?

Question 94-15 : 263 nm 185 nm 131 nm 59 nm

..under index set true track 120° under the center bore set tas 130 kt and with the rotative scale set wind 180°/40 kt . /com en/com033 444a jpg.drift is always measured from heading to track so turn to set true heading 120° + 17° = 137° . /com en/com033 444b jpg.gs out is 105 kt .proceed the same way to find ground speed home .under index set true track 300°.13° right drift true heading is 300° 13° = 287°.gs home is 146 kt .psr = time x gs out x gs home / gs out + gs home .psr = 4 3 x 105 x 146 / 105 + 146 .psr = 263 nm exemple 250 263 nm.263 nm.

The flight crew of a turbojet aeroplane prepares a flight using the following ?

Question 94-16 : The fuel transport operation is not recommended in this case 8 000 kg 22 000 kg 15 000 kg

.it's very simple fuel is cheaper at destination so fuel transport operation is not recommended in this case exemple 254 The fuel transport operation is not recommended in this case.The fuel transport operation is not recommended in this case.

At reference or see flight planning manual mrjt 1 figure 4 3 5 .the following ?

Question 94-17 : 3740 nm 3640 nm 3500 nm 3250 nm

.you have to use the graph backward you must go to the condition first 10 kt tailwind and then go to the ref line instead of the 'normal' way to proceed first the ref line and then the condition . /com en/com033 447 jpg. exemple 258 3740 nm.3740 nm.

Cas is 190 kt and altitude 9000 ft.temperature isa 10°c.true course ?

Question 94-18 : 203 nm 147 nm 183 nm 167 nm

.oat at fl90 = 15°c 9 x 2°c = 3°c .we are in isa 10°c thus oat = 13°c .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 exemple 262 203 nm.203 nm.

Which best describes the significant cloud if any forecast for the area ?

Question 94-19 : 5 to 7 oktas cu and cb base below fl100 tops fl180 5 to 7 oktas cu and cb base fl100 tops fl180 3 to 7 oktas cu and cb base below fl100 tops fl180 nil

. /com en/com033 450 jpg.the chart is from fl100 to fl450 exemple 266 5 to 7 oktas cu and cb base below fl100, tops fl180.5 to 7 oktas cu and cb base below fl100, tops fl180.

Given .distance from departure to destination 190 nm .safe endurance 2 4 h ?

Question 94-20 : 148 nm 95 nm 73 nm 44 nm

..set 120° true track on top with rotating scale mark the wind 030°/40 kt .you read 17° right drift .set on top 120° 17° = 103° .now read the drift 18° right .set on top 102° and you can read your 'ground speed out' of 123 5 kt .the wind is perpendicular to our track so gsh will also be 123 5 kt...point of safe return psr = endurance x homeward gs / outbound gs + homeward gs ..ground speed out = 123 5 kt.ground speed home = 123 5 kt..point of safe return psr = 2 4 x 123 5 / 123 5 + 123 5 .point of safe return psr = 296 4 / 247.point of safe return psr = 1 2 h..1 2 x 60 = 72 minutes..distance of the psr from the departure point at a speed of 123 5 kt .72 min x 123 5/60 = 148 nm exemple 270 148 nm.148 nm.

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

Question 94-21 : 30000 ft 25000 ft 21000 ft 27500 ft

. /com en/com033 458 jpg. exemple 274 30000 ft.30000 ft.

Given .distance from departure to destination 350 nm .true track 320 .w/v ?

Question 94-22 : Distance 210 nm time 122 min distance 139 nm time 54 min distance 123 nm time 74 min distance 139 nm time 81 min

exemple 278 Distance: 210 nm time: 122 minDistance: 210 nm time: 122 min

Given .distance from departure to destination 240 nm .safe endurance 3 5 h .tas ?

Question 94-23 : Distance 216 nm time 118 min distance 134 nm time 58 min distance 108 nm time 52 min distance 24 nm time 13 min

.point of safe return psr = endurance x gs home / gs out + gs home .outbound gs = 110 kt.homeward gs = 140 kt.endurance = 3 5 h.point of safe return psr = 3 5 x 140 / 110 + 140 .point of safe return psr = 1 96 h.1 96 h = 118 min.distance of the psr from the departure 1 96 x 110 = 215 6 nm exemple 282 Distance: 216 nm time: 118 minDistance: 216 nm time: 118 min

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

Question 94-24 : 271° 259° 247° 279°

.center your protractor at mid distance between a and c you will find 260° you must apply variation 11°w .average magnetic course is 260° + 11° = 271° exemple 286 271°.271°.

Given .distance from departure to destination 165 nm .true track 055 .w/v ?

Question 94-25 : 92 nm 73 nm 83 nm 132 nm

. /com en/com033 475a jpg. /com en/com033 475b jpg.ground speed out = 92 kt..proceed the same way to find ground speed home .true track 235° drift = 8° true heading = 243° gs home = 116 kt ..pet = distance x gsh / gso+ gsh .pet = 165 x 116 / 92 + 116 .pet = 19140 / 208 = 92 nm exemple 290 92 nm92 nm

Given .distance from departure to destination 140 nm .gs out 90 kt .gs home 80 ?

Question 94-26 : 66 nm 74 nm 70 nm 124 nm

exemple 294 66 nm.66 nm.

Given .distance from departure to destination 6340 nm .safe endurance 15 h ?

Question 94-27 : 3756 nm 2560 nm 1878 nm 2584 nm

exemple 298 3756 nm3756 nm

Route manual chart nap.the distance nm from a 64°n006°e to c 62°n020°w is . ?

Question 94-28 : 720 nm 690 nm 1590 nm 1440 nm

.report the track distance along latitude 63°n average latitude between a and c .you will count a little bit more of 26° separation between a and c .26° x 60 nm x cos 63° = 708 nm . /com en/com033 483 jpg.you can also use meridian 1° = 60 nm . maxscail .26° change in longitude 60 x 26 x cos 63 = 708 225.between 64 and 62 2 ° ==> 2 x 60 1°=60nm = 120.now use phytagore sqrt 708 225 + sqrt 120 = 15108 225.square root of 515982 905 = 718 32 nm exemple 302 720 nm.720 nm.

Given . course a to b 088° t .distance 1250 nm.mean tas 330 kt.mean w/v ?

Question 94-29 : 1 hour 42 minutes 1 hour 54 minutes 1 hour 39 minutes 2 hours 02 minutes

. /com en/com033 484a jpg. /com en/com033 484b jpg.ground speed out = 343 kt.proceed the same way to find ground speed home 306 kt .pet = distance x gsh / gso+ gsh .pet = 1250 x 306 / 343 + 306 .pet = 382500 / 649 = 589 nm.589 nm / 343 = 1 71 h 1 hour 42 minutes exemple 306 1 hour 42 minutes1 hour 42 minutes

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

Question 94-30 : 259° 247° 271° 079°

. /com en/com033 497 jpg.we are looking for average true course .with you protractor aligned on true north between a and c you will find a true course of 259° exemple 310 259°.259°.

Given .distance from departure to destination 340 nm.true track 320°.wind ?

Question 94-31 : 112 nm 228 nm 121 nm 219 nm

..under index set true track 320° centre dot on tas 110 kt with the rotative scale set wind 160°/40 kt you find a right drift of 5° . /com en/com033 500a jpg.now drift is always measured from heading to track .turn to set true heading 315° 320° 5° right drift under index you now read your ground speed out of 147 kt. /com en/com033 500b jpg.proceed in the same way to find the ground speed home of 72 kt . left drift of 10° true heading of 150° .ground speed out gso = 147 kt.ground speed home gsh = 72 kt.distance to pet = distance x gsh / gso + gsh .distance to pet = 340 x 72 / 147 + 72 .distance to pet = 24480 / 219 = 111 78 nm exemple 314 112 nm.112 nm.

Use route manual chart nap.on a direct great circle course from reykjavik ?

Question 94-32 : A 131° b 1095 nm a 311° b 1824 nm a 118° b 1095 nm a 140° b 1824 nm

.average true course . /com en/com033 510 jpg.cos mean latitude x difference of longitude x 60 nm .cos58 31 x 26 83° x 60 nm = 846 nm.difference of latitude 11°42' = 11 70° x 60 nm = 702 nm.distance between reykjavik and amsterdam = sqrt 702² + 846² = 1099 nm exemple 318 (a) 131° (b) 1095 nm(a) 131° (b) 1095 nm

Given .distance from departure to destination 330 nm .safe endurance 5 h .true ?

Question 94-33 : 302 nm 194 nm 150 nm 30 nm

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .on the computer .under index set true track 170° centre dot on tas 125 kt with the rotative scale set wind 140°/25 kt we read a right drift of 7° .drift is always measured from heading to track so turn to set true heading 163° 170° 7° right drift under index you read a ground speed out of 104 kt and a drift of 6º right .proceed on the same way for gs home you will find 140 kt .point of safe return psr = 5 x 140 / 104 + 140 .point of safe return psr = 700 / 244.point of safe return psr = 2 87 h.distance of the psr from the departure point at a speed of 104 kt .2 87 x 104 = 298 nm exemple 322 302 nm.302 nm.

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

Question 94-34 : 275° 267° 271° 262°

. /com en/com033 520 jpg.we are looking for initial magnetic course .with you protractor aligned on true north you have a true course of 271° add the 4°west magnetic variation = 275° exemple 326 275°.275°.

The maximum wind velocity °/kt immediately north of tunis 36°n010°e . err a ?

Question 94-35 : 190°/95 kt 280°/110 kt 250°/85 kt 180°/105 kt

. 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 521 jpg. exemple 330 190°/95 kt.190°/95 kt.

Given . distance a to b 3060 nm.mean groundspeed 'out' 440 kt.mean groundspeed ?

Question 94-36 : 5 hours 30 minutes 5 hours 45 minutes 3 hours 55 minutes 5 hours 20 minutes

exemple 334 5 hours 30 minutes.5 hours 30 minutes.

The approximate mean wind component kt at mach 0 78 along true course 270° at ?

Question 94-37 : 40 kt headwind component 55 kt headwind component 35 kt tailwind component 25 kt tailwind component

. /com en/com033 541 jpg. 55 kt + 55 kt + 30 kt / 3 = 46 kt .mean wind is 240°/46kt .headwind component = wind speed x cos angle between the wind and the course .headwind component = 46 kt x cos 30° = 40 kt exemple 338 40 kt headwind component.40 kt headwind component.

Given .distance from departure to destination 340 nm.gs out 150 kt.gs home 120 ?

Question 94-38 : 151 nm 189 nm 170 nm 272 nm

.ground speed out gso = 150 kt.ground speed home gsh = 120 kt.distance to pet = distance x gsh / gso + gsh .distance to pet = 340 x 120 / 150 + 120 .distance to pet = 40800 / 270 = 151 nm exemple 342 151 nm.151 nm.

Given .distance from departure to destination 470 nm.true track 237°.wind ?

Question 94-39 : 256 nm 214 nm 235 nm 205 nm

..under index set true track 237° centre dot on tas 125 kt with the rotative scale set wind 300°/25 kt you find a left drift of 11° .now drift is always measured from heading to track .turn to set true heading 248° 237° + 11° left drift under index you now read your ground speed out of 111 kt.proceed in the same way to find the ground speed home of 135 kt . right drift of 9° true heading of 048° .ground speed out gso = 111 kt.ground speed home gsh = 135 kt.distance to pet = distance x gsh / gso + gsh .distance to pet = 470 x 135 / 111 + 135 .distance to pet = 63450 / 246 = 258 nm closest answer is 256 nm exemple 346 256 nm256 nm

From which of the following would you expect to find details of the search and ?

Question 94-40 : Aip atcc broadcasts notam sigmet

exemple 350 AipAip


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