Question 121-1 : Given distance from departure to destination 3750 nm safe endurance 95 h true track 360 wv 36050 tas 480 ktwhat is the distance of the psr from the departure point ? [ Preparation civilian ]

Question 121-2 : Which best describes be maximum intensity of icing if any at fl150 in the vicinity of bucharest 45°n 026°e err a 033 430 ?

Moderate.

629 com encom033 430jpgfrom below the chart base to fl 200 icing is moderate over bucarest

Question 121-3 : Given distance from departure to destination 210 nmsafe endurance 35 h true track 310wv 27030tas 120 ktwhat is the distance of the psr from the departure point ?

200 nm.

Question 121-4 : What maximum surface windspeed kt is forecast for bordeauxmerignac at 1600 utc err a 033 434 ?

30 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

Question 121-5 : Which best describes the significant cloud forecast over toulouse 44°n001°e err a 033 435 ?

Broken accu base below fl100 tops fl150 embedded isolated cb base below fl100 tops fl270.

com encom033 435png

Question 121-6 : At reference or see flight planning manual mrjt 1 figure 44holding planningngm= nam x tas+ wind tas the fuel required for 45 minutes holding in a racetrack pattern at 5000 ft pressure altitude and a weight of 47000 kg is err a 033 437 ?

Question 121-7 : The wind °kt at 50°n 015°w is err a 033 438 ?

29075.

com encom033 438pngwind is coming from 290°number or pennants andor feathers correspond to speed pennants correspond to 50 ktfeathers correspond to 10 kthalf feathers correspond to 5 kt50+10+10+5 = 75 kt

Question 121-8 : At reference or see flight planning manual mrjt 1 figure 436 in order to get alternate fuel and time the twin jet aeroplane operations manual graph shall be entered with err a 033 443 ?

Question 121-9 : Given distance from departure to destination 320 nmsafe endurance 43 htrue track 120°wind 180°40 kttas 130 ktwhat is the distance of the psr from the departure point ?

263 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 encom033 444ajpgdrift is always measured from heading to track so turn to set true heading 120° + 17° = 137° com encom033 444bjpggs 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 = 43 x 105 x 146 105 + 146 psr = 263 nm

Question 121-10 : The flight crew of a turbojet aeroplane prepares a flight using the following data flight leg distance 3 500 nm flight level fl 310 true airspeed 450 kt headwind component at this level 5 kt initially planned take off mass without extra fuel on board 180 000 kg fuel price 035 us dollarsl at ?

The fuel transport operation is not recommended in this case.

It's very simple fuel is cheaper at destination so fuel transport operation is not recommended in this case

Question 121-11 : At reference or see flight planning manual mrjt 1 figure 435the following apply tail wind component 10 kttemperature isa +10°cbrake release mass 63000 kgtrip fuel available 20000 kgwhat is the maximum possible trip distance err a 033 447 ?

3740 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 encom033 447jpg

Question 121-12 : Cas is 190 kt and altitude 9000 fttemperature isa 10°ctrue course 350°wind 32040distance from departure to destination is 350 nmendurance 3 hours and actual time of departure is 1105 utcthe distance from departure to point of equal time pet ?

203 nm.

Oat at fl90 = 15°c 9 x 2°c = 3°cwe are in isa 10°c thus oat = 13°cconvert cas to tas on your computer com encom033 1156jpgcalculate the outbound and inbound groud speed start first with the wind 40 x cos30 = 34ktoutbound ground speed 215 34 = 181 ktinbound ground speed 215+34 = 249 ktpoint of equal time = 350x249 181+249 = 20267 nm

Question 121-13 : Which best describes the significant cloud if any forecast for the area southwest of bodo 67°n 014°e err a 033 450 ?

5 to 7 oktas cu and cb base below fl100 tops fl180.

com encom033 450jpgthe chart is from fl100 to fl450

Question 121-14 : Given distance from departure to destination 190 nm safe endurance 24 h true track 120° wind 030°40 kt tas 130 kt what is the distance of the psr from the departure point ?

148 nm.

Set 120° true track on top with rotating scale mark the wind 030°40 ktyou read 17° right driftset on top 120° 17° = 103°now read the drift 18° rightset on top 102° and you can read your 'ground speed out' of 1235 ktthe wind is perpendicular to our track so gsh will also be 1235 ktpoint of safe return psr = endurance x homeward gs outbound gs + homeward gs ground speed out = 1235 ktground speed home = 1235 ktpoint of safe return psr = 24 x 1235 1235 + 1235 point of safe return psr = 2964 247point of safe return psr = 12 h12 x 60 = 72 minutesdistance of the psr from the departure point at a speed of 1235 kt 72 min x 123560 = 148 nm

Question 121-15 : At reference or see flight planning manual mrjt 1 figure 42find the short distance cruise altitude for the twin jet aeroplanegiven brake release mass 40000 kgtemperature isa + 20°ctrip distance 150 nautical air miles nam err a 033 458 ?

30000 ft.

com encom033 458jpg

Question 121-16 : Given distance from departure to destination 350 nm true track 320 wv 35030 tas 130 kt what is the distance and time of the pet from the departure point ?

Distance 210 nm time 122 min.

Question 121-17 : Given distance from departure to destination 240 nm safe endurance 35 h tas 125 kt ground speed out 110 ktground speed home 140 ktwhat is the distance and time of the psr from the departure point ?

Distance 216 nm time 118 min.

Point of safe return psr = endurance x gs home gs out + gs home outbound gs = 110 kthomeward gs = 140 ktendurance = 35 hpoint of safe return psr = 35 x 140 110 + 140 point of safe return psr = 196 h196 h = 118 mindistance of the psr from the departure 196 x 110 = 2156 nm

Question 121-18 : Route manual chart napthe average magnetic course from a 64°n006°e to c 62°n020°w is err a 033 464 ?

271°.

Center your protractor at mid distance between a and c you will find 260° you must apply variation 11°waverage magnetic course is 260° + 11° = 271°

Question 121-19 : Given distance from departure to destination 165 nm true track 055 wv 36020 tas 105 kt what is the distance of the pet from the departure point ?

92 nm.

com encom033 475ajpg com encom033 475bjpgground speed out = 92 ktproceed the same way to find ground speed home true track 235° drift = 8° true heading = 243° gs home = 116 ktpet = distance x gsh gso+ gsh pet = 165 x 116 92 + 116 pet = 19140 208 = 92 nm

Question 121-20 : Given distance from departure to destination 140 nm gs out 90 kt gs home 80 kt what is the distance of the pet from the departure point ?

66 nm.

Question 121-21 : Given distance from departure to destination 6340 nm safe endurance 15 h true track 090 wv 270100 tas 520 ktwhat is the distance of the psr from the departure point ?

3756 nm.

Question 121-22 : Route manual chart napthe distance nm from a 64°n006°e to c 62°n020°w is err a 033 483 ?

720 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 c26° x 60 nm x cos 63° = 708 nm com encom033 483jpgyou can also use meridian 1° = 60 nm maxscail 26° change in longitude 60 x 26 x cos 63 = 708225between 64 and 62 2 ° ==> 2 x 60 1°=60nm = 120now use phytagore sqrt 708225 + sqrt 120 = 15108225square root of 515982905 = 71832 nm

Question 121-23 : Given course a to b 088° t distance 1250 nmmean tas 330 ktmean wv 340°60 ktthe time from a to the pet between a and b is ?

1 hour 42 minutes.

com encom033 484ajpg com encom033 484bjpgground speed out = 343 ktproceed 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 nm589 nm 343 = 171 h 1 hour 42 minutes

Question 121-24 : Route manual chart nap the average true course from a 64°n006°e to c 62°n020°w is err a 033 497 ?

259°.

com encom033 497jpgwe 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°

Question 121-25 : Given distance from departure to destination 340 nmtrue track 320°wind 160°40 kttas 110 ktwhat is the distance of the pet from the departure point ?

112 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 encom033 500ajpgnow 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 encom033 500bjpgproceed 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 ktground speed home gsh = 72 ktdistance to pet = distance x gsh gso + gsh distance to pet = 340 x 72 147 + 72 distance to pet = 24480 219 = 11178 nm

Question 121-26 : Use route manual chart napon a direct great circle course from reykjavik 64°10' n 022°00'w to amsterdam 52°32'n 004°50'e the a average true course and b distance are err a 033 510 ?

A 131° b 1095 nm.

Average true course com encom033 510jpgcos mean latitude x difference of longitude x 60 nm cos5831 x 2683° x 60 nm = 846 nmdifference of latitude 11°42' = 1170° x 60 nm = 702 nmdistance between reykjavik and amsterdam = sqrt 702² + 846² = 1099 nm

Question 121-27 : Given distance from departure to destination 330 nm safe endurance 5 h true track 170°wind 14025tas 125 ktwhat is the distance of the psr from the departure point ?

302 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 244point of safe return psr = 287 hdistance of the psr from the departure point at a speed of 104 kt 287 x 104 = 298 nm

Question 121-28 : Route manual chart nap the initial magnetic course from a 64°n006°e to c 62°n020°w is err a 033 520 ?

275°.

com encom033 520jpgwe 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°

Question 121-29 : The maximum wind velocity °kt immediately north of tunis 36°n010°e err a 033 521 ?

190°95 kt.

arrows feathers and pennants arrows indicate direction number or pennants andor feathers correspond to speedexample with a 270°115 kt wind com encom033 346ajpgpennants correspond to 50 ktfeathers correspond to 10 kthalf feathers correspond to 5 kt com encom033 521jpg

Question 121-30 : Given distance a to b 3060 nmmean groundspeed 'out' 440 ktmean groundspeed 'back' 540 ktsafe endurance 10 hoursthe time to the point of safe return psr is ?

5 hours 30 minutes.

Question 121-31 : The approximate mean wind component kt at mach 078 along true course 270° at 50°n from 000° to 010°w is err a 033 541 ?

40 kt headwind component.

com encom033 541jpg 55 kt + 55 kt + 30 kt 3 = 46 ktmean wind is 240°46ktheadwind component = wind speed x cos angle between the wind and the course headwind component = 46 kt x cos 30° = 40 kt

Question 121-32 : Given distance from departure to destination 340 nmgs out 150 ktgs home 120 ktwhat is the distance of the pet from the departure point ?

151 nm.

Ground speed out gso = 150 ktground speed home gsh = 120 ktdistance to pet = distance x gsh gso + gsh distance to pet = 340 x 120 150 + 120 distance to pet = 40800 270 = 151 nm

Question 121-33 : Given distance from departure to destination 470 nmtrue track 237°wind 30025 kt tas 125 ktwhat is the distance of the pet from the departure point ?

256 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 ktproceed 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 ktground speed home gsh = 135 ktdistance 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

Question 121-34 : From which of the following would you expect to find details of the search and rescue organisation and procedures sar ?

Aip.

Question 121-35 : Given distance from departure to destination 2380 nmgs out 420 ktgs home 520 ktwhat is the time of the pet from the departure point ?

188 minutes.

Ground speed out = 420 ktground speed home = 520 ktpet = distance x gsh gso + gsh pet = 2380 x 520 420 + 520 pet = 1237600 940 = 13166 nmtime of the pet from the departure point 13166 420 = 313 h313 x 60 = 188 minutes

Question 121-36 : Given distance from departure to destination 338 nm true track 045°wind 225°35 kt tas 120 ktwhat is the distance and time of the pet from the departure point ?

Distance 120 nm time 46 min.

Pet point of equal timedistance to pet = d x h o + hd = total track distanceh = groundspeed homeo = groundspeed outd = 338 nmh = 120 35 kt headwind = 85 kto = 120 + 35 kt tailwind = 155 ktpet distance = 338 x 85 155 + 85 distance = 1197 nm = 120 nmtime = 120 155 ground speed out x 60 = 46 minutes

Question 121-37 : Given distance from departure to destination 1500 nm safe endurance 45 h tas 450 kt ground speed out 480 kt ground speed home 410 kt what is the time of the psr from the departure point ?

124 min.

Point of safe return psr = endurance x homeward gs outbound gs + homeward gs ground speed out = 480 ktground speed home = 460 ktpoint of safe return psr = 45 x 410 480 + 410 point of safe return psr = 1845 890point of safe return psr = 207 h207 x 60 = 124 minutes

Question 121-38 : Over london 51°n 000°ew the lowest fl listed which is unaffected by cat is err a 033 567 ?

Fl230.

London is located in the cat area n°1this cat area extends from fl240 to fl350 thus two levels are out of this layer fl230 and fl360 the question asks for the lowest one

Question 121-39 : What lowest cloud conditions oktasft are forecast for 1900 utc at hamburg eddh err a 033 572 ?

5 to 7 at 500 ft.

Question 121-40 : Given distance from departure to destination 5000 nm safe endurance 10 h tas 450 kt ground speed out 500 kt ground speed home 400 kt what is the distance of the psr from the departure point ?

2222 nm.

Point of safe return psr = endurance x homeward gs outbound gs + homeward gs ground speed out = 500 ktground speed home = 400 ktpoint of safe return psr = 10 x 400 500 + 400 point of safe return psr = 4000 900point of safe return psr = 444 hdistance of the psr from the departure point at a speed of 500 kt 444 h x 500 = 2222 nm