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Question 118-1 : Use route manual chart e hi 1.an aircraft has to fly from glasgow 55°52'n 004°27'w to benbecula 57°29'n 007°22'w. cruising at 320 kt tas..assuming a headwind of 40 kt and cruise fuel consumption of 2300 kg/h. what is the forecast fuel used for this flight.. err a 033 6 ? [ Training professional ]
Question 118-2 : Given.distance from departure to destination 500 nm.safe endurance 4 h.tas 140 kt.ground speed out 150 kt.ground speed home 130 kt.what is the distance and time of the psr from the departure point ?
Distance 279 nm, time 111 min.
Point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 150 kt.ground speed home = 130 kt..point of safe return psr = 4 x 130 / 150 + 130.point of safe return psr = 65000 / 280.point of safe return psr = 1.86 h..1.86 x 60 = 111.6 min..distance of the psr from the departure point at a speed of 150 kt.111.6 min x 150/60 = 279 nm.
Question 118-3 : On a given path, it is possible to chose between four flight levels fl , each associated with a mandatory flight mach number m..the flight conditions, static air temperature sat and headwind component hwc are given below.fl 370 m = 0.80 ts = 60°c hwc = 15 kt.fl 330 m = 0.78 ts = 60°c hwc= 5 kt.fl ?
Fl 270.
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. 60° and mach 0.8 => tas is 451 kt.. 1137..fl 370 => tas = 451 kt => gs = 439 kt.fl 330 => tas = 440 kt => gs = 435 kt.fl 290 => tas = 456 kt => gs = 441 kt.fl 270 => tas = 445 kt => gs = 445 kt..the flight level allowing the highest ground speed is fl 270.
Question 118-4 : Given.. distance from departure to destination 410 nm. safe endurance 3.6 h. true track 055°. w/v 180/35. tas 120 kt..what is the distance of the psr from the departure point ?
203 nm.
First step, find the gs out.given. true track. true air speed. wind direction and velocity..required wind correction angle and ground speed.. computer solution.a set true track to true index...b turn the indicator to the wind direction, in this case using the black azimuth 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 angle at the same place. read the ground speed under the center bore from the scale on the axis of the slide...setting 055° to true index..set the indicator to 180° on the black azimuth circle being upwind. adjust the speed arc labelled 120 kt of the diagram slide to the wind speed 35 kt of the indicator scale...reading under the plotted point read the wind correction angle 14°. under the center bore read the ground speed 136 kt.. 1141..step two proceed on the same way for gs home you will find 97 kt...step three now, apply the psr formula.psr = time x gs out x gs home / gs out + gs home.psr = 3.6 x 136 x 97 / 136 + 97.psr = 203.8 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 118-5 : The surface weather system over england 53°n 002°w is.. err a 033 22 ?
An occluded front moving east.
. /com en/com022 33a.png..over england, the surface weather system is an an occluded front.. /com en/com033 22.jpg.the arrow attached to the occluded front indicates direction and speed 20 km/h.
Question 118-6 : Which describes the intensity of icing, if any, at fl 150 in the vicinity of toulouse 44°n 01°e .. err a 033 25 ?
Moderate or severe.
.its due to the presence of cbs around toulouse. always assume moderate or severe icing next to them.
Question 118-7 : Given.distance a to b 2050 nm.mean groundspeed 'on' 440 kt.mean groundspeed 'back' 540 kt.the distance to the point of equal time pet between a and b is ?
1130 nm.
.ground speed out gso = 440 kt.ground speed home gsh = 540 kt..distance to pet = distance x gsh / gso + gsh.distance to pet = 2050 x 540 / 440 + 540.distance to pet = 1107000 / 980 = 1130 nm.
Question 118-8 : Which is the heaviest type of precipitation, if any, forecast for bordeaux/merignac at 1000 utc .. err a 033 27 ?
Light rain.
. /com en/com033 27.jpg..at 1000 utc, if any , the heaviest type of precipitation forcast is light rain.
Question 118-9 : Given.distance from departure to destination 315 nm.true track 343°.w/v 015/15.tas 100 kt.what is the distance of the pet from the departure point ?
176 nm.
...under index, set true track 343°, centre dot on tas, 100 kt, with the rotative scale, set wind 015° / 15 kt...now, drift is always measured from heading to track.turn to set true heading 345° 343° + 2° left drift under index, you now read a ground speed out of 88 kt...proceed in the same way to find the ground speed home of 112 kt.. right drift of 4°, true heading of 159°...distance to pet = distance x gsh / gso + gsh.distance to pet = 315 x 112 / 88 + 112.distance to pet = 35280 / 200 = 176 nm.
Question 118-10 : What lowest cloud conditions oktas/ft are forecast for johannesburg/jan smuts at 0300 utc.. err a 033 31 ?
5 to 7 at 400 ft.
.ft0900...prob30 0305 3000 bcfg bkn004.....30% chance that temporarily, between 0300 utc and 0500 utc, there will be patches fog with 3000 m visibility and 5 to 7 oktas of cloud at 400 ft.
Question 118-11 : Given.distance from departure to destination 210 nm.safe endurance 2.5 h.true track 035.w/v 250/20.tas 105 kt.what is the distance of the psr from the departure point ?
127 nm.
.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..on the computer.under index, set true track 035°, centre dot on tas, 105 kt, with the rotative scale, set wind 250°/20 kt we read a right drift of 5°..drift is always measured from heading to track, so turn to set true heading 030° 035° 5° right drift under index you read a ground speed out of 122 kt...proceed on the same way for gs home.under index, set true track 215°, centre dot on tas, 105 kt, with the rotative scale, set wind 250°/20 kt we read a left drift of 4°..drift is always measured from heading to track, so turn to set true heading 219° 215° + 4° right drift under index you read a ground speed home of 88 kt....point of safe return psr = 2.5 x 88 / 122 + 88.point of safe return psr = 220 / 210.point of safe return psr = 1.04 h..distance of the psr from the departure point at a speed of 122 kt.1.04 x 122 = 127 nm.
Question 118-12 : Given the following.d = flight distance.x = distance to point of equal time.gso = groundspeed out.gsr = groundspeed return.the correct formula to find distance to point of equal time is ?
X = d x gsr / gso + gsr.
Question 118-13 : The flight crew of a turbojet aeroplane prepares a flight using the following data. flight leg air distance 2 700 nm. flight level fl 310, true airspeed 470 kt. tailwind component at this level 35 kt. initially planned take off mass without extra fuel on board 195 000 kg. fuel price 0.28 euro/l at ?
The fuel transport operation is not recommended in this case.
.the fuel is cheaper at destination... the fuel transport operation is not recommended in this case.
Question 118-14 : From which of the following would you expect to find information regarding known short unserviceability of vor, tacan, and ndb ?
Notam.
Question 118-15 : The wind velocity over italy is.. err a 033 54 ?
A maximum of 110 kt at fl380.
. /com en/com033 54.jpg.a jet is crossing france and italy at fl 380, with a maximum of 110 kt.
Question 118-16 : Assuming the following data.ground distance to be covered 2 000 nm.cruise flight level fl 330.cruising speed mach 0.82 true airspeed 470 kt.head wind component 30 kt..planned destination landing mass 160 000 kg..temperature isa..cg 37.total anti ice on. pack flow hi..time needed to carry out such a ?
4 h 43 min.
.nam = ngm x tas/gs = 2000 x 470/440 = 2136 nam... /com en/com033 56.jpg..between 2100 and 2200, we can find our answer 4 h 43 min... 210883 .i think that the answer is wrong because. 4.39 + 4.51 /2 = 4.45 h > 4h 27'.. 4h 43' > 4.75h so the correct answer, as i think is 4h 27'....no, times on the annex are shown as 4.39 for 4h39 and 4.51 for 4h51..it's written on the top, below fuel consumed kg time h.min.
Question 118-17 : When calculating the fuel required to carry out a given flight, one must take into account..1 the wind..2 foreseeable airborne delays..3 other weather forecasts..4 any foreseeable conditions which may delay landing..the combination which provides the correct statement is ?
1, 2, 3, 4.
Question 118-18 : Which of the following flight levels, if any, is forecast to be clear of significant cloud, icing and cat along the marked route from shannon 53°n 10°w to berlin 53°n 13°e .. err a 033 64 ?
Fl250.
. /com en/com033 64.jpg..there are isolated embedded cbs from below fl100 to fl220, so fl 210 is excluded..you have cat area n°2 from fl270 to fl400, so fl290 is excluded..only fl250 is forecast to be clear of significant cloud, icing and cat along the marked route from shannon to berlin.
Question 118-19 : Given.distance from departure to destination 215 nm.safe endurance 3.3 h.true track 005°.w/v 290/15.tas 125 kt.what is the distance of the psr from the departure point ?
205 nm.
.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..on the computer.under index, set true track 005°, centre dot on tas, 125 kt, with the rotative scale, set wind 290°/15 kt we read a right drift of 7°..drift is always measured from heading to track, so turn to set true heading 358° 005° 7° right drift under index you read a ground speed out of 120 kt...proceed on the same way for gs home.under index, set true track 185°, centre dot on tas, 125 kt, with the rotative scale, set wind 290°/15 kt we read a left drift of 7°..drift is always measured from heading to track, so turn to set true heading 192° 185° + 7° right drift under index you read a ground speed home of 128 kt....point of safe return psr = 3.3 x 128 / 120 + 128.point of safe return psr = 422.4 / 248.point of safe return psr = 1.70 h..distance of the psr from the departure point at a speed of 120 kt.1.7 x 120 = 204 nm.
Question 118-20 : Given.distance from departure to destination 200 nm.safe endurance 3 h.tas 130 kt.ground speed out 150 kt.ground speed home 110 kt.what is the distance psr from the departure point ?
190 nm.
.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 150 kt.ground speed home = 110 kt..point of safe return psr = 3 x 110 / 150 + 110.point of safe return psr = 330 / 260.point of safe return psr = 1.2692 h..distance of the psr from the departure point at a speed of 150 kt.1.2692 h x 150 = 190 nm.
Question 118-21 : What minimum visibility m is forecast for 0600 utc at london lhr egll .. err a 033 71 ?
1500 m.
Question 118-22 : Excluding rvsm an appropriate flight level for ifr flight in accordance with semi circular height rules on a course of 180° m is ?
Fl100.
.the question states excluding rvsm... /com en/com033 1228.jpg..a vfr fligh level is flxx5 or flxx5..below fl290, in accordance with semi circular rules, for a magnetic heading of 180°, we need a even level.
Question 118-23 : Given.distance from departure to destination 400 nm.safe endurance 2,5 h.tas 115 kt.ground speed out 130 kt.ground speed home 105 kt.what is the distance of the psr from the departure point ?
145 nm.
.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 130 kt.ground speed home = 105 kt..point of safe return psr = 2.5 x 105 / 130 + 105.point of safe return psr = 262.5 / 235.point of safe return psr = 1.12 h..distance of the psr from the departure point at a speed of 130 kt.1.12 h x 130 = 145 nm.
Question 118-24 : Given.distance from departure to destination 300 nm.safe endurance 4 h.tas 110 kt.ground speed out 120 kt.ground speed home 100 kt.what is the distance of the psr from the departure point ?
218 nm.
Point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 120 kt.ground speed home = 100 kt..point of safe return psr = 4 x 100 / 120 + 100.point of safe return psr = 400 / 220.point of safe return psr = 1.81 h..distance of the psr from the departure point at a speed of 120 kt.1.81 h x 120 = 218 nm.
Question 118-25 : What is the mean temperature deviation °c from the isa over 50°n 010°w.. err a 033 83 ?
2°c.
. /com en/com033 83.jpg.temperature are negative unless prefixed by ps..at fl300, isa temperature is.15°c 2°c x 30 = 45°c...oat is 47°c, deviation from isa is 2°.
Question 118-26 : Given.distance from departure to destination 4630 nm.safe endurance 12,4 h.true track 240.w/v 060/80.tas 530 kt.what is the distance of the psr from the departure point ?
3211 nm.
.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 530 80 kt = 450 kt.ground speed home = 530 + 80 kt = 610 kt..point of safe return psr = 12,4 x 610 / 450 + 610.point of safe return psr = 7564 / 1060.point of safe return psr = 7.135h..distance of the psr from the departure point at a speed of 450 kt.7.135 x 450 = 3211 nm.
Question 118-27 : Given.distance from departure to destination 1100 nm.true track 280°.w/v 100/80.tas 440 kt.what is the distance time of the pet from the departure point ?
Distance 450 nm, time 52 min.
.track 280°, wind from 100°, it's a tailwind of 80 kt..ground speed out gso = 440 + 80 = 520 kt....track 100°, wind from 100°, it's a headwind of 80 kt..ground speed home gsh = 440 80 = 360 kt....distance to pet = distance x gsh / gso + gsh..distance to pet = 1100 x 360 / 520 + 360..distance to pet = 396000 / 880 = 450 nm.....time of the pet from the departure point..450 nm / 520 = 0.865 h..0.865 x 60 min = 52 minutes.
Question 118-28 : Given.distance from departure to destination 150 nm.safe endurance 3.2 h.tas 90 kt.ground speed out 100 kt.ground speed home 80 kt.what is the distance and time of the psr from the departure point ?
Distance 142 nm, time 85 min.
.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 100 kt.ground speed home = 80 kt..point of safe return psr = 3.2 x 80 / 100 + 80.point of safe return psr = 256 / 180.point of safe return psr = 1.42 h..1.42 x 60 = 85 minutes...distance of the psr from the departure point at a speed of 100 kt.85 min x 100/60 = 142 nm.
Question 118-29 : Given.distance from departure to destination 270 nm.true track 030.w/v 120/35.tas 125 kt.what is the distance and time of the pet from the departure point ?
Distance 135 nm, time 68 min.
...under index, set true track 030°, centre dot on tas, 125 kt, with the rotative scale, set wind 120°/35 kt, you find a left drift of 15°... /com en/com033 95a.jpg..now, drift is always measured from heading to track.turn to set true heading 045° 030° + 15° left drift under index, you now read your ground speed out of 120 kt.. /com en/com033 95b.jpg..proceed in the same way to find the ground speed home of 120 kt.. right drift of 15°, true heading of 195°...ground speed out gso = 120 kt.ground speed home gsh = 120 kt..distance to pet = distance x gsh / gso + gsh.distance to pet = 270 x 120 / 120 + 120.distance to pet = 32400 / 240 = 135 nm...135 nm at a ground speed out of 120 kt = 135 x 60/120 = 67.5 minutes.
Question 118-30 : At reference or see flight planning manual sep 1 figure 2.5..given.fl 75.lean mixture and full throttle, 2300 rpm.take off fuel 444 lbs.take off from msl..find.endurance in hours and minutes... err a 033 100 ?
05 hours 12 minutes.
. /com en/com033 100.jpg..5.2h ==> 5h + 0.2h 0.2 x 60 = 12 minutes..5 h 12 minutes.
Question 118-31 : From which of the following would you expect to find facilitation information regarding customs and health formalities ?
Aip.
Question 118-32 : An aircraft flies at a tas of 380 kt. it flies from a to b and back to a. distance ab = 480 nm..when going from a to b, it experiences a headwind component = 60 kt. the wind remains constant..the duration of the flight will be ?
2 h 35 min.
.take care the wind remains constant..headwind from a to b becomes tailwind from b to a...tas is 380 kt, from a to b, ground speed is 380 60 = 320 kt..480 nm / 320 kt = 1.5 h 1h30...from b to a, ground speed is 380+60 = 440 kt..480 nm / 440 kt = 1.09 h 1h05...1h30 + 1h05 = 2 h 35 min.
Question 118-33 : Given.distance from departure to destination 350 nm.true track 320.w/v 350/30.tas 130 kt.what is the distance and time of the pet from the departure point ?
Distance 210 nm, time 121 min.
Under index, set true track 320°, centre dot on tas, 130 kt, with the rotative scale, set wind 350°/30 kt, you find a left drift of 7°..now, drift is always measured from heading to track..turn to set true heading 327° 320° + 7° left drift under index, you now read your ground speed out of 104 kt...proceed in the same way to find the ground speed home of 155 kt... right drift of 6°, true heading of 134°...ground speed out gso = 104 kt.ground speed home gsh = 155 kt..distance to pet = distance x gsh / gso + gsh.distance to pet = 350 x 155 / 104 + 155.distance to pet = 54250 / 259 = 210 nm...210 nm at a ground speed out of 104 kt = 210 x 60/104 = 121 minutes.
Question 118-34 : Given.distance from departure to destination 250 nm.gs out 130 kt.gs home 100 kt.what is the distance of the pet from the departure point ?
109 nm.
Ground speed out gso = 130 kt.ground speed home gsh = 100 kt..distance to pet = distance x gsh / gso + gsh.distance to pet = 250 x 100 / 130 + 100.distance to pet = 25000 / 230 = 109 nm.
Question 118-35 : Given.distance from departure to destination 550 nm.endurance 3,6 h.true track 200.w/v 220/15.tas 130 kt.what is the distance of the psr from the departure point ?
231 nm.
...under index, set true track 200°, centre dot on tas, 130 kt, with the rotative scale, set wind 220°/15 kt, you find a left drift of 3°...now, drift is always measured from heading to track.turn to set true heading 203° 200° + 3° left drift under index, you now read your ground speed out of 115 kt..proceed in the same way to find the ground speed home of 142 kt.. true track of 020°, right drift of 2°, true heading of 018°...ground speed out gso = 115 kt.ground speed home gsh = 142 kt..point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..point of safe return psr = 3.6 x 142 / 115 + 142.point of safe return psr = 511.2 / 257.point of safe return psr = 1.99 h..distance of the psr from the departure point at a speed of 115 kt.1.99 x 115 = 229 nm closest answer is 231 nm.
Question 118-36 : Given.distance from departure to destination 180 nm.endurance 2 h.tas 120 kt.ground speed out 135 kt.ground speed home 105 kt.what is the distance and time of the psr from the departure point ?
Distance 118 nm, time 53 min.
.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 135 kt.ground speed home = 105 kt..point of safe return psr = 2 x 105 / 135 + 105.point of safe return psr = 210 / 240.point of safe return psr = 0.875 h..0.875 x 60 = 52.5 minutes..distance of the psr from the departure point at a speed of 135 kt.52.5 min x 135/60 = 118.125 nm.
Question 118-37 : Given.distance from departure to destination 2500 nm.gs out 540 kt.gs home 470 kt.what is the time of the pet from the departure point ?
129 min
.pet = d x vsr / vsa + vsr.pet = 2500 x 470 / 540 + 470.pet = 1163 nm..1163 / 540 = 2.15 h.. 2 x 60 + 0.15 x 60 = 129 minutes.
Question 118-38 : Given.distance from departure to destination 875 nm.true track 240.wind 060/50 kt.tas 500 kt.what is the distance and time of the pet from the departure point ?
Distance 394 nm, time 43 min.
.true track 240°.wind 060/50 kt..wind is parallel to our course, thus..ground speed out 500 + 50 = 550 kt.ground speed home 500 50 = 450 kt..pet = d x gsh / gso + gsh.pet = 875 x 450 / 550 + 450 = 393.75 nm...393.75 / 550 = 0.716 minutes..60 x 0.716 = 43 minutes.
Question 118-39 : The forecast period covered by the paris/charles de gaulle tafs totals hours .. err a 033 122 ?
Question 118-40 : If cas is 190 kts.altitude 9000 ft.temp. isa 10°c.true course tc 350°.w/v 320/40.distance from departure to destination is 350 nm.endurance 3 hours.and actual time of departure is 1105 utc..the point of equal time pet is reached at ?
1213 utc
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...202.67/181 = 1.12 h...1.12x0.6 = 67 minutes or 1h07....11h05+01h07 = 12h33.
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