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Question 181-1 : What is the average track °t and distance between bal vor n5318.0 w00626.9 and cfn ndb n5502.6 w00820.4 .. err a 061 421 ? [ Question security ]

327° 124 nm.

exemple 281 327° - 124 nm. — .report the true north tick, center your protractor, you read an average true track of 327°.. /com en/com061 421.jpg.use the scale to find the distance 124 nm.

Question 181-2 : Given.sha vor n5243.3 w00853.1 radial 205°, crk vor n5150.4 w00829.7 radial 317°. what is the aircraft position.. err a 061 424 ?

N5210 w00910

exemple 285 N5210 w00910 — . /com en/com061 424.jpg..

Question 181-3 : Given.sha vor n5243.3 w00853.1 radial 129°, crk vor n5150.4 w00829.7 radial 047°. what is the aircraft position.. err a 061 426 ?

N5220 w00750

exemple 289 N5220 w00750 — . /com en/com061 426.jpg..

Question 181-4 : Given.sha vor/dme n5243.3 w00853.1 radial 120°/35 nm. what is the aircraft position.. err a 061 428 ?

N5230 w00800

exemple 293 N5230 w00800 — . /com en/com061 428.jpg..

Question 181-5 : Given.sha vor/dme n5243.3 w00853.1 radial 232°/32 nm. what is the aircraft position.. err a 061 430 ?

N5220 w00930

exemple 297 N5220 w00930

Question 181-6 : Given.sha vor/dme n5243.3 w00853.1 radial 048°/22 nm. what is the aircraft position.. err a 061 432 ?

N5300 w00830

exemple 301 N5300 w00830 — . /com en/com061 432.jpg..

Question 181-7 : Given.sha vor n5243.3 w00853.1, crk vor n5150.4 w00829.7..aircraft position n5230 w00820..which of the following lists two radials that are applicable to the aircraft position.. err a 061 434 ?

Sha 131°, crk 017°.

exemple 305 Sha 131°, crk 017°. — Debiassi.when two answers are quite close, try to line up the douglas protractor with two known values ie sha 052° and 115° radials for the airways. this really improves accuracy.

Question 181-8 : Given. sha vor n5243.3 w00853.1, con vor n5354.8 w00849.1, aircraft position n5330 w00800. which of the following lists two radials that are applicable to the aircraft position.. err a 061 436 ?

Sha 042° con 138°

exemple 309 Sha 042° con 138° — . /com en/com061 436.jpg..magnetic north is indicated over vors and ndbs.

Question 181-9 : Given.sha vor n5243.3 w00853.1 dme 50 nm, crk vor n5150.4 w00829.7 dme 41 nm, aircraft heading 270° m , both dme distances increasing..what is the aircraft position.. err a 061 438 ?

N5200 w00935

exemple 313 N5200 w00935 — . /com en/com061 438.jpg..you draw two circles use the vertical scale to define the distances..both dme distances increasing so this is the only valid junction position.

Question 181-10 : Given. crn vor n5318.1 w00856.5 dme 18 nm, sha vor n5243.3 w00853.1 dme 30 nm,. aircraft heading 270° m , both dme distances decreasing. what is the aircraft position.. err a 061 440 ?

N5310 w00830

exemple 317 N5310 w00830

Question 181-11 : Given. con vor n5354.8 w00849.1 dme 30 nm, crn vor n5318.1 w00856.5 dme 25 nm,. aircraft heading 270° m , both dme distances decreasing. what is the aircraft position.. err a 061 442 ?

N5330 w00820

exemple 321 N5330 w00820

Question 181-12 : Given. sha vor/dme n5243.3 w00853.1 , birr aerodrome n5304 w00754. what is the sha radial and dme distance when overhead birr aerodrome.. err a 061 444 ?

068° 41 nm

exemple 325 068° - 41 nm

Question 181-13 : Given. con vor/dme n5354.8 w00849.1 , castlebar aerodrome n5351 w00917 , what is the con radial and dme distance when overhead castlebar aerodrome.. err a 061 446 ?

265° 17 nm

exemple 329 265° - 17 nm

Question 181-14 : An aircraft departs from position a 04°10' s 178°22'w and flies northward following the meridian for 2950 nm..it then flies westward along the parallel of latitude for 382 nm to position b..the coordinates of position b are ?

45°00'n 172°38'e.

exemple 333 45°00'n 172°38'e. — Admin.it flies northward following the meridian for 2950 nm.1 nm on a meridian = 1'.2950 nm = 2950 minutes of arc..60' = 1°.2950'/60' = 49°10'...we are in south hemisphere and we are heading north, 4°10' to reach the equator, and after it remains 45° to fly. our position is now 45°00'n 178°22'w...now we are heading west, along latitude 45°n. we need to use the formula.distance = change of longitude x cos of latitude.382 nm = change of longitude x cos45°.change of longitude = 382 / 0.707 = 540.3'..540.3'/60' = 9°..178°22'w + 9°w = 187°22'.we cross the anti meridian, we are now at 172°38'e 360° 187°22'.

Question 181-15 : What is the time required to travel along the parallel of latitude 60°n between meridians 010°e and 030°w at a groundspeed of 480 kt ?

2 h 30.

exemple 337 2 h 30. — Admin.distance in longitude = 40°.at the equator, 1° = 60 nm..distance in nm = 40° x 60 nm x cos latitude.distance = 40° x 60 nm x cos 60°.distance = 1200 nm..1200 / 480 kt = 2 h 30.

Question 181-16 : Given the following.magnetic heading 060°, magnetic variation 8°w, drift angle 4° right.what is the true track ?

056°.

exemple 341 056°. — Admin. 1363.use this wonderful table for those questions...gomis01.060° menos 8°w son 052° como es heading ya tiene el ángulo de deriva incluido, por lo que solo tenemos que quitarselo. 052° 4° = 048°... .no, drift is always measured from heading to track.

Question 181-17 : An aircraft is following a true track of 048° at a constant tas of 210 kt..the wind velocity is 350° / 30 kt..the gs and drift angle are ?

192 kt, 7° right.

exemple 345 192 kt, 7° right. — Under index, set true track 048°, centre dot on tas, 210 kt, with the rotative scale, set wind.. /com en/com061 12.jpg..now, drift is always measured from heading to track.turn to set true heading 041° 048° 7° right drift under index, you now read a ground speed of 192 kt.

Question 181-18 : Given.fl 350, mach 0.80, oat 55°c..calculate the values for tas and local speed of sound lss ?

461 kt, lss 576 kt.

exemple 349 461 kt, lss 576 kt. — .lss varies with the square root of absolute temperature..formula lss = 39 x square root of oat+273..lss = 39 x square root of 218.lss = 576 kt...mach number is the ratio of true airspeed tas to the local speed of sound lss.formula mach number = tas/lss..0.80 = tas/576.tas = 576 x 0.8 = 461 kt.

Question 181-19 : Given.true heading = 180°.tas = 500 kt.w/v 225° / 100 kt.calculate the gs ?

435 kt.

exemple 353 435 kt. — Admin.on the computer, set 180° under index, centre dot on 500 kt, under the wind speed 100 kt on the rotating scale.. 1367.you read a groud speed of 435 kt.

Question 181-20 : Given.true heading = 310°, tas = 200 kt, gs = 176 kt, drift angle 7° right..calculate the w/v ?

270° / 33 kt.

exemple 357 270° / 33 kt. — Admin.use low speed scale on the computer, set 310° under index, centre dot on 200 kt, where right drift 7° crosses ground speed arc 176 kt, read the wind on the rotating scale. 1368.you read 270°/33kt.

Question 181-21 : Given.true heading = 090°, tas = 200 kt, wind = 220°/30 kt..calculate the ground speed ?

220 kt.

exemple 361 220 kt. — Admin. 1369

Question 181-22 : The reported surface wind from the control tower is 240°/35 kt. runway 30 300°..what is the cross wind component ?

30 kt.

exemple 365 30 kt. — Admin.angle between the wind and the direction of the runway 300° 240° = 60°..crosswind = sine of the angle between the wind and the direction of the runway x windspeed..crosswind = sin60° x 35 kt = 30.31 kt.

Question 181-23 : Given magnetic heading 311°, drift angle 10° left, relative bearing of ndb 270°. what is the magnetic bearing of the ndb measured from the aircraft ?

221°.

exemple 369 221°. — Admin.heading.....311°..bearing....+270°..total.......581°..581° 360° = 221° magnetic bearing of the ndb measured from the aircraft.

Question 181-24 : Given the following.true track 192°, magnetic variation 7°e, drift angle 5° left..what is the magnetic heading required to maintain the given track ?

190°.

exemple 373 190°. — Admin. 1393.use this wonderful table for those questions.

Question 181-25 : The angle between the plane of the ecliptic and the plane of equator is approximately ?

23.5°.

exemple 377 23.5°. — Admin. 2028

Question 181-26 : Given.tas = 485 kt.oat = isa +10°c..fl 410..calculate the mach number ?

0.825

exemple 381 0.825 — Admin. 56°c is considered to be the lowest isa temperature, and the question states that we are in outside air temperature of isa +10°c...at fl410, isa is 56°c, isa +10°c will be 46°c...on the computer. 1396..by calculation.mach = tas x a.a = 39square root t° in kelvin...a = 39 square root 227°k= 587,59.mach number= 485/ 587,59= 0,825.

Question 181-27 : At 1215 utc lajes vortac 38°46'n 027°05'w rmi reads 178°, range 135 nm..calculate the aircraft position at 1215 utc. 2484 ?

40°55'n 027°55'w.

exemple 385 40°55'n 027°55'w. — Admin.rmi reads 178°, it is the magnetic direction to reach the station vor tacan at lajes...for a vor, we must apply the variation at the station, on the chart, variation line is 15°w..178° variation west, magnetic best , so 'minus' 15° 178° 15° = 163°...from lajes, 163° + 180° = 343°.. 1398..use the latitude scale to find 135 nm.

Question 181-28 : At reference..1300 utc dr position 37°30'n 021°30'w alter heading port santo ndb 33°03'n 016°23'w.tas 450 kt, forecast w/v 360°/30kt..calculate the eta at port santo ndb.. err a 061 53 ?

1348

exemple 389 1348 — .true track is 136°..wind 360°/30 kt..drift is .true heading is ..ground speed is ...center dot on 450 kt, true index on 136°, indicate wind 360°/30 kt drift is 3° right...turn to put 133° 136° 3° right drift below true index, the groundspeed is under the wind mark 30 kt 470 kt gs....plot distance between points and measure along line of longitude.. /com en/com061 53.jpg..6°20' = 6.33 x 60 nm = 380 nm...380 / 470/60 = 48.5 minutes....1300 + 48 minutes 30 secondes = 13h48 30sec.

Question 181-29 : For a distance of 1860 nm between q and r, a ground speed 'out' of 385 kt, a ground speed 'home' of 465 kt and an endurance of 8 hours excluding reserves the distance from q to the point of safe return psr is ?

1685 nm.

exemple 393 1685 nm. — .point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 385 kt.ground speed home = 465 kt..point of safe return psr = 8 x 465 / 385 + 465.point of safe return psr = 3720 / 850.point of safe return psr = 4.3764 h..distance of the psr from the departure point at a speed of 385 kt.4.3764 h x 385 = 1685 nm.

Question 181-30 : Two points a and b are 1000 nm apart. tas = 490 kt..on the flight between a and b the equivalent headwind is 20 kt..on the return leg between b and a, the equivalent tailwind is +40 kt..what distance from a, along the route a to b, is the the point of equal time pet ?

530 nm.

exemple 397 530 nm. — Admin.ground speed out 490 20 = 470 kt.ground speed home 490 + 40 = 530 kt..pet = distance x gsh / gso + gsh.pet = 1000 x 530 / 470 + 530 = 530 nm.

Question 181-31 : Given ad = air distance, gd = ground distance, tas = true airspeed, gs = groundspeed. which of the following is the correct formula to calculate ground distance gd gone ?

Gd = ad x gs /tas

exemple 401 Gd = (ad x gs)/tas — .gd = gs x t but t = ad/tas so gd = gs x ad /tas.

Question 181-32 : What is the isa temperature value at fl 330 ?

51°c

exemple 405 -51°c — 15°c 2° x 33 = 51°c.

Question 181-33 : Given.tas 487kt, fl 330, temperature isa + 15..calculate the mach number ?

0.81

exemple 409 0.81 — Admin.isa temperature at fl 330 = 15º 2º x 33 = 51ºc...deviation is +15ºc then oat is 51°c + +15°c = 36ºc...with the computer.in the airspeed window put the temperature 36ºc in front of 'mach kt ' index..in front of 487 kt, on the inner scale, you read 0.81 mach number.. 1435

Question 181-34 : How many nm would an aircraft travel in 1 minute 45 secondes if gs is 135 kt ?

3.94 nm.

exemple 413 3.94 nm. — Admin.135 kt /60 minutes = 2.25 nm/minute..1 minute 45 secondes = 1.75 minute..2.25 x 1.75 = 3.9375 nm.

Question 181-35 : An aircraft travels 100 statute miles in 20 min, how long does it take to travel 215 nm ?

50 min.

exemple 417 50 min. — Admin.1 nm = 1.15 sm..215 nm = 215 x 1.15 = 247.25 sm..100 sm in 20 minutes = 300 sm in 60 minutes 1h...247.25/300 = 0.824 hour..0.824 x 60 = 50 min.

Question 181-36 : Given.tas = 220 kt.magnetic course = 212°.wind = 160 ° m / 50 kt.calculate the gs ?

186 kt.

exemple 421 186 kt. — Admin.the given wind has a headwind component so the groundspeed is going to be less than the given tas and all but one of the answers are more.

Question 181-37 : Given.fl250.oat 15 °c.tas 250 kt.calculate the mach number ?

0.40

exemple 425 0.40 — . 2485.. aviat 617.put temperature 15ºc in front of m kt index, in the airspeed window...go to tas 250 kt on outer scale and read mach number on the inner scale...by calculation..mach number = tas / lss..mach number = 250 / 39 sqrt 273 15..mach number = 0.399.

Question 181-38 : During a low level flight 2 parallel roads that are crossed at right angles by an aircraft..the time between these roads can be used to check the aircraft ?

Groundspeed.

Question 181-39 : Given.magnetic track = 315°, magnetic heading = 301°, variation = 5°w, tas = 225 kt..the aircraft flies 50 nm in 12 min..calculate the wind °t ?

190°/63 kt.

exemple 433 190°/63 kt. — Admin.true heading = 296° 301° 5°..true course = 310° 315 5°..ground speed = 60 x 50/12 = 250 kt...on nav computer.set tas 225 kt on center dot, under true index set true heading 296°...mark where drift 14º right crosses ground speed 250 kt...rotate to shift mark under the vertical speed line, you read 190°/63 kt.

Question 181-40 : Given.tas = 270 kt, true hdg = 270°, actual wind 205° t /30kt..calculate the drift angle and gs ?

6r 259kt

exemple 437 6r - 259kt — Admin.under index, set true heading 270°, centre dot on tas, 270 kt, with the rotative scale, set wind 205°/30 kt.read the drift and the ground speed 6°r 259 kt.

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