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Question 184-1 : Given .heading 265°.tas 290 kt.wind 210°/35 kt.calculate track and groundspeed ? [ Question security ]

271° and 272 kt

Admin .center dot on tas 290 kt .true heading 265° under index.put wind direction under the red compass rose under 35 kt your drift is 6° right giving a track of 271° and a groundspeed under the wind mark of 272 kt . 2036 exemple 284 271° and 272 kt.

Question 184-2 : An aircraft is flying at fl 200 oat is 0°c .when the actual air pressure on an airfield at msl is placed in the subscale of the altimeter the indicated altitude is 19300 ft .calculate the aircraft's true altitude ?

21 200 ft

Admin .isa at fl200 is 15°c 2°c x 20 = 25°c.oat is 0°c we are in isa +25°c .you can use either the computer or with the following rule of thumb called the '4% rule' .the altitude/height changes by 4% for each 10°c temperature deviation from isa .0 04 x 19300 x 2 5 = 1930 ft .true altitude = 19300 + 1930 = 21230 ft exemple 288 21 200 ft.

Question 184-3 : An aircraft must fly 2000 ft above an obstacle of which the elevation is 13 600 ft the qnh at the nearest airfield is 991 hpa the elevation is 1500 ft and the temperature is 20°c calculate the minimum altitude required ?

17 400 ft

.the aircraft must be at a true height above the airfield of 13600 + 2000 1500 = 14100 ft ..at the airfield isa temperature is 15°c 1 5 x2°c = 12°c .temperature is report to be 20°c so we are in isa 32°c.temperature correction formula 4° x 14 1 x 32° = 1805 ft..the minimum height above the airfield is 14100 + 1805 ft = 15905 ft .now adding the 1500 ft of the airfield to have an altitude 15905 + 1500 = 17405 ft ...note qnh is calculated from the qfe reduced to mean sea level msl assuming isa conditions there is no temperature error between airfield elevation qfe and mean sea level msl exemple 292 17 400 ft.

Question 184-4 : Consider the following factors that determine the accuracy of a dead rekoning position .1 the flight time since the last position update .2 the accuracy of the forecasted wind .3 the accuracy of the tas .4 the accuracy of the steered heading .using the list which of the above statements are correct ?

1 2 3 and 4

exemple 296 1, 2, 3 and 4.

Question 184-5 : An aircraft is flying at fl100 .the oat = isa 15°c the qnh given by a station at an elevation 3000 ft is 1035 hpa .calculate the approximate true altitude ?

10 200 ft

Admin .note 061 general navigation learning objectives states . for questions involving height calculation 30 ft/hpa is to be used unless another figure is specified in the question .you have to turn your altimeter subscale setting knob clockwise from 1013 to 1035 .indicated altitude will be increased by 22 hpa x 30 ft = 660 ft.10000 + 660 = 10660 ft .we now have to correct the temperature above the station .you can use either the computer or with the following rule of thumb called the '4% rule' .the altitude/height changes by 4% for each 10°c temperature deviation from isa .4 ft x 10660 3000 /1000 x 15 = 460 ft.true altitude = 10660 460 = 10200 ft exemple 300 10 200 ft.

Question 184-6 : An aircraft has to fly over a mountain ridge .the highest obstacle indicated in the navigation chart has an elevation of 9 800 ft .the qnh given by a meteorological station at an elevation of 6200 ft is 1022 hpa .the oat = isa +5°c .calculate the approximate indicated altitude to obtain a ?

11 700 ft

.9800 ft + 2000 ft = 11800 ft .we have to correct the temperature above the qfe datum .you can use either the computer or with the following rule of thumb called the '4% rule' .the altitude/height changes by 4% for each 10°c temperature deviation from isa . /com en/com061 613 jpg..deviation from isa = +5°c.4% x 5 6 x 5 = 112 ft..11800 ft 112 ft = 11688 ft .it is 'minus' 112 ft because air is hotter than standard true altitude is higher than indicated altitude exemple 304 11 700 ft.

Question 184-7 : An aircraft is flying from a to b a distance of 50 nm .the true course in the flight log is 270° the forecast wind is 045° t /15 kt and the tas is 120 kt .after 15 minutes of flying with the planned tas and true heading the aircraft is 3 nm south of the intended track and 2 5 nm ahead of the dead ?

17°

Admin . 1798.with the forecasted wind we will fly at 130 kt ground speed .at 130 kt and 15 minutes of flight we will be at 32 5 nm from a .the question states 2 5 nm ahead of the dead reckoning position so we are at 35 nm from a .use the one in sixty rule .track error angle from a = 3 nm x 60 / 35 nm = 5° . it's the drift to applied in order to correct the wind .track error angle to join b from our current position = 3 nm x 60 / 15 nm = 12° .to reach destination b from this position the correction angle on the heading should be 5° + 12° = 17° exemple 308 17°.

Question 184-8 : An aircraft is flying from a to b a distance of 50 nm .the true course in the flight log is 090° the forecast wind is 225° t /15kt and the tas is 120 kt .after 15 minutes of flying with the planned tas and true heading the aircraft is 3 nm north of the intended track and 2 5 nm ahead of the dead ?

17°

.draw the exercice . /com en/com061 623 jpg.without wind at 120 kt and 15 minutes of flight we are at 32 5 nm from a .the question states 2 5 nm ahead of the dead reckoning position so we are at 35 nm from a .use the one in sixty rule .track error angle from a = 3 nm x 60 / 35 nm = 5° . it's the drift to applied in order to correct the wind .track error angle to join b from our current position = 3 nm x 60 / 15 nm = 12° .to reach destination b from this position the correction angle on the heading should be 5° + 12° = 17° exemple 312 17°.

Question 184-9 : An aircraft is flying from a to b .the true course according to the flight log is 090° the estimated wind is 225° t /15 kt and the tas is 120 kt .after 15 minutes of flying with the planned tas and true heading the aircraft is 3 nm south of the intended track and 2 5 nm ahead of the dead ?

5°r

Frist step find the ground speed .place centre dot on 120 kt tas .place 225° wind direction under true index .make a wind mark on centre line 15 kt below centre dot at 105 kt .rotate to set 090° true track under true index .wind mark has moved to 5° left drift .rotate to lined up 090° with 5° left drift .wind mark has stayed at 5° left drift you find .true heading 095° .ground speed 130 kt . 130/60 x 15 min = 32 5 nm.actual aircraft position is 2 5 nm ahead dead reckoning position at 32 5 + 2 5 = 35 nm .track error angle = distance off track x 60 / distance along track.track error angle = 3 x 60 / 35.track error angle = 180 / 35 = 5° exemple 316 5°r

Question 184-10 : On a true heading of 090° the aircraft experiences drift of 5°right on a true heading of 180° the aircraft experiences no drift on both headings the tas is 200 kt and it is assumed that the wind is the same .what is the experienced wind speed and direction ?

360°/17 kt

..since on a true heading of 180° there is no drift the wind is coming from 180° or 360° .under index set true heading 090° centre dot on tas 200 kt with the rotative scale set the 5°right drift . /com en/com061 627 jpg.read the wind 360°/17 kt exemple 320 360°/17 kt.

Question 184-11 : An aircraft is flying from salco to berry head on magnetic track 007° tas 445 kt .the wind is 050° t /40 kt .variation 5°w deviation +2°.at 1000 utc the rb of locator py is 311° .at 1003 utc the rb of locator py is 266° .calculate the distance of the aircraft from locator py at 1003 utc ?

21 nm

.first step find the aircraft magnetic heading . /com en/com061 631 jpg.calculate the drift between our true track 002° and the true wind 050°/40 kt with your computer the drift is 4° left you have to apply a 4°right wind angle correction and also a ground speed of 415 kt ..from the aircraft at 1000 utc the locator is at 60° to the left 011° to 311° = 60° .from the aircraft at 1003 utc the locator is at 105° to the left 011° to 266° = 105° ..we have an isosceles triangle and in an isosceles triangle two sides are equal in length .in 45° in 3 minutes .415 kt/60 = 6 92 nm/min.3 min x 6 92 = 20 76 nm exemple 324 21 nm.

Question 184-12 : An aircraft is flying at fl250 oat = 45°c the qnh given by a station at msl is 993 2 hpa .calculate the approximate true altitude ?

23400 ft

.you have to turn your altimeter subscale setting knob counterclockwise from 1013 2 to 993 2 .indicated altitude will be decreased by 20 hpa x 30 ft = 600 ft .25000 ft 600 ft = 24400 ft.you can use either the computer or with the following rule of thumb called the '4% rule' .the altitude/height changes by 4% for each 10°c temperature deviation from isa .deviation from isa = 15° 2 x 25 = 35°c . 35°c to 45°c = 10°c we are in isa 10°c .0 04 x 24400 x 1 = 976 ft .true altitude = 24400 976 = 23424 ft .keep in mind that air is colder than standard thus the air column is contracted our true altitude is lower than our indicated altitude exemple 328 23400 ft.

Question 184-13 : A vor is situated at position n55°26' w005°42' .the variation at the vor is 9°w .the position of the aircraft is n60°00'n w010°00' .the variation at the aircraft position is 11°w .the initial true track angle of the great circle from the aircraft position to the vor is 101 5° .which radial ?

294°

. /com en/com061 638 jpg.first step apply convergency .convergency = difference of longitude x sin mean latitude .convergency = 10°w 5°42'w x sin 60 +55°26' /2 .convergency = 4 35° x sin 57 5 = 3 67°.second step find true track vor .true track at vor = 101 5° + convergency = 101 5 + 3 67° = around 105° t .third step we must apply variation at the vor .105 + 9°w = 114° magnetic .last step we are looking for a radial .114° + 180° = 294° exemple 332 294°.

Question 184-14 : An aircraft tracks radial 200° inbound to a vor station with a magnetic heading of 010° .after being overhead the vor station the aircraft tracks radial 090° outbound with a mh of 080° .the tas is 240 kt and the magnetic variation in the area is 5°w .the wind is ?

320°/50 kt

Admin .following radial 200° inbound magnetic track 020° with a magnetic heading of 010° we have a 10° right drift .our true heading is 005° we have to applied the 5°west magnetic variation since it is a vor .after having overfly the vor we fly outbound on radial 090° magnetic track 090° our magnetic heading is 080° so our true heading is 075° .on the computer set 240 kt under the center dot and 005° below true index draw a line along the right 10° drift line .rotate and put 075° below true index draw a line along the right 10° drift line .rotate to bring back the intersection of the lines under the central wind line . 1788 exemple 336 320°/50 kt.

Question 184-15 : The fix of the aircraft position is determined by radials from three vor stations the measurements contain small random errors known systematic errors and unknown systematic errors the measured radials are corrected for known systematic errors and are plotted on a navigation chart the result is ?

1

. /com en/com061 644 jpg.point 1 is always located on the right side of the radials exemple 340 1

Question 184-16 : An aircraft flies at fl 250 with an oat of 45°c the qnh given by a meteorological station with an elevation of 2830 ft is 1033 hpa .calculate the clearance above a mountain ridge with an elevation of 20410 ft ?

4 200 ft

Admin .we need to calculate our true altitude .first you have to turn your altimeter subscale setting knob clockwise from 1013 to 1033 .indicated altitude will be increased by 20 hpa x 30 ft = 600 ft .25000 + 600 = 25600 ft .next step we must correct for temperature .outside temperature is 45°c at fl250 .isa at fl250 is 15°c 25 x 2°c = 35°c .we are in isa 10°c .you can use either the computer or with the following rule of thumb called the '4% rule' .the altitude/height changes by 4% for each 10°c temperature deviation from isa .an altimeter set to airport qnh will read correctly when on the ground at the airport irrespective of temperature .any temperature error therefore occurs due to non isa temperature in the layer of atmosphere between airport elevation and aircraft in flight .therefore 25600 2830 = 22770 ft.22770 x 0 04 x 1 = 910 ft .our true altitude is 25600 910 = 24690 ft .clearance above the mountain ridge is 24690 20410 = 4280 ft exemple 344 4 200 ft.

Question 184-17 : Given .an aircraft is flying at fl100 oat = isa 15°c .the qnh given by a meteorological station with an elevation of 100 ft below msl is 1032 hpa 1 hpa = 27 ft .calculate the approximate true altitude of this aircraft ?

9900 ft

Admin .you have to turn your altimeter subscale setting knob clockwise from 1013 to 1032 .indicated altitude will be increased by 19 hpa x 27 ft = 513 ft .10000 + 513 = 10513 ft .now we must correct for temperature .we are in isa 15°c .you can use either the computer or with the following rule of thumb called the '4% rule' .the altitude/height changes by 4% for each 10°c temperature deviation from isa .an altimeter set to airport qnh will read correctly when on the ground at the airport irrespective of temperature .any temperature error therefore occurs due to non isa temperature in the layer of atmosphere between airport elevation and aircraft in flight .therefore 10513 100 = 10613 ft.10613 x 0 04 x 1 5 = 637 ft .10513 637 = 9876 ft closest answer is 9900 ft exemple 348 9900 ft.

Question 184-18 : The accuracy of the manually calculated dead reckoning position of an aircraft is among other things affected by ?

The accuracy of the forecasted wind

Admin .dead reckoning is the process of calculating one's current position by using a previously determined position or fix and advancing that position based upon known or estimated speeds over elapsed time and course you have to take the wind into account and the more accurate the wind information is the more accurate the manually calculated position will be exemple 352 The accuracy of the forecasted wind.

Question 184-19 : The accuracy of the manually calculated dead reckoning position of an aircraft is among other things affected by ?

The flight time since the last position update

Admin .dead reckoning is the process of estimating one's current position based upon a previously determined position or fix and advancing that position based upon known or estimated speeds over elapsed time and course therefore the accuracy is among other things affected by the flight time since the last position update exemple 356 The flight time since the last position update.

Question 184-20 : What may cause a difference between a dead rekoning position and a fix ?

The difference between the actual wind and the forecasted wind

exemple 360 The difference between the actual wind and the forecasted wind.

Question 184-21 : Cas is 320 kt.flight level 330.oat isa +15°c.tas is approximately . compressibility factor 0 939 ?

530 kt

Admin .oat is 15°c 15°c 2 x 33 = 36°c .on computer in airspeed window set press alt '33' in front of coat °c ' 36°c' on the outer scale in front of cas 320 kt you can read tas 565 kt .true air speed tas is obtained from calibrated air speed cas by correcting for compressibility and density .568 x 0 939 = 533 kt exemple 364 530 kt.

Question 184-22 : Given .mach numer 0 8.flight level 330.oat isa +15°c .tas is approximately . compressibility factor 0 94 ?

480 kt

Admin .temperature at fl330 = 51°c 33°c x 2 + 15 .isa +15°c so 51°c + 15°c = 36°c.tas = m*lss.lss = 38 95 x sqrtt°a t°a =273 36= 237°k .lss = 38 95 x sqrt237 = 599 63.tas = 0 8 x 599 63 = 480 kt .you need to apply compressibility factor if you want to go from cas to tas not from mach to tas .true air speed tas is obtained from calibrated air speed cas by correcting for compressibility and density exemple 368 480 kt.

Question 184-23 : The main purpose of dr dead reckoning is ?

To obtain with reasonable accuracy the aircraft's position between fixes or in the absence of fixes

exemple 372 To obtain with reasonable accuracy, the aircraft's position between fixes or in the absence of fixes.

Question 184-24 : An aircraft is flying at fl390 at a speed of mach 0 821 .oat isa 4°c.the compressibility factor is 0 942 .calculate the tas ?

467 kt

Admin .isa temperature at fl390 = 56 5°c 56 5°c is considered to be the lowest isa temperature .isa 4°c so oat = 60 5°c.tas = m*lss.lss = 39 x sqrtt°a t°a =273 60 5= 212 5°k .lss = 39 x sqrt212 5 = 568 5.tas = 0 821 x 568 5 = 466 7 kt .you need to apply compressibility factor if you want to go from cas to tas not from mach to tas .true air speed tas is obtained from calibrated air speed cas by correcting for compressibility and density exemple 376 467 kt.

Question 184-25 : An aircraft descends from fl240 to fl80 for the final approach .track = 070°.cas = 220 kt.oat = isa 10°c.the average tas in the descent is ?

276 kt

Admin .at the exam average tas used for descent problems is calculated at the altitude 1/2 of the descent altitude .at fl160 isa temperature = 15°c 2°c x 16 = 17°c .oat is isa 10°c thus oat is 27°c at fl160 .on the computer in airspeed window put 27ºc next to fl160 go to cas 220 kt on inner scale and read tas on outer scale 276 kt exemple 380 276 kt.

Question 184-26 : An aircraft is flying at fl 350 with cas = 300 kt .oat = isa + 4°c .the compressibility factor is 0 939 .calculate the tas ?

509 kt

Admin .isa temperature at fl350 = 15°c + 35 x 2 = 55°c.isa +4°c so oat = 51°c.on the computer in airspeed window put 51ºc next to fl350 go to cas 300 kt on inner scale and read tas on outer scale 542 kt .multiply 542 kt x 0 939 = 509 kt .you need to apply compressibility factor if you want to go from cas to tas not from mach to tas .true air speed tas is obtained from calibrated air speed cas by correcting for compressibility and density exemple 384 509 kt.

Question 184-27 : Given .track = 355°.tas = 190 kt.wind 270°/25 kt.after 30 minutes of flying with the planned tas and true heading the aircraft is 3 5 nm right of track and 4 5 nm ahead of the dead reckoning position .calculate the actual wind ?

254°/34 kt

Ecqb03 august 2016 exemple 388 254°/34 kt.

Question 184-28 : Given .fl 400.oat = 65°c.ias = 260 kt.instrument and position error to be neglected .compressibility factor = 0 935.calculate the true air speed taking compressibility into account ?

479 kt

Admin .calibrated airspeed cas is indicated airspeed ias corrected for instrument error and position error the question states instrument and position error to be neglected .therefore ias = cas.oat = 65°c.on the computer in airspeed window put 65ºc next to fl400 go to cas 260 kt on inner scale and read tas on outer scale 513 kt .true air speed tas is obtained from calibrated air speed cas by correcting for compressibility and density .513 x 0 935 = 479 kt exemple 392 479 kt.

Question 184-29 : Given .tas = 210 kt.cas = 190 kt.pressure altitude = 9000 ft.calculate mach number ?

0 34

Admin .using the computer align tas external ring & cas internal ring when done go to your airspeed case and read the one corresponding to the pressure altitude above the 9000 ft line you should read about 22°c .mach number = tas / lss.lss = 39*sqrt t in k° .here t° = 22°c = 22 + 273 = 251°k .hence lss = 39*sqrt 251 = 617 876.thus mach number = 210/617 876 = 0 339 = 0 34 exemple 396 0.34.

Question 184-30 : A dr position is to be found ?

On the desired track

exemple 400 On the desired track.

Question 184-31 : Which of the factors named hereafter should be considered by the pilot when selecting landmarks as visual reference points .1 possibility of identification.2 transmitted frequency.3 visibility.4 closeness to the track.the combination that regroups all of the correct statements is ?

1 3 4

exemple 404 1, 3, 4.

Question 184-32 : Given .fl 300.oat = 45°c.ias = 260 kt.instrument and position error to be neglected.compressibility factor = 0 96.calculate the true air speed taking compressibility into account ?

408 kt

exemple 408 408 kt.

Question 184-33 : On a mercator chart one minute on n55° parallel is 3 1 mm .the map scale at 40°n is ?

1 457 650

exemple 412 1 : 457 650

Question 184-34 : The nominal scale of a north stereopolar map is ?

At the north pole

exemple 416 At the north pole.

Question 184-35 : Given .tas = 140 kt true hdg = 302° w/v = 045° t /45kt .calculate the drift angle and gs ?

16°l 156 kt

Admin .under index set true heading 302° centre dot on tas 140 kt with the rotative scale set wind . 1448.read drift 16° left .ground speed is 156 kt exemple 420 16°l - 156 kt.

Question 184-36 : Given .tas = 290 kt.true hdg = 171°.w/v = 310° t /30kt .calculate the drift angle and gs ?

4°l 314 kt

Admin . 2528 exemple 424 4°l - 314 kt

Question 184-37 : Given .tas = 485 kt .true heading = 226° .true wind = 110°/95kt .calculate the drift angle and gs ?

9°r 533 kt

Admin .under index set true heading 226° centre dot on tas 485 kt with the rotative scale set wind . 1451.read drift 9° right .ground speed is 533 kt exemple 428 9°r - 533 kt.

Question 184-38 : Given .tas = 472 kt .true heading = 005° .true wind = 110°/50kt .calculate the drift angle and gs ?

6°l/490 kt

Admin .under index set true heading 005° centre dot on tas 472 kt with the rotative scale set wind . 1740.read drift 5 5° left .ground speed is 487 kt .closest answer 6°l/490 kt exemple 432 6°l/490 kt.

Question 184-39 : Given .tas = 375 kt true heading = 124° wind = 130°/55 kt .calculate the true track and gs ?

123° 320 kt

Admin . 1741 exemple 436 123° - 320 kt.

Question 184-40 : Given .tas = 198 kt.hdg °t = 180.w/v = 359/25 .calculate the track °t and gs ?

180° 223 kt

exemple 440 180° - 223 kt


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