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Given TAS = 190 kt HDG T = 355° W/V = 165/25kt Calculate the drift and GS ?

Formation > assignment

exemple reponse 284
under index set true heading 355° centre dot on tas 190 kt with rotative scale set wind img /com_en/com061 174 jpg read drift 1° left ground speed 214 kt (close enough the answer).



Given TAS = 250 ktHDG T = 029°W/V = 035/45ktCalculate the drift and GS ?

exemple reponse 285
Given tas = 250 kthdg t = 029°w/v = 035/45ktcalculate drift and gs under index set true heading 029° centre dot on tas 250 kt with rotative scale set wind read drift 1° left ground speed 205 kt.

Given TAS = 485 kt True Heading = 168° wind = 130/75 kt Calculate True Track and Ground Speed ?

exemple reponse 286
Given tas = 485 kt true heading = 168° wind = 130/75 kt calculate true track and ground speed under index set true heading 168° centre dot on tas 485 kt with rotative scale set wind img /com_en/com061 180 jpg read drift 6° right (168° + 6° = 174°) ground speed 430 kt (close enough the answer).

  • exemple reponse 287
    Given tas = 130 kttrack t = 003°w/v = 190/40 ktcalculate hdg °t and gs under index set true heading 168° centre dot on tas 485 kt with rotative scale set wind img /com_en/com061 180 jpg read drift 6° right (168° + 6° = 174°) ground speed 430 kt (close enough the answer).

  • exemple reponse 288
    Given tas = 227 kt track t = 316° w/v = 205/15kt calculate hdg °t and gs under index set true track 316° centre dot on tas 227 kt with rotative scale set wind now drift always measured from heading to track turn to set true heading 312° (316° 4° right drift) under index you now read a ground speed of 232 kt.

  • exemple reponse 289
    Given tas = 200 kt track t = 073° w/v = 210/20kt calculate hdg °t and gs under index set true track 316° centre dot on tas 227 kt with rotative scale set wind now drift always measured from heading to track turn to set true heading 312° (316° 4° right drift) under index you now read a ground speed of 232 kt.

  • exemple reponse 290
    Given tas = 270 kt track t = 260° wind = 275°/30kt calculate hdg °t and gs under index set true track 260° centre dot on tas 270 kt with rotative scale set wind img /com_en/com061 187 jpg now drift always measured from heading to track turn to set true heading 262° (260° + 2° left drift) under index you now read a ground speed of 241 kt.

  • Question 169-8

    Given true hdg = 233° tas = 480 kt track t = 240° gs = 523 kt calculate w/v true heading 233° true track 240° our drift 7° right img /com_en/com061 189 jpg wind 111°/77kt (closest answer 110/75kt).

  • Question 169-9

    Given true heading = 074° tas = 230 kt true track = 066° ground speed = 242 kt calculate wind true heading 074° true track 066° our drift 8° left img /com_en/com061 191 jpg where rotative scale crosses ground speed arc (242 kt) we read wind 180°/ 35 kt.

  • Question 169-10

    Given true heading = 054° tas = 450 kt true track = 059° gs = 416 kt calculate wind true heading 054° true track 059° our drift 5° right wind 010°/50 kt.

  • Question 169-11

    Given true heading = 002°tas = 130 kttrue track = 353°ground speed = 132 kt calculate wind true heading 002° true track 353° our drift 9° left img /com_en/com061 195 jpg wind 094°/22 kt (closest answer 095°/20 kt).

  • Question 169-12

    Given gs = 236 kt distance from a to b = 354 nm what the time from a to b 354 nm /(236 kt/60 min) = 90 minutes (1h30).

  • Question 169-13

    Given gs = 345 kt distance from a to b = 3560 nm what the time from a to b 3560 nm /(345 kt/60 min) = 619 minutes (10h19).

  • Question 169-14

    Given gs = 95 kt distance from a to b = 480 nm what the time from a to b 480 nm /(95 kt/60 min) = 303 minutes (5h03).

  • Question 169-15

    Given gs = 120 kt distance from a to b = 84 nm what the time from a to b 84 nm /(120 kt/60 min) = 42 minutes.

  • Question 169-16

    Given distance 'a' to 'b' 1973 nmground speed out 430 ktground speed back 385 ktsafe endurance 7 hr 20 minthe distance from 'a' to point of safe return psr point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 430 kt ground speed home = 385 kt point of safe return (psr) = 7 33 x 385 / (430 + 385) point of safe return (psr) = 2822 / 815 point of safe return (psr) = 3 46 h distance of psr from departure point at a speed of 430 kt 3 46 h x 430 = 1489 nm.

  • Question 169-17

    Given distance 'a' to 'b' 2346 nmground speed out 365 ktground speed back 480 ktthe time from 'a' to point of equal time pet between 'a' and 'b' ground speed out 365 kt ground speed home 480 kt pet = distance x gsh / (gso + gsh) pet = 2346 x 480 /(365 + 480) = 1332 nm 1332 nm / 365 kt = 3 65 h 3 65 h x 60 minutes = 219 minutes.

  • Question 169-18

    Given distance 'q' to 'r' 1760 nmground speed out 435 ktground speed back 385 ktthe time from 'q' to point of equal time pet between 'q' and 'r' ground speed out 435 kt ground speed home 385 kt pet = distance x gsh / (gso + gsh) pet = 1760 x 385 /(435 + 385) = 826 nm 826 nm / 435 kt = 1 9h 1 9 h x 60 minutes = 114 minutes.

  • Question 169-19

    Given distance 'q' to 'r' 1760 nmground speed out 435 ktground speed back 385 ktsafe endurance 9 hrthe distance from 'q' to point of safe return psr between 'q' and 'r' point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 435 kt ground speed home = 385 kt point of safe return (psr) = 9 x 385 / (435 + 385) point of safe return (psr) = 3465 / 820 point of safe return (psr) = 4 22 h distance of psr from departure point at a speed of 435 kt 4 22 h x 435 = 1838 nm.

  • Question 169-20

    An aeroplane flying at tas 180 kt on a track of 090° the w/v 045° / 50kt how far can aeroplane fly out from its base and return in one hour centre dot on tas 180 kt rotate to wind direction 045° come down from centre dot wind speed 50 kt mark end of wind vector at 130 kt rotate to outbound track under heading index note drift 14°starboard rotate track to drift note drift now 11°starboard rotate track to drift note drift still 11°starboard our outbound heading will be 079° to track 090° our ground speed outbound 141 kt proceed same way to find ground speed homebound (you will find 212 kt) pnr = t x gso x gsh / (gso + gsh) pnr = 1 x 141 x 212 / (141 + 212) pnr = 29892 / 353 pnr = 84 67 nm.

  • Question 169-21

    An aircraft maintaining a 5 2% gradient at 7 nm from runway on a flat terrain its height approximately 1 nm = 6080 ft aircraft at 7 nm from runway 7 nm x 6080 ft = 42560 ft 42560 x (5 2/100) = 2213 ft.

  • Question 169-22

    An aircraft descends from fl250 to fl100 the rate of descent 1000 ft/min gs 360 kt the flight path angle vertical speed = gradient (%)* gs 1000 = gradient (%) * 360 gradient(%)= 2 77 gradient(%)= angle * 100/60 2 77 = angle * 100/60 > angle = 1 662º.

  • Question 169-23

    The outer marker of an ils with a 3° glide slope located 4 6 nm from threshold assuming a glide slope height of 50 ft above threshold approximate height of an aircraft passing outer marker 4 6 nm x tan 3° = 0 24 nm 1 nm = approximately 6000 ft 0 24 x 6000 = 1440 ft 1440 ft + 50 ft = 1490 ft the approximate height of aircraft 1490 ft (close to 1450 ft) svandam in a 3° glide slope 1nm = 300 feet then 4 6 nm * 300 = 1380 ft 1380 ft + 50 ft = 1430 ft the approximate height of aircraft 1430 ft (close to 1450 ft) johanjog more precise calculation 4 6 nm*6080 ft/1nm = 27968 ft 1 60 rule 3º/60 = d/27968 d = 1398 4 ft 1398 4 + 50 ft = 1448 4 ft (1450 ft).

  • Question 169-24

    730 ft/min equals if you don't have nav compupter or formula how to calculate it directly you can help yourself with easy counts 730 / 3 28 (1 m = 3 28 ft) = 222 6 m/min to convert it to seconds simply divide it 60 so 222 6/60 = 3 7 m/sec.

  • Question 169-25

    How long will it take to fly 5 nm at a groundspeed of 269 kt 5 / (269/60) =1 115 minutes 0 115 x 60 = 7 secondes 1 min 7 sec.

  • Question 169-26

    An aircraft travels 2 4 statute miles in 47 seconds what its groundspeed 2 4/47 x 3600 = 184 statute miles per hour 1 statute mile = 1 609 km = 0 87 nm 184 x 0 87 = 160 kt.

  • Question 169-27

    The icao definition of eta the Estimated time of arrival at destination. 2 4/47 x 3600 = 184 statute miles per hour 1 statute mile = 1 609 km = 0 87 nm 184 x 0 87 = 160 kt.

  • Question 169-28

    Assuming zero wind what distance will be covered an aircraft descending 15000 ft with a tas of 320 kt and maintaining a rate of descent of 3000 ft/min Estimated time of arrival at destination. 15000 ft / 3000 ft/min = 5 minutes 5 min / 60 min = 0 0833 hour 0 0833 x 320 = 26 67 nm.

  • Question 169-29

    An island appears 30° to left of centre line on an airborne weather radar display what the true bearing of aircraft from island if at time of observation aircraft was on a magnetic heading of 276° with magnetic variation 12°w Estimated time of arrival at destination. magnetic heading 276º variation 12ºw true heading 264º island bearing 30ºleft true bearing of island from aircraft 234º true bearing of aircraft from island 234°+/ 180º = 054°.

  • Question 169-30

    An aircraft at fl370 required to commence descent at 120 nm from a vor and to cross facility at fl130 if mean gs the descent 288 kt minimum rate of descent required Estimated time of arrival at destination. 37000 ft 13000 ft = 24000 ft 120 nm / 288kt = 0 417h > 25 min (0 417 x 60) 24000 ft / 25 min = 960 ft/min.

  • Question 169-31

    An aircraft at fl310 m0 83 temperature 30°c required to reduce speed in order to cross a reporting point five minutes later than planned assuming that a zero wind component remains unchanged when 360 nm from reporting point mach number should be reduced to Estimated time of arrival at destination. set temperature ( 30°c) in airspeed window in front of 8 3 (mach 83 inner scale) on outer scale you read tas = 503 kt (360/503) x 60 = 43 minutes we are at 43 minutes from reporting point (360/ ?) x 60 = 48 minutes ? = (360 x 60) / 48 = 450 kt in airspeed window set outside temperature ( 30°c) in front of mach index go to 450 kt on outer scale you read 7 4 (mach 0 74) on inner scale.

  • Question 169-32

    A ground feature was observed on a relative bearing of 325° and five minutes later on a relative bearing of 280° the aircraft heading was 165° m variation 25°w drift 10°right and gs 360 kt when relative bearing was 280° distance and true bearing of aircraft from feature was Estimated time of arrival at destination. 5 minutes at 360 kt = (360 / 60) x 5 = 30 nm magnetic heading 165° variation 25°w true heading = 140° true track= 140° + 10° = 150° we have an isosceles triangle in an isosceles triangle two sides are equal in length relative bearing 280° true bearing of feature from aircraft = (140 + 280) 360 = 060° true bearing of aircraft from feature = 060 + 180 = 240°.

  • Question 169-33

    An aircraft at fl350 required to descend to cross a dme facility at fl80 maximum rate of descent 1800 ft/min and mean gs descent 276 kt the minimum range from dme at which descent should start Estimated time of arrival at destination. 35000 ft 8000 ft = 27000 ft 27000 ft / 1800 ft/min = 15 min > 0 25h (15 / 60) 0 25 x 276 kt = 69 nm.

  • Question 169-34

    An aircraft at fl120 ias 200kt oat 5° and wind component +30kt required to reduce speed in order to cross a reporting point 5 min later than planned assuming flight conditions do not change when 100 nm from reporting point ias should be reduced to Estimated time of arrival at destination. use nav computer to find tas ias 200 kt = tas 240 kt ground speed = 240 kt + 30 kt = 270 kt 100 nm at gs 270 kt = 22 minutes 100 nm in 27 minutes = 100 / 27 = 3 7 nm per minute 3 7 x 60 = 222 kt 222 kt 30 kt (wind) = 192 kt tas 192 kt ==> with nav computer ==> 159 kt ias.

  • Question 169-35

    An aircraft at fl350 required to cross a vor/dme facility at fl110 and to commence descent when 100 nm from facility if mean gs the descent 335 kt minimum rate of descent required Estimated time of arrival at destination. 24000 ft to descend 335/60 = 5 583 nm/min 100/5 583 = 18 minutes 24000 / 18 min = 1340 ft/min.

  • Question 169-36

    An aircraft at fl370 m0 86 oat 44°c headwind component 110 kt required to reduce speed in order to cross a reporting point 5 minutes later than planned if speed reduction were to be made 420 nm from reporting point what mach number required Estimated time of arrival at destination. tas 503 kt ground speed is 503 110 = 393 kt 420 nm at 393 kt = 1 07 h (1 07 x 60 = 64 minutes) atc asks you to cross a rportin poitn 5 minutes later in 64 + 5 minutes 69 / 60 = 1 15 h 420 / 1 15 = 365 kt 365 kt + 110 kt = 475 kt = m 0 81.

  • Question 169-37

    An aircraft at fl390 required to descend to cross a dme facility at fl70 maximum rate of descent 2500 ft/min mean gs during descent 248 kt what the minimum range from dme at which descent should commence Estimated time of arrival at destination. 39000 7000 = 32000 ft 32000 / 2500 = 12 8 minutes 248 kt / 60 minutes = 4 14 nm/minute 12 8 x 4 14 = 53 nm.

  • Question 169-38

    An aircraft at fl370 required to commence descent when 100 nm from a dme facility and to cross station at fl120 if mean gs during descent 396 kt minimum rate of descent required approximately Estimated time of arrival at destination. 37000 12000 = 25000 ft 396 kt / 60 minutes = 6 6 nm/minute 100 nm / 6 6 = 15 15 minutes before fly over dme 25000 / 15 = 1650 ft/min.

  • Question 169-39

    An aircraft at fl140 ias 210 kt oat 5°c and wind component minus 35 kt required to reduce speed in order to cross a reporting point 5 minutes later than planned assuming that flight conditions do not change when 150 nm from reporting point aircraft should reduce ias Estimated time of arrival at destination. given ias (cas) 210 kt fl140 oat 5°c on computer we find tas = 262 kt we can now find groundspeed 262 35 = 227 kt 150 nm at 227 kt = 39 6 minutes we must reduce speed to cross a point 5 minutes later 150 nm in 39 6 + 5 minutes = 202 kt we must reduce ground speed by 227 202 = 25 kt a tas of 262 25 = 237 kt required now back to computer tas 237 kt fl140 oat 5°c new ias 190 kt we should reduce ias by 210 190 = 20 kt.

  • Question 169-40

    At 0422 an aircraft at fl370 gs 320kt on direct track to vor 'x' 185 nm distant the aircraft required to cross vor 'x' at fl80 for a mean rate of descent of 1800 ft/min at a mean gs of 232 kt latest time at which to commence descent Estimated time of arrival at destination. fl370 to fl80 = 29000 ft 29000 / 1800 ft/min = 16 1 minutes during descent aircraft will cover (232/60) x 16 1 = 62 25 nm distance before top of descent (tod) 184 62 25 = 121 75 nm time before tod 121 75 / (320/60) = 22 8 minutes the latest time to commence descent is 04h22 + 23 minutes = 04h45.


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