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Given TAS = 370 kt True Heading = 181° Wind = 095°/35 ktCalculate true track and Ground Speed ?

Formation > assignment

exemple reponse 281
center dot on tas 370 kt true heading 181° under index put wind direction under red compass rose under 35 kt your drift 5° right giving a track of 186° a groundspeed (under wind mark) of 370 kt.



Given TAS = 125 kt True Heading = 355° True Wind = 320°/30 kt Calculate the True Track and Ground Speed ?

exemple reponse 282
Given tas = 125 kt true heading = 355° true wind = 320°/30 kt calculate true track and ground speed center dot on tas 125 kt true heading 355° under index put wind direction under red compass rose under 30 kt your drift 10° right giving a track of 005° a groundspeed (under wind mark) of 102 kt cmarzocchini all this questions doesnt work with cr3 wich used us in spain be carefull about this ie in this question just in this one correct answer using cr3 003/95 i only did this comment in this question but take it into account the reminder questions feel free to download cr3 instructions here http //www jeppesen com/download/misc/crinstructions pdf page 44 45 true course (track) and ground speed on top set speed '125 kt' over 'ta index over 'tc' index set '355°' locate wind dot finding 320° line on green scale making point where this line intersects green 30 kt circle reading directly up from wind dot we see that there a left crosswind component of 17 kt looking at outer scale find 17 opposite it read 8° crab angle since wind from left true heading must be left of true course therefore rotate top disc 8° to left (counter clockwise) now 'tc' index points to 003° looking directly above wind dot after above move you now find that crosswind component has changed to 22 kt instead of 17 kt locate 22 on outer scale find opposite crab angle of 10° it now appears that first crab angle of 8° was 2° too less therefore add 2° of previous tc adjustment making a true course reading of 005° you can read a headwind of 23kt therefore 125 kt 23 kt = 102 kt ground speed.

Given TAS = 225 kt HDG °T = 123° W/V = 090/60kt Calculate the Track °T and GS ?

exemple reponse 283
Given tas = 225 kt hdg °t = 123° w/v = 090/60kt calculate track °t and gs put 225 kt in center dot under true index set wind direction 090° mark wind on centre line at 165 kt (60 kt below centre dot) rotate to put heading 123° under true index you read a drift of 11° right a groundspeed of 178 kt add drift to your heading to find true track 134°.

  • exemple reponse 284
    Given tas = 480 kt true heading = 040° wind = 090°/60 kt calculate true track and ground speed center dot on tas 480 kt true heading 040° under index put wind direction under red compass rose under 60 kt your drift 6° left giving a track of 034° a groundspeed (under wind mark) of 445 kt.

  • exemple reponse 285
    Given tas = 170 kttrue heading = 100°wind = 350/30kt calculate true track and gs center dot on tas 170 kt true heading 100° under index put wind direction under red compass rose under 30 kt your drift 9° right giving a track of 109° a groundspeed (under wind mark) of 182 kt img /com_en/com061 168 jpg the answer 109° 182 kt.

  • exemple reponse 286
    Given tas = 235 kt hdg t = 076° w/v = 040/40kt calculate drift angle and gs under index set true heading 076° centre dot on tas 235 kt with rotative scale set wind img /com_en/com061 169 jpg read drift 7° right ground speed 205 kt (close enough to answer 204 kt).

  • exemple reponse 287
    Given tas = 440 kt true heading = 349° wind = 040/40kt calculate drift and ground speed under index set true heading 349° centre dot on tas 440 kt with rotative scale set wind 040/40kt read drift 4° left ground speed 415 kt.

  • Question 166-8

    Given tas = 95 kt hdg t = 075° w/v = 310/20kt calculate drift and gs under index set true heading 349° centre dot on tas 440 kt with rotative scale set wind 040/40kt read drift 4° left ground speed 415 kt.

  • Question 166-9

    Given tas = 230 kt hdg t = 250° wind = 205/10kt calculate drift and gs under index set true heading 250° centre dot on tas 230 kt with rotative scale set wind read drift 2° right ground speed 223 kt.

  • Question 166-10

    Given tas = 205 kt hdg t = 180° wind = 240/25kt calculate drift and gs under index set true heading 180° centre dot on tas 205 kt with rotative scale set wind img /com_en/com061 176 jpg read drift 6° left ground speed 194 kt.

  • Question 166-11

    Given tas = 132 kt true heading = 053° wind = 205°/15 kt calculate true track and ground speed under index set true heading 053° centre dot on tas 132 kt with rotative scale set wind img /com_en/com061 178 jpg read drift 3° left ground speed 145 kt.

  • Question 166-12

    Given tas = 90 kttrue heading = 355°wind = 120/20 ktcalculate true track and ground speed under index set true heading 355° centre dot on tas 90 kt with rotative scale set wind read drift 9° left ground speed 103 kt (close enough the answer).

  • Question 166-13

    Given tas = 155 kt track t = 305° w/v = 160/18kt calculate hdg °t and gs under index set true track 305° centre dot on tas 155 kt with rotative scale set wind img /com_en/com061 181 jpg now drift always measured from heading to track turn to set true heading 301° (305° 4° right drift) under index you now read a ground speed of 169 kt.

  • Question 166-14

    Given tas = 465 kt track t = 007° w/v = 300/80kt calculate hdg °t and gs under index set true track 007° centre dot on tas 465 kt with rotative scale set wind img /com_en/com061 184 jpg now drift always measured from heading to track turn to set true heading 358° (007° 9° right drift) under index you now read a ground speed of 428 kt.

  • Question 166-15

    Given tas = 200 kt track t = 110° w/v = 015/40 kt calculate hdg °t and gs under index set true track 110° centre dot on tas 200 kt with rotative scale set wind now drift always measured from heading to track turn to set true heading 099° (110° 11° right drift) under index you now read a ground speed of 198 kt cmarzocchini the answer wrong you have tail wind !!! correct answer using sin cos cr3 098/205 don't be so confident! read carefully explanation drift always measured from heading to track when you will be on your track with correct heading to counteract drift wind becomes a headwind.

  • Question 166-16

    Given true hdg = 307° tas = 230 kt track t = 313° gs = 210 kt calculate w/v true heading 307° true track 313° our drift 6° right wind 261°/30kt.

  • Question 166-17

    Given true hdg = 133° tas = 225 kt track t = 144° gs = 206 kt calculate w/v true heading 133° true track 144° our drift 11° right .

  • Question 166-18

    Given true heading = 206° tas = 140 kt true track = 207° gs = 135 kt calculate wind true heading 206° true track 207° our drift 1° right img /com_en/com061 192 jpg wind 180°/05 kt.

  • Question 166-19

    Given true heading = 145° tas = 240 kt true track = 150° gs = 210 kt calculate wind true heading 145° true track 150° our drift 5° right img /com_en/com061 194 jpg wind 115°/35 kt.

  • Question 166-20

    Given true hdg = 035° tas = 245 kt track t = 046° gs = 220 kt calculate w/v true heading 035° true track 046° our drift 11° right .

  • Question 166-21

    Given course required = 085° t forecast w/v 030/100kt tas = 470 kt distance = 265 nm calculate true hdg and flight time tas = 470 kt true course = 085° vw = 030°/100kt drift = ? gs = ? (a) set true track to true index (b) turn indicator to wind direction in this case using black azimuth graduation (the angle being upwind counting anti clockwise) (c) shift speed arc corresponding to true air speed so as to coincide with wind speed on indicator (d) read wind correction at same place read ground speed under center bore from scal on axis of slide setting set 85° to true index set indicator to 030° on black azimuth circle (being upwind) adjust speed arc labelled 470 of diagram slide to wind speed 10 (100 kt) of indicator scale reading under plotted point read wind correction angle 10° under center bore read ground speed 405 kt then true heading = true course drift = 085° 10° = 075° 405/60 = 6 75 nm/minutes 265/6 75 = 39 minutes.

  • Question 166-22

    For a landing on runway 23 227° magnetic surfacewind reported the atis 180/30 kt variation 13°e calculate cross wind component wind from tower already corrected variation the wind from tower refers to magnetic north wind angle = 227° 180° = 47° crosswind = windspeed x sin wind angle crosswind = 30 kt x sin 47° = 22 kt.

  • Question 166-23

    Given maximum allowable tailwind component landing 10 kt planned runway 05 047° magnetic the direction of surface wind reported atis 210° variation 17°e calculate maximum allowable windspeed that can be accepted without exceeding tailwind limit wind from tower (atis recorded the tower) already corrected variation the wind from tower refers to magnetic north this a tailwind our wind angle = (047°+180°) 210° = 17° tailwind = windspeed x cos 17° = 10 kt maximum allowable windspeed = 10 kt / cos 17° = 10 46 kt.

  • Question 166-24

    Given maximum allowable crosswind component 20 kt runway 06 rwy qdm 063° m wind direction 100° m calculate maximum allowable windspeed wind angle = 100° 063° = 37° crosswind = windspeed x sin 37° = 20 kt maximum allowable windspeed = 20 kt / sin 37° = 33 kt.

  • Question 166-25

    Given true course a to b = 250° distance a to b = 315 nm tas = 450 kt w/v = 200°/60kt etd a = 0650 utc what the eta at b set 250° under index center dot on tas 450 kt wind 200º/60kt drift 6° right now set heading 244° under index read ground speed 410 kt 315 nm / 410 kt = 0 768 hour = 46 minutes (0 768 x 60) etd at a 0650 utc + 46 minutes = 0736 utc.

  • Question 166-26

    Given gs = 510 kt distance a to b = 43 nm what the time from a to b 43 nm /(510 kt/60 min) = 5 minutes.

  • Question 166-27

    Given gs = 122 kt distance from a to b = 985 nm what the time from a to b 985 nm /(122 kt/60 min) = 484 minutes (8h04).

  • Question 166-28

    Given gs = 435 kt distance from a to b = 1920 nm what the time from a to b 1920 nm /(435 kt/60 min) = 265 minutes (4h25).

  • Question 166-29

    Given gs = 480 kt distance from a to b = 5360 nm what the time from a to b 5360 nm /(480 kt/60 min) = 670 minutes (11h10).

  • Question 166-30

    Given gs = 105 kt distance from a to b = 103 nm what the time from a to b 103 nm /(105 kt/60 min) = 59 minutes.

  • Question 166-31

    Given gs = 135 kt distance from a to b = 433 nm what the time from a to b 433 nm /(135 kt/60 min) = 192 minutes (3h12).

  • Question 166-32

    Given runway direction 083° m surface wwind 035/35 kt calculate effective headwind component angle between wind the direction of runway 083° 035° = 48° effective headwind = cos of angle between wind the direction of runway x windspeed effective headwind = cos 48° x 35 kt = 23 42 kt.

  • Question 166-33

    Given for take off an aircraft requires a headwind component of at least 10 kt and has a cross wind limitation of 35 kt the angle between wind direction and runway 60° calculate minimum and maximum allowable wind speeds crosswind = 35 kt maximum 35 = x sin 60 x = 35 / sin 60 = 40 kt headwind = 10 kt minimum 10 = x cos 60 x = 10 / cos 60 = 20 kt.

  • Question 166-34

    Given runway direction 230° t surface wind 280° t /40 kt calculate effective cross wind component angle between wind the direction of runway 280° 230° = 50° crosswind = sine of angle between wind the direction of runway x windspeed crosswind = sin50° x 40 kt = 30 64 kt.

  • Question 166-35

    Given runway direction 210° m surface w/v 230° m /30 kt calculate cross wind component Angle between wind the direction of runway 230° 210° = 20° crosswind = sine of angle between wind the direction of runway x windspeed crosswind = sin20° x 30 kt = 10 26 kt.

  • Question 166-36

    An aircraft obtains a relative bearing of 315° from an ndb at 08h30 at 08h40 relative bearing from same position 270° assuming no drift and a gs of 240 kt what the approximate range from ndb at 08h40 you have an isoceles triangle the angles are 45° 45° 90° the hypotonuse the distance from 0830 position to ndb two equal sides are distance travelled between 0830 0840 and distance from 0840 position to ndb in 10 minutes at 240 kt aircraft will travel 40 nm so this also distance from 0830 position the ndb you can also use 1 in 60 rule 315° 270° = 45° 240 kt / 60 min = 4° per minute 10 min x 4° = 40 nm.

  • Question 166-37

    The equivalent of 70 m/sec approximately 1 nm = 0 5 m/s 70 / 0 5 = 140 kt (closest to 136 kt than 145 kt ) if you want to find exact answer 70 m/s x 3600 secondes = 252000 m/h 252000 m/h = 252 km/h 252 / 1 852 = 136 kt.

  • Question 166-38

    Given runway direction 305° m surface w/v 260° m /30 kt calculate cross wind component angle between wind the direction of runway 305° 260° = 45° crosswind = sine of angle between wind the direction of runway x windspeed crosswind = sin45° x 30 kt = 21 2 kt.

  • Question 166-39

    The distance between positions a and b 180 nm an aircraft departs position a and after having travelled 60 nm its position pinpointed 4 nm left of intended track assuming no change in wind velocity what alteration of heading must be made in order to arrive at position b use one in sixty rule for small angles every 60 nm along 1 nm off track equivalent to 1° deviation 4 nm deviation after having travelled 60 nm means there 4° off track deviation caused drift in order to counteract drift you need to correct your heading 4° now to arrive at position b you have to recover those 4 nm deviation in remaining 120 nm 2 nm per 60 nm > 2° alteration 4° + 2° = 6° it's a right heading alteration since we are drifting left of intended track.

  • Question 166-40

    A flight to be made from 'a' 49°s 180°e/w to 'b' 58°s 180°e/w the distance in kilometres from 'a' to 'b' approximately you are travelling south along greenwich anti meridian from 49°s to 58°s which a 9° change of latitude 9° x 60 nm = 540 nm 540 nm x 1 852 = 1000 km.


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