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Formation > assignment : Given . tas = 235 kt hdg t = 076° w/v = 040/40kt .calculate the drift angle ?

Question 168-1 : 7r 204 kt 7l 269 kt 5l 255 kt 5r 207 kt

...under index set true heading 076° centre dot on tas 235 kt with the rotative scale set wind. /com en/com061 169 jpg..read drift 7° right .ground speed is 205 kt close enough to answer 204 kt exemple 268 7r - 204 kt.7r - 204 kt.

Given .tas = 440 kt true heading = 349° wind = 040/40kt .calculate drift and ?

Question 168-2 : 4l 415 kt 2l 420 kt 6l 395 kt 5l 385 kt

.under index set true heading 349° centre dot on tas 440 kt with the rotative scale set wind 040/40kt.read drift 4° left .ground speed is 415 kt exemple 272 4l - 415 kt4l - 415 kt

Given tas = 95 kt hdg t = 075° w/v = 310/20kt calculate the drift and gs ?

Question 168-3 : 9r 108 kt 10l 104 kt 9l 105 kt 8r 104 kt

exemple 276 9r - 108 kt9r - 108 kt

Given .tas = 230 kt hdg t = 250° wind = 205/10kt .calculate the drift and gs ?

Question 168-4 : 2r 223 kt 2l 224 kt 1l 225 kt 1r 221 kt

.under index set true heading 250° centre dot on tas 230 kt with the rotative scale set wind . 2490.read drift 2° right .ground speed is 223 kt exemple 280 2r - 223 kt.2r - 223 kt.

Given .tas = 205 kt hdg t = 180° wind = 240/25kt .calculate the drift and gs ?

Question 168-5 : 6l 194 kt 7l 192 kt 3l 190 kt 4l 195 kt

...under index set true heading 180° centre dot on tas 205 kt with the rotative scale set wind. /com en/com061 176 jpg..read drift 6° left .ground speed is 194 kt exemple 284 6l - 194 kt6l - 194 kt

Given .tas = 132 kt true heading = 053° wind = 205°/15 kt .calculate the true ?

Question 168-6 : 050° 145 kt 057° 144 kt 052° 143 kt 051° 144 kt

...under index set true heading 053° centre dot on tas 132 kt with the rotative scale set wind . /com en/com061 178 jpg..read drift 3° left .ground speed is 145 kt exemple 288 050° - 145 kt050° - 145 kt

Given .tas = 90 kt.true heading = 355°.wind = 120/20 kt.calculate true track ?

Question 168-7 : 346 102 kt 006 95 kt 358 101 kt 359 102 kt

.under index set true heading 355° centre dot on tas 90 kt with the rotative scale set wind . 2491.read drift 9° left .ground speed is 103 kt close enough for the answer exemple 292 346 - 102 kt346 - 102 kt

Given .tas = 155 kt track t = 305° w/v = 160/18kt .calculate the hdg °t and gs ?

Question 168-8 : 301 169 kt 305 169 kt 309 170 kt 309 141 kt

...under index set true track 305° centre dot on tas 155 kt with the rotative scale set wind. /com en/com061 181 jpg..now drift is 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 exemple 296 301 - 169 kt.301 - 169 kt.

Given .tas = 465 kt track t = 007° w/v = 300/80kt .calculate the hdg °t and gs ?

Question 168-9 : 358° 428 kt 001° 435 kt 017° 490 kt 357° 502 kt

...under index set true track 007° centre dot on tas 465 kt with the rotative scale set wind. /com en/com061 184 jpg..now drift is 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 exemple 300 358° - 428 kt.358° - 428 kt.

Given .tas = 200 kt track t = 110° w/v = 015/40 kt .calculate the hdg °t and ?

Question 168-10 : 099° 199 kt 121° 207 kt 121° 199 kt 097° 201 kt

.under index set true track 110° centre dot on tas 200 kt with the rotative scale set wind . 1775.now drift is 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 is wrong you have tail wind correct answer using sin and cos and cr3 098/205. .don't be so confident and read carefully the explanation drift is always measured from heading to track .when you will be on your track with the correct heading to counteract drift the wind becomes a headwind exemple 304 099° - 199 kt.099° - 199 kt.

Given .true hdg = 307° tas = 230 kt track t = 313° gs = 210 kt .calculate the ?

Question 168-11 : 260/30kt 257/35kt 255/25kt 265/30kt

.true heading is 307° true track is 313° our drift is 6° right . 2492.wind 261°/30kt exemple 308 260/30kt.260/30kt.

Given .true hdg = 133° tas = 225 kt track t = 144° gs = 206 kt .calculate the ?

Question 168-12 : 075/45kt 065/45kt 060/50kt 075/70kt

.true heading is 133° true track is 144° our drift is 11° right . 2493 exemple 312 075/45kt.075/45kt.

Given .true heading = 206° tas = 140 kt true track = 207° gs = 135 kt ?

Question 168-13 : 180°/05 kt 000°/05 kt 000°/10 kt 180°/10 kt

...true heading is 206° true track is 207° our drift is 1° right. /com en/com061 192 jpg..wind 180°/05 kt exemple 316 180°/05 kt.180°/05 kt.

Given .true heading = 145° tas = 240 kt true track = 150° gs = 210 kt ?

Question 168-14 : 115°/35 kt 360°/35 kt 180°/35 kt 295°/35 kt

...true heading is 145° true track is 150° our drift is 5° right. /com en/com061 194 jpg..wind 115°/35 kt exemple 320 115°/35 kt.115°/35 kt.

Given .true hdg = 035° tas = 245 kt track t = 046° gs = 220 kt .calculate the ?

Question 168-15 : 340/50kt 335/45kt 335/55kt 340/45kt

.true heading is 035° true track is 046° our drift is 11° right . 2520 exemple 324 340/50kt340/50kt

Given course required = 085° t forecast w/v 030/100kt tas = 470 kt distance = ?

Question 168-16 : 075° 39 min 095° 31 min 096° 29 min 076° 34 min

.tas = 470 kt.true course = 085°.vw = 030°/100kt..drift = .gs =. 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 at the same place read the ground speed under the center bore from the scal on the axis of the slide. setting .set 85° to true index set the indicator to 030° on the black azimuth circle being upwind adjust the speed arc labelled 470 of the diagram slide to the wind speed 10 100 kt of the indicator scale. reading .under the plotted point read the wind correction angle 10° under the center bore read the ground speed 405 kt . 1770.then true heading = true course drift = 085° 10° = 075°..405/60 = 6 75 nm/minutes..265/6 75 = 39 minutes exemple 328 075°, 39 min.075°, 39 min.

For a landing on runway 23 227° magnetic surface.wind reported by the atis is ?

Question 168-17 : 22 kt 26 kt 15 kt 20 kt

.wind from tower is already corrected for 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 exemple 332 22 kt.22 kt.

Given . maximum allowable tailwind component for landing 10 kt planned runway ?

Question 168-18 : 10 kt 8 kt 15 kt 18 kt

.wind from tower atis is recorded by the tower is already corrected for variation the wind from tower refers to magnetic north .this is a tailwind our wind angle is = 047°+180° 210° = 17°..tailwind = windspeed x cos 17° = 10 kt..maximum allowable windspeed = 10 kt / cos 17° = 10 46 kt exemple 336 10 kt.10 kt.

Given .maximum allowable crosswind component is 20 kt .runway 06 rwy qdm 063° ?

Question 168-19 : 33 kt 27 kt 25 kt 16 kt

.wind angle = 100° 063° = 37°..crosswind = windspeed x sin 37° = 20 kt..maximum allowable windspeed = 20 kt / sin 37° = 33 kt exemple 340 33 kt.33 kt.

Given .true course a to b = 250° .distance a to b = 315 nm .tas = 450 kt .w/v ?

Question 168-20 : 0736 utc 0730 utc 0810 utc 0716 utc

.set 250° under index center dot on tas 450 kt and wind 200º/60kt . 2522.drift is 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 is 0650 utc + 46 minutes = 0736 utc exemple 344 0736 utc.0736 utc.

Given .gs = 510 kt distance a to b = 43 nm .what is the time from a to b ?

Question 168-21 : 5 minutes 4 minutes 6 minutes 7 minutes

.43 nm / 510 kt/60 min = 5 minutes exemple 348 5 minutes.5 minutes.

Given .gs = 122 kt .distance from a to b = 985 nm .what is the time from a to b ?

Question 168-22 : 8 hr 04 min 7 hr 48 min 7 hr 49 min 8 hr 10 min

.985 nm / 122 kt/60 min = 484 minutes 8h04 exemple 352 8 hr 04 min.8 hr 04 min.

Given .gs = 435 kt distance from a to b = 1920 nm .what is the time from a to b ?

Question 168-23 : 4 hr 25 min 3 hr 25 min 3 hr 26 min 4 hr 10 min

.1920 nm / 435 kt/60 min = 265 minutes 4h45 exemple 356 4 hr 25 min.4 hr 25 min.

Given .gs = 480 kt distance from a to b = 5360 nm .what is the time from a to b ?

Question 168-24 : 11 hr 10 min 11 hr 06 min 11 hr 07 min 11 hr 15 min

.5360 nm / 480 kt/60 min = 670 minutes 11h30 exemple 360 11 hr 10 min.11 hr 10 min.

Given .gs = 105 kt distance from a to b = 103 nm .what is the time from a to b ?

Question 168-25 : 00 hr 59 min 00 hr 57 min 00 hr 58 min 01 hr 01 min

.103 nm / 105 kt/60 min = 59 minutes exemple 364 00 hr 59 min.00 hr 59 min.

Given .gs = 135 kt distance from a to b = 433 nm .what is the time from a to b ?

Question 168-26 : 3 hr 12 min 3 hr 25 min 3 hr 19 min 3 hr 20 min

.433 nm / 135 kt/60 min = 192 minutes 3h32 exemple 368 3 hr 12 min.3 hr 12 min.

Given .runway direction 083° m .surface wwind 035/35 kt .calculate the ?

Question 168-27 : 24 kt 27 kt 31 kt 34 kt

.angle between the wind and the direction of the runway 083° 035° = 48°..effective headwind = cos of the angle between the wind and the direction of the runway x windspeed..effective headwind = cos 48° x 35 kt = 23 42 kt exemple 372 24 kt.24 kt.

Given .for take off an aircraft requires a headwind component of at least 10 kt ?

Question 168-28 : 20 kt and 40 kt 18 kt and 50 kt 15 kt and 43 kt 12 kt and 38 kt

.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 exemple 376 20 kt and 40 kt.20 kt and 40 kt.

Given .runway direction 230° t .surface wind 280° t /40 kt .calculate the ?

Question 168-29 : 31 kt 36 kt 21 kt 26 kt

.angle between the wind and the direction of the runway 280° 230° = 50°..crosswind = sine of the angle between the wind and the direction of the runway x windspeed..crosswind = sin50° x 40 kt = 30 64 kt exemple 380 31 kt.31 kt.

Given .runway direction 210° m surface w/v 230° m /30 kt .calculate the ?

Question 168-30 : 10 kt 19 kt 16 kt 13 kt

Angle between the wind and the direction of the runway 230° 210° = 20°..crosswind = sine of the angle between the wind and the direction of the runway x windspeed..crosswind = sin20° x 30 kt = 10 26 kt exemple 384 10 kt.10 kt.

An aircraft obtains a relative bearing of 315° from an ndb at 08h30 at 08h40 ?

Question 168-31 : 40 nm 50 nm 60 nm 30 nm

. 2521.you have an isoceles triangle and the angles are 45° 45° and 90°.the hypotonuse is the distance from the 0830 position to the ndb the two equal sides are . the distance travelled between 0830 and 0840 .and. the distance from the 0840 position to the ndb .in 10 minutes at 240 kt the aircraft will travel 40 nm so this is also the distance from the 0830 position and the ndb.you can also use the 1 in 60 rule .315° 270° = 45°.240 kt / 60 min = 4° per minute.10 min x 4° = 40 nm exemple 388 40 nm.40 nm.

The equivalent of 70 m/sec is approximately ?

Question 168-32 : 136 kt 145 kt 210 kt 35 kt

.1 nm = 0 5 m/s..70 / 0 5 = 140 kt closest to 136 kt than 145 kt .if you want to find the exact answer .70 m/s x 3600 secondes = 252000 m/h.252000 m/h = 252 km/h.252 / 1 852 = 136 kt exemple 392 136 kt.136 kt.

Given .runway direction 305° m surface w/v 260° m /30 kt .calculate the cross ?

Question 168-33 : 21 kt 24 kt 27 kt 18 kt

.angle between the wind and the direction of the runway 305° 260° = 45°.crosswind = sine of the angle between the wind and the direction of the runway x windspeed..crosswind = sin45° x 30 kt = 21 2 kt exemple 396 21 kt.21 kt.

The distance between positions a and b is 180 nm an aircraft departs position a ?

Question 168-34 : 6° right 8° right 2° left 4° right

.use the one in sixty rule .for small angles for every 60 nm along 1 nm off track is equivalent to 1° deviation.4 nm deviation after having travelled 60 nm means there is 4° off track deviation caused by drift .in order to counteract drift you need to correct your heading by 4° .now to arrive at position b you have to recover those 4 nm deviation in the 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 the intended track exemple 400 6° right.6° right.

A flight is to be made from 'a' 49°s 180°e/w to 'b' 58°s 180°e/w the ?

Question 168-35 : 1000 km 1222 km 804 km 540 km

.you are travelling south along the greenwich anti meridian from 49°s to 58°s which is a 9° change of latitude .9° x 60 nm = 540 nm.540 nm x 1 852 = 1000 km exemple 404 1000 km.1000 km.

Given .distance a to b = 120 nm after 30 nm aircraft is 3 nm to the left of ?

Question 168-36 : 8° right 6° right 8° left 4° right

.use the one in sixty rule.track error angle from a = 3 nm x 60 / 30 nm = 6°.. 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 / 90 nm = 2°..to reach destination b from this position the correction angle on the heading should be 6° + 2° = 8° exemple 408 8° right.8° right.

An aircraft was over 'a' at 1435 hours flying direct to 'b' given .distance 'a' ?

Question 168-37 : 1657 1744 1846 1721

Ground speed out 470 + 55 = 525 kt.ground speed home 470 75 = 395 kt..pet = d x gsh / gso + gsh .pet = 2900 x 395 / 525 + 395 = 1245 nm.1245 nm / 525 kt = 2 37h..2 37 x 60 minutes = 142 minutes 2h42minutes.14h35 + 2h42 = 16h57 exemple 412 1657.1657.

Given .distance 'a' to 'b' 2484 nm.groundspeed 'out' 420 kt.groundspeed 'back' ?

Question 168-38 : 193 minutes 163 minutes 173 minutes 183 minutes

.ground speed out 420 kt.ground speed home 500 kt..pet = distance x gsh / gso + gsh .pet = 2484 x 500 / 420 + 500 = 1350 nm.1350 nm / 420 kt = 3 21 h..3 21 h x 60 minutes = 193 minutes exemple 416 193 minutes.193 minutes.

Given .distance 'a' to 'b' 2484 nm.ground speed out 420 kt.ground speed home ?

Question 168-39 : 1940 nm 1908 nm 1736 nm 1630 nm

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs.ground speed out = 420 kt.ground speed home = 500 kt..point of safe return psr = 8 5 x 500 / 420 + 500 .point of safe return psr = 4250 / 920.point of safe return psr = 4 62 h..distance of the psr from the departure point at a speed of 420 kt .4 62 h x 420 = 1940 nm exemple 420 1940 nm.1940 nm.

An aircraft was over 'q' at 1320 hours flying direct to 'r' given .distance 'q' ?

Question 168-40 : 1752 1756 1820 1742

.ground speed out 480 90 = 390 kt.ground speed home 480 + 75 = 555 kt..pet = d x gsh / gso + gsh .pet = 3016 x 555 / 390 + 555 = 1771 nm.1771 nm / 390 kt = 4 54 h..4 54 x 60 minutes = 272 minutes 4h32minutes.13h40 + 4h32 = 17h52 exemple 424 1752.1752.

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