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Given .tas = 125 kt true heading = 355° true wind = 320°/30 kt .calculate the ? [ Multiple protocol ]

Question 170-1 : 005° 102 kt 345° 100 kt 348° 102 kt 002° 98 kt

Admin .center dot on tas 125 kt .true heading 355° under index.put wind direction under the red compass rose under 30 kt your drift is 10° right giving a track of 005° and a groundspeed under the wind mark of 102 kt . 1457.cmarzocchini .all this questions doesnt work with cr3 wich is used for us in spain be carefull about this ie in this question just in this one the correct answer using cr3 is 003/95 i only did this comment in this question but take it into account for the reminder questions . .feel free to download the 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 'tas' index .over 'tc' index set '355°' .locate the wind dot by finding the 320° line on the green scale and making the point where this line intersects the green 30 kt circle .reading directly up from the wind dot we see that there is a left crosswind component of 17 kt .looking at the outer scale find 17 and opposite it read 8° crab angle . 2486.since the wind is from the left the true heading must be left of the true course therefore rotate the top disc 8° to the left counter clockwise now the 'tc' index points to 003° .looking directly above the wind dot after the above move you now find that the crosswind component has changed to 22 kt instead of 17 kt .locate 22 on the outer scale and find opposite crab angle of 10° . 2487.it now appears that the first crab angle of 8° was 2° too less therefore add 2° of the 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 exemple 270 005° - 102 kt.005° - 102 kt.

Given .tas = 225 kt .hdg °t = 123° .w/v = 090/60kt .calculate the track °t ?

Question 170-2 : 134° 178 kt 134° 188 kt 120° 190 kt 128° 180 kt

Admin .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 . 2488.rotate to put heading 123° under true index . 2489.you read a drift of 11° right and a groundspeed of 178 kt .add the drift to your heading to find the true track 134° exemple 274 134° - 178 kt.134° - 178 kt.

Given .tas = 480 kt true heading = 040° wind = 090°/60 kt .calculate true ?

Question 170-3 : 034° 445 kt 028° 415 kt 032° 425 kt 036° 435 kt

Admin .center dot on tas 480 kt .true heading 040° under index. 2026.put wind direction under the red compass rose under 60 kt your drift is 6° left giving a track of 034° and a groundspeed under the wind mark of 445 kt exemple 278 034° - 445 kt.034° - 445 kt.

Given .tas = 170 kt.true heading = 100°.wind = 350/30kt .calculate the true ?

Question 170-4 : 109° 182 kt 091° 183 kt 103° 178 kt 098° 178 kt

..center dot on tas 170 kt .true heading 100° under index.put wind direction under the red compass rose under 30 kt your drift is 9° right giving a track of 109° and a groundspeed under the wind mark of 182 kt . /com en/com061 168 jpg.the answer is 109° and 182 kt exemple 282 109° - 182 kt.109° - 182 kt.

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

Question 170-5 : 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 286 7r - 204 kt.7r - 204 kt.

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

Question 170-6 : 4l 415 kt 2l 420 kt 6l 395 kt 5l 385 kt

Admin .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 290 4l - 415 kt4l - 415 kt

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

Question 170-7 : 9r 108 kt 10l 104 kt 9l 105 kt 8r 104 kt

exemple 294 9r - 108 kt9r - 108 kt

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

Question 170-8 : 2r 223 kt 2l 224 kt 1l 225 kt 1r 221 kt

Admin .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 298 2r - 223 kt.2r - 223 kt.

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

Question 170-9 : 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 302 6l - 194 kt6l - 194 kt

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

Question 170-10 : 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 306 050° - 145 kt050° - 145 kt

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

Question 170-11 : 346 102 kt 006 95 kt 358 101 kt 359 102 kt

Admin .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 310 346 - 102 kt346 - 102 kt

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

Question 170-12 : 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 314 301 - 169 kt.301 - 169 kt.

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

Question 170-13 : 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 318 358° - 428 kt.358° - 428 kt.

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

Question 170-14 : 099° 199 kt 121° 207 kt 121° 199 kt 097° 201 kt

Admin .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 322 099° - 199 kt.099° - 199 kt.

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

Question 170-15 : 260/30kt 257/35kt 255/25kt 265/30kt

Admin .true heading is 307° true track is 313° our drift is 6° right . 2492.wind 261°/30kt exemple 326 260/30kt.260/30kt.

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

Question 170-16 : 075/45kt 065/45kt 060/50kt 075/70kt

Admin .true heading is 133° true track is 144° our drift is 11° right . 2493 exemple 330 075/45kt.075/45kt.

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

Question 170-17 : 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 334 180°/05 kt.180°/05 kt.

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

Question 170-18 : 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 338 115°/35 kt.115°/35 kt.

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

Question 170-19 : 340/50kt 335/45kt 335/55kt 340/45kt

Admin .true heading is 035° true track is 046° our drift is 11° right . 2520 exemple 342 340/50kt340/50kt

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

Question 170-20 : 075° 39 min 095° 31 min 096° 29 min 076° 34 min

Admin .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 346 075°, 39 min.075°, 39 min.

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

Question 170-21 : 22 kt 26 kt 15 kt 20 kt

Admin .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 350 22 kt.22 kt.

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

Question 170-22 : 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 354 10 kt.10 kt.

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

Question 170-23 : 33 kt 27 kt 25 kt 16 kt

Admin .wind angle = 100° 063° = 37°.crosswind = windspeed x sin 37° = 20 kt.maximum allowable windspeed = 20 kt / sin 37° = 33 kt exemple 358 33 kt.33 kt.

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

Question 170-24 : 0736 utc 0730 utc 0810 utc 0716 utc

Admin .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 362 0736 utc.0736 utc.

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

Question 170-25 : 5 minutes 4 minutes 6 minutes 7 minutes

.43 nm / 510 kt/60 min = 5 minutes exemple 366 5 minutes.5 minutes.

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

Question 170-26 : 8 hr 04 min 7 hr 48 min 7 hr 49 min 8 hr 10 min

Admin .985 nm / 122 kt/60 min = 484 minutes 8h04 exemple 370 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 170-27 : 4 hr 25 min 3 hr 25 min 3 hr 26 min 4 hr 10 min

Admin .1920 nm / 435 kt/60 min = 265 minutes 4h45 exemple 374 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 170-28 : 11 hr 10 min 11 hr 06 min 11 hr 07 min 11 hr 15 min

Admin .5360 nm / 480 kt/60 min = 670 minutes 11h30 exemple 378 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 170-29 : 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 382 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 170-30 : 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 386 3 hr 12 min.3 hr 12 min.

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

Question 170-31 : 24 kt 27 kt 31 kt 34 kt

Admin .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 390 24 kt.24 kt.

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

Question 170-32 : 20 kt and 40 kt 18 kt and 50 kt 15 kt and 43 kt 12 kt and 38 kt

Admin .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 394 20 kt and 40 kt.20 kt and 40 kt.

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

Question 170-33 : 31 kt 36 kt 21 kt 26 kt

Admin .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 398 31 kt.31 kt.

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

Question 170-34 : 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 402 10 kt.10 kt.

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

Question 170-35 : 40 nm 50 nm 60 nm 30 nm

Admin . 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 406 40 nm.40 nm.

The equivalent of 70 m/sec is approximately ?

Question 170-36 : 136 kt 145 kt 210 kt 35 kt

Admin .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 410 136 kt.136 kt.

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

Question 170-37 : 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 414 21 kt.21 kt.

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

Question 170-38 : 6° right 8° right 2° left 4° right

Admin .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 418 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 170-39 : 1000 km 1222 km 804 km 540 km

Admin .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 422 1000 km.1000 km.

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

Question 170-40 : 8° right 6° right 8° left 4° right

Admin .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 426 8° right.8° right.

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