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An aircraft at fl290 is required to commence descent when 50 nm from a vor and ? [ Multiple protocol ]

Question 174-1 : 1900 ft / min 1700 ft / min 1800 ft / min 2000 ft / min

exemple 274 1900 ft / min. — Admin .50 nm / 271 kt = 0 185h > 11 min.29000 ft 8000 ft = 21000 ft.21000 ft / 11 min = 1900 ft/min 1900 ft / min.

An aircraft at fl350 is required to commence descent when 85 nm from a vor and ?

Question 174-2 : 1800 ft/min 1900 ft/min 1600 ft/min 1700 ft/min

exemple 278 1800 ft/min. — Admin .fl350 fl080 = 27000 ft .85 nm / 340 kt = 0 25 h 15 minutes .27000 ft / 15 min = 1800 ft/min 1800 ft/min.

An aircraft is planned to fly from position 'a' to position 'b' distance 480 nm ?

Question 174-3 : 12 06 utc 11 57 utc 12 03 utc 11 53 utc

exemple 282 12:06 utc. — Admin .150 nm / 240 kt = 0 625 > 37 5 min + 2 min behind = 39 5 min > 0 658h .150 nm / 0 658h = 228 kt .480 nm 150 nm = 330 nm.330 nm / 228 kt = 1 447h > 86 8 min .86 8 min + 39 5 min = 126 3 min > 2h 06 min 18 seconds.10 00 + 2h 06 min = 12 06 utc 12:06 utc.

An aircraft is planned to fly from position 'a' to position 'b' distance 320 nm ?

Question 174-4 : 13 33 utc 13 40 utc 13 47 utc 14 01 utc

exemple 286 13:33 utc. — Admin .70 nm / 180 kt = 0 389 > 23 3min .as we are 3 minute ahead of planned time so our actual time is 20 3 min > 0 339h .70 nm / 0 339 = 207 kt.320 nm 70 nm = 250 nm.250 nm / 207 kt = 1 21h > 72 5 min.72 5 min + 20 3 min = 92 8 min > 1h 32 min 48 seconds .12 00 + 1h 33 min = 13 33 utc 13:33 utc.

An aircraft is planned to fly from position 'a' to position 'b' distance 250 nm ?

Question 174-5 : 11 15 utc 10 44 utc 10 50 utc 11 10 utc

exemple 290 11:15 utc. — Admin .75 nm / 115 kt = 0 652h > 39 min 0 652 x 60 .9 00 + 39 min = 9 39 + 1 5 min = 9 40 5 we would arrive at 9 39 but we are 1 5 minute behind so +1 5 min .40 5 / 60 = 0 675 h now count the speed with revised time .75 / 0 675 = 111 1 kt adjusted gs to 1 5 min behind planned of course we are slower as we arrived later .250 nm 75 nm = 175 nm the remaining distance to fly .175 / 111 1 = 1 575 h > 94 5 min 1 575 x 60 remaining distance divided by new gs .9 40 5 + 1h 34 5 min = 11 15 utc .stanley .250/75 = 3 33.3 33 x 1 5min = 5 min.250nm/115kt = 2 17h.2 17 x 60 = 130 min = 2h30min + 5 min = 2h35min > 9 00 +2h35' > 11 15 11:15 utc.

Given .distance 'a' to 'b' is 475 nm planned gs 315 kt atd actual time ?

Question 174-6 : 340 kt 360 kt 300 kt 320 kt

exemple 294 340 kt — Admin .475 nm / 315 = 1 51 h 1 51 x 60 minutes = 91 minutes or 1h31 .estimated time of arrival at 'b' is 10h00 + 1h31 = 11h31 .fix obtained along track at 10h40 shows a groud speed of 190 nm / 40 minutes = 4 75 nm/minutes.4 75 x 60 minutes = 285 kt instead of 315 kt planned .it remains 475 nm 190 nm = 285 nm.it remains 51 minutes to achieve the distance 10h40 to 11h31 .285 nm / 51 minutes =5 58 nm/minutes.5 6 x 60 = 336 kt 340 kt

Given distance 'a' to 'b' is 325 nm planned gs 315 kt atd 1130 utc 1205 utc ?

Question 174-7 : 355 kt 375 kt 395 kt 335 kt

exemple 298 355 kt. — Admin .325 nm / 315 kt = 1 03h > 62 min 1 03 x 60 .11 30 + 62 min = 12 32.12 32 12 05 = 27 min > 0 45h 27/60 .325 nm 165 nm = 160 nm so we need to make 160 nm in 27 min .160 nm / 0 45 = 355 kt 355 kt.

Given distance 'a' to 'b' is 100 nm fix obtained 40 nm along and 6 nm to the ?

Question 174-8 : 15° right 9° right 6° right 18° right

exemple 302 15° right. — Admin .tan 1 6/40 = 8 53.tan 1 6/60 = 5 71.total = 14 24° .or using formula .tke = 6 x 60/40 and 6 x 60/60.tke = 9° and 6°.ca = 9° + 6° = 15° as we are left of the course correction is to the right 15° right.

Given distance 'a' to 'b' is 90 nm fix obtained 60 nm along and 4 nm to the ?

Question 174-9 : 12° left 16° left 4° left 8° left

exemple 306 12° left. — Admin .to calculate the heading change at an off course fix to directly reach the next waypoint use the one in sixty rule . 4/60 x 60 = 4°. 4/30 x 60 = 8°.8° + 4° = 12° and left because we are to the right of the target course 12° left.

Complete line 1 of the 'flight navigation log'.positions 'a' to 'b' what is the ?

Question 174-10 : Hdg 268° eta 1114 utc hdg 282° eta 1128 utc hdg 282° eta 1114 utc hdg 268° eta 1128 utc

exemple 310 Hdg 268° - eta 1114 utc. — Admin . 2496 Hdg 268° - eta 1114 utc.

Complete line 2 of the 'flight navigation log' positions 'c' to 'd' what is the ?

Question 174-11 : Hdg 193° eta 1239 utc hdg 188° eta 1229 utc hdg 193° eta 1249 utc hdg 183° eta 1159 utc

exemple 314 Hdg 193° - eta 1239 utc. — Admin . 2496 Hdg 193° - eta 1239 utc.

Complete line 3 of the 'flight navigation log' positions 'e' to 'f' what is the ?

Question 174-12 : Hdg 105° eta 1205 utc hdg 095° eta 1155 utc hdg 106° eta 1215 utc hdg 115° eta 1145 utc

exemple 318 Hdg 105° - eta 1205 utc. — Admin . 2496 Hdg 105° - eta 1205 utc.

Complete line 4 of the 'flight navigation log' positions 'g' to 'h' what is the ?

Question 174-13 : Hdg 344° eta 1336 utc hdg 354° eta 1326 utc hdg 334° eta 1336 utc hdg 344° eta 1303 utc

exemple 322 Hdg 344° - eta 1336 utc. — Admin . 2496 Hdg 344° - eta 1336 utc.

Complete line 5 of the 'flight navigation log' positions 'j' to 'k' what is the ?

Question 174-14 : Hdg 337° eta 1422 utc hdg 320° eta 1412 utc hdg 337° eta 1322 utc hdg 320° eta 1432 utc

exemple 326 Hdg 337° - eta 1422 utc. — Admin . 2496 Hdg 337° - eta 1422 utc.

Complete line 6 of the 'flight navigation log' positions 'l' to 'm' what is the ?

Question 174-15 : Hdg 075° eta 1502 utc hdg 064° eta 1449 utc hdg 075° eta 1452 utc hdg 070° eta 1459 utc

exemple 330 Hdg 075° - eta 1502 utc. — Admin . 2496.you have to find tas .set 55°c in the airspeed window next to mkt .read on the outer main scale in front of 0 84 a tas of 485 kt .below center dot set tas 485 kt under index set true course 070° with the rotative scale set the wind 020°/60kt read below 60 kt a 6° right drift .true heading = 070° 6° = 064° .gs is 441 kt after having set the true heading on the computer .magnetic heading = 064° + 11° variation = 075° . 495/441 x60 = 67 minutes Hdg 075° - eta 1502 utc.

Given .tas = 197 kt true course = 240° w/v = 180/30kt .descent is initiated at ?

Question 174-16 : 1400 ft/min 800 ft/min 950 ft/min 1500 ft/min

exemple 334 1400 ft/min. — Admin .using flight computer you will get gs = 182 kt .distance = rate x time.39 nm = 182 x time.t = 39/182 = 0 214h > 12 8 min total time of descent .22000 ft 4000 ft = 18000 ft height to be flown in descent .18000 / 12 8 min = 1406 ft/min 1400 ft/min.

Given .ils glide path angle = 3 5° ground speed = 150 kt .what is the ?

Question 174-17 : 900 ft/min 350 ft/min 700 ft/min 300 ft/min

exemple 338 900 ft/min. — Admin .1 in 60 rule is a rule of thumb . 3 5 x 150 x 100 /60 = 875 ft/min 900 ft/min.

Given aircraft height 2500 ft ils gp angle 3° at what approximate distance ?

Question 174-18 : 8 3 nm 7 0 nm 13 1 nm 14 5 nm

exemple 342 8.3 nm. — Admin .distance= height x 60 / angle° .distance= 2500x60/3° .distance= 50000 ft then divide by 6080 as 1nm=6080 ft 8.3 nm.

An island appears 60° to the left of the centre line on an airborne weather ?

Question 174-19 : 046° 086° 226° 026°

exemple 346 046°. — Admin .276° mh + 10°e = 286° th .286° 60° = 226° th to the island to get from the island simply reverse it .226° 180° = 046° 046°.

An island appears 45° to the right of the centre line on an airborne weather ?

Question 174-20 : 059° 101° 239° 329°

exemple 350 059°. — Admin .215° mh 21°w = 194° th .194° + 45° = 239° th to the island to get from the island simply reverse it .239° 180° = 059° 059°.

An island appears 30° to the right of the centre line on an airborne weather ?

Question 174-21 : 220° 160° 130° 190°

exemple 354 220°. — Admin .355° mh + 15°e = 010° th .010° + 30° = 040° th to the island to get from the island simply reverse it .040° + 180° = 220° 220°.

An island appears 30° to the left of the centre line on an airborne weather ?

Question 174-22 : 145° 325° 205° 195°

exemple 358 145°. — Admin .020° mh 25°w = 355° th .island is on the left so 355° 30° = 325°.true bearing from the island so 325° 180° = 145° 145°.

Given an aircraft is flying a track of 255° m 2254 utc it crosses radial ?

Question 174-23 : The same as it was at 2254 utc greater than it was at 2254 utc randomly different than it was at 2254 utc less than it was at 2254 utc

exemple 362 The same as it was at 2254 utc — Admin . 1724.this is an isosceles triangle .the first angle at 22 54 is 255° 180° = 75°.the second angle at the vor is 30° .the last one = 180° 75°+30° = 75° The same as it was at 2254 utc

The distance between two waypoints is 200 nm to calculate compass heading the ?

Question 174-24 : 14 nm 7 nm 0 nm 21 nm

exemple 366 14 nm. — Admin .for each degree of error that you have at every 60 nm of travel you will be 1 nm off track .you have 200/60 3 33 nm off track for each degree of error .total error is 4° from 2°e to 2°w .4° x 3 33 nm = 13 33 nm .using goniometric functions .tan4° = / 200. = tan4° x 200. = 13 99 nm 14 nm.

Given .eta to cross a meridian is 2100 utc.gs is 441 kt.tas is 491 kt.at 2010 ?

Question 174-25 : 40 kt 90 kt 75 kt 60 kt

exemple 370 40 kt. — Admin .at 20h30 the airplane is at 50 minutes of the meridian with a speed of 441 kt thus at a distance of 367 5 nm 441/60 x 50 .now it has to travel 367 5 nm in 55 minutes . 367 5 / 55 x 60 = 401 kt .tas reduction is 441 401 = 40 kt 40 kt.

The flight log gives the following data 'true track drift true heading magnetic ?

Question 174-26 : 119° 3°l 122° 2°e 120° +4° 116° 115° 5°r 120° 3°w 123° +2° 121° 117° 4°l 121° 1°e 122° 3° 119° 125° 2°r 123° 2°w 121° 4° 117°

exemple 374 119°, 3°l, 122°, 2°e, 120°, +4°, 116° — Admin . 2498.use this wonderful table for those questions 119°, 3°l, 122°, 2°e, 120°, +4°, 116°

At 0020 utc an aircraft is crossing the 310° radial at 40 nm of a vor/dme ?

Question 174-27 : 085° 226 kt 090° 232 kt 080° 226 kt 088° 232 kt

exemple 378 085° - 226 kt. — Admin .draw the situation . 2499.it is a isosceles triangle with at least two equal sides .the angle at the vor is 90° and the other two angles are 45° .at 00 20 the bearing from the aircraft to the vor is 310° 180° = 130° .track is 130° 45° = 085° .it is not mandatory to calculate the groundspeed but you can use pythagoras .40² + 40² = distance between position at 00 20 and postion at 00 35 ².distance is = 56 567 nm.56 nm covered in 15 minutes . 56 567/15 x60 = 226 kt 085° - 226 kt.

Given .tas is 120 kt.ata 'x' 1232 utc.eta 'y' 1247 utc.ata 'y' is 1250 utc.what ?

Question 174-28 : 1302 utc 1257 utc 1300 utc 1303 utc

exemple 382 1302 utc. — Admin .x to y = 30 nm in 18 minutes from 12h32 to 12h50 = 100 kt ground speed .y to z = 20 nm at 100 kt = 12 minutes .12h50 + 12 minutes = 13h02 1302 utc.

Given .fl120 oat is isa standard cas is 200 kt track is 222° m heading is ?

Question 174-29 : 050° t / 70 kt 040° t / 105 kt 055° t / 105 kt 065° t / 70 kt

exemple 386 050°(t) / 70 kt. — Admin .oat at fl120 is 15° 2° x 12 = 9°c. 2501.magnetic heading is 215° variation is 15°w = true heading is 200° .our magnetic track is 222° minus 15°w = true track is 207° . 105 nm / 21 min x 60 = 300 kt ground speed .on the computer under centre dot set tas 239 kt under index set true heading 200° mark the point where drift 7°right crosses the ground speed 300 kt . 2502.wind is 050°/70 kt 050°(t) / 70 kt.

A useful method of a pilot resolving during a visual flight any uncertainty in ?

Question 174-30 : Set heading towards a line feature such as a coastline motorway river or railway fly the reverse of the heading being flown prior to becoming uncertain until a pinpoint is obtained fly expanding circles until a pinpoint is obtained fly reverse headings and associated timings until the point of departure is regained

exemple 390 Set heading towards a line feature such as a coastline, motorway, river or railway — Set heading towards a line feature such as a coastline, motorway, river or railway

An aircraft is descending down a 6% slope whilst maintaining a ground speed of ?

Question 174-31 : 1800 ft/min 10800 ft/min 3600 ft/min 900 ft/min

exemple 394 1800 ft/min. — Admin .1 nm = 6080 ft.6% ==> 0 06. 300 kt x 6080 ft / 60 min x 0 06 = 1824 ft/min 1800 ft/min.

An aircraft is flying according the flight log at the annex .after 15 minutes ?

Question 174-32 : 258° 292° 270° 253°

exemple 398 258°. — Admin . 2517.15 min / 60 = 0 25h.0 25 x 130 kt gs = 32 5 nm.32 5 nm + 2 5 nm ahead of dr = 35 nm.50 nm distance 35 nm = 15 nm.tke = 3 x 60/35 and correction angle = 3 x 60/15.tke = 5° and 12° ca = 5° + 12° = 17° .as we are north of 275° to get back we need to fly more to the south therefore minus 17° so 275° 17° = 258° 258°.

An island is observed to be 30° to the right of the nose of the aircraft the ?

Question 174-33 : 330° 270° 250° 310°

exemple 402 330°. — Admin .magnetic heading 290°.variation east magnetic least .magnetic heading = true heading 10°e.true heading = magnetic heading + 10°e = 300° .the true bearing from the aircraft to the island is 300° + 30°right = 330° 330°.

An aircraft follows a radial to a vor/dme station at 10 00 the dme reads 120 nm ?

Question 174-34 : 10 24 10 27 10 18 10 21

exemple 406 10:24. — Admin .15 nm in 3 minutes = 5 nm per minute.distance to station 105 nm.105 / 5 nm/min = 21 minutes .10 03 + 00 21 = 10 24 10:24.

You are departing from an airport which has an elevation of 2000 ft .the qnh is ?

Question 174-35 : 920 ft/min 1080 ft/min 590 ft/min 750 ft/min

exemple 410 920 ft/min. — Admin .total climb = 7500 ft 2000 ft = 5500 ft .10 nm at 100 kt = 10 nm/ 100/60 = 6 minutes .5500 ft / 6 min = 916 6 ft/min 920 ft/min.

At reference or see europe low altitude enroute chart e lo 1a.an aircraft is ?

Question 174-36 : 280 kt 385 kt 485 kt 180 kt

exemple 414 280 kt. — Admin .the distance between vor's is 59 nm.20 5 nm + 10 5 nm = 31 nm.59 nm 31 nm = 28 nm we need to make 28 nm in 6 minutes so .28 / 0 1 = 280 kt 6 minutes = 0 1h 280 kt.

The distance between point of departure and destination is 340 nm and wind ?

Question 174-37 : 1h and 49 min 1h and 30 min 1h and 37 min 1h and 21 min

exemple 418 1h and 49 min. — Admin .psr = e x h / o + h e endurance h gs home o gs out .using flight computer you will get 20 kt headwind so gsout = 120 kt gshome = 160 kt .psr = 190 x 160 kt/120 + 160 3h 10min > 190 min .psr = 108 6 min > 1h 49 min 1h and 49 min.

You are tracking the 200° radial inbound to a vor and your true heading is ?

Question 174-38 : 310°/65 320°/55 330°/50 300°/50

exemple 422 310°/65. — Admin .200° radial inbound = magnetic track 020° variation +5° = true track 025° .true heading = 010°.drift = 15° right.090° radial outbound = magnetic track 090° variation +5° = true track 095° .magnetic heading 080° variation +5° = true heading 085° .drift = 10° right .on the computer .centre dot on tas 240 kt put true heading 010° under the index and mark a line down the 15° right drift line .rotate to put true heading 085° under index and mark a line down the 10° right drift line .the point where these two lines intersect is the end of the wind vector rotate to position it under the centre dot and read the wind 310°/65 kt 310°/65.

An aircraft is flying according the flight log at the annex after 15 minutes of ?

Question 174-39 : 112° 080° 090° 107°

exemple 426 112°. — Admin .15 min / 60 = 0 25h.0 25 x 130 kt gs = 32 5 nm.32 5 nm + 2 5nm ahead of dr = 35 nm.50 nm dist 35 nm = 15 nm.tke = 3 x 60/35 and 3 x 60/15.tke = 5° and 12° ca = 5° + 12° = 17°.as we are north of 095° to get back we need to fly more to the south therefore add 17° so 095 + 17° = 112° 112°.

An aircraft is departing from an airport which has an elevation of 2000 ft and ?

Question 174-40 : 11 1 nm 10 3 nm 13 3 nm 16 6 nm

exemple 430 11.1 nm. — Admin .2000 ft 300 ft 10 hpa diff = 1700 ft.10000 ft 1700 ft = 8300 ft.8300 / 1000 ft/min = 8 3 min > 0 1383h .0 1383 x 80 kt gs = 11 1 nm 11.1 nm.

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