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Question 200-1 : Complete line 5 of the 'flight navigation log', positions 'j' to 'k'. what is the hdg° m and eta. 2497 ? [ Exam pilot ]

Hdg 337° eta 1422 utc.

. 2496 exemple 300 Hdg 337° - eta 1422 utc.

Question 200-2 : Complete line 6 of the flight navigation log , positions l to m. what is the hdg° m and eta. 2497 ?

Hdg 075° eta 1502 utc.

. 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. exemple 304 Hdg 075° - eta 1502 utc.

Question 200-3 : Given.tas = 197 kt, true course = 240°, w/v = 180/30kt..descent is initiated at fl 220 and completed at fl 40..distance to be covered during descent is 39 nm..what is the approximate rate of descent ?

1400 ft/min.

.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. exemple 308 1400 ft/min.

Question 200-4 : Given.ils glide path angle = 3.5°, ground speed = 150 kt..what is the approximate rate of descent ?

900 ft/min.

.1 in 60 rule is a rule of thumb. 3.5 x 150 x 100 /60 = 875 ft/min. exemple 312 900 ft/min.

Question 200-5 : Given aircraft height 2500 ft, ils gp angle 3°. at what approximate distance from thr can you expect to capture the gp ?

8.3 nm.

.distance= height x 60 / angle°..distance= 2500x60/3°..distance= 50000 ft then divide by 6080 as 1nm=6080 ft. exemple 316 8.3 nm.

Question 200-6 : An island appears 60° to the left of the centre line on an airborne weather radar display. what is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading mh of 276° with the magnetic variation 10°e. ?

046°.

.276° mh + 10°e = 286° th..286° 60° = 226° th to the island, to get from the island, simply reverse it..226° 180° = 046°. exemple 320 046°.

Question 200-7 : An island appears 45° to the right of the centre line on an airborne weather radar display. what is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading mh of 215° with the magnetic variation 21°w ?

059°.

.215° mh 21°w = 194° th..194° + 45° = 239° th to the island, to get from the island, simply reverse it..239° 180° = 059°. exemple 324 059°.

Question 200-8 : An island appears 30° to the right of the centre line on an airborne weather radar display. what is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading mh of 355° with the magnetic variation var 15°e ?

220°.

.355° mh + 15°e = 010° th..010° + 30° = 040° th to the island, to get from the island, simply reverse it..040° + 180° = 220°. exemple 328 220°.

Question 200-9 : An island appears 30° to the left of the centre line on an airborne weather radar display. what is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading mh of 020° with the magnetic variation var 25°w ?

145°.

.020° mh 25°w = 355° th..island is on the left so 355° 30° = 325°..true bearing from the island, so 325° 180° = 145°. exemple 332 145°.

Question 200-10 : Given an aircraft is flying a track of 255° m , 2254 utc, it crosses radial 360° from a vor station, 2300 utc, it crosses radial 330° from the same station. at 2300 utc, the distance between the aircraft and the station is ?

The same as it was at 2254 utc

. 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°. exemple 336 The same as it was at 2254 utc

Question 200-11 : The distance between two waypoints is 200 nm, to calculate compass heading, the pilot used 2°e magnetic variation instead of 2°w..assuming that the forecast w/v applied, what will the off track distance be at the second waypoint ?

14 nm.

.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. exemple 340 14 nm.

Question 200-12 : Given.eta to cross a meridian is 2100 utc.gs is 441 kt.tas is 491 kt.at 2010 utc, atc requests a speed reduction to cross the meridian at 2105 utc..the reduction to tas will be approximately ?

40 kt.

.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. exemple 344 40 kt.

Question 200-13 : The flight log gives the following data 'true track, drift, true heading, magnetic variation, magnetic heading, compass deviation, compass heading' the right solution, in the same order, is ?

119°, 3°l, 122°, 2°e, 120°, +4°, 116°

. 2498.use this wonderful table for those questions. exemple 348 119°, 3°l, 122°, 2°e, 120°, +4°, 116°

Question 200-14 : At 0020 utc an aircraft is crossing the 310° radial at 40 nm of a vor/dme station. at 0035 utc the radial is 040° and dme distance is 40 nm. magnetic variation is zero..the true track and ground speed are ?

085° 226 kt.

.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. exemple 352 085° - 226 kt.

Question 200-15 : Given.tas is 120 kt.ata 'x' 1232 utc.eta 'y' 1247 utc.ata 'y' is 1250 utc.what is eta 'z'. 2500 ?

1302 utc.

.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. exemple 356 1302 utc.

Question 200-16 : Given.fl120, oat is isa standard, cas is 200 kt, track is 222° m , heading is 215° m , variation is 15°w..time to fly 105 nm is 21 min..what is the wind ?

050° t / 70 kt.

.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. exemple 360 050°(t) / 70 kt.

Question 200-17 : A useful method of a pilot resolving, during a visual flight, any uncertainty in the aircraft's position is to maintain visual contact with the ground and ?

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

exemple 364 Set heading towards a line feature such as a coastline, motorway, river or railway

Question 200-18 : An aircraft is descending down a 6% slope whilst maintaining a ground speed of 300 kt..the rate of descent of the aircraft is approximately ?

1800 ft/min.

.1 nm = 6080 ft.6% ==> 0.06.. 300 kt x 6080 ft / 60 min x 0.06 = 1824 ft/min. exemple 368 1800 ft/min.

Question 200-19 : An aircraft is flying according the flight log at the annex..after 15 minutes of flying with the planned tas and true heading the aircraft is 3 nm north of the intended track and 2.5 nm ahead of the dead reckoning position..to reach destination b from this position the true heading should be. 2516 ?

258°.

. 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°. exemple 372 258°.

Question 200-20 : An island is observed to be 30° to the right of the nose of the aircraft. the aircraft heading is 290° m , variation 10° e. the bearing ° t from the aircraft to the island is ?

330°.

.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°. exemple 376 330°.

Question 200-21 : An aircraft follows a radial to a vor/dme station. at 10 00 the dme reads 120 nm. at 10 03 the dme reads 105 nm. the estimated time overhead the vor/dme station is ?

10 24.

.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. exemple 380 10:24.

Question 200-22 : You are departing from an airport which has an elevation of 2000 ft..the qnh is 1013 hpa..10 nm away there is a waypoint you are required to pass at an altitude of 7500 ft..given a groundspeed of 100 kt, what is the minimum rate of climb ?

920 ft/min.

.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. exemple 384 920 ft/min.

Question 200-23 : At reference or see europe low altitude enroute chart e lo 1a..an aircraft is flying from inverness vordme n57°32.6', w004°02.5 to aberdeen vordme n57°18.6', w002°16.0'..at 1000 utc the fix of the aircraft is determined by vordme inverness.radial = 114.dme distance = 20.5 nm..at 1006 utc the fix of ?

280 kt.

exemple 388 280 kt.

Question 200-24 : The distance between point of departure and destination is 340 nm and wind velocity in the whole area is 100°/25 kt. tas is 140kt, true track is 135° and safe endurance 3h and 10 min..how long will it take to reach the point of safe return ?

1h and 49 min.

.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. exemple 392 1h and 49 min.

Question 200-25 : You are tracking the 200° radial inbound to a vor and your true heading is 010°..at the vor you then track the 090° radial outbound and are showing a heading of 080°m.the variation is +5° and the tas is 240 kt...what is the wind °t has affected the aircraft ?

310°/65.

.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. exemple 396 310°/65.

Question 200-26 : An aircraft is flying according the flight log at the annex. after 15 minutes of flying with the planned tas and true heading, the aircraft is 3 nm north of the intended track and 2.5 nm ahead of the dead reckoning position..to reach destination b from this position, the true heading should be. 2504 ?

112°.

.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°. exemple 400 112°.

Question 200-27 : An aircraft is departing from an airport which has an elevation of 2000 ft and the qnh is 1023 hpa. the tas is 100 kt, the head wind component is 20 kt and the rate of climb is 1000 ft/min. top of climb is fl 100..at what distance from the airport will this be achieved ?

11.1 nm.

.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. exemple 404 11.1 nm.

Question 200-28 : At 10 15 the reading from a vor/dme station is 211°/ 90nm, at 10 20 the reading from the same vor/dme station is 211°/120nm..compass heading = 200°..variation in the area = 31°w..deviation = +1°..tas = 390 kt..the wind vector t is approximately ?

110°/70kt.

.200° ch + 1°e 31°w = 170° th.211° mc 31°w = 180° tc..from 10 15 to 10 20 5 minutes has passed and 30 nm have been flown..5 min > 0,0833h 30 / 0,0833 = 360 kt gs.390 kt 360 kt = 30 kt hw component...now use your flight computer with 390 kt tas, 180° tc , 30 kt hw component and 10°crab angle to get 110°/70kt. exemple 408 110°/70kt.

Question 200-29 : An aircraft is departing from an airport which has an elevation of 2000 ft and the qnh is 1003 hpa..the tas is 100 kt, the head wind component is 20 kt and the rate of climb is 1000 ft/min..top of climb is fl 100..at what distance from the airport will this be achived ?

10.3 nm.

.2000 ft is at qnh 1003 hpa. at 1013 hpa it is 2300 ft...to reach fl100 you must climb 7700 ft 10000 2300...rate of climb is 1000 ft/min.7700/1000 = 7.7 min..at a ground speed of 80 kt, it will take.7.7 x 80/60 = 10.26 nm. exemple 412 10.3 nm.

Question 200-30 : Two consecutive waypoints of a flight plan are stornoway vordme n58°12.4', w006°11.0' and glasgow vordme n55°52.2', w004°26.7'..during the flight the actual time over stornoway is 11 15 utc and the estimated time over glasgow is 11 38 utc..at 11 21 utc the fix of the aircraft is exactly over ?

11 36

.distance stornoway to glasgow = 151 nm.distance stornoway to ronar = 44 nm..11 21 11 15 = 6 min..6 min = 44 nm so 60 min = 440 kt nm.151 44 = 107 nm..107/440 = 0,243h... 0,243 x 60 = 14,6 min.11 21 + 14,6 min = 11 35,36 sec. exemple 416 11:36

Question 200-31 : An aircraft at fl360 is required to descent to fl120..the aircraft should reach fl120 at 40 nm from the next waypoint..the rate of descent is 2000 ft/min..the average gs is 420 kt..the minimum distance from the next waypoint at which descent should start is ?

124 nm.

.24000 ft to lose with 2000 ft/min. this means descending 24000 ft in 12 min...the plane is flying 7 nm/min. 12x7 84 nm...the plane needs 84 nm to reach fl120...it also need to be leveled 40 nm before the next waypoint...that means we should start the descent 84 + 40 = 124 nm before next waypont. exemple 420 124 nm.

Question 200-32 : The distance between a and b is 90 nm. at a distance of 15 nm from a the aircraft is 4 nm right of course. to reach destination b, the correction angle on the heading should be ?

19°.

.tke = distance off track x 60 / distance along track..tke = 4 nm x 60 / 15 nm..tke = 16°..to join back on our track..tke = distance off track x 60 / distance to go..tke = 4 nm x 60 /75 nm... 90 nm 15 nm = 75 nm..tke = 3°..correction angle 16° + 3° = 19° to the left as we are right off the course. exemple 424 19°.

Question 200-33 : After 15 minutes of flying with the planned tas and true heading the aircraft is 3 nm south of the intended track and 2.5 nm ahead of the dead reckoning position..to reach destination b from this position, the true heading should be. 2516 ?

292°.

exemple 428 292°.

Question 200-34 : An aircraft is flying from salco to berry head on magnetic track 007°, tas 445 kt..the wind is 050° t /40 kt..variation 5°w, deviation +2°.at 1000 utc the rb of locator py is 311°..at 1003 utc the rb of locator py is 266°..calculate the true bearing of locator py at 1003 utc from the aircraft. ?

272° t.

. /com en/com061 635.jpg..calculate the drift between our true track 002° and the true wind 050°/40 kt with your computer, the drift is 4° left, you have to apply a 4°right wind angle correction...true heading + relative bearing = true bearing of locator from the aircraft.006° + 266° = 272°. exemple 432 272° (t).

Question 200-35 : An aircraft is departing from an airport which has an elevation of 2000 ft and the qnh is 1003 hpa..the tas is 100 kt, the head wind component is 20 kt and the rate of climb is 1000 ft/min..top of climb is fl 050..at what distance from the airport will this be achived ?

3.6 nm.

.2000 ft is at qnh 1003 hpa. at 1013 hpa it is 2300 ft...to reach fl050 you must climb 2700 ft 5000 2300...rate of climb is 1000 ft/min.2700/1000 = 2.7 min..at a ground speed of 80 kt, it will take. 2.7 x 80/60 = 3.6 nm. exemple 436 3.6 nm.

Question 200-36 : During approach the following data are obtained.dme 12.0 nm, altitude 3000 ft.dme 9.8 nm, altitude 2400 ft.tas = 160 kt, groundspeed 125 kt.the rate of descent is ?

570 ft/min.

.12 nm 9,8 nm = 2,2 nm..3000 ft 2400 ft = 600 ft..2,2 nm / 125 kt = 0,0176 h >1,056 min..600 ft / 1,056 = 568 ft/min. exemple 440 570 ft/min.

Question 200-37 : The distance between a and b is 90 nm. at a distance of 75 nm from a the aircraft is 4 nm right of course. the track angle error tke is ?

3°r.

.use the one in sixty rule.track error angle from a = distance off track x 60 / distance along track.track error angle from a = 4 nm x 60 / 75 nm.track error angle from a = 3°r. exemple 444 3°r.

Question 200-38 : The true course according to the flight log is 270°, the forecast wind is 045° t /15 kt and the tas is 120 kt..after 15 minutes of flying with the planned tas and true heading the aircraft is 3 nm south of the intended track and 2.5 nm ahead of the dead reckoning position..the track angle error tke ?

5°l.

. 1798.with forecasted wind, our ground speed is 130 kt..at 130 kt and 15 minutes of flight, we will be at 32.5 nm from a..but the question states 2.5 nm ahead of the dead reckoning position , so we are at 35 nm from a...use the one in sixty rule.track error angle from a = 3 nm x 60 / 35 nm = 5°. exemple 448 5°l.

Question 200-39 : An aircraft flies from waypoint 7 63°00'n, 073°00'w to waypoint 8 62°00'n, 073°00'w. the aircraft position is 62°00'n, 073°10'w. the cross track distance in relation to the planned track is ?

4,7 nm right.

.1° longitude at equator = 60 nm..1° long. at 60°lat = 30 nm..10' off track is 5 nm 10' = 1/6 from 1h, so 1/6 from 30 nm is 5 nm 30 / 6..as we are heading along meridian from 63°n to 62°n out true course is 180° and as we have ended up at 73°10', this is right of the track, so 5 nm right...mathematically..distance nm = chlong. in minutes * coslat..distance = 10 x cos62°..distance = 4,7 nm. exemple 452 4,7 nm right.

Question 200-40 : An aircraft is departing from an airport which has an elevation of 2000 ft and the qnh is 1003 hpa..the tas is 100 kt, the headwind component is 20 kt and the rate of climb is 500 ft/min. top of climb is fl 050..at what distance from the airport will this be achived ?

7.2 nm.

.1013 hpa 1003 hpa = 10 hpa > 300 ft 10 x 30 ft..2000 ft + 300 ft = 2300 ft..5000 ft 2300 ft = 2700 ft 2700 ft to climb..2700 ft / 500 ft/min = 5,4 min > 0,09h 5,4 / 60..gs = 80 kt 100 kt tas 20 headwind component..80 kt x 0,09h = 7,2 nm. exemple 456 7.2 nm.


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