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Question 187-1 : 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 ? [ Question security ]

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 exemple 287 280 kt.

Question 187-2 : 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

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 exemple 291 1h and 49 min.

Question 187-3 : 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

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 exemple 295 310°/65.

Question 187-4 : 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°

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° exemple 299 112°.

Question 187-5 : 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

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 exemple 303 11.1 nm.

Question 187-6 : 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

Admin .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 307 110°/70kt.

Question 187-7 : 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

Admin .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 311 10.3 nm.

Question 187-8 : 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

Admin .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 315 11:36

Question 187-9 : 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

Admin .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 319 124 nm.

Question 187-10 : 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°

Admin .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 323 19°.

Question 187-11 : 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°

Admin .draw the exercice . 1798.the dead reckoning position was at 15 min from a with a gs of 130 kt . 130/60 x 15 = 32 5 nm from a .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° . 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 / 15 nm = 12° .to reach destination b from this position the correction angle on the heading should be 5° + 12° = 17° .current heading is 275° new heading is 275° + 17° = 292° exemple 327 292°.

Question 187-12 : 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 331 272° (t).

Question 187-13 : 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

Admin .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 335 3.6 nm.

Question 187-14 : 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

Admin .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 339 570 ft/min.

Question 187-15 : 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

Admin .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 343 3°r.

Question 187-16 : 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 ?

5°l

Admin . 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 347 5°l.

Question 187-17 : 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

Admin .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 351 4,7 nm right.

Question 187-18 : 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

Admin .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 355 7.2 nm.

Question 187-19 : You are departing from an airport which has an elevation of 1500 ft the qnh is 1003 hpa .15 nm away there is a waypoint you are required to pass at an altitude of 7500 ft .given a groundspeed of 120 kt what is the minimum rate of climb ?

800 ft/min

Admin .7500 1500 = 6000 ft.6000 / 7 5 = 800 ft/min exemple 359 800 ft/min.

Question 187-20 : An aircraft is flying at fl200 .the qnh given by a meteorological station at an elevation of 1300ft is 998 2 hpa .oat = 40°c .the elevation of the highest obstacle along the route is 8 000 ft .calculate the aircraft's approximate clearance above the highest obstacle on this route ?

10 500 ft

Admin .find the qnh altitude 1013 998 2 = 14 8 x 27 = 400 ft.altitude is 19600ft qnh .with aviat 617 computer .against altitude pressure = 20 put °c oat = 40 .then read in the inner circle the altitude 19600 the.on the outer circle 18400 true altitude .18400 8000ft = 10400ft = approximate clearance over the obstacle .for information 061 general navigation learning objectives states for questions involving height calculation 30 ft/hpa is to be used unless another figure is specified in the question exemple 363 10 500 ft.

Question 187-21 : The qnh given by a station at 2500 ft is 980hpa .the elevation of the highest obstacle along a route is 8 000 ft and the oat = isa 10°c .when an aircraft on route has to descend the minimum indicated altitude qnh on the subscale of the altimeter to maintain a clearance of 2000 ft will be ?

10 400 ft

Admin .we need to be at 10000 ft to avoid the obstacle by 2000 ft .temperature correction formula 4° x 10 x 10° = 400 ft.the altimeter over reading in cold air and if we flew exactly at 10000 ft indicated our true altitude would be 9600 ft .we need to cruise at 10000 + 400 = 10400 ft indicated in order to maintain a clearance of 2000 ft exemple 367 10 400 ft.

Question 187-22 : An aircraft is departing from an airport which has an elevation of 2000 ft .outside temperature is 0°c qnh = 1013 hpa .it is planned to climb to fl 320 where outside temperature is 60°c .the average cas during climb will be 200 kt compressibility is negligeable .mean tas during climb is ?

276 kt

Admin . by convention at the exam easa specification average tas used for climb problems is calculated at the altitude 2/3 of the cruising altitude .temperature is 0°c at 2000 ft .approximately 4° at sea level we 'gain' 2°c per 1000 ft while descending .21000 ft x 2°/1000ft = 42°c.4°+ 42° = 38°c.temperature is around 38°c at fl210 .on computer in airspeed window set press alt '21' in front of coat °c ' 38°c' on the outer scale in front of cas 200 kt you can read tas 272 kt exemple 371 276 kt.

Question 187-23 : During visual navigation in freezing conditions after heavy snowfall which of the following landmark will give the best reference for a visual checkpoint ?

A large river

Admin .after heavy snowfall roads will not have been cleared by snow ploughs neither country road or railway a large river may freeze but you will always be able to distinguish its path exemple 375 A large river.

Question 187-24 : During a climb at a constant cas below the tropopause in standard conditions ?

Both tas and mach number will increase

Admin .for those questions use the very simple 'ertm' diagram . 1037.the cas line is vertical because the question states climb at a constant calibrated airspeed cas . ertm for e as/ r as rectified air speed or cas / t as/ m ach exemple 379 Both tas and mach number will increase.

Question 187-25 : An aircraft is descending down a 12% slope whilst maintaining a gs of 540 kt the rate of descent of the aircraft is approximately ?

6500 ft/min

Admin .vertical speed = 12% gradient x 540 kt.vertical speed = 6480 ft/min exemple 383 6500 ft/min.

Question 187-26 : The departure airfield is at 2000 ft elevation temperature at the field is +20°c qnh 1013 hpa the plan is to climb to fl 290 where outside air temperature is 40°c .the average tas in the climb should be calculated using what fl and temperature ?

Fl 200 with temperature 20°c

Admin .by convention at the exam easa specification average tas used for climb problems is calculated at the altitude 2/3 of the cruising altitude .29000 2000 = 27000 ft.2/3 de 27000 = 18000 ft .18000 + 2000 = 20000 ft fl200 .température à 20000 ft = 20°c + 20 x 2°c = 20°c exemple 387 Fl 200 with temperature -20°c.

Question 187-27 : The departure is from an airfield at 2000 ft elevation temperature at the field is +20°c qnh 1013 hpa the plan is to climb to fl 290 where outside air temperature is 40°c the cas in the climb is 180 kt compressibility negligible .the average tas in the climb is ?

249 kt

Admin .1 29000 2000= 27000ft.2 27000* 2/3 = 18000.3 18000+2000 = 20000 your average alt in climb patern .4 if oat at 2000 alt is +20 so at 20000 will be +20 reference figure 2*18 18 height explanation 2*18 because temp decrease 2 deg per 1000 ft . .5 align temp 16 with 22000 ft in th air speed window cr 3 or iwa 11092 and read oposit 180 kt your 249 tas at outer scale.this is only the way to solve tasks like this exemple 391 249 kt.

Question 187-28 : Given .w/v at arrival aerodrome at 1000 ft amsl is 230°/15kt w/v at tod at fl 130 is 280°/45kt average track after tod is 220° isa conditions descent speed ias = 170 kt .find the gs during the descent ?

163 kt

Admin .at fl130 isa condition .ias=170kt => tas=206kt .we have w/v 280°/45kt => drift 12°l so mh=232° to get an average track of 220°.so gs= 178kt with computer.at 1000ft amsl isa condition .ias=170kt => tas=172kt .we have w/v 230°/15kt => drift 1°l so mh=221° to get an average track of 220°.so gs= 157kt with computer.as a result the average gs for the descent is 157+178 /2 = 167 5kt exemple 395 163 kt.

Question 187-29 : Given .w/v at arrival aerodrome at msl is 200°/20kt w/v at tod at fl 100 is 260°/50kt average track after tod is 190° .isa conditions descent speed ias = 150 kt .find the gs during the descent ?

135 kt

exemple 399 135 kt.

Question 187-30 : An aircraft is cruising in fl180 and thereafter descends to ground level the following wind information is given .ground level 260°/25 kt.fl030 270°/30 kt.fl060 270°/35 kt.fl090 270°/40 kt.fl120 280°/50 kt.fl150 285°/55 kt.fl180 290°/55 kt.the wind to be used for descent calculation is ?

270°/40 kt

Admin .by convention average wind velocity used for climb problems is wind velocity at the altitude 2/3 of the cruising altitude .average wind velocity used for descent problems is wind velocity at the altitude 1/2 of the descent altitude exemple 403 270°/40 kt.

Question 187-31 : The distance between two waypoints is 150 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 ?

10 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 150/60 2 5 nm off track for each degree of error .total error is 4° from 2°e to 2°w .4° x 2 5 nm = 10 nm .using goniometric functions .tan4° = / 150. = tan4° x 150. = 10 nm exemple 407 10 nm.

Question 187-32 : True track 085°.groundspeed 180 kt.wind 290°/30kt.variation 4°e .the aircraft is 1 5 nm left of track after 12 minutes .what is the track angle error tke ?

2 5° l

Admin .tke = distance off track x 60 / distance along track.tke = 1 5 nm x 60 / 36 nm = 2 5°.the aircraft has drifted to the left therefore tke is 2 5° left exemple 411 2.5° l.

Question 187-33 : With only a visual straight line as visual cue a canal for example this line of position must be selected ?

More or less perpendicular to our track

exemple 415 More or less perpendicular to our track.

Question 187-34 : Given .a descending aircraft flies in a straight line to a dme .dme 55 nm altitude 33000 ft.dme 43 9 nm altitude 30500 ft.m = 0 72 gs = 525 kt oat = isa .the descent gradient is ?

3 70%

Admin .33000 30500 = 2500 ft.55 nm 43 9nm = 11 1 nm.11 1 nm = 67488 ft.2500 = 67488 x x.x= 2500 / 67488 = 0 0370 3 7% exemple 419 3.70%.

Question 187-35 : The descent gradient of an aircraft with the following data is . 60 nm norths of vor xyz fl350. 10 nm south of vor xyz fl120 ?

5 4%

Admin .total ground distance is 60+10 = 70 nm .altitude difference is 35000 12000 = 23000 ft .gradient in % = altitude difference in ft x 100 / ground distance in ft..ground distance in feet 70 nm x 6080 ft = 425600 ft.gradient in % = 23000 x 100 / 425600 = 5 4% exemple 423 5.4%

Question 187-36 : The average tas climbing from 1500 ft to fl180 with a given temperature of isa +15°c a cas of 230 kt and qnh 1032 hpa is ?

283 kt

Admin . by convention at the exam easa specification average tas used for climb problems is calculated at the altitude 2/3 of the cruising altitude .1032 1013 = 19 hpa.19 hpa x 30 ft = 570 ft .18000 + 570 = 18570 ft.2/3 of 18570 = 12380 ft .to convert cas to tas 1% for each 600 ft and 0 2% for each degree of isa deviation.tas = cas x 12380/600 + 0 2% x 15°c = 230 x 20% + 3% = 282 9 kt exemple 427 283 kt.

Question 187-37 : An aircraft is turning on a final approach to intercept a 3° glide slope which is located at an altitude of 700 ft amsl assuming the turn is made at 4 nm from the threshold what is a suitable altitude to intercept the glide slope ?

1916 ft

Admin .1 in 60 rule is a rule of thumb . 3° x 4 nm /60 = 0 2 nm .0 2 nm x 6080 ft = 1216 ft .add 700 ft since we are looking for an altitude = 1216 + 700 = 1916 ft exemple 431 1916 ft.

Question 187-38 : Given .tas 220 kt.cruising level fl180.track during climb 080°.wind at msl 260°/25kt.wind at fl180 290°/55 kt.from msl to cruising level find the gs during climb ?

262 kt

Admin .we have to use the wind at the altitude 2/3 of the cruising altitude .fl180 x 2/3 = fl120.wind changes by 30° from ground to fl180 an speed increases from 30 kt .mean wind at fl120 is 260°+20° and 25kt+20kt = 280°/45kt .with your nav computer you will find 262 kt exemple 435 262 kt.

Question 187-39 : An aircraft climbs from ground level to fl180 the following wind information is given .ground level 260°/25 kt.fl030 270°/30 kt.fl060 270°/35 kt.fl090 270°/40 kt.fl120 280°/50 kt.fl150 285°/55 kt.fl180 290°/55 kt.the wind to be used to solve climb problems . e g the calculation of the gs ?

280°/50 kt

Admin .by convention average wind velocity used for climb problems is wind velocity at the altitude 2/3 of the cruising altitude .average wind velocity used for descent problems is wind velocity at the altitude 1/2 of the descent altitude exemple 439 280°/50 kt.

Question 187-40 : An aircraft descends from fl240 to fl040 for the final approach .cas = 220 kt.oat = isa +10°c.the average tas in the descent is ?

273 kt

Admin .at the exam average tas used for descent problems is calculated at the altitude 1/2 of the descent altitude.at fl120 isa temperature = 15°c 2°c x 12 = 9°c .oat is isa +10°c thus oat is +1°c at fl120 .on the computer in airspeed window put +1ºc next to fl120 go to cas 220 kt on inner scale and read tas on outer scale 273 kt exemple 443 273 kt.


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