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Question 199-1 : Given.gs = 120 kt. distance from a to b = 84 nm..what is the time from a to b ? [ Level reports ]

00 hr 42 min.

.84 nm / 120 kt/60 min = 42 minutes. exemple 299 00 hr 42 min.

Question 199-2 : Given.distance 'a' to 'b' 1973 nm.ground speed out 430 kt.ground speed back 385 kt.safe endurance 7 hr 20 min.the distance from 'a' to the point of safe return psr is ?

1490 nm.

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 430 kt.ground speed home = 385 kt..point of safe return psr = 7.33 x 385 / 430 + 385.point of safe return psr = 2822 / 815.point of safe return psr = 3.46 h..distance of the psr from the departure point at a speed of 430 kt.3.46 h x 430 = 1489 nm. exemple 303 1490 nm.

Question 199-3 : Given.distance 'a' to 'b' 2346 nm.ground speed out 365 kt.ground speed back 480 kt.the time from 'a' to the point of equal time pet between 'a' and 'b' is ?

219 min.

.ground speed out 365 kt.ground speed home 480 kt..pet = distance x gsh / gso + gsh.pet = 2346 x 480 / 365 + 480 = 1332 nm...1332 nm / 365 kt = 3.65 h..3.65 h x 60 minutes = 219 minutes. exemple 307 219 min.

Question 199-4 : Given.distance 'q' to 'r' 1760 nm.ground speed out 435 kt.ground speed back 385 kt.the time from 'q' to the point of equal time pet between 'q' and 'r' is ?

114 min.

.ground speed out 435 kt.ground speed home 385 kt..pet = distance x gsh / gso + gsh.pet = 1760 x 385 / 435 + 385 = 826 nm...826 nm / 435 kt = 1.9h..1.9 h x 60 minutes = 114 minutes. exemple 311 114 min.

Question 199-5 : Given.distance 'q' to 'r' 1760 nm.ground speed out 435 kt.ground speed back 385 kt.safe endurance 9 hr.the distance from 'q' to the point of safe return psr between 'q' and 'r' is ?

1838 nm.

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 435 kt.ground speed home = 385 kt..point of safe return psr = 9 x 385 / 435 + 385.point of safe return psr = 3465 / 820.point of safe return psr = 4.22 h..distance of the psr from the departure point at a speed of 435 kt.4.22 h x 435 = 1838 nm. exemple 315 1838 nm.

Question 199-6 : An aeroplane is flying at tas 180 kt on a track of 090°..the w/v is 045° / 50kt..how far can the aeroplane fly out from its base and return in one hour ?

85 nm.

.centre dot on tas, 180 kt, rotate to wind direction, 045°.come down from centre dot for wind speed, 50 kt mark end of wind vector at 130 kt...rotate to outbound track under heading index note drift.14°starboard rotate track to drift note drift now 11°starboard rotate track to drift note drift still 11°starboard..our outbound heading will be 079° to track 090° and our ground speed outbound is 141 kt. 1382.proceed same way to find ground speed homebound you will find 212 kt...pnr = t x gso x gsh / gso + gsh..pnr = 1 x 141 x 212 / 141 + 212..pnr = 29892 / 353..pnr = 84.67 nm. exemple 319 85 nm.

Question 199-7 : An aircraft is maintaining a 5.2% gradient is at 7 nm from the runway, on a flat terrain, its height is approximately ?

2210 ft

.1 nm = 6080 ft..aircraft is at 7 nm from the runway 7 nm x 6080 ft = 42560 ft...42560 x 5.2/100 = 2213 ft. exemple 323 2210 ft

Question 199-8 : An aircraft descends from fl250 to fl100..the rate of descent is 1000 ft/min, the gs is 360 kt..the flight path angle is ?

1.6°.

.vertical speed = gradient % * gs.1000 = gradient % * 360.gradient % = 2,77..gradient % = angle * 100/60..2,77 = angle * 100/60 > angle = 1,662º. exemple 327 1.6°.

Question 199-9 : The outer marker of an ils with a 3° glide slope is located 4.6 nm from the threshold. assuming a glide slope height of 50 ft above the threshold, the approximate height of an aircraft passing the outer marker is ?

1450 ft.

4.6 nm x tan 3° = 0.24 nm.1 nm = approximately 6000 ft.0.24 x 6000 = 1440 ft..1440 ft + 50 ft = 1490 ft...the approximate height of the aircraft is 1490 ft close to 1450 ft... svandam .in a 3° glide slope, 1nm = 300 feet.then 4.6 nm * 300 = 1380 ft..1380 ft + 50 ft = 1430 ft..the approximate height of the aircraft is 1430 ft close to 1450 ft... johanjog .more precise calculation...4.6 nm*6080 ft/1nm = 27968 ft..1 60 rule..3º/60 = d/27968 d = 1398.4 ft..1398.4 + 50 ft = 1448.4 ft 1450 ft. exemple 331 1450 ft.

Question 199-10 : 730 ft/min equals ?

3.7 m/sec

exemple 335 3.7 m/sec

Question 199-11 : How long will it take to fly 5 nm at a groundspeed of 269 kt ?

1 min 07 sec

.5 / 269/60 =1.115 minutes.0.115 x 60 = 7 secondes..1 min 7 sec. exemple 339 1 min 07 sec

Question 199-12 : An aircraft travels 2.4 statute miles in 47 seconds. what is its groundspeed ?

160 kt.

.2.4/47 x 3600 = 184 statute miles per hour...1 statute mile = 1.609 km = 0.87 nm..184 x 0.87 = 160 kt. exemple 343 160 kt.

Question 199-13 : The icao definition of eta is the ?

Estimated time of arrival at destination.

exemple 347 Estimated time of arrival at destination.

Question 199-14 : Assuming zero wind, what distance will be covered by an aircraft descending 15000 ft with a tas of 320 kt and maintaining a rate of descent of 3000 ft/min ?

26.7 nm

.15000 ft / 3000 ft/min = 5 minutes..5 min / 60 min = 0,0833 hour..0,0833 x 320 = 26,67 nm. exemple 351 26.7 nm

Question 199-15 : 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 of 276° with the magnetic variation 12°w ?

054°.

.magnetic heading 276º.variation 12ºw.true heading 264º.island bearing 30ºleft.true bearing of the island from the aircraft 234º.true bearing of the aircraft from the island 234°+/ 180º = 054°. exemple 355 054°.

Question 199-16 : An aircraft at fl370 is required to commence descent at 120 nm from a vor and to cross the facility at fl130. if the mean gs for the descent is 288 kt, the minimum rate of descent required is ?

960 ft/min.

.37000 ft 13000 ft = 24000 ft..120 nm / 288kt = 0,417h > 25 min 0.417 x 60..24000 ft / 25 min = 960 ft/min. exemple 359 960 ft/min.

Question 199-17 : An aircraft at fl310, m0.83, temperature 30°c, is required to reduce speed in order to cross a reporting point five minutes later than planned. assuming that a zero wind component remains unchanged, when 360 nm from the reporting point mach number should be reduced to ?

M0.74

. 1745.set temperature 30°c in airspeed window..in front of 8.3 mach.83 inner scale , on the outer scale, you read tas = 503 kt... 360/503 x 60 = 43 minutes.we are at 43 minutes from the reporting point... 360/ x 60 = 48 minutes.. = 360 x 60 / 48 = 450 kt...in airspeed window, set outside temperature 30°c in front of mach index, go to 450 kt on the outer scale and you read 7.4 mach 0.74 on the inner scale. exemple 363 M0.74

Question 199-18 : A ground feature was observed on a relative bearing of 325° and five minutes later on a relative bearing of 280°. the aircraft heading was 165° m , variation 25°w, drift 10°right and gs 360 kt..when the relative bearing was 280°, the distance and true bearing of the aircraft from the feature was ?

30 nm and 240°.

.5 minutes at 360 kt = 360 / 60 x 5 = 30 nm. 1415.magnetic heading 165°.variation 25°w.true heading = 140°..true track= 140° + 10° = 150°...we have an isosceles triangle, and in an isosceles triangle, two sides are equal in length...relative bearing 280°.true bearing of the feature from the aircraft = 140 + 280 360 = 060°..true bearing of the aircraft from the feature = 060 + 180 = 240°. exemple 367 30 nm and 240°.

Question 199-19 : An aircraft at fl350 is required to descend to cross a dme facility at fl80. maximum rate of descent is 1800 ft/min and mean gs for descent is 276 kt. the minimum range from the dme at which descent should start is ?

69 nm.

.35000 ft 8000 ft = 27000 ft..27000 ft / 1800 ft/min = 15 min > 0,25h 15 / 60..0,25 x 276 kt = 69 nm. exemple 371 69 nm.

Question 199-20 : An aircraft at fl120, ias 200kt, oat 5° and wind component +30kt, is required to reduce speed in order to cross a reporting point 5 min later than planned..assuming flight conditions do not change, when 100 nm from the reporting point ias should be reduced to ?

159 kt.

.use nav computer to find tas. 1390.ias 200 kt = tas 240 kt...ground speed = 240 kt + 30 kt = 270 kt..100 nm at gs 270 kt = 22 minutes.100 nm in 27 minutes = 100 / 27 = 3.7 nm per minute...3.7 x 60 = 222 kt...222 kt 30 kt wind = 192 kt...tas 192 kt ==> with nav computer ==> 159 kt ias. exemple 375 159 kt.

Question 199-21 : An aircraft at fl350 is required to cross a vor/dme facility at fl110 and to commence descent when 100 nm from the facility. if the mean gs for the descent is 335 kt, the minimum rate of descent required is ?

1340 ft/min.

.24000 ft to descend..335/60 = 5.583 nm/min...100/5.583 = 18 minutes...24000 / 18 min = 1340 ft/min. exemple 379 1340 ft/min.

Question 199-22 : An aircraft at fl370, m0.86, oat 44°c, headwind component 110 kt, is required to reduce speed in order to cross a reporting point 5 minutes later than planned. if the speed reduction were to be made 420 nm from the reporting point, what mach number is required ?

M0.81

. 2494.tas is 503 kt, ground speed is 503 110 = 393 kt...420 nm at 393 kt = 1.07 h 1.07 x 60 = 64 minutes...atc asks you to cross a rportin poitn 5 minutes later, in 64 + 5 minutes...69 / 60 = 1.15 h..420 / 1.15 = 365 kt.. 2495.365 kt + 110 kt = 475 kt = m 0.81. exemple 383 M0.81

Question 199-23 : An aircraft at fl390 is required to descend to cross a dme facility at fl70. maximum rate of descent is 2500 ft/min, mean gs during descent is 248 kt. what is the minimum range from the dme at which descent should commence ?

53 nm.

.39000 7000 = 32000 ft.32000 / 2500 = 12.8 minutes..248 kt / 60 minutes = 4.14 nm/minute..12.8 x 4.14 = 53 nm. exemple 387 53 nm.

Question 199-24 : An aircraft at fl370 is required to commence descent when 100 nm from a dme facility and to cross the station at fl120..if the mean gs during the descent is 396 kt, the minimum rate of descent required is approximately ?

1650 ft/min.

.37000 12000 = 25000 ft.396 kt / 60 minutes = 6.6 nm/minute..100 nm / 6.6 = 15.15 minutes before fly over dme..25000 / 15 = 1650 ft/min. exemple 391 1650 ft/min.

Question 199-25 : An aircraft at fl140, ias 210 kt, oat 5°c and wind component minus 35 kt, is required to reduce speed in order to cross a reporting point 5 minutes later than planned. assuming that flight conditions do not change, when 150 nm from the reporting point the aircraft should reduce ias by ?

20 kt.

.given ias cas 210 kt, fl140 and oat 5°c.on the computer, we find tas = 262 kt...we can now find the groundspeed.262 35 = 227 kt..150 nm at 227 kt = 39.6 minutes..we must reduce speed to cross a point 5 minutes later.150 nm in 39.6 + 5 minutes = 202 kt..we must reduce ground speed by 227 202 = 25 kt.a tas of 262 25 = 237 kt is required...now back to the computer.tas 237 kt, fl140 and oat 5°c.new ias is 190 kt...we should reduce ias by 210 190 = 20 kt. exemple 395 20 kt.

Question 199-26 : At 0422 an aircraft at fl370, gs 320kt, is on the direct track to vor 'x' 185 nm distant..the aircraft is required to cross vor 'x' at fl80..for a mean rate of descent of 1800 ft/min at a mean gs of 232 kt, the latest time at which to commence descent is ?

04h45.

.fl370 to fl80 = 29000 ft.29000 / 1800 ft/min = 16.1 minutes...during the descent, the aircraft will cover. 232/60 x 16.1 = 62.25 nm..distance before top of descent tod.184 62.25 = 121.75 nm..time before tod.121.75 / 320/60 = 22.8 minutes...the latest time to commence descent is.04h42 + 23 minutes = 04h45. exemple 399 04h45.

Question 199-27 : An aircraft at fl330 is required to commence descent when 65 nm from a vor and to cross the vor at fl100. the mean gs during the descent is 330 kt. what is the minimum rate of descent required ?

1950 ft / min.

.33000 10000 = 23000 ft.330 kt / 60 minutes = 5.5 nm/minute..65 nm / 5.5 = 11.8 minutes before fly over dme..23000 / 11.8 = 1950 ft/min. exemple 403 1950 ft / min.

Question 199-28 : An aircraft at fl290 is required to commence descent when 50 nm from a vor and to cross that vor at fl80. mean ground speed during descent is 271kt. what is the minimum rate of descent required ?

1900 ft / min.

.50 nm / 271 kt = 0,185h > 11 min..29000 ft 8000 ft = 21000 ft..21000 ft / 11 min = 1900 ft/min. exemple 407 1900 ft / min.

Question 199-29 : An aircraft at fl350 is required to commence descent when 85 nm from a vor and to cross the vor at fl80. the mean gs for the descent is 340 kt. what is the minimum rate of descent required ?

1800 ft/min.

.fl350 fl080 = 27000 ft...85 nm / 340 kt = 0.25 h 15 minutes..27000 ft / 15 min = 1800 ft/min. exemple 411 1800 ft/min.

Question 199-30 : An aircraft is planned to fly from position 'a' to position 'b', distance 480 nm at an average ground speed of 240 kt. it departs 'a' at 1000 utc. after flying 150 nm along track from 'a', the aircraft is 2 minutes behind planned time..using the actual gs experienced, what is the revised eta at 'b' ?

12 06 utc.

.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. exemple 415 12:06 utc.

Question 199-31 : An aircraft is planned to fly from position 'a' to position 'b',distance 320 nm, at an average gs of 180 kt. it departs 'a' at 1200 utc. after flying 70 nm along track from 'a', the aircraft is 3 minutes ahead of planned time..using the actual gs experienced, what is the revised eta at 'b' ?

13 33 utc.

.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. exemple 419 13:33 utc.

Question 199-32 : An aircraft is planned to fly from position 'a' to position 'b', distance 250 nm at an average gs of 115 kt. it departs 'a' at 0900 utc. after flying 75 nm along track from 'a', the aircraft is 1.5 minute behind planned time..using the actual gs experienced, what is the revised eta at 'b' ?

11 15 utc.

.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. exemple 423 11:15 utc.

Question 199-33 : Given.distance 'a' to 'b' is 475 nm, planned gs 315 kt, atd actual time departure 1000 utc..at 1040 utc a fix is obtained at 190 nm along track..what gs must be maintained from the fix in order to achieve planned eta at 'b' ?

340 kt

exemple 427 340 kt

Question 199-34 : Given distance 'a' to 'b' is 325 nm, planned gs 315 kt, atd 1130 utc, 1205 utc fix obtained 165 nm along track..what gs must be maintained from the fix in order to achieve planned eta at 'b' ?

355 kt.

.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. exemple 431 355 kt.

Question 199-35 : Given distance a to b is 100 nm, fix obtained 40 nm along and 6 nm to the left of course. what heading alteration must be made to reach 'b' ?

15° right.

.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. exemple 435 15° right.

Question 199-36 : Given distance 'a' to 'b' is 90 nm, fix obtained 60 nm along and 4 nm to the right of course..what heading alteration must be made to reach 'b' ?

12° left.

.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. exemple 439 12° left.

Question 199-37 : Complete line 1 of the 'flight navigation log'.positions 'a' to 'b'. what is the hdg° m and eta. 2497 ?

Hdg 268° eta 1114 utc.

. 2496 exemple 443 Hdg 268° - eta 1114 utc.

Question 199-38 : Complete line 2 of the 'flight navigation log', positions 'c' to 'd'. what is the hdg° m and eta. 2497 ?

Hdg 193° eta 1239 utc.

. 2496 exemple 447 Hdg 193° - eta 1239 utc.

Question 199-39 : Complete line 3 of the 'flight navigation log', positions 'e' to 'f'. what is the hdg° m and eta. 2497 ?

Hdg 105° eta 1205 utc.

. 2496 exemple 451 Hdg 105° - eta 1205 utc.

Question 199-40 : Complete line 4 of the 'flight navigation log', positions 'g' to 'h'. what is the hdg° m and eta. 2497 ?

Hdg 344° eta 1336 utc.

. 2496 exemple 455 Hdg 344° - eta 1336 utc.


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