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Given .true heading = 054° tas = 450 kt true track = 059° gs = 416 kt ? [ Multiple protocol ]

Question 172-1 : 010°/50 kt 005°/50 kt 010°/55 kt 010°/45 kt

exemple 272 010°/50 kt. — Admin .true heading is 054° true track is 059° our drift is 5° right . 2525.wind 010°/50 kt 010°/50 kt.

Given .true heading = 002°.tas = 130 kt.true track = 353°.ground speed = 132 ?

Question 172-2 : 095°/20 kt 090°/15 kt 090°/20 kt 095°/25 kt

exemple 276 095°/20 kt. — ..true heading is 002° true track is 353° our drift is 9° left . /com en/com061 195 jpg.wind 094°/22 kt closest answer is 095°/20 kt 095°/20 kt.

Given .gs = 236 kt distance from a to b = 354 nm .what is the time from a to b ?

Question 172-3 : 1 hr 30 min 1 hr 09 min 1 hr 10 min 1 hr 40 min

exemple 280 1 hr 30 min. — Admin .354 nm / 236 kt/60 min = 90 minutes 1h30 1 hr 30 min.

Given .gs = 345 kt distance from a to b = 3560 nm .what is the time from a to b ?

Question 172-4 : 10 hr 19 min 10 hr 05 min 11 hr 00 min 11 hr 02 min

exemple 284 10 hr 19 min. — Admin .3560 nm / 345 kt/60 min = 619 minutes 10h39 10 hr 19 min.

Given .gs = 95 kt distance from a to b = 480 nm .what is the time from a to b ?

Question 172-5 : 5 hr 03 min 4 hr 59 min 5 hr 00 min 5 hr 08 min

exemple 288 5 hr 03 min. — Admin .480 nm / 95 kt/60 min = 303 minutes 5h03 5 hr 03 min.

Given .gs = 120 kt distance from a to b = 84 nm .what is the time from a to b ?

Question 172-6 : 00 hr 42 min 00 hr 43 min 00 hr 44 min 00 hr 45 min

exemple 292 00 hr 42 min. — Admin .84 nm / 120 kt/60 min = 42 minutes 00 hr 42 min.

Given .distance 'a' to 'b' 1973 nm.ground speed out 430 kt.ground speed back ?

Question 172-7 : 1490 nm 1664 nm 1698 nm 1422 nm

exemple 296 1490 nm. — Admin .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 1490 nm.

Given .distance 'a' to 'b' 2346 nm.ground speed out 365 kt.ground speed back ?

Question 172-8 : 219 min 290 min 197 min 167 min

exemple 300 219 min. — Admin .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 219 min.

Given .distance 'q' to 'r' 1760 nm.ground speed out 435 kt.ground speed back ?

Question 172-9 : 114 min 110 min 106 min 102 min

exemple 304 114 min. — Admin .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 114 min.

Given .distance 'q' to 'r' 1760 nm.ground speed out 435 kt.ground speed back ?

Question 172-10 : 1838 nm 1313 nm 1467 nm 1642 nm

exemple 308 1838 nm. — Admin .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 1838 nm.

An aeroplane is flying at tas 180 kt on a track of 090° .the w/v is 045° / ?

Question 172-11 : 85 nm 56 nm 88 nm 176 nm

exemple 312 85 nm. — Admin .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 85 nm.

An aircraft is maintaining a 5 2% gradient is at 7 nm from the runway on a flat ?

Question 172-12 : 2210 ft 680 ft 1890 ft 3640 ft

exemple 316 2210 ft — Admin .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 2210 ft

An aircraft descends from fl250 to fl100 .the rate of descent is 1000 ft/min ?

Question 172-13 : 1 6° 2 8° 3 2° 2 4°

exemple 320 1.6°. — Admin .vertical speed = gradient % * gs.1000 = gradient % * 360.gradient % = 2 77.gradient % = angle * 100/60.2 77 = angle * 100/60 > angle = 1 662º 1.6°.

The outer marker of an ils with a 3° glide slope is located 4 6 nm from the ?

Question 172-14 : 1450 ft 1350 ft 1300 ft 1400 ft

exemple 324 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 1450 ft.

730 ft/min equals ?

Question 172-15 : 3 7 m/sec 5 2 m/sec 1 6 m/sec 2 2 m/sec

exemple 328 3.7 m/sec — Admin .if you don't have nav compupter or formula how to calculate it directly you can help yourself with easy counts .730 / 3 28 1 m = 3 28 ft = 222 6 m/min to convert it to seconds simply divide it by 60 so 222 6/60 = 3 7 m/sec 3.7 m/sec

How long will it take to fly 5 nm at a groundspeed of 269 kt ?

Question 172-16 : 1 min 07 sec 1 min 55 sec 2 min 30 sec 0 min 34 sec

exemple 332 1 min 07 sec — Admin .5 / 269/60 =1 115 minutes.0 115 x 60 = 7 secondes .1 min 7 sec 1 min 07 sec

An aircraft travels 2 4 statute miles in 47 seconds what is its groundspeed ?

Question 172-17 : 160 kt 183 kt 209 kt 131 kt

exemple 336 160 kt. — Admin .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 160 kt.

The icao definition of eta is the ?

Question 172-18 : Estimated time of arrival at destination actual time of arrival at a point or fix estimated time of arrival at an en route point or fix estimated time en route

exemple 340 Estimated time of arrival at destination. — Estimated time of arrival at destination.

Assuming zero wind what distance will be covered by an aircraft descending ?

Question 172-19 : 26 7 nm 19 2 nm 38 4 nm 16 0 nm

exemple 344 26.7 nm — Admin .15000 ft / 3000 ft/min = 5 minutes.5 min / 60 min = 0 0833 hour.0 0833 x 320 = 26 67 nm 26.7 nm

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

Question 172-20 : 054° 318° 234° 038°

exemple 348 054°. — Admin .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° 054°.

An aircraft at fl370 is required to commence descent at 120 nm from a vor and ?

Question 172-21 : 960 ft/min 860 ft/min 890 ft/min 920 ft/min

exemple 352 960 ft/min. — Admin .37000 ft 13000 ft = 24000 ft.120 nm / 288kt = 0 417h > 25 min 0 417 x 60 .24000 ft / 25 min = 960 ft/min 960 ft/min.

An aircraft at fl310 m0 83 temperature 30°c is required to reduce speed in ?

Question 172-22 : M0 74 m0 76 m0 80 m0 78

exemple 356 M0.74 — Admin . 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 M0.74

A ground feature was observed on a relative bearing of 325° and five minutes ?

Question 172-23 : 30 nm and 240° 40 nm and 110° 40 nm and 290° 30 nm and 060°

exemple 360 30 nm and 240°. — Admin .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° 30 nm and 240°.

An aircraft at fl350 is required to descend to cross a dme facility at fl80 ?

Question 172-24 : 69 nm 79 nm 49 nm 59 nm

exemple 364 69 nm. — Admin .35000 ft 8000 ft = 27000 ft.27000 ft / 1800 ft/min = 15 min > 0 25h 15 / 60 .0 25 x 276 kt = 69 nm 69 nm.

An aircraft at fl120 ias 200kt oat 5° and wind component +30kt is required to ?

Question 172-25 : 159 kt 174 kt 165 kt 169 kt

exemple 368 159 kt. — Admin .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 159 kt.

An aircraft at fl350 is required to cross a vor/dme facility at fl110 and to ?

Question 172-26 : 1340 ft/min 1240 ft/min 1390 ft/min 1290 ft/min

exemple 372 1340 ft/min. — Admin .24000 ft to descend.335/60 = 5 583 nm/min .100/5 583 = 18 minutes .24000 / 18 min = 1340 ft/min 1340 ft/min.

An aircraft at fl370 m0 86 oat 44°c headwind component 110 kt is required to ?

Question 172-27 : M0 81 m0 75 m0 79 m0 73

exemple 376 M0.81 — Admin . 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 M0.81

An aircraft at fl390 is required to descend to cross a dme facility at fl70 ?

Question 172-28 : 53 nm 58 nm 63 nm 68 nm

exemple 380 53 nm. — Admin .39000 7000 = 32000 ft.32000 / 2500 = 12 8 minutes.248 kt / 60 minutes = 4 14 nm/minute.12 8 x 4 14 = 53 nm 53 nm.

An aircraft at fl370 is required to commence descent when 100 nm from a dme ?

Question 172-29 : 1650 ft/min 1550 ft/min 2400 ft/min 1000 ft/min

exemple 384 1650 ft/min. — Admin .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 1650 ft/min.

An aircraft at fl140 ias 210 kt oat 5°c and wind component minus 35 kt is ?

Question 172-30 : 20 kt 15 kt 25 kt 30 kt

exemple 388 20 kt. — Admin .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 20 kt.

At 0422 an aircraft at fl370 gs 320kt is on the direct track to vor 'x' 185 nm ?

Question 172-31 : 04h45 04h54 04h51 04h48

exemple 392 04h45. — Admin .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 04h45.

An aircraft at fl330 is required to commence descent when 65 nm from a vor and ?

Question 172-32 : 1950 ft / min 1750 ft / min 1650 ft / min 1850 ft / min

exemple 396 1950 ft / min. — Admin .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 1950 ft / min.

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

Question 172-33 : 1900 ft / min 1700 ft / min 1800 ft / min 2000 ft / min

exemple 400 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 172-34 : 1800 ft/min 1900 ft/min 1600 ft/min 1700 ft/min

exemple 404 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 172-35 : 12 06 utc 11 57 utc 12 03 utc 11 53 utc

exemple 408 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 172-36 : 13 33 utc 13 40 utc 13 47 utc 14 01 utc

exemple 412 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 172-37 : 11 15 utc 10 44 utc 10 50 utc 11 10 utc

exemple 416 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 172-38 : 340 kt 360 kt 300 kt 320 kt

exemple 420 340 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 172-39 : 355 kt 375 kt 395 kt 335 kt

exemple 424 355 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 172-40 : 15° right 9° right 6° right 18° right

exemple 428 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.

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