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Multiple > protocol : Given tas = 200 kt track t = 073° w/v = 210/20kt calculate the hdg °t and gs ?

Question 171-1 : 077 214 kt 079 211 kt 075 213 kt 077 210 kt

exemple 271 077 - 214 kt077 - 214 kt

Given .tas = 270 kt track t = 260° wind = 275°/30kt .calculate the hdg °t ?

Question 171-2 : 262° 241 kt 262° 237 kt 264° 241 kt 264° 237 kt

...under index set true track 260° centre dot on tas 270 kt with the rotative scale set wind. /com en/com061 187 jpg..now drift is always measured from heading to track .turn to set true heading 262° 260° + 2° left drift under index you now read a ground speed of 241 kt exemple 275 262° - 241 kt.262° - 241 kt.

Given .true hdg = 233° tas = 480 kt track t = 240° gs = 523 kt .calculate the ?

Question 171-3 : 110/75kt 115/70kt 110/80kt 105/75kt

...true heading is 233° true track is 240° our drift is 7° right. /com en/com061 189 jpg..wind 111°/77kt closest answer 110/75kt exemple 279 110/75kt.110/75kt.

Given .true heading = 074° tas = 230 kt true track = 066° ground speed = 242 ?

Question 171-4 : 180/35 kt 180/30 kt 185/35 kt 180/40 kt

...true heading is 074° true track is 066° our drift is 8° left . /com en/com061 191 jpg.where the rotative scale crosses the ground speed arc 242 kt we read the wind 180°/ 35 kt exemple 283 180/35 kt180/35 kt

Given .true heading = 054° tas = 450 kt true track = 059° gs = 416 kt ?

Question 171-5 : 010°/50 kt 005°/50 kt 010°/55 kt 010°/45 kt

.true heading is 054° true track is 059° our drift is 5° right . 2525.wind 010°/50 kt exemple 287 010°/50 kt.010°/50 kt.

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

Question 171-6 : 095°/20 kt 090°/15 kt 090°/20 kt 095°/25 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 exemple 291 095°/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 171-7 : 1 hr 30 min 1 hr 09 min 1 hr 10 min 1 hr 40 min

.354 nm / 236 kt/60 min = 90 minutes 1h30 exemple 295 1 hr 30 min.1 hr 30 min.

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

Question 171-8 : 10 hr 19 min 10 hr 05 min 11 hr 00 min 11 hr 02 min

.3560 nm / 345 kt/60 min = 619 minutes 10h39 exemple 299 10 hr 19 min.10 hr 19 min.

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

Question 171-9 : 5 hr 03 min 4 hr 59 min 5 hr 00 min 5 hr 08 min

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

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

Question 171-10 : 00 hr 42 min 00 hr 43 min 00 hr 44 min 00 hr 45 min

.84 nm / 120 kt/60 min = 42 minutes exemple 307 00 hr 42 min.00 hr 42 min.

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

Question 171-11 : 1490 nm 1664 nm 1698 nm 1422 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 311 1490 nm.1490 nm.

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

Question 171-12 : 219 min 290 min 197 min 167 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 315 219 min.219 min.

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

Question 171-13 : 114 min 110 min 106 min 102 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 319 114 min.114 min.

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

Question 171-14 : 1838 nm 1313 nm 1467 nm 1642 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 323 1838 nm.1838 nm.

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

Question 171-15 : 85 nm 56 nm 88 nm 176 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 327 85 nm.85 nm.

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

Question 171-16 : 2210 ft 680 ft 1890 ft 3640 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 331 2210 ft2210 ft

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

Question 171-17 : 1 6° 2 8° 3 2° 2 4°

.vertical speed = gradient % * gs.1000 = gradient % * 360.gradient % = 2 77..gradient % = angle * 100/60..2 77 = angle * 100/60 > angle = 1 662º exemple 335 1.6°.1.6°.

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

Question 171-18 : 1450 ft 1350 ft 1300 ft 1400 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 339 1450 ft.1450 ft.

730 ft/min equals ?

Question 171-19 : 3 7 m/sec 5 2 m/sec 1 6 m/sec 2 2 m/sec

.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 exemple 343 3.7 m/sec3.7 m/sec

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

Question 171-20 : 1 min 07 sec 1 min 55 sec 2 min 30 sec 0 min 34 sec

.5 / 269/60 =1 115 minutes.0 115 x 60 = 7 secondes .1 min 7 sec exemple 347 1 min 07 sec1 min 07 sec

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

Question 171-21 : 160 kt 183 kt 209 kt 131 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 351 160 kt.160 kt.

The icao definition of eta is the ?

Question 171-22 : 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 355 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 171-23 : 26 7 nm 19 2 nm 38 4 nm 16 0 nm

.15000 ft / 3000 ft/min = 5 minutes..5 min / 60 min = 0 0833 hour..0 0833 x 320 = 26 67 nm exemple 359 26.7 nm26.7 nm

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

Question 171-24 : 054° 318° 234° 038°

.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 363 054°.054°.

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

Question 171-25 : 960 ft/min 860 ft/min 890 ft/min 920 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 367 960 ft/min.960 ft/min.

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

Question 171-26 : M0 74 m0 76 m0 80 m0 78

. 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 371 m0.74m0.74

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

Question 171-27 : 30 nm and 240° 40 nm and 110° 40 nm and 290° 30 nm and 060°

.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 375 30 nm and 240°.30 nm and 240°.

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

Question 171-28 : 69 nm 79 nm 49 nm 59 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 379 69 nm.69 nm.

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

Question 171-29 : 159 kt 174 kt 165 kt 169 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 383 159 kt.159 kt.

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

Question 171-30 : 1340 ft/min 1240 ft/min 1390 ft/min 1290 ft/min

.24000 ft to descend..335/60 = 5 583 nm/min.100/5 583 = 18 minutes.24000 / 18 min = 1340 ft/min exemple 387 1340 ft/min.1340 ft/min.

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

Question 171-31 : M0 81 m0 75 m0 79 m0 73

. 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 391 m0.81m0.81

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

Question 171-32 : 53 nm 58 nm 63 nm 68 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 395 53 nm.53 nm.

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

Question 171-33 : 1650 ft/min 1550 ft/min 2400 ft/min 1000 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 399 1650 ft/min.1650 ft/min.

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

Question 171-34 : 20 kt 15 kt 25 kt 30 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 403 20 kt.20 kt.

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

Question 171-35 : 04h45 04h54 04h51 04h48

.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 407 04h45.04h45.

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

Question 171-36 : 1950 ft / min 1750 ft / min 1650 ft / min 1850 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 411 1950 ft / min.1950 ft / min.

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

Question 171-37 : 1900 ft / min 1700 ft / min 1800 ft / min 2000 ft / min

.50 nm / 271 kt = 0 185h > 11 min..29000 ft 8000 ft = 21000 ft..21000 ft / 11 min = 1900 ft/min exemple 415 1900 ft / min.1900 ft / min.

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

Question 171-38 : 1800 ft/min 1900 ft/min 1600 ft/min 1700 ft/min

.fl350 fl080 = 27000 ft.85 nm / 340 kt = 0 25 h 15 minutes.27000 ft / 15 min = 1800 ft/min exemple 419 1800 ft/min.1800 ft/min.

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

Question 171-39 : 12 06 utc 11 57 utc 12 03 utc 11 53 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 423 12:06 utc.12:06 utc.

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

Question 171-40 : 13 33 utc 13 40 utc 13 47 utc 14 01 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 427 13:33 utc.13:33 utc.

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