.center dot on tas 135 kt..true heading 278° under index.put wind direction under the red compass rose, under 20 kt, your drift is 5° right, giving a track of 283° and a groundspeed under the wind mark of 150 kt.. 1739
Question 185-2 : Given.tas = 155 kt.true heading = 216°.wind = 090°/60 kt.calculate the true track and gs ?
231° 196 kt.
.center dot on tas 155 kt..true heading 216° under index.put wind direction under the red compass rose, under 60 kt, your drift is 14.5° right, giving a track of 230.5° and a groundspeed under the wind mark of 195 kt.. 2527.the closest answer is 231° and 196 kt.
...under index, set true heading 124°, centre dot on tas, 465 kt, with the rotative scale, set wind.. /com en/com061 171.jpg..read drift 8° left..ground speed is 415 kt.
Question 185-4 : Given tas = 140 kt, hdg t = 005°, w/v = 265/25kt. calculate the drift and gs ?
10r 146 kt
.under index, set true heading 005°, centre dot on tas, 140 kt, with the rotative scale, set wind. 1716.read drift 10° right..ground speed is 146 kt.
Question 185-5 : Given.tas = 190 kt, hdg t = 355°, w/v = 165/25kt..calculate the drift and gs ?
1l 215 kt
...under index, set true heading 355°, centre dot on tas, 190 kt, with the rotative scale, set wind.. /com en/com061 174.jpg..read drift 1° left..ground speed is 214 kt close enough for the answer.
Question 185-6 : Given.tas = 250 kt.hdg t = 029°.w/v = 035/45kt.calculate the drift and gs ?
1l 205 kt
.under index, set true heading 029°, centre dot on tas, 250 kt, with the rotative scale, set wind. 2526.read drift 1° left..ground speed is 205 kt.
...under index, set true heading 168°, centre dot on tas, 485 kt, with the rotative scale, set wind. /com en/com061 180.jpg..read drift 6° right, 168° + 6° = 174°..ground speed is 430 kt close enough for the answer.
Question 185-8 : Given.tas = 130 kt.track t = 003°.w/v = 190/40 kt..calculate the hdg °t and gs ?
001° 170 kt.
Question 185-9 : Given.tas = 227 kt, track t = 316°, w/v = 205/15kt..calculate the hdg °t and gs ?
312° 232 kt.
.under index, set true track 316°, centre dot on tas, 227 kt, with the rotative scale, set wind. 2021.now, drift is always measured from heading to track.turn to set true heading 312° 316° 4° right drift under index, you now read a ground speed of 232 kt.
Question 185-10 : Given tas = 200 kt, track t = 073°, w/v = 210/20kt. calculate the hdg °t and gs ?
077 214 kt
Question 185-11 : Given.tas = 270 kt, track t = 260°, wind = 275°/30kt..calculate the hdg °t and gs ?
262° 241 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.
Question 185-12 : Given.true hdg = 233°, tas = 480 kt, track t = 240°, gs = 523 kt..calculate the w/v ?
110/75kt.
...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.
...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.
...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.
Question 185-16 : Given.gs = 236 kt. distance from a to b = 354 nm..what is the time from a to b ?
1 hr 30 min.
.354 nm / 236 kt/60 min = 90 minutes 1h30.
Question 185-17 : Given.gs = 345 kt. distance from a to b = 3560 nm..what is the time from a to b ?
10 hr 19 min.
.3560 nm / 345 kt/60 min = 619 minutes 10h39.
Question 185-18 : Given.gs = 95 kt. distance from a to b = 480 nm..what is the time from a to b ?
5 hr 03 min.
.480 nm / 95 kt/60 min = 303 minutes 5h03.
Question 185-19 : Given.gs = 120 kt. distance from a to b = 84 nm..what is the time from a to b ?
00 hr 42 min.
.84 nm / 120 kt/60 min = 42 minutes.
Question 185-20 : 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.
Question 185-21 : 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.
Question 185-22 : 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.
Question 185-23 : 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.
Question 185-24 : 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.
Question 185-25 : 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.
Question 185-26 : 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 ?
Question 185-27 : 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.
Question 185-28 : 730 ft/min equals ?
3.7 m/sec
Question 185-29 : 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.
Question 185-30 : 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.
Question 185-31 : The icao definition of eta is the ?
Estimated time of arrival at destination.
Question 185-32 : 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.
Question 185-33 : 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°.
Question 185-34 : 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.
Question 185-35 : 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.
Question 185-36 : 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°.
Question 185-37 : 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.
Question 185-38 : 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.
Question 185-39 : 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.
Question 185-40 : 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.
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