Given SHA VOR/DME N5243 3 W00853 1 radial 232°/32 NM What is the aircraft position ERR _a_061 430 ?
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Given SHA VOR/DME N5243 3 W00853 1 radial 048°/22 NM What is the aircraft position ERR _a_061 432 ?
Given SHA VOR N5243 3 W00853 1 CRK VOR N5150 4 W00829 7 Aircraft position N5230 W00820 Which of the following lists two radials that are applicable to the aircraft position ERR _a_061 434 ?
Given sha vor n5243 3 w00853 1 con vor n5354 8 w00849 1 aircraft position n5330 w00800 which of following lists two radials that are applicable to aircraft position err _a_061 436 Sha 3 ° crk 7°. img /com_en/com061 436 jpg magnetic north indicated over vors ndbs. 
Given sha vor n5243 3 w00853 1 dme 50 nm crk vor n5150 4 w00829 7 dme 41 nm aircraft heading 270° m both dme distances increasing what the aircraft position err _a_061 438 Sha 3 ° crk 7°. img /com_en/com061 438 jpg you draw two circles (use vertical scale to define distances) both dme distances increasing so this the only valid junction position. 
Given crn vor n5318 1 w00856 5 dme 18 nm sha vor n5243 3 w00853 1 dme 30 nm aircraft heading 270° m both dme distances decreasing what the aircraft position err _a_061 440 Sha 3 ° crk 7°. img /com_en/com061 438 jpg you draw two circles (use vertical scale to define distances) both dme distances increasing so this the only valid junction position. 
Given con vor n5354 8 w00849 1 dme 30 nm crn vor n5318 1 w00856 5 dme 25 nm aircraft heading 270° m both dme distances decreasing what the aircraft position err _a_061 442 Sha 3 ° crk 7°. img /com_en/com061 438 jpg you draw two circles (use vertical scale to define distances) both dme distances increasing so this the only valid junction position. Question 165-8
Given sha vor/dme n5243 3 w00853 1 birr aerodrome n5304 w00754 what the sha radial and dme distance when overhead birr aerodrome err _a_061 444 Sha 3 ° crk 7°. img /com_en/com061 438 jpg you draw two circles (use vertical scale to define distances) both dme distances increasing so this the only valid junction position. Question 165-9
Given con vor/dme n5354 8 w00849 1 castlebar aerodrome n5351 w00917 what the con radial and dme distance when overhead castlebar aerodrome err _a_061 446 Sha 3 ° crk 7°. img /com_en/com061 438 jpg you draw two circles (use vertical scale to define distances) both dme distances increasing so this the only valid junction position. Question 165-10
An aircraft departs from position a 04°10' s 178°22'w and flies northward following meridian 2950 nm it then flies westward along parallel of latitude 382 nm to position b the coordinates of position b are Sha 3 ° crk 7°. it flies northward following meridian 2950 nm 1 nm on a meridian = 1' 2950 nm = 2950 minutes of arc 60' = 1° 2950'/60' = 49°10' we are in south hemisphere we are heading north 4°10' to reach equator after it remains 45° to fly our position now 45°00'n 178°22'w now we are heading west along latitude 45°n we need to use formula distance = change of longitude x cos of latitude 382 nm = change of longitude x cos45° change of longitude = 382 / 0 707 = 540 3' 540 3'/60' = 9° 178°22'w + 9°w = 187°22' we cross anti meridian we are now at 172°38'e (360° 187°22'). Question 165-11
What the time required to travel along parallel of latitude 60°n between meridians 010°e and 030°w at a groundspeed of 480 kt Sha 3 ° crk 7°. distance in longitude = 40° at equator 1° = 60 nm distance in nm = 40° x 60 nm x cos latitude distance = 40° x 60 nm x cos 60° distance = 1200 nm 1200 / 480 kt = 2 h 30. Question 165-12
Given following magnetic heading 060° magnetic variation 8°w drift angle 4° rightwhat the true track Sha 3 ° crk 7°. use this wonderful table those questions gomis01 060° menos 8°w son 052° como es heading ya tiene el ángulo deriva incluido por lo que solo tenemos que quitarselo 052° 4° = 048° no drift always measured from heading to track. Question 165-13
An aircraft following a true track of 048° at a constant tas of 210 kt the wind velocity 350° / 30 kt the gs and drift angle are Sha 3 ° crk 7°. Under index set true track 048° centre dot on tas 210 kt with rotative scale set wind img /com_en/com061 12 jpg now drift always measured from heading to track turn to set true heading 041° (048° 7° right drift) under index you now read a ground speed of 192 kt. Question 165-14
Given fl 350 mach 0 80 oat 55°c calculate values tas and local speed of sound lss Sha 3 ° crk 7°. lss varies with square root of absolute temperature formula lss = 39 x square root of (oat+273) lss = 39 x square root of 218 lss = 576 kt mach number the ratio of true airspeed (tas) to local speed of sound (lss) formula mach number = tas/lss 0 80 = tas/576 tas = 576 x 0 8 = 461 kt. Question 165-15
Given true heading = 180°tas = 500 ktw/v 225° / 100 ktcalculate gs Sha 3 ° crk 7°. on computer set 180° under index centre dot on 500 kt under wind speed 100 kt on rotating scale you read a groud speed of 435 kt. Question 165-16
Given true heading = 310° tas = 200 kt gs = 176 kt drift angle 7° right calculate w/v Sha 3 ° crk 7°. use low speed scale on computer set 310° under index centre dot on 200 kt where right drift 7° crosses ground speed arc 176 kt read wind on rotating scale you read 270°/33kt. Question 165-17
Given true heading = 090° tas = 200 kt wind = 220°/30 kt calculate ground speed Sha 3 ° crk 7°. . Question 165-18
The reported surface wind from control tower 240°/35 kt runway 30 300° what the cross wind component Sha 3 ° crk 7°. angle between wind the direction of runway 300° 240° = 60° crosswind = sine of angle between wind the direction of runway x windspeed crosswind = sin60° x 35 kt = 30 31 kt. Question 165-19
Given magnetic heading 311° drift angle 10° left relative bearing of ndb 270° what the magnetic bearing of ndb measured from aircraft Sha 3 ° crk 7°. heading 311° bearing +270° total 581° 581° 360° = 221° magnetic bearing of ndb measured from aircraft. Question 165-20
Given following true track 192° magnetic variation 7°e drift angle 5° left what the magnetic heading required to maintain given track Sha 3 ° crk 7°. use this wonderful table those questions. Question 165-21
The angle between plane of ecliptic and plane of equator approximately Sha 3 ° crk 7°. . Question 165-22
Given tas = 485 ktoat = isa +10°c fl 410 calculate mach number Sha 3 ° crk 7°. 56°c considered to be lowest isa temperature the question states that we are in outside air temperature of isa +10°c at fl410 isa 56°c isa +10°c will be 46°c on computer by calculation mach = tas x a a = 39square root t° in kelvin a = 39 square root 227°k= 587 59 mach number= 485/ 587 59= 0 825. Question 165-23
At 1215 utc lajes vortac 38°46'n 027°05'w rmi reads 178° range 135 nm calculate aircraft position at 1215 utc 2484 Sha 3 ° crk 7°. rmi reads 178° it the magnetic direction to reach station (vor tacan) at lajes for a vor we must apply variation at station on chart variation line 15°w 178° (variation west magnetic best) so 'minu 15° 178° 15° = 163° from lajes 163° + 180° = 343° use latitude scale to find 135 nm. Question 165-24
At reference 1300 utc dr position 37°30'n 021°30'w alter heading port santo ndb 33°03'n 016°23'w tas 450 kt forecast w/v 360°/30kt calculate eta at port santo ndb err _a_061 53 Sha 3 ° crk 7°. true track is 136° wind 360°/30 kt drift is ? true heading is ? ground speed is ? center dot on 450 kt true index on 136° indicate wind 360°/30 kt drift 3° right turn to put 133° (136° 3° right drift) below true index groundspeed under wind mark 30 kt 470 kt gs plot distance between points measure along line of longitude img /com_en/com061 53 jpg 6°20' = 6 33 x 60 nm = 380 nm 380 / (470/60) = 48 5 minutes 1300 + 48 minutes 30 secondes = 13h48 30sec. Question 165-25
For a distance of 1860 nm between q and r a ground speed 'out' of 385 kt a ground speed 'home' of 465 kt and an endurance of 8 hours excluding reserves distance from q to point of safe return psr Sha 3 ° crk 7°. point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 385 kt ground speed home = 465 kt point of safe return (psr) = 8 x 465 / (385 + 465) point of safe return (psr) = 3720 / 850 point of safe return (psr) = 4 3764 h distance of psr from departure point at a speed of 385 kt 4 3764 h x 385 = 1685 nm. Question 165-26
Two points a and b are 1000 nm apart tas = 490 kt on flight between a and b equivalent headwind 20 kt on return leg between b and a equivalent tailwind +40 kt what distance from a along route a to b the point of equal time pet Sha 3 ° crk 7°. ground speed out 490 20 = 470 kt ground speed home 490 + 40 = 530 kt pet = distance x gsh / (gso + gsh) pet = 1000 x 530 /(470 + 530) = 530 nm. Question 165-27
Given ad = air distance gd = ground distance tas = true airspeed gs = groundspeed which of following the correct formula to calculate ground distance gd gone Sha 3 ° crk 7°. gd = gs x t but t = ad/tas so gd = (gs x ad)/tas. Question 165-28
What the isa temperature value at fl 330 Sha 3 ° crk 7°. 15°c (2° x 33) = 51°c. Question 165-29
Given tas 487kt fl 330 temperature isa + 15 calculate mach number Sha 3 ° crk 7°. isa temperature at fl 330 = 15º (2º x 33) = 51ºc deviation +15ºc then oat 51°c + (+15°c) = 36ºc with computer in airspeed window put temperature 36ºc in front of 'mach (kt)' index in front of 487 kt on inner scale you read 0 81 mach number . Question 165-30
How many nm would an aircraft travel in 1 minute 45 secondes if gs 135 kt Sha 3 ° crk 7°. 135 kt /60 minutes = 2 25 nm/minute 1 minute 45 secondes = 1 75 minute 2 25 x 1 75 = 3 9375 nm. Question 165-31
An aircraft travels 100 statute miles in 20 min how long does it take to travel 215 nm Sha 3 ° crk 7°. 1 nm = 1 15 sm 215 nm = 215 x 1 15 = 247 25 sm 100 sm in 20 minutes = 300 sm in 60 minutes (1h) 247 25/300 = 0 824 hour 0 824 x 60 = 50 min. Question 165-32
Given tas = 220 ktmagnetic course = 212°wind = 160 ° m / 50 ktcalculate gs Sha 3 ° crk 7°. the given wind has a headwind component so groundspeed going to be less than given tas all but one of answers are more. Question 165-33
Given fl250oat 15 °ctas 250 ktcalculate mach number Sha 3 ° crk 7°. aviat 617 put temperature ' 15ºc' in front of m (kt) index in airspeed window go to tas 250 kt on outer scale read mach number on inner scale by calculation mach number = tas / lss mach number = 250 / 39 sqrt (273 15) mach number = 0 399. Question 165-34
During a low level flight 2 parallel roads that are crossed at right angles an aircraft the time between these roads can be used to check aircraft Sha 3 ° crk 7°. aviat 617 put temperature ' 15ºc' in front of m (kt) index in airspeed window go to tas 250 kt on outer scale read mach number on inner scale by calculation mach number = tas / lss mach number = 250 / 39 sqrt (273 15) mach number = 0 399. Question 165-35
Given magnetic track = 315° magnetic heading = 301° variation = 5°w tas = 225 kt the aircraft flies 50 nm in 12 min calculate wind °t Sha 3 ° crk 7°. true heading = 296° (301° 5°) true course = 310° (315 5°) ground speed = 60 x(50/12) = 250 kt on nav computer set tas (225 kt) on center dot under true index set true heading 296° mark where drift (14º right) crosses ground speed 250 kt rotate to shift mark under vertical speed line you read 190°/63 kt. Question 165-36
Given tas = 270 kt true hdg = 270° actual wind 205° t /30kt calculate drift angle and gs Sha 3 ° crk 7°. under index set true heading 270° centre dot on tas 270 kt with rotative scale set wind 205°/30 kt read drift the ground speed 6°r 259 kt. Question 165-37
Given tas = 270 kt true hdg = 145° actual true wind = 205°/30kt calculate drift angle and gs Sha 3 ° crk 7°. under index set true track 145° centre dot on tas 270 kt with rotative scale set wind read drift the ground speed 6°l 256 kt. Question 165-38
Given tas = 470 kt true heading = 317° wind = 045° t /45 kt calculate drift angle and gs Sha 3 ° crk 7°. under index set true heading 317° centre dot on tas 470 kt with rotative scale set wind read drift 5° left ground speed 470 kt. Question 165-39
Given tas = 190 kt true heading = 085° true wind = 110°/50kt calculate drift angle and gs Sha 3 ° crk 7°. under index set true heading 085° centre dot on tas 190 kt with rotative scale set wind read drift 8° left ground speed 147 kt closest answer 8°l 146 kt. Question 165-40
Given tas = 132 kt true hdg = 257° true wind = 095°/35 kt calculate drift angle and gs Sha 3 ° crk 7°. under index set true heading 257° centre dot on tas 132 kt set wind (095°/35kt) on rotative scale you read 4 5°right 164 kt.
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