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Question 181-1 : What is the average track °t and distance between bal vor n5318 0 w00626 9 and cfn ndb n5502 6 w00820 4 . err a 061 421 ? [ Question security ]

327° 124 nm

.report the true north tick center your protractor you read an average true track of 327° . /com en/com061 421 jpg.use the scale to find the distance 124 nm exemple 281 327° - 124 nm.

Question 181-2 : Given .sha vor n5243 3 w00853 1 radial 205° crk vor n5150 4 w00829 7 radial 317° what is the aircraft position . err a 061 424 ?

N5210 w00910

. /com en/com061 424 jpg. exemple 285 N5210 w00910

Question 181-3 : Given .sha vor n5243 3 w00853 1 radial 129° crk vor n5150 4 w00829 7 radial 047° what is the aircraft position . err a 061 426 ?

N5220 w00750

. /com en/com061 426 jpg. exemple 289 N5220 w00750

Question 181-4 : Given .sha vor/dme n5243 3 w00853 1 radial 120°/35 nm what is the aircraft position . err a 061 428 ?

N5230 w00800

. /com en/com061 428 jpg. exemple 293 N5230 w00800

Question 181-5 : Given .sha vor/dme n5243 3 w00853 1 radial 232°/32 nm what is the aircraft position . err a 061 430 ?

N5220 w00930

exemple 297 N5220 w00930

Question 181-6 : Given .sha vor/dme n5243 3 w00853 1 radial 048°/22 nm what is the aircraft position . err a 061 432 ?

N5300 w00830

. /com en/com061 432 jpg. exemple 301 N5300 w00830

Question 181-7 : 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 ?

Sha 131° crk 017°

Debiassi .when two answers are quite close try to line up the douglas protractor with two known values ie sha 052° and 115° radials for the airways this really improves accuracy exemple 305 Sha 131°, crk 017°.

Question 181-8 : Given . sha vor n5243 3 w00853 1 con vor n5354 8 w00849 1 aircraft position n5330 w00800. which of the following lists two radials that are applicable to the aircraft position . err a 061 436 ?

Sha 042° con 138°

. /com en/com061 436 jpg.magnetic north is indicated over vors and ndbs exemple 309 Sha 042° con 138°

Question 181-9 : 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 is the aircraft position . err a 061 438 ?

N5200 w00935

. /com en/com061 438 jpg.you draw two circles use the vertical scale to define the distances .both dme distances increasing so this is the only valid junction position exemple 313 N5200 w00935

Question 181-10 : 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 is the aircraft position . err a 061 440 ?

N5310 w00830

exemple 317 N5310 w00830

Question 181-11 : 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 is the aircraft position . err a 061 442 ?

N5330 w00820

exemple 321 N5330 w00820

Question 181-12 : Given . sha vor/dme n5243 3 w00853 1 birr aerodrome n5304 w00754 what is the sha radial and dme distance when overhead birr aerodrome . err a 061 444 ?

068° 41 nm

exemple 325 068° - 41 nm

Question 181-13 : Given . con vor/dme n5354 8 w00849 1 castlebar aerodrome n5351 w00917 what is the con radial and dme distance when overhead castlebar aerodrome . err a 061 446 ?

265° 17 nm

exemple 329 265° - 17 nm

Question 181-14 : An aircraft departs from position a 04°10' s 178°22'w and flies northward following the meridian for 2950 nm .it then flies westward along the parallel of latitude for 382 nm to position b .the coordinates of position b are ?

45°00'n 172°38'e

Admin .it flies northward following the meridian for 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 and we are heading north 4°10' to reach the equator and after it remains 45° to fly our position is now 45°00'n 178°22'w .now we are heading west along latitude 45°n we need to use the 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 the anti meridian we are now at 172°38'e 360° 187°22' exemple 333 45°00'n 172°38'e.

Question 181-15 : What is the time required to travel along the parallel of latitude 60°n between meridians 010°e and 030°w at a groundspeed of 480 kt ?

2 h 30

Admin .distance in longitude = 40°.at the 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 exemple 337 2 h 30.

Question 181-16 : Given the following .magnetic heading 060° magnetic variation 8°w drift angle 4° right.what is the true track ?

056°

Admin . 1363.use this wonderful table for those questions .gomis01 .060° menos 8°w son 052° como es heading ya tiene el ángulo de deriva incluido por lo que solo tenemos que quitarselo 052° 4° = 048° . .no drift is always measured from heading to track exemple 341 056°.

Question 181-17 : An aircraft is following a true track of 048° at a constant tas of 210 kt .the wind velocity is 350° / 30 kt .the gs and drift angle are ?

192 kt 7° right

Under index set true track 048° centre dot on tas 210 kt with the rotative scale set wind . /com en/com061 12 jpg.now drift is 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 exemple 345 192 kt, 7° right.

Question 181-18 : Given .fl 350 mach 0 80 oat 55°c .calculate the values for tas and local speed of sound lss ?

461 kt lss 576 kt

Admin .lss varies with the 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 is the ratio of true airspeed tas to the local speed of sound lss .formula mach number = tas/lss.0 80 = tas/576.tas = 576 x 0 8 = 461 kt exemple 349 461 kt, lss 576 kt.

Question 181-19 : Given .true heading = 180°.tas = 500 kt.w/v 225° / 100 kt.calculate the gs ?

435 kt

Admin .on the computer set 180° under index centre dot on 500 kt under the wind speed 100 kt on the rotating scale . 1367.you read a groud speed of 435 kt exemple 353 435 kt.

Question 181-20 : Given .true heading = 310° tas = 200 kt gs = 176 kt drift angle 7° right .calculate the w/v ?

270° / 33 kt

Admin .use low speed scale on the computer set 310° under index centre dot on 200 kt where right drift 7° crosses ground speed arc 176 kt read the wind on the rotating scale . 1368.you read 270°/33kt exemple 357 270° / 33 kt.

Question 181-21 : Given .true heading = 090° tas = 200 kt wind = 220°/30 kt .calculate the ground speed ?

220 kt

Admin . 1369 exemple 361 220 kt.

Question 181-22 : The reported surface wind from the control tower is 240°/35 kt runway 30 300° .what is the cross wind component ?

30 kt

Admin .angle between the wind and the direction of the runway 300° 240° = 60°.crosswind = sine of the angle between the wind and the direction of the runway x windspeed.crosswind = sin60° x 35 kt = 30 31 kt exemple 365 30 kt.

Question 181-23 : Given magnetic heading 311° drift angle 10° left relative bearing of ndb 270° what is the magnetic bearing of the ndb measured from the aircraft ?

221°

Admin .heading 311°.bearing +270°.total 581°.581° 360° = 221° magnetic bearing of the ndb measured from the aircraft exemple 369 221°.

Question 181-24 : Given the following .true track 192° magnetic variation 7°e drift angle 5° left .what is the magnetic heading required to maintain the given track ?

190°

Admin . 1393.use this wonderful table for those questions exemple 373 190°.

Question 181-25 : The angle between the plane of the ecliptic and the plane of equator is approximately ?

23 5°

Admin . 2028 exemple 377 23.5°.

Question 181-26 : Given .tas = 485 kt.oat = isa +10°c .fl 410 .calculate the mach number ?

0 825

Admin . 56°c is considered to be the lowest isa temperature and the question states that we are in outside air temperature of isa +10°c .at fl410 isa is 56°c isa +10°c will be 46°c .on the computer . 1396.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 exemple 381 0.825

Question 181-27 : At 1215 utc lajes vortac 38°46'n 027°05'w rmi reads 178° range 135 nm .calculate the aircraft position at 1215 utc . 2484 ?

40°55'n 027°55'w

Admin .rmi reads 178° it is the magnetic direction to reach the station vor tacan at lajes .for a vor we must apply the variation at the station on the chart variation line is 15°w .178° variation west magnetic best so 'minus' 15° 178° 15° = 163° .from lajes 163° + 180° = 343° . 1398.use the latitude scale to find 135 nm exemple 385 40°55'n 027°55'w.

Question 181-28 : 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 the eta at port santo ndb . err a 061 53 ?

1348

.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 is 3° right .turn to put 133° 136° 3° right drift below true index the groundspeed is under the wind mark 30 kt 470 kt gs ..plot distance between points and measure along line of longitude . /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 exemple 389 1348

Question 181-29 : 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 the distance from q to the point of safe return psr is ?

1685 nm

.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 the psr from the departure point at a speed of 385 kt .4 3764 h x 385 = 1685 nm exemple 393 1685 nm.

Question 181-30 : Two points a and b are 1000 nm apart tas = 490 kt .on the flight between a and b the equivalent headwind is 20 kt .on the return leg between b and a the equivalent tailwind is +40 kt .what distance from a along the route a to b is the the point of equal time pet ?

530 nm

Admin .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 exemple 397 530 nm.

Question 181-31 : Given ad = air distance gd = ground distance tas = true airspeed gs = groundspeed which of the following is the correct formula to calculate ground distance gd gone ?

Gd = ad x gs /tas

.gd = gs x t but t = ad/tas so gd = gs x ad /tas exemple 401 Gd = (ad x gs)/tas

Question 181-32 : What is the isa temperature value at fl 330 ?

51°c

15°c 2° x 33 = 51°c exemple 405 -51°c

Question 181-33 : Given .tas 487kt fl 330 temperature isa + 15 .calculate the mach number ?

0 81

Admin .isa temperature at fl 330 = 15º 2º x 33 = 51ºc .deviation is +15ºc then oat is 51°c + +15°c = 36ºc .with the computer .in the airspeed window put the temperature 36ºc in front of 'mach kt ' index .in front of 487 kt on the inner scale you read 0 81 mach number . 1435 exemple 409 0.81

Question 181-34 : How many nm would an aircraft travel in 1 minute 45 secondes if gs is 135 kt ?

3 94 nm

Admin .135 kt /60 minutes = 2 25 nm/minute.1 minute 45 secondes = 1 75 minute.2 25 x 1 75 = 3 9375 nm exemple 413 3.94 nm.

Question 181-35 : An aircraft travels 100 statute miles in 20 min how long does it take to travel 215 nm ?

50 min

Admin .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 exemple 417 50 min.

Question 181-36 : Given .tas = 220 kt.magnetic course = 212°.wind = 160 ° m / 50 kt.calculate the gs ?

186 kt

Admin .the given wind has a headwind component so the groundspeed is going to be less than the given tas and all but one of the answers are more exemple 421 186 kt.

Question 181-37 : Given .fl250.oat 15 °c.tas 250 kt.calculate the mach number ?

0 40

Admin . 2485. aviat 617 .put temperature ' 15ºc' in front of m kt index in the airspeed window .go to tas 250 kt on outer scale and read mach number on the inner scale .by calculation .mach number = tas / lss.mach number = 250 / 39 sqrt 273 15 .mach number = 0 399 exemple 425 0.40

Question 181-38 : During a low level flight 2 parallel roads that are crossed at right angles by an aircraft .the time between these roads can be used to check the aircraft ?

Groundspeed


Question 181-39 : Given .magnetic track = 315° magnetic heading = 301° variation = 5°w tas = 225 kt .the aircraft flies 50 nm in 12 min .calculate the wind °t ?

190°/63 kt

Admin .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 the vertical speed line you read 190°/63 kt exemple 433 190°/63 kt.

Question 181-40 : Given .tas = 270 kt true hdg = 270° actual wind 205° t /30kt .calculate the drift angle and gs ?

6r 259kt

Admin .under index set true heading 270° centre dot on tas 270 kt with the rotative scale set wind 205°/30 kt .read the drift and the ground speed 6°r 259 kt exemple 437 6r - 259kt


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