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An aircraft departs from position a 04°10' s 178°22'w and flies northward ? [ Formation assignment ]

Question 168-1 : 45°00'n 172°38'e 53°20'n 169°22w 45°00'n 169°22w 53°20'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 268 45°00'n 172°38'e.45°00'n 172°38'e.

What is the time required to travel along the parallel of latitude 60°n ?

Question 168-2 : 2 h 30 1 h 15 1 h 45 5 h 00

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 272 2 h 30.2 h 30.

Given the following .magnetic heading 060° magnetic variation 8°w drift angle ?

Question 168-3 : 056° 064° 048° 072°

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 276 056°.056°.

An aircraft is following a true track of 048° at a constant tas of 210 kt .the ?

Question 168-4 : 192 kt 7° right 200 kt 3 5° right 192 kt 7° left 225 kt 7° left

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 280 192 kt, 7° right.192 kt, 7° right.

Given .fl 350 mach 0 80 oat 55°c .calculate the values for tas and local speed ?

Question 168-5 : 461 kt lss 576 kt 237 kt lss 296 kt 490 kt lss 461 kt 461 kt lss 296 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 284 461 kt, lss 576 kt.461 kt, lss 576 kt.

Given .true heading = 180°.tas = 500 kt.w/v 225° / 100 kt.calculate the gs ?

Question 168-6 : 435 kt 600 kt 535 kt 450 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 288 435 kt.435 kt.

Given .true heading = 310° tas = 200 kt gs = 176 kt drift angle 7° right ?

Question 168-7 : 270° / 33 kt 360° / 33 kt 090° / 33 kt 180° / 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 292 270° / 33 kt.270° / 33 kt.

Given .true heading = 090° tas = 200 kt wind = 220°/30 kt .calculate the ?

Question 168-8 : 220 kt 230 kt 180 kt 200 kt

Admin . 1369 exemple 296 220 kt.220 kt.

The reported surface wind from the control tower is 240°/35 kt runway 30 300° ?

Question 168-9 : 30 kt 24 kt 27 kt 21 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 300 30 kt.30 kt.

Given magnetic heading 311° drift angle 10° left relative bearing of ndb ?

Question 168-10 : 221° 208° 211° 180°

Admin .heading 311°.bearing +270°.total 581°.581° 360° = 221° magnetic bearing of the ndb measured from the aircraft exemple 304 221°.221°.

Given the following .true track 192° magnetic variation 7°e drift angle 5° ?

Question 168-11 : 190° 194° 204° 180°

Admin . 1393.use this wonderful table for those questions exemple 308 190°.190°.

The angle between the plane of the ecliptic and the plane of equator is ?

Question 168-12 : 23 5° 25 3° 27 5° 66 5°

Admin . 2028 exemple 312 23.5°.23.5°.

Given .tas = 485 kt.oat = isa +10°c .fl 410 .calculate the mach number ?

Question 168-13 : 0 825 0 9 0 85 0 87

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 316 0.8250.825

At 1215 utc lajes vortac 38°46'n 027°05'w rmi reads 178° range 135 nm ?

Question 168-14 : 40°55'n 027°55'w 40°50'n 027°30'w 41°00'n 028°10'w 41°05'n 027°50'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 320 40°55'n 027°55'w.40°55'n 027°55'w.

At reference .1300 utc dr position 37°30'n 021°30'w alter heading port santo ?

Question 168-15 : 1348 1344 1341 1354

.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 324 13481348

For a distance of 1860 nm between q and r a ground speed 'out' of 385 kt a ?

Question 168-16 : 1685 nm 1532 nm 930 nm 1865 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 328 1685 nm.1685 nm.

Two points a and b are 1000 nm apart tas = 490 kt .on the flight between a and ?

Question 168-17 : 530 nm 455 nm 500 nm 470 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 332 530 nm.530 nm.

Given ad = air distance gd = ground distance tas = true airspeed gs = ?

Question 168-18 : Gd = ad x gs /tas gd = ad tas /tas gd = ad x gs tas /gs gd = tas/ gs x ad

.gd = gs x t but t = ad/tas so gd = gs x ad /tas exemple 336 Gd = (ad x gs)/tasGd = (ad x gs)/tas

What is the isa temperature value at fl 330 ?

Question 168-19 : 51°c 56°c 66°c 81°c

15°c 2° x 33 = 51°c exemple 340 -51°c-51°c

Given .tas 487kt fl 330 temperature isa + 15 .calculate the mach number ?

Question 168-20 : 0 81 0 76 0 78 0 84

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 344 0.810.81

How many nm would an aircraft travel in 1 minute 45 secondes if gs is 135 kt ?

Question 168-21 : 3 94 nm 2 36 nm 39 nm 3 25 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 348 3.94 nm.3.94 nm.

An aircraft travels 100 statute miles in 20 min how long does it take to travel ?

Question 168-22 : 50 min 100 min 90 min 80 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 352 50 min.50 min.

Given .tas = 220 kt.magnetic course = 212°.wind = 160 ° m / 50 kt.calculate ?

Question 168-23 : 186 kt 290 kt 246 kt 250 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 356 186 kt.186 kt.

Given .fl250.oat 15 °c.tas 250 kt.calculate the mach number ?

Question 168-24 : 0 40 0 42 0 44 0 39

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 360 0.400.40

During a low level flight 2 parallel roads that are crossed at right angles by ?

Question 168-25 : Groundspeed drift position track

exemple 364 Groundspeed.Groundspeed.

Given .magnetic track = 315° magnetic heading = 301° variation = 5°w tas = ?

Question 168-26 : 190°/63 kt 355°/15 kt 195°/61 kt 195°/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 368 190°/63 kt.190°/63 kt.

Given .tas = 270 kt true hdg = 270° actual wind 205° t /30kt .calculate the ?

Question 168-27 : 6r 259kt 6l 256kt 6r 251kt 8r 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 372 6r - 259kt6r - 259kt

Given .tas = 270 kt true hdg = 145° actual true wind = 205°/30kt .calculate ?

Question 168-28 : 6°l 256 kt 6°r 251 kt 8°r 261 kt 6°r 259 kt

Admin .under index set true track 145° centre dot on tas 270 kt with the rotative scale set wind . 1446.read the drift and the ground speed 6°l 256 kt exemple 376 6°l - 256 kt6°l - 256 kt

Given .tas = 470 kt true heading = 317° wind = 045° t /45 kt .calculate the ?

Question 168-29 : 5°l 470 kt 3°r 470 kt 5°l 475 kt 5°r 475 kt

Admin .under index set true heading 317° centre dot on tas 470 kt with the rotative scale set wind . 1447.read drift 5° left .ground speed is 470 kt exemple 380 5°l - 470 kt5°l - 470 kt

Given .tas = 190 kt .true heading = 085° .true wind = 110°/50kt .calculate ?

Question 168-30 : 8°l 146 kt 7°l 156 kt 4°l 168 kt 4°l 145 kt

Admin .under index set true heading 085° centre dot on tas 190 kt with the rotative scale set wind . 1452.read drift 8° left .ground speed is 147 kt .closest answer 8°l 146 kt exemple 384 8°l - 146 kt.8°l - 146 kt.

Given .tas = 132 kt true hdg = 257° true wind = 095°/35 kt .calculate the ?

Question 168-31 : 4°r 165 kt 2°r 166 kt 4°l 167 kt 3°l 166 kt

Admin .under index set true heading 257° centre dot on tas 132 kt set wind 095°/35kt on the rotative scale . 1453.you read 4 5°right and 164 kt exemple 388 4°r - 165 kt.4°r - 165 kt.

Given .tas = 370 kt true heading = 181° wind = 095°/35 kt.calculate true ?

Question 168-32 : 186° 370 kt 176° 370 kt 192° 370 kt 189° 370 kt

Admin .center dot on tas 370 kt .true heading 181° under index. 1454.put wind direction under the red compass rose under 35 kt your drift is 5° right giving a track of 186° and a groundspeed under the wind mark of 370 kt exemple 392 186° - 370 kt.186° - 370 kt.

Given .tas = 125 kt true heading = 355° true wind = 320°/30 kt .calculate the ?

Question 168-33 : 005° 102 kt 345° 100 kt 348° 102 kt 002° 98 kt

Admin .center dot on tas 125 kt .true heading 355° under index.put wind direction under the red compass rose under 30 kt your drift is 10° right giving a track of 005° and a groundspeed under the wind mark of 102 kt . 1457.cmarzocchini .all this questions doesnt work with cr3 wich is used for us in spain be carefull about this ie in this question just in this one the correct answer using cr3 is 003/95 i only did this comment in this question but take it into account for the reminder questions . .feel free to download the cr3 instructions here .http //www jeppesen com/download/misc/crinstructions pdf. page 44 45 true course track and ground speed .on top set speed '125 kt' over 'tas' index .over 'tc' index set '355°' .locate the wind dot by finding the 320° line on the green scale and making the point where this line intersects the green 30 kt circle .reading directly up from the wind dot we see that there is a left crosswind component of 17 kt .looking at the outer scale find 17 and opposite it read 8° crab angle . 2486.since the wind is from the left the true heading must be left of the true course therefore rotate the top disc 8° to the left counter clockwise now the 'tc' index points to 003° .looking directly above the wind dot after the above move you now find that the crosswind component has changed to 22 kt instead of 17 kt .locate 22 on the outer scale and find opposite crab angle of 10° . 2487.it now appears that the first crab angle of 8° was 2° too less therefore add 2° of the previous tc adjustment making a true course reading of 005° .you can read a headwind of 23kt therefore 125 kt 23 kt = 102 kt ground speed exemple 396 005° - 102 kt.005° - 102 kt.

Given .tas = 225 kt .hdg °t = 123° .w/v = 090/60kt .calculate the track °t ?

Question 168-34 : 134° 178 kt 134° 188 kt 120° 190 kt 128° 180 kt

Admin .put 225 kt in center dot under true index set wind direction 090° mark wind on centre line at 165 kt 60 kt below centre dot . 2488.rotate to put heading 123° under true index . 2489.you read a drift of 11° right and a groundspeed of 178 kt .add the drift to your heading to find the true track 134° exemple 400 134° - 178 kt.134° - 178 kt.

Given .tas = 480 kt true heading = 040° wind = 090°/60 kt .calculate true ?

Question 168-35 : 034° 445 kt 028° 415 kt 032° 425 kt 036° 435 kt

Admin .center dot on tas 480 kt .true heading 040° under index. 2026.put wind direction under the red compass rose under 60 kt your drift is 6° left giving a track of 034° and a groundspeed under the wind mark of 445 kt exemple 404 034° - 445 kt.034° - 445 kt.

Given .tas = 170 kt.true heading = 100°.wind = 350/30kt .calculate the true ?

Question 168-36 : 109° 182 kt 091° 183 kt 103° 178 kt 098° 178 kt

..center dot on tas 170 kt .true heading 100° under index.put wind direction under the red compass rose under 30 kt your drift is 9° right giving a track of 109° and a groundspeed under the wind mark of 182 kt . /com en/com061 168 jpg.the answer is 109° and 182 kt exemple 408 109° - 182 kt.109° - 182 kt.

Given . tas = 235 kt hdg t = 076° w/v = 040/40kt .calculate the drift angle ?

Question 168-37 : 7r 204 kt 7l 269 kt 5l 255 kt 5r 207 kt

..under index set true heading 076° centre dot on tas 235 kt with the rotative scale set wind . /com en/com061 169 jpg.read drift 7° right .ground speed is 205 kt close enough to answer 204 kt exemple 412 7r - 204 kt.7r - 204 kt.

Given .tas = 440 kt true heading = 349° wind = 040/40kt .calculate drift and ?

Question 168-38 : 4l 415 kt 2l 420 kt 6l 395 kt 5l 385 kt

Admin .under index set true heading 349° centre dot on tas 440 kt with the rotative scale set wind 040/40kt .read drift 4° left .ground speed is 415 kt exemple 416 4l - 415 kt4l - 415 kt

Given tas = 95 kt hdg t = 075° w/v = 310/20kt calculate the drift and gs ?

Question 168-39 : 9r 108 kt 10l 104 kt 9l 105 kt 8r 104 kt

exemple 420 9r - 108 kt9r - 108 kt

Given .tas = 230 kt hdg t = 250° wind = 205/10kt .calculate the drift and gs ?

Question 168-40 : 2r 223 kt 2l 224 kt 1l 225 kt 1r 221 kt

Admin .under index set true heading 250° centre dot on tas 230 kt with the rotative scale set wind . 2490.read drift 2° right .ground speed is 223 kt exemple 424 2r - 223 kt.2r - 223 kt.

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