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Formation > assignment : Given . crn vor n5318 1 w00856 5 dme 18 nm sha vor n5243 3 w00853 1 dme 30 nm . ?

Question 167-1 : N5310 w00830 n5252 w00923 n5307 w00923 n5355 w00825

exemple 267 n5310 w00830n5310 w00830

Given . con vor n5354 8 w00849 1 dme 30 nm crn vor n5318 1 w00856 5 dme 25 nm . ?

Question 167-2 : N5330 w00820 n5343 w00925 n5335 w00925 n5337 w00820

exemple 271 n5330 w00820n5330 w00820

Given . sha vor/dme n5243 3 w00853 1 birr aerodrome n5304 w00754 what is the ?

Question 167-3 : 068° 41 nm 248° 42 nm 060° 42 nm 240° 41 nm

exemple 275 068° - 41 nm068° - 41 nm

Given . con vor/dme n5354 8 w00849 1 castlebar aerodrome n5351 w00917 what is ?

Question 167-4 : 265° 17 nm 077° 18 nm 257° 17 nm 086° 18 nm

exemple 279 265° - 17 nm265° - 17 nm

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

Question 167-5 : 45°00'n 172°38'e 53°20'n 169°22w 45°00'n 169°22w 53°20'n 172°38'e

.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 283 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 167-6 : 2 h 30 1 h 15 1 h 45 5 h 00

.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 287 2 h 30.2 h 30.

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

Question 167-7 : 056° 064° 048° 072°

. 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 291 056°.056°.

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

Question 167-8 : 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 295 192 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 167-9 : 461 kt lss 576 kt 237 kt lss 296 kt 490 kt lss 461 kt 461 kt lss 296 kt

.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 299 461 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 167-10 : 435 kt 600 kt 535 kt 450 kt

.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 303 435 kt.435 kt.

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

Question 167-11 : 270° / 33 kt 360° / 33 kt 090° / 33 kt 180° / 33 kt

.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 307 270° / 33 kt.270° / 33 kt.

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

Question 167-12 : 220 kt 230 kt 180 kt 200 kt

. 1369 exemple 311 220 kt.220 kt.

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

Question 167-13 : 30 kt 24 kt 27 kt 21 kt

.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 315 30 kt.30 kt.

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

Question 167-14 : 221° 208° 211° 180°

.heading 311°..bearing +270°..total 581°..581° 360° = 221° magnetic bearing of the ndb measured from the aircraft exemple 319 221°.221°.

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

Question 167-15 : 190° 194° 204° 180°

. 1393.use this wonderful table for those questions exemple 323 190°.190°.

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

Question 167-16 : 23 5° 25 3° 27 5° 66 5°

. 2028 exemple 327 23.5°.23.5°.

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

Question 167-17 : 0 825 0 9 0 85 0 87

. 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 331 0.8250.825

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

Question 167-18 : 40°55'n 027°55'w 40°50'n 027°30'w 41°00'n 028°10'w 41°05'n 027°50'w

.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 335 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 167-19 : 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 339 13481348

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

Question 167-20 : 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 343 1685 nm.1685 nm.

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

Question 167-21 : 530 nm 455 nm 500 nm 470 nm

.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 347 530 nm.530 nm.

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

Question 167-22 : 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 351 gd = (ad x gs)/tasgd = (ad x gs)/tas

What is the isa temperature value at fl 330 ?

Question 167-23 : 51°c 56°c 66°c 81°c

15°c 2° x 33 = 51°c exemple 355 -51°c-51°c

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

Question 167-24 : 0 81 0 76 0 78 0 84

.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 359 0.810.81

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

Question 167-25 : 3 94 nm 2 36 nm 39 nm 3 25 nm

.135 kt /60 minutes = 2 25 nm/minute..1 minute 45 secondes = 1 75 minute..2 25 x 1 75 = 3 9375 nm exemple 363 3.94 nm.3.94 nm.

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

Question 167-26 : 50 min 100 min 90 min 80 min

.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 367 50 min.50 min.

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

Question 167-27 : 186 kt 290 kt 246 kt 250 kt

.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 371 186 kt.186 kt.

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

Question 167-28 : 0 40 0 42 0 44 0 39

. 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 375 0.400.40

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

Question 167-29 : Groundspeed drift position track

exemple 379 groundspeed.groundspeed.

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

Question 167-30 : 190°/63 kt 355°/15 kt 195°/61 kt 195°/63 kt

.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 383 190°/63 kt.190°/63 kt.

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

Question 167-31 : 6r 259kt 6l 256kt 6r 251kt 8r 259kt

.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 387 6r - 259kt6r - 259kt

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

Question 167-32 : 6°l 256 kt 6°r 251 kt 8°r 261 kt 6°r 259 kt

.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 391 6°l - 256 kt6°l - 256 kt

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

Question 167-33 : 5°l 470 kt 3°r 470 kt 5°l 475 kt 5°r 475 kt

.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 395 5°l - 470 kt5°l - 470 kt

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

Question 167-34 : 8°l 146 kt 7°l 156 kt 4°l 168 kt 4°l 145 kt

.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 399 8°l - 146 kt.8°l - 146 kt.

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

Question 167-35 : 4°r 165 kt 2°r 166 kt 4°l 167 kt 3°l 166 kt

.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 403 4°r - 165 kt.4°r - 165 kt.

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

Question 167-36 : 186° 370 kt 176° 370 kt 192° 370 kt 189° 370 kt

.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 407 186° - 370 kt.186° - 370 kt.

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

Question 167-37 : 005° 102 kt 345° 100 kt 348° 102 kt 002° 98 kt

.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 411 005° - 102 kt.005° - 102 kt.

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

Question 167-38 : 134° 178 kt 134° 188 kt 120° 190 kt 128° 180 kt

.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 415 134° - 178 kt.134° - 178 kt.

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

Question 167-39 : 034° 445 kt 028° 415 kt 032° 425 kt 036° 435 kt

.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 419 034° - 445 kt.034° - 445 kt.

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

Question 167-40 : 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 423 109° - 182 kt.109° - 182 kt.

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