Question 198-1 : The accuracy of the manually calculated dead reckoning position of an aircraft is among other things affected by ? [ Level reports ]

Question 198-2 : The accuracy of the manually calculated dead reckoning position of an aircraft is among other things affected by ?

The flight time since the last position update.

Question 198-3 : What may cause a difference between a dead rekoning position and a fix ?

The difference between the actual wind and the forecasted wind.

Question 198-4 : Cas is 320 ktflight level 330oat isa +15°ctas is approximately compressibility factor 0939 ?

530 kt.

Oat is 15°c 15°c 2 x 33 = 36°con computer in airspeed window set press alt '33' in front of coat °c ' 36°c' on the outer scale in front of cas 320 kt you can read tas 565 kttrue air speed tas is obtained from calibrated air speed cas by correcting for compressibility and density 568 x 0939 = 533 kt

Question 198-5 : Given mach numer 08flight level 330oat isa +15°ctas is approximately compressibility factor 094 ?

480 kt.

Temperature at fl330 = 51°c 33°c x 2 + 15 isa +15°c so 51°c + 15°c = 36°ctas = m*lsslss = 3895 x sqrtt°a t°a =273 36= 237°k lss = 3895 x sqrt237 = 59963tas = 08 x 59963 = 480 ktyou need to apply compressibility factor if you want to go from cas to tas not from mach to tastrue air speed tas is obtained from calibrated air speed cas by correcting for compressibility and density

Question 198-6 : The main purpose of dr dead reckoning is ?

To obtain with reasonable accuracy the aircraft's position between fixes or in the absence of fixes.

Question 198-7 : An aircraft is flying at fl390 at a speed of mach 0821oat isa 4°cthe compressibility factor is 0942calculate the tas ?

467 kt.

Isa temperature at fl390 = 565°c 565°c is considered to be the lowest isa temperature isa 4°c so oat = 605°ctas = m*lsslss = 39 x sqrtt°a t°a =273 605= 2125°k lss = 39 x sqrt2125 = 5685tas = 0821 x 5685 = 4667 ktyou need to apply compressibility factor if you want to go from cas to tas not from mach to tastrue air speed tas is obtained from calibrated air speed cas by correcting for compressibility and density

Question 198-8 : An aircraft descends from fl240 to fl80 for the final approach track = 070°cas = 220 ktoat = isa 10°cthe average tas in the descent is ?

276 kt.

At the exam average tas used for descent problems is calculated at the altitude 12 of the descent altitude at fl160 isa temperature = 15°c 2°c x 16 = 17°coat is isa 10°c thus oat is 27°c at fl160on the computer in airspeed window put 27ºc next to fl160 go to cas 220 kt on inner scale and read tas on outer scale 276 kt

Question 198-9 : An aircraft is flying at fl 350 with cas = 300 ktoat = isa + 4°cthe compressibility factor is 0939calculate the tas ?

509 kt.

Isa temperature at fl350 = 15°c + 35 x 2 = 55°cisa +4°c so oat = 51°con the computer in airspeed window put 51ºc next to fl350 go to cas 300 kt on inner scale and read tas on outer scale 542 ktmultiply 542 kt x 0939 = 509 ktyou need to apply compressibility factor if you want to go from cas to tas not from mach to tastrue air speed tas is obtained from calibrated air speed cas by correcting for compressibility and density

Question 198-10 : Given track = 355°tas = 190 ktwind 270°25 ktafter 30 minutes of flying with the planned tas and true heading the aircraft is 35 nm right of track and 45 nm ahead of the dead reckoning positioncalculate the actual wind ?

254°34 kt.

Ecqb03 august 2016

Question 198-11 : Given fl 400oat = 65°cias = 260 ktinstrument and position error to be neglectedcompressibility factor = 0935calculate the true air speed taking compressibility into account ?

479 kt.

Calibrated airspeed cas is indicated airspeed ias corrected for instrument error and position error the question states instrument and position error to be neglected therefore ias = casoat = 65°con the computer in airspeed window put 65ºc next to fl400 go to cas 260 kt on inner scale and read tas on outer scale 513 kttrue air speed tas is obtained from calibrated air speed cas by correcting for compressibility and density 513 x 0935 = 479 kt

Question 198-12 : Given tas = 210 ktcas = 190 ktpressure altitude = 9000 ftcalculate mach number ?

034.

Using the computer align tas external ring & cas internal ring when done go to your airspeed case and read the one corresponding to the pressure altitude above the 9000 ft line you should read about 22°cmach number = tas lsslss = 39*sqrt t in k° here t° = 22°c = 22 + 273 = 251°khence lss = 39*sqrt 251 = 617876thus mach number = 210617876 = 0339 = 034

Question 198-13 : A dr position is to be found ?

On the desired track.

Question 198-14 : Which of the factors named hereafter should be considered by the pilot when selecting landmarks as visual reference points 1 possibility of identification2 transmitted frequency3 visibility4 closeness to the trackthe combination that regroups all of the correct statements is ?

1 3 4.

Question 198-15 : Given fl 300oat = 45°cias = 260 ktinstrument and position error to be neglectedcompressibility factor = 096calculate the true air speed taking compressibility into account ?

408 kt.

Question 198-16 : On a mercator chart one minute on n55° parallel is 31 mmthe map scale at 40°n is ?

1 457 650.

Question 198-17 : The nominal scale of a north stereopolar map is ?

At the north pole.

Question 198-18 : Given tas = 140 kt true hdg = 302° wv = 045° t 45ktcalculate the drift angle and gs ?

16°l 156 kt.

Under index set true heading 302° centre dot on tas 140 kt with the rotative scale set wind 1448read drift 16° leftground speed is 156 kt

Question 198-19 : Given tas = 290 kttrue hdg = 171°wv = 310° t 30ktcalculate the drift angle and gs ?

4°l 314 kt.

2528

Question 198-20 : Given tas = 485 kttrue heading = 226°true wind = 110°95ktcalculate the drift angle and gs ?

9°r 533 kt.

Under index set true heading 226° centre dot on tas 485 kt with the rotative scale set wind 1451read drift 9° rightground speed is 533 kt

Question 198-21 : Given tas = 472 kttrue heading = 005°true wind = 110°50ktcalculate the drift angle and gs ?

6°l490 kt.

Under index set true heading 005° centre dot on tas 472 kt with the rotative scale set wind 1740read drift 55° leftground speed is 487 ktclosest answer 6°l490 kt

Question 198-22 : Given tas = 375 kt true heading = 124° wind = 130°55 ktcalculate the true track and gs ?

123° 320 kt.

1741

Question 198-23 : Given tas = 198 kthdg °t = 180wv = 35925calculate the track °t and gs ?

180° 223 kt.

Question 198-24 : Given tas = 135 kt true heading = 278° true wind = 140°20 ktcalculate the true track and ground speed ?

283° 150 kt.

Center dot on tas 135 kttrue heading 278° under indexput 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 198-25 : Given tas = 155 kttrue heading = 216°wind = 090°60 ktcalculate the true track and gs ?

231° 196 kt.

Center dot on tas 155 kttrue heading 216° under indexput wind direction under the red compass rose under 60 kt your drift is 145° right giving a track of 2305° and a groundspeed under the wind mark of 195 kt 2527the closest answer is 231° and 196 kt

Question 198-26 : Given tas = 465 kt true heading = 124° wind = 170°80 ktcalculate drift and ground speed ?

8l 415 kt.

Under index set true heading 124° centre dot on tas 465 kt with the rotative scale set wind com encom061 171jpgread drift 8° leftground speed is 415 kt

Question 198-27 : Given tas = 140 kt hdg t = 005° wv = 26525kt 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 1716read drift 10° rightground speed is 146 kt

Question 198-28 : Given tas = 190 kt hdg t = 355° wv = 16525ktcalculate 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 encom061 174jpgread drift 1° leftground speed is 214 kt close enough for the answer

Question 198-29 : Given tas = 250 kthdg t = 029°wv = 03545ktcalculate 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 2526read drift 1° leftground speed is 205 kt

Question 198-30 : Given tas = 485 kt true heading = 168° wind = 13075 ktcalculate true track and ground speed ?

174° 428 kt.

Under index set true heading 168° centre dot on tas 485 kt with the rotative scale set wind com encom061 180jpgread drift 6° right 168° + 6° = 174° ground speed is 430 kt close enough for the answer

Question 198-31 : Given tas = 130 kttrack t = 003°wv = 19040 ktcalculate the hdg °t and gs ?

001° 170 kt.

Question 198-32 : Given tas = 227 kt track t = 316° wv = 20515ktcalculate 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 2021now 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 198-33 : Given tas = 200 kt track t = 073° wv = 21020kt calculate the hdg °t and gs ?

077 214 kt.

Question 198-34 : Given tas = 270 kt track t = 260° wind = 275°30ktcalculate 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 encom061 187jpgnow 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 198-35 : Given true hdg = 233° tas = 480 kt track t = 240° gs = 523 ktcalculate the wv ?

11075kt.

True heading is 233° true track is 240° our drift is 7° right com encom061 189jpgwind 111°77kt closest answer 11075kt

Question 198-36 : Given true heading = 074° tas = 230 kt true track = 066° ground speed = 242 ktcalculate the wind ?

18035 kt.

True heading is 074° true track is 066° our drift is 8° left com encom061 191jpgwhere the rotative scale crosses the ground speed arc 242 kt we read the wind 180° 35 kt

Question 198-37 : Given true heading = 054° tas = 450 kt true track = 059° gs = 416 ktcalculate the wind ?

010°50 kt.

True heading is 054° true track is 059° our drift is 5° right 2525wind 010°50 kt

Question 198-38 : Given true heading = 002°tas = 130 kttrue track = 353°ground speed = 132 ktcalculate the wind ?

095°20 kt.

True heading is 002° true track is 353° our drift is 9° left com encom061 195jpgwind 094°22 kt closest answer is 095°20 kt

Question 198-39 : Given gs = 236 kt distance from a to b = 354 nmwhat is the time from a to b ?

1 hr 30 min.

354 nm 236 kt60 min = 90 minutes 1h30

Question 198-40 : Given gs = 345 kt distance from a to b = 3560 nmwhat is the time from a to b ?

10 hr 19 min.

3560 nm 345 kt60 min = 619 minutes 10h19