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Question 198-1 : The accuracy of the, manually calculated, dead reckoning position of an aircraft is, among other things, affected by ? [ Level reports ]
The flight time since the last position update.
Question 198-2 : What may cause a difference between a dead rekoning position and a fix ?
The difference between the actual wind and the forecasted wind.
Question 198-3 : Cas is 320 kt.flight level 330.oat isa +15°c.tas is approximately. compressibility factor 0.939 ?
530 kt.
.oat is 15°c 15°c 2 x 33 = 36°c...on 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 kt...true air speed tas is obtained from calibrated air speed cas by correcting for compressibility and density.568 x 0.939 = 533 kt.
Question 198-4 : Given.mach numer 0.8.flight level 330.oat isa +15°c..tas is approximately. compressibility factor 0.94 ?
480 kt.
.temperature at fl330 = 51°c 33°c x 2 + 15..isa +15°c, so 51°c + 15°c = 36°c..tas = m*lss..lss = 38,95 x sqrtt°a t°a =273 36= 237°k..lss = 38,95 x sqrt237 = 599,63..tas = 0,8 x 599,63 = 480 kt...you need to apply compressibility factor if you want to go from cas to tas, not from mach to tas..true air speed tas is obtained from calibrated air speed cas by correcting for compressibility and density.
Question 198-5 : 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 198-6 : An aircraft is flying at fl390 at a speed of mach 0.821..oat isa 4°c.the compressibility factor is 0.942..calculate the tas ?
467 kt.
.isa temperature at fl390 = 56.5°c 56.5°c is considered to be the lowest isa temperature..isa 4°c, so oat = 60.5°c..tas = m*lss..lss = 39 x sqrtt°a t°a =273 60.5= 212.5°k..lss = 39 x sqrt212.5 = 568.5..tas = 0,821 x 568.5 = 466.7 kt...you need to apply compressibility factor if you want to go from cas to tas, not from mach to tas..true air speed tas is obtained from calibrated air speed cas by correcting for compressibility and density.
Question 198-7 : An aircraft descends from fl240 to fl80 for the final approach.track = 070°.cas = 220 kt.oat = isa 10°c..the average tas in the descent is ?
276 kt.
.at the exam, average tas used for descent problems is calculated at the altitude 1/2 of the descent altitude.at fl160, isa temperature = 15°c 2°c x 16 = 17°c...oat is isa 10°c, thus oat is 27°c at fl160...on 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 198-8 : An aircraft is flying at fl 350 with cas = 300 kt..oat = isa + 4°c..the compressibility factor is 0.939..calculate the tas ?
509 kt.
.isa temperature at fl350 = 15°c + 35 x 2 = 55°c..isa +4°c, so oat = 51°c..on 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 kt...multiply 542 kt x 0.939 = 509 kt...you need to apply compressibility factor if you want to go from cas to tas, not from mach to tas..true air speed tas is obtained from calibrated air speed cas by correcting for compressibility and density.
Question 198-9 : Given.track = 355°.tas = 190 kt.wind 270°/25 kt.after 30 minutes of flying with the planned tas and true heading, the aircraft is 3.5 nm right of track and 4.5 nm ahead of the dead reckoning position..calculate the actual wind ?
254°/34 kt.
Ecqb03, august 2016
Question 198-10 : Given.fl 400.oat = 65°c.ias = 260 kt.instrument and position error to be neglected..compressibility factor = 0.935.calculate 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 = cas..oat = 65°c..on 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 kt...true air speed tas is obtained from calibrated air speed cas by correcting for compressibility and density.513 x 0.935 = 479 kt.
.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°c...mach number = tas / lss.lss = 39*sqrt t in k°..here, t° = 22°c = 22 + 273 = 251°k..hence, lss = 39*sqrt 251 = 617.876.thus, mach number = 210/617.876 = 0.339 = 0.34.
Question 198-12 : A dr position is to be found ?
On the desired track.
Question 198-13 : Which of the factors named hereafter should be considered by the pilot when selecting landmarks as visual reference points.1 possibility of identification.2 transmitted frequency.3 visibility.4 closeness to the track..the combination that regroups all of the correct statements is ?
1, 3, 4.
Question 198-14 : Given.fl 300.oat = 45°c.ias = 260 kt.instrument and position error to be neglected.compressibility factor = 0.96.calculate the true air speed taking compressibility into account ?
408 kt.
Question 198-15 : On a mercator chart, one minute on n55° parallel is 3.1 mm...the map scale at 40°n is ?
1 457 650
Question 198-16 : The nominal scale of a north stereopolar map is ?
At the north pole.
Question 198-17 : Given.tas = 140 kt, true hdg = 302°, w/v = 045° t /45kt..calculate 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. 1448.read drift 16° left..ground speed is 156 kt.
Question 198-18 : Given.tas = 290 kt.true hdg = 171°.w/v = 310° t /30kt..calculate the drift angle and gs ?
4°l 314 kt
. 2528
Question 198-19 : Given.tas = 485 kt,.true heading = 226°,.true wind = 110°/95kt..calculate 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. 1451.read drift 9° right..ground speed is 533 kt.
Question 198-20 : Given.tas = 472 kt,.true heading = 005°,.true wind = 110°/50kt..calculate the drift angle and gs ?
6°l/490 kt.
.under index, set true heading 005°, centre dot on tas, 472 kt, with the rotative scale, set wind. 1740.read drift 5.5° left..ground speed is 487 kt..closest answer 6°l/490 kt.
.center dot on tas 135 kt..true heading 278° under index.put 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 198-24 : Given.tas = 155 kt.true heading = 216°.wind = 090°/60 kt.calculate the true track and gs ?
231° 196 kt.
.center dot on tas 155 kt..true heading 216° under index.put wind direction under the red compass rose, under 60 kt, your drift is 14.5° right, giving a track of 230.5° and a groundspeed under the wind mark of 195 kt.. 2527.the closest answer is 231° and 196 kt.
...under index, set true heading 124°, centre dot on tas, 465 kt, with the rotative scale, set wind.. /com en/com061 171.jpg..read drift 8° left..ground speed is 415 kt.
Question 198-26 : Given tas = 140 kt, hdg t = 005°, w/v = 265/25kt. 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. 1716.read drift 10° right..ground speed is 146 kt.
Question 198-27 : Given.tas = 190 kt, hdg t = 355°, w/v = 165/25kt..calculate 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 en/com061 174.jpg..read drift 1° left..ground speed is 214 kt close enough for the answer.
Question 198-28 : Given.tas = 250 kt.hdg t = 029°.w/v = 035/45kt.calculate 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. 2526.read drift 1° left..ground speed is 205 kt.
...under index, set true heading 168°, centre dot on tas, 485 kt, with the rotative scale, set wind. /com en/com061 180.jpg..read drift 6° right, 168° + 6° = 174°..ground speed is 430 kt close enough for the answer.
Question 198-30 : Given.tas = 130 kt.track t = 003°.w/v = 190/40 kt..calculate the hdg °t and gs ?
001° 170 kt.
Question 198-31 : Given.tas = 227 kt, track t = 316°, w/v = 205/15kt..calculate 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. 2021.now, 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 198-32 : Given tas = 200 kt, track t = 073°, w/v = 210/20kt. calculate the hdg °t and gs ?
077 214 kt
Question 198-33 : Given.tas = 270 kt, track t = 260°, wind = 275°/30kt..calculate 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 en/com061 187.jpg..now, 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 198-34 : Given.true hdg = 233°, tas = 480 kt, track t = 240°, gs = 523 kt..calculate the w/v ?
110/75kt.
...true heading is 233°, true track is 240° our drift is 7° right... /com en/com061 189.jpg..wind 111°/77kt closest answer 110/75kt.
...true heading is 074°, true track is 066° our drift is 8° left.. /com en/com061 191.jpg.where the rotative scale crosses the ground speed arc 242 kt , we read the wind 180°/ 35 kt.
