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An aircraft has to fly over a mountain ridge .the highest obstacle indicated in ? [ Multiple protocol ]

Question 172-1 : 11 700 ft 11 100 ft 12 000 ft 11 200 ft

.9800 ft + 2000 ft = 11800 ft .we have to correct the temperature above the qfe datum .you can use either the computer or with the following rule of thumb called the '4% rule' .the altitude/height changes by 4% for each 10°c temperature deviation from isa . /com en/com061 613 jpg..deviation from isa = +5°c.4% x 5 6 x 5 = 112 ft..11800 ft 112 ft = 11688 ft .it is 'minus' 112 ft because air is hotter than standard true altitude is higher than indicated altitude exemple 272 11 700 ft.11 700 ft.

An aircraft is flying from a to b a distance of 50 nm .the true course in the ?

Question 172-2 : 17° 12° 5° 14°

Admin . 1798.with the forecasted wind we will fly at 130 kt ground speed .at 130 kt and 15 minutes of flight we will be at 32 5 nm from a .the question states 2 5 nm ahead of the dead reckoning position so we are at 35 nm from a .use the one in sixty rule .track error angle from a = 3 nm x 60 / 35 nm = 5° . it's the drift to applied in order to correct the wind .track error angle to join b from our current position = 3 nm x 60 / 15 nm = 12° .to reach destination b from this position the correction angle on the heading should be 5° + 12° = 17° exemple 276 17°.17°.

An aircraft is flying from a to b a distance of 50 nm .the true course in the ?

Question 172-3 : 17° 12° 5° 10°

.draw the exercice . /com en/com061 623 jpg.without wind at 120 kt and 15 minutes of flight we are at 32 5 nm from a .the question states 2 5 nm ahead of the dead reckoning position so we are at 35 nm from a .use the one in sixty rule .track error angle from a = 3 nm x 60 / 35 nm = 5° . it's the drift to applied in order to correct the wind .track error angle to join b from our current position = 3 nm x 60 / 15 nm = 12° .to reach destination b from this position the correction angle on the heading should be 5° + 12° = 17° exemple 280 17°.17°.

An aircraft is flying from a to b .the true course according to the flight log ?

Question 172-4 : 5°r 12°r 17°l 6°l

Frist step find the ground speed .place centre dot on 120 kt tas .place 225° wind direction under true index .make a wind mark on centre line 15 kt below centre dot at 105 kt .rotate to set 090° true track under true index .wind mark has moved to 5° left drift .rotate to lined up 090° with 5° left drift .wind mark has stayed at 5° left drift you find .true heading 095° .ground speed 130 kt . 130/60 x 15 min = 32 5 nm.actual aircraft position is 2 5 nm ahead dead reckoning position at 32 5 + 2 5 = 35 nm .track error angle = distance off track x 60 / distance along track.track error angle = 3 x 60 / 35.track error angle = 180 / 35 = 5° exemple 284 5°r5°r

On a true heading of 090° the aircraft experiences drift of 5°right on a true ?

Question 172-5 : 360°/17 kt 180°/17 kt 360°/34 kt 180°/34 kt

..since on a true heading of 180° there is no drift the wind is coming from 180° or 360° .under index set true heading 090° centre dot on tas 200 kt with the rotative scale set the 5°right drift . /com en/com061 627 jpg.read the wind 360°/17 kt exemple 288 360°/17 kt.360°/17 kt.

An aircraft is flying from salco to berry head on magnetic track 007° tas 445 ?

Question 172-6 : 21 nm 24 nm 23 nm 22 nm

.first step find the aircraft magnetic heading . /com en/com061 631 jpg.calculate the drift between our true track 002° and the true wind 050°/40 kt with your computer the drift is 4° left you have to apply a 4°right wind angle correction and also a ground speed of 415 kt ..from the aircraft at 1000 utc the locator is at 60° to the left 011° to 311° = 60° .from the aircraft at 1003 utc the locator is at 105° to the left 011° to 266° = 105° ..we have an isosceles triangle and in an isosceles triangle two sides are equal in length .in 45° in 3 minutes .415 kt/60 = 6 92 nm/min.3 min x 6 92 = 20 76 nm exemple 292 21 nm.21 nm.

An aircraft is flying at fl250 oat = 45°c the qnh given by a station at msl is ?

Question 172-7 : 23400 ft 24000 ft 25500 ft 26100 ft

.you have to turn your altimeter subscale setting knob counterclockwise from 1013 2 to 993 2 .indicated altitude will be decreased by 20 hpa x 30 ft = 600 ft .25000 ft 600 ft = 24400 ft.you can use either the computer or with the following rule of thumb called the '4% rule' .the altitude/height changes by 4% for each 10°c temperature deviation from isa .deviation from isa = 15° 2 x 25 = 35°c . 35°c to 45°c = 10°c we are in isa 10°c .0 04 x 24400 x 1 = 976 ft .true altitude = 24400 976 = 23424 ft .keep in mind that air is colder than standard thus the air column is contracted our true altitude is lower than our indicated altitude exemple 296 23400 ft.23400 ft.

A vor is situated at position n55°26' w005°42' .the variation at the vor is ?

Question 172-8 : 294° 276° 278° 296°

. /com en/com061 638 jpg.first step apply convergency .convergency = difference of longitude x sin mean latitude .convergency = 10°w 5°42'w x sin 60 +55°26' /2 .convergency = 4 35° x sin 57 5 = 3 67°.second step find true track vor .true track at vor = 101 5° + convergency = 101 5 + 3 67° = around 105° t .third step we must apply variation at the vor .105 + 9°w = 114° magnetic .last step we are looking for a radial .114° + 180° = 294° exemple 300 294°.294°.

An aircraft tracks radial 200° inbound to a vor station with a magnetic ?

Question 172-9 : 320°/50 kt 310°/60 kt 300°/50 kt 330°/50 kt

Admin .following radial 200° inbound magnetic track 020° with a magnetic heading of 010° we have a 10° right drift .our true heading is 005° we have to applied the 5°west magnetic variation since it is a vor .after having overfly the vor we fly outbound on radial 090° magnetic track 090° our magnetic heading is 080° so our true heading is 075° .on the computer set 240 kt under the center dot and 005° below true index draw a line along the right 10° drift line .rotate and put 075° below true index draw a line along the right 10° drift line .rotate to bring back the intersection of the lines under the central wind line . 1788 exemple 304 320°/50 kt.320°/50 kt.

The fix of the aircraft position is determined by radials from three vor ?

Question 172-10 : 1 4 2 3

. /com en/com061 644 jpg.point 1 is always located on the right side of the radials exemple 308 11

An aircraft flies at fl 250 with an oat of 45°c the qnh given by a ?

Question 172-11 : 4 200 ft 4 600 ft 3 500 ft 3 000 ft

Admin .we need to calculate our true altitude .first you have to turn your altimeter subscale setting knob clockwise from 1013 to 1033 .indicated altitude will be increased by 20 hpa x 30 ft = 600 ft .25000 + 600 = 25600 ft .next step we must correct for temperature .outside temperature is 45°c at fl250 .isa at fl250 is 15°c 25 x 2°c = 35°c .we are in isa 10°c .you can use either the computer or with the following rule of thumb called the '4% rule' .the altitude/height changes by 4% for each 10°c temperature deviation from isa .an altimeter set to airport qnh will read correctly when on the ground at the airport irrespective of temperature .any temperature error therefore occurs due to non isa temperature in the layer of atmosphere between airport elevation and aircraft in flight .therefore 25600 2830 = 22770 ft.22770 x 0 04 x 1 = 910 ft .our true altitude is 25600 910 = 24690 ft .clearance above the mountain ridge is 24690 20410 = 4280 ft exemple 312 4 200 ft.4 200 ft.

Given .an aircraft is flying at fl100 oat = isa 15°c .the qnh given by a ?

Question 172-12 : 9900 ft 9400 ft 10600 ft 11200 ft

Admin .you have to turn your altimeter subscale setting knob clockwise from 1013 to 1032 .indicated altitude will be increased by 19 hpa x 27 ft = 513 ft .10000 + 513 = 10513 ft .now we must correct for temperature .we are in isa 15°c .you can use either the computer or with the following rule of thumb called the '4% rule' .the altitude/height changes by 4% for each 10°c temperature deviation from isa .an altimeter set to airport qnh will read correctly when on the ground at the airport irrespective of temperature .any temperature error therefore occurs due to non isa temperature in the layer of atmosphere between airport elevation and aircraft in flight .therefore 10513 100 = 10613 ft.10613 x 0 04 x 1 5 = 637 ft .10513 637 = 9876 ft closest answer is 9900 ft exemple 316 9900 ft.9900 ft.

The accuracy of the manually calculated dead reckoning position of an aircraft ?

Question 172-13 : The accuracy of the forecasted wind the accuracy of the actual wind the accuracy of the adjustment of the position lines for the motion of the aircraft between the last fix and the dr position the accuracy of the adjustment of the position lines for the motion of the aircraft between the last and the new dr position

Admin .dead reckoning is the process of calculating one's current position by using a previously determined position or fix and advancing that position based upon known or estimated speeds over elapsed time and course you have to take the wind into account and the more accurate the wind information is the more accurate the manually calculated position will be exemple 320 The accuracy of the forecasted wind.The accuracy of the forecasted wind.

The accuracy of the manually calculated dead reckoning position of an aircraft ?

Question 172-14 : The flight time since the last position update the accuracy of the actual wind the accuracy of the adjustment of the position lines for the motion of the aircraft between the last fix and the dr position the accuracy of the adjustment of the position lines for the motion of the aircraft between the last and the new dr position

Admin .dead reckoning is the process of estimating one's current position based upon a previously determined position or fix and advancing that position based upon known or estimated speeds over elapsed time and course therefore the accuracy is among other things affected by the flight time since the last position update exemple 324 The flight time since the last position update.The flight time since the last position update.

What may cause a difference between a dead rekoning position and a fix ?

Question 172-15 : The difference between the actual wind and the forecasted wind the difference between no wind and the actual wind the difference between no wind and the forecasted wind the difference between the magnetic and the true wind direction

exemple 328 The difference between the actual wind and the forecasted wind.The difference between the actual wind and the forecasted wind.

Cas is 320 kt.flight level 330.oat isa +15°c.tas is approximately . ?

Question 172-16 : 530 kt 560 kt 265 kt 340 kt

Admin .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 exemple 332 530 kt.530 kt.

Given .mach numer 0 8.flight level 330.oat isa +15°c .tas is approximately . ?

Question 172-17 : 480 kt 420 kt 450 kt 265 kt

Admin .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 exemple 336 480 kt.480 kt.

The main purpose of dr dead reckoning is ?

Question 172-18 : To obtain with reasonable accuracy the aircraft's position between fixes or in the absence of fixes to monitor an inertial navigation system to obtain without reasonable accuracy the aircraft's position between fixes to improve gps accuracy

exemple 340 To obtain with reasonable accuracy, the aircraft's position between fixes or in the absence of fixes.To obtain with reasonable accuracy, the aircraft's position between fixes or in the absence of fixes.

An aircraft is flying at fl390 at a speed of mach 0 821 .oat isa 4°c.the ?

Question 172-19 : 467 kt 439 kt 459 kt 433 kt

Admin .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 exemple 344 467 kt.467 kt.

An aircraft descends from fl240 to fl80 for the final approach .track = ?

Question 172-20 : 276 kt 288 kt 268 kt 282 kt

Admin .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 exemple 348 276 kt.276 kt.

An aircraft is flying at fl 350 with cas = 300 kt .oat = isa + 4°c .the ?

Question 172-21 : 509 kt 540 kt 535 kt 479 kt

Admin .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 exemple 352 509 kt.509 kt.

Given .track = 355°.tas = 190 kt.wind 270°/25 kt.after 30 minutes of flying ?

Question 172-22 : 254°/34 kt 251°/21 kt 246°/21 kt 248°/21 kt

Ecqb03 august 2016 exemple 356 254°/34 kt.254°/34 kt.

Given .fl 400.oat = 65°c.ias = 260 kt.instrument and position error to be ?

Question 172-23 : 479 kt 470 kt 512 kt 533 kt

Admin .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 exemple 360 479 kt.479 kt.

Given .tas = 210 kt.cas = 190 kt.pressure altitude = 9000 ft.calculate mach ?

Question 172-24 : 0 34 0 54 0 62 0 44

Admin .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 exemple 364 0.34.0.34.

A dr position is to be found ?

Question 172-25 : On the desired track closeness to the destination perpendicular to the desired track at a distance not less than 50 km perpendicular to the desired track at a distance not less than 100 km

exemple 368 On the desired track.On the desired track.

Which of the factors named hereafter should be considered by the pilot when ?

Question 172-26 : 1 3 4 1 2 3 2 3 4

exemple 372 1, 3, 4.1, 3, 4.

Given .fl 300.oat = 45°c.ias = 260 kt.instrument and position error to be ?

Question 172-27 : 408 kt 400 kt 435 kt 424 kt

exemple 376 408 kt.408 kt.

On a mercator chart one minute on n55° parallel is 3 1 mm .the map scale at ?

Question 172-28 : 1 457 650 1 779 880 1 447 320 1 797 890

exemple 380 1 : 457 6501 : 457 650

The nominal scale of a north stereopolar map is ?

Question 172-29 : At the north pole at the south pole at the equator the 45°north parallel

exemple 384 At the north pole.At the north pole.

Given .tas = 140 kt true hdg = 302° w/v = 045° t /45kt .calculate the drift ?

Question 172-30 : 16°l 156 kt 9°r 143 kt 9°l 146 kt 18°r 146 kt

Admin .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 exemple 388 16°l - 156 kt.16°l - 156 kt.

Given .tas = 290 kt.true hdg = 171°.w/v = 310° t /30kt .calculate the drift ?

Question 172-31 : 4°l 314 kt 4°r 310 kt 4°r 314 kt 4°l 310 kt

Admin . 2528 exemple 392 4°l - 314 kt4°l - 314 kt

Given .tas = 485 kt .true heading = 226° .true wind = 110°/95kt .calculate ?

Question 172-32 : 9°r 533 kt 7°r 531 kt 9°r 433 kt 8°l 435 kt

Admin .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 exemple 396 9°r - 533 kt.9°r - 533 kt.

Given .tas = 472 kt .true heading = 005° .true wind = 110°/50kt .calculate ?

Question 172-33 : 6°l/490 kt 7°r/491 kt 7°l/487 kt 7°r/491 kt

Admin .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 exemple 400 6°l/490 kt.6°l/490 kt.

Given .tas = 375 kt true heading = 124° wind = 130°/55 kt .calculate the true ?

Question 172-34 : 123° 320 kt 125° 322 kt 126° 320 kt 125° 318 kt

Admin . 1741 exemple 404 123° - 320 kt.123° - 320 kt.

Given .tas = 198 kt.hdg °t = 180.w/v = 359/25 .calculate the track °t and gs ?

Question 172-35 : 180° 223 kt 179° 220 kt 181° 180 kt 180° 183 kt

exemple 408 180° - 223 kt180° - 223 kt

Given .tas = 135 kt true heading = 278° true wind = 140°/20 kt .calculate the ?

Question 172-36 : 283° 150 kt 279° 152 kt 272° 121 kt 275° 150 kt

Admin .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 exemple 412 283° - 150 kt.283° - 150 kt.

Given .tas = 155 kt.true heading = 216°.wind = 090°/60 kt.calculate the true ?

Question 172-37 : 231° 196 kt 224° 175 kt 222° 181 kt 226° 186 kt

Admin .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 exemple 416 231° - 196 kt.231° - 196 kt.

Given .tas = 465 kt true heading = 124° wind = 170°/80 kt .calculate drift ?

Question 172-38 : 8l 415 kt 3l 415 kt 4l 400 kt 6l 400 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 exemple 420 8l - 415 kt8l - 415 kt

Given tas = 140 kt hdg t = 005° w/v = 265/25kt calculate the drift and gs ?

Question 172-39 : 10r 146 kt 9r 140 kt 11r 142 kt 11r 140 kt

Admin .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 exemple 424 10r - 146 kt10r - 146 kt

Given .tas = 190 kt hdg t = 355° w/v = 165/25kt .calculate the drift and gs ?

Question 172-40 : 1l 215 kt 1l 225 kt 1r 175 kt 1r 165 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 exemple 428 1l - 215 kt1l - 215 kt

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