An aircraft is flying at FL100 The OAT = ISA 15°C The QNH given by a station at an elevation 3000 ft is 1035 hPa Calculate the approximate True Altitude ?
Formation > assignment
An aircraft has to fly over a mountain ridge The highest obstacle indicated in the navigation chart has an elevation of 9 800 ft The QNH given by a meteorological station at an elevation of 6200 ft ?
An aircraft is flying from A to B a distance of 50 NM The True Course in the flight log is 270° the forecast wind is 045° T /15 kt and the TAS is 120 kt After 15 minutes of flying with the planned ?
An aircraft flying from a to b a distance of 50 nm the true course in flight log 090° forecast wind 225° t /15kt and tas 120 kt after 15 minutes of flying with planned tas and true heading aircraft 3 nm north of intended track and 2 5 nm ahead of dead reckoning position to reach destination b from this position correction angle on heading should be draw exercice img /com_en/com061 623 jpg without wind at 120 kt 15 minutes of flight we are at 32 5 nm from a the question states 2 5 nm ahead of dead reckoning position so we are at 35 nm from a use one in sixty rule track error angle from a = 3 nm x 60 / 35 nm = 5° (it's drift to applied in order to correct 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 correction angle on heading should be 5° + 12° = 17°. 
An aircraft flying from a to b the true course according to flight log 090° estimated wind 225° t /15 kt and tas 120 kt after 15 minutes of flying with planned tas and true heading aircraft 3 nm south of intended track and 2 5 nm ahead of dead reckoning position the track angle error tke Frist step find 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 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°. 
On a true heading of 090° aircraft experiences drift of 5°right on a true heading of 180° aircraft experiences no drift on both headings tas 200 kt and it assumed that wind the same what the experienced wind speed and direction since on a true heading of 180° there no drift wind coming from 180° or 360° under index set true heading 090° centre dot on tas 200 kt with rotative scale set 5°right drift img /com_en/com061 627 jpg read wind 360°/17 kt. 
An aircraft flying from salco to berry head on magnetic track 007° tas 445 kt the wind 050° t /40 kt variation 5°w deviation +2°at 1000 utc rb of locator py 311° at 1003 utc rb of locator py 266° calculate distance of aircraft from locator py at 1003 utc first step find aircraft magnetic heading img /com_en/com061 631 jpg calculate drift (between our true track 002° the true wind 050°/40 kt) with your computer drift 4° left you have to apply a 4°right wind angle correction (and also a ground speed of 415 kt ) from aircraft at 1000 utc locator at 60° to left (011° to 311° = 60°) from aircraft at 1003 utc locator at 105° to left (011° to 266° = 105°) we have an isosceles triangle 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. Question 168-8
An aircraft flying at fl250 oat = 45°c the qnh given a station at msl 993 2 hpa calculate approximate true altitude you have to turn your altimeter subscale setting knob counterclockwise from 1013 2 to 993 2 indicated altitude will be decreased 20 hpa x 30 ft = 600 ft 25000 ft 600 ft = 24400 ft you can use either computer or with following rule of thumb called '4% rule' the altitude/height changes 4% 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 colder than standard thus air column contracted our true altitude lower than our indicated altitude. Question 168-9
A vor situated at position n55°26' w005°42' the variation at vor 9°w the position of aircraft n60°00'n w010°00' the variation at aircraft position 11°w the initial true track angle of great circle from aircraft position to vor 101 5° which radial the aircraft on img /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 vor 105 + 9°w = 114° magnetic last step we are looking a radial 114° + 180° = 294°. Question 168-10
An aircraft tracks radial 200° inbound to a vor station with a magnetic heading of 010° after being overhead vor station aircraft tracks radial 090° outbound with a mh of 080° the tas 240 kt and magnetic variation in area 5°w the wind following radial 200° inbound (magnetic track 020°) with a magnetic heading of 010° we have a 10° right drift our true heading 005° (we have to applied 5°west magnetic variation since it a vor) after having overfly vor we fly outbound on radial 090° (magnetic track 090°) our magnetic heading 080° so our true heading 075° on computer set 240 kt under center dot 005° below true index draw a line along right 10° drift line rotate put 075° below true index draw a line along right 10° drift line rotate to bring back intersection of lines under central wind line . Question 168-11
The fix of aircraft position determined radials from three vor stations the measurements contain small random errors known systematic errors and unknown systematic errors the measured radials are corrected known systematic errors and are plotted on a navigation chart the result shown at reference what the most probable position of aircraft err _a_061 644 img /com_en/com061 644 jpg point 1 always located on right side of radials. Question 168-12
An aircraft flies at fl 250 with an oat of 45°c the qnh given a meteorological station with an elevation of 2830 ft 1033 hpa calculate clearance above a mountain ridge with an elevation of 20410 ft 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 20 hpa x 30 ft = 600 ft 25000 + 600 = 25600 ft next step we must correct temperature outside temperature 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 computer or with following rule of thumb called '4% rule' the altitude/height changes 4% each 10°c temperature deviation from isa an altimeter set to airport qnh will read correctly when on ground at airport irrespective of temperature any temperature error therefore occurs due to non isa temperature in layer of atmosphere between airport elevation 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 mountain ridge is 24690 20410 = 4280 ft. Question 168-13
Given an aircraft flying at fl100 oat = isa 15°c the qnh given a meteorological station with an elevation of 100 ft below msl 1032 hpa 1 hpa = 27 ft calculate approximate true altitude of this aircraft you have to turn your altimeter subscale setting knob clockwise from 1013 to 1032 indicated altitude will be increased 19 hpa x 27 ft = 513 ft 10000 + 513 = 10513 ft now we must correct temperature we are in isa 15°c you can use either computer or with following rule of thumb called '4% rule' the altitude/height changes 4% each 10°c temperature deviation from isa an altimeter set to airport qnh will read correctly when on ground at airport irrespective of temperature any temperature error therefore occurs due to non isa temperature in layer of atmosphere between airport elevation aircraft in flight therefore 10513 ( 100) = 10613 ft 10613 x 0 04 x 1 5 = 637 ft 10513 637 = 9876 ft (closest answer 9900 ft). Question 168-14
The accuracy of manually calculated dead reckoning position of an aircraft among other things affected The accuracy of forecasted wind. dead reckoning the process of calculating one's current position using a previously determined position or fix advancing that position based upon known or estimated speeds over elapsed time course you have to take wind into account the more accurate wind information is more accurate manually calculated position will be. Question 168-15
The accuracy of manually calculated dead reckoning position of an aircraft among other things affected The flight time since last position update. dead reckoning the process of estimating one's current position based upon a previously determined position or fix advancing that position based upon known or estimated speeds over elapsed time course therefore accuracy is among other things affected the flight time since last position update. Question 168-16
What may cause a difference between a dead rekoning position and a fix The difference between actual wind the forecasted wind. dead reckoning the process of estimating one's current position based upon a previously determined position or fix advancing that position based upon known or estimated speeds over elapsed time course therefore accuracy is among other things affected the flight time since last position update. Question 168-17
Cas 320 ktflight level 330oat isa +15°ctas approximately compressibility factor 0 939 The difference between actual wind the forecasted wind. oat 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 outer scale in front of cas 320 kt you can read tas 565 kt true air speed (tas) obtained from calibrated air speed (cas) correcting compressibility density 568 x 0 939 = 533 kt. Question 168-18
Given mach numer 0 8flight level 330oat isa +15°c tas approximately compressibility factor 0 94 The difference between actual wind the forecasted wind. 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) obtained from calibrated air speed (cas) correcting compressibility density. Question 168-19
The main purpose of dr dead reckoning To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. 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) obtained from calibrated air speed (cas) correcting compressibility density. Question 168-20
An aircraft flying at fl390 at a speed of mach 0 821 oat isa 4°cthe compressibility factor 0 942 calculate tas To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. isa temperature at fl390 = 56 5°c ( 56 5°c considered to be 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) obtained from calibrated air speed (cas) correcting compressibility density. Question 168-21
An aircraft descends from fl240 to fl80 the final approach track = 070°cas = 220 ktoat = isa 10°cthe average tas in descent To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. at exam average tas used descent problems calculated at altitude 1/2 of descent altitude at fl160 isa temperature = 15°c (2°c x 16) = 17°c oat isa 10°c thus oat is 27°c at fl160 on computer in airspeed window put 27ºc next to fl160 go to cas 220 kt on inner scale read tas on outer scale 276 kt. Question 168-22
An aircraft flying at fl 350 with cas = 300 kt oat = isa + 4°c the compressibility factor 0 939 calculate tas To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. isa temperature at fl350 = 15°c +(35 x ( 2)) = 55°c isa +4°c so oat = 51°c on computer in airspeed window put 51ºc next to fl350 go to cas 300 kt on inner scale 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) obtained from calibrated air speed (cas) correcting compressibility density. Question 168-23
Given track = 355°tas = 190 ktwind 270°/25 ktafter 30 minutes of flying with planned tas and true heading aircraft 3 5 nm right of track and 4 5 nm ahead of dead reckoning position calculate actual wind To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. isa temperature at fl350 = 15°c +(35 x ( 2)) = 55°c isa +4°c so oat = 51°c on computer in airspeed window put 51ºc next to fl350 go to cas 300 kt on inner scale 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) obtained from calibrated air speed (cas) correcting compressibility density. Question 168-24
Given fl 400oat = 65°cias = 260 ktinstrument and position error to be neglected compressibility factor = 0 935calculate true air speed taking compressibility into account To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. calibrated airspeed (cas) indicated airspeed (ias) corrected instrument error position error the question states instrument position error to be neglected therefore ias = cas oat = 65°c on computer in airspeed window put 65ºc next to fl400 go to cas 260 kt on inner scale read tas on outer scale 513 kt true air speed (tas) obtained from calibrated air speed (cas) correcting compressibility density 513 x 0 935 = 479 kt. Question 168-25
Given tas = 210 ktcas = 190 ktpressure altitude = 9000 ftcalculate mach number To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. using computer align tas (external ring) & cas (internal ring) when done go to your airspeed case read one corresponding to pressure altitude above 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 168-26
A dr position to be found To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. using computer align tas (external ring) & cas (internal ring) when done go to your airspeed case read one corresponding to pressure altitude above 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 168-27
Which of factors named hereafter should be considered the pilot when selecting landmarks as visual reference points 1 possibility of identification2 transmitted frequency3 visibility4 closeness to trackthe combination that regroups all of correct statements To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. using computer align tas (external ring) & cas (internal ring) when done go to your airspeed case read one corresponding to pressure altitude above 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 168-28
Given fl 300oat = 45°cias = 260 ktinstrument and position error to be neglectedcompressibility factor = 0 96calculate true air speed taking compressibility into account To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. using computer align tas (external ring) & cas (internal ring) when done go to your airspeed case read one corresponding to pressure altitude above 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 168-29
On a mercator chart one minute on n55° parallel 3 1 mm the map scale at 40°n To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. using computer align tas (external ring) & cas (internal ring) when done go to your airspeed case read one corresponding to pressure altitude above 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 168-30
The nominal scale of a north stereopolar map To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. using computer align tas (external ring) & cas (internal ring) when done go to your airspeed case read one corresponding to pressure altitude above 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 168-31
Given tas = 140 kt true hdg = 302° w/v = 045° t /45kt calculate drift angle and gs To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. under index set true heading 302° centre dot on tas 140 kt with rotative scale set wind read drift 16° left ground speed 156 kt. Question 168-32
Given tas = 290 kttrue hdg = 171°w/v = 310° t /30kt calculate drift angle and gs To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. . Question 168-33
Given tas = 485 kt true heading = 226° true wind = 110°/95kt calculate drift angle and gs To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. under index set true heading 226° centre dot on tas 485 kt with rotative scale set wind read drift 9° right ground speed 533 kt. Question 168-34
Given tas = 472 kt true heading = 005° true wind = 110°/50kt calculate drift angle and gs To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. under index set true heading 005° centre dot on tas 472 kt with rotative scale set wind read drift 5 5° left ground speed 487 kt closest answer 6°l/490 kt. Question 168-35
Given tas = 375 kt true heading = 124° wind = 130°/55 kt calculate true track and gs To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. . Question 168-36
Given tas = 198 kthdg °t = 180w/v = 359/25 calculate track °t and gs To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. . Question 168-37
Given tas = 135 kt true heading = 278° true wind = 140°/20 kt calculate true track and ground speed To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. center dot on tas 135 kt true heading 278° under index put wind direction under red compass rose under 20 kt your drift 5° right giving a track of 283° a groundspeed (under wind mark) of 150 kt . Question 168-38
Given tas = 155 kttrue heading = 216°wind = 090°/60 ktcalculate true track and gs To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. center dot on tas 155 kt true heading 216° under index put wind direction under red compass rose under 60 kt your drift 14 5° right giving a track of 230 5° a groundspeed (under wind mark) of 195 kt the closest answer 231° 196 kt. Question 168-39
Given tas = 465 kt true heading = 124° wind = 170°/80 kt calculate drift and ground speed To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. under index set true heading 124° centre dot on tas 465 kt with rotative scale set wind img /com_en/com061 171 jpg read drift 8° left ground speed 415 kt. Question 168-40
Given tas = 140 kt hdg t = 005° w/v = 265/25kt calculate drift and gs To obtain with reasonable accuracy aircraft's position between fixes or in absence of fixes. under index set true heading 005° centre dot on tas 140 kt with rotative scale set wind read drift 10° right ground speed 146 kt.
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