Given Distance A to B = 120 NM After 30 NM aircraft is 3 NM to the left of course What heading alteration should be made in order to arrive at point 'B' ?
Formation > assignment
An aircraft was over 'A' at 1435 hours flying direct to 'B' Given Distance 'A' to 'B' 2900 NMTrue airspeed 470 ktMean wind component 'out' +55 ktMean wind component 'back' 75 kt The ETA for reaching ?
Given Distance 'A' to 'B' 2484 NMGroundspeed 'out' 420 ktGroundspeed 'back' 500 ktThe time from 'A' to the Point of Equal Time PET between 'A' and 'B' is ?
Given distance 'a' to 'b' 2484 nmground speed out 420 ktground speed home 500 ktsafe endurance 08 h 30 minutes the distance from 'a' to point of safe return psr point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 420 kt ground speed home = 500 kt point of safe return (psr) = 8 5 x 500 / (420 + 500) point of safe return (psr) = 4250 / 920 point of safe return (psr) = 4 62 h distance of psr from departure point at a speed of 420 kt 4 62 h x 420 = 1940 nm. 
An aircraft was over 'q' at 1320 hours flying direct to 'r' given distance 'q' to 'r' 3016 nmtrue airspeed 480 ktmean wind component 'out' 90 ktmean wind component 'back' +75 kt endurance 10 hwhat the eta at point of equal time pet ground speed out 480 90 = 390 kt ground speed home 480 + 75 = 555 kt pet = d x gsh / (gso + gsh) pet = 3016 x 555 /(390 + 555) = 1771 nm 1771 nm / 390 kt = 4 54 h 4 54 x 60 minutes = 272 minutes (4h32minutes) 13h20 + 4h32 = 17h52. 
Given distance 'a' to 'b' 1973 nmgroundspeed 'out' 430 ktgroundspeed 'back' 385 ktthe time from 'a' to point of equal time pet between 'a' and 'b' ground speed out 430 kt ground speed home 385 kt pet = distance x gsh / (gso + gsh) pet = 1973 x 385 /(430 + 385) = 932 nm 932 nm / 430 kt = 2 16 h 2 16 h x 60 minutes = 129 6 minutes. 
Given distance 'a' to 'b' 2346 nmground speed out 365 ktground speed back 480 ktsafe endurance 8 h 30 minutesthe time from 'a' to point of safe return point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 365 kt ground speed home = 480 kt point of safe return (psr) = 8 5 x 480 / (365 + 480) point of safe return (psr) = 4080 / 845 point of safe return (psr) = 4 83 h the time from 'a' to point of safe return is 4 83 x 60 minutes = 290 minutes. Question 167-8
Given distance 'a' to 'b' 3623 nmgroundspeed 'out' 370 ktgroundspeed 'back' 300 ktthe time from 'a' to point of equal time pet between 'a' and 'b' ground speed out 370 kt ground speed home 300 kt pet = distance x gsh / (gso + gsh) pet = 3623 x 300 /(370 + 300) = 1622 nm 1622 nm / 370 kt = 4 38h 4 38 h x 60 minutes = 262 8 minutes. Question 167-9
Given magnetic track = 075°magnetic heading = 066°variation = 11°etas = 275 kt aircraft flies 48 nm in 10 min calculate true wind 48 nm in 10 minutes > gs = 288 kt on computer under index set true heading 077° in center dot tas 275 kt our true track 086° so drift 9° right mark point where 9° right drift crosses ground speed 288 kt on rotating scale you can read a wind of 335°/45kt . Question 167-10
Given magnetic track = 210°magnetic heading = 215°variation = 15°etas = 360 ktaircraft flies 64 nm in 12 minutes calculate true w/v 64 nm in 12 minutes > gs = 320 kt on computer under index set true heading 230° over center dot tas 360 kt our true track 225° so drift 5° left mark point where 5° left drift crosses ground speed 320 kt on rotating scale you can read a wind of 265°/50 kt . Question 167-11
Given aircraft at fl150 overhead an airport elevation of airport 720 ft qnh 1003 hpa oat at fl150 5°c what the true altitude of aircraft assume 1 hpa = 27 ft at fl150 isa = 15°c (2°c x 15) = 15°c oat 5°c we are in air mass 10°c warmer than isa changing subscale from 1013 hpa to 1003 hpa means that indicated altitude on altimeter will decrease 270 ft (10 hpa) 15000 270 = 14730 ft temperature correction 4 x 15 x 10 = 600 ft 14730 + 600 = 15330 ft ('plu 600 ft because air mass warmer than isa) maxscail how do you find 4x15x10=600ft it a formula? yes this the rule of thumb formula called '4% rule' the altitude/height changes 4% each 10°c temperature deviation from isa this an official formula given easa altitude/height calculations at exam. Question 167-12
An aircraft takes off from aerodrome of brioude altitude 1483 ft qfe = 963 hpa temperature = 32°c five minutes later passing 5000 ft on qfe second altimeter set on 1013 hpa will indicate approximately difference between 963 hpa 1013 hpa 50 hpa 50 hpa x 30 ft/hpa = 1500 ft 5000 + 1500 = 6500 ft your altimeter indicates your pressure altitude not your true altitude this the reason why we do not correct temperature (we only want to know reading of altimeter) note 061 general navigation learning objectives states for questions involving height calculation 30 ft/hpa to be used unless another figure specified in question. Question 167-13
Given distance a to b 360 nm wind component a b 15 kt wind component b a +15 kt tas 180 kt what the distance from equal time point to b babar350 e*o*h / ( o+h ) e endurance o gs outnbound h gs inbound e = distance * o eoh/(o+h) = 2 18*165*195/ 360 giving distance from etp from a 195 nm the question etp from b so 360 195 = 165 nm. Question 167-14
Given half way between two reporting points navigation log gives following information tas 360 kt w/v 330°/80kt compass heading 237° deviation on this heading 5° variation 19°w what the average ground speed this leg first step find true heading 237° 5° (deviation on this heading) = magnetic heading 232° 232° 19° (variation west) = true heading 213° put 360 kt in center dot under index set true heading 213° on rotating scale set wind 330°/80kt you read a ground speed of 403 kt the answer will be more accurate on a real computer (such as aviat 617 example). Question 167-15
Given an aircraft on final approach to runway 32r 322° the wind velocity reported the tower 350°/20 kt tas on approach 95 kt in order to maintain centre line aircraft's heading °m should be babar350 maximum drift (60/95) x 20 = 12 63 actual drift max drift x sin (350 322) = 12 63 x sin 28 = 5 92° mainting centre line 322°+5 92 = 328°. Question 167-16
An aircraft takes off from an airport 2 hours before sunset the pilot flies a track of 090° t w/v 130°/ 20 kt tas 100 kt in order to return to point of departure before sunset furthest distance which may be travelled resolve this question as a point of safe return question point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) outbound gs on computer when you start with a track under index you must first apply drift before read ground speed! drift=8° left set heading 098° (090° + 8°) under index read outbound gs 86 kt proceed on same way to finde homeward gs 116 kt point of safe return (psr) = (2h x 116) / (86 + 116) point of safe return (psr) = 1 148 h 1 148 x 86 kt = 98 7 nm. Question 167-17
From departure point distance to point of equal time Inversely proportional to sum of ground speed out ground speed back. distance to point of equal time = d x gsh/ (gso + gsh) where d = distance between departure arrival gso = ground speed out gsh = ground speed home. Question 167-18
Given required course 045° m variation 15°ew/v 190° t /30 ktcas 120 kt at fl 55 in standard atmosphere what are heading °m and gs Inversely proportional to sum of ground speed out ground speed back. at flight level 55 temperature in standard atmosphere is 15° (2° x 5 5) = 4°c on computer in airspeed window put 4ºc next to fl55 go to cas 120 kt on inner scale read tas on outer scale 131 kt now centre dot on tas 131 kt under true index put 190° (wind direction) mark wind speed 30 kt below at 101 kt then rotate to put true track 060° (045° + variation east) under true index drift 9° left it means that on this heading 060° aircraft would be on a true track of 051° rotate to put true track 060° under drift note that drift has changed to 10° left (as we turn drift changes on a heading of 070° we will have a 10° left drift) magnetic heading = 070° minus variation east = 055° ground speed under wind mark 147 kt. Question 167-19
Given airport elevation 1000 ft qnh 988 hpa what the approximate airport pressure altitude assume 1 hpa = 30 ft Inversely proportional to sum of ground speed out ground speed back. 1013 988 = 25 hpa 25 hpa x 30 ft = 750 ft 1000 + 750 = 1750 ft. Question 167-20
Given true altitude 9000 ftoat 32°ccas 200 ktthe true air speed tas Inversely proportional to sum of ground speed out ground speed back. you have first to convert true altitude in pressure altitude 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°c (2 x 9) = 3°c 3°c to 32°c = 29°c we are in isa 29°c 4% x 9 x 29 = 1044 ft pressure altitude = 9000 + 1044 = 10044 ft (10000 ft) now in airspeed window set 29°c in front of 10000 ft pressure alitude read cas 200 kt on inner scale corresponding tas 220 kt on outer scale. Question 167-21
Given course 040° t tas 120 kt wind speed 30 kt maximum drift angle will be obtained a wind direction of Inversely proportional to sum of ground speed out ground speed back. maximum drift obtained when wind at a right angle from our course 040° + 90° = 130° or 040° 90° = 310°. Question 167-22
Given cas 120 kt fl 80 oat +20°c what the tas Inversely proportional to sum of ground speed out ground speed back. in airspeed window set tempertaure +20° in front of pressure altitude fl80 on outside scale in front of cas 120 kt you read tas=141 kt cas = ias + correction position instrument error instrument error an eventual error of airspeed indicator itself position error the error produced from airflow around static port wherever that located around probe at low speed we consider cas = ias. Question 167-23
Route 'a' 44°n 026°e to 'b' 46°n 024°e forms an angle of 35° with longitude 026°e variation at a 3°e what the initial magnetic track from a to b Inversely proportional to sum of ground speed out ground speed back. draw situation img /com_en/com061 399 jpg initial true track from a to b will be 360° 35° = 325° variation 3°e 325° 3° = 322° (variation west magnetic best variation east magnetic least). Question 167-24
Given compass heading 090° deviation 2°w variation 12°e tas 160 kt whilst maintaining a radial 070° from a vor station aircraft flies a ground distance of 14 nm in 6 min what the wind °t Inversely proportional to sum of ground speed out ground speed back. first step search the ground speed 14 nm in 6 minutes = (14/6)x60 = 140 kt next step calculate drift img /com_en/com061 484a jpg drift (x) 18° left (100° 082°) put 160 kt (tas) on center dot under true index set 100° (heading) img /com_en/com061 484 jpg where ground speed crosses 18° left drift line you read on red scale a wind coming from 160° 50 kt. Question 167-25
Given m 0 80 oat 50°c fl 330 gs 490 kt variation 20°w magnetic heading 140° drift 11° right calculate true wind Inversely proportional to sum of ground speed out ground speed back. tas 462 kt magnetic heading 140° variation 20°w = true heading 120° read wind direction force on rotative scale. Question 167-26
Given pressure altitude 29000 ft oat 55°c calculate density altitude Inversely proportional to sum of ground speed out ground speed back. img /com_en/com061 486 jpg . Question 167-27
An aircraft flying at fl180 and outside air temperature 30°c if cas 150 kt what the tas Inversely proportional to sum of ground speed out ground speed back. in airspeed window put 30°c next to fl180 in front of 150 kt on inner scale read tas (195 kt) on outer scale. Question 167-28
Calibrated airspeed cas indicated airspeed ias corrected Instrument error position error. in airspeed window put 30°c next to fl180 in front of 150 kt on inner scale read tas (195 kt) on outer scale. Question 167-29
An aircraft was over 'q' at 1320 hours flying direct to 'r' given distance 'q' to 'r' 3016 nmtrue airspeed 480 ktmean wind component out 90 ktmean wind component back +75 ktsafe endurance 10 h 00 the distance from 'q' to point of safe return psr 'q' Instrument error position error. Point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 480 90 = 390 kt ground speed home = 480 + 75 = 555 kt point of safe return (psr) = 10 x 555 / (390 + 555) point of safe return (psr) = 5555 / 945 point of safe return (psr) = 5 878h distance of psr from departure point at a speed of 390 kt 5 878h x 390 = 2292 nm. Question 167-30
An aircraft flying at fl150 with an outside air temperature of 30° above an airport where elevation 1660 ft and qnh 993 hpa calculate true altitude assume 30 ft = 1 hpa Instrument error position error. you have to turn your altimeter subscale setting knob counterclockwise from 1013 to 993 indicated altitude will be decreased 20 hpa x 30 ft = 600 ft 15000 600 = 14400 ft now we must correct temperature outside temperature 30°c at fl150 isa at fl150 is 15°c (15 x 2°c) = 15°c 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 14400 1660 = 12740 ft 12740 x 0 04 x 1 5 = 764 ft 14400 764 = 13636 ft. Question 167-31
Given true track 239°true heading 229°tas 555 ktg/s 577 ktcalculate wind velocity Instrument error position error. 229° to true index 555 kt to center bore with a true headin of 229° a true track of 239° we have 10° right drift pencil mark intersection of 10° right drift with ground speed arc 577 kt read wind speed wind direction 130°/100 kt. Question 167-32
Given true track 245°drift 5° rightvariation 3°ecompass heading 242° calculate deviation Instrument error position error. use this wonderful table those questions. Question 167-33
Given true heading 090°tas 180 ktgs 180 ktdrift 5° rightthe wind Instrument error position error. set heading 090° undex index center dot on tas 180 kt where right drift 5° crosses ground speed arc 180 kt read wind on rotating scale 005°/15kt. Question 167-34
Given magnetic heading = 255°variation = 40°wgs = 375 ktw/v = 235° t / 120 ktcalculate drift angle Instrument error position error. the wind coming in front of us our true air speed will be more than 375 kt set wind index under true index set 400 kt (for example) in center dot mark wind at 120 kt below center dot on rotating indicator turn to put heading 215° (255° 40°) under true index shift speed arc under wind speed mark drift 6° left you also notice a tas of 490 kt . Question 167-35
Given true track = 095° tas = 160 kt true heading = 087° gs = 130 kt calculate wind Instrument error position error. the wind coming in front of us our true air speed will be more than 375 kt set wind index under true index set 400 kt (for example) in center dot mark wind at 120 kt below center dot on rotating indicator turn to put heading 215° (255° 40°) under true index shift speed arc under wind speed mark drift 6° left you also notice a tas of 490 kt . Question 167-36
Given true track 245°drift 5° rightvariation 3°ecompass heading 242° calculate magnetic heading Instrument error position error. use this wonderful table those questions note compass heading (242°) given nothing. Question 167-37
Given heading 265°tas 290 ktwind 210°/35 ktcalculate track and groundspeed Instrument error position error. center dot on tas 290 kt true heading 265° under index put wind direction under red compass rose under 35 kt your drift 6° right giving a track of 271° a groundspeed (under wind mark) of 272 kt . Question 167-38
An aircraft flying at fl 200 oat 0°c when actual air pressure on an airfield at msl placed in subscale of altimeter indicated altitude 19300 ft calculate aircraft's true altitude Instrument error position error. isa at fl200 is 15°c (2°c x 20) = 25°c oat 0°c we are in isa +25°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 0 04 x 19300 x 2 5 = 1930 ft true altitude = 19300 + 1930 = 21230 ft. Question 167-39
An aircraft must fly 2000 ft above an obstacle of which elevation 13 600 ft the qnh at nearest airfield 991 hpa elevation 1500 ft and temperature 20°c calculate minimum altitude required Instrument error position error. the aircraft must be at a true height above airfield of 13600 + 2000 1500 = 14100 ft at airfield isa temperature 15°c (1 5 x2°c) = 12°c temperature report to be 20°c so we are in isa 32°c temperature correction formula 4° x 14 1 x 32° = 1805 ft the minimum height above airfield is 14100 + 1805 ft = 15905 ft now adding 1500 ft of airfield to have an altitude 15905 + 1500 = 17405 ft note qnh calculated from qfe reduced to mean sea level (msl) assuming isa conditions there no temperature error between airfield elevation (qfe) mean sea level (msl). Question 167-40
Consider following factors that determine accuracy of a dead rekoning position 1 the flight time since last position update 2 the accuracy of forecasted wind 3 the accuracy of tas 4 the accuracy of steered heading using list which of above statements are correct Instrument error position error. the aircraft must be at a true height above airfield of 13600 + 2000 1500 = 14100 ft at airfield isa temperature 15°c (1 5 x2°c) = 12°c temperature report to be 20°c so we are in isa 32°c temperature correction formula 4° x 14 1 x 32° = 1805 ft the minimum height above airfield is 14100 + 1805 ft = 15905 ft now adding 1500 ft of airfield to have an altitude 15905 + 1500 = 17405 ft note qnh calculated from qfe reduced to mean sea level (msl) assuming isa conditions there no temperature error between airfield elevation (qfe) mean sea level (msl).
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