An aircraft is departing from an airport which has an elevation of 2000 ft and the QNH is 1023 hPa The TAS is 100 kt the head wind component is 20 kt and the rate of ?
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At 10 15 the reading from a VOR/DME station is 211°/ 90NM at 10 20 the reading from the same VOR/DME station is 211°/120NM Compass Heading = 200° Variation in the area = 31°W Deviation = +1° TAS ?
An aircraft is departing from an airport which has an elevation of 2000 ft and the QNH is 1003 hPa The TAS is 100 kt the head wind component is 20 kt and the rate of climb is 1000 ft/min Top of climb ?
Two consecutive waypoints of a flight plan are stornoway vordme n58°12 4' w006°11 0' and glasgow vordme n55°52 2' w004°26 7' during flight actual time over stornoway 11 15 utc and estimated time over glasgow 11 38 utc at 11 21 utc fix of aircraft exactly over reporting point ronar what the revised utc over glasgow based on this last fix 2505 distance stornoway to glasgow = 151 nm distance stornoway to ronar = 44 nm 11 21 11 15 = 6 min 6 min = 44 nm so 60 min = 440 kt(nm) 151 44 = 107 nm 107/440 = 0 243h 0 243 x 60 = 14 6 min 11 21 + 14 6 min = 11 35 36 sec. 
An aircraft at fl360 required to descent to fl120 the aircraft should reach fl120 at 40 nm from next waypoint the rate of descent 2000 ft/min the average gs 420 kt the minimum distance from next waypoint at which descent should start 24000 ft to lose with 2000 ft/min this means descending 24000 ft in 12 min the plane flying 7 nm/min 12x7 84 nm the plane needs 84 nm to reach fl120 it also need to be leveled 40 nm before next waypoint that means we should start descent 84 + 40 = 124 nm before next waypont. 
The distance between a and b 90 nm at a distance of 15 nm from a aircraft 4 nm right of course to reach destination b correction angle on heading should be tke =(distance off track x 60) / distance along track tke =(4 nm x 60) / 15 nm tke = 16° to join back on our track tke =(distance off track x 60) / distance to go tke =(4 nm x 60)/75 nm (90 nm 15 nm = 75 nm) tke = 3° correction angle 16° + 3° = 19° (to left as we are right off course). 
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 to reach destination b from this position true heading should be 2516 draw exercice the dead reckoning position was at 15 min from a with a gs of 130 kt (130/60) x 15 = 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° current heading 275° new heading 275° + 17° = 292°. Question 171-8
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 true bearing of locator py at 1003 utc from aircraft img /com_en/com061 635 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 true heading + relative bearing = true bearing of locator from aircraft 006° + 266° = 272°. Question 171-9
An aircraft departing from an airport which has an elevation of 2000 ft and qnh 1003 hpa the tas 100 kt head wind component 20 kt and rate of climb 1000 ft/min top of climb fl 050 at what distance from airport will this be achived 2000 ft at qnh 1003 hpa at 1013 hpa it 2300 ft to reach fl050 you must climb 2700 ft (5000 2300) rate of climb 1000 ft/min 2700/1000 = 2 7 min at a ground speed of 80 kt it will take 2 7 x (80/60) = 3 6 nm. Question 171-10
During approach following data are obtained dme 12 0 nm altitude 3000 ftdme 9 8 nm altitude 2400 fttas = 160 kt groundspeed 125 ktthe rate of descent 12 nm 9 8 nm = 2 2 nm 3000 ft 2400 ft = 600 ft 2 2 nm / 125 kt = 0 0176 h >1 056 min 600 ft / 1 056 = 568 ft/min. Question 171-11
The distance between a and b 90 nm at a distance of 75 nm from a aircraft 4 nm right of course the track angle error tke use one in sixty rule track error angle from a = distance off track x 60 / distance along track track error angle from a = 4 nm x 60 / 75 nm track error angle from a = 3°r. Question 171-12
The true course according to flight log 270° forecast wind 045° 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 with forecasted wind our ground speed 130 kt at 130 kt 15 minutes of flight we will be at 32 5 nm from a but 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°. Question 171-13
An aircraft flies from waypoint 7 63°00'n 073°00'w to waypoint 8 62°00'n 073°00'w the aircraft position 62°00'n 073°10'w the cross track distance in relation to planned track 1° longitude at equator = 60 nm 1° long at 60°lat = 30 nm 10' off track 5 nm (10' = 1/6 from 1h so 1/6 from 30 nm 5 nm (30 / 6)) as we are heading along meridian from 63°n to 62°n out true course 180° as we have ended up at 73°10' this right of track so 5 nm right mathematically distance(nm) = chlong (in minutes) * coslat distance = 10 x cos62° distance = 4 7 nm. Question 171-14
An aircraft departing from an airport which has an elevation of 2000 ft and qnh 1003 hpa the tas 100 kt headwind component 20 kt and rate of climb 500 ft/min top of climb fl 050 at what distance from airport will this be achived 1013 hpa 1003 hpa = 10 hpa > 300 ft (10 x 30 ft) 2000 ft + 300 ft = 2300 ft 5000 ft 2300 ft = 2700 ft (2700 ft to climb) 2700 ft / 500 ft/min = 5 4 min > 0 09h (5 4 / 60) gs = 80 kt (100 kt (tas) 20(headwind component)) 80 kt x 0 09h = 7 2 nm. Question 171-15
You are departing from an airport which has an elevation of 1500 ft the qnh 1003 hpa 15 nm away there a waypoint you are required to pass at an altitude of 7500 ft given a groundspeed of 120 kt what the minimum rate of climb 7500 1500 = 6000 ft 6000 / 7 5 = 800 ft/min. Question 171-16
An aircraft flying at fl200 the qnh given a meteorological station at an elevation of 1300ft 998 2 hpa oat = 40°c the elevation of highest obstacle along route 8 000 ft calculate aircraft's approximate clearance above highest obstacle on this route find qnh altitude 1013 998 2 = 14 8 x 27 = 400 ft altitude 19600ft qnh with aviat 617 computer against altitude pressure = 20 put °c oat = 40 then read in inner circle altitude 19600 the on outer circle 18400 true altitude 18400 8000ft = 10400ft = approximate clearance over obstacle for information 061 general navigation learning objectives states for questions involving height calculation 30 ft/hpa to be used unless another figure specified in question. Question 171-17
The qnh given a station at 2500 ft 980hpa the elevation of highest obstacle along a route 8 000 ft and oat = isa 10°c when an aircraft on route has to descend minimum indicated altitude qnh on subscale of altimeter to maintain a clearance of 2000 ft will be we need to be at 10000 ft to avoid obstacle 2000 ft temperature correction formula 4° x 10 x ( 10°) = 400 ft the altimeter over reading in cold air if we flew exactly at 10000 ft indicated our true altitude would be 9600 ft we need to cruise at 10000 + 400 = 10400 ft indicated in order to maintain a clearance of 2000 ft. Question 171-18
An aircraft departing from an airport which has an elevation of 2000 ft outside temperature 0°c qnh = 1013 hpa it planned to climb to fl 320 where outside temperature 60°c the average cas during climb will be 200 kt compressibility negligeable mean tas during climb by convention at exam (easa specification) average tas used climb problems calculated at altitude 2/3 of cruising altitude temperature 0°c at 2000 ft approximately 4° at sea level (we 'gain' 2°c per 1000 ft while descending) 21000 ft x 2°/1000ft = 42°c 4°+ ( 42°) = 38°c temperature around 38°c at fl210 on computer in airspeed window set press alt '21' in front of coat °c ' 38°c' on outer scale in front of cas 200 kt you can read tas 272 kt. Question 171-19
During visual navigation in freezing conditions after heavy snowfall which of following landmark will give best reference a visual checkpoint after heavy snowfall roads will not have been cleared snow ploughs neither country road or railway a large river may freeze but you will always be able to distinguish its path. Question 171-20
During a climb at a constant cas below tropopause in standard conditions Both tas mach number will increase. for those questions use very simple 'ertm' diagram the cas line vertical because question states climb at a constant calibrated airspeed (cas) ertm e as/r as(rectified air speed or cas)/t as/m ach. Question 171-21
An aircraft descending down a 12% slope whilst maintaining a gs of 540 kt the rate of descent of aircraft approximately Both tas mach number will increase. vertical speed = 12% gradient x 540 kt vertical speed = 6480 ft/min. Question 171-22
The departure airfield at 2000 ft elevation temperature at field +20°c qnh 1013 hpa the plan to climb to fl 290 where outside air temperature 40°c the average tas in climb should be calculated using what fl and temperature Fl 2 with temperature 2 °c. by convention at exam (easa specification) average tas used climb problems calculated at altitude 2/3 of cruising altitude 29000 2000 = 27000 ft 2/3 27000 = 18000 ft 18000 + 2000 = 20000 ft (fl200) température à 20000 ft = 20°c + (20 x 2°c) = 20°c. Question 171-23
The departure from an airfield at 2000 ft elevation temperature at field +20°c qnh 1013 hpa the plan to climb to fl 290 where outside air temperature 40°c the cas in climb 180 kt compressibility negligible the average tas in climb Fl 2 with temperature 2 °c. 1 29000 2000= 27000ft 2 27000*(2/3)= 18000 3 18000+2000 = 20000 (your average alt in climb patern) 4 if oat at 2000 alt +20 so at 20000 will be +20 (reference figure) 2*18( 18 height) explanation ( 2*18 because temp decrease 2 deg per 1000 ft) 5 align temp 16 with 22000 ft in th air speed window (cr 3 or iwa 11092 ) read oposit 180 kt your 249 tas at outer scale this is only the way to solve tasks like this. Question 171-24
Given w/v at arrival aerodrome at 1000 ft amsl 230°/15kt w/v at tod at fl 130 280°/45kt average track after tod 220° isa conditions descent speed ias = 170 kt find gs during descent Fl 2 with temperature 2 °c. at fl130 isa condition ias=170kt => tas=206kt we have w/v 280°/45kt => drift 12°l so mh=232° to get an average track of 220° so gs= 178kt with computer at 1000ft amsl isa condition ias=170kt => tas=172kt we have w/v 230°/15kt => drift 1°l so mh=221° to get an average track of 220° so gs= 157kt with computer as a result average gs the descent is (157+178)/2 = 167 5kt. Question 171-25
Given w/v at arrival aerodrome at msl 200°/20kt w/v at tod at fl 100 260°/50kt average track after tod 190° isa conditions descent speed ias = 150 kt find gs during descent Fl 2 with temperature 2 °c. at fl130 isa condition ias=170kt => tas=206kt we have w/v 280°/45kt => drift 12°l so mh=232° to get an average track of 220° so gs= 178kt with computer at 1000ft amsl isa condition ias=170kt => tas=172kt we have w/v 230°/15kt => drift 1°l so mh=221° to get an average track of 220° so gs= 157kt with computer as a result average gs the descent is (157+178)/2 = 167 5kt. Question 171-26
An aircraft cruising in fl180 and thereafter descends to ground level the following wind information given ground level 260°/25 ktfl030 270°/30 ktfl060 270°/35 ktfl090 270°/40 ktfl120 280°/50 ktfl150 285°/55 ktfl180 290°/55 ktthe wind to be used descent calculation Fl 2 with temperature 2 °c. by convention average wind velocity used climb problems wind velocity at altitude 2/3 of cruising altitude average wind velocity used descent problems wind velocity at altitude 1/2 of descent altitude. Question 171-27
The distance between two waypoints 150 nm to calculate compass heading pilot used 2°e magnetic variation instead of 2°w assuming that forecast w/v applied what will off track distance be at second waypoint Fl 2 with temperature 2 °c. for each degree of error that you have at every 60 nm of travel you will be 1 nm off track you have 150/60 2 5 nm off track each degree of error total error 4° (from 2°e to 2°w) 4° x 2 5 nm = 10 nm using goniometric functions tan4° = ? / 150 ? = tan4° x 150 ? = 10 nm. Question 171-28
True track 085°groundspeed 180 ktwind 290°/30ktvariation 4°e the aircraft 1 5 nm left of track after 12 minutes what the track angle error tke Fl 2 with temperature 2 °c. tke = distance off track x 60 / distance along track tke = 1 5 nm x 60 / 36 nm = 2 5° the aircraft has drifted to left therefore tke 2 5° left. Question 171-29
With only a visual straight line as visual cue a canal example this line of position must be selected More or less perpendicular to our track. tke = distance off track x 60 / distance along track tke = 1 5 nm x 60 / 36 nm = 2 5° the aircraft has drifted to left therefore tke 2 5° left. Question 171-30
Given a descending aircraft flies in a straight line to a dme dme 55 nm altitude 33000 ftdme 43 9 nm altitude 30500 ftm = 0 72 gs = 525 kt oat = isa the descent gradient More or less perpendicular to our track. 33000 30500 = 2500 ft 55 nm 43 9nm = 11 1 nm 11 1 nm = 67488 ft 2500 = 67488 x x x= 2500 / 67488 = 0 0370 (3 7%). Question 171-31
The descent gradient of an aircraft with following data 60 nm norths of vor xyz fl350 10 nm south of vor xyz fl120 More or less perpendicular to our track. total ground distance 60+10 = 70 nm altitude difference 35000 12000 = 23000 ft gradient in % = altitude difference in ft x 100 / ground distance in ft ground distance in feet 70 nm x 6080 ft = 425600 ft gradient in % = 23000 x 100 / 425600 = 5 4%. Question 171-32
The average tas climbing from 1500 ft to fl180 with a given temperature of isa +15°c a cas of 230 kt and qnh 1032 hpa More or less perpendicular to our track. by convention at exam (easa specification) average tas used climb problems calculated at altitude 2/3 of cruising altitude 1032 1013 = 19 hpa 19 hpa x 30 ft = 570 ft 18000 + 570 = 18570 ft 2/3 of 18570 = 12380 ft to convert cas to tas 1% each 600 ft 0 2% each degree of isa deviation tas = cas x (12380/600) + (0 2% x 15°c) = 230 x (20% + 3%) = 282 9 kt. Question 171-33
An aircraft turning on a final approach to intercept a 3° glide slope which located at an altitude of 700 ft amsl assuming turn made at 4 nm from threshold what a suitable altitude to intercept glide slope More or less perpendicular to our track. 1 in 60 rule a rule of thumb (3° x 4 nm)/60 = 0 2 nm 0 2 nm x 6080 ft = 1216 ft add 700 ft since we are looking an altitude = 1216 + 700 = 1916 ft. Question 171-34
Given tas 220 ktcruising level fl180track during climb 080°wind at msl 260°/25ktwind at fl180 290°/55 ktfrom msl to cruising level find gs during climb More or less perpendicular to our track. we have to use wind at altitude 2/3 of cruising altitude fl180 x 2/3 = fl120 wind changes 30° from ground to fl180 an speed increases from 30 kt mean wind at fl120 260°+20° 25kt+20kt = 280°/45kt with your nav computer you will find 262 kt. Question 171-35
An aircraft climbs from ground level to fl180 the following wind information given ground level 260°/25 ktfl030 270°/30 ktfl060 270°/35 ktfl090 270°/40 ktfl120 280°/50 ktfl150 285°/55 ktfl180 290°/55 ktthe wind to be used to solve climb problems e g calculation of gs from tas and track in climb More or less perpendicular to our track. by convention average wind velocity used climb problems wind velocity at altitude 2/3 of cruising altitude average wind velocity used descent problems wind velocity at altitude 1/2 of descent altitude. Question 171-36
An aircraft descends from fl240 to fl040 the final approach cas = 220 ktoat = isa +10°cthe average tas in descent More or less perpendicular to our track. at exam average tas used descent problems calculated at altitude 1/2 of descent altitude at fl120 isa temperature = 15°c (2°c x 12) = 9°c oat isa +10°c thus oat is +1°c at fl120 on computer in airspeed window put +1ºc next to fl120 go to cas 220 kt on inner scale read tas on outer scale 273 kt. Question 171-37
When flying a visual navigation exercise in controlled airspace it confirmed aircraft exactly on track atc instructions are to turn left 30 degrees to avoid conflicting traffic after two minutes they advise 'you are now two miles left of your original track turn right to regain track in 30 miles ' what the heading change required to comply with atc instructions More or less perpendicular to our track. Question issue ae this question asks about on in sixty rule this rule states that after 60 nm a drift of 1° corresponds to an off track distance of 1 nm (2° = 2 nm 3° = 3 nm etc ) the question mentions an off track distance of 2 nm if you cut flight distance into half (30 nm instead of 60 nm) angle off track distance must double 30 nm => 4° => 4 nm the question states that controller wants us to regain our original track after 30 nm so correction angle should be 4° + 30° = 34°. Question 171-38
An aircraft in cruise at fl120 cleared to descend to 3000 ft the distance to go 25 nm calculate descent gradient More or less perpendicular to our track. we have to descend 9000 ft 25 nm in ft is 25 nm x 6000 ft/nm = 151900 ft (9000 / 150000) x 100 = 6%. Question 171-39
Given descent from 15000 ft to 3000 ft mslglide path angle during descent 3°ground speed 180 kt in 15000 ftground speed 150 kt in 3000 ftcalculate rate of descent It decreases from 9 ft/min to 75 ft/min. rate of descent (3°) = ground speed (kt) x 10/2 at 15000 ft with a ground speed of 180 kt rate of descent = 180 x 10/2 = 900 ft/min approaching 3000 ft with a decelerating speed to reach 150 kt rate of descent = 150 x 10/2 = 750 ft/min. Question 171-40
Which formula can be used to calculate rate of climb/descent rate of climb/descent ft/min = Groundspeed (kt) x gradient (ft/nm) / 6. we must know groundspeed to calculate a climb/descent gradient calculate rate of descent (rod) on a given glide path angle or gradient using following rule of thumb formulae rod (ft/min) = gp degrees x gs (nm/min) x 100 or rod (ft/min) = gp per cent x gs (kt) calculate climb/descent gradient (ft/nm per cent degrees) gs or vertical speed according to following formula vertical speed (ft/min) = (gs (kt) x gradient (ft/nm)) / 60.
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