Spring and autumn equinox are the moments at which the sun reaches ?
MCQ > aircraft
An aircraft at latitude 02°20'N tracks 360° T for 685 km On completion of the flight the latitude will be ?
Given A 50°N 070°W and B 50°N 080°W and the position of one of the vertices of the great circle between A and B as being equal to 50°06 4'N 075°00 0'W what is the position of the other vertex ?
Sunrise in dublin 53°29'n 006°15'w 06 23 lmt calculate sunrise at bremen airport 53°09'n 008°45'e in lmt (5 ° 6 4s 5° 'e). difference in longitude 006°15'+008°45' = 15° the earth moves around sun at a rate of 4 minutes per degree of longitude (15°/h) latitude remains same only longitude changes sun rises at bremen airport 1h before dublin but sun rises at same local mean time. 
Calculate difference in lmt between dublin 53°29'n 006°15'w and bremen airport 53°09'n 008°45'e (5 ° 6 4s 5° 'e). difference in longitude 006°15'+008°45' = 15° the earth moves around sun at a rate of 4 minutes per degree of longitude (15°/h) 4x15 = 60 minutes. 
Calculate approximate distance from waypoint dbu 53°29 0'n 000°28 6'w to a waypoint 20 nm north of bremen airport the coordinates of bremen airport are 53°09 0'n 008°45 0'e (5 ° 6 4s 5° 'e). latitude difference 53°29 0'n 53°09 0'n = 20' (20'x1°/60'=0 33°) if (0 33°x60nm/1°=20nm) it could be our waypoint longitude difference 008°45 0'e + 000°28 6'w = 9 22° distance = change longitude x 60nm x lat wpt = 9 22° x 60nm x cos53 48° = 329 2 nm andresmarina it said 20 nm of north of bremen airport as you know 1 degree = 60 mins 1 degree = 60 nm so 1 min = 1 nm because of that waypoint located at same latitude than dbu waypoint in this case we should use rhumb line formula d = (change in long) * 60* cos (lm) d = 9 2266 * 60 * cos 53 5 = 329 2939 nm. 
Calculate approximate distance from dublin 53°29 0'n 006°15 3'w to a waypoint 20 nm north of bremen airport the coordinates of bremen airport are 53°09 0'n 008°45 0'e (5 ° 6 4s 5° 'e). dublin > bremen = 15° 1°=60nm on equator 1°=60nm x cos (mean latitude) 1°=60nm x cos (53 33°) 1°=35 844nm 15° x 35 844 = 537 66 nm the closest answer 535 7 nm. Question 158-8
Route a 53°24'n 015°54'e to b 32°00'n 052°51'w distance flight plan 3150 nm average gs 450 kt difference between standard time a and utc 1 hour difference between standard time b and utc 4 hours estimated time of arrival eta b 10 00 st b on 5 august calculate estimated time of departure etd a expressed in standard time (5 ° 6 4s 5° 'e). flight time (3150x60)/450=7h eta (b) 10 00 st (5 aug) change to utc (utc+4)=10+4=14 00 utc(5 aug) departure time 14 00 7h(flight time) =07 00 st(5 aug) etd(a) 07 00 st(5 aug) change to utc (utc+1)=08 00 utc(5 aug). Question 158-9
In producing chart projections following projection surfaces may be used Plane cylinder cone. flight time (3150x60)/450=7h eta (b) 10 00 st (5 aug) change to utc (utc+4)=10+4=14 00 utc(5 aug) departure time 14 00 7h(flight time) =07 00 st(5 aug) etd(a) 07 00 st(5 aug) change to utc (utc+1)=08 00 utc(5 aug). Question 158-10
The term 'oblique' in relation to map projections means that The axis of cylinder or cone neither parallel to or perpendicular to earth's axis of rotation. flight time (3150x60)/450=7h eta (b) 10 00 st (5 aug) change to utc (utc+4)=10+4=14 00 utc(5 aug) departure time 14 00 7h(flight time) =07 00 st(5 aug) etd(a) 07 00 st(5 aug) change to utc (utc+1)=08 00 utc(5 aug). Question 158-11
A day at a place as measured in local mean time starts When mean sun transits anti meridian of place in question. Ecqb03 december 2016. Question 158-12
Standard time The time enforced the legal authority to be used in a country or an area. Ecqb03 december 2016. Question 158-13
Daylight saving time summer time 1 used to extend sunlight period in evening2 introduced setting standard time forward one hour3 used in some countries The time enforced the legal authority to be used in a country or an area. Ecqb03 december 2016. Question 158-14
The countries having a standard time slow on utc Will generally be located at western longitudes. Ecqb03 december 2016. Question 158-15
The international date line located At 8 ° e/w meridian or in vicinity of this meridian. Ecqb03 december 2016. Question 158-16
Consider following statements on sunset Sunset the time when observer at sea level sees last part of sun disappear below horizon. Ecqb03 december 2016. Question 158-17
Atmospheric refraction Causes sunrise to occur earlier the sunset to occur later. Ecqb03 december 2016. Question 158-18
Consider following statements on sunrise and sunset At equator sunrise sunset occur at quite regular times throughout year. Ecqb03 december 2016. Question 158-19
'true north' The direction along any meridian toward true north pole. Ecqb03 december 2016. Question 158-20
In international aviation following units shall be used horizontal distance Metres kilometres nautical miles. Ecqb03 december 2016. Question 158-21
When dealing with heights and altitudes in international aviation we use following units Metres kilometres nautical miles. Ecqb03 december 2016. Question 158-22
'kilometre' defined as A / part of meridian length from equator to pole. Ecqb03 december 2016. Question 158-23
How long 25 kilometres at 60°n A / part of meridian length from equator to pole. Ecqb03 december 2016. Question 158-24
Position a 50°00 0'n 138°30 0'w st a = utc 9 h position b 50°00 0'n 175°45 0'e st b = utc + 12 h the ground distance between a and b 1736 nm on 4 february at 08 00 st a an aircraft exactly above a at moment aircraft arrives at position b air distance between a and b 1636 nm the average tailwind component from a to b was 25 kt calculate time of arrival at b A / part of meridian length from equator to pole. 1st nothing to do with coordinates! 2nd select required facts at point a 08 00st (04th feb) a 9h b +12h a >b 1736nm (ground) a >b 1636nm (air) wind 25kt st time at b? 3rd calculate ground air distance 100nm use logic 100nm/25kt = 4h the flight time! at point a 08 00st +9h > 17 00utc the time at departure at point b 17 00utc +12h > 05 00st the time at departure but 05th feb!!! the aircraft arrives 4h later so at b 05 00st +4h > 09 00st 5th feb. Question 158-25
Estimated time of departure a 15°15 0'n 072°06 0'w on 12 march 01 00 st st = utc 5 h estimated time of arrival b 55°18 0'n 005°45 0'e 16 15 st on same date st = utc + 1 h according to jeppesen table sunset at b occurs at 18 57 calculate flight time from a to b and time between arrival and sunset at b A / part of meridian length from equator to pole. 1st nothing to do with coordinates! 2nd select required facts at point a 08 00st (04th feb) a 9h b +12h a >b 1736nm (ground) a >b 1636nm (air) wind 25kt st time at b? 3rd calculate ground air distance 100nm use logic 100nm/25kt = 4h the flight time! at point a 08 00st +9h > 17 00utc the time at departure at point b 17 00utc +12h > 05 00st the time at departure but 05th feb!!! the aircraft arrives 4h later so at b 05 00st +4h > 09 00st 5th feb. Question 158-26
What meant the term 'polar circle' It the parallel at lowest latitude at which an observer can see sun 24 hours above horizon. 1st nothing to do with coordinates! 2nd select required facts at point a 08 00st (04th feb) a 9h b +12h a >b 1736nm (ground) a >b 1636nm (air) wind 25kt st time at b? 3rd calculate ground air distance 100nm use logic 100nm/25kt = 4h the flight time! at point a 08 00st +9h > 17 00utc the time at departure at point b 17 00utc +12h > 05 00st the time at departure but 05th feb!!! the aircraft arrives 4h later so at b 05 00st +4h > 09 00st 5th feb. Question 158-27
Grivation 56w when It the parallel at lowest latitude at which an observer can see sun 24 hours above horizon. grivation = convergence + variation westerly grivation = negative value easterly grivation = positive value (just like variation) grid heading = magnetic heading ± grivation grid heading = magnetic heading w grivation (or + e grivation) 103° (gh) = 159° (mh) 56° (grivation). Question 158-28
The term 'ellipsoi may be used to describe It the parallel at lowest latitude at which an observer can see sun 24 hours above horizon. grivation = convergence + variation westerly grivation = negative value easterly grivation = positive value (just like variation) grid heading = magnetic heading ± grivation grid heading = magnetic heading w grivation (or + e grivation) 103° (gh) = 159° (mh) 56° (grivation). Question 158-29
By 'ecliptic' meant The apparent yearly path of sun around earth. grivation = convergence + variation westerly grivation = negative value easterly grivation = positive value (just like variation) grid heading = magnetic heading ± grivation grid heading = magnetic heading w grivation (or + e grivation) 103° (gh) = 159° (mh) 56° (grivation). Question 158-30
The duration of civil twilight the time Between sunset when centre of sun 6° below celestial horizon. the term 'civil twilight' generally used to refer to a certain period in morning it refers to that period of time when center of sun less than 6 degrees below celestial horizon it ends at sunrise during civil twilight natural illumination supposed to be sufficient to see objects outdoors well enough to get along without artificial lighting. Question 158-31
An aircraft over position ho 55°30'n 060°15'w where yyr vor 53°30'n 060°15'w can be received the magnetic variation 31°w at ho and 28°w at yyr what the radial from yyr Between sunset when centre of sun 6° below celestial horizon. ho postion yyr vor are on same meridian (060°15'w) spaced 2° of latitude ho position located north of yyr variation applied at vor (at position or aircraft a ndb) «variation west magnetic best» aircraft on radial 028° (360° + 28°w) from yyr vor. Question 158-32
An aircraft in northern hemisphere making an accurate rate one turn to right if initial heading was 135° after 30 seconds direct reading magnetic compass should read Between sunset when centre of sun 6° below celestial horizon. in case of a turning speed of 3° per second (2 minutes bend or rate one turn) following rules are to be observed to a course north turn will be ended when compass indicates 30° less than north (undershoot) to a course south turn will be ended when compass indicates 30° past south (overshoot) to a course east or west turn will be ended when compass indicates desired course to a course between east or west north or south an undershoot or overshoot between 30° 0° used depending on whether desired course closer or less close to north or south general rule of thumb never see the north ever see the south for 30 seconds will turn through 90° the aircraft should be on a heading of 135° + 90° = 225° the compass will over read when turning through south in northern hemisphere read more than 225°. Question 158-33
When accelerating on a westerly heading in northern hemisphere compass card of a direct reading magnetic compass will turn Anti clockwise giving an apparent turn towards north. cmarzocchini westerly mean to west the correct answer should be clockwise to north! if your direct reading magnetic compass indicates a turn to north when on a westerley heading it turns anti clockwise because your heading on compass will increase img /com_en/com061 70a jpg if your heading increases from 270° to 300° while accelerating 270° moves anti clockwise the 300° will replace it img /com_en/com061 70b jpg it even more understable on a vertical card compass. Question 158-34
A ground feature appears 30° to left of centre line of crt of an airborne weather radar if heading of aircraft 355° m and magnetic variation 15° east true bearing of aircraft from feature Anti clockwise giving an apparent turn towards north. if we were in an area with no magnetic variation our magnetic heading our true heading will be 355° in that case bearing of ground feature will be 355° 30° = 325° from ground feature our aircraft bearing will be 325° 180° = 145° but magnetic variation 15° east ! thus our magnetic heading 355° magnetic variation 15° east our true heading 355°+15° = 010° 010° 30°= 340° 340° 180° = 160°. Question 158-35
When decelerating on a westerly heading in northern hemisphere compass card of a direct reading magnetic compass will turn Clockwise giving an apparent turn toward south. acceleration errors these are caused inertia on east west headings because centre of gravity of compasse under pivot point accelerating makes bulk of compass lag behind machine dispace centre of gravity aft of pivot point img /com_en/com061 350 jpg if you were just going north south all you would get extra dip but because you are going east or west north bit of compass pointing to side of aircraft the displaced centre of gravity not being vertically in line with pivot point goes towards north to create a couple that makes compass turn clockwise to read less than 90° during turn a deceleration has oppposite effect to south (in northern hemisphere) acceleration errors are maximum on east/west headings near magnetic poles nil on north/south headings at equator the watchword here ands northern hemisphere (accelerate north decelerate south) or sand southern hemisphere (south accelerate north decelerate) img /com_en/com061 77 jpg example in northern hemisphere flying east if you accelerate needle will deflect to nearest pole (north an easterly deviation) to south when decelerate during deceleration after landing on runway 18 example a compass in northern hemisphere would indicate no apparent turn during deceleraton after landing in an easterly direction a magnetic compass in northern hemisphere indicates an apparent turn south during deceleraton after landing in an westerly direction a magnetic compass in southern hemisphere indicates an apparent turn north. Question 158-36
When turning right from 330° c to 040° c in northern hemisphere reading of a direct reading magnetic compass will Under indicate turn liquid swirl will increase effect. for turning errors northern hemisphere 'uno underturn/under read through north overturn/over read through south southern hemisphere 'onu overturn/over read through north underturn/under read through south. Question 158-37
When accelerating on an easterly heading in northern hemisphere compass card of a direct reading magnetic compass will turn Clockwise giving an apparent turn toward north. acceleration errors these are caused inertia on east west headings because centre of gravity of compasse under pivot point accelerating makes bulk of compass lag behind machine dispace centre of gravity aft of pivot point img /com_en/com061 350 jpg if you were just going north south all you would get extra dip but because you are going east or west north bit of compass pointing to side of aircraft the displaced centre of gravity not being vertically in line with pivot point goes towards north to create a couple that makes compass turn clockwise to read less than 90° during turn a deceleration has oppposite effect to south (in northern hemisphere) acceleration errors are maximum on east/west headings near magnetic poles nil on north/south headings at equator the watchword here ands northern hemisphere (accelerate north decelerate south) or sand southern hemisphere (south accelerate north decelerate) img /com_en/com061 77 jpg example in northern hemisphere flying east if you accelerate needle will deflect to nearest pole (north an easterly deviation) to south when decelerate during deceleration after landing on runway 18 example a compass in northern hemisphere would indicate no apparent turn during deceleraton after landing in an easterly direction a magnetic compass in northern hemisphere indicates an apparent turn south during deceleraton after landing in an westerly direction a magnetic compass in southern hemisphere indicates an apparent turn north. Question 158-38
A direct reading compass should be swung when There a large permanent change in magnetic latitude. a compass swing should be done on installation of compasse in first place as per maintenance schedules whenever there any doubt about accuracy after a shock to airframe or a lightning strike if aircraft has been left standing some time or has moved to a significantly different latitude if there a large permanent change in magnetic latitude (angle of dip) if a major component or electrical installation has changed august 2020 this question also exists with right answer a period of 12 months has passed during which aircraft has remained stationary on ground. Question 158-39
The direct reading magnetic compass made aperiodic dead beat Keeping magnetic assembly mass close to compass point by using damping wires. aperiodicity the ability to settle quickly after a disturbance without overshooting or oscillating which helped the (transparent) suspension liquid a wire spider assembly two magnets keep mass of assembly near pivot reducing inertia light alloys reduce inertia even more thus mass of assembly kept close to compass point damping wires are used. Question 158-40
Given true track 348° drift 17° left variation 32°w deviation 4°e what the compass heading Keeping magnetic assembly mass close to compass point by using damping wires. use this wonderful table those questions.
Exclusive rights reserved. Reproduction prohibited under penalty of prosecution.
6279 Free Training Exam