Causes the sunrise to occur earlier and the sunset to occur later.
Question 188-2 : Consider the following statements on sunrise and sunset ?
At equator sunrise and sunset occur at quite regular times throughout the year.
Question 188-3 : 'true north' is ?
The direction along any meridian toward the true north pole.
Question 188-4 : In international aviation the following units shall be used for horizontal distance ?
Metres, kilometres and nautical miles.
Question 188-5 : When dealing with heights and altitudes in international aviation, we use the following units ?
Metre and feet.
Question 188-6 : 'kilometre' is defined as ?
A 1/10000 part of the meridian length from equator to the pole.
Question 188-7 : How long is 25 kilometres at 60°n ?
13.5 nautical mile.
Question 188-8 : 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 is 1736 nm..on 4 february at 08 00 st a an aircraft is exactly above a. at the moment the aircraft arrives at position b the air distance between a and b is ?
09 00 st b 05/02.
.1st nothing to do with coordinates..2nd select the 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 is the flight time..at point a 08 00st +9h > 17 00utc is the time at departure...at point b 17 00utc +12h > 05 00st is the time at departure. but 05th feb ..the aircraft arrives 4h later. so,at b 05 00st +4h >..09 00st 5th feb.
Question 188-9 : Estimated time of departure a 15°15.0'n, 072°06.0'w on 12 march is 01 00 st st = utc 5 h..estimated time of arrival b 55°18.0'n, 005°45.0'e 16 15 st on the same date st = utc + 1 h..according to the jeppesen table sunset at b occurs at 18 57..calculate the flight time from a to b and the time ?
09h35m and 02h42m.
Question 188-10 : What is meant by the term 'polar circle' ?
It is the parallel at the lowest latitude at which an observer can see the sun for 24 hours above the horizon.
Question 188-11 : Grivation is 56w when ?
Gh is 103° and mh is 159°.
.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 188-12 : The term 'ellipsoid' may be used to describe ?
The shape of the earth.
Question 188-13 : By 'ecliptic' is meant ?
The apparent yearly path of the sun around the earth.
Question 188-14 : Given.ils glide path angle = 3.5°, ground speed = 150 kt..what is the approximate rate of descent ?
875 ft/min.
.1 in 60 rule is a rule of thumb. 3.5 x 150 x 100 /60 = 875 ft/min.
Question 188-15 : For a given track, the following data is provided.wind component = +45 kt.drift angle = 15° left.true air speed tas = 240 kt.what is the wind component on the reverse track ?
65 kt.
Since the initial wind component is +45 knots on the original track, the reverse track will have the wind component as 45 knots.
Question 188-16 : The duration of civil twilight is the time ?
Between sunset and when the centre of the sun is 6° below the celestial horizon.
Question 188-17 : An aircraft is over position ho 55°30'n 060°15'w , where yyr vor 53°30'n 060°15'w can be received..the magnetic variation is 31°w at ho and 28°w at yyr. what is the radial from yyr ?
028°.
.ho postion and yyr vor are on the same meridian 060°15'w , spaced by 2° of latitude..ho position is located north of yyr...variation is applied at the vor at the position or the aircraft for a ndb . variation west, magnetic best , aircraft is on the radial 028° 360° + 28°w from the yyr vor.
Question 188-18 : An aircraft in the northern hemisphere is making an accurate rate one turn to the right. if the initial heading was 135°, after 30 seconds the direct reading magnetic compass should read ?
More than 225°.
.in case of a turning speed of 3° per second 2 minutes bend or rate one turn , the following rules are to be observed.. to a course north, the turn will be ended when the compass indicates 30° less than north undershoot... to a course south, the turn will be ended when the compass indicates 30° past south overshoot... to a course east or west, the turn will be ended when the compass indicates the desired course... to a course between east or west and north or south, an undershoot or overshoot between 30° and 0° is used, depending on whether the desired course is closer or less close to the north or south...general rule of thumb.never see the north.ever see the south..for 30 seconds will turn through 90° and the aircraft should be on a heading of 135° + 90° = 225°...the compass will over read when turning through south in the northern hemisphere and read more than 225°.
Question 188-19 : When accelerating on a westerly heading in the northern hemisphere, the compass card of a direct reading magnetic compass will turn ?
Anti clockwise giving an apparent turn towards the north.
Cmarzocchini.westerly mean to the west. the correct answer should be clockwise to the north...if your direct reading magnetic compass indicates a turn to the north when on a westerley heading it turns anti clockwise because your heading on the compass will increase.. /com en/com061 70a.jpg.if your heading increases from 270° to 300° while accelerating, the 270° moves anti clockwise and the 300° will replace it... /com en/com061 70b.jpg.it is even more understable on a vertical card compass.
Question 188-20 : A ground feature appears 30° to the left of the centre line of the crt of an airborne weather radar. if the heading of the aircraft is 355° m and the magnetic variation is 15° east, the true bearing of the aircraft from the feature is ?
160°.
.if we were in an area with no magnetic variation, our magnetic heading and our true heading will be 355°..in that case, the bearing of the ground feature will be 355° 30° = 325°, and from the ground feature, our aircraft bearing will be 325° 180° = 145°... but magnetic variation is 15° east thus..our magnetic heading is 355°, magnetic variation is 15° east, our true heading is 355°+15° = 010°...010° 30°= 340°...340° 180° = 160°.
Question 188-21 : When decelerating on a westerly heading in the northern hemisphere, the compass card of a direct reading magnetic compass will turn ?
Clockwise giving an apparent turn toward the south.
.acceleration errors.these are caused by inertia on east west headings..because the centre of gravity of the compasse is under the pivot point, accelerating makes the bulk of the compass lag behind the machine and dispace the centre of gravity aft of the pivot point... /com en/com061 350.jpg..if you were just going north south, all you would get is extra dip, but because you are going east or west, the north bit of the compass is pointing to the side of the aircraft, and the displaced centre of gravity, not being vertically in line with the pivot point, goes towards north to create a couple that makes the compass turn clockwise to read less than 90° during the turn. a deceleration has the oppposite effect to the south in the northern hemisphere...acceleration errors are maximum on east/west headings and near the magnetic poles, and nil on north/south headings, and at the equator...the watchword here is ands for northern hemisphere accelerate north, decelerate south , or sand for southern hemisphere south accelerate, north decelerate... /com en/com061 77.jpg..example.in the northern hemisphere flying east, if you accelerate, the needle will deflect to the nearest pole north, for an easterly deviation and to the south when decelerate... during deceleration after landing on runway 18 for example, a compass in the northern hemisphere would indicate no apparent turn... during deceleraton after landing in an easterly direction, a magnetic compass in the northern hemisphere indicates an apparent turn south... during deceleraton after landing in an westerly direction, a magnetic compass in the southern hemisphere indicates an apparent turn north.
Question 188-22 : When turning right from 330° c to 040° c in the northern hemisphere, the reading of a direct reading magnetic compass will ?
Under indicate the turn and liquid swirl will increase the effect.
.for turning errors.northern hemisphere unos underturn/under read through north, overturn/over read through south..southern hemisphere onus overturn/over read through north, underturn/under read through south.
Question 188-23 : When accelerating on an easterly heading in the northern hemisphere, the compass card of a direct reading magnetic compass will turn ?
Clockwise giving an apparent turn toward the north.
.acceleration errors.these are caused by inertia on east west headings..because the centre of gravity of the compasse is under the pivot point, accelerating makes the bulk of the compass lag behind the machine and dispace the centre of gravity aft of the pivot point... /com en/com061 350.jpg..if you were just going north south, all you would get is extra dip, but because you are going east or west, the north bit of the compass is pointing to the side of the aircraft, and the displaced centre of gravity, not being vertically in line with the pivot point, goes towards north to create a couple that makes the compass turn clockwise to read less than 90° during the turn. a deceleration has the oppposite effect to the south in the northern hemisphere...acceleration errors are maximum on east/west headings and near the magnetic poles, and nil on north/south headings, and at the equator...the watchword here is ands for northern hemisphere accelerate north, decelerate south , or sand for southern hemisphere south accelerate, north decelerate... /com en/com061 77.jpg..example.in the northern hemisphere flying east, if you accelerate, the needle will deflect to the nearest pole north, for an easterly deviation and to the south when decelerate... during deceleration after landing on runway 18 for example, a compass in the northern hemisphere would indicate no apparent turn... during deceleraton after landing in an easterly direction, a magnetic compass in the northern hemisphere indicates an apparent turn south... during deceleraton after landing in an westerly direction, a magnetic compass in the southern hemisphere indicates an apparent turn north.
Question 188-24 : A direct reading compass should be swung when ?
There is a large, and permanent, change in magnetic latitude.
.a compass swing should be done. on installation of the compasse in the first place.. as per maintenance schedules.. whenever there is any doubt about accuracy.. after a shock to the airframe or a lightning strike.. if the aircraft has been left standing for some time or has moved to a significantly different latitude.. if there is a large and permanent change in magnetic latitude angle of dip.. if a major component or electrical installation has changed...august 2020 this question also exists with the right answer. a period of 12 months has passed during which the aircraft has remained stationary on the ground.
Question 188-25 : The direct reading magnetic compass is made aperiodic dead beat by ?
Keeping the magnetic assembly mass close to the compass point and by using damping wires.
.aperiodicity the ability to settle quickly after a disturbance, without overshooting or oscillating, which is helped by the transparent suspension liquid and a wire spider assembly. two magnets keep the mass of the assembly near the pivot, reducing inertia. light alloys reduce inertia even more. thus, the mass of the assembly is kept close to the compass point, and damping wires are used.
Question 188-26 : Given.true track is 348°, drift 17° left, variation 32°w, deviation 4°e..what is the compass heading ?
033°.
. 1429.use this wonderful table for those questions.
Question 188-27 : Which of the following statements is correct concerning the effect of turning errors on a direct reading compass ?
Turning errors are greatest on north/south headings, and are greatest at high latitudes
.turning errors are caused by the dip of the compass needle which increases with latitude..turning errors are greatest on north/south headings and they are zero when turning on east/west headings.
Question 188-28 : Which of the following is an occasion for carrying out a compass swing on a direct reading compass ?
After an aircraft has passed through a severe electrical storm, or has been struck by lightning.
.a compass swing should be done. on installation of the compasse in the first place.. as per maintenance schedules.. whenever there is any doubt about accuracy.. after a shock to the airframe or a lightning strike.. if the aircraft has been left standing for some time or has moved to a significantly different latitude.. if there is a large and permanent change in magnetic latitude angle of dip.. if a major component or electrical installation has changed.
Question 188-29 : When is the magnetic compass most effective ?
About midway between the magnetic poles.
. 1430.a direct reading compass has a pivoted magnet that is free to aligne itself with the horizontal component of the earth's magnetic field. 1945.the compass is pendulous and symmetrical and will sit level if there is no magnetic field or if the field is horizontal with no vertical component of the earth's magnetic force, as on the magnetic equator...magnetic compass is the most effective about midway between the magnetic poles.
Question 188-30 : When an aircraft on a westerly heading on the northern hemisphere accelerates, the effect of the acceleration error causes the magnetic compass to ?
Indicate a turn towards the north.
.acceleration/deceleration errors are false compass indications of a swing to the north or south during speed changes of the airplane. this error is most pronounced when flying on a heading of east or west and decreases when flying closer to a north or south heading. in a direct north or south heading this error does not occur..in the northern hemisphere, the compass swings towards the north during acceleration and towards the south during deceleration. in the southern hemisphere this error occurs the other way round. when the speed stabilises, the error disappears...learn this for northern hemisphere ands. a cceleration gives apparent turn to n orth.. d eceleration gives apparent turn to s outh.
. 1443.use this wonderful table for those questions.
Question 188-36 : Deviation applied to magnetic heading gives ?
Compass heading.
Learn the composition of this table.. /com en/com061 141.jpg..
Question 188-37 : Given.true course from a to b = 090°, tas = 460 kt, w/v = 360/100kt, average variation = 10°e, deviation = 2°..calculate the compass heading and gs ?
070° 450 kt.
.put 460 kt tas on center dot, under true index set true course 090°. on the rotative scale, under the wind mark 360°/100 kt, you read 12° right drift.. /com en/com061 198a.jpg..turn to set under true index the true heading 078° 090° 12° drift , you now read a ground speed under the wind mark of 450 kt... /com en/com061 198b.jpg..your true heading is 078°, minus 10°e variation, your magnetic heading is 068°..for compass heading, add 2°w deviation 068° + 2° = 070°.
Question 188-38 : An island is observed by weather radar to be 15° to the left. the aircraft heading is 120° m and the magnetic variation 17°w. what is the true bearing of the aircraft from the island ?
268°.
.our magnetic heading is 120°, magnetic variation is 17° west, our true heading is 120° 17° = 103°...true bearing of the island from the aircraft 103° 15°= 088°...true bearing of the aircraft from the island 088° 180° = 268°.
Question 188-39 : The main reason for mounting the detector unit of a remote reading compass in the wingtip of an aeroplane is ?
To minimise the amount of deviation caused by aircraft magnetism and electrical circuits
Question 188-40 : A ground feature was observed on a relative bearing of 315° and 3 minutes later on a relative bearing of 270°. the wind is calm..aircraft gs 180 kt..the minimum distance between the aircraft and the ground feature is ?
9 nm.
. 2524.we have an isosceles triangle, and in an isosceles triangle, two sides are equal in length..3 minutes at 180 kt = 9 nm...this question has been seen in february 2023 with a relative bearing of 225° instead of 315°. the answer remains the same, the aircraft is flying south instead of north.
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