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Question 174-1 : What is 'vertex' ? [ Practice protocol ]
The point on the great circle with greatest latitude.
. 1718.a great circle which has an inclination of 33° to the equator reach a latitude of 33° n/s..the point on the great circle with greatest latitude is called vertex..every great circle has two vertices, one in the northern hemisphere and the other in the southern hemisphere..at the 'vertex', the great circle track is 090° or 270°, and it is the closest point to the pole and the farther from the equator... notice this question also exists with the following statement.'which statement about the vertex of a great circle is true.in the vertex the great circle reaches its highest latitude'.
Question 174-2 : Which statement about the vertex of a great circle is always true ?
In the vertex the true track is 090° or 270°.
. 1718.a great circle which has an inclination of 33° to the equator reach a latitude of 33° n/s..the point on the great circle with greatest latitude is called vertex..every great circle has two vertices, one in the northern hemisphere and the other in the southern hemisphere..at the 'vertex', the great circle track is 090° or 270°, and it is the closest point to the pole and the farther from the equator.
Question 174-3 : A small circle ?
Has a plane that does not pass through the centre of the earth.
.the centre of small circles does not coincide with the centre of the earth.
Question 174-4 : Spring and autumn equinox are the moments at which the sun reaches ?
A declination of 0°.
Ecqb03, august 2016....declination of the sun is the angular distance of the sun north or south of the celestial equator.. /com en/com061 682a.jpg..the earth's equator is tilted 23.5 degrees with respect to the plane of the earth's orbit around the sun, so at various times during the year, as the earth orbits the sun, declination varies from 23.5 degrees north to 23.5 degrees south.. /com en/com061 682b.jpg..this gives rise to the seasons. around december 21, the northern hemisphere of the earth is tilted 23.5 degrees away from the sun, which is the winter solstice for the northern hemisphere and the summer solstice for the southern hemisphere. around june 21, the southern hemisphere is tilted 23.5 degrees away from the sun, which is the summer solstice for the northern hemisphere and winter solstice for the southern hemisphere. on march 21 and september 21 are the spring and autumn equinoxes when the sun is passing directly over the equator a declination of 0°. note that the tropics of cancer and capricorn mark the maximum declination of the sun in each hemisphere.
Question 174-5 : An aircraft at latitude 02°20'n tracks 360° t for 685 km. on completion of the flight the latitude will be ?
08°29'n.
.on a meridian, 1° = 60 nm..we are heading north..685 km / 1.852 = 370 nm..370 / 60 = 6.16°..0.16° x 60 = 9.6 min..02°20'n + 6°09' = 08°29'n.
Question 174-6 : 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 of this great circle ?
50°06.4s, 105°00.0'e.
. 1718.a great circle which has an inclination of 33° to the equator reach a latitude of 33° n/s..the point on the great circle with greatest latitude is called vertex..every great circle has two vertices, one in the northern hemisphere and the other in the southern hemisphere...for this question, latitude is 50°06.4n, therefore the antipode is 50°06.4s..concerning the longitude, we are at + or 180° so 075°w + or 180° = 105°e.
Question 174-7 : Sunrise in dublin 53°29'n, 006°15'w is 06 23 lmt. calculate the sunrise at bremen airport 53°09'n,.008°45'e in lmt ?
06 23.
.difference in longitude 006°15'+008°45' = 15°...the earth moves around the sun at a rate of 4 minutes per degree of longitude 15°/h. latitude remains the same, only the longitude changes, the sun rises at bremen airport 1h before dublin, but the sun rises at the same local mean time.
Question 174-8 : Calculate the difference in lmt between dublin 53°29'n, 006°15'w and bremen airport 53°09'n, 008°45'e ?
01h 00m.
Question 174-9 : Calculate the 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 ?
329.4 nm.
.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 is said 20 nm of north of bremen airport...as you know 1 degree = 60 mins and 1 degree = 60 nm. so 1 min = 1 nm...because of that, the waypoint is located at the same latitude than dbu waypoint. in this case we should use the rhumb line formula...d = change in long * 60* cos lm..d = 9.2266 * 60 * cos 53.5 = 329.2939 nm.
Question 174-10 : Calculate the 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 ?
535.7 nm.
.dublin > bremen = 15°..1°=60nm on the 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 174-11 : Route a 53°24'n, 015°54'e to b 32°00'n, 052°51'w..distance flight plan is 3150 nm, average gs is 450 kt..difference between standard time a and utc is 1 hour, difference between standard time b and utc is 4 hours..estimated time of arrival eta b is 10 00 st b on 5 august.. ?
08 00 st a 05/08.
.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 174-12 : In producing chart projections, the following projection surfaces may be used ?
Plane, cylinder, cone.
Question 174-13 : The term 'oblique' in relation to map projections means that ?
The axis of the cylinder or cone is neither parallel to or perpendicular to the earth's axis of rotation.
Question 174-14 : A day at a place as measured in local mean time starts ?
When the mean sun transits the anti meridian of the place in question.
Ecqb03, december 2016
Question 174-15 : Standard time is ?
The time enforced by the legal authority to be used in a country or an area.
Question 174-16 : Daylight saving time summer time is..1 used to extend the sunlight period in the evening.2 introduced by setting the standard time forward by one hour.3 used in some countries ?
1, 2, 3.
Question 174-17 : The countries having a standard time slow on utc ?
Will generally be located at western longitudes.
Question 174-18 : The international date line is located ?
At the 180° e/w meridian, or in the vicinity of this meridian.
Question 174-19 : Consider the following statements on sunset ?
Sunset is the time when the observer at sea level sees the last part of the sun disappear below the horizon.
Question 174-20 : Atmospheric refraction ?
Causes the sunrise to occur earlier and the sunset to occur later.
Question 174-21 : Consider the following statements on sunrise and sunset ?
At equator sunrise and sunset occur at quite regular times throughout the year.
Question 174-22 : 'true north' is ?
The direction along any meridian toward the true north pole.
Question 174-23 : In international aviation the following units shall be used for horizontal distance ?
Metres, kilometres and nautical miles.
Question 174-24 : When dealing with heights and altitudes in international aviation, we use the following units ?
Metre and feet.
Question 174-25 : 'kilometre' is defined as ?
A 1/10000 part of the meridian length from equator to the pole.
Question 174-26 : How long is 25 kilometres at 60°n ?
13.5 nautical mile.
Question 174-27 : 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. ?
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 174-28 : 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.. ?
09h35m and 02h42m.
Question 174-29 : 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 174-30 : 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 174-31 : The term 'ellipsoid' may be used to describe ?
The shape of the earth.
Question 174-32 : By 'ecliptic' is meant ?
The apparent yearly path of the sun around the earth.
Question 174-33 : 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 174-34 : The duration of civil twilight is the time ?
Between sunset and when the centre of the sun is 6° below the celestial horizon.
Question 174-35 : 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 174-36 : 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 174-37 : 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 174-38 : 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 174-39 : 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 174-40 : 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.
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