WorksheetsIOP. General Navigation
Total questions: 233
Worksheet time: 58hrs 15mins
The term 'ellipsoid' may be used to describe:
a) a great circle on the celestial sphere.
b) the shape of the ecliptic.
c) the movement of the Earth around the Sun.
d) the shape of the Earth.
The circumference of the Earth is approximately:
a) 43.200 NM
b) 10.800 NM
c) 21.600 NM
d) 5.400 NM
The Earth is:
a) a sphere whose centre is equidistant (the same distance) from the Poles and the
Equator.
b) a sphere which has a larger polar circumference than equatorial circumference.
c) none of the above statements is correct.
d) considered to be a perfect sphere as far as basic (simple) navigation is
concerned.
Seasons are due to the:
a) Earth's rotation on its polar axis.
b) Earth's elliptical orbit around the Sun.
c) variable distance between Earth and Sun.
d) inclination of the polar axis with the ecliptic plane.
The poles on the surface of the Earth may be defined as:
a) the points on the surface of the Earth where all meridians intersect at right angles.
b) the points where the Earth's axis of rotation penetrates the surface of the
Earth.
c) the points at which the vertical lines runs through the centre of the Earth.
d) the points from where the distance to the equator is equal.
When the Sun's declination is northerly:
a) midnight Sun may be observed at the south pole.
b) it is winter on the northern hemisphere.
b) it is winter on the northern hemisphere.
d) the daylight period is shorter in the southern hemisphere than the northern.
c) the sunrise occurs earlier at southern latitudes than the northern latitudes.
Consider the positions (00°N/S, 000°E/W) and (00°N/S, 180°E/W) on the ellipsoid.
Which statement about the distances between these positions is correct?
a) The route via the equator is shorter than the route via the South Pole.
b) The route via the North Pole is shorter than the route along the equator.
c) The route via either pole and the route via the equator are of equal length.
d) The route via the South pole is shorter than the route via the North Pole.
The maximum difference between geocentric and geodetic latitude occurs at about:
a) 60° North and South.
b) 90° North and South.
c) 0° North and South (equator).
d) 45° North and South.
What is difference in latitude from 30°39'S-20°20'E to 45°23'N-40°40'E:
a) 14°44'N
b) 76°2'N
c) 76°2'S
d) 14°44'S
Given: A (56°N , 145°E); B (57°N , 165°W). What is the difference in longitude
between A and B?
a) 020°
b) 050°
c) 001°
d) 130°
11.Which statement about the duration of daylight is true?
a) Close to the solstices the influence of latitude on the duration of daylight is at its
smallest.
b) In summer the length of the period of daylight decreases with increasing latitude.
c) Close to the equinoxes the influence of latitude on the duration of daylight is
at its smallest.
d) On September 10th the duration of daylight is longer on the Southern Hemisphere
than on the Northern Hemisphere.
The angle between True North and Magnetic North is known as:
a) deviation.
b) alignment error.
c) variation.
d) dip.
The definition of True North for any observer is:
a) the direction of the Greenwich meridian to the North Pole.
b) the direction of the observer's Magnetic North corrected for local variation.
c) the direction of the observer's meridian to the North Pole.
d) the reading of the observer's compass corrected for deviation and local variation.
The purpose of establishing a grid is:
a) to provide a system for directions where a great circle has a constant
direction, even if its true direction varies.
b) minimise the errors introduced when making calculations involving variation.
c) make a chart covering high latitudes that has the same qualities as the equatorial
Mercator chart.
d) to make the system of latitude and longitude available on a gridded map.
Compass deviation is defined as the angle between:
a) Magnetic North and compass North.
b) the horizontal and the total intensity of the Earth's magnetic field.
c) True North and compass North.
d) True North and Magnetic North.
The evaluation of your plotting work shows a WCA +3° and a drift 3° left:
a) your actual position is on the intended track.
b) the GS was exactly calculated.
c) the expected W/V and the actual W/V coincide.
d) the track error is 6°.
You should follow a track due North taking account of a northwesterly wind. You
calculated a WCA -8°.
a) The drift will be 8° left.
b) A track error of -2° (Ieft) shows a WCA of only -6°.
c) The drift will be 8° right.
d) A track error of 2° (right) shows a drift of 10° right.
Given: Course: 040°(T); TAS: 120 kts; Wind speed: 30 kts. Maximum drift angle will
be obtained for a wind direction of:
a) 120°
b) 145°
c) 130°
Given: TAS: 132 kts; True HDG: 257°; W/V: 095°(T)/35 kts. Calculate the drift
angle and GS.
a) 2°R - 166 kts.
b) 4°R - 165 kts.
c) 3°L - 166 kts.
d) 4°L - 167 kts.
Given: TAS: 472 kts; True HDG: 005°; W/V: 110°(T)/50 kts; Calculate the drift
angle and GS.
a) 6°L - 487 kts.
b) 7°R - 487 kts.
c) 7°L - 491 kts.
d) 7°R - 491 kts
1 nautical mile equals:
a) 5 280 feet.
b) 1 852 metres.
c) 3 081 yards.
d) 0,896 statute mile.
In international aviation the following units shall be used for horizontal distance:
a) metres, statute miles and nautical miles.
b) metres, kilometres and nautical miles.
c) kilometres, statute miles and nautical miles.
d) kilometres, feet and nautical miles.
What are the initial true course and distance between positions 58°00'N-013°00'W
and 66°00'N-002°00'E?
a) 032° - 470 NM.
b) 036° - 638 NM.
c) 042° - 635 NM.
d) 029° - 570 NM.
The distance between positions A and B is 180 NM. An aircraft departs position A
and after having traveled 60 NM, its position is pinpointed 4 NM left of the intended
track. Assuming no change in wind velocity, what alteration of heading must be made in
order to arrive at position B?
a) 8° right.
b) 6° right.
c) 4° right.
d) 2° left.
Given: AD = Air distance; GD = Ground distance; TAS = True Airspeed; GS =
Groundspeed. Which of the following is the correct formula to calculate ground distance
(GD) gone?
a) GD = (AD - TAS)/TAS
b) GD = AD X (GS -TAS)/GS
c) GD = (AD X GS)/TAS
d) GD = TAS/(GS X AD)
Given: AD: Air distance; GD: Ground distance; TAS: True airspeed; GS: Ground
speed. Which of the following is the correct formula to calculate ground distance (GD)
gone?
a) GD = (AD _ GS) ÷ TAS
b) GD = TAS ÷ (GS x AD)
c) GD = AD x (GS -TAS) ÷ GS
d) GD = (AD - TAS) ÷ TAS
An aircraft is flying at FL180 and the outside air temperature is -30°C. If the CAS is
150 kt, what is the TAS?
a) 115 kt
b) 180 kt
c) 195 kt
d) 145 kt
Given: CAS: 230 kts; FL120; OAT: -10°C. What is the TAS?
a) 266 kts
b) 280 kts
c) 273 kts
d) 287 kts
Given: Mach number: 0.8; Flight level: 330; OAT: ISA +15°C; TAS is approximately
(compressibility factor 0.94):
a) 420 kts
b) 450 kts
c) 265 kts
d) 480 kts
You are flying at FL80 and air temperature is ISA +15. What CAS is required to
make TAS 240 kts?
a) 214 kts
b) 208 kts
c) 226 kts
d) 220 kts
An aircraft is climbing at a constant CAS in ISA conditions. What will be the effect
on TAS and Mach number?
a) TAS increases and Mach No decreases.
b) Both increase.
c) TAS decreases and Mach No increases.
d) Both decrease.
Given: Half way between two reporting points the navigation log gives the following
information: TAS 360 kt, W/V 330°/80kt, Compass heading 237°, Deviation on this
heading -5°, Variation 19°W. What is the average ground speed for this leg?
a) 354 kt
b) 373 kt
c) 403 kt
d) 360 kt
The Sun moves from East to West at a speed of 15° longitude an hour. What ground
speed will give you the opportunity to observe the Sun due South at all times at 60°00'N?
a) 300 kts
b) 450 kts
c) 520 kts
d) 780 kts
You start from P (70°00'N-015°00'E) and fly westward along the parallel of latitude
for 2 hours at ground speed 220 kts. What is your position after two hours flight?
a) 006°26'W
b) 021°26'W
c) 007°40'E
d) 006°44'W
Given: Actual HDG: 290°; TAS: 250 kts; Wind: 135/75 kts. What is the ground
speed?
a) 300 kts
b) 175 kts
c) 320 kts
An aircraft travels 2,4 statute miles in 47 seconds. What is its ground speed?
a) 183 kts
b) 160 kts
c) 131 kts
d) 209 kts
An aircraft is planned to fly from position A to position B, distance 320 NM, at an
average GS of 180 kts. It departs A at 12:00 UTC. After flying 70 NM along track from
A, the aircraft is 3 min ahead of planned time. Using the actual GS experienced, what is
the revised ETA at B?
a) 1401 UTC
b) 1333 UTC
c) 1340 UTC
d) 1347 UTC
Given: A descending aircraft flies in a straight line to a DME; DME 55,0 NM,
altitude 33.000 ft; DME 43,9 NM, altitude 30.500 ft; M = 0.72, GS = 525 kts, OAT =
ISA. The descent gradient is:
a) 3,70%
b) 3,90%
c) 3,50%
d) 4,10%
An aircraft is maintaining a 5,2% gradient at 7 NM from the runway, on a flat terrain
its height is approximately:
a) 2.210 ft
b) 3.640 ft
c) 1.890 ft
d) 680 ft
You are departing from an airport which has an elevation of 2.000 ft. The QNH is
1013 hPa. 10 NM away there is a waypoint you are required to pass at an altitude of
7.500 ft. Given a groundspeed of 100 kts, what is the minimum rate of climb?
a) 920 ft/min
b) 590 ft/mins
c) 750 ft/min
d) 1.080 ft/min
An aircraft is departing from an airport which has an elevation of 2.000 ft and the
QNH is 1023 hPa. The TAS is 100 kts, the head wind component is 20 kts and the rate of
climb is 1.000 ft/min. Top of climb is FL100. At what distance from the airport will this
be achieved?
a) 16,6 NM
b) 11,1 NM
c) 10,3 NM
d) 13,3 NM
An aircraft at FL350 is required to commence descent when 85 NM from a VOR and
to cross the VOR at FL80. The mean GS for the descent is 340 kts. What is the minimum
rate of descent required?
a) 1.900 ft/min.
b) 1.600 ft/min.
c) 1.800 ft/min.
d) 1.700 ft/min.
An aircraft at FL390 is required to descend to cross a DME facility at FL70.
Maximum rate of descent is 2.500 ft/min, mean GS during descent is 248 kts. What is the
minimum range from the DME at which descent should commence?
a) 63 NM
b) 58 NM
c) 68 NM
d) 53 NM
The outer marker of an ILS with a 3° glide slope is located 4,6 NM from the
threshold. Assuming a glide slope height of 50 ft above the threshold, the approximate
height of an aircraft passing the outer marker is (use the 1:60 rule):
a) 1.400 ft
b) 1.450 ft
c) 1.300 ft
d) 1.350 ft
Construct the triangle of velocities on a piece of paper, showing the following data:
True Heading: 305°
TAS: 135 kts
W/V: 230°/40
Period of time: from 11:30 to 11:45
What is the track in this period of time?
a) 290°
b) 322°
c) 316°
d) 310°
Given: True course from A to B: 090°; TAS: 460 kts; W/V: 360/100 kts; Average
variation: 10°E; Deviation: -2°. Calculate the compass heading and GS.
a) 070° - 450 kts.
b) 102° - 450 kts.
c) 068° - 460 kts.
d) 078° - 450 kts.
Given: TAS = 375 kt, True HDG = 124°, W/V = 130°(T)/55kt. Calculate the true
track and GS?
a) 125 - 322 kt
b) 123 - 320 kt
c) 125 - 318 kt
True Heading of an aircraft is 265° and TAS is 290 kt. If W/V is 210°/35kt, what is
True Track and GS?
a) 259° and 272kt
b) 271° and 272kt
c) 259° and 305kt
d) 260° and 315kt
Given: TAS: 135 kts; True Heading: 278°; W/V: 140/20 kts. Calculate the True track
and GS.
a) 279° - 152 kts.
b) 283° - 150 kts.
c) 275° - 150 kts.
d) 272° - 121 kts.
50.The DR position represents:
a) the estimated position taking account of the estimated TAS and wind
condition.
b) the air position corrected by the track error.
c) the actual position corrected by the track error.
d) the estimated position in no wind condition.
Given: Required course: 045°(T); W/V: 190/30; FL55; ISA; Variation: 15°E; CAS:
120 knots. What is the magnetic heading and GS?
a) 038°(M) 154 kts
b) 038°(M) 113 kts
c) 068°(M) 154 kts
d) 053°(M) 154 kts
Given: True HDG: 035°; TAS: 245 kts; Track: 046° (T); GS: 220 kts. Calculate the
W/V.
a) 340/50 kts
b) 340/45 kts
c) 335/45 kts
d) 335/55 kts
The True course in the flight log is 270°, the forecast wind is 045°(T)/15kts and the
TAS is 120 kts. After 15 minutes of flying with the planned TAS and TH the aircraft is 3
NM South of the intended track and 2,5 NM ahead of the dead reckoning position. The
track angle error (TKE) is:
a) 5°L
b) 6°R
c) 3°R
d) 2°L
An aircraft is flying from A to B. The true course according to the flight log is 090°,
the estimated wind is 225°(T)/15 kts and the TAS is 120 kts. After 15 minutes of flying
with the planned TAS and TH the aircraft is 3 NM South of the intended track and 2,5
NM ahead of the dead reckoning position. The Track angle error (TKE) is:
a) 5°R
b) 6°L
c) 17°L
d) 12°R
After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM South
of the intended track and 2,5 NM ahead of the dead reckoning position. To reach
destination B from this position, the TH should be:
a) 292°
b) 258°
c) 280°
d) 287°
What is the average TAS climbing from 2.000 ft up to FL120 at standard
temperatures, given a CAS 185 kts and QNH 1013?
a) 210 kts
b) 221 kts
c) 197 kts
d) 188 kts
What is the average TAS climbing from 1.500 ft up to FL180, given a temperature
ISA +15 °C, a CAS 230 kts and QNH 1032 ?
a) 270 kts
b) 263 kts
c) 309 kts
d) 283 kts
An aircraft is following a true track of 048° at a constant TAS of 210 kt.
The wind velocity is 350° / 30 kt.
The GS and drift angle are:
a) 192 kt, 7° right
b) 225 kt, 7° left
c) 192 kt, 7° left
d) 200 kt, 3.5° right
Given: An aircraft is on final approach to runway 32R (322°); The wind velocity
reported by the tower is 350°/20 kt.; TAS on approach is 95 kt. In order to maintain the
centre line, the aircraft's heading (°M) should be:
a) 316°
b) 322°
c) 326°
d) 328°
Given: TAS = 197 kt, True course = 240°, W/V = 180/30kt. Descent is initiated at FL
220 and completed at FL 40. Distance to be covered during descent is 39 NM. What is
the approximate rate of descent?
a) 950 FT/MIN
b) 800 FT/MIN
c) 1500 FT/MIN
d) 1400 FT/MIN
You are departing from an airport which has an elevation of 2000 ft. The QNH is
1013 hPa. 10 NM away there is a waypoint you are required to pass at an altitude of 7500
ft. Given a groundspeed of 100 kt, what is the minimum rate of climb?
a) 590 ft/min
b) 1080 ft/min
c) 750 ft/min
d) 920 ft/min
An aircraft is departing from an airport which has an elevation of 2.000 ft and the
QNH is 1003 hPa. The TAS is 100 kts, the head wind component is 20 kts and the rate of
climb is 500 ft/min. Top of climb is FL050. At what distance from the airport will this be
achieved?
a) 7,2 NM
b) 10,8 NM
c) 8,8 NM
d) 6,6 NM
During approach the following data are obtained: DME: 12,0 NM, altitude 3.000 ft;
DME: 9,8 NM, altitude 2.400 ft; TAS: 160 kts; GS: 125 kts. The rate of descent is:
a) 600 ft/min
b) 570 ft/min
c) 730 ft/min
d) 700 ft/min
Given: ILS GP angle: 3.5°; GS: 150 kts. What is the approximate rate of descent?
1.0 ft/min.
a) 900 ft/min.
b) 800 ft/min.
c) 700 ft/min.
You are required to descend from FL230 to FL50 over a distance of 32 NM in 7
minutes. What will the glideslope be when you expect a Wind Component (WC) of -25
kts during the descend?
a) 6,25°
°
b) 4,07
c) 4,68°
d) 5,29°
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:
a) 220°
b) 310°
c) 160°
d) 130°
Consider the following statements on sunset:
a) at sunset the centre of the Sun is at the observers horizon.
b) for positions at the same longitude, sunset will occur simultaneously at all
latitudes.
c) night-flying regulations start at the time of sunset.
d) sunset is the time when the observer at sea level sees the last part of the Sun
disappear below the horizon.
A map is conformal when:
a) the meridians are straight lines and the scale is constant.
b) the variation information is printed on the map as isogonals.
c) when it conforms to the specifications.
d) the meridians and the parallels of latitude intersects at right angles and when
the scale from any selected point is the same in all directions.
Parallels of latitude, except the equator are:
a) rhumb lines.
b) are neither rhumb lines nor great circles.
c) great circles.
d) both rhumb lines and great circles.
Consider the following statements on rhumb lines:
a) most rhumb lines will run as spirals from the one pole to another.
b) the true direction of a rhumb line on northern hemisphere will increase in true
direction, while on southern hemisphere it will decrease.
c) a rhumb line and a great circle will never have the same true direction for some
distance.
d) a rhumb line will never cross a great circle.
The shortest distance between 2 points on the surface of the Earth is:
a) half the rhumb line distance.
b) the arc of a small circle.
c) rhumb line.
d) a great circle.
The convergence of meridians:
a) is the distance between the meridians in degrees, minutes, and seconds.
b) is independent of latitude and longitude.
d) is greater using rhumb line track than using greater circle.
c) is the angular difference between the meridians.
You are flying from A (50°N 10°W) to B (58°N 02°E). What is the convergence
between A and B?
a) 9,7°
d) 6,5°
c) 10,2°
b) 6,8°
The convergence factor of a Lambert conformal conic chart is quoted as 0,78535. At
what latitude on the chart is Earth convergence correctly represented?
a) 38°15'
b) 51°45'
c) 80°39'
d) 52°05'
What is the value of the convergence factor on a Polar Stereographic chart?
a) 0,866
b) 1
c) 0
d) 0,5
What is the convergence at 50°00'N between the meridians 105°00';W and 145°00'W
on the Earth?
a) 32,1°
b) 30,6°
c) 40,0°
d) 50,0°
Radio bearings:
a) are rhumb lines.
b) are great circles.
c) are lines of fixed direction.
d) cut all meridians at the same angle.
Which one of the following, concerning great circles on a Direct Mercator chart, is
correct?
a) They approximate to straight lines between the standard parallels
b) They are all curves concave to the equator
c) They are all curves convex to the equator
d) With the exception of meridians and the equator, they are curves concave to
the equator
Conversions angle is:
a) convergency.
b) 0,5 convergency.
c) 4 times convergency.
d) twice convergency.
The great circle bearing of position B from position A in the Northern Hemisphere is
040°. If the conversion angle is 4°, what is the great circle bearing of A from B?
a) 212°
b) 228°
c) 224°
d) 220°
On a Lambert conformal chart the distance between two parallels of latitude
(difference of latitude = 2°), is measured to be 112 mm. The distance between two
meridians, spaced 2° longitude, according to the chart is 70 NM. The parallel of origin
(selected parallel) runs through the middle of the described square. What is the
convergence for a d-long of 15° on this map?
a) 14,56°
b) 9,23°
c) 7,50°
d) 12,18°
On a Lambert conformal chart the distance between two parallels of latitude
(difference of latitude = 2°), is measured to be 112 mm. The distance between two
meridians, spaced 2° longitude, according to the chart is 70 NM. The parallel of origin
(selected parallel) runs through the middle of the described square. What is the
convergence for a d-long of 15° on this map?
a) The parallel of origin is the parallel at which the scale reaches its maximum value.
b) The parallel of origin is the only parallel at which the chart is conformal.
c) The parallel of origin is the parallel at which the scale reaches its minimum
value.
On a Mercator's projection the distance between (17°N, 035°E) and (17°N, 040°E) is
5 cm. The scale at 57°N is approximately:
a) 1 : 10 626 460
b) 1 : 18 658 470
c) 1 : 6 052 030
d) 1 : 5 556 000
On a direct Mercator projection, the distance measured between two meridians spaced
5° apart at latitude 60°N is 8 cm. The scale of this chart at latitude 60°N is
approximately:
a) 1 : 3 500 000
b) 1 : 6 000 000
c) 1 : 7 000 000
d) 1 : 4 750 000
The polar Stereographic projection is:
a) a cylinder projection.
b) a conical projection.
c) a plane projection.
d) a variable cone projection.
Which map projection is described as follows: - Meridians are straight lines. - The
scale vary with latitude. - Most rhumb lines are curved lines.
a) Lambert conformal projection.
b) Equatorial Mercator projection.
c) Polar stereographic projection.
d) Lambert conformal or a polar stereographic projection.
The chart that is generally used for navigation in polar areas is based on a:
a) direct Mercator projection.
b) stereographical projection.
c) Lambert conformal projection.
d) gnomonic projection.
The Polar Stereographic projection is:
a) a cylinder projection.
b) a variable cone projection.
c) a plane projection.
d) a conical projection.
On a polar stereographic map, a straight line is drawn from position A (70°N 102°W)
to position B (80°N 006°E). The point of highest latitude along this line occurs at
longitude 035°W. What is the initial straight-line track angle from A to B, measured at
A?
a) 049°
b) 023°
c) 229°
d) 077°
A straight line from A (75°S, 120°E) to B (75°S, 160°E) is drawn on a Polar
Stereographic chart. When passing the meridian 155°E, the True Track is:
a) 075°
b) 255°
c) 095°
d) 105°
Given: Direct Mercator chart with a scale of 1: 200 000 at equator; Chart length from
'A' to 'B', in the vicinity of the equator, 11 cm. What is the approximate distance from 'A'
to 'B'?
a) 21 NM
b) 12 NM
c) 14 NM
d) 22 NM
What is the chart distance between longitudes 179°E and 175°W on a direct Mercator
chart with a scale of 1 : 5 000 000 at the equator?
a) 133 mm
b) 72 mm
c) 167 mm
d) 106 mm
93. On a Lambert chart (standard parallels 37°N and 65°N), with respect to the straight
line drawn on the map the between A (49°N 030°W) and B (48°N 040°W), the:
a) great circle and rhumb line are to the North.
b) great circle is to the North, the rhumb line is to the South.
c) rhumb line is to the North, the great circle is to the South.
d) great circle and rhumb line are to the South.
The standard parallels of a Lambert's conical orthomorphic projection are 07°40'N
and 38°20'N. The constant of the cone for this chart is:
a) 0,6
b) 0,92
c) 0,39
d) 0,42
The constant of the cone in a Lambert chart is 0,8666500. The angle between the
north directions of the meridian in position A (65°00'N, 018°00'W) and the meridian of
position B (75°00'N, 023°00'W) on the chart is:
a) 5,0°
b) 4,3°
c) 5,8°
d) 10,0°
A straight line from A (53°N, 155°W) to B (53°N, 170°E) is drawn on a Lambert
Conformal conical chart with standard parallels at 50°N and 56°N. When passing the
meridian 175°E, the True Track is:
a) 100,0°
b) 102,5°
c) 257,5°
d) 260,0°
A Lambert's conical conformal chart has standard parallels at 63°N and 41°N. What
is the constant of the cone?
a) 0,891
b) 0,788
c) 0,707
d) 0,656
On a Lambert conformal chart, the distance between two parallels of latitude
(difference of latitude = 2°), is measured to be 112 mm. The distance between two
meridians, spaced 2° longitude, according to the chart is 70 NM. What is the latitude in
the centre of the described square?
a) 54°
b) 42°
c) 38°
d) 49°
Which of the following lists all the aeronautical chart symbols shown at position
N5150.4 W00829.7?
a) VOR: DME: NDB:compulsory reporting point
b) civil airport: VOR: non-compulsory reporting point
c) VOR: DME: NDB: ILS
d) civil airport: VOR: DME: compulsory reporting point
Which of the following lists all the aeronautical chart symbols shown at position
N5318.1 W00856.5?
a) civil airport: VOR: DME: non-compulsory reporting point
b) VOR: DME: NDB: compulsory reporting point
c) VOR: DME: NDB: compulsory reporting point
d) civil airport: NDB: DME: non-compulsory reporting point
Which of the aeronautical chart symbols indicates a VOR/DME?
a) 2
b) 6
c) 1
d) 7
What is the meaning of aeronautical chart symbol No. 15?
a) Aeronautical ground light
b) Lighthouse
c) Hazard to aerial navigation
d) Visual reference point
What is the meaning of aeronautical chart symbol No. 16
a) Shipwreck showing above the surface at low tide
b) Off-shore helicopter landing platform
c) Off-shore lighthouse
d) Lightship
A pilot receives the following signals from a VOR DME station: Radial: 180°+/- 1°;
Distance: 200 NM. What is the approximate error?
a) ± 1 NM
b) ± 3,5 NM
c) ± 7 NM
d) ± 2 NM
At 10:00 hours an aircraft is on the 310° radial from a VOR/DME, at 10 nautical
miles range. At 10:10 the radial and range are 040°/10 NM. What is the aircraft's track
and ground speed?
a) 080 / 85 knots
b) 085 / 85 knots
c) 085 / 90 knots
d) 080 / 80 knots
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?
a) 028°
b) 332°
c) 031°
d) 208°
At 0020 UTC an aircraft is crossing the 310° radial at 40 NM of a VOR/DME
station. At 0035 UTC the radial is 040° and DME distance is 40 NM. Magnetic variation
is zero. The true track and ground speed are:
a) 090° - 232 kt
b) 085° - 226 kt
c) 088° - 232 kt
d) 080° - 226 kt
What is the radial and DME distance from CON VOR/DME (N5354.8 W00849.1)
to position N5430 W00900?
a) 214° - 26 NM
b) 358° - 36 NM
c) 169° - 35 NM
d) 049° - 45 NM
What is the radial and DME distance from BEL VOR/DME (N5439.7 W00613.8) to
position N5410 W00710?
a) 223° - 36 NM
b) 320° - 44 NM
c) 236° - 44 NM
d) 333° - 36 NM
Which statement is correct about the apparent solar day?
a) The duration of the apparent solar day is constant throughout a year due to the
constant velocity of the earth in its orbit around the sun.
b) The duration of the apparent solar day is constant throughout a year due to the
constant rotational speed of the earth around its axis.
c) The apparent solar day is the period between two successive transits of the mean
sun through the same meridian.
d) The apparent solar day is the period between two successive transits of the
true sun through the same meridian.
For 1st February the Air Almanac lists the following data: Latitude: 66°00'N;
Morning civil twilight: 07:56; Sunrise: 09:00; Sunset: 15:28; Evening civil twilight:
16:32. The duration of morning twilight at 66°00'N is:
a) 7 hours 56 minutes and starts at 09:00 UTC.
b) 1 hour 4 minutes and starts at 09:00 UTC.
c) 1 hour 4 minutes and starts at 07:56 LMT.
d) 8 hours 36 minutes and starts at 07:56 UTC.
Position 'Elephant Point' is situated at (58°00'N, 135°30'W). Standard time for this
location is listed in the Air Almanac as UTC -8. If sunset occurs at 00:57 UTC on 21st
January, what is the time of Sunset in LMT?
a) 15:55 on January 20th.
.
b) 16:57 on January 20th
c) 09:59 on January 21st.
d) 08:57 on January 21st.
On 27 Feb, at 52°10'S 040°00'E, the sunrise is at 02:30 UTC. On the same day, at
52°10'S 035°00'W, the sunrise is at:
a) 05:10 UTC
b) 02:30 UTC
c) 21:30 UTC
d) 07:30 UTC
Daylight Saving Time:
a) is used to extend the sunlight period in the evening.
b) is used in some countries.
c) is introduced by setting the standard time forward by one hour.
d) all answers are correct.
Position "Elephant Point" is situated at (58°00'N, 135°30'W). Standard time for this
location is listed in the Air Almanac as UTC -8. If sunset occurs at 00:57 UTC on 21st
January, what is the time of sunset in LMT?
a) 15:55 on January 20th.
b) 09:59 on January 21st.
c) 16:57 on January 20th.
d) 08:57 on January 21st.
The countries having a standard time slow on UTC:
a) will generally be located at westerly longitudes.
b) will often experience sunrise earlier than the sunrise occurs at the greenwich
meridian.
c) will generally be located at easterly longitudes.
d) will often have an earlier standard date than the UTC date.
Refer to almanac: The UTC of sunrise at (66°48'N, 095°26'W) on 27th of January is?
a) 1541 UTC
b) 0927 UTC
c) 1549 UTC
d) 0814 UTC
On 4th February the Air Almanac lists 19:41 as the time of sunset at 50°00'S. An
observer registers sunset at 21:13 UTC this day. What is the observers position?
a) 50°00'S 022°00'E.
b) 50°00'S 010°35'W.
c) 50°00'S 010°35'E.
d) 50°00';S 023°00'W.
You have calculated point of no return (PNR) on a flight, having all negative WCs
in the flight plan. During the flight you experience that the W/V is stronger but coming
from the same direction as in the flight plan. Considered the following statements.
a) the PNR will not change because the neither TAS nor fuel flow has changed.
b) a recalculated PNR will move toward the place of departure.
c) the PNR will, if recalculated, move toward the no-wind PNR.
d) you will arrive at the PNR at a later time than flight planned.
The distance from A to B is 2.368 NM. If outbound ground speed is 365 kts and
homebound ground speed is 480 kts and safe endurance is 8 hrs 30 min, what is the time
to the PNR?
a) 290 minutes.
b) 190 Minutes.
c) 219 Minutes.
d) 209 Minutes.
Given: Distance A to B: 1.973 NM; Ground speed OUT: 430 kts; Ground speed
BACK: 385 kts. The time from A to the point of equal time (PET) between A and B is:
a) 130 min.
b) 162 min.
c) 181 min.
d) 145 min.
Given: Distance A to B: 2.484 NM; Mean ground speed OUT: 420 kts; Mean
ground speed BACK: 500 kts; Safe endurance 08 hrs 30 min. The distance from A to the
point of safe return (PSR) A is:
a) 1.940 NM
b) 1.736 NM
c) 1.630 NM
d) 1.908 NM
An aircraft was over 'A' at 1435 hours flying direct to 'B'. Given: Distance 'A' to 'B'
2900 NM; True airspeed 470 kt; Mean wind component 'out' +55 kt; Mean wind
component 'back' -75 kt; Safe endurance 9 HR 30 MIN. The distance from 'A' to the
Point of Safe Return (PSR) 'A' is:
a) 1759 NM
b) 1611 NM
c) 2844 NM
d) 2141 NM
An aircraft is planned to fly from position 'A' to position 'B', distance 480 NM at an
average GS of 240 kt. It departs 'A' at 1000 UTC. After flying 150 NM along track from
'A', the aircraft is 2 MIN behind planned time. Using the actual GS experienced, what is
the revised ETA at 'B'?
a) 1153
b) 1203
c) 1157
d) 1206
Complete line 1 of the 'FLIGHT NAVIGATION LOG'; positions 'A' 'to 'B'. What is
the HDG°(M) and ETA?
a) 282° - 1128 UTC
b) 268° - 1114 UTC
c) 268° - 1128 UTC
d) 282° - 1114 UTC
Complete line 2 of the 'FLIGHT NAVIGATION LOG', positions'C' to 'D'. What is
the HDG°(M) and ETA?
a) HDG 193° - ETA 1249 UTC
b) HDG 188° - ETA 1229 UTC
c) HDG 183° - ETA 1159 UTC
d) HDG 193° - ETA 1239 UTC
Complete line 3 of the 'FLIGHT NAVIGATION LOG', positions 'E' to 'F'. What is
the HDG°(M) and ETA?
a) HDG 095° - ETA 1155 UTC
b) HDG 105° - ETA 1205 UTC
c) HDG 115° - ETA 1145 UTC
d) HDG 106° - ETA 1215 UTC
Complete line 4 of the 'FLIGHT NAVIGATION LOG', positions 'G' to 'H'. What is
the HDG°(M) and ETA?
a) HDG 354° - ETA 1326 UTC
b) HDG 344° - ETA 1336 UTC
c) HDG 344° - ETA 1303 UTC
d) HDG 034° - ETA 1336 UTC
Complete line 5 of the 'FLIGHT NAVIGATION LOG', positions 'J' to 'K'. What is
the HDG°(M) and ETA?
a) HDG 320° - ETA 1412 UTC
b) HDG 337° - ETA 1322 UTC
c) HDG 337° - ETA 1422 UTC
d) HDG 320° - ETA 1432 UTC
Complete line 6 of the 'FLIGHT NAVIGATION LOG', positions 'L' to 'M'. What is
the HDG°(M) and ETA?
a) HDG 075° - ETA 1452 UTC
b) HDG 064° - ETA 1449 UTC
c) HDG 070° - ETA 1459 UTC
d) HDG 075° - ETA 1502 UTC
Given: TAS = 472 kt; True HDG = 005°; W/V = 110°(T)/50k. Calculate the drift
angle and GS.
a) 6°R/490 kt
b) 6°L/490 kt
c) 6°R/462 kt
d) 6°L/402 kt
Given: True Track 245°; Drift 5° right; Variation 3° E; Compass Hdg 242°;
Calculate the deviation.
a) 1° E
b) 5° E
c) 5° W
d) 11° E
Given: TAS: 150 kts; Actual HDG: 270°; Wind: 245/12 kts. What is the wind
correction angle?
2° to the left.
2° to the right.
4° to the left.
4° to the right.
Given: TAS: 470 kts; True HDG: 317°; W/V: 045°(T)/45 kts. Calculate the drift
angle and GS.
3°R - 470 kts.
5°L - 470 kts.
5°R - 475 kts.
5°L - 475 kts.
Given: TAS: 270 kts; True HDG: 270°; Actual wind: 205° (T)/30 kts. Calculate the
drift angle and GS.
6°L - 256 kts.
6°R - 259 kts.
6°R - 251 kts.
8°R - 259 kts.
An aircraft equipped with an Inertial Navigation System (INS) flies with INS 1
coupled with autopilot 1. Both inertial navigation systems are navigating from way-point
A to B. The inertial systems' Central Display Units (CDU) shows:- XTK on INS 1 = 0; -
XTK on INS 2 = 8L (XTK = cross track). From this information it can be deduced that:
a) only inertial navigation system No. 2 is drifting
b) only inertial navigation system No. 1 is drifting
c) at least one of the inertial navigation systems is drifting
d) the autopilot is unserviceable in NAV mode
An aircraft is flying with the aid of an inertial navigation system (INS) connected to
the autopilot. The following two points have been entered in the INS computer: WPT 1:
60°N 030°W; WPT 2: 60°N 020°W. When 025°W is passed the latitude shown on the
display unit of the inertial navigation system will be:
a) 60°05.7';N
b) 60°11.0'N
c) 59°49.0'N
d) 60°00.0'N
As the INS position of the departure aerodrome, coordinates 35°32.7'N; 139°46.3'W
are input instead of 35°32.7'N-139°46.3'E. When the aircraft subsequently passes point
52°N 180°W, the longitude value shown on the INS will be:
a) 099° 32.6'W
b) 080° 27.4'W
c) 080° 27.4'E
d) 099° 32.6'E
The following points are entered into an inertial navigation system (INS). WPT 1:
60°N 30°W; WPT 2: 60°N 20°W; WPT 3: 60°N 10°W. The inertial navigation system is
connected to the automatic pilot on route (1-2-3).The track change when passing WPT 2
will be approximately:
a) zero
b) a 9° increase
c) a 9° decrease
d) a 4° decrease
An aircraft has a TAS of 300 kts and a safe endurance of 10 hrs. If the wind
component on the outbound leg is 50 kts head, what is the distance to the point of safe
endurance?
a) 1.458 NM
b) 1.544 NM
c) 1.622 NM
d) 1.500 NM
An aircraft at position 27°00'N-170°00'W travels 3000 km on a track of 180° (T),
then 3000 km on a track of 090° (T), then 3000 km on a track of 000° (T), then 3000 km
on a track of 270° (T). What is its final position?
a) 27°00'N 170°00'W
b) 27°00'N 173°18'W
c) 00°00'N/S 170°00'W
d) 27°00'N 143°00'W
Given: Runway direction: 083°(M); Surface W/V: 035/35 kts. Calculate the
effective headwind component.
a) 31 kts
b) 27 kts
c) 34 kts
d) 24 kts
If the headwind component is 50 kts, the FL is 330, temperature ISA -7 °C and the
ground speed is 495 kts, what is the Mach number?
a) 0,78
b) 0,98
c) 0,75
d) 0,95
Given: Pressure Altitude: 27.000 feet; OAT: -35 °C; Mach number: 0,45; W/V:
270°/85; Track: 200°(T). What is drift and ground speed?
a) 15R / 235 knots
b) 18L / 285 knots
c) 17R / 287 knots
d) 17L / 228 knots
An aircraft at FL370, M 0,86, OAT -44 °C, headwind component 110 kts, is
required to reduce speed in order to cross a reporting point 5 min later than planned. If
the speed reduction were to be made 420 NM from the reporting point, what Mach
number is required?
a) M 0,81
b) M 0,75
c) M 0,73
d) M 0,79
By what amount must you change your rate of descent given a 10 knot increase in
headwind on a 3° glideslope:
a) 50 feet per minute increase.
b) 30 feet per minute increase
c) 50 feet per minute decrease.
d) 30 feet per minute decrease.
An island is observed to be 15° to the left. The aircraft heading is 120° (M),
variation 17°W. The bearing from the aircraft to the island is:
a) 268° (T)
b) 122° (T)
c) 302° (T)
d) 088° (T)
Given: TAS: 197 kts; True course: 240°; W/V: 180/30 kts; Descent is initiated at
FL220 and completed at FL40. Distance to be covered during descent is 39 NM. What is
the approximate rate of descent?
a) 800 ft/min
b) 1.400 ft/min
c) 1.500 ft/min
d) 950 ft/min
Given: Distance A to B: 475 NM; Planned GS: 315 kt; ATD: 1000 UTC; 1040 UTC
- fix obtained 190 NM along track. What GS must be maintained from the fix in order to
achieve planned ETA at B?
a) 300 kt
b) 360 kt.
c) 340 kt
d) 320 kt.
What is the effect on the Mach number and TAS in an aircraft that is climbing with
constant CAS?
a) Mach number decreases; TAS decreases.
b) Mach number increases; TAS increases.
c) Mach number increases; TAS remains constant.
d) Mach number remains constant; TAS increases.
A ground feature was observed on a relative bearing of 315° and 3 min later on a
relative bearing of 270°. The W/V is calm; aircraft GS 180 kts. What is the minimum
distance between the aircraft and the ground feature?
a) 3 NM
b) 12 NM
c) 9 NM
d) 6 NM
Complete line 5 of the FLIGHT NAVIGATION LOG, positions 'J' to 'K'. What is
the HDG° (M) and ETA?
a) HDG 320° - ETA 1412 UTC.
b) HDG 337° - ETA 1422 UTC.
c) HDG 320° - ETA 1432 UTC.
d) HDG 337° - ETA 1322 UTC.
An aircraft at FL330 is required to commence descent when 65 NM from a VOR
and to cross the VOR at FL100. The mean GS during the descent is 330 kts. What is the
minimum rate of descent required?
a) 1.750 ft/min.
b) 1.650 ft/min.
c) 1.850 ft/min.
d) 1.950 ft/min.
Complete line 3 of the FLIGHT NAVIGATION LOG, positions 'E' to 'F'. What is
the HDG° (M) and ETA?
a) HDG 105°; ETA 1205 UTC.
b) HDG 115°; ETA 1145 UTC.
c) HDG 106°; ETA 1215 UTC.
d) HDG 095°; ETA 1155 UTC.
Given: Aircraft height: 2.500 ft; ILS GP: angle 3°. At what approximate distance
from THR can you expect to capture the GP?
a) 14,5 NM
b) 8,3 NM
c) 13,1 NM
d) 7,0 NM
The flight log gives the following data: True track, Drift, True heading, Magnetic
variation, Magnetic heading, Compass deviation, Compass heading. The right solution, in
the same order, is:
a) 119°; 3°L; 122°; 2°E; 120°; +4°; 116°.
b) 117°; 4°L; 121°; 1°E; 122°; -3°; 119°.
c) 115°; 5°R; 120°; 3°W; 123°; +2°; 121°.
d) 125°; 2°R; 123°; 2°W; 121°; -4°; 117°.
Given: true track is 348°, drift 17° left, variation 32° W, deviation 4°E. What is the
compass heading?
a) 033°
b) 337°
c) 359°
d) 007°
Given:True course from A to B = 090°, TAS = 460 KT, W/V = 360/100 KT,
Average variation = 10°E, Deviation = -2°. Calculate the compass heading and GS?
a) 068° - 460 KT
b) 070° - 450 KT
c) 073° - 453 KT
d) 078° - 450 KT
Given: True track 180°; Drift 8°R; Compass heading 195°; Deviation -2°. Calculate
the variation.
a) 25°W
b) 21°W
c) 9°W
d) 5°W
Given:True course 300°; drift 8°R; variation 10°W; deviation -4°. Calculate the
compass heading.
a) 322°
b) 306°
c) 278°
d) 294°
Given: true track 352°; variation 11° W; deviation is -5°; drift 10°R. Calculate the
compass heading?
a) 346°
b) 018°
c) 358°
d) 025°
265 US-GAL equals? (Specific gravity 0.80)
a) 862 kg
b) 895 kg
c) 803 kg
d) 940 kg
730 fpm equals:
a) 1.6 m/s
b) 5.2 m/s
c) 2.2 m/sec
d) 3.7 m/s
An aircraft travels 2.4 statute miles in 47 seconds. What is its groundspeed?
a) 183 kt
b) 209 kt
c) 160 kt
d) 131 kt
Given: True Heading = 180°; TAS = 500 kt; W/V 225° / 100 kt. Calculate the GS?
a) 600 kt
b) 435 kt
c) 450 kt
d) 535 kt
Given: True heading = 310°; TAS = 200 kt; GS = 176 kt; Drift angle 7° right.
Calculate the W/V?
a) 090° / 33 kt
b) 360° / 33 kt
c) 180° / 33 kt
d) 270° / 33 kt
The following information is displayed on an Inertial Navigation System: GS 520 kt,
True HDG 090°, Drift angle 5° right, TAS 480 kt. SAT (static air temperature) -51°C.
The W/V being experienced is:
a) 225° / 60 kt
b) 220° / 60 kt
c) 320° / 60 kt
d) 325° / 60 kt
Given: TAS = 270 kt, True HDG = 270°, Actual wind 205°(T)/30kt, Calculate the
drift angle and GS?
a) 6R - 251kt
b) 6L - 256kt
c) 8R - 259kt
d) 6R - 259kt
Given: TAS = 270 kt, True HDG = 145°, Actual wind = 205°(T)/30kt. Calculate the
drift angle and GS?
a) 8°R - 261 kt
b) 6°R - 251 kt
c) 6°R - 259 kt
d) 6°L - 256 kt
Given: TAS = 198 kt, HDG (°T) = 180, W/V = 359/25. Calculate the Track(°T) and
GS?
a) 181 - 180 kt
b) 179 - 220 kt
c) 180 - 183 kt
d) 180 - 223 kt
Given: TAS = 135 kt, HDG (°T) = 278, W/V = 140/20kt Calculate the Track (°T)
and GS?
a) 279 - 152 kt
b) 272 - 121 kt
c) 283 - 150 kt
d) 275 - 150 kt
Given: TAS = 205 kt, HDG (T) = 180°, W/V = 240/25kt. Calculate the drift and
GS?
a) 7L - 192 kt
b) 6L - 194 kt
c) 4L - 195 kt
d) 3L - 190 kt
Given: TAS = 250 kt, HDG (T) = 029°, W/V = 035/45kt. Calculate the drift and
GS?
a) 1L - 205 kt
b) 1R - 295 kt
c) 1L - 265 kt
d) 1R - 205 kt
Given: course required = 085° (T), Forecast W/V 030/100kt, TAS = 470 kt,
Distance = 265 NM. Calculate the true HDG and flight time?
a) 075°, 39 MIN
b) 095°, 31 MIN
c) 096°, 29 MIN
d) 076°, 34 MIN
For a landing on runway 23 (227° magnetic) surface; W/V reported by the ATIS is
180/30kt. VAR is 13°E. Calculate the cross wind component?
a) 15 kt
b) 26 kt
c) 20 kt
d) 22 kt
Given: For take-off an aircraft requires a headwind component of at least 10 kt and
has a cross-wind limitation of 35 kt. The angle between the wind direction and the
runway is 60°, Calculate the minimum and maximum allowable wind speeds?
a) 12 kt and 38 kt
b) 20 kt and 40 kt
c) 18 kt and 50 kt
d) 15 kt and 43 kt
Given: FL 350, Mach 0.80, OAT -55°C. Calculate the values for TAS and local
speed of sound (LSS)?
a) 461 kt , LSS 576 kt
b) 461 kt , LSS 296 kt
c) 490 kt, LSS 461 kt
d) 237 kt, LSS 296 kt
Given: FL250, OAT -15 °C, TAS 250 kt. Calculate the Mach No.?
a) 0.44
b) 0.42
c) 0.39
d) 0.40
Given: True altitude 9000 FT, OAT -32°C, CAS 200 kt. What is the TAS?
a) 215 kt
b) 200 kt
c) 220 kt
d) 210 kt
Given: A NORTHERN polar stereographic chart whose grid is aligned with the zero
meridian. Grid track 344°, Longitude 115°00'W, Calculate the true course?
a) 099°
b) 229°
c) 049°
d) 279°
A Mercator chart has a scale at the equator = 1: 3 704 000. What is the scale at
latitude 60° S?
a) 1: 3 208 000
b) 1: 7 408 000
c) 1: 185 200
d) 1: 1 852 000
The total length of the 53°N parallel of latitude on a direct Mercator chart is 133 cm.
What is the approximate scale of the chart at latitude 30°S?
a) 1: 30 000 000
b) 1: 25 000 000
c) 1: 21 000 000
d) 1: 18 000 000
Assume a North polar stereographic chart whose grid is aligned with the Greenwich
meridian. An aircraft flies from the geographic North pole for a distance of 480 NM
along the 110°E meridian, then follows a grid track of 154° for a distance of 300 NM. Its
position is now approximately:
a) 78°45'N 087°E
b) 80°00'N 080°E
c) 70°15'N 080°E
d) 79°15';N 074°E
For a distance of 1860 NM between Q and R, a ground speed 'out' of 385 kt, a
ground speed 'back' of 465 kt and an endurance of 8 HR (excluding reserves) the distance
from Q to the point of safe return (PSR) is:
a) 1685 NM
b) 1865 NM
c) 930 NM
d) 1532 NM
Two points A and B are 1000 NM apart. TAS = 490 kt. On the flight between A and
B the equivalent wind is -20 kt. On the return leg between B and A, the equivalent wind
is +40 kt. What distance from A, along the route A to B, is the Point of Equal Time
(PET)?
a) 455 NM
b) 470 NM
c) 500 NM
d) 530 NM
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 the point of departure before
sunset, the furthest distance which may be travelled is:
a) 97 NM
b) 84 NM
c) 105 NM
d) 115 NM
A ground feature was observed on a relative bearing of 325° and five minutes later
on a relative bearing of 280°. The aircraft heading was 165°(M), variation 25°W, drift
10°Right and GS 360 kt. When the relative bearing was 280°, the distance and true
bearing of the aircraft from the feature was:
a) 40 NM and 060°
b) 30 NM and 240°
c) 30 NM and 060°
d) 40 NM and 240°
Assuming zero wind, what distance will be covered by an aircraft descending 15000
FT with a TAS of 320 kt and maintaining a rate of descent of 3000 FT/MIN?
a) 26.7 NM
b) 16.0 NM
c) 38.4 NM
d) 19.2 NM
An aircraft at FL370 is required to commence descent at 120 NM from a VOR and
to cross the facility at FL130. If the mean GS for the descent is 288 KT, the minimum
rate of descent required is:
a) 960 fpm
b) 920 fpm
c) 890 fpm
d) 860 fpm
An aircraft at FL370 is required to commence descent when 100 NM from a DME
facility and to cross the station at FL120. If the mean GS during the descent is 396 kt, the
minimum rate of descent required is approximately:
a) 1000 FT/MIN
b) 2400 FT/MIN
c) 1550 FT/MIN
d) 1650 FT/MIN
At 0422 an aircraft at FL370, GS 320 KT, is on the direct track to VOR 'X' 185 NM
distant. The aircraft is required to cross VOR 'X' at FL80. For a mean rate of descent of
1800 fpm at a mean GS of 232 KT, the latest time at which to commence descent is:
a) 0451
b) 0448
c) 0454
d) 0445
Given: aircraft height 2500 FT, ILS GP angle 3°. At what approximate distance
from THR can you expect to capture the GP?
a) 13.1 NM
b) 7.0 NM
c) 14.5 NM
d) 8.3 NM
An aircraft at FL370, M0.86, OAT -44°C, headwind component 110 kt, is required
to reduce speed in order to cross a reporting point 5 MIN later than planned. If the speed
reduction were to be made 420 NM from the reporting point, what Mach Number is
required?
a) M0.75
b) M0.73
c) M0.79
d) M0.81
Given: Distance 'A' to 'B' is 100 NM, Fix obtained 40 NM along and 6 NM to the
left of course. What heading alteration must be made to reach 'B'?
a) 9° Right
b) 6° Right
c) 15° Right
d) 18° Right
Given: Distance 'A' to 'B' is 90 NM, Fix obtained 60 NM along and 4 NM to the
right of course. What heading alteration must be made to reach 'B'?
a) 16° Left
b) 12° Left
c) 8° Left
d) 4° Left
Given: Fuel flow: 42 US Gal/hr; Specific gravity: 0,72; TAS: 210 kts. What is the
specific fuel consumption?
a) 0,757 kgM air distance.
b) 1,052 kgM air distance.
c) 0,144 kgM air distance.
d) 0,545 kgM air distance.
Given: Fuel flow: 28 Imp Gal/hr; Specific gravity: 0,72; TAS: 154 MPH. What is
the specific range?
a) 1,67 NM air distance / kg.
b) 0,68 NM air distance / kg.
c) 1,46 NM air distance / kg.
d) 2,0 NM air distance / kg.
If the TAS exceeds the CAS by 20% at FL100, the OAT should be:
a) +15 °C.
b) is not defined.
c) +5 °C.
d) -5 °C.
Given: CAS: 130 kts; PA: 1.000 ft; TAS: 127 kts; What is the OAT?
a) +20 °C
b) +10 °C
c) -8 °C
d) 0 °C
On a Lambert conformal conic chart the convergence of the meridians:
a) varies as the secant of the latitude
b) is zero throughout the chart
c) equals earth convergency at the standard parallels
d) is the same as earth convergency at the parallel of origin
The constant of the cone, on a Lambert chart where the convergence angle between
longitudes 010°E and 030°W is 30°, is:
a) 0.50
b) 0.40
c) 0.64
d) 0.75
The angular difference, on a Lambert conformal conic chart, between the arrival and
departure track is equal to:
a) Earth convergence
b) conversion angle
c) chart convergence
d) difference in longitude
Scale on a Lambert's conformal chart is:
a) constant over the whole chart.
b) constant along a meridian of longitude.
c) varies with latitude and longitude.
d) constant along a parallel of latitude.
On a Lamberts conformal chart the distance between two parallels of latitude
(difference of latitude = 2°), is measured to be 112 mm. The distance between two
meridians, spaced 2° longitude, according to the chart is 70 NM. The parallel of origin
(selected parallel) runs through the middle of the described square. What is the
convergence for a d-long of 15° on this map?
a) 9.23°
b) 12.18°
c) 14.56°
d) 7.50°
Given: Great circle from P to Q measured at P = 095°; Southern hemisphere;
Conversion angle P - Q = 7°; What is the rhumb line track P - Q?
a) 102°
b) 081°
c) 109°
d) 088°
An approximate equation for calculation conversion angle is:
a) CA = dlong x sin (mean lat) x sin (long)
b) CA = (dlong-dlat) x 0,5
c) CA = 0,5 x dlong x sin (mean lat)
d) CA = 0,5 x dlat x sin (mean lat)
What is 'conversion angle':
a) the angular difference between the rhumb line and the great circle between
two positions, measured at any of the two positions.
b) the angle at which speech from another person enters the ear.
c) the angle used to convert from true to compass directions.
d) the difference between the rhumb line and the great circle directions.
Conversion angle is:
a) convergency.
b) 4 times convergency.
c) 0,5 convergency.
d) twice convergency.
What is the final position after the following rhumb line tracks and distances have
been followed from position 60°00'N'-030°00'W? South for 3.600 NM East for 3.600 NM
North for 3.600 NM West for 3.600 NM. The final position of the aircraft is:
a) 59°00' 090°00'W
b) 60°00'N 030°00'E
c) 60°00'N 090°00'W
d) 59°00'N 060°00'W
Assuming the Earth being a perfect sphere:
a) distances will vary, dependent on the latitude.
b) distances will vary, dependent on their directions.
c) a 1 minute arc of a Great Circle measured on the surface of the Earth will be
equally long wherever it is measured.
d) all answers are correct.
An aircraft flies the following rhumb line tracks and distances from position
04°00'N 030°00'W: 600 NM South, then 600 NM East, then 600 NM North, then 600
NM West. The final position of the aircraft is:
a) 04°00';N 030°02'W.
b) 04°00'N 029°58'W.
c) 03°58'N 030°02'W.
d) 04°00'N 030°00'W.
Given: Position A: 60°N 020°W; Position B: 60°N 021°W; Position C: 59°N
020°W. What are, respectively, the distances from A to B and from A to C?
a) 52 NM and 60 NM.
b) 60 NM and 30 NM.
c) 60 NM and 52 NM.
d) 30 NM and 60 NM.
Given: CON VOR (53°54.8'N 008°49.1'W) DME 30 NM; CRN DME (53°18.1'N
008°56.5'W) DME 25 NM; Aircraft heading is 270°(M) and both DME distances are
decreasing. What is the aircraft position (map extract from E(LO)1, Jeppesen)?
a) 53°43'N 009°25'W
b) 53°30'N 008°20'W
c) 53°37'N 008°20'W
d) 53°35'N 009°25'W
Given: SHA VOR (52°43,3'N 008°53,1'W) DME 50 NM; CRK VOR (51°50,4'N
008°29,7'W) DME 41 NM; Aircraft heading is 270° (M) and both DME distances are
increasing. What is the aircraft position?
a) 52°00'N 009°35'W
b) 52°35'N 007°50'W
c) 52°15'N 009°40'W
d) 52°15'N 007°45'W
Given: Position NDB (55°10'N, 012°55'E); DR Position (54°53'N, 009°58'E); NDB
on the RMI reads 090°; Magnetic variation 10°W. The position line has to be plotted on a
Lamberts conformal chart with standard parallels at 40°N and 48°N. Calculate the
direction (T) of the bearing to be plotted from the NDB.
a) 272°
b) 258°
c) 262°
d) 265°
Given: True altitude: 9.000 ft; OAT: -32°C; CAS: 200 kts. What is the TAS?
a) 215 kts
b) 200 kts
c) 220 kts
d) 210 kts
The data that needs to be inserted into an Inertial Reference System in order to
enable the system to make a successful alignment for navigation is:
a) aircraft heading
b) the position of an in-range DME
c) aircraft position in latitude and longitude
d) airport ICAO identifier
The automatic flight control system is coupled to the guidance outputs from an
inertial navigation system. Which pair of latitudes will give the greatest difference
between initial track read-out and the average true course given, in each case, a difference
of longitude of 10°?
a) 60°N to 50°N
b) 60°N to 60°N
c) 30°S to 25°S
d) 30°S to 30°N
Given: True HDG: 133°; TAS: 225 kts; Track: 144° (T); GS: 206 kts; Calculate the
W/V.
a) 070/45 kts
b) 070/40 kts
c) 075/50 kts
d) 075/45 kts
Given: Magnetic track: 075°; HDG: 066° (M); VAR: 11°E; TAS: 275 kts. Aircraft
flies 48 NM in 10 min. Calculate the true W/V.
a) 210° / 15 kts.
b) 300° / 50 kts.
c) 180° / 45 kts.
d) 340° / 45 kts.
A great circle track joins position A (59°S 141°W) and B (61°S 148°W). What is the
difference between the great circle track at A and B?
a) It increases by 3°
b) It decreases by 6°
c) It decreases by 3°
d) It increases by 6°
The Great Circle bearing from A (70°S 030°W) to B (70°S 060°E) is approximately:
a) 048°(T)
b) 090°(T)
c) 132°(T)
d) 312°(T)
An aircraft flies a great circle track from 56° N 070° W to 62° N 110° E. The total
distance travelled is?
a) 5420 NM
b) 3720 NM
c) 2040 NM
d) 1788 NM
Parallels of latitude, except the equator, are:
a) Great circles
b) both Rhumb lines and Great circles
c) Rhumb lines
d) are neither Rhumb lines nor Great circles
Consider the following statements on rhumb lines:
a) a rhumb line and a great circle will never have the same true direction for some
distance.
b) a rhumb line will never cross a great circle.
.
c) the true direction of a rhumb line on the northern hemisphere will increase in true
direction, while on southern hemisphere it will decrease
d) most rhumb lines will run spirals from the one polar to another.
The ICAO definition of ETA is the:
a) actual time of arrival at a point or fix
b) estimated time en-route
c) estimated time of arrival at destination
d) estimated time of arrival at an en-route point or fix
'Mean time' has been introduced in order to:
a) compensate for the irregularities of the speed of rotation of the Earth around it's
axis.
b) save energy as the problem of adjusting our watches entering every leap year is
eliminated.
c) introduced a constant measurement of time, independent of the daily
variations in the movement of the Sun as observed from the Earth.
d) have one fixed time to be used within the border of a country.
Civil Twilight is defined by:
a) Sun's altitude is 6° below the celestial horizon.
b) Sun's altitude is 18° below the celestial horizon.
d) Suns altitude is 12° below the celestial horizon.
c) Sun's upper egde tangential to horizon.
The 'duration of twilight':
a) will in the period around the equinoxes increase as you approach the equator from
North or South.
b) is generally longer in positions at high latitudes than in positions at lower
altitudes.
c) is independently of the Sun's declination, and only depends on the observers
latitude and longitude.
d) is longer in the morning than in the evening because of the refraction in the
atmosphere.
When approaching the International Date line from the East, you:
a) should be prepared to decrease your date by 1.
b) should increase your date by an extra date at the first midnight you experience.
c) should not change date at the first midnight you experience.
d) should be prepared to increase your date by 1
The International Date Line is located:
a) at the apparent sun's anti meridian.
b) at the 180° E/W meridian, or in the vicinity of this meridian.
c) at latitudes on the 180° E/W meridian.
d) at the Greenwich meridian
Which of the following alternatives is correct when you cross the international date
line?
a) The date will increase if you are crossing on a westerly heading.
b) If you are crossing from westerly longitude to easterly longitude the date will
remain the same.
c) The date will always be the same.
d) The date will increase if you are crossing on a easterly heading.
You are required to descend from FL230 to FL50 over a distance of 32 NM in 7
minutes. What is the required TAS when you expect a Wind Component (WC) of -25 kts
during the descent?
a) 317 kts
b) 300 kts
c) 308 kts
d) 329 kts
