ატფლ Flashcards
1
Q
5 HR 20 MIN 20 SEC corresponds to a longitude difference of ;
A
80°05’
2
Q
265 US-GAL equals ; ( Specific gravity 0.80 )
A
803 kg.
3
Q
730 ft/min equals ;
A
3.7 m/sec
4
Q
1215 UTC LAJES VORTAC ( 38°46’N 027°05’W ) RMI reads 178° , range 135 NM. Calculate the aircraft position at 1215 UTC ?
A
40°55’N 027°55’W
5
Q
1300 UTC DR position 37°30’N 021°30’W alter heading PORT SANTO NDB ( 33°03’N 016°23’W ) , TAS 450 kt , Forecast W/V 360°/30 kt. Calculate the ETA at PORT SANTO NDB ?
A
1348
6
Q
A chart has the scale 1 : 1 000 000. From A to B on the chart measures 1.5 inches ( one inch equals 2.54 centimetres ) , the
distance from A to B in NM is ;
A
20.6
7
Q
A chart has the scale 1 : 1 000 000. From A to B on the chart measures 3.8 cm , the distance from A to B in NM is ;
A
20.5
8
Q
A course of 120°( T ) is drawn between “X” ( 61°30’N ) and “Y” ( 58°30’N ) on a Lambert Conformal conic chart with a scale of 1 : 1 000 000 at 60°N. The chart distance between “X” and “Y” is ;
A
66.7 cm
9
Q
A definition of a Magnetic Track angle is ;
A
The direction of a line referenced to Magnetic North.
10
Q
A direct Mercator graticule is based on a projection that is ;
A
cylindrical.
11
Q
A direct reading compass should be swung when ;
A
there is a large , and permanent , change in magnetic latitude.
12
Q
A flight is to be made from “A” 49°S 180’E/W to “B” 58°S 180’E/W. The distance in kilometres from “A” to “B” is approximately ;
A
1000.
13
Q
A great circle intersects the equator in 030°W with a great circle direction of 035°( T ). An aircraft tracking the great circle will reach the maximum Northern/Southern latitude in position ;
A
( 55°N , 060°E ).
14
Q
A great circle on the Earth running from the North Pole to the South Pole is called ;
A
a meridian.
15
Q
A great circle track crosses the equator at 30°W has an initial track of 035°T. It’s highest or lowest North/South point is ;
A
55ºN 060ºE.
16
Q
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 6°.
17
Q
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
160°
18
Q
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 kt. What is the minimum distance between the aircraft and the ground feature ?
A
9 NM
19
Q
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
30 NM and 240°.
20
Q
A Lambert conformal conic chart has a constant of the cone of 0.75. The initial course of a straight line track drawn on this chart from A ( 40°N 050°W ) to B is 043°( T ) at A , course at B is 055°( T ). What is the longitude of B ?
A
34°W
21
Q
A Lambert conformal conic chart has a constant of the cone of 0.80. A straight line course drawn on this chart from A ( 53°N
004°W ) to B is 080° at A , course at B is 092° ( T ). What is the longitude of B ?
A
011°E
22
Q
A Lambert conformal conic projection , with two standard parallels ;
A
the scale is only correct along the standard parallels.
23
Q
A line drawn on a chart which joins all points where the value of magnetic variation is zero is called an ;
A
agonic line.
24
Q
A Mercator chart has a scale at the equator = 1 : 3 704 000. What is the scale at latitude 60° S ?
A
1 : 1 852 000
25
A nautical mile is ;
1852 metres.
26
A negative ( westerly ) magnetic variation signifies that ;
True North is East of Magnetic North.
27
A pilot accidently turning OFF the IRS in flight , and then turns it back ON a few moments later. Following this incident
it can only be used for attitude reference.
28
A rhumb line from a position 86°N 30°W has an initial track of 085°T is it ;
a spiral to the North pole
29
A Rhumb line is ;
a line on the surface of the earth cutting all meridians at the same angle.
30
A rhumb line on a Direct Mercator chart appears as a ;
straight line.
31
A ring laser gyro is ;
a device which measures angular movements.
32
A route is flown from ( 80°S , 100°W ) to ( 80°S , 140°E ). At 160°W the Grid Track ( GT ) and True Track ( TT ) on a
Polar Stereographic chart with a grid orientated on the 180º meridian are respectively ;
290º(G) and 270º(T).
33
A route is flown from ( 80°S , 100°W ) to ( 80°S , 140°E ). At 180°E/W the Grid Track ( GT ) and True Track ( TT ) on a Polar Stereographic chart , whose grid is aligned with the Greenwich meridian , are respectively ;
110º(G) and 290º(T).
34
A route is flown from ( 85°S , 100°E ) to ( 85°S , 140°W ). At 160°E the Grid Track ( GT ) and True track ( TT ) on a Polar Stereographic chart with a grid orientated on the 180º meridian are respectively ;
070º(G) and 090º(T).
35
A route is flown from ( 85°S , 100°E ) to ( 85°S , 140°W ). At 180°E/W the Grid Track ( GT ) and True Track ( TT ) on a Polar Stereographic chart , whose grid is aligned with the Greenwich meridian , are respectively ;
250º(G) and 070º(T).
36
A straight line drawn on a chart measures 4.63 cm and represents 150 NM. The chart scale is ;
1 : 6 000 000
37
A straight line from A ( 53ºN , 155ºW ) to B ( 53ºN , 170ºE ) is drawn on a Lambert Conformal conical chart with standard parallels at 50ºN and 56ºN. When passing the meridian 175ºE , the True Track is ;
260.0º
38
A straight line from A ( 53ºS , 155ºE ) to B ( 53ºS , 170ºW ) is drawn on a Lambert Conformal conical chart with standard parallels at 50ºS and 56ºS. When passing 175ºW , the True Track is ;
078.0º
39
A straight line from A ( 75ºN , 120ºE ) to B ( 75ºN , 160ºE ) is drawn on a Polar Stereographic chart. When passing the
meridian 155ºE , the True Track is ;
105º
40
A straight line from A ( 75ºS , 120ºE ) to B ( 75ºS , 160ºE ) is drawn on a Polar Stereographic chart. When passing the meridian 155ºE , the True Track is ;
075º
41
A straight line is drawn on a Lamberts conformal conic chart between two positions of different longitude. The angular difference between the initial true track and the final true track of the line is equal to ;
chart convergency
42
A straight line on a chart 4.89 cm long represents 185 NM. The scale of this chart is approximately ;
1 : 7 000 000
43
A straight line on a Lambert Conformal Projection chart for normal flight planning purposes ;
is approximately a Great Circle
44
A useful method of a pilot resolving , during a visual flight , any uncertainty in the aircraft’s position is to maintain visual contact with the ground and ;
set heading towards a line feature such as a coastline , motorway , river or railway.
45
A VOR is situated at position ( N55°26' , W005°42' ). The variation at the VOR is 9°W. The position of the aircraft is
( N60°00'N , W010°00' ). The variation at the aircraft-position is 11°W. The initial TT-angle of the great circle from the aircraft position to the VOR is 101.5°. Which radial is the aircraft on ?
294
46
A VOR is situated at position ( 74ºN , 094ºW ) , local variation is 50ºW. A Polar Stereographic chart supplied with a
Greenwich grid is used for navigation. To proceed along ( magnetic ) radial 238 inbound an aircraft has to follow a Grid Track of ;
103º
47
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 ;
292°
48
After alignment of the stable platform of an Inertial Navigation System , the output data from the platform is ;
acceleration north/south and east/west , attitude and true heading.
49
After alignment of the stable platform of the Inertial Navigation System , the output data from the INS computer to the platform is
rate corrections to the gyros.
50
An aeronautical chart is conformal when ;
at any point the scale over a short distance in the direction of the parallel is equal to the scale in the direction of the meridian and the meridians are perpendicular to the parallels.
51
An aeroplane flies from A ( 59°S 142°W ) to B ( 61°S 148°W ) with a TAS of 480 kt. The autopilot is engaged and coupled with an Inertial Navigation System in which AB track is active. On route AB , the true track ;
increases by 5°.
52
An aeroplane is flying at TAS 180 kt on a track of 090°. The W/V is 045° / 50kt. How far can the aeroplane fly out from its base and return in one hour ?
85 NM
53
An Agonic line is a line that connects ;
positions that have 0° variation.
54
An aircraft at FL120 , IAS 200 kt , OAT –5° and wind component +30 kt , is required to reduce speed in order to cross a reporting point 5 MIN later than planned. Assuming flight conditions do not change , when 100 NM from the reporting point IAS
should be reduced to ;
159 kt.
55
An aircraft at FL140 , IAS 210 kt , OAT –5°C and wind component minus 35 kt , is required to reduce speed in order to cross a reporting point 5 MIN later than planned. Assuming that flight conditions do not change , when 150 NM from the reporting point the IAS should be reduced by ;
20 kt.
56
An aircraft at FL290 is required to commence descent when 50 NM from a VOR and to cross that VOR at FL80. Mean GS during descent is 271 kt. What is the minimum rate of descent required ?
1900 FT/MIN
57
An aircraft at FL310 , M0.83 , temperature –30°C , is required to reduce speed in order to cross a reporting point five minutes later than planned. Assuming that a zero wind component remains unchanged , when 360 NM from the reporting point Mach Number should be reduced to ;
M0.74
58
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 kt. What is the minimum rate of descent required ?
1950 FT/MIN
59
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 kt. What is the minimum rate of descent required ?
1800 FT/MIN
60
An aircraft at FL350 is required to cross a VOR/DME facility at FL110 and to commence descent when 100 NM from the facility. If the mean GS for the descent is 335 kt , the minimum rate of descent required is ;
1340 FT/MIN
61
An aircraft at FL350 is required to descend to cross a DME facility at FL80. Maximum rate of descent is 1800 FT/MIN and mean GS for descent is 276 kt. The minimum range from the DME at which descent should start is ;
69 NM
62
An aircraft at FL360 is required to descent to FL120. The aircraft should reach FL120 at 40 NM from the next waypoint. The rate of descent is 2000 ft/min. The average GS is 420 kt. The minimum distance from the next waypoint at which descent should start
124 NM
63
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 ;
960 FT/MIN
64
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 ;
1650 FT/MIN
65
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
M0.81
66
An aircraft at FL390 is required to descend to cross a DME facility at FL70. Maximum rate of descent is 2500 FT/MIN ,
mean GS during descent is 248 kt. What is the minimum range from the DME at which descent should commence ?
53 NM
67
An aircraft at latitude 02°20'N tracks 180°( T ) for 685 km. On completion of the flight the latitude will be ;
03°50'S
68
An aircraft at latitude 10°North flies south at a groundspeed of 445 km/hr. What will be its latitude after 3 HR ?
02°00'S
69
An aircraft at latitude 10°South flies north at a GS of 890 km/hr. What will its latitude be after 1.5 HR ?
02°00'N
70
An aircraft at position 60°N 005°W tracks 090°( T ) for 315 km. On completion of the flight the longitude will be ;
000°40'E
71
An aircraft departing A ( N40º00' E080º00' ) flies a constant true track of 270º at a ground speed of 120 kt. What are the coordinates of the position reached in 6 HR ?
N40º00' E064º20'
72
An aircraft departs from position A ( 04°10'S 178°22'W ) and flies northward following the meridian for 2950 NM. It then flies westward along the parallel of latitude for 382 NM to position B. The coordinates of position B are ?
45°00'N 172°38'E
73
```
An aircraft departs from Schiphol airport and flies to Santa Cruz in Bolivia ( South America ) via Miami in Florida. The
departure time ( off blocks ) is 07:45 ST at the 10th of November , taxi time before take off at Schiphol is 25 minutes. The flight time to Miami over the Atlantic Ocean is 09 h 20 m. The total taxi time in Miami to and from the gate is 25 minutes. The time spend at the gate is 02 h 40 m. From Miami to Santa Cruz the airborne time is 06 h 30 m. Calculate the time and date of touch down in Santa Cruz in ST Bolivia if the difference between ST and UTC is 5 hours ;
```
21:05 10th Nov.
74
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 ;
at least one of the inertial navigation systems is drifting.
75
An aircraft flies a great circle track from 56°N 070°W to 62°N 110°E. The total distance travelled is ?
3720 NM
76
An aircraft flies at FL 250. OAT = –45°C. The QNH , given by a meteorological station with an elevation of 2830 ft , is 1033 hPa. Calculate the clearance above a mountain ridge with an elevation of 20410 ft ;
4200 ft
77
An aircraft flies from waypoint 7 ( 63°00'N , 073°00'W ) to waypoint 8 ( 62°00'N , 073°00'W ). The aircraft position is
( 62°00'N , 073°10'W ). The cross track distance in relation to the planned track is ;
4.7 NM R
78
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 ;
04°00'N 029°58'W
79
An aircraft follows a great circle in the Northern Hemisphere. At a certain moment the aircraft is in the position on the great
circle where the great circle direction is 270°(T). Continuing on the great circle the ;
track angle will decrease and the latitude will decrease.
80
An aircraft follows a radial to a VOR/DME station. At 10:00 the DME reads 120 NM. At 10:03 the DME reads 105 NM. The estimated time overhead the VOR/DME station is ;
10:24
81
An aircraft has to fly over a mountain ridge. The highest obstacle , indicated in the navigation chart , has an elevation of 9800 ft. The QNH , given by a meteorological station at an elevation of 6200 ft , is 1022 hPa. The OAT = ISA+5ºC. Calculate the approximate indicated altitude to obtain a clearance of 2000 ft.
Between 11500 ft and 11700 ft.
82
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°
83
An aircraft in the northern hemisphere makes an accurate rate one turn to the right/starboard. If the initial heading was 330° , after 30 seconds of the turn the direct reading magnetic compass should read ;
less than 060°
84
An aircraft is at position ( 53ºN , 006ºW ) and has a landmark at position ( 52º47'N , 004º45'W ) , with a relative bearing of 060º. Given ;
Compass Heading = 051º
Variation = 16ºW
Deviation = 2ºE
What is the true bearing of the position line to be plotted from the landmark to the aircraft on a Lambert chart with standard parallels at 37ºN and 65ºN ?
278º
85
An aircraft is departing from an airport which has an elevation of 2000 ft and the QNH is 1003 hPa. The TAS is 100 kt , the
head wind component is 20 kt and the rate of climb is 500 ft/min. Top of climb is FL 050. At what distance from the airport will this be achived
7.2 NM
86
An aircraft is departing from an airport which has an elevation of 2000 ft and the QNH is 1003 hPa. The TAS is 100 kt , the head wind component is 20 kt and the rate of climb is 1000 ft/min. Top of climb is FL 050. At what distance from the airport will this be achived
3.6 NM
87
An aircraft is departing from an airport which has an elevation of 2000 ft and the QNH is 1003 hPa. The TAS is 100 kt , the head wind component is 20 kt and the rate of climb is 1000 ft/min. Top of climb is FL 100. At what distance from the airport will this be achived ?
10.3 NM
88
An aircraft is departing from an airport which has an elevation of 2000 ft and the QNH is 1023 hPa. The TAS is 100 kt , the
head wind component is 20 kt and the rate of climb is 1000 ft/min. Top of climb is FL 100. At what distance from the airport will this
be achieved ?
11.1 NM
89
An aircraft is descending down a 6% slope whilst maintaining a G/S of 300 kt. The rate of descent of the aircraft is approximately ;
1800 FT/MIN
90
An aircraft is flying according the flight log at the Annex. After 15 minutes of flying with the planned TAS and TH the
aircraft is 3 NM North 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 ;
(refer to image 5)
112º
91
An aircraft is flying according the flight log at the Annex. After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM North 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 ; (refer to image 9)
258º
92
An aircraft is flying at FL100. The OAT = ISA – 15°C. The QNH given by a station at an elevation 3000 ft is 1035 hPa. Calculate the approximate True Altitude ;
10 200 ft
93
An aircraft is flying at FL150 , with an outside air temperature of –30° , above an airport where the elevation is 1660 ft and the QNH is 993 hPa. Calculate the true altitude ; ( Assume 30 ft = 1 hPa )
13 660 ft
94
An aircraft is flying at FL180 and the outside air temperature is –30°C. If the CAS is 150 kt , what is the TAS ?
195 kt
95
An aircraft is flying at FL 200. OAT = 0°C. When the actual air pressure on an airfield at MSL is placed in the subscale of the altimeter the indicated altitude is 19300 ft. Calculate the aircraft’s true altitude ;
21.200 ft
96
An aircraft is flying at FL200. The QNH , given by a meteorological station at an elevation of 1300 ft is 998.2 hPa.
OAT = –40ºC. The elevation of the highest obstacle along the route is 8 000 ft. Calculate the aircraft’s approximate clearance above the highest obstacle on this route ;
10.500 ft
97
An aircraft is flying at FL250 , OAT = –45°C. The QNH , given by a station at MSL , is 993.2 hPa. Calculate the
approximate True Altitude
23400 ft
98
An aircraft is flying from A to B a distance of 50 NM. The True course in the flight log is 090º , the forecast wind is 225º(T)/15 kt and the TAS is 120 kt. 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 correction angle on the heading should be ;
17º
99
An aircraft is flying from A to B a distance of 50 NM. The True Course in the flight log is 270º , the forecast wind is 045º(T)/15 kt and the TAS is 120 kt. 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 correction angle on the heading should be
17º
100
An aircraft is flying from A to B. The true course according to the flight log is 090º , the estimated wind is 225º(T)/15 kt and the TAS is 120 kt. 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 ;
5ºR
101
An aircraft is flying from Inverness VORDME ( N57°32.6'' , W004°02.5W ) to Aberdeen VORDME ( N57°18.6'' , W002° 16.0''W ). At 1000 UTC the fix of the aircraft is determined by VORDME Inverness ; radial = 114 , DME-distance = 20.5 NM. At 1006 UTC the fix of the aircraft is determined by VORDME Aberdeen ; radial = 294 , DME-distance = 10.5 NM. What is the average GS of the aircraft between 1000 UTC and 1006 UTC ?
( For this question see Europe Low Altitude Enroute Chart E(LO) 1A )
280 kt
102
An aircraft is flying from SALCO to BERRY HEAD on Magnetic Track 007º , TAS 445 kt. The wind is 050º(T)/40 kt. Variation 5ºW , deviation +2º. At 1000 UTC the RB of locator PY is 311º. At 1003 UTC the RB of locator PY is 266º. Calculate the distance of the aircraft from locator PY at 1003 UTC ;
21 NM
103
An aircraft is flying from SALCO to BERRY HEAD on Magnetic Track 007º , TAS 445 kt. The wind is 050º(T)/40 kt.
Variation 5ºW , deviation +2º. At 1000 UTC the RB of locator PY is 311º. At 1003 UTC the RB of locator PY is 266º. Calculate the
True bearing of locator PY at 1003 UTC from the aircraft ;
272º(T)
104
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 ;
60°05.7'N
105
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 ;
192 kt , 7° right
106
An aircraft is following the 45°N parallel of latitude. The track followed is a ;
rhumb line.
107
An aircraft is in the position ( 86°N , 020°E ). When following a rhumb line track of 085°( T ) it will ;
fly via a spiral to the North Pole.
108
An aircraft is maintaining a 5.2% gradient is at 7 NM from the runway , on a flat terrain , its height is approximately ;
2210 ft.
109
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°
110
An aircraft is planned to fly from position “A” to position “B” , distance 250 NM at an average GS of 115 kt. It departs “A”
at 0900 UTC. After flying 75 NM along track from “A” , the aircraft is 1.5 MIN behind planned time. Using the actual GS experienced , what is the revised ETA at “B” ?
1115 UTC
111
An aircraft is planned to fly from position “A” to position “B” , distance 320 NM , at an average GS of 180 kt. It departs “A” at 1200 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” ?
1333 UTC
112
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” ?
1206
113
An aircraft must fly 2000 ft above an obstacle of which the elevation is 13 600 ft. The QHN at the nearest airfield is 991 hPa , the elevation is 1500 ft and the temperature is –20°C. Calculate the minimum altitude required.
17 400 ft
114
An aircraft obtains a relative bearing of 315° from an NDB at 0830. At 0840 the relative bearing from the same NDB is 270°.
Assuming no drift and a GS of 240 kt , what is the approximate range from the NDB at 0840 ?
40 NM
115
An aircraft passes position A ( 60°00'N 120°00'W ) on route to position B ( 60°00'N 140°30'W ). What is the great circle track on departure from A ?
279°
116
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 ;
97 NM.
117
An aircraft takes off from the aerodrome of BRIOUDE ( altitude 1483 ft , QFE = 963 hPa , temperature = 32°C ). Five minutes later , passing 5 000 ft on QFE , the second altimeter set on 1013 hPa will indicate approximately ;
6 500 ft.
118
An aircraft travels 2.4 statute miles in 47 seconds. What is its groundspeed ?
160 kt
119
An aircraft travels 100 statute miles in 20 MIN , how long does it take to travel 215 NM ?
50 MIN
120
An aircraft travels from point A to point B , using the autopilot connected to the aircraft’s inertial system. The coordinates of A ( 45°S 010°W ) and B ( 45°S 030°W ) have been entered. The true course of the aircraft on its arrival at B to the nearest degree , is ;
277°
121
```
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 ;
```
2141 NM
122
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
The ETA for reaching the Point of Equal Time ( PET ) between “A” and “B” is ;
1657
123
An aircraft was over “Q” at 1320 hours flying direct to “R”. Given ;
Distance “Q” to “R” 3016 NM
True airspeed 480 kt
Mean wind component “out” –90 kt
Mean wind component “back” +75 kt
Safe endurance 10:00 HR
The distance from “Q” to the Point of Safe Return ( PSR ) “Q” is ;
2290 NM
124
An aircraft was over “Q” at 1320 hours flying direct to “R”.
Given ;
Distance “Q” to “R” 3016 NM
True airspeed 480 kt
Mean wind component “out” –90 kt
Mean wind component “back” +75 kt
The ETA for reaching the Point of Equal Time ( PET ) between “Q” and “R” is ;
1752
125
An island appears 30° to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading ( MH ) of 020° with the magnetic variation ( VAR ) 25°W ?
145°
126
An island appears 30° to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading of 276° with the magnetic variation
12°W ?
054°
127
An island appears 30° to the right of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading ( MH ) of 355° with the magnetic variation ( VAR ) 15°E ?
220°
128
An island appears 45° to the right of the centre line on an airborne weather radar display. What is the true bearing of the
aircraft from the island if at the time of observation the aircraft was on a magnetic heading ( MH ) of 215° with the magnetic variation
( VAR ) 21°W ?
059°
129
An island appears 60° to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading ( MH ) of 276° with the magnetic variation ( VAR ) 10°E ?
046°
130
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°
131
An island is observed to be 15° to the left. The aircraft heading is 120° ( M ) , variation 17° ( W ). The bearing ( °T ) from the aircraft to the island is ;
088
132
```
An island is observed to be 30° to the right of the nose of the aircraft. The aircraft heading is 290° ( M ) , variation 10° ( E ).
The bearing ( °T ) from the aircraft to the island is ;
```
330°
133
An NDB is located at position ( N55°26' , W005°42' ). The variation at the NDB is 9°W. The position of the aircraft is
( 56°00'N , 010°00'W ). The variation at the aircraft position is 11°W. The initial TT of the great circle from the aircraft position to the NDB position , is 101.5°. What is the Magnetic Bearing of the NDB from the aircraft ?
112.5º
134
An Oblique Mercator projection is used specifically to produce ;
charts of the great circle route between two points.
135
An observer is situated on the parallel of 23.5°S. Which statement about the passage of the apparent sun in relation to this
position is correct ?
It passes through the zenith once a year around December 22nd.
136
Approximately how many nautical miles correspond to 12 cm on a map with a scale of 1 : 2 000 000 ?
130
137
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 ;
099° 32.6'W
138
Assume a Mercator chart. The distance between positions A and B , located on the same parallel and 10° longitude apart , is 6 cm. The scale at the parallel is 1 : 9 260 000. What is the latitude of A and B ?
60° N or S
139
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 ;
( For this question use annex 061-1818A )
80°00'N 080°E
140
Assuming mid-latitudes ( 40° to 50°N/S ). At which time of year is the relationship between the length of day and night , as well as the rate of change of declination of the sun , changing at the greatest rate ?
Spring equinox and autumn equinox.
141
Assuming zero wind , what distance will be covered by an aircraft descending 15 000 FT with a TAS of 320 kt and maintaining a rate of descent of 3000 FT/MIN ?
26.7 NM
142
At 00:00 Local Mean Time of an observer ;
the mean sun is in transit with the observer’s anti-meridian.n.
143
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 ;
085° - 226 kt
144
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 FT/MIN at a mean GS of 232 kt , the latest time at which to commence descent is ;
0445
145
At 10:15 the reading from a VOR/DME station is 211°/ 90 NM , at 10:20 the reading from the same VOR/DME station is 211°/120 NM.
Compass Heading = 200º
Variation in the area = 31ºW
Deviation = +1º
TAS = 390 kt
The wind vector ( T ) is approximately ;
110º/70 k
146
At 47° North the chart distance between meridians 10° apart is 5 inches. The scale of the chart at 47° North approximates ;
1 : 6 000 000
147
At 47° North the chart distance between meridians 10° apart is 12.7 cm. The scale of the chart at 47° North approximates ;
1 : 6 000 000
148
At ( 54°N , 020°W ) the sun rises on November 28th at 09:01 UTC. At ( 44°N , 020°W ) the sun will rise ;
earlier since the latter position lies further South.
149
At 60° N the scale of a direct Mercator chart is 1 : 3 000 000. What is the scale at the equator ?
1 : 6 000 000
150
At a specific location , the value of magnetic variation ;
varies slowly over time.
151
At latitude 60°N the scale of a Mercator projection is 1 : 5 000 000. The length on the chart between “C” N60° E008° and “D” N60° W008° is ;
17.8 cm
152
At reference. Magnetic heading of an aircraft is 040 degrees. On the airborne weather radar ( AWR ) display the relative bearing of the distance to the must southerly part of Lands End , ( approximate position : 50 03 N 005 40 W ) are 030 degrees R and 80 NM. What is the position of the aircraft based on these observations ? The slant range correction and the map convergency between aircraft position and Lands End may be neglected.( Refer to Image 1 )
( 49 25 N 007 30 W )
153
At the magnetic equator , when accelerating after take off on heading West , a direct reading pivot suspended compass ;
indicates the correct heading.
154
At what approximate date is the earth closest to the sun ( perihelion ) ?
Beginning of January.
155
At what approximate date is the earth furthest from the sun ( aphelion ) ?
Beginning of July.
156
ATT Mode of the Inertial Reference System ( IRS ) is a back-up mode providing ;
only attitude and heading information.
157
Calculate the constant of the cone on a Lambert Chart given chart convergency between 010°E and 030°W as being 30° ;
0.75
158
Calibrated Airspeed ( CAS ) is Indicated Airspeed ( IAS ) corrected for ;
instrument error and position error.
159
Compass deviation is defined as the angle between ;
Magnetic North and Compass North.
160
Complete line 1 of the “FLIGHT NAVIGATION LOG” , positions “A” to “B”. What is the HDG° ( M ) and ETA ?( For this question use annex 061-9437A )
268° - 1114 UTC
161
Complete line 2 of the “FLIGHT NAVIGATION LOG” , positions “C” to “D”. What is the HDG° ( M ) and ETA ? For this question use annex 061-9437A )
HDG 193° - ETA 1239 UTC
162
Complete line 3 of the “FLIGHT NAVIGATION LOG” , positions “E” to “F”. What is the HDG° ( M ) and ETA ? For this question use annex 061-9437A )
HDG 105° - ETA 1205 UTC
163
Complete line 4 of the “FLIGHT NAVIGATION LOG” , positions “G” to “H”. What is the HDG° ( M ) and ETA ? For this question use annex 061-9437A )
HDG 344° - ETA 1336 UTC
164
Complete line 5 of the “FLIGHT NAVIGATION LOG” , positions “J” to “K”. What is the HDG° ( M ) and ETA ?
For this question use annex 061-9437A )
HDG 337° - ETA 1422 UTC
165
Complete line 6 of the “FLIGHT NAVIGATION LOG” , positions “L” to “M”. What is the HDG° ( M ) and ETA ? For this question use annex 061-9437A )
HDG 075° - ETA 1502 UTC
166
Complete the following statement regarding magnetic variation. The charted values of magnetic variation on earth normally change annually due to
magnetic pole movement causing numerical values at all locations to increase or decrease.
167
Concerning direct reading magnetic compasses , in the northern hemisphere , it can be said that ;
on an Easterly heading , a longitudinal acceleration causes an apparent turn to the North.
168
Consider the following factors that determine the accuracy of a DR position ;
1. The flight time since the last position update.
2. The accuracy of the forecasted wind.
3. The accuracy of the TAS.
4. The accuracy of the steered heading.
Using the list above which of the following contains the most complete answer ?
1 , 2 , 3 and 4.
169
Consider the positions ( 00ºN/S , 000ºE/W ) and ( 00ºN/S , 180ºE/W ) on the ellipsoid. Which statement about the distances between these positions is correct ?
The route via the North Pole is shorter than the route along the equator.
170
Contour lines on aeronautical maps and charts connect points ;
having the same elevation above sea level
171
Deviation applied to magnetic heading gives ;
compass heading.
172
Deviation on the standby compass is
dependent on the heading of the aircraft.
173
Double integration of the output from the east/west accelerometer of an inertial navigation system ( INS ) in the NAV
MODE give ;
distance east/west.
174
During a low level flight 2 parallel roads that are crossed at right angles by an aircraft. The time between these roads can be used to check the aircraft ;
groundspeed.
175
```
During approach the following data are obtained ;
DME 12.0 NM , altitude 3000 ft
DME 9.8 NM , altitude 2400 ft
TAS = 160 kt , GS = 125 kt
The rate of descent is ;
```
570 ft/min
176
During the initial alignment of an inertial navigation system ( INS ) the equipment ;
will not accept a 10° error in initial latitude but will accept a 10° error in initial longitude.
177
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 ;
1685 NM
178
For a landing on runway 23 ( 227° magnetic ) surface W/V reported by the ATIS is 180 / 30 kt. VAR is 13°E. Calculate the cross wind component ;
22 kt
179
```
From Rakovnik ( 50° 05.9' N , 013° 41.5' E ) to Frankfurt FFM ( 50° 05.9' N , 008° 38.3' E ) the True Track of departure
along the straight line is 272.0°. The constant of the cone of this Lambert conformal projection is ;
```
0.79
180
From the departure point , the distance to the point of equal time is ;
inversely proportional to the sum of ground speed out and ground speed back.
181
Fuel flow per HR is 22 US-GAL , total fuel on board is 83 IMP GAL. What is the endurance ?
4 HR 32 MIN
182
Geodetic latitude and geocentric latitude coincide ;
at the Poles and on the equator.
183
```
Given ;
A polar stereographic chart whose grid is aligned with the zero meridian.
Grid track 344°
Longitude 115°00'W
Calculate the true course ?
```
229°
184
```
Given ;
ETA to cross a meridian is 2100 UTC
GS is 441 kt
TAS is 491 kt
At 2010 UTC , ATC requests a speed reduction to cross the meridian at 2105 UTC. The reduction to TAS will be approximately ;
```
40 kt
185
Given ;
An aircraft is flying a track of 255°( M )
2254 UTC , it crosses radial 360° from a VOR station
2300 UTC , it crosses radial 330° from the same station
At 2300 UTC , the distance between the aircraft and the station is ;
the same as it was at 2254 UTC.
186
Given ;
Aircraft height 2500 ft
ILS GP angle 3°
At what approximate distance from THR can you expect to capture the GP ?
8.3 NM
187
Given ;
An aircraft is flying at FL100
OAT = ISA – 15ºC
The QNH given by a meteorological station with an elevation of 100 ft below MSL is 1032 hPa
1 hPa = 27 ft
Calculate the approximate True Altitude of this aircraft ;
9.900 ft
188
```
Given ;
True course 300°
drift 8°R
variation 10°W
deviation –4°
Calculate the compass heading ;
```
306°
189
```
Given ;
true track 070°
variation 30°W
deviation +1°
drift 10°R
Calculate the compass heading ;
```
089°
190
```
Given ;
true track 352°
variation 11° W
deviation is –5°
drift 10°R
Calculate the compass heading ?
```
358°
191
```
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 ;
```
070° - 450 kt
192
Given ;
Runway direction 210° ( M )
Surface W/V 230° ( M ) / 30 kt
Calculate the cross-wind component ;
10 kt
193
Given ;
Runway direction 305° ( M )
Surface W/V 260° ( M ) / 30 kt
Calculate the cross-wind component ?
21 kt
194
Given ;
Pressure Altitude 29000 ft
OAT –55°C
Calculate the Density Altitude ?
27500 ft.
195
```
Given ;
True Track 245°
Drift 5° right
Variation 3°E
Compass Hdg 242°
Calculate the deviation ;
```
5°W
196
```
Given ;
TAS = 95 kt
HDG ( T ) = 075°
W/V = 310 / 20 kt
Calculate the drift and GS ;
```
9R - 108 kt
197
```
Given ;
TAS = 140 kt
HDG ( T ) = 005°
W/V = 265 / 25kt
Calculate the drift and GS
```
10R - 146 kt
198
```
Given ;
TAS = 190 kt
HDG ( T ) = 355°
W/V = 165 / 25kt
Calculate the drift and GS ;
```
1L - 215 kt
199
```
Given ;
TAS = 205 kt
HDG ( T ) = 180°
W/V = 240 / 25 kt
Calculate the drift and GS ;
```
6L - 194 kt
200
```
Given ;
TAS = 230 kt
HDG ( T ) = 250°
W/V = 205 / 10 kt
Calculate the drift and GS ;
```
2R - 223 kt
201
```
Given ;
TAS = 250 kt
HDG ( T ) = 029°
W/V = 035 / 45 kt
Calculate the drift and GS ;
```
1L - 205 kt
202
```
Given ;
TAS = 440 kt
HDG ( T ) = 349°
W/V = 040 / 40 kt
Calculate the drift and GS ;
```
4L - 415 kt
203
```
Given ;
TAS = 465 kt
HDG ( T ) = 124°
W/V = 170 / 80 kt
Calculate the drift and GS ;
```
8L - 415 kt
204
```
Given ;
Magnetic heading = 255°
VAR = 40°W
GS = 375 kt
W/V = 235°( T ) / 120 kt
Calculate the drift angle ?
```
6° left
205
```
Given ;
TAS = 132 kt
True HDG = 257°
W/V = 095° ( T ) / 35 kt
Calculate the drift angle and GS ;
```
4°R - 165 kt
206
```
Given ;
TAS = 140 kt
True HDG = 302°
W/V = 045°( T ) / 45 kt
Calculate the drift angle and GS ;
```
16°L - 156 kt
207
```
Given ;
TAS = 190 kt
True HDG = 085°
W/V = 110°( T ) / 50 kt
Calculate the drift angle and GS ;
```
8°L - 146 kt
208
```
Given ;
TAS = 235 kt
HDG ( T ) = 076°
W/V = 040 / 40kt
Calculate the drift angle and GS ;
```
7R - 204 kt
209
```
Given ;
TAS = 270 kt
True HDG = 145°
Actual wind = 205° ( T ) / 30 kt
Calculate the drift angle and GS ;
```
6°L - 256 kt
210
```
Given ;
TAS = 270 kt
True HDG = 270°
Actual wind 205°( T ) / 30 kt
Calculate the drift angle and GS ;
```
6°R - 259 kt
211
```
Given ;
TAS = 290 kt
True HDG = 171°
W/V = 310°( T ) / 30 kt
Calculate the drift angle and GS ;
```
4°L - 314 kt
212
```
Given ;
TAS = 470 kt
True HDG = 317°
W/V = 045°( T ) / 45 kt
Calculate the drift angle and GS ;
```
5°L - 470 kt
213
```
Given ;
TAS = 472 kt
True HDG = 005°
W/V = 110°( T ) / 50 kt
Calculate the drift angle and GS ;
```
6°L - 490 kt
214
```
Given ;
TAS = 485 kt
True HDG = 226°
W/V = 110°( T ) / 95 kt
Calculate the drift angle and GS ;
```
9°R - 533 kt
215
Given ;
Runway direction 230°( T )
Surface W/V 280°( T ) / 40 kt
Calculate the effective cross-wind component ;
31 kt
216
Given ;
Runway direction 083°( M )
Surface W/V 035 / 35 kt
Calculate the effective headwind component ;
24 kt
217
```
Given ;
TAS = 220 kt
Magnetic course = 212 º
W/V 160º( M ) / 50kt
Calculate the GS ?
```
186 kt
218
```
Given ;
True Heading = 090°
TAS = 200 kt
W/V = 220° / 30 kt
Calculate the GS ?
```
220 kt
219
```
Given ;
True Heading = 180°
TAS = 500 kt
W/V 225° / 100 kt
Calculate the GS ?
```
435 kt
220
```
Given ;
TAS = 130 kt
Track ( T ) = 003°
W/V = 190 / 40 kt
Calculate the HDG ( °T ) and GS ;
```
001 - 170 kt
221
```
Given ;
TAS = 155 kt
Track ( T ) = 305°
W/V = 160 / 18 kt
Calculate the HDG ( °T ) and GS ;
```
301 - 169 kt
222
```
Given ;
TAS = 200 kt
Track ( T ) = 073°
W/V = 210 / 20 kt
Calculate the HDG ( °T ) and GS ;
```
077 - 214 kt
223
```
Given ;
TAS = 200 kt
Track ( T ) = 110°
W/V = 015 / 40 kt
Calculate the HDG ( °T ) and GS ;
```
099 - 199 kt
224
```
Given ;
TAS = 227 kt
Track ( T ) = 316°
W/V = 205 / 15 kt
Calculate the HDG ( °T ) and GS ;
```
312 - 232 kt
225
```
Given ;
TAS = 270 kt
Track ( T ) = 260°
W/V = 275 / 30 kt
Calculate the HDG ( °T ) and GS ;
```
262 - 241 kt
226
```
Given ;
TAS = 465 kt
Track ( T ) = 007°
W/V = 300 / 80 kt
Calculate the HDG ( °T ) and GS ;
```
358 - 428 kt
227
```
Given ;
FL250
OAT –15 ºC
TAS 250 kt
Calculate the Mach No. ?
```
0.40
228
```
Given ;
TAS = 485 kt
OAT = ISA +10°C
FL 410
Calculate the Mach Number ?
```
0.825
229
Given ;
TAS 487 kt , FL 330 , Temperature ISA + 15
Calculate the MACH Number ?
0.81
230
```
Given ;
True Track 245°
Drift 5° right
Variation 3° E
Compass Hdg 242°
Calculate the Magnetic Heading ;
```
237°
231
Given ;
Maximum allowable crosswind component is 20 kt
Runway 06 , RWY QDM 063°( M )
Wind direction 100°( M )
Calculate the maximum allowable wind speed ;
33 kt
232
Given ;
Maximum allowable tailwind component for landing
10 kt. Planned runway 05 ( 047° magnetic )
The direction of the surface wind reported by ATIS 210° Variation is 17°E
Calculate the maximum allowable wind speed that can be accepted without exceeding the tailwind limit ;
10 kt
233
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 ;
20 kt and 40 kt.
234
```
Given ;
TAS = 90 kt
HDG ( T ) = 355°
W/V = 120 / 20 kt
Calculate the Track ( °T ) and GS ;
```
346 - 102 kt
235
```
Given ;
TAS = 132 kt
HDG ( T ) = 053°
W/V = 205 / 15 kt
Calculate the Track ( °T ) and GS
```
050 - 145 kt
236
```
Given ;
TAS = 135 kt
HDG ( °T ) = 278
W/V = 140 / 20 kt
Calculate the Track ( °T ) and GS ;
```
283 - 150 kt
237
```
Given ;
TAS = 155 kt
HDG ( T ) = 216°
W/V = 090 / 60 kt
Calculate the Track ( °T ) and GS ;
```
231 - 196 kt
238
```
Given ;
TAS = 170 kt
HDG ( T ) = 100°
W/V = 350 / 30 kt
Calculate the Track ( °T ) and GS ;
```
109 - 182 kt
239
```
Given ;
TAS = 198 kt
HDG ( °T ) = 180
W/V = 359 / 25
Calculate the Track ( °T ) and GS ;
```
180 - 223 kt
240
```
Given ;
TAS = 225 kt
HDG ( °T ) = 123°
W/V = 090 / 60kt
Calculate the Track ( °T ) and GS ;
```
134 - 178 kt
241
```
Given ;
TAS = 480 kt
HDG ( °T ) = 040°
W/V = 090 / 60kt
Calculate the Track ( °T ) and GS ;
```
034 - 445 kt
242
```
Given ;
TAS = 485 kt
HDG ( T ) = 168°
W/V = 130 / 75 kt
Calculate the Track ( °T ) and GS ;
```
174 - 428 kt
243
```
Given ;
Course required = 085°( T )
Forecast W/V 030 / 100 kt
TAS = 470 kt
Distance = 265 NM
Calculate the true HDG and flight time ;
```
075° , 39 MIN
244
```
Given ;
TAS = 125 kt
True HDG = 355°
W/V = 320°( T ) / 30kt
Calculate the true track and GS ;
```
005 - 102 kt
245
```
Given ;
TAS = 370 kt
True HDG = 181°
W/V = 095°( T ) / 35 kt
Calculate the true track and GS ;
```
186 - 370 kt
246
```
Given ;
TAS = 375 kt
True HDG = 124°
W/V = 130°( T ) / 55 kt
Calculate the true track and GS ;
```
123 - 320 kt
247
```
Given ;
M 0.80
OAT –50°C
FL 330
GS 490 kt
VAR 20°W
Magnetic heading 140°
Drift is 11° Right
Calculate the true W/V ?
```
020° / 95 kt
248
```
Given ;
Magnetic track = 075°
HDG = 066°( M )
VAR = 11°E
TAS = 275 kt
Aircraft flies 48 NM in 10 MIN
Calculate the true W/V ?
```
340° / 45 kt
249
```
Given ;
Magnetic track = 210°
Magnetic HDG = 215°
VAR = 15°E
TAS = 360 kt
Aircraft flies 64 NM in 12 MIN
Calculate the true W/V ?
```
265° / 50 kt
250
Given ;
FL 350 , Mach 0.80 , OAT –55°C
Calculate the values for TAS and local speed of sound ( LSS ) ?
461 kt , LSS 576 kt
