new_navigation.pdf

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NAVIGATION

NAVIGATION

NAVIGATION

BASICS OF NAVIGATION - The Solar System



Assuming mid-latitudes (40o to 50o N/S). At which time of year is the relationship between the length of day and night, as well as the rate ofchange of declination of the sun, changing at the greatest rate?

What is the approximate date of perihelion, when the Earth is nearest to the Sun?

At what approximate date is the earth furthest from the sun (aphelion)?

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Summer solstice and spring equinox

Spring equinox and autumn equinox

Summer solstice and winter solstice

Winter solstice and autumn equinox

Beginning of January

End of December

Beginning of July

End of March

Beginning of July

End of December


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BASICS OF NAVIGATION - The Earth

At what approximate date is the earth furthest from the sun (aphelion)?

Seasons are due to the:

A great circle track joins position A (59oS 141oW) and B (61oS 148oW). What is the difference between the great circle track at A and B?

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End of September

Earth's elliptical orbit around the Sun

inclination of the polar axis with the ecliptic plane

Earth's rotation on its polar axis

variable distance between Earth and Sun

it increases by 6o

it decreases by 6o

it increases by 3o

it decreases by 3o

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The angle between the plane of the Equator and the plane of the Ecliptic is:

Given: Value for the ellipticity of the Earth is 1/297. Earth's semi-major axis, as measured at the equator, equals 6378.4 km. What is thesemi-minor axis (km) of the earth at the axis of the Poles?

At what approximate latitude is the length of one minute of arc along a meridian equal to one NM (1852 m) correct?

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66.5 deg

23.5 deg

25.3 deg

65.6 deg

6 356.9

6 378.4

6 367.0

6 399.9

45o

0o

90o

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At what approximate latitude is the length of one minute of arc along a meridian equal to one NM (1852 m) correct?

The maximum difference between geocentric and geodetic latitude occurs at about:

What is the UTC time of sunrise in Vancouver, British Columbia, Canada (49N 123 30W) on the 6th December?

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30o

90o North and South

60o North and South

45o North and South

0o North and South (equator)

2324 UTC

0724 UTC

1552 UTC

0738 UTC

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The circumference of the Earth is approximately:

In order to fly from position A (10o00N, 030o00W) to position B (30o00N), 050o00W), maintaining a constant true course, it is necessary to fly:

The diameter of the Earth is approximately:

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43200 nm

10800 nm

21600 nm

5400 nm

the great-circle route

the constant average drift route

a rhumb line track

a straight line plotted on a Lambert chart

18 500 km

6 350 km

12 700 km

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The diameter of the Earth is approximately:

At what approximate date is the earth closest to the sun (perihelion)?

In which two months of the year is the difference between the transit of the Apparent Sun and mean Sun across the Greenwich Meridian thegreatest?

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40 000 km

End of June

End of March

Beginning of July


March and September

February and November

June and December

April and August

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What is a line of equal magnetic variation?

The circumference of the parallel of latitude at 60oN is approximately:

Parallels of latitude, except the equator are:

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An isocline

An isogonal

An isogriv

An isovar

10 800 NM

18 706 NM

20 000 NM

34 641 NM

both Rhumb lines and Great circles

Great circles

Rhumb lines

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Parallels of latitude, except the equator are:

The angle between the plane of the ecliptic and the plane of equator is approximately:

Given:The coordinates of the heliport at Issy les Moulineaux are:N48o50 E002o16.5The coordinates of the antipodes are:

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are neither Rhumb lines nor Great circles

27.5o

25.3o

23.5o

66.5o

S41o10 W177o43.5

S48o50 E177o43.5

S48o50 W177o43.5

S41o10 E177o43.5

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An aircraft at latitude 02o20N tracks 180o(T) for 685 km. On completion of the flight the latitude will be:

An aircraft departing A(N40o 00¿E080o00¿) flies a constant true track of 270o at a ground speed of 120 kt. What are the coordinates of theposition reached in 6 HR?

If an aeroplane was to circle around the Earth following parallel 60oN at a ground speed of 480 kt. In order to circle around the Earth alongthe equator in the same amount of time, it should fly at a ground speed of:

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03o50S

04o10S

04o30S

09o05S

N40o 00¿ E068o 10¿

N40o 00¿ E064o 20¿

N40o 00¿ E070o 30¿

N40o 00¿ E060o 00¿

550 kt

240 kt

960 kt

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If an aeroplane was to circle around the Earth following parallel 60oN at a ground speed of 480 kt. In order to circle around the Earth alongthe equator in the same amount of time, it should fly at a ground speed of:

The angle between the true great-circle track and the true rhumb-line track joining the following points: A (60oS 165oW) B (60oS 177oE), at theplace of departure A, is:

An aircraft flies the following rhumb line tracks and distances from position 04o00N 030o00W: 600 NM South, then 600 NM East, then 600 NM North,then 600 NM West. The final position of the aircraft is:

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480 kt

7.8o

9o

15.6o

5.2o

04o00N 029o58W

04o00N 030o02W

04o00N 030o00W

03o58N 030o02W

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Which of the following statements concerning the earth's magnetic field is completely correct?

You are flying from A (50n 10W) to B (58N 02E). What is the Convergency between A and B?

Radio bearings:

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Dip is the angle between total magnetic field and vertical field component

The blue pole of the earth's magnetic field is situated in North Canada

At the earth's magnetic equator, the inclination varies depending on whether the geographic equator is north or south of themagnetic equatorThe earth's magnetic field can be classified as transient semi-permanent or permanent

6.5o

9.7o

10.2o

6.8o

are Rhumb lines

cut all meridians at the same angle

are Great circles

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Radio bearings:

How many nm are equivalent to 1o of arc of latitude:

The earth may be referred to as:

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are lines of fixed direction

1 nm

15 nm

60 nm

600 nm

round

an oblate spheroid

a globe

elliptical

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What is the standard formula for convergency?

A line which cuts all meridians at the same angle is called a:

A Parallel of Latitude is a:

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Convergency = dial x sin mean latitude

Convergency = dial x cos mean latitude

Convergency = diong x cos mean latitude

Convergency = diong x sin mean latitude

Line of variation

Great circle

Rhumb line

Agonic line

Great circle

Rhumb line

Small circle

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A Parallel of Latitude is a:

The shortest distance between 2 point of the surface of the earth is:

Generally what line lies closer to the pole?

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kj---900038

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Meridian of tangency

a great circle

the arc of a great circle

half the rhumb line distance

Rhumb line

Rhumb line

Orthodromic line

Equator

The rhumb line or great circle depending on the chart used

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BASICS OF NAVIGATION - Time and time conversions

The Earth is:

At what time of the year is the Earth at its furthest point from the sun (aphelion)?

(Refer to figure 061-14)When it is 1000 Standard Time in Kuwait, the Standard time in Algeria :

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A sphere which has a larger polar circumference than equatorial circumference

A sphere whose centre is equidistant (the same distance) from the Poles and the Equator

Considered to be a perfect sphere as far as navigation is concerned

None of the above statements is correct

Early July

Late December

Early January

Mid-June

0700

1200

1300

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(Refer to figure 061-14)When it is 1000 Standard Time in Kuwait, the Standard time in Algeria :

The duration of civil twilight is the time:

(Refer to figures 061-13 and 061-15)An aircraft takes off from Guam at 2300 Standard Time on 30 April local date. After a flight of 11 HR 15MIN it lands at Los Angeles (California). What is the Standard Time and local date of arrival (assume summer time rules apply)?

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0800

between sunset and when the centre of the sun is 12o below the true horizon

agreed by the international aeronautical authorities which is 12 minutes

needed by the sun to move from the apparent height of 0o to the apparent height of 6o

between sunset and when the centre of the sun is 6o below the true horizon

1715 on 30 April

1215 on 1 May

1315 on 1 May

1615 on 30 April

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What is the definition of EAT?

In which months is the difference between apparent noon and mean noon the greatest?

Civil Twilight occurs between:

2

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Estimated on-blocks arrival time

Estimated time overhead the destination airfield

Estimated initial approach fix time

Estimated final approach fix time

November and February

January and July

March and September

June and December

sunset and 6 deg below the horizon

6 deg and 12 deg below the horizon

12 deg and 18 deg below the horizon

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Civil Twilight occurs between:

Which is the highest latitude listed below at which the sun will rise above the horizon and set every day?

On the 27th of February, at 52oS and 040oE, the sunrise is at 0243 UTC. On the same day, at 52oS and 035oW, the sunrise is at:

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sunrise and sunset

62o

68o

72o

66o

2143 UTC

0243 UTC

0743 UTC

0523 UTC

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(Refer to figures 061-13 and 061-15)At 1200 Standard Time on the 10th of July in Queensland, Australia, what is the Standard Time in Hawaii,USA?

What is the local mean time, position 65o25N 123o45W at 2200 UTC?

The main reason that day and night, throughout the year, have different duration is due to the:

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kj---900050

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1200 ST 10 July

1000 ST 10 July

1600 ST 09 July

0200 ST 10 July

1345

2200

0615

0815

inclination of the ecliptic to the equator

earth's rotation

relative speed of the sun along the ecliptic

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The main reason that day and night, throughout the year, have different duration is due to the:

The Local Mean Time at longitude 095o20W at 0000 UTC, is:

What is the meaning of the term standard time?

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gravitational effect of the sun and moon on the speed of rotation of the earth

1738:40 same day

0621:20 same day

1738:40 previous day

0621:20 previous day

It is the time zone system applicable only in the USA

It is an expression for local mean time

It is another term for UTC

It is the time set by the legal authorities for a country or part of a country

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Civil twilight is defined by:

(Refer to figure 061-12)The UTC of sunrise on 6 December at WINNIPEG (Canada) (49o 50' N 097o 30'W) is:

(Refer to figure 061-04) Given: TAS is 120 kt ATA ‘X’ 1232 UTC ETA ‘Y’ 1247 UTC ATA ‘Y’ is 1250 UTC What is ETA ‘Z’?

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sun altitude is 12o below the celestial horizon


sun upper edge tangential to horizon


0930

0113

2230

1413

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(Refer to figure 061-04) Given: TAS is 120 kt ATA ‘X’ 1232 UTC ETA ‘Y’ 1247 UTC ATA ‘Y’ is 1250 UTC What is ETA ‘Z’?

In 8 hours and 8 minutes the mean sun has moved how many degrees (o) along the celestial equator?

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1257 UTC

1302 UTC

1300 UTC

1303 UTC

18o

148o

122o

56o

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NAVIGATION

NAVIGATION

NAVIGATION




Morning Civil twilight begins when:

When the time is 1400 LMT at 90o West, it is:

When the time is 2000 UTC, it is:

2

2

2

Complexity

Complexity

Complexity

1

1

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kj---900058

kj---900059

kj---900060

Ref

Ref

Ref

the sun's upper edge is tangential to the celestial horizon

the centre of the sun is 12o below the celestial horizon



1400 LMT at 90o East

1200 LMT at 120o West


0600 LMT at the Prime meridian



1200 LMT at 60o East

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NAVIGATION

NAVIGATION

NAVIGATION




When the time is 2000 UTC, it is:

Which of the following alternatives is correct when you cross the international date line?

On 27 Feb, at S5210.0 E04000.0, the sunrise is at 0230 UTC. On the same day, at S5210.0 W03500.0, the sunrise is at:

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

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kj---900060

kj---900061

kj---900062

Ref

Ref

Ref

Ref

0800 LMT at the Prime meridian

The date will increase if you are crossing on a westerly heading

The date will increase if you are crossing on a easterly heading

The date will always be the same

If you are crossing from westerly longitude to easterly longitude the date will remain the same

0230 UTC

0510 UTC

0730 UTC

2130 UTC

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NAVIGATION

NAVIGATION

NAVIGATION




The UTC of the end of Evening Civil Twilight in position N51000' W008000' on 15 August is:

(Refer to figures 061-13 and 061-15)If it is 1200 Standard Time on 10th July in Queensland, Australia, the Standard Time in Hawaii, USA is:

The months in which the difference between apparent noon and mean noon is greatest are:

2

2

2

Complexity

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kj---900063

kj---900064

kj---900065

Ref

Ref

Ref

1928 UTC

1944 UTC

2000 UTC

2032 UTC

1200ST 10 July

1000ST 10 July

1600ST 09 July

0200ST 10 July

February and November

January and July

March and September

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NAVIGATION

NAVIGATION

NAVIGATION




The months in which the difference between apparent noon and mean noon is greatest are:

(Refer to figures 061-13 and 061-15)What is the Standard Time in Hawaii when it is 0600 ST on the 16th July in Queensland, Australia?

If it is 0700 hours Standard Time in Kuwait, what is the Standard Time in Algeria?

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

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kj---900065

kj---900066

kj---900067

Ref

Ref

Ref

Ref

June and December

1000 ST 15th July

2000 ST 15th July

1000 ST 16th July

1000 ST 17th July

0500

0900

1200

0300

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NAVIGATION

NAVIGATION

NAVIGATION

BASICS OF NAVIGATION - Distance



Isogrives are lines that connect positions that have:

The lines on the earth's surface that join points of equal magnetic variation are called:

An aircraft is following the 45oN parallel of latitude. The track followed is a:

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900068

kj---900070

kj---900072

Ref

Ref

Ref

the same horizontal magnetic field strength

the same grivation

the same variation

0o magnetic dip

isogrives

isoclines

isogonals

isotachs

constant-heading track

rhumb line

great circle

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft is following the 45oN parallel of latitude. The track followed is a:

How does the convergency of any two meridians on the Earth change with varying latitude?

How many small circles can be drawn between any two points on a sphere?

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

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kj---900072

kj---900073

kj---900074

Ref

Ref

Ref

Ref

constant-drift track

It changes as cosine of latitude

It changes as sine of latitude

It increases with decrease of latitude

It is of constant value and does not change with latitude

One

None

An unlimited number

Two

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NAVIGATION

NAVIGATION

NAVIGATION




If you are flying along a parallel of latitude, you are flying:

In which occasions does the rhumb line track and the great circle track coincide on the surface of the Earth?

When flying on a westerly great circle track in the Southern Hemisphere you will:

2

2

2

Complexity

Complexity

Complexity

1

1

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kj---900075

kj---900076

kj---900077

Ref

Ref

Ref

a great circle track

on a north-south track

on a track which is constantly changing direction

a rhumb line track

On East-West tracks in polar areas

On high latitude tracks directly East-West

On East-West tracks in the northern hemisphere north of the magnetic equator

On tracks directly North-South and on East-West tracks along the Equator

fly a spiral and finally end up at the south pole

experience an increase in the value of true track

always have the rhumb line track between the departure point and the destination to the left of your great circle track

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NAVIGATION

NAVIGATION

NAVIGATION




When flying on a westerly great circle track in the Southern Hemisphere you will:

How many feet are there in 1 sm?

How many feet are there in a nm?

2

2

2

Complexity

Complexity

Complexity

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1

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kj---900077

kj---900078

kj---900079

Ref

Ref

Ref

Ref

experience a decrease in the value of true track

3.280 ft

5.280 ft

6.080 ft

1.000 ft

3.280 ft

5.280 ft

6.080 ft

1.000 ft

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NAVIGATION

NAVIGATION

NAVIGATION




How many feet are there in a km?

How many centimetres are equivalent to 36.25 inches?

How many feet are equivalent to 9.5 km?

2

2

2

Complexity

Complexity

Complexity

1

1

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kj---900080

kj---900081

kj---900082

Ref

Ref

Ref

3.280 ft

5.280 ft

6.080 ft

1.000 ft

92.08 cm

0.014 m

14.27 cm

11.05 cm

31.160 ft

50.160 ft

57.760 ft

1

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NAVIGATION

NAVIGATION

NAVIGATION




How many feet are equivalent to 9.5 km?

The International Nautical Mile defined by ICAO is equivalent to ___ m.

A nautical mile is:

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

1

kj---900082

kj---900083

kj---900084

Ref

Ref

Ref

Ref

9.500 ft

1.582m

1.652m

1.852m

1.962m

1609 metres

1852 metres

1012 metres

1500 metres

1

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NAVIGATION

NAVIGATION

NAVIGATION




The distance along a meridian between 63o55’N and 13o47’S is:

What is the length of one degree of longitude at latitude 60o South?

What is the rhumb line distance, in nautical miles, between two positions on latitude 60oN, that are separated by 10o of longitude?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900085

kj---900086

kj---900087

Ref

Ref

Ref

3008 NM

7702 NM

5008 NM

4662 NM

30 NM

52 NM

60 NM

90 NM

300 NM

520 NM

600 NM

1

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NAVIGATION

NAVIGATION

NAVIGATION


MAGNETISM AND COMPASSES - General Principles


What is the rhumb line distance, in nautical miles, between two positions on latitude 60oN, that are separated by 10o of longitude?

What is the dip angle at the South Magnetic Pole?

What is the value of magnetic dip at the South Magnetic Pole?

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

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kj---900087

kj---900088

kj---900089

Ref

Ref

Ref

Ref

866 NM

0 deg

90 deg

180 deg

64 deg

360o

180o

090o

0o

1

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NAVIGATION

NAVIGATION

NAVIGATION




The angle between True North and Magnetic North is known as:

Isogonic lines connect positions that have:

At a specific location, the value of magnetic variation:

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900090

kj---900091

kj---900092

Ref

Ref

Ref

deviation

variation

alignment error

dip

the same variation

0o variation

the same elevation

the same angle of magnetic dip

depends on the true heading

depends on the type of compass installed

depends on the magnetic heading

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NAVIGATION

NAVIGATION

NAVIGATION




At a specific location, the value of magnetic variation:

What is the definition of magnetic variation?

The horizontal component of the earth's magnetic field:

2

2

2

Complexity

Complexity

Complexity

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1

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kj---900092

kj---900093

kj---900094

Ref

Ref

Ref

Ref

varies slowly over time

The angle between the direction indicated by a compass and Magnetic North

The angle between True North and Compass North

The angle between Magnetic North and True North

The angle between Magnetic Heading and Magnetic North

is approximately the same at all magnetic latitudes less than 60o

weakens with increasing distance from the magnetic poles

weakens with increasing distance from the nearer magnetic pole

is approximately the same at magnetic latitudes 50oN and 50oS

1

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NAVIGATION

NAVIGATION

NAVIGATION




Isogonals converge at the:

If variation is West; then:

Complete the following statement regarding magnetic variation. The charted values of magnetic variation on earth normally change annually dueto:

2

2

2

Complexity

Complexity

Complexity

1

1

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kj---900095

kj---900096

kj---900097

Ref

Ref

Ref

Magnetic equator

North and South geographic and magnetic poles

North magnetic pole only

North and South magnetic poles only

True North is West of Magnetic North

Compass North is West of Magnetic North

True North is East of Magnetic North

Magnetic North is West of Compass North

a reducing field strength causing numerical values at all locations to decrease

magnetic pole movement causing numerical values at all locations to increase

magnetic pole movement causing numerical values at all locations to increase or decrease

1

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NAVIGATION

NAVIGATION

NAVIGATION




Complete the following statement regarding magnetic variation. The charted values of magnetic variation on earth normally change annually dueto:

Which of these is a correct statement about the Earth's magnetic field?

When turning right from 330o (C) to 040o (C) in the northern hemisphere, the reading of a direct reading magnetic compass will:

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

1

kj---900097

kj---900098

kj---900099

Ref

Ref

Ref

Ref

an increasing field strength causing numerical values at all locations to increase

It acts as though there is a large blue magnetic pole in Northern Canada

The angle of dip is the angle between the vertical and the total magnetic force

It may be temporary, transient, or permanent

It has no effect on aircraft deviation

over-indicate the turn and liquid swirl will decrease the effect

under-indicate the turn and liquid swirl will increase the effect

under-indicate the turn and liquid swirl will decrease the effect

over-indicate the turn and liquid swirl will increase the effect

1

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NAVIGATION

NAVIGATION

NAVIGATION




Isogonals are lines of equal:

A negative (westerly) magnetic variation signifies that:

An aircraft is over position HO (55o30N 060o15W), where YYR VOR (53o30N 060o15W) can be received. The magnetic variation is 31oW at HO and 28oWat YYR. What is the radial from YYR?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900100

kj---900101

kj---900102

Ref

Ref

Ref

compass deviation

magnetic variation

pressure

wind velocity

True North is East of Magnetic North

True North is West of Magnetic North

Compass North is East of Magnetic North

Compass North is West of Magnetic North

031o

208o

028o

1

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft is over position HO (55o30N 060o15W), where YYR VOR (53o30N 060o15W) can be received. The magnetic variation is 31oW at HO and 28oWat YYR. What is the radial from YYR?

A line drawn on a chart which joins all points where the value of magnetic variation is zero is called an:

The angle between True North and Magnetic North is called:

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

1

kj---900102

kj---900103

kj---900104

Ref

Ref

Ref

Ref

332o

isogonal

aclinic line

agonic line

isotach

compass error

deviation

variation

drift

1

1

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1

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1

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NAVIGATION

NAVIGATION

NAVIGATION




The value of magnetic variation on a chart changes with time. This is due to:

Given: True track is 348o Drift 17o left Variation 32oW Deviation 4oE What is the compass heading?

The agonic line:

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900105

kj---900106

kj---900107

Ref

Ref

Ref

movement of the magnetic poles, causing an increase

increase in the magnetic field, causing an increase

reduction in the magnetic field, causing a decrease

movement of the magnetic poles, which can cause either an increase or a decrease

007o

033o

359o

337o

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NAVIGATION

NAVIGATION

NAVIGATION




The agonic line:

Isogonal lines converge as follows:

Which of the following statements concerning earth magnetism is completely correct?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900107

kj---900108

kj---900109

Ref

Ref

Ref

is midway between the magnetic North and South poles

follows the geographic equator

is the shorter distance between the respective True and Magnetic North and South poles

Follows separate paths out of the North polar regions, one currently running through Western Europe and the other through the USA

at the North Magnetic Pole

at the North and South Magnetic and Geographical Poles

at the North and South Magnetic poles

at the Magnetic equator

An isogonal is a line which connects places with the same magnetic variation; the agonic line is the line of zero magnetic dip

An isogonal is a line which connects places with the same magnetic variation; the aclinic is the line of zero magnetic dip

An isogonal is a line which connects places of equal dip; the aclinic is the line of zero magnetic dip

1

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NAVIGATION

NAVIGATION

NAVIGATION




Which of the following statements concerning earth magnetism is completely correct?

An Agonic line is a line that connects:

The Earth can be considered as being a magnet with the:

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

1

kj---900109

kj---900110

kj---900111

Ref

Ref

Ref

Ref

An isogonal is a line which connects places with the same magnetic variation; the aclinic connects places with the same magneticfield strength

positions that have the same variation

positions that have 0o variation

points of equal magnetic dip

points of equal magnetic horizontal field strength

blue pole near the north pole of the earth and the direction of the magnetic force pointing straight up from the earth's surface

red pole near the north pole of the earth and the direction of the magnetic force pointing straight down to the earth's surface

blue pole near the north pole of the earth and the direction of the magnetic force pointing straight down to the earth's surface

red pole near the north pole of the earth and the direction of the magnetic force pointing straight up from the earth's surface

1

1

1

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NAVIGATION

NAVIGATION

NAVIGATION




The value of magnetic variation:

Where is a compass most effective?

A magnetic compass will be most effective at:

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900113

kj---900114

kj---900115

Ref

Ref

Ref

varies between maximum values of 45oE and 45oW

is a maximum of 180o

is always 0o at the magnetic equator

is never greater than 90o

About midway between the earth's magnetic poles

In the region of the magnetic South pole

In the region of the magnetic North pole

On the geographic equator

a position roughly half way between the magnetic poles

the South Magnetic Pole

the North Magnetic Pole

1

1

1

1

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4

1

2

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1

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NAVIGATION

NAVIGATION

NAVIGATION




A magnetic compass will be most effective at:

When accelerating on a westerly heading in the northern hemisphere, the compass card of a direct reading magnetic compass will turn:

When a magnetized compass needle is freely suspended in the Earth's magnetic field, when free from extraneous magnetic influence, it will alignitself with:

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

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kj---900115

kj---900116

kj---900117

Ref

Ref

Ref

Ref

the Equator

clockwise giving an apparent turn towards the north

clockwise giving an apparent turn towards the south

anti-clockwise giving an apparent turn towards the north

anti-clockwise giving an apparent turn towards the south

true North

magnetic North

absolute North

relative North

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NAVIGATION

NAVIGATION

NAVIGATION




When a magnetized compass needle is freely suspended in the Earth's magnetic field, and affected by extraneous magnetic influence, it will alignitself with:

When is Magnetic North Pole is East of the True North Pole variation is:

When the Magnetic Pole is West of the True North pole variation is:

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900118

kj---900119

kj---900120

Ref

Ref

Ref

true North

magnetic North

compass North

relative North

+ and easterly

- and easterly

- and westerly

+ and westerly

+ and easterly

- and easterly

- and westerly

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NAVIGATION

NAVIGATION

NAVIGATION




When the Magnetic Pole is West of the True North pole variation is:

An isogonal is:

The agonic line is:

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

1

kj---900120

kj---900121

kj---900122

Ref

Ref

Ref

Ref

+ and westerly

a line of equal wind speed

a line of equal magnetic deviation

a line of zero magnetic variation

a line of equal magnetic variation

a line of zero magnetic deviation

a line of equal magnetic deviation

a line of zero magnetic variation

a line of equal magnetic variation

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NAVIGATION

NAVIGATION

NAVIGATION




What is deviation?

Deviation is:

The force acting on the needle of a direct reading compass varies:

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900123

kj---900124

kj---900125

Ref

Ref

Ref

The angle between magnetic North and compass North

The angle between magnetic North and True North

The angle between True North and compass North

The angle between True North and magnetic North

an error to be added to magnetic headings

a correction to be added to magnetic heading to obtain compass heading

a correction to be added to compass heading to obtain magnetic heading

an error to be added to compass heading to obtain magnetic heading

directly with the horizontal component of the earth's magnetic field

directly with the vertical component of the earth's magnetic field

inversely with both vertical and horizontal components of the earth's magnetic field

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NAVIGATION

NAVIGATION

NAVIGATION




The force acting on the needle of a direct reading compass varies:

The horizontal component of the earth's magnetic field:

The lines on a chart joining places of equal magnetic dip are called:

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

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kj---900125

kj---900126

kj---900127

Ref

Ref

Ref

Ref

inversely with the horizontal component of the earth's magnetic field

weakens with increasing distance from the nearer magnetic pole

weakens with increasing distance from the magnetic poles

is stronger closer to the magnetic equator

is approximately the same at all magnetic latitudes less than 60o

Aclinic lines

Isogonals

Isoclinals

Agonic lines

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft is accelerating on a westerly heading in the Northern Hemisphere; the effect on a Direct Reading Compass will result in:

When should a DRC be 'swung'?

An aircraft, in the Northern Hemisphere, turns right from 330(C) in a Rate 1 Turn for 30 secs. As the aircraft rolls out, does the compassoverread or underread and will liquid swirl increase or decrease the error:

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900128

kj---900129

kj---900130

Ref

Ref

Ref

An apparent turn to the West

An indication of a turn to the North

A decrease in the indicated reading

An indication of a turn to the South

Every 6 months

Following a change of magnetic latitude

For night use

After flying in an area where lightning is visible

Underread Decrease

Underread Increase

Overread Decrease

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft, in the Northern Hemisphere, turns right from 330(C) in a Rate 1 Turn for 30 secs. As the aircraft rolls out, does the compassoverread or underread and will liquid swirl increase or decrease the error:

An aircraft is accelerating on a westerly heading in the Northern Hemisphere. The effect on a Direct Reading Magnetic Compass is:

What is the maximum possible value of Dip Angle?

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

1

kj---900130

kj---900131

kj---900132

Ref

Ref

Ref

Ref

Overread Increase

Underreads North

Underreads South

Overreads North

Overreads South

66o

180o

90o

45o

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NAVIGATION

NAVIGATION

NAVIGATION

MAGNETISM AND COMPASSES - Aircraft Magnetism



Given: True Track = 352 deg Variation = 11W Deviation = .5 Drift = 10R What is Heading (C)?

When decelerating on a westerly heading in the Northern Hemisphere, the compass card of a direct reading magnetic compass will turn:

When an aircraft on a westerly heading on the northern hemisphere accelerates, the effect of the acceleration error causes the magnetic compassto:

2

2

2

Complexity

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1

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kj---900133

kj---900134

kj---900135

Ref

Ref

Ref

078 C

346 C

358 C

025 C

clockwise giving an apparent turn toward the south

anti-clockwise giving an apparent turn towards the south

clockwise giving an apparent turn towards the north

anti-clockwise giving an apparent turn towards the north

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NAVIGATION

NAVIGATION

NAVIGATION




When an aircraft on a westerly heading on the northern hemisphere accelerates, the effect of the acceleration error causes the magnetic compassto:

In Northern Hemisphere, during an acceleration in an easterly direction, the magnetic compass will indicate:

Concerning direct reading magnetic compasses, in the northern hemisphere, it can be said that:

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900135

kj---900136

kj---900137

Ref

Ref

Ref

lag behind the turning rate of the aircraft

indicate a turn towards the north

indicate a turn towards the south

to turn faster than the actual turning rate of the aircraft

a decrease in heading

an increase in heading

an apparent turn to the South

a heading of East

on an Easterly heading, a longitudinal acceleration causes an apparent turn to the South

on an Easterly heading, a longitudinal acceleration causes an apparent turn to the North

on a Westerly heading, a longitudinal acceleration causes an apparent turn to the South

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NAVIGATION

NAVIGATION

NAVIGATION




Concerning direct reading magnetic compasses, in the northern hemisphere, it can be said that:

The angle between Magnetic North and Compass North is called:

You are in the Northern hemisphere, heading 135C on a Direct Reading Magnetic Compass. You turn right in a Rate 1 turn for 30 seconds. Do youroll out on an indicated heading of:

2

2

2

Complexity

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Complexity

Complexity

1

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kj---900137

kj---900138

kj---900139

Ref

Ref

Ref

Ref

on a Westerly heading, a longitudinal deceleration causes an apparent turn to the North

magnetic variation

compass error

compass deviation

alignment error

greater than 225

less than 225

equal to 225

not possible to determine

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NAVIGATION

NAVIGATION

NAVIGATION




When turning right from 330o(C) to 040o(C) in the northern hemisphere the reading of a direct reading magnetic compass will:

Compass deviation is defined as the angle between:

The value of variation:

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900140

kj---900141

kj---900142

Ref

Ref

Ref

over-indicate the turn and liquid swirl will decrease the effect

under-indicate the turn and liquid swirl will increase the effect

under-indicate the turn and liquid swirl will decrease the effect

over-indicate the turn and liquid swirl will increase the effect

True North and Magnetic North

Magnetic North and Compass North

True North and Compass North

The horizontal and the total intensity of the earth's magnetic field

is zero at the magnetic equator

has a maximum value of 180 deg

has a maximum value of 45E or 45W

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NAVIGATION

NAVIGATION

NAVIGATION




The value of variation:

Deviation applied to magnetic heading gives:

At the magnetic equator, when accelerating after take off on heading West, a direct reading compass:

2

2

2

Complexity

Complexity

Complexity

Complexity

1

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kj---900142

kj---900143

kj---900144

Ref

Ref

Ref

Ref

cannot exceed 90 deg

magnetic course

true heading

compass heading

magnetic track

underreads the heading

overreads the heading

indicates the correct heading

indicates a turn to the south

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft in the northern hemisphere makes an accurate rate one turn to the right/starboard. If the initial heading was 330o after 30 secondsof the turn the direct reading magnetic compass should read:

When accelerating on an easterly heading in the Northern hemisphere, the compass card of a direct reading magnetic compass will turn:

You are turning from 330o to 040o in the Northern hemisphere using timing. You stop the turn at the correct time. Before the direct indicatingmagnetic compass settles down, does it over-read or under-read, and does the effect of liquid swirl increase or decrease?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900145

kj---900146

kj---900147

Ref

Ref

Ref

060o

less than 060o

more than 060o

more or less than 060o depending on the pendulous suspension used

anti-clockwise giving an apparent turn toward the south

clockwise giving an apparent turn toward the south

anti-clockwise giving an apparent turn toward the north

clockwise giving an apparent turn toward the north

Under-read; increase

Over-read; decrease

Under-read; decrease

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NAVIGATION

NAVIGATION

NAVIGATION




You are turning from 330o to 040o in the Northern hemisphere using timing. You stop the turn at the correct time. Before the direct indicatingmagnetic compass settles down, does it over-read or under-read, and does the effect of liquid swirl increase or decrease?

Which of the following statements is correct concerning the effect of turning errors on a direct reading compass?

Permanent magnetism in aircraft arises chiefly from:

2

2

2

Complexity

Complexity

Complexity

Complexity

1

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1

kj---900147

kj---900148

kj---900149

Ref

Ref

Ref

Ref

Over-read; increase

Turning errors are greatest on north/south headings, and are least at high latitudes

Turning errors are greatest on east/west headings, and are least at high latitudes

Turning errors are greatest on north/south headings, and are greatest at high latitudes

Turning errors are greatest on east/west headings, and are greatest at high latitudes

exposure to the earth's magnetic field during normal operation

hammering, and the effect of the earth's magnetic field, whilst under construction

the combined effect of aircraft electrical equipment and the earth's magnetic field

the effect of internal wiring and exposure to electrical storms

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NAVIGATION

NAVIGATION

NAVIGATION


MAGNETISM AND COMPASSES - Principles; Direct


One purpose of a compass calibration is to reduce the difference, if any, between:

In a remote indicating compass system the amount of deviation caused by aircraft magnetism and electrical circuits may be minimised by:

The main advantage of a remote indicating compass over a direct reading compass is that it:

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900150

kj---900151

kj---900152

Ref

Ref

Ref

compass north and magnetic north

compass north and true north

true north and magnetic north

compass north and the lubber line

positioning the master unit in the centre of the aircraft

the use of repeater cards

mounting the detector unit in the wingtip

using a vertically mounted gyroscope

is able to magnify the earth's magnetic field in order to attain greater accuracy

has less moving parts

requires less maintenance

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NAVIGATION

NAVIGATION

NAVIGATION




The main advantage of a remote indicating compass over a direct reading compass is that it:

The purpose of compass check swing is to:

Which of the following is an occasion for carrying out a compass swing on a Direct Reading Compass?

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

1

kj---900152

kj---900153

kj---900154

Ref

Ref

Ref

Ref

senses, rather than seeks, the magnetic meridian

cancel out the horizontal component of the earth's magnetic field

cancel out the vertical component of the earth's magnetic field

measure the angle between Magnetic North and Compass North

cancel out the effects of the magnetic fields found on board the aeroplane

After an aircraft has passed through a severe electrical storm, or has been struck by lightning

Before an aircraft goes on any flight that involves a large change of magnetic latitude

After any of the aircraft radio equipment has been changed due to unserviceability

Whenever an aircraft carries a large freight load regardless of its content

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NAVIGATION

NAVIGATION

NAVIGATION




Why are the detector units of slaved gyro compasses usually located in the aircraft wingtips?

A direct reading compass should be swung when:

The direct reading magnetic compass is made aperiodic (dead beat) by:

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900155

kj---900156

kj---900157

Ref

Ref

Ref

With one detector unit in each wingtip, compass deviations are cancelled out

To isolate the detector unit from the aircraft deviation sources

To isolate the detector unit from the Earth's magnetic field

To reduce turning and acceleration errors

there is a large, and permanent, change in magnetic latitude

there is a large change in magnetic longitude

the aircraft is stored for a long period and is frequently moved

the aircraft has made more than a stated number of landings

using the lowest acceptable viscosity compass liquid

keeping the magnetic assembly mass close to the compass point and by using damping wires

using long magnets

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NAVIGATION

NAVIGATION

NAVIGATION




The direct reading magnetic compass is made aperiodic (dead beat) by:

The main reason for usually mounting the detector unit of a remote indicating compass in the wingtip of an aeroplane is to:

The annunciator of a remote indicating compass system is used when:

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

1

kj---900157

kj---900158

kj---900159

Ref

Ref

Ref

Ref

pendulous suspension of the magnetic assembly

facilitate easy maintenance of the unit and increase its exposure to the Earth's magnetic field

reduce the amount of deviation caused by aircraft magnetism and electrical circuits

place it is a position where there is no electrical wiring to cause deviation errors

place it where it will not be subjected to electrical or magnetic interference from the aircraft

synchronising the magnetic and gyro compass elements

compensating for deviation

setting local magnetic variation

setting the heading pointer

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NAVIGATION

NAVIGATION

NAVIGATION




Which one of the following is an advantage of a remote reading compass as compared with a standby compass?

An aircraft's compass must be swung:

The sensitivity of a direct reading magnetic compass is:

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900160

kj---900161

kj---900162

Ref

Ref

Ref

It senses the magnetic meridian instead of seeking it, increasing compass sensitivity

It is lighter than a direct reading compass because it employs, apart from the detector unit, existing aircraft equipment

it eliminates the effect of turning and acceleration errors by pendulously suspending the detector unit

It is more reliable because it is operated electrically and power is always available from sources within the aircraft

if the aircraft has been in the hangar for a long time and has been moved several times

if the aircraft has been subjected to hammering

every maintenance inspection

after a change of theatre of operations at the same magnetic latitude

inversely proportional to the horizontal component of the earth's magnetic field

proportional to the horizontal component of the earth's magnetic field

inversely proportional to the vertical component of the earth's magnetic field

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NAVIGATION

NAVIGATION

NAVIGATION




The sensitivity of a direct reading magnetic compass is:

What is the advantage of the remote indicating compass (slaved gyro compass) over the direct reading magnetic compass?

The main reason for mounting the detector unit of a remote reading compass in the wingtip of an aeroplane is:

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

1

kj---900162

kj---900163

kj---900164

Ref

Ref

Ref

Ref

inversely proportional to the vertical and horizontal components of the earth's magnetic field

It is lighter

It is connected to a source of electrical power and so is more accurate

It senses the earth's magnetic field rather than seeks it, so is more sensitive

It is not affected by aircraft deviation

to ensure that the unit is in the most accessible position on the aircraft for ease of maintenance

by having detector units on both wingtips, to cancel out the deviation effects caused by the aircraft structure

to minimise the amount of deviation caused by aircraft magnetism and electrical circuits

to maximise the units exposure to the earth's magnetic field

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NAVIGATION

NAVIGATION

NAVIGATION




The sensitivity of a direct reading compass varies:

If compass HDG is 340o and deviation +3, what is magnetic heading?

If true HDG is 165o and variation -3 what is magnetic heading?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900165

kj---900166

kj---900167

Ref

Ref

Ref

inversely with the vertical component of the earth's magnetic

directly with the horizontal component of the earth's magnetic field

directly with the vertical component of the earth's magnetic field

inversely with both vertical and horizontal components of the earth's magnetic field

Deviation is plus therefore East, so compass is least, so magnetic is 343o

Deviation is plus therefore West, so compass is least, so magnetic is 343o

Deviation is plus therefore East, so compass is best, so magnetic is 337o

Deviation is plus therefore East, so compass is best, so magnetic is 343o

Variation is minus therefore West, so magnetic is best, so magnetic is 168o

Variation is minus therefore West, so magnetic is least, so magnetic is 162o

Variation is plus therefore East, so magnetic is best, so magnetic is 162o

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NAVIGATION

NAVIGATION

NAVIGATION




If true HDG is 165o and variation -3 what is magnetic heading?

In still air, you wish to fly a true of 315o. Variation is 4oW. Deviation is 2oE. What Compass heading should you fly?

Magnetic compass calibration is carried out to reduce:

2

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Complexity

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1

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kj---900167

kj---900168

kj---900169

Ref

Ref

Ref

Ref

Variation is plus therefore East, so magnetic is best, so magnetic is 168o

321

313

317

309

deviation

variation

parallax error

acceleration errors

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NAVIGATION

NAVIGATION

NAVIGATION



CHARTS - General properties of miscellaneous types of projections

You are in the northern hemisphere, heading West, and the aircraft is accelerating. Will a direct reading magnetic compass over-read or under-read and is the compass indicating a turn to the north or to the south:

Concerning a Direct Reading Compass in the Northern Hemisphere, it can be said:

The standard parallels of a Lamberts conical orthomorphic projection are 07o40N and 38o20N. The constant of the cone for this chart is:

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kj---900170

kj---900171

kj---900172

Ref

Ref

Ref

over-reads north

overreads south

under-reads north

under-reads south

On an easterly heading, a lateral acceleration produces an apparent turn to the South

On an easterly heading, a longitudinal acceleration produces an apparent turn to the North

On a westerly heading, a lateral acceleration produces an apparent turn to the North

On a westerly heading, a longitudinal acceleration produces an apparent turn to the South

0.60

0.39

0.92

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NAVIGATION

NAVIGATION

NAVIGATION




The standard parallels of a Lamberts conical orthomorphic projection are 07o40N and 38o20N. The constant of the cone for this chart is:

On a Lambert Conformal Conic chart earth convergency is most accurately represented at the:

An Oblique Mercator projection is used specifically to produce:

2

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Complexity

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Complexity

Complexity

1

1

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kj---900172

2kj---900174

kj---900175

Ref

Ref

Ref

Ref

0.42

north and south limits of the chart

parallel of origin

standard parallels

equator

plotting charts in equatorial regions

radio navigational charts in equatorial regions

topographical maps of large east/west extent

charts of the great circle route between two points

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NAVIGATION

NAVIGATION

NAVIGATION




The main use for an Oblique Mercator chart would be:

Scale on a Lamberts conformal chart is:

On a transverse Mercator chart, with the exception of the Equator, parallels of latitude appear as:

2

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Complexity

Complexity

Complexity

1

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kj---900176

kj---900177

kj---900178

Ref

Ref

Ref

for countries with large changes in latitude but small changes in longitude

route charts for selected great circle routes

better topographical coverage of polar regions

topographical coverage of equatorial regions

constant along a parallel of latitude

constant along a meridian of longitude

constant over the whole chart

varies with latitude and longitude

hyperbolic lines

straight lines

ellipses

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NAVIGATION

NAVIGATION

NAVIGATION




On a transverse Mercator chart, with the exception of the Equator, parallels of latitude appear as:

The two standard parallels of a conical Lambert projection are at N10o40 and N41o20. The cone constant of this chart is approximately:

The constant of the cone, on a Lambert chart where the convergence angle between longitudes 010oE and 030oW is 30o, is:

2

2

2

Complexity

Complexity

Complexity

Complexity

1

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kj---900178

kj---900179

kj---900180

Ref

Ref

Ref

Ref

parabolas

0.18

0.90

0.66

0.44

0.40

0.75

0.50

0.64

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NAVIGATION

NAVIGATION

NAVIGATION




The chart that is generally used for navigation in polar areas is based on a:

A Mercator chart has a scale at the equator = 1:3 704 000. What is the scale at latitude 60o S?

A Lambert conformal conic projection, with two standard parallels:

2

2

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Complexity

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1

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kj---900181

kj---900182

kj---900183

Ref

Ref

Ref

Stereographical projection

Direct Mercator projection

Gnomonic projection

Lambert conformal projection

1 : 1 852 000

1 : 7 408 000

1 : 3 208 000

1 : 185 200

shows lines of longitude as parallel straight lines

shows all great circles as straight lines

the scale is only correct at parallel of origin

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NAVIGATION

NAVIGATION

NAVIGATION




A Lambert conformal conic projection, with two standard parallels:

The convergence factor of a Lambert conformal conic chart is quoted as 0.78535. At what latitude on the chart is earth convergency correctlyrepresented?

The nominal scale of a Lambert conformal conic chart is the:

2

2

2

Complexity

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kj---900183

kj---900184

kj---900185

Ref

Ref

Ref

Ref

the scale is only correct along the standard parallels

38o15

51o45

52o05

80o39

scale at the equator

scale at the standard parallels

mean scale between pole and equator

mean scale between the parallels of the secant cone

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NAVIGATION

NAVIGATION

NAVIGATION




The constant of cone of a Lambert conformal conic chart is quoted as 0.3955. At what latitude on the chart is earth convergency correctlyrepresented?

On a direct Mercator projection, the distance measured between two meridians spaced 5o apart at latitude 60oN is 8 cm. The scale of this chartat latitude 60oN is approximately:

At 60o N the scale of a direct Mercator chart is 1:

2

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kj---900186

kj---900187

kj---900188

Ref

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68o25

21o35

23o18

66o42

1 : 4 750 000

1 : 7 000 000

1 : 6 000 000

1 : 3 500 000

1 : 3 000 000

1 : 3 500 000

1 : 1 500 000

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NAVIGATION

NAVIGATION

NAVIGATION




At 60o N the scale of a direct Mercator chart is 1:

Transverse Mercator projections are used for:

A direct Mercator graticule is based on a projection that is:

2

2

2

Complexity

Complexity

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kj---900188

22kj---900189

kj---900190

Ref

Ref

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Ref

1 : 6 000 000

maps of large north/south extent

maps of large east/west extent in equatorial areas

radio navigation charts in equatorial areas

plotting charts in equatorial areas

spherical

concentric

cylindrical

conical

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NAVIGATION

NAVIGATION

NAVIGATION




What is the value of the convergence factor on a Polar Stereographic chart?

The Earth has been charted using:

A straight line is drawn on a Lamberts conformal conic chart between two positions of different longitude. The angular difference between theinitial true track and the final true track of the line is equal to:

2

2

2

Complexity

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kj---900191

kj---900192

kj---900193

Ref

Ref

Ref

0.866

0.5

0.0

1.0

WGP84

WGS84

GD84

GPS84

earth convergency

chart convergency

conversion angle

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NAVIGATION

NAVIGATION

NAVIGATION




A straight line is drawn on a Lamberts conformal conic chart between two positions of different longitude. The angular difference between theinitial true track and the final true track of the line is equal to:

How does the chart convergency change with latitude in a Lambert Conformal projection?

How does the scale vary in a Direct Mercator chart?

2

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kj---900193

kj---900194

kj---900195

Ref

Ref

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Ref

difference in longitude

It changes with sine of latitude

It changes with cosine of latitude

It increases with increase of latitude

It is constant and does not change with latitude

The scale increases with increasing distance from the Equator

The scale decreases with increasing distance from the Equator

The scale is constant

The scale increases south of the Equator and decreases north of the Equator

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NAVIGATION

NAVIGATION

NAVIGATION




On a chart a straight line is drawn between two points and has a length of 4.63 cm. What is the chart scale if the line represents 150 NM?

What is the constant of the cone for a Lambert conic projection whose standard parallels are at 50oN and 70oN?

Isogrivs on a chart indicate lines of:

2

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kj---900196

kj---900197

kj---900198

Ref

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Ref

1 : 1 000 000

1 : 6 000 000

1 : 3 000 000

1 : 5 000 000

0.500

0.941

0.866

0.766

Zero magnetic variation

Equal magnetic tip

Equal horizontal directive force

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NAVIGATION

NAVIGATION

NAVIGATION


CHARTS - Representation of meridians, parallel, great circles



On a Lambert conformal conic chart the convergence of the meridians:

On a Direct Mercator chart a great circle will be represented by a:

2

2

2

Complexity

Complexity

Complexity

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1

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kj---900198

kj---900199

kj---900200

Ref

Ref

Ref

Ref

Equal grivation 1 min

is the same as earth convergency at the parallel of origin

is zero throughout the chart

varies as the secant of the latitude

equals earth convergency at the standard parallels

complex curve

curve concave to the equator

curve convex to the equator

straight line

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NAVIGATION

NAVIGATION

NAVIGATION




On a Direct Mercator chart at latitude 15oS, a certain length represents a distance of 120 NM on the earth. The same length on the chart willrepresent on the earth, at latitude 10oN, a distance of:

On a Direct Mercator chart at latitude of 45oN, a certain length represents a distance of 90 NM on the earth. The same length on the chart willrepresent on the earth, at latitude 30oN, a distance of:

The parallels on a Lambert Conformal Conic chart are represented by:

2

2

2

Complexity

Complexity

Complexity

1

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2kj---900202

kj---900203

kj---900204

Ref

Ref

Ref

122.3 NM

117.7 NM

124.2 NM

118.2 NM

45 NM

73.5 NM

78 NM

110 NM

parabolic lines

straight lines

arcs of concentric circles

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NAVIGATION

NAVIGATION

NAVIGATION




The parallels on a Lambert Conformal Conic chart are represented by:

On a Lambert Conformal Conic chart great circles that are not meridians are:

On a Direct Mercator chart, a rhumb line appears as a:

2

2

2

Complexity

Complexity

Complexity

Complexity

1

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1

kj---900204

kj---900205

kj---900206

Ref

Ref

Ref

Ref

hyperbolic lines

curves concave to the parallel of origin

straight lines

curves concave to the pole of projection

straight lines within the standard parallels

straight line

small circle concave to the nearer pole

spiral curve

curve convex to the nearer pole

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NAVIGATION

NAVIGATION

NAVIGATION




On a Direct Mercator chart, great circles are shown as:

A Rhumb line is:

Which one of the following, concerning great circles on a Direct Mercator chart, is correct?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900207

kj---900208

kj---900209

Ref

Ref

Ref

curves convex to the nearer pole

straight lines

rhumb lines

curves concave to the nearer pole

the shortest distance between two points on a Polyconic projection

a line on the surface of the earth cutting all meridians at the same angle

any straight line on a Lambert projection

a line convex to the nearest pole on a Mercator projection

They are all curves convex to the equator

They are all curves concave to the equator

They approximate to straight lines between the standard parallels

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NAVIGATION

NAVIGATION

NAVIGATION




Which one of the following, concerning great circles on a Direct Mercator chart, is correct?

How does scale change on a normal Mercator chart?

Which one of the following describes the appearance of rhumb lines, except meridians, on a Polar Stereographic chart?

2

2

2

Complexity

Complexity

Complexity

Complexity

1

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kj---900209

kj---900210

kj---900211

Ref

Ref

Ref

Ref

With the exception of meridians and the equator, they are curves concave to the equator

Expands as the secant2 (1/2 co-latitude)

Expands directly with the secant of the latitude

Correct on the standard parallels, expands outside them, contracts within them

Expands as the secant of the E/W great circle distance

Straight lines

Ellipses around the Pole

Curves convex to the Pole

Curves concave to the Pole

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NAVIGATION

NAVIGATION

NAVIGATION




A straight line on a Lambert Conformal Projection chart for normal flight planning purposes:

On a Lambert chart (standard parallels 37oN and 65oN), with respct to the straight line drawn on the map the between A (N49o W030o) and B (N48oW040o), the:

Which one of the following statements is correct concerning the appearance of great circles, with the exception of meridians, on a PolarStereographic chart whose tangency is at the pole?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900212

kj---900213

kj---900214

Ref

Ref

Ref

can only be a parallel of latitude

is a Loxodromic line

is a Rhumb line

is approximately a Great Circle

great circle is to the north, the rhumb line is to the south

great circle and rhumb line are to the north

great circle and rhumb line are to the south

rhumb line is to the north, the great circle is to the south

The higher the latitude the closer they approximate to a straight line

Any straight line is a great circle

They are complex curves that can be convex and/or concave to the Pole

1

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NAVIGATION

NAVIGATION

NAVIGATION




Which one of the following statements is correct concerning the appearance of great circles, with the exception of meridians, on a PolarStereographic chart whose tangency is at the pole?

On a Direct Mercator, rhumb lines are:

On which of the following chart projections is it NOT possible to represent the north or south poles?

2

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Complexity

Complexity

Complexity

1

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kj---900214

kj---900215

kj---900216

Ref

Ref

Ref

Ref

They are curves convex to the Pole

straight lines

curves concave to the equator

ellipses

curves convex to the equator

Lamberts conformal

Direct Mercator

Transverse Mercator

Polar stereographic

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NAVIGATION

NAVIGATION

NAVIGATION




On a Lambert conformal conic chart, with two standard parallels, the quoted scale is correct:

Parallels of latitude on a Direct Mercator chart are:

The scale on a Lambert conformal conic chart:

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900217

kj---900218

kj---900219

Ref

Ref

Ref

along the prime meridian

along the two standard parallels

in the area between the standard parallels

along the parallel of origin

parallel straight lines equally spaced

arcs of concentric circles equally spaced

straight lines converging above the pole

parallel straight lines unequally spaced

is constant along a meridian of longitude

is constant across the whole map

varies slightly as a function of latitude and longitude

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NAVIGATION

NAVIGATION

NAVIGATION




The scale on a Lambert conformal conic chart:

On a Lambert conformal conic chart the distance between parallels of latitude spaced the same number of degrees apart:

What is the Rhumb line (RL) direction from 45oN 14o12W to 45oN 12o48E?

2

2

2

Complexity

Complexity

Complexity

Complexity

1

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1

kj---900219

kj---900220

kj---900221

Ref

Ref

Ref

Ref

is constant along a parallel of latitude

expands between, and reduces outside, the standard parallels

is constant throughout the chart

reduces between, and expands outside, the standard parallels

is constant between, and expands outside the standard parallels

270o (T)

090o (T)

090o (M)

270o (M)

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NAVIGATION

NAVIGATION

NAVIGATION




A rhumb line on a Direct Mercator chart appears as a:

Where on a Direct Mercator projection is the chart convergency correct compared to the earth convergency?

The rhumb line distance between points C (N6000.0 E00213.0) and D (N60000.0 W 00713.0) is:

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straight line

complex curve

curve convex to the nearer pole

small circle concave to the nearer pole

All over the chart

At the two parallels of tangency

At the poles

At the equator

300 nm

520 nm

150 nm

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NAVIGATION

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The rhumb line distance between points C (N6000.0 E00213.0) and D (N60000.0 W 00713.0) is:

An aircraft starts at position 0411.0S 17812.2W and heads True North for 2950nm, then turns 90o left maintaining a rhumb line track for 314 km.The aircraft's final position is:

The appearance of a rhumb line on a Mercator chart is:

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600 nm

5500.0N 17412.2W

4500.0N 17412.2W

5500.0N 17713.8E

4500.0N 17713.8E

A small circle concave to the nearer pole

A straight line

A curved line

A spiral curve

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NAVIGATION

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The distance on a Lambert's chart, between two parallels of latitude the same number of degrees apart:

The scale quoted on a Lamberts chart is:

On a conformal chart, scale is:

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is constant all over the chart

is constant between the Standard Parallels and expands outside them

Expands between the Standard Parallels, but reduces outside them

Reduces between the Standard Parallels, but expands outside them

The scale at the Standard Parallels

The scale at the Equator

The mean scale between the Pole and the Equator

The mean scale at the Parallel of the Secant of the Cone

Constant

Constant along a meridian of longitude

Variable: it varies as a function of latitude and longitude

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NAVIGATION

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On a conformal chart, scale is:

On a Transverse Mercator chart scale is correct at:

A pilot navigates from A to B on 7000.0N on a Polar Stereographic chart. A is at 6000.0W, B is at 6000.0E; the initial track at A is:

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Constant along a parallel of latitude

The 180o meridian

The False Meridian

The Great Circle of Tangency

The Meridian of Tangency

030o

150o

350o

210o

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NAVIGATION

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Which of the following differences in latitude will give the biggest difference in the initial Great Circle track and the mean Great Circletrack between two points separated by 10o change of longitude?

In which of the following projections does a plane surface touch the Reduced Earth at one of the Poles?

On a Polar Stereographic map, a straight line is drawn from position A (70N 102W) to position B (80N 006E). The point of highest latitude alongthis line occurs at longitude 035W. What is the initial straight-line track angle from A to B, measured at A?

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60N and 60S

60N and 55N

30S and 30N

30S and 25S

Gnomic

Stereographic

Lambert's

Direct Mercator

049

077

229

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NAVIGATION

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CHARTS - The use of current aeronautical charts

On a Polar Stereographic map, a straight line is drawn from position A (70N 102W) to position B (80N 006E). The point of highest latitude alongthis line occurs at longitude 035W. What is the initial straight-line track angle from A to B, measured at A?

The initial straight track from A (75N 60E) to B (75N 60W) on a Polar Stereographic chart is:

On Lambert Conformal chart the distance between meridians 5o apart along latitude 37o North is 9 cm. The scale of the chart at that parallelapproximates:

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023

030o

360o

060o

330o

1 : 3 750 000

1 : 5 000 000

1 : 2 000 000

1 : 6 000 000

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NAVIGATION

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A straight line is drawn on a North Polar Stereographic chart joining Point A (7000N 06000W) to Point B (7000N 06000E). What is the initialtrack direction (going eastwards) of the line at A?

(Refer to Jeppesen Student Manual - chart (E(LO)1 or figure 061-11)What is the radial and DME distance from BEL VOR/DME (N5439.7 W00613.8) toposition N5410 W00710?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Which of the following lists all the aeronautical chart symbols shown atposition N5318.0 W00626.9?

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kj---900242

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090 T

030 T

120 T

330 T

223o - 36 NM

236o - 44 NM

320o - 44 NM

333o - 36 NM

VOR: DME: danger area

Civil airport: VOR: DME

Military airport: VOR: NDB

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:CRK VOR/DME (N5150.4 W00829.7)Kerry aerodrome (N5210.9 W00931.4)What isthe CRK radial and DME distance when overhead Kerry aerodrome?

On a Mercator chart, at latitude 60oN, the distance measured between W002o and E008ois 20 cm. The scale of this chart at latitude 60oN isapproximately:

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Military airport: VOR: DME

307o - 43 NM

119o - 44 NM

127o - 45 NM

299o - 42 NM

1 : 5 560 000

1 : 278 000

1 : 780 000

1 : 556 000

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:SHA VOR/DME (N5243.3 W00853.1)Birr aerodrome (N5304 W00754)What is theSHA radial and DME distance when overhead Birr aerodrome?

On a Mercator chart, the scale:

An aircraft starts at position 0410S 17822W and heads true north for 2950 nm, then turns 90 degrees left, and maintains a rhumb line track for314 kilometers. What is its final position?

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068o - 41 NM

248o - 42 NM

060o - 42 NM

240o - 41 NM

varies as 1/cosine of latitude (1/cosine=secant)

varies as the sine of the latitude

is constant throughout the chart

varies as ½ cosine of the co-latitude

5500N 17422W

4500N 17422W

5500N 17738E

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NAVIGATION

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An aircraft starts at position 0410S 17822W and heads true north for 2950 nm, then turns 90 degrees left, and maintains a rhumb line track for314 kilometers. What is its final position?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the average track (oT) and distance between CRN NDB (N5318.1 W00856.5)and EKN NDB (N5423.6 W00738.7)?

Given: Direct Mercator chart with a scale of 1: 200 000 at equatorChart length from A to B, in the vicinity of the equator, 11 cmWhat is theapproximate distance from A to B?

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kj---900249

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4500N 17738E

044o - 82 NM

042o - 83 NM

036o - 81 NM

035o - 80 NM

21 NM

12 NM

22 NM

14 NM

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NAVIGATION

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Given that: A is N55 E/W 000 B is N54 E 010 If the true great circle track from A to B is 100T, what is the true Rhumb Line track at A?

(Refer to figure 061-10)What are the average magnetic course and distance between position N6000 W02000 and Sumburg VOR (N5955 W 00115)?

On a Polar Stereographic chart, the initial great circle course from A 70oN 060oW to B 70oN 060oE is approximately:

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kj---900252

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096

107

104

100

105o - 562 NM

091o - 480 NM

091o - 562 NM

105o - 480 NM

030o (T)

330o (T)

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NAVIGATION

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On a Polar Stereographic chart, the initial great circle course from A 70oN 060oW to B 70oN 060oE is approximately:

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:SHA VOR (N5243.3 W00853.1) DME 50 NMCRK VOR (N5150.4 W00829.7) DME 41NMAircraft heading 270o(M)Both DME distances increasingWhat is the aircraft position?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the average track (oT) and distance between BAL VOR (N5318.0 W00626.9)and CFN NDB (N5502.6 W00820.4)?

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kj---900254

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150o (T)

210o (T)

N5215 W00745

N5215 W00940

N5200 W00935

N5235 W00750

335o - 128 NM

327o - 124 NM

325o - 126 NM

320o - 127 NM

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:SHA VOR N5243.3 W00853.1CON VOR N5354.8 W00849.1Aircraft position N5320W00950Which of the following lists two radials that are applicable to the aircraft position?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:SHA VOR (N5243.3 W00853.1) radial 129oCRK VOR (N5150.4 W00829.7) radial047oWhat is the aircraft position?

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Civil airport: VOR: DME: non-compulsory reporting point

VOR: DME: NDB: compulsory reporting point

Civil airport: NDB: DME: non-compulsory reporting point


SHA 325o CON 235o

SHA 137o CON 046o

SHA 317o CON 226o

SHA 145o CON 055o

N5205 W00755

N5215 W00755

N5210 W00750

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NAVIGATION

NAVIGATION

NAVIGATION





Given: Chart scale is 1: 850 000 The chart distance between two points is 4 centimetres Earth distance is approximately:

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What feature is shown on the chart at position N5351 W009017?

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kj---900259

kj---900260

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N5220 W00750

4 NM

74 NM

100 NM

40 NM

Connaught aerodrome

Castlebar aerodrome

Connemara aerodrome

Brittas Bay aerodrome

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(Refer to Jeppesen Student Manual chart E(LO)1 or figure 061-11)What is the average track (oM) and distance between CRN NB (N5318.1 W00856.5)and BEL VOR (N5439.7 W00613.8)?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the average track (oM) and distance between KER NDB (N5210.9 W00931.5)and CRN NDB (N5318.1 W00856.5)?

On a direct Mercator projection, at latitude 45o North, a certain length represents 70 NM. At latitude 30o North, the same length representsapproximately:

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kj---900262

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229o - 125 NM

089o - 95 NM

057o - 126 NM

237o - 130 NM

025o - 70 NM

197o - 71 NM

205o - 71 NM

017o - 70 NM

57 NM

86 NM

70 NM

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NAVIGATION

NAVIGATION

NAVIGATION




On a direct Mercator projection, at latitude 45o North, a certain length represents 70 NM. At latitude 30o North, the same length representsapproximately:

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:SHA VOR/DME (N5243.3 W00853.1)Radial 165o/36 NMWhat is the aircraftposition?

Approximately how many nautical miles correspond to 12 cm on a map with a scale of 1: 200 000?

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kj---900264

kj---900265

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81 NM

N5210 W00830

N5208 W00840

N5315 W00915

N5317 W00908

130

150

329

43

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NAVIGATION

NAVIGATION

NAVIGATION




A course of 120o(T) is drawn between X(61o30N) and Y(58o30N) on a Lambert Conformal conic chart with a scale of 1: 1 000 000 at 60oN. The chartdistance between X and Y is:

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:SHA VOR N5243.3 W00853.1CRK VOR N5150.4 W00829.7Aircraft position N5230W00820Which of the following lists two radials that are applicable to the aircraft position?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:CRN VOR (N5318.1 W00856.5) DME 18 NMSHA VOR (N5243.3 W00853.1) DME 30NMAircraft heading 270o(M)Both DME distances decreasingWhat is the aircraft position?

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kj---900267

kj---900268

kj---900269

Ref

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33.4 cm

66.7 cm

38.5 cm

36.0 cm

SHA 131o CRK 017o

SHA 304o CRK 189o

SHA 312o CRK 197o

SHA 124o CRK 009o

N5252 W00923

N5310 N00830

N5307 W00923

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NAVIGATION

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:CRN VOR (N5318.1 W00856.5) DME 18 NMSHA VOR (N5243.3 W00853.1) DME 30NMAircraft heading 270o(M)Both DME distances decreasingWhat is the aircraft position?

On a chart, the distance along a meridian between latitudes 45oN and 46oN is 6 cm. The scale of the chart is approximately:

The following waypoints are entered into an inertial navigation system (INS) WPT 1: 60N 30W WPT 2: 60N 20W WPT 3: 60N 10W The intertial navigation is connected to the automatic pilot on the route WP1-WP2-WP3. The track change on passing WPT:

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kj---900269

kj---900270

kj---900271

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N5355 W00825

1 : 1 000 000

1 : 850 000

1 : 185 000

1 : 18 500 000

1 9 deg increase

1 4 deg decrease

zero

a 9 deg decrease

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NAVIGATION




(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:SHA VOR (N5243.3 W00853.1) radial 205oCRK VOR (5150.4 W00829.7) radial317oWhat is the aircraft position?

(Refer to Jeppesen Student manual - chart E(LO)1 or figure 061-11)What feature is shown on the chart at position NS311 W00637?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the radial and DME distance from CRK VOR/DME (N5150.4 W00829.7) toposition N5140 W00730?

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kj---900272

kj---900273

kj---900274

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N5210 W00910

N5118 W00913

N5205 W00915

N5215 W00917

Clonbullogue aerodrome

Connemara aerodrome

KERRY/Farranfore aerodrome

Punchestown aerodrome

106o - 38 NM

104o - 76 NM

293o - 39 NM

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NAVIGATION

NAVIGATION





The chart distance between meridians 10o apart at latitude 65o North is 3.75 inches. The chart scale at this latitude approximates:

(Refer to Jeppesen Student Manual - chart (E(LO)1 or figure 061-11)Which of the following lists all the aeronautical chart symbols shown atposition N5150.4 W00829.7?

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113o - 38 NM

1 : 6 000 000

1 : 5 000 000

1 : 2 500 000

1 : 3 000 000

Civil airport: VOR: non-compulsory reporting point

Civil airport: VOR: DME: compulsory reporting point


VOR: DME: NDB: ILS

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An aircraft at position 6000N 00522WS flies 165 km due East. What is the new position?

Two positions plotted on a polar stereographic chart, A (80oN 000o) and B (70oN 102oW) are joined by a straight line whose highest latitude isreached at 035oW. At point B, the true course is:

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TUSKAR ROCK LT.H. NDB

WTD NDB



6000N 00820E

6000N 00224WS

6000N 00108E

6000N 00108W

247o

023o

203o

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Two positions plotted on a polar stereographic chart, A (80oN 000o) and B (70oN 102oW) are joined by a straight line whose highest latitude isreached at 035oW. At point B, the true course is:

Given:An aircraft is flying a track of 255o(M). At 2254 UTC, it crosses radial 360o from a VOR station. At 2300 UTC, it crosses radial 330ofrom the same station. At 2300 UTC, the distance between the aircraft and the station is:

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the average track (oT) and distance between WTD NDB (N5211.3 W00705.0)and FOY NDB (N5234.0 W00911.7)?

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305o

the same as it was at 2254 UTC

greater than it was at 2254 UTC

randomly different that it was at 2254 UTC

less than it was at 2254 UTC

075o - 81 NM

294o - 80 NM

286o - 81 NM

277o - 83 NM

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the radial and DME distance from CON VOR/DME (N5354.8 W00849.1) toposition N5430 W00900?

Given: Waypoint 1.60oS 030oW Waypoint 2.60oS 020oW What will be the approximate latitude shown on the display unit of an inertial navigation system at longitude 025oW?

On a chart, 49 nautical miles is represented by 7.0 centimetres. What is the scale?

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049o - 45 NM

214o - 26 NM

358o - 36 NM

169o - 35 NM

060o 11’S

059o 49'S

060o 00'S

060o 06'S

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On a chart, 49 nautical miles is represented by 7.0 centimetres. What is the scale?

(refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the radial and DME distance from SHA VOA/DME (N5243.3 W00853.1) toposition N5210 W00920?


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1/700,000

½,015,396

1/1,296,400

1/1,156,600

346o - 34 NM

354o - 34 NM

198o - 37 NM

214o - 37 NM

N5205 W00805

N5155 W00810

N5210 W00800

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NAVIGATION

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NAVIGATION





The distance measured between two points on a navigation map is 42 mm (millimetres). The scale of the chart is 1:1 600 000. The actualdistance between these two points is approximately:

What is the chart distance between longitudes 179oE and 175oW on a direct Mercator chart with a scale of 1:5 000 000 at the equator?

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N5200 W00800

3.69 NM

370.00 NM

67.20 NM

36.30 NM

133 mm

106 mm

167 mm

72 mm

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An aircraft is at 5530N 03613W where the variation is 15W. It is tuned to a VOR located at 5330N 03613W where the variation is 12W. What VORradial is the aircraft on?


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039o - 48 NM

024o - 43 NM

023o - 48 NM

017o - 43 NM

348

012

165

015

N5230 W00800

N5225 W00805

N5220 W00750

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NAVIGATION

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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 Bin NM is:


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N5240 W00750

44.5

38.1

20.6

54.2

048o - 40 NM

030o - 33 NM

014o - 33 NM

220o - 40 NM

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the average track (oM) and distance between WTD NDB (N5211.3 W00705.0)and KER NDB (N5210.9 W00931.5)?

A straight line drawn on a chart measures 4.63 cm and represents 150 NM. The chart scale is:

Route A (44oN 026oE) to B (46oN 024oE) forms an angle of 35o with longitude 026oE. Average magnetic variation between A and B is 3oE. What isthe average magnetic course from A to B?

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270o - 89 NM

090o - 91 NM

278o - 90 NM

098o - 90 NM

1 : 3 000 000

1 : 6 000 000

1 : 5 000 000

1 : 1 000 000

322o

328o

032o

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NAVIGATION

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NAVIGATION




Route A (44oN 026oE) to B (46oN 024oE) forms an angle of 35o with longitude 026oE. Average magnetic variation between A and B is 3oE. What isthe average magnetic course from A to B?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:SHA VOR (N5243.3 W00853.1) DME 41 NMCRK VOR (N5150.4 W00829.7) DME 30NMAircraft heading 270o(M)Both DME distances decreasingWhat is the aircraft position? - A - N5225 W00810 - B - N5205 W00915 - C - N5215 W00805- D - N5215 W00915 Ref: all Ans: C 9817. 5 hours 20 minutes and 20 seconds hours time difference is equivalent to which change of longitude:

An aircraft flies a great circle track from 56o N 070o W to 62o N 110o E. The total distance travelled is:

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038o

81o 30

78o 15

79o 10

80o 05

2040 NM

1788 NM

5420 NM

3720 NM

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NAVIGATION

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NAVIGATION




An aircraft departs a point 0400N 17000W and flies 600 nm South, followed by 600 nm East, then 600 nm North, then 600 nm West. What is itsfinal position?

A Lambert conformal conic chart has a constant of the cone of 0.80. A straight line course drawn on this chart from A (53oN 004oW) to B is 080oat A; course at B is 092o(T). What is the longitude of B?

On a polar stereographic projection chart showing the South pole, a straight line joins position A (70oS 065oE) to position B (70oS 025oW). Thetrue course on departure from position A is approximately:

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0400N 17000W

0600S 17000W

0400N 16958.1W

0400N 17001.8W

011oE

009o36E

008oE

019oE

250o

225o

135o

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NAVIGATION

NAVIGATION

NAVIGATION




On a polar stereographic projection chart showing the South pole, a straight line joins position A (70oS 065oE) to position B (70oS 025oW). Thetrue course on departure from position A is approximately:


Assume a Mercator chart. The distance between positions A and B located on the same parallel and 10o longitude apart, is 6 cm. The scale atthe parallel is 1: 9 260 000. What is the latitude of A and B?

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315o

N5230 W00800

N5300 W00945

N5225 W00805

N5250 W00950

45o N or S

30o N or S

0o

60o N or S

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Magnetic heading 311o Drift angle 10o left Relative bearing of NDB 270o What is the magnetic bearing of the NDB measured from the aircraft?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the radial and DME distance from SHA VOR/DME (N5243.3 W00853.1) toposition N5310 W00830?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)At position 5211N 00931W, which of the following denotes all the symbols?

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211o

208o

221o

180o

019o - 31 NM

070o - 58 NM

207o - 31 NM

035o - 30 NM

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NAVIGATION

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)At position 5211N 00931W, which of the following denotes all the symbols?

A straight line on a chart 4.89 cm long represents 185 NM. The scale of this chart is approximately:

(Refer to figure 061-10)What are the average magnetic course and distance between INGO VOR (N6350 W01640) and Sumburg VOR (N5955 W 00115)?

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Military airport, ILS, NDB

Civil airport, VOR, ILS

Military airport, VOR, ILS

Civil airport, ILS, NDB

1 : 5 000 000

1 : 3 500 000

1 : 6 000 000

1 : 7 000 000

131o - 494 NM

118o - 440 NM

117o - 494 NM

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NAVIGATION

NAVIGATION

NAVIGATION




(Refer to figure 061-10)What are the average magnetic course and distance between INGO VOR (N6350 W01640) and Sumburg VOR (N5955 W 00115)?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the radial and DME distance from CON VOR/DME (N5354.8 W00849.1) toposition N5340 W00820)?

On a particular Direct Mercator wall chart, the 180W to 180E parallel of latitude at 53N is 133 cm long. What is the scale of the chart at 30S?

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kj---900310

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130o - 440 NM

311o - 22 NM

119o - 42 NM

140o - 23 NM

240o - 24 NM

1 : 3 000 000

1 : 18 000 000

1 : 21 000 000

1 : 25 000 000

1

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NAVIGATION

NAVIGATION

NAVIGATION




(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the average track (oT) and distance between SLG NDB (N5416.7 W00836.0)and CFN NDB (N5502.6 W00820.4)?

The total length of the 53oN parallel of latitude on a direct Mercator chart is 133 cm. What is the approximate scale of the chart at latitude30oS?

In a navigation chart a distance of 49 NM is equal to 7 cm. The scale of the chart is approximately:

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kj---900311

kj---900312

kj---900313

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191o - 45 NM

020o - 46 NM

348o - 46 NM

011o - 47 NM

1 : 25 000 000

1 : 30 000 000

1 : 18 000 000

1 : 21 000 000

1 : 130 000

1 : 700 000

1 : 1 300 000

1

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NAVIGATION

NAVIGATION

NAVIGATION




In a navigation chart a distance of 49 NM is equal to 7 cm. The scale of the chart is approximately:

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the average track (oT) and distance between WTD NDB (N5211.3 W00705.0)and SLG NDB (N5416.7 W00836.0)?

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(40oN 050oW) to B is 043o(T) at A; course at B is 055o(T). What is the longitude of B?

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kj---900313

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kj---900315

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1 : 7 000 000

344o - 139 NM

336o - 137 NM

156o - 136 NM

164o - 138 NM

41oW

36oW

38oW

34oW

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NAVIGATION

NAVIGATION

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the average track (oT) and distance between CON VOR (N5354.8 W00849.1)and BEL VOR (N5439.7 W00613.8)?

At latitude 60oN the scale of a Mercator projection is 1:5 000 000. The length on the chart between C N60o W008o and D N60o W008o is:

At 47o North the chart distanced between meridians 10o apart is 5 inches. The scale of the chart at 47o North approximates:

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kj---900316

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293o - 98 NM

071o - 100 NM

113o - 97 NM

063o - 101 NM

19.2 cm

16.2 cm

35.6 cm

17.8 cm

1: 2 500 000

1 : 8 000 000

1 : 3 000 000

1

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NAVIGATION

NAVIGATION

NAVIGATION




At 47o North the chart distanced between meridians 10o apart is 5 inches. The scale of the chart at 47o North approximates:

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:SHA VOR/DME (N5243.3 W00853.1)Connemara aerodrome (N5314 W00928)What isthe SHA radial and DME distance when overhead Connemara aerodrome?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the average track (oM) and distance between WTD NDB (N5211.3 W00705.0)and BAL VOR (N5318.0 WS00626.9)?

1

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Complexity

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Complexity

1

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kj---900318

kj---900319

kj---900320

Ref

Ref

Ref

Ref

1 : 6 000 000

333o - 37 NM

154o - 38 NM

326o - 37 NM

146o - 38 NM

206o - 71 NM

018o - 153 NM

026o - 71 NM

198o - 72 NM

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the average track and distance between BAL VOR (N5318.0 W00626.9) andCRN NDB (N5318.1 W00856.5)?


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139o - 35 NM

129o - 46 NM

132o - 36 NM

212o - 26 NM

278o - 89 NM

270o - 90 NM

268o - 91 NM

272o - 89 NM

VOR, DME, NDB non-compulsory reporting point

VOR, DME, NDB, compulsory reporting point

civil airport, VOR, DME, non-compulsory reporting point

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Waypoint 1 is 60N 30W. Waypoint 2 is 60N 20W. The aircraft autopilot is coupled to the INS steer. What is the latitude on passing 25W?

Position A is at 70S 030W, position B is 70S 060E. What is the Great Circle track of B from A, measured at A?

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civil airport, NDB, DME, compulsory reporting point

6005N

6011N

6032N

5949M

132T

048T

090T

228T

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the radial and DME distance from CON VOR/DME (N5354.8 W00849.1) toposition N5330 W00930?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Which of the following lists all the aeronautical chart symbols shown atposition N5211 W00705?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the radial and DME distance from BEL VOR/DME (N5439.7 W00613.8) toposition N5500 W00700?

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165o - 27 NM

233o - 35 NM

335o - 43 NM

025o - 38 NM

NDB: ILS

VOR: NDB

civil airport: ILS

civil airport: NDB

126o - 33 NM

296o - 65 NM

315o - 34 NM

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the radial and DME distance from BEL VOR/DME (N5439.7 W00613.8) toposition N5500 W00700?

(Refer to figure 061-10)What are the initial true course and distance between positions N5800 W01300 and N6600 E00200?


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222o - 48 NM

032o - 470 NM

029o - 570 NM

042o - 635 NM

036o - 638 NM

N5303 W00810

N5305 W00815

N5228 W00935

N5220 W00930

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(Refer to Jeppesen Student Manual chart - E(LO)1 or figure 061-11)Given:SHA VOR (N5243.3 W00853.1) radial 223oCRK VOR (5150.4 W00829.7) radial322oWhat is the aircraft position:

An aircraft at latitude 0220N tracks 180T for 685 kilometres. What is its latitude at the end of the flight?

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057o - 27 NM

309o - 33 NM

293o - 33 NM

324o - 17 NM

N5220 W00920

N5230 W00910

N5210 W00910

N5210 W00930

0350S

0250S

0210S

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An aircraft at latitude 0220N tracks 180T for 685 kilometres. What is its latitude at the end of the flight?

Contour lines on aeronautical maps and charts connect points:

(Refer to figure 061-10)An aircraft on radial 315o at a range of 150 NM from MYGGENES NDB (N6206 W00732) is at position?

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0850S

of equal latitude

with the same variation

having the same longitude

having the same elevation above sea level

N6320 W01205

N6020 W00405

N6345 W01125

N6040 W00320

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:CON VOR/DME (5354.8 W00849.1)Castlebar aerodrome (N5351 W00917)What isthe CON radial and DME distance when overhead Castlebar aerodrome


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Waterford NDB

Connemara aerodrome

Punchestown aerodrome

265o - 17 NM

077o - 18 NM

257o - 17 NM

086o - 18 NM

Cammore aerodrome

Belmullet aerodrome

EAGLE ISLAND LT.H. NDB

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:CRN VOR (N5318.1 W00856.5) DME 34 NMSHA VOR (N5243.3 W00853.1) DME 26NMAircraft heading 090o(M)Both DME distances increasingWhat is the aircraft position?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:SHA VOR/DME (N5243.3 W00853.1) radial 048o/22 NMWhat is the aircraftposition?

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N5255 W00815

N5250 W0030

N5305 W00930

N5310 W00820

N5228 00920

N5300 W0830

N5258 W00825

N5225 W00917

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:SHA VOR N5243.3 W00853.1CRK VOR N5150.4 W00829.7Aircraft position N5230W00930Which of the following lists two radials that are applicable to the aircraft position?

(Refer to figure 061-09)At 1215 UTC LAJES VORTAC (38o 46¿N 027o 05¿W) RMI reads 178o, range 135 NM. Calculate the aircraft position at 1215UTC:

(Refer to figure 061-07)Assume a North polar stereographic chart whose grid is aligned with the Greenwich meridian. An aircraft flies from thegeographic North pole for a distance of 480 NM along the 110oE meridian, then follows a grid track of 154o for a distance of 300 NM. Itsposition is now approximately:

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SHA 068o CRK 145o

SHA 060o CRK 138o

SHA 240o CRK 137o

SHA 248o CRK 325o

41o 00¿N 028o 10¿W

41o 05¿N 027o 50¿W

40o 55¿N 027o 55¿W

40o 50¿N 027o 40¿W

70o 15¿N 080o E

80o 00¿N 080oE

78o 45¿N 087oE

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(Refer to figure 061-07)Assume a North polar stereographic chart whose grid is aligned with the Greenwich meridian. An aircraft flies from thegeographic North pole for a distance of 480 NM along the 110oE meridian, then follows a grid track of 154o for a distance of 300 NM. Itsposition is now approximately:

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Given:CON VOR/DME (N5354.8 W00849.1)Abbey Shrule aerodrome (N5335 W00739)Whatis the CON radial and DME distance when overhead Abbey Shrule aerodrome?

(Refer to figure 061-10)An aircraft on radial 110o at a range of 120 NM from SAXAVORD VOR (N6050 W00050) is at position:

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79o 15¿N 074oE

296o - 46 NM

304o - 47 NM

124o - 46 NM

116o - 47 NM

N6127 W00443

N6010 E00255

N6109 E00255

N6027 E00307

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the average track (oT) and distance between SHA VOR (N5243.3W000853.1) and CON VOR (5354.8 W00849.1)?

(Refer to figure 061-06)Complete line 5 of the 'FLIGHT NAVIGATION LOG' positions 'J' to 'K'. What is the HDGo (M) and ETA?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the average track (oM) and distance between BAL VOR (N5318.0 W00626.9)and SLG NDB (N5416.7 W00836.0)?

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010o - 71 NM

358o - 72 NM

006o - 71 NM

002o - 72 NM

HDG 337o - ETA 1422 UTC




262o - 86 NM

128o - 99 NM

308o - 98 NM

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the average track (oM) and distance between BAL VOR (N5318.0 W00626.9)and SLG NDB (N5416.7 W00836.0)?

(Refer to figure 061-06)Complete line 4 of the 'FLIGHT NAVIGATION LOG' positions 'G' to 'H'. What is the HDGo (M) and ETA?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the radial and DME distance from CRK VOR/DME (N5150.4 WS00829.7) toposition N5210 W00920?

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316o - 96 NM





350o - 22 NM

295o - 38 NM

170o - 22 NM

311o - 38 NM

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(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the average track (oM) and distance between RK VOR (N5150.4 W00829.7)and CRN NDB (N5318.1 W00856.5)?

(Refer to figure 061-03)Which of the aeronautical chart symbols indicates a VOR/DME?

(Refer to figure 061-03)Which of the aeronautical chart symbols indicates a DME?

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177o - 92 NM

357o - 89 NM

169o - 91 NM

349o - 90 NM

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(Refer to figure 061-03)Which of the aeronautical chart symbols indicates a DME?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the average track (oM) and distance between CRN NDB (N5318.1 00856.5)and WTD NDB (N5211./3 W00705.0)?

(Refer to figure 061-03)Which of the aeronautical chart symbols indicates a VOR?

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135o - 96 NM

322o - 95 NM

142o - 95 NM

315o - 94 NM

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(Refer to figure 061-03)Which of the aeronautical chart symbols indicates an NDB?

(Refer to figure 061-03)Which of the aeronautical chart symbols indicates a basic, non-specified, navigation aid?

(Refer to figure 061-03)Which of the aeronautical chart symbols indicates a VORTAC?

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(Refer to figure 061-03)Which of the aeronautical chart symbols indicates a VORTAC?

(Refer to figure 061-01)Which aeronautical chart symbol indicates a flight Information Region (FIR) boundary?

(Refer to figure 061-01)Which aeronautical chart symbol indicates a Control Zone boundary?

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5

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(Refer to figure 061-03)Which of the aeronautical chart symbols indicates a TACAN?

(Refer to figure 061-01)Which aeronautical chart symbol indicates a group of unlighted obstacles?

(Refer to figure 061-01)Which aeronautical chart symbol indicates a group of lighted obstacles? A - 9B - 10C - 11D - 12Ref: allAns: D21695.(Refer to figure 061-01)Which aeronautical chart symbol indicates an exceptionally high unlighted obstacle?

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6

7

3

1

9

10

11

12

13

6

9

1

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(Refer to figure 061-01)Which aeronautical chart symbol indicates a group of lighted obstacles? A - 9B - 10C - 11D - 12Ref: allAns: D21695.(Refer to figure 061-01)Which aeronautical chart symbol indicates an exceptionally high unlighted obstacle?

(Refer to figure 061-01)Which aeronautical chart symbol indicates an exceptionally high lighted obstacle?

(Refer to figure 061-01)What is the meaning of aeronautical chart symbol number 16?

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12

14

13

12

16

Shipwreck showing above the surface at low tide

Off-shore lighthouse

Lightship

Off-shore helicopter landing platform

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(Refer to figure 061-01)Which aeronautical chart symbol indicates a lightship?

(Refer to figure 061-01)Which aeronautical chart symbol indicates a lighted obstacle?

(Refer to figure 061-01)Which aeronautical chart symbol indicates an unlighted obstacle?

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12

10

15

16

9

10

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16

9

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8

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(Refer to figure 061-01)Which aeronautical chart symbol indicates an unlighted obstacle?

(Refer to figure 061-01)Which aeronautical chart symbol indicates a Waypoint?

(Refer to figure 061-01)Which aeronautical chart symbol indicates a compulsory reporting point?

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kj---900371

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Ref

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15

15

6

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8

8

15

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Sequence N

Sequence N

Answer

Answer

Right/Wrong

Right/Wrong

1234567

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NAVIGATION

NAVIGATION

NAVIGATION




(Refer to figure 061-01)Which aeronautical chart symbol indicates a non-compulsory reporting point?

(Refer to figure 061-01)Which aeronautical chart symbol indicates the boundary of advisory airspace?

(Refer to figure 061-01)Which aeronautical chart symbol indicates an uncontrolled route?

3

3

3

Complexity

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Complexity

0

0

0

kj---900372

kj---900373

kj---900374

Ref

Ref

Ref

15

6

7

8

2

3

4

5

3

4

5

1

1

1

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1

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NAVIGATION

NAVIGATION

NAVIGATION




(Refer to figure 061-01)Which aeronautical chart symbol indicates an uncontrolled route?

(Refer to figure 061-01)What is the meaning of aeronautical chart symbol number 15?

(Refer to figure 061-01)Which aeronautical chart symbol indicates an aeronautical ground light?

3

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3

Complexity

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Complexity

Complexity

0

0

0

kj---900374

kj---900375

kj---900376

Ref

Ref

Ref

Ref

2

Aeronautical ground light

Visual reference point

Hazard to aerial navigation

Lighthouse

14

16

10

15

1

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NAVIGATION

NAVIGATION

NAVIGATION




Determine the distance between points A (N4500.0 E01000.0) and B (N4500.0 W00500.0) is:

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)SHA VOR (5243N 00853W) 205o radialCRK VOR (5150N 00829W) 317o radialWhat isthe position of the aircraft?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)SHA VOR (5243N 00853W) DME 41 nmCRK VOR (5150N 00829W) DME 30 nmWhat is theposition of the aircraft?

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kj---900377

kj---900378

kj---900379

Ref

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Ref

300 nm

636.4 nm

900 nm

212.1 nm

5210N 00910W

5118N 00913W

5205N 00915W

5215N 00917W

5215N 00805W

5205N 00915W

5215N 00915W

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NAVIGATION

NAVIGATION

NAVIGATION




(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)SHA VOR (5243N 00853W) DME 41 nmCRK VOR (5150N 00829W) DME 30 nmWhat is theposition of the aircraft?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the radial and DME distance from CRK VOR (5151N 00830W) to position5220N 00910W?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the radial and DME distance from SHA VOR (5243N 00853W) to BIRRairport (5311N 00754W)?

3

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0

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0

kj---900379

kj---900382

kj---900383

Ref

Ref

Ref

Ref

5225N 00810W

322 M 39 nm

330 M 41 nm

330 M 39 nm

322 M 41 nm

068 M 42 nm

060 M 40 nm

068 M 40 nm

060 M 42 nm

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NAVIGATION

NAVIGATION

NAVIGATION



DEAD RECKONING NAVIGATION (DR) - Basics of dead reckoning

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the lat and long of the SHA VOR (5243N 00853W) 239M/36nmradial/range?

(Refer to figure 061-10)Which of the following beacons is 185 NM from AKRABERG (N6124 W00640)?

Given: A is N55o 000o B is N54o E010o The average true course of the great circle is 100o. The true course of the rhumbline at point A is:

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kj---900384

kj---900385

kj---900386

Ref

Ref

Ref

5215N 00930W

5220N 00937W

5212N 00930W

5212N 00915W

KIRKWALL (N5858 W00254)

STORNOWAY (N5815 W00617)

SUMBURGH (N5955 W00115)

SAXAVORD (N6050 W00050)

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NAVIGATION

NAVIGATION

NAVIGATION




Given: A is N55o 000o B is N54o E010o The average true course of the great circle is 100o. The true course of the rhumbline at point A is:

The rhumb-line distance between points A (60o00N 002o30E) and B (60o00N 007o30W) is:

An aircraft is climbing at a constant CAS in ISA conditions. What will be the effect on TAS and Mach No?

2

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Complexity

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kj---900386

kj---900387

kj---900388

Ref

Ref

Ref

100o

096o

104o

107o

150 NM

450 NM

600 NM

300 NM

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft is climbing at a constant CAS in ISA conditions. What will be the effect on TAS and Mach No?

Heading is 156oT, TAS is 320 knots, W/V is 130o/45. What is your true track?

The ICAO definition of ETA is the:

2

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1

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kj---900388

kj---900389

kj---900390

Ref

Ref

Ref

TAS increases and Mach No decreases

Both increase

Both decrease

TAS decreases and Mach No increases

160

152

104

222

actual time of arrival at a point or fix

estimated time of arrival at destination

estimated time of arrival at an en-route point or fix

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NAVIGATION

NAVIGATION

NAVIGATION




The ICAO definition of ETA is the:

Given: True track: 192o Magnetic variation: 7oE Drift angle: 5o left What is the magnetic heading required to maintain the given track?

Given: True course A to B = 250o Distance A to B = 315 NM TAS = 450 kt W/V = 200o/60 kt ETD A – 0650 UTC What is the ETA at B?

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kj---900390

kj---900391

kj---900392

Ref

Ref

Ref

estimated time en route

190o

194o

204o

180o

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NAVIGATION

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NAVIGATION




Given: True course A to B = 250o Distance A to B = 315 NM TAS = 450 kt W/V = 200o/60 kt ETD A – 0650 UTC What is the ETA at B?

An aircraft passes position A (60o00N 120o00W) on route to position B (60o00N 140o30W). What is the great circle track on departure from A?

Given: Position A N60 W020 Position B N60 W021

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kj---900392

kj---900393

kj---900394

Ref

Ref

Ref

0730 UTC

0736 UTC

0810 UTC

0716 UTC

261o

288o

279o

270o

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Position A N60 W020 Position B N60 W021 Position C N59 W020 What are, respectively, the distances from A to B and from A to C?

What is the longitude of a position 6 NM to the east of 58o42N 094o00W?

Given: Position A is N00o E100o Position B is 240o(T), 200 NM from A What is the position of B?

2

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kj---900394

kj---900395

kj---900396

Ref

Ref

Ref

60 NM and 30 NM

52 NM and 60 NM

30 NM and 60 NM

60 NM and 52 NM

093o53.1W

093o54.0W

093o48.5W

094o12.0W

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Position A is N00o E100o Position B is 240o(T), 200 NM from A What is the position of B?

The rhumb line track between position A (45o00N, 010o00W) and position B (48o30N, 015o00W) is approximately:

What is the time required to travel along the parallel of latitude 60oN between meridians 010oE and 030oW at a ground speed of 480 kt?

2

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1

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kj---900396

kj---900397

kj---900398

Ref

Ref

Ref

S01o40 E101o40

N01o40 E097o07

S01o40 E097o07

N01o40 E101o40

345

300

330

315

1

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NAVIGATION

NAVIGATION

NAVIGATION




What is the time required to travel along the parallel of latitude 60oN between meridians 010oE and 030oW at a ground speed of 480 kt?

Given the following: Magnetic heading: 060o Magnetic variation: 8oW Drift angle: 4o right What is the true track?

An aircraft has a TAS of 300 knots and is over a stretch of water between 2 airfields 500 nm apart. If the wind component is 60 knots head,what is the distance from the first airfield to the critical point?

2

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Complexity

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1

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kj---900398

kj---900399

kj---900400

Ref

Ref

Ref

1 HR 45 MIN

1 HR 15 MIN

2 HR 30 MIN

5 HR 00 MIN

048o

064o

056o

072o

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft has a TAS of 300 knots and is over a stretch of water between 2 airfields 500 nm apart. If the wind component is 60 knots head,what is the distance from the first airfield to the critical point?

An aircraft departs from position A (04o10S 178o22W) and flies northward following the meridian for 2950 NM. It then flies westward along theparallel of latitude for 382 NM to position B. The co-ordinates of position B are?

An aircraft in the northern hemisphere is making an accurate rate one turn to the right. If the initial heading was 135o, after 30 seconds thedirect reading magnetic compass should read:

2

2

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1

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kj---900400

kj---900401

kj---900402

Ref

Ref

Ref

250 nm

200 nm

300 nm

280 nm

53o20 N 172o38 E

45o00 N 172o38 E

53o20 N 169o22 W

45o00 N 169o22 W

225o

less than 225o

more or less than 225o depending on the pendulous suspension used

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft in the northern hemisphere is making an accurate rate one turn to the right. If the initial heading was 135o, after 30 seconds thedirect reading magnetic compass should read:

5 HR 20 MIN 20 SEC corresponds to a longitude difference of:

What is the ration between the litre and the US gallon?

2

2

2

Complexity

Complexity

Complexity

Complexity

1

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kj---900402

kj---900403

kj---900404

Ref

Ref

Ref

Ref

more than 225o

75o00

78o45

80o05

81o10

1 US-GAL equals 4.55 litres

1 litre equals 4.55 US-GAL

1 US-GAL equals 3.78 litres

1 litre equals 3.78 US-GAL

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Answer

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NAVIGATION

NAVIGATION

NAVIGATION




What is the ISA temperature value at FL 330?

An aircraft leaves 0oN/S 45oW and flies due south for 10 hours at a speed of 540 kts. What is its position?

An aircraft leaves 0oN/S 45oW and flies due south for 10 hours at a speed of 540 kts. What is its position as a true bearing from the southpole?

2

2

2

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Complexity

Complexity

1

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kj---900405

kj---900406

kj---900407

Ref

Ref

Ref

-56oC

-66oC

-81oC

-51oC

South pole

North pole

30oS

45oS

30oT

000oT

45oT

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Answer

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft leaves 0oN/S 45oW and flies due south for 10 hours at a speed of 540 kts. What is its position as a true bearing from the southpole?

You are flying from A (50N 10W) to B (58N 02E). If initial Great circle track is 047oT what is Final Great circle track?

You are flying from A (30S 20E) to B (30S 20W). What is the RL track from A to B?

2

2

2

Complexity

Complexity

Complexity

Complexity

1

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1

kj---900407

kj---900408

kj---900409

Ref

Ref

Ref

Ref

60oT

57o

52o

43o

29o

250o (T)

270o (T)

290o (T)

300o (T)

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NAVIGATION

NAVIGATION

NAVIGATION




You are flying from A (30S 20E) to B (30S 20W). What is the initial GC track?

An aircraft is flying at FL 180 and the outside air temperature is -30oC. If the CAS is 150 kt, what is the TAS?

Calibrated Airspeed (CAS) is indicated Airspeed (IAS) corrected for:

2

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1

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kj---900410

kj---900411

kj---900412

Ref

Ref

Ref

260o (T)

270o (T)

290o (T)

300o (T)

115 kt

195 kt

180 kt

145 kt

density

temperature and pressure error

compressibility error

1

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Answer

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1234567

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NAVIGATION

NAVIGATION

NAVIGATION




Calibrated Airspeed (CAS) is indicated Airspeed (IAS) corrected for:

If the Compass Heading is 265o variation is 33oW and deviation is 3oE, what is the True Heading?

If the chart scale is 1 : 500 000, what earth distance would be represented by 7 cm on the chart?

2

2

2

Complexity

Complexity

Complexity

Complexity

1

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kj---900412

kj---900413

kj---900414

Ref

Ref

Ref

Ref

instrument error and position error

229o

235o

301o

295o

35 NM

3.5 km

35 000 m

0.35 km

1

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Sequence N

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Answer

Answer

Right/Wrong

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1234567

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NAVIGATION

NAVIGATION

NAVIGATION




In the Northern Hemisphere the rhumb line track from position A to B is 230o, the convergency is 6o and the difference in longitude is 10o.What is the initial rhumb line track from B to A?

On a Direct Mercator projection a particular chart length is measured at 30oN. What earth distance will the same chart length be if measured at60oN?

The Great Circle bearing from A (70oS 030oW) to B (70oS 060oE) is approximately:

2

2

2

Complexity

Complexity

Complexity

1

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kj---900415

kj---900416

kj---900417

Ref

Ref

Ref

050o

053o

056o

047o

A larger distance

Twice the distance

The same distance

A smaller distance

090o (T)

048o (T)

132o (T)

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Answer

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NAVIGATION

NAVIGATION

NAVIGATION




The Great Circle bearing from A (70oS 030oW) to B (70oS 060oE) is approximately:

The great circle bearing of position B from position A in the Northern Hemisphere is 040o. If the Conversion Angle is 4o, what is the greatcircle bearing of A from B?

The great circle track measured at A (45o00-N 010o00-W) from A to B (45o00-N 019o00-W) is approximately:

2

2

2

Complexity

Complexity

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Complexity

1

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kj---900417

kj---900418

kj---900419

Ref

Ref

Ref

Ref

312o (T)

228o

212o

220o

224o

270o

090o

273o

093o

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1234567

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NAVIGATION

NAVIGATION

NAVIGATION




The initial great circle track from A to B is 080o and the rhumb line track is 083o. What is the initial great circle track from B to A and inwhich Hemisphere are the two positions located?

A flight is planned from A (N37000- E/W000000-) to B (N46000- E/W000000-). The distance in kilometres from A to B is approximately:

Given: Variation 7oW Deviation 4oE If the aircraft is flying a Compass heading of 270, the True and Magnetic Headings are:

2

2

2

Complexity

Complexity

Complexity

1

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kj---900420

kj---900421

kj---900422

Ref

Ref

Ref

266o and in the northern hemisphere

260o and in the southern hemisphere

260o and in the northern hemisphere

266o and in the southern hemisphere

540

794

1000

1771

1

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1234567

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Variation 7oW Deviation 4oE If the aircraft is flying a Compass heading of 270, the True and Magnetic Headings are:

Given: True track 140o Drift 8oS Variation 9oW Deviation 2oE What is the compass heading?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900422

kj---900423

kj---900424

Ref

Ref

Ref

274o (T) 267o (M)

267o (T) 274o (M)

277o (T) 281o (M)

263o (T) 259o (M)

147o (C)

155o (C)

139o (C)

125o (C)

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1234567

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NAVIGATION

NAVIGATION

NAVIGATION



DEAD RECKONING NAVIGATION (DR) - Use of the navigational computer

On a chart, 49 nm is represented by 7.0 cm; the scale of the chart is:

The distance Q to R is 3016 nm; TAS is 480 kts. Flying outbound Q to R the head wind component is calculated as 90 kts and the tail windcomponent R to Q is 75 kts. Leaving Q at 1320 UTC, what is the ETA at the point of Equal Time:

Airfield elevation is 1000 feet. The QNH is 988. Use 27 feet per millibar. What is pressure altitude?

2

2

2

Complexity

Complexity

Complexity

1

1

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kj---900424

kj---900425

kj---900426

Ref

Ref

Ref

1:700 000

1:2 015 396

1:1 296 400

1: 156 600

1631 UTC

1802 UTC

1702 UTC

1752 UTC

675

325

1675

1

1

1

1

2

3

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3

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1234567

1234567

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NAVIGATION

NAVIGATION

NAVIGATION




Airfield elevation is 1000 feet. The QNH is 988. Use 27 feet per millibar. What is pressure altitude?

Given: True course 300o Drift 8oR Variation 10oW Deviation -4o Calculate the compass heading?

265 US-GAL equals? (Specific gravity 0.80)

2

2

2

Complexity

Complexity

Complexity

1

1

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kj---900426

kj---900427

kj---900428

Ref

Ref

Ref

825

306o

322o

294o

278o

862 kg

803 kg

895 kg

1

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NAVIGATION

NAVIGATION

NAVIGATION




265 US-GAL equals? (Specific gravity 0.80)

Ground speed is 540 knots. 72 nm to go. What is time to go?

The relative bearing to a beacon is 270oR. Three minutes later, at a ground speed of 180 knots, it has changed to 225oR. What was the distanceof the closest point of approach of the aircraft to the beacon?

2

2

2

Complexity

Complexity

Complexity

Complexity

1

1

1

kj---900428

kj---900429

kj---900430

Ref

Ref

Ref

Ref

940 kg

8 mins

9 mins

18 mins

12 mins

45 nm

18 nm

9 nm

3 nm

1

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1234567

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NAVIGATION

NAVIGATION

NAVIGATION




An aeroplane flying at 180 kts TAS on a track of 090o. The wind is 045o/50 kts. The distance the aeroplane can fly out and return in one houris:

Given: GS = 122 kt Distance from A to B = 985 NM What is the time from A to B?

Given: IAS 120 kt FL 80 OAT +20oC What is the TAS?

2

2

2

Complexity

Complexity

Complexity

1

1

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kj---900431

kj---900432

kj---900433

Ref

Ref

Ref

88 NM

85 NM

56 NM

176 NM

7 HR 48 MIN

8 HR 04 MIN

7 HR 49 MIN

8 HR 10 MIN

1

1

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1234567

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NAVIGATION

NAVIGATION

NAVIGATION




Given: IAS 120 kt FL 80 OAT +20oC What is the TAS?

The wind velocity is 359/25. An aircraft is heading 180 at a TAS of 198 knots. (All directions are True). What is its track and ground speed?


2

2

3

Complexity

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Complexity

1

1

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kj---900433

kj---900434

kj---900435

Ref

Ref

Ref

132 kt

102 kt

120 kt

141 kt

180.223

179.220

180.220

179.223

1

1

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NAVIGATION

NAVIGATION

NAVIGATION





You are flying 090oC heading. Deviation is 2oW and Variation is 12E. Your TAS is 160 knots. You are flying the 070 radial outbound from a VORand you have gone 14 nm in 6 minutes. What is the W/V?


3

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900435

kj---900436

kj---900437

Ref

Ref

Ref

11 HR 07 MIN

11 HR 06 MIN

11 HR 10 MIN

11 HR 15 MIN

158oT/51

060oT/50

340oT/25

055oT/25

1

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NAVIGATION

NAVIGATION

NAVIGATION





Given: GS = 510 kt Distance A to B = 43 NM What is the time (MIN) from A to B?

Given: GS = 120 kt Distance from A to B = 84 NM

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900437

kj---900438

kj---900439

Ref

Ref

Ref

1 HR 09 MIN

1 HR 30 MIN

1 HR 10 MIN

1 HR 40 MIN

6

4

5

7

1

1

1

2

3

4

1

2

3

4

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1234567

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NAVIGATION

NAVIGATION

NAVIGATION





On a particular take-off, you can accept up to 10 knots tailwind. The runway QDM is 047, the variation is 17E and the ATIS gives the winddirection as 210. What is the maximum wind strength you can accept?

Given: Course 040o(T) TAS is 120 kt Wind speed 30 kt Maximum drift angle will be obtained for a wind direction of:

2

2

2

Complexity

Complexity

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1

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kj---900439

kj---900440

kj---900441

Ref

Ref

Ref

00 HR 42 MIN

00 HR 43 MIN

00 HR 44 MIN

00 HR 45 MIN

18 knots

11 knots

8 knots

4 knots

1

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Course 040o(T) TAS is 120 kt Wind speed 30 kt Maximum drift angle will be obtained for a wind direction of:

G/S = 240 knots, Distance to go = 500 nm. What is time to go?

Pressure Altitude is 27,000 feet, OAT = -35C, Mach No = 0.45 W/V = 270/85, Track = 200T What is drift and ground speed?

2

2

2

Complexity

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Complexity

1

1

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kj---900441

kj---900442

kj---900444

Ref

Ref

Ref

120o

145o

115o

130o

20 minutes

29 minutes

2 h 05 m

2 h 12 m

1

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NAVIGATION

NAVIGATION

NAVIGATION




Pressure Altitude is 27,000 feet, OAT = -35C, Mach No = 0.45 W/V = 270/85, Track = 200T What is drift and ground speed?

If the true track from A to B is 090o, TAS is 460 knots, wind velocity is 360o/100 kts, variation is 10oE and deviation is -20; calculate thecompass heading and ground speed.

. Given: True track 180o Drift 8oR Compass heading 195o Deviation -2o Calculate the variation?

2

2

2

Complexity

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Complexity

1

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kj---900444

kj---900445

kj---900446

Ref

Ref

Ref

18L/252 knots

15R/310 knots

17L/228 knots

17R/287 knots

069o and 448 kts

068o and 460 kts

078o and 450 kts

070o and 453 kts

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NAVIGATION

NAVIGATION

NAVIGATION




. Given: True track 180o Drift 8oR Compass heading 195o Deviation -2o Calculate the variation?


2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900446

kj---900447

kj---900448

Ref

Ref

Ref

25oW

21oW

5oW

9oW

10 HR 19 MIN

10 HR 05 MIN

11 HR 00 MIN

11 HR 02 MIN

1

1

1

2

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft travels 2.4 statute miles in 47 seconds. What is the ground speed?

At 1000 hours an aircraft is on the 310 radial from a VOR/DME, at 10 nautical miles range. At 1010 the radial and range are 040/10 nm. What isthe aircraft¿s track and ground speed?

How long will it take to fly 5 NM at a ground speed of 269 kt?

2

2

2

Complexity

Complexity

Complexity

1

1

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kj---900448

kj---900449

kj---900450

Ref

Ref

Ref

183 kt

160 kt

209 kt

131 kt

080 / 85 knots

085 / 85 knots

080 / 80 knots

085 / 90 knots

1 MIN 07 SEC

1 MIN 55 SEC

2 MIN 30 SEC

1

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1234567

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NAVIGATION

NAVIGATION

NAVIGATION




How long will it take to fly 5 NM at a ground speed of 269 kt?

. Given: GS = 135 kt Distance from A to B = 433 NM What is the time from A to B?

An aircraft is landing on runway 23 (QDM 227o), surface wind 180o/30 kts from ATIS; variation is 13oE. The cross wind component on landingis:

2

2

2

Complexity

Complexity

Complexity

1

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kj---900450

kj---900451

kj---900452

Ref

Ref

Ref

0 MIN 34 SEC

3 HR 20 MIN

3 HR 25 MIN

3 HR 19 MIN

3 HR 12 MIN

26 kts

23 kts

20 kts

15 kts

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NAVIGATION

NAVIGATION

NAVIGATION




Given: GS = 105 kt Distance from A to B = 103 NM Whatis the time from A to B?

You leave A to fly to B, 475 nm away, at 1000 hours. Your ETA at B is 1130. At 1040 you are 190 nm from A. What ground speed is required toarrive on time at B?

An aircraft travels 100 statute miles in 20 MN, how long does it take to travel 215 NM?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900453

kj---900455

kj---900456

Ref

Ref

Ref

01 HR 01 MIN

00 HR 57 MIN

00 HR 58 MIN

00 HR 59 MIN

317 knots

330 knots

342 knots

360 knots

50 MIN

1

1

1

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3

4

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1234567

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft travels 100 statute miles in 20 MN, how long does it take to travel 215 NM?

The equivalent of 70 m/sec is approximately:

Given: Required course 045o(M) Variation is 15oE W/V is 190o(T)/30 kt CAS is 120 kt at FL 55 in standard atmosphere What are the heading (oM) and GS?

2

2

2

Complexity

Complexity

Complexity

1

1

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kj---900456

kj---900457

kj---900458

Ref

Ref

Ref

100 MIN

90 MIN

80 MIN

145 kt

136 kt

210 kt

35 kt

036o and 151 kt

1

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Required course 045o(M) Variation is 15oE W/V is 190o(T)/30 kt CAS is 120 kt at FL 55 in standard atmosphere What are the heading (oM) and GS?


Given: True track: 352o

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900458

2kj---900459

kj---900460

Ref

Ref

Ref

055o and 147 kt

052o and 154 kt

056o and 137 kt

4 HR 10 MIN

3 HR 25 MIN

3 HR 26 MIN

4 HR 25 MIN

1

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1234567

1234567

NAVIGATION

NAVIGATION



Given: True track: 352o Variation 11oW Deviation is -5o Drift 10oR Calculate the compass heading?

Given: True course from A to B = 090o TAS = 460 kt W/V = 360/100 kt Average variation = 10oE Deviation = -2o Calculate the compass heading and GS?

2

2

Complexity

Complexity

Complexity

1

1

kj---900460

kj---900461

Ref

Ref

Ref

358o

346o

018o

025o

078o - 450 kt

068o - 460 kt

069o - 448 kt

070o - 453 kt

1

1

1

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1234567

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1234567

NAVIGATION

NAVIGATION

NAVIGATION




730 FT/MIN equals:

Fuel flow per HR is 22 US-GAL, total fuel on board is 83 IMP GAL. What is the endurance?

How many NM would an aircraft travel in 1 MIN 45 SEC if GS is 135 kt?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900462

kj---900463

kj---900464

Ref

Ref

Ref

3.7 m/sec

5.2 m/sec

1.6 m/sec

2.2 m/sec

4 HR 32 MIN

3 HR 12 MIN

3 HR 53 MIN

2 HR 15 MIN

39.0

2.36

3.25

1

1

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3

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Sequence N

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Answer

Answer

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1234567

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NAVIGATION

NAVIGATION

NAVIGATION




How many NM would an aircraft travel in 1 MIN 45 SEC if GS is 135 kt?

Given: FL 250 OAT -15oC TAS 250 kt Calculate the Mach No?

Given: TAS = 225 kt HDG (oT) – 123o W/V – 090/60 kt Calculate the Track (oT) and GS?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900464

kj---900465

kj---900466

Ref

Ref

Ref

3.94

0.44

0.40

0.39

0.42

134 - 178 kt

1

1

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4

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1234567

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NAVIGATION

NAVIGATION

NAVIGATION




Given: TAS = 225 kt HDG (oT) – 123o W/V – 090/60 kt Calculate the Track (oT) and GS?

Given: TAS = 170 kt HDG (T) = 100o W/V – 350/30 kt Calculate the Track (oT) and GS?

Given: TAS – 230 kt

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900466

kj---900467

kj---900468

Ref

Ref

Ref

134 - 188 kt

120 - 190 kt

123 - 180 kt

098 - 178 kt

109 - 182 kt

091 - 183 kt

103 - 178 kt

1

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1234567

1234567

1234567

NAVIGATION

NAVIGATION

NAVIGATION




Given: TAS – 230 kt HDG (T) – 250o W/V m 205/10 kt Calculate the drift and GS?

Given: True HDG = 145o TAS – 240 kt Track (T) – 150o GS – 210 kt Calculate the W/V?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900468

kj---900469

kj---900470

Ref

Ref

Ref

1L - 225 kt

1R - 221 kt

2R - 223 kt

2L - 224 kt

360/35 kt

180/35 kt

295/35 kt

115/35 kt

1

1

1

2

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4

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1234567

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NAVIGATION

NAVIGATION

NAVIGATION




Given: True altitude 9000 FT OAT -32oC CAS 200 kt What is the TAS?

Given: True Track = 095o TAS = 160 kt True Heading = 087o GS = 130 kts Calculate W/V

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900470

kj---900471

kj---900472

Ref

Ref

Ref

215 kt

200 kt

210 kt

220 kt

124o/36 kt

237o/36 kt

307o/36 kt

057o/36 kt

1

1

1

2

3

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NAVIGATION

NAVIGATION



Given: True Track 239o True Heading 229o TAS 555 kt G/S 577 kt Calculate the wind velocity.

Given: True Track 245o Drift 5o right Variation 3o E Compass Hdg 242o Calculate the deviation.

2

2

Complexity

Complexity

Complexity

1

1

kj---900472

kj---900473

Ref

Ref

Ref

300o/100 kt

310o/100 kt

130o/100 kt

165o/100 kt

11o E

1o E

5o E

5o W

1

1

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1234567

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1234567

NAVIGATION

NAVIGATION

NAVIGATION




True Heading of an aircraft is 265o and TAS is 290 kt. If W/V is 210o/35kt, what is True Track and GS?

Course 040oT, TAS 120 kt, Wind speed 30 knots. From which direction will the wind give the greatest drift:

Required course 045oT, W/V = 190/30, FL 55, ISA, Variation 15oE, CAS 120 knots. What is the magnetic heading and G/S?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900474

kj---900475

kj---900476

Ref

Ref

Ref

259o and 305 kt

259o and 272 kt

260o and 315 kt

271o and 272 kt

215o

230oT

235oT

240oT

052oM 154

067oM 154

037oM 154

1

1

1

1

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4

1

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Sequence N

Sequence N

Answer

Answer

Answer

Right/Wrong

Right/Wrong

Right/Wrong

1234567

1234567

1234567

NAVIGATION

NAVIGATION

NAVIGATION




Required course 045oT, W/V = 190/30, FL 55, ISA, Variation 15oE, CAS 120 knots. What is the magnetic heading and G/S?

Given: Pressure Altitude = 5000 ft OAT = +35C What is true altitude:

Given: Pressure Altitude 29 000 ft OAT -55oC What is the density altitude:

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kj---900476

kj---900477

kj---900478

Ref

Ref

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037oM 113

4550 ft

5550 ft

4290 ft

5320 ft

27 500 ft

31 000 ft

33 500 ft

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NAVIGATION

NAVIGATION

NAVIGATION



DEAD RECKONING NAVIGATION (DR) - The Triangle of velocities; methods of solution

Given: Pressure Altitude 29 000 ft OAT -55oC What is the density altitude:

If the headwid component is 50 kt, the FL is 330, temperature JSA -7oC and the ground speed is 496 kt, the Mach No. is:

Given: True Heading = 090o TAS = 180 kt GS = 180 kt Drift 5o right Calculate the W/V?

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kj---900478

kj---900479

kj---900480

Ref

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36 000 ft

0.98

0.78

0.95

0.75

360o / 15 kt

190o / 15 kt

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NAVIGATION

NAVIGATION

NAVIGATION




Given: True Heading = 090o TAS = 180 kt GS = 180 kt Drift 5o right Calculate the W/V?

Given: FL 120 OAT is ISA standard CAS is 200 kt Track is 222o (M) Heading is 215o(M) Variation is 15oW Time to fly 105 NM is 21 MIN. What is the W/V?

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kj---900480

kj---900481

kj---900482

Ref

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010o / 15 kt

180o / 15 kt

050o(T) / 70 kt

040o(T) / 105 kt

055o(T) / 105 kt

065o(T) / 70 kt

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NAVIGATION

NAVIGATION

NAVIGATION




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 winddirection and the runway is 60o. Calculate the minimum and maximum allowable wind speeds?

Given: Maximum allowable tailwind component for landing 10 kt Planned runway 05 (047o magnetic) The direction of the surface wind reported by ATIS 210o Variation is 17oE Calculate the maximum allowable windspeed that can be accepted without exceeding the tailwind limit?

Given: Runway direction 083o(M)

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kj---900482

kj---900483

kj---900484

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12 kt and 38 kt

20 kt and 40 kt

15 kt and 43 kt

18 kt and 50 kt

15 kt

18 kt

8 kt

11 kt

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Runway direction 083o(M) Surface W/V 035/35 kt Calculate the effective headwind component?

An aircraft is following a true track of 048o at a constant TAS of 210 kt. The wind velocity is 350o/30 kt. The GS and drift angle are:

Given: Runway direction 230o(T) Surface W/V 280o(T)/40 kt Calculate the effective cross-wind component?

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kj---900484

kj---900485

kj---900486

Ref

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24 kt

27 kt

31 kt

34 kt

192 kt, 7o left

200 kt - 3.5o right

195 kt, 7o right

225 kt, 7o left

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Runway direction 230o(T) Surface W/V 280o(T)/40 kt Calculate the effective cross-wind component?

Given: TAS = 485 kt True HDG = 226o W/V = 110o(T)/95 kt Calculate the drift angle and GS?

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kj---900486

kj---900487

kj---900488

Ref

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21 kt

36 kt

31 kt

26 kt

7oR - 531 ktg

9oR - 533 kt

9oR - 433 kt

8oL - 435 kt

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NAVIGATION

NAVIGATION

NAVIGATION




Given: TAS = 198 kt HDG (oT) = 180 W/V = 359/25 Calculate the Track (oT) and GS?

Given: True HDG = 307o TAS = 230 kt Track (T) = 313o GS = 210 kt Calculate the W/V?

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kj---900488

kj---900489

kj---900490

Ref

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180 - 223 kt

179 - 220 kt

181 - 180 kt

180 - 183 kt

255/25 kt

257/35 kt

260/30 kt

265/30 kt

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NAVIGATION

NAVIGATION

NAVIGATION



Given: Magnetic track = 210o Magnetic HDG = 215o VAR = 15oE TAS = 360 kt Aircraft flies 64 NM in 12 MIN Calculate the true W/V?


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kj---900490

kj---900491

kj---900492

Ref

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265o/50 kt

195o/50 kt

235o/50 kt

300o/30 kt

8oL- 146 kt

7oL - 156 kt

4oL - 168 kt

4oL - 145 kt

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NAVIGATION

NAVIGATION

NAVIGATION




Given: TAS = 95 kt HDG (T) = 075o W/V = 310/20 kt Calculate the drift and GS?


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kj---900492

kj---900493

kj---900494

Ref

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9R - 108 kt

10L - 104 kt

9L - 105 kt

8R - 104 kt

2oR - 166 kt

4oR - 165 kt

4oL - 167 kt

3oL - 166 kt

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NAVIGATION

NAVIGATION

NAVIGATION




Given: TAS = 250 kt HDG (T) = 029o W/V = 035/45kt Calculate the drift and GS?


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kj---900494

kj---900495

kj---900496

Ref

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1L - 205 kt

1R - 205 kt

1L - 265 kt

1R - 295 kt

088/15 kt

095/20 kt

088/20 kt

093/25 kt

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NAVIGATION

NAVIGATION

NAVIGATION




Given: TAS = 235 kt HDG (T) = 076o W/V = 040/40kt Calculate the drift angle and GS?


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kj---900496

kj---900497

kj---900498

Ref

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Ref

5R- 207 kt

7L - 269 kt

5L - 255 kt

7R - 204 kt

7L - 192 kt

6L - 194 kt

3L - 190 kt

4L - 195 kt

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NAVIGATION

NAVIGATION

NAVIGATION




Given: TAS = 90 kt HDG (T) = 355o W/V = 120/20 kt Calculate the Track (oT) and GS?

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kj---900498

kj---900499

kj---900500

Ref

Ref

Ref

070/40 kt

075/45 kt

070/45 kt

075/50 kt

006 - 95 kt

346 - 102 kt

358 - 101 kt

359 - 102 kt

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NAVIGATION

NAVIGATION



Given: True Heading = 090o TAS = 200 kt W/V = 220o/30 kt Calculate the GS?

Given: Compass Heading 090o Deviation 2oW Variation 12oE TAS 160 kt Whilst maintaining a radial 070o from a VOR station, the aircraft flies a ground distance of 14 NM in 6 MIN. What is the W/V (oT)?

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kj---900500

kj---900501

Ref

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Ref

180 kt

230 kt

220 kt

200 kt

165o/25 kt

340o/25 kt

340o/98 kt

160o/50 kt

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NAVIGATION

NAVIGATION

NAVIGATION






2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900502

kj---900503

kj---900504

Ref

Ref

Ref

11R - 140 kt

9R - 140 kt

11R - 142 kt

10R - 146 kt

9oR - 143 kt

16oL - 156 kt

9oL - 146 kt

18oR - 146 kt

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NAVIGATION

NAVIGATION

NAVIGATION



Given: TAS = 465 kt Track (T) = 007o W/V = 300/80 kt Calculate the HDG (oT) and GS?

Given: Maximum allowabl crosswind component is 20 kt Runway 06 RWY QDM 063o(M) Wind direction 100o(M) Calculate the maximum allowable windspeed?

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kj---900504

kj---900505

kj---900506

Ref

Ref

Ref

000 - 430 kt

001 - 432 kt

358 - 428 kt

357 - 430 kt

26 kt

31 kt

33 kt

25 kt

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Course required = 085o (T) Forecast W/V 030/100 kt TAS = 470 kt Distance = 265 NM Calculate the true HDG and flight time?

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Complexity

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kj---900506

kj---900507

kj---900508

Ref

Ref

Ref

6oL - 487 kt

7oR - 491 kt

7oL - 491 kt

7oR - 487 kt

096o, 29 MIN

076o, 34 MIN

075o, 39 MIN

095o, 31 MIN

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NAVIGATION

NAVIGATION

NAVIGATION





Given: TAS = 132 kt HDG (T) = 053o W/V = 205/15 kt Calculate the track (oT) and GS?

2

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Complexity

Complexity

Complexity

1

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kj---900508

kj---900509

kj---900510

Ref

Ref

Ref

077 - 214 kt

079 - 211 kt

075 - 213 kt

077 - 210 kt

057 - 144 kt

050 - 145 kt

052 - 143 kt

051 - 144 kt

1

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NAVIGATION

NAVIGATION

NAVIGATION




Given: TAS = 270 kt True HDG = 145o Actual wind = 205o(T)/30 kt Calculate the drift angle and GS?


2

2

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Complexity

Complexity

Complexity

1

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1

kj---900510

kj---900511

kj---900512

Ref

Ref

Ref

8oR - 261 kt

6oR - 251 kt

6oL - 256 kt

6oR - 259 kt

313 - 235 kt

311 - 230 kt

312 - 232 kt

310 - 233 kt

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NAVIGATION

NAVIGATION

NAVIGATION




Given: TAS = 220 kt Magnetic course = 212o W/V 160o (M)/50 kt Calculate the GS?

Given: Runway direction 305o(M) Surface W/V 260o(M)/30 kt Calculate the cross-wind component?

Given: TAS = 470 kt

2

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Complexity

Complexity

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kj---900512

kj---900513

kj---900514

Ref

Ref

Ref

186 kt

290 kt

246 kt

250 kt

18 kt

24 kt

27 kt

21 kt

1

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NAVIGATION

NAVIGATION

NAVIGATION




Given: TAS = 470 kt True HDG = 317o W/V = 045o(T)/45 kt Calculate the drift angle and GS

Given: True HDG = 074o TAS = 230 kt Track (T) = 066o GS = 242 kt Calculate the W/V

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900514

kj---900515

kj---900516

Ref

Ref

Ref

3oR - 470 kt

5oL - 270 kt

5oL - 475 kt

5oR - 475 kt

180/30 kt

180/35 kt

185/35 kt

180/40 kt

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1234567

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NAVIGATION

NAVIGATION

NAVIGATION




Given: TAS = 155 kt HDG (T) = 216o W/V = 090/60 kt Calculate the track (oT) and GS?

An aeroplane is flying at TAS 180 kt on a track of 090o. The W/V is 045o/50 kt. How far can the aeroplane fly out from its base and return inone hour?

Given: TAS = 370 kt True HDG = 181o W/V = 095o(T)/35 kt Calculate the true track and GS?

2

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Complexity

Complexity

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kj---900516

kj---900517

kj---900518

Ref

Ref

Ref

224 - 175 kt

231 - 196 kt

222 - 181 kt

226 - 186 kt

56 NM

88 NM

85 NM

176 NM

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NAVIGATION

NAVIGATION

NAVIGATION





Given: Magnetic heading = 255o VAR = 40oW GS = 375 kt W/V = 235o(T)/120 kt Calculate the drift angle?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900518

kj---900519

kj---900520

Ref

Ref

Ref

186 - 370 kt

176 - 370 kt

192 - 370 kt

189 - 370 kt

7o left

7o right

9o left

16o right

1

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1234567

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NAVIGATION

NAVIGATION

NAVIGATION





Given: TAS = 485 kt HDG (T) = 168o W/V = 130/75 kt Calculate the Track (oT) and GS?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900520

kj---900521

kj---900522

Ref

Ref

Ref

010/55 kt

005/50 kt

010/50 kt

010/45 kt

175 - 432 kt

173 - 424 kt

175 - 420 kt

174 - 428 kt

1

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Magnetic track = 315o HDG = 301o (M) VAR = 5oW TAS = 225 kt The aircraft flies 50 NM in 12 MIN. Calculate the W/V (oT)?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900522

kj---900523

kj---900524

Ref

Ref

Ref

1R - 165 kt

1L - 225 kt

1R - 175 kt

1L - 215 kt

195o/63 kt

355o/15 kt

195o/61 kt

190o/63 kt

1

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NAVIGATION

NAVIGATION

NAVIGATION





Given: True Heading = 180o TAS = 500 kt W/V 225o/100 kt Calculate the GS?

2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900524

kj---900525

kj---900526

Ref

Ref

Ref

301 - 169 kt

305 - 169 kt

309 - 170 kt

309 - 141 kt

450 kt

600 kt

535 kt

435 kt

1

1

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1234567

1234567

1234567

NAVIGATION

NAVIGATION

NAVIGATION




For a given track the: Wind component = 45 kt Drift angle = 15o left TAS = 240 kt What is the wind component on the reverse track?

An aircraft is on final approach to runway 32R (322o). The wind velocity reported by the tower is 350o/20 kt. TAS on approach is 95 kt. Inorder to maintain the centre line, the aircrafts heading (oM) should be:

Given: TAS = 270 kt Track (T) = 260o W/V = 275/30 kt

2

2

2

Complexity

Complexity

Complexity

1

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1

kj---900526

kj---900527

kj---900528

Ref

Ref

Ref

-55 kt

-65 kt

-45 kt

-35 kt

322o

328o

316o

326o

1

1

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1234567

1234567

1234567

NAVIGATION

NAVIGATION

NAVIGATION






2

2

2

Complexity

Complexity

Complexity

1

1

1

kj---900528

kj---900529

kj---900530

Ref

Ref

Ref

264 - 237 kt

262 - 237 kt

264 - 241 kt

262 - 241 kt

115/70 kt

110/75 kt

110/80 kt

105/75 kt

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Runway direction 210o(M) Surface W/V 230o (M)/30 kt Calculate the crosswind component?


2

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kj---900530

kj---900531

kj---900532

Ref

Ref

Ref

19 kt

10 kt

16 kt

13 kt

335/55 kt

335/45 kt

340/50 kt

340/45 kt

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Magnetic track = 075o HDG = 066o(M) VAR = 11oE TAS = 275 kt Aircraft flies 48 NM in 10 MIN. Calculate the true W/V?

Given: TAS = 480 kt HDG (oT) = 040o W/V = 090/60 kt Calculate the Track (oT) and GS?

2

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kj---900532

kj---900533

kj---900534

Ref

Ref

Ref

340o/45 kt

320o/50 kt

210o/15 kt

180o/45 kt

032 - 425 kt

028 - 415 kt

034 - 445 kt

036 - 435 kt

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NAVIGATION

NAVIGATION

NAVIGATION






2

2

2

Complexity

Complexity

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1

1

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kj---900534

kj---900535

kj---900536

Ref

Ref

Ref

002 - 98 kt

345 - 100 kt

348 - 102 kt

005 - 102 kt

180/10 kt

000/05 kt

000/10 kt

180/05 kt

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NAVIGATION

NAVIGATION

NAVIGATION



. Given: True heading = 310o TAS = 200 kt GS = 176 kt Drift angle 7o right Calculate the W/V?

Given: TAS = 135 kt HDG = (oT) = 278 W/V = 140/20 kt Calculate the Track (oT) and GS?

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kj---900536

kj---900537

kj---900538

Ref

Ref

Ref

090o/33 kt

360o/33 kt

270o/33 kt

180o/33 kt

279 - 152 kt

283 - 150 kt

282 - 148 kt

275 - 150 kt

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NAVIGATION

NAVIGATION

NAVIGATION






2

2

2

Complexity

Complexity

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kj---900538

kj---900539

kj---900540

Ref

Ref

Ref

8L - 415 kt

3L - 415 kt

4L - 400 kt

6L - 400 kt

002 - 173 kt

001 - 170 kt

359 - 166 kt

357 - 168 kt

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NAVIGATION

NAVIGATION

NAVIGATION






2

2

2

Complexity

Complexity

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1

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kj---900540

kj---900541

kj---900542

Ref

Ref

Ref

125 - 322 kt

123 - 320 kt

126 - 320 kt

125 - 318 kt

097 - 201 kt

121 - 207 kt

121 - 199 kt

099 - 199 kt

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NAVIGATION

NAVIGATION

NAVIGATION




. Given: TAS = 270 kt True HDG = 270o Actual wind 205o(T)/30 kt Calculate the drift angle and GS?

The following information is displayed on an Inertial Navigation System: GS 520 kt. True HDG 090o, Drift angle 5o right, TAS 480 kt SAT (staticair temperature) -51oC. The W/V being experienced is:

Given: TAS = 440 kt HDG (T) = 349o W/V = 040/40 kt

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kj---900542

kj---900544

kj---900545

Ref

Ref

Ref

6R - 259 kt

6L - 256 kt

6R - 251 kt

8R - 259 kt

225o/60 kt

320o/60 kt

220o/60 kt

325o/60 kt

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NAVIGATION

NAVIGATION




Given: M 0.80 OAT -50oC FL 330 GS 490 kt VAR 20oW Magnetic heading 140o Drift is 11o Right Calculate the true W/V?

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kj---900545

kj---900546

Ref

Ref

Ref

4L - 415 kt

2L - 420 kt

6L - 395 kt

5L - 385 kt

200o/95 kt

025o/47 kt

020o/95 kt

025o/45 kt

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NAVIGATION

NAVIGATION

NAVIGATION




The reported surface wind from the control tower is 240o/35 kt.Runway 30 (300o). What is cross-wind component?

For a landing on runway 23 (227o magnetic) surface W/V reported by the ATIS is 180/30 kt. VAR is 13oE. Calculate the cross wind component?

How long will it take to travel 284 nm at a speed of 526 KPH?

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kj---900547

kj---900548

kj---900549

Ref

Ref

Ref

30 kt

24 kt

27 kt

21 kt

20 kt

22 kt

26 kt

15 kt

1.6 h

1.9 h

45 min

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NAVIGATION

NAVIGATION

NAVIGATION



DEAD RECKONING NAVIGATION (DR) - List elements required for establishing DR position

How long will it take to travel 284 nm at a speed of 526 KPH?

If it takes 132.4 mins to travel 840 nm, what is your speed in kmh?

A useful method of a pilot resolving, during a visual flight, any uncertainty in the aircraft's position is to maintain visual contact with theground and:

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kj---900549

kj---900550

kj---900551

Ref

Ref

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Ref

1 h

705 kmh

290 kmh

120 kmh

966 kmh

set heading towards a line feature such as a coastline, motorway, river or railway

fly the reverse of the heading being flown prior to becoming undertain until a pinpoint is obtained

fly expanding circles until a pinpoint is obtained

fly reverse headings and associated timings until the point of departure is regained.

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NAVIGATION

NAVIGATION

NAVIGATION




Position A is located on the equator at longitude 130o00E. Position B is located 100 NM from A on a bearing of 225o(T). The co-ordinates ofposition B are:

Given: Position A 45oN, ?oE Position B 45oN, 45o15E Distance A-B = 280 NM B is to the East of A Required: longitude of position A?

The Great Circle bearing of B (70oS 060oE), from A (70oS 030oW), is approximately?

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kj---900552

kj---900553

kj---900554

Ref

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Ref

01o11N 128o 49E

01o11S 128o 49E

01o11N 131o 11E

01o11S 131o 11E

38o39E

49o57E

51o51E

40o33E

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NAVIGATION

NAVIGATION

NAVIGATION




The Great Circle bearing of B (70oS 060oE), from A (70oS 030oW), is approximately?

What is the final position after the following rhumb line tracks and distances have been followed from position 60o00N 030o00W? South for 3600 NM East for 3600 NM North for 3600 NM West for 3600 NM The final position of the aircraft is:

An aircraft at positon 60oN 005oW tracks 090o(T) for 315km. On completion of the flight the longitude will be:

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kj---900554

kj---900555

kj---900556

Ref

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Ref

150o (T)

090o (T)

318o (T)

135o (T)

59o00N 090o00W

60o00N 090o00W

60o00N 030o00E

59o00N 060o00W

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft at positon 60oN 005oW tracks 090o(T) for 315km. On completion of the flight the longitude will be:

The departure between positions 60oN 160oE and 60sN x is 900 NM. What is the longitude of x?

An aircraft at latitude 10o South flies north at a GS of 890 km/HR. What will its latitude be after 1.5 HR?

2

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kj---900556

kj---900557

kj---900558

Ref

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Ref

002o 10W

000o 15E

000o 40E

005o 15E

170oW

140oW

145oE

175oE

22o00N

03o50N

02o00N

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft at latitude 10o South flies north at a GS of 890 km/HR. What will its latitude be after 1.5 HR?

You are flying from A (30S 20E) to B (30S 20W). At what longitude will the GC track equal the RL track?

What is the Chlong (in degrees and minutes) from A (45N 1630E) to B (45N 15540W)?

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kj---900558

kj---900559

kj---900561

Ref

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Ref

12o15N

10oE

10oW

0oE/W

20oW

38o05E

38o50W

38o05W

38o50E

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NAVIGATION

NAVIGATION

NAVIGATION



DEAD RECKONING NAVIGATION (DR) - Calculate DR elements

Given: True Track 245o Drift 5o right Variation 3oE Compass Hdg 242o Calculate the Magnetic Heading:

Grid heading is 299o, grid convergency is 55o West and magnetic variation is 90o West. What is the corresponding magnetic heading?

OAT = +35oC Pressure alt = 5000 feet What is true alt?

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1

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kj---900562

kj---900563

kj---900564

Ref

Ref

Ref

247o

243o

237o

253o

084o

334o

154o

264o

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NAVIGATION

NAVIGATION

NAVIGATION




OAT = +35oC Pressure alt = 5000 feet What is true alt?

Given: Airport elevation is 1000 ft QNH is 988 hPa What is the approximate airport pressure altitude? (Assume 1 hPa = 27 FT

Your pressure altitude is FL 55, the QNH is 998, and the SAT is +30C. What is Density Altitude?

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1

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kj---900564

kj---900565

kj---900566

Ref

Ref

Ref

4550 feet

5550 feet

4290 feet

5320 feet

680 FT

320 FT

1680 FT

-320 FT

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NAVIGATION

NAVIGATION

NAVIGATION




Your pressure altitude is FL 55, the QNH is 998, and the SAT is +30C. What is Density Altitude?

You are on ILS 3o glideslope which passes over the runway threshold at 50 feet. Your DME range is 25 nm from the threshold. What is yourheight above the runway threshold elevation? (Use the 1 in 60 rule and 6000 feet = 1 nautical mile)

Given: FL 350 Mach 0.80 OAT -55oC Calculate the values for TAS and local speed of sound (LSS)?

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kj---900566

kj---900567

kj---900568

Ref

Ref

Ref

6980 feet

7750 feet

8620 feet

10020 feet

8010 feet

7450 feet

6450 feet

7550 feet

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NAVIGATION

NAVIGATION

NAVIGATION




Given: FL 350 Mach 0.80 OAT -55oC Calculate the values for TAS and local speed of sound (LSS)?

The pressure alt is 29000 feet and the SAT is -55C. What is density altitude?

You are flying at a True Mach No of 0.82 in a SAT of -45oC. At 1000 hours you are 100 nm from the POL DME and your ETA at POL is 1012. ATC askyou to slow down to be at POL at 1016. What should your new TMN be if you reduce speed at 100 nm distance to:

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kj---900568

kj---900569

kj---900570

Ref

Ref

Ref

461 kt, LSS 296 kt

237 kt, LSS 296 kt

490 kt, LSS 461 kt

461 kt, LSS 576 kt

27500 feet

26000 feet

30000 feet

31000 feet

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NAVIGATION

NAVIGATION

NAVIGATION




You are flying at a True Mach No of 0.82 in a SAT of -45oC. At 1000 hours you are 100 nm from the POL DME and your ETA at POL is 1012. ATC askyou to slow down to be at POL at 1016. What should your new TMN be if you reduce speed at 100 nm distance to:

Given: TAS = 485 kt OAT = ISA +10oC FL 410 Calculate the Mach Number?

Given: TAS 487 kt FL 330 Temperature ISA + 15 Calculate the Mach Number?

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1

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kj---900570

kj---900571

kj---900572

Ref

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Ref

M .76

M .72

M .68

M .61

0.85

0.90

0.825

0.87

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NAVIGATION

NAVIGATION

NAVIGATION




Given: TAS 487 kt FL 330 Temperature ISA + 15 Calculate the Mach Number?

A flight is to be made from A 49oS 180oE/W to B 58oS, 180oE/W. The distance is kilometres from A to B is approximately:

An aircraft is at 10N and is flying South at 444 km/hour. After 3 hours the latitude is:

2

2

2

Complexity

Complexity

Complexity

1

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kj---900572

kj---900573

kj---900574

Ref

Ref

Ref

0.81

0.84

0.76

0.78

1222

1000

540

804

1

1

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft is at 10N and is flying South at 444 km/hour. After 3 hours the latitude is:

Given: Aircraft at FL 150 overhead an airport elevation of airport 720 ft QNH is 1003 hPa OAT at FL 150 -5oC What is the true altitude of the aircraft? (Assume 1 hPa = 27 ft)

An aircraft takes off from the aerodrome of BRIOUDE (altitude 1 483 ft, QFE = 963 hPa, temperature = 32oC). Five minutes later, passing 5,000ft on QFE, the second altimeter set on 1,013 hPa will indicate approximately:

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kj---900574

kj---900575

kj---900576

Ref

Ref

Ref

10S

02N

02S

0N/S

15,840 ft

15,280 ft

14,160 ft

14,720 ft

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft takes off from the aerodrome of BRIOUDE (altitude 1 483 ft, QFE = 963 hPa, temperature = 32oC). Five minutes later, passing 5,000ft on QFE, the second altimeter set on 1,013 hPa will indicate approximately:

An aircraft maintaining a 5.2% gradient is at 7 NM from the runway, on a flat terrain; its height is approximately:

Given: Pressure Altitude 29,000 ft, OAT -55C. Calculate the Density Altitude?

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kj---900576

kj---900577

kj---900578

Ref

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Ref

6,900 ft

6,400 ft

6,000 ft

4,000 ft

680 ft

2210 ft

1890 ft

3640 ft

27,500 ft

31,500 ft

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Pressure Altitude 29,000 ft, OAT -55C. Calculate the Density Altitude?

An aircraft leaves point A (75N 50W) and flies due North. At the North Pole it flies due south along the meridian of 65o50E unit reaches 75N(point B). What is the total distance covered?

Your true altitude is 5500 feet, the QNH is 995, and the SAT is +30oC. What is Density Altitude:

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kj---900578

kj---900579

kj---900580

Ref

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Ref

33,500 ft

26,000 ft

1,650 nm

2,000 nm

2,175 nm

1,800 nm

7080 feet

8120 feet

9280 feet

9930 feet

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1234567

1234567

1234567

NAVIGATION

NAVIGATION

NAVIGATION



DEAD RECKONING NAVIGATION (DR) - Construct DR position

Given: Pressure Altitude = 29,000 ft OAT = -50o Calculate the Density Altitude

Given: M0.9 FL370 OAT -70C Determine CAS:

(Refer to figure 061-09)1300 UTC DR position 37o30N 021o30W alter heading PORTO SANTO NDB (33o03N 016o23W) TAS 450 kt, Forecast W/V 360o/30 kt.Calculate the ETA at PORTO SANTO NDB:

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26,000 ft

27,000 ft

31,000 ft

33,500 ft

500 kts

281 kts

293 kts

268 kts

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NAVIGATION

NAVIGATION

NAVIGATION




(Refer to figure 061-09)1300 UTC DR position 37o30N 021o30W alter heading PORTO SANTO NDB (33o03N 016o23W) TAS 450 kt, Forecast W/V 360o/30 kt.Calculate the ETA at PORTO SANTO NDB:

You are flying from A (50N 10W) to B (58N 02E). At what longitude will the Great Circle track equal the Rhumb Line (RL) track between A and B:

At N6010.0 on a Mercator chart the scale is 1:5 000 000; the length of a line on the chart between C N6010.0 E00810.0 and D N6010.0 W00810.0 is:

3

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0

1

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kj---900586

kj---900587

kj---900588

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1341

1344

1348

1354

06oW

0oW

04oW

04oE

19.2 cm

16.2 cm

17.8 cm

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NAVIGATION

NAVIGATION

NAVIGATION



DEAD RECKONING NAVIGATION (DR) - Name range specifics of maximum range and radius of action

At N6010.0 on a Mercator chart the scale is 1:5 000 000; the length of a line on the chart between C N6010.0 E00810.0 and D N6010.0 W00810.0 is:

Given: Aircraft position S8000.0 E14000.0 Aircraft tracking 025o(G) If the grid is aligned with the Greenwich Anti-Meridian, the True track is:

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:

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kj---900588

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kj---900590

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35.6 cm

245o

205o

165o

065o 1

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NAVIGATION

NAVIGATION

NAVIGATION




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:

An aircraft was over A at 1435 hours flying direct to B. Given: Distance A to B 2,900 NM True airspeed 470 kt Mean wind component OUT +55 ktMean wind component BACK -75 kt. The ETA for reaching the Point of Equal Time (PET) between A and B is:

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2

2

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kj---900590

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1820

1756

1752

1742

1721

1744

1846

1657

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Distance A to B 2346 NM Groundspeed OUT 365 kt Groundspeed BACK 480 kt Safe endurance 8 HR 30 MIN The time from A to the Point of Safe Return (PSR) A is:

Two points A and B are 1000 NM apart. TAS = 490 kt. On the flight between A and B the equivalent headwind is -20 kt. On the return legbetween B and A, the equivalent headwind is +40 kt. What distance from A, along the route A to B, is the Point of Equal Time (PET)?

An aircraft was over A at 1435 hours flying direct to B Given: Distance A to B 2900 NM

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197 min

219 min

290 min

209 min

470 NM

530 NM

455 NM

500 NM

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NAVIGATION

NAVIGATION



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:

Given: Distance A to B 1973 NM Groundspeed OUT 430 kt Groundspeed BACK 385 kt The time from A to the Point of Equal Time (PET) between A and B is:

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2

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kj---900594

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2844 NM

1611 NM

1759 NM

2141 NM

145 min

130 min

162 min

181 min

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NAVIGATION

NAVIGATION



Given: Distance A to B 2484 NM Mean groundspeed out 420 kt Mean groundspeed back 500 kt Safe endurance 08 Hr 30 min The distance from A to the Point of Safe Return (PSR) A is:

Given: Distance Q to R 1760 NM Groundspeed out 435 kt Groundspeed back 385 kt The time from Q to the Point of Equal Time (PET) between Q and R is:

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2

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kj---900596

kj---900597

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1908 NM

1940 NM

1736 NM

1630 NM

110 min

114 min

106 min

102 min

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NAVIGATION

NAVIGATION

NAVIGATION




From the departure point, the distance to the point of equal time is:

Given: Distance A to B 2484 NM Groundspeed OUT 420 kt Groundspeed BACK 500 kt The time from A to the Point of Equal Time (PET) between A and B is:

Given: AD = Air distance GD = Ground distance TAS = True airspeed

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kj---900598

kj---900599

kj---900600

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proportional to the sum of ground speed out and ground speed back

inversely proportional to the sum of ground speed out and ground speed back

inversely proportional to the total distance to go

inversely proportional to ground speed back

173 min

163 min

193 min

183 min

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NAVIGATION

NAVIGATION



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?

Given: Distance A to B is 360 NM Wind component A – B is -15 kt Wind component B – A is +15 kt TAS is 180 kt What is the distance from the equal-time-point to B?

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kj---900600

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GD = (AD X GS)/TAS

GD = (AD - TAS)/TAS

GD = AD X (GS - TAS)/GS

GD = TAS/(GS X AD)

170 NM

195 NM

180 NM

165 NM

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NAVIGATION

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NAVIGATION




. Given: Distance A to B 3623 NM Groundspeed out 370 kt Groundspeed back 300 kt The time from a to the Point of Equal Time (PET) between A and B is:

An aircraft has a TAS of 300 knots and a safe endurance of 0 hours. If the wind component on the outbound leg is 50 knots head, what is thedistance to the point of safe endurance?

The distance from A to B is 2368 nautical miles. If outbound groundspeed in 365 knots and homebound groundspeed is 480 knots and safe enduranceis 8 hours 30 minutes, what is the time to the PNR?

2

2

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323 min

288 min

263 min

238 min

1500 nm

1458 nm

1544 nm

1622 nm

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NAVIGATION

NAVIGATION

NAVIGATION




The distance from A to B is 2368 nautical miles. If outbound groundspeed in 365 knots and homebound groundspeed is 480 knots and safe enduranceis 8 hours 30 minutes, what is the time to the PNR?

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 (excludingreserves) the distance from Q to the point of safe return (PSR) is:

Given: Distance Q to R 1760 NM Groundspeed out 435 kt Groundspeed back 385 kt Safe endurance 9 hr The distance from Q to the Point of Safe Return (PSR) between Q and R is:

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kj---900604

kj---900605

kj---900606

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290 minutes

209 minutes

219 minutes

190 minutes

930 NM

1532 NM

1685 NM

1865 NM

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NAVIGATION

NAVIGATION



Given: Distance Q to R 1760 NM Groundspeed out 435 kt Groundspeed back 385 kt Safe endurance 9 hr The distance from Q to the Point of Safe Return (PSR) between Q and R is:

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:

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2

Complexity

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kj---900606

kj---900607

Ref

Ref

1313 NM

1838 NM

1467 NM

1642 NM

2370 NM

2290 NM

1310 NM

1510 NM

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NAVIGATION

NAVIGATION

NAVIGATION



DEAD RECKONING NAVIGATION (DR) - Miscellaneous DR uncertainties and practical means of correction

An aircraft takes off from an airport 2 hours before sunset. The pilot flies a track of 090o(T). W/V 130o/20 kt, TAS 100 kt. In order toreturn to the point of departure before sunset, the furthest distance which may be travelled is:

The distance between point of departure and destination is 340 NM and wind velocity in the whole area is 100o/25 kt. TAS is 140 kt. True Trackis 135o and safe endurance 3 hr and 10 min. How long will it take to reach the Point of Safe Return?

Calculate the dlat from N 001 15 E090 00 to S090 00:

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kj---900608

kj---900609

kj---900610

Ref

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97 NM

115 NM

105 NM

84 NM

1 hr and 44 min

1 hr and 37 min

1 hr and 21 min

5 hr and 30 min

91o15N

88o45N

91o15S

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NAVIGATION

NAVIGATION

NAVIGATION



IN-FLIGHT NAVIGATION - Use of visual observations and application to in-flight navigation

Calculate the dlat from N 001 15 E090 00 to S090 00:

Calculate the dlong from N001 15 E090 00 to N001 15 E015 15:

(Refer to figure 061-01)What is the symbol for an unlighted obstacle?

2

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1

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kj---900610

kj---900611

kj---900612

Ref

Ref

Ref

Ref

268o15N

74o45E

74o15E

74o45W

105o15N

10

14

15

9

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NAVIGATION

NAVIGATION

NAVIGATION




An island appears 45o to the right of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from theisland if at the time of observation the aircraft was on a magnetic heading (MH) of 215o with the magnetic variation (VAR) 21oW?

An island is observed to be 15o to the left. The aircraft heading is 120o(M), variation 17o(W). The bearing (oT) from the aircraft to theisland is:

A ground feature was observed on a relative bearing of 315o and 3 min later on a relative bearing of 270o. The W/V is calm; aircraft GS 180 kt. What is the minimum distance between the aircraft and the ground feature?

2

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kj---900613

kj---900614

kj---900615

Ref

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Ref

101o

059o

239o

329o

122

088

268

302

3 NM

12 NM

9 NM

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NAVIGATION

NAVIGATION

NAVIGATION




A ground feature was observed on a relative bearing of 315o and 3 min later on a relative bearing of 270o. The W/V is calm; aircraft GS 180 kt. What is the minimum distance between the aircraft and the ground feature?

An island is observed by weather radar to be 15o to the left. The aircraft heading is 120o(M) and the magnetic variation 17oW. What is thetrue bearing of the aircraft from the island?

(Refer to figure 061-01)What is the chart symbol for a lightship?

2

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kj---900615

kj---900616

kj---900617

Ref

Ref

Ref

Ref

6 NM

122o

302o

088o

268o

15

16

10

12

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NAVIGATION

NAVIGATION

NAVIGATION




An island appears 30o to the right of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from theisland if at the time of observation the aircraft was on a magnetic heading (MH) of 355o with the magnetic variation (VAR) 15oE?

(Refer to figure 061-01)Which of the following is the symbol for an exceptionally high (over 1000 feet AGL) lighted obstruction?

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 checkthe aircraft:

2

3

2

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kj---900618

kj---900619

kj---900620

Ref

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Ref

160o

130o

220o

190o

13

10

14

12

groundspeed

position

track

1

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NAVIGATION

NAVIGATION

NAVIGATION




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 checkthe aircraft:

An island appears 60o to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from theisland if at the time of observation the aircraft was on a magnetic heading (MH) of 276o with the magnetic variation (VAR) 10oE?

An island appears 30o to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from theisland if at the time of observation the aircraft was on a magnetic heading (MH) of 020o with the magnetic variation (VAR) 25o W?

2

2

2

Complexity

Complexity

Complexity

Complexity

1

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kj---900620

kj---900621

kj---900622

Ref

Ref

Ref

Ref

drift

046o

086o

226o

026o

145o

195o

205o

325o

1

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NAVIGATION

NAVIGATION

NAVIGATION



IN-FLIGHT NAVIGATION - Navigation in climb and descent

You are flying a VFR route and have become uncertain of your position. Which is the best course of action?

A ground feature appears 30o to the left of the centre line of the CRT of an airborne weather radar. If the heading of the aircraft is 355o (M)and the magnetic variation is 15o East, the true bearing of the aircraft from the feature is:

Given: Aircraft height 2500 ft ILS GP angle 3o At what approximate distance from TRH can you expect to capture the GP?

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kj---900623

kj---900624

kj---900625

Ref

Ref

Ref

Set heading towards a line feature - coastline, river, or motorway

Turn round and fly your flight plan tracks in reverse until you see something you recognised before

Fly a series of ever-expanding circles from your present position till you find your next check point

Turn round and fly your flight plan in reverse back to base

160o

220o

310o

130o

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Aircraft height 2500 ft ILS GP angle 3o At what approximate distance from TRH can you expect to capture the GP?

An aircraft is descending down a 12% slope whilst maintaining a GS of 540 kt. The rate of descent of the aircraft is approximately:

On a 12% glide slope, your ground speed is 540 knots. What is your rate of descent?

2

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Complexity

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kj---900625

kj---900626

kj---900627

Ref

Ref

Ref

14.5 NM

7.0 NM

13.1 NM

8.3 NM

650 ft/min

6500 ft/min

4500 ft/min

3900 ft/min

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NAVIGATION

NAVIGATION

NAVIGATION




On a 12% glide slope, your ground speed is 540 knots. What is your rate of descent?

An aircraft at FL 350 is required to commence descent when 85 NM from a VOR and to cross the VOR at FL 80. The mean GS for the descent is 340kt. What is the minimum rate of descent required?

An aircraft at FL 330 is required to commence descent when 65 NM from a VOR and to cross the VOR at FL 100. The mean GS during the descent is330 kt. What is the minimum rate of descent required?

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kj---900627

kj---900628

kj---900629

Ref

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Ref

6550 feet/min

4820 feet/min

8740 feet/min

3120 feet/min

1900 ft/min

1800 ft/min

1600 ft/min

1700 ft/min

1950 ft/min

1650 ft/min

1750 ft/min

1

1

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft at FL 330 is required to commence descent when 65 NM from a VOR and to cross the VOR at FL 100. The mean GS during the descent is330 kt. What is the minimum rate of descent required?

An aircraft at FL 350 is required to descend to cross a DME facility at FL80. Maximum rate of descent is 1800 ft/min and mean GS for descent is276 kt. The minimum range from the DME at which descent should start is:

What is the effect on the Mach number and TAS in an aircraft that is climbing with constant CAS?

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kj---900629

kj---900630

kj---900631

Ref

Ref

Ref

Ref

1850 ft/min

79 NM

69 NM

49 NM

59 NM

Mach number decreases; TAS decreases

Mach number remains constant; TAS increases

Mach number increases; TAS increases

Mach number increases; TAS remains constant

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NAVIGATION

NAVIGATION

NAVIGATION




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 of3000 ft/min?

At 65 nm from a VOR you commence a descent from FL 330 in order to arrive over the VOR at FL 100. Your mean groundspeed in the descent is 240knots. What rate of descent is required?

An aircraft at FL 370 is required to commence descent when 100 NM from a DME facility and to cross the station at FL 120. If the mean GS duringthe descent is 396 kt, the minimum rate of descent required is approximately:

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kj---900632

kj---900633

kj---900634

Ref

Ref

Ref

26.7 NM

19.2 NM

38.4 NM

16.0 NM

1420 feet/min

1630 feet/min

1270 feet/min

1830 feet/min

1650 ft/min

2400 ft/min

1000 ft/min

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft at FL 370 is required to commence descent when 100 NM from a DME facility and to cross the station at FL 120. If the mean GS duringthe descent is 396 kt, the minimum rate of descent required is approximately:

Given: ILS GP angle = 3.5o GS = 150 kt What is the approximate rate of descent?

At 0422 an aircraft at FL 370, GS 320 kt, is on the direct track to VOR X 185 NM distant. The aircraft is required to cross VOR X at FL 80.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:

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kj---900634

kj---900635

kj---900636

Ref

Ref

Ref

1550 ft/min

1000 ft/min

700 ft/min

900 ft/min

800 ft/min

0448

0445

0451

0454

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft at FL 350 is required to cross a VOR/DME facility at FL110 and to commence descent when 100 NM from the facility. If the mean GSfor the descent is 335 kt, the minimum rate of descent required is:

An aircraft at FL 390 is required to descend to cross a DME facility at FL 70. Maximum rate of descent is 2500 ft/min, mean GS during descentis 248 kt. What is the minimum range from the DME at which descent should commence?

An aircraft at FL 290 is required to commence descent when 50 NM from a VOR and to cross that VOR at FL 80. Mean GS during descent is 271 kt.What is the minimum rate of descent required?

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kj---900637

kj---900638

kj---900639

Ref

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Ref

1390 ft/min

1340 ft/min

1240 ft/min

1290 ft/min

53 NM

58 NM

63 NM

68 NM

1700 ft/min

2000 ft/min

1900 ft/min

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft at FL 290 is required to commence descent when 50 NM from a VOR and to cross that VOR at FL 80. Mean GS during descent is 271 kt.What is the minimum rate of descent required?

An aircraft at FL 370 is required to commence descent at 120 NM from a VOR and to cross the facility at FL 130. If the mean GS for the descentis 288 kt, the minimum rate of descent required is:

Given: TAS = 197 kt True course = 240o W/V = 180/30 kt 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?

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kj---900639

kj---900640

2kj---900641

Ref

Ref

Ref

1800 ft/min

960 ft/min

860 ft/min

890 ft/min

920 ft/min

800 ft/min

1400 ft/min

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NAVIGATION

NAVIGATION

NAVIGATION




Given: TAS = 197 kt True course = 240o W/V = 180/30 kt 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?

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:

The outer marker of an ILS with a 3o glide slope is located 4.6 NM from the threshold. Assuming a glide slope height of 50 ft above thethreshold, the approximate height of an aircraft passing the outer marker is:

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2kj---900641

kj---900642

kj---900643

Ref

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Ref

950 ft/min

1500 ft/min

1800 ft/min

10800 ft/min

3600 ft/min

900 ft/min

1400 ft

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NAVIGATION

NAVIGATION

NAVIGATION



IN-FLIGHT NAVIGATION - Navigation in Cruising Flight, Use of Fixes to Revise Navigation Data

The outer marker of an ILS with a 3o glide slope is located 4.6 NM from the threshold. Assuming a glide slope height of 50 ft above thethreshold, the approximate height of an aircraft passing the outer marker is:

By what amount must you change your rate of descent given a 10 knot increase in headwind on a 3o glideslope:


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kj---900643

kj---900644

kj---900645

Ref

Ref

Ref

1450 ft

1350 ft

1300 ft

40 feet per minute increase



50 feet per minute decrease

zero magnetic variation

equal magnetic dip

equal horizontal directive force

equal grivation

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NAVIGATION

NAVIGATION

NAVIGATION




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:

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the radial and DME distance from Connaught VOR/DME (CON, 5355N 00849W)to overhead Abbey Shrule aerodrome (5336N 00739W)?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)You are on a heading of 105o, deviation 3 E WTD NDB (5211.3N 00705.0W) bears013R, CRK VOR (5150.4N 00829.7W) QDM is 211. What is your position?

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kj---900646

kj---900647

kj---900648

Ref

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60 kt

90 kt

75 kt

40 kt

304 47 nm

124 47 nm

296 46 nm

116 46 nm

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NAVIGATION

NAVIGATION

NAVIGATION




(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)You are on a heading of 105o, deviation 3 E WTD NDB (5211.3N 00705.0W) bears013R, CRK VOR (5150.4N 00829.7W) QDM is 211. What is your position?

An aircraft at FL 140, IAS 210 kt, OAT -5oC and wind component minus 35 kt, is required to reduce speed in order to cross a reporting point 5min later than planned. Assuming that flight conditions do not change, when 150 NM from the reporting point the IAS should be reduced by:

An island appears 30o to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from theisland if at the time of observation the aircraft was on a magnetic heading of 276o with the magnetic variation 12oW?

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kj---900648

kj---900649

kj---900650

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Ref

5245N 00757W

5228N 00802W

5412N 00639W

5217N 00745W

25 kt

20 kt

30 kt

15 kt

318o

054o

234o

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NAVIGATION

NAVIGATION

NAVIGATION




An island appears 30o to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from theisland if at the time of observation the aircraft was on a magnetic heading of 276o with the magnetic variation 12oW?

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. Afterflying 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?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)Kerry (5210.9N 00932.0W) is 41 nm DME, Galway 5318.1N 00856.5W) is 50 nm DME.What is your position?

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kj---900650

kj---900651

kj---900652

Ref

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038o

1203

1206

1153

1157

5242N 00827W

5230N 00834W

5255N 00819W

5219N 00809W

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Half way between two reporting points the navigation log gives the following information: TAS 360 kt W/V 330o/80 kt Compass heading 237o Deviation on this heading -5o Variation 19oW What is the average ground speed for this leg?

An aircraft at FL 310, M0.83, temperature -30oC, is required to reduce speed in order to cross a reporting point five minutes later thanplanned. Assuming that a zero wind component remains unchanged, when 360 NM from the reporting point Mach Number should be reduced to:

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kj---900653

kj---900654

kj---900655

Ref

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Ref

360 kt

354 kt

373 kt

403 kt

M 0.76

M 0.74

M 0.78

M 0.80

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NAVIGATION

NAVIGATION

NAVIGATION




TAS = 240 knots The relative bearing from an NDB is 315R at 1410. At 1420 the bearing has changed to 270R. What is your distance from the NDB at 1420?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)An aircraft is on the 025 radial from Shannon VOR (SHA, 5243N 00853W) at 49DME. What is its position?

An aircraft at position 2700N 17000W travels 3000 km on a track of 180T, then 3000 km on a track of 090T, then 3000 km on a track of 000T, then3000 km on a track of 270T. What is its final position?

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kj---900655

kj---900656

kj---900657

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40 nm

50 nm

60 nm

70 nm

5329N 00930W

5239N 00830W

5229N 00930W

5329N 00830W

2700N 17000W

0000N 17000W

2700N 17318W

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft at position 2700N 17000W travels 3000 km on a track of 180T, then 3000 km on a track of 090T, then 3000 km on a track of 000T, then3000 km on a track of 270T. What is its final position?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the mean true track and distance from the BAL VOR (5318N 00627W) toCFN NDB (5520N 00820W)?

A pilot receives the following signals from a VOR DME station: radial 180o+/- 1o, distance = 200 NM. What is the approximate error?

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kj---900657

kj---900658

kj---900659

Ref

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Ref

2700N 14300W

328o 125

148o 125

328o 134

148o 134

+/- 3.5 NM

+/- 1 NM

+/- 2 NM

+/- 7 NM

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft at FL 120, IAS 200 kt, OAT -5o and wind component +30 kt, is required to reduce speed in order to cross a reporting point 5 minlater than planned. Assuming flight conditions do not change, when 100 NM from the reporting point IAS should be reduced to:

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. Afterflying 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?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)You are at position 5340N 00840W. What is the QDR from the SHA VOR (5243N00853W)?

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kj---900660

kj---900661

kj---900662

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Ref

169 kt

165 kt

159 kt

174 kt

1401 UTC

1333 UTC

1347 UTC

1340 UTC

217

037

209

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NAVIGATION

NAVIGATION

NAVIGATION




(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)You are at position 5340N 00840W. What is the QDR from the SHA VOR (5243N00853W)?

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. Afterflying 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?

Given: Distance A to B = 120 NM After 30 NM aircraft is 3 NM to the left of course What heading alteration should be made in order to arrive at point B?

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kj---900662

kj---900663

kj---900664

Ref

Ref

Ref

029

1110 UTC

1115 UTC

1044 UTC

1050 UTC

8o left

6o right

4o right

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NAVIGATION

NAVIGATION

NAVIGATION




Given: Distance A to B = 120 NM After 30 NM aircraft is 3 NM to the left of course What heading alteration should be made in order to arrive at point B?

What is the Rhumb Line track from A (4500N 01000W) to B (4830N 01500W)?

A ground feature was observed on a relative bearing of 325o and five minutes later on a relative bearing of 280o. The aircraft heading was165o(M), variation 25oW, drift 10o right and GS 360 kt. When the relative bearing was 280o the distance and true bearing of the aircraft fromthe feature was:

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kj---900664

kj---900665

kj---900666

Ref

Ref

Ref

8o right

315 T

330 T

215 T

150 T

30 NM and 240o

40 NM and 110o

40 NM and 290o

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NAVIGATION

NAVIGATION

NAVIGATION




A ground feature was observed on a relative bearing of 325o and five minutes later on a relative bearing of 280o. The aircraft heading was165o(M), variation 25oW, drift 10o right and GS 360 kt. When the relative bearing was 280o the distance and true bearing of the aircraft fromthe feature was:

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the approximate course (T) and distance between Waterford NDB (WTD,5212N 00705W) and Sligo NDB (SLG, 5417N 00836W)?

An aircraft obtains a relative bearing of 315o from an NDB at 0830. At 0840 the relative bearing from the same position is 270o. Assuming nodrift and a GS of 240 kt, what is the approximate range from the NDB at 0840?

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kj---900666

kj---900667

kj---900668

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Ref

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Ref

30 NM and 060o

344o 139 nm

164o 138 nm

156o 136 nm

336o 137 nm

50 NM

40 NM

60 NM

30 NM

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NAVIGATION

NAVIGATION

NAVIGATION




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?

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?

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kj---900669

kj---900670

kj---900671

Ref

Ref

Ref

6o Right

9o Right

15o Right

18o Right

4o Left

16o Left

12o Left

8o Left

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NAVIGATION

NAVIGATION

NAVIGATION




The distance between positions A and B is 180 NM. An aircraft departs position A and after having travelled 60 NM, its position is pinpointed 4NM 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?

The distance between two waypoints is 200 NM. To calculate compass heading, the pilot used 2oE magnetic variation instead of 2oW. Assumingthat the forecast W/V applied, what will the off track distance be at the second waypoint?

An aircraft at FL 370, M0.86, OAT -44oC, headwind component 110 kt, is required to reduce speed in order to cross a reporting point 5 min laterthan planned. If the speed reduction were to be made 420 nm from the reporting point, what Mach Number is required?

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kj---900671

kj---900672

kj---900673

Ref

Ref

Ref

6o Right

8o Right

2o Left

4o Right

0 NM

7 NM

14 NM

21 NM

M 0.79

M 0.73

M 0.75

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft at FL 370, M0.86, OAT -44oC, headwind component 110 kt, is required to reduce speed in order to cross a reporting point 5 min laterthan planned. If the speed reduction were to be made 420 nm from the reporting point, what Mach Number is required?

Given: Distance A to B is 475 NM, Planned GS 315 kt, ATD 1000 UTC, 1040 UTC - fix obtained 190 NM along track. What GS must be maintained fromthe fix in order to achieve planned ETA at B?

As the INS position of the departure aerodrome, co-ordinates 35o32.7N 139o46.3W are input instead of 35o32.7N 139o46.3E. When the aircraftsubsequently passes point 52o N 180oW, the longitude value show on the INS will be:

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kj---900673

kj---900674

kj---900675

Ref

Ref

Ref

Ref

M 0.81

320 kt

360 kt

300 kt

340 kt

080o27.4W

099o32.6W

099o32.6 E

080o27.4 E

1

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NAVIGATION

NAVIGATION



Given: Distance A to B 1973 NM Groundspeed out 430 kt Groundspeed back 385 kt Safe endurance 7 hr 20 min The distance from A to the Point of Safe Return (PSR) A is:

Given: Distance A to B 2346 NM Groundspeed out 365 kt Groundspeed back 480 kt The time from A to the Point of Equal Time (PET) between A and B is:

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1664 nm

1698 nm

1422 nm

1490 nm

167 min

219 min

260 min

197 min

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NAVIGATION

NAVIGATION

NAVIGATION




You are flying from A (30S 20E) to B (30S 20W). What is the final GC track?

Refer to figure 061-02)What is the True bearing of point A from point B?

An aircraft at latitude 10o North flies south at a groundspeed of 445 km/hr. What will be its latitude after 3 hrs?

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250o (T)

270o (T)

280o (T)

300o (T)

000o

090o

270o

360o

03o 50¿S

02o 00¿S

12o 15¿S

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft at latitude 10o North flies south at a groundspeed of 445 km/hr. What will be its latitude after 3 hrs?

(Refer to figure 061-11)Given:CON VOR (N5354.8 W00849.1) DME 30 NMCRN VOR (N 5318.1 W00856.5) DME 25 NMAircraft heading 270o(M)Both DMEdistances decreasingWhat is the aircraft position?

An island is observed to be 30o to the right of the nose of the aircraft.The aircraft heading is 290o(M), variation 10o(E)The bearing (oT) fromthe aircraft to the island is:

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kj---900680

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22o 00¿S

N5330 W00820

N5343 W00925

N5335 W00925

N5337 W00820

330

270

250

310

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NAVIGATION

NAVIGATION

NAVIGATION




You are heading 080oT when you get a range and bearing fix from your AWR on a headland at 185 nm 30o left of the nose. What true bearing do youplot on the chart?

An aircraft starts from (S0400.0 W17812.2) and flies north for 2950 nm along the meridian, then west for 382 nm along the parallel of latitude.What is the aircraft's final position?

An aircraft at latitude S0612.0 tracks 000oT for 1667 km. On completion of the flight the latitude will be:

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050 from the headland, using the headland's meridian

050 from the headland, using the aircraft's meridian

230 from the headland, using the headland's meridian

230 from the headland, using the aircraft's meridian

N45100 E172138

N53120 W169122

N45100 W169122

N53120 E172138

S2112.0

N2112.5

N0848.0

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft at latitude S0612.0 tracks 000oT for 1667 km. On completion of the flight the latitude will be:

An airraft departs from N0212.0 E0450.0 on a track of 180oT and flies 685 km. On completion of the flight the latitude will be:

A is at S4500.0 W01000.0 B is at S4500.0 W03000.0 The true course of an aircraft on its arrival at B, to the nearest degree is:

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N0914.0

S1112.5

S0813.0

S0357.0

S0910.5

263o

270o

277o

284o

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NAVIGATION

NAVIGATION

NAVIGATION



IN-FLIGHT NAVIGATION - Flight Log

An aircraft at position 6010.0N 00512.2W flies 165 km due East. The aircraft's new position is:

An aircraft at position 0000N/S 16327W flies a track of 225oT for 70 nm. What is its new position?

You are heading 345M, the variation is 20E, and you take a radar bearing of 30 left of the nose from an island. What bearing do you plot?

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6010.0N 00812.0E

6010.0N 00212.4W

6010.0N 00110.8E

6010.0N 00110.8W

0049N 16238W

0049S 16238W

0049N 16416W

0049S 16416W

160T

155T

140T

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NAVIGATION

NAVIGATION

NAVIGATION




You are heading 345M, the variation is 20E, and you take a radar bearing of 30 left of the nose from an island. What bearing do you plot?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What are the symbols at Galway Carnmore (5318.1N 00856.5W)?

(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)The airport at 5211N 00932W is:

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180T

VOR, NDB, DME, compulsory reporting point

Civil airport, NDB, DME, non-compulsory reporting point

Civil airport, VOR, DME, non-compulsory reporting point

VOR, NDB, DME, non-compulsory reporting point

Kerry

Cork

Shannon

Waterford

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NAVIGATION

NAVIGATION

NAVIGATION




The flight log gives the following data: True track, Drift, True heading, Magnetic variation, Magnetic heading, Compass deviation, Compassheading. The right solution, in the same order, is:

(Refer to figure 061-06)Complete line 6 of the FLIGHT NAVIGATION LOG, positions L to M. what is the HDGo (M) and ETA?

(Refer to figure 061-06)Complete line 3 of the FLIGHT NAVIGATION LOG, positions E to F. What is the HDGo (M) and ETA?

2

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125o, 2oR, 123o, 2oW, 121o, -4o, 117o

115o, 5oR, 120o, 3oW, 123o, +2o, 121o

117o, 4oL, 121o, 1oE, 122o, -3o, 119o

119o, 3oL, 122o, 2oE, 120o, +4o, 116o








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NAVIGATION

NAVIGATION

NAVIGATION




(Refer to figure 061-06)Complete line 3 of the FLIGHT NAVIGATION LOG, positions E to F. What is the HDGo (M) and ETA?

(Refer to figure 061-06)Complete line 2 of the FLIGHT NAVIGATION LOG, positions C to D. What is the HDGo (M) and ETA?

(Refer to figures 061-06 and 061-05)Complete line 1 of the FLIGHT NAVIGATION LOG; positions A to B. What is the HDGo (M) and ETA?

2

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268o - 1114 UTC

282o - 1128 UTC

282o - 1114 UTC

268o - 1128 UTC

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NAVIGATION

NAVIGATION

NAVIGATION

IN-FLIGHT NAVIGATION - Purposes of (FMS) Flight Management Systems



Which of the following lists the first three pages of the FMC/CDU normally used to enter data on initial start-up of te B737-400 ElectronicFlight Intrument System?

In the B737-400 Flight Management System the CDUs are used during pre-flight to:

In which of the following situations is the FMC present position of a B737-400 Electronic Flight Instrument System likely to be least accurate?

11

6

11

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kj---900698

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IDENT - RTE - DEPARTURE

POS INIT - RTE - IDENT

IDENT - POS INIT - RTE

POS INIT - RTE - DEPARTURE

manully initialise the IRSs and FMC with dispatch information

automatically initialise the IRSs and FMC with dispatch information

manually initialise the Flight Director System and FMC with dispatch information

manually initialise the IRSs, FMC and Autothrotle with dispatch information

At top of descent

At top of climb

Just after take-off

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NAVIGATION

NAVIGATION

NAVIGATION




In which of the following situations is the FMC present position of a B737-400 Electronic Flight Instrument System likely to be least accurate?

How is the radio position determined by the FMC in the B737-400 Electronic Flight Instrument System?

What is the validity period of the permanent data base of aeronautical information stored in the FMC in the B737-400 Flight Management System?

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kj---900700

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On final approach

DME ranges and/or VOR/ADF bearings

DME/DME or VOR/DME

DME/DME

VOR/DME range and bearing

28 days

One calendar month

3 calendar months

14 days

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NAVIGATION

NAVIGATION

NAVIGATION




Which component of the B737-400 Flight Management System (FMS) is used to enter flight plan routeing and performance parameters?

Given: Distance A to B is 325 NM Planned GS 315 kt ATD 1130 UTC 1205 UTC – fi obtained 165 NM along track What GS must be maintained from the fix in order to achieve planned ETA at B?

The purpose of the Flight Management System (FMS) as for example installed in the B737-400 is to provide:

6

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kj---900703

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Flight Management Computer

Multi-Function Control Display Unit

Inertial Reference System

Flight Director System

335 kt

375 kt

395 kt

355 kt

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NAVIGATION

NAVIGATION

NAVIGATION




The purpose of the Flight Management System (FMS) as for example installed in the B737-400 is to provide:

Which of the following can all be stored as five letter waypoint identifiers through the CDU of a B737-400 Electronic Flight Instrument System?

Which FMC/CDU page normally appears on initial power application to the B737-400 Electronic Flight Instrument System?

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kj---900705

kj---900706

kj---900707

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both manual navigation guidance and performance management

manual navigation guidance and automatic performance management

continuous automatic navigation guidance as well as manual performance management

continuous automatic navigation guidance and performance management

Waypoint names; navaid frequencies; runway codes; airport ICAO identifiers

Airway names; navaid identifiers; airport names; waypoint code numbers

Waypoint names; navaid identifiers; runway numbers; airport ICAO identifiers

Waypoint names; navaid positions; airport ICAO identifiers; airport names

IDENT

INITIAL

POS INIT

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NAVIGATION

NAVIGATION

NAVIGATION




Which FMC/CDU page normally appears on initial power application to the B737-400 Electronic Flight Instrument System?

What are the levels of message on the Boeing 737-400 FMC?

Which of the following lists all the methods that can be used to enter Created Waypoints into the CDU of a B737-400 Electronic Flight InstrumentSystem?

6

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kj---900707

kj---900708

kj---900709

Ref

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PERF INIT

Urgent and Routine

Priority and Alerting

Alert and Advisory

Urgent and Advisory

Identifier bearing/distance; place bearing/place bearing; latitude and longitude;waypoint name

Identifier bearing/distance; place bearing/place distance; along/across track displacement; latitude and longitude

Identifier bearing/distance; place distance/place distance; along track displacement; latitude and longitude

Identifier bearing/distance; place distance/place distance; along-track displacement; latitude and longitude

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NAVIGATION

NAVIGATION

NAVIGATION




What indication, if any, is given in the B737-400 Flight Management System if radio updating is not available?

What are, in order of highest priority followed by lowest, the two levels of message produced by the CDU of the B737-400 Electronic FlightInstrument System?

An aeroplane flies from A (59oS 142oW) to B (61oS 148oW) with a TAS of 480 kt. The autopilot is engaged and coupled with an Inertial NavigationSystem in which AB track is active. On route AB, the true track:

6

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kj---900710

kj---900711

kj---900712

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A warning message is displayed on the IRS displays

A warning message is displayed on the EHSI and MFDU

A warning message is displayed on the Flight Director System

No indication is given so long as the IRS positions remain within limits

Priority and Alerting

Urgent and Routine

Alerting and Advisory

Urgent and Advisory

varies by 10o

decreases by 6o

varies by 4o

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NAVIGATION

NAVIGATION

NAVIGATION




An aeroplane flies from A (59oS 142oW) to B (61oS 148oW) with a TAS of 480 kt. The autopilot is engaged and coupled with an Inertial NavigationSystem in which AB track is active. On route AB, the true track:

In a Flight Management System (FMS), control Display Units (CDUs) are used pre-flight to

When can a pilot change the data in the FMS data base?

11

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kj---900712

kj---900713

kj---900714

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increases by 5o


automatically initialise the IRSs and FMC with dispatch information

manually initialise the IRSs and FMC with dispatch information

manually initialise the IRSs, FMC and Air Data Computer with dispatch information

Every 28 days

When deemed necessary

When there is a fault

He can't; for the pilot the FMS data base is read only

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NAVIGATION

NAVIGATION

NAVIGATION



INERTIAL NAVIGATION SYSTEMS (INS) - Principles and Practical application

The FMC position is:

Which of the following can be input to the FMC using a maximum of 5 alphanumerics:

What does the sensor of an INS/IRS measure?

11

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6

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kj---900715

kj---900716

kj---900717

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Ref

The average of the IRS positions

The average of the IRS and radio navigation positions

Computer generated from the IRS and radio navigation positions

Computer generated from the radio navigation positions

Waypoints, latitude and longitude and SIDs/STARs

ICAO aerodrome indicators, navigation facilities and SIDs/STARs

Waypoints, airway designators and latitude and longitude

Navigation facilities, reporting points and airway designators

Velocity

Precession

Horizontal Earth Rate

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NAVIGATION

NAVIGATION

NAVIGATION




What does the sensor of an INS/IRS measure?

An INS platform is kept at right angles to local gravity by applying corrections for the effects of: i. Aircraft manoeuvres ii. Earth rotation iii. Transport wander iv. Coriolis v. Gyroscopic inertia

The term drift refers to the wander of the axis of a gyro in:

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kj---900717

kj---900718

6kj---900719

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Acceleration

i, iii and v

ii, iii and v

ii, iv and v

ii, iii and iv

the vertical and horizontal plane

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NAVIGATION

NAVIGATION

NAVIGATION




The term drift refers to the wander of the axis of a gyro in:

What additional information is required to be input to an Inertial Navigation System (INS) in order to obtain an W/V readout?

In an IRS:

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6kj---900719

kj---900720

kj---900721

Ref

Ref

Ref

the vertical plane

the horizontal plane

any plane

Mach Number

IAS

Altitude and OAT

TAS

the accelerometers are strapped down but the platform is gyro stabilised

the platform is strapped down but the accelerometers are gyro-stabilised

accelerometers and platform are both gyro-stabilised

accelerometers and platform are both strapped down

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NAVIGATION

NAVIGATION

NAVIGATION




In order to maintain an accurate vertical using a pendulous sytem, an aircraft inertial platform incorporates a device:

With reference to inertial navigation systems, a TAS input is:

In what plane is gyro wander known as drift?

6

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1

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kj---900722

kj---900723

kj---900724

Ref

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without damping and a period of 84.4 min

with damping and a period of 84.4 min

without damping and a period of 84.4 sec

with damping and a period of 84.4 sec

not required

required to provide a W/V read out

required for Polar navigation

required for rhumb line navigation

Horizontal

Vertical

Horizontal and vertical

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NAVIGATION

NAVIGATION

NAVIGATION




In what plane is gyro wander known as drift?

In a ring laser gyro, the purpose of the dither motor is to:

An Inertial Navigation System, what is the output of the first stage North/South integrator?

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kj---900724

kj---900725

kj---900726

Ref

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Ref

Neither - it is a separate phenomenon

enhance the accuracy of the gyro at all rotational rates

overcome laser lock

compensate for transport wander

stabilise the laser frequencies

Groundspeed

Latitude

Velocity along the local meridian

Change of latitude

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NAVIGATION

NAVIGATION

NAVIGATION




What measurement is used to carry out alignment of an Inertial Navigation System?

The resultant of the first integration of the output from the east/west accelerometer of an inertial navigation system (INS) in NAV MODE is:

What is the name given to an Inertial Reference System (IRS) which has the gyros and accelerometers as part of the units fixture to the aircraftstructure?

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kj---900727

kj---900728

kj---900729

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Acceleration sensed by the east gyro horizontal accelerometer

Acceleration sensed by the north gyro horizontal accelerometer

Acceleration sensed by the north gyro ertical accelerometer

Difference in magnitude of the value of gravity compared with the gravity at the last known position

velocity along the local parallel of latitude

change of longitude

vehicle longitude

departure

Solid state

Rigid

Strapdown

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NAVIGATION

NAVIGATION

NAVIGATION




What is the name given to an Inertial Reference System (IRS) which has the gyros and accelerometers as part of the units fixture to the aircraftstructure?

One of the errors inherent in a ring laser gyroscope occurs at low input rotation rates tending towards zero when a phenomenon known as lock-inis experienced. What is the name of the technique, effected by means of a piezo- electric motor, that is used to correct this error?

In an Inertial Navigation System (INS), Ground speed (GS) is calculated:

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kj---900729

kj---900730

kj---900731

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Ref

Ring laser

Dither

Cavity rotation

Zero drop

Beam lock

from TAS and W/V from RNAV data

from TAS and W/V from Air Data Computer (ADC)

by integrating measured acceleration

by integrating gyro precession in N/S and E/W directions respectively

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NAVIGATION

NAVIGATION

NAVIGATION




IRS differs from INS in that it:

Double integration of the output from the east/west accelerometer of an inertial navigation system (INS) in the NAV MODE give:

Some inertial reference systems are known as strapdown. This means:

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kj---900733

kj---900734

kj---900735

Ref

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has a longer spin-up time and is not affected by vertical accelerations due to gravity

has a shorter spin-up time and suffers from laser lock

does not need to correct for coriolis and central acceleration

does not experience Schuler errors as accelerometers are strapped down and are not rotated by a V/R feedback loop

distance north/south

vehicle longitude

distance east/west

velocity east/west

the system is mounted on a stabilised platform

the system is mounted and fixed to the aircraft structure

the accelerometers are fixed bu the gyros are stabilised

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NAVIGATION

NAVIGATION

NAVIGATION




Some inertial reference systems are known as strapdown. This means:

Some inertial reference and navigation systems are known as strapdown. This means that:

The principle of Schuler Tuning as applied to the operation of inertial Navigation Systems Inertial Reference Systems is applicable to:

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kj---900735

kj---900736

kj---900737

Ref

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Ref

the gyros are fixed but the accelerometers are stabilised

only the gyros and not the accelerometers, become part of the units fixture to the aircraft structure

gyros, and accelerometers are mounted on a stabilised platform in the aircraft

gyros and accelerometers need satellite information input to obtain a vertical reference

the gyroscopes and accelerometers become part of the units fixture to the aircraft structure

both gyro-stabilised platform and strapdown systems

only gyro-stabilised systems

both gyro-stabilised and laser gyro systems but only when operating in the non strapdown mode

only to strapdown laser gyro systems

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NAVIGATION

NAVIGATION

NAVIGATION




After alignment of the stable platform of an Inertial Navigation System, the output data from the platform is:

In an Inertial Reference System, accelerations are measured in relation to:

Inertial Reference System sensors include:

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kj---900738

kj---900739

kj---900740

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acceleration north/south and east/west and true heading

latitude, longitude and attitude

acceleration north/south and east/west, attitude and true heading

latitude, longitude and true heading

the direction of true north

WGS 84 Earth co-ordinates

local vertical at the aircraft position

aircraft axis

A – one east-west and one north-south gyro; one east-west and one north- south accelerometer

B – accelerometers mounted in the direction of the aircraft axis

C – laser gyros mounted in the direction of the aircraft axis

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NAVIGATION

NAVIGATION

NAVIGATION




Inertial Reference System sensors include:

The platform of an inertial navigation system (INS) is maintained at right angles to the local vertical by applying corrections for the effectsof:

The purpose of the TAS input, from the air data computer, to the Inertial Navigation System is for:

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kj---900740

kj---900741

kj---900742

Ref

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Ref

D – accelerometers, and laser gyros, mounted in the direction of the aircraft axis

aircraft manoeuvres, earth rotation, transport wander and coriolis

gyroscopic inertia, earth rotation and real drift

vertical velocities, earth precession, centrifugal forces and transport drift

movement in the yawing plane, secondary precession and pendulous oscillation

position update in Attitude mode

the calculation of wind velocity

position update in Navigation mode

the calculation of drift

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NAVIGATION

NAVIGATION

NAVIGATION


INERTIAL NAVIGATION SYSTEMS (INS) - Alignment Procedures


A laser reference system (IRS), as compared to a gyro reference system (INS):

Which of the following statements concerning the aircraft positions indicated on a triple fit Inertial Navigation System (INS)/InertialReference System (IRS) on the CDU is correct?

After alignment, is it possible to update IRS positions?

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kj---900743

kj---900744

kj---900745

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is not strapped down and is adversely affected by g-forces

is strapped down and is not adversely affected by g-forces

the platform is strapped down but the accelerometers are not

the accelerometers are strapped down but the platform is not

The positions will only differ if one of the systems has been decoupled because of a detected malfunction

The positions will be the same because they are an average of three difference positions

The positions are likely to differ because they are calculated from different sources

The positions will only differ if an error has been made when inputting the present position at the departure airport

Yes by operation of the TO/GA switch, the runway threshold co- ordinates are inserted into the IRS

No

Yes, the pilots can insert updates

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NAVIGATION

NAVIGATION

NAVIGATION




After alignment, is it possible to update IRS positions?

Alignment of INS and IRS equipments can take place in which of the following modes?

Which of the following statements concerning the loss of alignment by an Inertial Reference System (IRS) in flight is correct?

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kj---900745

kj---900746

kj---900747

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Yes, theprocess is automatic in flight from the DMEs

ATT and ALIGN

NAV and ALIGN

ALIGN and ATT

NAV and ATT

It is not usable in any mode and must be shut down for the rest of the flight

The IRS has to be coupled to the remaining serviceable system and a realignment carried out in flight

The mode selector has to be rotated to ATT then back through ALIGN to NAV in order to obtain an in-flight realignment

The navigation mode, including present position and ground speed outputs, in inoperative for the remainder of the flight

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NAVIGATION

NAVIGATION

NAVIGATION




During initial alignment an inertial navigation system is north aligned by inputs from:

During the initial alignment of an inertial navigation system (INS) the equipment:

When initial position is put into an FMS, the system:

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kj---900748

kj---900749

kj---900750

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horizontal accelerometers and the east gyro

the aircraft remote reading compass system

computer matching of measured gravity magnitude to gravity magnitude of initial alignment

vertical accelerometers and the north gyro

will accept a 10o error in initial latitude but will not accept a 10o error in initial longitude

will not accept a 10o error in initial latitude but will accept a 10o error in initial longitude

will accept a 10o error in initial latitude and initial longitude

will not accept a 10o error in initial latitude or initial longitude

rejects initial latitude error, but it will accept longitude error

rejects initial longitude error, but it will accept latitude error

rejects initial latitude or longitude error

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NAVIGATION

NAVIGATION

NAVIGATION




When initial position is put into an FMS, the system:

Which of the following statements is correct concerning gyro-compassing of an inertial navigation system (INS)?

The alignment time, at mid-latitudes, for an Inertial Reference System using laser ring gyros is approximately:

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kj---900750

kj---900751

kj---900752

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cannot detect input errors, and accepts whatever is put in

Gyro-compassing of an INS is possible in flight because it can differentiate between movement induced and misalignment inducedaccelerationsGyro-compassing of an INS is not possible in flight because it cannot differentiate between movement induced and misalignmentinduced accelerationsGyro-compassing of an INS is possible in flight because it cannot differentiate between movement induced and misalignment inducedaccelerationsGyro-compassing of an INS is not possible in flight because it can differentiate between movement induced and misalignment inducedaccelerations

5 min

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10 min

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NAVIGATION

NAVIGATION

NAVIGATION




A pilot accidently turning OFF the INS in flight, and then turns it back ON a few moments later. Following this incident:

Which of the following statements concerning the alignment procedure for Inertial Navigation Systems (INS/Inertial Reference Systems (IRS) atmid-latitudes is correct?

When and where are IRS positions updated?

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kj---900753

kj---900754

kj---900755

Ref

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Ref

everything returns to normal and is usable

no useful information can be obtained from the INS

it can only be used for attitude reference

the INS is usable in NAV MODE after a position update

INS/IRS can only be aligned in the ALIGN mode

INS/IRS can be aligned in either the ALIGN or NAV mode

INS/IRS can be aligned in either the ALIGN or ATT mode

INS/IRS can only be aligned in NAV mode

During all phases of flight

Only on the ground during the alignment procedure

When the FMS is in IRS ONLY NAV operation

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NAVIGATION

NAVIGATION

NAVIGATION




When and where are IRS positions updated?

After alignment of the stable platform of the Inertial Navigation System, the output data from the INS computer to the platform is:

The data that needs to be inserted into an Inertial Reference System in order to enable the system to make a successful alignment for navigationis:

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kj---900755

kj---900756

kj---900757

Ref

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Ref

When the VHF Nav Radios are selected to AUTO

rate corrections to the gyros

accelerations from the accelerometers

attitude

latitude and longitude

airport ICAO identifier

aircraft heading

the position of an in-range DME

aircraft position in latitude and longitude

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NAVIGATION

NAVIGATION

NAVIGATION


INERTIAL NAVIGATION SYSTEMS (INS) - Accuracy, reliability, errors and coverage of INS/IRS


The full alignment of the stable platform on an Inertial Navigation System:

The drift of the azimuth gyro on an inertial unit induces an error in the position given by this unit. T being the elapsed time. The totalerror is:

The azimuth gyro of an inertial unit has a drift of 0.01o/hr. After a flight of 12 hrs with a ground speed of 500 kt, the error on theaeroplane position is approximately:

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kj---900758

kj---900759

kj---900760

Ref

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Ref

may be carried out on the ground or when in straight and level flight

may be carried out during any phase of flight

is only possible on the ground when the aircraft is at a complete stop

may be carried out at any time so long as an accurate position is inserted into the system

sinusoidal

proportional to the square of time, t?

proportional to t/2

proportional to t

6 NM

1 NM

12 NM

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NAVIGATION

NAVIGATION

NAVIGATION




The azimuth gyro of an inertial unit has a drift of 0.01o/hr. After a flight of 12 hrs with a ground speed of 500 kt, the error on theaeroplane position is approximately:

The platform of an inertial navigation system (INS) is maintained at right angles to the local vertical by applying corrections for the effectsof:

Which of the following lists, which compares an Inertial Reference System that utilises Ring Laser Gyroscopes (RLG) instead of conventionalgyroscopes, is completely correct?

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kj---900760

kj---900761

kj---900762

Ref

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Ref

60 NM

gyroscopic inertia, earth precession and pendulous oscillation

vertical velocities, earth precession, centrifugal forces and transport drift

movements in the yawing plane, secondary precession and pendulous oscillation

aircraft manoeuvres, earth rotation, transport wander and coriolis

The platform iskept stable relative to the earth mathematically rather than mechanically but it has a longer spin up time

It does not suffer from lock in error and it is insensitive to gravitational (g) forces

There is little or no spin up time and it does not suffer from lock in error

There is little or no spin up time and it is insensitive to gravitational (g) forces

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NAVIGATION

NAVIGATION

NAVIGATION



INERTIAL NAVIGATION SYSTEMS (INS) - Flight deck equipment and operation

Comparing the Present Position display on the Boeing 737-400 FMC, you note that there is a 10-mile difference between the left IRS and the rightIRS positions. This means that:

Within the platform levelling loop of an earth-vertical referenced INS:

An aircraft equipped with an Inertial Navigation System (INS) flies with INS 1 coupled with autopilot 1. Both inertial navigation systems are navigating from waypoint A to B. The inertial systems Central Display Units (CDU) shows: -XTK on INS 1 = 0 –XTK on INS 2 = BL (XTK = cross track) From this information it can be deduced that:

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kj---900763

kj---900764

kj---900765

Ref

Ref

Ref

One system is in IRS ONLY NAV operation and the other has the VHF Nay Radios selected to AUTO

No special significance this is normal

At least one of the IRS is drifting

One position has been computer generated from radio nay positions whilst the other is raw IRS

The levelling signals are unbounded, with a period of 84.4 seconds

The levelling signals are bounded with a period of 84.4 minutes

The levelling signals are unbounded, with a period of 84.4 minutes

The levelling signals are bounded, with a period of 84.4 seconds

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NAVIGATION

NAVIGATION

NAVIGATION




An aircraft equipped with an Inertial Navigation System (INS) flies with INS 1 coupled with autopilot 1. Both inertial navigation systems are navigating from waypoint A to B. The inertial systems Central Display Units (CDU) shows: -XTK on INS 1 = 0 –XTK on INS 2 = BL (XTK = cross track) From this information it can be deduced that:

ATT Mode of the Inertial Reference System (IRS) is a back-up mode providing:


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kj---900765

kj---900766

kj---900767

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the autopilot is unserviceable in NAV mode

only inertial navigation system No. 2 is drifting

only inertial navigation system No. 1 is drifting

at least one of the inertial navigation systems is drifting

only attitude and heading information

only attitude information

navigation information

altitude, heading and position information

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NAVIGATION

NAVIGATION

NAVIGATION





In the Boeing 737-400 FMS, the CDU is used to:

The period of validity of an FMS database is:

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kj---900767

kj---900768

kj---900769

Ref

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The positions will only differ if one of the systems has been decoupled because of a detected malfunction

The positions will be the same because they are an average of three different positions

The positions are likely to differ because they are calculated from different sources

The positions will only differ if an error has been made when inputting the present position at the departure airport

manually initialise the IRS and FMC with dispatch information

automatically initialise the IRS and FMC with dispatch information


manually initialise the Flight Director System, FMC and Autothrottle with dispatch information

56 days

one week

28 days

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NAVIGATION

NAVIGATION

NAVIGATION




The period of validity of an FMS database is:

What are the positions (in the order left to right) on the Boeing 737-400 IRS MSU mode selector?

With reference to an inertial navigation system (INS) the initial great circle track between computer inserted waypoints will be displayed whenthe control display unit (CDU) is selected to:

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kj---900769

kj---900770

kj---900771

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varies depending on the area of operational cover

OFF STBY ALIGN NAV

OFF ON ALIGN NAV

OFF STBY ATT NAV

OFF ALIGN NAV ATT

TK/GS

HDG/DA

DSRTK/STS

XTK/TKE

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NAVIGATION

NAVIGATION

NAVIGATION




On a triple-fit IRS system, present positions on the CDU:

Aircraft position determined by radio navigation in an FMC is derived from:

Waypoints can be entered in an INS memory in different formats. In which of the following formats can waypoints be entered into all INSs?

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will only differ if one IRS has been decoupled due to a detected malfunction

will only differ if an initial input error of aircraft position has been made

are likely to differ as the information comes from different sources

will not differ as the information is averaged

VOR/DME

DME ranges and/or VOR/ADF bearings

VOR/ADF

DME only

Bearing and distance

Geographic co-ordinates

Hexadecimal

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NAVIGATION

NAVIGATION

NAVIGATION




Waypoints can be entered in an INS memory in different formats. In which of the following formats can waypoints be entered into all INSs?

When is the last point at which an INS or IRS may be selected to NAV mode?

Which of the following correctly lists the order of available selections of the Mode Selector switches of an inertial reference system (IRS)mode panel?

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By waypoints name

After passengers and freight are aboard

Immediately prior to push back or taxi from the gate

At the holding point

On operation of the TOGA switch when opening the throttles for the take- off

OFF - ON - ALIGN - NAV

OFF - ALIGN - NAV - ATT

OFF - STBY - ALIGN - NAV

OFF - ALIGN - ATT - NAV

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NAVIGATION

NAVIGATION

NAVIGATION




On the IRS, selection of ATT mode gives?

On an INS, what is the output of the E/W second-stage integrator?

Gyro-compassing of an inertial reference system (IRS) is accomplished with the mode selector switched to:

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attitude and heading

altitude, heading, and groundspeed

altitude, attitude, and heading

attitude information only

Velocity N/S

Distance N/S

Distance E/W

Velocity E/W

ATT/REF

STBY

ALIGN

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NAVIGATION

NAVIGATION

NAVIGATION




Gyro-compassing of an inertial reference system (IRS) is accomplished with the mode selector switched to:

Which of the following statements concernikng the operation of an Inertial Navigation System (INS)/Inertial Reference System (IRS) is correct?

What method of entering waypoints can be used on all INS equipments?

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ON

NAV mode must be selected prior to movement of the aircraft off the gate

NAV mode must be selected on the runway just prior to take-off

NAV mode must be selected prior to the loading of passengers and/or freight

NAV mode must be selected when the alignment procedure is commenced

Distance and bearing

Waypoint name

Navaid identifier

Latitude and longitude

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NAVIGATION

NAVIGATION

NAVIGATION

INERTIAL NAVIGATION SYSTEMS (INS) - INS operation



Gyro-compassing in an INS:

In what formats can created waypoints be entered into the scratch pad of the B737-400 FMS?

The following points are entered into an inertial navigation system (INS).WPT 1:60oN 30oW; WPT 2:60oN 20oW; WPT 3:60oN 10oWThe inertialnavigation system is connected to the automatic pilot on route (1-2-3). The track change when passing WPT 2 will be approximately:

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is possible in flight as the gyros can differentiate between acceleration due to aircraft movement and initial alignment errors

is not possible in flight as the gyros can differentiate between acceleration due to aircraft movement and initial alignment errors

is not possible in flight as the gyros cannot differentiate between acceleration due to aircraft movement and initial alignmenterrorsis possible in flight as the gyros cannot differentiate between acceleration due to aircraft movement and initial alignment errors

Place Bearing/Distance, Place Distance/Place Distance, Along-Track Displacement, Latitude and Longitude

Place Bearing/Distance, Place Bearing/Place Bearing, Across-Track Displacement, Latitude and Longitude

Place Bearing/Distance, Place Bearing/Place Bearing, Along-Track Displacement, Latitude and Longitude

Place, Place Bearing/Distance, Along-Track Displacement, Latitude and Longitude

a 9o increase

zero

a 9o decrease

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NAVIGATION

NAVIGATION

NAVIGATION




The following points are entered into an inertial navigation system (INS).WPT 1:60oN 30oW; WPT 2:60oN 20oW; WPT 3:60oN 10oWThe inertialnavigation system is connected to the automatic pilot on route (1-2-3). The track change when passing WPT 2 will be approximately:

The automatic flight control system (AFCS) in an aircraft is coupled to the guidance outputs from an inertial navigation system (INS). Theaircraft is flying between inserted waypoints No. 3 (55o00N 020o00W) and No. 4 (55o00N 030o00W). With DSRTK/STS selected on the CDU, to thenearest whole degree, the initial track read-out from waypoint No. 3 will be:

The sensors of an INS measure:

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a 4o decrease

278o

274o

266o

270o

precession

velocity

the horizontal component of the earth's rotation

acceleration

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NAVIGATION

NAVIGATION

NAVIGATION




What is the source of magnetic variation information in a Flight Management System (FMS)?

Where and when are the IRS positions updated?

The automatic flight control system is coupled to the guidance outputs from an inertial navigation system. Which pair of latitudes will givethe greatest difference between initial track read-out and the average true course given, in each case, a difference of longitude of 10o?

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Magnetic variation is calculated by each IRS based on the respective IRS position and the aircraft magnetic heading

The main directional gyro which is coupled to the magnetic sensor 9flux valve) positioned in the wing-tip

The FMS calculates MH and MT from the FMC position

Magnetic variation information is stored in each IRS memory; it is applied to the true heading calculated by the respective IRS

During flight IRS positions are automatically updated by the FMC

Only on the ground during the alignment procedure

IRS positions are updated by pressing the Take-off/Go-around button at the start of the take-off roll

Updating is normally carried out by the crew when over-flying a known position (VOR station or NDB)

30oS to 25oS

60oN to 50oN

30oS to 30oN

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NAVIGATION

NAVIGATION

NAVIGATION




The automatic flight control system is coupled to the guidance outputs from an inertial navigation system. Which pair of latitudes will givethe greatest difference between initial track read-out and the average true course given, in each case, a difference of longitude of 10o?

An aircraft travels from point A to point B, using the autopilot connected to the aircraft's inertial system. The co-ordinates of A (45oS010oW) and B (45oS 030oW) have been entered. The true course of the aircraft on its arrival at B, to the nearest degree, is:

The automatic flight control system (AFCS) in an aircraft is coupled to the guidance outputs from an inertial navigation system (INS) and theaircraft is flying from waypoint No. 3 (60o00S 070o00W) to No. 3 (60o00S 080o00W). Comparing the initial track (oT) at 070o00W and the finaltrack (oT) at 080o00W, the difference between them is that the initial track is approximately:

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60oN to 80oN

277o

284o

263o

270o

9o greater than the final one

5o greater than the final one

9o less than the final one

5o less than the final one

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NAVIGATION

NAVIGATION

NAVIGATION




Which of the following statements concerning the position indicated on the Inertial Reference System (IRS) display is correct?

Which of the following statements concerning the operation of an Inertial Navigation System (INS) /Inertial Reference System (IRS) is correct?

An aircraft is flying with the aid of an inertial navigation system (INS) connected to the autopilot. The following two points have beenentered in the INS computer:WPT 1: 60oN 030oWWPT 2: 60oN 020oWWhen 025oW is passed the latitude shown on the display unit of the inertialnavigation system will be:

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It is updated when go-around is selected on take-off

It is constantly updated from information obtained by the FMC

It is not updated once the IRS mode is set to NAV

The positions from the two IRSs are compared to obtain a best position which is displayed on the IRS

NAV mode must be selected prior to movement of the aircraft off the gate

NAV mode must be selected on the runway just prior to take-off

NAV mode must be selected prior to the loading of passengers and/or freight

NAV mode must be selected when the alignment procedure is commenced

60o 00.0N

59o 49.0N

60o 11.0N

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NAVIGATION

NAVIGATION

NAVIGATION



BASICS OF NAVIGATION

An aircraft is flying with the aid of an inertial navigation system (INS) connected to the autopilot. The following two points have beenentered in the INS computer:WPT 1: 60oN 030oWWPT 2: 60oN 020oWWhen 025oW is passed the latitude shown on the display unit of the inertialnavigation system will be:

What is the sequence of pages on start-up of the Boeing 737-400 FMS?

What is the highest latitude listed below at which the sun will reach an altitude of 90o above the horizon at some time during the year?

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60o 05.7N

POS INIT, IDENT, DEPARTURES

IDENT, POS INIT, RTE

POS INIT, RTE, IDENT

IDENT, POS INIT, DEPARTURES

0o

45o

66o

23o

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1234567 NAVIGATION

BASICS OF NAVIGATION

What is the initial great circle direction from 45oN 14o12W to 45oN 12o48E?

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86.5o (T)

80.4o (T)

090o (M)

270o (M)

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new_navigation.pdf

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