Cpl Navigation

CPL

INDEX

CPL NAVIGATION

1. The Earth

01

2. Charts

25

3. Relative Velocity

55

4. Solar System & Time

59

5. Navigation Computer

83

6. Plotting

95

Annex A

Sample Exams

137

Annex B

Answers to Questions

149

Copyright



2001 Flight Training College of Africa

All Rights Reserved. No part of this manual may be reproduced in any manner

whatsoever including electronic, photographic, photocopying, facsimile, or stored in a

CHAPTER 1

THE EARTH

The earth is not a perfect sphere, there is a slight bulge at the Equator and a flattening at the Poles. The earth's shape is described as an oblate spheroid. The polar diameter is 6860.5 nm which is 23.2 nm shorter than the average equatorial diameter of 6883.7 nm. This gives a compression ratio of 1/2967 which for all practical purposes can be ignored. Cartographers and Inertial Navigation systems will take the true shape of the earth into account.

PARALLELS OF LATITUDE

Parallels of Latitude are small circles that are parallel to the Equator. They lie in a 090 and 270 Rhumb Line direction as they cut all Meridians at 90.

LATITUDE

The Latitude of a point is the arc of a Meridian from the Equator to the point. It is expressed in degrees and minutes North or South of the Equator. It can be presented in the following

LONGITUDE

The Longitude of a point is the shorter arc of the Equator measured East or West from the Greenwich Meridian. It can be presented in the following forms.

FINDING PLACES WITH LAT/LONG Example 1

At 26:34 S / 26:16.5 E what do you find? Did you find a town?

GREAT CIRCLE (GC)

A Great Circle is a circle drawn on the surface of a sphere whose centre and radius are those of the sphere itself. A Great Circle divides the sphere into two halves. The Equator is a Great Circle dividing the earth into the Northern and Southern Hemispheres. On a flat surface the shortest distance between TWO points is a straight line. On a sphere the shortest distance between two points is the shorter arc of a Great Circle drawn through the two points. To fly from Europe to the West Coast of America the shortest distance is of course a Great Circle which usually takes the least time and fuel used. A Great Circle cuts all Meridians at different angles.

RHUMB LINE (RL)

A Rhumb Line is a curved line drawn on the surface of the earth which cuts all Meridians at the same angle. An aircraft steering a constant heading of 065(T) with zero wind will be flying a Rhumb Line.

MERIDIANS

Meridians are Great semi-circles that join the North and South Poles. Every Great Circle passing through the poles forms a Meridian and its Anti-Meridian. All Meridians indicate True North or 000(T) and 180(T). As Meridians have a constant direction they are Rhumb Lines as well as Great Circles.

EQUATOR

The Equator cuts all Meridians at 90 providing a True East-West or 090(T) and 270(T) erection. As the Equator cuts all Meridians at 90 it is a Rhumb Line as well as a Great

SMALL CIRCLE

A Small Circle is a circle drawn on a sphere whose centre and radius are not those of the sphere itself.

DIRECTION

True North

True North is the direction of the Meridian passing through a position. True Direction

Aircraft Heading or Track is measured clockwise from True North. It is usually expressed in degrees and decimals of a degree, e.g. 092(T) 107.25GC 265.37 RL

Magnetic North

Magnetic North is the direction in the horizontal plane indicated by a freely suspended magnet influenced by the earth's magnetic field only.

Variation

Variation is the angular difference between True North and Magnetic North

Magnetic Direction (M)

Aircraft Magnetic Heading or Magnetic Track is measured clockwise from Magnetic

North, which is sometimes referred to as the Magnetic Meridian, e.g. 100



(M)

Compass North (C)

Compass North is the direction indicated by the compass needle in an aircraft. Magnetic Fields in the aircraft will attract the compass needle away from Magnetic North causing Compass Deviation.

Deviation

The angular difference between Compass North and Magnetic North.

Deviation is Westerly when Compass North is to the West of Magnetic North Deviation is Easterly when Compass North is to the East of Magnetic North

DEVIATION EAST COMPASS LEAST DEVIATION WEST COMPASS BEST Heading l00(C) Dev+4e 104(M) Heading 100(C) Dev-3w 096(M)

Deviation West is Negative (-) Deviation East is Positive (+) Deviation is a correction to Compass Heading to give Magnetic Heading

CONVERGENCE AND CONVERSION ANGLE

CONVERGENCE

Meridians are Semi Great Circles joining the North and South Poles. They are parallel at the Equator. As the meridians leave the Equator either Northwards or Southwards they converge and meet at the Poles.

Convergence is defined as the angle of inclination between two selected meridians measured at a given Latitude.

Considering the two meridians shown above, one at 20W and the other at 20E. The Change of Longitude (Ch. Long) or Difference in Longitude (D Long) between the two meridians is 40.

At the Equator (Latitude 0) they are parallel, the angle of convergence is 0. At the Poles (Latitude 90) they meet, and the angle of convergence is the Difference of Longitude, 40.

At any intermediate Latitude the angle of inclination between the same two meridians

will between 0



and 40



depending on the Latitude.

Example 1. Calculate the value of Convergence between A (N 45:25 E 025:36) and B (N 37:53 E042:17).

A N 45:25 E 025:36

B N 37:53 E 042:17

N 41:39 Mean Latitude 16:41 Change of Longitude

Convergence = Ch. Long x Sin Mean Latitude = 1641' x Sin 41 39'

= 16.6833 x Sin 41.65 = 11.0874

NOTE Both Mean Latitude and Change of Longitude must be changed into decimal notation.

CONVERGENCE = CHANGE OF LONGITUDE x SIN MEAN LATITUDE

CONVERGENCE = DIFFERENCE BETWEEN INITIAL AND FINAL GC TRACKS

Example 1 A and B are in the same hemisphere The Great Circle Track from A to B is 062 The Great Circle Track from B to A is 278 (a) In which hemisphere are A and B?

(b) What is the value of Convergence between A and B?

Example 2 C and D are in the same hemisphere

The Great Circle bearing of D from C is 136 (brg of D measured at C) The Great Circle bearing of C from D is 262 (brg of C measured at D) (a) In which hemisphere are C and D?

CONVERSION ANGLE (CA)

CONVERSION ANGLE = DIFFERENCE BETWEEN GREAT CIRCLE AND RHUMB LINE Conversion Angle (CA) is used to change Great Circle bearings and tracks into Rhumb Line bearings and tracks or vice versa.

THE GREAT CIRCLE IS ALWAYS NEARER THE POLE THE RHUMB LINE IS ALWAYS NEARER THE EQUATOR

CONVERSION ANGLE CONVERGENCE CONVERGENCE CONVERSION ANGLE CONVERSION ANGLE CONVERGENCE = = = = = = ½ CONVERGENCE

TWICE CONVERSION ANGLE

CHANGE OF LONGITUDE x SIN MEAN LATITUDE ½ CHANGE OF LONGITUDE x SIN MEAN LATITUDE

DIFFERENCE BETWEEN GREAT CIRCLE AND RHUMB LINE DIFFERENCE BETWEEN INITIAL AND FINAL GREAT CIRCLES

The Rhumb Line is a constant direction. If the Rhumb Line track from A to B is 100º, then the Rhumb Line track from B to A is 280º. You can always take the reciprocal of a Rhumb Line, NEVER A GC.

Initial GC track A to B is 080° GC, initial GC track B to A is 300° GC (Conv. angle 20°) Example 3 The Great Circle bearing of A from B is 255 GC

The Rhumb Line bearing of B from A is 084 RL

Example 4 The Great Circle bearing of X from Y is 072 GC The Rhumb Line bearing of Y from X is 259 RL What is the great circle bearing of Y from X?

DISTANCE Kilometre (KM.)

A Kilometre is 1/10 000 th. part of the average distance from the Equator to either Pole. It is generally accepted to equal 3280 feet.

Statute Mile (SM)

Defined in British law as 5280 feet. Nautical Mile (NM)

A Nautical Mile is defined as the distance on the surface of the earth of one minute of arc at the centre of the earth. As the earth is not a perfect sphere the distance is variable.

At the Equator I NM is 6046.4 feet At the pole 1 NM -Is 6078 feet For navigation purposes the Standard Nautical Mile is 6080 feet (South Africa and UK)

ICAO 1 NM = 1852 metres or 6076.1 feet

Most navigational electronic calculators use I NM = 6076.1 feet. To answer questions in the CAA examinations any of the following may be used :-

1 NM = 6080 feet or 1852 metres Conversion Factors 1 Foot = 12 inches

1 Inch = 2.54 Centimetres

As one minute of arc is 1 NM, then Great Circle distance along a Meridian can be calculated. One minute of Latitude is 1 NM and 1Degree of Latitude is 60 NM.

The Great Circle distance from N75:30 E065:45 to N82:15 W114:15 is:-

As W114:15 is the anti-meridian of E065:45 the Great Circle distance is along a Meridian over the Pole where 1 of Latitude equals 1 nm.

N 75:30 to the Pole = 1430' change of Latitude (14=x 60 = 840 nm+30 nm) = 870nm

Pole to N 82:15 = 745' change of Latitude (7 x 60 = 420nm + 45nm) = 465nm + 870nm

= 1335 nm

CHANGE OF LONGITUDE (CH. LONG) or DEPARTURE DISTANCE

Departure is the distance in Nautical Miles along a parallel of Latitude in an East-West direction. At the Equator, two meridians (5W and 5E) have a change of Longitude of 10 of arc. As the Equator is a Great Circle, 10 of arc equals 600 nautical miles. As Latitude

The departure between any 2 points is thus a function of their latitudes and the change of longitude, and the relationship is given by

Where mean lat = lat A + lat B 2

E 032:45 W 067:25 Both East or West SUBTRACT E 021:15 E 027:30 One East & One West ADD 11:30Ch.Long 94:55 Ch. Long

DEPARTURE = CHANGE of LONGITUDE (in minutes) x COSINE LATITUDE Example 1 The distance from A (N 20:10 E 005:00) to B (N 20:10 W 005:00) is :-

Departure = Ch. Long x cos Lat = 10 x 60 x cos2010' = 600 x cos 20.1667 = 563.2163 nm

Example 2 An aircraft leaves A (E 012:30) and flies along the parallel of S 29:30 in an Easterly direction. After flying 1050 nm its Longitude is :-

Departure = Ch. Long x cos Lat 1050nm = Ch. Long xcos2930' Ch Long = 1050 nm

Example 3 An aircraft in the Northern Hemisphere flies around the world in an Easterly direction at an average groundspeed of 515 Kts in 14 hours. The Latitude at which the aircraft flew was :-

Departure = Ch. Long cos Latitude GS 515 x 14 Hrs. = 360 x 60 x cos Lat

7210

21600 = cos Lat = 70 30’ N

DISTANCE ALONG A PARALLEL OF LATITUDE IS DEPARTURE DISTANCE ALONG A MERIDIAN IS CHANGE OF LATITUDE

As a Meridian is a Great Circle, then the arc of Change of Latitude can be converted into nautical miles.

Example 4 The shortest distance from A (N 78:15 W 027:13) in B (N82:30 E 152:47) is :- As E 152:47 is the anti-meridian of W 027:13, A to B is the arc of a Great Circle.

N 78:15 to the North Pole = 11:45 Change of Latitude North Pole to N 82:30 = 7:30 Change of Latitude

_____

19:15 Change of Latitude 19° x 60 = 1140nm + 15 minutes = 1155nm shortest (GC) distance A to B

Example 5 An aircraft departs A (N 25:13 W017:25) and flies a track of 090°(T) at GS 360 for 1 hour 35 minutes. Then the aircraft flies a track of l80° (T) for I hour 55 minutes and arrives at position;

Departure = Ch. Long x cos Latitude N 25:13

W017:25;

Track 180°

Change of Latitude

Departure = Ch. Long x cos Latitude Departure

cos Lat = Ch. Long GS360 x 1:35

cos 25:13 = 630 minutes of Longitude = 10°30-East of W 017:25 = W006:55 GS360 x 1:55 = 690nm =11°30 t= N 13:43

RADIO BEARINGS

VHF D/F VERY HIGH FREQUENCY - DIRECTION FINDING VDF

Major airports in South Africa have a VDF service, it is usually on the Approach frequency and will provide radio bearings to aircraft on request. The aircraft transmits on the appropriate frequency and direction finding equipment at the airport will sense the direction of the incoming radio wave. The bearing will be passed to the aircraft in Q-code form.

Q CODE QTE

QDR QUJ QDM

TRUE bearing FROM the VDF station MAGNETIC bearing FROM the VDF station TRUE track TO the VDF station

MAGNETIC track TO the VDF station

Take the shortest route to change one bearing to another QTE ± 180 = QUJ

QDR ± 180 = QDM

QDR ± Variation = QTE

VOR VOR Radials are Magnetic bearings from the VOR = QDR RMI Readings are Magnetic tracks to the VOR = QDM

RMI BEARINGS (VOR & ADF)

ADF BEARINGS

ADF Relative bearings are measured from the Fore and Aft axis of the aircraft.

ADF Relative bearings must be converted into True Bearings (QTE) before they can be plotted on a chart.

RELATIVE BEARING + TRUE HEADING = QUJ  180 = QTE

MAGNETIC VARIATION AT THE AIRCRAFT IS ALWAYS USED WITH ADF BEARINGS Lets demonstrate it to you..

ADF bearing 095 Relative Heading (T) + 057

QUJ 152 (T) TO NDB

 180

QTE 332 (T) FROM NDB

ADF bearing 200 Relative Heading (T) 318 QUJ 518 Subtract 360 QUJ 158 (T) TO NDB  180 QTE 338 (T) FROM NDB

QUESTIONS

1. The Initial Great Circle track from A to B is 067 The Initial Great Circle track from B to A is 263 The Rhumb Line track from A to B is:-

(a) 059 (b) 075 (c) 083

2. The Initial Great Circle track from C to D is 097 The Initial Great Circle track from D to C is 263 The Rhumb Line track from D to C is :-

(a) 256  (b) 262 (c) 270

3. The Great Circle bearing of A from B is 255 The Rhumb Line bearing of B from A is 084 The Great Circle bearing of B from A is :- (a) 093

(b) 096 (c) 099

4. The Great Circle bearing of X from Y is 072 The Rhumb Line bearing of Y from X is 259 The Great Circle bearing of Y from X is :- (a) 262

(b) 266 (c) 270

5. The initial great circle track from A (S 30:45 E 045:15) to B(S 30:45 E 062:38) is (a) 085.6

(b) 094.4 (c) 098.9

6. The initial great circle track from A (S 28:30 W 015:15) to B is 099 If A and B are on the same parallel of latitude the longitude of B is :-

7. The latitude where the value of convergency is half the value of convergency at 60 N is

(a) 30 00’ N (b) 27 52’ N (c) 25 39’ N

8. The Initial Great Circle Track from A (S 27:30 E 017:45) to B (S 27:30 E 029:15)is: (a) 092.65 GC

(b) 087.35 GC (c) 095.31 GC

9. A and B are in the Northern Hemisphere. The Great Circle bearing of B from A is 068.

If Conversion angle is 6 the Great Circle bearing of A from B is :- (a) 242 GC

(b) 254 GC (c) 260 GC

10. A and B are in the Southern Hemisphere. The Great Circle bearing of A from B is 245

If Conversion Angle is 7 the Great Circle bearing of B from A is :- (a) 079 GC

(b) 065 GC (c) 051 GC

11. A and B are in the same hemisphere.

The Rhumb Line bearing of A from B is 100 The Great Circle bearing of B from A is 275 The Great Circle bearing of A from B is :- (a) 105 GC

(b) 100 GC (c) 095 GC

12. A and B are in the same hemisphere. A bears 080 GC from B

B bears 255 GC from A

The Rhumb Line Track from B to A is :- (a) 077.5 RL

13. A and B are in the same hemisphere. The bearing of B from A is 143 GC The bearing of A from B is 319 GC The Rhumb Line Track from A to B is :- (a) 141 RL

(b) 145 RL (c) 147 RL

14. The Latitude where the Convergency between two meridians is twice the value of their Convergency at 20 N is :-

(a) N 42:45 (b) N 43:10 (c) N 43:16

15. A and B are in the same hemisphere. The Great Circle bearing of A from B is 080 The Great Circle bearing of B from A is 270 The Rhumb Line Track from B to A is :- (a) 075RL

(b) 080RL (c) 085RL

16. A and B are in the same hemisphere. The Great Circle bearing of A from B is 290 The Great Circle bearing of B from A is 118 The Rhumb Line bearing of A from B is :- (a) 294RL

(b) 298RL (c) 302RL

17. A and B are in the same hemisphere. The bearing of A from B is 280 GC The bearing of B from A is 095 RL The Great Circle bearing of B from A is (a) 090GC

(b) 100GC (c) 105GC

18. Two positions on the same parallel of Latitude are in the Northern Hemisphere. A at 171E and B at 173 W have a 8 angle of Convergency between them. The Initial Great Circle Track from A to B is :-

(a) 086GC (b) 090GC (c) 094GC

19.

Two positions on the same parallel of Latitude are in the Northern

Hemisphere.

A at 171 E and B at 173 W have a 8 angle of Convergency between them. The Latitude of A is :-

(a) 25N (b) 30N (c) 35N

20. The position of A is S 30:00 W 010:00. Position B is on the same parallel of Latitude. The Initial Great Circle Track from A to B is 256.

The Longitude of B is :- (a) 56W

(b) 61W (c) 66W

21. The position of A is N 42:13 W 158:24. Position B is on the same parallel of Latitude. The Great Circle bearing of B from A is 278.

The Longitude of B is :- (a) E 175:13

(b) E 182:13 (c) E 177:47

22. The distance from A (S 27:43 W 005:15) to B(S 27:43 E 018:29) is :- (a) 703 nm

(b) 1261 nm (c) 1452 nm

23. An aircraft departs C (N 45:17 E 025:52) on a track of 270° (T) and arrives at D after a flight of 456 nm. The Longitude of D is :-

(a) E 015:04 (b) E 016:32 (c) E 017:25

24. An aircraft in the Southern Hemisphere flies around the world in 16 hours 35 minutes at GS 478. The Latitude at which the aircraft flew was :-

25. An aircraft leaves ORLANDO. FLORIDA (N 28:32 W 081 :20) at 08:20 Z, GS 375. The ETA at TENERIFE. CANARIES (N 28:32 W 016:16) is :-

(a) 17:29 Z (b) 17:44 Z (c) 17:57 Z

26. An aircraft departed ABU ‘DHABI (N 24:23 E 054:44) at 06:52 Z and arrived at KARACHI (N 24:23 E 067:11) at 08:17 Z.

The average groundspeed for the flight was :- (a) 470 Kts

(b) 480 Kts (c) 490 Kts

27. An aircraft leaves X (N 57:42 E 030:15) on a Rhumb Line track of 270° (T) After flying 1037 nm the Longitude of the aircraft is :-

(a) W 002:06 (b) W 003:38 (c) W 004:29

28. An aircraft flies 425 nm along the parallel of Latitude N 46:52. The change of Longitude is

(a) 10°51’ (b) 10°36’ (c) 10°22’

29. The length of one nautical mile is :- (a) constant

(b) maximum at the poles (c) maximum at the equator

30. The shortest distance from A (N 75:39 E 123:17) to B(N 78:27 W 056:43) is :- (a) 1554 nm

(b) 1672 nm (c) 1739 nm

31. An aircraft leaves X (S 34:58 E 018:24) at 06:30 Z, track 360° (T), GS 300 Kts.

At 07:55 Z the aircraft turns right onto track 090° (T). The longitude of the aircraft at 09:05 is

(a) E 024:30 (b) E 025:00

32. The Latitude where the distance between two meridians is half the distance between the same two meridians at 25° N is :-

(a) N 62:08 (b) N 62:31 (c) N 63:03

33.

An aircraft leaves P (N 32:27 E 027:56) at 17:30 Z. track 270° (T), GS 390 Kts. At E 021:00 it flies due South and passes abeam of Q(N 20:20 E 021:56) at 20:12 Z. The groundspeed on the second leg was :-

(a) 384 Kts (b) 394 Kts (c) 404 Kts

34. An aircraft departs X (S 27:34 W 034:15) at 09:00 Z, track 090°(T). GS 455 Kts. At W 015:00 it flies due North at GS 422 Kts. The ETA abeam of Y (S 19:35 W 013:45) is:

(a) 12:23 (b) 12:33 (c) 12:43

35. Aircraft A and aircraft B depart the same position (60ºN 150ºW). Aircraft A flies a track of 090ºT. Aircraft B flies directly north to the pole then a track of 180ºT so as to intercept aircraft A. The groundspeeds are the same for A and B. At what longitude will aircraft A and B intercept each other?

(a) 90E

(b) 30W (c) 30E

CHAPTER 2

CHART PROJECTION THEORY

The original problem of map making is still with us even in the 21st century, how can you represent the curved surface of the earth on a flat piece of paper without distortion???

The answer is IT CANNOT BE DONE!! It’s the same as trying to flatten out a Orange peel, it too cannot be done.

Charts which are produced by conic projections are used widely in aviation – mainly because conic projections “

1. preserve true shapes

2. preserve angular relationships (called conformal or orthomorphic) 3. have a reasonably constant scale over the whole chart

4. show great circle as straight lines..

Lets now look at the chart projections and properties that we as pilots are interested in: ORTHOMORPHISM

Orthomorphism means true shape. In theory a cartographer starts with a 'reduced earth' which is the earth reduced by the required scale. The 'reduced earth' is a true undistorted representation of the earth. Details, such as Parallels of Latitude, Meridians and topographical features are 'projected' from the reduced earth onto a cylinder (Mercator's Projection), a cone (Lambert's Projection) or a flat sheet of paper (Polar Stereographic Projection). The ideal chart would possess the following features.

 Scale, both correct and constant  Bearings correct

 Shapes correctly shown  Areas correctly shown

 Parallels of Latitude and Meridians will intersect at 90

Unfortunately to reproduce a spherical surface on a flat sheet of paper is impossible. Distortions will occur. Only one of the above features can be shown correctly.

If shapes and areas are approximately correct to enable map reading, then slight distortions can be tolerated.

The 1 nm square on the reduced earth is correct, the diagonal of a square is 45



and

bearings are correct.

The 1 nm square of the reduced earth projected onto a cylinder becomes a rectangle. Bearings are no longer correct. The scale has been expanded in the North/South direction to a greater degree than the East/West case. To overcome this problem the scale expansion North/South is reduced mathematically to equal the scale expansion East/West. The rectangle becomes a square and the diagonal is 45 Bearings are now correct. Meridians and Parallels of Latitude intersect at 90 Scale is expanded, but by the same amount in all directions over short distances. Shapes and areas are approximately correct and the chart is orthomorphic. On the Mercator, Lambert and Polar Stereographic charts the Parallels of Latitude are adjusted in the above manner. Bearings are correct but the scale is variable. SCALE

Scale is the ratio of a line drawn on a chart to the corresponding distance on the surface of the earth.

Statement In Words 1 inch equals 40 nm

Usually found on radio facility charts. 1 inch on the chart equals 40 nm.

Graduated Scale Line

0 10 20 30 40 50 60 70 80 90 100 1_____1_____1_____1_____!_____1_____1_____1_____1______1

Representative Fraction

1

1000 000 or 1/1 000 000 or 1:1 000 000  1 Unit on the chart equals 1 000 000 units on the earth

 1 Centimetre on the chart equals 1 000 000 centimetres on the earth .  1 Inch on the chart equals 1 000 000 inches on the earth

SCALE FACTOR

Due to the inherent difficulty of presenting a spherical object (the earth) on a flat sheet of paper. there is no such thing as a constant scale chart. Scale expansion or contraction will occur. Usually scale will be correct at a certain Latitude but expands else where. For example :-

Mercator Chart Scale 1:1 000 000 at the Equator

What is the scale at 40N with a Scale factor of 1.3054 1

_______ x Scale factor 1.3054 1 000 000

Scale at 40N = 1: 766 049

Example 1 A chart has a scale of 1:2 500 000. How many nautical miles are represented by 4 cm on the chart?

Chart Length (CL) 1 4 cm Scale = ________________ ________ ______

Earth Distance (ED) 2 500 000 ED ED = 2 500 000 x 4 cms

2 500 000 x 4 cms Divide by 2.54 = Inches ______________ = 53.96nm Divide by 12 = Feet

2.54 x 12 x 6080 Divide by 6080 = Nautical Mile; Example 2 32 centimetres on a chart represents 468 nm. The scale of the chart is :

CL 32 cms 1 Scale = _________________________ = _______

Example 3 The scale of a chart is 1: 3 500 000. The length of a line that represents 105 nm is :- CL 1 CL Scale = ___ = ________ = __________________________ ED 3 500 000 105 nm x 6080 x 12 x 2.54 = 3 500 000 x CL = 105 nm x 6080 x 12 x 2.54 105 nm x 6080 x 12 x 2.54 CL = _____________________ = 5.56 cms 3 500 000

Example 4 Chart A has a scale of 1:2 500 000 Chart B has a scale of 1:1 750 000 Which chart has the larger scale?

1 1 Chart B has the larger scale ___  ___

2 4

LAMBERT CONFORMAL CONIC CHART

The Lambert's chart was developed from the Simple Conic chart. Simple Conic

A cone is placed over a reduced earth so it is tangential to a selected parallel of latitude. The apex of the cone is above the pole. A light source at the centre of the reduced earth projects details onto the cone. The cone is opened to give a simple conic projection.

The scale is correct at the parallel of tangency (45N) and expands north and south of 45N. Due to the scale expansion the chart is not suitable for navigation.

The Meridians are straight lines converging on the nearer pole and the value of convergence is constant throughout the chart.

SIMPLE CONIC CONVERGENCE

When the cone is opened, 360 of Longitude is represented by the angular extent of the chart which is 254.5584. The angular extent of the chart is controlled by the latitude chosen to be the parallel of tangency.

Angular extent of the chart 254.5584

______________________________ = 0.7071 Constant of the Cone or 'n' factor Change of Longitude 360

Two Meridians 1 apart have a convergence 0.7071

and this is called the: CHART CONVERGENCE FACTOR (CCF)

Parallel of Tangency 45 Sine 45 = 0.7071 = CCF = Constant of the Cone = 'n' factor LAMBERT CONFORMAL CONIC CHART

The Lambert's chart is based on the simple conic and is produced mathematically from it. Firstly, the scale is reduced throughout the chart. Since scale on the simple conic is correct only on the parallel of tangency and expands either side, the reduction will give two Standard Parallels (SP) on which scale is correct, one on either side of the simple conic parallel of tangency which is renamed the Parallel of Origin (// 0). Further mathematical modification is applied by adjusting the radius of the parallels of latitude to produce an orthomorphic projection.

The above can be shown be lowering the simple conic cone so that it cuts the earth at the two Standard Parallels instead of the original parallel of tangency of the simple conic.

Lambert's Chart Properties

PARALLELS OF LATITUDE Arcs of circles, radius the Pole, unequally spaced. MERIDIANS Straight lines converging towards the nearer Pole SCALE Correct at the two Standard Parallels

Expands outside the Standard Parallels Contracts between the Standard Parallels

Scale variation throughout 1:1 000 000 and 1:500 000 charts is negligible and can be considered constant if the band of Latitude projected is small and the Standard Parallels are positioned according to the one sixth rule. That is one sixth of that Latitude band from the top and bottom of the chart. Charts of the North Atlantic with a scale of 1:5 600 000 have a marked scale variation and care must be taken when measuring distances.

CONVERGENCE Constant throughout the chart Correct at the Parallel of Origin Chart Convergence Ch. Long x sin Parallel of Origin

Chart Convergence Ch. Long x CCF (Chart Convergence Factor) Chart Convergence Ch. Long x 'n'

Chart Convergence Ch. Long x Constant of the Cone SHAPES and AREAS Slight distortion

CHART FIT Charts of the same scale and Standard Parallels will fit N/S and E/W. Charts with different SP will not fit.

Lambert's Chart - Tracks

For all practical purposes the Great Circle is a straight line.

The Rhumb Line track is parallel to the mean Great Circle track at the Mid Meridian between two positions

The difference between the Great Circle and the Rhumb Line is :

Chart Conversion Angle (CCA)

The difference between the Initial Great Circle track and the Final Great Circle track is Chart Convergence (CC)

NB: For examination purposes

Unless otherwise stated in a question, the Great Circle is taken to be the straight line and Chart Convergence (CC) is used.

 Where a question asks for 'the most accurate value of the Great Circle' or 'the true Great Circle' then Earth Convergence (EC) is used.

 The Parallel of Origin of a Lamberts chart is mid way between the two Standard Parallels

 If the Standard Parallels (SP) are 20S and 40S, then the Parallel of Origin (// 0) is 30S

 If one SP is 20S and the // 0 is 30S - Then the other SP is 40S

 Chart Convergence (CC) = Change of Longitude x sine Parallel of Origin  Chart Convergence (CC) = Change of Longitude x Chart Convergence Factor  Sine Parallel of Origin = Chart Convergence Factor (CCF)

e.g. a Lamberts chart has a chart convergence of 5 between the meridians of 10E and 20E) then the Parallel of Origin can be calculated (CC 5 = ch. long 10 x sin 30) and the CCF = 0.5 As convergence is proportional to the CCF, convergence between any two meridians is easily found.

Lets look at some problems that can arise in the examination..

Q1. On a North Hemisphere Lamberts chart (SP 20N & 40 N) the initial GC track from A (10E) to B (42E) is 065. The GC track at B is:

CC = Ch Long X sin // O = 32 X sin 30 = 16

65 + 16  = 81 GC at B

Q2. The Chart Convergency factor of a Lamberts chart is .5. The Great Circle track from C (20N 10E) to D (45N 30 W) measures 316 at C. The Rhumb Line Track from C to D is:

CC = CH Long X CCF Track C to D 316 GC

CC = 40 X 0.5 CCA 10

CC = 20 Track C TO D 306

CCA = 10

Q3 The CCF of a Lambert's chart is 0.5

If one Standard Parallel (SP) is 25S then the Latitude of the other Standard Parallel D C GC RL A B 65 _65 CC 16 10E 42E

Lambert's Chart Plotting Radio Bearings

Radio bearings are Great Circles. Straight Lines on a Lambert's chart are Great Circles and plotting radio bearings is simple.

VOR & VDF

VOR and VDF bearings are determined at the station, that is measured from the Meridian of the station.

VOR Radial (QDR) Correct for VOR station Variation only and plot from VOR Meridian VOR RMI reading RMI reading is a QDM VOR Variation 180 = QTE

Plot QTE from VOR Meridian (do not apply compass deviation) VDF QDM = Station Variation 180 = QTE

QDR Station Variation = QTE QUJ ± 180 = QTE

Plot QTE from VDF station Meridian Plotting ADF Bearings

A problem arises with the plotting of ADF bearings due to the bearing being measured at the aircraft's Meridian and plotted from the NDB's Meridian which differ by the amount of Convergence between the two positions.

ADF QUJ ± 180 = Bearing to plot from Aircraft Meridian paralleled through NDB ADF QUJ ± 180 ± CC = Bearing to plot from NDB Meridian

Scale Problems

Lambert's scale 1:2 500 000, SP20 S and 40S. The scale is correct at the two Standard Parallels Scale 20S = Scale at 40S

Some problems that may arise..

Example 1 A Lambert's chart has Standard Parallels of 30N and 50 N

The Rhumb Line distance from A (50N 30E)to B (50N 10E) is 13.75 inches. The scale at 30N is :- CL 13.75 inches 1 Scale = __ = ________________________________ = ________ ED 20 Ch. Long x 60 x cos 50 x 6080 x 12 4 092 898 (Departure in nm)

Example 2 On a Lambert's chart the Standard Parallel of 35S measures 58.4 cms. The other Standard Parallel measures 43.9 cms.

The Latitude of the second Standard Parallel is :-

CL 58.4 cms CL 43.9 cms Scale at 35S = _________________ Scale at 2nd SP = _______________

ED Ch. Long x cos 35 ED Ch.Long x cos Lat The scales are equal.

As CH. Long is the same in both equations it disappears

58.4 cms 43.9 cms ___________ = ____________ cos 35 cos Lat

cos lat = 0.6158

lat = (.6158)COS-1 lat = 52S

Lambert's Questions

1. The convergence between the meridians of 37E and 59E at 30S on a Lamberts chart (Standard Parallels 29S and 41S) is :-

(a) 11.0 (b) 11.8 (c) 12.6

2. Chart convergency on a Lamberts chart between the meridians of 10E and 10W is 12. If one standard parallel is S 30:20 the other standard parallel is at :-

(a) S 40:48 (b) S 41:52 (c) S 43:24

3. On a Lamberts chart the standard parallel of 32N measures 77.5 cms. The other standard parallel measures 72 cms. The latitude of the second standard parallel is :- (a) 37N

(b) 38N (c) 39N

4. A Lamberts chart has standard parallel of 20S and 40S. The rhumb line distance from A (20S 15E) to B (20S 37E) is 76.62 cms. The scale of the chart at 40S is :-

(a) 1:3 000 000 (b) 1:3 500 000 (c) 1:4 000 000

5. A Lamberts chart has standard parallels of 25S and 45S. The rhumb line distance from X (45S 17W) to Y (45S 05E) is 80.5 cms.

The rhumb line distance from 25S 17W to 25S 05E is :- (a) 97 cms

(b) 100 cms (c) 103 cms

6. A Lamberts chart has standard parallels of 30N and 50N.

The initial great circle track from A (32N 63W) to B (45N 15W) is 056(T). The longitude at which the great circle track becomes 090(T) is :-

(a) W 012:37 (b) W 010:06 (c) W 008:28

7. A northern hemisphere Lamberts chart has a CCF of 0.57.

A straight line from A (27W) to C (32E) passes through B (09E). The initial great circle track at A is 062(T). The great circle track at B is :-

(a) 082.5 (b) 087.2 (c) 093.8

8. An aircraft heading 321(T) in the northern hemisphere receives an ADF bearing of 094 relative. If chart convergency between the aircraft and the NDB is 5 the bearing to plot from the NDB on a Lamberts Chart is :-

(a) 230 (b) 235  (c) 240

9. An aircraft heading 112(T) in the southern hemisphere receives an ADF bearing of 156 relative. If chart convergency between the aircraft and the NDB is 4 the bearing to plot from the NDB on a Lamberts Chart is :-

(a) 084 (b) 092 (c) 096

10. A Lambert's chart has a chart convergency of 12.04 between the Meridians of 70E and 84E. If One Standard Parallel is 64N, the Latitude of the other SP is :-

(a) N59:19 (b) N57:15 (c) N54:38

11. On a Lambert's chart, a straight line from A (45N 045E) to B (47N 035E) cuts the 40E Meridian at 75. Convergency between A and B is 8. The Rhumb Line track from A to B is :-

(a) 255 (b) 270 (c) 285

12. On a Lambert's chart, Convergency between A(50S 005E)and B (52S 015E) is 8. The Great Circle track from A to C (50 S 020 E) is :-

(a) 084  (b) 090 (c) 096

13. A bearing obtained from a NDB is 273 relative. Aircraft heading 330 (T). d.long between aircraft and NDB is 14, mean Latitude 26S. Parallel of Origin 30S. The bearing to plot on a Lambert's chart is :-

(a) 063 (b) 070 (c) 077

14. Two Meridians. 175W and 171E, in the Northern Hemisphere, have a convergency of 7 on a Lambert's chart. The GC track from X (169E) to Y (175W) is 080(T) at X. The Rhumb Line track from Y to X is :-

(a) 256 (b) 260 (c) 264

15. A Northern Hemisphere Lambert's chart has a CCF of 0.75. The straight line from A (20E) to C (60E) passes through B (044E). The direction of the track at A is 100 (T). The Great Circle track at B is :-

(a) 109 (b) 115  (c) 118

16. A Lambert's chart has Standard Parallels of 24S and 46S. Position X (46S 045W) and Y (46S 025W) are plotted on the chart. The Great Circle bearing of Y from X is :-

(a) 095.74 (b) 096.94 (c) 097.19

17. The chart convergency factor of a Lambert's chart is 0.50. A straight line is drawn from X (20N 010E) to Y (40N 030W) and measures 305 (T) at X. The RL track from X to Y is :-

(a) 295 (b) 305 (c) 315

18. An aircraft heading 173 (T) obtains a relative bearing of 313 from a NDB. Convergency between aircraft and NDB is 14. What bearings would you plot from the NDB on a Lambert's chart in:

1. Northern Hemisphere? 2. Southern Hemisphere?

19. Positions X and Y are in the Southern Hemisphere. An aircraft at X. heading 240 (T). obtains a relative bearing of 198 from a NDB at Y. When plotted as a straight line from the NDB on a Lambert's chart, the bearing measures 250 (T).

The RL track from X to Y is :- (a) 074

(b) 078 (c) 082

20. The Standard Parallels on a Lambert's chart are 10N and 30 N. A straight line is drawn from Position A (30 N 160W) to B (10 N 150 E) on a Mercator and a Lambert's chart. The straight line cuts the 180 Meridian at 244 on the Mercator. The GC tracks from A to B on the Lambert's chart at 160 W and 170 E are :-.

160 W 170 E (a) 249.67 (b) 252.55 (c) 254.05 (a) 241.41 (b) 242.29 (c) 243.84

21. Positions X (45 N 010 W) and Y (20 N 060 E) are joined by a straight lines on a Mercator and Lambert's chart. Lambert's: SPs 20 N and 45 N. Mercator track X to Y is 124 (T) at 40 E. The Longitude that the track on the Lambert's chart equals 124 (T) is :-

(a) 23E (b) 25E (c) 27E

22. On a Northern Hemisphere Lambert's chart the Initial Great Circle track from A to B is 082(T). An aircraft leaves A steering a constant heading of 082(T) in zero wind conditions. The aircraft will pass :-

(a) North of B (b) Overhead B (c) South of B

23. A Lambert's chart has Standard Parallels of 15S and 35S. The difference between Chart Convergency and Earth Convergency of Meridians 22 apart at 34S is :-

(a) 2 (b) 3 (c) 4

24. A Lambert's chart has Standard Parallels of 25N and 45N. A straight line is drawn on the chart from X (42N 15W) to Y (43N 15E). The true Great Circle between X

25. Parallels of Latitude on a Lambert's chart are :- (a) Parallel straight lines unequally spaced (b) Arcs of circles equally spaced

(c) Arcs of circles unequally spaced

26. Lambert's charts of the same scale and Standard Parallels will fit :- (a) N/S only

(b) E/W only

(c) both N/S and E/W

27. A Lambert's chart has a scale of 1:2 500 000. The chart length of 1 of Latitude varies as follows :- Latitude 70N 65N 60N 55N 50N 45N

Chart length of 1 of Latitude 4.471 cms 4.441 cms 4.422 cms 4.441 cms 4.471 cms 4.516 cms

The Standard Parallels are :-

(a) 54N & 66N (b) 55N & 67N (c) 56N & 68N

MERCATOR CHART

Before the advent of Inertial Navigation, and GPS computers aircraft flew constant headings. They flew Rhumb Lines. The Mercator chart was constructed so that Rhumb Lines are straight lines and the headings flown were easily plotted.

A cylinder is positioned over the reduced earth tangential to the Equator. A light source at the centre of the reduced earth projects details of the reduced earth onto the cylinder and we have a Geometric Cylindrical Projection. After adjusting the Parallels of Latitude so that the scale expansion North/South equals the scale expansion East/West it becomes a Mercator chart.

MERCATOR CHART PROPERTIES

POINT OF PROJECTION Centre of the reduced earth POINT OF TANGENCY Equator

PARALLELS OF LATITUDE Parallel straight lines, unequally spaced MERIDIANS Parallel straight lines, equally spaced

CONVERGENCY Constant

Value Zero

Correct at the Equator

SCALE Correct at the Equator

Expands as the secant of the Latitude RHUMB LINES Straight Lines

GREAT CIRCLES Complex curves towards the nearer Pole Convex to the Pole, Concave to the Equator

SHAPES & AREAS Approximately correct, excellent between 12N and 12S becoming distorted with increasing Latitude. The chart has a limit of 70N and 70S.

CHART FIT Charts of the same equatorial scale will fit N/S. E/W and diagonally.

USES Plotting and Met charts

Topographical maps between 12N and 12S ADVANTAGES Rhumb Lines are straight lines - plotting easy DISADVANTAGES Great Circles (radio bearings) are complex curves.

Great care must be taken measuring distances due to rapidly changing scale.

SCALE

Scale is correct at the Equator and expands North and South as the secant of the Latitude. Every Parallel of Latitude has its own scale. (SF x cos LAT)

Equator 1:2 000 000 5S 1:1 992 389 10S 1:1 969 615 30S 1:1 732 051 60S 1:1 000 000

Great care must be taken when measuring distances on a Mercator chart due to the variable scale. Use the Latitude scale at the mid point between the two positions.

SCALE PROBLEMS

Scale problems are easily solved by use of ABBA

SCALE DENOMINATOR A x COS B = SCALE DENOMINATOR B x COS A

Example 1 The scale of a Mercator chart is l:2500000 at 15S. 15S = A What is the scale at 45N? 45N = B

SCALE DENOMINATOR A x COS B = SCALE DENOMINATOR B x COS A 2 500 000 x cos 45 = Scale B x cos 15

2 500 000 x cos 45

cos 15 = 1 830 127

Scale at 45N 1:1 830 127 Example 2 The scale of a Mercator chart is 1:3 500 000 at 10N 10N = A At what Latitude is the scale 1:2 500 000? Lat X = B SCALE DENOMINATOR A x COS B = SCALE DENOMINATOR B x COS A

3 500 000 x cos X = 2 500 000 x cos 10 cos X = 2 500 000 x cos 10 3 500 00 = 0:7034 X = (0.7034)cos -1 = 45 17'49" N/S

Example 3 The Meridian spacing on a Mercator chart is 2.7 cms. The scale at 30S is If ABBA cannot solve the problem, then revert to:-

CL 2.7 cms Scale = __ = _______________________________ ED 1 Long x 60 x cos 30 x 6080 x 12 x 2.54 = 2.7 (scale is 1/xxxxxx) 9 629 426 = 1:3 566 454

PLOTTING RADIO BEARINGS VDF & VOR

Radio bearings are Great Circle bearings. They have to be converted into Rhumb Line bearings by applying Conversion Angle before they can be plotted.

Both VDF and VOR bearings are measured at the station, thus station variation must be applied. Conversion angle is also applied where the bearing was measured, that is the VDF or VOR station.

VDF & VOR APPLY STATION VARIATION APPLY CA TO QTE

BEARING VARIATION HEMISPHERE CA PLOT RL

QDM 100 10 W NORTHERN 2

QDR 258 16 W SOUTHERN 3

NOTE

VOR RMI readings are QDM's. Apply VOR station variation, but not compass deviation.

Answers to VOR/VDF bearings

QDM 100 VAR 10 W QUJ 090 180 GC QTE 270 N CA -2 RL QTE 268 QDR 258 VAR 16 W GC QTE 242 S CA +3 RL QTE 245 QDM 113 VAR 15 E QUJ 128 180 GC QTE 308 N CA -4 RL QTE 304 QDM 088 VAR 11 W QUJ 077 180 GC QTE 257 S CA +2 RL QTE 259 QDR 129 VAR 20 E GC QTE 149 N CA +3 RL QTE 152 QDM 285 VAR 14 W QUJ 271 180 GC QTE 091 S CA -4 RL QTE 087 QUJ 131 180 GC QTE 311 N CA -1 RL QTE 310 PLOTTING ADF/NDB BEARINGS

ADF bearings are presented to the pilot by either a RELATIVE BEARING INDICATOR (RBI) or by a RADIO MAGNETIC INDICATOR (RMI).

RELATIVE BEARING INDICATOR (RBI)

ADF bearings are measured clockwise from the fore and aft axis of the aircraft and are termed RELATIVE BEARINGS, that is relative to the aircraft's fore and aft axis. ADF Relative bearings must be converted into True Bearings (QTE) before they can be plotted on a chart,

RELATIVE BEARING + TRUE HEADING = GC QUJ ± CA = RL QUJ ± 180 = RL QTE NB The GC QUJ must be converted into a RL QUJ before the reciprocal is taken. The

reciprocal of a Rhumb Line can always be taken, never the reciprocal of a Great Circle

RADIO MAGNETIC INDICATOR (RMI)

The RMI is a remote gyro compass on which radio bearings (both ADF and VOR) are shown. As it is a compass, the heading index is heading compass and it may suffer from deviation, for which a correction must be made to ADF bearings but not VOR bearings. The sharp end of the pointers are referred to as RMI readings or QDM. The opposite or blunt end of the needle will be a QDR.

ADF QDM ± Deviation ± Aircraft Variation = GC QUJ ± CA RL QUJ ± 180= RL QTE

ADF BEARINGS

HEADING DEVIATION VARIATION BEARING HEMISPHERE CA

016 (C) 2 W 13 E 071 REL NORTHERN 4

065 (C) 3 E 18 W 214 REL SOUTHERN 5

225 (C) 4 W 21 W 069 REL NORTHERN 3

Hdg 016 ( C ) Hdg 065 ( C ) Hdg 225 ( C ) RMI 123 Dev 2 W Dev 3 E Dev 4 W Dev 2 E Hdg 014 (M) Hdg 068 (M) Hdg 221 (M) QDM 125 Var 13 E Var 18 W Var 21 W Var 19 W Hdg 027 (T) Hdg 050 (T) Hdg 200 (T) QUJ 106 GC ADF 071 Rel ADF 214 Rel ADF 069 Rel CA - 2 S QUJ 098 GC QUJ 264 GC QUJ 269 GC QUJ 104 RL CA + 4 N CA +5 S CA - 3 N ± 180 QUJ 102 RL QUJ 269 RL QUJ 266 RL QTE 284 RL  180 180 180

QTE 282 RL QTE 089 RL QTE 086 RL

NB

VOR & VDF APPLY STATION VARIATION DO NOT APPLY DEVIATION APPLY CA TO QTE

RMI READING = QDM

ADF NDB APPLY AIRCRAFT DEVIATION & VARIATION GC QUJ  CA = RL QUJ  180 = RL QTE Station West of ACFT Station East of ACFT

Mercator Questions

1. A Mercator chart has a scale of 1:2 500 000 at 20N. The scale at 50N is:- (a) 1:1 710 100

(b) 1:1 835 000 (c) 1:1 972 060

2. A Mercator chart has a scale of 1:2 000 000 at 20N. The latitude where the scale would be 1:1 500 000 is :-

(a) N 44:11 (b) N 45:11 (c) N 46:1 I

3. The distance between meridians on a Mercator chart 1 apart is 2.7 centimetres The scale of the chart at 33S is :-

(a) 1:3 247 500 (b) 1:3 453 800 (c) 1:3 694 400

4. An aircraft at position S 22:35 E 029:45 obtains an ADF QDM of 095. If variation is 14W and conversion angle 2 the bearing to plot on a Mercator chart is :-

(a) 263 (b) 261 (c) 259

5. An aircraft obtains a QDM of 275 from a VHF D/F station in the Southern Hemisphere. Aircraft Variation is 16W and VHF D/F station Variation is 18W. lf Convergency between the aircraft and the VHF D/F station is 6 the bearing to be plotted on a Mercator chart is :-

(a) 080 (b) 074 (c) 071

6. An aircraft at position S 23:15 E 027:32 (variation 15W) heading 267(T) obtains a relative bearing of 346 Relative from a NDB at S 25:27 E 019:32 (variation 17W). CA 2. The bearing to plot on a Mercator chart is :-

(a) 060 (b) 071 (c) 075

7. On a Mercator chart the distance between two meridians 5º apart is 156 millimetres. The scale of the chart at 32N is :-

(a) 1:3 022 286 (b) 1:3 127 654 (c) 1:3 226 327

8. Two straight lines of equal length are drawn East/West on a Mercator chart. One at 15S and the other at 45N.

(a) The line at 15S represents a greater distance than the line at 45N (b) The line at 45N represents a greater distance than the line at 15S (c) Both line represent the same distance

9. Two straight lines representing 200 nm are drawn East/West on a Mercator chart. One at 22S and the other at 48N.

(a) The line at 22S is longer than the line at 48N (b) The line at 48N is longer than the line at 22S (c) The lines are of equal length

10. The Meridian spacing on a Mercator chart is 3 cms. The Latitude where the scale is 1:2 500 000 is

(a) 42 45' (b) 45 15' (c) 47 35'

11. The Meridians on a Mercator chart are 5,75 cms apart. The ratio of nautical miles to the centimetre at 52N is :-

(a) 6.42 (b) 7.53 (c) 8.64

12. The Parallels of Latitude on a Mercator chart are :- (a) Parallel straight lines equally spaced

(b) Parallel straight lines unequally spaced. (c) Arcs of circles radius the nearer Pole 13. The Meridians on a Mercator chart are :-

(a) Parallel straight lines equally spaced (b) Parallel straight lines unequally spaced. (c) Straight lines converging on the nearer Pole 14. A Great Circle on a Mercator chart is :-

15. Mercator charts of the same scale at the Equator will fit :- (a) N/S only

(b) E/W only (c) in all directions

CHAPTER 3

RELATIVE VELOCITY

Relative Velocity is the comparison of aircraft speeds or the speed of one aircraft relative to another.

The calculations can be broken down into 3 main areas: Aircraft meeting Aircraft overtaking Speed adjustment Meeting Aircraft A GS 240 Kts Aircraft B GS 300 Kts SPEED OF CLOSING 540 Kts Overtaking





Aircraft A GS 340 Kts Aircraft B GS 250 Kts

Aircraft A being faster will overtake Aircraft B SPEED OF CLOSING 90 Kts





Aircraft A GS 220 Kts Aircraft B GS 290 Kts Aircraft B being faster than Aircraft A will fly further ahead of B

Example 1. Aircraft A is overhead NDB PY at 0900 Z enroute to VOR CN. GS 240 Kts Aircraft B is overhead VOR CN at 0920 Z enroute to NDB PY. GS 300 Kts Distance PY to CN is 1150 nm

1150 nm

Note: Times have been rounded off to the nearest minute

Example 2 Aircraft A. GS 180 Kts, passes overhead X at 1200 Z bound for Y Aircraft B, GS 270 Kts, passes overhead X at 1225 Z bound for Y At what time will aircraft B overtake aircraft A?

556nm 595nm

PY

CN

Y

X

B

Example 3 Two aircraft at the same Flight Level following the same route are approaching a VOR.

Aircraft A, GS 350 Kts, is 260 nm from the VOR at 0800 Z. Aircraft B, GS 450 Kts, is 390 nm from the VOR at 0750 Z. At what time must aircraft B reduce to GS 350 in order to :- (a) ensure a 50 nm separation at the VOR?

(b) ensure a 5 minute separation at the VOR?

As aircraft B reduces speed to the same speed as aircraft A it is a 'speed of closing' problem. If aircraft B reduces speed to a different speed than aircraft A it is a 'delay' problem.

390 nms 315nms 260 nms

B B A

0750 0800 0800

5 mins @ 350 kts =29 nms

(a) Speed of closing 100 kts (b) Speed of closing 100 kts Distance to close (55-50) 5 nm Distance to close (55-29) 26 nms

Time to close 3 mins Time to close 16 mins

Reduce speed at 0803Z Reduce speed at 0816 Example 4 An aircraft, GS 450 Kts, estimates overhead 'Delta' at 0915 Z.

ATC requests the aircraft to cross 'Delta' at 0920 Z.

To accomplish this the aircraft reduces speed to 390 Kts at time :-

Delay x Old GS x New GS 5 x 450 x 390

Distance = ______________________ = ___________ = 243.75 nm Difference in GS x 60 60 x 60

GS 450 Dist 243.75 nm Time 32½ mins ETA 0915 - 32½ mins = 0842½ GS 390 Dist 243.75 nm Time 37½ mins ETA 0920 - 37½ mins = 0842½' Alternative solution DISTANCE = SPEED x TIME

At the point where speed is reduced, the aircraft is 'D nm' from Delta. At GS 450 D = 450 x T At GS 390 D = 390 x (T+5) As D is common, then 450 T = 390 (T + 5)

450 T = 390 T + 1950 450 T-390T = 1950

Example 5 A/c A, GS 180 Kts passes over NDB PB 5 minutes ahead of a/c B A/c B. GS 260 Kts. passes over VOR CPL 8 minutes ahead of a/c A. The distance from NDB PB to VOR CPL is :-

As aircraft B overtakes aircraft A. the times are added.

Delay x Old GS x New GS 13 x 180 x 260

Distance = ______________________ = ______________ = 126.75 nm Difference in GS x 60 80x60

Example 6 Aircraft A. GS 250 Kts, passes NDB DN 14 minutes ahead of aircraft B. GS 315 Kts.

Aircraft A then passes VOR PON 5 minutes ahead of aircraft B The distance from NDB DN to VOR PON is :-

As aircraft B does not overtake aircraft A the times are subtracted Delay x Old OS x New GS 9 x 250 x 315

Distance = _____________________ = _______________ = 181.73 nm Difference in GS x 60 65 x 60

QUESTIONS

Q1. An aircraft is at FL 330, at M.85 with IOAT -35ºC W/V 250/45. The aircraft Estimates overhead PWV at 1011Z, track to PWV is 61ºT the local variation is 21ºE. ATC requests that the aircraft change GS to 200 so as to arrive overhead PWV at 1020Z. At what time must the aircraft be slowed down.

a. 1015Z

b. 1005Z c. 1001Z

d. 1008Z

Q2. Aircraft A is overhead PNV at 1010Z FL 240 enroute to JWV, GS 250kts. Aircraft B is overhead JWV at 1020Z FL 330 enroute to PNV, GS 420 KTS. Distance PNV to JWV is 998 NMS. What time will they cross?

a. 11:49Z b. 11:43Z c. 11:46Z d. 11:52Z

Q3. An aircraft flying at FL120, IAS 200 knots, temperature –5° C, wind component +30 knots. At a position 100 nm from the next reporting point the aircraft is ordered to delay arrival by 5 minutes. The immediate reduction in IAS to comply with this order is:

CHAPTER 4

THE SOLAR SYSTEM & TIME

The measurement of the passage of time is based upon observations of events occurring at regular intervals. The two repetitive events which most influence life on Earth are the rotation of the Earth on its axis. Causing day and night, and the movement of the Earth in its orbit around the Sun, causing the seasons.

THE EARTH’S ORBIT

The orbit of a planet around the Sun conforms with Kepler’s Laws of Planetary Motion which state :-

1. The orbit of a planet is an ellipse, with the Sun at one of the foci.

2. The line joining the planet to the Sun, known as the radius vector, sweeps out equal areas in equal in equal intervals of time.

In the above sketch the planet (P) moves anticlockwise in its orbit and is at its closest position to the Sun at position A which is called PERIHELION. At Perihelion the Earth is about 91½ million miles from the Sun and occurs on January 4.

At position C the planet is furthest from the Sun and is known as APHELION. At Aphelion the Earth is about 94½ million miles from the Sun and occurs on July 4.

The mean distance of the Earth from the Sun is about 93 million miles.

According to Kepler’s Law the radius vector sweeps out equal areas in equal intervals of time. If the area SAX equals the area SYC then as the distance AX is greater than the

SAX SYC

A

X

C Y

The Earth completes one orbit around the Sun in about 365.25 days. The plane of the orbit is called the plane of the Ecliptic, and the N/S axis of the Earth is inclined to this plane at an angle of 66½. The plane of The Ecliptic is at an angle of 23½ º to the Earth’s Equator and this angle is known as the obliquity of the ecliptic.

THE SEASONS

One effect of the tilt of the Earth’s axis is the annual cycle of seasons. As the Earth moves around the Sun, on or near 23rd of December the North Pole is inclined away from the Sun, which is vertically above Latitude 23½°N. This is known as winter solstice and is midwinter in the Northern Hemisphere and midsummer in the Southern Hemisphere.

As the Earth travels around its orbit, being a gyro. Its axis will always point in the same direction relative to space and will reach a point at the summer solstice, on or about 22nd June, when the Sun is vertically overhead Latitude 23½°N. It is then midsummer in the Northern Hemisphere and midwinter in the Southern Hemisphere.

Between these dates the Sun. will be overhead the Equator. These events occur on 21st March which is the spring or vernal equinox, and 23rd September which is the autumn equinox. Jan 4 Mar 21 Jun 22 July 4 Sep 23 Dec 23 Perihelion

Vernal or Spring Equinox Summer Solstice

Aphelion

Autumn Equinox Winter Solstice

Sun 91½ million miles

Sun overhead Equator Declination 00:N/S

Sun overhead Tropic of Cancer Declination 23½°N Sun 94½ million miles

Sun overhead Equator Declination 00:N/S Sun overhead Topic of Capricorn 23½°S

MEASUREMENT OF TIME – THE DAY

The rotation of the Earth on its axis is used as a basis for the measurement of the length of a day. The length of time taken for the Earth to complete one revolution on its axis can be found by taking the time between two successive transits of a fixed point in space over a particular meridian.

Sidereal Day (23 hours 56 minutes 4 seconds)

As stars are at immense distances from the Earth, they can be considered to be at infinity and rays of light from stars can be considered parallel regardless of the position of the Earth in its orbit round the Sun. The time interval between two successive transits of a star or a fixed point in space over a meridian is called a SIDEREAL DAY and is constant at 23 hours 56 minutes and 4 seconds.

Apparent Solar Day

The time interval between two successive transits of the True Sun over a meridian is an Apparent Solar Day.

The Sun and a star are in transit overhead a meridian. After 23 hours 56 minutes and 4 seconds the star is in transit For a second time (a Sidereal Day), rays of light from a star being parallel. Due to the Earth’s orbital speed (approximately 58 000 Kts) it has moved some 1 400 000 nm along its orbit and the Earth has to rotate ‘X’ degrees before the Sun is in transit for a second time. This of course takes time thus an Apparent Solar Day is always longer than a Sidereal Day.

An average of 365 Apparent Solar Days is taken and termed a Mean Solar Day which is 24 hours.

Mean Solar Day

The 24 hour day is based on the Mean Sun. When the Mean Sun is overhead a meridian it is 12:00 Local Mean Time (LMT). Each and every meridian has its own LMT.

The Siderial Day

Because of the relative proximity of the earth to the sun, attempts to measure the

length of the day (one revolution of the earth) are contaminated by the movement of

the earth in its orbit relative to the sun.

To solve this problem, a fixed point in space is chosen which is so enormously distant that the movement of the earth in its orbit relative to this point is basically zero. This point in space is called the Siderial point or the first point of Aries.

The Siderial day then, is defined as two successive transits of the Siderial point at the same meridian. The Siderial day is of constant duration : 23 hours 56 mins 4 seconds.

The Earth rotates on its axis from West to East. It is more convenient to imagine the Earth stationary with the Sun rising in the East and setting in the West.

At the Greenwich Meridian the sun is rising at 06:00 LMT. At 90ºE the sun is overhead at 12:00 LMT.

At 180ºE/W the sun is setting at 18:00 LMT.

At 90ºW it is midnight 24:00 LMT on the 5th LD Local Date or 00:00 LMT on the 6 th LD. The Local Date changes at midnight and also at the International Date Line.

ARC TO TIME

The Earth rotates through 360 in 24 hours. 90 in 6 hours, or 15° per hour, there is a direct relationship between Longitude and LMT. The Conversion of Arc to Time table is available in the Navigation Tables booklet provided in the examination.

The first six columns are degrees of Longitude on the left with the corresponding time in hours and minutes on the right.

10º 0:40 15º 1:00 134º 8:56 314º 20:56 The right hand column gives the time equivalent for minutes of Longitude. 10' Long 40 seconds 16' Long 1 minute 04 seconds

UNIVERSAL CO-ORDINATED TIME (UTC)

UTC is the LMT at the Greenwich Meridian and is used as the standard reference from time keeping for aviation. UTC is the same as GMT (Greenwich Mean Time).

CONVERSION OF LMT TO UTC

LONGITUDE EAST UTC LEAST LONGITUDE WEST UTC BEST

(Tables used for these following questions start on page 65)

Example 1. At position A (N 45:05 E 065:30) it is 13:15 LMT on 23rd March. The UTC at this position is :-

A 13:15 LMT 23 March

E 065:30 Arc to Time 4:22

A 08:53 UTC 23 March

Longitude East - UTC Least UTC must be an earlier time than LMT

Example 2. The time is 06:45 UTC on 21st May GD (Greenwich Date). At position B (S 28:37 W 092:20) the LMT is :-

B 06:45:00 UTC 21°May GD°

W 092:20 Arc to Time 6:09:20

B 00:35:40 LMT 21" May LD Longitude West - UTC Best UTC must be a later time than LMT

Example 3. If the UTC is 15:30 on the 22nd June GD and the LMT at position X is 09:45 on 22nd June, LD the Longitude of X is :-

15:30 UTC 22nd June 09:45 LMT 22nd June

Time difference 5:45 Time to Arc = W 086° 15' Longitude

Example 4. An aircraft departs C (N 45:35 E 010:15) at 15:30 LMT on 15th May LD. Flight time to D (42:37 E 135:45) is 11 hours 18 minutes.

The ETA in LMT is :-

C ETD 15:30 LMT 15th May LD

E 010:15 Arc to Time 0:41

C ETD 14:49 UTC 15th May GD

Flight Time 11:18

LOCAL STANDARD TIME

As every Meridian has a different LMT, LMT is not suitable for civil time keeping. Durban has a different LMT to Johannesburg. Each country has its own standard time factor which is applied to UTC to give local standard time. Standard Time tables appear on page 67 onwards. For GMT (Greenwich Mean Time) read UTC.

List 1 Mainly countries with Easterly Longitude (including Spain & Portugal which are Westerly Long.)

List 2 Countries normally keeping GMT or UTC. List 3 Countries with Westerly Longitude

Apply Standard Times in the same manner as LMT (Long East - UTC Least & Long West - UTC Best) or apply as given at the top of each list. Ignore summer time.

INTERNATIONAL DATE LINE

The International Date Line roughly follows the 180 E/W meridian, with some divergences to accommodate certain groups of South Sea Islands and regions of Eastern Siberia.

GOING EAST 1 DAY LEAST (lose a day) GOING WEST 1 DAY BEST (gain a day)

TIME QUESTIONS

1. An aircraft departs X(S 23:46E 023:45) at 21:00Z on 3 July and arrives at Y(S 29:13 W 066:30) at 05:58 LMT on 4 July. If the distance from X to Y is 4732 nm the average groundspeed was :-

(a) 353 Kts (b) 376 Kts (c) 396 Kts

2. An aircraft departs Ascension (S 07:58 W 014:30) at 22:15 LMT on 15th June LD enroute for Johannesburg (S 26:08 E 028:15). If the flight time is 9 hours 14 minuses the ETA FAJS is :-

(a) 08:31 SAST (b) 10:20 SAST (c) 10:27 SAST

3. An aircraft departs Perth (S31:57 E 115:57) at 0930 LMT on a flight to Mauritius (S20:26 E 057:40). Distance 3207 nm, average groundspeed 427 kts. The LMT of arrival at Mauritius is :-

(a) 13:00 (b) 13:07 (c) 13:14

4. At 08:15 LMT on 19 th Sept Local Date an aircraft leaves A (N27:00 E 035:15). After a flight of 7 hours 27 minutes the aircraft arrives at B (N32:00 W 028:45). The LMT of arrival at B is :-

(a) 11:26 (b) 13:21 (c) 15:16

5. An aircraft leaves Prestwick (N55:00 W 005:00) at 1 115 LMT on 23rd June LD. Flight time to San Francisco (N37:30 W 122:00) is 11 hours 15 minutes. The LMT of arrival at San Francisco is :-

(a) 14:02 (b) 14:22 (c) 14:42

6. An aircraft is to fly from Wellington. New Zealand (S 41:10 E 174:45) to Tahiti. The Standard Time Factor New Zealand is 12 hours) Arrival at Tahiti (S 17:29 W 149:29) must not be later than Sunset 1823 LMT on 5th March LD. If the flight time is 5 hours, the latest local mean time and date at which the aircraft must leave Wellington is :- (a) 10:55 5 th

(b) 11:00 6 th (c) 11:05 5 th