ATPL Navigation

ATP Navigation & Plotting FLIGHT TRAINING COLLEGE ATP DOC 2 & 6 Revision : 1/1/2001 Version 4

ATP NAVIGATION

ATP Navigation & Plotting FLIGHT TRAINING COLLEGE ATP DOC 2 & 6 Revision : 1/1/2001 Version 4

INDEX

ATP NAVIGATION

1. The Earth 01 2. Charts 21 3. Relative Velocity 77 4. Solar System & Time 89 5. The 1 in 60 Rules and General Maths 115 6. Navigational Computer 119 7. Basic Plotting on the Lamberts Chart 133 8. Plotting on the British Isles Chart 157 9. Advanced Plotting 163

Annex A Sample Exams 173 Annex B Answers to Questions 187

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 retrieval system,

without the prior permission of Flight Training College of Africa.

ATP NAVIGATION & PLOTTING FLIGHT TRAINING COLLEGE ATP DOC 2 & 6 Revision : 1/1/2001 Page 1 Version 4

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 forms.

N 27:30 27:30 N 27°30'N 35°25'45"S 35:25:45S


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.

E032:15 32°15' E 32:15 E 65°24W 65°24’38”W 65:24:38 W


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 Circle. 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


CONVERGENCY AND CONVERSION ANGLE

CONVERGENCY 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.

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

Considering the two meridians shown above, one at 20W and the other at 20E. 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.


This is a sine relationship, convergence varies as Sine Mean Latitude. Convergency also varies as the Change of Longitude between the two meridians. The greater the Ch. Long, the greater the convergency.

Convergency = Ch. Long x Sine Mean Latitude Ex 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 E042:17

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

Convergency = 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.

THE MERIDIANS CONVERGE TOWARDS THE NEARER POLE


CONVERGENCY = CHANGE OF LONGITUDE x SIN MEAN LATITUDE CONVERGENCY = DIFFERENCE BETWEEN INITIAL AND FINAL GC TRACKS


Q 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?

Q2. C and D are in the same hemisphere

The Great Circle bearing of D from C is l36° (bearing of D measured at C) The Great Circle bearing of C from D is 262° (bearing of C measured at D) (a) In which hemisphere are C and D? (b) What is the value of Convergency between 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 = ½ CONVERGENCEY CONVERGENCY = TWICE CONVERSION ANGLE CONVERGENCY = CHANGE OF LONGITUDE° x SIN MEAN LATITUDE CONVERSION ANGLE = ½ CHANGE OF LONGITUDE° x SIN MEAN LATITUDE CONVERSION ANGLE = DIFFERENCE BETWEEN GREAT CIRCLE AND RHUMB LINE CONVERGENCY - 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 (Conversion angle 20°)


Q3 The Great Circle bearing of A from B is 255° GC The Rhumb Line bearing of B from A is 084° RL

Q4 The Great Circle bearing of X from Y is 072° GC The Rhumb Line bearing of Y from X is 259° RL


THE CALCULATION OF RHUMB LINE TRACKS AND DISTANCES Departure must be used when determining rhumb line tracks and distances. Calculate the rhumb line track and distance between A (00° N and 010° W) and B (° N 010° E).

In order to express the dLAT in nm's : dLAT = 30° = 1800' = 1800 nm (No Departure) In order to express the dLONG in nm's, DEP (nm) = dL' x COS MID LAT = 1200' x COS 15° = 1159 nm

To determine angle A :

TAN ∅ = 1159 nm1800 nm

TAN ∅ = 0.6438 ∅ = 32.8° (rhumb line track A - B)


To determine distance x, use Pythagoras: x² = 1800² + 1159² x² = 4583281 nm x = 4583281 nm x = 2141 nm (rhumb line distance A - B)

It is important to note that this method of determining rhumb line tracks and distances is very limited in terms of its accuracy DISTANCE KILOMETRE (KM.) A Kilometre is 1/10 000 th. part of the average distance from the Equator to either Pole It 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 1 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 1 NM = 6076.1 feet. To answer questions in the CAA examinations any of the following may be used :-

1 NM = 6080 feet or 1853 metres 1 NM = 6076.1 feet: or I852 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 = 1335nm 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 increases, either to the North or to the South, the meridians converge, and the distance between them decreases, until they meet at the Poles where the distance between them is zero. Departure (nm) = ch long (mins) x cos mean lat: 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 Q1 The distance from A (N 20:10 E 005:00) to B (N 20:10 \V 005:00) is :-

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

Q2 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' 1050 nm

= 1206.4 minutes of Longitude cos 29.5°

60 = 20° 06' 24" Easterly

+ 12° 30' 032° 36' 24" E


Q3 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

= cos Lat = 70° 30' N 21600

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. Q4 The shortest distance from A (N 78:15 W 027:13) in B (SS3:30 E 15.2:4-) 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 Q5 An aircraft departs A (N 25:13 W017:25) and flies a track of 090°(T) at GS 360

for I 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 W 017:25 ;

Track 180° Change of Latitude

Departure = Ch. Long x cos Latitude Departure _________ = Ch. Long

cos Lat

GS360 x 1:35 ____________ = 630 minutes of Longitude = 10°30-East of W 017:25 = W 006:55 cos 25:13 Track 180° = Change of Latitude Old Latitude N 25:13 11:30 GS360 x 1:55 = 690nm = 11°30 Southern-Change of Latitude = position N 13:43 W 006:55


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

QDM ±180° QDR

±Variation = QUJ

±180°

±Variation = QTE

Take the shortest route to change one bearing to another

VOR VOR Radials are Magnetic bearings QDR RMI Readings are Magnetic tracks to the VOR QDM RMI BEARINGS (VOR & ADF) Usually termed RMI READING which is QDM (for ADF RMI ± DEV= QDM)

QDM ±Variation QUJ ±180° ±180° QDR =Variation QTE


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

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 great circle bearing from A to B is 260°. Convergency 12°. Southern hemisphere.

i) What is the rhumb line bearing from B to A?

ii) What is the great circle bearing from B to A?

2. Positions A and B are in the same hemisphere. The great circle bearing from A to B is 140°. The great circle bearing from B to A is 330°.

i) In which hemisphere are A and B?

ii) What is the rhumb line bearing from B to A?

3. At what latitude on earth is the convergency twice the value of convergency at 25°N?

4. Position A (40° N 170° E). Position B is on the same parallel of latitude. The great circle bearing from A to B is 082°.

What is the longitude of position B?

5. What is the rhumb line distance from A (30° N 070° E) to B (30° N 085°E)?

6. An aircraft flies around the world on a rhumb line track of 090° at a ground speed of 480

Kts. The flying time if 19 hours.

At what latitude did the aircraft fly?

7. An aircraft (G/S 480 Kts) departs position A (20° N 010°E) on a track of 360° for 3 HRS. It then turns onto a track of 270° for 2 HRS 30. It then turns onto a track of 180° for 4 hours.

What is the position of the aircraft at the end of the 3rd leg?

8. What is the shortest distance between A (065° N 13° 30’ W) and B (78° N 166° 30’E)?



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. Bearings and scale must be correct, but we cannot have both.


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_____i___________1


REPRESENTATIVE FRACTION 1 __________ or 1/1000000 or 1:1000000

1000 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 elsewhere. For example :- Mercator Chart Scale 1:1 000 000 at the Equator Scale factor 1.3054 at 40°N

1 1

________ x Scale factor 1.3054 = Scale at 40°'N _______ 1 000 000 766 049 Q1 A chart has a scale of 1:2 500 000. How many nautical miles are represented by 4 cm on

the chart?

CL Chart Length 1 4 cm Scale = ________________ ________ ______

ED Earth Distance 2 500 000 ED

ED = 2500000 x 4 cms

2500000 x 4 cms Divide by 2.54 = Inches ______________ = 53.96nm Divide by 12 = Feet

2.54 x 12 x 6080 Divide bv 6080 = Nautical Mile; Q2 32 centimetres on a chart represents 468 nm. The scale of the chart is :

CL 32 cms 1 Scale = _________________________ = _______

ED 468 nm x 6080 x 12 x 2.54 2710282


Q3 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 3500000 105 nm x 6080 x 12 x 2.54

3500000 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 Q4 Chart A has a scale of 1:2 500 000

Chart B has a scale of 1:1750 000

Which chart has the larger scale? 1 1

Chart B has the larger scale ___ > ___ 2 4

The smaller denominator is the larger scale (half a cake is larger than quarter of a cake)

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.

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 Q1 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 Q2 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

________________ = 0:7034 3 500 000

(0,7034)cos-1 = 45° 17'49" N/S Q3 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 = __ = _______________________________ = 1:3566454

ED 1 Long x 60 cos 30 x 6080 x 12 x 2.54 (Departure) Q4 The scale at 200 N is 1: 250000 What is the scale at 50 N?

Always work latitude - equator - latitude.

SCALE AT O° - 1 COS 200 x 250000 1

COS 20° 250000

= 1

266044


SCALE AT 5O°N = 1 SEC 50 ْ = X

266044 1

1 1 = x

266044 COS 500 = 1

171010 This particular problem can also be solved in one step:

SCALE AT 50° N - SCALE AT 20° N COS 50° x

COS 2° ْ 1 250000 COS 50°

x COS 200 1 = 1

171010

CALCULATING DISTANCE AND dLONG ON A MERCATOR CHART When calculating distance and dLONG on a Mercator chart, remember that between any

two given meridians: - the chart length remains the same regardless of latitude change. - the dLONG remains the same regardless of latitude change. - the scale varies with latitude (use the Mercator scale formula). - the earth distance varies with latitude (use the departure formula). EXAMPLE: Two meridians at latitude 30° N measure 13 cm apart on a Mercator chart. What is the

dLONG between these two meridians if the scale is 1

250000 at 30° N?


At 30° N : SC =

CLED

1

250000 = 13 CM

ED

ED = 13 CM x 250000 ED = 3250000 CM ED = 17.539 nm DEP (nm's) = dLONG' x COS LAT 17.539 nm's = dLONG' x COS 30°

dLONG' = 17.539 nmCOS 30°

dLONG' = 20.252' Because chart length is constant regardless of latitude and dLONG is constant regardless of latitude, this question could also have been calculated at the equator, or any other latitude, provided that the scale is calculated at that latitude. At the equator:

SCALE AT 0° = 1

250000 COS 30

1×

°

= 1

288675

At 0° N : SC = CLED

1

288675 = 13 CM

ED

ED = 13 CM x 288675 ED = 3752777 CM ED = 20.252 nm There is no departure at the equator, therefore 20.252 nm = 20.252' dLONG.


EXAMPLE: Two meridians at 30° N are 27 cm apart. What is the earth distance between these two

meridians if the scale at 60° N is 1

500000?

Again, apply the scale at the latitude where the work is being done:

SCALE AT 30° N = SCALE AT 60 N

COS 60 COS 30

1°

°×

°

= 1

866025

SC = CLED

1

866025 = 27 CM

ED

ED = 27 CM x 866025 ED = 23382686 CM ED = 126 nm

Note that this question must be solved at 30° N. The earth distance at 30° N is required and, unlike dLONG, earth distance is not a constant regardless of latitude.


PLOTTING RADIO BEARINGS ON A MERCATOR When plotting radio bearings, the final goal is always to plot a QTE, and on a Mercator chart, specifically a rhumb line QTE, because this is a straight line. STEPS TO PLOTTING ON A MERCATOR CHART a) Always draw a sketch. b) Orientate the hemisphere (to determine which way the great circle will

curve). c) Take the given information and make it true. d) Plot this great circle. e) Measure the bearing where the work was done. f) Apply the conversion angle to convert the GC to a RL. VDF BEARING EXAMPLE ATC passes an aircraft a QDM of 060°. The variation at the station is 15° W. The variation at the aircraft is 20° W. The deviation is 5° E. The convergency between the aircraft and the station is 10°. Southern hemisphere.


RBI EXAMPLE Aircraft compass heading 200°. Relative bearing to an NDB station 40°. Aircraft variation 20°W. Station variation 15° W. Deviation 5° E. Convergency between the aircraft and the station is 8°. Northern hemisphere.

VOR NEEDLE ON THE RMI EXAMPLE The VOR needle at the RMI indicates a radial of 270° (tail of the needle). The variation at the aircraft is 20° W. The variation at the station is 15° W. The deviation is 5° E. The convergency between the aircraft and the station is 14°. Southern hemisphere.


MERIDIONAL PARTS INTRODUCTION In essence, meridional parts solves the rhumb line track and distance problem. Given the following question, from A (00° N 010° W) to B (30° N 010° E), determine the rhumb line track and distance.

Thus far, the suggested method to solve this question has been to convert the dLat into nm's, convert the dLONG into nm's using departure and the cosine of the mid-latitude, and then apply trigonometry to solve the rhumb line track and distance. Unfortunately this method is only accurate for distances up to 600 nm's, mainly due to the fact that the cosine of the mid-latitude is being used to express the dLONG in nm's. Another possible solution is to physically measure the distance A - B, but due to the continually changing scale on the Mercator chart, this is also not accurate. The solution is to use meridional parts. MERIDIONAL PARTS A meridional part is equal to a minute of longitude. The meridional parts tables indicate “how many times one minute of longitude will fit into a particular change of latitude”. For example, if you look up 30° (latitude) on the table, you will find 1876.67 meridional parts. This means that one minute of longitude will fit into the dLat 0° - 30° 1876.67 times.


PRACTICAL APPLICATION

Although expressing the dLONG in nm's by using departure and the COS MID latitude is doubtful in terms of its accuracy, there is absolutely no doubt that the dLONG is 1200'. Expressing the dLat in nm's is accurate, but trigonometry can't be applied because the sides of the triangle would have different units. If we express the LAT in meridional parts, however we can proceed with trigonometry. The sides of the triangle are in the same units because one minute of longitude is equal to one meridional part. (The meridional parts tables do correct for the effect of the earth's compression). Now, using trigonometry:

TAN ∅ = 1200

1876.67 MP

∅ = 32.6° (track A - B)

Now transfer the track angle to the triangle labelled nm's. Expressing the dLONG in nm's wouldn't be accurate, but expressing the dLat in nm's certainly is. Now using trigonometry:

COS 32.6° = 1800 nm

x

x = 1800 nm

COS 32.6°

x = 2137 nm (rhumb line distance) RECOMMENDED TECHNIQUES


a) Always draw two sketches, one for MINS LONG/MP's and another for nm's.

b) Sometimes angle ∅ is not the track. In the following sketch, the track is ∅ + 90°.

c) When working from one latitude to another, neither of which is the equator, the

latitude side of the triangle will be the difference in meridional parts (DMP), or the sum of meridional parts (SMP) if changing hemispheres.


QUESTIONS The vast majority of meridional parts questions fall into one of five categories. An example of each follows below, with a heading for each to assist with identification. a) Determine the rhumb line track and distance flown. As per previous example. b) Determine the aircraft's position, given rhumb line track and distance flown.

An aircraft leaves position A (18° N 047° E) on a rhumb line track of 047°. What is its position after flying for 1246 nm's?

32° 10' N = 2027.73 MP COS 47° =

x1246 nm

18° N = 1090.99 MP x = 1246 nm x COS 47 dLAT = 936.74 DMP x = 849.8 nm

TAN 47° = x

936.74 LAT B =

849.860

x = 936.74 x TAN 47° = 14° 10' + 18° N x = 1004.53 MIN LONG = 32° 10' N x = 16° 45' LONG B = 16° 45' + 047° E LONG B = 063° 45'E


c) At which latitude will an aircraft cross a given meridian?

An aircraft departs A (12° S 063° W) on a track of 125°. At which latitude will the aircraft cross the meridian 043° W?

dLONG = 063° W - 043° W

= 20° W = 1200'

TAN 35° = x

1200'

x = 1200' x TAN 35° x = 840.25 MINS or MP/s 12° S = 720.46 MP DMP = 840.25 MP NEW LAT = 1560.71 MP NEW LAT = 25° 19' S


d) At which meridian will an aircraft cross a given latitude?

An aircraft departs position A (14° N 025° E) on a track of 295°. At which meridian will the aircraft cross the latitude 22° N?

22° N = 1344.92 MP 14° N = 842.83 MP 502.09 DMP

TAN 25° = 502.09

x

x = 502.09

TAN 25°

x = 1076.74 MP's or MINS LONG dLONG = 1076.74' = 17° 57' NEW LONG = 25° - 17° 57' NEW LONG = 007° 03'E

e) Meridional parts scale


As can be seen from any Mercator chart, the chart length of one minute of longitude has a constant value chart length, regardless of latitude. The exact value of the chart length of course depends on the scale of the chart at that point. Because the chart length between any two meridians is constant throughout the chart, the scale at any latitude may be used.

If one minute of longitude is equal to one meridional part, then it stands to reason that 1 MP must also have a constant value chart length throughout the chart.

EXAMPLE:

A mercator chart has a scale of 1

1000000 at the equator. What is the chart length of 1 MP

in cm's?

SC = CL

ED (1 MP / 1 MIN LONG)

1

1000000 =

CL1NM

1

1000000 =

CL185300 CM

CL = 1853001000000

CL = 0.1853 CM (CL of 1 MP/1 MIN LONG)

As previously stated, because the CL of 1 MIN LONG is constant throughout the chart, the chart length may be calculated at any latitude, provided the scale at that latitude is used.

Calculation of the same question, but at 60°N.

SCALE AT 60° N = 1

1000000 1

COS 60×

°

= 1

500000

Calculate the earth distance of 1 MIN LONG at 60° N. DEP (nm's) = dLONG' x COS LAT = 1' x COS 60° = 0.5 nm's


SC = CL

ED (1 MP / 1 MIN LONG)

1

500000 =

CL0.5 NM

1

500000 =

CL92650 CM

CL = 92650

500000

CL = 0.1853 CM (CL of 1 MP/1 MIN LONG)

What is the CL in CM's between A (12° N 006° E) and B (18° S 024° E) if the scale at 52° N is

1400000

?

Determine the CL of 1 MP/1 MIN LONG

SC = CLED

1

400000 =

CL1' COS 52 185300× ° ×

1

400000 =

CL114082 CM

CL = 114082400000

CL = 0.2852 CM (CL of 1 MP/1 MIN LONG) 12° N = 720.46 MP 18° S = 1090.99 MP SMP = 1811.45 MP


dLONG = 024° E - 006°E dLONG = 18° E dLONG = 1080' Using Pythagoras: x² = 1811.45² + 1080² x² = 4447751 x = 4447751 x = 2109 MP at 0.2952 cm per MP x = 601.5 cm











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 (45N) and expands north and south of 45N. 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. Parallels of Latitude are arcs of circles radius the Pole. 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 convergency 0.7071° is called the: CHART CONVERGENCY 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. 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. RHUMB LINES Curves concave to the Pole and convex to the Equator GREAT CIRCLES A straight line joining two positions on the Parallel of origin,

Curves slightly concave to the Parallel of Origin.


CONVERGENCY Constant throughout the chart Correct at the Parallel of Origin

Chart Convergency Ch. Long x sin Parallel of Origin

Chart Convergency 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. THE ADVANTAGES OF THE CHART a) Constant scale.

b) Radio bearings are great circles and on this chart, great circles are straight lines, which means that radio bearings can be easily plotted.

THE DISADVANTAGES OF THE CHART

a) The grid is not rectangular. b) Light aircraft generally fly rhumb line tracks, but the rhumb line is a curved line on

this chart and therefore cannot be accurately plotted.


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 Convergency (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 Convergency (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 ║O is 30°S - Then the other SP is 40°S Chart Convergency (CC) = Change of Longitude x sine Parallel of Origin Chart Convergency (CC) = Change of Longitude x Chart Convergency Factor Sine Parallel of Origin = Chart Convergency Factor (CCF)


If a statement regarding convergency is given :- (e.g. a Lamberts chart has a chart convergency 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 convergency is proportional to the CCF, convergency between any two meridians is easily found.

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 is :- The Parallel of Origin (║O) is midway between the two Standard Parallels CCF 0.5 = sin║O = 30°S

SP25°S Parallel of Origin 30°S Other SP35°S


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. The final goal when plotting radio bearings on the Lambert's chart is to plot a QTE, and specifically the great circle QTE, because this is a straight line. STEPS TO PLOTTING ON A LAMBERT'S CHART a) Always draw a sketch. b) Orientate the hemisphere correctly. c) Take the given information and make it true. d) Plot this great circle. e) Measure this bearing where the work was done. f) Apply convergency if required to obtain the great circle QTE. VDF BEARING EXAMPLE ATC passes an aircraft a QDM of 060°. The variation at the station is 15° W. The variation at the aircraft is 20° W. The deviation is 5° E. The convergency between the aircraft and the station is 10°. Southern hemisphere.


RBI EXAMPLE Aircraft compass heading 200°, relative bearing to an NDB station 040°. Aircraft variation 20° W. Station variation 15° W. Deviation 5° E. Convergency between the aircraft and the station is 8°. Northern hemisphere.

THE VOR NEEDLE ON THE RMI EXAMPLE The VOR needle on the RMI indicates a radial of 270° (tail of the needle). The variation at the aircraft is 20° W. The variation at the station is 15° W. The deviation is 5° E. The convergency between the aircraft and the station is 14°. Southern hemisphere.


PLOTTING RHUMB LINE TRACKS On the Lambert's chart, a rhumb line is a curved line and cannot actually be plotted. Join points A and B on the chart with a straight line (great circle). Measure the track of the great circle at the mid-meridian. If the aircraft departs from position A and maintains this track, it will fly the equivalent rhumb line track. At the mid-meridian, the great circle and the rhumb line are parallel.

MEASURING RHUMB LINE DISTANCES The rhumb line (curved) is never actually plotted, thus its distance cannot be measured. Instead, measure the great circle (straight line) distance to obtain the equivalent rhumb line distance. To obtain the greatest degree of accuracy, measure this distance: a) Along a meridian scale.

b) Across the mid latitude of the track


SCALE PROBLEMS Lambert's scale 1:2 500 000, SP20° Sand40°S. The scale is correct at the two Standard Parallels

Scale 20°S = Scale at 40°S Q1. 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) Q2 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 = 0.6158 = cos52°S


THE POLAR STEREOGRAPHIC CHART THE CONSTRUCTION OF THE CHART A model earth is constructed in glass with a light source at one of the poles. A flat piece of paper is then placed on top of the pole to be constructed, and opposite to the light source. When the light is switched on, the data is projected onto the flat piece of paper. When the piece of paper is removed, a polar stereographic chart has been created.

SOUTH POLAR STEREOGRAPHIC CHART THE PROPERTIES OF THE CHART THE MERIDIANS The meridians are straight lines radiating from the pole. THE PARALLELS The parallels are concentric circles. The spacing between the parallels increases away from the pole. The formula for determining the chart length of the radius from the pole to a particular parallel of latitude is: r = 2 R tan ½ co-lat Where R is the radius of the model earth and co-lat is the difference between 90º and the latitude in question.


THE POINT OF TANGENCY The point of tangency is the north or south pole. THE POINT OF PROJECTION The point of projection is a light source at the opposite pole. SCALE The scale is correct at the point of tangency (the pole). Elsewhere on the chart, the scale expands with movement away from the pole or contracts with movement towards the pole.

The formula for determining scale expansion away from the pole is:

1

SCALE AT LATITUDE = 1

SCALE AT POLE SEC CO - LAT

1

2 12 ×

RHUMB LINES Rhumb lines curve towards the equator and cut successive meridians at the same angle. GREAT CIRCLES Great circles may be considered to be straight lines and will cut successive meridians at different angles. (In truth the great circle is slightly concave to the pole.)


ORTHOMORPHIC The chart is orthomorphic because a) Meridians and parallels cut at 90º.

b) The scale expands at the same rate in all directions over short distances. CONVERGENCY On this chart, convergency is correct at the pole. CONVERGENCYº = dLONGº However, convergency is constant throughout the chart because the meridians are straight lines. Therefore convergency all over the chart is simply calculated with the formula : CONVERGENCYº = dLONGº SHAPES AND AREAS The nearer the pole, the more accurate the representation of shapes and areas.


Measuring Directions On The Charts Remember that direction true is always measured clockwise and relative to true north. On a North polar stereographic chart, the North pole is at the centre of the chart. On a South polar stereographic chart, the South pole is at the centre of the chart and true North is 180° away from true South. Remember also that a parallel of latitude runs E/W.

Plotting Radio Bearings On The Polar Stereographic Chart The final goal when plotting radio bearings on the polar stereographic chart is to plot a QTE, because this is a straight line. Steps To Plotting On A Polar Stereographic Chart a) Draw a sketch. b) Orientate the hemisphere. c) Take what is given and make it true. d) Plot this great circle. e) Measure this bearing where the work was done. f) Apply convergency if required to obtain the great circle QTE. VDF Bearing EXAMPLE ATC passes the aircraft a QDM of 060°. The variation at the station is 15° W. The variation at the aircraft is 20° W. The deviation is 5° E. The station is at position 70° S 090° E. The aircraft is at position 70°S 010° E. Southern hemisphere. The QTE to plot is?


RBI EXAMPLE Aircraft compass heading 270°. Relative bearing to an NDB station 040°. Station variation 15° W. Aircraft variation 20° W. Deviation 10° E. The station is at 70° N 030° W. The aircraft is at 70° N 030° E. Northern hemisphere.

VOR Needle On The RMI EXAMPLE The VOR needle on the RMI indicates a radial of 165°. The variation at the station is 15° W. The variation at the aircraft is 20° W. The deviation is 12° E. The station is at position 70° S 040° E. The aircraft is at position 70° S 160° E. Southern hemisphere.


Determining The Radius Of A Parallel Of Latitude

The scale of the model earth is 1

8000000. The radius of the real earth is 3438 nm. On a polar

stereographic chart of the north pole, calculate the chart length between 70°N and 60° N in cm's.

a. Calculate the radius of the model earth in cm's.

SC

CL

= CLED

= CL3438 185300

= 6370614008000000

18000000 ×

CL = 79.6 cm (radius of the model earth)

b. Calculate the radius from 90° N to 60° N. r = 2 R tan ½ co-lat = 2 x 79.6 x tan ½ (90° - 60°) = 2 x 79.6 x tan 15° = 42.7 cm

c. Calculate the radius from 90° N to 70° N. r = 2 R tan ½ co-lat = 2 x 79.6 x tan 10° = 28.1 cm The chart length between 70° N and 60° N. 42.7 cm - 28.1 cm = 14.6 cm


Determining Scale On The Polar Stereographic Chart

On a polar stereographic chart of the north pole, the scale at 60° N is 1

1000000. What is the scale

at 70° N. Take the scale from 60° N to 90° N.

1

SCALE AT 90 N° =

11000000

COS co - lat1

2 12×

= COS co - lat1000000

2 12

= COS 151000000

2 °

= 1

1071797

Take the scale from 90° N to 70° N

1

SCALE AT 70 N° =

11071797

SEC co - lat1

2 12×

= 1

1071797 1

COS co - lat2 12

×

= 1

1071797 1

COS 102×°

= 1

1039478

By ABBA :

Scale A x {cos (½co-lat)B}² = Scale B x {cos (½co-lat)A}² Gee that was a short and noisy landing


GRID NAVIGATION One of the problems associated with the polar stereographic chart is that if you were at the north pole, it would be impossible to plot a course anywhere, because every single direction is south. Similarly, if you were at the south pole, every single direction is north. Certainly less serious, but also warranting improvement is the Lambert's chart. Flying great circle tracks is ideal, but care must be taken when plotting these tracks, because they cut each meridian at a different angle. The solution to both of these problems is grid navigation. A square grid is placed over the applicable chart, grid north is always at the top of the chart and direction is now referenced to grid north rather than true north. Direction will always be constant relative to grid north because the grid is square.


THE POLAR STEREOGRAPHIC GRID On the polar stereographic grid, the datum meridian (the meridian on the chart with which the grid is lined up) is always the Greenwich meridian / anti-meridian of Greenwich. THE LAMBERT'S GRID

On the Lambert's grid, the datum meridian (the meridian on the chart with which the grid is lined up) can vary, and is normally positioned at a meridian closest to where the chart will be used. CONVERGENCE Convergence is defined as being the angular difference between grid north and true north.

If convergency is the angular difference between any two meridians, then convergence is the angular difference, not between any two meridians, but between the datum meridian and another meridian. Convergence and convergency thus always have the same numerical value. On the polar stereographic chart :

CONVERGENCE° = CONVERGENCY° = dLONG° On the Lambert's chart

CONVERGENCE° = CONVERGENCY° = dLONG° x SIN LAT//of O EXAMPLE:


Grid heading 080°. Convergence 20° W. What is the true heading?

Grid heading 080°. Convergence 20° E. What is the true heading?

RULE: CONVERGENCE WEST - TRUE IS BEST CONVERGENCE EAST - TRUE IS LEAST BEWARE: On the polar stereographic grid, although CONVERGENCE° = CONVERGENCY° = dLONG °, an easterly convergence does not necessarily mean that the aircraft is in the eastern hemisphere. Similarly, a westerly convergence does not necessarily mean that the aircraft is in the western hemisphere. This will be the case on a south polar chart, but will not be the case on a north polar chart. Always draw a sketch. GRID VARIATION (GRIVATION) Grivation is defined as the angular difference between grid north and magnetic north. It is thus the algebraic sum of convergence and variation.


ISOGRIVS Isogrivs are defined as being lines joining places of equal grivation.


QUESTIONS PART 1 1. On a mercator chart, the scale at 18° N is 1:1000000 What is the scale at 36° S. 2. On a Mercator chart, a line 21 cm long is drawn along the parallel 36° S. What change in longitude does this line represent if the scale of the chart is 1:400 000 at

50° S? 3. What earth distance is represented by a line 18” long drawn along the parallel 27° N if the

scale on the Mercator chart is1:250 000 at 60 N? 4. Two lines of equal length are drawn on a Mercator chart, one at the equator and the other

at 60° N. Which of these two lines represents the greater earth distance? 5. With the needle centralised, the VOR CDI indicates 145° TO. The variation at the aircraft

position is 10° W. The variation at the station position is 15° W The deviation is 5° W.

What bearing should be plotted on a Mercator chart of the northern hemisphere if the convergency between the aircraft and the station is 10°?

6. The ADF bearing on an RMI is 060°. The variation at the aircraft position is 10° W. The

variation at the station position is 20° W. the deviation is 5° E the convergency between the aircraft and the station is 12°.

What bearing should be plotted on a Mercator chart of the southern hemisphere?

7. Using meridional parts, calculate the rhumb line track and distance from A (08°N 016° 30’

W) to B (16° 27’ S 004° 18’ E). 8. An aircraft leaves position A (27 ْ27’ S 014° 28’ E) on a rhumb line track of 205° and flies for

a distance of 4087 nm. What is the aircraft’s final position? 9. An aircraft departs from position A (10 18’ S 002° 03’E) on a rhumb line track of 040°. At which meridian will the aircraft cross the equator? 10. An aircraft departs position A (21° 37’ N 012° 12’ W) on a rhumb line track of 137°. At which latitude will the aircraft cross the Greenwich meridian?


11. On a flight from A (22° N 165° E) to B (37 N 178° W), what is the chart length in cm’s of the

rhumb line distance if the scale of the chart is at 60° 1:1 000 000 N?

If the northernmost latitude of this chart is 60 ْN and the north/south length of the chart is 150 cm’s, what is the southernmost latitude?

12. On a flight along the 50th parallel, the measured distance between fixes A and B is 22,5 cm

on a Mercator chart of the northern hemisphere. The scale of the chart is 1:2 500 000 at 20°N. What is the aircraft’s groundspeed if the time between fixes was 17 minutes?

13. A Mercator chart has a scale of 1:2 000 000 at latitude 30°N. At what latitude will the scale

be 1: 1 500 000? 14. On a Mercator chart, the perpendicular distance between parallels 37°N and 39°N is 4 cms.

What is the scale of the chart at 30°N? Part 2 1. The chart convergency factor on a Lambert's chart is 0.766. On standard parallel is at 40°

N. What is the latitude of the other standard parallel? 2. The great circle track from A (40° S 015° E) to B (20° S 015° W) cuts the Greenwich

meridian at an angle of 45°. The P of O is at 30° S. i) What is the great circle track measured at A? ii) What is the great circle track measured at B? iii) What is the rhumb line track from B - A? 3. The ADF needle on an RMI indicates an QDM of 040°. The variation at the station position

is 20° W. The variation at the aircraft position is 15º W. The deviation is 5° E. The dLONG between the aircraft position and station position is 60°. The parallel of Origin is at 30° N.

What is the bearing to plot on a Lambert's chart of the northern hemisphere? 4. With the needle on the VOR CDI centralised, the indication is 360° TO. The variation at the

aircraft position is 15° W. The variation at the station position is 20° W. The deviation is 5° E. The convergency between the aircraft and the station is 10°.

What is the bearing to plot on a Lambert's chart of the southern hemisphere? 5. On a Lambert's chart of the northern hemisphere, the standard parallel of 30° N has a chart

length of 50 cm's. The other standard parallel measures 38 cm's. What is the latitude of the other standard parallel? 6. On a Lambert’s chart in the Northern hemisphere, a straight line is drawn from X to Y. The

track measured at X is 60°T. If an aircraft leaves X on a constant heading of 60°T in zero wind conditions, will it pass:


a) North of Y. b) Overhead Y. c) South of Y. 7. A Lambert’s chart has standard parallels 20°N and 60°N. The initial great circle track form

a 27°N 061°W to B 47°N 017°W is 52° (T). The longitude at which the great circle track becomes 084° is ...? PART 3 1. On a polar stereographic chart, a flight is planned from A (70° N 035° E) to B (70° N 043°

W). i) What is the great circle track from A - B? ii) What is the great circle track from B - A? 2. On a polar stereographic chart, a flight is planned from A - D. A - B great circle track 041°. B - C great circle track 059°. C - D great circle track 064°.

What is the great circle track from A direct to D if all these positions lie on the parallel 75° N?

3. On a polar stereographic chart, a flight is planned from A (75° S 168° E) to B (75° S x°W). The great circle track from A - B is 120°. i) What is the longitude of position B?

ii) What will the great circle track be when crossing the anti-meridian of Greenwich? 4. On a polar stereographic chart, a flight is planned from A (72° N 032° W) to B (72° N 098°

W). i) What is the great circle heading at A if the drift is 5° right? ii) What is the highest latitude which this line will attain?


5. On a polar stereographic chart, a flight is planned from A (75° S 047° E) to B (75° S 063°

W). i) What is the great circle track from A - B?

ii) If there was an NDB station at B, what would the QDM be when the aircraft crosses the prime meridian assuming zero deviation and variation 15° W?

6. A Mercator chart and a polar stereographic chart have a rolling fit at 70°N. The scale of the

Mercator chart is 1:1 000 000 at the Equator. What is the scale of the polar stereographic chart at 90°N?

Part 4 1. Aircraft heading 231° G. Convergence 15° W. What is the aircraft's true heading. 2. A grid is superimposed on a polar stereographic chart of the north pole. An aircraft has a

heading of 060° T and 130° G. What is the aircraft's longitude? 3. A grid is superimposed on a polar stereographic chart of the south pole. An aircraft has a

heading of 210° T and 160° G. What is the aircraft's longitude? 4. On a north polar grid chart, an aircraft at position 70° N 040° E, has a heading of 060° G. What is the aircraft's heading ° T? 5. On a south polar grid chart, an aircraft at position 75° S 060° W, has a heading of 160° T. What is the aircraft's heading ° G? 6. A grid is superimposed on a Lambert's chart of the northern hemisphere with the datum

meridian at 030° W. The CCF is 0.5. An aircraft at position 45° N 010° W has a heading of 080° T.

What is the aircraft's grid heading? 7. A grid is superimposed on a Lambert's chart of the southern hemisphere with the datum

meridian at 060° E. The n factor is 0.5. An aircraft at position 20° S 020° E has a grid heading of 160° G.

If the variation is 15° W, what is the aircraft's magnetic heading? 8. On a Lambert's conformal/Grid chart of the northern hemisphere, an aircraft at position 30°

N 050° E has a true heading of 060° T and a grid heading of 100° G. The parallel of origin is at 30° N.


What is the datum meridian used on this chart? 9. On a Lambert's conformal/Grid chart of the southern hemisphere, an aircraft at position 25°

S 030° W has a magnetic heading of 120° M. The grid heading is 090°G. The variation is 15° W. The CCF is 0.5.

What is the datum meridian used on this chart?


CHAPTER 3

RELATIVE VELOCITY

Relative Velocity is the comparison of aircraft speeds or the speed of one aircraft relative to another. PROBLEM TYPES The easiest way to solve relative velocity problems is to first identify the problem type. TYPE 1

a) Formula : TIME = RELATIVE DISTANCE

RELATIVE SPEED

b) Identifiable by : - 2 aircraft. - neither aircraft changes speed or - one aircraft changes speed to a new, unknown speed or - one aircraft changes speed to the same speed as the other aircraft. TYPE 2 a) Formula:

DIST FROM DEST = DELAY OLD G / S NEW G / S

DIFFERENCE IN G / S× ×

b) Identifiable by: - 1 aircraft.

- 2 aircraft, one of which reduces speed to a new, but known speed, which is not the same as the other aircraft's speed.

TYPE 3

a) Formula : a

SIN A = b

SIN B = c

SIN C

b) Identifiable by: - 2 aircraft on converging tracks.


PROBLEM EXAMPLES The following examples demonstrate the types of relative velocity problems. In each case: - Draw a sketch. - Get both aircraft to the same time - Identify the problem type. - Write down the appropriate formula

- Solve. FIRST WE TACKLE CPL LEVEL QUESTIONS: Q 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 Q2. 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?


Q3 Two aircraft at the same Flight Level following the same route are approaching a VOR.

Aircraft A, GS 390 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 390 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.

(a) Speed of closing 60Kts (b) Speed of closing 60Kts

Distance to close (55 - 50) 5 nm Distance to close (55-32.5) 22.5 nm Time to close 5 mins Time to close 22.5mn Reduce speed at 0805 Z Reduce speed at 0822.5

Q4. 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 GS450 Dist243.75mn Time 32½ mins ETA 0915-32½ mins = 0842½ GS 390 Dist 243.75 nm Time 37½ mins ETA 0920 - 37½ mins = 0842½' Q4 Alternative solution DISTANCE = SPEED x TIME

At the point where speed is reduced, the aircraft is 'D nm' from Delta.

AtGS450 D = 450 x T AtGS390 D-390 x (T+5) As D is common, then 450 T = 390 (T + 5)

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

60 T = 1950 T = 32½ mins

At GS4 50 ETA 0915-T 32½ mins = 0842½ At GS 390 ETA 0920-(T+5) 37½ mins - 0842½

DTC 55 nms


Q5 Aircraft A, GS 180 Kts. passes over NDB PB 5 minutes ahead of aircraft B. Aircraft B. GS 260 Kts. passes over VOR CPL 8 minutes ahead of aircraft 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 Q6 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 GS x New GS 9 x 250 x 315 Distance = _____________________ = _______________ = 181.73 nm

Difference in GS x 60 65 x 60


QUESTIONS 1. At 1205, aircraft A is 76 nm behind aircraft B. Aircraft A 180 Kts. Aircraft B 140 Kts. a) At what time will aircraft A pass aircraft B? b) At what time will aircraft A be 42 nm ahead of aircraft B?

76 nm

A - 180 Kts1205

B - 140 Kts1205

a) T = RDRS

= 76 - 0

180 - 140

= 7640

= 1 HR 54 after 1205 = 1359 (aircraft A will pass aircraft B)

b) T = RDRS

= 76 + 42

180 - 140

= 2 HR 57 after 1205 = 1502 (aircraft A will be 42 nm ahead of B) 2. Route X - Y 500 nm. Aircraft A (G/S 320 Kts) passes X at 1205.

Aircraft B (G/S 360 Kts) passes X at 1215.

a) At what time must aircraft B reduce speed to 300 Kts to ensure a 20 nm separation when aircraft A reaches Y?

b) At what time must aircraft B reduce speed to 300 Kts to ensure arrival at Y, 10 minutes after A?

60 nm

B - 360 Kts1205

A - 320 Kts1205

20 nm

B - 360 Kts1215

X Y

Z

500 nm


a) Time for A to reach Y

= 500320

= 1 H 34 + 1205 = 1339 (also time for B to reach Z) B's original estimate for Z

= 60 + (500 - 20)

360

= 1 H 30 + 1205 = 1335 B must delay his arrival by 1339 - 1335 = 4 mins


DIFF IN G / S× ×

= 0.067 360 300

60× ×

= 120 nm from Z (speed reduction point) B's time to reach speed reduction point

T = 60 + (500 - 20 - 120)

360

= 1 HR 10 + 1205 = 1315 b) Time for A to reach Y = 1339 Time for B to reach Y = 1349 B's original estimate for Y

T = 60 + 500

360

= 1 HR 33 + 1205 = 1338


B must delay arrival by 1349 - 1338 = 11 mins


DIFF IN G / S× ×

= 0.183 360 300

60× ×

= 330 nm from Y (speed reduction point) Time to reach speed reduction point

T = 60 + (500 - 330)

360

= 38 min + 1205 = 1243 3. At 1205, aircraft A and B are 75 nm's apart and are on a collision course. Aircraft A 330

Kts. Aircraft B 360 Kts. The relative bearing from A to B is 075° a) What is the relative bearing from B to A? b) At what time will the two aircraft collide?

c) At what time will the aircraft first see each other if the in-flight visibility is 20 nm.

aSIN A

= bSIN B

= cSIN C

Please note that the sides of the triangle must have constant units, either all 3 sides speed or all 3 sides distance, but not a mixture of both.


a) SIN B

b = SIN A

a

SIN B = SIN A b

a×

= SIN 75 330

360×

= 62.3° - 360° = 297.7° b) Calculate the time when the two aircraft will collide. Calculate angle C. A + B + C = 180° C = 180° - 75° - 62.3° C = 42.7° Calculate side C (relative speed)

c

SIN C = a

SIN A

C = a SIN C

A×SIN

= 360 SIN 42.7

75× °

°SIN

= 252.75 Kts Calculate the time when the aircraft will meet.

T = RDRS

= 75

252.75

= 18 mins after 1205 = 1223


c) Calculate the time when the aircraft will first see each other if the in-flight visibility is

20 nm.

T = RDRS

= 75 - 20252.75

= 13 mins after 1205 = 1218


QUESTIONS 1. Positions X and Y are 960 nm's apart. At 1205, aircraft A leaves position X en-route to Y at

G/S 370 Kts. At 1225, aircraft B leaves position Y en-route to X at G/S 290 Kts. i) At what time will the aircraft pass each other? ii) When the aircraft pass each other, what is their distance from position X? 2. Positions X and Y are 290 nm's apart. At 1205, aircraft A is overhead position X G/S 350

Kts. At 1230, aircraft B is overhead position X G/S 450 Kts. i) At what time will the separation between the two aircraft be 130 nm's? ii) When aircraft A reaches position Y, the separation between the two aircraft must be

130 nm's. What reduction in speed must B make at position X to ensure this separation?

3. Aircraft A and B are both bound for position Y. Aircraft A G/S 350 Kts is 200 nm's from Y at

1225. Aircraft B G/S 450 Kts is 300 nm's from Y at 1215. i) At what time must aircraft B reduce speed to 350 Kts to ensure a 20 nm separation

when aircraft A reaches position Y? ii) At what time must aircraft B reduce speed to 350 Kts to ensure a two minute

separation when aircraft A reaches position Y? 4. Aircraft A G/S 250 Kts passes position X 10 minutes ahead of aircraft B G/S 300 Kts.

Some time later, aircraft B passes position Y 10 minutes ahead of aircraft A. What is the distance from X to Y? 5. Aircraft A G/S 420 Kts is inbound to position Y. ATC requests that the aircraft delay its

arrival over position Y by 15 minutes. At what distance from Y must aircraft A reduce speed to 370 Kts to ensure this delay? 6. Position X and Y are 700 nm's apart. Aircraft A G/S 350 Kts passes position X at 1200 en-

route to Y. Aircraft B G/S 470 Kts passes position X at 1215 en-route to Y. i) At what time must aircraft B reduce speed to 320 Kts to ensure a 50 nm separation

when aircraft A passes position Y? ii) At what time must aircraft B reduce speed to 320 Kts to ensure an arrival over point

Y 10 minutes after aircraft A?


7. At 1305, aircraft A and B are on a collision course, 250 nm's apart. Aircraft A G/S 280 Kts. Aircraft B G/S 250 Kts. The relative bearing from B to A is 320°.

i) What is the relative bearing from A to B? ii) At what time will the aircraft collide? iii) At what time will the aircraft first see each other if the in-flight visibility is 20 nm's?



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 ±3. 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 3. 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 distance CY and the orbital speed of the planet is faster at Perihelion than at Aphelion. The orbital speed of the Earth is variable. 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.

SAX SYC

A

X

C

Y


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. Approximate dates 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

The seasons apply to the Northern Hemisphere and reversed in the Southern Hemisphere. 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 Sun 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 EQUATION OF TIME The equation of time is the time difference between the apparent solar day and the mean solar day and is of varying duration. 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 90E the sun is overhead at 12:00 LMT. At 180°E/W the sun is setting at 18:00 LMT. At 90°W 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. UTC UNIVERSAL CO-ORDINATED TIME 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). ARC TO TIME As 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 on the next page is also 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.

10° 0:40 15° 1:00 79° 5:16 161° 10:44 The right hand column gives the time equivalent for minutes of Longitude.

28' Long 1 minutes 52 seconds 42' Long 2 minutes 48 seconds 127° 37'E Arc to Time 127° = 8:28 37' = 2:28 127° 37' = 8:30:28



Q1. At position A (N 45:05 E 065:30) it is 13:15 LMT on 23rd March.

The LJTC at this position is :-

A 13:15 LMT 23rd March E 065:30 Arc to Time 4:22 A 08:53 UTC 23rd March

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

Q2. 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

Q3. 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

Q4. 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 D ETA 26:07 UTC 15 th May GD

ETA 02:07 UTC 16 th May GD E 135:45 Arc to Time 9:03 D ETA 11:10 LMT 16 th May LD

NOTE: In flight the time standard is UTC. always work in UTC.


ZONE TIME (ZT) UTC is a method of co-ordination, not time keeping. LMT could be a method of time keeping, but is not practical because each and every meridian would have its own time. To solve this problem, the earth is divided into zones, 15º wide. The time in each zone is the same and is referenced to the mid-meridian for that zone. To convert UTC to zone time or vice versa, add or subtract the zone number (hours). Remember, east is later and west is earlier. Rule for the international date line: When heading east, subtract 24 hours. When heading west, add 24 hours.



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 times appear on the next four pages. 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.

OOPS






SUNRISE, SUNSET AND TWILIGHT SUNRISE The time at which the upper rim of the sun just becomes visible above the horizon. This time can be extracted from one of the following sources: a) The Air Almanac (LMT) b) The Jeppesen (LMT)

c) The Aerad (UTC) d) The SA AIP for the major airports

The times extracted have been corrected for atmospheric refraction. SUNSET The time at which the upper rim of the sun just disappears below the horizon. This time can be extracted from one of the following sources: a) The Air Almanac (LMT) b) The Jeppesen (LMT) c) The Aerad (UTC) d) The SA AIP for the major airports The time extracted have been corrected for atmospheric refraction. TWILIGHT The beginning of morning civil twilight. The time at which the sun is 6° below the horizon on its way up. This time can be extracted from the Air Almanac (LMT). The end of evening civil twilight The time at which the sun is 6° below the horizon on its way down. This time can be extracted from the Air Almanac (LMT). IMPORTANT NOTES a) Sunrise, sunset and twilight do not occur at the same LMT for places on the same

meridian. Due to the inclination of the earth's axis, the time for sunrise, sunset and twilight varies with latitude and date. Sunrise, sunset and twilight do however occur at the same LMT for all places on the same latitude for a particular date.

b) The duration of morning civil twilight is determined by subtracting the time for

sunrise (end of morning civil twilight) and the time for (beginning of) morning civil twilight.


c) The duration of evening civil twilight is determined by subtracting the time by sunset (beginning of evening civil twilight) and the time for (the end of) evening civil twilight.





MOONRISE AND MOONSET

The moon rotates anti-clockwise around the earth in a nearly circular orbit at an average distance of 250 000 miles. The point where the moon is closest to the earth in its orbit is called PERIGEE. The point where the moon is furthest from the earth in its orbit is called APOGEE. THE SYNODIC PERIOD This is the time it takes for the moon to make one complete orbit around the earth relative to the sun i.e. the time interval between two new moons. The synodic period is + 29½ days and forms the origin of the month. THE SIDERIAL PERIOD This is the time it takes for the moon to make one complete orbit around the earth relative to a fixed point in space. The Siderial period is 27 days 7 hours 43 minutes + 3 minutes. THE PHASES OF THE MOON


THE MOON'S DAILY LAG The earth rotates approximately 360º in 24 hours. If the moon were stationary, the time interval between successive moonrises/moonsets would be 24 hours. Due to the fact that the moon rotates anti-clockwise as the earth is revolving anti-clockwise, the time interval between successive moonrises/moonsets is more than 24 hours, in fact about 24 hours and 50 minutes depending on latitude and date. Thus if moonrise occurs at 20h00 on day 1, it will occur again at about 20h50 on day 2.

This fact is referred to as the moon's daily lag.

MOONRISE AND MOONSET TIMES Much like sunrise, sunset and twilight, moonrise and moonset times vary with latitude and date. Unlike sunrise, sunset and twilight however, moonrise and moonset do not occur at the same LMT for places on the same parallel of latitude. If the moon's daily lag is + 50 minutes, and moonrise at 30° N on the Greenwich meridian occurs at 20h00, it will occur at 20h25 at 30° N 180° E/W (half

the daily difference 180360

).

The Air Almanac lists the LMT at Greenwich of moonrise and moonset at different latitudes and different dates. Next to the time of moonrise and moonset is a column labelled DIFF which indicates half the daily lag in minutes for that particular latitude and date. Associated with the moonrise and moonset tables is a table labelled F4. This table simply indicates what fraction of the half daily difference applies to a particular longitude. E.G.: If the moon's daily lag for a particular latitude and date is 60 minutes, the diff column would indicate 30 minutes (½ daily lag).

If we were at a longitude of 90°, the F4 table would indicate 15 minutes (90

180 = 1

2 = 15

30).

This 15 minutes would then be added to or subtracted from the moonrise/moonset LMT at Greenwich to derive the LMT at 90° longitude. Whether to add or subtract the 15 minutes depends on whether the longitude is east


SPECIAL NOTES

i) Once the LMT for moonrise/moonset has been derived, a particular question may require that this time be expressed in UTC by adding or subtracting arc to time.

If, for example, a particular question required the UTC of moonrise for JAN 1, and in the process of subtracting arc to time, we arrived at a UTC for moonrise, but on DEC 31st, add a full day (24 hours) to get to the 1st AND add the FULL daily difference.

In the same way, if in the process of adding arc to time, we arrived at the UTC for moonrise, but on the 2nd JAN, subtract a full day (24 hours) to get to the 1st AND subtract the FULL daily difference.

ii) Because the moon's daily lag obviously has a cumulative effect, there will be one

day per month when there will be no moonrise (near the last quarter) and one day per month when there will be no moonset (near the first quarter). Now, instead of leaving the table blank for that day, the table will indicate the time for moonrise/moonset, but for the next day.

E.G.: JAN 1 moonrise 2445 is actually: JAN 2 moonrise 0045. EXAMPLE What is the UTC for moonrise for position 30° N 060°E on 1st JAN? 1 JAN 30° N at Greenwich moonrise 0010 LMT 1st DIFF 29 on F4 table against 60° E - 10 1 JAN 30° N 060° E moonrise 0000 LMT 1st ARC to time -400 UTC moonrise 30° N 060° E 2000 UTC 31st Add 24 hours plus full daily diff (2 x 29) 2458 UTC moonrise 30° N 060° E 1 JAN 2058 UTC 1st





QUESTIONS Part 1 1. An aircraft leaves position A (25° N 067° 19' E) at 1407 LMT on the 3rd on a flight to B (23°

N 014° 27' W). The flight time is 9 H 12. What is the arrival time at B in LMT? 2. The aircraft arrives at position B (63° N 168° W) at 2208 ST on the 2nd. The standard

factor at B is 11 hours. The aircraft departed from A (71° N 174° E) and the flight time was 7 H 06.

What was the departure time in LMT? 3. An aircraft arrives at position B (12° N 008° E) at 1823 ST on the 4th. The standard factor

at B is 1 hour. The aircraft departed position A (08° N x° W) at 0838 LMT on the 4th and the flying time was 6H33.

What is the longitude of position A? 4. An aircraft flies a rhumb line track of 270° for 3 hours and covers 1580 nm. If the LMT of departure is the same as the LMT of arrival, what was the

parallel of l altitude which the aircraft followed? Part 2 1. An aircraft is to fly from A (12° S 22° 08' W) to B (10° S 63° 47' W). The aircraft must arrive

at B no later than the end of evening civil twilight on 11 JAN. What is the latest standard LMT that the aircraft can depart from A if the flying time is

4H20? 2. An aircraft departs from position A (60° N 015° E) at 0715 LMT on JAN 2nd. the aircraft

flies a rhumb line track of 090° at a ground speed of 300 Kts. The aircraft arrives at destination on sunset of the same day.

What is the destination position? 3. An aircraft departs from position A (60°N 098°30’E) at 0800Z on Jan 15 flying due west at

a mean groundspeed of 300 Kts and lands at sunset the same day, at destination longitude?


Part 3 1. What is the UTC of moonrise for position 62° N 090° E on the 1st JAN? 2. What is the UTC of moonset for position 50° S 120° W on the 1st JAN? 3. What is the UTC of moonrise for position 52° S 060° W on the 1st JAN?


CHAPTER 5

THE 1 IN 60 RULE AND GENERAL MATHEMATICS

Every CAA ATP Navigation paper will normally have one or two questions requiring the use of the 1 in 60 rule or general mathematics in order to derive an answer. THE 1 IN 60 RULE The 1 in 60 rule is a simplified way to calculate an aircraft’s drift angle in flight. If an aircraft has drifted 1 nm. off track after 60 nm’s, its drift angle is 1°.

FORMULA

ERROR

DISTANCE 60

1× = DRIFT ANGLE

1

60 60

1× = 1°

EXAMPLE A to B 476 nm's. Track 090° T. The aircraft departs position A and maintaining a heading

of 090° T. After 157 nm's, the aircraft is 11 nm's left of track. i) What is the new heading to steer to regain track at point B? ii) What is the new track to point B?

11

157 60

1× = 4.2°

11319

601

× = 2°

i) The new heading to B is : 090° + 4.2° + 2° = 096.2° T ii) The new track to B is : 090° + 2° = 092° T


GENERAL MATHEMATICS The questions requiring the use of general mathematics normally involve the use of the COSINE RULE and/or the formulae for determining the RADIUS, DIAMETER or CIRCUMFERENCE of a CIRCLE. THE COSINE RULE The COSINE RULE is used in NON-RIGHT ANGLED TRIANGLES when given the length of two sides and one angle and the unknown is the length of the side opposite the known angle or when given the length of all three sides and the unknown is any angle. FORMULA a² = b² + c² - 2bc COS A Naturally, this formula can be arranged in any other fashion to isolate the unknown. EXAMPLE

Solve the length of Side a. a² = b² + c² - 2bc COS A a² = 3² + 7² - (2 × 3 × 7 × COS 40) a² = 9 + 49 - 32,17 a² = 25,83 a = 25 83, a = 5,08 UNITS


THE CIRCLE Various questions may be asked relating to the radius, diameter or circumference of a circle. FORMULA d (diameter) = 2r c (circumference) = 2πr c (circumference) = πd EXAMPLE If the radius of a circle is 7 units, determine its circumference? c = 2πr = 2 × 3,14 × 7 = 43,96 UNITS


QUESTIONS 1. An aircraft departs from position A on a heading of 247° M in order to fly track 229°T. After

132 nm's, the aircraft is 11 nm's right of the intended track. What is the magnetic heading to steer to return to A? (Variation 15° W). 2. With the aircraft's weather radar in the MAP mode, the following observations were made of

a ground feature: UTC RELATIVE BEARING RANGE 1205 067º 76 1215 107º 83 TAS 300 Kts. Heading 143° C. Deviation 3° W. Variation 15° W. i) What is the aircraft's groundspeed? ii) What is the W/V? 3. An aircraft passes overhead point A on a heading of 270° and commences a rate on turn to

the left for 7 minutes W/V 360/25. TAS 300 Kts. What is the position of the aircraft as a bearing and distance from the station at the end of

the 7 minutes? 4. An airship overhead the equator flies west around the world in 280 HRS. An aircraft also

flies west around the world, but in a time of 70 HRS and at 3 times the airships speed. At what latitude did the aircraft fly? The effect of the altitude of both the airship and the aircraft is negligible. 5. The satellite Oculus follows a polar orbit around the earth at a speed of 2900 km/h at a

constant radius from the centre of the earth of 6800 km's. Oculus crosses the equator at longitude 60° W at 1400 LMT on 27 JULY on its northbound passage.

At what longitude does Oculus cross the equator on its southbound passage and what is

the LMT at this point?


CHAPTER 5

Navigational computer VECTOR TRIANGLE (TRIANGLE OF VELOCITIES) Navigation plotting is based around the Vector Triangle which comprises of three vectors.

NOTE All directions are TRUE DIRECTIONS (measured from TRUE NORTH) The length of each Vector is the value for ONE HOUR. (TAS 240 = 240 nm)(W/V 340/30 = 30 nm) The AIR VECTOR (TAS & True Heading) has one arrow and is called the AIR PLOT. The GROUND VECTOR (True Track & Groundspeed) has two arrows and is called the TRACK PLOT. The W/V has three arrows. W/V 340/30 is the direction from which the wind blows at 30 Kts. In the above sketch the Drift angle is 7° Right. The Wind blows from the Air Vector to the Ground Vector. The units cannot be interchanged. The Air Vector is TAS and True Heading only (never TAS & Track) If four of the six values are known, the other two can be calculated. NAVIGATIONAL COMPUTER


Prior to flight the Heading and GS must be known as well as the fuel required for the flight and the time intervals between enroute points. This can all be found by using simple calculations from the flight computer or Whiz Wheel. There are many wide and varied versions of the Whiz Wheel, but basically they can all do the same thing in the same way. There are two methods of working with the wind side:

TAS under the grommet (center) wind down. or

GS under the grommet (center) wind up – Jeppesen method. The first method can solve all 3 common triangle of velocity problems, method 2 can only solve 2. Therefore method one will be used in this chapter. In this method the wind is plotted down from the grommet. WIND EXAMPLES Example 1 HDG 330º TAS 150kts W/V 040/25 Find: Track made good The groundspeed Solution: Step1 Plot wind down, then set HDG 330º under index on top. Step2 Read off the drift 10º left, the TRK is therefore 320º Step3 Read off the GS 144 kts Example 2 Required Track 150° TAS 100kt W/V 360°/30 Find the Hdg and GS Step 1 Place the W/V on the plotting disk: Step 2


Move the circular scale to have the Track under the Index mark. Step 3 The drift is noted to be 7ºR, adjust the disk so that 143º (150º-7) is under the index. Observe that the drift has changed and is now 8ºR. Further adjust the disk until the difference between the required track and the HDG under the index equals the drift. Note that if this is done correctly HDG 141 is under the index and the drift will be 9ºR. The TRK will equal 150º which is what we require. Now read off the GS at the end of the wind vector 125 kts. Example 3 If TAS is 174kt, Track is 290°, the wind velocity is 240°/40. Find the Heading and GS. ANSWER: Approx 280°, 145kt Example 4 The in-flight type of problem…finding wind You know the following figures, find out the Wind Velocity. HDG 138° TAS 120kts TRK 146° GS 144kts

Step 1 Place the HDG under index and TAS in the middle on the whiz wheel:


Step 2 (diagram to the right) Now in your head work out the drift, and its found to be 8° right, so now draw in a straight line along the 8° right drift. The GS is 144kt, draw a line along the 144 line so as to intersect the 8º drift line. Draw a line from the grommet to the intersection of 144kt and 8° drift and you have drawn in the wind vector.

Step 3 (diagram below) Rotate the grid until the wind vector blows straight down. Under the Index mark you can read the direction, in this case its 360°, and from the 120kts under the grommet at the beginning of the wind vector to its tail is the strength of the wind, in this case its 30kt. So the answer is 360/30


THE CALCULATOR SIDE OF THE COMPUTER This side of the computer can do many weird and wonderful calculations, but we are only concerned with the GS/Dist/Time and the Fuel Qty/Fuel Flow/Time problems. In order to make things simple we shall use the whiz wheel in the same manner as you would a electronic calculator, in that we use the following methods for the equations: Example 1 If the aircraft has a GS of 154kts, and the Distance for the leg it 77nm, what is the EET for the leg?

Answer = 30minutes (found under the Arrow head)

DISTANCE

GS

TIME

TIME

Fuel flow

Fuel Qtty


Example 2 At a GS of 147kt, how far will you travel in 11minutes? ANSWER 27nm Example 3 A leg is 25 minutes long, and the fuel flow is 32 lph, what is the fuel burn for this leg of the flight?

ANSWER = 13.3 litres, say 14 litres.. Example 4 If you burn 24 litres per hour, and the duration of the leg is 88minutes, what will be the fuel burn? ANSWER = 35 litres


MULTIPLE DRIFT W/V USINE THE WHIZ WHEEL Given:- TAS 190Kts Heading 040° + Drift 8° Right Heading 085° + Drift 8½° Right Heading 355° + Drift 1° Right Method: Set TAS 190 Kts at the CENTRE. Set HEADING 040° against TRUE INDEX, draw 8° RIGHT DRIFT LINE. Set HEADING 085° against TRUE INDEX, draw 8½° RIGHT DRIFT LINE. Set HEADING 355° against TRUE INDEX, draw 1° RIGHT DRIFT LINE. Place the intersection of the THREE DRIFT LINES on the CENTRE LINE below the CENTRE CIRCLE. Read off WIND DIRECTION 348° against the TRUE INDEX. Read off WIND SPEED 30 Kts along the CENTRE LINE

2 3

41

5


TRACK & GROUNDSPEED W/V (DOPPLER W/V) Given Heading 126°(T) TAS 156Kts Doppler Drift 10° Right or Track 136° Doppler GS 142 Kts Method: Set HEADING 126° against TRUE INDEX Set TAS 155 Kts at the centre circle Draw 10° Right DRIFT LINE. Draw arc of GROUNDSPEED 142 Kts. Position the intersection of the DRIFT and GROUNDSPEED lines BELOW the CENTRE CIRCLE. Read off WIND DIRECTION 070° against the TRUE INDEX. Read off WIND SPEED 30 Kts along the CENTRE LINE MULTIPLE DRIFT W/V PRACTICE PROBLEMS TAS 230 Kts Heading 195° Heading 257° Heading 332° W/V 135/30

Drift 7° Right Drift 6° Right Drift 2° Left

TAS 200 Kts Heading 045° Heading 090° Heading 340° W/V 313/32

Drift 10° Right Drift 6° Right Drift 5° Right

DOPPLER W/V PRACTICE PROBLEMS Heading 045° 225° 352°

TAS 240 300 420

Drift 10° Right 7° Left 12° Right

Groundspeed 275 285 465

W/V 282/57 289/39 242/103

The DOPPLER DRIFT may be given on one heading and the DOPPLER GROUNDSPEED on another. In this case the W/V can only be solved by the manual nav computer. Given: 1000 z Heading 055° (T) TAS 250 Kts Doppler Drift 10° Right 1012 Z Heading 010° (T) Doppler GS 235 Kts Method Set TAS 250 Kts at CENTRE Set HEADING 055° at TRUE INDEX Draw 10 Right DRIFT LINE Set HEADING 010° at TRUE INDEX Draw arc of GROUNDSPEED 235 Kts Position the intersection of the DRIFT and GROUNDSPEED lines BELOW the CENTRE CIRCLE. Read off WIND DIRECTION 303° against the TRUE INDEX. Read off WIND SPEED 50 Kts along the CENTRE LINE.


Given: 1800 Z Heading 120° (T) TAS 200 Kts Doppler Drift 12 Left 1812 Z Heading 055° (T) Doppler GS 250 Kts 1825 Z The Groundspeed on Heading 335° is :- Method: Calculate W/V 232/50 as above Set Heading 335° at TRUE INDEX and TAS 200 Kts at CENTRE Read off DRIFT 13° Right and GROUNDSPEED 218 Kts MEAN W/V The following winds are forecast for a climb to cruising altitude:- 045/25 080/45 120/55 The mean W/V for the climb is :- Method: Select a vacant area on the chart and start from the intersection of a Meridian and Parallel of Latitude. This can best be done on normal ruled paper using the lines as reference and a suitable scale. Draw the three wind vectors to scale, from head to tail. Join the end (tail) of the third wind vector to the starting point (head) and measure the wind direction. Measure the length of the vector and divide by the number of W/V’s to give the wind speed. The above method is used to calculate the mean W/V at cruising altitude when several W/V are given for a route. 091º/36.7


MEAN W/V (whiz wheel) A mean W/V can be found on the square side of the slide rule. Given: W/V 180/20 135/16 100/14 Proceed as follows: a. Set 180º against the true index and plot 20 kts down from the centre spot.

b. Set 135 against the true index and plot 16 kts down from the end of the previous wind

c. Set 100 against the true index and plot 14 kts down from the end of the previous wind


d. Rotate the disk to align the end of the last vector with the lubber line and read off the direction under true index –143º.

e. The length of the mean wind vector of 42 nms is divided by the number of winds (3) to give the mean wind speed -(14kts) Sample problem W/V 025/20 W/V 090/30 W/V 160/30 Mean W/V = 103º/15


ELECTRONIC NAVIGATION COMPUTERS TO CALCULATE TRACK & GROUNDSPEED (GS) ON THE PATHFINDER Given Heading 090° TAS240Kts W/V 010/60 Method: Select WIND function Enter HEADING 090° Enter WIND SPEED 60 Kts as GROUNDSPEED Enter TAS 240Kts Enter WIND DIRECTION 010° as CRS (TRACK) Computed W/V 104/237 WIND DIRECTION 104° is the TRACK WIND SPEED 237 Kts is the GROUNDSPEED TO CALCULATE MEAN W/V ON PATHFINDER: Method: USE Req TAS function 1. Enter 1st W/V in W Dir W Spd 2. Enter 2nd W/V in Crs GS Answer Hdg = Wind Direction ] remember this TAS = Wind Spd ] 3. Enter 3st W/V in W Dir W Spd Enter Hdg (from 2) in Crs Enter TAS (from 2) in GS ANSWER Mean W/V Hdg = Mean Wind Direction TAS = Mean Wind Speed (÷ by number of winds)


TAKE OFF WIND COMPONENTS

In order to take-off, an aircraft needs a headwind component of at least 20 kts. The maximum permitted crosswind component is 25 kts. The wind direction is 40° from the runway direction.

a) Determine the maximum and minimum wind speed acceptable for take-off.

Proceed as follows:

1) Set 040° against the true index and draw a vertical line representing the W/V.

2) Set 360° against the true index. This is the difference between the runway

and the wind direction. 3) Along the headwind component of 20 read off the required minimum wind

velocity of 25 kts. 4) Along the crosswind component of 25 read off the maximum wind velocity of

40 kts. TAS CALCULATION Given: RAS 140 Kts. Pressure Altitude 8000 feet. OAT +20°C Using the AIRSPEED WINDOW set pressure Altitude 8000 feet against OAT +20°C Against RAS 140 on the INSIDE SCALE, read off TAS 164 Kts on the OUTSIDE SCALE.


AIRSPEED COMPRESSIBILITY CORRECTION ELECTRONIC CALCULATORS Electronic calculators correct for the compressibility error at high speeds. Given OAT or Corrected OAT use PLANNED TAS Given Indicated OAT use ACTUAL TAS MANUAL FLIGHT COMPUTER Given: Pressure Altitude 20 000 feet OAT -23C RAS 320kts Using the AIRSPEED WINDOW set Pressure Altitude 20 000 feet against OAT -23°C Against RAS 320 on the INSIDE SCALE Read off TAS 440 kts on the OUTSIDE SCALE. RAS 320 and TAS 440 kts are too high due to compressibility, use the correction factor from the table below.

ALTERNATIVE METHOD (EQUIVALENT AIR SPEED EAS) RAS 320 x 0.97 = EAS 310.4 Using the AIRSPEED WINDOW set Pressure Altitude 20 000 feet against OAT -23°C Against EAS 310.4 on the INSIDE SCALE Read off TAS 427 kts on the OUTSIDE SCALE. ELECTRONIC CALCULATORS If given OAT or COAT use PLAN TAS or PLAN MACH If given IOAT use ACT TAS or ACT MACH


CHAPTER 7

BASIC PLOTTING ON THE LAMBERT'S CONFORMAL CONIC CHART

The charts used for the South African Airline Transport Pilot’s plotting examination are the Lambert’s Conformal Conic of Southern Africa. Scale 1:5 000 000 with Standard Parallels S 20:20 and S 33:40 and the Lambert’s Conformal Conic of the British Isles with Scale 1:2 500 000 and Standard Parallels N30:00 N60:00. Chart convergency is measured at S27:00 and N45:00 for the Southern African and British charts respectively, The Chart Convergency Factor is 0.45° being the sine of the Parallel of Origin S 27:00. Straight lines drawn on the chart are considered to be GREAT CIRCLES for all practical purposes, which is the prime advantage of the chart, especially when plotting radio bearings. The other main advantages of the chart is that Great Circle tracks can be flown which are shorter than Rhumb Line tracks. This is marvellous when using Great Circle navigation systems such as INS, Omega, Loran and satellite GPS. The main disadvantages of the chart is when Rhumb Line navigation is used (flying constant headings). This is overcome by splitting the Great Circle track into short segments of 200 to 300 nautical miles or perhaps 5° of Longitude and using the MID-MERIDIAN technique. MEASUREMENT OF DISTANCES Use the VERTICAL LATITUDE SCALE near the Latitude of the plot. 1° of Latitude = 60 Nautical Miles. 1° of Latitude has 12 increments of 5 Nautical Miles each. RADIO BEARINGS All radio bearings MUST be converted into QTE True Bearings and plotted from the nearest Meridian to the radio facility. VHF D/F and VOR use station variation. ADF / NDB use aircraft variation. For the ATP examination the chart convergency (CCF 0.45°) which should be applied to ADF / NDB bearings in Southern African Plotting problems may be ignored as the value is small. This is not the case, however, for European plots.


MID MERIDIAN TECHNIQUE - GREAT CIRCLE NAVIGATION Draw the straight line from A to B, this is the TRUE GREAT CIRCLE TRACK which cuts all meridians at a different angle. Highlight the nearest MERIDIAN to the mid point along track, this is the MID MERIDIAN between A and B. Typical exam questions are: 1. What is the INITIAL magnetic heading from A to B? Measure the INITIAL TRUE TRACK at the nearest meridian at A. Calculate the TRUE HEADING using the nav computer. Apply the variation of the first isogonal along track. 2. What is the MEAN magnetic heading from A to B? Measure the MEAN TRUE TRACK at the mid meridian between A and B. Calculate the TRUE HEADING using the nav computer. Apply the variation at the mid meridian. 3. What is the FINAL magnetic heading from A to B? Measure the FINAL TRUE TRACK at meridian closest to B. Calculate the TRUE HEADING using the nav computer. Apply the variation at B. The above procedures are to be used for ATP exams whenever a heading (True or Magnetic) has to be calculated.


MID MERIDIAN TECHNIQUE - RHUMB LINE NAVIGATION Draw the straight line from A to B, this is the TRUE GREAT CIRCLE TRACK which cuts all meridians at a different angle.

Highlight the nearest MERIDIAN to the mid point along track, this is the mid meridian between A and B. Measure the mean great circle track at the mid meridian. At the mid meridian the mean great circle track is parallel to the rhumb line track so in effect we now have the rhumb line track from A to B. The HDG (T) is calculated by the nav computer, variation applied and the HDG (M) is flown between A and B. This is the basis of rhumb line navigation on a Lambert’s chart. All plotting of true headings between A and B is referenced to the mid meridian. Plotting or measuring wind direction is referenced to the Mid Meridian. This is the same as superimposing a rectangular grid similar to the meridians of a Mercator chart over the Lambert’s chart, the grid being parallel to the Mid Meridian. ATP exam - use the above technique. THE DEAD RECKONING (DR) POSITION The DR position is the calculated ground position of the aircraft. It may be plotted at any time during the flight by either the AIR PLOT or the TRACK PLOT method.


AIR PLOT METHOD Overhead position A at 0800. Heading 080° (T) TAS 180 Kts W/V 020/30

Plot the heading 080° (T) using the nearest meridian AHEAD of the aircraft. This is the AIR VECTOR. The length of the Air Vector is proportional to the TAS and time, TAS 180 Kts for 45 minutes = 120 NM. Plot the AIR POSITION + 120 NM along the AIR VECTOR. From the 0845 Air position plot the WIND VECTOR DOWNWIND. The wind is blowing FROM 020° (T). The length of the wind vector is also proportional to the wind speed and time. Wind Speed 30 Kts for 45 mins = 20 NM. Plot the DR POSITION 20 NM along the WIND VECTOR. This is the DR POSITION ∆ of the aircraft at 0845. The AIR PLOT method is flexible, alterations of heading and/or TAS may be made at any time.

The heading at 0830 is plotted from the first meridian ahead of the aircraft. As the heading is changed at 0852 and 0910 the headings are plotted from the first meridians ahead of the respective air positions. Note that the length of the wind vector will be 1 hour 03 mins of the wind speed (the Air Plot has been running from 0830 to 0933).


TRACK AND GROUND SPEED METHOD 0900 Overhead position A Heading 080° (T) 0930 Overhead position B

A line joining the two fixes is the TMG TRACK MADE GOOD, which is the actual track followed by the aircraft over the ground. Measure the distance A - B, say 120 NM which is computed against time between fixes of 30 mins and gives a groundspeed of 240 Kts. Extend the TMG and plot the DR position for the required time. Groundspeed 240 Kts for 12 mins = 48 NM. Note : This method can be used when there has been no alteration of heading or groundspeed

between the two fixes. WIND FINDING A flight plan is based on a met forecast that can be many hours old. The forecast W/Vs may be accurate, but if the pressure pattern changes unexpectedly, the W/Vs can be very different thus affecting the navigation of the aircraft. It is strongly recommended that the W/V is found in flight to improve navigation, update the ETA at checkpoints and monitor the fuel flight plan. Air Plot Method

Overhead A at 0900, Heading 082° (T), TAS 180 Kts. Fix at 0945. Plot the heading 082° (T) from the first meridian ahead of the aircraft in the direction of flight. The length of the AIR VECTOR is proportional to time, TAS 180 Kts for 45 mins, which is 135 NM and give the AIR POSITION + at 0945. This would be the position of the aircraft in zero wind conditions, BUT the aircraft is South of track BECAUSE the actual W/V is different from the planned W/V.


Join the AIR POSITION to the FIX which is the WIND VECTOR, annotate with 3 arrows, the wind blowing from AIR to GROUND. The wind direction is measured from the nearest meridian to the fix, in this case about 025° (T). The wind speed is proportional to the time that the AIR PLOT has been running. Wind Vector 33 NM for 45 mins = Wind Speed 44 Kts. W/V 025/44. The above method is suitable for short distances between fixes in the order of less than 250 NM. The heading may be altered at any time with no loss of accuracy.

As the heading has been changed at 0925 then the meridian at 0925 is used to plot the new heading. Note: For distances greater than 250 NM (ATP exam) the mid meridian between the two fixes is

used and all headings and the wind direction is measured from the mid meridian. Track And Groundspeed Method

This method is used when a single heading has been flown between fixes. Heading 082° (T), TAS 190 Kts. The TMG (Track Made Good) is measured at the mid meridian between the fixes - say 090° (T). The GS (Groundspeed) is proportional to the time between the fixes - 168 NM in 45 mins = GS 224 Kts. Using the nav computer: Heading 082° TAS 190 Kts TMG 090° GS 224 Kts W/V 306/45 This method is accurate over any distance.


POSITION LINES (P/L) or LINES OF POSITION (LOP) A Position Line is a line along which the aircraft is known to be at a particular time. The main source of Position Lines are radio bearings but coastlines, railway lines, rivers etc. may be used. Radio Bearings must be converted into TRUE BEARINGS (QTE) before they can be plotted from the meridian passing through the radio facility. QDM + Variation QUJ + 180° + 180° QDR + Variation QTE Follow the shortest route to convert one bearing to another. QDM MAGNETIC TRACK TO the station. QDR MAGNETIC BEARING FROM the station (VOR Radial) QUJ TRUE TRACK TO the station. QTE TRUE BEARING FROM the station. LAMBERTS CONFORMAL CHART As straight lines on the chart are GREAT CIRCLES and Radio Bearings are also GREAT CIRCLES no corrections are necessary to: VOR and VDF (VHF D/F) bearings Simply convert into a QTE using magnetic variation at the station as the bearing was measured at the station and plot from the Meridian passing through the station. NDB / ADF Bearings ADF bearings are measured at the aircraft, thus are measured with reference to the meridian passing through the aircraft and a correction for CHART CONVERGENCY has to be made before the bearing can be plotted from the meridian passing through the NDB. For the ATP plotting exam Chart Convergency in Southern Africa can be ignored over short distances (3° of Longitude change) as the CCF for the South African chart is 0.45°. QDM and QDR bearings - convert to QTE using aircraft variation.


RELATIVE BEARINGS RELATIVE + TRUE HEADING = QUJ + 180° = QTE USE OF SINGLE POSITION LINES Groundspeed check and revised ETA A position line at right angles to track can be used as a Ground speed check and to revise the ETA.

0800 Overhead A en route to B 0824 NDB XY QTE 182 What is the revised ETA at B? The distance along track from A to where the P/L cuts track is measured and computed against time to give the actual groundspeed. 70 NM run in 24 min = GS 175 From the P/L to B is 96 NM at GS 175 = 33 mins ETA 0857 Position Lines that cut the track within 15° of the perpendicular to the track are acceptable for groundspeed checks. Note that a P/L has single arrows at either end and a time. Track check A bearing from a radio facility that the aircraft has over flown is an indication of the TMG (Track Made Good). Example:

1205 Overhead NDB ABC, Heading 095° (T) 1221 NDB ABC bears 172 Relative


172 Relative + Heading 095 = QUJ 267 - 180 = QTE 087 TMG 087 Heading 095 Drift 8 Left Multi position lines fix Two or more position lines can be used to construct a fix. The ideal situation is that two position lines are obtained at the same time preferably at 90° to each other. 0956 Z VOR JSV Radial 090, DME range 105 NM JSV Variation 16 W Plot QTE 074 (T)

THE RUNNING FIX - TRANSFER OF POSITION LINES A position line is usually a bearing of the aircraft from a radio facility. If the radio facility were moved on a track parallel to that of the aircraft and at the same groundspeed, the bearing of the aircraft from the radio facility would remain constant.

Aircraft Track 090° (T) Groundspeed 240 Kts The aircraft at position A at 0900 Z obtains a QTE of 320° from VDF station X. At 0912 the aircraft will have flown 48 NM along track to position B. If the VDF station is imagined to travel from X to Y at the same speed of the aircraft, then XY is equal and parallel to AB and the line joining Y to B will be an imaginary position line parallel to AX. (The distance AB or XY is known as the run). The line BY drawn through the aircraft’s position at 0912 is known as a transferred position line and has two arrows and no time.


Transfer of circular (DME) position lines

0810 Overhead A, Heading 083° (T), TAS 195 Kts Track 090° (T), GS 180 Kts 0830 JSV VOR/DME Range 78 nm 0842 JSV VOR/DME Range 95 nm What is the position of the aircraft at 0842? The transfer of DME position lines is achieved by the Track & GS method of moving the DME station along a line parallel to the aircraft’s track at the aircraft’s groundspeed and plotting the original range of 78 nm. THE AIR PLOT TRANSFER OF POSITION LINES A situation may arise where heading changes are being made during a running plot which makes constant monitoring of a DR track difficult or even impossible. The AIR PLOT may then be used to transfer position lines and hence obtain a fix on the ground. Such a question may well be expected in the ATP exam although only for direct radio bearings (VOR, ADF, RMI) and not DME arcs. The techniques involves using time intervals relating to an air plot from a specific fix - the method cannot be accurately applied to an air plot from a DR position.


METHOD: i. Plot the air positions from the fix for the times of the bearings. ii. Plot the QTE's for each bearing. iii. Joint the 1027 air position (B) to any point E to give a convenient vector for the 22 minutes

flown since the fix at 1005 (A). iv. Repeat the above procedure between C and F for 29 minutes the same unit scale as that

used at 1027. v. From the 1040 air position (D) draw two lines, one parallel to line be (Line DH) and the

other parallel to CF (Line DI). Mark off an arc or 35 units radius to correspond to the time interval of 35 minutes from the last fix (A). The same scale must again be used.

vi. Redraw the position lines from point H and point I to establish the fix at 1040 (G). By

joining the 1040 air position to the fix, the mean wind can be established since 1005. TYPICAL THREE POSITION LINE FIX EXAMPLE 1 1200 Overhead A en route to B TAS 200 Kts, Heading 081° (T), DR Groundspeed 192 Kts 1223 X QTE 327 1230 X QTE 358 1240 X QTE 035

1. Plot the 3 QTE’s. 2. Inspect the 3 bearings to see if one crosses the track at approximately 90°. 3. If so, then discard the DR Groundspeed and perform a GS check. From 1200 to 1230 is

120 NM in 30 mins = GS 240 Kts. 4. Transfer the 1223 P/L 17 mins at GS 240 = 68 nm along track. 5. Transfer the 1230 P/L 10 mins at GS 240 = 40 nm along track. NOTE: If a GS check cannot be made then the DR groundspeed may be used.


EXAMPLE 2 The following VOR Radials are obtained from TIMBUKTOO. Aircraft track 045° (T), Groundspeed 180 Kts 0815 VOR TIM 288 0824 VOR TIM 321 0835 VOR TIM 348

As there is no fix from which to start the plot, the track of 045° is drawn to cut the 3 P/Ls. It is parallel to the actual track of the aircraft. A Groundspeed check cannot be made so the DR GS 180 is used. Transfer the 0815 P/L 20 mins at GS 180 = 60 nm along track. Transfer the 0824 P/L 11 mins at GS 180 = 33 nm along track. EXAMPLE 3 0700 Overhead PORT ELIZABETH VORTAC PEV, FL 230, RAS 225 kts, OAT -35° C, set

heading 016° (M). 0723 BURGERSDORP VOR BDV RADIAL 226 0734 BURGERSDORP NDB JF 326° relative by ADF.

a. Give the position at 0742.

b. Give the mean W/V experience since 0700. Use the air plot transfer of position lines to find the fix at 0742. Join the fix to the air position at 0742 to find the wind direction and speed. Remember that only 42 minutes has elapsed since 0700 and therefore the wind speed will be the

measured vector x 6072

knots.



THE CLIMB CONSTANT SPEED (RAS) CLIMB This is the normal climb technique used by commercial aircraft. Navigation wise the most important parameter required for a climb to cruising altitude is the mean TAS. This will occur at the mid point of the climb in TIME and not ALTITUDE. This applies to both piston and jet aircraft. The mid point of the climb in TIME occurs at approximately two thirds of the climb. Example 1 Climbing from Sea Level to FL 240 at RAS 175 Kts and a mean rate of climb of 800 ft/min, temperature ISA + 10°C. 24 000 ft x 2/3 = 16 000 ft mean climb altitude. ISA at Sea Level + 15° C 16 000 ft x 2° C/1 000 ft = - 32° ISA at 16 000 ft = - 17° C ISA + 10° = - 7° C By computer FL 160, temp - 7° C, RAS 175 Kts, TAS 227 Kts Example 2 Climbing from 6 000 ft to FL 270 at RAS 225 and a mean rate of climb of 1 250 ft/min, temperature ISA + 13° C. 27 000 ft 6 000 ft 21 000 ft x 2/3 = 14 000 ft 6 000 ft initial climb altitude 20 000 ft mean climb altitude ISA at Sea Level = + 15° C 20 000 ft x 2°/1 000 ft = - 40° C ISA at FL 200 = - 25° C ISA + 13°C = - 12°°C By computer FL 200, temp - 12° C, RAS 225 Kts, TAS 311. The above calculations were made with an electronic calculator which corrects for compressibility. If a manual nav computer is used the compressibility correction must be made according to the table below.


PRESSURE

ALTITUDE IN FT RECTIFIED AIR SPEED IN KT

200 250 300 350 400 450 500 550 10000 1.0 1.0 0.99 0.99 0.98 098 0.97 0.97 20000 0.99 0.98 0.97 0.97 0.96 0.95 0.94 0.93 30000 0.97 0.96 0.95 0.94 0.94 0.91 0.90 0.89 40000 0.96 0.94 0.92 0.90 0.90 0.87 0.87 0.86 50000 0.93 0.90 0.87 0.86 0.86 0.84 0.84 0.84

Ex. 2 FL 100, temp -12° C, TAS 225, TAS 316 x 0.985 = TAS 311 THE DESCENT As a CONSTANT RATE OF DESCENT is normally used, the temperature at the mid altitude is used to calculate the TAS and the W/V also at the mid altitude is used to calculate the groundspeed. Example 1 An aircraft cruising at FL 370 at GS 495 Kts obtains a fix at 1000 Z which gives a distance of 230 nm to go to destination. Descent details: Rate of descent 2 000 ft/min Mean descent GS 360 Kts Plan a descent to arrive overhead destination at FL 90. GS DIST TIME ETA 1000 FIX 495 146 18 1018 230 1018 TOD 360 84 14 1032 Calculate the descent distance and subtract from distance to go and complete the cruise sector. AIR PLOT ON THE CLIMB As the climb is of short duration, seldom longer than 25 mins, the INITIAL track is used to calculate the heading and groundspeed. At the TOC (Top of Climb) position which theoretically is on track, the Mid Meridian method is then used. The Air Position at the TOC is plotted, the mean climb W/V applied to give the TOC DR position. The heading for the cruise is plotted from the TOC Air Position and not the DR position. A heading, Air Vector or Air Plot can only be started from a positive fix, NEVER a DR POSITION. It may be continued from an Air Position.


QUESTIONS 1. 1000 Overhead BURGERSDORP, set heading 260(M), FL 100. RAS l72kts, OAT +6C, W/V 120/30. 1026 Alter heading 202(M). 1050 Alter heading GEORGE.

a) What is the magnetic heading to steer and ETA GEORGE? 2. 0800 Overhead UPINGTON, set heading 180((M), TAS l90kts, W/V 010/30. 0840 Alter heading 135((M). 0900 Alter heading 220((M). 0925 Alter heading for CTV.

a) What is the DR position at 0925? b) What is the magnetic heading to CTV? c) What is the ETA for CTV?

3. 0835 Overhead Cape Town VOR CTV, set heading for VICTORIA WEST (VWV), FL 100, OAT +2(C,

RAS l45kts, W/V 275/25. 0926 SUTHERLAND VOR Radial 147, DME 35.

a) What has been the mean W/V since 0835? b) What is the revised ETA VICTORIA WEST? e) What relative bearing should be maintained on the ADF in order to

Home on VICTORIA WEST NDB, VW? 4. 1513 Overhead DURBAN VOR DNV, heading 260((C), deviation 3(E, FL 125. OAT -3(C, RAS l64kts. 1555 Overhead UMTATA NDB, alter heading for EAST LONDON.

a) What is the ETA for EAST LONDON? b) What radial should be maintained on EAST LONDON VOR ELV?

5. 1300 Overhead DNV, pressure altitude 12000ft, OAT +5(C, RAS l50kts.Magnetic

heading 300(. 1345 Alter heading 230((M). 1420 Alter heading 175((M). 1435 Visual fix 3200S 02800E.

a) Determine the W/V since 1300. b) What is the mean magnetic heading to PEV? c) What is the ETA for PEV?


6. 0700 Overhead CTV, pressure altitude 15000ft, OAT -5(C, RAS l65kts, magnetic heading 090((M).

0745 Overhead SLV. a) What has been the mean W/V since 0700? From SLV alter heading for EAST LONDON, TAS l80kts. b) What is the initial magnetic heading to steer? 0805 QTE GGV 350( 0815 QTE GGV 008( 0825 QTE GGV 030(

c) What is the position at 0825? d) What has been the mean W/V since 0745?

0825 Alter heading for ELV.

e) What is the revised ETA for ELV?

f) What radial should be maintained to ELV? 7. 1530 Overhead GGV, pressure altitude 13000ft, OAT +3(C, RAS l55kts,

W/V 290/35, set heading for BLV. 1645 VW NDB relative bearing 225(. 1705 BDV radial 325(. 1715 JF NDB relative bearing 320(.

a) What is the position at 1715? b) What has been the mean W/V since 1530?

1715 Using the W/V found, alter heading for BLV.

c) What is the mean magnetic heading to BLV? d) What is the ETA for BLV? e) What relative bearing must be maintained on the ADF to home on BL

NDB? 8. 1723 Overhead KMV, set heading for PEV, FL 150, OAT -7(C, RAS l50kts,

W/V 225/20, compass deviation 2(E.

a) What is the mean compass heading for PEV?

1745 BLV DME 116mn 1810 BDV radial 280(. 1820 COOKHOUSE NDB CH RMI reading 158(.

b) What is the position at 1820? c) What has been the mean W/V since 1723?


1828 Alter heading for PEV?

d) What is the mean compass heading to steer? e) What is the ETA for PEV?

9. 1213 Overhead BULAWAYO VOR, set heading for HARARE, FL 110,

OAT +4(C, RAS l42kts, forecast WV 285/20. 1249 THORNHILL DME 35nm 1256 NORTON NDB relative bearing 333(.

a) What has been the mean W/V since 1213? b) What is the radial to maintain on the HARARE VOR?

10. 0740 Overhead KEETMANSHOOP VOR KTV, set heading for KMV, FL 115 OAT 0(C. RAS l42kts. As the forecast W/V is not available, steer the

track. 0825 UPINGTON NDB UP RMI reading 168(. 0843 UPV radial 030(. 0900 UPV DME 67nm.

a) What is the position at 0900? b) What has been the mean W/V since 0740?

0906 Alter heading for KMV.

c) What is the ETA for KMV? d) What bearing must be maintained on the RMI in order to home on KM

NDB? 11. 0537 Take off PIETERMARITZBURG for MAPUTO.

0540 Overhead PM at 3000ft. climbing to FL 130, constant RAS l38kts. mean OAT +8(C, mean forecast W/V 350/20, mean rate of climb 500ft/min. You are at TOC on ETA. Cruise RAS l62kts, OAT -2(C, forecast W/V 325/35.

0607 ESHOWE NDB ES RMI 166(.

0617 VRYHEID NDB VHD relative 280(. 0628 VRYHEID NDB VHD relative 246(.

a) What has the mean W/V been since 0540? Using the W/V found at 0628, determine the mean magnetic heading

To steer from the fix at 0628 to MAPUTO. 12. 0637 Overhead CHIREDZI NDB CZ, FL 120, TAS l97kts, heading 253((M). 0722 Overhead GREEFSWALD VOR, alter heading for ELLISRAS.

A descent is planned to arrive overhead ELLISRAS at 5000ff. Constant rate Of descent 500ft/min, mean TAS l73kts, mean forecast W/V 270/20.

a) What is the ETA overhead ELLISRAS?


13. 1020 Overhead DURBAN VOR DNV at 3000ft climbing to FL 140, maintaining radial 280.

Mean climb information; Constant RAS l40kts, rate of climb 500ft/min, temperature deviation ISA +5(C, W/V 300/30.

a) What is the mean magnetic heading for the climb? b) Give the DR position for TOC. c) Give the ETA for the TOC position.

1045 You reach the TOC position. Set heading 280(M), RAS l80kts. temp -10(C,

forecast W/V 280/30. 1130 MASERU VORTAC MZV DME 75nm. 1130 BURGERSDORP VOR BDV RMI reading 270(.

d) Determine the position at 1130. e) What has been the mean W/V since 1020?

1136 Divert to PORT ELIZABETH via COOKHOUSE NDB CH using a

forecast W/V of 360/45.

f) What is the ETA for CH? g) What relative bearing must be maintained on the ADF to home on CH?

14. 0545 Overhead BURGERSDORP VOR BDV, set heading for CAPE TOWN,

FL 160, OAT -5(C, RAS l45kts, W/V 035/15. 0650 VICTORIA WEST VORTAC VWV DME 85nm 0710 GEORGE VOR GGV radial 025.

a) What is the position at 0710? b) What has been the mean W/V experienced since 0545?

0720 Alter heading for CTV

c) What is the DR position at 0720?

A descent is planned to arrive overhead CTV at 3000ft, constant rate of Descent 500ft/min, mean temperature +10(C, mean WV 300/25 and mean RAS l60kts.

d) Give the mean magnetic heading to steer to CTV from TOD. e) What is the ETA for TOD and CTV?

15. 0630 Fix at 2900S 02730E, set heading for CAPE TOWN, FL 310, Mach

0.82, temperature -42(C, forecast W/V 300/70. a) What is the initial magnetic heading to steer? 0635 Doppler observation: Drift 2( left, GS 440kts. b) Give the spot wind being experienced.


The following readings are obtained from the BURGERSDORP VOR BDV: 0640 RMI reading 201(. 0647 RMI reading 163(. 0658 RMI reading 121(. c) What is the position at 0658?

0704 Fix 31°04S 023° 04E, FL 310, OAT -40°C, Mach 0.83, alter heading for CTV.

0727 Crossing SUTHERLAND VOR SLV radial 167.

A descent is planned to arrive overhead CTV at FL 050. Constant rate of descent 2000ft/min, mean TAS 390kts, mean W/V 260/40.

d) Determine the latest time to commence the descent. e) What is the ETA overhead CTV at FL 050?

16. 0643 Overhead CAPE TOWN VORTAC CTV at 3000ft, climbing to FL 290

on heading 007°(M) and at a constant RAS 220kts. Mean climb W/V 240/70, temperature deviation +3°C.

0705 Level FL 290, temp. dev. +5°C, Mach 0.79, forecast W/V 280/90, set heading for KEETMANSHOOP VOR KTV

a) Give the mean magnetic heading and ETA KTV

0710 NIEUWOUDVILLE VOR NVV RMI reading 109°.

0720 NVV RMI reading 180°. 0728 KLEINSEE NDB KZ RMI reading 253°. 0730 ALEXANDER BAY VOR ABV RMI reading 288°.

b) Give the position at 0730. c) What mean W/V has been experienced since 0643?

At 0730 the aircraft commences a descent at a constant rate while

Maintaining the present radial from KTV and arrives overhead KTV at FL 050 at 0803.

Assume for the descent: mean W/V 300/50, temp. dev. +10°C.

d) Give the radial to maintain to KTV. e) Give the magnetic heading to maintain the above radial f) What is the rate of descent required? g) What is the required mean RAS during descent?

17. 0315 Overhead SALISBURY VOR VSB, FL 070, set heading for MMV, climbing to

FL 180, mean rate of climb 600ft/min, constant RAS l6Okts. Forecast W/V 100/30, temp. dev. ISA + 10°C.

a) What is the initial magnetic heading to steer? b) What is the ETA TOC?


At TOC, FL 180, the RMI indication of VSB is 034 temperature -11°C, RAS 200kts.

c) Using the DRIFT experienced, alter heading to parallel the TRACK

Give the initial new magnetic heading to steer.

0343 THORNHILL VOR RMI reading 310°. 0400 FRANCISTOWN NDB FT relative bearing 055°. 0405 BULAWAYO VOR VBU RMJ reading 344° 0414 MESSINA NDB RMI reading 122°.

d) What is the position at 0414? e) What mean W/V has been experienced since 0315?

0414 Alter heading for MMV using W/V 140/35, FL 180, temperature -14°C,

RAS 208kts.

f) Give the mean magnetic heading and ETA for MMV. 18. 0610 Overhead KMV, FL 180, temperature -18°C, RAS 208kts, forecast W/V 260/60. Set heading CTV.

a) What is the mean magnetic heading and ETA CTV? 0654 KMV radial 250°. 0654 VWV radial 337°. b) Determine the W/V experienced since 0610.

0700 Using the above W/V alter heading for CTV. c) What is the initial heading (M) and revised ETA CTV?

You are instructed by ATC as follows: Enter the CAPE TOWN CTA (50nm) at FL 100 and commence descent at 500ft/min. as late as possible. Mean W/V for the descent is 200/30, RAS 180, Temp. dev. ISA + 8°C.

d) What is the mean pressure altitude, temperature, TAS and magnetic

heading for the descent? e) What is the ETA for TOD? f) What is the ETA for the CAPE TOWN CTA?

19. 1025 Overhead EAST LONDON, set heading for UPIINGTON, altitude

2000ft climbing to FL 310 at a constant RAS of 230kts, mean rate of climb 1200ft/min, forecast W/V 040/35, mean temp -22°C.

a) Give the initial magnetic heading for the climb. b) At TOC, estimate what QDR to expect from VICTORIA WEST VOR


1050 Level at FL 310, alter heading for UPINGTON, Mach 0.72, temperature -36°C, forecast W/V 010/65.

c) Give the mean magnetic heading and ETA UPINGTON. 1053 VICTORIA WEST VOR VWV QDR 113°

1107 VICTORIA WEST VOR VWV QDR 071° 1110 BLOEMFONTEIN DME range 132nm. d) Give the position at 1110 and the mean W/V experienced since 1025. 1116 Alter heading 340(M) TAS 434kts. 1126 SISHEN NDB SI RMI reads 020°. Doppler GS 390kts and drift 4° left. e) Give the spot W/V.

1126 UPINGTON DME range 100nm Using the spot wind found, alter heading for UPV.

f) Give the initial magnetic heading and ETA UPV. g) What relative bearing must be maintained on UP NDB?

20. 1030 Airborne LANSERIA. 1045 Overhead GRASMERE VOR GAV at FL 140, maintaining radial 245,

RAS l65kts, OAT -2°C.

1120 Abeam WELKOM NDB WM. Alter heading 220°(M) to avoid a thunderstorm. The drift observed on this heading 5° left.

1132 Alter heading 290°(M). The drift on this heading is 8° left. 1147 Alter heading 240°(M). The drift on this heading 7° left. 1200 Alter heading for SUTHERLAND.

a) What was the mean W/V experienced since 1120? b) What is the DR position at 1200? c) What is the mean magnetic heading to SUTHERLAND? d) What is the ETA for SUTHERLAND?

21. 1103 Airborne DURBAN, radar vectors to radial 150 DNV, 10 DME. 1109 DNV radial 150, DME range 10nm FL 050, climbing to FL 260, mean

rate of climb 700ft/min, RAS l85kts, mean temperature -11°C, forecast W/V 120/20.

Set heading on flight plan heading 221°(C), deviation °2W. a) Give the true heading, ETA and DR position for TOC. 1137 Level at FL 260, temperature -25°C, RAS 190, forecast WV 180/30. Alter heading 275°(M).

b) Give the ETA abeam COOKHOUSE.


1150 Crossing the coast inbound. 1203 EAST LONDON NDB bears 287° relative. 1215 EAST LONDON NDB bears 225° relative.

c) Give the aircraft’s position at 1215. d) What mean W/V has been experienced since 1109? 1227 Alter heading 250°(M), TAS 300kts, forecast W/V 200/30.

1233 Alter heading 323°(M) PORT ELIZABETH VOR PEV bears 181( by RMI. 1245 Alter heading 274( (M) GEORGE VOR GGV bears 248 by RMI 1300 SUTHERLAND NDB SL bears 310 ْby RMI.

e) Give the position at 1300. f) What mean W/V has been experienced since 1215? 1300 Set heading for WOLSELEY. CAPE TOWN Control requests your arrival at WOLSELEY at FL 100

not before 1348. From the 1300 position to TOD the W/V is forecast as 200/30, temperature -30(C. Descend at 800ft/min, mean TAS 240kts and mean descent W/V 140/30

g) Give the mean magnetic heading and ETA for TOD. h) Give the RAS while at FL260 that will ensure arrival at WOLSELEY at

1348. i) Give the TOD position. j) What will be the time in the descent?



CHAPTER 8

BRITISH ISLES CHART PLOTTING EXERCISES

1. 1000 Overhead EAGLE VOR EGL, FL 100, temperature -5oC, RAS 155kts,

W/V 050/30, set heading for STRUMBLE VOR STU. a) Give the mean magnetic heading and ETA STRUMBLE. 1045 EAGLE VOR EGL radial 140(. 1047 RUSH VOR RSH RMI reading 048o. 1047 KNIGHTON NDB KNI bears 337o relative by ADF. b) Give the position at 1047. c) What mean W/V has been experienced since 1000. 1100 Alter heading for STRUMBLE VOR STU using the W/V found. d) Give the mean magnetic heading and ETA STRUMBLE. 2. 1000 OLNO (50:35N 005:43E), altitude 3000ft, temp. dev. +7oC, set heading for

NEWCASTLE 311o(M), climbing to FL 240 at constant RAS 184kts. Mean W/V for the climb is 210/45.

1030 Level at FL 240, temperature -48oC, RAS 205kts, forecast W/V 220/45, alter heading

NEWCASTLE. a) Give the mean TAS for the climb. b) Give the TAS for the cruise at FL 240. c) Give the DR position at 1030. 1033 Fix at 52:25N 003:00E. 1043 CLACTON NDB CL RMI reading 214o. d) What is the true bearing to plot from CL? 1059 POLE HILL NDB PH RMI reading 280o. e) What is the true bearing to plot from PH? 1104 OTTRINGHAM NDB OT bears 279o relative by ADF. f) What is the true bearing to plot from OT? g) What is the position at 1104? h) Give the mean W/V experienced since 1033.


3. 0930 At position 56:00N 010:00W FL 350, temperature ISA, Mach 0.83, W/V

constant at 260/120. The autopilot is coupled to the INS en route to WPT 2, 59:00N 021:00W.

a) Whilst maintaining the required track, what is the aeroplane’s true heading at

longitude 12oW? b) What is the true heading at longitude 15oW? c) What is the true heading at longitude 21oW? 4. 2116 INS WPT 3 (60:00N 020:00W) enroute to MACKNISH VOR MAC WPT 4.

The following observations are recorded: Heading 145o(M), GS 561kts, ETA Scottish FIR/10oW at 2156, FL 410, Mach 0.81, temperature -52oC.

a) Give the W/V at 2116.

2136 Due to an electrical fault, the coupled INS ceases to function. The heading noted is 144o(M). Alter heading for MACKNISH VOR.

b) Give the aircraft’s most probable position (MPP) at 2136.

c) Using the W/V found at 2116, give the mean magnetic heading to reach MACKNISH and the ETA.

2145 TIREE NDB TI bears 339o relative by ADF. 2150 STORNOWAY NDB SN bears 299o relative by ADF. 2155 TORY ISLAND NDB TY bears 013o relative by ADF. d) Give the position at 2155. e) What mean W/V has been experienced since 2116?

2200 Alter heading for MACKNISH VOR using forecast W/V 270/100, Mach 0.84, temperature -50oC.

f) Give the mean heading (M) and ETA MACKNISH VOR. 5. 1012 Overhead SPIJKERBOOR VOR SPY, FL 200, temperature -15oC,

RAS 215kts, set heading for PRESTWICK VOR PWK, W/V 260/50. a) Give the mean magnetic heading and ETA PWK. 1036 CLACTON NDB CL bears 272o relative by ADF. 1046 OTTRINGHAM NDB OT bears 322o relative by ADF. 1101 OTTRINGHAM NDB OT bears 231o relative by ADF. b) Give the position at 1101. c) What mean W/V has been experienced since 1012?

1107 FL 200, temperature -20oC, RAS 220kts. Using forecast W/V 280/60, alter heading for PRESTWICK VOR PWK.

d) Give the mean magnetic heading to PWK. e) Give the revised ETA PWK. f) What radial must be maintained to reach PWK?


6. 1220 Overhead PRESTWICK VOR PWK, FL 310, temp. dev. +5oC, Mach

0.80, forecast W/V 270/120. Set heading for ROCKALL ISLAND (57:33N 013:48W).

a) Give the mean magnetic heading and ETA ROCKALL. 1235 TIREE VOR TIR RMI reading 036o.

1235 With the VOR PWK selected and the OBS set to 306 FROM, the CDI indication on a five dot indicator shows three dots fly right.

b) What mean W/V has been experienced since leaving PWK? 1241 Using a mean W/V of 310/80 alter heading for ROCKALL ISLAND. c) What is the mean magnetic heading and revised ETA?

d) Assuming that the VORTAC BEN at BENBECULA is within range on ETA ROCKALL, what will be the RMI reading and DME range of BEN?

7. 1235 WPT 8 (53:00N 015:00W) set heading for RUSH VOR RSH, FL 370,

temperature -57oC, Mach 0.80, W/V 320/80. a) Give the mean magnetic heading and ETA RUSH. 1240 EAGLE NDB EG RMI reading 078o. 1245 SHANNON VOR SNN RMI reading 122o. 1250 EAGLE NDB EG relative bearing 325o. b) Give the position at 1250. c) What mean W/V has been experienced since 1235?

1256 Alter heading for WPT 9 (53:30N 004:00W), FL 370, temperature -53oC, Mach 0.80, forecast W/V 260/80.

d) Give the mean magnetic heading and ETA WPT 9. 1306 RUSH VOR RSH radial 277°. 1309 BELFAST VOR BEL radial 202°. 1311 RUSH VOR RSH radial 134°. e) Give the position at 1311. f) Give the mean W/V experienced since 1250.

g. Using the W/V found, give the new mean magnetic heading and revised ETA for WPT 9.


8. 0720 Overhead SOLA at FL 310, temp. dev. ISA - 4oC, Mach 0.74, forecast

W/V 030/60, set heading for BENBECULA. a) Give the mean magnetic heading and ETA BENBECULA. 0804 ABERDEEN VOR AND radial 293°. 0808 TIREE NDB TI bears 337o relative by ADF. 0814 STORNOWAY ground DF station gives a QDM of 354o, class A. b) Give the position at 0814. c) What mean W/V has been experienced since 0720?

0820 Alter heading for BENBECULA VOR BEN using the W/V found at 0814, FL 310, temp. dev. ISA - 6oC, Mach 0.80.

d) What is the mean magnetic heading and revised ETA BENBECULA?

9. 0935 Overhead ABBEVILLE VOR ABB, FL 50, set heading for position A

(53:40N 000E/W) climbing to FL310 at constant RAS 186kts. Mean W/V 290/30, mean temp -21°C, mean rate of climb 2000ft/min.

a) Give the initial magnetic heading.

b) What is the DR position for TOC?

0950 Overhead DOVER VOR DVR, level at FL 310, RAS 220kts, temp. -48oC, forecast W/V 310/70, alter heading for position A.

c) Give the mean magnetic heading. 1008 DOVER VOR DVR RMI reads 172o. 1014 LICHFIELD VOR LIC RMI reads 257o. 1017 OTTRINGHAM VOR OTR RMI reads 306o. d) Give the position at 1017. e) What mean W/V has been experienced since 0950?

1023 Alter heading for ABERDEEN VOR AND using forecast W/V 270/100, FL 310, RAS 224kts, temperature -51oC.

f) Give the mean magnetic heading. 1051 St. ABBS VOR SAB radial 074°. 1051 ADN VOR radial 168°. g) Give the position at 1051. h) What mean W/V has been experienced since 1017?


10. 1119 Overhead STRUMBLE VOR STU, FL 100, temperature 0oC, RAS

160kts, W/V 360/35. Set heading for COMBRAI VOR CMB. a) Give the ETA abeam MIDHURST VOR MID. b) Give the mean magnetic heading to COMBRAI. c) Give the initial magnetic heading. d) Give the ETA COMBRAI. 1132 BERRYHEAD VOR BHD RMI reading 180o. 1142 BERRYHEAD VOR BHD RMI reading 205o. 1200 BERRYHEAD VOR BHD RMI reading 244o. e) Give the position at 1200. f) What mean W/V has been experienced since STRUMBLE? 1208 Alter heading for COMBRAI using forecast W/V 300/40.

g) Give the radial to maintain to CMB.



CHAPTER 9

ADVANCED PLOTTING CRITICAL POINT (CP) The PET/CP is a point along the track from which it would take equal time to continue to destination or to return to the point of departure given certain performance criteria. It is important to measure all distances from the same point on the chart using the appropriate mid-latitude scale. THE CP CONSTRUCTION

Construct a perpendicular bisector between departure point X and destination Y. The bisector cuts the track at the midpoint distance (Z). Z will be the PET in zero wind conditions. Measure all distances with reference to the meridian closest to this point. Plot the appropriate W/V at Y, blowing towards the track. Remember that wind always blows from the AIR POSITION to the GROUND POSITION. From W scribe an arc of the TAS to cut the track at A. Parallel heading WA through Y to cut the bisector at B. From B parallel the wind vector WY to cut the track at the PET. Measure the Lat. and Long. of this position.


THE CRITICAL POINT (one engine) CONSTRUCTION To plot the position of the CP, use the one engine inoperative TAS for the construction described above. To calculate the ETA for the CP, the normal multi-engine G/S is used. POINT OF NO RETURN (PNR) AND RADIUS OF ACTION (RA) The PNR is that point along the track beyond which it is calculated that an aircraft cannot continue and still return to its point of departure, given the forecast W/V. For practical reasons, fuel is usually kept in reserve and under such conditions, the PNR is referred to as the Radius of Action. PNR - Maximum endurance (dry tanks) RA - Safe endurance (reserves) THE PNR/RA CONSTRUCTION

(a) A dry tank position (maximum range) or safe range position A is calculated by: ENDURANCE x G/S OUT

b) Construct a perpendicular between X and A. c) Plot the W/V at A, blowing towards track. d) From W draw an arc of TAS to cut the track at B. e) Parallel the heading WB through A to cut the bisector at C.


f) From C parallel the wind vector WA to cut the track at the PNR. g) Measure the Lat. and Long. of this position. PET/CP WITH ALTERNATE DESTINATION An aircraft will usually fly from X to Y with alternate Z. A PET may therefore be required which includes the time to fly to the alternate destination in the construction. THE PET WITH ALTERNATE CONSTRUCTION

a) Construct a perpendicular bisector between Y and Z. b) Plot the W/V at Y, blowing towards the track. c) From W draw an arc of TAS to cut XY at A. d) Parallel the heading WA through Y to cut the bisector at B. Note: If the construction shows that by paralleling WA through Y, the bisector cannot be

intercepted or the PET does not lie on the line XY then it will be quicker to fly direct to the destination than to the alternate and there is no PET.

e) From B parallel the wind vector to cut the track at the PET. f) Measure the Lat. and Long. of the position.

g) A second PET between X and Y can be constructed. When the aircraft is between the two PET’s, it will be quicker to divert to Z than to continue to Y or return to X.


POINT OF NO ALTERNATE (PNA) The aircraft may not be able to carry enough fuel to fly from X to Y and then divert to Z. In this case a PNA will have to be constructed. The PNA is the latest time and position at which an aircraft may depart from track in order to arrive over an alternate airfield either with or without reserves as the case may be. THE PNA CONSTRUCTION

a) Calculate a dry tank or safe endurance and extend the track XY to position A by:

ENDURANCE x G/S OUT b) Construct a perpendicular bisector between A and Z. c) Plot the W/V at A, blowing towards the track. d) From W draw an arc of TAS to cut the track at B. e) Parallel the heading WB through A to cut the bisector at C. f) From C parallel the wind vector to cut the track at the PNA. g) Measure the Lat. and Long. of the position.


ALTERNATIVE METHOD (Only Track and G/S of one aircraft known) In most cases, the aircraft to be intercepted is following a known track at a specific ground speed but the heading and TAS may be unknown to the intercepting aircraft. An alternative construction is therefore necessary to accomplish the interception.

a) Update positions of both aircraft to a common time A and B. b) Join A to B. AB is the LINE OF CONSTANT BEARING. c) Plot AC, A’s TRACK and G/S for one hour. d) Parallel AB through C. Line A1B1. e) From B draw in the W/V, blowing AWAY from B, line BD. f) From O draw an arc at B’s TAS to cut A1B1 at E.

g) Join B to E to cut AC at F and hence reveal the track, G/S and position for aircraft B to intercept aircraft A. DE is then B’s heading for the intercept.

EXAMPLE 1 1700z Aircraft A: TAS 240kts, heading 090o(T). 1715z Aircraft B: TAS 250kts, bears 170o(T) and 80nm from A. W/V 340/25


DETERMINE: a) The track that B must make good to intercept A. b) The time that the aircraft will be 25nm apart. (hint: use the line of constant bearing) SOLUTION:

* Update position of A to 1715 * Complete construction as previously described. * Construct track from B to H and measure at mid meridian.

* Scribe an arc 25nm from A along the line of constant bearing. Parallel aircraft A’s heading from point I to J. The point along BG at which the aircraft will be 25nm apart will be at J.

* Using B’s TAS, calculate the time to reach J and add to 1715. NB! Remember that the wind vector is related in magnitude to the time flown. EXAMPLE 2 0900z Aircraft A: G/S 190kts, track 080o(T), drift 5o right. 0900z Aircraft B: TAS 210kts, bears 110nm and 80o relative to A. W/V 290/30 DETERMINE: a) The heading B must steer to intercept A. b) The time required for B to intercept A. c) The distance that the aircraft will be apart at 0930. d) The closing speed of B on A.


SOLUTION:

* Calculate and draw in B’s true position relative to A at 0900. (QTE = 80o - 5o + 80o = 155o)

* Continue with construction as previously described to find B’s heading and time for the intercept.

* Determine A or B’s position along their respective tracks, corresponding to the time

of 0930. Parallel the line of constant bearing through the 0930 position and measure the separation distance between aircraft.

* Divide the original distance that the aircraft were apart (110nm) by the time taken to

make the intercept to determine the closing speed.


INTERCEPTION An aircraft, in the course of its operation, may be required to carry out an interception mission on another aircraft. It follows that a plotting construction method is necessary in order to predict a particular heading and time for one aircraft to intercept the other. THE INTERCEPT CONSTRUCTION (HDG and TAS of both aircraft known)

a) Update the aircraft positions to a common time A and B. b) Join A and B. AB is known as the LINE OF CONSTANT BEARING. c) Plot AC, A’s TAS and heading for one hour. d) Parallel AB through point C, line A1B1. e) From B draw an arc of B’s TAS to cut A1B1 at F. f) BF is the INTERCEPT HEADING (Note: not track) g) G is the AIR POSITION (+) of interception. h) Calculate the intercept time: A to G at A’s TAS = Time to intercept = B to G at B’s TAS i) Apply the W/V to point G to find the GROUND POSITION of

interception. NOTE: The wind velocity vector must be relative in its magnitude to the time to intercept. e.g. If

the interception took 45 minutes then the wind vector would be 0.75 x the forecast velocity in length.


ADVANCED PLOTTING EXERCISES 1. Plan a flight from JAN SMUTS to PORT ELIZABETH. The TAS is 150kts and the forecast

W/V is 180/30.

Determine the position along track where the time to PEV will be equal to the time to return to JAN SMUTS.

2. A flight is made from KEETMANSHOOP KTV to BURGERSDORP BDV. Fuel available

excluding reserves is 1400lbs. Fuel consumption is 340lbs./hr, TAS 170kts, W/V 200/30. Determine the radius of action in time and distance from KEETMANSHOOP. 3. A flight is made from FRANCISTOWN to GROOTFONTEIN with GHANZI as alternate. The

TAS is 350kts and the W/V is 210/40.

Determine the position along track that is equidistant in time from FRANCISTOWN and GHANZI.

1. 1215 UTC, overhead ALEXANDER BAY, heading 345o(T), TAS 250kts and W/V 040/20.

Determine the latest time and position at which to depart from track in order to arrive overhead KEETMANSHOOP AT 1435.

5. 1200 UTC, overhead GROOTFONTEIN on a direct flight to UPINGTON. Any diversion will

be to ALEXANDER BAY routing via KEETMANSHOOP. The TAS is 300kts. Forecast W/V between GROOTFONTEIN and UPINGTON is 135/30. Forecast W/V between KEETMANSHOOP and ALEXANDER BAY is 150/40.

Determine the latest time and position at which the aircraft must depart from the present track to arrive overhead ALEXANDER BAY at 1432.

6. Overhead HARARE at 1036. Set heading for PIETERSBURG. Pressure altitude 16000ft,

temp dev. ISA - 8oC, RAS 226kts, forecast W/V 070/55. Endurance excluding reserves 01:30.

Determine the following:

a) The furthest point along track from which a diversion to BULAWAYO may be made leaving the reserve fuel intact.

b) The co-ordinates of the PNA. c) The mean magnetic heading and ETA at the PNA. d) The mean magnetic heading and ETA from the PNA to BULAWAYO.


7. 1240 UTC, FIX at 29:00S 026:10E on a direct flight to GEORGE, endurance 01:50, excluding reserves. Any diversion will be to alternate PORT ELIZABETH via COOKHOUSE NDB CH. Forecast W/V for the entire area is 250/30.

If the TAS throughout the flight is 240kts, determine the latest time and position at which to depart from the direct track in order to arrive over PORT ELIZABETH with reserves unused.

8. Aircraft A leaves GROOTFONTEIN at 0800 on track to MMABATHO, G/S 140kts. Aircraft B leaves UPINGTON at 0830 to intercept aircraft A, TAS 250kts. W/V affecting both aircraft is 090/30. Determine the following: a) The time of interception. b) The position of interception. c) The true heading and G/S of aircraft B while intercepting A. d) The closing speed of B on A. 9. Aircraft A leaves CAPE TOWN at 1800 on a track 030o(T), TAS 300kts. Aircraft B leaves VICTORIA WEST at 1825, TAS 280kts to intercept aircraft A. W/V affecting both aircraft is 060/30. Determine the following: a) The time and position of interception. b) The true heading and G/S of aircraft B to intercept aircraft A. c) The closing speed. d) The time and position at which B will first be within 25nm of A. 10. Aircraft X is overhead LIVINGSTON NDB LI on a track to WINDHOEK NDB WH, TAS

300kts. At the same time aircraft Z, TAS 420kts, is overhead PIETERSBURG to intercept aircraft X. The W/V for the entire area is 045/50 and the visibility is 30nm. Determine the following: a) The time interval for the interception. b) The mean magnetic heading for Z to steer to intercept X. c) The distance from PIETERSBURG to the intercept point. d) The closing speed of the two aircraft. e) The estimated time to the first possible visual contact.


ANNEX A

SAMPLE QUESTIONS


QUESTIONS 1. It is 1823 ST on 10 October at Norfolk Island. What is the Standard Time (ST) at Haiti? a) 2053 on 09 October; b) 1523 on 09 October; c) 0153 on 10 October.

(2) 2. A spy ship in keeping Zone Time (ZT) at 115° 23’W. A revolutionary coup has started in

Guyana (42° 20’W) at 0635 ST on 12 September. What is the GMT of the coup? a) 0935 on 12 September; b) 0335 on 12 September; c) 1235 on 12 September.

(2) 3. With reference to the above question, what is the LMT on the ship at the time of the coup? a) 1718 on 12 September; b) 0242 on 12 September; c) 0152 on 12 September.

(2) 4. With reference to the above question, what is the Zone Time on the ship at the time of the

coup? a) 1735 on 12 September; b) 0135 on 12 September; c) 0235 on 12 September.

(2) 5. What is the Rhumb Line (RL) track and distance from Seychelles (S 04° 40’ E 055° 32’) to

Mauritius (S 20° 35’ E 057° 39’)? a) 174° / 951 NMS; b) 176° / 943 NMS; c) 172° / 963 NMS.

(4) 6. An aircraft takes 18 minutes 24 seconds to cover a distance of 7.3 cm between A and B on

a chart. The scale of the chart is 1:2000 000. What is the aircraft’s groundspeed? a) 257 KTS; b) 265 KTS; c) 275 KTS.

(2)


7. An aircraft flying from A to B (distance 422 nm) is 10 nm to the right of track with 122 nm to

go to B. What alteration of heading is required to fly to B? a) 2° right b) 5° left c) 7° left

(2) 8. The position of A is N 38° 14’ W 132º 20’. B is on the same parallel of latitude. The great

circle bearing of B from A is 278º. What is the longitude of B? a) W 106º 29’; b) W 158º 11’; c) E 179º 10’.

(3) 9. The Standard Parallels (SP’s) on a Lambert’s Chart are 20º N and 40º N. The RL track

from A (40º N 160º W) to B (20º N 150º E) is 245º. What is the GC track from A to B at the anti meridian of Greenwich?

a) 249.5º b) 252.5º c) 247.5º

(2) 10. On a globe representing the earth, the circumference of the 54º S parallel measures 48 cm.

What is the distance in cm, measured along a meridian from the 54º S parallel to the equator on the globe?

a) 10.02 cm; b) 14.87 cm; c) 12.25 cm.

(3) 11. An aircraft leaves Buenos Aires (S 34º 35’ W 058º 20’) on a track of 026º T and flies a

distance of 3898nm. What is its final position? a) N25º 50’ W 025º 33’; b) N23º 49’ W 028º 33’ c) N20º 57’ W 018º 47’

(4) 12. The scale of a Mercator Chart is 1:3000 000 at the Equator. An aircraft flies from a (S 43º

32’) W 008º 00’) to B S 43º 32’ E 009º 37’). The aircraft is overhead A at 0900 and over the Greenwich Meridian at 0940. TAS and W/V are constant throughout the flight. What is the ETA at B?

a) 1018; b) 1008; c) 1028.

(2)


13. You have to land at Cape Town (E 018º 30’) before last light at 1806 LMT on 10 October.

The Flying time is 7 hours 45 minutes. What is the latest time to get airborne from Argentina to land at Cape Town? Give your answer in Argentinean ST.

a) 0607; b) 0907; c) 1207.

(3) 14. What is the GMT for first light at Stockholm (N 59º 40’ E 017º 55’) on 10 Jan ? a) 0759 b) 0647 c) 0747

(2) 15. A southern hemisphere Lambert’s Chart (n=0,8) has a datum meridian at 40º W. An

Aircraft’s heading is 300º M and 330º Grid (G). Variation is 10º E. What is the aircraft’s longitude?

a) 015º W; b) 022º W; c) 005º W.

(3) 16. A track represented by a straight line is drawn on a PS chart from position A (S 80º 00 E

150º 00’) to B S 75º 00’) W 150º 00). The track is 080º T at longitude 180º E/W. What is the initial track at A?

a) 117º T; b) 110º T; c) 097º T.

(2) 17. With reference to the previous question, what is the final track at B? a) 050º T; b) 054º T; c) 046º T.

(2) 18. With reference to the previous question, the scale of the chart is 1:992 047 at 80º S. What

is the scale at 75º S? a) 1:989 763; b) 1:982 609; c) 1:980 098.

(3)


19. An aircraft flying East along the 10º S parallel covers 13.335 cm in 30 mins as measured on

a Mercator Chart. The scale of the chart is 1:1 000 000 at 25º S. What is the ground speed of the aircraft?

a) 156 KTS; b) 200 KTS; c) 176 KTS.

(3) 20. Aircraft heading is 281º C. Variation is 21º E. Deviation is 2º W. The aircraft obtains a

relative bearing of 131º in the Northern hemisphere CA is 2º. What is the QTE to plot? a) 171º; b) 253º; c) 176º.

(2) 21. On a westerly flight along the 60º N parallel, the 030º W Meridian is crossed 2 hours after

the Greenwich Meridian is crossed. What is the aircraft’s groundspeed in KPH? a) 450 KPH; b) 1688 KPH; c) 834 KPH.

(2) 22. The position of A is N 42º 13’ W 158º 24’. B is on the same parallel of latitude. The great

Circle bearing of B from A is 278º T. What is the longitude of B? a) W 179º 13’; b) E 177º 47’; c) E 175º 57’.

(3) 23. An aircraft leaves A (S 15º 35’ E 017º 52’) on a rhumb line track of 137º T. What is the

position of the aircraft after flying for 1500 NMS? a) S 33º 52’ E 036º 39’; b) S 33º 19’ E 036º 55’; c) S 34º 02’ E 037º 06’;

(5) 24. A Mercator Chart is drawn to a scale of 1:1 000 000 at 56º N. The position of A is N 52º 00’

E 005º 00’. Position B is approximately South West at A at a chart length of 58 cm. The change of longitude between A and B is 8º. What is the position of B?

a) N 48º 56’ W 003º 00’; b) N 50º 04’ W 003º 04’; c) N 52º 00’ W 003º 00’.

(4) 25. On a P.S. Chart of the Southern Hemisphere, the scale at the pole is 1:5 000 000. The

radius of the earth is 250 000 000 inches. What is the chart length in CMS between 84º S and 71º S?


a) 28,7839; b) 29,1934; c) 33,8406.

(4) 26. The following positions are plotted on a P.S. Chart of the Northern Hemisphere. Position A

N 75º 00’ W 100º 00’ position B N 75º 00’ W 030º 00’. The scale of the chart is 1:3 450 000 at 82º N. What is the Great Circle track from A to B?

a) 065º T; b) 055º T; c) 045º T.

(2) 27. With reference to the previous question, what is the Great Circle track from B to A? a) 305º T; b) 055º T; c) 290º T;

(2) 28. With reference to the previous question, what is the longitude of the most Northerly point

along the track? a) 065º W; b) 135º W; c) 005º E.

(2) 29. What is the Zone number for longitude E 088º? a) +6; b) -6; c) -5.

(1) 30. Aircraft A : Mach 0.76 OAT -40º C W/C -80 KTS ETA for overhead VOR ABC is 0250 Aircraft B : Mach 0.80 OAT -40º C W/C -60 KTS ETA for overhead VOR ABC is 0235


At what time should aircraft A reduce speed to Mach 0.70 in order to arrive at VOR ABC 20

minutes after aircraft B? a) 0203; b) 0208; c) 0212.

(3) 31. Aircraft P (G/S 240 KTS) arrives at X at 0900 en-route to Y. Aircraft Q (G/S 320 KTS)

arrives at X at 0918 and passes aircraft P at Y. What is the distance X to Y? a) 120 nms; b) 288 nms; c) 300 nms.

(3) 32. Solve by any method: Aircraft A track 030º T. G/S 200 KTS. Aircraft B track 300º T G/S

250 KTS. What is the relative velocity of B with respect to A? a) 300º / 300 KTS; b) 245º / 315 KTS; c) 260º / 320 KTS.

(3) 33. An aircraft (G/S 280 KTS) estimates overhead VOR ABC at 0630 Z. ATC requests the

aircraft to cross VOR ABC at 0642 Z, due to traffic. How far from VOR ABC will the aircraft have to reduce speed by 60 KTS to arrive at 0642 Z?

a) 205 nms; b) 56 nms ; c) 44 nms .

(3) 34. Two aircraft at the same flight level are approaching a VOR. Aircraft A (G/S 390 KTS) is

260 nms from the VOR at 0350 Z. At what time must aircraft B reduce groundspeed from 634 KTS to 390 KTS in order to ensure a 5-minute separation at the VOR?

a) 0417; b) 0422; c) 0427.

(2) 35. The following details apply to an aircraft:

Track 075º T. G/S 180 KTS. At 10h00, the VOR needle on the RMI indicates a QDM of 020º. AT 10h10, the VOR needle on the RMI indicates a QDM of 290º. The local variation is 10º E. What is the aircraft distance from the station at 10h10?

a) 18.5 nms ; b) 21,2 nms ; c) 15.3 nms .

(4)


36. At 10h26, an aircraft at FL 250, temperature –35º C, M 0.81, headwind 30 KTS is estimating position Charlie at 11h16. To reach position Charlie at 11h20, the aircraft must reduce speed to M 0.70 at?

a) 1052; b) 1056; c) 1100

(3) 37. The Rhumb Line track from position A (N 32º 22’ W 064º 42’) to B (N 16º 42’ W 022º 57’)

crosses the Meridian of 040º W at which latitude? a) N 23º 31’; b) N 25º 07’; c) N 26º 54’.

(3) 38. The duration of evening civil twilight at Boston USA (N 42º 22’ W 071º 00’) on JAN 27 is? a) 20 mins; b) 25 mins; c) 30 mins.

(3) 39. 100 nms is equal to 30 cms on a chart. What is the chart length in cms of a straight line

representing 80 kms? a) 12,97 CMS; b) 24,0 CMS; c) 1,297 CMS.

(2) 40. On a south P.S. chart, position A (S 75º 00’ E 136º 00’) and position B (S 75º 00’ W 152º

00’), are joined by a straight line. What is the initial G.C. track from A-B? a) 126º T; b) 234º T; c) 306º T.

(3) 41. A grid is aligned with the Greenwich Meridian on a south P.S. chart. At position A (S 75º

00’ W 040º 00’), the grid heading is 220º G. What is the true heading? a) 180º; b) 220º; c) 260º.

(3)


42. The following positions are plotted on a south P.S. chart. A (S 85º 00’ E 140º 00’) and B (S 75º 00’ W 120º 00’). The initial G.C track from A to B is 125º T. What is the track at the anti-meridian of Greenwich?

a) 085º T; b) 095º T; c) 165º T.

(2) 43. The aircraft is at DR position S 78º 00’ E 170º 00’. The aircraft is steering a heading of

280º T and obtains a relative bearing of 299º to an NDB situated at S 80º 00’ E 160º 00’. The bearing to plot on a south P.S. chart is?

a) 029º T; b) 039º T; c) 049º T.

(3) 44. Position A and B, both at latitude N75: 00 are plotted on a Polar Stereographic Chart, the

initial great circle track from A at N75: 00 W 107:00 to B is 326º (T), and the longitude of B is:

a) E 141: 00; b) W034:00; c) E 039:00.

(2) 45. In order to arrive at position S20:00 W172:00, 15 minutes before the end of the evening

civil twilight on JAN 14th, an aeroplane on a six hour flight must leave position S08:00 E178:00 at:

(The end of Evening Civil Twilight for this position and date is 18 HRS 12 LMT). a) 1230 LMT JAN 15th; b) 1215 LMT JAN 15th; c) 1222 LMT JAN 14th.

(3) 46. An aeroplane departs from A N60:00 E118:45 at 0600Z on January 2nd flying due west at a

mean groundspeed of 300 KTS and lands at Sunset the same day, at destination in longitude?

a) E083:15; b) E071:39; c) E101:00.

(4) 47. The scale of Polar Stereographic Chart is 1:992 047 at S80:00 and at S75:00 the scale is? a) 1:999 640; b) 1:1 016 966; c) 1:982 609.

(4)


48. On a Conical Orthomorphic Chart, one standard parallel is N40:00 and measures 63,88 cm. The length of the other standard parallel is 41,7 centimetres at latitude:

a) N20:00; b) N55:00; c) N60:00.

(3) 49. On a Mercator Chart the perpendicular distance between parallels N34:00 and N35:00 is 2

centimetres and the scale at N30:00 is: a) 1:5 815 000; b) 1:5 559 500; c) 1:5 432 000.

(4) 50. A Mercator Chart is constructed to a scale of 1:5 000 000 at the Equator, and the width of

the graticule between the Greenwich meridian and E030:00 is: a) 26.25 CM; b) 69,49 CM; c) 66,65 CM.

(3) 51. During a flight along the N48:00 parallel, the measured distance between fixes X and Y 25

minutes apart is 17,9 CM on a Mercator chart. The scale of the chart is 1:3 000 000 at N15:00, and the aeroplanes ground speed is therefore:

a) 434 KTS; b) 450 KTS; c) 482 KTS.

(3) 52. An aircraft over flies position A on a heading of 067º ( C), variation 21W, deviation 3E,

intending to maintain a track of 058º (T). A fix is obtained 105 NM from A which places the aircraft 7 NM to the left of the intended track. The heading (M) to be steered to return to A from the fix is:

a) 239º (M); b) 257º (M); c) 260º (M);

(2) 53. The initial great circle track from A (S41:27 W171:24), to B (S41:27 E161:24) is: a) 288º (T); b) 279º (T); c) 261º (T).

(2)


54. On a Lambert’s Chart in the Northern Hemisphere, a straight line is drawn from A to B, the

track measured at A is 070º (T). An aircraft leaves A on a constant heading of 070º (T), zero wind conditions, and will pass:

a) north of B; b) overhead of B; c) south of B.

(2) 55. Aircraft A, GS 280KTS, passes point X 7 minutes ahead of aircraft B, GS 326 KTS. Aircraft

B passes the point Y 6 minutes ahead of aircraft A. The distance from X to Y is: a) 429,7 NM; b) 411,9 NM; c) 400,7 NM.

(3) 56. An aircraft flies from A (N48:47 E169:20). On a Rhumb Line track of 090º (T) to B, the

distance is 450 NM. The longitude of B is: a) W 179:17; b) E 179:43; c) W 179:24.

(2) 57. The position of A is N40:00 E008:00, B is on the same parallel. The great circle track from

A to B is 276º. The longitude of B is: a) W001:20; b) W010:40; c) E026:40.

(2) 58. An aircraft obtains a QDM of 073º from a VDF station in position N30:00 W020:00, variation

at the aircraft is 17E and 19E at the VDF. Convergency 4º. The bearing to plot on a Lambert’s chart is:

a) 272º; b) 270º; c) 268º.

(2) 59. Two meridians, 174 W and 172 E, in the Southern Hemisphere, have a convergency of 7º

on a Lambert’s chart. The great circle track from A (W168:00) to B (E172:00) measures 250º (T) at A. The Rhumb Line track from B to A is:

a) 075,0º; b) 073,5º; c) 080,0º.

(3) 60. Two aeroplanes at the same flight level estimate the same position at the same time.

Aeroplane A, GS 200 KTS is flying due North. Aeroplane B, GS 300 KTS is flying due East. Aeroplane A’s velocity relative to B is:


a) 303º at 360 KTS; b) 056º at 360 KTS; c) 034º at 400 KTS.

(3) 61. The pilot of an aeroplane TAS 360 KTS, heading 309º (T), makes the following

observations using Airborne Search Radar (ASR): Time Relative bearing Range NM 1036 015 47 1042 055 20 Using the navigational computer the local W/V is: a) 248/48 KTS; b) 055/47 KTS; c) 010/44 KTS.

(3) 62. An aeroplane leaves position X (S30:00 W 168:45) at sunrise on January 18th for position Y

(S30:00 E 165:18), reaching Y at sunset on January 19th. The average GS for the flight is: a) 126 KTS; b) 98 KTS; c) 101 KTS.

(4) 63. An aeroplane passes overhead NDB A at 1023Z, TAS 260 KTS, heading 060º C (M).

Variation is 17E. At 1046Z the radio compass is tuned to the NDB A, which gives a steady relative bearing of 178. Simultaneously the aeroplane passes over NDB B, 110NM from A. The wind velocity is:

a) 335/35; b) 277/29; c) 237/29.

(4) 64. The following wind velocities are forecast for the climb to cruising altitude. W/V 270/15; W/V 240/27 W/V 180/42. The mean W/V for the climb is: a) 230/28; b) 214/23; c) 195/27. (3) 65. A flight of 2 152 KM is flown along the 69th parallel on a track of 270º (T). The LMT of

arrival at the destination is the same as the LMT of departure, after a flight of: a) 1 hour 12 minutes; b) 3 hours 36 minutes;


c) 6 hours 40 minutes. (4)

66. An aeroplane departs from P (S22:55 E008:10) for Q following a Rhumb line track of 340º

(T) and crosses the equator in longitude; a) W009:00; b) W000:21; c) W008:19.

(5) 67. The great circle bearing of A from B in 094º (T), the Rhumb line bearing of B from A is 270º

(T), the change of longitude between A and B is 10 degrees and the mean latitude between A and B is approximately:

a) S36:52; b) N36:52; c) S53:08.

(2) 68. Two aeroplanes on the same track at the same flight level are approaching Waypoint 4.

The first at a groundspeed of 240 KTS, is 280 NM from WPT 4 at 0620Z. The second at groundspeed 300 KTS, is 315 NM from WPT4 at 0626Z. In order to ensure a 50 NM separation at WPT 4, the second aeroplane must reduce groundspeed to 240 KTS at:

a) 0630Z; b) 0635Z; c) 0645Z.

(5) 69. A track is drawn from N70:00 E100:00 to N70:00 W140:00 on a Polar Stereographic chart.

At its most northerly point this track will be: a) at a tangent to N80:00 latitude; b) North of N80:00; c) south of N80:00.

(4) 70. The track of an aeroplane represented by a straight line on a Polar Stereographic chart

measures 328º (T) at position N74:00 E015:20, and re-crosses latitude N74:00 at: a) W079:20; b) W116:20; c) W100:40.

(2)


71. A Lambert’s chart (Northern Hemisphere n factor 0,8) has an overprinted grid based on

W40:00. The grid heading of an aeroplane crossing E30:00 (variation 10 degrees west) is 250 (G) and the heading magnetic is:

a) 268; b) 316; c) 300.

(3) 72. On a Lambert’s projection the meridians through A S50:00 E008:30 and B S48:00 W006:30

converge at an angle of 10,095º and the parallel of origin of this projection is: a) S42:18; b) S49:12; c) S55:40.

(3) 73. At 1530Z an aeroplane, track 190º (M), groundspeed 360 KTS observes a QDM of 240 to

an NDB. At 1540Z the QDM to the NDB was 330. The closest distance of the aeroplane from the NDB was:

a) 20 NM; b) 30 NM; c) 40 NM.

(4) 74. The Standard Time of moonrise at Cape Town S33:58 E018:36 on January 2nd was:

a) 0002; b) 0048; c) 0116.

(3)


ANNEX B ANSWERS TO QUESTIONS


Chapter 1 1.

i) 086° ii) 092° 2.

i) Northern hemisphere

ii) 325°


3. CONV° = dLONG° x SIN LAT = 1° x SIN 25° = 0.42° CONV° = dLONG° x SIN LAT 0.42° x 2 = 1° x SIN LAT

0 84

2. °

= SIN LAT

57° 42' N/S = LAT 4.

CONV° = dLONG° x SIN LAT 16° = dLONG° x SIN 40°

16

SIN 40°°

= dLONG°

24° 53' = dLONG° 24° 53' east of 170° E = 165° 07' W


5. DEP (nm's) = dLONG' x COS LAT = 15° x 60' x COS 30° = 900' x COS 30° = 779.4 nm's 6. DEP (nm's) = dLONG' x COS LAT 480 Kts x 19 HRS = 360° x 60' x COS LAT 9120 nm's = 21600' x COS LAT

9120

21600 = COS LAT

65° = LAT 7. Track 360° : 480 Kts x 3 HRS = 1440 nm's Track 270° : 480 Kts x 2 H 30 = 1200 nm's Track 180° : 480 Kts x 4 HRS = 1920 nm's Track 360° : 1440 nm's = 1440' = 24° Track 270° : DEP (nm's) = dLONG' x COS LAT 1200 nm's = dLONG' x COS 20° N + 24° N

1200

COS 44° = dLONG'

1668' = dLONG' 27° 48' = dLONG°


Track 180° : 1920 nm's = 1920' ÷ 60 = 32° 8. The sum of two longitudes 13° 30' and 166° 30' = 180°. Therefore, A and B are on anti-

meridians. The shortest distance will therefore be along the meridian, over the north pole, and along the anti-meridian. There is no departure involved.

90° - 65° = 25° 90° - 78° = 12° 25° = 1500' 12° = 720' 1500' = 1500 nm's 720' = 720 nm's DISTANCE A TO B = 1500 nm's + 720 nm's = 2220 nm's


CHAPTER 2

PART 1 1. Take the scale from 18° N to the equator.

1

SCALE OF EQUATOR =

1SCALE AT 18 N

COS 181°

×°

= 1 COS 18

11000000×

= COS 181000000

°

= 1

1051462

Take the scale from the equator to 36° S.

1

SCALE AT 36° =

1SCALE OF EQUATOR

SEC 361

×

= 1 1

COS 361051462×

= 1

850651

2.

This question may be calculated at any latitude, provided that the scale is used at the latitude. The scale is given at 50° S, work the problem there.


SC = CLED

1

400000 =

21 cmED

ED = 21 cm x 400000 ED = 840000 cm ED = 45.33 nm's DEP (nm's) = dLONG' x COS LAT 45.33 nm's = dLONG' x COS 50°

dLONG' = 45.33 nm's

COS 50'

dLONG' = 70.5' dLONG° = 1° 10.5 ' 3. Calculate the scale at 27° N

1

SCALE AT 27° =

1SCALE AT 60

COS 601

1COS 27°

×°

×°

= 1 COS 60

1 1

COS 27250000×

°×

°

= 1

445503

SC = CLED

1

445503 =

18"ED

ED = 18" x 445503 ED = 8019059" ED = 109.9 nm's

4. The scale at the equator is smaller, the scale at 60° N is larger. 1

e.g. Equator 1:1000000 60N 1:500000


The smaller the scale, the larger the earth distance e.g a l0cm long line

At the equator ED = 1000000 x 10 cm ED = 53.96 nm’s

At 60 N ED = 500000 x 10cm ED = 26.98 nm’s

The line at the equator represents the larger earth distance.

5.


6.

7.

08° N = 478.31 MP 08° N + 16° 27' S 16° 27'S = 994.22 MP = 24° 27' 1472.53 SMP = 1467 nm

16° 30' W + 004° 18' E COS 40.3° = 1467 nm

x

= 20° 48' x = 1467 nm

COS 40.3°

= 1248' x = 1923.5 nm's Rhumb line distance

tan ∅ = 1248'

MP1472 53.

= 40.3° Thumb line track = 180° - 40.3° = 139.7°


8.

27° 27' = 1703.09 MP COS 25° =

x4087 nm

89° 11' = 16972.37 MP x = 4087 x COS 25° 15269.38 MP x = 3704 nm x = 3704' x = 61° 44' + 27° 27' x = 89° 11' S

tan 25° = x

15269.28

x = tan 25° x 15269.38 x = 7120.18' dLONG x = 118° 40' dLONG + 14° 28' E x = 104° 12' W


9.

10° 18' S = 617.17 MP

tan 40° = x

617.17

x = 617.17 x tan 40° x = 517.9' x = 8° 38' dLONG 8° 38' + 002° 03' = 010° 41' E 10.


dLONG = 12° 12' dLONG = 732'

tan 43° = 732'

x

x = 732'

tan 43°

x = 785' or 785 MP's 21° 37' N = 1320.29 MP = 785.00 - NEW LAT = 535.39 MP NEW LAT = 08° 57' N 11. Determine the chart length of 1 min long or 1 MP

SC = CLED

1

1000000 =

CL1' COS 60 185300× ° ×

CL = 92650 cm1000000

CL = 0.09265 cm (per MP/MIN LONG)


37° N = 2378.54 MP 22° N = 1344.92 MP = 1033.62 DMP dLONG = 165° E to 178° W = 17° = 1020' x² = a² + b² x² = 1033.62² + 1020² x = 1452.16 MP 1452.16 MP at 0.09265 cm's per MP = 134.5 cm 150 cm's can accommodate 1618,99 MPS at 0.09265 cm/MP 60° N = 4507.08 MP 1618,99 MP 2888.09 MP 2888.09 MP = 43° 30' N 12. Find the scale at 50°N.

1

2 500 000 COS 20

1 1

COS 50 = 1

1 710 101× ×

Find the earth distance at 50°N.

SC = CLED

1

1 710 101 = 22,5 CM

ED

ED = 207,6 NM’s Find the G/S

G / S DT

=

= 733 KTS


13. On a Mercator chart, the scale expands with latitude and 1:15 000 000 is a larger scale than 1:2 000 000, therefore:

1

2 000 000 COS 30

1 = 1

2 309 401 (EQUATOR SCALE) ×

1

2 309 401 1

COS x = 1

1 500 000 ×

1

COS x 1

1 500 000 = 2 309 401

1 ×

1

COS x2 309 4011 500 000

=

COS x =1 500 0002 309 401

COS x = 0,64952 x = 49° 30’ N 14. This is a Merdional Parts scale question. 39°N = 2 530.20 MP’s 37°N = 2 378.54 MP’s 151.66 DMP’s = 4 cm

4 cm

151.66 MP's0,02637 cm / MP=

But an MP is equal to 1 minute of longitude which at the equator is equal to 1 NM or 185

300 cms. Calculate the scale at the equator.

SC = CLED

= 0,02637 cm185 300 cm

= 17 025 649

Calculate the scale at 30°N.

1

7 025 649 1

COS 30 = 1

6 084 391×


Part 2 1. The chart convergency factor = SIN LAT P of O 0.766 = SIN LAT P of O 50° N = LAT P of O The P of O is halfway between the standard parallels. If one standard parallel is at 40 ° N,

the other must be at 60° N. B.

i) The great circle track at A = 315° - convergency (dLONG° x SIN LAT P of O) = 315° - (15° x SIN 30°) = 315° - 7.5° = 307.5°


ii) The great circle track at B = 315° - convergency (dLONG° x SIN 30°) = 315° - (15° x SIN 30°) = 315° - 7.5° = 322.5° iii) The rhumb line track from B to A = 142.5° - CA = 142.5° - (½ dLONG° x SIN LAT P of O) = 142.5° - (½ 30° x SIN 30°) = 142.5° - 7.5° = 135° 3.


4.

5.

Whatever the two meridians on this sketch are, they are the same two meridians at 30°N

and at X°N. Therefore, the dLong at 30°N and at X°N is the same. 30°N 30°N DEP (CM) = dLONG’ × COS 30° DEP (CM) = dLONG’ × COS 30°

dLONG’ = DEP(CM)COS 30

dLONG’ = οXCOSDEP(CM)

But, the two dLONG’s are the same, therefore:

DEP(CM)COS 30

= 38 CMCOS X

50 CM

COS 30= 38 CM

COS X

COS X = 38 CM COS 3050 CM×

COS X° = 0,658 X° = 48° 50’N


6. Remember that if the aircraft leaves X on a constant heading of 060°T, it will be flying a

rhumb line and cut each meridian at 060°T.

The aircraft will pass North of position Y. 7. 011° `13’ W 7. The great circle is a straight line and the only reason why a G.C. track of 52°T should

become anything else (in this case 84°T) is because of the convergency of the Meridians. What we are looking for is what dLONG would give us a convergency of 84° - 52° =

32°. CONV° = dLONG° × SIN LAT P of O 32° = dLONG° × SIN 40°

dLONG° = 32

SIN 40

dLONG° = 49° 47’ (East of 061°W) = 11° 13’ W

61°W 17°W

27°N

47°N32° conv

52° 84°


Part 3 1.

i) Great circle track A to B = 270° (RL) + CA = 270° + 39° = 309° ii) Great circle track B to A = 090° (RL) - CA = 090° - 39° = 051° 2.

Great circle track A to D = 270° (RL) + CA = 270° + 74° = 344°


3.

i) The longitude of position B is 132° W. ii) The great circle track at the anti-meridian of Greenwich is: 120° - CONV 120° - 12° 108° 4.

i) The great circle track A to B = 270° + CA = 270° + 33 = 303° If the drift is 5° R, the great circle heading A - B = 303° - drift = 303° - 5° = 298° ii) To determine the highest latitude which this line will attain, take the triangle out of

the circle.


The highest latitude which this track will reach will be indicated by the shortest line

between the apex and the base of the triangle, or the perpendicular bisector.

SIN 57° =

x18°

x = 18° x SIN 57° x = 15° 06' The highest latitude is 90° - 15° 06' = 74° 54' N NOTE: In truth, the highest latitude which this track reaches will in fact be south of

74° 54' N, because once this triangle is put back onto the chart, we would see the effect of the parallel spacing increasing as we move away from the pole.


5.

i) The great circle track from A - B = 270° - CA = 270° - 55° = 215° ii) Determine the QUJ at the Prime meridian. The great circle track (QUJ) = 215° + CONV = 215° + 47° = 262° QUJ Zero deviation. Variation + 15° W = 277° QDM 6. The term ‘ROLLING FIT’ means that at 70°N, the Mercator and Polar stereographic share the same scale. Calculate the Mercator scale at 70°N.

11 000 000

1COS 70

= 1342 020

×

(This is the scale on the Polar Stereographic at 70°N.) Calculate the Polar Stereographic scale at 90°N.

1342 020

COS CO - LAT1

2 2×

1

1342 020

COS (90 - 70)1

2 2×

1

1

342 020 COS 10

1 = 1

352 645

2

×


Part 4 1.

2.

Now superimpose this drawing onto the polar stereographic chart.


Longitude 70° W Convergence = convergency = dLONG (diagram solves east or west) 3.

Now superimpose this drawing onto the polar stereographic chart.


Longitude = 50° W Convergence = convergency = dLONG (diagram solves east or west) 4.

The aircraft heading is 100° T. 5.

Aircraft heading 100° G.


6.

Convergency = convergency = dLONG° x CCF = 20° x 0.5° = 10° This grid heading is 070° G. 7.

Convergence = convergency = dLONG x n = 40° x 0.5 = 20° The true heading is 180°, therefore the magnetic heading is 195°.


8.

Convergence = convergency = dLONG x SIN LAT P of O 40° = dLONG x 0.5

dLONG = 400 5°

.

dLONG = 80° The datum meridian is at 130° E.


9.

Convergence = convergency = dLONG x CCF 15° = dLONG x 0.5

dLONG = 150 5°

.

dLONG = 30° The datum meridian is 0° E/W.


CHAPTER 3 1.

X Y

A - 370 Kts1205

290 Kts - B1225

290 Kts - B1205

960 nm 97 nm

Move B back to 1205. 20 minutes x 290 Kts = 97 nm's

i) T = RDRS

= 960 + 97370 + 290

= 1 HR 36 + 1205 = 1341 ii) Aircraft A flies for 1 HR 36 from position X before passing B. 1 HR 36 x 370 Kts = 592 nm's from X 2.

290 nm

X Y

A - 350 Kts1205

B - 450 Kts1230

Move aircraft A on to 1230.

146 144

B - 450 Kts1230

A - 350 Kts1230

X Y

25 minutes x 350 Kts = 146 nm's

i) T = RDRS

= 146 - 30450 - 350

= 10 minutes + 1230 = 1240


ii) Calculate the time for aircraft A to reach Y.

T = 144 nm350 Kts

T = 25 minutes

i) T = RDRS

RS = RDT

= 146 - 130

25 mins

= 38 Kts Aircraft B must be 38 Kts faster than aircraft A to close from 146 nm to 130 nm in 25

minutes. 350 Kts + 38 Kts = 388 Kts (new speed for B) 450 Kts - 388 Kts = 62 Kts (reduction in speed)


3.

100 nm 200 nm

B - 450 Kts1215

A - 350 Kts1225

Y

Move aircraft B on to 1225. 10 mins x 450 Kts = 75 nm.

25 nm 200 nm

B - 450 Kts1225

A - 350 Kts1225

Y

i) T = RDRS

= 25 - 20

450 - 350

= 3 minutes + 1225 = 1228 ii) When aircraft A reaches position Y, aircraft B must be 2 minutes behind at 350 Kts. 2 mins x 350 Kts = 12 nm

T = RDRS

= 25 - 12

450 - 350

= 8 minutes + 1225 = 1233


4.

10 mins50 nm X Y

B - 300 Kts A - 250 Kts A - 250 Kts B - 300 Kts

10 mins42 nm

T = RDRS

= 50 + 42

300 - 250

= 1 HR 50 (1 HR 50 x 250 Kts) + 42 nm = 502 nm X - Y

5. DIST FROM DEST = DELAY OLD G / S NEW G / S

DIFF IN G / S× ×

= 0.25 420 370

50× ×

= 777 nm from Y 6.

117 nm 700 nm

B - 470 Kts1200

A - 350 Kts1200

YX

B - 470 Kts1215

Move B back to 1200. 15 minutes x 470 Kts = 117 nm


i) When aircraft A reaches position Y, aircraft B must be at position Z, 50 nm before Y. This will occur at time.

T = DS

= 700350

= 2 HRS + 1200 = 1400 (A at Y and B at Z) Aircraft B's original estimate for position Z would have been:

T = DS

= 117 + (700 - 50)

470

= 1 HR 38 + 1200 = 1338 Aircraft B must therefore delay its arrival by 1400 - 1338 = 22 minutes.


DIFF IN G / S× ×

= 0.36 470 320

150× ×

= 368 nm from Z (speed reduction point)


T = DS

= 117 + (700 - 50 - 368)

470

= 51 minutes + 1200 = 1251 ii) Aircraft A's arrival over point Y is at 1400. Aircraft B must therefore arrive over point

Y at 1410. B's original estimate for Y was at time:

T = DS

= 117 + 700

470

= 1 HR 44 + 1200 = 1344 Aircraft B must therefore delay its arrival over position Y by 1410 - 1344 = 26 minutes.


DIFF IN G / S× ×

= 0.433 470 320

150× ×

= 434 nm from Y (speed reduction point) B's estimate for the speed reduction point is:

T = DS

= 117 + (700 - 434)

470

= 0 HR 49 + 1200 = 1249


7.

i) SIN A

a = SIN B

b

SIN A = SIN B a

b×

= SIN 40 250

280° ×

= 35° (relative bearing from A - B) Angle C = 180° - 40° - 35° = 105° ii) Solve for Side C (relative speed)

c

SIN C = b

SIN B

C = b SIN C

SIN B×

C = 421 Kts

T = RDRS

= 250421

= 36 minutes + 1305 = 1341


iii) T = RDRS

= 250 - 20

421

= 33 minutes + 1305 = 1338


Chapter 4 Part 1 1. DeP 1407 LMT 3rd 429 - A/T DEP 0938 UTC 3rd 912 + FT ARR 1850 UTC 3rd 058 - A/T ARR 1752 LMT 3rd

2. ARR 2208 ST 2nd 1100 + SF ARR 0908 UTC 3rd 706 - FT DEP 0202 UTC 3rd 1136 + A/T DEP 1338 LMT 3rd

3. ARR 1823 ST 4th 100 - SF ARR 1723 UTC 4th 633 - FT DEP 1050 UTC 4th X - A/T DEP 0838 LMT 4th i) 1050 - x = 0838 x = 1050 - 0838 x = 2 H 12 2 H 12 x 15° PER HOUR = 33° W 4. If the LMT of departure and arrival are the same, the aircraft obviously covers the same

dLONG per hour as the sun which is 15°n per hour. Flight time 3 hours x 15° per our = 45° or 2700' DEP = dLONG' x COS LAT 1580 nm = 2700' x COS LAT

COS LAT = 15802700

LAT = 54° 11' N/S


Part 2 1. End of evening civil twilight on JAN 11 10° S is 1850 LMT. ARR 1850 ST 11th 415 + A/T ARR 2305 UTC 11th 420 - FT DEP 1845 UTC 11th 129 - A/T DEP 1716 LMT 11th 2. Arrival is at sunset 60° N JAN 2nd 1506 LMT Departure is at 0715 LMT 7 HR 51

What is the speed of the sun at 60° N? (15° per hour) DEP = dLONG' x COS LAT = 15° x 60' x COS 60 = 900 x COS 60 = 450 Kts How far does the sun travel in 7 HR 51? 7 HR 51 x 450 Kts = 3532,5 nm's. This distance is being covered at the closing speed of both aircraft and sun (300 Kts +

450 Kts).

T = RDRS

= 3532.5 nm's

750 Kts

= 4 HRS 43 How far does the aircraft fly in this time? 4 HR 43 x 300 Kts = 1412 nm's DEP = dLONG' x COS LAT 1412 = dLONG' x COS 60


dLONG' = 1413

COS 60

dLONG' = 2826 ' dLONG° = 47° 06' EAST OF 15° LONG = 62° 06' EAST If you struggle to conceptualise this problem, imagine the departure of the aircraft at 1200

LMT (sun overhead the aircraft). How long until sunset? How far does the sun travel in this time? At the combined speed of aircraft and sun, how long does it take to open this distance? How far does the aircraft travel in this time?

3. Sunset for 60°N on JAN 15 is: 15H31 LMT. Departure time is: 0800Z + 6H34 (A/T) = 14H34 LMT The time between departure and arrival is: 15H31 - 14H34 = 0H57 At 60°N the speed of the sun is 450 Kts. 450 Kts × 0H57 = 427.5 NMS. If the sun and the aircraft are both going west, the relative velocity is: 450 Kts - 300 Kts = 150 Kts

T RDRS

=

=427.5 NMS150 KTS

= 2H51 How far does the aircraft fly in this time? 300 Kts × 2H51 = 855 NM DEP (NM) = dlong’ × COS LAT 855 NM = dlong’ × COS 6O°N 1710 = dlong’ 28° 30’ = dlong° West of 098° 30’E is 70°E


Part 3 1. Moonrise at 62° N 0° E/W 0047 LMT 1st DIFF 47 MINS to F4 table - 23 Moonrise at 62° N 090° E 0024 LMT 1st -600 A/T 1824 UTC 3rd ADD 24 HOURS + FULL DAILY DIFF (47 x 2) 2534 1958 UTC 1st 2. Moonset at 50° S 0° E/W 1303 LMT 1st DIFF 38 MINS to F4 table + 25 Moonset at 50° S 120° W 1328 LMT 1st +800 A/T 2128 UTC 1st 3. Moonrise at 52° S 0° E/W 2334 LMT 1st DIFF 6 MINS to F4 table + 2 Moonrise 52° S 060° W 2336 LMT 1st +400 A/T 0336 UTC 2nd Subtract 24 hours + FULL DAILY DIFF (2 x6) -2412 0324 UTC 1st Chapter 5 1. Calculate the aircraft's drift and its track made good (TMG)

11

132 60

1× = 5° RIGHT

TMG = 234° The aircraft steered 3° right (232° T - 229° T). The aircraft drifted 5° right (234° T - 229° T). The wind is responsible for 2° right drift (5° - 3°) To return: 234° T - 180° = 054° T 054° T + 2° Right for wind = 056° T 056° T + 15° W variation = 071° M


2. HDG 143° C HDG 125° HDG 125° DEV 3°- RB 67°+ RB 107°+ 140° M QUJ 192° QUJ 232° VAR 15°- 180°- 180°- HDG 125° T QTE 012° QTE 052°

a² = b² + c² = 2bc COS A a² = 83² + 76² - (2 x 83 x 76 x COS 40º) a² = 6889 + 5776 - (9664) a = 54.78 nm 54,78 nm in 10 mins = 328.6 Kts

SIN B

b =

SIN Aa

SIN B = SIN A b

a×

B = 76.9° 192° - 76.9° = 115.1° (aircraft track) TRACK 115.1° T HDG 125° T W/V 239/61 G/S 328.6 Kts TAS 300 Kts


3. Rate one turn = 360° in 2 minutes, therefore the aircraft does 3½ turns. First do an air plot, then apply the wind for 7 minutes. CIRC = 300 Kts x 2 mins = 10 nm CIRC = π d

d = CIRCπ

d = 3.18 nm D = 7 mins x 25 Kts D = 292 nm From point A : Bearing 180° : Distance 6.1 nm 4. Airship : DEP = dLONG' x COS LAT = 360° x 60' x COS 0 = 21600 nm

S = DT

= 21600280

= 77.14 Kts Aircraft : S = 77.14 x 3 = 231.42 Kts D = S x T = 231.42 x 70 HRS = 16199.4 nm DEP = dLONG' x COS LAT 16199.4 = 21600' x COS LAT

COS LAT = 16199 421600

.

LAT = 41° 25' N/S


5. C = 2πr = 2 x π x 6800 km = 42726 km

½ C = 42726

2

= 21363 km

T = DS

= 213632900

= 7 H 22 If the earth were stationary, Oculus would cross the equator at 60° W and 120° E. By the time Oculus reaches the equator, the earth has rotated in an anti-clockwise direction 7 H 22 at 15°/HR = 110° 30'. 120° - 110° 30' = 09° 30' E DEP A 1400 LMT 27th 400 + A/T DEP A 1800 UTC 27th 722 + FT ARR B 0122 UTC 28th 038 + A/T ARR B 0200 LMT 28th


Chapter 7 1. a) HDG 251°(M); ETA 1125 2. a) DR position 32°46S 022°30 E

b) HDG 280°(M) c)ETA 1027 3. a) W/V314/34

b) ETA 1011 c) RB 011°

4. a) ETA 1620

b) Radial 049 ELV 5. a) W/V 323/12

b) HDG 253° c) ETA 1530

6. a) W/V 127/70

b) INI HDG 109°(M) c) FIX 32°35’S 023° 17’E d) W/V 2 18/48 e) Revised ETA 0936 f) Radial 300 inbound

7. a) FIX 2952S 02526E

b) W/V 348/33 c) HDG 057(M) d) ETA1738 e) RB OO8° on BL

8. a) HDG 196°(C)

b) FIX 31°45’S 024°41’E c) W/V 114/14 d) HDG 174°(C) e) ETA 1907

9. a) W/V 293/30

b) Radial 227 on VSB 10. a) FIX 27° 53’ S 022° 22 E

b) W/V264/13 c) ETA 0952

d) RMI 1320 on KM 11. a) W/V 303/44

b) HDG 023°(M) 12. a) ETA 0807


13. a) HIDG 288°(M)

b) DR 30°05’ 030°00’E c) ETA TOC 1042 d) FIX 30°40’S 027°18’E e) W/V287/39 f) ETA 1207

g) RB 355° on CH 14. a) FIX 33°13’S 022°23’E

b) W/V 306/39 c) DR 33°30’S 021°55’E d) HDG 289°(M) e) ETA TOD 0756 ETA CTV 0822

15. a) INT HDG 262°(M)

b) W/V 261/49 c) FIX 30°41’S 024°10’E d) ETA TOD 0737 e) ETA CTV 0750

16. a) HDG 005°(M) ETA 0749

b) FIX 28°33’ S 018°17’E c) W/V 247/88 d) Radial 193 KTV e) HDG 004°(M) f) ROD 727ft/min. g) Mean RAS l90kts

17. a) INI HDG 212°(M)

b) ETA TOC 0333 c) INI HDG 219°(M) d) FIX21°58’S 028°55’E e) W/V015/31 f) HDG 225°(M); ETA 0521

18. a) HDG 254°(M) ETA 0809 b) W/V 251/44 c) INT HDG 247°(M);Revised ETA 0804 d) PA. 14000FT; Temp. -5°C; TAS 225kts; HDG 242°(M) e) ETA TOD 0738 f) ETA CTA 0754

19. a) INI HDG 335°(M)

b) QDR 120( on VWV c) HDG 337((M); ETA 1137 d) FIX 30(18’S 024(11’E ;W/V358/45 e) W/V352/53 f) INI HDG 321((M); ETA 1141 g) RB 354( on UP


20. a) W/V 340/30

b) FIX 29(06’S 024(45’E c) HDG 255((M)

d) ETA 1319 21. a) HDG 198((T); ETA TOC 1139; DR Pos. 31(51’S 030(13’E

b) ETA 1227 c) FIX 32(27’S 026(46’E

d) W/V 147/20 e) FIX 32(51’S 022(37’E f) W/V 111/20 g) HDG 273((M); ETA TOD 1328 h) RAS l45kts i) DR Pos. TOD 33(10’S 020(49’E j) Time on descent 20min.


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Chapter 8 NOTE: The plotting method of transferring the aircraft’s meridian to the NDB is correct,

however when the question asks for the QTE to plot the convergency must be calculated and applied to the bearing to get the QTE to plot using the NDB’s meridian. Both methods give the same bearing plotted.

1. a) HDG 128°(M); ETA 1119

b) FTX 52°49’N 007°04’W c) W/V 040/37

d) HDG 114°(M); ETA 1118 2. a) TAS 24l kts

b) TAS 286kts c) DR Pos.52°08’N 003°24’E d) QTE 028° from CL e) QTE 092° from PH f) QTE 043° from OT g) F1X 53°55’N 000°17’E h) W/V 258/49

3. a) HDG 290°(T)

b) HDG 288°(T) c) HDG 285°(T)

4. a) W/V 258/130

b) MPP 58°33’N 014°33’W c) HDG 146°(M); ETA 2214 d) FIX 56°59’N 010°13’W e) W/V 261/108

f) HDG 136°(M); ETA 2214 5. a) HDG 296°(M); ETA 1141

b) FIX 54°25’N 000°30’W c) W/V 235/51 d) HDG 295°(M) e) Revised ETA 1140 f) Radial 117° From PWK

6. a) HDG 299°(M); ETA 1314

b) W/V 276/93 c) HDG 312°(M); Revised ETA 1310 d) BEN RMI 104°; DME 206nm

7. a) HDG 088°(M); ETA 1313

b) FIX 53°33’N 011°07’W c) W/V 252/122 d) HDG 105°(M); ETA 1318 e) FIX 53°24’N 005°49’W f) W/V 290/86 g) HDG 087°(M); Revised ETA 1318


8. a) HDG 272°(M); ETA 0814

b) FIX 57°11’N 005°32’W c) W/V 336/83 d) HDG 340°(M); Revised ETA 0826

9. a) INI HDG343°(M)

b) DR Pos.TOC 50°59’N 001°25’E c) HDG 342°(M) d) FIX 53°29’N 000°26’E e) W/V 282/75 f) HDG 328°(M) g) FIX 56°05’N 001°28’W

h) W/V 275/88 10. a) ETA 1212

b) HDG 105°(M) c) INI HDG 104°(M) d) ETA 1259 e) FIX 51°I9’N 001°06’W f) W/V 318/43

g) Radial 294° inbound to CMB.











278° 028° QTE’s plotted using the NDB stations meridian and calculated convergency


294° 255° 328° QTE’s plotted using the NDB stations meridian and calculated convergency


248° 222° QTE’s plotted using the NDB stations meridian and calculated convergency


058° QTE’s plotted using the NDB stations meridian and calculated convergency



Chapter 9 1. PET: 30°53’S026°41’E

2. RA: 02:07 EET DIST: 341nm from KTV

3. PET.with alternate: 20°02’S 020°19’E

4. ETA: 13:35 P0S: 23°24’S 015°04’E

5. ETA: 13:36 P0S: 26°30’S 020°31’E

6. a) (Practical plotting) b) PNA: 22°34’S 029°48’E

c) HDG: 198°(M) ETA: 11:32 d) HDG: 359°(M) ETA: 12:06

7. PNA: 14:11 Pos: 32°20’S 023°40’E

8. a) 09:54

b) INT. 22°38’S 021°31’E c) HDG: 009°(T) G/S: 247kts d) Closing at 355kts

9. a) INT. 18:50 Pos: 30°39’S 020°50’E

b) HDG: 296°(T) G/S: 298kts c) Closing at 439kts d) 25nm apart at 18:47 Pos: 32°52’S 020°42’E

10. a) INT. EET: 01:22

b) HDG: 300°(M) c) DIST: 605nm d) Closing at 305kts e) First possible visible contact at 01:16 elapsed time.












Sample Exam Questions 1. C 26. B 51. C 2. A 27. A 52. C 3. C 28. A 53. C 4. B 29. A 54. A 5. C 30. A 55. A 6. A 31. B 56. A 7. C 32. C 57. B 8. B 33. A 58. A 9. C 34. B 59. A 10. C 35. B 60. A 11. B 36. A 61. C 12. C 37. A 62. B 13. A 38. C 63. C 14. B 39. A 64. B 15. A 40. A 65. B 16. B 41. C 66. B 17. A 42. A 67. C 18. B 43. C 68. B 19. A 44. A 69. A 20. B 45. B 70. C 21. C 46. A 71. B 22. B 47. C 72. A 23. A 48. C 73. B 24. A 49. A 74. B 25. B 50. C


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