ATPL Navigation

ATP NAVIGATIONATP Navigation & Plotting ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Version 4 INDEX ATP NAVIGATION 1. The Earth 2. Charts 3. Relative Velocity 4. Solar System & Time 5. The 1 in 60 Rules and General Maths 6. Navigational Computer 7. Basic Plotting on the Lamberts Chart 8. Plotting on the British Isles Chart 9. Advanced Plotting Annex A Sample Exams Annex B Answers to Questions 01 21 77 89 115 119 133 157 163 173 187 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 ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE 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 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 1 Version 4 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 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 2 Version 4 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. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 3 Version 4 e.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. 100°(M) COMPASS NORTH (C) Compass North is the direction indicated by the compass needle in an aircraft.g. Magnetic Fields in the aircraft will attract the compass needle away from Magnetic North causing Compass Deviation. which is sometimes referred to as the Magnetic Meridian. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 4 Version 4 . 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 Heading l00°(C) Dev+4°e 104°(M) DEVIATION WEST COMPASS BEST 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 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 5 Version 4 . Convergency is defined as the angle of inclination Between two selected meridians measured at a given Latitude. Long) or Difference in Longitude (D Long) between the two meridians is 40°. Considering the two meridians shown above.CONVERGENCY AND CONVERSION ANGLE CONVERGENCY Meridians are Semi Great Circles joining the North and South Poles. The Change of Longitude (Ch. At any intermediate Latitude the angle of inclination between the same two meridians will between 0° and 40° depending on the Latitude. At the Poles (Latitude 90°) they meet. As the meridians leave the Equator either Northwards or Southwards they converge and meet at the Poles. They are parallel at the Equator. 40°. the angle of convergence is 0°. and the angle of convergence is the Difference of Longitude. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 6 Version 4 . one at 20W and the other at 20E. At the Equator (Latitude 0°) they are parallel. the greater the convergency. convergence varies as Sine Mean Latitude.This is a sine relationship. Long. 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). Convergency = Ch.65° = 11. Convergency also varies as the Change of Longitude between the two meridians. Long° x Sin Mean Latitude = 16°41' x Sin 41° 39' = 16. The greater the Ch.6833°x Sin 41.0874° NOTE Both Mean Latitude and Change of Longitude must be changed into decimal notation. A B N 45:25 N 37:53 N 41:39 Mean Latitude E 025:36 E042:17 16:41 Change of Longitude Convergency = Ch. THE MERIDIANS CONVERGE TOWARDS THE NEARER POLE ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 7 Version 4 . CONVERGENCY = CHANGE OF LONGITUDE x SIN MEAN LATITUDE CONVERGENCY = DIFFERENCE BETWEEN INITIAL AND FINAL GC TRACKS ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 8 Version 4 . C and D are in the same hemisphere The Great Circle bearing of D from C is l36° The Great Circle bearing of C from D is 262° (bearing of D measured at C) (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? ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 9 Version 4 .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. 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 . If the Rhumb Line track from A to B is 100º. You can always take the reciprocal of a Rhumb Line. initial GC track B to A is 300° GC (Conversion angle 20°) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 10 Version 4 .DIFFERENCE BETWEEN INITIAL AND FINAL GREAT CIRCLES The Rhumb Line is a constant direction. then the Rhumb Line track from B to A is 280º. Initial GC track A to B is 080° GC. NEVER A GC. 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 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 11 Version 4 . 8° (rhumb line track A . 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.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). DEP (nm) = = = dL' x COS MID LAT 1200' x COS 15° 1159 nm To determine angle A : TAN ∅ TAN ∅ ∅ = = = 1159 nm 1800 nm 0.6438 32.B) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 12 Version 4 . 1 feet ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 13 Version 4 . As the earth is not a perfect sphere the distance is variable. part of the average distance from the Equator to either Pole It generally accepted to equal 3280 feet.) A Kilometre is 1/10 000 th. use Pythagoras: x² x² x x = = = = 1800² + 1159² 4583281 nm 4583281 nm 2141 nm (rhumb line distance A .To determine distance x.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. 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. 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. As Latitude increases.54 Centimetres As one minute of arc is 1 NM. until they meet at the Poles where the distance between them is zero. either to the North or to the South. To answer questions in the CAA examinations any of the following may be used :1 NM = 6080 feet or 1853 metres Conversion Factors 1 NM = 6076. 10 of arc equals 600 nautical miles.1 feet: or I852 metres 1 Foot = 12 inches 1 Inch = 2. Long Both East or West SUBTRACT One East & One West ADD ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 14 Version 4 .Most navigational electronic calculators use 1 NM = 6076. then Great Circle distance along a Meridian can be calculated. At the Equator. and the distance between them decreases. 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. As the Equator is a Great Circle. two meridians (5W and 5E) have a change of Longitude of 10 of arc. the meridians converge.1 feet. 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. 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. LONG) or DEPARTURE DISTANCE Departure is the distance in Nautical Miles along a parallel of Latitude in an East-West direction.Long W 067:25 E 027:30 94:55 Ch. One minute of Latitude is 1 NM and 1Degree of Latitude is 60 NM. and the relationship is given by Where mean lat = lat A + lat B 2 E 032:45 E 021:15 11:30Ch. 4 minutes of Longitude cos 29.1667° 563. Long xcos29°30' = 1206. After flying 1050 nm its Longitude is :Departure 1050nm 1050 nm = = Ch. Long x cos Lat Ch. Long x cos Lat 10° x 60 x cos20°10' 600 x cos 20.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.2163 nm Q2 An aircraft leaves A (E 012:30) and flies along the parallel of S 29:30 in an Easterly direction.5° 60 = 20° 06' 24" Easterly + 12° 30' 032° 36' 24" E ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 15 Version 4 . then the arc of Change of Latitude can be converted into nautical miles. Departure = Ch. Then the aircraft flies a track of l80° (T) for I hour 55 minutes and arrives at position. Track 180° Change of Latitude Departure = Ch.2:4-) is :- As E 152:47 is the anti-meridian of W 027:13. Long x cos Latitude N 25:13 W 017:25 . The Latitude at which the aircraft flew was :Departure GS 515 x 14 Hrs 7210 = 21600 cos Lat = 70° 30' N = = Ch. Long x cos Latitude Departure _________ cos Lat = Ch. Long cos Latitude 360° x 60 x cos Lat DISTANCE ALONG A PARALLEL OF LATITUDE IS DEPARTURE DISTANCE ALONG A MERIDIAN IS CHANGE OF LATITUDE As a Meridian is a Great Circle. Long GS360 x 1:35 ____________ = cos 25:13 Track 180° = 630 minutes of Longitude = 10°30-East of W 017:25 = W 006:55 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 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 16 Version 4 . Q4 The shortest distance from A (N 78:15 W 027:13) in B (SS3:30 E 15.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. A to B is the arc of a Great Circle. N 78:15 to the North Pole North Pole to N 82:30 = = 11:45 Change of Latitude 7:30 Change of Latitude _____ 19:15 Change of Latitude 19° x 60 Q5 = 1140nm+ 15 minutes = 1155nm shortest (GC) distance A to B An aircraft departs A (N 25:13 W017:25) and flies a track of 090°(T) at GS 360 for I hour 35 minutes. RADIO BEARINGS VHF D/F VERY HIGH FREQUENCY . The bearing will be passed to the aircraft in Q-code form. 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.DIRECTION FINDING VDF Major airports in South Africa have a VDF service. Q CODE QTE QDR QUJ QDM QDM ±180° QDR TRUE bearing FROM the VDF station MAGNETIC bearing FROM the VDF station TRUE track TO the VDF station MAGNETIC track TO the VDF station ±Variation = QUJ ±180° ±Variation = QTE Take the shortest route to change one bearing to another QDM ±180° QDR ±Variation =Variation QUJ ±180° QTE VOR VOR Radials are Magnetic bearings RMI Readings are Magnetic tracks to the VOR QDR QDM RMI BEARINGS (VOR & ADF) Usually termed RMI READING which is QDM (for ADF RMI ± DEV= QDM) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 17 Version 4 . ADF Relative bearings must be converted into True Bearings (QTE) before they can be plotted on a chart.ADF BEARINGS ADF Relative bearings are measured from the Fore and Aft axis of the aircraft. 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 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 18 Version 4 . The great circle bearing from A to B is 082°. It then turns onto a track of 180° for 4 hours. Positions A and B are in the same hemisphere. What is the longitude of position B? 5. i) ii) What is the rhumb line bearing from B to A? What is the great circle bearing from B to A? 2. The great circle bearing from A to B is 140°. What is the rhumb line distance from A (30° N 070° E) to B (30° N 085°E)? 6. At what latitude on earth is the convergency twice the value of convergency at 25°N? 4. The great circle bearing from A to B is 260°. An aircraft flies around the world on a rhumb line track of 090° at a ground speed of 480 Kts.QUESTIONS 1. What is the position of the aircraft at the end of the 3rd leg? 8. An aircraft (G/S 480 Kts) departs position A (20° N 010°E) on a track of 360° for 3 HRS. Position A (40° N 170° E). The great circle bearing from B to A is 330°. Southern hemisphere. Position B is on the same parallel of latitude. At what latitude did the aircraft fly? 7. What is the shortest distance between A (065° N 13° 30’ W) and B (78° N 166° 30’E)? ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 19 Version 4 . The flying time if 19 hours. i) ii) In which hemisphere are A and B? What is the rhumb line bearing from B to A? 3. It then turns onto a track of 270° for 2 HRS 30. Convergency 12°. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 20 Version 4 . CHAPTER 2 CHART PROJECTION THEORY The original problem of map making is still with us even in the 21st century. a cone (Lambert's Projection) or a flat sheet of paper (Polar Stereographic Projection). 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. Details. 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. but we cannot have both. Meridians and topographical features are 'projected' from the reduced earth onto a cylinder (Mercator's Projection). Bearings and scale must be correct. Lets now look at the chart projections and properties that we as pilots are interested in: ORTHOMORPHISM Orthomorphism means true shape. 1. 3. Charts which are produced by conic projections are used widely in aviation – mainly because conic projections. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 21 Version 4 . 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. In theory a cartographer starts with a 'reduced earth' which is the earth reduced by the required scale. it too cannot be done. The ideal chart would possess the following features. 4. The 'reduced earth' is a true undistorted representation of the earth. such as Parallels of Latitude. Distortions will occur. 2. preserve true shapes preserve angular relationships (called conformal or orthomorphic) have a reasonably constant scale over the whole chart show great circle as straight lines. STATEMENT IN WORDS 1 inch equals 40 nm Usually found on radio facility charts. On the Mercator. Lambert and Polar Stereographic charts the Parallels of Latitude are adjusted in the above manner. GRADUATED SCALE LINE 0 10 20 30 40 50 60 70 80 90 100 1_____1_____1_____1_____!_____1_____1_____1_____i___________1 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 22 Version 4 . 1 inch on the chart equals 40 nm. SCALE Scale is the ratio of a line drawn on a chart to the corresponding distance on the surface of the earth. The rectangle becomes a square and the diagonal is 45° Bearings are now correct. but by the same amount in all directions over short distances. Meridians and Parallels of Latitude intersect at 90° Scale is expanded. To overcome this problem the scale expansion North/South is reduced mathematically to equal the scale expansion East/West. Bearings are correct but the scale is variable. Bearings are no longer correct. Shapes and areas are approximately correct and the chart is orthomorphic. the diagonal of a square is 45° and bearings are correct.The 1 nm square on the reduced earth is correct. The 1 nm square of the reduced earth projected onto a cylinder becomes a rectangle. The scale has been expanded in the North/South direction to a greater degree than the East/West case. there is no such thing as a constant scale chart. 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.54 x 12 x 6080 Q2 Divide by 2.96nm 2. Usually scale will be correct at a certain Latitude but expands elsewhere.3054 = Scale at 40°'N A chart has a scale of 1:2 500 000. Scale expansion or contraction will occur.REPRESENTATIVE FRACTION 1 __________ 1000 000 or 1/1000000 or 1:1000000 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 . The scale of the chart is : Scale = CL 32 cms _________________________ ED 468 nm x 6080 x 12 x 2.54 = Inches Divide by 12 = Feet Divide bv 6080 = Nautical Mile. 32 centimetres on a chart represents 468 nm. For example :Mercator Chart Scale 1:1 000 000 at the Equator 1 ________ 1 000 000 Q1 Scale factor 1. How many nautical miles are represented by 4 cm on the chart? Scale CL Chart Length = ________________ ED Earth Distance 1 ________ 2 500 000 4 cm ______ ED ED = 2500000 x 4 cms 2500000 x 4 cms ______________ = 53.54 = 1 _______ 2710282 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 23 Version 4 .3054 at 40°N 1 _______ 766 049 x Scale factor 1. 54 105 nm x 6080 x 12 x 2.Q3 The scale of a chart is 1: 3 500 000. After adjusting the Parallels of Latitude so that the scale expansion North/South equals the scale expansion East/West it becomes a Mercator chart. A cylinder is positioned over the reduced earth tangential to the Equator. The length of a line that represents 105 nm is :Scale CL ___ ED 1 ________ 3500000 CL __________________________ 105 nm x 6080 x 12 x 2.56 cms 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. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 24 Version 4 . The Mercator chart was constructed so that Rhumb Lines are straight lines and the headings flown were easily plotted.54 CL = _____________________ 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? Chart B has the larger scale 1 ___ 2 1 > ___ 4 = 5. 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. and GPS computers aircraft flew constant headings. They flew Rhumb Lines.54 3500000 x CL = 105 nm x 6080 x 12 x 2. unequally spaced Parallel straight lines. equally spaced Constant Value Zero Correct at the Equator Correct at the Equator Expands as the secant of the Latitude Straight Lines SCALE RHUMB LINES ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 25 Version 4 .MERCATOR CHART PROPERTIES POINT OF PROJECTION POINT OF TANGENCY PARALLELS OF LATITUDE MERIDIANS CONVERGENCY Centre of the reduced earth Equator Parallel straight lines. Every Parallel of Latitude has its own scale. Charts of the same equatorial scale will fit N/S. Use the Latitude scale at the mid point between the two positions. excellent between 12°N and 12°S becoming distorted with increasing Latitude. The chart has a limit of 70°N and 70°S. Equator 5°S 10°S 30°S 60°S 1:2 000 000 1:1 992 389 1:1 969 615 1:1 732 051 1:1 000 000 Great care must be taken when measuring distances on a Mercator chart due to the variable scale. Concave to the Equator SHAPES & AREAS Approximately correct.GREAT CIRCLES Complex curves towards the nearer Pole Convex to the Pole. Plotting and Met charts topographical maps between 12°N and 12°S Rhumb Lines are straight lines .plotting easy Great Circles (radio bearings) are complex curves great care must be taken measuring distances due to rapidly changing scale. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 26 Version 4 . E/W and diagonally. CHART FIT USES ADVANTAGES DISADVANTAGES SCALE Scale is correct at the Equator and expands North and South as the secant of the Latitude. 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. What is the scale at 45°N? 15°S = A 45°N = B SCALE DENOMINATOR A x COS B = SCALE DENOMINATOR B x COS A 2 500 000 x cos 45 2 500 000 x cos 45 cos 15 Q2 = = Scale B x cos 15 1 830 127 Scale at 45°N 1:1 830 127 10°N = A Lat X = B The scale of a Mercator chart is 1:3 500 000 at 10°N At what Latitude is the scale 1:2 500 000? SCALE DENOMINATOR A x COS B = SCALE DENOMINATOR B x COS A 3 500 000 x cos X cos X (0.latitude.7034)cos-1 = 45° 17'49" N/S Q3 = = 2 500 000 x cos 10 2 500 000 x cos 10 ________________ 3 500 000 = 0:7034 The Meridian spacing on a Mercator chart is 2.54 (Departure) What is the scale at 50 N? = 1:3566454 Q4 The scale at 200 N is 1: 250000 Always work latitude .equator . then revert to:Scale = CL __ = ED 2. The scale at 30°S is :If ABBA cannot solve the problem.7 cms. SCALE AT O° 1 x 250000 COS 20° 250000 = 1 266044 1 COS 200 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 27 Version 4 .7 cms _______________________________ 1 Long x 60 cos 30 x 6080 x 12 x 2. remember that between any two given meridians: the chart length remains the same regardless of latitude change. the scale varies with latitude (use the Mercator scale formula). What is the dLONG between these two meridians if the scale is 1 at 30° N? 250000 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 28 Version 4 .SCALE AT 5O°N = = 1 X 266044 1 SEC 50ْ 1 1 x COS 500 = 266044 = 1 171010 This particular problem can also be solved in one step: SCALE AT 50° N SCALE AT 20° N x COS 2°ْ 250000 x COS 200 = 1 171010 1 1 COS 50° COS 50° CALCULATING DISTANCE AND dLONG ON A MERCATOR CHART When calculating distance and dLONG on a Mercator chart. EXAMPLE: Two meridians at latitude 30° N measure 13 cm apart on a Mercator chart. the dLONG remains the same regardless of latitude change. the earth distance varies with latitude (use the departure formula). ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 29 Version 4 .252 nm There is no departure at the equator.539 nm dLONG' x COS LAT dLONG' x COS 30° 17. provided that the scale is calculated at that latitude. this question could also have been calculated at the equator. At the equator: SCALE AT 0° = = At 0° N : SC = 1 288675 ED ED ED = = = = 1 250000 1 288675 CL ED 13 CM ED × COS 30° 1 13 CM x 288675 3752777 CM 20.252' Because chart length is constant regardless of latitude and dLONG is constant regardless of latitude.252' dLONG.539 nm COS 30° 20.At 30° N : SC = 1 250000 ED ED ED DEP (nm's) 17.539 nm's dLONG' dLONG' = = = = = = = = CL ED 13 CM ED 13 CM x 250000 3250000 CM 17. or any other latitude.252 nm = 20. therefore 20. earth distance is not a constant regardless of latitude. unlike dLONG.EXAMPLE: Two meridians at 30° N are 27 cm apart. The earth distance at 30° N is required and. apply the scale at the latitude where the work is being done: SCALE AT 30° N = SCALE AT 60° N COS 60° 1 866025 × COS 30° 1 = SC = CL ED 27 CM ED 27 CM x 866025 23382686 CM 126 nm 1 866025 ED ED ED = = = = Note that this question must be solved at 30° N. What is the earth distance between these two meridians if the scale at 60° N is 1 ? 500000 Again. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 30 Version 4 . The convergency between the aircraft and the station is 10°. Plot this great circle. Measure the bearing where the work was done. The deviation is 5° E.PLOTTING RADIO BEARINGS ON A MERCATOR When plotting radio bearings. Take the given information and make it true. the final goal is always to plot a QTE. specifically a rhumb line QTE. The variation at the aircraft is 20° W. Apply the conversion angle to convert the GC to a RL. The variation at the station is 15° W. Southern hemisphere. STEPS TO PLOTTING ON A MERCATOR CHART a) b) c) d) e) f) Always draw a sketch. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 31 Version 4 . because this is a straight line. and on a Mercator chart. VDF BEARING EXAMPLE ATC passes an aircraft a QDM of 060°. Orientate the hemisphere (to determine which way the great circle will curve). Relative bearing to an NDB station 40°. Southern hemisphere. Convergency between the aircraft and the station is 8°. Northern hemisphere. The convergency between the aircraft and the station is 14°. 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 station is 15° W. Aircraft variation 20°W.RBI EXAMPLE Aircraft compass heading 200°. Deviation 5° E. The variation at the aircraft is 20° W. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 32 Version 4 . The deviation is 5° E. Station variation 15° W. MERIDIONAL PARTS INTRODUCTION In essence. For example. This means that one minute of longitude will fit into the dLat 0° . Another possible solution is to physically measure the distance A . MERIDIONAL PARTS A meridional part is equal to a minute of longitude. but due to the continually changing scale on the Mercator chart.67 meridional parts.B. Thus far.67 times. The meridional parts tables indicate “how many times one minute of longitude will fit into a particular change of latitude”. determine the rhumb line track and distance.30° 1876. from A (00° N 010° W) to B (30° N 010° E). Unfortunately this method is only accurate for distances up to 600 nm's. if you look up 30° (latitude) on the table. Given the following question. convert the dLONG into nm's using departure and the cosine of the mid-latitude. the suggested method to solve this question has been to convert the dLat into nm's. you will find 1876. and then apply trigonometry to solve the rhumb line track and distance. mainly due to the fact that the cosine of the mid-latitude is being used to express the dLONG in nm's. meridional parts solves the rhumb line track and distance problem. this is also not accurate. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 33 Version 4 . The solution is to use meridional parts. there is absolutely no doubt that the dLONG is 1200'. but expressing the dLat in nm's certainly is.6° = 1800 nm x 1800 nm COS 32. (The meridional parts tables do correct for the effect of the earth's compression). using trigonometry: TAN ∅ = ∅= 1200 1876.67 MP 32. Now.6° (track A . If we express the LAT in meridional parts. Expressing the dLat in nm's is accurate.B) Now transfer the track angle to the triangle labelled nm's. The sides of the triangle are in the same units because one minute of longitude is equal to one meridional part.PRACTICAL APPLICATION Although expressing the dLONG in nm's by using departure and the COS MID latitude is doubtful in terms of its accuracy. Now using trigonometry: COS 32. but trigonometry can't be applied because the sides of the triangle would have different units.6° 2137 nm (rhumb line distance) x x = = RECOMMENDED TECHNIQUES ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 34 Version 4 . Expressing the dLONG in nm's wouldn't be accurate. however we can proceed with trigonometry. Sometimes angle ∅ is not the track. c) When working from one latitude to another. the latitude side of the triangle will be the difference in meridional parts (DMP). In the following sketch. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 35 Version 4 . neither of which is the equator. one for MINS LONG/MP's and another for nm's.a) b) Always draw two sketches. or the sum of meridional parts (SMP) if changing hemispheres. the track is ∅ + 90°. 99 MP = 936. a) Determine the rhumb line track and distance flown. An example of each follows below. b) Determine the aircraft's position. An aircraft leaves position A (18° N 047° E) on a rhumb line track of 047°. with a heading for each to assist with identification.74 DMP COS 47° x x = = = x 1246 nm 1246 nm x COS 47 849. given rhumb line track and distance flown. As per previous example. What is its position after flying for 1246 nm's? 32° 10' N 18° N dLAT = 2027.8 nm TAN 47° x x x LONG B LONG B = x 936.53 MIN LONG = 16° 45' = 16° 45' + 047° E = 063° 45'E 14° 10' + 18° N 32° 10' N ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 36 Version 4 .74 x TAN 47° = 1004.73 MP = 1090.QUESTIONS The vast majority of meridional parts questions fall into one of five categories.74 LAT B = = = 849.8 60 = 936. 46 MP 840. At which latitude will the aircraft cross the meridian 043° W? dLONG = = = 063° W .25 MP 1560.71 MP 25° 19' S ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 37 Version 4 .c) At which latitude will an aircraft cross a given meridian? An aircraft departs A (12° S 063° W) on a track of 125°.25 MINS or MP/s 720.043° W 20° W 1200' TAN 35° x x 12° S DMP NEW LAT NEW LAT = = = = = = = x 1200' 1200' x TAN 35° 840. 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°.09 DMP TAN 25° = 502.17° 57' 007° 03'E x x dLONG = = = = NEW LONG NEW LONG e) Meridional parts scale = = ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 38 Version 4 .92 MP 842.83 MP 502.74 MP's or MINS LONG 1076.09 x 502. At which meridian will the aircraft cross the latitude 22° N? 22° N = 14° N = 1344.09 TAN 25° 1076.74' 17° 57' 25° . the scale at any latitude may be used. EXAMPLE: A mercator chart has a scale of in cm's? 1 at the equator. Calculation of the same question.5 nm's ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 39 Version 4 .As can be seen from any Mercator chart. provided the scale at that latitude is used. but at 60°N. the chart length may be calculated at any latitude. because the CL of 1 MIN LONG is constant throughout the chart. SCALE AT 60° N = 1 1000000 1 500000 × 1 COS 60° = Calculate the earth distance of 1 MIN LONG at 60° N. What is the chart length of 1 MP 1000000 SC = CL (1 MP / 1 MIN LONG) ED CL 1NM 1 1000000 = 1 1000000 CL CL = CL 185300 CM 185300 1000000 0. regardless of latitude. then it stands to reason that 1 MP must also have a constant value chart length throughout the chart. Because the chart length between any two meridians is constant throughout the chart. DEP (nm's) = = = dLONG' x COS LAT 1' x COS 60° 0. The exact value of the chart length of course depends on the scale of the chart at that point. If one minute of longitude is equal to one meridional part. the chart length of one minute of longitude has a constant value chart length.1853 CM (CL of 1 MP/1 MIN LONG) = = As previously stated. 46 MP 1090.45 MP FLIGHT TRAINING COLLEGE Page 40 Version 4 1 400000 1 400000 CL CL = = = = 12° N = 18° S = SMP = ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 .99 MP 1811.2852 CM (CL of 1 MP/1 MIN LONG) 720.5 NM CL 92650 CM 1 500000 1 500000 CL CL = = = = 92650 500000 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 1 ? 400000 Determine the CL of 1 MP/1 MIN LONG SC = CL ED CL 1' × COS 52 ° × 185300 CL 114082 CM 114082 400000 0.SC = CL (1 MP / 1 MIN LONG) ED CL 0. 006°E 18° E 1080' 1811.5 cm ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 41 Version 4 .45² + 1080² 4447751 4447751 2109 MP at 0.dLONG dLONG dLONG Using Pythagoras: x² x² x x x = = = = = = = = 024° E .2952 cm per MP 601. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 42 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 43 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 44 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 45 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 46 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 47 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 48 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 49 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 50 Version 4 . LAMBERT CONFORMAL CONIC CHART The Lambert's chart was developed from the Simple Conic chart. The cone is opened to give a simple conic projection. A light source at the centre of the reduced earth projects details onto the cone. The apex of the cone is above the pole. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 51 Version 4 . SIMPLE CONIC A cone is placed over a reduced earth so it is tangential to a selected parallel of latitude. Due to the scale expansion the chart is not suitable for navigation.5584° ______________________________ Change of Longitude 360° = 0.7071 = CCF = Constant of the Cone = 'n' factor FLIGHT TRAINING COLLEGE Page 52 Version 4 . 360° of Longitude is represented by the angular extent of the chart which is 254.The scale is correct at the parallel of tangency (45N) and expands north and south of 45N.7071° is called the: CHART CONVERGENCY FACTOR (CCF) Parallel of Tangency 45° ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 Sine 45° = 0. The angular extent of the chart is controlled by the latitude chosen to be the parallel of tangency.5584°. Parallels of Latitude are arcs of circles radius the Pole. The Meridians are straight lines converging on the nearer pole and the value of convergence is constant throughout the chart. Angular extent of the chart 254.7071° Constant of the Cone or 'n' factor Two Meridians 1° apart have a convergency 0. SIMPLE CONIC CONVERGENCE When the cone is opened. Curves slightly concave to the Parallel of Origin. 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. Since scale on the simple conic is correct only on the parallel of tangency and expands either side. which is. RHUMB LINES GREAT CIRCLES Curves concave to the Pole and convex to the Equator A straight line joining two positions on the Parallel of origin. one on either side of the simple conic parallel of tangency.LAMBERT CONFORMAL CONIC CHART The Lambert's chart is based on the simple conic and is produced mathematically from it. the reduction will give two Standard Parallels (SP) on which scale is correct. renamed the Parallel of Origin. Further mathematical modification is applied by adjusting the radius of the parallels of latitude to produce an orthomorphic projection. unequally spaced. radius the Pole. LAMBERT'S CHART PROPERTIES PARALLELS OF LATITUDE MERIDIANS SCALE Arcs of circles. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 53 Version 4 . 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. the scale is reduced throughout the chart. Straight lines converging towards the nearer Pole 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. Firstly. THE ADVANTAGES OF THE CHART a) b) Constant scale. Long x Constant of the Cone Slight distortion Charts of the same scale and Standard Parallels will fit N/S and E/W. but the rhumb line is a curved line on this chart and therefore cannot be accurately plotted. Light aircraft generally fly rhumb line tracks. which means that radio bearings can be easily plotted. Long x 'n' Ch. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 54 Version 4 . great circles are straight lines. Long x sin Parallel of Origin Ch. Radio bearings are great circles and on this chart.CONVERGENCY Chart Convergency Chart Convergency Chart Convergence Chart Convergence SHAPES and AREAS CHART FIT Constant throughout the chart Correct at the Parallel of Origin Ch. THE DISADVANTAGES OF THE CHART a) b) The grid is not rectangular. Long x CCF (Chart Convergence Factor) CH. Charts with different SP will not fit. LAMBERT'S CHART .Then the Parallel of Origin (// 0) is 30°S If one SP is 20°S and the ║O is 30°S . 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.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) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 55 Version 4 .TRACKS For all practical purposes the Great Circle is a straight line. 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 . Q3 The CCF of a Lambert's chart is 0. 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.5 = sin║O = 30°S SP25°S Parallel of Origin 30°S Other SP35°S ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 56 Version 4 .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.If a statement regarding convergency is given :(e. long 10° x sin 30°) and the CCF = 0.g.5°. convergency between any two meridians is easily found. As convergency is proportional to the CCF. LAMBERT'S CHART PLOTTING RADIO BEARINGS Radio bearings are Great Circles. The deviation is 5° E. because this is a straight line. The final goal when plotting radio bearings on the Lambert's chart is to plot a QTE. VDF BEARING EXAMPLE ATC passes an aircraft a QDM of 060°. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 57 Version 4 . Take the given information and make it true. Southern hemisphere. STEPS TO PLOTTING ON A LAMBERT'S CHART a) b) c) d) e) f) Always draw a sketch. The convergency between the aircraft and the station is 10°. Measure this bearing where the work was done. and specifically the great circle QTE. Plot this great circle. The variation at the station is 15° W. Straight Lines on a Lambert's chart are Great Circles and plotting radio bearings is simple. The variation at the aircraft is 20° W. Orientate the hemisphere correctly. Apply convergency if required to obtain the great circle QTE. Northern hemisphere. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 58 Version 4 . Deviation 5° E. Convergency between the aircraft and the station is 8°. The variation at the station is 15° W. The variation at the aircraft is 20° W. Aircraft variation 20° W. The convergency between the aircraft and the station is 14°. The deviation is 5° E. THE VOR NEEDLE ON THE RMI EXAMPLE The VOR needle on the RMI indicates a radial of 270° (tail of the needle). Southern hemisphere. Station variation 15° W.RBI EXAMPLE Aircraft compass heading 200°. relative bearing to an NDB station 040°. thus its distance cannot be measured. Join points A and B on the chart with a straight line (great circle). Instead. the great circle and the rhumb line are parallel. a rhumb line is a curved line and cannot actually be plotted. Measure the track of the great circle at the mid-meridian. If the aircraft departs from position A and maintains this track. Across the mid latitude of the track ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 59 Version 4 .PLOTTING RHUMB LINE TRACKS On the Lambert's chart. At the mid-meridian. To obtain the greatest degree of accuracy. measure this distance: a) b) Along a meridian scale. measure the great circle (straight line) distance to obtain the equivalent rhumb line distance. MEASURING RHUMB LINE DISTANCES The rhumb line (curved) is never actually plotted. it will fly the equivalent rhumb line track. 75 inches = ________________________________ 20° Ch. The Rhumb Line distance from A (50°N 30°E)to B (50°N 10°E) is 13. Long x 60 x cos 50° x 6080 x 12 (Departure in nm) = 1 ________ 4 092 898 The other Q2 On a Lambert's chart the Standard Parallel of 35°S measures 58.4 cms Scale at 35°S= _________________ ED Ch. SP20° Sand40°S.75 inches.9 cms ____________ cos Lat = 0. The Latitude of the second Standard Parallel is :CL 58. The scale at 30°N is :Scale = CL __ ED 13.4 cms ___________ cos 35 = 43. A Lambert's chart has Standard Parallels of 30°N and 50° N. Long is the same in both equations it disappears 58. Long x cos Lat = cos52°S ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 60 Version 4 . Standard Parallel measures 43.9 cms ED Ch.9 cms.4 cms. Long x cos 35 The scales are equal. The scale is correct at the two Standard Parallels Scale 20°S = Scale at 40°S Q1.SCALE PROBLEMS Lambert's scale 1:2 500 000. As CH.6158 Scale at 2nd SP = CL 43. THE PARALLELS The parallels are concentric circles. 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 spacing between the parallels increases away from the pole. and opposite to the light source. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 61 Version 4 . When the piece of paper is removed. A flat piece of paper is then placed on top of the pole to be constructed.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 polar stereographic chart has been created. When the light is switched on. SOUTH POLAR STEREOGRAPHIC CHART THE PROPERTIES OF THE CHART THE MERIDIANS The meridians are straight lines radiating from the pole. the data is projected onto the flat piece of paper. GREAT CIRCLES Great circles may be considered to be straight lines and will cut successive meridians at different angles. SCALE The scale is correct at the point of tangency (the pole).THE POINT OF TANGENCY The point of tangency is the north or south pole. The formula for determining scale expansion away from the pole is: 1 SCALE AT LATITUDE RHUMB LINES = 1 SCALE AT POLE × SEC 2 1 2 CO . THE POINT OF PROJECTION The point of projection is a light source at the opposite pole. the scale expands with movement away from the pole or contracts with movement towards the pole. Elsewhere on the chart. (In truth the great circle is slightly concave to the pole.LAT 1 Rhumb lines curve towards the equator and cut successive meridians at the same angle.) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 62 Version 4 . ORTHOMORPHIC The chart is orthomorphic because a) b) Meridians and parallels cut at 90º. The scale expands at the same rate in all directions over short distances. Therefore convergency all over the chart is simply calculated with the formula : CONVERGENCYº = dLONGº SHAPES AND AREAS The nearer the pole. CONVERGENCYº = dLONGº However. CONVERGENCY On this chart. convergency is correct at the pole. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 63 Version 4 . the more accurate the representation of shapes and areas. convergency is constant throughout the chart because the meridians are straight lines. The aircraft is at position 70°S 010° E. Steps To Plotting On A Polar Stereographic Chart a) b) c) d) e) f) Draw a sketch.Measuring Directions On The Charts Remember that direction true is always measured clockwise and relative to true north. The station is at position 70° S 090° E. Take what is given and make it true. VDF Bearing EXAMPLE ATC passes the aircraft a QDM of 060°. 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. The QTE to plot is? ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 64 Version 4 . Southern hemisphere. Apply convergency if required to obtain the great circle QTE. Measure this bearing where the work was done. The deviation is 5° E. because this is a straight line. the South pole is at the centre of the chart and true North is 180° away from true South. Orientate the hemisphere. Plot this great circle. the North pole is at the centre of the chart. The variation at the aircraft is 20° W. On a North polar stereographic chart. Remember also that a parallel of latitude runs E/W. On a South polar stereographic chart. The variation at the station is 15° W. Southern hemisphere. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 65 Version 4 . Deviation 10° E.RBI EXAMPLE Aircraft compass heading 270°. VOR Needle On The RMI EXAMPLE The VOR needle on the RMI indicates a radial of 165°. Relative bearing to an NDB station 040°. The variation at the aircraft is 20° W. The station is at position 70° S 040° E. The variation at the station is 15° W. Station variation 15° W. The deviation is 12° E. The station is at 70° N 030° W. The aircraft is at 70° N 030° E. The aircraft is at position 70° S 160° E. Northern hemisphere. Aircraft variation 20° W. 28. r = = = = 2 R tan ½ co-lat 2 x 79. calculate the chart length between 70°N and 60° N in cm's. 1 .6 cm ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 66 Version 4 .6 x tan ½ (90° .7 cm .1 cm = 14. Calculate the radius from 90° N to 70° N. The radius of the real earth is 3438 nm.Determining The Radius Of A Parallel Of Latitude The scale of the model earth is stereographic chart of the north pole. 42. a.7 cm c.6 x tan 10° 28. r = = = 2 R tan ½ co-lat 2 x 79. On a polar 8000000 SC = 1 8000000 CL CL b.60°) 2 x 79. CL ED = CL 3438 × 185300 637061400 8000000 = 79.6 cm (radius of the model earth) = Calculate the radius from 90° N to 60° N.6 x tan 15° 42.1 cm The chart length between 70° N and 60° N. Calculate the radius of the model earth in cm's. lat 1000000 = COS 2 15° 1000000 1 1071797 = Take the scale from 90° N to 70° N 1 SCALE AT 70° N = 1 1071797 1 1071797 1 1071797 1 1039478 × × × SEC 2 1 2 co . What is the scale 1000000 1 SCALE AT 90° N = 1 1000000 × COS 2 1 2 co . Take the scale from 60° N to 90° N.lat 1 1 COS 2 1 2 = co . the scale at 60° N is at 70° N.lat = 1 COS 2 10° = 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 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 67 Version 4 . 1 .lat 1 = COS2 12 co .Determining Scale On The Polar Stereographic Chart On a polar stereographic chart of the north pole. Certainly less serious. it would be impossible to plot a course anywhere. Similarly. if you were at the south pole. but also warranting improvement is the Lambert's chart. A square grid is placed over the applicable chart. every single direction is north. The solution to both of these problems is grid navigation. but care must be taken when plotting these tracks. Flying great circle tracks is ideal. grid north is always at the top of the chart and direction is now referenced to grid north rather than true north. because every single direction is south. Direction will always be constant relative to grid north because the grid is square. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 68 Version 4 .GRID NAVIGATION One of the problems associated with the polar stereographic chart is that if you were at the north pole. because they cut each meridian at a different angle. If convergency is the angular difference between any two meridians. 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 POLAR STEREOGRAPHIC GRID On the polar stereographic grid. then convergence is the angular difference. On the polar stereographic chart : CONVERGENCE° = CONVERGENCY° = dLONG° On the Lambert's chart CONVERGENCE° = CONVERGENCY° = dLONG° x SIN LAT//of O EXAMPLE: ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 69 Version 4 . but between the datum meridian and another meridian. not between any two meridians. CONVERGENCE Convergence is defined as being the angular difference between grid north and true north. and is normally positioned at a meridian closest to where the chart will be used. THE LAMBERT'S GRID On the Lambert's grid. Convergence and convergency thus always have the same numerical value. the datum meridian (the meridian on the chart with which the grid is lined up) can vary. Always draw a sketch. It is thus the algebraic sum of convergence and variation.Grid heading 080°. What is the true heading? Grid heading 080°. a westerly convergence does not necessarily mean that the aircraft is in the western hemisphere. although CONVERGENCE° = CONVERGENCY° = dLONG °. but will not be the case on a north polar chart. an easterly convergence does not necessarily mean that the aircraft is in the eastern hemisphere. This will be the case on a south polar chart. Similarly. Convergence 20° W. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 70 Version 4 .TRUE IS LEAST BEWARE: On the polar stereographic grid. Convergence 20° E. GRID VARIATION (GRIVATION) Grivation is defined as the angular difference between grid north and magnetic north. What is the true heading? RULE: CONVERGENCE WEST .TRUE IS BEST CONVERGENCE EAST . ISOGRIVS Isogrivs are defined as being lines joining places of equal grivation. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 71 Version 4 . 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 scale at 18° N is 1:1000000 What is the scale at 36° S. 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? Two lines of equal length are drawn on a Mercator chart. 4.QUESTIONS PART 1 1. What is the aircraft’s final position? 9. Using meridional parts. 2. On a Mercator chart. An aircraft departs position A (21° 37’ N 012° 12’ W) on a rhumb line track of 137°. At which meridian will the aircraft cross the equator? 10. one at the equator and the other at 60° N. At which latitude will the aircraft cross the Greenwich meridian? ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 72 Version 4 . An aircraft leaves position A (27ْ 27’ S 014° 28’ E) on a rhumb line track of 205° and flies for a distance of 4087 nm. the VOR CDI indicates 145° TO. With the needle centralised. calculate the rhumb line track and distance from A (08°N 016° 30’ W) to B (16° 27’ S 004° 18’ E). An aircraft departs from position A (10 18’ S 002° 03’E) on a rhumb line track of 040°. the deviation is 5° E the convergency between the aircraft and the station is 12°. Which of these two lines represents the greater earth distance? 5. 8. What bearing should be plotted on a Mercator chart of the southern hemisphere? 7. On a mercator chart. The variation at the station position is 20° W. The ADF bearing on an RMI is 060°. 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. The variation at the aircraft position is 10° W. The variation at the station position is 15° W The deviation is 5° W. The variation at the aircraft position is 10° W. The variation at the station position is 20° W. The deviation is 5° E.5 cm on a Mercator chart of the northern hemisphere. 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 60ْ N and the north/south length of the chart is 150 cm’s. What is the scale of the chart at 30°N? 13. At what latitude will the scale be 1: 1 500 000? On a Mercator chart. i) ii) iii) 3.A? The ADF needle on an RMI indicates an QDM of 040°. If an aircraft leaves X on a constant heading of 60°T in zero wind conditions. the measured distance between fixes A and B is 22. The other standard parallel measures 38 cm's. The parallel of Origin is at 30° N. What is the latitude of the other standard parallel? 6.11. What is the bearing to plot on a Lambert's chart of the southern hemisphere? 5. will it pass: FLIGHT TRAINING COLLEGE Page 73 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 . The chart convergency factor on a Lambert's chart is 0.766. Part 2 1. On a Lambert’s chart in the Northern hemisphere. The variation at the station position is 20° W. The dLONG between the aircraft position and station position is 60°. The track measured at X is 60°T. the standard parallel of 30° N has a chart length of 50 cm's. What is the bearing to plot on a Lambert's chart of the northern hemisphere? 4. The deviation is 5° E. what is the southernmost latitude? 12. 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°. On a flight along the 50th parallel. The variation at the aircraft position is 15° W. What is the latitude of the other standard parallel? 2. 14. a straight line is drawn from X to Y. What is the aircraft’s groundspeed if the time between fixes was 17 minutes? A Mercator chart has a scale of 1:2 000 000 at latitude 30°N. The variation at the aircraft position is 15º W. What is the great circle track measured at A? What is the great circle track measured at B? What is the rhumb line track from B . With the needle on the VOR CDI centralised. On a Lambert's chart of the northern hemisphere. the indication is 360° TO. The scale of the chart is 1:2 500 000 at 20°N. On standard parallel is at 40° N. The P of O is at 30° S. the perpendicular distance between parallels 37°N and 39°N is 4 cms. On a flight from A (22° N 165° E) to B (37 N 178° W). The convergency between the aircraft and the station is 10°. .D.A? On a polar stereographic chart. Overhead Y. a flight is planned from A . The great circle track from A . The initial great circle track form a 27°N 061°W to B 47°N 017°W is 52° (T).a) b) c) 7. The longitude at which the great circle track becomes 084° is . i) ii) 2. B .B great circle track 041°.D great circle track 064°. On a polar stereographic chart. a flight is planned from A (72° N 032° W) to B (72° N 098° W). What is the great circle track from A . On a polar stereographic chart. C . What is the great circle track from A direct to D if all these positions lie on the parallel 75° N? 3. A Lambert’s chart has standard parallels 20°N and 60°N. a flight is planned from A (70° N 035° E) to B (70° N 043° W). i) ii) What is the great circle heading at A if the drift is 5° right? What is the highest latitude which this line will attain? ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 74 Version 4 .B is 120°.C great circle track 059°.. a flight is planned from A (75° S 168° E) to B (75° S x°W). South of Y.? PART 3 1. North of Y. A . i) ii) What is the longitude of position B? What will the great circle track be when crossing the anti-meridian of Greenwich? 4.B? What is the great circle track from B . On a polar stereographic chart. 5. On a polar stereographic chart, a flight is planned from A (75° S 047° E) to B (75° S 063° W). i) ii) What is the great circle track from A - B? 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. FLIGHT TRAINING COLLEGE Page 75 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 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? ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 76 Version 4 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) b) TYPE 2 a) Formula: DIST FROM DEST = 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. Formula : TIME = 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. RELATIVE DISTANCE RELATIVE SPEED DELAY × OLD G / S × NEW G / S DIFFERENCE IN G / S TYPE 3 a) b) Formula : a SIN A = b SIN B = c SIN C Identifiable by: 2 aircraft on converging tracks. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 77 Version 4 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? ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 78 Version 4 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) (b) ensure a 50 nm separation at the VOR? 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. DTC 55 nms (a) Speed of closing 60Kts Distance to close (55 - 50) Time to close Reduce speed at 5 nm 5 mins 0805 Z (b) Speed of closing 60Kts Distance to close (55-32.5) Time to close Reduce speed at 22.5 nm 22.5mn 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 := 5 x 450 x 390 ___________ 60 x 60 = 243.75 nm Delay x Old GS x New GS Distance = ______________________ Difference in GS x 60 GS450 GS 390 Q4 Dist243.75mn Dist 243.75 nm Time 32½ mins Time 37½ mins ETA 0915-32½ mins = 0842½ ETA 0920 - 37½ mins = 0842½' Alternative solution DISTANCE = SPEED x TIME At the point where speed is reduced, the aircraft is 'D nm' from Delta. AtGS450 D = 450 x T AtGS390 450 T 450 T 450T-390T 60 T T D-390 x (T+5) = = = = = 390 (T + 5) 390 T - 1950 1950 1950 32½ mins As D is common, then At GS4 50 At GS 390 ETA 0915-T 32½ mins = 0842½ ETA 0920-(T+5) 37½ mins - 0842½ ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 79 Version 4 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 Distance = ______________________ = Difference in GS x 60 13 x 180 x 260 ______________ = 126.75 nm 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 = _____________________ = _______________ Difference in GS x 60 65 x 60 = 181.73 nm ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 80 Version 4 QUESTIONS 1. At 1205, aircraft A is 76 nm behind aircraft B. Aircraft A 180 Kts. Aircraft B 140 Kts. a) b) At what time will aircraft A pass aircraft B? At what time will aircraft A be 42 nm ahead of aircraft B? 76 nm A - 180 Kts 1205 B - 140 Kts 1205 a) T = RD RS 76 - 0 180 - 140 76 40 1 HR 54 after 1205 1359 (aircraft A will pass aircraft B) = = = = b) T = RD RS 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) b) At what time must aircraft B reduce speed to 300 Kts to ensure a 20 nm separation when aircraft A reaches Y? At what time must aircraft B reduce speed to 300 Kts to ensure arrival at Y, 10 minutes after A? 60 nm B - 360 Kts 1205 X 500 nm A - 320 Kts 1205 B - 360 Kts 1215 Y Z 20 nm ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 81 Version 4 120) 360 1 HR 10 + 1205 1315 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 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 82 Version 4 .067 × 360 × 300 60 120 nm from Z (speed reduction point) = = B's time to reach speed reduction point T = = = b) 60 + (500 .20) 360 1 H 30 + 1205 1335 B must delay his arrival by 1339 .20 .a) Time for A to reach Y = = = 500 320 1 H 34 + 1205 1339 (also time for B to reach Z) B's original estimate for Z = = = 60 + (500 .1335 = 4 mins DIST FROM DEST = DELAY × OLD G / S × NEW G / S DIFF IN G / S 0. 60 + (500 .330) 360 38 min + 1205 1243 At 1205.B must delay arrival by 1349 . Aircraft B 360 Kts. but not a mixture of both. either all 3 sides speed or all 3 sides distance. Aircraft A 330 Kts. a SIN A = b SIN B = c SIN C Please note that the sides of the triangle must have constant units.1338 = 11 mins DIST FROM DEST = DELAY × OLD G / S × NEW G / S DIFF IN G / S 0. The relative bearing from A to B is 075° a) b) c) What is the relative bearing from B to A? At what time will the two aircraft collide? At what time will the aircraft first see each other if the in-flight visibility is 20 nm. aircraft A and B are 75 nm's apart and are on a collision course.183 × 360 × 300 60 330 nm from Y (speed reduction point) = = Time to reach speed reduction point T = = = 3. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 83 Version 4 . 7 ° SIN 75° 252.75° . Calculate angle C.360° 297.7° Calculate side C (relative speed) c SIN C C = = a SIN A a × SIN C SIN A 360 × SIN 42.3° C = 42. T = RD RS 75 252.62.75 18 mins after 1205 1223 = = = ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 84 Version 4 .3° .a) SIN B b SIN B = SIN A a = SIN A × b a SIN 75 × 330 360 62. A + B + C = 180° C = 180° .7° = = = b) Calculate the time when the two aircraft will collide.75 Kts = = Calculate the time when the aircraft will meet. c) Calculate the time when the aircraft will first see each other if the in-flight visibility is 20 nm. T = RD RS = = = 75 .75 13 mins after 1205 1218 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 85 Version 4 .20 252. Aircraft A G/S 420 Kts is inbound to position Y. what is their distance from position X? Positions X and Y are 290 nm's apart. At what time will the aircraft pass each other? When the aircraft pass each other. At 1205. Aircraft A G/S 350 Kts passes position X at 1200 enroute to Y. aircraft A is overhead position X G/S 350 Kts. Aircraft A and B are both bound for position Y. What reduction in speed must B make at position X to ensure this separation? 3. i) ii) 2. At 1230. Aircraft A G/S 250 Kts passes position X 10 minutes ahead of aircraft B G/S 300 Kts. ATC requests that the aircraft delay its arrival over position Y by 15 minutes. Positions X and Y are 960 nm's apart. i) ii) At what time must aircraft B reduce speed to 320 Kts to ensure a 50 nm separation when aircraft A passes position Y? At what time must aircraft B reduce speed to 320 Kts to ensure an arrival over point Y 10 minutes after aircraft A? ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 86 Version 4 . 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. aircraft B is overhead position X G/S 450 Kts. At what distance from Y must aircraft A reduce speed to 370 Kts to ensure this delay? 6. i) ii) At what time will the separation between the two aircraft be 130 nm's? When aircraft A reaches position Y. aircraft B passes position Y 10 minutes ahead of aircraft A. At 1225. What is the distance from X to Y? 5. At 1205. Some time later.QUESTIONS 1. Position X and Y are 700 nm's apart. the separation between the two aircraft must be 130 nm's. aircraft B leaves position Y en-route to X at G/S 290 Kts. i) ii) At what time must aircraft B reduce speed to 350 Kts to ensure a 20 nm separation when aircraft A reaches position Y? 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 B G/S 470 Kts passes position X at 1215 en-route to Y. aircraft A leaves position X en-route to Y at G/S 370 Kts. Aircraft A G/S 280 Kts. 250 nm's apart. Aircraft B G/S 250 Kts. The relative bearing from B to A is 320°. At 1305. i) ii) iii) What is the relative bearing from A to B? At what time will the aircraft collide? At what time will the aircraft first see each other if the in-flight visibility is 20 nm's? ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 87 Version 4 .7. aircraft A and B are on a collision course. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 88 Version 4 . THE EARTH'S ORBIT The orbit of a planet around the Sun conforms with Kepler's Laws of Planetary Motion which state :1. and the movement of the Earth in its orbit around the Sun. The orbit of a planet is an ellipse. causing the seasons. 2. 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. A Y SAX SYC C X 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. sweeps out equal areas in equal in equal intervals of time. with the Sun at one of the foci. 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 orbital speed of the Earth is variable. According to Kepler's Law the radius vector sweeps out equal areas in equal intervals of time. The Earth completes one orbit around the Sun in about 365. At Perihelion the Earth is about 91½ million miles from the Sun and occurs on January ±3.CHAPTER 4 THE SOLAR SYSTEM & TIME The measurement of the passage of time is based upon observations of events occurring at regular intervals. The mean distance of the Earth from the Sun is about 93 million miles. The line joining the planet to the Sun. Causing day and night. 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. At position C the planet is furthest from the Sun and is known as APHELION.25 days. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 89 Version 4 . known as the radius vector. At Aphelion the Earth is about 94½ million miles from the Sun and occurs on ±July 3. The two repetitive events which most influence life on Earth are the rotation of the Earth on its axis. 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. and ±23rd September which is the autumn equinox. 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. on or near 23rd of December the North Pole is inclined away from the Sun. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 90 Version 4 . As the Earth moves around the Sun. It is then midsummer in the Northern Hemisphere and midwinter in the Southern Hemisphere. its axis will always point in the same direction relative to space and will reach a point at the summer solstice. 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. on or about 22nd June. when the Sun is vertically overhead Latitude 23½°N. will be overhead the Equator.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 SEASONS One effect of the tilt of the Earth's axis is the annual cycle of seasons. These events occur on ±21st March which is the spring or vernal equinox. As the Earth travels around its orbit. Between these dates the Sun. MEASUREMENT OF TIME . being a gyro. This is known as winter solstice and is midwinter in the Northern Hemisphere and midsummer in the Southern Hemisphere. SIDEREAL DAY (23 hours 56 minutes 4 seconds) As stars are at immense distances from the Earth. which is vertically above Latitude 23½°N. Approximate dates Perihelion Jan 4 Vernal or Spring Equinox Mar 21 Summer Solstice Jun 22 Aphelion July 4 Autumn Equinox Sep 23 Winter Solstice Dec 23 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. 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. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 91 Version 4 . stationary with the Sun rising in the East and setting in the West. 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.APPARENT SOLAR DAY The time interval between two successive transits of the True Sun over a meridian is an Apparent Solar Day. An average of 365 Apparent Solar Days is taken and termed a Mean Solar Day which is 24 hours. Each and every meridian has its own LMT. When the Mean Sun is overhead a meridian it is 12:00 Local Mean Time (LMT). MEAN SOLAR DAY The 24 hour day is based on the Mean Sun. After 23 hours 56 minutes and 4 seconds the star is in transit for a second time (a Sidereal Day). The Siderial day then. THE SIDERIAL DAY Because of the relative proximity of the earth to the sun. It is more convenient to imagine the Earth. This of course takes time thus an Apparent Solar Day is always longer than a Sidereal Day. rays of light from a star being parallel. The Siderial day is of constant duration : 23 hours 56 mins 4 seconds The Earth rotates on its axis from West to East. 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. 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. The Sun and a star are in transit overhead a meridian. To solve this problem. is defined as two successive transits of the Siderial point at the same meridian. This point in space is called the siderial point or the first point of Aries. The Conversion of Arc to Time table on the next page is also available in the Navigation Tables booklet provided in the examination. or 15° per hour there is a direct relationship between Longitude and LMT. ARC TO TIME As the Earth rotates through 360 in 24 hours. The Local Date changes at midnight and also at the International Date Line. 90 in 6 hours. The first six columns are degrees of Longitude on the left with the corresponding time in hours and minutes on the right. At 90E the sun is overhead at 12:00 LMT. 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. 28' Long 127° 37'E 1 minutes 52 seconds 127° = 8:28 42' Long 37' = 2:28 2 minutes 48 seconds Arc to Time 127° 37' = 8:30:28 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 92 Version 4 . UTC is the same as GMT (Greenwich Mean Time). At 180°E/W the sun is setting at 18:00 LMT. 10° 0:40 15° 1:00 79° 5:16 161° 10:44 The right hand column gives the time equivalent for minutes of Longitude.At the Greenwich Meridian the sun is rising at 06:00 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. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 93 Version 4 . The LJTC at this position is :A E 065:30 Arc to Time A 13:15 LMT 23rd March 4:22 08:53 UTC 23rd March Longitude East . Flight time to D (42:37 E 135:45) is 11 hours 18 minutes.UTC Least UTC must be an earlier time than LMT Q2. always work in UTC.Q1. The time is 06:45 UTC on 21st May GD (Greenwich Date). ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 94 Version 4 . At position A (N 45:05 E 065:30) it is 13:15 LMT on 23rd March. At position B (S 28:37 W 092:20) the LMT is :B W 092:20 Arc to Time B 06:45:00 UTC 21°May GD° 6:09:20 00:35:40 LMT 21" May LD Longitude West . 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 Q4.UTC Best UTC must be a later time than LMT Q3. The ETA in LMT is :C E 010:15 Arc to Time C Flight Time D E 135:45 Arc to Time D ETD ETD ETA ETA ETA 15:30 LMT 15th May LD 0:41 14:49 UTC 15th May GD 11:18 26:07 UTC 15 th May GD 02:07 UTC 16 th May GD 9:03 11:10 LMT 16 th May LD NOTE: In flight the time standard is UTC. 5:45 Time to Arc = W 086° 15' Longitude An aircraft departs C (N 45:35 E 010:15) at 15:30 LMT on 15th May LD. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 95 Version 4 . To convert UTC to zone time or vice versa. add 24 hours.ZONE TIME (ZT) UTC is a method of co-ordination. the earth is divided into zones. LMT could be a method of time keeping. 15º wide. To solve this problem. Rule for the international date line: When heading east. Remember. but is not practical because each and every meridian would have its own time. When heading west. east is later and west is earlier. not time keeping. subtract 24 hours. add or subtract the zone number (hours). The time in each zone is the same and is referenced to the mid-meridian for that zone. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 96 Version 4 . Standard times appear on the next four pages. List 3 Countries with Westerly Longitude Apply Standard Times in the same manner as LMT (Long East .UTC Least & Long West .LOCAL STANDARD TIME As every Meridian has a different LMT. Durban has a different LMT to Johannesburg.) List 2 Countries normally keeping GMT or UTC.UTC Best) or apply as given at the top of each list. OOPS ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 97 Version 4 . For GMT (Greenwich Mean Time) read UTC. Each country has its own standard time factor which is applied to UTC to give local standard time. Ignore summer time. List 1 Mainly countries with Easterly Longitude (including Spain & Portugal which are Westerly Long. LMT is not suitable for civil time keeping. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 98 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 99 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 100 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 101 Version 4 . This time can be extracted from the Air Almanac (LMT). b) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 102 Version 4 . the time for sunrise. This time can be extracted from one of the following sources: a) b) c) d) The Air Almanac (LMT) The Jeppesen (LMT) The Aerad (UTC) The SA AIP for the major airports The time extracted have been corrected for atmospheric refraction. 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. sunset and twilight varies with latitude and date. Sunrise. SUNSET AND TWILIGHT SUNRISE The time at which the upper rim of the sun just becomes visible above the horizon. IMPORTANT NOTES a) Sunrise. sunset and twilight do however occur at the same LMT for all places on the same latitude for a particular date. The end of evening civil twilight The time at which the sun is 6° below the horizon on its way down. SUNSET The time at which the upper rim of the sun just disappears below the horizon.SUNRISE. This time can be extracted from one of the following sources: a) b) c) d) The Air Almanac (LMT) The Jeppesen (LMT) The Aerad (UTC) The SA AIP for the major airports The times extracted have been corrected for atmospheric refraction. This time can be extracted from the Air Almanac (LMT). The time at which the sun is 6° below the horizon on its way up. TWILIGHT The beginning of morning civil twilight. Due to the inclination of the earth's axis. sunset and twilight do not occur at the same LMT for places on the same meridian. 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. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 103 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 104 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 105 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 106 Version 4 . 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.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 PHASES OF THE MOON ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 107 Version 4 . The point where the moon is furthest from the earth in its orbit is called APOGEE. The point where the moon is closest to the earth in its orbit is called PERIGEE. The synodic period is + 29½ days and forms the origin of the month. the time interval between two new moons. 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.e. the F4 table would indicate 15 minutes ( This 15 minutes would then be added to or subtracted from the moonrise/moonset LMT at Greenwich to derive the LMT at 90° longitude. This fact is referred to as the moon's daily lag. in fact about 24 hours and 50 minutes depending on latitude and date. it will occur again at about 20h50 on day 2. the time interval between successive moonrises/moonsets is more than 24 hours. 360 The Air Almanac lists the LMT at Greenwich of moonrise and moonset at different latitudes and different dates. Thus if moonrise occurs at 20h00 on day 1. 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. Due to the fact that the moon rotates anti-clockwise as the earth is revolving anti-clockwise. 30 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 108 Version 4 . sunset and twilight however.G. This table simply indicates what fraction of the half daily difference applies to a particular longitude. E. Unlike sunrise. moonrise and moonset times vary with latitude and date. If we were at a longitude of 90°. Whether to add or subtract the 15 minutes depends on whether the longitude is east 90 = 180 1 2 = 15 ). it will occur at 20h25 at 30° N 180° E/W (half the daily difference 180 ). the time interval between successive moonrises/moonsets would be 24 hours. sunset and twilight. the diff column would indicate 30 minutes (½ daily lag). If the moon were stationary. MOONRISE AND MOONSET TIMES Much like sunrise. and moonrise at 30° N on the Greenwich meridian occurs at 20h00.THE MOON'S DAILY LAG The earth rotates approximately 360º in 24 hours.: If the moon's daily lag for a particular latitude and date is 60 minutes. If the moon's daily lag is + 50 minutes. moonrise and moonset do not occur at the same LMT for places on the same parallel of latitude. In the same way. for example. but for the next day. add a full day (24 hours) to get to the 1st AND add the FULL daily difference. the table will indicate the time for moonrise/moonset. if in the process of adding arc to time. 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 . If. and in the process of subtracting arc to time. we arrived at the UTC for moonrise.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) UTC moonrise 30° N 060° E 1 JAN 2458 2058 UTC 1st ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 109 Version 4 . but on the 2nd JAN.: JAN 1 moonrise 2445 is actually: JAN 2 moonrise 0045. ii) Because the moon's daily lag obviously has a cumulative effect. Now. 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). subtract a full day (24 hours) to get to the 1st AND subtract the FULL daily difference.SPECIAL NOTES i) Once the LMT for moonrise/moonset has been derived. we arrived at a UTC for moonrise. E. but on DEC 31st. a particular question may require that this time be expressed in UTC by adding or subtracting arc to time.G. instead of leaving the table blank for that day. a particular question required the UTC of moonrise for JAN 1. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 110 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 111 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 112 Version 4 . the aircraft flies a rhumb line track of 090° at a ground speed of 300 Kts. An aircraft arrives at position B (12° N 008° E) at 1823 ST on the 4th. what was the parallel of l altitude which the aircraft followed? Part 2 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 aircraft departed from A (71° N 174° E) and the flight time was 7 H 06. The aircraft arrives at destination on sunset of the same day. The flight time is 9 H 12.QUESTIONS Part 1 1. If the LMT of departure is the same as the LMT of arrival. 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. What is the destination position? 3. What is the latest standard LMT that the aircraft can depart from A if the flying time is 4H20? 2. An aircraft is to fly from A (12° S 22° 08' W) to B (10° S 63° 47' W). What is the arrival time at B in LMT? 2. at destination longitude? ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 113 Version 4 . The aircraft departed position A (08° N x° W) at 0838 LMT on the 4th and the flying time was 6H33. An aircraft departs from position A (60° N 015° E) at 0715 LMT on JAN 2nd. The aircraft must arrive at B no later than the end of evening civil twilight on 11 JAN. What is the longitude of position A? 4. The standard factor at B is 1 hour. The standard factor at B is 11 hours. An aircraft flies a rhumb line track of 270° for 3 hours and covers 1580 nm. The aircraft arrives at position B (63° N 168° W) at 2208 ST on the 2nd. What was the departure time in LMT? 3. Part 3 1. 2. 3. What is the UTC of moonrise for position 62° N 090° E on the 1st JAN? What is the UTC of moonset for position 50° S 120° W on the 1st JAN? What is the UTC of moonrise for position 52° S 060° W on the 1st JAN? ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 114 Version 4 . off track after 60 nm’s. FORMULA ERROR DISTANCE 1 60 EXAMPLE × = 60 1 1° = DRIFT ANGLE × 60 1 A to B 476 nm's. 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. After 157 nm's. i) ii) What is the new heading to steer to regain track at point B? What is the new track to point B? 11 157 i) ii) × 60 1 = 4. its drift angle is 1°. the aircraft is 11 nm's left of track.2° T : 090° + 2° = 092° T ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 115 Version 4 .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.2° 11 319 × 60 1 = 2° The new heading to B is The new track to B is : 090° + 4.2° + 2° = 096. The aircraft departs position A and maintaining a heading of 090° T. Track 090° T. a² a² a² a² a a = = = = = = b² + c² . DIAMETER or CIRCUMFERENCE of a CIRCLE.17 25.32. FORMULA a² = b² + c² .83 5.08 UNITS FLIGHT TRAINING COLLEGE Page 116 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 .(2 × 3 × 7 × COS 40) 9 + 49 . 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.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.2bc COS A Naturally. EXAMPLE Solve the length of Side a. this formula can be arranged in any other fashion to isolate the unknown.2bc COS A 3² + 7² .83 25. 96 UNITS = = = 2r 2πr πd ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 117 Version 4 . determine its circumference? c = = = 2πr 2 × 3. FORMULA d (diameter) c (circumference) c (circumference) EXAMPLE If the radius of a circle is 7 units.THE CIRCLE Various questions may be asked relating to the radius.14 × 7 43. diameter or circumference of a circle. but in a time of 70 HRS and at 3 times the airships speed. Deviation 3° W. 5. After 132 nm's. the aircraft is 11 nm's right of the intended track. Heading 143° C. What is the aircraft's groundspeed? What is the W/V? 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. An airship overhead the equator flies west around the world in 280 HRS.QUESTIONS 1. Oculus crosses the equator at longitude 60° W at 1400 LMT on 27 JULY on its northbound passage. i) ii) 3. With the aircraft's weather radar in the MAP mode. At what longitude does Oculus cross the equator on its southbound passage and what is the LMT at this point? ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 118 Version 4 . An aircraft departs from position A on a heading of 247° M in order to fly track 229°T. What is the position of the aircraft as a bearing and distance from the station at the end of the 7 minutes? 4. At what latitude did the aircraft fly? The effect of the altitude of both the airship and the aircraft is negligible. the following observations were made of a ground feature: UTC 1205 1215 RELATIVE BEARING 067º 107º RANGE 76 83 TAS 300 Kts. 2. TAS 300 Kts. Variation 15° W. An aircraft also flies west around the world. What is the magnetic heading to steer to return to A? (Variation 15° W). 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. (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 other two can be calculated. W/V 340/30 is the direction from which the wind blows at 30 Kts. The Air Vector is TAS and True Heading only (never TAS & Track) If four of the six values are known. The GROUND VECTOR (True Track & Groundspeed) has two arrows and is called the TRACK PLOT. The W/V has three arrows. The units cannot be interchanged. NOTE All directions are TRUE DIRECTIONS (measured from TRUE NORTH) The length of each Vector is the value for ONE HOUR.CHAPTER 5 Navigational computer VECTOR TRIANGLE (TRIANGLE OF VELOCITIES) Navigation plotting is based around the Vector Triangle which comprises of three vectors. NAVIGATIONAL COMPUTER ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 119 Version 4 . The Wind blows from the Air Vector to the Ground Vector. In the above sketch the Drift angle is 7° Right. WIND EXAMPLES Example 1 HDG TAS W/V 330º 150kts 040/25 Find: Track made good The groundspeed Solution: Step1 Plot wind down. Therefore method one will be used in this chapter. but basically they can all do the same thing in the same way. Step2 Read off the drift 10º left. There are two methods of working with the wind side: TAS under the grommet (center) wind down. 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 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 120 Version 4 . or GS under the grommet (center) wind up – Jeppesen method.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. then set HDG 330º under index on top. The first method can solve all 3 common triangle of velocity problems. method 2 can only solve 2. In this method the wind is plotted down from the grommet. There are many wide and varied versions of the Whiz Wheel. Note that if this is done correctly HDG 141 is under the index and the drift will be 9ºR. Further adjust the disk until the difference between the required track and the HDG under the index equals the drift. The TRK will equal 150º which is what we require. the wind velocity is 240°/40. Observe that the drift has changed and is now 8ºR. Find the Heading and GS.Move the circular scale to have the Track under the Index mark. ANSWER: Approx 280°. Track is 290°. HDG TAS TRK GS 138° 120kts 146° 144kts Step 1 Place the HDG under index and TAS in the middle on the whiz wheel: ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 121 Version 4 . find out the Wind Velocity. Example 3 If TAS is 174kt. Now read off the GS at the end of the wind vector 125 kts. adjust the disk so that 143º (150º-7) is under the index. Step 3 The drift is noted to be 7ºR. 145kt Example 4 The in-flight type of problem…finding wind You know the following figures. so now draw in a straight line along the 8° right drift. Draw a line from the grommet to the intersection of 144kt and 8° drift and you have drawn in the wind vector. in this case its 360°.Step 2 (diagram to the right) Now in your head work out the drift. and its found to be 8° right. Under the Index mark you can read the direction. The GS is 144kt. So the answer is 360/30 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 122 Version 4 . and from the 120kts under the grommet at the beginning of the wind vector to its tail is the strength of the wind. draw a line along the 144 line so as to intersect the 8º drift line. Step 3 (diagram below) Rotate the grid until the wind vector blows straight down. in this case its 30kt. and the Distance for the leg it 77nm. but we are only concerned with the GS/Dist/Time and the Fuel Qty/Fuel Flow/Time problems.THE CALCULATOR SIDE OF THE COMPUTER This side of the computer can do many weird and wonderful calculations. what is the EET for the leg? Answer = 30minutes (found under the Arrow head) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 123 Version 4 . in that we use the following methods for the equations: DISTANCE TIME GS Fuel Qtty TIME Fuel flow Example 1 If the aircraft has a GS of 154kts. In order to make things simple we shall use the whiz wheel in the same manner as you would a electronic calculator. 3 litres. Example 4 If you burn 24 litres per hour. what is the fuel burn for this leg of the flight? ANSWER = 13. what will be the fuel burn? ANSWER = 35 litres ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 124 Version 4 .. and the fuel flow is 32 lph. how far will you travel in 11minutes? ANSWER 27nm Example 3 A leg is 25 minutes long. and the duration of the leg is 88minutes. say 14 litres.Example 2 At a GS of 147kt. draw 8½° RIGHT DRIFT LINE. draw 8° RIGHT DRIFT LINE. draw 1° RIGHT DRIFT LINE. Read off WIND SPEED 30 Kts along the CENTRE LINE 2 3 Heading 040° + Drift 8° Right 4 1 5 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 125 Version 4 . Set HEADING 085° against TRUE INDEX. Set HEADING 040° against TRUE INDEX. Read off WIND DIRECTION 348° against the TRUE INDEX. Set HEADING 355° against TRUE INDEX. Place the intersection of the THREE DRIFT LINES on the CENTRE LINE below the CENTRE CIRCLE.MULTIPLE DRIFT W/V USINE THE WHIZ WHEEL Given:TAS 190Kts Heading 085° + Drift 8½° Right Heading 355° + Drift 1° Right Method: Set TAS 190 Kts at the CENTRE. In this case the W/V can only be solved by the manual nav computer. Position the intersection of the DRIFT and GROUNDSPEED lines BELOW the CENTRE CIRCLE. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 126 Version 4 . 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. Draw arc of GROUNDSPEED 142 Kts. Read off WIND SPEED 50 Kts along the CENTRE LINE. Read off WIND DIRECTION 070° against the TRUE INDEX. Given: 1012 Z 1000 z Heading 055° (T) TAS 250 Kts Heading 010° (T) Doppler GS 235 Kts Doppler Drift 10° Right 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.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. Join the end (tail) of the third wind vector to the starting point (head) and measure the wind direction. from head to tail. Measure the length of the vector and divide by the number of W/V’s to give the wind speed. Draw the three wind vectors to scale. The above method is used to calculate the mean W/V at cruising altitude when several W/V are given for a route.7 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 127 Version 4 . This can best be done on normal ruled paper using the lines as reference and a suitable scale. 091º/36.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. Given: 135/16 100/14 W/V 180/20 Proceed as follows: a.MEAN W/V (whiz wheel) A mean W/V can be found on the square side of the slide rule. Set 100 against the true index and plot 14 kts down from the end of the previous wind FLIGHT TRAINING COLLEGE Page 128 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 . Set 135 against the true index and plot 16 kts down from the end of the previous wind c. Set 180º against the true index and plot 20 kts down from the centre spot. b. 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 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 129 Version 4 .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. ELECTRONIC NAVIGATION COMPUTERS TO CALCULATE TRACK & GROUNDSPEED (GS) ON THE PATHFINDER Given 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 1. Enter 3st W/V in ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 130 Version 4 . Req TAS function Enter 1st W/V in Enter 2nd W/V in W Dir W Spd Crs GS ] remember this ] 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) Heading 090° TAS240Kts W/V 010/60 Answer Hdg = Wind Direction TAS = Wind Spd 3. 2. Along the headwind component of 20 read off the required minimum wind velocity of 25 kts.TAKE OFF WIND COMPONENTS In order to take-off. Pressure Altitude 8000 feet. The wind direction is 40° from the runway direction. Proceed as follows: 1) 2) 3) 4) Set 040° against the true index and draw a vertical line representing the W/V. This is the difference between the runway and the wind direction. an aircraft needs a headwind component of at least 20 kts. The maximum permitted crosswind component is 25 kts. OAT +20°C Using the AIRSPEED WINDOW set pressure Altitude 8000 feet against OAT +20°C Against RAS 140 on the INSIDE SCALE. TAS CALCULATION Given: RAS 140 Kts. Set 360° against the true index. Along the crosswind component of 25 read off the maximum wind velocity of 40 kts. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 131 Version 4 . read off TAS 164 Kts on the OUTSIDE SCALE. a) Determine the maximum and minimum wind speed acceptable for take-off. 4 Using the AIRSPEED WINDOW set Pressure Altitude 20 000 feet against OAT -23°C Against EAS 310. use the correction factor from the table below. RAS 320 and TAS 440 kts are too high due to compressibility.97 = EAS 310. 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. ELECTRONIC CALCULATORS If given OAT or COAT use PLAN TAS or PLAN MACH If given IOAT use ACT TAS or ACT MACH ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 132 Version 4 .4 on the INSIDE SCALE Read off TAS 427 kts on the OUTSIDE SCALE.AIRSPEED COMPRESSIBILITY CORRECTION ELECTRONIC CALCULATORS Electronic calculators correct for the compressibility error at high speeds. ALTERNATIVE METHOD (EQUIVALENT AIR SPEED EAS) RAS 320 x 0. VHF D/F and VOR ADF / NDB use station variation. however. The main disadvantages of the chart is when Rhumb Line navigation is used (flying constant headings).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. for European plots. 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 N60:00.45° being the sine of the Parallel of Origin S 27:00. Chart convergency is measured at S27:00 and N45:00 for the Southern African and British charts respectively. For the ATP examination the chart convergency (CCF 0. This is marvellous when using Great Circle navigation systems such as INS. Straight lines drawn on the chart are considered to be GREAT CIRCLES for all practical purposes. 1° of Latitude has 12 increments of 5 Nautical Miles each. which is the prime advantage of the chart. 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. This is not the case. The Chart Convergency Factor is 0. 1° of Latitude = 60 Nautical Miles. Loran and satellite GPS. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 133 Version 4 . The other main advantages of the chart is that Great Circle tracks can be flown which are shorter than Rhumb Line tracks. especially when plotting radio bearings. Omega.45°) which should be applied to ADF / NDB bearings in Southern African Plotting problems may be ignored as the value is small. MEASUREMENT OF DISTANCES Use the VERTICAL LATITUDE SCALE near the Latitude of the plot. use aircraft variation. RADIO BEARINGS All radio bearings MUST be converted into QTE True Bearings and plotted from the nearest Meridian to the radio facility. The above procedures are to be used for ATP exams whenever a heading (True or Magnetic) has to be calculated. this is the MID MERIDIAN between A and B.MID MERIDIAN TECHNIQUE . Highlight the nearest MERIDIAN to the mid point along track. Apply the variation at the mid meridian.GREAT CIRCLE NAVIGATION Draw the straight line from A to B. Calculate the TRUE HEADING using the nav computer. Apply the variation of the first isogonal along track. Calculate the TRUE HEADING using the nav computer. Apply the variation at B. What is the INITIAL magnetic heading from A to B? Measure the INITIAL TRUE TRACK at the nearest meridian at A. Typical exam questions are: 1. What is the FINAL magnetic heading from A to B? Measure the FINAL TRUE TRACK at meridian closest to B. this is the TRUE GREAT CIRCLE TRACK which cuts all meridians at a different angle. 2. What is the MEAN magnetic heading from A to B? Measure the MEAN TRUE TRACK at the mid meridian between A and B. 3. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 134 Version 4 . Calculate the TRUE HEADING using the nav computer. This is the basis of rhumb line navigation on a Lambert’s chart.use the above technique. The HDG (T) is calculated by the nav computer. THE DEAD RECKONING (DR) POSITION The DR position is the calculated ground position of the aircraft. 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.RHUMB LINE NAVIGATION Draw the straight line from A to B. Plotting or measuring wind direction is referenced to the Mid Meridian. this is the TRUE GREAT CIRCLE TRACK which cuts all meridians at a different angle. variation applied and the HDG (M) is flown between A and B. Measure the mean great circle track at the mid meridian. It may be plotted at any time during the flight by either the AIR PLOT or the TRACK PLOT method. 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. this is the mid meridian between A and B. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 135 Version 4 . ATP exam .MID MERIDIAN TECHNIQUE . All plotting of true headings between A and B is referenced to the mid meridian. Highlight the nearest MERIDIAN to the mid point along track. From the 0845 Air position plot the WIND VECTOR DOWNWIND. The length of the Air Vector is proportional to the TAS and time. The AIR PLOT method is flexible. The heading at 0830 is plotted from the first meridian ahead of the aircraft. This is the DR POSITION ∆ of the aircraft at 0845. 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). As the heading is changed at 0852 and 0910 the headings are plotted from the first meridians ahead of the respective air positions. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 136 Version 4 . 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. TAS 180 Kts for 45 minutes = 120 NM. The wind is blowing FROM 020° (T). Plot the DR POSITION 20 NM along the WIND VECTOR. Plot the AIR POSITION + 120 NM along the AIR VECTOR.AIR PLOT METHOD Overhead position A at 0800. Wind Speed 30 Kts for 45 mins = 20 NM. alterations of heading and/or TAS may be made at any time. The length of the wind vector is also proportional to the wind speed and time. The length of the AIR VECTOR is proportional to time.TRACK AND GROUND SPEED METHOD 0900 Overhead position A 0930 Overhead position B Heading 080° (T) A line joining the two fixes is the TMG TRACK MADE GOOD. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 137 Version 4 . TAS 180 Kts for 45 mins. Heading 082° (T). Note : This method can be used when there has been no alteration of heading or groundspeed between the two fixes. Fix at 0945. which is 135 NM and give the AIR POSITION + at 0945. which is the actual track followed by the aircraft over the ground. update the ETA at checkpoints and monitor the fuel flight plan. Measure the distance A . The forecast W/Vs may be accurate. TAS 180 Kts. BUT the aircraft is South of track BECAUSE the actual W/V is different from the planned W/V. It is strongly recommended that the W/V is found in flight to improve navigation. Groundspeed 240 Kts for 12 mins = 48 NM.B. This would be the position of the aircraft in zero wind conditions. WIND FINDING A flight plan is based on a met forecast that can be many hours old. but if the pressure pattern changes unexpectedly. Extend the TMG and plot the DR position for the required time. say 120 NM which is computed against time between fixes of 30 mins and gives a groundspeed of 240 Kts. Plot the heading 082° (T) from the first meridian ahead of the aircraft in the direction of flight. Air Plot Method Overhead A at 0900. the W/Vs can be very different thus affecting the navigation of the aircraft. The wind speed is proportional to the time that the AIR PLOT has been running. W/V 025/44.Join the AIR POSITION to the FIX which is the WIND VECTOR. The heading may be altered at any time with no loss of accuracy. The wind direction is measured from the nearest meridian to the fix. Wind Vector 33 NM for 45 mins = Wind Speed 44 Kts. TAS GS 190 Kts 224 Kts ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 138 Version 4 . Track And Groundspeed Method This method is used when a single heading has been flown between fixes. The TMG (Track Made Good) is measured at the mid meridian between the fixes . annotate with 3 arrows. in this case about 025° (T). As the heading has been changed at 0925 then the meridian at 0925 is used to plot the new heading. Heading 082° (T). The GS (Groundspeed) is proportional to the time between the fixes . the wind blowing from AIR to GROUND.168 NM in 45 mins = GS 224 Kts. TAS 190 Kts. 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. The above method is suitable for short distances between fixes in the order of less than 250 NM.say 090° (T). Using the nav computer: Heading TMG 082° 090° W/V 306/45 This method is accurate over any distance. QDM and QDR bearings .convert to QTE using aircraft variation. may be used. CIRCLES and Radio Bearings are also GREAT NDB / ADF Bearings ADF bearings are measured at the aircraft. rivers etc. TRUE BEARING FROM the station. QDM QDR QUJ QTE MAGNETIC TRACK TO the station. 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. LAMBERTS CONFORMAL CHART As straight lines on the chart are 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. MAGNETIC BEARING FROM the station (VOR Radial) TRUE TRACK TO the station. The main source of Position Lines are radio bearings but coastlines.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. 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. QDM + 180° QDR + Variation + Variation QUJ + 180° QTE Follow the shortest route to convert one bearing to another.45°. railway lines. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 139 Version 4 . Radio Bearings must be converted into TRUE BEARINGS (QTE) before they can be plotted from the meridian passing through the radio facility. Note that a P/L has single arrows at either end and a time. Heading 095° (T) NDB ABC bears 172 Relative FLIGHT TRAINING COLLEGE Page 140 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 . Example: 1205 1221 Overhead NDB ABC. 0800 0824 Overhead A en route to B 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. Track check A bearing from a radio facility that the aircraft has over flown is an indication of the TMG (Track Made Good).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. 0956 Z VOR JSV Radial 090.172 Relative + Heading 095 = QUJ 267 . 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).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. If the VDF station is imagined to travel from X to Y at the same speed of the aircraft. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 141 Version 4 . Aircraft Track 090° (T) Groundspeed 240 Kts The aircraft at position A at 0900 Z obtains a QTE of 320° from VDF station X. JSV Variation 16 W Plot QTE 074 (T) DME range 105 NM THE RUNNING FIX - TRANSFER OF POSITION LINES A position line is usually a bearing of the aircraft from a radio facility. the bearing of the aircraft from the radio facility would remain constant. 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. At 0912 the aircraft will have flown 48 NM along track to position B. If the radio facility were moved on a track parallel to that of the aircraft and at the same groundspeed. the method cannot be accurately applied to an air plot from a DR position. 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. Such a question may well be expected in the ATP exam although only for direct radio bearings (VOR. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 142 Version 4 . The techniques involves using time intervals relating to an air plot from a specific fix .Transfer of circular (DME) position lines 0810 0830 0842 Overhead A. Heading Track 083° (T). The AIR PLOT may then be used to transfer position lines and hence obtain a fix on the ground. RMI) and not DME arcs. 090° (T). ADF. TAS 195 Kts GS 180 Kts JSV VOR/DME Range 78 nm 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. 4. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 143 Version 4 . If a GS check cannot be made then the DR groundspeed may be used. 3. The same scale must again be used. 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). Heading 081° (T). Plot the air positions from the fix for the times of the bearings. Inspect the 3 bearings to see if one crosses the track at approximately 90°.METHOD: i. From the 1040 air position (D) draw two lines. vi. ii. Transfer the 1223 P/L 17 mins at GS 240 = 68 nm along track. 5. From 1200 to 1230 is 120 NM in 30 mins = GS 240 Kts. NOTE: Plot the 3 QTE’s. If so. Transfer the 1230 P/L 10 mins at GS 240 = 40 nm along track. 2. By joining the 1040 air position to the fix. v. iii. then discard the DR Groundspeed and perform a GS check. DR Groundspeed 192 Kts X X X QTE 327 QTE 358 QTE 035 1. Redraw the position lines from point H and point I to establish the fix at 1040 (G). the mean wind can be established since 1005. one parallel to line be (Line DH) and the other parallel to CF (Line DI). iv. Mark off an arc or 35 units radius to correspond to the time interval of 35 minutes from the last fix (A). TYPICAL THREE POSITION LINE FIX EXAMPLE 1 1200 1223 1230 1240 Overhead A en route to B TAS 200 Kts. Plot the QTE's for each bearing. Repeat the above procedure between C and F for 29 minutes the same unit scale as that used at 1027. b. BURGERSDORP VOR BDV RADIAL 226 BURGERSDORP NDB JF 326° relative by ADF. Use the air plot transfer of position lines to find the fix at 0742. Groundspeed 180 Kts 0815 0824 0835 VOR VOR VOR TIM 288 TIM 321 TIM 348 As there is no fix from which to start the plot. Give the position at 0742. RAS 225 kts. Give the mean W/V experience since 0700. It is parallel to the actual track of the aircraft. a. Transfer the 0815 P/L 20 mins at GS 180 = 60 nm along track. Aircraft track 045° (T). 72 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 144 Version 4 . FL 230.EXAMPLE 2 The following VOR Radials are obtained from TIMBUKTOO. A Groundspeed check cannot be made so the DR GS 180 is used. EXAMPLE 3 0700 0723 0734 Overhead PORT ELIZABETH VORTAC PEV. Transfer the 0824 P/L 11 mins at GS 180 = 33 nm along track. Join the fix to the air position at 0742 to find the wind direction and speed. OAT -35° C. the track of 045° is drawn to cut the 3 P/Ls. set heading 016° (M). Remember that only 42 minutes has elapsed since 0700 and therefore the wind speed will be the measured vector x 60 knots. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 145 Version 4 . 12°°C ISA at Sea Level 20 000 ft x 2°/1 000 ft ISA at FL 200 ISA + 13°C By computer FL 200. RAS 225 Kts. RAS 175 Kts. 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 = = = = + 15° C .12° C. The mid point of the climb in TIME occurs at approximately two thirds of the climb.25° C .32° . temp .7° C. Navigation wise the most important parameter required for a climb to cruising altitude is the mean TAS. ISA at Sea Level 16 000 ft x 2° C/1 000 ft ISA at 16 000 ft ISA + 10° = = = + 15° C . This will occur at the mid point of the climb in TIME and not ALTITUDE. This applies to both piston and jet aircraft.40° C . temp . If a manual nav computer is used the compressibility correction must be made according to the table below. The above calculations were made with an electronic calculator which corrects for compressibility. Example 1 Climbing from Sea Level to FL 240 at RAS 175 Kts and a mean rate of climb of 800 ft/min.7° C By computer FL 160. 24 000 ft x 2/3 = 16 000 ft mean climb altitude. temperature ISA + 10°C. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 146 Version 4 . 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.THE CLIMB CONSTANT SPEED (RAS) CLIMB This is the normal climb technique used by commercial aircraft.17° C . TAS 311. temperature ISA + 13° C. 94 0.86 450 098 0. seldom longer than 25 mins.84 500 0. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 147 Version 4 .90 0. NEVER a DR POSITION. Descent details: Rate of descent 2 000 ft/min Mean descent GS 360 Kts Plan a descent to arrive overhead destination at FL 90. the INITIAL track is used to calculate the heading and groundspeed.90 0. AIR PLOT ON THE CLIMB As the climb is of short duration.91 0.96 0. TAS 316 x 0.99 0. It may be continued from an Air Position. TAS 225. the mean climb W/V applied to give the TOC DR position.PRESSURE ALTITUDE IN FT 10000 20000 30000 40000 50000 Ex.87 0. At the TOC (Top of Climb) position which theoretically is on track.98 0. the Mid Meridian method is then used.95 0.97 0.0 0.92 0.93 250 1. 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.89 0. temp -12° C. A heading.97 0. 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.97 0.95 0.94 0.84 550 0. GS 1000 FIX 1018 TOD 495 360 DIST 146 230 84 TIME 18 14 ETA 1018 1032 Calculate the descent distance and subtract from distance to go and complete the cruise sector.97 0. The heading for the cruise is plotted from the TOC Air Position and not the DR position.94 0.86 400 0. The Air Position at the TOC is plotted.84 FL 100.985 = TAS 311 THE DESCENT As a CONSTANT RATE OF DESCENT is normally used.87 0. 2 200 1.98 0. Air Vector or Air Plot can only be started from a positive fix.94 0.90 RECTIFIED AIR SPEED IN KT 300 0.96 0.99 0.90 0.93 0.97 0.87 350 0.99 0.86 0.0 0.96 0. 1513 1555 a) b) 5. Overhead UMTATA NDB. 1000 Overhead BURGERSDORP. 0800 0840 0900 0925 a) b) c) 3. OAT +5(C. Alter heading 135((M). heading 260((C). deviation 3(E.Magnetic heading 300(. FL 100. set heading 180((M). Alter heading 220((M). Visual fix 3200S 02800E. 1050 Alter heading GEORGE. What is the magnetic heading to steer and ETA GEORGE? Overhead UPINGTON. What is the mean magnetic heading to PEV? What is the ETA for PEV? 4. 1300 1345 1420 1435 a) b) c) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 148 Version 4 . FL 125. RAS l64kts. a) 2. What is the ETA for EAST LONDON? What radial should be maintained on EAST LONDON VOR ELV? Overhead DNV. RAS l72kts. set heading for VICTORIA WEST (VWV). What is the DR position at 0925? What is the magnetic heading to CTV? What is the ETA for CTV? 0835 Overhead Cape Town VOR CTV. What has been the mean W/V since 0835? What is the revised ETA VICTORIA WEST? What relative bearing should be maintained on the ADF in order to Home on VICTORIA WEST NDB. OAT -3(C. 1026 Alter heading 202(M). RAS l45kts. Alter heading 230((M). set heading 260(M). pressure altitude 12000ft. VW? Overhead DURBAN VOR DNV. W/V 010/30. OAT +6C. OAT +2(C.QUESTIONS 1. FL 100. W/V 275/25. Alter heading 175((M). RAS l50kts. TAS l90kts. 0926 a) b) e) SUTHERLAND VOR Radial 147. alter heading for EAST LONDON. W/V 120/30. Alter heading for CTV. Determine the W/V since 1300. DME 35. JF NDB relative bearing 320(. VW NDB relative bearing 225(. b) 0805 0815 0825 c) d) 0825 e) f) 7. pressure altitude 15000ft. What is the revised ETA for ELV? What radial should be maintained to ELV? Overhead GGV. W/V 225/20. 0700 Overhead CTV. magnetic heading 090((M). What has been the mean W/V since 0700? From SLV alter heading for EAST LONDON. FL 150.6. set heading for BLV. What is the position at 1820? What has been the mean W/V since 1723? ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 149 Version 4 . alter heading for BLV. BDV radial 325(. OAT -5(C. RAS l65kts. What is the mean compass heading for PEV? BLV DME 116mn BDV radial 280(. OAT +3(C. 0745 a) Overhead SLV. RAS l55kts. OAT -7(C. What is the position at 1715? What has been the mean W/V since 1530? Using the W/V found. TAS l80kts. compass deviation 2(E. W/V 290/35. 1530 1645 1705 1715 a) b) 1715 c) d) e) 8. set heading for PEV. pressure altitude 13000ft. 1723 a) 1745 1810 1820 b) c) What is the initial magnetic heading to steer? QTE GGV 350( QTE GGV 008( QTE GGV 030( What is the position at 0825? What has been the mean W/V since 0745? Alter heading for ELV. COOKHOUSE NDB CH RMI reading 158(. RAS l50kts. What is the mean magnetic heading to BLV? What is the ETA for BLV? What relative bearing must be maintained on the ADF to home on BL NDB? Overhead KMV. constant RAS l38kts. 0537 0540 0607 0617 0628 a) 12. set heading for HARARE.1828 d) e) 9. RAS l42kts. Cruise RAS l62kts. mean TAS l73kts. TAS l97kts. What has the mean W/V been since 0540? Using the W/V found at 0628. Alter heading for PEV? What is the mean compass heading to steer? What is the ETA for PEV? Overhead BULAWAYO VOR. mean forecast W/V 270/20. Constant rate Of descent 500ft/min. THORNHILL DME 35nm NORTON NDB relative bearing 333(. climbing to FL 130. 0637 0722 A descent is planned to arrive overhead ELLISRAS at 5000ff. Overhead GREEFSWALD VOR. mean OAT +8(C. VRYHEID NDB VHD relative 246(. Overhead CHIREDZI NDB CZ. alter heading for ELLISRAS. Overhead PM at 3000ft. a) b) 0906 c) d) What is the position at 0900? What has been the mean W/V since 0740? Alter heading for KMV. As the forecast W/V is not available. forecast WV 285/20. 1213 1249 1256 a) b) 10. set heading for KMV. You are at TOC on ETA. OAT +4(C. OAT -2(C. mean forecast W/V 350/20. ESHOWE NDB ES RMI 166(. mean rate of climb 500ft/min. a) What is the ETA overhead ELLISRAS? ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 150 Version 4 . determine the mean magnetic heading To steer from the fix at 0628 to MAPUTO. FL 110. FL 115 OAT 0(C. heading 253((M). VRYHEID NDB VHD relative 280(. 0900 UPV DME 67nm. forecast W/V 325/35. 0825 UPINGTON NDB UP RMI reading 168(. What has been the mean W/V since 1213? What is the radial to maintain on the HARARE VOR? 0740 Overhead KEETMANSHOOP VOR KTV. What is the ETA for KMV? What bearing must be maintained on the RMI in order to home on KM NDB? Take off PIETERMARITZBURG for MAPUTO. RAS l42kts. 0843 UPV radial 030(. steer the track. 11. FL 120. MASERU VORTAC MZV DME 75nm. What is the initial magnetic heading to steer? Doppler observation: Drift 2( left. Set heading 280(M). ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 151 Version 4 . a) b) c) 1045 1130 1130 d) e) 1136 f) g) What is the mean magnetic heading for the climb? Give the DR position for TOC. W/V 035/15. maintaining radial 280. Constant RAS l40kts. mean WV 300/25 and mean RAS l60kts. 1020 Overhead DURBAN VOR DNV at 3000ft climbing to FL 140. set heading for CAPE TOWN. constant rate of Descent 500ft/min. RAS l45kts. Give the ETA for the TOC position. Determine the position at 1130.82. What is the ETA for CH? What relative bearing must be maintained on the ADF to home on CH? Overhead BURGERSDORP VOR BDV. set heading for CAPE TOWN. OAT -5(C. FL 310. Give the spot wind being experienced. VICTORIA WEST VORTAC VWV DME 85nm GEORGE VOR GGV radial 025. W/V 300/30. d) e) 15. What is the ETA for TOD and CTV? Fix at 2900S 02730E. FL 160. 0545 0650 0710 a) b) 0720 c) A descent is planned to arrive overhead CTV at 3000ft. forecast W/V 280/30. Mean climb information. temp -10(C. temperature deviation ISA +5(C. temperature -42(C. You reach the TOC position.13. BURGERSDORP VOR BDV RMI reading 270(. forecast W/V 300/70. What has been the mean W/V since 1020? Divert to PORT ELIZABETH via COOKHOUSE NDB CH using a forecast W/V of 360/45. What is the position at 0710? What has been the mean W/V experienced since 0545? Alter heading for CTV What is the DR position at 0720? 14. 0630 a) 0635 b) Give the mean magnetic heading to steer to CTV from TOD. Mach 0. GS 440kts. rate of climb 500ft/min. mean temperature +10(C. RAS l80kts. RMI reading 163(. set heading for MMV. Mach 0. mean TAS 390kts. temperature deviation +3°C. dev. alter heading for CTV. d) e) f) g) 17. dev. What is the initial magnetic heading to steer? What is the ETA TOC? a) b) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 152 Version 4 . ISA + 10°C. Forecast W/V 100/30. Crossing SUTHERLAND VOR SLV radial 167. climbing to FL 290 on heading 007°(M) and at a constant RAS 220kts. RMI reading 121(. FL 310. 0643 Determine the latest time to commence the descent. NVV RMI reading 180°. temp. FL 070. +5°C. c) What is the position at 0658? 0704 0727 Fix 31°04S 023° 04E. Constant rate of descent 2000ft/min. temp. forecast W/V 280/90. Mean climb W/V 240/70. 0315 Give the radial to maintain to KTV.83. ALEXANDER BAY VOR ABV RMI reading 288°. What is the ETA overhead CTV at FL 050? Overhead CAPE TOWN VORTAC CTV at 3000ft. climbing to FL 180. +10°C.79. KLEINSEE NDB KZ RMI reading 253°. What mean W/V has been experienced since 0643? 0705 a) 0710 0720 0728 0730 b) c) 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. mean rate of climb 600ft/min. set heading for KEETMANSHOOP VOR KTV Give the mean magnetic heading and ETA KTV NIEUWOUDVILLE VOR NVV RMI reading 109°. Assume for the descent: mean W/V 300/50. Level FL 290. Give the position at 0730. dev. constant RAS l6Okts. d) e) 16.The following readings are obtained from the BURGERSDORP VOR BDV: 0640 0647 0658 RMI reading 201(. temp. Mach 0. Give the magnetic heading to maintain the above radial What is the rate of descent required? What is the required mean RAS during descent? Overhead SALISBURY VOR VSB. mean W/V 260/40. OAT -40°C. A descent is planned to arrive overhead CTV at FL 050. 0700 Using the above W/V alter heading for CTV. set heading for UPIINGTON. Temp. At TOC. 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. temperature -18°C. TAS and magnetic heading for the descent? What is the ETA for TOD? What is the ETA for the CAPE TOWN CTA? Overhead EAST LONDON. Overhead KMV. forecast W/V 040/35. RAS 208kts. estimate what QDR to expect from VICTORIA WEST VOR a) b) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 153 Version 4 . Set heading CTV. ISA + 8°C. alter heading to parallel the TRACK Give the initial new magnetic heading to steer. 0343 0400 0405 0414 d) e) 0414 f) 18. 0610 a) 0654 0654 b) THORNHILL VOR RMI reading 310°. RAS 200kts. the RMI indication of VSB is 034 temperature -11°C. Determine the W/V experienced since 0610. FRANCISTOWN NDB FT relative bearing 055°. as late as possible. What is the position at 0414? What mean W/V has been experienced since 0315? Alter heading for MMV using W/V 140/35. c) Using the DRIFT experienced. RAS 208kts. RAS 180. mean rate of climb 1200ft/min. temperature -14°C. Mean W/V for the descent is 200/30. altitude 2000ft climbing to FL 310 at a constant RAS of 230kts. Give the mean magnetic heading and ETA for MMV. FL 180. d) e) f) 19. What is the mean magnetic heading and ETA CTV? KMV radial 250°. VWV radial 337°. BULAWAYO VOR VBU RMJ reading 344° MESSINA NDB RMI reading 122°.At TOC. Give the initial magnetic heading for the climb. temperature. mean temp -22°C. dev. 1025 What is the mean pressure altitude. FL 180. forecast W/V 260/60. FL 180. temperature -36°C. RAS l65kts. alter heading for UPV. climbing to FL 260. 10 DME. Alter heading for SUTHERLAND. Give the position at 1110 and the mean W/V experienced since 1025.72. SISHEN NDB SI RMI reads 020°. forecast W/V 010/65. The drift on this heading 7° left. 1030 1045 1120 Level at FL 310. VICTORIA WEST VOR VWV QDR 113° VICTORIA WEST VOR VWV QDR 071° BLOEMFONTEIN DME range 132nm. OAT -2°C. DME range 10nm FL 050. mean rate of climb 700ft/min. Alter heading 240°(M). maintaining radial 245. alter heading for UPINGTON. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 154 Version 4 . RAS l85kts. Alter heading 340(M) TAS 434kts. What relative bearing must be maintained on UP NDB? Airborne LANSERIA. Set heading on flight plan heading 221°(C). Abeam WELKOM NDB WM. 1103 1109 Alter heading 290°(M). The drift observed on this heading 5° left. Mach 0. deviation °2W. Give the spot W/V. DNV radial 150.1050 c) 1053 1107 1110 d) 1116 1126 e) 1126 f) g) 20. RAS 190. What was the mean W/V experienced since 1120? What is the DR position at 1200? What is the mean magnetic heading to SUTHERLAND? What is the ETA for SUTHERLAND? Airborne DURBAN. forecast W/V 120/20. The drift on this heading is 8° left. radar vectors to radial 150 DNV. temperature -25°C. a) 1137 b) Give the true heading. 1132 1147 1200 a) b) c) d) 21. Doppler GS 390kts and drift 4° left. forecast WV 180/30. ETA and DR position for TOC. Overhead GRASMERE VOR GAV at FL 140. UPINGTON DME range 100nm Using the spot wind found. Alter heading 220°(M) to avoid a thunderstorm. Give the ETA abeam COOKHOUSE. mean temperature -11°C. Level at FL 260. Give the mean magnetic heading and ETA UPINGTON. Alter heading 275°(M). Give the initial magnetic heading and ETA UPV. forecast W/V 200/30. Alter heading 274( (M) GEORGE VOR GGV bears 248 by RMI SUTHERLAND NDB SL bears 310ْ by RMI. temperature -30(C. From the 1300 position to TOD the W/V is forecast as 200/30. EAST LONDON NDB bears 225° relative. Give the RAS while at FL260 that will ensure arrival at WOLSELEY at 1348. What mean W/V has been experienced since 1109? Alter heading 250°(M). Descend at 800ft/min. Give the aircraft’s position at 1215. CAPE TOWN Control requests your arrival at WOLSELEY at FL 100 not before 1348. What mean W/V has been experienced since 1215? Set heading for WOLSELEY. EAST LONDON NDB bears 287° relative. What will be the time in the descent? g) h) i) j) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 155 Version 4 .1150 1203 1215 c) d) 1227 1233 1245 1300 e) f) 1300 Crossing the coast inbound. TAS 300kts. Give the TOD position. Alter heading 323°(M) PORT ELIZABETH VOR PEV bears 181( by RMI. mean TAS 240kts and mean descent W/V 140/30 Give the mean magnetic heading and ETA for TOD. Give the position at 1300. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 156 Version 4 . FL 100. RAS 155kts. set heading for NEWCASTLE 311o(M). forecast W/V 220/45. 1000 Overhead EAGLE VOR EGL. Give the TAS for the cruise at FL 240. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 157 Version 4 . Give the mean TAS for the climb. alter heading NEWCASTLE. Give the mean magnetic heading and ETA STRUMBLE. EAGLE VOR EGL radial 140(.CHAPTER 8 BRITISH ISLES CHART PLOTTING EXERCISES 1. 1000 a) 1045 1047 1047 b) c) 1100 d) 2. dev. Fix at 52:25N 003:00E. 1030 a) b) c) 1033 1043 d) 1059 e) 1104 f) g) h) What is the true bearing to plot from OT? What is the position at 1104? Give the mean W/V experienced since 1033. temperature -48oC. RUSH VOR RSH RMI reading 048o. Mean W/V for the climb is 210/45. Alter heading for STRUMBLE VOR STU using the W/V found. What is the true bearing to plot from CL? POLE HILL NDB PH RMI reading 280o. Level at FL 240. CLACTON NDB CL RMI reading 214o. Give the DR position at 1030. Give the position at 1047. What is the true bearing to plot from PH? OTTRINGHAM NDB OT bears 279o relative by ADF. altitude 3000ft. temp. temperature -5oC. W/V 050/30. RAS 205kts. set heading for STRUMBLE VOR STU. Give the mean magnetic heading and ETA STRUMBLE. KNIGHTON NDB KNI bears 337o relative by ADF. What mean W/V has been experienced since 1000. OLNO (50:35N 005:43E). +7oC. climbing to FL 240 at constant RAS 184kts. Give the revised ETA PWK. Mach 0. STORNOWAY NDB SN bears 299o relative by ADF. 2116 2136 b) c) 2145 2150 2155 d) e) 2200 f) 5. TORY ISLAND NDB TY bears 013o relative by ADF. temperature -52oC. ETA Scottish FIR/10oW at 2156. temperature ISA. What radial must be maintained to reach PWK? FLIGHT TRAINING COLLEGE Page 158 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 .81. Give the mean magnetic heading to PWK. 59:00N 021:00W. Mach 0. 0930 At position 56:00N 010:00W FL 350. W/V 260/50. what is the aeroplane’s true heading at longitude 12oW? What is the true heading at longitude 15oW? What is the true heading at longitude 21oW? INS WPT 3 (60:00N 020:00W) enroute to MACKNISH VOR MAC WPT 4. What mean W/V has been experienced since 2116? Alter heading for MACKNISH VOR using forecast W/V 270/100. What mean W/V has been experienced since 1012? FL 200.83. RAS 215kts. Using forecast W/V 280/60. CLACTON NDB CL bears 272o relative by ADF. RAS 220kts. a) b) c) 4. OTTRINGHAM NDB OT bears 231o relative by ADF.84. The autopilot is coupled to the INS en route to WPT 2. Overhead SPIJKERBOOR VOR SPY. give the mean magnetic heading to reach MACKNISH and the ETA. Give the position at 2155. Give the position at 1101. FL 410. Using the W/V found at 2116. temperature -20oC. the coupled INS ceases to function. 1012 a) 1036 1046 1101 b) c) 1107 d) e) f) Due to an electrical fault. The following observations are recorded: Heading 145o(M). alter heading for PRESTWICK VOR PWK. temperature -15oC. Give the aircraft’s most probable position (MPP) at 2136. TIREE NDB TI bears 339o relative by ADF. GS 561kts. W/V constant at 260/120. temperature -50oC. set heading for PRESTWICK VOR PWK. a) Give the W/V at 2116. Alter heading for MACKNISH VOR. Give the mean heading (M) and ETA MACKNISH VOR. Whilst maintaining the required track.3. Give the mean magnetic heading and ETA PWK. The heading noted is 144o(M). FL 200. Mach 0. OTTRINGHAM NDB OT bears 322o relative by ADF. temperature -57oC. EAGLE NDB EG relative bearing 325o. With the VOR PWK selected and the OBS set to 306 FROM. dev. RUSH VOR RSH radial 134°. W/V 320/80. Mach 0. What mean W/V has been experienced since 1235? Alter heading for WPT 9 (53:30N 004:00W). 1220 Overhead PRESTWICK VOR PWK. +5oC. forecast W/V 270/120. Mach 0. What is the mean magnetic heading and revised ETA? Assuming that the VORTAC BEN at BENBECULA is within range on ETA ROCKALL. Using the W/V found. RUSH VOR RSH radial 277°. Give the mean magnetic heading and ETA ROCKALL.80. Give the mean magnetic heading and ETA RUSH. forecast W/V 260/80. a) 1235 1235 b) 1241 c) d) 7. Mach 0. What mean W/V has been experienced since leaving PWK? Using a mean W/V of 310/80 alter heading for ROCKALL ISLAND. Give the position at 1250. temperature -53oC. Give the mean magnetic heading and ETA WPT 9. the CDI indication on a five dot indicator shows three dots fly right. TIREE VOR TIR RMI reading 036o.6. what will be the RMI reading and DME range of BEN? WPT 8 (53:00N 015:00W) set heading for RUSH VOR RSH. temp. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 159 Version 4 . 1235 a) 1240 1245 1250 b) c) 1256 d) 1306 1309 1311 e) f) g. BELFAST VOR BEL radial 202°. Set heading for ROCKALL ISLAND (57:33N 013:48W). give the new mean magnetic heading and revised ETA for WPT 9. SHANNON VOR SNN RMI reading 122o. EAGLE NDB EG RMI reading 078o. Give the position at 1311. FL 370.80. Give the mean W/V experienced since 1250. FL 310. FL 370.80. forecast W/V 030/60. temp. ISA . 0935 a) b) 0950 c) 1008 1014 1017 d) e) 1023 f) 1051 1051 g) h) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 160 Version 4 . Give the mean magnetic heading. forecast W/V 310/70. What mean W/V has been experienced since 0950? Alter heading for ABERDEEN VOR AND using forecast W/V 270/100. Give the position at 1051. DOVER VOR DVR RMI reads 172o. dev. What mean W/V has been experienced since 1017? 9. temperature -51oC. ABBS VOR SAB radial 074°. ADN VOR radial 168°. FL 50. What is the mean magnetic heading and revised ETA BENBECULA? Overhead ABBEVILLE VOR ABB. Give the position at 1017. Give the mean magnetic heading. Mach 0. What is the DR position for TOC? Overhead DOVER VOR DVR. level at FL 310. LICHFIELD VOR LIC RMI reads 257o. -48oC. Give the position at 0814. ABERDEEN VOR AND radial 293°. temp. TIREE NDB TI bears 337o relative by ADF. Mean W/V 290/30. class A.6oC. mean rate of climb 2000ft/min.80. alter heading for position A. FL 310. 0720 a) 0804 0808 0814 b) c) 0820 d) Overhead SOLA at FL 310.74. set heading for position A (53:40N 000E/W) climbing to FL310 at constant RAS 186kts. RAS 220kts. St. Give the mean magnetic heading and ETA BENBECULA. RAS 224kts. ISA .4oC. Give the initial magnetic heading. dev. temp. STORNOWAY ground DF station gives a QDM of 354o. What mean W/V has been experienced since 0720? Alter heading for BENBECULA VOR BEN using the W/V found at 0814. Mach 0.8. mean temp -21°C. set heading for BENBECULA. OTTRINGHAM VOR OTR RMI reads 306o. FL 310. BERRYHEAD VOR BHD RMI reading 180o. temperature 0oC. BERRYHEAD VOR BHD RMI reading 205o. RAS 160kts. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 161 Version 4 . FL 100. Give the position at 1200. BERRYHEAD VOR BHD RMI reading 244o.10. g) Give the radial to maintain to CMB. 1119 a) b) c) d) 1132 1142 1200 e) f) Overhead STRUMBLE VOR STU. What mean W/V has been experienced since STRUMBLE? 1208 Alter heading for COMBRAI using forecast W/V 300/40. Give the ETA abeam MIDHURST VOR MID. W/V 360/35. Give the ETA COMBRAI. Give the mean magnetic heading to COMBRAI. Give the initial magnetic heading. Set heading for COMBRAI VOR CMB. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 162 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 163 Version 4 . The bisector cuts the track at the midpoint distance (Z). and Long. Measure the Lat. THE CP CONSTRUCTION Construct a perpendicular bisector between departure point X and destination Y. Z will be the PET in zero wind conditions. blowing towards the track. Remember that wind always blows from the AIR POSITION to the GROUND POSITION. It is important to measure all distances from the same point on the chart using the appropriate midlatitude scale. From W scribe an arc of the TAS to cut the track at A. Plot the appropriate W/V at Y. From B parallel the wind vector WY to cut the track at the PET. Measure all distances with reference to the meridian closest to this point. Parallel heading WA through Y to cut the bisector at B.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. of this position. blowing towards track.Maximum endurance (dry tanks) . Plot the W/V at A. 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. use the one engine inoperative TAS for the construction described above.THE CRITICAL POINT (one engine) CONSTRUCTION To plot the position of the CP. the normal multi-engine G/S is used. For practical reasons. the PNR is referred to as the Radius of Action. To calculate the ETA for the CP. From W draw an arc of TAS to cut the track at B.Safe endurance (reserves) THE PNR/RA CONSTRUCTION (a) b) c) d) e) A dry tank position (maximum range) or safe range position A is calculated by: ENDURANCE x G/S OUT Construct a perpendicular between X and A. PNR RA . Parallel the heading WB through A to cut the bisector at C. fuel is usually kept in reserve and under such conditions. given the forecast W/V. FLIGHT TRAINING COLLEGE Page 164 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 . of the position. Parallel the heading WA through Y to cut the bisector at B. of this position. A second PET between X and Y can be constructed. Plot the W/V at Y. Measure the Lat.f) g) From C parallel the wind vector WA to cut the track at the PNR. PET/CP WITH ALTERNATE DESTINATION An aircraft will usually fly from X to Y with alternate Z. THE PET WITH ALTERNATE CONSTRUCTION a) b) c) d) Construct a perpendicular bisector between Y and Z. A PET may therefore be required which includes the time to fly to the alternate destination in the construction. 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. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 165 Version 4 . blowing towards the track. When the aircraft is between the two PET’s. From W draw an arc of TAS to cut XY at A. it will be quicker to divert to Z than to continue to Y or return to X. Note: If the construction shows that by paralleling WA through Y. Measure the Lat. e) f) g) From B parallel the wind vector to cut the track at the PET. and Long. and Long. of the position. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 166 Version 4 . In this case a PNA will have to be constructed. and Long. blowing towards the track. THE PNA CONSTRUCTION a) b) c) d) e) f) g) Calculate a dry tank or safe endurance and extend the track XY to position A by: ENDURANCE x G/S OUT Construct a perpendicular bisector between A and Z. Parallel the heading WB through A to cut the bisector at C. 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. Measure the Lat.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. From C parallel the wind vector to cut the track at the PNA. Plot the W/V at A. From W draw an arc of TAS to cut the track at B. DE is then B’s heading for the intercept. blowing AWAY from B. a) b) c) d) e) f) g) EXAMPLE 1 Update positions of both aircraft to a common time A and B. AB is the LINE OF CONSTANT BEARING. line BD. bears 170o(T) and 80nm from A. W/V 340/25 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 167 Version 4 . Plot AC. From O draw an arc at B’s TAS to cut A1B1 at E. Parallel AB through C. 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. From B draw in the W/V. Join A to B. An alternative construction is therefore necessary to accomplish the interception.ALTERNATIVE METHOD (Only Track and G/S of one aircraft known) In most cases. heading 090o(T). 1700z Aircraft A: 1715z Aircraft B: TAS 240kts. Line A1B1. G/S and position for aircraft B to intercept aircraft A. Join B to E to cut AC at F and hence reveal the track. TAS 250kts. A’s TRACK and G/S for one hour. Scribe an arc 25nm from A along the line of constant bearing.DETERMINE: a) b) The track that B must make good to intercept A. TAS 210kts. The time required for B to intercept A. * NB! Remember that the wind vector is related in magnitude to the time flown. The point along BG at which the aircraft will be 25nm apart will be at J. (hint: use the line of constant bearing) SOLUTION: * * * * Update position of A to 1715 Complete construction as previously described. Parallel aircraft A’s heading from point I to J. Using B’s TAS. The closing speed of B on A. Construct track from B to H and measure at mid meridian. calculate the time to reach J and add to 1715. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 168 Version 4 . EXAMPLE 2 0900z Aircraft A: 0900z Aircraft B: G/S 190kts. bears 110nm and 80o relative to A. The time that the aircraft will be 25nm apart. W/V 290/30 DETERMINE: a) b) c) d) The heading B must steer to intercept A. drift 5o right. The distance that the aircraft will be apart at 0930. track 080o(T). Divide the original distance that the aircraft were apart (110nm) by the time taken to make the intercept to determine the closing speed. corresponding to the time of 0930. Determine A or B’s position along their respective tracks. (QTE = 80o . * ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 169 Version 4 . Parallel the line of constant bearing through the 0930 position and measure the separation distance between aircraft.SOLUTION: * Calculate and draw in B’s true position relative to A at 0900.5o + 80o = 155o) * * Continue with construction as previously described to find B’s heading and time for the intercept. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 170 Version 4 .INTERCEPTION An aircraft. From B draw an arc of B’s TAS to cut A1B1 at F.75 x the forecast velocity in length. If the interception took 45 minutes then the wind vector would be 0. THE INTERCEPT CONSTRUCTION (HDG and TAS of both aircraft known) a) b) c) d) e) f) g) h) Update the aircraft positions to a common time A and B. Join A and B. A’s TAS and heading for one hour. Parallel AB through point C. in the course of its operation. BF is the INTERCEPT HEADING (Note: not track) G is the AIR POSITION (+) of interception. Plot AC.g. NOTE: The wind velocity vector must be relative in its magnitude to the time to intercept. e. 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. line A1B1. AB is known as the LINE OF CONSTANT BEARING. 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. may be required to carry out an interception mission on another aircraft. Determine the position along track where the time to PEV will be equal to the time to return to JAN SMUTS. Determine the latest time and position at which to depart from track in order to arrive overhead KEETMANSHOOP AT 1435. The co-ordinates of the PNA. Overhead HARARE at 1036. 2. 1. Endurance excluding reserves 01:30. 1200 UTC. The mean magnetic heading and ETA at the PNA. 3. temp dev. Fuel available excluding reserves is 1400lbs. A flight is made from KEETMANSHOOP KTV to BURGERSDORP BDV./hr. Forecast W/V between GROOTFONTEIN and UPINGTON is 135/30. heading 345o(T). Any diversion will be to ALEXANDER BAY routing via KEETMANSHOOP. Plan a flight from JAN SMUTS to PORT ELIZABETH. TAS 250kts and W/V 040/20. The mean magnetic heading and ETA from the PNA to BULAWAYO. 6. forecast W/V 070/55.ADVANCED PLOTTING EXERCISES 1.8oC. Forecast W/V between KEETMANSHOOP and ALEXANDER BAY is 150/40. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 171 Version 4 . A flight is made from FRANCISTOWN to GROOTFONTEIN with GHANZI as alternate. 1215 UTC. Set heading for PIETERSBURG. Fuel consumption is 340lbs. Determine the position along track that is equidistant in time from FRANCISTOWN and GHANZI. The TAS is 150kts and the forecast W/V is 180/30. Determine the radius of action in time and distance from KEETMANSHOOP. overhead ALEXANDER BAY. The TAS is 300kts. Determine the latest time and position at which the aircraft must depart from the present track to arrive overhead ALEXANDER BAY at 1432. overhead GROOTFONTEIN on a direct flight to UPINGTON. TAS 170kts. 5. ISA . Determine the following: a) b) c) d) The furthest point along track from which a diversion to BULAWAYO may be made leaving the reserve fuel intact. RAS 226kts. Pressure altitude 16000ft. W/V 200/30. The TAS is 350kts and the W/V is 210/40. 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) b) c) d) The time of interception. The position of interception. The true heading and G/S of aircraft B while intercepting A. 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) b) c) d) The time and position of interception. The true heading and G/S of aircraft B to intercept aircraft A. The closing speed. 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) b) c) d) e) The time interval for the interception. The mean magnetic heading for Z to steer to intercept X. The distance from PIETERSBURG to the intercept point. The closing speed of the two aircraft. The estimated time to the first possible visual contact. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 172 Version 4 ANNEX A SAMPLE QUESTIONS ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 173 Version 4 QUESTIONS 1. It is 1823 ST on 10 October at Norfolk Island. What is the Standard Time (ST) at Haiti? a) b) c) 2. 2053 on 09 October; 1523 on 09 October; 0153 on 10 October. (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) b) c) 3. 0935 on 12 September; 0335 on 12 September; 1235 on 12 September. (2) With reference to the above question, what is the LMT on the ship at the time of the coup? a) b) c) 4. 1718 on 12 September; 0242 on 12 September; 0152 on 12 September. (2) With reference to the above question, what is the Zone Time on the ship at the time of the coup? a) b) c) 5. 1735 on 12 September; 0135 on 12 September; 0235 on 12 September. (2) 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) b) c) 6. 174° / 951 NMS; 176° / 943 NMS; 172° / 963 NMS. (4) 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) b) c) 257 KTS; 265 KTS; 275 KTS. (2) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 174 Version 4 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) b) c) 2° right 5° left 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) b) c) W 106º 29’; W 158º 11’; 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) b) c) 249.5º 252.5º 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) b) c) 10.02 cm; 14.87 cm; 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) b) c) N25º 50’ W 025º 33’; N23º 49’ W 028º 33’ 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) b) c) 1018; 1008; 1028. (2) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 175 Version 4 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) b) c) 0607; 0907; 1207. (3) 14. What is the GMT for first light at Stockholm (N 59º 40’ E 017º 55’) on 10 Jan ? a) b) c) 0759 0647 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) b) c) 015º W; 022º W; 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) b) c) 117º T; 110º T; 097º T. (2) 17. With reference to the previous question, what is the final track at B? a) b) c) 050º T; 054º T; 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) b) c) 1:989 763; 1:982 609; 1:980 098. (3) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 176 Version 4 (4) 25. E 177º 47’. The position of A is N 52º 00’ E 005º 00’. N 50º 04’ W 003º 04’. An aircraft flying East along the 10º S parallel covers 13. The great Circle bearing of B from A is 278º T. The radius of the earth is 250 000 000 inches. Position B is approximately South West at A at a chart length of 58 cm. the scale at the pole is 1:5 000 000. 176º. On a westerly flight along the 60º N parallel. The position of A is N 42º 13’ W 158º 24’. S 34º 02’ E 037º 06’.S. Aircraft heading is 281º C. The aircraft obtains a relative bearing of 131º in the Northern hemisphere CA is 2º. 1688 KPH. the 030º W Meridian is crossed 2 hours after the Greenwich Meridian is crossed. Chart of the Southern Hemisphere. (2) 21. What is the position of B? a) b) c) N 48º 56’ W 003º 00’. Variation is 21º E. The change of longitude between A and B is 8º. 200 KTS.335 cm in 30 mins as measured on a Mercator Chart. 834 KPH. On a P. A Mercator Chart is drawn to a scale of 1:1 000 000 at 56º N. 253º.19. (5) 24. (2) 22. (3) 23. Deviation is 2º W. What is the ground speed of the aircraft? a) b) c) 156 KTS. An aircraft leaves A (S 15º 35’ E 017º 52’) on a rhumb line track of 137º T. What is the aircraft’s groundspeed in KPH? a) b) c) 450 KPH. What is the QTE to plot? a) b) c) 171º. N 52º 00’ W 003º 00’. The scale of the chart is 1:1 000 000 at 25º S. B is on the same parallel of latitude. What is the chart length in CMS between 84º S and 71º S? FLIGHT TRAINING COLLEGE Page 177 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 . What is the longitude of B? a) b) c) W 179º 13’. S 33º 19’ E 036º 55’. (3) 20. 176 KTS. E 175º 57’. What is the position of the aircraft after flying for 1500 NMS? a) b) c) S 33º 52’ E 036º 39’. 80 OAT -40º C W/C -60 KTS ETA for overhead VOR ABC is 0235 30.7839. Aircraft A Aircraft B : ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 178 Version 4 . (4) The following positions are plotted on a P. (1) : Mach 0. 290º T. (2) 28. What is the Zone number for longitude E 088º? a) b) c) +6.a) b) c) 26.76 OAT -40º C W/C -80 KTS ETA for overhead VOR ABC is 0250 Mach 0. (2) 29. 005º E. 29. What is the Great Circle track from A to B? a) b) c) 065º T. Chart of the Northern Hemisphere. 045º T. Position A N 75º 00’ W 100º 00’ position B N 75º 00’ W 030º 00’. what is the longitude of the most Northerly point along the track? a) b) c) 065º W. -6. With reference to the previous question. 33. With reference to the previous question. The scale of the chart is 1:3 450 000 at 82º N. 055º T. 28. 055º T.1934.8406. (2) 27. 135º W.S. -5. what is the Great Circle track from B to A? a) b) c) 305º T. At what time should aircraft A reduce speed to Mach 0. 0427. How far from VOR ABC will the aircraft have to reduce speed by 60 KTS to arrive at 0642 Z? a) b) c) 34. 205 nms. 300º / 300 KTS. 44 nms . (2) The following details apply to an aircraft: Track 075º T. G/S 200 KTS. 288 nms. G/S 180 KTS. AT 10h10. What is the aircraft distance from the station at 10h10? a) b) c) 18. 56 nms . 0208. At 10h00. Aircraft A (G/S 390 KTS) is 260 nms from the VOR at 0350 Z. (3) An aircraft (G/S 280 KTS) estimates overhead VOR ABC at 0630 Z. 260º / 320 KTS. What is the distance X to Y? a) b) c) 32. 0212. (3) Aircraft P (G/S 240 KTS) arrives at X at 0900 en-route to Y. 21.70 in order to arrive at VOR ABC 20 minutes after aircraft B? a) b) c) 31. 15. The local variation is 10º E. 0203. (3) Solve by any method: Aircraft A track 030º T. (3) Two aircraft at the same flight level are approaching a VOR. What is the relative velocity of B with respect to A? a) b) c) 33.3 nms . 245º / 315 KTS. (4) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 179 Version 4 . 300 nms. 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) b) c) 35. Aircraft Q (G/S 320 KTS) arrives at X at 0918 and passes aircraft P at Y. Aircraft B track 300º T G/S 250 KTS. 0422. the VOR needle on the RMI indicates a QDM of 290º. the VOR needle on the RMI indicates a QDM of 020º. due to traffic. ATC requests the aircraft to cross VOR ABC at 0642 Z. 120 nms.5 nms . 0417.2 nms . 100 nms is equal to 30 cms on a chart. (3) 41.70 at? a) b) c) 1052. 1056.297 CMS. an aircraft at FL 250. N 25º 07’. 260º.97 CMS. 306º T. the aircraft must reduce speed to M 0. What is the initial G. position A (S 75º 00’ E 136º 00’) and position B (S 75º 00’ W 152º 00’).S. 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) b) c) N 23º 31’. 1100 (3) 37. N 26º 54’. 1. At 10h26. 25 mins. chart. At position A (S 75º 00’ W 040º 00’). M 0. headwind 30 KTS is estimating position Charlie at 11h16. are joined by a straight line. The duration of evening civil twilight at Boston USA (N 42º 22’ W 071º 00’) on JAN 27 is? a) b) c) 20 mins. (3) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 180 Version 4 . On a south P. What is the true heading? a) b) c) 180º. 234º T. temperature –35º C. the grid heading is 220º G. To reach position Charlie at 11h20. (2) 40. track from A-B? a) b) c) 126º T.C. 24.S. chart. 220º.81.36. (3) 38. 30 mins. What is the chart length in cms of a straight line representing 80 kms? a) b) c) 12.0 CMS. A grid is aligned with the Greenwich Meridian on a south P. (3) 39. at destination in longitude? a) b) c) E083:15. (3) 44. (3) 46. The aircraft is at DR position S 78º 00’ E 170º 00’. W034:00. The following positions are plotted on a south P. 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’. (2) 43. 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 (S 85º 00’ E 140º 00’) and B (S 75º 00’ W 120º 00’). 1215 LMT JAN 15th. 095º T. both at latitude N75: 00 are plotted on a Polar Stereographic Chart. E071:39. 1222 LMT JAN 14th.S.42. (4) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 181 Version 4 . 165º T. (2) 45. 039º T. E 039:00. a) b) c) 1230 LMT JAN 15th.C track from A to B is 125º T. 1:1 016 966. 1:982 609. The initial G. and the longitude of B is: a) b) c) E 141: 00. The scale of Polar Stereographic Chart is 1:992 047 at S80:00 and at S75:00 the scale is? a) b) c) 1:999 640. the initial great circle track from A at N75: 00 W 107:00 to B is 326º (T). Position A and B. The bearing to plot on a south P. 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. (4) 47.S. chart. What is the track at the anti-meridian of Greenwich? a) b) c) 085º T. 049º T. chart is? a) b) c) 029º T. In order to arrive at position S20:00 W172:00. E101:00. 15 minutes before the end of the evening civil twilight on JAN 14th. (3) 51. (3) 49. (4) 50. and the aeroplanes ground speed is therefore: a) b) c) 434 KTS.65 CM.25 CM. A fix is obtained 105 NM from A which places the aircraft 7 NM to the left of the intended track. variation 21W. deviation 3E. A Mercator Chart is constructed to a scale of 1:5 000 000 at the Equator. (3) 52. (2) 53.9 CM on a Mercator chart. 69. and the width of the graticule between the Greenwich meridian and E030:00 is: a) b) c) 26. On a Conical Orthomorphic Chart.88 cm. 450 KTS. (2) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 182 Version 4 . 1:5 559 500. one standard parallel is N40:00 and measures 63. 482 KTS. An aircraft over flies position A on a heading of 067º ( C). During a flight along the N48:00 parallel. N55:00. 260º (M). The heading (M) to be steered to return to A from the fix is: a) b) c) 239º (M). intending to maintain a track of 058º (T).49 CM. The initial great circle track from A (S41:27 W171:24). The length of the other standard parallel is 41. 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) b) c) 1:5 815 000. 257º (M). the measured distance between fixes X and Y 25 minutes apart is 17. 261º (T). N60:00. 279º (T). 1:5 432 000. to B (S41:27 E161:24) is: a) b) c) 288º (T). The scale of the chart is 1:3 000 000 at N15:00.48. 66.7 centimetres at latitude: a) b) c) N20:00. 7 NM. The Rhumb Line track from B to A is: a) b) c) 075. W 179:24. a straight line is drawn from A to B. south of B. variation at the aircraft is 17E and 19E at the VDF. have a convergency of 7º on a Lambert’s chart. The longitude of B is: a) b) c) W 179:17. GS 280KTS. the distance is 450 NM. The position of A is N40:00 E008:00. Aeroplane A. passes point X 7 minutes ahead of aircraft B. (3) 56. W010:40. Two aeroplanes at the same flight level estimate the same position at the same time. Aircraft A.0º. The distance from X to Y is: a) b) c) 429. 268º. (2) 57. (2) 58. and will pass: a) b) c) north of B. (2) 59. 073. GS 326 KTS. An aircraft leaves A on a constant heading of 070º (T). 400. GS 200 KTS is flying due North. GS 300 KTS is flying due East. On a Rhumb Line track of 090º (T) to B. 270º. E 179:43. 411. E026:40. The great circle track from A to B is 276º. An aircraft flies from A (N48:47 E169:20). Aeroplane A’s velocity relative to B is: FLIGHT TRAINING COLLEGE Page 183 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 .54. The bearing to plot on a Lambert’s chart is: a) b) c) 272º.9 NM. Aeroplane B. (2) 55. 080. On a Lambert’s Chart in the Northern Hemisphere. B is on the same parallel. The great circle track from A (W168:00) to B (E172:00) measures 250º (T) at A. The longitude of B is: a) b) c) W001:20. Two meridians. (3) 60. overhead of B.7 NM. in the Southern Hemisphere. An aircraft obtains a QDM of 073º from a VDF station in position N30:00 W020:00.0º. the track measured at A is 070º (T). zero wind conditions. Aircraft B passes the point Y 6 minutes ahead of aircraft A. Convergency 4º. 174 W and 172 E.5º. (3) The pilot of an aeroplane TAS 360 KTS. heading 309º (T). 195/27. heading 060º C (M). 034º at 400 KTS. 335/35. Simultaneously the aeroplane passes over NDB B. The wind velocity is: a) b) c) 64. 303º at 360 KTS. which gives a steady relative bearing of 178. TAS 260 KTS. 214/23. 126 KTS. FLIGHT TRAINING COLLEGE Page 184 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 . 101 KTS. W/V 270/15. 277/29. W/V 240/27 W/V 180/42. 230/28. (4) An aeroplane passes overhead NDB A at 1023Z. 056º at 360 KTS. 237/29.a) b) c) 61. 010/44 KTS. 055/47 KTS. The LMT of arrival at the destination is the same as the LMT of departure. 98 KTS. 110NM from A. (3) A flight of 2 152 KM is flown along the 69th parallel on a track of 270º (T). The mean W/V for the climb is: a) b) c) 65. The average GS for the flight is: a) b) c) 63. reaching Y at sunset on January 19th. (3) An aeroplane leaves position X (S30:00 W 168:45) at sunrise on January 18th for position Y (S30:00 E 165:18). 3 hours 36 minutes. At 1046Z the radio compass is tuned to the NDB A. after a flight of: a) b) 1 hour 12 minutes. makes the following observations using Airborne Search Radar (ASR): Time 1036 1042 Relative bearing 015 055 Range NM 47 20 Using the navigational computer the local W/V is: a) b) c) 62. 248/48 KTS. (4) The following wind velocities are forecast for the climb to cruising altitude. Variation is 17E. (5) 67. 6 hours 40 minutes. 0645Z. N36:52. At its most northerly point this track will be: a) b) c) at a tangent to N80:00 latitude. is 280 NM from WPT 4 at 0620Z. North of N80:00. The second at groundspeed 300 KTS. south of N80:00. and re-crosses latitude N74:00 at: a) b) c) W079:20. Two aeroplanes on the same track at the same flight level are approaching Waypoint 4. (4) 70. the Rhumb line bearing of B from A is 270º (T). The track of an aeroplane represented by a straight line on a Polar Stereographic chart measures 328º (T) at position N74:00 E015:20. The first at a groundspeed of 240 KTS. (5) 69. (2) 68. 0635Z. (4) 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. S53:08. W008:19. W116:20.c) 66. W100:40. (2) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 185 Version 4 . the second aeroplane must reduce groundspeed to 240 KTS at: a) b) c) 0630Z. A track is drawn from N70:00 E100:00 to N70:00 W140:00 on a Polar Stereographic chart. a) b) c) W009:00. is 315 NM from WPT4 at 0626Z. the change of longitude between A and B is 10 degrees and the mean latitude between A and B is approximately: a) b) c) S36:52. The great circle bearing of A from B in 094º (T). In order to ensure a 50 NM separation at WPT 4. W000:21. (4) 74. 0048. 30 NM. groundspeed 360 KTS observes a QDM of 240 to an NDB. A Lambert’s chart (Northern Hemisphere n factor 0. (3) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 186 Version 4 . The Standard Time of moonrise at Cape Town S33:58 E018:36 on January 2nd was: a) b) c) 0002. 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. At 1540Z the QDM to the NDB was 330. S49:12. The grid heading of an aeroplane crossing E30:00 (variation 10 degrees west) is 250 (G) and the heading magnetic is: a) b) c) 268.8) has an overprinted grid based on W40:00. 0116.71. (3) 73. At 1530Z an aeroplane. track 190º (M). (3) 72. The closest distance of the aeroplane from the NDB was: a) b) c) 20 NM. S55:40.095º and the parallel of origin of this projection is: a) b) c) S42:18. 300. 316. 40 NM. ANNEX B ANSWERS TO QUESTIONS ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 187 Version 4 . Chapter 1 1. i) ii) 2. 086° 092° i) ii) Northern hemisphere 325° ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 188 Version 4 . 42° x 2 = = = = 0.3. 84 ° 2 57° 42' N/S 4. CONV° 16° = = = = dLONG° x SIN LAT dLONG° x SIN 40° dLONG° dLONG° 16° SIN 40° 24° 53' 24° 53' east of 170° E = 165° 07' W ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 189 Version 4 . CONV° = = = dLONG° x SIN LAT 1° x SIN 25° 0.42° dLONG° x SIN LAT 1° x SIN LAT SIN LAT LAT CONV° 0. Track 360° Track 270° Track 180° Track 360° Track 270° : : : : : 480 Kts x 3 HRS = 1440 nm's 480 Kts x 2 H 30 = 1200 nm's 480 Kts x 4 HRS = 1920 nm's 1440 nm's = 1440' = 24° DEP (nm's) 1200 nm's = = = = = dLONG' x COS LAT dLONG' x COS 20° N + 24° N dLONG' dLONG' dLONG° 1200 COS 44 ° 1668' 27° 48' ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 190 Version 4 . DEP (nm's) = = = = dLONG' x COS LAT 15° x 60' x COS 30° 900' x COS 30° 779.5. DEP (nm's) 480 Kts x 19 HRS 9120 nm's 9120 21600 65° 7.4 nm's = = = = = dLONG' x COS LAT 360° x 60' x COS LAT 21600' x COS LAT COS LAT LAT 6. 65° = 25° 25° = 1500' 1500' = 1500 nm's DISTANCE A TO B = = 90° . There is no departure involved. The shortest distance will therefore be along the meridian. over the north pole.78° = 12° 12° = 720' 720' = 720 nm's 1500 nm's + 720 nm's 2220 nm's ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 191 Version 4 . A and B are on antimeridians.Track 180° 8. : 1920 nm's = 1920' ÷ 60 = 32° The sum of two longitudes 13° 30' and 166° 30' = 180°. and along the anti-meridian. 90° . Therefore. 1 SCALE OF EQUATOR = 1 SCALE AT 18° N × COS 18° 1 = 1 COS 18 × 1000000 1 COS 18° 1000000 1 1051462 = = Take the scale from the equator to 36° S. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 192 Version 4 .CHAPTER 2 PART 1 1. 1 SCALE AT 36° = 1 SEC 36 × SCALE OF EQUATOR 1 1 1 × 1051462 COS 36 1 850651 = = 2. The scale is given at 50° S. This question may be calculated at any latitude. Take the scale from 18° N to the equator. provided that the scale is used at the latitude. work the problem there. SC = CL ED 21 cm ED 21 cm x 400000 840000 cm 45.g. = = 18" ED 18" x 445503 = 8019059" = 109.33 nm's dLONG' dLONG' dLONG° 3.5' 1° 10. Equator 1:1000000 60N 1:500000 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 193 Version 4 .33 nm's dLONG' x COS LAT dLONG' x COS 50° 1 400000 ED ED ED DEP (nm's) 45.33 nm' s COS 50' 70.5 ' Calculate the scale at 27° N 1 SCALE AT 27 ° = 1 COS 60° 1 × × COS 27 ° 1 SCALE AT 60° = 1 COS 60° 1 × × 250000 1 COS 27 ° 1 445503 CL ED = SC = 1 445503 ED ED ED 4. the scale at 60° N is larger.9 nm's The scale at the equator is smaller. = = = = = = = = = 45. 1 e. g a l0cm long line At the equator ED ED At 60 N ED ED = = = = 1000000 x 10 cm 53. 5.96 nm’s 500000 x 10cm 26. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 194 Version 4 . the larger the earth distance e.The smaller the scale.98 nm’s The line at the equator represents the larger earth distance. 31 MP 16° 27'S = 994.6.53 SMP 16° 30' W + 004° 18' E = 20° 48' = 1248' tan ∅ = = 1467 nm x 1467 nm x = COS 40. 7.3° = = 180° . 53 MP 40.3° = x = 1923.40.3° 139.7° Thumb line track ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 195 Version 4 . 08° N = 478.3° COS 40.5 nm's Rhumb line distance 08° N + 16° 27' S = 24° 27' = 1467 nm 1248' 1472.22 MP 1472. 27° 27' = 1703.8.38 MP tan 25° = x x x x = = = = x 15269.28 tan 25° x 15269.37 MP 15269.09 MP COS 25° = x x x x x = = = = = x 4087 nm 4087 x COS 25° 3704 nm 3704' 61° 44' + 27° 27' 89° 11' S 89° 11' = 16972.38 7120.18' dLONG 118° 40' dLONG + 14° 28' E 104° 12' W ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 196 Version 4 . 10° 18' S = 617.17 617.9. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 197 Version 4 .17 x tan 40° 517.17 MP tan 40° x x x = = = = x 617.9' 8° 38' dLONG 8° 38' + 002° 03' = 010° 41' E 10. Determine the chart length of 1 min long or 1 MP SC = CL ED CL 1' × COS 60° × 185300 92650 cm 1000000 0.dLONG = 12° 12' dLONG = 732' tan 43° = 732' x 732' tan 43° 785' or 785 MP's 21° 37' N = 1320.00 535.39 MP 08° 57' N x x = = 11.29 MP = NEW LAT NEW LAT = = 785.09265 cm (per MP/MIN LONG) 1 1000000 CL CL = = = ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 198 Version 4 . = 1 1 710 101 SC 1 1 710 101 ED Find the G/S = = = CL ED 22.99 MP 1 COS 20 1 × × 2 500 000 1 COS 50 Find the earth distance at 50°N.5 CM ED 207.54 MP 1344.09 MP = 43° 30' N 12. = 4507.09 MP 2888.62² + 1020² 1452.62 DMP 165° E to 178° W 17° 1020' x² x² x = = = a² + b² 1033.09265 cm's per MP = 150 cm's can accommodate 1618.99 MPS at 0.6 NM’s G /S = = D T 733 KTS ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 199 Version 4 .92 MP 1033.09265 cm/MP 60° N 2888.08 MP 1618. Find the scale at 50°N.16 MP 134.16 MP at 0.5 cm 1452.37° N 22° N dLONG = = = = = = 2378. 13. the scale expands with latitude and 1:15 000 000 is a larger scale than 1:2 000 000.64952 49° 30’ N This is a Merdional Parts scale question.20 MP’s 2 378.54 MP’s 151. = = = = 2 309 401 1 500 000 1 500 000 2 309 401 0. 1 1 = × 7 025 649 COS 30† ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 200 Version 4 .02637 cm 185 300 cm 1 7 025 649 1 6 084 391 Calculate the scale at 30°N. SC = = = CL ED 0.66 MP' s 0. Calculate the scale at the equator.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. On a Mercator chart. 39°N 37°N = = 2 530. therefore: 1 COS 30 = × 2 000 000 1 1 1 × 2 309 401 COS x = 1 2 309 401 (EQUATOR SCALE) 1 1 500 000 2 309 401 1 1 1 = × COS x 1 500 000 1 COS x COS x COS x x 14.66 DMP’s = 4 cm 4 cm = 151. B. If one standard parallel is at 40 ° N.5° 307.5° ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 201 Version 4 . i) The great circle track at A = = = = 315° .766 50° N = = SIN LAT P of O LAT P of O The P of O is halfway between the standard parallels.Part 2 1. the other must be at 60° N.(15° x SIN 30°) 315° .7. The chart convergency factor = SIN LAT P of O 0.convergency (dLONG° x SIN LAT P of O) 315° . (15° x SIN 30°) 315° .5° iii) The rhumb line track from B to A = = = = = 142.(½ dLONG° x SIN LAT P of O) 142.5° 135° 3.5° .convergency (dLONG° x SIN 30°) 315° .ii) The great circle track at B = = = = 315° .7.5° .5° .CA 142.(½ 30° x SIN 30°) 142.7.5° 322. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 202 Version 4 .5° . 30°N 30°N DEP (CM) = dLONG’ × COS 30° DEP (CM) = dLONG’ × COS 30° dLONG’ = DEP(CM) COS 30† = = = = = dLONG’ = DEP(CM) COS X ο But.4. they are the same two meridians at 30°N and at X°N. the two dLONG’s are the same.658 48° 50’N ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 203 Version 4 . 5. Therefore. therefore: DEP(CM) COS 30† 50 CM COS 30† COS X† COS X° X° 38 CM COS X† 38 CM COS X† 38 CM × COS 30† 50 CM 0. the dLong at 30°N and at X°N is the same. Whatever the two meridians on this sketch are. 52° = 32°.6. Remember that if the aircraft leaves X on a constant heading of 060°T. 011° `13’ W 32° conv 52° 27°N 84° 47°N 61°W 17°W 7. CONV° 32° dLONG° dLONG° = = = = = dLONG° × SIN LAT P of O dLONG° × SIN 40° 32† SIN 40† 49° 47’ (East of 061°W) 11° 13’ W ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 204 Version 4 . The aircraft will pass North of position Y. The great circle is a straight line and the only reason why a G. it will be flying a rhumb line and cut each meridian at 060°T. 7. What we are looking for is what dLONG would give us a convergency of 84° . track of 52°T should become anything else (in this case 84°T) is because of the convergency of the Meridians.C. 39° 051° ii) Great circle track B to A 2. i) Great circle track A to B = = = = = = 270° (RL) + CA 270° + 39° 309° 090° (RL) .CA 090° .Part 3 1. Great circle track A to D = = = 270° (RL) + CA 270° + 74° 344° ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 205 Version 4 . CONV 120° . The great circle track at the anti-meridian of Greenwich is: 120° . i) ii) The longitude of position B is 132° W.3. i) The great circle track A to B = = = 270° + CA 270° + 33 303° If the drift is 5° R.5° = 298° ii) To determine the highest latitude which this line will attain.drift = 303° . FLIGHT TRAINING COLLEGE Page 206 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 . the great circle heading A-B = 303° .12° 108° 4. take the triangle out of the circle. SIN 57° x x = = = x 18° 18° x SIN 57° 15° 06' The highest latitude is 90° . we would see the effect of the parallel spacing increasing as we move away from the pole. the highest latitude which this track reaches will in fact be south of 74° 54' N.15° 06' = 74° 54' N NOTE: In truth. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 207 Version 4 .The highest latitude which this track will reach will be indicated by the shortest line between the apex and the base of the triangle. because once this triangle is put back onto the chart. or the perpendicular bisector. Variation + = 215° + CONV 215° + 47° 262° QUJ 15° W 277° QDM 6. i) The great circle track from A .55° 215° ii) Determine the QUJ at the Prime meridian.LAT 1 1 2 (90 . the Mercator and Polar stereographic share the same scale. The great circle track (QUJ) = = = Zero deviation. 1 1 × = 1 000 000 COS 70† 1 342 020 (This is the scale on the Polar Stereographic at 70°N.B = = = 270° . The term ‘ROLLING FIT’ means that at 70°N. Calculate the Mercator scale at 70°N. COS 2 1 × 342 020 COS 2 1 × 342 020 1 2 CO .) Calculate the Polar Stereographic scale at 90°N.5.CA 270° .70) 1 1 COS2 10 = × 342 020 1 1 352 645 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 208 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 209 Version 4 . Now superimpose this drawing onto the polar stereographic chart.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. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 210 Version 4 . The aircraft heading is 100° T. 5.Longitude = 50° W Convergence = convergency = dLONG (diagram solves east or west) 4. Aircraft heading 100° G. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 211 Version 4 . 5° 10° Convergence = convergency = = = dLONG x n 40° x 0. therefore the magnetic heading is 195°.6. 7. = dLONG° x CCF 20° x 0. Convergency = convergency = = This grid heading is 070° G.5 20° The true heading is 180°. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 212 Version 4 . = = = = dLONG x SIN LAT P of O dLONG x 0. 5 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 213 Version 4 . Convergence = convergency 40° dLONG dLONG The datum meridian is at 130° E.8.5 80° 40° 0. 5 = 30° The datum meridian is 0° E/W. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 214 Version 4 .5 dLONG dLONG = 15° 0.9. Convergence = convergency = dLONG x CCF 15° = dLONG x 0. 960 nm Y 290 Kts .350 Kts 1205 B . 290 nm X A .B 1205 20 minutes x 290 Kts = 97 nm's i) T = = = = ii) 1 HR 36 + 1205 1341 RD RS 960 + 97 370 + 290 Aircraft A flies for 1 HR 36 from position X before passing B.450 Kts 1230 146 A .370 Kts 1205 Move B back to 1205.350 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 215 Version 4 .350 Kts 1230 144 Y 25 minutes x 350 Kts = 146 nm's i) T = = = = 10 minutes + 1230 1240 RD RS 146 .30 450 .CHAPTER 3 1.450 Kts 1230 Y Move aircraft A on to 1230. X B .B 1225 97 nm 290 Kts . X A . 1 HR 36 x 370 Kts = 592 nm's from X 2. 388 Kts = 62 Kts (reduction in speed) ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 216 Version 4 . T T = = 144 nm 350 Kts 25 minutes i) T = RD RS RD T 146 .130 25 mins 38 Kts RS = = = 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 .ii) Calculate the time for aircraft A to reach Y. 2 mins x 350 Kts = 12 nm T = RD RS = = = 25 .450 Kts 1225 A .20 450 .350 3 minutes + 1225 1228 = = = ii) When aircraft A reaches position Y.450 Kts 1215 A .350 8 minutes + 1225 1233 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 217 Version 4 .12 450 .3. aircraft B must be 2 minutes behind at 350 Kts. 10 mins x 450 Kts = 75 nm. 25 nm B .350 Kts 1225 200 nm Y i) T = RD RS 25 .350 Kts 1225 200 nm Y Move aircraft B on to 1225. 100 nm B . 250 Kts Y B .470 Kts 1215 700 nm Y Move B back to 1200. 10 mins 50 nm B .250 1 HR 50 (1 HR 50 x 250 Kts) + 42 nm = 502 nm X .250 Kts 10 mins 42 nm A .300 Kts T = RD RS = = 50 + 42 300 .350 Kts 1200 B .25 × 420 × 370 50 777 nm from Y 6.300 Kts X A .470 Kts 1200 X A . 117 nm B . DIST FROM DEST = DELAY × OLD G / S × NEW G / S DIFF IN G / S = = 0.4. 15 minutes x 470 Kts = 117 nm ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 218 Version 4 .Y 5. 50) 470 1 HR 38 + 1200 1338 = = = Aircraft B must therefore delay its arrival by 1400 . aircraft B must be at position Z. 50 nm before Y.36 × 470 × 320 150 368 nm from Z (speed reduction point) = = ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 219 Version 4 .1338 = 22 minutes. DIST FROM DEST = DELAY × OLD G / S × NEW G / S DIFF IN G / S 0.i) When aircraft A reaches position Y. This will occur at time. T = D S 700 350 2 HRS + 1200 1400 (A at Y and B at Z) = = = Aircraft B's original estimate for position Z would have been: T = D S 117 + (700 . T = D S 117 + (700 . DIST FROM DEST = DELAY × OLD G / S × NEW G / S DIFF IN G / S = = 0.368) 470 51 minutes + 1200 1251 = = = ii) Aircraft A's arrival over point Y is at 1400.50 . Aircraft B must therefore arrive over point Y at 1410. B's original estimate for Y was at time: T = D S 117 + 700 470 1 HR 44 + 1200 1344 = = = Aircraft B must therefore delay its arrival over position Y by 1410 .433 × 470 × 320 150 434 nm from Y (speed reduction point) B's estimate for the speed reduction point is: T = D S = = = 0 HR 49 + 1200 1249 117 + (700 .1344 = 26 minutes.434) 470 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 220 Version 4 . 35° 105° SIN B × a b SIN 40° × 250 280 Angle C ii) = = Solve for Side C (relative speed) c SIN C C C T = = = = = = = 421 Kts b SIN B b × SIN C SIN B 36 minutes + 1305 1341 RD RS 250 421 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 221 Version 4 . i) SIN A a SIN A = = = = SIN B b 35° (relative bearing from A .7.B) 180° .40° . iii) T = RD RS 250 .20 421 33 minutes + 1305 1338 = = = ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 222 Version 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. DeP 1407 LMT 3rd 429 .FT DEP 1050 UTC 4th X .FT DEP 0202 UTC 3rd 1136 + A/T DEP 1338 LMT 3rd 3.x = 0838 x = 1050 .A/T DEP 0838 LMT 4th i) 1050 . Flight time 3 hours x 15° per our = 45° or 2700' DEP 1580 nm COS LAT LAT = = = = dLONG' x COS LAT 2700' x COS LAT 1580 2700 54° 11' N/S ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 223 Version 4 . ARR 2208 ST 2nd 1100 + SF ARR 0908 UTC 3rd 706 . ARR 1823 ST 4th 100 .A/T DEP 0938 UTC 3rd 912 + FT ARR 1850 UTC 3rd 058 .0838 x = 2 H 12 2 H 12 x 15° PER HOUR = 33° W 4.SF ARR 1723 UTC 4th 633 .Chapter 4 Part 1 1.A/T ARR 1752 LMT 3rd 2. This distance is being covered at the closing speed of both aircraft and sun (300 Kts + 450 Kts). End of evening civil twilight on JAN 11 10° S is 1850 LMT.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 1412 = = dLONG' x COS LAT dLONG' x COS 60 FLIGHT TRAINING COLLEGE Page 224 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 . T = RD RS 3532.5 nm's. 1850 ST 11th 415 + A/T 2305 UTC 11th 420 .Part 2 1. ARR ARR DEP DEP 2.A/T 1716 LMT 11th 1506 LMT 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 Arrival is at sunset 60° N JAN 2nd Departure is at How far does the sun travel in 7 HR 51? 7 HR 51 x 450 Kts = 3532.FT 1845 UTC 11th 129 . Sunset for 60°N on JAN 15 is: 15H31 LMT. the relative velocity is: 450 Kts . How long until sunset? How far does the sun travel in this time? At the combined speed of aircraft and sun. Departure time is: 0800Z + 6H34 (A/T) = 14H34 LMT The time between departure and arrival is: 15H31 . If the sun and the aircraft are both going west.300 Kts = 150 Kts T = = = RD RS 427.5 NMS 150 KTS 2H51 How far does the aircraft fly in this time? 300 Kts × 2H51 = 855 NM DEP (NM) 855 NM 1710 28° 30’ = = = = dlong° dlong’ × COS LAT dlong’ × COS 6O°N dlong’ West of 098° 30’E is 70°E ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 225 Version 4 .dLONG' dLONG' dLONG° LONG = = = = 1413 COS 60 2826 ' 47° 06' EAST OF 15° 62° 06' EAST If you struggle to conceptualise this problem. how long does it take to open this distance? How far does the aircraft travel in this time? 3. imagine the departure of the aircraft at 1200 LMT (sun overhead the aircraft).14H34 = 0H57 At 60°N the speed of the sun is 450 Kts. 450 Kts × 0H57 = 427.5 NMS. 229° T). Moonset at 50° S 0° E/W DIFF 38 MINS to F4 table Moonset at 50° S 120° W 3.3°) To return: 234° T . Moonrise at 62° N 0° E/W DIFF 47 MINS to F4 table Moonrise at 62° N 090° E ADD 24 HOURS + FULL DAILY DIFF (47 x 2) 0047 .23 0024 -600 1824 2534 1958 1303 + 25 1328 +800 2128 LMT 1st LMT 1st A/T UTC 3rd UTC 1st LMT 1st LMT 1st A/T UTC 1st 2.229° T). The wind is responsible for 2° right drift (5° . The aircraft drifted 5° right (234° T .Part 3 1. Moonrise at 52° S 0° E/W DIFF 6 MINS to F4 table Moonrise 52° S 060° W Subtract 24 hours + FULL DAILY DIFF (2 x6) 2334 LMT 1st + 2 2336 LMT 1st +400 A/T 0336 UTC 2nd -2412 0324 UTC 1st Chapter 5 1. Calculate the aircraft's drift and its track made good (TMG) 11 132 TMG × = 60 1 234° = 5° RIGHT The aircraft steered 3° right (232° T .180° = 054° T 054° T + 2° Right for wind = 056° T 056° T + 15° W variation = 071° M ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 226 Version 4 . 1° (aircraft track) TRACK 115.9° = 115. HDG DEV VAR HDG 143° C 3°140° M QUJ 15°125° T HDG RB 192° QTE 125° 67°+ 180°012° HDG RB QUJ QTE 125° 107°+ 232° 180°052° a² a² a² a = = = = b² + c² = 2bc COS A 83² + 76² .6 Kts HDG 125° T TAS 300 Kts W/V 239/61 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 227 Version 4 .76.2.6 Kts SIN B b SIN B B = SIN A a SIN A × b a 76.(2 x 83 x 76 x COS 40º) 6889 + 5776 .1° T G/S 328.78 nm in 10 mins = 328.78 nm 54.(9664) 54.9° = = 192° . 42 Kts S x T 231. Airship : DEP S = D T = = Aircraft : S = = D = = = 21600 280 77.4 COS LAT LAT dLONG' x COS LAT = 21600' x COS LAT = = 41° 25' N/S 16199 . then apply the wind for 7 minutes. First do an air plot. CIRC = = CIRC = d d D D = = = = 300 Kts x 2 mins 10 nm πd 3.4 nm DEP = 16199.14 Kts 77.1 nm = = = dLONG' x COS LAT 360° x 60' x COS 0 21600 nm CIRC π From point A : : 4. Rate one turn = 360° in 2 minutes.18 nm 7 mins x 25 Kts 292 nm Bearing 180° Distance 6. therefore the aircraft does 3½ turns.3. 4 21600 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 228 Version 4 .14 x 3 231.42 x 70 HRS 16199. 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 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 229 Version 4 . By the time Oculus reaches the equator. the earth has rotated in an anti-clockwise direction 7 H 22 at 15°/HR = 110° 30'. Oculus would cross the equator at 60° W and 120° E. C = = = 2πr 2 x π x 6800 km 42726 km ½C = = 42726 2 21363 km T = D S 21363 2900 7 H 22 = = If the earth were stationary.5. 120° . 12. 7. ETA 1125 DR position 32°46S 022°30 E HDG 280°(M) c)ETA 1027 W/V314/34 ETA 1011 RB 011° ETA 1620 Radial 049 ELV W/V 323/12 HDG 253° ETA 1530 W/V 127/70 INI HDG 109°(M) FIX 32°35’S 023° 17’E W/V 2 18/48 Revised ETA 0936 Radial 300 inbound FIX 2952S 02526E W/V 348/33 HDG 057(M) ETA1738 RB OO8° on BL HDG 196°(C) FIX 31°45’S 024°41’E W/V 114/14 HDG 174°(C) ETA 1907 W/V 293/30 Radial 227 on VSB FIX 27° 53’ S 022° 22 E W/V264/13 ETA 0952 RMI 1320 on KM W/V 303/44 HDG 023°(M) ETA 0807 4. 9. 11.Chapter 7 1. 10. 2. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 230 Version 4 . 8. 5. 3. a) a) b) a) b) c) a) b) a) b) c) a) b) c) d) e) f) a) b) c) d) e) a) b) c) d) e) a) b) a) b) c) d) a) b) a) HDG 251°(M). 6. a) b) c) d) e) f) g) a) b) c) d) e) a) b) c) d) e) a) b) c) d) e) f) g) a) b) c) d) e) f) a) b) c) d) e) f) a) b) c) d) e) f) g) HIDG 288°(M) DR 30°05’ 030°00’E ETA TOC 1042 FIX 30°40’S 027°18’E W/V287/39 ETA 1207 RB 355° on CH FIX 33°13’S 022°23’E W/V 306/39 DR 33°30’S 021°55’E HDG 289°(M) ETA TOD 0756 ETA CTV 0822 INT HDG 262°(M) W/V 261/49 FIX 30°41’S 024°10’E ETA TOD 0737 ETA CTV 0750 HDG 005°(M) ETA 0749 FIX 28°33’ S 018°17’E W/V 247/88 Radial 193 KTV HDG 004°(M) ROD 727ft/min. -5°C. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 231 Version 4 . 14000FT.W/V358/45 W/V352/53 INI HDG 321((M). ETA 1137 FIX 30(18’S 024(11’E . Temp. ETA 0521 HDG 254°(M) ETA 0809 W/V 251/44 INT HDG 247°(M). 17. ETA 1141 RB 354( on UP 14. Mean RAS l90kts INI HDG 212°(M) ETA TOC 0333 INI HDG 219°(M) FIX21°58’S 028°55’E W/V015/31 HDG 225°(M).Revised ETA 0804 PA. HDG 242°(M) ETA TOD 0738 ETA CTA 0754 INI HDG 335°(M) QDR 120( on VWV HDG 337((M). 19. 15. 18. TAS 225kts.13. 16. ETA TOD 1328 RAS l45kts DR Pos. 21. a) b) c) d) a) b) c) d) e) f) g) h) i) j) W/V 340/30 FIX 29(06’S 024(45’E HDG 255((M) ETA 1319 HDG 198((T). TOD 33(10’S 020(49’E Time on descent 20min. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 232 Version 4 .20. DR Pos. 31(51’S 030(13’E ETA 1227 FIX 32(27’S 026(46’E W/V 147/20 FIX 32(51’S 022(37’E W/V 111/20 HDG 273((M). ETA TOC 1139. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 233 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 234 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 235 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 236 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 237 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 238 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 239 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 240 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 241 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 242 Version 4 ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 243 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 244 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 245 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 246 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 247 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 248 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 249 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 250 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 251 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 252 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 253 Version 4 . 3. 6. ETA 1314 W/V 276/93 HDG 312°(M). ETA 1119 FTX 52°49’N 007°04’W W/V 040/37 HDG 114°(M). a) b) c) d) a) b) c) d) e) f) g) h) a) b) c) a) b) c) d) e) f) a) b) c) d) e) f) a) b) c) d) a) b) c) d) e) f) g) HDG 128°(M). ETA 1318 FIX 53°24’N 005°49’W W/V 290/86 HDG 087°(M). ETA 1141 FIX 54°25’N 000°30’W W/V 235/51 HDG 295°(M) Revised ETA 1140 Radial 117° From PWK HDG 299°(M). 2. 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.52°08’N 003°24’E QTE 028° from CL QTE 092° from PH QTE 043° from OT F1X 53°55’N 000°17’E W/V 258/49 HDG 290°(T) HDG 288°(T) HDG 285°(T) W/V 258/130 MPP 58°33’N 014°33’W HDG 146°(M). ETA 2214 HDG 296°(M). ETA 1118 TAS 24l kts TAS 286kts DR Pos. 4. 5. Revised ETA 1310 BEN RMI 104°. Both methods give the same bearing plotted. 7. Revised ETA 1318 1. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 254 Version 4 . DME 206nm HDG 088°(M). ETA 1313 FIX 53°33’N 011°07’W W/V 252/122 HDG 105°(M). ETA 2214 FIX 56°59’N 010°13’W W/V 261/108 HDG 136°(M).Chapter 8 NOTE: The plotting method of transferring the aircraft’s meridian to the NDB is correct. ETA 0814 FIX 57°11’N 005°32’W W/V 336/83 HDG 340°(M). 10.8. 9. a) b) c) d) a) b) c) d) e) f) g) h) a) b) c) d) e) f) g) HDG 272°(M).TOC 50°59’N 001°25’E HDG 342°(M) FIX 53°29’N 000°26’E W/V 282/75 HDG 328°(M) FIX 56°05’N 001°28’W W/V 275/88 ETA 1212 HDG 105°(M) INI HDG 104°(M) ETA 1259 FIX 51°I9’N 001°06’W W/V 318/43 Radial 294° inbound to CMB. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 255 Version 4 . Revised ETA 0826 INI HDG343°(M) DR Pos. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 256 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 257 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 258 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 259 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 260 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 261 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 262 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 263 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 264 Version 4 . 278° 028° QTE’s plotted using the NDB stations meridian and calculated convergency ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 265 Version 4 . 294° 255° 328° QTE’s plotted using the NDB stations meridian and calculated convergency ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 266 Version 4 . 248° 222° QTE’s plotted using the NDB stations meridian and calculated convergency ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 267 Version 4 . 058° QTE’s plotted using the NDB stations meridian and calculated convergency ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 268 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 269 Version 4 . PET: 30°53’S026°41’E RA: 02:07 EET DIST: 341nm from KTV alternate: P0S: P0S: 20°02’S 020°19’E PET. 4. 3. 9. 22°38’S 021°31’E HDG: 009°(T) G/S: 247kts Closing at 355kts INT. 8. 10. 6. 5. PNA: 14:11 a) b) c) d) a) b) c) d) a) b) c) d) e) 09:54 INT. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 270 Version 4 . EET: 01:22 HDG: 300°(M) DIST: 605nm Closing at 305kts First possible visible contact at 01:16 elapsed time.Chapter 9 1.with ETA: ETA: a) b) c) d) 13:35 13:36 23°24’S 015°04’E 26°30’S 020°31’E (Practical plotting) PNA: 22°34’S 029°48’E HDG: 198°(M) ETA: 11:32 HDG: 359°(M) ETA: 12:06 Pos: 32°20’S 023°40’E 7. 18:50 Pos: 30°39’S 020°50’E HDG: 296°(T) G/S: 298kts Closing at 439kts 25nm apart at 18:47 Pos: 32°52’S 020°42’E INT. 2. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 271 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 272 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 273 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 274 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 275 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 276 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 277 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 278 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 279 Version 4 . ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 280 Version 4 . 73. 56. 43. 60. 44. 40. 74. 4. 11. 54. 3. 71. 37. 57. 50. 24. 66. 13. 7. 36.Sample Exam Questions 1. 67. 58. 48. C A C B C A C B C C B C A B A B A B A B C B A A B 26. 15. 10. 12. 68. 41. 2. 20. 31. 53. 62. 55. 8. 22. 29. 59. 38. 64. C C C A A A B A A A C B C B B B C B A C B A B B ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 281 Version 4 . 34. B A A A A B C A B B A A C A A C A C A B A C C A C 51. 47. 5. 27. 72. 33. 45. 52. 25. 21. 32. 6. 16. 46. 23. 19. 39. 61. 9. 70. 63. 35. 42. 49. 30. 14. 65. 18. 28. 69. 17. ATP NAVIGATION & PLOTTING ATP DOC 2 & 6 Revision : 1/1/2001 FLIGHT TRAINING COLLEGE Page 282 Version 4 .