PRACTICAL NAVIGATION NUTSHELL SERIES BOOK | BY CAPT. H. SUBRAMANIAM EXTRA MASTER, FRMAS, MILLN, FNL, ECMMLLy MUMAETeehs MAMS, VIJAYA PUBLICATIONSPRACTICAL NAVIGATION NUTSHELL SERIES BOOK 1 BY CAPT. H. SUBRAMANIAM Extra Master, FR.MetS., MALIN. FINI, FCMMI, MMarTech, M.LMeLS. Principal Emeritus, L.B.S. College of Advanced Maritime Studies & Resear Mumbai. General Manager & Principal, Eurasia Centre for Advanced Learning, Mumbai. VIJAYA PUBLICATIONS 9122.25217044; e-mail: subramaniam.harry@ gmail.com 2 CHAITRA, 550 ELEVENTH ROAD, CHEMBUR, MUMBAI 400 071First edition: Jun 1976 Second edition Aug 1978 Reprinted: Sep 80, Mar 83, Mar 85, Oct 88, Dec 89, July 91, Aug 92, Jun 93, Jan 95, Third edition: Dec 1995 Reprinted: May 96, Apr 97, Feb 98, May 99, ‘Aug 00, Dec 01, Jan 03 Fourth edition Feb 2004 Reprinted: Dec 04, Oct 05, Jun 06 Jan 2007, Aug 07 Copyright Alll rights reserved Price in India: Rs. 300/- Including "Selected pages of the 1992 Nautical Almanac’ Prine and published by Mrs Vijaya Harry for Vijaya Publications of 2 Chaitra, $50 11 Road, Chembur, Mumbai 400 071 at the Book Centre Lid. {6* Road, Sion East, Mumbai 400 022. Dedicated to my mother, without whose patient and constant encouragement. this book would not have been possible.INDIAN NATIONAL SHIPOWNERS ASSOCIATION Grams: *Hindships’ Scindia House, Ballard Estate, Capt. J. C. ANAND ‘Bombay 400038 President, 18th May 1976 FOREWORD Indian Merchant Marine and the other allied fields offer to our national talent ample opportunities 10 satisfy the ambitions for self-advancement and self-fulfilment without having to subordinate personality or the desire for dedication 10 the national service. Today our maritime technical personnel dare hailed everywhere in the world as the best, and form the backbone of our national shipping. We achieved this success ‘because of the ready availability of trained technical personne! for efficient manning and operation of the shipping industry for which one cannot but thank the silent service effectively being rendered by the LBS Nautical & Engg College, the TS Rajendra (which has replaced the 1.8. Dufjerin), the Directorate of Marine Engineering Training and the ratings training establishments to produce the most competetitiand disciplined floating personnel we have today - navigating and engineering, cofficers and men - who are a pride of the nation. Capt. H. Subramaniam, the author of this book, has to hhis eredit a brilliant ‘Dufferin’ career and subsequent service with the Scindia Steam Navigation Company. This book presents 10 the students, the deep knowledge of the subject gained by his Jong personal experience whilst at sea and whilst teaching in the Nautical College. It is the first text book in his. ‘Nutshell Series’ for nautical students and is intended to prepare students for any grade of M.0.T. examination from Mate (Home Trade) to First Mate (Foreign Going). Its special and distinctive feature is the facility 1 affords 10 the students for study while at sea, so that they can utilise thetr leisure time on board more gainfully towards self advancement and save on the examination leave ume The excellent treatment and layout of the subject gives scope jor firm grounding in the science of navigation, which 1s 50 very necessary for the navigator as he has in his care, safe conduct of a modern sophisticated vessel costing crores of Rupees of capital investment and carrying valuable cargoes worth many more crores, besides the reputation of the owners and safety of several lives on board. He has to operate on the hugh seas amidst all nature of currents and swell, shallow or deep water, narrow or wide sea lanes, straight or sharp courses, thin or thick traffic, different conditions of weather and visibility, etc. Therefore, the need for human expertise, skill and care remain paramount. Navigation of modern ships ts a science calling for theoretical and practical knowledge of the highest order to enable the navigator to mect all eventualities and emergencies at sea The responsibility carries with it the need for instantaneous, instinctive and correct reactions to meet the situations as they arrtve. That is where the student's responsibility while under training, comes in and Capt. Subramaniam’s ‘Practical Navigation’ offer a useful aid to this end. It is @ ploneering project that is mast commendable and in line with the noble calling of teaching that he is pursuing at great monetary sacrifice I would like to offer him congratulations from the shipping industry and my own self for this most worthy effort Welcoming this first book of the sertes planned by him, I wish this and subsequent publications brilliant success as envisaged above, Bh {. fe (.CAniand), saePREFACE TO THE FOURTH EDITION ‘The overwhelming response, in India and abroad, to the “Nutshell Series” of books, especially this first one ~ Practical Navigation - has encouraged me to keep thinking of various ways ‘which improvements could be made with each subsequent In the third edition, sufficient theory was added to make the student understand the subject of Practical Navigation better than was possible with the previous editions. ‘The use of a simple, scientific, electronic calculator was illustrated and encouraged throughout, though working with the help of Nautical Tables was retained. Since the subject of “Spherical Trigonometry’ has been adequately covered by ‘Nutshell Series’ book mumber eight, the descriptions of various steps in the chapters on Great Circle and ‘Composite Circle Sailing have been suitably reduced, During a voyage in command from India to Europe and ‘back in mid 2003, I discovered that the combination of sun sights ‘at noon was not being done efficiently by officers of today. What ‘was routine in the mid sixties now seems to be an awkward procedure being done to mainly satisfy ISM requirements! Hence | decided to add Chapter 26, “Noon position by Sun’ to this book In this fourth edition, the only improvement made to the third edition is the addition of Chapter 26, Mumbai Ist February 2004 H, Subramaniam. waune 10 u 12 13 14 15 CONTENTS Recommended terms and abbreviations Terrestrial references Celestial references Figure drawing Preliminary calculations Plane and parallel sailing, Exercise 1 - Plane & parallel sailing Use of Traverse Tables Exercise 2 - Use of Traverse Tables ‘The Mercator Chart Mercator sailing Exereise 3 - Mereator Sailing Correction of altitudes - theory Correction of altitudes - practical Days work Chronometer time Exercise 4 - Chronometer error Theory of astronomical position lines SUN - worked examples and exercises 14.1 ~ Meridian altitude 14.2 - Azimuth 14,3 - Amplitude 14.4 - Longitude by chronometer 14.5 = Intercept 1466 - Ex-meridian STARS - worked examples and exercises 15.1 - Meridian altitude 15.2 - Azimuth 15.3 - Longitude by chronometer 15.4 - Intercept 15.5 - Exemeridian 15.6 - Polaris, Page 1s 20 25 30 31 39 40, 44 a7 4B 52 37 68 4 1 84 87 89 1 95 9 103 105 108 112 1s 1816 7 19 2 2 2B 4 25, 26 PLANETS - worked examples and exercises 16.1 Meridian altitude 16.2- Azimuth 16.3 - Longitude by chronometer 16.4 Intercept 16.5 - Ex-meridian MOON - worked examples and exercises 17.1 - Amplitude 172 - Meridian altitude 173 - Azimuth 174 - Longitude by chronometer 175 = Intercept 176 - Ex-meridian Graphical combination of sights ‘A ~ Simultancous observations B - Staggered observations Correction of simple errors in sights Exercise 28 - Errors in sights Computation of altitudes Exercise 29 - altitude computation Star identification Exercise 30 - star identification Great Circle Sailing Exercise 31 - great circle sailing ‘Composite Circle Sailing, Exereise 32 - composite circle sailing Altitude above and below the pole Stars suitable for observation Exercise 33 - suitable stars Noon position by Sun Answers Summary of formulae for use by calculator 121 124 126 129 133 136 139 140 142 146 149 153 165 175 182 183 187 188 192 193 208 209 212 213 221 224 25 233 242 [ABBREVIATIONS] RECOMMENDED TERMS AND ABBREVIATIONS TERMS: (@) Dead reckoning (or DR) position is that obtained by allowing for courses and distances only. (b) Estimated position (or EP) is that obtained by allowing for courses and distances and also for estimated leeway and current. ifany. (©) Chosen position (or CP) is that position, nearest to the ‘observer, chosen so that the latitude is an integral degree and the longitude is such that the local hour angle of the body at the time of the observation is also an integral degree (@) Sextamt altitude is that read off a sextant. (©) Observed altitude is the sextant altitude corrected for index error, if any. ABBREVIATIONS Latitude lat Course Co Cortatitude Coat Gyro GMean latitude Diff of lat Longitude Dif of long Meridian Meridian passage Meridional parts Difference of MP DR position Estimated position Geogr. position Sextant altitude Observed altitude Apparent altitude True altitude Tabulated altitude Calculated ZD Int terminal point ‘Chronometer Local mean time Greenwich mean time Greenwich hour angle Local hour angle Sidereal hour angle First Point of Aries Parallax in alt Horizontal parallax ‘Azimuth Position line Amplitude Traverse tables Nautical almanac Northern hemisphere Southern hemisphere mat lat along mer ‘mer pass MP DMP DR EP or Sext alt Obs alt App alt Tak tab alt czD re LMT GMT GHA SHA Phx in alt HP PL TT NA NH SH [ABBREVIATIONS] Magnetic ™ ‘Compass r True , Deviation Dev Variation Var Compass error CE ‘Bearing brs Distance dist ‘Nautical miles = M Knows Kn Altitude alt ‘Departure dep Days " Hours h Minutes m Seconds s Zenith distance —_ ZD True ZD 1D Intercept Int Zone time 2 Hour angle HA Declination dec Polar distance PD Rational horizon RH Star * Lower limb LL Upper limb UL Semi-diameter SD Index error IE Refraction refe Height ofeye HE Correction Corm Metres m Kilometres kam [1- TERRESTRIAL REFERENCES] TERRESTRIAL REFERENCES GEOGRAPHIC POLES ‘The two points where the axis of rotation cuts the surface of the arth are called the geographic poles - the upper one is called the North pole and the lower, the South Pole. EQUATOR ‘The equator is a great circle which is equidistant from, and therefore 90° away from, the geographic poles. The equator divides the earth into two hemispheres - the Northen Hemisphere and the Southern Hemisphere PARALLELS OF LATITUDE Parallels of latitude are small circles parallel to the equator, MERIDIANS OF LONGITUDE Meridians are great circles that pass through the geographic ppoles. Meridians cross the equator and all parallels of latitude at Tight angles[]- TERRESTRIAL REF ENCES] [Figure showing axis, poles, equator, parallels of lat, meridians | AXIS OF ROTATION —————_> LATITUDE Latitude of a place is the arc of a meridian, or the angle at the centre of the earth, measured between the equator and the parallel of latitude passing through that place. Latitude is expressed in degrees and minutes North or South of the equator. The latitude ‘of a place can have any value between 0® and 90° N or S. The latitude of the North Pole is 90°N and that of the South Pole, 90°S. [1- TERRESTRIAL REFERENCES] [Figure showing latitude) PARALLEL OF LATITUDE a |e. | PRIME MERIDIAN ‘The meridian which passes through Greenwich is called the Prime Meridian and has the value of 0° of longitude LONGITUDE Longitude of a place is the are of the equator, or the angle at the geographic pole, contained between the Prime Meridian and the ‘meridian passing through that place. Longitude is expressed in degrees and minutes East or West of Greenwich. Longitude of a place can have any value between 0° and 180°, Longitude 180°E ‘and 180°W refer to the same meridian([1- TERRESTRIAL REFERENCES] [Plan view showing 0° - 180° longitude E & W} 180° WESTERN EASTERN 90° W NoRTH | POLE 7 POSITION OF OBSERVER ‘The position of any place or person is indicated by Latitude and Longitude, For example 18° 58.2'N. 172° 52.7. D'LAT 4 Difference of latitude or d'lat between two places is the arc of a ‘meridian, or the angle at the centre of the earth, contained ‘between the parallels of latitude passing through those two places 6 [1- TERRESTRIAL REFERENCES) [Figure showing dlat & d’long] D’'LONG Difference of longitude or dong between two places is the arc of the equator, or the angle at the geographic pole, contained between the meridians passing through those two places. M'LAT Mean latitude or milat between two places is that parallel of latitude which lies midway between the parallels of latitude of hose two places.esananaae [2-CELESTIAL REFERENCES} 2 CELESTIAL REFERENCES ‘THE CELESTIAL SPHERE For the purposes of astronomical navigation, the earth is assumed to be surrounded by a concentric sphere, of infinite radius, called the Celestial Sphere. All astronomical bodies such as the Sun, Moon, stars and planets are assumed to lie on the surface of the celestial sphere. Since all calculations are based on angular measurements, the radius of the celestial sphere is irrelevant. The earth is assumed to be stationary while the celestial bodies are assumed to move on the surface of the celestial sphere. If the carth were to be transparent, and a source of light were to be installed at its contre. the projections of the various reference points/lines on the earth would have equivalent points/lnes on the surface of the celestial sphere, For example: EARTH passing through that body. Declination is expressed in degrees and minutes North or South of the equinoctial. The declination of a celestial body must have a value between 0° and 90° N or S. GHA Greenwich Hour Angle or GHA of a celestial body is the are of the equinoctial, or the angle at the celestial pole, measured 9([2-CELESTIAL REFERENCES] ‘westerly from the celestial meridian of Greenwich to the celestial ‘meridian passing through that body. GHA is expressed in degrees ‘and minutes from 0° to 360°. GHA of a celestial body would increase steadily from 0° to 360°. GHA would be 0° when the body is on the Greenwich Meridian, increase steadily until it is 360° (ic. 0°) when it would again be on the Greenwich meridian POSITION OF A CELESTIAL BODY ‘At any instant, the position of a celestial body is indicated by its declination and GHA at that instant LHA Local Hour Angle or LHA of a celestial body is the arc of the ‘equinoctial, or the angle at the celestial pole, measured westerly from the observer's celestial meridian to the celestial meridian passing through that body. Like GHA, LHA of a celestial body ‘would increase steadily from 0° to 360°. LHA would be 0° when the body is on the observer's meridian and steadily increase until it is 360° (ic. 0°) when it would again be on the observer's meridian. At any instant, for any celestial body : LHA = GHA + Longitude of Observer ‘The actuai calculations involving GHA, LHA and longitude are ‘explained later on in this book, ‘THE SUN'S ORBIT As mentioned earlier, itis assumed that the earth is stationary and that all celestial bodies move along the surface of the celestial sphere. The Sun's orbit is inclined to the equinoctial by about 23° 26.5'. Maximum Northerly declination (about 23° 26.5'N) occurs around June 22nd and the maximum Southerly declination (about 10 [2-CELESTIAL REFERENCES] [Figure showing Aries, Libra, ete. ] AXIS OF e 23° 26,5'S), around December 22nd. There are two specific reference points, on the Sun's orbit, called the First point of Aries and the First Point of Libra. They are also referred to as the ‘equinoctial points.([2-CELESTIAL REFERENCES} FIRST POINT OF ARIES ‘The First Point of Aries is that point on the surface of the celestial sphere where the sun's path crosses the equinoctial from SOUTH to NORTH. This occurs around 21st March and is called Vernal Equinox. The Sun's declination at this point is 00° 00". The First Point of Aries is used very frequently in practical navigation and is represented by the Greek letter y (gamma), FIRST POINT OF LIBRA The First Point of Libra is that point on the surface of the celestial sphere where the sun's path crosses the equinoctial from NORTH to SOUTH. This occurs around 23rd Sept and is called ‘Autumnal equinox. The Sun's declination at this point is 00° 00" SHA Stars are so far away from the earth that they are, for all practical purposes, fixed objects in space. So when we consider them to be on the surface of the celestial sphere, their GHA would increase at a uniform rate and their change of declination per day ‘would be negligible. If the GHA and dec of each star were to be tabulated every hour of the day, the Almanac would be unnecessarily bulky, costly and awkward to consult. Hence the introduction of a value called SHA for each star. ‘SHA is the arc of the equinoctial, or the angle at the celestial pole, measured Westerly from the First Point of Arics to the celestial meridian passing through the star. SHA and dec are tabulated in the Almanac, once every three days, for cach star. The GHA of Aries is tabulated, for each day, for very hour of GMT. GHA * =GHAy + SHA * (2-CELESTIAL REFERENCES) GEOGRAPHICAL POSITION The line joining a celestial body and the centre of the earth would cut the carth's surface at a point called the geographical position or GP of that body at that instant. GP is expressed in latitude and longitude. The value of dectination is the latitude of the GP and the value of GHA, converted to longitude, would be the longitude of the GP. (GHA is expressed in degrees and minutes west of Greenwich from 0° to 360° whereas longitude is expressed from 0° to 180° E or W of Greenwich). RELATIONSHIP BETWEEN ARC AND TIME, ‘When the earth rotates on its axis once, one day has elapsed. This ‘means that: ARC TIME 360° 24 hours ise 1 hour r 4 minutes Yr 4 seconds A table for converting arc to time and vice versa is given in the Nautical Almanac. GHA and LHA, though usually expressed in degrees and minutes of arc, may also be expressed in hours, minutes and seconds of time. The symbols for degroes and minutes of arc are ° and ' ‘whereas the symbols for time are h, m and s. cmr ‘When the sun is on the Greenwich meridian, GHA = 0° but the time is said to be 1200 hours GMT. So GMT is ahead of GHA by 12 hours. GMT is the number of hours, minutes and seconds since the sun crossed the inferior meridian of Greenwich. (The 13([2-CELESTIAL REFERENCES] inferior meridian is one which is 180° away ie. on the opposite side of the earth/celestial sphere) LMT When the sun is on the observer's meridian, LHA = 0° but the time is said to be 1200 hours LMT. So LMT is ahead of LHA by 12 hours. LMT is the number of hours, minutes and seconds since the sun crossed the inferior meridian of the observer. (The inferior ‘meridian is one which is 180° away i. on the opposite side of the carth/celestial sphere). RELATIONSHIP BETWEEN GMT & LMT It is obvious from the foregoing that the difference of time between GMT and LMT is only longitude expressed in units of time. Longitude in time = GMT + LMT ‘The actual calculations involving GMT, LMT and longitude are ‘explained later on in this book. “4 (3-FIGURE DRAWING] 3 FIGURE DRAWING FOR ASTRO CALCULATIONS Imagine a globe, representing the earth, as shown in the following figure A . NP & SP are the north and south geographic poles respectively. WQE is the equator and Il is the parallel of latitude of the observer Z. The N-S diametric line is the meridian of the observer Ifthe globe is now tilted slightly towards us, the points P, Z and Q would move downwards and both, the equator and the parallel of latitude, would appear as curves, as shown in figure B. The south geographic pole would have disappeared from view We have to bear in mind that P, Z and Q are on the surface of the lobe and not on a flat surface. If the globe is gradually tilted more and more, until Z is in transit with the centre of the globe, the globe would appear as shown in figure C. Since all the points, P, Z and Q have apparently moved only because of the tilting of the globe, each thas shifted out of its original position by the same amount i¢., NP = ZQ (see figure C). But ZQ = latitude. So NP = ZQ = latitude, This can be proved mathematically as follows: 15[3-FIGURE DRAWING] (3-FIGURE DRAWING](3-FIGURE DRAWING] In figure C, NZ = PQ (each = 90°) But NZ = NP + PZ and PQ = PZ +ZQ SoNP +PZ=PZ+ZQ Subtracting PZ from both sides, ‘NP = ZQ (and each = latitude), In figure C, we notice that Z (on the surface of the globe) is equidistant from all parts of the circle NESW. Z is hence the pole of the great circle NESW. If the globe now represents the celestial sphere, and Z, the observer's zenith, NESW is callled the Rational Horizon of the observer. Figure C can now be explained as under: NESW - rational horizon or RH P = celestial pole Z_— = zenith of observer PZS_~—- __ part of observer's meridian NP - part of observer's inferior meridian NP = latof observer =ZQ WZE - prime vertical circle NZS principal vertical circle. NOTE: Any great circle on the celestial sphere, passing through the Zenith of the observer, is called a vertical circle. The N-S vertical circle is called the principal vertical circle and the E-W ‘one, the prime vertical circle. Il, the parallel of latitude of the observer, was necessary for the explanation of the figure but may now be omitted, [3-FIGURE DRAWING] In case the observer is in the southern hemisphere, the globe would be tilted with the north end going away from us, as shown in figures G, H and I The position of a celestial body can be inserted in the figure knowing its declination and LHA ‘The declination circle of the body can be drawn knowing, that it would be a cirele, around the pole, with a radius equal to its polar distance (PD). If the observer's lat was N, we would be considering the north pole. Ifdec was N, PD = 90 - dec. If dee was S, PD = 90 + dec. If the observer's lat was S, we would be considering the south pole. Ifdec was $, PD = 90 - dec. If dec was N, PD = 90 + dec. NOTE: The Polar Distance of the equinoctial is 90°. Having drawn the declination circle, the position of the body can be inserted knowing its LHA at that time, When rising, the celestial body would be on the rational horizon cast of Z ‘When setting, it would be on the RH west of Z. When at maximum altitude, the body would be on the meridian (PZQS in figure C). Various possibilities of latitude and declination are shown in figures D, E, F, G, H and 1, wherein the suffix R indicates the point of rising, M indicates the point of meridian passage and suffix S, the point of setting 19[4-PRELIM CALCULATIONS] PRELIMINARY CALCULATIONS 4.1 To obtain d'lat between two given latitudes: If the latitudes are of the SAME name, SUBTRACT, as illustrated below. If the latitudes are of DIFFERENT names, ADD, as illustrated below. @) (b) ©. (d) From 21° 18N 36° 44'S 43°27N 32° 09'S To 45° 10N 10°27'S 1230S 30° O1'N Dat 23°52N 26°17N 55°S7S_—«6210'N ‘The name of the d'lat is obtained by comparing the departure and arrival positions - If going from N lat to more N lat:- d'lat N ‘OR from S fai to less $ lat:- lat N OR from $ lat to N lat d'lat N If going from S lat to more S lat:- dat S OR from N lat to less N lat’ d'lat S OR from N lat to S lat d'lat S. 20 [4-PRELIM CALCULATIONS] 4,2 To obtain the m'lat between two places: If the latitudes are of the SAME name, add, divide by two and retain the name. If the latitudes are of DIFFERENT names, subtract, divide by two and retain the name of the larger one, as illustrated below :- @) tb) ©) @ LatA 10°26N 46° 52'S._ 20° 19N_ 37° 48'S. LatB 30°SON = 30° 10'S 42'S SHE SGN Milt 20°38'N38°31'S_«10°0S'S._—07°54'N 4.3 To obtain d'long between two longitudes: If the longitudes are of the SAME name, SUBTRACT, as illustrated below. Jf the longitudes are of DIFFERENT names, ADD, as illustrated below. @) () © (@) From 75° 10'E 124° 1'W_ 13° O6E_— 175° 30'E To 69° O8'E 136° 14'W 05° 33°W 170° 20.W Dilong 06°.02"°W 12° 03'W -18°39'W 345° S0'W* ie. 14°10 ‘The name of the dlong is obtained by comparing the departure and arrival positions - If going from E long to more E long:- long E. OR from W long to less W long:- dlong E. ‘OR from W long to E long:- long E. I going from W long to more W long:- d'long W. OR from E long to less E long:- dlong W. OR from E long to W tong:- dong W. 2[4-PRELIM CALCULATIONS] *¥elre it is necessary to add, and the d'long so obtained is greater than 180°, first put down the value of d'long as it is and name it as above. Then subtract it from 360° and CHANGE THE NAME. This is because dong can never be sreater than 180° 4.4 Connection between GHA, LHA and long: Where any two'of the above are known, the third can be found by the slogan Longitude EAST, GHA LEAST Longitude WEST, GHA BEST. ie, GHA+Elong=LHA and GHA - W long = LHA Example jiven GHA & long, to find LHA: fa) (b) ©). (d) GHA 049° 11.6 097° 148 343° 18.7" 012° 13.7 Long 36° 10.0E _20°40.0W 36° 20.0E 48° 56,0W LHA 085° 21.6" 076°34.8' 379° 38.7" 323° 17 7@ ie. 019° 38.7" * Since LHA cannot be greater than 360%, subtract 360° and the balance is the answer. @ Where direct subtraction is not possible, add 360° to GHA ‘and then subtract. Example B: Given'LHA & long, to find GHA: @ ), ©). «@). LHA 076° 18.2" 27°17." 341° 33,8' 013° 17.6 Long —_30° L1.0E _50°49,0W 63° 10,0W 48° 04.0E GHA — 046°07.2' 278069 404° 43.8% 325° 13.6@ ie, 044° 43.8) * Since GHA cannot be greater than 360° subtract 360° and the balance is the answer. 2 [4-P REL IM CALCULATIONS] (@ Where direct subtraction is not possible, add 360° to LHA and then subtract. Example C: Given GHA & LHA, to find long: Subtract the smaller one from the larger one and name the longitude according to the slogan longitude East GHA least, longitude west GHA best. In NO case must GHA and LHA be added, @) (b) © @ GHA 087° 11.4 320°.47.2" 012° 18.3" 327° 18.3" LHA 110° 19.7' 290173" 341°52.5' 040° 32.8 Long 23° 08.3E 30° 29.9W 329°34.2E 286° 45. SW ie. 30°25.8W 073° 14.5E ‘Where the longitude obtained by direct subtraction is greater than 180°, put it down as it is and name it. Then subtract it from 360° and CHANGE THE NAME. This is because ongitude can never be greater than 180°. 4.5 Connection between GMT, LMT & LIT: ‘When any two of the above are known, the third can be found bby applying the slogan - Long East GMT least, Long west GMT best. ie, GMT+Elong=LMT and GMT-Wlong=LMT Example A: Given GMT & long, to find LMT: (a) ) () @ dhms dhms dhms dhms GMT 18041006 05234004 30 200419 120418 12 LIT E.061836 E 050056 W 080542 W.06 1236 LMT 18 102842 06044100 30115837 112205 36 23(4-PRELIM CALCULATIONS] Example B: Given LMT & long, to find GMT: @) (b) © @ dhms dhms dhms dhms LMT 21162210 14061219 30050950 1916.44 10 LIT E 101216 E080912 W050416 W073604 GMT 21060954 13220307 30101406 20002014 Example C: Given GMT & LMT, to find long: Subtract the smaller one from the larger one and name it as per slogan, In NO case should GMT and LMT be added. (@) ) © @ dhms dhms dhms dhms GMT 18 161030 28041026 06082712 19205716 LMT 18100750 27182740 06111506 20080408 LIT 060240 094246 = 02.4754 1106 52 Jong 90°40W 145° 41.5W 41° 58.5°E 166° 43'E 4.6 SHA of stars: Find the GHA and dee of SIRIUS at 18h 24m 10s on Jan 20th 92. Referring to page of 20th Jan in Almanac: GHA y 20d 18h 029° 22.61 Incr 24m 10s 6° 03.5' GHA y for GMT 035° 26.1" SHA > SIRIUS 258°.47.4 GHA & SIRIUS for GMT 294° 13.5' Dec % SIRIUS for 20d 16° 42.4'S -000- 24 [S-PLANE & PARALLEL SAILING] PLANE AND PARALLEL SAILING ‘THE PLANE SAILING TRIANGLE: ‘Small portions of the earth may be considered to be a flat, or plane surface without appreciable error. Where the distance between two places is less than 600 miles or so, plane trigonometry can be used to :~ (a) compute the course the distance between two positions (b) calculate the position arrived given the starting position, ‘course and distance steamed. ‘To facilitate easy calculation, ready solutions of right angled triangles are available in the form of Traverse Tables. The use of Traverse Tables is illustrated in Chapter 6 of this book In order to obtain a plane sailing triangle, the course is, first converted from three figure notation into quadrantal notation (NE, SE, SW and NW). The actual distance between two places is split up into its N-S component and its E-W component, as illustrated below. In each case, the plane sailing triangle will be similar to one of the four triangles shown below, depending on the quadrantal course - NE, SE, SW or NW. 2[S-PLANE & PARALLEL SAILING] “The N-S component is the d'lat between the two places ‘and the E-W component, the departure (dep). All the three sides ~ dist, d'lat and dep - are expressed in M (nautical miles). One M = 1852 m or 1.852 km. Also one km = 0.54 M. « DEP-—-» <«—— DEP —-» ‘THE PARALLEL SAILING FORMUL, To convert the dep into long, or vice versa, the parallel sailing formula is used: Dep_ = Cos mat Diong 26 (5-PLANE & PARALLEL SAILING] Note: ‘Technically speaking, the right side of the above formula should read mid lat instead of mean lat. The application of a very ‘small correction to mean fat wouid convert it into mid lat. This is because the earth is an oblate spheroid and not a perfect sphere: In practical navigation, for the sake of convenience, mean lat is used instead of mid lat as the resultant error is insignificant. 5.1 To find the position arrived, given the starting position, course and distance: . Worked example Given the starting position to be 20° LI’N 72° 52'W, course (032°(T) and distance 238 M, find the position arrived. 032°(T) = N32°E, Dist = 238 M. dilat = dist. Cos co = 238.Cos 32° Log 238 2.37658 latlet 20° 11.0'N Log Cos 32° -140,92842 dat 3221.8'N Log dtat 2.30500 latarrd 23° 32.8'N dilat = 201.84" = 3°21.8'N milat 21°52" ist . Sin Co = 238 . Sin 32° “ss —» Log 238 2.37658 a LogSin32° “14072421 f «Log dep 2.10079 | dep = 126.12E & s S. dilong = dep . Sec m'lat 5 Ss 26.12. See 21°52" D> Log 126.12 2.10079 Y Log Sec 21°52. Log d'long, 2 nIS-PLANE & PARALLEL SAILING} dilong = 135.9 = 02° 159 long left 7 s2.0W Jong arr’d 70° 36.1'W Pos arrived: 23° 32.8'N 70° 36.1'W Answer. 5.2 To find the course and distance between two positions: ‘Worked example Find Co & Dist from 20° 10'N 179° 40'W to 13° 40'N 178° 10'E. From: lat 20° 10'N long 179° 40'W J To: tat 13°40N tong 178° 10 Gilat 6° 30'S d'long 357° S0’E s ie, 2° 10W | dilat =390°S, d'long = 130°W, and sftat 16°55" dep = d'long . Cos m'lat Tan Co= dep = 124.38 = 130. Cos 16° 53! dat 390 ‘Log 130 2.11304 Log 124,38 2.09473 Log Cos 16°55 -1+0,98079 Log 390 2.59105 Log dep 2.09473 Log Tan Co -1+0.50367 dep 124.386 | Course $ 17° 41'W Answer (i). | dist = dilat . Sec Co = 390. See 17°41" Log 390 2.59106 Log See 17°41" 0.02102, Log dist 2.61208 Distance = 409.34 M Answer (ii). 8 {S-PLANE & PARALLEL SAILING] 5.3 To find the set and drift of current: Set and drift of current is the course and distance FROM the DR position TO the fix. Drift is expressed in miles. Where required, the rate of current can be obtained in knots by dividing, the drift by the number of hours during which the drift occurred, ‘The working is similar to that explained earlier in 5.2. Hence no ‘worked example is shown here separately. 29[5-PLANE & PARALLEL SAILING] EXERCISE 1 (Plane & parallels 2) In the following cases, find the position arrived: STARTING POSITION Latitude Longitude Course Distance 1 10°20N (060° 20°F 155° 300M 2 10° 12'S 120° 11'W 260° 458M 3 00° 10S 179° 40'W, 340° 510M 4 60° 112N 120° 18.6E 250° 3124M 5 30°48 168° 12 046712’ 4268M In the following cases, find the course and distance: FROM TO Latiude Longitude Latitude Longitude 6 20°300'N 179° 36.0'E 167 18.0'N-178°32.0'W 7 03° 120'N 004° 113" 02°30.4'S_ 002° 10.0;W ¥ $6 125'S 046° 12.5°W SOPLL3'S 044° 14. W. 9 36° TN 075° 12.6'E 40° IRG'N O80" LI S'E 10 GO°I16'N 076° 443 $5°10.3N 080° 16 'W In the following cases, find the set and drift of current: DR FIX Latitude Longitude Latitude Longitude 46°44.3N 076° 363'E 47° 00.0'N 076" 104. 12 30° 168'S 087°49.'E 31° 00.75 058° 20.4'E 3 00° H16N —179°50.2W 00° 40.38 178° 40.'E ‘S 175° S4.9°W N 000° 20.1'W 60° 20.6 176° 18'W 6001 15 50°16.3'N 000° 123"E 49°50. (6- TRAVERSE TABLES) THE USE OF TRAVERSE TABLES Traverse Tables are ready solutions of right angled ‘triangles to facilitate navigational calculations. In this chapter, the illustrations are limited to the solution of problems in plane and parallel sailing, 6.1 To find the position arrived given the starting position, course and distance: Worked example 1 Starting position 20° 11'N 72° 52'W, Co 032° (1) dist 238 M, To find the position arrived, (1) Convert the three figare notation into a quadrantal one - 032° becomes N37°E, (2) If the quadramtal Co 1s less If the quadrantal Co 1s more than 45°. enter TT from than 45°, enter TT from the top and use the column below and use the column headings from the top of the headings from bottom of the pave. as shown in eapital page, as. shown in. capital letters, nthe following letters, im the following diagram ~ diagram(6- TRAVERSE TABLES) (G) From ‘TT, for Co = 32° and dist = 238 M, extract values of dilat and dep. Since Co is'N32°E, lat must be Nand dep ‘must be E. Values are d'lat = 201.8'N and dep = 126.1'E. (4) Apply:dllat to the departure latitude and obtain the latitude arrived. (5) Obtain the m‘lat (to the nearest decimal of a degree) between the departure and arrival latitudes. In this case, m'lat = 21.9°. (6) Enter TT again, using mat in place of course. If lat is less than 45°, enter TT from the top of the page and use the columns shown in italics in the foregoing figure {flat is more than 45°, enter TT from bottom of the page and use the columns shown in italies inthe foregoing figure. (7) Locate 126.1 in the dep column and obtain dong from its ‘column, interpolating as necessary. D'long will always have the same name as dep, 2 [6- TRAVERSE TABLES) In this case, interpolation is as follows. milat 21°, dep 126.1, dlong = 135.1" milat 22° dep 126.1, d'long = 136.0' milat 1° changes dlong by 0.9" so 0.1° changes dlong by 01° For mllat 21.9° dep 126.1, d'long = 135.9° ie, d’long = 2° 15.9E (8) Apply d'long to the longitude left and obtain the longitude arrived, ‘Method of showing written work: Left lat 20° 11.0'N long 72° 52.0'W Co N32E 238M dilat 3° _21.8N dilong 2° 159° dep =126.1'E Arrd lat 23° 32.8'N long 70° 36.1'W milat = 21.9°N Worked example 2 From position 36° 18'S 178° 46°W, course of 294° (T) was ‘steamed for 177 M. Find the position arrived. 294° = N66°W. Since Co> 45°, enter FT from below. Loft lat 36° 18.0'S long 178°°46,0W CoN66W 177M lat 1° 12.0'N d'long 3° 19.1'W dep = 161.7W Ard lat 35% 06.0S long 182° 05.1'W mlat= 35.7°S ie, 177 54.9E Note: Since the long. arrived exceeds 180°, subtract it from 360° and change its name, as explained in Chapter 4 33[6 - TRAVERSE TABLES} Worked example 3 From a position 00° 10'N 178° 57'E, a course of 156° 18' (T) and ‘a distance of 321.3 M were made good. Find the position arrived. Co 156° 18' = $23° 42'E = $23.7°E From TT: Co $23°E dist 321.3! d'lat 2958'S dep 125.5'E Co $24°E dist 321.3 dilat 293.5'S dep 130.7. Interpolating between Co 23° & Co 24° for 23.7°, dilat = 294.2'S Interpolating between Co 23° & Co 24° for 23.7°, dep = 129.1'E. ‘Method of showing written work: Left lat 00° 10.0'N long 178° 57.0'E $23.7°E 321.3M dilat 04° $4.2'S long 002° 09.2"E dep = 129.1'E Arrd lat 04° 44,2'S Jong 181° 06.2'E milat = 2.3°S ie, 178° 538W 6.2 To find the course & distance between two given po: (1) Write down the two given positions and obtain the dat and {long beeen them. Name them FROM the positon left TO the position arrived. (2) Calculate the mat to the nearest decimal of a degree. (3) Enter TT using the m'at & d'long and obtain the dep If the m'lat is less than 45°, If the m'let is more than 45°, enter TT from the top and enter TT from below and use use the column headings the column headings from from the top of the page as bottom of the page as shown shown here in italies'~ hore im italies:~ u [6~ TRAVERSE TABLES) | : | DIST | DEP | DLAT LONG, __| DEP 58° (4) Compare the d'lat and dep as follows: (a) If dlat & dep are equal, course = 45°. So if they are nearly equal, course must be near about 45°. If they differ by a large amount, course will be far away from 45° (closer to 0 or 90°) (b) If diat is more than dep, Co If dep is more than dat, Co must be less than 45° (as must be more than 45° (as illustrated below) so enter illustrated below) so enter TT TTT from above using capital from below using capital leter letter headings headings DEP a5[6 - TRAVERSE TABLES} (5) Bearing in mind the foregoing points, enter TT using the capital letter columns shown in the figure and locate the page hhaving the values of dilat and dep closest to the required values Interpolate for the Co and dist as shown in the following worked example:- Worked example Find the course & distance from 20° 10'N 179° 40'W to 13° 40'N 178° 10. From lat 20° 10'N long 179° 40'W To lat 13° 40N long 178° 10E Gilat 06° 30'S dilong 357° 50E ie, 2° 10W dilat =390'S dilong = 130'W milat = 16.9°N From TT, mat 16.9° and dong 130°W, dep = 124.4°W Bearing in mind the points mentioned in 6.2 (4): ‘Step number 1:- Step number 2:- To obtain Co from TT: To get dist from TT: Co diat_ dep Co dat dist M3900 1244 17 390.0 2? 18° 390.0. 126.7 18° 390.0 410.1 7 __390,0_119.2 7390.0 407.8 60° changes 7.5 60° changes 23 8 eee 23 8! > 0.7 177 390.0 1244 177% 390.0 409.4 Course = $17° 42" W Distance = 409.4M [6- TRAVERSE TABLES} 6.3 To calculate the set and drift of current: Set and drift of current is the course and distance FROM the DR. position TO the Fix. The working is similar to that explained in ‘6.2 but the interpolation may be simpler, because of the small distance involved, as illustrated below:~ Worked example ‘At 0800 hours the DR position of a ship being 10° 12.5'S 47° 42'W, a fix was obtained and found to be 10° 04.3'S 47° 56.6W. Find the set and drift of current, DR lat 10° 12.5'S long 47° 42.0'W milat 10.1°S Fix lat 10° 043'S long 47° 56.6W Diat—082'N dong —(14.6'W mlat dilong dep 10.1" “6M NOTE: By using 146 instead of 14.6 to enter TT, interpolation is rich faster. The decimal point in the dep so obtained can be inserted later. milat long dep 10° 146 143.8 e 146 143.3 10,1° 46 143.8 101° 146 4 So d'lat = 8.2'N and dep = 14.4. Step number 1:~ Step number 2:- To obtain Co from TT: To get dist from TT Co dep lat Co dep dist 2 Ma 82 603° 1442? 37[6- TRAVERSE TABLES] NOTE: By using 144 & 82 instead of 14.4 & 8.2 to enter TT. interpolation is much faster. The decimal points can be inserted after interpolation. Step mumber 1:- Step number 2:- ‘To obtain Co from TT: To get dist from TT: Co dep dat Co dep dist oo 14483 60° 144166 61° 14480 61° 144168 60° changes 3 60° changes 20 eee 20_23- 03 603° 14482 603° 144 165.7 603 144 82 603° 144 166 Course (set) = N60? 20° W Dist (drift) = 16.6M Note: The above interpolation has been written elaborately in ‘order to explain the method to the student The actual interpolation can be done mentally very quickly and the answer written down, -000- 8 [6- TRAVERSE TABLES) EXERCISE 2 (Use of Traverse Tables) Using ONLY Traverse Tables, find the position arrived: STARTING POSITION Latitude Longitude Course Distance 1 10°20 060° 205, 155° 300M 2 10°12 120° 1'W. 260° 438M 3 00° 10S 179° 40W 340° 510M 4 60° 1L2N 120° 18.6E (250° 3124M 5 30° 14'S 168° 12;W 46°12" 4268 M. Using ONLY Traverse Tables, find the course and distance: FROM TO Latitude Longitude Latitude_——-Longitude 6 20°300'N 179° 36.0 16° 18.0'N_ 178° 32.0W 7 03° 120N 0047 11.3E —02°30.4'S 002° 10.0/W 8 56° 125'S 046° 12.5;W SOP1N3'S 044° 14.8;W 9 36°11 7N 075° 126E 40° I8.6'N 080° LSE 10 60° 11.6'N 076° 44.3'W 55° 10.3'N__ 080° 16.8'W Using ONLY Traverse Tables, find the set & drift of curren DR FIX Latitude Longitude Latitude ‘Longitude 11 46°44.3'N 076°363'E 47° 00.6'N 076" 10.4E 12 30°16.8'S 057°49.3E 31° 00.78 058° 20.4E 13. 00° 116N —179°S02';W 00°40,3'S._ 178° 40.1'E 14 60°206S 176° 184'W_ 60° 01.7175" 54.9'W 15 50°16.3'N 000° 123 49°50.4'N 000° 20.1'W 39[7-MERCATOR CHART] THE MERCATOR CHART ‘The three dimensional, curved surface of the earth has to be represented on a two dimensional flat surface of a chart for the sake of convenience in navigation. Mercator solved this problem by using the cylindrical projection. Imagine that the earth is a ‘transparent globe with parallels of latitude and meridians of longitude drawn on it. Imagine that a cylinder, having the same radius as that of the globe, is placed enclosing the globe, tangential to the equator. Imagine that a source of light is placed at the centre of the globe and the projections of the parallels of latitude and meridians of longitude on the inside of the cylinder get marked permanently. Ifthe cylinder is now cut along its length, opened out and placed on a flat table, it would represent a Mercator Chart «wherein the following characteristics will be apparent: - 40 (7-MERCATOR CHART] On the Mercator Chart (a) The equator appears as a straight line (b) All meridians appear as straight tines parallel to, and equidistant from, each other and cross the equator at right angles. (©) Parallels of latitude would appear as straight lines parallel to each other but the distance between consecutive parallels ‘would increase as latitude increases, (@) Polar regions do not appear 41[7-MERCATOR CHART] (©) One minute of d'long is the same size in all parts of the chart (as mentioned in (b) above) (®) One minute of d'lat increases in size steadily as we go away from the equator (as mentioned in (c) above). Distances On a Mercator chart the distance, in any direction, ‘measured in minutes of latitude is the distance in Nautical Miles. Note: Since the size of one minute of latitude on a Mercator chart increases as latitude inereases, the dividers should be used against the latitude in which the distance lies. For example, if place A is in latitude 24°N, and place B is in latitude 26°N, one leg of the dividers should be placed on A, the other on B and that represents the distance AB on that chart. To read off the value of the distance, the dividers should be placed N-S along the latitude markings on the chart such that the centre of the dividers lies on latitude 25°N and then the d'lat contained between the legs read off {in minutes. Suppose the d'at so obtained is 180', the distance AB is 180 Nautical Miles, Dep and d'long On a Mercator chart, the E-W distance between two places, measured along the latitude scale (in minutes of latitude), is the departure between them in Nautical Miles but, if’ measured along the longitude scale (in minutes of longitude), itis the dlong, ‘between them. D'lat and DMP On a Mercator chart, the N-S distance between two places, measured along the latitude scale, is the dat between them (in minutes or Nautical Miles) but, if measured along the longitude 42 : h [7-MERCATOR CHART] scale (in minutes of LONGITUDE), it is called the DMP (Gifference of meridional parts) between them MERIDIONAL PARTS, Meridional Parts for any latitude is the arc of the meridian, or the angle at the centre of the carth, contained between the equator and that parallel of latitude, expressed in minutes of LONGITUDE. In other words, the N-S distance from the equator to any parallel of latitude, expressed in minutes of LONGITUDE, is called the Meridional Parts (MP) for that latitude, MP for each latitude is given in Nautical Tables DMP Difference of Meridional Parts or DMP between any two parallels of latitude is the are of the meridian, or the angle at the centre of the earth, contained between those two parallels of latitude, expressed in minutes of LONGITUDE. In other words, the N-S distance between the two parallels of latitude, expressed in ‘minutes of LONGITUDE, is called the DMP between them. DMP between position A and position B is the difference between the MP for latitude A and that for latitude B, both of which are extracted from Nautical Tables. The rule, whether to ‘add or subtract, is similar to that for obtaining d'lat between two places - same names subtract, different names add ‘RHUMB LINE, ‘A rhumb line is one that cuts all meridians at the same angle. It would appear as a straight line on a Mercator chart but, ‘on the surface of the earth, a rhumb line would be curved. This 1s because, on a Mereator Chart, meridians appear parallel to cach other whereas, on the surface of the earth, they are not 4B[$-MERCATOR SAILING] MERCATOR SAILING ‘Where the distance exceeds 600 M, itis recommended that the calculation be done by Mercator Sailing in order to get more ‘accurate results than by plane and parallel sailing methods. Mercator Sailing Formula: Dilong = Tan Co DMP. For the sake of convenience, the above formula can be superimposed on the plane sailing triangle, as shown here Worked example 1 Find the rhumb line course and distance from starting position P 02°12'S 160°18'W to final position Q 10°19'N 140°40'W. [MERCATOR SAILING] From Tables latP 02°12'S_ longP 160° 18'W) MPof P_ 131.18 lat Q JongQ 140° 40'W MP of Q: 618,2N dilat dlong «19° 38E DMP 749.3 te, ie, 1178'E Note: To get DMP, apply same rule as for dilat - same names subtract, different names add Tan Co=dilong = 1178 DMP 749.3 Course = NS7° 32'E Answer (i) Log 1178 3.07115 Log 749.3 2.87466 Log Tan Co 0.19649 To get distance, use the plane sailing formula: dist = d'lat . Sec Co ‘Log d'lat 2.87564 = 751. Sec 57° 32! Log Sec Co 9.27018 distance = 1399.0 M Answer (ii). Log dist 3.14582 Worked Example 2 Find by Mercator’s Principle, the position arrived if a ship sailed a course of 301° (T) for 1408 M_ from position 00° 04'S 178° 20'W. 301° = NS9°W. dilat = dist . Cos Co = 1408 . Cos 59° Log 1408 3.14860 Log Cos 59° -1+0.71184 Log dat 2.86044 dat = 725,16" = 12° 05.2'N 45[S-MERCATOR SAILING] lat left 00°.04.0'S dilat 2° 05.2'N MP forlatlet 4.0 MP for lat arr'd 721.7N lat ard 12° 01.2"N DMP 757 diong = Tan€o or long = DMP. Tan Co DMP long = 752.7 Tan 59° Log 725.7 2.86076 Log Tan 59° 0.22123 long = 1207 8.W_ Log dong 3.08199 long left 178° 20.0'W dlong —-20°078'W Jong arr’d 198° 27.8'W ie, 161° 3228 Pos arrived :- lat 12° 01.2'N long 161° 32.2'E Answer. [8-MERCATOR SAILING] EXERCISE 3 (Mercator Sailing) Find, by Mercator Sailing, the course and distance: FROM TO Latitude Longitude Latitude Longitude 24°00N O74 SW 46° 00'N. 053° 45'W 06° 00'N 079° 00W 38° 00'S_——«179° 00 40° 18N —100°20'W_ 68° 00'N 140° 10E 70°20N 010° 22;W$2°S0'N 009° 4S'E 32°29'S 064° OO 49°S50'S 005° IS'E Find, by Mercator's Principle, the position arrived:- Lacie Lnginde Came Dian 36°48'N08S°S3'W 241? 1897M. 06° 10S 176° 4TW 333° 4450M. 38° 18'S 00S? LI'W 124° 3256M. 44° 1I'N 140°20W 056° 222M 18°S8N 072° 52265? 7126M. "7[9-ALTITUDES- THEORY] 9 CORRECTION OF ALTITUDES - THEORY In the following figure, C is the centre of the earth & also that of the celestial sphere. B__ isa celestial body on the celestial sphere. © is the position of the observer on the carth's surface. E__ isthe eye of the observer, OE is the height of the observer's eye above the sea surface. It includes his own height and that of the bridge deck above the sea, WV is the visible horizon of the observer. ES isthe sensible horizon, Z isthe observer's zenith. RH is the rational horizon of observer. It is a great circle, of the celestial sphere, of which Z is the pole. BEV is the observed altitude of the body above the visible SEV is the angle of ‘dip’. It depends on the height of the observer's eye above the sea surface, [9-ALTITUDES- THEORY) BES is the apparent altitude of the body. It is obvious from the figure that BES = BEV - SEVie., App alt = Obs alt - dip. EBC is the angle of parallax, or parallactic angle, which is the angle subtended at the body by the radius of the earth, OF is negligible compared to CB, CE & EB and is hence diregarded for this purpose. Stars and planets are so far away that the radius of the earth does not subtend a finite angle at that distance, Hence there is no parallax for stars and planets. Parallax is significant for the Sun and the Moon, It will be noticed from the figure that the value of parallax is maximum when the body is on the rational ‘horizon and nil when it at the zenith of the observer. Hence[9-ALTITUDES- THEORY] the correction for parallax is tabulated against apparent altitude, EHC is the horizontal parallax or the parallax when the body is on the observer's rational horizon. BCH is the true altitude of the body i.c., the altitude, at the centre of the earth, of the body above the rational horizon. From the figure it is obvious that: BCH = BFS (because ES is parallel to CH). But BFS = BES + EBC (exterior angle = sum of interior opposite angles), Hence BCH = BES+EBC, or Truc alt = App alt + Parallax. Correction for refraction: While passing through the atmosphere, the rays of light ‘coming from a celestial body get refracted. Since the density of the air gets less and less as height above the earth's surface increases, rofraction causes the body to appear at a higher altitude than the real altitude, Hence the correction for refraction is always negative, as shown in the following figure wherein the body at B appears to be at Bi The correction for refraction is maximum when the altitude ig nil and gradually decreases in value until it is mil when the body is overhead. This is because, when the altitude is nil or low, the angle of incidence is very high whereby the rays pass obliquely through the atmosphere. When the altitude is 90°, the angle of ineidenc is nil whereby the rays are normal to the layers ‘of atmosphere. Mence’ the correction for refraction is tabulated against apparent altitude {9-ALTITUDES. THEORY) Bi Correction for Semi-diameter: It is not possible to obtain the altitude of the centre of the Sun or ‘Moon directly with a sextant because they appear as large discs, Hence we observe the altitude of either the lower limb (LIL) or the ‘upper limb (UL) and then apply a correction. When the LL is observed, the centre of the dise would be higher than the altitude observed and hence semi-diameter has to be added. Similarly, when observing the UL, the semi-diameter correction is subtractive st[HO-ALTITUDE CORRECTION} 10 CORRECTION OF ALTITUDES - PRACTICAL 10.1 Correction of altitudes of all celestial bodies except the MOON:- ‘This is illustrated by a worked example with comments where necessary ‘On 16th March 1992, the sextant altitude of the Sun's lower limb was 56° 11.4" Ifthe index error of the sextant was 2.8' off the arc and the height of eye was 12m, find the true altitude. Sext alt 56° 11.4" iE (28) [Of = (+), On=()] Obs alt 56° 14.2" Dip (HE 12m) (-)_06.1' [Always (] Appalt 56° 081° Tot corm — (4) 15.6 [LL (+), UL Talt 36° 23.7 [O-ALTITUDE CORRE NOTES 1: Corrections for altitude may be taken from Nautical Tables or from the Nautical Almanac, 2: Dip is tabulated against height of eye and is always subtracted from Obs alt to get App alt. 3: App alt is obtained in the above manner for all celestial bodies. 4: Total correction (Tot. corr) is tabulated against App alt, 5: FOR THE SUN, total correction is given in two columas - one for October to March and the other for April to September. Different columns are also given for upper limb (UL) and for ower limb (LL). Total correction should be applied according ‘to the sign given therein. However, it will be observed that total correction is always (+) when the LLL is used for observation and (-) when the UL is used. 6; UL or LL is only applicable for the Sun and the Moon because they appear as large discs to the observer. He cannot judge the location of the contre of the visible disc, Hence he has to observe the altitude of the UL or LL and then apply suitable correction to obtain the altitude of the centre of the Sun or ‘Moon, Stars and planets are so far away that they only appear ‘as pin points to the observer, Hesce LL or UL is not applicable for stars and plancts. 7: For STARS & PLANETS, total correction is given in a separate table from the Sun and is always to be subtracted. 8: In the case of Venus and Mars, one more correction called “Additional correction’ is to be applied. This is given in the ‘Nautical Almanac tabulated against altitude. 33[10-ALTITUDE CORRECTION] 10.2 Correction of altitudes of the MOON:- ‘The apparent altitude is obtained as explained in 10.1 otal correction for the Moon is given in the inside part of the back cover of the Nautical Almanac, The method of use of those tables is clearly illustrated therein and two solved examples will illustrate this fully Given: Sext alt of Moon's LL Given: Sext alt of Moon's UL 42° 24.6, IE 2.4' on the arc, 56° 27.1', IE 3.8' off the arc, HE 12m, HP $4.6' Find the HE 15m, HP 56.0. Find the ‘true altitude. true altitude. Sext alt 42° 24.6" Sext alt 56° 27.1' IE (2) _2.4" rE 38 Obs alt 42° 22.2° Obs alt 56° 30.9" Dip (HE 12m) )_6.1° Dip (HE 15m) )_6.8' App alt 42° 16.1 App alt 56° 24.1" ‘Main corr @) 52.2" ‘Main corr () 419" HP corm (54.6) (+) 01.9 HP corm (56.0) (+) 03.2' ‘Taltitude 43° 10,2" 57°09." 2.30.0 Taltitude 5° 39.2! NOTES 1: HP stands for Horizontal Parallax of the Moon. It is obtained ‘rom that day's page of the Almanac, alongside the declination. 2: Both main and HP corrections are always positive. 3: For UL, 30* has to be subtracted from the total to get true altitude. 4 {[10-ALTITUDE CORRECTION] 10.3 Back angle observations: ‘Sometimes, the horizon under the celestial body may not be clearly visible owing to the presence of land, haze, mist, rain, etc, In such cases, it may be possible to observe the altitude of the body in the opposite direction i.e, if the azimuth (bearing of the celestial body) is 080°, the altitude is taken while facing 260° Since the observer would then have his back to the celestial body, such a measurement is called a 'Back angle’ observation. Back angle observations will always be greater than 90° However, the sextant cannot measure angles greater than 120° ‘Hence back angle observations will be between 90° and 120°. Worked example 1 - Back angle - stars & planets: Given: On 20th Dec 1992, Sext alt of Venus was 118° 52.3'. IE 3.6' off the arc, HE 20m. Required: The true altitude. Sext alt 118° 52.3 iE @)_3.6 [As usual] Obs alt 118° 55.97 Dip (HE 20m) (2) _07.9' [As usual] App alt 8° 48.0 ‘61° 12.0" (Subtract from 180°} Tot. corm —_(-)_00.5' [As usual] ole 1st ‘Addl corm (+) 00,1" [Mars & Venus only} Tait erale ‘Worked example 2 - Back angle Sun: ‘Total correction cannot be used here, The individual corrections ~ Dip, Refraction, Semi- diameter and Parallax have to be applied separately. Slogan for easy remembrance: Don't Rob Spare Parts 55[10-ALTITUDE CORRECTION] Given: Sext alt of Sun's LL was 119° 11.61. IE 2.8! on the are, HE 12m, date 30th Nov 1992. Required: True altitude, Sext alt 119° 11.6 IE (-)_2,8' [As usual, on are (-)) Obs alt 119° 08.8" Dip (HE 12m) ()_06.1' [As usual, always (-)] App alt 9° 02.7 (60° 57.3" [Subtract from 180°] Refraction (-) 00.5! [As usual, always (-)] 61° 56.8 Semi-diam* (-)_16.2* {Opposite, LL ()]* 61° 40.6 Parallax (4)_00.1"[As usual, always (+)] ‘True alt 6140.7, NOTES 1: The correction for refraction can be obtained from the Nautical ‘Almanac or from Nautical Tables. Since stars and planets do fot have any semi-diameter or parallax, the total correction ‘tabulated in the Nautical Almanac consists entirely of refraction. So to get the refraction correction for the Sun, as in this case, look up the total correction for stars. This is tabulated against apparent altitude in the Nautical Almanac 2: The value of the semi-diameter of the Sun is given atthe foot of that day's page in the Nautical Almanac. In back angle ‘observations, correction for SD is applied opposite - Normally LL +, UL -, but in this case, LL -, UL +. 3: The correction for parallax is given in Nautical Tables. -000- [L-DAY'S WORK] 11 DAY’S WORK Day’s Work is a summary of the ship's passage from 1200 hours on one day to 1200 hours the next day (ship's time). ‘The working is illustrated by worked examples with explanatory notes where required Worked example 1 ‘At noon on 14th Sept., a ship in position 40° 12°N 76° 46°W set course of 250°(C) - Dev 4°W, Var 6°W ~ at an engine speed of 16 knots. At 1600, course was altered to 287%(C) - Dev 2°E, Var 6°W - and the speed was decreased to 14 knots. At 2200, course was again altered to 340°(C) - Dev 5°E, Var S°W - and speed of 14 knots was maintained till 0600 next morning, when ‘course was altered to 277°(C) - Dev 2°W, Var 5°W - and speed increased to 15 knots. This course and speed was maintained till noon on 15th Sept. Find the EP on 15th noon, the course and the distance made good, if a current was setting 027°(T) at 2 knots throughout, ‘Note: This is called the narrative or linear form of question. All ‘courses should be tabulated and corrected first, as follows: 37 sussnatanemeateseesiesn[11-DAY'S WORK] 1200 1600 2200-0600 to to to to 1600 __2200 ___0600 __1200__Current Compass Co 250°C) 287(C) 340°C) 277°) Deviation aw 2E SE 2w ‘Magnetic Co 246°(M) 289°(M) 345°(M) 275°(M) Variation == WWW SOW. TmeCo ——240°(T)_ 283°) 340°CT)_270°(T) 027°C) ssorw N77°>W N20°W WO N2T*E Distance 64M 84M 112M 90M 48M Note: Set of current is taken as a course and the drift for the whole 24 hour period is taken as distance. All the true courses are ‘converted into their quadrantal figures so that TT can be entered. Quadrantal figures also give the names of d'lat and d'long. The distance ran on each course is entered in the lowest line just beneath the corresponding. course in quadrantal figures. ‘The last two rows in the foregomg table should be used to enter ‘TT and the information tabulated as follows: Course Dist lat dep wohl Me Nees SB Wa, S6o°W 64 320 554 NIPW 84 18.9. 818 N20°W 112 105.2 383 Wot 0 90.0 NOTE 48 _428 _218 _,, Total 166.9 320 218 265.5 32.0 218 Resultant 134.9 243.7 Note: After obtaining total d'lat and total dep in each direction, the smaller figure of dilat is transferred under the larger figure of dlat and the ‘Resultant «lat is obtained by subtraction. ‘Resultant dep’ (LL-DAY'S WORK) is also obtained similarly. Entering TT with resultant dat and resultant dep, the course and distance made good, noon 10 noon, is obtained (as explained in Chapter 6.2). From TT: dat dep CoCo dist. 1349 243.3 61° 61° 278.2 134.9. 253.8 62° 62° 287.2 105->60' 60" > 9.0 fie) bovO spRh. ue sab O. 1349 243.7 2782 (Co MG = N61° 02'W and dist = 278.5 M Answer (i). “The EP next noon is obtained by TT'as explained in Chapter 6.1 NIN pos 14th lat 40° 12.0°'N- long 76°46.0'W milat 413° Resultant dlat 02°14.9'N dilong S°24.6W dep 243.7 NIN EP [Sth lat 42° 26.0 long 82° 10.6'W d'long 324.6'W ‘Answer (i). Worked example 2 On 16th Jan a ship in position 00° 10'N 68° 09'E set ‘course as follows: Time Co(C) Dev Var — L'way Wind Log 1200 126 2E 4E 3° SW we 1800 149° 3°E 4*E 2° NE 89 we 2300 210° IW S°E 3° SE 168 ae 0700 240° 2°W SE Nil W290 ae 1200 270° Nil 4E 3 ON 368 ‘A current set the vessel 183°(T) at 1.5 knots throughout Find the EP next noon and the course and distance made good. 9[1-DAY'S WORK} NOTES 1: This is called the tabular form or Log extract form of presentation. It further differs from example 1 in that leeway is included. Since course actually steered is given, it is necessary to apply leeway and obtain the effective course (course made ‘g0od with wind only) for each leg of run 2: If the wind is coming from the starboard side of the ship, the ship would be blown off-course to port. So the effective course will be less than the true course steered (when expressed in the three figure notation), and vice versa - that is, if the wind is ‘coming from the port side of the ship, the ship would be blown off-course to starboard. So the effective course will be more than the true course steered (in the three figure notation) 3: The distance steamed on each course is obtained by the difference of log readings taken at the beginning and at the end of each course. 4; The alteration of course to 270° shown against 1200 hours on 17th Jan would be effective from 1200 onwards and is hence not to be included in the working from 1200 on 16th Jan to 1200 on 17th Jan. The log reading at that time is, however, necessary to compute the distance run from 0700 to 1200 on the 17th 1200 1800-2300 0700 to to to to 300 2300 __0700__1200__Current Compass Co 126%(C) 149°(C) 210°C) 240°(C) Deviation 2B SE lw zw ‘Magnetic Co 128°(M) 152°(M) 209°(M) 238°(M) Variation Ee 4 SE SE TrueCo ——132°(T) 156°) 214°) 243°CT) Lecway = 2 zz Nil Effective Co 129° (T) 158°) 217°(T) 243°(T) 183°(T) SSISE $22°F S37°W S63°W S03°;W Distance 89M = 79M. 12M 78M_—-36M (1-DAY'S WORK} ‘The rest of the working is similar to that of example 1. Effective Dist dilat ep w SSI°E 89 56.0 69.2 S22°E p 732 296 S37°W 122 974 BA ‘$63°W 78 35.4 69.5 )3°W 36. 36.0 Total 298.0 988 1448 00.0 98.8 Resultant 298.0 46.0 From TT, Co MG = $08° 47°W, dist MG = 301.5 M Answer (I). NIN pos 16th lat 00° 10.0'N long 68°09.0'E milat_ 02.3°S Resultant dlat 04° 58.0'S d'long 0°46.0W dep 46.0W NIN EP 17th lat 04°48.0'S long 67°23.0E dlong 46.0'W ‘Answer (i), ‘Note: Since m’lat is only 2.3°, TT shows dep = d’long = 46", ‘Worked example 3 ‘On 6th March a ship in position 46° 36'S 175° 34'E steamed the following courses:~ Time Co(C) Dev Lway Wind Speed 1200 150° 5°E SWxW 8 Kn 1600 140° 4°E sw 8 2000 120° 3° sw 75 2400 120° 3°E SW 65 0400 100° 1°E BC 0800 095° Nil Ss 8 1200 095° Nil Biv vd. a[11-DAY'S WORK] ‘Var 10°E throughout. Find the DR position at noon on 7th March ‘and, if the observed position then was 48° 14.3'S 178° 06.5°E, find the set and drift of current. Note 1: The DR next noon should be worked out as shown in examples | and 2. Since Co and dist MG are not asked, they need rot be calculated. It is implied that the alteration of course, in each case, is at the hour mentioned alongside. For example: 140°(C) ‘was set at 1600 hours. 120°(C) was set at 2000 hours, and so on, Note 2: There is no alteration of course at 2400 hours but the ‘speed has altered. So from 2000 to 2400, the distance will be 30M and from 2400 to 0400 it will be 26M. Hence from 2000 to 0400, the distance will be $6 M on a course of 120°(C). 1200 1600-2000» 04000800 to to to to to _____ 1600 _-2000___0400__0800__1200 Compass Co 150°C) 140°C) 120%(C) 100%) _095%(C) Deviation SE 4B BE LE Nil Magnetic Co 155°(M) 144°(M) 123%(M) 101°(M) 095°(M) Variation — 10°R_— 10°E 10° 10°E 10°E TrueCo ——165%(T)154°(T) 1331) 111°CT) —105°(T) Leeway zx a Nil Nil Nil Effective Co 162° (T) 149°(T) 1331) 111°(T) —105°(T) SISE S31°E S47°E = $69°E S1S°E Distance 32M 32M S6M_— 32M. 32M Effective Dist dat dep ‘Co(T) Be Om NOS JUNE CoccTW S18°E 32 304 09.9 S31°E 32 m4 165 S47°E 56 382 410 S69°E 32 115 299 ‘S7S°E 32 083309 Total and resultant 158 128.2 [I-DAY'S WORK] NIN pos 06th lat 46° 36.0'S long 175°34.0'E milat 47. Resultant dlat 01° 55.8'S dilong 03° 10.0 dep 128.2'E NIN DR O7th lat 48° 38'S long 178° 44.0E dilong 190.0 DRNN 07th lat 48° 31.8°S long 178° 44,0°E Using TT :- Fix NIN 07th lat 48° 14.3°S long 178° 06,5°E m'lat 48.4°S d'lat "17.5'Nd’long —37.5°W dep 24.9" W Set = NSS°00'W and drift = 30.5M. ‘Worked example 4 At noon on 14th Dec a light house in lat 05° 56N 80° 36'E bore 000°(C), error 4°W, distance 10 M. Course was then set to 220%C), Dey 1°E, Var 2°W, log 0. At 2000, engines broke down and the log, showing 82, was hauled in, At 2200, engines ‘were repaired and course was resct to 200°%(C) Dev nil, Var 2°W, og 0, Ship maintained this course till noon next day when log showed 140, A current was estimated to set 350°(T) at 2 knots throughout. Find the EP at noon on 15th Dec. NOTES 1: There are two differences in this question from those illustrated carlicr. Firstly, the noon position on starting is not given directly and secondly, an engine stoppage is given. 2: Since the current was constant, even during engine stoppage, it is allowed as a course and distance for the whole day as usual. First obtain the noon position on 14th Dec as follows Peenet ace fe 000° (C) Bearing of ship from Lt Ho: ime ie, Course $04°E & distance 10 M. 3(11-DAY'S WORK] 3: Sometimes the bearing of the ship from the light house is given. In that case do not reverse the bearing, but use the iven bearing as the course, as illustrated below: Pos of LtHo lat 0S°56.0'N long 80° 36.0°E milat $.9°N SO4°E-10M dlat _100'S dong _O.7E dep 0.7E Noon pos 14th lat 05° 46,0'N long 80° 36.7E 1200 2000 2200 to to to ates 2000 1200 Compass Co 220°(C) 200%C) Deviation JE Nil Magnetic Co —_-221°(M) 200%(M) Variation Zw 2w True Co 219°(1) 198°(1)_350°(T) s30°W ‘Sis°w NIO°W Distance 82M 140M 48M ‘The working of the rest of the problem is similar to those illustrated earlier. The student may work it out on his own, Answer: EP 15th noon lat 03° 16.5°N long 78° 53.0°B. Worked example S At noon on 20th July, Pargo Point (32° 48°N 17° 16°W) bore 080° (C) 10 M off while steaming 219° (C), Dev 3° E Var 18° W. Ship maintained this course at a steady speed of 15 knots till noon next day. Clocks were retarded 12 minutes at 0200 hours. Find the DR at 1200 hours on 21st July. If the fix then was 27° 12'N 20° 05°W, find the set and drift of current. Note: This question is different from the previous ones only in one aspect - alteration of clocks. The steaming time from 1200 on 20th July to 1200 on 21st July becomes 24h 12m because clocks were (11-DAY'S WORK] retarded 12 minutes. If clocks were advanced, the steaming time would be reduced accordingly Distance at 15 knots, for 24h 12m steaming, is 363M. The rest of the working is similar to that of example 3. Answ DR 27° 12.2°N 20° 17.4°W - Set 091° (T), drift 11 M. Worked example 6 AA ship steams the following courses by Gyro compass, error nil 13th Jan 1200 - Co 236° (G) speed 15.0 knots ale 1700 - Co 284° (G) speed 16.0 knots We 2000 - Co 262° (G) speed 15.0 knots \4th a/e 0400 - Co 198° (G) speed 14.5 knots and continued thus till noon, 14th Jan At 1800 on 13th Jan, a light house in 48° 28.5°N 67° 0S°E bore 4 Points on the starboard bow and at 1845 it was abeam. Find the Mth noon DR, the course and distance made good noon to noon, Note: This is very simple if approached systematically. First the course and distance made good noon to noon can be got as usual Since the courses are true, with no leeway, working is simplified Course Dist dlat dep m BE NBER gh aat Rost ray, 1200-1700 ssw 75 419 62.2 1700-2000 N76°W 48116 46.6 2000-0400 S82°W 120 16.8 118.8, 0400-1200 S18°>W 116 1103 358 Total 6 169.0 263.4 ibs Nil Resultant 157.4 263.4 6s[AL-DAY'S WORK] From Traverse Tables:- d'lat dep Co Co dist 1574 262.0 59° 59° 305.6 157.4 272.6 60° 60° 314.3 10.660" 60" > 92 1428 8 > 12 157.4 263.4 > 59°08" -» 306.8 Answer (i): Course MG = $59° 08°W Dist MG = 306.8 M. To calculate the pos of the ship at 1845 hrs:- ‘Cojbetween 1800 and 1845 284°(T) Starboard beam +4902 1845 beam bearing of Lt from ship = 374°(T) = 014°(T) 1845 bearing of ship from Light = 194°(T) = S14°W eae By the principle of doubling the ‘angle on the bow, distance steamed from 1800 to 1845 is 284 § the distance off abeam at 1845. Ms Distance for 45 minutes at 16 knots = 12 M = distance off at eas 1845 on a bearing of $14°W 1800 from the light house. Pos of LtHo lat 48°28.5'N long 67°05.0'E milat 48.4°N S14°W-12M diat__11.6'S dilong _4.3;W dep 2.9'W 1845 position lat 48°16.9'N long 67° 00.7E NOTES 1: From 1845 to the next a/c at 2000 Co = 284°(T) & dist = 20M. 2: D'lat and dep for the run on each leg after 2000 can be copied from the earlier table made for the noon to noon Co and dist. 66 LIL-DAY'S WORK] Course Dist d’lat dep ) M N 8 Boy 1845-2000 N76°W 2004.8 194 2000-0400 S82°;W 120 168 118.8 0400-1200 S18°W 116 110.3 35.8 Total 048 127.1 174.0 048 Nil Resultant 1223 1740 1845 pos of ship 48° 16.9'N long 067° 00.3'E milat 47.25°N Resultant d'lat 02°.02.3'S dong 04° 16.3°;W dep 174.0W NIN DR léth lat 46° 14.6'N long 062°44.4'E dllong 256.3'E ‘Answer (ii) Note: Owing to the large number of worked examples included here on Day's Work, no separate exercise is given in this chapter, Students desiring to practice Day's Work can work out the same Jot without copying the working shown here and compare the answers. 67 mapesensensensemenaen[12-CHRONOMETER TIME] 12 CHRONOMETER TIME 12A: THE AMBIGUITY OF CHRONOMETER TIME ‘Most chronometers are graduated like ordinary clocks i from 0 to 12 hours only. So if chronometer shows 4h, it could be 04h or 16h, creating a twelve hour ambiguity. So whenever chronometer time is given in a navigation problem, it should be checked before commencing the calculation. If GMT is given, no 12h ambiguity exists as GMT is ‘expressed in the 24 hour notation - 4h means 04h NOT 16h, Remember: GMT hours - aceept. Chron hours - suspect ‘On modern ships, GMT for celestial observations is taken from the navigation satellite receiver instead of the chronometer. In ‘such cases, the date and exact GMT in the 24-hour notation would bbe known directly without any ambiguity ‘The reason why the ambiguity in chronometer time is stil retained in the examination system is that the student is constantly reminded that he should check the input data before plunging into calculations. GMT is also called UTC-Universal Time Coordinate 68 [12-CHRONOMETER TIME] ‘The date given in a problem usually refers to the date on the ship NOT the GMT date or chronometer date, Hence this has to be checked as the GMT date may possibly be one day less or ‘more than the date kept on the ship. This is illustrated below:~ On 24th June at 0500 IST in Bombay, GMT would be 05h 30m less ie., 234 23h 30m. Yet, one hour later, at 0600 IST on 24th, GMT would be 24d 00h 30m. In E long, GMT may be same date as ship or one day behind. In W long, GMT may be same date as ship or one day ahead. The procedure for solving the am is as follows: (1) Write down the two possibilities of chronometer time @) Apply LIT (longitude in time) to each and obtain both the possibilities of LMT hours. (3) An indication whether it is A.M or P.M on the ship would be given/known, Insert the date on the ship alongside appropriatc LMT and, working backwards, insert the date of GMT. Example: On 13th Sept, PM at ship in DR 27° 16°N 75° 00°W, the chron showed 03h 18m 12s, error 02m 05s fast. Find GMT. ho ms hom s Chron 03 18 12 or 15 18 12 Error 0 -) 02 05 (02 05 GMT (i4)* 03:16. 07 15 16 07 LIT) ©0500 09 © 05.00 00 LMT (13)* 22:16 07 10 16 07 GMT: 14d 03h 16m 075 (1 Application of error is described in 12B, * Inserted last A lf direct subtraction not possible, 24h to be added. 69[12-CHRONOMETER TIME] In problems in practical navigation, AM or PM at ship, or the approximate ship's time, MUST be given. However, as a point of interest, itis possible to resolve the 12 hour ambiguity by other ‘means which have fondly been called navigational gymnastics:- (a) For azimuth, long by chron and intercept - SUN:- ‘The correct LMT must be after sunrise and before sunset, both ‘of which are obtained from that day's page in the nautical almanac. (b) For ex-meridian alt - SUN:- The correct LMT must be very near 1200 hrs (between 1030 and 1330 or so). (c) For azimuth - STARS AND PLANETS:- The correct LMT must be at night ie, after sunsot and before ‘sunrise (both obtained from that day's page of the nautical almanac) as they are not visible during daytime. (@) For sextant observations - STARS & PLANETS:- From that day's page of the nautical almanac, extract the civil and nautical twilight timings and tabulate them thas :~ ‘The correct LMT will. be AM PM between Civil and Nautical twi- CryiL. 05401905 light of either AM or PM NAUT 0510 1930 only. Jf, in one chance in a million, A.M and P.M are both possible, the question would involve "superior navigational ‘gymnastics. (©) For ex-meridian alt - STAR:~ Calculate the approx. LMT mer pass star as follows :~ LMT mer pass: 03h 17.6m —_ |(From the corresponding; SHA * in time (-) 09h 16.3m [page of the Almanac). LMT mer pass * 18h 013m The correct LMT must be very close to the time calculated above. The time so calculated will give only an approximate idea of the LMT meridian passage of the star and cannot be used for any other purpose other than to sort out ambiguity of chronometer time. 70 [12-CHRONOMETER TIME] (® For ex-mer alt - PLANETS AND MOON: ‘The approx LMT of mer pass is given in that day's page of the ‘Nautical Almanac and the correct LMT must be near that (g) For azimuth, long by chron and intercept - MOON:- Correct LMT must be between approx. LMT of moonrise and moonset, both of which are given in that day's page of the Nautical Almanac. (h) For ex-mer alt - MOON:- ‘Approx. LMT mer pass moon is given in that day's page of the Nautical Almanac. Correct LMT must be close to this. 12 B: PROBLEMS INVOLVING DAILY RATE OF CHRON ‘The difference between chronometer time and GMT, at any instant, is called the chron error. Chron error is said to be FAST if the chron time is ahead of GMT. Fast error is, therefore, to be subtracted from chron time to obtain GMT. Chron error is said to be SLOW if the chron time is behind GMT, Slow error is, therefore, to be added to chron time to obtain GMT. ‘The words FAST and SLOW only indicate whether the chronometer is ahead of GMT or behind it. They do NOT indicate the speed of the chronometer. The words GAINING or LOSING are used to indicate the speed of the chronometer. ‘A chronometer is said to be GAINING in the following cases: (1) If the chron is correct on GMT one day and is fast next day (2) Ifit is fast one day and more fast the next day. (3) Ifit is slow one day and less slow next day. (4) Iftis slow one day and fast next day. ‘A chronometer is said to be LOSING in the following cases: (1) If the chron is correct on GMT one day and is slow next day, (2) If'it is slow one day and more slow the next day. @) If it is fast one day and less fast next day. (4) If tis fast one day and slow next day. nm[12-CHRONOMETER TIME} ‘The daily rate of the chronometer is the number of seconds gained or lost by it in one day. It is usually taken as an average over a number of days to overcome observational errors and thereby obtain an accurate result, A good, reliable chronometer may not keep perfect time but will have a constant daily rate by which GMT can be computed at any required instant. Over a period of time, the change of chron error (caused by daily rate) is called the accumulated rate, Accumulated rate is obtained by multiplying the daily rate by the interval between the time of the last time signal and the required date and time as illustrated in the following worked examples. If the daily rate is gaining, the accumulated rate would be called fast and if the daily rate is losing, accumulated rate would be called slow. Accumulated rate, applied to the old chron error, ‘would give the chron error at the required instant Worked Example 1 ‘At 1200 GMT on 23rd June 1992, a chronometer was 6m 20s fast, Find the chron error at 0700 GMT on 4th August if the daily rate was 3s losing. 23rd June 1200 GMT to 30th June 1200GMT: 07d 00h, 30th June 1200 GMT to 31st July 1200GMT: 31d_——00h 31st July 1200 GMT to 04th Aug 0700GMT: 03d _19h 23rd June 1200 GMT to04th Aug 0700GMT: 41d 19h Interval: 41.8d* Accumulated rate = daily rate x interval = 3.x 41.8 = 125s 4 * Squaring off to one decimal of a day would suffice. 4 Squaring of to nearest second would suffice. ([12-CHRONOMETER TIME] Accumulated rate = 02m 05s (Slow) Old chronometer error = 06m 20s (Fast Error at GMT Aug 04d 07h 00m = 04m 15s (Fast) ‘The interval ean be calculated in an easier way by using page 5 of the Nautical Almanac as follows:~ Odth Aug 0700GMT = 217th day of 1992 0700 GMT. 23rd Jun 1200 GMT 175th day of 1992 1200 GMT Interval = 4idays 1Shours 418 days. In problems in practical navigation, it is essential to first obtain the correct GMT date and hours of the astronomical ‘observation (as shown earlier in this chapter) so that the correct interval is obtained in days and hours. It would be considered an error in principle if this was not done first. This is illustrated in worked example 2. Worked example 2 On 29th Nov, AM at ship in DR 26° 27'N 130° 27°W, an astronomical observation was made when the chron showed 0Sh 46m 20s. If at 0600 GMT on Oct 14th the chron was 00m 10s fast, and its daily rate was 4s losing, find the correct GMT of the observation. dh oms ahoms Chron 05 46 20° OR 17 46 20 Error Not yet calculated Approx GMT 05 46 20 29:17 46 20 LUT(w) (>) 08 41 48 (© 0841 48 LMT 21 04 32 29 09 04 32 Correct GMT d & h: 29d 17h B[12-CHRONOME. TIME] 29thNov 1746GMT = 334th dayof 1992 1746 GMT 14th Oct 0600 GMT 288th day of 1992 9600 GMT Interval 46 days 1h 46m 46.5 days. “Accumulated rate = daily rate x interval = 4 x 46.5 = 18€3 SLOW Accumulated rate = 03m 06s (Slow) (Old chronometer error = 00m 1 Error at GMT Nov 294 17h 46m = 02m 56s (Slow) Chron time of observation 29:17 46 20 Chron error (Slow) (4.02.56 Correct GMT of observation 29 1749.16 EXERCISE 4 (Accumulated rate of chronometer) 1 On 8th March 1992, AM at ship in DR 50° 12°N 176° 18°E, the cchron showed 06h Olm 10s, If at 1200 GMT on 20th Feb chron error was 3m 12s fast and its daily rate was 5s gaining, find the correct GMT of the observation 2.At about 1820 at ship in DR 20° 187S 120° 11°W on 18th April 1992, find the carrect GMT when chron showed 02h 21m 06s, Chron error on 30th March at 0600 GMT was 00m 40s slow and its daily rate was 2s gaining. 3.On Jan 4th 1992, PM at ship in DR 16° 01°N 84? 00°F, an astronomical observation was made when chron showed 01h 15m 10s. If chron error at 1800 GMT on 20th Dec 1991 was 10m 16s slow, and its daily rate was 6s gaining, find the correet GMT. (Note: Page 5 of the 1992 almanac cannot be used here). 14 [13 - ASTRO PL - THEORY} 13 THEORY OF ASTRONOMICAL POSITION LINES ‘An astronomical position line is really a part of a large circle drawn on the surface of the earth, with its centre at the Geographical Position (GP) of the body and having a radius equal to the Zenith Distance. (GP and ZD have been described in Chapters 2 & 3), All observers on the circumference of this circle ‘will observe that the body has the same ZD at the same time, Since this cirele has a large radius, any small part of the circumference of this position circle may be considered to be a straight line and called @ position fine (PL) or line of position (LOP). The azimuth (true bearing) of the celestial body, from the observer, represents the direction of the radius. Since the circumference of a circle and its radius mect at right angles, we say that the celestial PL. (or LOP) is at right angles to the azimuth ‘The azimuth of the body is easily calculated with the help of ABC tables using the DR position. The direction of the celestial PL (or LOP) is obtained by adding. and subtracting 90° to the 18[13 - ASTRO PL - THEORY] azimuth - eg., if azimuth was calculated to be 040°, the direction ‘of the PL (or LOP) would be 130° - 310°. One astronomical observation can only give one PL (or LOP) and a position through which to draw it. Another PL (or LOP) is necessary to intersect the first PL (or LOP) in order to obtain a *Fix’, There are five common methods of obtaining the position through which to draw the astronomical PL (or LOP) - (1) Latitude by meridian altitude, (2) Longitude by chronometer, (3) Intercept, (4) Ex-meridian and (8) Altitude of Polaris. Each of these is briefly described here:~ 16 [13 - ASTRO PL - THEORY] (J) Latitude by meridian altitude: Here the altitude of a celestial body is taken when the body is on the observer's meridian, Its ‘azimuth at that time would be either 000° or 180°, ‘The PL (or LOP) will, in both cases, be 090° <> 270° (ic., 000° + 90° = 090° <> 270° and 180° + 90° = 090° <> 270°), Since the PI. (or LOP) runs E - W, it will coincide with a parallel of latitude. The actual value of that latitude is obtained by a simple caleulation. Such an observation will give a definite value of the ship's latitude, irrespective of any error in the ship’s DR position. ‘The actual calculations are shown and explained in the worked ‘exaniples under the appropriate heading in subsequent chapters. (2) Longitude by chronometer: Using the Haversine formula, we calculate the Observed Longitude (longitude where the position cirele cuts the DR latitude). The azimuth is calculated by use of ‘ABC tables and the direction of the PL (or LOP) is obtained by adding and subtracting 90° to it. This PL (or LOP) is drawn through the DR latitude and Observed longitude, The DR longitude is of no consequence in this calculation, Actual calculation by this method is shown under the appropriate heading in subsequent chapters. DR lat[13 - ASTRO PL - THEORY] (3) Intercept or Marcq St. Hilaire method: The angular distance, between the DR position and the GP of the body, called the Calculated Zenith Distance (CZD), is calculated by use of the Haversine formula. CZD is then compared with the TZD (True Zenith Distance) which is obtained with the help of a sextant. The difference between them is called the intercept. Intercept is ‘expressed in minutes of arc, taken to be Nautical Miles. The azimuth is calculated by using the DR position and ABC tables. ‘The direction of the PL (or LOP) is obtained by adding and subtracting 90° to the azimuth. “The position circle always has the GP as its centre and ZD as its radius. The CZD is the radius of the position circle which passes through the DR position. The TZD is the radius of the position circle that passes through the actual position of the ship. Since we know the DR position of the ship, we imagine ourselves standing on the DR and going by the shortest distance to the true position circle. If TZD < CZD, we note that we will be going towards the GP (ic., with azimuth as course and intercept as distance), If TZD > CZD, we will be going away from the GP (ie., with a course equal to azimuth + 180° and with the intercept as distance. This is illustrated in the following figures. ‘The solution of the intercept method of work should state the DR latitude, DR longitude, True Azimuth, the value of the intercept in minutes of are, whether intercept is towards or away and finally the direction of the PL (or LOP), A rough figure (not necessarily to scale) should also be drawn as shown in the following examples | and 2. ‘The Intercept Terminal Point (ITP) is that point on the PL (or LOP) which is closest to the DR position. Calculation of the actual latitude and longitude of the TTP is not normally necessary and is done only if specifically asked to so. [13 - ASTRO PL - THEORY] 0[13 - ASTRO PL - THEORY] Example 1 DR position: latitude: 36° 18°N a DRiat longitude: 72° 56’°W_ Intercept 3.8 AWAY Az 042° (T) PL 312° - 132° Example 2 DRiet DR position: latitude: 16° 37°S longitude: 142° 11°E Intercept 4.5” TOW ‘Az 130° (T) PL 220° - 040° ‘Both methods - Long by chron and Intercept - give equally ‘accurate results and take the same time for working. In the ‘examination, the choice of method is often left to the student but hhe should be thoroughly familiar with both these methods. The intercept method is much simpler and faster if several simultaneous observations are to be calculated and plotted to obtain a "Fix’ ‘The next figure illustrates the similarity of result of working by Long by chron and by Intercept. Using DR position “D’, Obs long °B’ and PL were obtained and plotted on a chart. If the ‘same astronomical observation was worked by the intercept method, an intercept of DP would be obtained by calculation and, in the case illustrated, it would be "Away". 80 [13 - ASTRO PL - THEORY) __ As the celestial body is so far away, afew miles difference in the position of the observer would not appreciably affect the azimuth. So whether the azimuth was worked using the DR long or the Obs long, the answer would be the same and hence the direction of the PL (or LOP) of both, Long by chron and Intercept ‘methods, would be the same. When the PL’s (or LOP’s) of both the methods are plotted on a chart, it will be observed that they ‘coincide as shown in the foregoing figure. Actual working by the intercept method is shown under the appropriate heading in ‘subsequent chapters.[13 - ASTRO PL. - THEORY] (4) Ex-meridian method: Sometimes, it may not be possible to ‘obtain the meridian altitude of a celestial body (especially the sun) ‘owing to clouds, hazy horizon, rain or other causes, In such cases, it may be possible, under certain conditions (given below), to ‘observe the body ‘near the meridian’ and, by use of Nautical “Tables, calculate the latitude where the PL (or LOP) euts the DR longitude (Observed latitude or Obs lat). The obs lat is not necessarily the latitude of the ship. The direction of the PL (or LOP) is obtained in the usual manner - calculating azimuth by ABC Tables and applying + 90° to it. “The circumstances under which a sight may be worked by the ex-meridian method is governed by the values and names of the latitude and the declination. The value of the cx-meridian limit (EML) is obtained from Nautical Tables. In the case of Burton's ‘Tables, EML is obtained by entering Ex-meridian Table IV. In the case of Norie’s Tables, we have to consult Ex-meridian Tables I & TV to obtain EML. Ex-meridian Table [is in two parts ~ one for lat & dec having same name and the other for lat & dec having ‘opposite names. ‘The EML so obtained will be in minutes of time. If the EML is found to be 16 minutes, it means that an ex-meridian sight may be taken upto 16 minutes on either side of the time of ‘meridian passage, as illustrated below:~ Ship's time of meridian passage 1206 EML (from nautical tables) £16 Excmer sights may be taken betwicen ship’s time:- 1150 and 1222 ‘A rough thumb rule is available to obtain the approximate value of EML without Nautical Tables:- calculate the L~D (Lat and Dec same names subtract and opposite names add), square off to the nearest whole number of degrees and just call that value the EML in minutes of time. 82 [13 - ASTRO PL - THEORY] ‘Example: DR lat 20°16°N dec 20°30'S. Approximate EML = ? L-D=41 whole degrees. Approximate EML = 41 minutes of time For examination purposes, when a body is said to be °Near the meridian’, it is expected to be worked by the ex-meridian method. The solution to an ex-meridian problem should state the Obs lat, the DR long and the direstion of the PL (LOP), Actual working of an ex-meridian problem is shown under the appropriate heading.in subsequent chapters, (5) Polaris:- is a star of very high declination (around 89° 09") and hhence its azimuth is very nearly North at all times, resulting in a PL very nearly E-W. By applying special corrections (given in the ‘Nautical Almanac) to the true altitude, the Obs lat (latitude where the PL cuts the DR longitude) is obtained. The true azimuth also is obtained from the same page of the Nautical Almanac, The direction of the PL is obtained as usual, by applying 90° to the azimuth. The solution to an altitude of Polaris should specify the Obs lat, the DR long and the direction of the PL, similar to the solution of an ex-meridian sight. ‘Note: The Obs lat obtained by an observation of Polaris is not necessarily the latitude of the vessel as the PL is nearly E-W not necessarily exactly E-W. Polaris cannot be seen from the Southem Hemisphere or ‘from very low latitudes in the Northem Hemisphere. In the former ‘case it will be below the horizon and in the latter case, it will be above but too close to the horizon for observation by a sextant. ‘The actual working of an altitude of Polaris is shown ‘under the appropriate heading later on in this book 8(4-SUN} 14 WORKED EXAMPLES AND EXERCISES - SUN 14.1 Latitude by meridian altitude - Sun:- On 23rd Sept 1992, in DR 23° 40°N 161°S6°E, the sextant meridian altitude of the Sun's lower limb was 66° 10.6”. If TE was 2.3” on the are and HE was 10.5m, find the latitude and the PL. From Almanac Sext Alt 66° 10.6" LMT mer pass 23d 11h 52m 00s IE (on) -) 02.3" LITE) 10 47 44 Obs alt 66° 08.3" GMT mer pase 23d 01h Of Is ip (10.5m) )_05.7' Dee 00° 06.1" Appak 66026 dda) GOO Tot corm LL (+) 15.5’ Deo 90° 06,2" S Talt 662 18.1 S > (Named same as azimuth) MZzD 23° 41.9°N_ > (Named opposite to T alt) Dec 00° 06,2" -» (See note | below). Latitude 23°.35.7°N PL: E-W 84 [14- SUN] Notes (1) MZD and dec are of same name so add and retain name. If of opposite names, subtract the smaller one from the larger one and retain the name of the larger one, (2) Corrections to sext alt are explained in Chapters 9 and 10 @) Im the above example, DR tat is 23° 40°N while dec is Practically nil. Hence the sun’s Az at mer pass must be S and so the true altitude is named $ and MZD is named N. This naming system is a quick and sure method of finding out whether MZD and dec are to be added or to be subtracted to obtain the latitude without the necessity of drawing a figure to reason this out. (4) Whether MZD and dec are to added or subtracted can, in many cases, be decided by knowing that the answer must be very close to the DR lat. However, if either MZD or dec is very small, both addition and subtraction would give results reasonably clase to DR lat, as in the foregoing worked ‘example, and the unwary student would be puzzled as to what to do, A mistake of this sort is considered a "Principle Error’ in the examination, resulting in heavy loss of marks. The student is therefore advised to get used to the naming system shown here, (5) In some cases, DR lat is not given at all but then an indication of the Az at mer pass would be given. By using the above ‘naming system, the answer is obtained easily and quickly, (6) Whether “a” correction is to be added or subtracted is known by inspecting the dec for the required hour and the next hour in the almanac. If dee is increasing, d correction is additive and vive versa 85[4- SUN] EXERCISE - 5 (Latitude by meridian altitude SUN) (1) On 21st Jan 1992, in DR 24° 36'S 110° 20°W, the sextant altitude of the Sun’s LL on the meridian was 85° 05.5". If IE was 1.6° off the arc and HE was 10m, find the latitude and state the direction of the PL (LOP). (2) On Ist Sept 1992, DR equator 50° 27°E, the sextant meridian altitude of the Sun’s UL was 82° 10.4". If IE was 2.4” on the are and HE was 17m, required the latitude and the PL (LOP), (3) On Ist May 1992, in DR longitude 179° 58°E, the observed altitude of the Sun’s LL on the meridian was 64° 35.9" South of the observer. If HE was 15m, find the latitude and the PL. (4) On 14th Sept 1992, in DR longitude 116° 27°W, the sextant meridian altitude of the Sun's UL North of the observer was 70° 29.8". If1E was 3.2’ off the arc and HE was 12m, find the latitude and state the direction of the PL (LOP). (5) On Ist Dec 1992, in DR 06° 35’N 64° 18°W, owing to a hazy hhorizon to the South, a back angle observation of the Sun’s LL. fon the meridian was made and the sextant altitude was found to be 118° 11.8”. If HE was 14m, and IE was 2.4” on the arc, roquired the latitude and the direction of the PL . (Note: Back angle altitude has been explained in Chapter 10) 86 [14-SUN] 14.2 Azimuth - Sun:- On 13th Sept 1992, in DR 23° 21°S 47° 18'W, the ‘observed azimuth of the Sun was 046° (C) when the chron showed O1b 08m 10s. If chron error was 02m 12s slow and Var (variation) was 3°W, find the Dev (deviation) of the compass. dhms dhms Chron 01 08 10 or 13-08 10 Error wo 2 woe oT O1 10 22 13 13 10 22 LIT(w) (03 09 12 ©) 03 09 12 LMT 22 01 10 13 10 oF 10 GMT 134 13h 10m 22s GHA 016° 03.4" dec 03° 34.5°N Incr 902° 35.5" 4 (1.0) 00.2" GHA 018° 38.9" dec 03°34.3°N Long(W) (-) 047° 18.0° tat 23°21.0°S LHA 331°20.9° From Nautical Tables TAz 049.8°(T) A 079N CAz 046.08 (C) B O1BN Error 3.8° E © 092N Var _3.0° W Dev 68° E Notes (1) The actual procedure of entering ABC Tables is not explained here as this is done in the Nautical Tables themselves. (2) Whilst using Norie’s Tables: “A” is normally named opposite to lat (i.e., when LHA is between 270° & 090°) and same as Jat in unusual cases (i.c., when LHA is between 090° & 270°), °B’ is named the same as declination. For naming °C”: If A & B are of same name, add and retain name; if of contrary names, subtract and retain name of larger one. The Az is named as follows: The prefix N or S is the name of °C’ 87[14 - SUN] whereas the suffix E or W is depends on the value of LHA. If LIA is between 0° and 180°, the body lies to the West and if it is between 180° and 360°, the body lies to the East. (3) When the latitude is 00° 00.0°, it may be named N or S. (4) The calculation of deviation is a matter of elementary chart- work and is not explained here. However, many students get stumped as to whether variation is to be added or subtracted to ‘compass error to obtain deviation. A simple solution: TAz 049.8°(1) Var = _3.0° W MAz 052.8 (M) CAz 046.0° (C) Dey «68° E ‘Calculation of Azimuth by scientific calculator: A=Tan lat B= Tan dec Tan Az=__1 Tan P Sin P C.. Cos tat If LHA < 180°, P = LHA. If LHA > 180°, P = (360 - LHA). ‘Naming of A, B and Azimuth is as stated before. When P > 90°, the minus sign obtained for the value of A is to be ignored as itis taken care of by changing name of A (sce foregoing note 2). A= Tanlat =Tan 23° 21,0" =0.7901031N TanP Tan 28°39.1° B= Tan dec = Tan 03° 34.3" =0.1301780N SinP Sin 28°39.1° © =0,9202811N Tan Az= __J__-= 11835568 C (Cos 23° 21") [14-SUN] EXERCISE 6 (Azimuth - SUN) (21) On 29th Nov 1992, AM at ship in DR 26° 27°N 130° 27°W, the azimuth of the Sun was 130°(C) when chron showed Sh 49m 20s. If chron error was Olm 31s fast and variation was 3°E, find the deviation for the ship’s head, (2) On 22nd Sept 1992, PM at ship in DR 18° 20°N 85° 40°B, the azimuth of the Sun was 265°(C) when chron showed 10h 03m 20s, If chron error was 06m 18s slow and variation was 2°E, calculate the deviation of the compass. G) On Jan 19th 1992, in DR 40° 16'S 175° 31°F, the azimuth of the Sun was 267°(C) at 1618 ship’s time. If the ship’s time difference was 11h 30m from GMT and variation was 2.3°E, find the deviation for the ship's head (4) On 30th April 1992, in DR 00° 00.0° 60° 12°W, the Sun bore 080°(C) when chron showed 10h 57m 43s, If chron error was (03m 09s slow and variation was 1°W, find the deviation of the ‘compass. (5) On 31st Aug 1992, in DR 10° 11°S 000° 00”, the azimuth of the Sun was 282°(C) when chron showed 05h 10m 25s. If chron error was 00m 05s fast and variation was 3°E, find the ‘deviation for the ship’s head. 143 Amplitude - SUN:~ On Sth March 1992, in DR 32° 12°N 178° 16°E, the rising Sun bore 100°(C), If variation was 3°E, find the deviation of the compass 89[14- SUN] From Almanac d oh m s dec 06° 08.0°S LMT Sunrise 05 06 22 (1.0) () 00.5" ‘LIT (E) (-) 11 53 04 dec 06° 07.5°S GMT Sunrise 04:18 28 56 lat 32° 12.0°N From Nautical Tables, True amplimde = £7.2°S True azimuth 097.2° (1) Compass azimuth = 100,0°(C) ‘Compass error 28° W Variation = 30° E Deviation Ft EW. Notes ()) LMT of sunrise or sunset from the almanac needs only be interpolated and taken in whole minutes of time. (2) When entering Nautical Tables for true amplitude, great care ‘must be taken during interpolation to ensure accurate results. (3) The prefix of true amplitude is E if the body is rising and W if the body is setting. Its suffix is the same as that of declination, (4) Amp £20°N means: Sun’s bearing is 20° to N of E - 070°(T). Similarly, True amplitude E10°S = True bearing of 100° WIS°N = True bearing of 285° W9.5°S = Truc bearing of 260.5° (5) Sometimes the time of observation of the amplitude is given in the problem. Then given time should be used for calculation. It is, however, necessary to remove any ambiguity of chron time as explained in Chapter 12. (© If the latitude is 00° 00°, the value of the amplitude is equal to the value of the declination. Naming of the amplitude is done as usual. For example: If lat is 00° 00° and dec is 05° 11°S, the rising amplitude = B5.2°S ic., 095.2°(T) and the setting amplitude = W5.2°S i.c,, 264.8°(T). (7) Amplitude can also be worked out by the formuta:- oun Sin amp = (Sin dec . Sec lat) 90 [14-SUN] EXERCISE 7 (Amplitude - SUN) (1) On 2nd Sept 1992, in DR 40° 28°N 64° 20°E, the rising Sun bore 090°(C). If variation was 5°W, find the deviation of the ‘compass. (2) On Ist May 1992, in DR 30° 06°N 179° 45°W, the setting Sun bore 285%(C). If variation was 2°W, find the deviation for the ship's head, (3) Jan 20th 1992, in DR 54° 20°S 46° 27°W, the Sun set bearing 234°(C), If variation was 3°W, find the deviation of the ‘compass. (4) Nov 30th 1992, the Sun rose bearing 110°(C) in DR 49° 18°N 110° 12°E at 00h 10m 00s chron time. If chron error was 0Sm 10s slow and variation was 10°E, find the deviation of the ‘compass. (5) Sept 23rd 1992, the Sun rose bearing 094°(C) in DR 00° 00° 120° 27°W when chron showed OLh 54m, If chron error was (03m 12s fast and variation was 2.7°W, find the deviation for the ship’s head. 14.4 Longitude by chronometer - SUN:- On 29th Nov 1992 in DR 26° 27°N 130° 27°W, the sextant altitude of the Sun’s UL East of the meridian was 28° 11.0" when chron (error O1m 31s fast) showed 05h 49m 20s. If HE was 10m and IE was 2.3” off the arc, calculate the direction of the PL and a position through which it passes. a1[14 - SUN] dhms dhms Chron 05 49 20 or 17 49 20 Error 013 (01 3 GMT 05 47 49 29 17-47 49 LIT) 08 41 48 08 41 48 LMT 21-06 01 29 09 06 01 GMT 29d 17h 47m 49s Solution by Nautical Tables: Hay LHA = Sec L.. See D [Hav ZD - Hay (L~ D)] GHA om 52.1" Sextalt 28° 11.0" Incr ous s77° TE (of) = (4) 023° GHA ggs° 49.4" Obs alt 28° 13.3" dec 21° 36.1°S Dip (10m) ()_05.6° a@4) 003" Appalt 28° 07.7 dec 21° 36.4°S Tot Corm UL (-)_17.8° lat 262 27.0°N Tal 2P 499° (L~D) 48° 03.4" Tz, 62° 10)" NatHavZD 0.26656 ‘Log Hav diff 9.00329 Nat Hal ~D) 0.16580 Log SecL 0.04802 Nat Hav diff 0.10076 Log SecD 0.03164 (See note I below to obtain LHA) —> Leg Hav LHA 9.08295 LHA 319° 168° A 0588 GHA 039° 49.4" B Osis Obs ong ‘130° 32.6°;W c 119s DR iat 26°27.0'N Az 136.7°(T)- Az 43.3°E. PL 046.7° - 226,7° through DR lat and Obs long. Answer. [14 -SUN] Notes (1) Before meridian passage, LHA will be between 180° and 360° - obtain LHA from the bottom of the Haversine Table. After meridian passage, LHA will be between 000° and 180° - obtain LHA from the top of the Haversine Table. (2) Correction of altitudes has been explained in Chapters 9 & 10. (3) To obtain (L ~ D), same names subtract, opposite names add. (4) The answer must clearly state the direction of the PL, and that it is to be drawn through DR lat and Obs long, Solution by scientific calculator: Cos P= Sin T altF Sin lat. Sin dec Cos lat . Cos dee Note:- If Lat and dec are same name (-), contrary names (+) ‘Hence in the formula, (-) has been put above the (+) T at=27 499° lat = 26°27.0°N dec = 21° 36.4°S. Cos P = Sin 27° 49.9°(+)Sin 26° 27.0", Sin 21° 36.4’ = 0,757907 Cos 26° 27.0" . Cos 21° 36.4” P= 40° 43.2". Before mer pass so LHA = 360° - P = 319°16.8° GHA =039°49.4" ‘Obs long = 130°32.6°W A= Tan lat B=Tandec Tanaris ‘Tan P Sin P C. Cos lat If LHA < 180°, P = LHA. If LHA > 180°, P = (360 - LHA), ‘Naming of A, B and Azimuth is as stated before. When P > 90°, the minus sign obtained for the value of A is to be ignored as it is taken care of by changing name of A (see note 2, Chap. 14.2).[14-SUN] ‘Tan lat = Tan 26° 27.0" = 057798018 TanP — Tan 40° 43.2" Tan dec = Tan 21° 36.4” = 0.6071199S SinP Sin 40°.43.2° C = 118510008 TanAz= ___1___ = 0.9424646 Ar=S433°E C (Cos 26° 27°) ie, 136.7°(T) PL 046,7° - 226.7° through the DR lat and the Obs long. Answer. EXERCISE 8 (Longitude by chron - SUN) (1) On 31st Aug 1992, PM at ship in DR 10° 11°S 000° 00’, the Sextant altitude of the Sun’s LL was 39° 15° when the chron (crtor 01m 30s fast) showed 03h 11m 20s, If IE was 2.5” on the are and HE was 17m, find the direction of the PL and a position through which to draw it. (2) On 30th April 1992, in DR 00° 20°N 60° 12'W, the sextant altimde of the Sun’s UL East of the meridian was 44° 13.4” when chron (error 03m 09s slow) showed 00h $7m 43s. If TE ‘was 3.1” off the arc and HE was 20m, find the direction of the PL and the longitude where it crosses the DR lat (3) On 19th Jan 1992, at about 1530 at ship in DR 40° 16°S 175° 31’E, the sextant altitude of the Sun's LL was 43° 27.4" when the chron (error 02m 12s fast) showed 03h SOm 12s. If HE 94 [14-SUN] was 22m and IE was 1.5" on the arc, required the direction of the PL and a position through which it passes, (4) On 22nd Sept 1992, PM at ship in DR 48° 20°N. 85° 40°F, the Sextant altitude of the Sun’s UL was 20° 14,8” when the chron (error 06m 18s slow) showed 10h 03m 20s, If IE was 2.2’ on the arc and HE was 25m, find the direction of the PL and a position through which it passes. (5) On Sth March 1992, AM at ship in DR 38° 11°S 151° 10°E, ‘the sextant altitude of the Sun’s LL was 35° 59.1" when the chron showed 10h 54m 54s. The chron was Olm 20s slow at 12h GMT on 25th Feb 1992 and gaining 4s daily. If IE was 13° off the arc and HE was 30m, find the direction of the PL. and a position through which it passes. 14.5 Intercept - SUN:- On 29th Nov 1992 in DR 26° 27'N 130° 27°W, the sextant altitude of the Sun's UL East of the meridian was 28° 11’, ‘when the chron (error 01m 31s fast) showed 0Sh 49m 20s. If HE was 10m and IE was 2.3” off the arc, calculate the direction of the PPL and a position through which it passes. dhms dhms ‘Chron 05 49 20 or 1749 20 Error (-) 01 3! () 01 31 GMT 05 47 49 29 1747 49 LIT) 08 41 48 08 41 48 LMT 21 06 OL 29 09 06 O1 GMT 29d 17h 47m 49s 95:(14 -SUN] ([14- SUN} Solution by Nautical Tables: Solution by scientific caleulator Hay CZD = (Hav LHA . Cos L Cos D) + Hav (L~ D) Cos CZD = Cos P . Cos lat . Cos dec + Sin lat. Sin dee GHA 77? 52.1" Sext alt 28° 11.0° ; Note:- If Lat and dec are same name (+), contrary names (-) Incr oe 57.3 TE(f) = (+)_02. Hence in the formula, the (+) has been put above the (-). GHA 089° 49.4" Obsatt 28° 13.3" ‘The signs are opposite to that for long by chron formula, Long W 30° 27.0" Dip (10m) 05.6" LHA 319° 22.4" Appalt. 26° 017" IELHA lies between 0° and 180°, P = LHA dec 21° 36.1°S Tot Corm UL(-)_17.8° If LHA is between 180° & 360°, P = 360-LHA doa) @) 003° Tak 2 49.9 | . dec : 21° 36.4’S TzpD 62° 10.1" P = 40° 37.6" lat = 26° 27.0°N dec = 21° 36.4°S lat 26227.0°N Hay LHA 9.08104 @~D) 48°03.4 iesn 9.95198 Cos CZD = Cos 40° 37.6". Cos 26° 27.0" . Cos 21° 36.4 LogCosD 9.96836 ©) Sin 26° 27.0" . Sin 21° 36.4” Log Hav sum 9.00138 = 04677575 CZD =62°06.7° ‘Nat Hav Sum 0.10032 TZD = 62° id ‘Nat Hav(L ~ D)0.16580_ ' Int (AWAY) = 3.4" NatHav CZD 0.26612 1, czD 62° 06.7" AMS Daa bt: B= Tan dec Tan Az=__1 Mey Ria 72D sao 1 TanP Sin P © Gos at Int (AWAY) 034 q _ i. ens % DRiat 26° 27.0°N A= Tan lat = Tan 26° 27.0° = 0.5798880 S ? DRiong 130° 27.0°W TanP = Tan 40° 37.6" AOS8S AzS$43.3°E & Shas. inert B= Tandec=Tan21°364° = 060827198 C1198 PL 046.7- 226.7 SinP Sin 40° 37.6" C = 118815998 Notes TanAz=___1__—- = 0.9400374 =$432°E (1) A rough sketch (not necessarily to scale) should be drawn to f C (Cos 26°27') ie, 136.8°(T) show that the student knows where the PL is to be drawn. (2) Calculation of the latitude and longitude of the ITP (intercept DR lat 26° 27.0°N DR long 130° 27.0°W Intercept 3.4” AWAY ‘terminal point) is not necessary in such a problem. Azimuth 136.8°(T) PL 046.8° - 226.8° Answer. | 96 ”7[14 SUN) EXERCISE 9 (Intercept - SUN) (1) On Sth March 1992, AM at ship ia DR 38° 11’S 151° 10°E, the sextant altitude of the Sun's LL was 35° 59.1’ when the chron showed 10h $4m 54s. The chron was 01m 20s slow at 12h GMT on 25th Feb 1992 and gaining 4s daily. If IE was 1.3° off the are and HE was 30m, find the direction of the PL. and a position through which it passes. (2) On 22nd Sept 1992, PM at ship in DR 48° 20°N 85° 40°E, the Sextant altitude of the Sun’s UL was 20° 14.8" when the chron (error 06m 18s slow) showed 10h 03m 20s. If IE was 2.2” on the are and HE was 25m, find the direction of the PL and a position through which it passes. (3) On 19th Jan 1992, at about 1530 at ship in DR 40° 16°S 175° 31°E, the sextant altitude of the Sun's LL was 43° 27.4” when the chron (error 02m 12s fast) showed 03h 50m 12s. If HE ‘was 22m and IE was 1.5” on the arc, required the direction of the PL and a position through which it passes. (4) On 30th April 1992, in DR 00° 20°N 60° 12°W, the sextant altitude of the Sun’s UL Bast of the meridian was 44° 13.4” ‘when chron (error 03m 09s slow) showed 00h $7m 43s. If TE ‘was 3.1” off the arc and HE was 20m, find the intercept and the direction of the PL. (5) On 3ist Aug 1992, PM at ship in DR 10° 11°S_ 000° 00°, the sextant altitude of the Sun's LL was 39° 15° when the chron (error 0m 30s fast) showed 03h’ 11m 20s, If IE was 2.5° on the arc and HE was 17m, find the direction of the PL and a position through which to draw it 8 [14-SUN] 14,6 Ex-meridian - SUN:~ ‘On 4th March 1992, DR 27° 18°N 168° 11'W, the sextant altitude of the Sun's LLL near the meridian was 56° 19.8" when the chron showed 11h 13m 24s, If chron error was 01m 20s slow, HE ‘was 12m and IE was 2.8" on the are, tind the direction of the PL. and a position through which it passes. Note: Near the meridian ‘means that working is to be by the ex-meridian method, dhms dhoms Chron 11 13 24 or 23:13 24 Enoor (01 20 ) 01 20 mT 1 14 44 04 23:14 44 ur) =). 2a i244 LMT 00 02 00 04-12 02 00 GMT 044 23h 14m 44s Solution by Nautical Tables: GHA 162° 06.4" Sext alt 56° 19.8" Incr 003° 410° IE(on) —@)_02.8" GHA 165° 47.4" Obs alt 56° 17.0" Long W 168°_11.0° Dip (12m) (-)_06.1" LHA 357° 36.4" Appalt 56° 10.9" doe 06° 03.2°S ‘Tot Corm LL (+) 15.6" z Talt dec TZ Reduction @)_04.9"* A 12.408 MZD 33° 28.6'N B02538 dec 06° 03.0'S. 14.93 Obs lat 27° 25.6'N AzS4.3°E DR long — 168° 11.0°W ie, 175.7° (1) PL 085,7°- 265.7° 99.[14- SUN] * Working steps to obtain Reduction from Nautical Tables: From Table I - value F (Burton's) or value A (Norie’s) = 3.20 From Table I, First Correction = 4.9" From Table Ill, Second Correction = (-) 0,0" Reduction tothe TZD observed = 4,9" Notes: (1) In this problem, no hint of ship's time is given but it is noticed the possibilities of LMT are 00h or 12h. Since mer pass of ‘Sun must be close to 12h only, 00h is impossible. Q) By entering ex-meridian table I with lat and dec, we get A (Norie’s Tables) or F (Burton’s Tables). There are separate tables for lat & dec having same name and contrary names. (3) By entering, ex-meridian table II with LHA and A (or F), we ‘get the First Correction which is in minutes of arc. (4) By entering ex-meridian Table III with first correction and true alt, we get the Second Correction. (8) Second Correction is always subtracted from First Correction to get the Reduetion (or R) in minutes of arc. (©) The value of R is to be subtracted from TZD to get MZD. (7) Some books say “Add reduction to true altitude to get meridian alt. Then subtract 90° to get MZD’. Though this is mathematically correct, this method cannot be practically reasoned and explained to the student. Reduction should reduce something, not increase. Also, the student can easily remember that, except for Polaris, sextant altitude should be converted into TZD in all problems in Practical Navigation. (8) Ex-meridian table IV is not actually used in working the problem. It is used only to ascertain the ex-meridian limits (EML) as explained in Chapter 13 (4). (9) Whether MZD and dec are to be added or subtracted, to obtain Obs lat, is decided by naming the true alt and MZD as explained earlier in this Chapter under “Lat by mer alt - Sun’ (10) MZD can also be obtained by the Haversine formula:- Hay MZD = Hay TZD - (Hav LHA . Cos L . Cos D). 100 [14-SUN] ‘Solution by scientific caleulator:- Cos MZD = Cos TZD + {(1 - Cos P) . Cos DR lat . Cos dec} Note; TZD 33° 33.5’, P 2° 23.6", DR lat 27° 18°N, dec 06° 03'S = Cos 33°33.5’ + [(I- Cos 2°23.6") . Cos 27°18" . Cos 06°03" = 0.8340942. Hence MZD, by scientific calculator, = 33° 28.7" MZD = 33°28.7°N A=Tanlat = 12,34903408 dec 06° 03.0" S Tan P Obs tnt = 27° 25.77 N B= Tandec = 02,5380303 § DR long = 168° 11° W SieR ow OT C = 14.8870640 Tan Az= 1. = 0.075592. Az=$04.3°E = 175.7° (1) ©. Cos lat PL = 085.7° ~ 265.7° -000- 101 ausnetesneanaseasassererrarreeemees[14- SUN] Exercise 10 (Ex-meridian altitude - SUN) (1) On Ist Sept 1992, in DR 23° 18'N 165° 02°E, the sextant altitude of the Sun’s UL near the meridian was 75° 01.7’ when ‘chron (error 03m 2s slow) showed 00h 45m Sis. If IE was 3.2’ off the arc and HE was 20m, find the direction of the PL. and a position through which it passes. (2) On 2nd May 1992, in DR 15° 36'S 080° 11°W, the sextant altitude of the Sun’s LL near the meridian was 58° 25.6” when chron (error 02m 18s fast) showed 05h 40m 06s, If IE was 1.6° on the arc and HE was 15m, find the direction of the PL and the lat where it cuts the DR long, (3) On 6th March 1992, in EP 52° 12’N 170° 40°E, the sextant altitude of the Sun's UL near the meridian was 31° $9.8” when chron (error O1m 50s fast) showed 01h 29m 20s. If TE was 2.3” on the arc and HE was 40m, find the direction of the PL. and a position through which to draw it. (4) On 2ist Jan 1992, in DR 00° 00° 97° 48°W, the sextant altitude of the Sun’s LL near the meridian was 69° 28.7° when chron (error 12m 05s slow) showed 06h 13m 27s. If IE was 2.0" off the arc and HE was 12m, find the direction of the PL ‘and a position through which it passes. (5) On 23rd Sept 1992, in EP 48° 20°S 158° 46°E, the sextant altitude of the Sun's LL near the meridian was 41° 36.7" at Oth [1m 16s chron time. If chron error was 11m 04s fast, TE ‘was 3.4” on the are and HE was 17m, find the direction of the PL and the lat where it cuts the DR long, 102 [15 - STARS] 15 WORKED EXAMPLES AND EXERCISES - STARS 15.1 Latitude by meridian altitude - STAR ‘On Ist Dec 1992, AM at ship in DR 45° 20°S 75°00'E, the sextant meridian altitude of the star PROCYON was 39° 28.8, IfTE was 1.5” off the are and HE was 25m, find the latitude and the PL ‘Sext Alt 39° 28.8" IE (off) OLS Obs alt 39° 303° Dip (25m) —(-)_08.8° App alt 39° 21.5" ‘Tot corm (-)_ 01.2" Talt 39°_20,3"N + (Named same as azimuth) MzD 50° 39.7°S ~» (Named opposite to T alt) Dec 05° _14.6°N — (Contrary names subtract) Latitude: 45° 25.8 PL:- E-W 103,{15 - STARS] Sometimes, the question also says “State the GMT of ‘meridian passage’. In such a case, procced as follows: GHA & +E (or - W) long = LHA * Note GHA * = GHA y + SHA *, At mer pass %, LHAx = 360° SHA + is obtained from that day’s page in the almanac. Hence at mer pass, GHA y + SHA % + E (or - W) long = 360° GHA y + 245° 15.5" + 75° 00'=360° or GHAy=039°44.5° ‘From almanac, for GMT December 01d 2h GHA y= 025°57.9° ‘Sm 57s < Increment < 013°46.6" Note: In referring to the almanac, it has been assumod that the date of GMT would be the same as that of ship ic., Ist December. However, it may be possible that the date of GMT may be one day earlier or later than that assumed. It is therefore necessary to check up by applying LIT to the assumed GMT, as shown below, and alterations made, if necessary, to the GMT date and time, OMT mer pass: Dec Old 21h S4m 57s LITE) G)_05 0000 LMT 02. 02 54 57 But reqd date is Ist Dec Q)_23_56 _04 (one sidereal day)* Correct LMT O1 0258 53 LIT (E) (05 00 00 Correct GMT Nov 30 21 58 53 Answer. * The time interval between any two successive meridian passages ‘of 7 is one sidereal day which is equal to 23h 56m 04s. Hence cone sidereal day may be added ot subtracted as appropriate to ‘ensure that the meridian passage is correct for ship's date. 104 [15 - STARS] EXERCISE - 11 (Latitude by meridian altitude - STARS) (1) On 4th March 1992, in DR 45° 10°N 120° 30°W, the sextant meridian altitude of the star ANTARES was 18° 26.2’. If IE was 3.2’ off the arc and HE was 10m, find the latitude and state the direction of the PL (LOP). (2) On 12th Sept 1992, DR 43° 05'S 72° 20°E, the sextant meridian altitude of the star ALDEBARAN was 30° 40.2’. If TE was nil and HE was 18m, required the latitude and PL (LOP) and state the GMT of meridian passage. @) On 22nd Sept 1992, in DR longitude 90° 06°E, the observed altitude of the star RIGEL on the meridian was 73° 242° ‘North of the observer. If HE was 15m, find the latitude and the PL. State the GMT of meridian passage. (4) On 20th Jaf 1992, in DR 57° 31°N 164° 20°W, the sextant meridian altitude of the star DIPHDA was 14° 33.7". If IE ‘was 0,6" on the arc and HE was 14.5m, find the latitude and the direction of the PL. State the GMT of meridian passage (5) On Ist May 1992, in DR 30° 18°N 135° 02°W, a back angle sextant meridian altitude of the star REGULUS was 108° 165°. IF HE was 14m, and IE was 2.4” on the arc, required the GMT of mer pass, the latitude and the direction of the PL. 15.2 Azimuth - STAR:- On 21st Sept 1992, PM at ship in DR 43° 18°S 140° 11, the star CANOPUS bore 150° (C) at 07h 281i 52s chron time (error 02m 12s slow). If Var was 3°E, find dev of compass los[15 - STARS} dh ms d bms Chron, 07 28 52 or 19 28 $2 Error 02 12 +) 02 12 GMT 22 07 31 04 19 31 04 LIT(W) ~—)_09 20 44 (09 20 44 LMT 21 2 10 20 10 10 20 GMT 224 07h 31m O45 GHAy 106° 23.7" From Nautical Tables Incr 007°473" A 0588 GHAY neo B 1558 Long(W) (-) 1409110" C 2.138 ——_Altemative LHAy 334°00.0° TAz 147.2°(T) TAz 147.2°(1) SHA* 264903." CAz_150.0°(C) Var__3.0° E LHA*& 238° 03.1" Error 2.8° W MAz1442° M dec 52°41.2°S Var __3.0° E CAz 150.0°(C) lat 43° 18.0°S Dev 58° W Dev 5.8° W ‘Note: Working is very similar to that of azimuth - Sun except that GHA y and SHA have been used (see Chapters 2 & 4.6). ‘When looking up GHA y and its increment care must be taken to consult the y columns and not those of the Sun. ‘Calculation of Azimuth by scientific calculator: A=Tanat B= Tan dec ‘Tan P ‘Sin P Tan Az=__1 C. Cos lat If LHA < 180°, P = LHA. If LHA > 180°, P = (360 - LHA). Naming of A, B and Azimuth is as stated under Az Sun (Chapter 14.2), When P > 90°, the minus sign obtained for the value of A is to be ignored as it is taken care of by changing name of A (sce note 2 under Chapter 14.2). 106 [15 - STARS] = Tanlat =Tan 43° 18.0' = 0.587661 S TanP Tan 121° 56.9° B= Tandec = Tan 52° 41,2" = 1,5462752S SinP Sin 121° 56.9" © =2.1339413 Tan Az= 1 =0,6439049 Ar=S32.8°E C (Cos 43° 18") ie, 147.2°(1) EXERCISE 12 (Azimuth - STARS) (1) On 6th March 1992, AM at ship in DR 24° 12°S 83° 46°E, the azimuth of the star ALTAIR was 078%C) at 10h 38m 40s cchron time. If chron error was 03m 24s fast and variation was 3°W, find the deviation for the ship's head. (2) On 30th Nov 1992, PM at ship in DR 48° 57°N 173° 18°W, the azimuth of the star VEGA was 296°(C) at 07h 27m 12s chron time. If chron error was 12m 10s slow and variation was 3°W, calculate the deviation of the compass. (3) On Ist May 1992, AM at ship in DR 62° 11°S 179° 58°E, the azimuth of the star SPICA was 312(C) at 01h 03m 16s chron time (error 03m O8s fast). If variation was 10°E, find the deviation for the ship’s head. (4) On 13th Sept 1992, in DR 30° 46°N 90° 36°W, the star RASALHAGUE bore 275%(C) at 04h 36m GMT. If variation ‘was 5°W, find the deviation of the compass. (Caution: Date given is ship's date not GMT date) 107[15 - STARS] (5) On 2ist Jan 1992, at 0320 ship's time, in DR 64° 12'N 112° 18°E, the star DENEB bore 034.5°(C). If variation was 4°E, ‘and ship's time was 7h ahead of GMT, find the deviation. 15,3 Longitude by chronometer - STAR:- On 23rd Aug 1992 PM at ship in DR 34° 31°S 03° 30°W, the sextant altitude of the star SPICA was 45° 27.2’ when chron (error 02m 19s slow) showed 06h 15m 00s. If HE was 11m and IE was 2.1” on the arc, calculate the direction of the PL and a position through which it passes. dh ms d hms ‘Chron 06 15 00 or 18 15 00 Error G02 19 G02 19 GMT 06 17 19 23 18 17 19 LIT(w) G00 40 00 21.00 4000 LMT 05.37 19 23.17 37.19 GMT 23d 18h 17m 19s Solution by Nautical Tables: Hay LHA = Sec L. Sec D {Hay ZD - Hav (L ~D)} GHAy 242° 16.6" Sextalt 45°272° Incr 904° 20.5" TE (on) O01 GHAY 246° 37.1" Obs alt — 45° 25.1" SHA * 58°.47.8° Dip (11m) ) 05.8" GHA* 045° 24.9" Appa 45° 19.3" dec 11°07.4°S Tot Corm — (-) 01.0" lat 34°31.0°S Talt 45° 18.3" @-D) — -23°23.6 Tz 44° 41.7" 108 [IS - STARS] NatHavZD 0.14457 Log Hav diff 9.01479 ‘Nat Hav(L~D) 0.04110 Log Sec 0.08409 Nat Hav diff 0.10347 Log SecD 0.00824 (See note 1 below to obtain LHA)—> Log Hav LHA 9.10712 LHA* 41° 55.4” A 0.76N GHA * 045° 24.9" B 0298 Obs long 003° 29.5°W © 047N DR lat 34°31.0°S Az 291.2° (T}<-AzN 68.8° W PL 021.2° -201,2° through DR lat and Obs long. Answer. Notes (Q) When converting Log Hav LHA into calculated LHA * a small doubt arises whether LHA should be a small number or a large one i.¢., 041° or 319° in this problem. This doubt is removed by applying DR long to GHA % and getting approximate LHA 5. The calculated LHA will be very close to the approximate LHA. (2) Working is similar to long by chron Sun except that GHA. and SHA % are to be used. G) For entering ABC Tables, use the caleulated LHA. (4) The answer must clearly state the direction of the PL. and that itis tobe drawn through DR lat and Obs long, Solution by scientific caleulator: Cos P = Sin T alt Sin at. Sin dec Cos lat . Cos dee = If Lat and dec are same name (-), contrary names (+). Hence in the formula, (-) has been put above the (+). Entire numerator is Sin and entire denominator is Cos. 109[1S - STARS] T alt=45° 18,3" lat= 34°31.0°S — dec= 11° 07.48 Cos P = Sin 45° 18,3"()Sin 34° 31,0" Sin 11° 07,4" = 0.744039 Cos 34° 31.0" . Cos 11° 07.4" P= 41° 55.4" (Sce foregoing note no:1) LEA = P = 041°55.4° GHA = 045°24,9° Obs long = 003°29.5°W A= Tan lat B= Tan dee Tan Az=—_l___ TanP Sin P Cos lat If LHA < 180°, P = LHA. If LHA > 180°, P = (360 - LHA) ‘Naming of A, B and Azimuth is as stated before. When P > 90°, ‘the minus sign obtained for the value of A is to be ignored as itis ‘taken care of by changing name of A (see note 2, Chap. 14.2). A= Tanlat =Tan34°31,0° =0.7658369N TanP = Tan 41° 55.4" B= Tandec=Tan 11°074° =0.2942743 S SinP Ps © =0.4715626N Tan Az= = 2,5736755 Az=N68.8°W C (Cos 34° 31°) , 291.2°(T) M0 [15 - STARS} EXERCISE 13 (Longitude by chron - STAR) (1) On 29th Noy 1992, AM at ship in DR 25° 30°S_ 107° 20°W, the sextant altitude of the star RIGEL was 35° 10.3° when the chron (error 02m 50s fast) showed I1h 32m 10s. If TE was 2.8° on the are and HE was 12m, find the direction of the PL. and a position through which to draw it. (2) On 22nd Sept 1992, PM at ship in DR 60° 10°N 92° 27°E, the sextant altitude of the star ARCTURUS was 25° 01° when chron (error 05m 01s slow) showed 00h 46m 31s. If LE was 0.2" on the are and HE was 17m, find the direction of the PL and the longitude where it crosses the DR lat. (3) On 19th Jan 1992, at about 1900 at ship in DR 00° 02°N 170° 50°E, the sextant altitude of the star BETELGEUSE was 43° 11.1" when the chron (error O1m 18s fast) showed 07h 35m 02s, If HE was 18m and IE was 1.3’ off the arc, required the direction of the PL and a position through which it passes. (4) On 30th April 1992, PM at ship in DR 34? 18°S 40° 20°W, the observed altitude of the star SIRIUS was 57° 50.7” at 08h Sm 385 chron time. The chron error was 00m 10s slow at 12 GMT on 22nd April and its daily rate was 3s gaining, If HE ‘was 21m, find the direction of the PL and a position through ‘which it passes. (5) On 31st Aug 1992, AM at ship in DR 40° 30°N 64° 56°E, the sextant altitude of the star DIPHDA was 21° 23.4” when the chron (error nil) showed 00h 20m 26s. If IE was 0.9° off the are and HE was 9m, find the direction of the PL and a position through which it passes. a[15 - STARS] 15.4 Intercept - STAR:- On 23rd Aug 1992 in DR 34° 31°S 003° 30°W, at about 1800 hours at ship, the sextant altitude of the star SPICA was 45° 27.2’, when the chron (error 02m 19s slow) showed 06h 15m 00s, If HE was 11m and IE was 2.1" on the arc, calculate the direction ‘of the PL and a position through which it passes. dhms dhms Chron, 06 15 00 or 1815 00 Error (+) 02 19 (+) 02 19 GMT 06 17 19 23:18 17:19 LT) @_00 14 00 00 14 00 LMT 06 03:19 23 18 03 19 GMT. 23d 18h 17m 195 ‘Solution by Nautical Tables: Hav CZD = (Hav LHA . Cos L . Cos D) +Hav (L ~D) GHAY 242° 16.6" ‘Sext alt 45° 27.2" Incr 03° 205° E@ or GHA y (246° 37,1" Obs alt 45° 251° Long W 003° 30,0°W Dip (iim) ()_05.8° LHAY 243° 07.1" App alt, 45° 19.3" SHA* [58° 47,8" Tot Corm (-) O10" LHA* 041° 54.9" Talt 45° 18.3" TZD 44° 41.7" Log Hav LHA 9.10698 dee 11° 07.4'S Log Cos L’ 9.91591 {at 342 310s LogCosD 9.99176 «-D) 2 Be Log Hav sum 9.01465 [15 - STARS) Nat Hav Sum 0,10343 as A ‘Nat Hav(L ~ D)0.04110 DR lat Nat Hav CZD 0.14453 © czD 44°.41,3° wD ar 417° £ at Int (AWAY) 00.4" | DRiat 34°31.0°S DRiong 03° 30.0°W A0.76N B IS AN 688° C047 ie, 291.2° (1) “ M PL 021.2°- 201.2° (1) Notes (1) A rough sketch (not necessarily to scale) should be drawn to show that the student knows where the PL is to be drawn. (2) Calculation of the latitude and longitude of the ITP intercept ‘terminal point is not necessary in such a problem ‘Solution by scientific calculator Cos CZD = Cos P . Cos lat . Cos dec + Sin lat . Sin dec Note:~ If Lat and dec are same name (+), contrary names (-) Hence in the formula, the (+) has been put above the (-) ‘The signs are opposite to that for long by chron formula. TPLHA < 180°, P= LHA & if > 180°, P = (360 - LHA). P =41°549° lat= 34°31.0°S dec = 11°07.4°S Cos CZD = Cos 41° 54.9", Cos 34° 31.0". Cos 1 (¥) Sin 34° 31.0". Sin 11° 07.4° = 0.7109397 CzD = 44° 41.3" TZD = 44° 41.7" Int (AWAY) = 04° ona 113[15 - STARS] A=Tantat B= Tan dec TanAz=__] Tan P Sin P € Cos lat = Tanlat = Tan 34°31,0° = 0.7660610N TanP Tan 41° 54.9" B= Tandec=Tan 11°07.4" = 0,2943219S SinP Sin 41°54.9° © = 04717391 N ‘Tan Az = 1 -=2:5727125 Az=N68.8°W © (Cos 34° 31°) ie, 291.2° (1) DR lat 34° 31.0°S DR long 003° 30.0°W Intercept 0.4° AWAY ‘Azimuth 291.2°CT) PL 021.2° - 201.2° (T) Answer. EXERCISE 14 (Intercept - STARS) (1) On 30th April 1992, PM at ship in DR 34° 18°S 40° 20°W, the observed altitude of the star SIRIUS was 57° 50.7" at 08h ‘52m 05s chron time, The chron error was Olm 40s fast at 06 GMT on 16th April and its daily rate was 4s losing. If HE was 21m, find the direction of the PL. and a position through which it passes. (2) On 19th Jan 1992, at about 1900 at ship in DR 00° 02'N 170° '50°E, the sextant altitude of the star BETELGEUSE was 43° 11.1” when the chron (error 01m 185 fast) showed 07h 35m 02s, If HE was 18m and IE was 1.3° off the arc, required the direction of the PL and a position through which it passes. ua [15 - STARS} (3) On 29th Nov 1992, AM at ship in DR 25° 30°S 107° 20°W, the sextant altitude of the star RIGEL was 35° 10.3” when the chron (error 02m 50s fast) showed 11h 32m 10s. If IE was 2.8" on the arc and HE was 12m, find the direction of the PL. and a position through which to draw it (4) On 31st Aug 1992, AM at ship in DR 40° 30°N 64° S6°E, the sextant altitude of the star DIPHDA was 21° 23.4” when the cchron (error 01m 06s fast) showed 00h 21m 32s, If IE was 0.9" off the arc and HE was 9m, find the direction of the PL and a position through which it passes. (5) On 22nd Sept 1992, PM at ship in DR 60° 10°N 92° 27°, the sextant altitude of the star ARCTURUS was 25° 01° when chron (error 05m O1s slow) showed 00h 46m 31s. IF TE was 0.2° on the are and HE was 17m, find the direction of the PL and the longitude where it crosses the DR lat. 15.5 Ex-meridian - STAR:- On 2nd March 1992, PM at ship in DR 16° 12°N 92° 10°E, the sextant altitude of the star CAPELLA near the meridian was 60° 29.4° at 00h 30m 12s chron time (error 02m 06s slow). If HE was 48m and IF was 2.0" off the arc, find the direction of the PL and a position through which it passes. Note: Near the ‘meridian means that working is to be by the ex-meridian method. dhms dh ms Chron 00 30 12 or 12 30 12 Error +) 02 06 (4) 02 06 GMT 00 32-18 02 12 32 18 LIT(E) (4). 06 08 40 (+) 06 08 40 LMT 06 40 58 02-18 40 58 GMT 02d 12h 32m 18s us[15 - STARS) Solution by Nautical Tables: GHAy 340° 31.7" Sextalt 60° 29.4 Incr 008° 05.8" IE (off) 4)02.0° GHAy 348° 375° Obsalt —60°31.4" LongE 092° 10.0° Dip (48m) 122° LHAy 080°. 475° App alt 60° 19.2" SHA* 280° 57.9" Tot Corm —) 00.6” LHA*® 01" 45.4" Talt ‘60° 18.6" N dee 45° 59.6°N 1zD 2414S Reduction ()_2.1 409.568 MzD 29° 39.3°S B33.77N dec 43° 59.6 N C2421N Obs lat 16°20.3°N AzN2.5°W DRiong 92° 10.0°F ie, 357.5° (1) PL (087.5°- 267.5° ‘* Working steps to obtain Reduction from Nautical Tables:~ From Table I - value F (Burton's) or value A (Norie’s) = 2.63 From Table Il, First Correction = 2.1" ‘From Table Ill, Second Correction = (-) 0.0" ‘Reduction to the TZD observed = = 2.1" Notes (1) Working is similar to Ex-meridian altitude Sun (Chapter 14.6) ‘except that GHA y and SHA » have to be used, (2) Ex-meridian star ean also be worked by the Haversine formula ‘as mentioned in Chapter 14.6 (note: 10). 116 [15 - STARS) Solution by scientific calculator:- Cos MZD = Cos TZD + {(1 - Cos P) . Cos DR lat . Cos dec} Note: TZD 29° 41.4", P 1° 45.4’, lat 16° 12°N, dec 45° 59.6°N = Cos 29°41.4” + [(1-Cos 1°45.4°) Cos 16°12" Cos 45°59 6") = 0,8690315. Hence MZD, by scientific calculator, = 29° 39.2° MzD 29° 30.2°S A=Tanlat = 09,5593685 $ dec 45°59.6°N Tan P Obs lat = 16° 20.4°N B=Tandec = 33.7724840N DR long = 92° 10.0" E Sin P we C = 24.2131155N Tan Az= = 0.043038 Az=NO2S°W = 357,5°(T) C.. Cos lat PL = 087.5° - 267,5° Exercise 15 (Ex-meridian altitude - STARS) (1) On 12th Sept 1992, AM at ship in DR 00° 30°S 160° 20°W, the sextant altitude of the star ALDEBARAN near the meridian was 73° 09.5” at 03h 59m 29s chron time (error 0Sm 03s fast). If IE was 1.2” on the arc and HE was 9m, find the direction of the PL and a position through which it passes, (2) On 2nd May 1992, PM at ship in DR 44° 11'S 102° 40°E, the sextant altitude of the star POLLUX near the meridian was 17° 14.6” at 10h 52m 08s chron time (error 02m 12s slow). If TE was 3.6° off the arc and HE was 12m, find the direction of ‘the PL and the latitude where it cuts the DR longitude, 117[15 - STARS] (3) On 20th Jan 1992, during morning twilight in DR 44° 07°N 64° 47°E, the sextant altitude of the star ARCTURUS near the meridian was 64° 58.8” at 02h 04m 54s by chron (error Lm 58s slow). IE 3.1" off the are; HE 18m, Find the direction of the PL and a position through which to draw it. (4) On Sth March 1992, at about 1835 hours at ship in DR 66° 40N 86° SO°W, the sextant altitude of the star BETELGEUSE near the meridian was 30° 35.2’ when chron (error 18m 04s fast) showed 00h 40m 30s. If IE was 0.7” off the are and HE was 14m, find the direction of the PL and a position through which it passes. (5) On 2st Sept 1992, AM at ship in DR 20° 50°S 62° 30°E, the observed altitude of the star CAPELLA near the meridian was 23° 07.1" at O1h 15m 06s chron time. If chron error was 00m 04s slow and HE was 10m, find the direction of the PL and the Jat where it cuts the DR long. 15,6 Observation of Polaris:~ ‘On Ist Sept 1992, AM at ship in DR 18° 00°N 178° 11”E, the sextant altitude of the Pole star was 18° 47.4” at 0Sh 21m 08s by chronometer (error 01m 18s slow). HE = 12.5m. IE = 1.6" on the are. Find the direction of the PL and a position through which to draw it, If the azimuth was 001° (C), and var was 1.3° E, find the deviation. dh ms dhms Chron 05 21 08 on 17:21 08 Enor 10118 (+) O1 18 GMT 05 22 26 ‘Aug31_17 22 26 LIT) G41 52 44 G4 32 44 LMT 17 15 10 Sep 01 05 15 10 GMT 31d 17h 22m 26s 18 [15 STARS] Sextalt 18° 47.4" GHAy 235° 073° TE (on) Q_oL6" Incr 005° 37,4" Obs alt 18% 45.8" GHAy 240° 44.7 Dip (12.5m) (2) 06.2" LongE 178° 11.0° Appalt 18° 39.6" LHAY — 058° 55.7° Tot Corm —(-)_02.8" Tal 18° 368° True Az 359.7° (1) 20 00° 16.5" Comp Az 001.0° (C) at 00° 00.6" Comp Err 1.3°(W) a 00° _00.3° Variation 13° (E) ‘Sum 18° 54.2" Deviation 2.6° (W) Ci Raaae Obslat 17° $4.2'N DRlong 178° 11.0'E PL 089.79. 269.7° NOTES (1) Values of ao, at and a2 are obtained from the Pole Star Tables siven in the Nautical Almanac. Q) The latitude obtained by an observation of Polaris will be the correct latitude ONLY if the PL is exactly E - W. If not, the fatitude obtained will be the latitude where the PL. cuts the DR Jongitude, as in the foregoing worked example, (3) Sometimes, the DR lat is not given in the question, For the purpose of obtaining ai correction, the true altitude may be ‘used, in place of DR latitude, to enter the table. (@) The latitude obtained by an observation of Polaris must necessarily be N because the Pole Star is not visible from the ‘Southern Hemisphere. -000- 119[15 - STARS] Exercise 16 (Altitude and Azimuth - POLARIS) (1) On the morning of Ist Dec 1992, in DR 47° 16°N 143° 26°E, the sextant altitude of the Polestar was 46° 50.7" at 08h S1m 15s chron time (error 05m Is slow). If TE was 2.1" offthe are and HE was 17m, find the direction of the PL and a position ‘through which it passes. (2) In the evening of 22nd Sept 1992, a ship in DR long 160° 12°W, found the sextant altitude of Polaris to be 36° 18.6” at 05h 23m 175 chron time (error 02m 09s fast). If TE was 2.8” on the arc and HE was 10m, find the direction of the PL and the latitude where it cuts the DR longitude. (3) At about 0330 ship's time on Ist May 1992, in DR longitude 150°E, the observed altitude of Polaris was 50° 46.8" bearing 005° (C) at OSh 30m 30s by chron (error nil). HE 14m var 1°E, Find the deviation of the compass, the direction of the PL and a position through which to draw it. (4) On 6th March 1992, at 0200 ship's time in DR 20° 37°N 00° 00, the Pole star bore 356° (C). If variation was 3.7°W, find the deviation of the compass. (5) On 13th Sept 1992, PM at ship in DR 37° 26°N 72° 46°E, the Pole star bore 350° (C) at 03h 59m 03s by chron (error 10m 03s slow). If variation was 10°B, find the deviation for the ship’s head. 120 [16 - PLANETS} 16 WORKED EXAMPLES AND XERCISES - PLANETS 16.1 Latitude by meridian altitude - PLANET:- On 15th June 1992, in DR 45° 00°S 91°10°E, the sextant ‘meridian altitude of SUPITER was 35° 14.2’. If IE was 0.5” on the are and HE was 9m, find the latitude and the PL, dhms Sext Alt 35° 14.2’ Approx LMT mer pass 15 17 02 00 TE(on) — @) 00.5" LIT (E) 06 04 40 Obs alt 35° 13.7" Approx GMT mer pass 15 10 57.20 Dip (9m) (-)_05.3° declination 09° 48.0°N, App alt 35° 08.4" d(-0.1) 00° 00.1" Tot corm (2) 01.4 declination Tak 352 07.0" N > (Named same as azimuth) MZD —_ 54° 53,0' S —» (Named opposite to T alt) Dec (09° 47,9" N—> (Contrary names subtract) Latitude 45° 05.1 S PL: E-W 121[16 - PLANETS} Notes (1) The LMT of mer pass of the planet, taken directly from the almanac without any corrections is approximate only. It may bbe out by about three or four minutes. However, since the ‘change of declination per hour (value of d) for planets is very small, no appreciable error creeps in by taking the LMT directly as tabulated in the almanac. (2) Sometimes the chronometer time of the meridian altitude may be given. Then that time should be used after taking the usual precaution of obtaining the correct date and hours of GMT as explained in Chapter 12. (G) In correcting the altitude of Venus and Mars, an additional correction (given in the almanac) has to be applied. Sometimes, the question may also say “State the correct, GMT of meridian passage’. In such a case, proceed as follows:- ‘At mer pass, GHA. +E (or=W) long = LHA = 360° GHA + 91° 10° = 360° or GHA = 268°50.0° From almanac, for GMT 1992 June 15d 10h, GHA = 254°13.4" Approx. 58m < Increment «~ 014°36.6" v correction (+2.2) for 58m -¥ 000°02,1° S8m 18s < Increment < 014°34.5" Precise GMT of mer pass = Jun 15d 10h 58m 18s Answer, Notes: (1) While extracting GHA of a planct from the almanac, v correction has to be applied. The value of v correction is applied to GHA just the same way d correction is applied to declination. V correction is always positive except for Venus where it could sometimes be plus and sometimes minus - this is indicated in the almanac. 122 [16 - PLANETS] (2) In referring to the almanac, it has been assumed that the date of GMT would be the same as that of ship ie, 15th June. However, it may be possible that the date of GMT may be one day earlier or later than that assumed, It is therefore necessary to check up by applying LIT to the assumed GMT, as shown below, and alterations made, if necessary, to the GMT date and time. GMT mer pass: Jun 15d 10h 58m 18s uT@) +) 06 04 40 LMT mer pass:Jun 15 17 02 58 If the LMT so calculated fell on 14th or 16th June in this, case, the entire computation of GMT would have to be repeated for the new date of GMT, EXERCISE - 17 (Latitude by meridian altitude - PLANETS) (1) On Sth May 1992, in DR 50° 10'S 64° 15°W, the observed ‘meridian altitude of SATURN was 56° 00.3”. If HE was 10m, find the latitude and the direction of the PL. Also find, to the nearest second, the GMT of meridian passage. 2) On 17th Jan 1992, in DR longitude 36° 40°B, the sextant ‘meridian altitude of JUPITER was 37° 43.5" bearing N. If TE was 0.3” on the arc and HE was 12m, required the latitude and PL. (3) On 14th Oct 1992, in DR longitude 110° 20°W, the sextant altitude of MARS on the meridian was 61° 14.5” South of the observer, If HE was 17m and IE was 3.6" off the arc, find the latitude and the PL, 123 i sreereseseeseninmmanaitbassndiedassnanattasiaataaasssssiaeasteeeteeeniamtiiaiodl[16 - PLANETS} (@) On 30th Nov 1992, in DR 56° 07°N 120° 04°E, the sextant meridian altitude of SATURN was 16° 24.0". If IE was 0.6° ‘on the are and HE'was 12m, find the latitude and the direction of the PL. (5) On 3rd May 1992, in DR 35° 07°N 68° 30°W, the sextant meridian altitude of JUPITER was 66° 05.2’ at 0h 07m 26s (error 06m 02s fast). If HE was 10m, and IE was 0.7" off the arc, required the latitude and the direction of the PL. 16.2 Azimuth - PLANET:- ‘On Ist Dec 1992, PM at ship in, DR 36° 27°N 144° 44°E, VENUS bore,235° (C) at 09h 18m 08s chiron time (error 10m 04s fast), If Var was 2.5°E, find the deviation fos the ship’s head, dhms dhms Chron 09 18 08 or 21 18 08 Error 10 04 10 04 omT 01 09 08 04 21 08 04 LITE) (+) 09 38 56 (4.09 38 56 LMT 01 18 47 00 06 47 00 GMT 01d 09h 08m O43 GHA 272° 02.3" From Nautical Tables Ince 002° 01.07 A 045S vO) 001) B 0528 GHA 274° 032" C 0978 —_—_Altemative Long(E) (+) 144° 44.0" T Az 232,1°(T) TAz 232.1°(T) LHA 058° 47.2" CAz 235.0°(C) Var __2.5° E dec 24° 03.5'S Error 2.9° W M Az 229.6° M 4(-04) 0° 00.1" Var 25° CAz 235.0°(C) doc 24° 03.4'S Dev 54° W Dev 54° W 12a [16 - PLANETS} Note: Working is very similar to that of azimuth - Sun except that ¥ correction has to be applied to GHA just like the d correction for declination. V correction is always plus ‘except in the case of Venus when it may sometimes be minus - this is indicated in the almanac Calculation of Azimuth by scientific calculator: A=Tanlat B= Tan dec Tan Az=___1 Tan P Sin P €. Cos lat If LHA < 180°, P = LHA. If LHA > 180°, P = (360 - LHA) ‘Naming of A, B and Azimuth is as stated under Az Sun (Chapter 14.2). When P > 90°, the minus sign obtained for the value of A is to be ignored as it is taken care of by changing name of A (sce ‘note 2.under Chapter 14.2), A= Tan lat = Tan 36° 27.0" =0.4475539 TanP Tan 58° 47.2" B= Tandec= Tan 24° 03.4" =0.5219732§ SinP Sin 58°47.2° C =0,9695271S Tan Az = 1__ =1,2822749 Az =852.1°W C (Cos 36° 27°) ie, 232.1°() EXERCISE 18 (Azimuth - PLANETS) (1) On 23rd Sept 1992, at about 0019 at ship in DR 36° 08'S. 78° 50°W, SATURN bore 286°(C). If ship's time difference was ‘Sh from GMT and if variation was 3°W, find the deviation 125[16 - PLANETS] (2) On Ist May 1992, AM at ship in DR 40° 26°N 60° 40°E, ‘Mars bore 096%(C) at 11h Sim 14s by chron (error 04m 06s slow). Variation was 3.7°W. Calculate the deviation of the compass. (3) On 18th Jan 1992, in DR 00° 00° 62° 40°E, VENUS bore 120°(C) at 0310 ship's time (4h from GMT). If variation was 2°W, find the compass error and the deviation, 16.3 Longitude by chronometer - PLANET:~ ‘On 31st Aug 1992 in DR 60° 06°N 66° 18°W, the sextant altitude of MARS was 41° 32.4” at 08h 1Sm 02s GMT. If HE ‘was 10m and IE was 2.1” on the arc, calculate the direction of the PL and a position through which it passes. Note: GMT is given so there GMT 31 08 15 02 is mo ambiguity in the hours LIT(W) @) 04 25 12 but the date of GMT is to be LMT 31 03 49 50 verified GMT 31 08 15 02 Solution by Nautical Tables: Hav LHA = Sec L , See D (Hav ZD - Hav (L ~ D)) GHA 017° 22.6" Sextalt 41° 32.4" Incr 003° 45.5" TE (on) 202" v(+0.8) 00.2" Obs alt 41° 30.3" GHA 212 08.3" Dip (om) @)_05.6° fae 22° 56.9°N Appalt 41° 24.7" d(+0.1) 90° 00.0" ‘Tot Corm ol dec 22° 56.9°N ‘Addl corm (4).00.1° lat 602_06,0°N Talt 4° 23.7" (L~D) 37° 09.1" ‘1ZD 48° 363° 126 ([16- PLANETS} NatHayZD 0.16938 Log Hav diff 8.83187 NatHav(L~D) 0.10148 LogSeck —_0,30235 Nat Hay diff 0.06790 Log SecD 0.03581 Log Hav LHA 9.17003, LHA — 314°45.7° <(see note 1, Chapter 15.3) A L738 - GHA 021°. 08,3" B 0.60N Obs long 066° 22.6°W c 13s DRlat 60° 06.0°N Az 119.4 (I) Az $ 60.6? E PL 029.4° - 209.4° through DR lat and Obs long. Answer, Solution by scientific calculator: Cos P= Sin Talt_Sin at, Sin dec Cos lat . Cos dec Note:- If Lat and dee are same name (-), contrary names (+) Hence in the formula, (-) has been put above the (+). Entire numerator is Sin and entire denominator is Cos. T at=41° 23.7 lat = 60°06.0°N dec = 22° 56.9°N Cos P = Sin 41° 23.7'()Sin 60° 06,0", Sin 22° $6.9 = 0.704172 Cos 60° 06.0" . Cos 22° 56.9 P= 45° 14.2" (see note 1, Chapter 15.3) LHA = 3149458" GHA = 021° 083" Obs long = 066° 22.5°W To calculate azimuth by scientific calculator: A=Tanlat B=Tandec Tan Az=__1. Tan P ‘Sin P C. Cos lat 127