TEXT-BOOK ON NAVIGATION ' AND NAUTICAL ASTRONOMY TEXT-BOOK ON NAVIGATION AND NAUTICAL ASTRONOMY BY J. GILL, F.R.A.S. * r REVISED AND ENLARGED BY W. V. MERRIFIELD, B.A., F.R.A.S. HEAD-MASTER OF THE LIVERPOOL CORPORATION NAUTICAL COLLEGE NEW EDITION LONGMANS, GREEN AND CO 39 PATERNOSTER ROW, LONDON FOURTH AVENUE & 30 T H STREET, NEW YORK BOMBAY, CALCUTTA, AND MADRAS IQl8 All rights reserved d PREFACE TO NEW EDITION THIS work is designed to meet the special requirements of officers of the Merchant Service, and to be a complete Text- Book on Navigation, etc., for the general student. Candidates for Board of Trade Certificates will find in it all that is needed to cope with the new regulations for the Examinations of Mates and Masters, from the lowest to the highest grades, as regards both practice and principles. The new features in this edition are (1) Rearrangement, by which the proofs of the various problems are inserted at the beginning of the separate chapter instead of being collected in one chapter at the end of the book. This gives the student a better chance of comprehending each problem. (2) All the astronomical examples are fresh, the necessary rules are given for working them, and, where possible, an example is worked direct from the figure, without using special formula; this will be found of great assistance to those con- templating going in for the " extra " examination. (3) The necessary elements from the "Nautical Almanac" are embodied in the book in the form found in the "N.A.," and so will afford as good practice in extracting the data as the possession of the Almanac itself. (4) The chapters on TRIGONOMETRY have been entirely re- written, and the student is shown how to construct both Plane and Spherical Triangles to scale. 642615 VI PREFACE. For further information in this important subject he is recommended to consult a recognised School Text-Book. (5) New chapters have been introduced on the Projections, by which any figure may be drawn to scale, and on the Cor- rection of Altitudes. (6) The chapters on the Tides and Tidal Soundings have been rewritten to suit the recent Tide Tables, extracts from which are found at the end of the book. The candidate is warned that the abridged "Nautical Almanac " is in use at the examination room, and so must be on his guard when correcting all data, as the entries are arranged in such a manner that they may be extracted at sight. For the number of page in the new edition of Norie corre- sponding to the number of table as given in this book, the student is referred to the Index of that work. Chart work and the Construction of Charts have been more fully dealt with, and it is hoped that the alterations made will ensure the appreciation of the student. W. V. MERRIFIELD. LIVERPOOL, 1918. CONTENTS PART I. CHAPTER PAQE I. PROJECTIONS . . i II. DEFINITIONS AND INSTRUMENTS 7 III. LOGARITHMS 33 IV. SOLUTION OF PLANE TRIANGLES 43 V. SOLUTION OF SPHERICAL TRIANGLES 60 VI. THE SAILINGS 71 VII. TIDES 93 VIII. REDUCTION OF SOUNDINGS 97 IX. TIME 107 X. ELEMENTS FROM THE " NAUTICAL ALMANAC " 117 XI. SIDEREAL AND SOLAR TIME 127 XII. CORRECTION OF ALTITUDES 139 XIII. LATITUDE BY MERIDIAN ALTITUDE 145 XIV. LATITUDE BY EX-MERIDIAN ALTITUDE 157 XV. COMPASS ERROR BY AMPLITUDES 168 XVI. COMPASS ERROR BY AZIMUTHS 173 XVII. LONGITUDE BY CHRONOMETER m 183 XVIII. SUMNER'S METHOD OF PROJECTION 195 XIX. THE CHART 201 XX. USE OF NAPIER'S DIAGRAM 208 XXI. GREAT CIRCLE SAILING 216 Vlll CONTENTS. CHAPTER PAQB XXII. LATITUDE BY DOUBLE ALTITUDES 227 XXIII. FINDING ERROR OF CHRONOMETER 241 XXIV. CALCULATION OP ALTITUDES 246 XXV. LONGITUDE BY LUNAR DISTANCES 250 PART II. XXVI. CONSTRUCTION OP CHARTS 263 XXVII. LAWS OF STORMS 271 XXVIII. MAGNETISM AND DEVIATION OF THE COMPASS 278 XXIX. SYLLABUS 299 XXX. MISCELLANEOUS PROBLEMS 331 APPENDIX , . 343 ANSWERS 403 TIDE TABLES . . at end 1-12 PART I. CHAPTER I. PROJECTIONS. A WELL-DRAWN figure is at all times a great help to a student in the solution of a problem, and for a clear understanding of new terms made use of in a fresh subject. For this some knowledge of projection is required. Projection is the art of representing a solid body on any surface by means of lines drawn according to definite laws. In Nautical Astronomy the solid is a sphere, and the surface a plane. These projections are either Natural or Artificial : Natural when the drawings are such as they would appear to the eye situated at a given point ; Artificial when they are not perspective representations. According to the relative positions of the eye, sphere and plane of projection or primitive, the methods employed in Natural Projections receive different names, of which the most important are the Orthographic, Gnomonic, and Stereographic. In the Orthographic, the eye is supposed so distant from the sphere that the visual rays are parallel to one another, the primitive is perpendicular to this direction. In the Gnomonic, the eye is supposed to be at the centre of the sphere, and the primitive is a tangent plane to the sphere. In the Stereographic, the eye is on the circumference of the sphere, and the primitive is a plane cutting the sphere in a great circle, which has the eye as its pole. Since the primitive may be parallel to the Equator, Meridian, or Horizon, the above projections may be further distinguished as Equatorial, Meridional) or Horizontal. The one used throughout this work is the Horizontal Stereo- graphic Projection, and its important properties are : (1) all circles on the sphere project into either circles or straight lines ; (2) any figure on the sphere projects into a similar figure on the primitive. (a) Projection on the Plane of the Equinoctial. Here the primitive is the equinoctial, and the pole P the projecting point. The concentric circles are parallels of declination. Hour circles B 7' TEXT-BOOK ON NAVIGATION. are represented;]^ stfatigSit lines, since their planes pass through the projecting jDoihl ' 'NPg is the celestial meridian, and EPW th^sfe^JteJ^pWr cJrele; EEW is the ecliptic (Fig. 1). ' A' projection of th'e* *&& on the plane of the equator shows FIG. 1. FIG. 2. the meridians of longitude as straight lines, and the parallels of latitude as concentric circles whose common centre is the pole. This is useful for showing the course of a rhumb line, also for mapping the circumpolar regions (Fig. 2). (b) Projection on the Plane of the Meridian (Figs. 3 and FIG. 3. 4). The primitive circle represents the celestial meridian, the projecting point O being the east or west point of the horizon; Z and N are the zenith and nadir ; P and PI, the north and south poles respectively ; HR (at right angles to ZN) is the horizon ; and EQ (at right angles to PPi) is the equinoctial. The elevation of the upper pole P above the horizon, viz. the arc RP, is equal to the latitude of the observer's position. The arc EZ also equals the PROJECTIONS. latitude. Supposing R to be the north point of the horizon, then P is the north pole (Fig. 3), and the latitude is north. For south latitude, the south pole PI must be above the horizon (Fig. 4). Hour circles are represented by arcs from P to PI, and vertical or altitude circles by arcs from Z to N. (c) Projection on the Plane of the Horizon (Fig. 5.) Here the primitive circle is the hori- zon, N., E., S., W. being the cardi- nal points. The zenith (Z) is the projecting point, hence all verti- cal circles appear as straight lines passing through Z, NZS being the celestial 'meridian, and WZE the prime vertical. P is the north pole, at a distance from N. equal to the latitude. The arc ZQ is also equal to the latitude, and PZ is the co-lati- tude ; WQE is the equator, cutting the horizon in the E. and W. points; Pp b Pp 2 , etc., are hour circles. The projection on the plane of the horizon has some advantages over that on the plane of the meridian for illustrating most of the problems of nautical astronomy, and is therefore more commonly employed. To draw a figure to scale, a " Gunter " rule is necessary ; the following illustration explains the connection between the lines marked Rum., Cho., Sin., Tan., and S.T. on the rule (Fig. 6). With centre 0, distance OA, describe a semicircle. AOB is Fio. 6. 4 TEXT-BOOK ON NAVIGATION. a vertical diameter ; OX, a horizontal line through centre ; BY, a tangent at B. OX divides the figure into two quadrants ; the upper one is divided into eight equal parts, and with A as centre and distance the different chords from A to each point; the distances are brought back on the chord of 90, forming the line of Rhumbs (RuM.). A3 is thus the chord of a circle described with radius OA, which subtends an angle of three points at IB C FIG. 7. centre 0. The line of Chords (Cno.) is similarly formed on the lower quadrant, but the quadrant is here divided into nine equal parts instead of eight, and thus gives divisions to 10, which can be further subdivided. We can now see that the chord of 60 is the radius that must be used to describe the circle. For the line of Tangents (TAN.) through and points of division in lower quadrant, draw straight lines meeting BY in required points. For the line of Semi-Tangents (S.T.) through A, and PROJECTIONS. points of division in lower quadrant, and also corresponding points in upper quadrant, draw lines to cut OX in required points. Example. Draw a figure from following data : Latitude 40 N. ; western hour angle 2 h 48 m ; declination 10 N. ; and take off the Amplitude at setting, Azimuth and Altitude (Fig. 7). With centre Z, and radius the chord of 60, describe a circle N.E.S.W. ; draw SZNB vertically, and lay off E. and W. as the extremities of a horizontal diameter through Z. Make ZM = 40 (Lat.) from scale of tangents, and through M draw a line perpendicular to MZ ; in this line are the centres of all hour circles. Make NO = chord of 40 (Lat.), where the line OW cuts the vertical line is the elevated pole P. At P to the right make an angle = complement hour angle that is 48 ; this line cuts the line through M in D, and is the centre for describing the hour circle required PXB. For the equinoctial, add the colat to 90 = 140 ; subtract the colat from 90 = 40, and from scale of S.T. make ZB = 140, and ZQ = 40 ; bisect BQ in 1, this is centre for describing equinoctial WQE. For the declination parallel, add colat to polar distance = 130, subtract colat from polar distance = 30 ; from S.T. scale make ZC = 130, and ZL = 30; bisect CL in 2 ; this is centre for describing the parallel TXL, where the parallel cuts the hour circle is X, the position of the object. Draw ZXA the vertical circle through the object. Measure WT on the scale of chords for the Amplitude = 15, or W. 15 N. Measure SA on the scale of chords for the Azimuth = 63, or S. 63 W. Measure ZX on the S-.T. scale for the Zenith Distance = 49, or Altitude = 41. For South Latitude set off ZM to the north, and con- struct the figure downwards instead of upwards. The only Artificial Projection to be considered is Mercator's (Fig. 8). Mercator's Projection. This is the kind of projection used in the construction of Navigating Charts, and represents the Earth's surface as a plane. The meridians of longitude are drawn as parallel straight lines, and the divisions of latitude are expanded in the same proportion as the meridian distances are lengthened. The effect of this is to make the rhumb course a straight line, which is of great advantage to the navigator. The distance from the equator of any given parallel of latitude, and the amount of expansion of a degree of latitude, can be found from the table of Meridional Parts; for example, the parallel of 30 is 1800 geographical miles from the equator on the globe, but the meridional parts for 30 FIG. 8. 1888, 6 TEXT-BOOK ON NAVIGATION. which means that on a Mercator's chart drawn to the same scale tliis parallel would be 1888 geographical miles from the equator. Again, the meridional parts for 31 = 1958, and 1958 - 1888 = 70 miles, which would be the length of the expanded degree from 30 to 31. The meridional parts for any given latitude may be calcu- lated by adding together the secants of all minutes of latitude up to the given latitude ; thus sec 0' + sec 1' + sec 2' + sec 3', etc. or they may be calculated independently from the formula log of mer. parts = 3'8984895 + log (log cot J co-lat. - 10) CHAPTER II. DEFINITIONS AND INSTRUMENTS. 1. Great Circles. Circles whose planes pass through the centre of a sphere or globe. 2. Vertex of a Great Circle. That point in it which is farthest from the Equator. 3. Small Circles. Circles whose planes do not pass through the centre of the sphere. AYB and CV'D are great circles. V and V are vertices. EOF, a email circle. FIG. 9. 4. Right Angle. An angle measuring 90, or one quarter of a circle. 5. Oblique Angle. An angle not a right angle. 6. Obtuse Angle. An angle greater than a right angle. 7. Acute Angle. An angle less than a right angle. A OB FIG. 10. F. FIG. 11. FIG. 12. AOC and BOO are right angles. DEF and XYZ are oblique angles. DEF is an obtuse angle. XYZ is an acute angle. 8. Spherical Angle, An angle included between two great circles of a sphere (Fig. 13). JC AOC and BOG are spherical angles. FIG. 13. TEXT-BOOK ON NAVIGATION. 9. Arc, A part of the circumference of a circle. 10. Complement of an Arc or Angle. Its difference from 90. 11. Supplement of an Arc or Angle. Its difference from 180. O A FIG. 14. AB, BC, CD are arcs. BOG is complement of BOA. BC is complement of BA. COD is complement of AOD. R Q P FIG. 15. PS and SK are arcs. SR is supplement of PS. EQS is supplement of SQP. 12. The Poles, The points where the earth's axis meets the surface. 13. Equator. A great circle of the earth 90 distant from the poles. 14. A Meridian. Half a great circle extending from pole to pole of the earth. 15. Parallels of Latitude. Circles whose planes are parallel to the plane of the Equator. 16. Tropics. Two parallels of latitude distant 23 28' on each side of the Equator : Cancer on the north, and Capricorn on the South. 17. Difference of Latitude. The arc of a meridian intercepted between the parallels of latitude on which the places stand. 18. Prime Meridian. The meridian from which longitudes are measured ; in this country it is the meridian of Greenwich. 19. Longitude. Distance in arc east or west of the meridian of Greenwich. It is measured on the Equator, or on a graduated parallel, vp * \%t> 20. Difference of Longitude. An arc of the Equator intercepted between two meridians of longitude (Fig. 16). P and P' are the poles. EQ is the equator. PAP' and PBP' are meridians. POP' the prime meridian. LL', TT', RR' are parallels of latitude. TT is tropic of Cancer. RR' is tropic of Capricorn. XY is difference of latitude when places are on same side of equator, on parallels LL' and TT'. XZ is difference of latitude when places are on opposite sides of equator, on parallels LL' and RR'. OA and OB are Longitudes of places on meridians PAP' and PBP'. AB is difference of longitude, between places on meridians PAP' and PBP' DEFINITIONS AND INSTRUMENTS. 9 21. Meridional Parts. The distance from the Equator in geo- graphical miles of a given latitude on Mercator's Chart (Fig. 17). ab = parallel on globe. cd same parallel on chart. ec = meridional parts for lat. ea. FIG. 17. 22. Departure. The distance in geographical miles between two places in the same latitude. In the traverse table, it is the number of miles of easting or westing made on a given course and distance. 23. Nautical Nile. The one-sixtieth part of a degree measured on a meridian. It varies slightly with the latitude, and its mean value is taken at 6080 feet. 24 Rhumb Line. A line which makes equal angles with all the meridians it crosses. 25. Zenith. The point in the heavens vertically over the observer's head. 26. Nadir. The point in the heavens diametrically opposite to the Zenith. 27. Equinoctial. A great circle in the heavens 90 from the celestial poles. 28. Ecliptic. A great circle in the heavens representing the sun's annual apparent path. 29. First Point of Aries. The point where the sun crosses the Equinoctial about the 21st of March. 30. Celestial Meridian of Observer. The vertical circle which cuts the Horizon in the true North and South points (Fig. 18). NESW., Plane of horizon. Z is zenith. EQW is equinoctial. CvC' is ecliptic. rp is first point of Aries. NPZS is celestial meridian of observer. 31. Azimuth. The bearing of an elevated object reckoned from the N. or S. point. io TEXT-BOOK ON NAVIGATION. 32. Amplitude. The bearing of an object when rising or setting, reckoned from the E. or W. point. 33. Vertical Circles. Circles passing through the Zenith and Nadir, and therefore perpendicular to the Horizon. They are also called " altitude circles " and " azimuth circles." 34. Prime Vertical. The vertical circle which cuts the Horizon in the true East and West points. 35. Declination. Distance in arc north or south from the Equinoctial measured on an Hour Circle. 36. Polar Distance, The distance in arc from the elevated pole, and therefore the complement of the Declination, measured on an Hour Circle. 37. Hour Circles. Halves of great circles which pass through the celestial poles, one hour being equal to 15 (Fig. 19). NESW, Plane of horizon. NPZS, Celestial meridian. P, Elevated pole. X, a heavenly body on parallel of declination LL'. NZA or arc NA is Azimuth of X. WL, Amplitude of X at setting. ZXA, Vertical circle through X. WZE, Prime vertical. XD, Declination of X. PX, Polar distance of X. PXD, Hour circle through X. 38. Right Ascension, Distance eastward from the first point of Aries measured on the Equinoctial and expressed in Time. 89. Hour Angle. The angle at the pole between the observer's Celestial Meridian and an Hour Circle. 40. Civil Time. The time used in civil life. The day begins at midnight, and is divided into two parts of twelve hours each, a,m. from midnight to noon, and p.m. from noon to midnight. 41. Astronomical Time, Time used for astronomical calcula- tions. The day begins at noon, and the hours count from to 24. 42. Mean Time. The Westerly hour angle of the Mean Sun, it is the time shown by a correctly going timepiece. The Mean Sun is an imaginary body which is supposed to move along the equinoctial with the average velocity of the True Sun in the ecliptic. 43. Apparent Time. The Westerly hour angle of the true Sun. 44. Equation of Time. The difference between Mean and Apparent Times ; and therefore used to convert one time to the other. The Equation of Time is least about April 15th, June 15th, August 31st, and December 24th; and greatest about February 10th, May 14th, July 26th, and November 2nd in each year. DEFINITIONS AND INSTRUMENTS. n 45. Sidereal Time. The Westerly hour angle of the first point of Aries (Fig. 20). Figure as before on horizon piano. OrC' is Ecliptic. T is first point of Aries. M is mean Sun. T is true Sun. X is any heavenly object. Then rA is right ascension of X. QPA is easterly hour angle of X. QPM is westerly hour angle of mean sun, or Mean Time. QPT is westerly hour angle of true sun, or Apparent Time. QPT is westerly hour angle of T, or Sidereal Time. TPM is Equation of Time. 4G. Visible Horizon. The circle which bounds the observer's view at sea. 47. Sensible Horizon. A plane touching the Earth at the observer's position and extended to the Heavens. 48. Rational Horizon. A plane passing through the Earth's Centre parallel to the Sensible Horizon and extended to the Heavens. 49. Artificial Horizon. A horizontal reflecting surface used for observing altitudes in the absence of a good sea-horizon. 50. Dip. The angular depression of the Visible horizon below the Sensible, due to the observer's elevated position. 51. Observed Altitude. The angular height of an object above the Visible horizon as measured by a Sextant. 52. Apparent Altitude. The angular height of an object above the Sensible horizon, found by correcting the observed altitude for Index Error and Dip. 53. True Altitude. The angular height of an object above the Rational Horizon, found by correcting the Apparent Altitude for Refraction and Parallax (Fig. 21). AB ia visible horizon. MOA = angle OA makes with sensible horizon is Dip. A OX = observed altitude. MOX = apparent altitude. XCN = true altitude. FIG. 21. 12 TEXT-BOOK ON NAVIGATION. 54. Refraction, The error of altitude caused by the bending of rays of light in passing through the atmosphere (Fig. 22). FIG. 22. O, Observer's position. S, True position of star. Sj, Apparent position of star. SS 1} Refraction. OH, Horizon. HS,, Apparent altitude of a star. KS, True altitude of a star. 55. Parallax is the angle at the centre of the object subtended by the earth's radius at the position of the observer. It is a correction for reducing an observation to the earth's centre (Fig. 23). O, Observer's position. Q H, Sun's position on horizon. S, Sun's position in altitude. L OHC, Horizontal parallax. L OSC, Parallax in alti- Sa. tude. & > Z, The zenith of ob- server.. Bs 2 , True altitude ol sun. Fid. 23. DEFINITIONS AND INSTRUMENTS. 13 56. Semidiameter. The angle subtended at the earth's centre by the radius of the sun, moon, or planet. 57. Augmentation of Moon's Semidiameter. An apparent increase in the moon's Semidiameter due to the fact that her distance from the observer decreases as her altitude increases (Fig. 24). FIG. 24. E, Centre of earth. O, Observer's position. MA, Position of moon on horizon. MB, MO, Other positions of moon in altitude : her centre being at fixed distance from B. As she rises her distance from O decreases. MO A, Semidiameter of moon on horizon. MOB and MOC, Semidiameters when in altitude. MOB-MOA and MOC-MOA, Augmentation of Semidiameter when in altitude. 58. Reduction of Horizontal Parallax. A correction necessary on account of the shortening of the earth's radii towards the poles (Fig, 25), O FIG O, Observer's position on longest diameter of earth. O,, Position on a shorter diameter. O C C, Equatorial horizontal parallax. O,, C C, Horizontal parallax for latitude of observer, which is less because subtended by a shorter radius. TEXT-BOOK ON NAVIGATION. 59. Zenith Distance. The angular distance from the Zenith, or the complement of the true altitude measured on a vertical circle. 60. Magnetic Meridian at a place is a great circle in whose plane a compass-needle lies when under the earth's influence only. 61. Variation. The angle between the True and Magnetic meridians, measured from True North. 62. Deviation. The angle between the Compass- needle and the Magnetic meridian. 63. Error of the Compass, The angle between the Compass- needle and the True meridian (Fig. 26). T, True north. MN, Magnetic meridian. C and Cj, Different directions of compass north. TM, Variation (west). MO, Deviation (east). MC,, (west). TC, Error of compass (east). TO,, (west). CD or CjD, Compass course from north. TD, True course from north. MD, Magnetic course from north. FIG. 2G. 64. Leeway. The angle between the ship's keel and her "wake"( Fi S- 27). FIG. 27. NOTE. The foregoing comprise the definitions required by the Board of Trade Regulations for the examination of mates and masters. 65. Latitude. Distance in arc north or south from the Equator measured on a Meridian, 66. Arctic Circle. The parallel of 66 32' N. lat. 67. Antarctic Circle. The parallel of 66 32' S. lat, 68. Obliquity of Ecliptic. The angle between the planes of the Ecliptic and Equinoctial. DEFINITIONS AND INSTRUMENTS. 15 69. Equinoxes. The points where the Ecliptic intersects the Equinoctial. The Vernal Equinox is where the sun crosses the Equinoctial in March, and the Autumnal where it crosses in September. 70. Precession. A slow motion of the Equinoxes along the Equinoctial westwards, amounting to about 50" annually. 71. Solstices. The points on the Ecliptic farthest from the Equinoctial. N.B. The terms " Equinoxes " (Spring and Autumn) and " Solstices " (Summer and Winter) are also applied to the times of the year when the sun is on the Equinoctial or at the Solstices. 72. Celestial Latitude. Distance in arc north or south from the Ecliptic. 73. Celestial Longitude. Distance in arc from the " first point of Aries " measured on the Ecliptic. 74. Circumpolar Stars. Stars whose polar distance is less than the latitude of the place, and consequently never set. QUESTIONS ON VARIATION AND DEVIATION, FROM THB BOARD or TRADE REGULATIONS. 1. Q. Does the variation change with time ? A. Yes ; it slowly changes in amount. At London it now decreases about 9' annually. 2. Q. Is the variation the same all over the world ? A. No ; all amounts from to 180 may be found. 3. Q. Where do you find the variation ? A. On a variation chart. 4. Q. Does the deviation change ? if so, when ? A. It changes with a change of course ; also, for the same course with lapse of time, because the magnetic condition of the ship is liable to change, and from change of cargo. 5. Q. Where is the north magnetic pole situated ? A. In about 70 1ST. lat. and 92 W. long. 6. Q. Where is the south magnetic pole situated ? A. In about lat. 75 S., and long. 150 E. 7. Q. Should the compass-needle point to the magnetic or the true pole of the earth 1 A. To the magnetic pole. 8. Q. When is the altitude of an object most seriously affected by refraction ? A. When nearest to the horizon. 9. Q. Whore is the pole star situated ? A. At the distance of 1 12' from the true celestial pole. 10. Q. Which is the most favourable time for determining the hour angle of a celestial body, and thence the longitude, and state the reason why? A. When it is on the prime vertical, because the hour angle is then least affected by errors in altitude or latitude. i6 TEXT-BOOK ON NAVIGATION. INSTRUMENTS. Mariner's Compass. The Mariner's Compass consists of three parts : Bowl, Card, and Needles. The Bowl is a hollow hemi- sphere of brass, weighted at its lowest part to keep it horizontal. On its inside a vertical line is painted called the " lubber " line, which indicates on the card the Course of the ship. From the centre of the bottom of the bowl springs a support which holds a sharp pivot tipped with iridium to render it hard and non- rustable. The pivot fits into a brass cap having an agate or ruby centre to work on. From the cap by means of silk threads the card is suspended. The card is circular, made of thin paper with the centre part cut out to lessen the weight, and having a band of aluminium round its circumference to strengthen it, and is graduated to quarter-points and degrees. FIG. 29. From the card hang the magnetised needles, which are generally 8 in number, arranged symmetrically, parallel to one another, and to the N. and S. points of the card (Fig. 28). The needles are about the size of a large knitting-needle, the longest pair with a 10" compass card being 3" long. The whole instrument is mounted on gimbals, with strong wire supports to take off the vibration of the ship as much as possible, and is placed in a binnacle (Fig. 29). Two brackets at the side support cast-iron correctors, and provision is made inside the binnacle for compensating magnets. A brass cylindrical box on the outside is provided for a Flinders Bar. A Clinometer shows the angle of " heeling." Azimuth Compass. The Azimuth Compass is like an ordinary Steering-compass in the mounting of the needles and card. There is a movable ring on the bowl, provided with sight- vanes at opposite ends of a diameter. One of these should have DEFINITIONS AND INSTRUMENTS. 17 a reflecting prism for reading the graduations on the card, and the other a mirror for reflecting elevated objects (Fig. 30). It is mounted in a box or on a tripod, and in using it the observer looks through the narrow slit, and moves the ring round until the object is bisected by the wire in the other sight-vane, and then reads off the bearing by means of the prism. Sometimes a vertical style is fixed on the centre of the glass cover of an ordinary compass, for showing the sun's compass bearing by means of the shadow. FIG. 30. Pelorus. The " Pelorus " resembles an Azimuth Compass, but has neither magnets nor bowl. The " card," usually of metal, is not suspended, but is free to move round its centre, and is mounted on gimbals and weighted to keep it horizontal. A bar carrying the two sight-vanes is pivoted at the centre of the card, and may be clamped in any position. ' Four " lubber- points," 90 distant, are usually marked outside the edge of the "card" (Fig. 31). The uses of the " Pelorus " are (a) To find the Deviation on any course. The sight- vane being clamped to the known correct magnetic bearing of a distant object, the card is turned round until, on looking through the slit, the object is seen bisected by the wire. The correct magnetic course is shown at the lubber-point, and the difference between this course and the course by compass is the Deviation. (6) To place the ship's head in any desired direction for the purpose of compensating the compasses. The sight-vanes being clamped to a known correct magnetic bearing, and the card turned round until the desired direction of ship's head is at the lubber-point, then the ship is swung until C iS TEXT-BOOK ON NAVIGATION. the object is seen bisected by the wire. Her head is now on the desired point, and the magnets or correctors may be placed in position so that the Compass shall show the same course as the Pelorus. (c) For taking Bearings when the Standard Compass is not available. Care must be taken that the ship is upright, and that a line FIG. 31. through the centre of the card and the lubber-point shall be parallel to the line of the keel. Barometers. The atmosphere surrounds the earth to a height of two hundred miles or more, and presses on the surface with an average weight of nearly 15 Ibs. on a square inch, and its average pressure at sea-level is nearly 1000 millibars. A cubic foot of atmospheric air under ordinary circumstances weighs approximately an ounce and a quarter and its density increases one per cent, with an increase of ten millibars of pressure, and decreases one per cent, with an increase of temperature of 5 F. (see " Barometer Manual "). The pressure varies with changes of temperature and moisture and from other causes, and decreases from the sea-level upwards. A Barometer is an instrument for measuring this pressure. Barometers are generally of two kinds, viz. " Aneroid," meaning without liquid, and "Mercurial." The Aneroid Barometer has a shallow metal cylinder with a flexible cover, and partially exhausted of air. The tendency to collapse in consequence of the partial vacuum is balanced by a DEFINITIONS AND INSTRUMENTS. FIG. 32. strong spring attached to the cover and fastened to the bottom of the containing box. As the pressure of the atmosphere varies, the movements of the flexible cover inwards and outwards are, by a mechanical contriv- ance, conveyed to a pointer which shows on a dial the pressure of the atmosphere (Fig. 32). The readings from an aneroid cannot be depended upon, and should be com- pared from time to time with those taken from a mercurial barometer, and adjusted by means of a screw at the back of the instrument, and index error recorded. It is very portable, convenient for reading, and the readings recorded require no correction for temperature or latitude, but must be corrected for height above sea-level and index error. The dial of the barometer is graduated on three concentric circles, the outer showing the reading in millibars, the centre in inches and tenths, and the inner in millimetres. The converters for the three scales may be obtained from the fact that 100 centibars or 1000 millibars correspond to 29 '53 inches of mercury at 32 Fahr., and 45 lat. and to 750 millimetres at C. and 45 lat. A table for this purpose is found in the " Seaman's Hand Book on Meteorology," and may be obtained by the student as follows : One inch of mercury corresponds to 33*86 millibars. Hence to convert " inches " to " mbars." multiply reading in inches by 33'86. To convert " millibars " to " inches," divide reading in " mbars." by 33-86. Again '03937 inch equals 1 millimetre. Hence to convert "inches" to "mms." divide reading in inches by '03937. To convert " mms." to inches, multiply reading in " mms." by -03937. And since 1000 mbars. correspond to 750 mms., To convert mbars. to "mms.," subtract one-quarter of reading in mbars. for corresponding " mms." To convert "mms." to mbars. add one-third of reading in " mms." for corresponding " mbars." The Mercurial Barometer consists of a glass tube about 33 in. 20 TEXT-BOOK ON NAVIGATION. long (whose open end dips into a quantity of mercury contained in a trough or " cistern "), enclosing a column of mercury of varying height. The height of this column measured from the surface of the mercury in the cistern at any moment is the measure of the pressure of the atmosphere on a surface equal to the section of the tube. This pressure is indicated in inches of mercury and mbars. on scales near the top. Between the top of the column, and the top of the tube there is a space called the Torricellian vacuum, because of the experi- ments made in 1643 by an Italian called Torricelli (v, Fig. 34). The portable barometers (Fig. 33) used at sea have the glass tube protected by a casing of wood or iron, and the mercury in the cistern is prevented from escaping by a cover or bag of washleather, which, how- ever, does not interfere with the air-pressure. In this form of barometer the " inches " on the scale are shortened, so as to compensate for the changing level of the mercury in the cistern as the column rises or falls. For scientific purposes, corrections of the barometer-readings are made for temperature height above sea-level, and latitude. Thermometer. A Thermometer is an in- strument for measuring temperatures. The common Thermometer is a glass tube of exceedingly small bore, blown out at one end into a bulb, which is filled with mercury, the other end being hermetically sealed after the air is driven out. As the mercury in the bulb expands with heat and contracts with cold to a greater degree than the glass envelope, this causes the column in the tube to advance or retire. The degree of the scale indicated by the end of the column is the temperature. FIG. 33. FIG. 34. Three different scales of temperature are in use. (a) The Fahrenheit thermometer 1 (Fig. 35, a), which is the one generally used in this country, marks the freezing-point of water 32 and the boiling-point 212, when the barometer reads 29-905 inches at 32 Fahr., a difference of 180. Zero (0) on this scale is the temperature of a mixture of salt and snow. 1 So called because Fahrenheit, a German, in 1721, first suggested this method of graduation. DEFINITIONS AND INSTRUMENTS. 21 (6) The Centigrade or Celsius l thermometer (Fig. 35, b) marks the freezing-point zero (0) and the boiling-point 100. (c) The Reaumur thermometer 2 (Fig. 35, c) marks freezing- point zero (0) and boiling-point 80. The attached thermometer in the barometer (T, Fig. 33) has two scales, 212 -\ wo one graduated to give the temperature Fahrenheit, the other graduated in Centi- grade degrees, showing the temperature reckoned from absolute zero, which is -273 C. or -459 Fahr., and is the temperature at which permanent gases following the ordinary rate of contrac- tion on decrease of temperature would occupy no volume and meet no pressure. Mercury is the liquid generally in 3? use in thermometers because of its low freezing-point ( 40 C.), high boiling- point (358 C.), and its steady rate of a be expansion through long ranges of tern- Fio. 35. perature ; other advantages are its opacity and the fact that it does not wet the glass envelope. For lower temperatures alcohol is used instead of mercury, as it solidifies at - 130 C. and boils at 78 C., the alcohol being coloured to render it visible. The number of divisions on the three scales between freezing and boiling-points are therefore ^in the ratio of 180, 100, and 80, or 9, 5, 4, from which it is easy to compare the readings of any one of these instruments with each of the other two. Examples. 1. What are the readings of the Centigrade and the Re'aumur scales corresponding to 72, 18 and -40 Fahr. 2 72 40 40 Freezing 32 5 4 Above freezing 40 18 Fahr. 32 Below freezing 14 C 9)200 22-2 Cent. 273 295-2 A. 14 5 9)70 -77 Cent. 273 9)160 177 Reau. 14 4 9)56 -6'2Reau. 265'3 A. 1 Named after Celsius, a Swede, who, about 1743, suggested this method of graduation. 2 Called after its proposer, Reaumur, a Frenchman, about 1743. 22 TEXT-BOOK ON NAVIGATION. -40Fahr. 72 72 32 5 4 Below freezing -72 9)360 9)288 -40 Cent. -32Reau. 273 233 A. The amounts marked "A." being the readings when the thermometer graduations commence at absolute zero. 2. Convert 32 Cent, and 20 Reaumur into the corresponding Fahrenheit temperatures. 32 Cent. 20 Reau. 9 9 5)288 4)180 57'6 above freezing. 45 above freezing. + 32 +32 89-6 Fahr. 77 Fahr. 3. Convert -25 Cent, and 268 A. to Fahr. -25 Cent. 268 A. 9 273 5)225 -5 Cent. 9 45 below freezing. + 32 5)45 - 13 Fahr. -9 below freezing. +32 23 Fahr. Chronometer (Fig. 36). A chronometer is an accurately made and compensated timepiece with its balance so constructed that any change of temperature produces the smallest change in its error from day to day; this difference of error is called its daily rate. Most commonly it is made to work for two days, and on its face is an index hand showing how long since the instrument was wound. The winding should take place every day at the same hour in order that the operation may become a habit, and as a guarantee that a constant interval elapses between the operations, and is performed by six or seven half- turns of the key. The keyhole at the back is protected by a sliding cap. The instrument is mounted on gimbals in a mahogany case having a glass roof and outer cover. In the case is a clamp by means of which it can be prevented from swinging in the gimbals when conveyed from ship to shore. DEFINITIONS AND INSTRUMENTS. 23 Its use is to obtain correct mean time at Greenwich (the error and rate being known), and thence the longitude. FIG. 36. Hydrometer. The Hydrometer is an instrument for measur- ing Density or Specific Gravity (Fig. 37). Specific gravity of a solid or liquid is its weight compared with the weight of an equal bulk or volume of distilled water at the temperature of 62 Fahr., or, as some scientists prefer, 39'2 Fahr., which is the point of maximum density. The hydrometer used on board ship for obtaining the density of sea- water, and the water in docks or rivers, is a glass or metal stem with two bulbs, the lower of which is weighted with mercury or small shot to make the instrument float upright. The scale on the stem reads from downwards. It is seen that 24 TEXT-BOOK ON NAVIGATION. in using this hydrometer we displace constant weights of different volumes : the instrument is graduated so that the S.G. may be read directly from the scale, which is effected by shorten- ing the divisions from downwards. When in distilled water (temp. 62), the level of the water is at 0, but in sea- water at about 26. In brackish water it will be somewhere between those marks, according to the degree of saltness. One thousand must be added to the figures on the scale when stating the S.G. Hence the S.G. of distilled water (temp. 62 Fahr.) is 1000, and the S.G. of sea-water 1025 or 1026, whilst the S.G. of the water in a river or estuary or dock where fresh water enters might be anything between these two limits. These facts have an important bearing on the loading of ships, since the law fixes the load-line for seagoing vessels. It is obvious that a ship may be loaded deeper than her " load-mark " in fresh or brackish water, for she will " rise to her marks " when she goes to sea. The S.G. of the water at the loading-place may be obtained by a Hydrometer, and, her draught being known, a simple proportion will give the " sea draught " nearly. Also when the "draught" at sea is known, FIG. 37. the " draught " in dock or river can be found. Examples. 1. The S.G. of the water in the dock is 1015, and the ship's draught 23 ft. 6 in. : required her draught at sea. Taking S.G. of sea- water as 1025 1025 : 1015 : : 23 : 5 : 23-3 nearly 23-5 5075 3045 2030 1025)23852-5(23-27 ft. nearly 2050 3352 3075 2775 2050 7250 7175 75 Answer, 23 "3 ft. nearly. DEFINITIONS AND INSTRUMENTS. 25 2. At sea a ship draws 26 ft. 9 in. : what will her draught be in dock where the S.G. is 1010? ft. In. ft. in. 1010 : 1025 : : 26 9 : 27 2 12 321 in. 1025 1605 642 321 1010)329025(326 inches nearly 3030 = 27 ft. 2 in. Answer. 2602 2020 5825 3. The number of inches between the F.W. mark and centre of disc is given on the load-line certificate as 7'5 : how much may the Plimsoll mark be submerged in dock when Hydrometer reads 9 ? In diagram density at F.W. is at centre of disc 25 : vessel floats in 9. Hence for a differ- ence in saltness of 25, the mark is put down O i FW 7 '5 inches : how much will it be put down for a _ difference in saltness of 25 - 9 = 16 ? ^ 7'5 x 16 . = -^ inches 2o = 4*8 inches, and F.W. mark would be out of water 2*7 inches. 25 4. The tons per inch in sea water is 40 : what would it be in water where Hydrometer Fia. 37A. reads 16 ? As the water is lighter, fewer tons will sink her one inch. 40 x 1016 1025 = 39'65 tons per inch. The Station Pointer. The Station Pointer is a graduated circle having three radial arms, one of which is fixed, and the other two moveable, the latter being provided with verniers and clamp and tangent screws (Fig. 38). The use of the Station Pointer is to locate the ship's position on a chart from the angles subtended by three lighthouses or other fixed objects on shore. With a sextant, the horizontal angles between the middle object and each of the other two are measured, and the moveable arms are set to these angles. Then the instrument is laid on the chart with the bevelled edge of the fixed arm on the middle object, and moved about 26 TEXT-BOOK ON NAVIGATION. until the bevelled edges of the other two arms fall on the other two objects. The centre of the circle gives the ship's position. FIG. 38. The Sextant (Fig. 39). The "full" Sextant for sea use consists of a metal frame in the form of a Sector of about 60, with an arc or limb graduated to degrees and 10' spaces. An index bar, pivoted at the centre of the circle, carries the silvered Index Glass, and at the other end a Vernier or Nonius for subdividing to 10". The index bar is provided with clamp and tangent screws, and carries a reading microscope. The Horizon Glass half of which is silvered stands on the frame. Both glasses are provided with screens for taking off the glare of a strong light when the instrument is used. A handle at the back of the frame is convenient for holding, and a collar, which is moveable towards or from the frame by means of a milled- head screw, serves to hold a plain tube or a telescope. When in correct adjustment, the index and horizon glasses are perpendicular to the plane of the instrument (the upper surface of the frame), and parallel to each other when the pointer on the vernier indicates on the arc. The axis of the tele- scope when screwed into the collar is parallel to the plane of the instru- ment. The first two adjust- F IQ . 39. ~ ments are essential, but if the glasses are slightly out of parallelism, it is of little consequence, for the "index error " can be easily determined and applied. Screws are pro- vided for making all the adjustments. DEFINITIONS AND INSTRUMENTS. 27 QUESTIONS ON THE ADJUSTMENTS OF THE SEXTAIST IN BOARD OF TRADE EXAMINATIONS FOR MATES AND MASTERS. 1. What is the first adjustment of the sextant ? 2. How do you make that adjustment ? 3. What is the second adjustment ? 4. Describe how to make that adjustment. 5. What is the third adjustment 1 6. How would you make the third adjustment 1 7. In the absence of a screw, how would you proceed ? 8. How would you find the index error by the horizon ? 9. How is it to be applied ? 10. Place the index at error of minutes to be added, clamp it and leave it NOTE. The examiner will see that it is correct. 11. The examiner will then place the zero of the vernier on the arc, not near any of the marked divisions, and the candidate will read it. NOTE. In all cases the applicant will name or otherwise point out the screws used in the various adjustments. 12. How do you find the index error by the sun ? 13. How is the same applied ? Suppose the reading on the arc to be , and the reading off the arc to be , find the index error. 14. What proof have you that those measurements or angles have beep taken with tolerable accuracy ? 15. Describe the fourth adjustment. ANSWERS TO THE QUESTIONS ON THE SEXTANT. 1. To set the Index Glass perpendicular to the plane of the sextant. 2. The index should be clamped at about 60, and the sextant held face upwards with the limb from the observer ; then, on looking into the index glass towards the limb, if the reflected and real arcs do not coincide, turn the adjusting screws at the back of the glass until they do coincide. 3. To set the Horizon Glass perpendicular to the plane of the sextant. 4. The index being clamped at 0, the sextant should be held horizontally, face upwards then, on looking through the telescope collar towards the horizon, if the reflected and real horizons do not coincide, turn the upper adjusting screw at the back of the glass until they do coincide. By night, look at a star and see if its reflection passes up and down through the star, if it does the horizon glass is truly perpendicular, if not, use the upper screw to make it do so. 5. To set the Horizon Glass parallel to the Index Glass when the index is at on the arc. 6. The index being clamped at 0, the sextant should be held vertically ; then, on looking through the tube or telescope at the horizon, if the reflected and real horizons do not coincide, turn the lower adjusting screw at the back of the glass until they do coincide. 7. Find the Index Error. 8. Placing the index at 0, and holding the sextant as for the third adjustment, make the reflected and real horizons coincide by means of the Tangent Screw. By night, if when the index is at 0, the true and reflected stars do not coincide, make them do so by means of the tangent screw. The reading " On " or " Off" is the Index Error. 9. It is subtractive if " on," and additive if " off." 10 and 11 are practical and require no written answer. 12. Clamping the index at 40' "on" the arc and looking through the sight-tube or telescope at the sun, make the limbs of the reflected and real suns touch by means of the tangent screw, and note the reading ; then clamp at 40' "off," and again make the contact and note the reading. Half the difference of the readings is the Index Error. 28 TEXT-BOOK ON NAVIGATION. 13. If the greater reading is "on," the error is subtraclive : if "off," it is additive. 14. One-fourth the sum of the readings should be the Sun's Semi- diameter, as in the Nautical Almanac for the given day. 15. The Fourth Adjustment is to set the line of collimation or axis of the telescope parallel to the plane of the instrument. The adjustment is made as follows : Screw on the Inverting Telescope, and, getting it in correct focus, turn the eye-piece round until two parallel wires in the tube are parallel to the plane of the sextant. Then select two celestial objects not less than 90 apart, and holding the sextant with its face in the plane of the objects, move the index forward until the reflected image of one object comes into view in the centre of the field of the telescope with the direct image of the other. Bring both images in contact on the wire nearest the frame, and by slightly moving the instrument get them on the other wire : if they remain in contact, the adjustment is perfect ; if they separate, the object end inclines towards the frame ; but if they overlap, it declines from it. The adjust- ment is made by means of two screws in the collar which holds the telescope. The effect of the collimation error is to make the observed angle too'grcat. Any error of centering is an essential defect of the instrument, and may be detected by comparing the distance of two stars as measured by the sextant, and as measured by a standard instrument, or deduced from calculation. Any error due to eccentricity of the limb may be determined by Lord Ellenborough's method. Select two stars of nearly equal zenith distances on opposite sides of the zenith, and from their altitudes calculate the latitudes. If the latitudes agree there is no error, otherwise half the difference of the latitudes is the error, additive when the latitude from the star on the equatorial side is greater, and sultractive if less. For suitable stars to observe see "Stars and Sextants," by J. A. Sprigge and others. O N Principle of the Sex- \ v tant. The principle of the sextant is based on the optical fact that the angle of incidence of a ray of light upon a mirror is equal to the angle of reflection, and hence " The angle between the first and last directions of a ray which has suffered two reflections in one plane is equal to twice the inclination of the re- flecting surfaces to one another." Let I and H repre- sent the index and hori- zon glasses of a sextant. Light from S falling on I is reflected at an equal angle to H, T, H IG. DEFINITIONS AND INSTRUMENTS. 29 and thence reflected to the eye at E, so that the object S would be seen by reflection in the direction Er, and if r were the horizon, the object's image would be seen on the horizon. IO represents the index bar when the mirrors are parallel, and IP when it has been moved to P. Produce PI to a, PH to 6, and EH to r. Now, since L HIP = L Sla = L EIP(Euc. 1. 15), the L HIE is bisected by IP, making L HIP = L EIP. Also, since L. 6HI = Z.EHP = L 6Hr (Euc. I. 15), the L rHI is bisected by 6H, making L rUb = L 6HL The L rHI = L HIE + L HEI (Euc. I. 32); therefore LrUb = i Z.HIE + J Z.HEI = Z.HIP + Z.HEI. Again, LrSb (= 6HI) = Z.HIP + L HPI; therefore HIP Taking away the common angle HIP, there remains L HPI = J Z. HEI ; and as Z. OIP = Z. HPI, because 10 and HP are parallel, it follows that the angle through which the idex bar IP has moved from zero is half the angle described by the image of the object. For this reason an arc of J on the limb is read as 1, and an arc of 60 will measure 120. ART. 87. Principle of the Vernier, or Nonius. The value of the smallest division on an ordinary sextant is 10, but the 10 ARC 10 8 FIG. 41. vernier attached to the index bar subdivides into single minutes and 10, 20, 30, etc. seconds. This is accomplished by taking 59 divisions of the limb for the length of the vernier, and dividing it into 60 equal parts ; Thus, 60V = 59L ,.IV = gL = L-^ that is, one division on the vernier is J of 10' = 10" shorter than one on the limb. Six of these vernier divisions are marked 1, and the whole length reads 10'. When the pointer or on the vernier coincides with a cut on the limb, the next vernier cut falls short of the one above it on the limb by 10", the second one by 20", the third by 30", etc. So that if the index is moved forward until the first, second, 3 o TEXT-BOOK ON NAVIGATION. third, etc., cut coincides with the one above it on the limb, the pointer will have moved 10", 20", 30", etc., and so on to the end of the vernier scale. When the 10' cut coincides, the pointer will have moved a whole division of the limb, i.e. 10'. There- fore, to find how far the pointer or has moved from the last limb division, it is only necessary to see what cut on the vernier coincides with a cut on the limb. The same rule holds good when the reading is "off" the arc, only in this case both scales must be read from left to right. Star Maps and Planispheres. A knowledge of the Star-groups or Constellations and the names of the principal bright Stars is essential to the Navigator who would make use of these objects to find the Latitude and Longitude at sea. Many Star-maps and "Planispheres" are in use, but it is unnecessary to mention here more than the following : Proctor's Star Atlas. Philip's Planisphere. Poole's Celestial Planisphere. On a Star-map or a Celestial Globe, the position of a star can be found from its R.A. and Dec. in the same manner as a place is located on a globe or chart by the Latitude and Longitude. Planispheres are designed to show the principal star-groups above the Horizon at a given hour. Philip's Planisphere consists of a moveable disc turning in a leather frame about the pole as centre. On the disc is a map of the heavens, hour circles being drawn for every two hours, the equinoctial, ecliptic, and circles of declination are also shown. On the circumference of the disc the days of the month are printed. Out of the leather frame an oval is cut, and around the edge of the frame are printed the hours of the day and night. To find the stars visible at any time, turn the disc until the date is abreast the hour ; the stars visible in the oval are those above the horizon at the time, and in the place for which the planisphere is constructed. A wire stretched across the middle of the ellipse represents the Meridian. The Planisphere gives by inspection the approximate Time of Meridian Passage of a star as well, as the times of Rising and Setting. It can also be used to find by inspection what bright stars are within a given time from the Meridian at any given hour of Greenwich or local Time. Tables of Meridian Passage of Stars. Table XLIV. of Norie gives the apparent times of Mer. Pass, of the principal stars at intervals of five days, and as a star crosses the Meridian about 4 m earlier on each successive day, it is easy to find the time for intermediate days. Table 27 of Eaper gives the time of Mer. Pas. of fifty-five bright stars on the first day of each month, and by subtract- DEFINITIONS AND INSTRUMENTS. 31 ing 4 m for each day elapsed, as before, the required time is approximately found for any day, or correction found in Table 27A. The Meridian Altitude of a selected star being computed from its Dec. and the Approximate Latitude, the sextant can be set to this angle ; and as the approximate time of Mer. Pass, is known, the observer can be in readiness to measure the altitude at the moment the star is on the Meridian, and so obtain the Correct Latitude. EXERCISES. 1. The meridional parts for lat. 38 25' are 2500 '08. Calculate those for 38 31'. Aiis. 2507*6. 2. The meridional parts for lat. 57 46' are 4267*97. Calculate those for 57 50'. Am. 4275*5. 3. The vernier of a sextant, the limb of which is divided to 10', has 40 divisions, coinciding with 119 divisions on the limb. With what accuracy may the instrument be read 1 Ans. 15". 4. The limb of a sextant is divided to 15', and 59 divisions on the limb are formed into CO divisions on the vernier. With what accuracy may it be read ? Ans. 15". 5. If every inch on a barometer scale be divided into 10 parts, show how to form a vernier such that the reading of the instrument may be taken to 005 of an inch. Ans. Take 9 divisions and divide into 20 equal parts. 6. Describe Philip's Planisphere. 7. Where may the Meridian Passage of a star be found 1 Is it given in mean or apparent time ? 8. On the arc readings, 34' 50", 34' 20", 34' 20" ; off the arc readings, 29' 40", 29' 0", 29' 20" ; find the Index Error and Semidiameter. Ans. - 2' 35" : 15' 57 '5". 9. On the arc reading, 28' 50" ; off the arc, 34' 40" ; find Index Error and Semidiameter. Ans. + 2' 55" : 15' 52 '5". 10. When the hydrometer marks 16, a ship's draught is 25 ft. 8 in. : find her draught at sea. Ans. 25 ft. 5 in. 11. At sea a ship draws 18 ft. 10 in. : what will her draught be in a dock where the reading on the hydrometer is 12 1 Ans. 19 ft. 1 in. 12. Between the F.W. mark and centre of disc, the distance is 6 inches. How much can the Plimsoll mark be submerged in dock, where hydrometer reads 9 ? How much is F. W. mark above water level ? Ans. 3-8 in. : 2'2 in. 13. Ship's displacement is 10,000 tons. Her tons per inch in salt water 40. What will be her change in draught on entering dock where hydrometer reads 10? Ans. 37 ins. 14. The area of a ship at load water level is 15,120 sq. ft. How many tons will sink her 1 inch in salt water ? Ans. 36 tons. 15. Barometer reads 29*92 in. : find what pressure per sq. in. this is equivalent to, taking the Specific Gravity of mercury to be 13 '598. Ans. 14-72 Ibs. 16. Barometer reads 29*65 inches : find corresponding readings in mm. and mbars. Ans. 753 '1 mm. ; 1003' 9 mbars. 17. Express a barometrical reading of 745 '6 mm. in inches and mbars. Ans. 29-35 ins. ; 994" 1 mbars. 32 TEXT-BOOK ON NAVIGATION. 18. Convert a reading of 983*4 mbars. to mm. and ins. Am. 29-04 ins. ; 737 '6 mm. 19. The temperature of the maximum density of water is 39 '4 F., and at which salt water freezes is 26 F. Express these as Cent, and Absolute readings. Ans. 4'1 C. and 277'1 A. ; - 3J C., 269| A 20. Alcohol boils at 68 C. ; absolute zero is - 273 C. What are the corresponding temperatures Fahr. ? Ans. 154'4F. ; 459-4 F. CHAPTER III. LOGARITHMS. ART. 1. In the expression N = a* ; " N " is the number obtained by raising "a" to the power of "a;"; keeping "a" fixed in value, and finding all the values of " N " obtained by performing the operation a x : where "05" may be anything fractional or integral, results in a table of logarithms : the fixed, or constant, number is called the " Base," the power chosen the " Logarithm " : thus a Logarithm is the index to which the Base must be raised to give the number. In common logs the base is always 10, hence the logs used in our problems represent the power to which 10 must be raised to equal the number. Logs are chiefly fractional, since very few numbers are exact powers of 10 : the whole number, or integral part of the log., is called the index, and the decimal part the mantissa. It is the decimal part that is found on pages 2-16 in Norie's Tables, or any other nautical tables. The index is added by the computer, and is found by subtracting unity, " 1," from the number of figures in the integral part of the quantity; this is evident from the following considerations : 10 = 1, 10 1 = 1, 10 2 = 100, 10 3 = 1000. Thus from definition log 1 =0 log 10 =1 log 100 = 2 log 1000 = 3 ; and so on. Hence the logs of all numbers between 1 and 10, those which only have one digit in the whole number part, fall between and 1, and so are wholly decimal, having for index. Again, the logs of all numbers between 100 and 1000, which are ex- pressed by three digits in the whole number part, fall between 2 and 3, and so always have 2 for the index. The index of a wholly decimal number is negative, and is found by adding unity, " 1," to the number of cyphers im- mediately after the decimal point, and before the first significant figure, and putting a minus sign ( ) over it. D 34 TEXT-BOOK ON NAVIGATION. Logs of all numbers which are expressed by the same figures, but differ only in the position of the decimal point, have the same mantissa. A minus index is made positive by adding 10 to it. ART 2. Finding the Logarithms of Numbers. In Table, pp. 2-16, Norie, the first column is headed " No." ; here are entered all numbers from 100 to 999. The other columns are headed to 9 ; and the end one " Diff." : the first three figures of the number whose log is required are found under " No." ; the fourth figure represents the column to be used; the number found in this column abreast the figures under "No." is the mantissa of log required. The index must be placed as indicated above. Should there be fewer than four figures in the number, add cyphers to make up four and take out mantissa, using index corresponding to original number. Examples. Numbers. Logarithms. 1. 6 ... . 0-778151 2. 34 3. 420 4. 7-856 5. 38270 6. -5 7. '065 8. -0013 9. -00000671 ... 1-531479 2-623249 0-895201 4-582858 1-698970 2-812913 3-113943 6-826723 NOTE. The negative index is generally_changed into a positive for Nautical Computations by adding 10, so that 1 becomes 9, 2 becomes 8, etc. When the number contains more than four figures, take out the log. of the first four ; then find the correction for the remain- ing figures by multiplying the "Diff." by these figures, and cutting off as many to the right. The remaining quantity is the correction to be added. Number. Logarithms. Diff. 185 1. 2-340401 ...... 0-369216 i Examples. ber. Logari 0401 ...... 0-369 74 T^ ---- 740 0-369290 2.55300-017 ...... 4-742725 ^J! 017 -- 546 4742725 _^_ ,1326 3. 86430000 ...... 7'936664 NOTE. There is no correction for O's at the end of a number. LOGARITHMS. 35 EXERCISES. 1. Find the Logarithms of the following numbers : 3409 3 47 50 100 275 57-29 305-74 1-0009 75230 62-7014 175-008. 23002-7 164-3019 21550-007 23460 7-5 163-45 704-009 32636-04 5 049 0036 001204 006728 00028041 0000753 0346009 075604 3070007 ART. 3. Finding the Numbers corresponding to Given Logarithms. When searching for the number corresponding to a log., no account is taken of the index, the mantissa is looked out in the tables, and will be found to either correspond exactly with an entry there, or fall between two entries. In the event of exact coincidence, the figures in the " No." column and at the top of the column where the log. is found, give the first four figures of the Number required: the place of the decimal point depends on the Index, and is found by counting off from the left one more than the number of figures in the index. Should the index call for more than four digits in the Number, cyphers are added to make up the required quantity. Should the log. not agree, as is more often the case, the next less entry is taken from the tables and subtracted from the given log., cyphers are added to this difference and then divided by the number abreast in the " Diff." column. The figures in the dividend are extra ones to be placed to the right after the four figures taken from the tables corresponding to the less entry used. Examples. Given logs. Numbers. 1. 0-113943 1-3 2. 1-791480 ... ... 61-87 3. 2-724931 530'8 "When the Index exceeds three, and more than four figures are required, the log. next less in the tables is taken out, and the remainder divided (as in Division of Decimals) by the "Diff." gives the additional figures required. 36 TEXT-BOOK ON NAVIGATION. Examples. Given logs. Numbers. 1. 6-490379 3092993 490239 Diff. 141)140-0(993 1269 1310 1269 410 423 2. 5-820006 660703 820004 Diff. 66)2 00(03 198 2 3. 4-875003 74990 When the index is negative, and the number therefore a decimal, the position of the first figure (of value) is indicated by the index ; thus, 1 shows it will occupy the first place after the decimal point, 2 requires the first figure (of value) in the second place, etc. Examples. . Logs. Numbers. 1-432488 -2707 2-715339 -05192 4-934550 -0008601. NOTE. These index numbers when increased by 10 become 9, 8, 6. EXERCISES. 2. Find the numbers of which the following are the Logarithms : 3-203577 2-145818 1-700271 0-715920 0-004382 4-681484 5-690196 6-148900 7-040203 4-806044 5-175801 6-531481 7-000435 5-361917 4-041393 6-049606 0-000434 3-670154 7-276003 5-380032 NOTE. Since logarithms are merely the powers of "10," which would make the numbers, to multiply quantities, we do as in Algebra, add the powers. To divide quantities, subtract the powers. To perform Involution, multiply the log. by the power. To perform Evolution, divide the log. by the root. LOGARITHMS. 37 ART. 4. Multiplication by Logarithms. Eule. Add together the logarithms of the numbers, and find the number corresponding to the sum of logarithms. Examples. 1. Multiply 234 by 1*76 by common logarithms. 234 log 2-369216 176 log 0-245513 Ans. 411-8 log 2-614729 2. Multiply 32640 by '00876. 32640 log 4-513750 00876 log 7-942504 Ans. 285-9 log 2'456254 3. Multiply together 317, 24'781, and -067. 3-17 log 0-501059 24-781 log 1-394119 067 log 8-826075 Ans. 5-263 log 0*721253 4. Multiply 760-56 by '0001877. 760-56 log 2-881133 0001877 log 6-273464 Ans. -1427 log 9-154597 5. Multiply -89472 by -003708. 89472 log 9-951687 003708 log 7-569140 003318 log 7-520827 Remark. It is evident that the product of a whole number by a whole number is a whole number, and the product of a decimal by a decimal is a decimal ; but when a whole number and a decimal are multiplied together, the answer contains a whole number if the sum of indexes amounts to ten or more (in which case ten is rejected) ; but if the sum is less than ten, the answer is a decimal. EXERCISES. 1. Multiply 479 by 41. 2. 4672 by 5. 3. 1037- 5 by 269. 4. 35890 by 10. 5. 16-536 by 42-90. 0. 12-45 by 3'84. 7. 7-5642 by 279-008. 8. 2-3467 by 123'006. 9. 164-3019 by 21550'007. TEXT-BOOK ON NAVIGATION. 10. Multiply 356-8 by '0167. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 001204 by -006728. 32415-5 by -0002. 046009 by -00028041, 1000012 by -00259. 100 by -005. 7642-005 by 3-26507- 999999 by -1. 0867 by 235-6004. 98421 by '234. 00765 by -23461. 28550 by -0444. 30000000 by -00765. 23. Find the product of 2 '36 x 19-742 x '3897. 24. 104-26 x -018 x 95-41 x '8. ART. 5. Division by Logarithms, Rule. Subtract the log of the Divisor from the log of the Dividend, and then find the corresponding number. Examples. 1. Divide 37640 by 4705. 37640 log 4-575650 4705 log 3-672560 Ans. 8-000 log 903090 2. Divide 246 '78 by -0272. 246-78 log 2-392310 0272 log 8-434569 Ans. 9073 log 3-957741 3. Divide -14 by -09876. 14 log 9-146128 09876 log 8-994581 Ans. 1-418 log 0-151547 4. Divide 207 -4 by 3709-1. 207-4 log 2-316809 3709-1 log 3-569269 Ans. -05592 log 8-747540 6. Divide '0076438 by -92. 0076438 log 7-883310 92 log 9-963788 Ans. -008308 log 7 '919522 Remark. The answer always contains whole numbers (one more than the index) when the Divisor is less than the Dividend, but is all decimals when the Divisor is the greater. LOGARITHMS. 39 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. EXERCISES. Divide 32636-04 by 704-009. 687 -31 by 2'34. 45-2 by '005. 985712 by -0056. 9 23 by 2 001. 64238 by '026. 99 by -0000745. 35161-9 by 34'24. 746202 by '52496. 197 '65773 by '00139. 35 by 34-24063. 9 by 874303. 16192 -2 by '01086. 1-3 by '0000000290009. 001 by 1000. 000876 by '0000438. 012 by -009847. 76591 by '13. Find the value of 2-13 x 856-4 087 * 00175 x 3872-4 7'69 x -043* * 8-32 x '096 x 248-109 1-74 x 23'42 x l4lF6 r * 100-108 x -36749 00984 x 468-06 ' 0756 x -00998 x -247 -0001 x '0137 x -142 ' ART. 6. Involution by Logarithms. Rule. Multiply the log of the number by the required Power, and then find the corresponding number. Example. Required the 7th power of 1 '09. 1-09 log 0-037426 7 Ans. 1-828 log 0-261982 EXERCISES. 1. Required the square or 2nd power of 23 '4. 2. ,, cube or 3rd power of 1'876. 3. ,, 5th power of '08967. A 9th power of '1039. 10 ^ power of '367. ART. 7. Evolution by Logarithms. Rule. Divide the log of the given number by the required Root, and then find the corresponding number. 40 TEXT-BOOK ON NAVIGATION. Examples. 1. Required the square root of 10816. 10816 log 2)4 034066 Ans. 104 log 2-017033 2. Required the cube root of 42875. 42875 log 3)4-632204 Ans. 35 log 1-544068 3. Required the 5th root of 9 '618. 9-618 log 5)0-983085 Ans. 1-573 log 0-196617 4. Required the 7th root of 234-68. 234-68 log 7)2-370476 Ans. 2-181 log 0-338639 5. Required the cube (or 3rd) root of _ '004913. 004913 log 3)3-691347 Ans. -17 log 1-230449 6. Required the cube root of '06174. 06174 log 2-790567 or - 3 + 1 '790567 3) -3 + 1-790567 Ans. -3952 log f-596856 7. Required the 5th root of '007652. 007652 log 5)3-883775 (3 = 5 + 2) Ans. -3774 log 1-576755 EXERCISES. 1. Required the square root of 306 '25. 2. cube ,, 46-656. 3. 5th -002435. 4. cube 12167000. 5. ,, 7th -00000002097152. Eemark. With a Table of Logarithms containing only six places of figures, such as is used for Navigation, the answers in the preceding exercises cannot be expected to be correct to more than four or five places. Of course, it must be understood that whenever logs are added in the calculations of problems in Navigation and Nautical Astronomy, it means that the terms are multiplied together, and when logs are subtracted, it means division. This is important to remember when one of the terms is zero (or 0). LOGARITHMS. 41 ADDITIONAL EXERCISES. Find by logs the values of the following : 1. V20675; 2. (29-865) 2 and (98 7652) 2 . 3. (91-543)' x V8-794W; V8794-08 4. V276509 ART. 8. To find the Logarithm of a Given Ratio (Sin, Cos, Tan, etc.). In Table xxv.,pp.264-353,Norie, are entered the logarithms of the Sines, Cosines, Tangents, etc., of the various angles from to 90, given to every half-minute, two pages being reserved for each degree : on the top of the page the degrees from to 44 in consecutive order, the minutes and half -minutes being entered in the column on the left-hand side, the names of the ratios at the top ; and in columns between two ratios the differences for each second of arc to 30". At the bottom of the page the degrees from 45 to 89 are entered, the names of the ratios, and on the righi^hand side reading upwards the minutes and half-minutes of arc. In order that there shall be no minus indices, 10 is added to each index before tabulation. Pages 240-263 at the beginning of the table are devoted to the Sines of small angles from to 4 to each second of arc, and to the Cosines of large angles from 86 to 90. To find a log, turn up the number of degrees on the top of the page if the angle is less than 44, at the bottom of the page if the angle exceeds 44 ; find the minutes on the left if the degrees are at the top, and on the right if the degrees are at the bottom, and abreast the minutes in the column corresponding to the ratio required is the log. Examples. Lsin 23 25' = 9'599244 Lsin 68 36' = 9'931229 L sin 27 431' = 9-667666 Lsin 6414|'= 9'954549 L cosec 75 56 f = 10-013222 L cot 37 28' = 10-115412 Lcos 40 36' = 9-880397 Ltan 47 28' =10*037440 Lcos 65 22V = 9-619800 L sec 14 19 r = 10-013701 L cosec 51 40' =10-105454 Ltan 13 47' = 9'389724 If the angle exceeds 90, subtract the amount from 180 and take out the same ratio of the remainder ; or take 90 from the angle and take out the complementary ratio of the remainder. Examples. L sin 102 47' = L sin 77 13' = L cos 12 47' = 9 '989100 L cosec 111 15' = L cosec 68 45' = L sec 21 15' = 10-030580 L cot 98 32' = L cot 81 28' = L tan 8 32' = 9 '176224 To correct for odd seconds in an angle, find the amount in 42 TEXT-BOOK ON NAVIGATION. the proper " Diff." column corresponding to the seconds required, add to the log for sines, tangents, and secants, and subtract for cosines, cotangents, and cosecants. Remark. The beginning part of this table gives the L Sines of small angles (0 to 4) to seconds of arc : or the L Cos of large angles (86 to 90) : seeing that sines and cosecants are comple- mentary, and that the sines and tangents, also the cosecants and cotangents of small angles are approximately equal, by sub- tracting the L Sines from 20 we have the corresponding L Cosec, and when the angle is very small (0 to 30') the L Cot. ; this avoids using the very large differences tabulated for 30". Examples. L sin 56 37' 25" = 9 '921726 corr. for 25" = +35 Lcos 37 26' 47"= 9'899779 17" = -27 L sec 55 14' 38" .= 10'244061 8" = +24 L cosec 104 17' 20" = 10-013648 10" = - 5 L cos 88 12' 26" = 8'495333 (large " cos " table). ART. 9. To find the Angle corresponding to a Logarithm. Reverse the above process, looking out the log in its proper column, and taking the next less log to subtract from the given log in the case of sines, sees, and tans, and the next greater one in the case of cosines, cosecs, and cots, to subtract the given log from ; this difference referred to the corresponding " Diff." column gives the extra seconds. Examples. 9 -764318 = L sin 35 32' 3" diff. of 10 gives 3" 10124397 = L tan 53 5' 45" ,, 65 15" 9 -914386 = L cos 34 48' 25" 36 25" 8-452450 = L sin 1 37' 27" (small " sine " table) 9-796824 = L tan 32 3' 41" diff. of 51 gives 11" 10-495387 = L cot 17 43' 27" 195 27" 9 -918645 = Lain 56 0' 50" 28 20" CHAPTER IV. SOLUTION OF PLANE TRIANGLES. ART. 10. Trigonometrical Ratios. Comparisons can only be made between two quantities which refer to the same kind of thing. A Ratio is the comparison of two quantities, and in Trigonometry the quantities compared are the lengths of the sides, or functions of the angles. A Ratio may be expressed by writing the two quantities with two dots (:) between them, or by forming a vulgar or decimal fraction of them. In every triangle there are three sides, and since two are used to form the ratio, it follows that these can be arranged in "six" different ways ; and hence there are " six " Trigonometrical Ratios. Each one has received a definite name ; thus there are sines, cosines, tangents, cotangents, secants, and cosecants. In any right-angled triangle the side opposite the right angle is always called the HYPOTENUSE. The side which with the hypotenuse forms the angle is called the BASE. The side opposite the angle under consideration is called the PERPENDICULAR. The ratio is called "sine" or shortly "sin." hyp base cosne tangent cotangent " "secant" " cosecant " hyp base " hyp perp Thus in triangle PQR, right angled at Q, the hypotenuse is always PR ; and when the ratios of P are considered, PQ is the base, RQ is the perpendicular. .But when the ratios of R are under consideration, then RQ is the base, PQ is the perpendicular. "tan." " cot." "sec." "cosec." R FIG. 42. 44 TEXT-BOOK ON NAVIGATION. Thus from definition "-s -*-a . -'-s -'-S -'-a *~F-|| Hence and - B - S c,R,g T/^V tan R = RQ RQ PQ PR RQ cosec R = 15 cotR sec R sin P = cos R tan P = cot R sec P = cosec R and since P and R are complementary, this is expressed by saying that The sine of an angle is equal to the cosine of its complement, and vice versa. The cosine of an angle is equal to the sine of its complement. And so with tan and cot : sec and cosec. It will be seen that since cosec = sin sec cos cot = _ tan R 2 the cosec is the reciprocal of the sin, sec of the cos, and cot of the tan. This fact is useful, as it enables a divisor to be replaced by a multiplier in a formula; the same result being obtained whether we divide by sin, cos, or tan ; or multiply by their reciprocals cosec, sec, or cot. ART. 11. Changes in Sign and Magnitude of Angles. Consider two lines POQ and ROT meeting in O at right angles. It is necessary, in order to know in what direction a line is drawn, that some differ- ence should be made when a line is drawn perpendicularly to TR up or down, or when perpendicu- larly to PQ towards R or T ; this is done by the use of the signs -f and . Lines drawn perpen- dicularly up from TR are con- sidered -f,down . Lines drawn perpendicularly to PQ to the right are considered +, to the left . T M 6 FIG. 43. SOLUTION OF PLANE TRIANGLES. 45 Let a line with centre O revolve from the position OR, in a direction against the hands of a clock. In going from OR to OP, it traces out all angles from to 90, and in going from OP to OT all angles from 90 to 180 ; larger values need not be considered, as the angles and arcs dealt with in this book do not exceed 180. Suppose the revolving line to take up positions indicated by RI, R2, Ra, RI, RS, Re- Draw perpendiculars from these points meeting TR in MI, M 2 , M 3 , M 4 , M 5 , M 6 ; Then sin sin Since ORi = OR^ = . . . OR 6 , it follows that the above fractions, which represent the values the sin has as the angle increases, also increase as the line revolves from OR to OP, and decrease as the line revolves from OP to OT. In the position OR or OT the perpendicular vanishes, consequently sin = sin 180 = Also, when the revolving line coincides with OP, the perpendicular coincides with and is equal to it, or sin 90 = 1 All the time, however, the perpendicular has been above TR and is reckoned +. We arrive at the following conclusions : As the angle increases from to 90, the sine takes all values from to 1. As the angle increases from 90 to 180, the sine takes all values from 1 to 0. In both the above cases the sine is positive. Again cos cos R 4 OM 4 = The value of the cosine depends on the base, and it is seen from the figure that as the angle increases from to 90 the base gradually diminishes, and the cosine with it ; also when the angle increases from 90 to 180 the base increases, but is now 46 TEXT-BOOK ON NAVIGATION. drawn in a contrary direction, and so of opposite sign. Hence in this case As the angle increases from to 90, the cosine takes all values from 1 to 0. As the angle increases from 90 to 180, the cosine takes all values from to 1. The tangent depending on the ratio of perpendicular to base follows the same rule as the sine in increasing as the angle increases ; and follows the cosine by becoming a negative value in the quadrant from 90 to 180. Reciprocals of ratios follow the same rule as the ratios themselves, only as regards sign. ART. 12. Supplementary Angles. Make angle TOS = angle ROM, where OR = OS ; ST and RM ^ perpendiculars. The angles ROM and SOM are S R supplementary. O M .'. sin ROM = sin SOM. PIG. 44. Again /. cos ROM = -cos SOM; or sine of an angle equals the sine of its supplement ; but cosine of an angle equals the cosine of its supplement with sign prefixed. ART. 13. Summary. As angles increase, the sines, tangents, and secants increase. As angles increase, the cosines, cotangents, and cosecants decrease. All ratios are positive (-f) when angles fall between and 90". When angles fall between 90 and 180 all ratios except sines and cosecants are negative ( ); these two are positive (+). Sin A = cos (comp. A) Cos A = sin (comp. A) Tan A = cot (comp. A) Cot A = tan (comp. A) Sin A = sin (supp. A) Cos A = cos (supp. A) CosecA= - - sin A SecA= COS A CotA= _!_. tan A SOLUTION OF PLANE TRIANGLES. 47 ART. 14. Solution of Right-angled Triangles. In all triangles there are six parts (three sides and three angles). In a plane triangle the sum of the three angles is 180 ; if three parts be given (one given part must be a side), the other parts may be determined. In right-angled triangles a knowledge of the ratios enables any one to effect the solution; one of the given parts being a right angle, the other two parts which are given can only be (1) two sides, or (2) one side and an angle. The tables are needed, because in them are tabulated the logarithms of all ratios of angles from to 90; consequently, when an angle is given, the log of any ratio of that angle may be found ; or if an angle is required, knowing the log of the ratio, the angle may be looked up. Considering the first case where two sides are given. These sides put in the form of a fraction will make four different ratios of the two acute angles, and there are four different ways of starting the solution, all equally correct. Remark. All plane triangles should be drawn to scale. To do this the student requires a pen or pencil bow and a six-inch boxwood protractor. The advantage is that the problem may be solved graphically, and the answer obtained is a useful check on the numerical calculation, and so serious mistakes avoided. Angles are laid off or measured as with a Field's parallel rule, by placing the plain edge on the line, and the centre point at the point at which the angle with the line is to be made or measured. On the space at disposal depends the scale of the figure, and this scale should be entered at the side. Thus if we wish to represent a length of 215 inches, and choose a scale of -fa inch to 10 inches, the total length of the line would be 21 J tenths of inches, or 2*15 inches. Again, to represent a length of 425 miles on a scale of ^j inch to 25 miles, the length would ^ e 4 A 5 ~ * 1*0 inches = 1*7 inches : or if in such a figure the length 2'6 X 25 of a side is 2*6 inches, its actual size would be .. miles = 650 miles. Example 1. In triangle ABC, B = 90, AB = 368, BC = 215. Find the angles, and AC. Graphical Solution. Draw horizontal line AB and make it Yo 1 mm - = 36-8 mm. in length : at B draw BC perpendicular to AB and make it Zfif- = 21-5 mm. in length. Join AC. Measure AC = 42 '5 mm. /. AC = 425 BAC = 30 BCA = 60 4 8 TEXT-BOOK ON NAVIGATION. Trigonometrical Solution. Since AB and BC are known, we may say- Scale 1 mm. =10 ^,C AB AB -J77T = cot A or ^77^ = tan C 1 90' FIG. 45. BC _ , . BC Using the first ratio , . AB 368 cofcA = BC = 2i5 Apply logs, and adding 10 to top index, since tabular cotangent log has been increased by 10 AB ... 368 log = 12-565848 BC ... 215 log = 2-332438 90 30 17' 42" L cot = 10-233410 /. A = 30 17' 42" and C = 59 42 18 To find AC, we form a fraction, putting side wanted in the numerator and side known in the denominator, and writing down the name of ratio so formed. This gives rise to several ways of solution, as we may say AC AC AC AC -po = sec A or gp = cosec A or ^-~ = cosec C or g^ = sec C .'. AC = AB x sec A Apply logs, rejecting 10 from index, added to make a tabular secant. AB ... 368 log = 2-565848 A 30 17' 42" L sec = 10-063768 AC 426-2 log = 3,2-629616 Ans. A = 30 17' 42" ; C = 59 42' 18" ; AC = 426-2. Example 2. In triangle PQR, Q = 90, PR = 17 '8, PQ = 15'5. Solve. Graphical Solution. Draw a horizontal line RQ ; at Q make QP perpen- dicular to RQ and -^- of an inch in length. Take j - of an inch as a radius, and with P as centre, cut RQ in R. Then RPQ is triangle. Measure RQ = 8 -8 sixteenths /. RQ = 8 '8 RPQ = 30 PRQ = 60. Trigonometrical Solution. Apply logs. PQ PR 15-5 log = 11-190332 17-8 log = 1-250420 R ... 60 33' 1" L sin = 9-939912 P = 90 - R = 29 26' 59" Again- FIG. 46. /. RQ = PQ x cot R SOLUTION OF PLANE TRIANGLES. 49 Apply logs. PQ 15-5 log = 1-190332 R ... 60 33' 1" L cot = 9-751752 RQ 8-751 log = ^0-942084 Ans. P = 29 26' 59" ; R = 60 33' 1" ; RQ = 8 "751. The second case, where one side and an angle are given, is solved by remembering that the ratio to work from is formed by placing the side wanted in the numerator and the side known in the denominator. As before, there are four different ways, all equally right, of starting. Example 3. In triangle LMN, M = 90, N = 55 15', LM = 1342. Solve. Graphical Solution. Draw a horizontal line NM : at M draw ML perpendicular to MN and make it equal to^}^ inch = 1*342 inch. At any point O in MN make an angle = 55 15' : through L draw LN parallel to this line. Then LNM = 55 15'. Measure NL = 1'62 inch /. NL = 1620 N M MN = *92 /. MN = 920 Scale r-iooo NLM = 35 Fm. 17. Trigonometrical Solution LN M ,MN V^M. = cosec N, and ^^ = cot N JjiVl JjlVl /. LN = LM x cosec N, and MN = LM x cot N Apply logs. LM...1342 log= 3-127752 log = 3-127752 N 55 15' L cosec = 10*085315 L cot = 9-841187 1633 log = 1^3-213067 931 log = ^2-968939 L = 90 - N = 34 45' Ans. L = 34 45' ; LN = 1633 ; MN = 931. EXERCISES. RIGHT-ANGLED TRIANGLES (PLANE). 1. In triangle ABC, AB = 503, AC = 357, C = 90. Find A, B, and BC. 2. In triangle PQR, PQ = 371, QR = 204, Q = 90. Find P, R, and PR. 3. In triangle OPB, OP = 125*7, PB = 695, O = 90. Find P, B, and OB. 4. In triangle ABC, A = 64 41', C = 90, AB = 1037. Find AC and BC. 5. In triangle ABC, B = 34 17', = 90, BC = 905. Find AB and AC. 6. In triangle ABC, AB = 147, BC = 116, C = 90. Find A, B, and AC. 7. In triangle PQR, P = 33 14', PQ = 4372, Q = 90. Find PR and RQ. 8. In triangle ABC, C = 90, AC = 706, BC = 981. Find A, B, and AB. 9. At 145 ft. from the base of a flagstaff its angle of elevation is observed to be 32 16'. Find the height. 10. The difference of latitude made good is 236 -5 miles south, the departure 215*7 miles west. Find the course, and distance sailed. 11. The length of the shadow cast by a vertical post is 39 ft. 6 in. when the sun has an elevation of 21 32'. Find the height of the post. 12. From a ship a promontory was observed to be 3^ points on the bow, E 50 TEXT-BOOK ON NAVIGATION. and after making 7 miles on her course, it was found to be abeam. What distance was the vessel from the point at the two bearings ? 13. A lighthouse, whose top is 200 ft. above the sea-level, marks a danger situated at three cables' length from the shore. Required the vertical "danger angle." 14. From a reef awash at low water, the angle of elevation to the summit of a hill distant 2050 yards in a horizontal direction is 1 12' ; height of the eye, 6 ft. ; tidal rise, 21 ft. Required the height of the hill above high water. ART. 15. Solution of Oblique-angled Plane Triangles. Triangles which do not contain a right angle are called oblique-angled. Facts to be remembered in connection with triangles are The sum of the angles is 180. Equal angles are subtended by equal sides. The greater side is opposite the greater angle. Any two sides are together greater than the third side. According to the parts given the solution falls into three different cases. 1. Given two sides and an angle opposite to one of the known sides. Or, given two angles and a side. 2. Given three sides. 3. Given two sides and the angle contained by the known sides. CASE I. For the first case, the following property is used : Sides are proportional to the sines of the opposite angles. ACB is a triangle : draw CD at right r angles to AB. Then ^ = sinA I. AC and ^ = sin B II. BO Divide I. by II. . CD _._ CD _ sinA FIG. is. '''AC * BC~sm"B ,, , ^^- v BC sin A that -^ X 75^ = -j - ~6i3 sin B or, cancelling CD, BC _ sin A AC sin B Now, if a perpendicular were dropped from A on BC, it might be proved in a similar manner that AC = sin B AB sin C Hence the proposition is established. SOLUTION OF PLANE TRIANGLES. Example 1. Given PQ = 350, PR = 367, and Q = 54 30'. Solve. Draw a horizontal line RQ, and at Q make angle 54 30', also on a scale of I'l inch to 350 make PQ I'l inch long. Find length of PR on our scale _ 1-1 x 367 _ ^ 15 inchj take thig length in 350 Scale 1-1" = 350 P compass, and with P as centre draw an arc cutting QR in R. Then PQR is triangle. PRQ = 51 RPQ = 75 RQ = 1-38 inch RQ = 439. Two sides and an angle opposite to a given side ; hence the part to be found is the angle opposite to the other given side ; here we form a fraction having for its numerator the sine of the angle wanted, and for its de- nominator the sine of the given angle. This fraction is equal to one formed from preceding proof. FIG. 49. Thus sinR _ PQ sin Q PR Apply logs. PQ Q PR PR ... 350 log= 2-544068 54 30' L sin = 9-910686 12-454754 367 lo* = 2-564666 50 55' 57" L sin = 9 890033 Q = 54 30' 0" R = 50 55 57 R'=105 25 57 180 P = 74 34 3 Again, to find the third side RQ : form a fraction with the side wanted (RQ) as numerator, and one of the given sides (PR) as denominator ; this fraction equals one formed by placing sine of angle opposite RQ (sin P) in numerator and sine of angle opposite PR (sin Q) in denominator. RQ _ sinP PR ~ sin Q . -r,-. PR x sin P " RQ= sinQ = PR X sin P x co?ec Q In order to make all the logs added, sin Q as a divisor is changed into cosec Q as a multiplier since cosec Q = sin Q TEXT-BOOK ON NAVIGATION. PR P Q 3G7 log = 2-564666 74 34' 3" L sin = 9-984052 54 30' L cosec = 10-089314 RQ 434-5 log = $2-638032 Ans. P = 74 34' 3" ; R = 50 55' 57" ; RQ = 434-5. Example 2. Given C = 118 30', B = 21 22', and AC = 67 -5. Solve. Graphical Solution. Draw horizontal line AC of length inches = -68 inch. At C draw CB making an angle of 118 30' with AC, and through A draw AD parallel to CB. At A draw AB making an angle 21 22' with AD, to meet CB in B. Then ABC is triangle. Measure A = 40 AB = 1-6 inch .'. AB = 160 BC = 1-2 /. BC = 120 Scale 100 67*5 FIG. 50. Trigonometrical Solution. A + B + C = 180 /. A = 180 - (B + C) = 180 - 139 52' = 40 8' To find BC BC _sinA AC BC = sinB AC x sin A AC A B BC sinB 67-5 log = 1-829304 40 8' L sin = 9-809269 21 22' L cosec = 10 -438499 = AC x sin A x cosec B To find AB 119-4 log = $2-077072 AB-_ AC ~ AB = AC C B AB sinC sin B AC x sin C sinB ... 67'5 log = 1-829304 118 30' L sin = 9'943899 21 22' L cosec = 10-438499 x sinC x cosecB (Use sin 61 30', the supplement.) 162-8 log = $2-211702 Ans. A = 40 8' ; AB = 162-8 ; BC = 119-4. Example 3. Given AP = 1134 ; PM = 875 ; A = 35 20'. Solve. Here is given the angle opposite to the smaller side, and when the triangle is drawn to scale, it is found that two triangles can be drawn from the data ; there are thus two solutions, and it is known as the ambiguous Graphical Solution. Lay off a horizontal line of unlimited length, and at one end A make an angle equal to the given angle, viz. MAP = 35 20'. SOLUTION OF PLANE TRIANGLES. 53 Scale 1 =1000 Along AP measure the longer side = 1134, that is 1*13 inch ; and with centre P and distance equal to the shorter side 875, that is '87 inch ; describe an arc cutting AM in M and M'. Thus both PM and PM' = 875. Hence there are two triangles, PAM and PAM', which satisfy the given conditions. Measure AM = 1-5 inch, that is 1504 AM' = -35 inch, that is 350. Trigonometrical Solution. To find M in triangle APM sin M AP FIG. 51. sin A sin M = AP A PM M A PM AP x sin A PM 1134 log = 35 20' L sin = 3-054613 9762177 875 12-816790 log= 2-942008 48 32' 55" L sin = 9 -874782 35 20 A + M 83 52 55 180 7 5 P ... 96 To find AM in triangle AMP AM _ sin P PM sin A .-. AM = PM x s ! n P = PM x sin P x cosec A PM P A sin A 875 log = 2-942008 96 7' 5" L sin = 9*997519 35 20' L cosec = 10'237823 1504 log = ^3-177350 AM To find M' in APM' Since PM = PM', angle PMM' = angle PM'M Hence M' = 180 - M = 131 27' 5" A= 35 20 M' + A = 166 47 5 180 P = 13 12 55 To find AM' in triangle APM' AM' _ sin P PM' sin A AM' ^ sin A = pM , x sln p x cosec A M 54 TEXT BOOK ON NAVIGATION. PM' P A 875 log= 2942008 13 12' 55" L sin = 9 "359096 35 20' L cosec = 10 '237823 cos AM' 345-9 log = $2-538927 Ans. M = 48 32' 55", P = 96 7' 5", AM = 1504 or 131 27 5,- or 13 12 55, or 345 "9 CASE IL Given three sides, to find the angles. The half sum of the three sides is denoted by the letter s. If ABC is a triangle, the formula for finding an angle is A . / 8 . (g - BC) V AB.AC where s = J(AB + AC + BC) This is proved in any " Trigonometry," and may be expressed in writing thus Cosine of half an angle is equal to the square root of the amount obtained by dividing the product of the half sum of sides and the remainder by the product of the sides which contain the angles. The remainder is obtained by subtracting the side opposite the angle required from the half sum of the sides. Example 4. Given AB = 362, BC = 314, AC = 276. Solve. Graphical Solution. Draw a horizontal line AB in length = inch = 1'21 inch. With centre A, radius = = '92 inch, describe an arc, and with centre B, radius = f J = 1'05 inch, describe another arc cutting first arc in C. Then ABC is the triangle. A measures 57| B 47* C 74 Fio. 52. Trigonometrical Solution. To find C - AH- 2 V A AC.BC AB = 362 AC = 276 ... ... log BC = 314 ...... log 2)952 s = 476 log 2-677607 AB = 114 log 2-056905 2-440909 2-496930 24-734512 4-937839 4-937839 2)19-796673 37 41' 37" L cos = 9-898336 2 C = 75 23 14 SOLUTION OF PLANE TRIANGLES. Again, to find B T> / p . ~ 38 2 = V AB . BC~ AB = 362 log = 2-558709 BC = 314 ... ... log = 2-496930 s = 476 log - 2-677607 s - AC = 200 log = 2-301030 55 23 46' 15" 2 24-978637 5-055639 2)19-922998 L cos = 9-961499 6-055639 B = 47 32 30 A = 180 - (B + C) = 180 - 122 55' 44" = 57 4' 16" Ans. A = 57 4' 16" ; B = 47 32' 30" ; C = 75 23' 14". Case III. Given two sides and the contained angle; to solve the triangle. The solution depends on the following property. In any triangle, the tangent of half the difference of two angles bears the same ratio to the tangent of half the sum of the same two angles as the difference of the sides opposite the angles bears to the sum of the same sides. This is expressed as a formula, thus tan i (A - B) _ EC -AC tan~f~(A + B) ~ BC + AC A - B BC - AC and since tan tan 2 A + B 2 A + B BC + AC = cot 2 tan B and the formula used for the solution reduces to tan A - B BC - AC BC + AC , . cot x- Example 5. Given BC = 145 '5, AC = 98 -1, and C = 87 20'. Solve the triangle. Graphical Solution. Draw a horizontal line BC = 1-46 inch ; at C make angle = 87 20' by drawing QQ.-l CA = ~ = -98 inch long. Measure A = 58 B = 35 AB = 171 inch = 171 TEXT-BOOK ON NAVIGATION. Trigonometrical Solution. A + B = g()0 _ C = 90 - 43 40' = 46 20' tan A - B BG - AC BC + AC ' C 2 145-5 - 98-1 40 - cot 43 40' 47-4 log = 1-675778 .. 43 40' L cot = 10-020220 11-695998 ... 243-6 log= 2-386677 11 31' 20" L tan = 9 '309321 46 20 A B To find AB AB AC sin C sinB 57 51 20 J34,48 40 . _ AC x sin C . ~ . /* .*. AB = r ~ = AC x sin C x cosec B AC C B sin B 98-1 log = 1-991669 87 20' L sin = 9 '999529 34 48' 40" L cosec = 10-243461 AB 171-7 log = ^2-234659 Ans. A = 57 61' 20" ; B = 34 48' 40" ; AB = 171 '7. ART. 16. The proofs of the rules used in the two last cases are inserted here for the benefit of any student who may not have a " Trigonometry " handy. To express the cosine of an angle in terms of the sides. C D FIG. 54. B B FIG. 55. Let ABC be a triangle ; one figure represents B acute, the other B obtuse. Draw CD at right angles to AB or AB produced. By Euclid II. 13 and 12 AC 2 = AB 2 + BC 2 - 2AB . BD or AC 2 = AB 2 + BC 2 + 2AB.BD using small letters for sides and observing that SOLUTION OF PLANE TRIANGLES. 57 BD . BD = cos B also ~- = cos CBD .-. BD = BC . cos B = a . cos B /. BD = BC . cos CBD = a . cos B Insert these values for BD .-. 6 2 = c 2 + a, 2 - 2ca cos B or 6 2 = c 2 + a 2 - 2c . a cos B c 2 + a 2 - & 2 Hence cos B = -- - Similarly for cos A and cos C. To show that cos ^ = x A- (-<*) 2 V i c Now- cos A = l l (b + c + a)(6 + c - a) Let s = ^(l + c + a); then s a = |(fr 4- c a) TT T A ^S Hence 1 f cos A = 2 bo A 2 4=v^. Again A- B To show that tan L ft . sin A a Since - ^ = r sin > 6 . sin A , a + !=? + ! I sin A + sin B or and sin B 6 sin A __ .. __ a __ -j sin B b sin A sin B a b or : ^ j sin B 6 58 TEXT-BOOK ON NAVIGATION. Divide II. by I. TT sin A - sin B a - b Hence ; =- = - 7 sin A + sin B a + b EXEECISES. OBLIQUE-ANGLED PLANE TRIANGLES. If ABC be any triangle, complete the solution in the following cases : 1. a = 1056, 6 = 359, c = 1267. 2. a = 364, b = 217, c = 494. 3. a = 96, 6 = 80, c = 63. 4. A = 37 54', B = 74 25', a = 104-6. 5. B = 14 20', C = 101 13', 6 = 296. 6. A = 97 34', C = 50 16', c = 16. 7. a = 6954, 6 = 4865, C = 39 40'. 8. a = 267, c = 341, B = 68 46'. 9. 6 = 61, c = 49, A = 97 18'. 10. a = 865, 6 = 742, A = 52 26'. 11. 6 = 469, c = 348, C = 39 4'. 12. a = 1065, c = 897, A = 49 36'. 13. a = 587, b = 925, A = 21 23'. 14. a = 7, b = 8, c = 9. 15. a = 600, b = 500, c = 1000. 16. a = 50, 6 = 60, c = 70. 17. A = 18 20', B = 150, b = 10,000. 18. A = 78 40', C = 39 55', c = 135. 19. b = 237-09, c = 130-96, A = 57 59'. 20. c = 2265-4, a = 1779, B = 58 17'. 21. a = 1300, b = 500, A - B = 30 22'. 22. a = 1586-6, b - 5374-5, A = 15 11'. 23. a = 200, b = 300, A = 35 15'. 24. The angular elevation of a tower at a place A due south of it is 45, and at a place B, due west of A, the elevation is 30 ; AB = 300 ft. Find height of tower. 25. AB is a line 250 ft. long, in the same horizontal plane as the foot D of a tower CD ; the angles DAB and DBA are respectively 61 23' and 47 14' j the angle of elevation CAD is 34 50'. Find height of tower. MISCELLANEOUS PROBLEMS, 1. The difference between the base arid perpendicular of a right-angled triangle is one-fifth of their sum. Find the angles. 2. P and Q are two stations 1000 yards apart on a straight stretch of sea- shore which bears east and west. At P a rock bears S. 42 W., at Q it bears S. 35 E. Find distance of rock from shore. 3. A person on a ship, which is sailing on a straight course observes two lighthouses, one of which, L, is due north of the other, M. When M is due SOLUTION OF PLANE TRIANGLES. 59 east, L bears 45 to the east of north, and when L is due east, M bears 60 to the east of south. Find direction of ship's course. 4. AB is a vertical pole 50 feet high, A being above B ; BC is inclined upward at an angle of 20 to the horizontal line, BD, so that ABC is an angle of 70 ; the shadow of AB on BC is 50 feet long. If the shadow fell on BD, what would be its length ? 5. From the top of a vertical cliff 100 ft. high, forming one bank of a river, the angles of depression of the top and bottom of a vertical cliff, which forms the opposite bank of the river, are 28 40' and 64 30' respectively. Find height of opposite cliff and width of river. 6. The summit of a spire is vertically over the middle point of a horizontal square enclosure, side = 1000 ft. When the shadow of the spire just reaches a corner of the square the altitude of the sun is 27 29' 50". Find height of spire. 7. A tower near the coast is 197 feet high. From a rock the angle of elevation of the top of the tower is 8 15'. A ship steams so as to clear the rock by 1*5 cables. What is the angle of elevation of tower from ship ? 8. At two boats 600 yds. apart, the angles of elevation of a hydroplane flying directly between the boats are observed to be 57 and 65 30'. Find the height of the hydroplane. 9. Two ships, A and B, leave harbour together. A sails S. 12 W. at a rate of 10'5 knots, and B S. 67 E. at 8 knots. How far apart will they be in 2^ hours, and what will be the bearing of A from B ? 10. From a ship the bearings of two objects, A and B, are N. 30 E., and N. 20 W. After the ship has sailed 5 miles due N., the bearings are N. 60 E. and N. 40 W. Find distance and bearing of A from B. 11. The foot, Q, of a tow^r and two stations, A and B, are in the same horizontal plane. The angular elevation of the tower at A is 60, and at B is 45, the distance from A to B is 100 feet, and the angle AQB is 60. Find height of tower. 12. From P the bearing and distance of A is NE. 5 miles, and of B, S. 70 E. From Q, due east of P, the bearing of A is N. 25 E. If B is due south of A, what is the bearing of B from Q 1 13. The hypotenuse of a right-angled triangle is 1000 feet long, and the difference between the other two sides is 240 feet : calculate the other sides and angles. 14. The height of a tower is 67'375 yds. At P, which is 5 '35 yds. above the level of its base, the angle subtended by the tower is 74 40'. Find the angle of elevation of the top of the tower as seen from P, and horizontal distance of P from tower. CHAPTER V. SOLUTION OF SPHEKICAL TRIANGLES. ART. 17. A spherical triangle is a portion of the surface of a sphere bounded by arcs of three great circles, and, like a plane triangle, has three sides and three angles. The sides, being arcs of circles, are measured like angles, in degrees, minutes, and seconds ( ' ") When one of the angles is 90, the triangle is right-angled ; when one side is 90, it is quadrantal ; and when no side or angle is 90, it is an oblique-angled triangle. Unlike a plane triangle, a spherical triangle may have two or three right angles, or one, two, or three sides of 90, but no angle or side can be as much as 180. The sum of any two sides is greater than the third side. The sum of the three angles of a spherical triangle is always more than 180, but less than 540, and the sum of the three sides is less than 360. The greater angle is opposite to the greater side, as in plane triangles. The angles at the base of an isosceles triangle are equal, as in plane triangles. ART. 18. Solution of Right-angled Spherical Triangles. Here one of the given parts is the right angle ; any other two parts being given, the triangle can be solved. In order to simplify the solution, instead of using the five remaining parts of the triangle, Napier devised rules which refer only to his " Circular Parts." Napier's Circular Parts, These are five in number, and are the two sides which contain the right angle, the complement of the hypotenuse, and the complements of the two angles. Thus in triangle ABC, right-angled at A, the circular parts are AB ; AC ; Co. C ; Co. BC ; Co. B These are split into three groups, consisting of 1 middle, 2 adjacents, 2 opposites. Any part may be taken as middle. SOLUTION OF SPHERICAL TRIANGLES. 61 Thus Middle. Adjacents. Opposites. AC Co. C Co.BC AB ; co. C AC ; co. BC Co. B ; co. C Co. B ; co. BC Co. B ; AB AB; AC Again Napier's Rules. I. Sine mid part = tangents of adjacents multiplied. II. Sine mid part = cosines of opposites multiplied. Each of these rules connects three parts. When used it must be arranged that two are given parts; also that the given parts are the only ones used to find the others. To do this, always start by considering whether the given parts are adjacents or opposites ; then take as middle each given part in succession, and use the Rule which will connect the given parts. Hence sin AC = tan AB . tan co. C = tan AB . cot C and sin AC = cos co. B . cos co. BC = sin B . sin BC sin co. BC = tan co. B . tan co. C or cos BC = cot B . cot C and sin co. BC = cos AB . cos AC or cos BC = cos AB . cos AC ART 19. Hints on construction of Spherical Triangles. Draw a circle as a primitive, with radius of sphere, we shall use one inch; but the larger the radius the better the result will be ; with the Gunter rule it would be two inches. Let C be the centre, then any diameter drawn will represent a great circle at right angles to the primitive : thus ACB (Fig. 57) is a right circle. Draw DCE at right angles to AB. Then D and E are the Poles of ACB. The centres of all oblique circles through A and B lie on the line DCE. Thus with any centre F and radius FA describe an arc of a circle, then AGB is an oblique circle. To find its pole, join BG and produce to b, make be equal to EA (90), join eB cutting DE in P : then P is the pole of AGB. N.B. Great circles at right angles to AGB must pass through its pole P. Hence great circles can be drawn through a point on or within the primitive, or through two points, one on, the other FIG. 57. 62 TEXT-BOOK ON NAVIGATION. inside the primitive. Thus in Fig. 58 let A and B be two points, A on, and B inside. Draw diameter ACD, and a line through C, as CO, at right angles. Join AB, and bisect it by a line cutting CO in X. Then X is centre of great circle through A and B. If an oblique circle is to be drawn through any two points inside the primitive as through A and B, Fig. 59, proceed FIG. 58. FIG. 59. thus Draw diameter AC, and CD through C at right angles to AC. Join AD, make ADE a right angle, cutting AC in E : join AB, BE, bisect these by lines at right angles intersecting in F. F will be the centre for describing oblique circle required. To draw a great circle at right angles to a given oblique circle through a given point, first find the pole of the given oblique circle; then by the preceding examples draw a great circle through the given point and this pole; this will be at right angles to the given oblique circle, the given point may be either on the primitive, or on the oblique circle. A great circle is drawn making a given angle with the primitive (say 35) through a given point as follows : From centre with radius = tan 35 describe a small arc, and with given point as centre and radius = sec 35 : draw another arc, this intersects the first arc at the centre from which the great circle is to be drawn with sec 35 as radius. N.B. With a radius of one inch, the tan and sec of any angle may be obtained from the traverse table thus Take angle (say 35) as course and radius (I'OO) in lat. col., then sec (1-22) is found in dist. col. and tan (O f 7) in dep. col., the semi-tangent = tangent of half the angle = 0'31. In Fig. 60, A is given point, and o is centre of required great circle. If oC be drawn as a diameter, the arc ab measures the angle : Ca = 0*52 ; in traverse table find 1 in lat. 0*52 in dep. ; angle = 27 J, hence Ca on scale of semi-tangents = 55, or ab = 35. A small circle parallel to the primitive and at any distance from it is drawn from the centre of primitive, with radius = semi-tangent of complement of required distance. SOLUTION OF SPHERICAL TRIANGLES. A small circle through any point parallel to a right circle at a given distance is drawn thus : let A be given point, and 35 given distance; then with sec (Co. 35), that is sec 55, as radius sweep an arc from centre C, where this arc intersects an arc drawn from A as centre and tan 55 as radius is the centre for describing the small circle (Fig. 61). On scale Sec 55 = 1-74" Tan 55 = 1'43". FIG. 60. FIG. 61. To measure any arc of an oblique circle, say arc AB (Fig. 62). Find pole P of oblique circle, join PA, PB, and produce to meet primitive in a and 6, then ab measured on the scale of chords is the length of AB. Similarly, to measure an angle at the intersection of two great circles, first find their poles, join the point of intersection of the FIG. 62. FIG. 63. circles by lines passing through the poles, and produced to cut the primitive. The measure of this arc of the primitive on the scale of chords is the measure of the angle. Example 1. Given A = 64 30', B = 56 2G', = 90. Solve. Construction (Fig. 63). Draw primitive, AD and EF, diameters at right angles. At A make angle of 64 30' with primitive, centre of circle is on EF produced, radius = sec 64 30' = 2 -32. Find pole a of circle ACD. Then 64 TEXT-BOOK ON NAVIGATION. with N as centre and radius = semi-tangent of angle B 56 25' = 0*53 describe an arc cutting ACD in b, this point is pole of circle required. Draw diameter &N and BG at right angles. Since C is to be a right angle, the circle drawn through BaG- is one required ; this cuts ACD at C. ABC is triangle. By measurement AB = 71 BC = 60 AC = 51 Calculation. Here the given parts are opposites. We therefore use co. A with its opposites, and then co. B with its opposites. Finally, to find AB, we use adjacents. Sin (co. A) = cos (co. B) . cos BC or cos A = sin B . cos BC T-,/-, COS A -r> . . cos BC = = cos A . cosec B . . . I. sin B Again, sin (co. B) cos (co. A) . cos AC or cos B = sin A . cos AC .*. cos AC = = cos B . cosec A ... II. sin A Now, sin (co. AB) = tan (co. A) . tan (co. B) or cos AB = cot A . cot B III. A 64 30' cos = 9-633984 cosec = 10-044512 cot = 9-678496 B 56 25 cosec = 10' 079312 cos = 9*742842 cot = 9-822154 58 53' 3" cos = 9713296 52 12' 15" cos = 9787354 71 32' 11" cos = 9-500650 Ans. BC = 58 53' 3", AC = 52 12' 15", AB = 71 32' 11". Example 2. Given A = 90, B = 117 20', BC = 113 40'. Solve. Construction (Fig. 64). Draw primitive, diameters BD and EF at right angles. At B make an angle 117 20' with primitive ; sec = 2*21 as radius, draw circle BAD, find its pole b. Make BC = 113 40', through C draw diameter CG and through C6G de- scribe a circle cutting BAD in A. Then ABC is triangle. Pole of C6G is at a. By measurement radius for C&G = 1-63 = sec. 52 C = 52 AB = 45 AC = 127 FIG. 64. Calculation. Here the given parts are adjacents. We therefore use co. B with its adjacents ; then co. BC with its adjacents. Lastly, to find AC, we use opposites. Sin (co. B) = tan AB . tan (co. BC) or cos B = tan AB . cot BC + cos B .'. tan AB = cot BG = cos B . tan BC Again, sin (co. BC) = tan (co. B) . tan (co. C) or cos BC = cot B . cot C + cos BC .*. cot C = g- = cos BC . tan B. + And sin AC = cos (co. B) . cos (co. BC) = sin B . sin BC. SOLUTION OF SPHERICAL TRIANGLES. B = 117 20' cos = 9-661970 BC = 113 40 tan = 10-358253 tan = 10-286614 cos = 9-603594 sin = 9-948584 sin = 9-961846 46 20' 1" tan = 10-020223 52 9' 59' cot 9-890208 54 27' 10" sin = 9-910430 In the case of AB and C, notice that they are marked + , so these angles are in the first quadrant, and the values are those obtained direct from the tables ; in the case of AC, which is marked +, the value might be the tabular value or its supplement. As angle B opposite to it exceeds 90, use the supplement. Ans. AB = 46 20' 1" ; C = 52 9' 59" ; AC = 125 32' 50". ART 20. In solving these triangles, observe 1. Angles and opposite sides are of the same affection, that is, are both over, or both under, 90. 2. The hypotenuse is under 90 when the angles or sides are of the same affection. The hypotenuse is over 90 when the angles or sides are of contrary affection. 3. When an angle and opposite side are given, the triangle admits of two solutions ; this is called the Ambiguous case, and may be understood from the following figure. Let AOB and AMB be two arcs of great circles intersecting at A and B. The angles at A and B are equal, and each arc is a semi-circle. Now let OM, another great circle arc, be drawn at right angles to AMB. Thus we have two right-angled triangles AOM and BOM, in which angle A and arc OM are equal to angle B and arc OM. Hence with angle = A and arc = OM given, there is nothing to indicate which of the triangles AOM or BOM is the required one. EXERCISES. RIGHT-ANGLED SPHERICAL TRIANGLES. Solve the following triangles : 1.0 = 90, AC = 65 12', BC = 51 36'. 2. A = 78 11', B = 39 49', C = 90. 3. AB = 84 13', BC = 61 44, C = 90. 4. AB = 79 27', A = 51 30', C = 90. 5. BC = 32 51', B = 56 17', C = 90. 6. BC = 54 37', A = 67 53', C = 90. 7. B = 90, AB = 70 29', BC = 102 38'. 8. B = 90, AB = 110 17', BC = 98 46'. 9. B = 90, C = 114 51', A = 67 32'. 10. B = 90, C = 92 31', A = 118 44'. 11. B = 90, AC = 107 19', AB = 69 8'. 12. B = 90, AC = 88 36', BC = 101 25'. 13. A = 90, AB = 97 20', BC = 94 13'. 14. A = 90, BC = 113 10', C = 74 32'. FIG. 65. 66 TEXT-BOOK ON NAVIGATION. 15 A = 90, BC = 83 12', C = 98 15'. 16. A = 90, AB = 95 52', B = 76 13'. 17. A = 90, AC = 106 27', C = 94 8'. 18. A = 90, AB = 114 22', C = 108 19'. ART. 21. Solution of Oblique-angled Spherical Triangles. The object of this section is to give the student an insight into the solution of those triangles which refer particularly to the ordinary problems of Nautical Astronomy. Two cases only will be considered. CASE I. Given the three sides, to find the angles. If ABC be a triangle, where the sides AB, AC, and BC are known. The formulae used are cos- =A / sin a sin ( ~ OB) 2 V sin AB. sin AC cos C ? =A /si" 2 V i ( - AC) AB. FIG. 66. C /sin s . sin (s AB) COS 2" = V ~ sin AC . sin Example l.-In PQR, given PQ = 84 30', QR = 108 44', Pll = 72 20'. Find the angles. Construction (Fig. 67). Draw primitive, lay off PQ = 84 30'. Through P draw diameter, and another AB at right angles. Around P as pole describe small circle pRp distant 72 20' from it or 17 40' from AB : the centre for this small circle is on NP produced at distance sec 72 20' = 3 '24" from N, radius = tan 72 20' = 3 '08". Through Q draw diameter, and another CD at right angles. Around S as pole de- scribe small circle qRq distant supplement 108 44' or 71 16' from it or 18 44' from CD : the centre for this small circle is on NS produced at distance sec 71 16' from N = 3-07", and radius is tan 71 16' = 2'9". Where these two small circles intersect is R. Through PR and QR draw oblique circles. PQR is triangle required. By measurement QPR = 112 semi-tangent = 0'19 or 22. PQR = 70 semi-tangent = 0'18 or 20. PRQ - 78. Calculation. FIG. 67. cos ? = / sin s . sin (s - QR) 2 V sin PQ . sin PR ain 6- . sin (.s PR) sin PQ . sin QR R _ /sin s . sin (s PQ) 2 V ""sin PR . sin QR SOLUTION OF SPHERICAL TRIANGLES. 67 To make all the logs added, " sines" in denominators are changed to their reciprocals " cosecs." 'QR = 108 44' cosec = 10-023639 cosec = 10 '023639 PQ = 84 30 cosec = 10 '002004 cosec = 10 '002004 PR = 72 20 cosec = 10-020981 cosec = 10-020981 2)265 34 s = 132 47 sin = 9 '865653 sin = 9-865653 sin s - QR = 24 3 sin = 9 '610164 ,s-PR= 60 27 ......... sin = 9'939482 s-PQ= 48 17 .................. sin 2)19-498802 2)19-830778 = 9 '865653 = 9-872998 2)19-783271 55 50' 9"cos= 9-749401 2 34 36' 53" cos = 9-915389 2 38 48' 53" cos = 9-891635 P = 111 40 18 2 Q = 69 13 46 R = 77 37 46 Ans. P = 111 40' 18" ; Q = 69 13' 46" ; R = 77 37' 46". CASE II. Given two sides and the contained angle. In the above triangle ABC, let AB, BC be known and angle B ; sup- pose AB greater than BC. The solution depends on Napier's Analogies. sn - BC) , B These by addition and subtraction determine C and A ; the third side is obtained from Gauss' theorem. AC cosJ(AB COS -^- = -- i/n , A \ 2 cos J(C -f A) sm B Note that the reciprocals of the ratios in the denomi- nators are used to make the logs added, and that the half sum of angles is of the same affection as the half sum of sides. Example 2. In XYZ, given XY = 88 20', YZ = 60_50', angle Y = 75 30'. Solve. Construction. Draw primitive, measure on it YX = 88 20'. Draw the diameter through Y : and AB at right angles to it. At Y make angle with primitive = 75 30' by drawing oblique circle YZ. Around Y as pole, draw a small circle aZa distant 60 50' from it at distance sec = 2 -06", and radius tan = 1'8" : this cuts the oblique circle in Z. XYZ is triangle required. FIG. 68. 68 TEXT-BOOK ON NAVIGATION. By measurement angle X = 60 ,, semi -tangent = 0'26 or 30 angle Z = 96 XZ = 77 Calculation. tan Z - X) = COS ZX cosi(XY -- - i/ry . vx cos J(Z + X) Y . cot XY= 88 20' YZ = 60 50 XY + YZ = 149 10 XY-YZ= 27 30 i(XY -I- YZ) = 74 35 sec = 10-575385 cosec = 10-015915 cos = 9*424615 i(XY-YZ) = 13 45 cos= 9-987372 sin = 9-376003 ~ = 37 45 cot = 10-111100 2 cot = 10-111100 sin = 9-786906 i(Z + X) = 78 2' 8" tan = 10-673857 sec = 10-683391 i-(Z - X) = 17 39 48 tan = 9-503018 Z = 95 41 56 X = 60 22 20 38 16' 20" cos = 9-894912 2 76 32 40 Ana. X = 60 22' 20" ; Z = 95 41' 56" ; XZ = 76 32' 40". ART. 22. To express the cosine of an angle of a spherical triangle in terms of the sines and cosines of the sides. Let be the centre of the sphere, and PQE, a spherical triangle. Let tangent at P to arc PR meet OR produced in T; and let tangent at P to arc PQ meet OQ produced in S. Join ST. Thus angle TPS is angle P of spherical triangle, and TOS measures the side RQ ; SOP the side PQ ; and TOP the side PR. In triangle SPT FIG. 69. ST 2 = SP 2 + PT 2 - 2SP . PT cos P I. SOLUTION OF SPHERICAL TRIANGLES. 69 Again, in triangle TOS ST 2 = SO 2 + OT 2 - 2SO . OT cos SOT = SP 2 + PO 2 + PT 2 + PO 2 - 2SO . OT . cos RQ since OPS and OPT are right angles = 2P0 2 + SP 2 + PT 2 - 2SO . OT cos RQ II. Subtract I. from II. .-. = 2P0 2 + 2SP . PT cos P - 2SO . OT cos RQ Hence cos RQ = g > PO + PTg cosp = cos TOP. cos POS + sin TOP . sin POS . cos P = cos PR . cos PQ + sin PR . sin PQ . cos P. This is the fundamental formula, and, where ABC is the triangle and sides denoted by a, b } c, is expressed cos a = cos b . cos c + sin b . sin c . cos A. The formula used in Case I. is derived from this in the following manner : cos a = cos b . cos c + sin b . sin c . cos A cos a cos b . cos c .'. cos A = sin b . sin c cos a cos b . cos c Hence 1 + cos A = 1 + sin 6 . sm c Q A sin 6 . sin e + cos a cos b . cos c Or Z COS = ; = ; 2 sin b . sin c _ cos a cos (6 + c) sin b . sin c __ 2 sin %(a 4- b + c) . sin %(b -f c - a) sin 6 . sin c Let s = -|-(a -f 6 + c) /. s a = ^(b -f c a) 9 A sin s . sin (s a) PO;^ v / V^VJO o ^ . 7 2 sin o . sm c or cos t = ^ms. S m(s-a) v sin 6 . sin c EXERCISES. OBLIQUE-ANGLED SPHERICAL TRIANGLES. 1. If BC = 82 26', AC = 92 2', AB = 104 28'. Solve. 2. If BC = 140, AC = 70, AB = 80. Solve. 3. If BC = 113 2', AC = 82 40', and C = 138 50'. Solve. 4. If AC = 60, AB = 84 22', and A = 95. Find BC. 5. If BC = 115 20', AB = 84 42', and B = 65 30'. Find A and C. 0. If BC = 90, AC = 75 29', AB = 84 13'. Find A. 7. Sun's R.A. and Dec. being l h 30 ra 10 s , and 9 26' 30" N. Find his long. 8. The R.A. of a star is 16 h 49 ra 48% and its Dec. is 7 23' 10" N. ; find its long, and lat., the obliquity of the ecliptic being taken to be 23 27' 30". 70 TEXT-BOOK ON NAVIGATION. 9. In a right-angled spherical triangle C = 90, show tan a = tan c . cos B = tan A . sin b ; also if B = &, show triangle must be isosceles. 10. Two ports A and B, both in northern hemisphere, are in long. 90 W. and 10 W. Two ships leave A and B respectively, proceeding each upon the great circle which touches the parallel of the port at the point of departure. Ship from A steams to the eastward at uniform rate of 12 knots, that from B to the westward at 17 knots ; they meet each other in 5 days. Find lat. and long, of point of meeting. CHAPTER VI. THE SAILINGS. ART. 23. Parallel Sailing. In parallel sailing, the ship's course is supposed to continue on the same parallel of latitude. There are three things concerned, viz. the Latitude, the Departure or Meridian Distance, and the Difference of Longi- tude. Definitions. (1) Latitude : the distance of the ship north or south from the Equator. (2) Departure (or Meridian Distance) : the dis- tance in geographical miles between two places on -, the same parallel of latitude. (3) Difference of Longitude: the arc of the Equator corresponding to the Meridian Dis- tance. Illustration. AB and CD are Departures at different distances from the Equator; EQ is the D. long., or corresponding Arc of the Equator. Remark. It is seen that the Meridian Distance decreases as the Latitude increases, and vanishes altogether at the Poles. The formulfe which express the relation between the three terms are 1. D. long. = dep. x sec lat. 2. Dep. = diff. long. X cos lat. Fio. 70. 3. Sec lat. = diff. long, dep. These are the three cases in Parallel Sailing, and show how the D. long, or Dep. or Lat. may be found when the other two terms are given. Let P be the pole, PE and PQ meridians of longitude, EQ TEXT-BOOK ON NAVIGATION. an arc of the Equator, DL an arc of a parallel of latitude, and C the centre of the Earth. Then QL or ED will represent the lat. of the parallel DL, EQ is the diff. long., and DL the departure. Join CL, and draw FL perpendicular to PC. It is evident that FL is a radius of the small circle of which DL is a part, and CQ a radius of the great circle of which EQ is a part; that is, a radius of the Earth. And since the arcs are proportional to the radii of the two circles : are EQ : : rad. FL : rad. CQ : EQ : : FL : CL FIG. 71. ArcDL therefore DL because CL is also a radius of the Earth ; or = = sin FCL, or cos QCL because CFL is a plane right-angled triangle, and QCL is the complement of FCL. Now, the angle QCL is measured by the arc QL, which is the latitude of the parallel DL therefore _ fP' = cos lat (1) D. long. from which are derived the formulas Dep. = D. long, x cos lat (2) or dep. X sec lat. (3) Now the formula ^ P' = cos lat. expresses the relation D. long. between the base, hypotenuse, and angle at the base of a right-angled plane triangle. Such a triangle may therefore be used to derive the formulae of parallel sailing, by calling the base departure, the hypotenuse diff. of longitude, and the angle at the base latitude. dep. Then cos lat. FIG. 72. D. long. , D. long. , and 3 = sec lat. . . . dep. From (1) dep. = D. long, x cos lat. from (2) D. long. = dep. X sec lat. . which are the same as those derived from the sphere. (2) (3) (4) 22- * THE SAILINGS. 73 ART. 24. Middle-latitude Sailing. This is a combination of plane and parallel sailing, the dep. being com- mon to both, but the latitude used is the " middle lat.," or mean of the two latitudes, for in this case the ship does not sail on a parallel of latitude. It is evident that the formulae for right- angled triangles will solve every case, but this method is only made use of in modern navigation for the purpose of finding the D. long, in the " day's work," the formula em- ployed being D. long. = dep. X sec of middle lat. The mean is not the true "middle lat. /'being too small, but is sufficiently near for short distances. Examples. 1. In lat. 46 45' the Departure made good was 110*5 miles : required the Difference of Longitude. Construction. In Fig. 74, scale j\j inch = 10 miles. C Draw horizontal line AB = 1*1 inches for departure, make angle BAG = 46 45', and draw BC perpendicular to AB : then AC represents D. long. Measure AC = 1'6 inches or 160 miles. FIG. 73. Dep. 110-5 Lat. 46 45' log 2-043362 sec 10-164193 B FlG 74 - D. long. = 161-3 ... log 2-207555 2. In lat. 53 14' the D. long, being 3 16', required the Meridian Distance or Departure. In Fig. 75 draw horizontal line AB. Make angle BAG = 53 14', and AC = 0'98 inch ; scale ^ inch = 20 miles. From C draw CB perpendicular to AB : then AB represents Dep. Measure AB = 0'6 inch or 120 miles. D. long. 196' ...... log 2-292256 Lat. 53 14' ...... cos 9*777106 Dep. = 117-3 ...... log 2 -069362 3. On what parallel of latitude would a Meridian Dis- tance of 245 miles give a difference of longitude of 5 54' ? In Fig. 76, scale ^ inch = 20 miles, draw AB = 1-2 long, at B draw a perpendicular to AB, and with A as centre, radius = 1"77 inch, draw an arc, cutting this perpendicular in G : then BAG represents lat. Measure BAG = 46. D. long. 354 ............ log 2-549003 Dep. 245 ............ log 2-389166 Lat. 46 12' ............ sec 10 -159837 Remark. In the first two cases 10 is rejected from the sum of indexes, and in the third case 10 is added to the upper index, because in the Table of Log Sines, etc., the proper index is always increased by 10. 74 TEXT-BOOK ON NAVIGATION. EXERCISES. 1. In lat. 16 30' N. the dep. made good was 167 '6 miles : what will the difference of. longitude be by parallel sailing ? 2. The difference of longitude between two places in lat. 63 19' 30" is 4 20' : required the distance as measured along the parallel. 3. In lat. 11 56' N. the dep. made good was 365 miles : required the difference of longitude in minutes. 4. In what latitude is 200 miles of dep. equal to 600 minutes of longitude ? 5. In lat. 35 40' S. the dep. made good was 976 : required the difference of longitude measured on the parallel. 6. What is the difference of latitude between two parallels of latitude on different sides of the Equator, when between the same two meridians the dep. on the northern parallel is 250 miles, on the southern 350 miles, and on the Equator 470 miles ? 7. On the parallel of 48 12' ran true west 1005 miles : if the longitude left was 9 30' W. , what was the longitude in 1 8. From two ports in lat. 40 20' N., distant 300 minutes of longitude, two ships sail N. for 300 miles : how many miles apart will they be, and in what latitude? 9. In lat. 33 15' N. ran east 10*95 miles : how many minutes of longitude will that represent ? 10. If two ships in lat. 51 S. should sail directly north until the departure changes from 400 to 500 minutes, what latitude will they be in 1 11. In lat. 72 59' the dep. was 370*5 miles : convert that into minutes of longitude. 12. In what latitude does 300 miles of dep. correspond to 600 minutes of longitude ? 13. The latitude by Mer. alt. was 36 35' N. ; longitude by after- noon sights, 137 38' E. ; run since noon due east, 39 miles : find noon longitude. 14: A vessel sailed due east along the parallel of 45 24' S. for 29 miles from long. 43 40' E. : what is her longitude in ? 15. What distance must a vessel sail on the parallel 48 25' N. to change her longitude 2 25' ? 16. A vessel sailing due east finds her difference of longitude to be 1 19', and distance made good t>0' : what is her latitude 1 17. How far on the parallel of 38 S. must a vessel sail from long. 47 29' E. to reach long. 49 27' E. ? 18. For every 1 diff. long. I have to sail 45 miles due west : along what parallel am I sailing ? ART. 25.. Plane and Traverse Sailings. For short distances, the surface of the Earth may be regarded as a plane. In plane and traverse sailings no notice is taken of the curvature, and all cases may be solved by the formulae for right-angled triangles. Thus in sailing from A to B, the hypotenuse represents the distance, the perpendicular the departure, and the base the D. lat., whilst the angle which B z ^rr 'C .AP makes with AC (lying true N. and is the course. In the Traverse Table the values of D. lat. and De^. in miles are given for every degree of course THE SAILINGS. 75 up to 89 and all distances up to 600 miles, being computed from the formulae ' * ' = cos course dist. -^?- = sin course dist. whence D. lat. = dist. x cos course^ .... (1) and dep. = dist. x sin course .... (2) When D. lat. and dep. are given, the course and distance made good may be found from the formulae tan course = 35^ (3) distance = D. lat. X sec course ... (4) ART. 25. The Day's Work and Traverse Sailing. The purpose of the Day's Work is to find the position of the ship at noon by " Dead reckoning," that is, from the courses and distances sailed since the preceding noon. The direct -course and distance made good are also determined. The courses given are by compass as per ship's Log Book, but the true courses must be found for computing a " Traverse Table." Directions for finding the True Courses. (a) Allow the leeway to the Eight (R.) when the wind blows on the left or Port Side of the ship, and to the Left (L.) when on the Right or Starboard Side, (b) Allow E. deviation and variation (or total correction) to the Right (R.), and W. to the Left (L.). NOTE. The Leeway may be determined by towing the log-ship astern and observing the angle the line makes with the keel, but it is often merely estimated. The Deviation is found by reference to the Deviation Card, and the Variation from a chart. When the Departure is taken from a known point of land or light, the opposite course to the bearing of the point or light from the ship is called the " Departure Course," and the distance off is the distance belonging to that course. The " Set " and " Drift " of the current are found by chart, and are treated as a course and distance made by the ship, because the effect is the same on the Day's Work. NOTE. Should the "Set" and "Drift" of the current be required, it is the course and distance calculated by Mercator from the D.R. position found by Traverse to the observation position given in the question. They may also be found by turning D. long, into Dep. : then with the D. lat. and Dep. find Co. and Dist. in Traverse Table, for Set required. 7 6 Illustrations and Minor Calculations. TEXT-BOOK ON NAVIGATION. Example 1. Opposite course to bearing. (D S. 45 C Corr. 11 S. 56 R. (2) N. 65 L. Leeway 20 L. N. 85 L. Corr. 3 R. N. 82 L. (3) N. 77 R. 17 R. N. 94 R. 12 K. N. 106 R. 180 (4) S. 74 E. N. 57 L. 16 L. N. 73 2 (5) (6) N. 75 L. S. 90 L. 19 R. S. P 71 L. 21 R. 8. 50 L. N. 52 L. 14 L. N. 66 L. Corr. N. 66 L. 1 Course steered. 1 5 5 5 6 6 5 5 5 1 Wind. Lee- way. Devia- tion. Remarks. 1 2 3 4 N. 65 W. 6 5 8 1 2 9 7 2 N. by E. 20 12 W A point of land in lat. 46 3' S., long. 146 15' E., bore N.E. from ship, dis- tant 12 miles, ship's head being W., end Deviation 4 W. Variation 15 E. 5 6 7 8 9 10 11 12 N. 77 E. N. by E. 17 3W. N. 57 W. 5 5 5 5 2 3 5 N. by E. | E. 16 17 W. 1 2 3 4 5 6 7 8 9 10 11 12 E. 5 6 6 7 7 3 8 2 2 3 6 9 1 2 7 N.N.E. 19 6E. A current set N. magnetic 24 miles during the 24 hours from the time of de- parture to the end of the day. N. 52 W. 7 7 7 7 N.N.B. 14 15 W S. 80 E. 8 8 8 8 N.E.byN.^N. 21 14 E. Correct the courses, and find the course and distance made good ; also the Latitude and Longitude in by inspection. Traverse Table. True courses. Distance. D. Lat. Dep. N. S. E. W. S. 56 W. N. 82 W. S. 74 E. N. 75 W. S. 50 E. N. 66 W. S. 30 E. N. 15 E. 12 23 23 21 26 30 33 24 3-2 5-4 12-2 23-2 67 6-3 16-7 28-6 221 199 16-5 6-2 9-9 22-8 20-3 274 44-0 58-3 44-0 64-7 80-4 64'7 D. lat. 14-3 Dop. 15-7 The D. lat. and dep. are found nearest together under the course 48 THE SAILINGS. 77 and dist. 21 miles, which are the course and distance made good (course Minor O. 48 W.). Calculate Lat. left 46 3' S. Long, left 146 15' E. D. lat. 14' S. D. long. 23' W. ^A (7) S ' Lat. in 46 17' S. 2)92 20' Long, in 145 52' E. / sec 10-159541 / log 1-195900 / l 1 >ZKRAA-\ n / 8. 59 29 8.30 (8) Currer N. Oo 15 D N. 16 Mid. lat. 46 10' Dep. 157 Diff. long. 22-67 By inspection : The M. lat. 46 as a course with dep. 15 ? in Lab. column gives 23' long, in Dist. column. In Fig. 77 A, scale 1 inch = 10 miles, let ABC represent portion of meridian. AB = 1-43 inch is D. lat. BD = 1-57 inch is Dep. where ABD is 90. Lay off BDC = 46 Mid. lat. Then BAD is course = 48. AD is Dist. =2-1 inch. = 21 miles. DC is D. long = 2'3 inch. = 23 miles. Example 2. FIG. 77A. Hour. Course steered. Knts. lOths. Wind. Lee- way, Devia- tion. Remarks. 1 2 3 4 aw. 8 8 9 9 2 4 2 2 S.E. by S. Pts. H 6 W. A point of land in lat. 49 57' N., long. 5 12' 5 6 7 8 &W. by W. 8 8 7 7 9 5 6 S.S.E. if 3 W. W., bore by compass N.W. \ N., distant 20 miles, ship's headbeinjrS.W., 9 10 11 12 S.W. by S. 6 6 6 5 7 5 2 6 S.E. i* 2E. per log. Variation 19 W. 1 2 3 4 S.SE. 7 7 8 9 5 9 5 1 S.W. 2 7E. 5 6 7 8 S. by W. $ W. 10 10 10 10 1 3 6 W. If 4E. A current set E. by N. \ N. (corr. mag.) 30 miles, during the 24 hours. 9 10 11 12 MB. 11 11 11 12 3 6 7 4 W. by N. \ 2E. Eequired the course and distance made good, and the Latitude and Longitude in. TEXT-BOOK ON NAVIGATION. Minor Calculations. 1) S. 39 L. Corr. 25 L. S. 64 L. !) S. 62 R. Corr. 25 L. S. 37 R. I) S. 76 R. 22 L. S. 54 R. ) S. 51 L. 17 L. S. 34 R. ) S. 45 L. 12 L. S. 57 L. ) S. 3 3 L. 15 L. S. 18 L.' ) S. 8L. 17 L. S. 25 L. .rreut N. 73 R. Var. 19 L. N. 54 R. Traverse Table. True courses. Distance. D. Lat. Dep. N. s. E. W. S. 64 E. S. 37 W. S. 54 W. S. 34 W. S. 57 E. S. 18 E. S. 25 E. N. 54 E. 20 35 32 25 33 41 47 30 17-6 8-8 28-0 18-8 20-7 18-0 39-0 42-6 18-0 27-7 12-7 19-9 24-3 21-1 25-9 14-0 17'6 175-9 17-6 102-6 61-0 61-0 158-3 41-6 Course and dist. made good S. 15 E. 164 miles. Lat. left 49 57' N. D. lat. 2 38' S. Lat. in 47 19' N. 2)97 16' Mid. lat. 48 38' Dep. 41-6 Long, left 5 12' W. D. long. 1 3' E. Long in 4 9' W. sec 10-179880 log 1-619093 D. long. 62-95 log 1'798973 In Fig. 78 scale 1 inch = 100 miles, ABC represents meridian. AB = 1-58 inch is D. lat. BD = 0-42 inch is Dep. ABD = 90. Make BDC = 48| = Mid. lat. Then BAD is course = 15. AD is Dist. = 1-64 inch. = 164 miles. CD is D. long = 0'63 inch. = 63 miles. NOTE 1. The D. long, can also be found by inspection thus : Take the mid. lat. as a course and find dep. in D. lat. column ; the dist. corresponding is the D. long. NOTE 2. When course and leeway are in * ' points, " it is most convenient to allow the leeway before changing points into degrees, then apply the correction. NOTE 3. To calculate course and distance made good. D. lat. 158-3 S., dep. 41-6 E. In triangle ABC, C = 90, AC = 158-3, BC = 41'6. Find A = course AB = distance. Now tan A = BC 41-6 AC A = 14 43' 26" 41-6 log = 11-619093 158-3 log = 2-199481 tan = 9-419612 , AB and AC = sec A .-. AB = AC x sec A 2-199481 sec = 10-014501 163-7 log = 2-213982 Course S. 14 43' 26" E., distance 163*7 miles. THE SAILINGS. EXERCISES. 79 1. Hour Course steered. Knts. lOths. Wind. Lee- way. Devia- tion. Remarks. 1 2 3 4 N.N.W. 17 16 14 13 1 2 3 4 W. \ S. PtB. 1 2 7E. A point of land in lat. 34' S., long. 17' W., bearing by com- pass S. f E., distant 49 miles, with the ship's head at N.N.W. Deviation as per log. Variation 30 W. 5 6 7 8 N. by W. 13 15 17 16 2 5 6 7 N.W. by W. | W. 36 E. 9 10 11 12 N.N.E. 15 10 12 16 8 9 1 2 East 3 19 W. 1 2 3 4 N.E. by E. \ E. 13 17 10 15 3 4 5 8 S.E. by E. 3J 5W. A current set cor- rect magnetic South 4 miles from the time departure was taken to the end of the day. 5 6 7 8 9 10 11 12 S.W. J W. 14 13 15 17 9 3 9 9 S. by E. \ E. i 4E. B.S.W. 11 14 13 16 4 9 8 9 West 2 5E. Correct the courses for Deviation, Variation, and Leeway, and find the course and distance from the given point, and the Latitude and Longitude in. 2. Hour. Course steered. Knts. lOths. Wind. Lee- way. Devi i- tion. Remarks. 1 N. 85 W. 12 6 S.S.W. 3" 14 E. 2 12 7 3 12 4 A point in lat. 55 4 12 14' N., long. 8 5 12 2 20' W., bearing 6 12 8 by compass S. 23 7 12 5 K., ship's head 8 12 4 N.85W. Devi- 9 12 5 ation as per log., 10 12 6 distant 7 miles. 11 12 3 12 12 . (Continued on next page,} 8o TEXT-BOOK ON NAVIGATION. Hour. Course steered. Knts. lOths. Wind. Lee- way. Devia- tion. Remarks. 1 2 N. 82 W. 11 11 9 8 S.W. 12 E. Variation during first 8 hours, 28 4 11 9 W. ; during se- cond 8 hours, 30 5 G 7 8 9 10 11 12 12 12 12 12 12 12 11 11 1 4 5 3 2 9 4 W.; during third 8 hours, 32 W. A current set cor- rect magnetic S.S.E. | E. 2J miles per hour for all day. Find the course and distance made good ; Latitude and Longitude of the ship. 3. Hour Course steered. Knts. lOths. Wind. Lee- way. Devia- tion. Remarks. 1 2 3 4 N. by E. i E. 9 9 9 8 1 3 9 7 S.E. by E. Pts. \ 18 E. A point of land in lat. 18 59' S., long. 179 53' W., bearing by compass E.S.E., distant 19 miles, with the ship's head at N. by E. \ E. Devia- tion as per log. Variation 9 E. 5 6 7 8 9 10 11 12 S.E. by E. 6 n 6 6 6 5 5 5 4 1 3 2 S. by W. 1* 23 W. E. by N. \ N. 9 8 3 N. J E. If 3 W. 1 2 3 4 S.W. f W. 5 5 5 5 3 1 2 4 S.S.E. 2* A current set cor- rect magnetic S.S.E. f E., 15 miles from the time departure was taken to the end of the day. 5 6 7 8 s.s.w. 7 8 8 , 9 9 2 4 5 W. by N. 1 11 W. 9 10 11 12 N.N.E. 9 10 10 10 6 3 1 N.W \ W. n 16 E. Correct the courses for Deviation, Variation, and Leeway, and find the course and distance from the given point, and the Latitude and Longitude in. THE SAILINGS. 81 4. Hour. Course steered. Knts. lOths. Wind. Lee- way. Devia- tion. Remarks. 1 9 3 4 S. by W. J W. 16 17 10 9 3 1 2 4 W. JS. Pts. 11 7E. A point of land in lat. 58 30' S., long. 64 54' W., bearing by com- pass N.W. \ N., distant 60 miles, with the ship's head at S. by W. \ W. Devia- tion as per log. Variation 21 E. 5 6 7 8 W. by S. \ S. 11 12 13 14 9 1 N.W. by N. 1 9E. 9 10 11 12 S.W. J W. 16 14 12 13 9 8 8 5 S.S.E \ 8W. 1 2 3 4 5 G 7 8 N.E. by N. 10 11 12 13 9 8 8 5 w.s.w. 11 W. A current set cor- rect magnetic W. | S. 39 miles, from the time departure was taken to the end of the day. N. by E. 14 15 16 17 8 8 8 6 N.W. by W. i I 12 W. 9 10 11 12 W. by N. 9 1 2 6 7 7 7 9 S.W. by S. 41 W. Correct the courses for Deviation, Variation, and Leeway, and find the course and distance from the given point, and the Latitude and Longitude in. 5. Hour. 1 2 3 4 5 6 7 8 "IT 10 11 12 Course steered. Knts. lOths. Wind. Lee- way. Devia- tion. Remarks. N.N.W. W. 4 3 3 1 1 3 1 5 N.E. Pts. If 2 2W. A point in lat. 29 59' N., long. 132 54' E., bearing by compass N.N.E. J E., ship's head N.W. by W. De- viation 6 W., distant 15 miles. E.S.E. 2 3 3 4 I 1 1 7 3 9 4 4 N.E. 7E. S. f E. 6 6 5 6 E.S.E. 2J 2 W. (.Continued on next page.) TEXT-BOOK ON NAVIGATION. Hour. Course steered. Knts. lOtbs. Wind. Lee- way. Devia- tion. Remarks. Pts. 1 N.E. \ N. 9 7 E.S.E. 1* 8E. 2 7 4 3 2 4 2 7 5 W. N. 3 5 S.S.W. \ W. M D W. 6 4 3 Variation l c E. 7 3 6 8 3 6 9 N. by E. 8 5 E. by N. 5 6E. 10 1 1 11 8 1 12 9 3 The position by observation was found to be lat. 30 25' N., long. 133- 7' E. Find the set and drift of the current. Hour. Course steered. Knots lOtbs. Winds. Lee- way. Devia- tion. Remarks. 1 2 3 4 S. by W. 12 13 11 12 1 2 3 4 W. by S. ipt- 9W. A point in lat. 54 22' N., long. 178 26' E., bearing by compass N.W. \ W., ship's head N. ] E. Devia- tion as per log, distant 8 miles. Variation 12 E. 5 6 7 8 N. | E. 12 13 14 11 9 8 7 6 East Jpt 6E. 9 10 11 12 1 2 3 4 East 11 12 12 13 2 2 2 4 S.S.E. iPt- 19 E. S.E. 9 11 12 10 3 4 7 6 S.S.W. Jpt. 11 E. 5 6 7 8 9 10 11 12 S.S.W. \ W. 12 11 14 15 6 2 9 3 N.W. 12 W. E.$N. 13 12 11 13 9 9 9 3 North Jpt. 18 E. The ship by observation was found in lat. 51 59' N., long. 178 5' W. Find set and drift of current. THE SAILINGS. 83 ART. 27. Mercator's Sailing. The problem in Mercator's Sailing commonly used, and the only one in the Board of Trade Examination for Mates and Masters, is to find the Course and Distance from place to place, the latitudes and longitudes of both places being given. The data used in the calculation are The difference of latitude. The difference of longitude. The meridional difference of latitude. Definitions. (1) D. lat. : The sum or difference of the two latitudes (according as they are of different names or of the same name) reduced to miles. (2) D. long. : The sum or difference of the two longitudes (according as they are of different names or of the same name) reduced to minutes of arc. N.B. When the sum of longitudes exceeds 180, subtract from 360. (3) Mer. D. lat. : The sum or difference of the meridional parts for the two latitudes. Formulce. (1) To find the Course Tangent of course = ^ ^' mer. D. lat. (2) To find the Distance- Distance == secant of course x D. lat. It may be graphically represented by two similar triangles, DEC and ABC, the angle at C being the course, which is common to both triangles. The sides of the smaller triangle represent the elements of plane sailing. The extended meridian CB represents the meridional D. lat. found from the meridional parts for the two latitudes ; then the side AB will be the D. long, in its correct proportion to the departure. The object aimed at | in Mercator's method is to find the A D.Ldr rhumb course from point to point. For FlQ this purpose, the diff. lat. is extended by means of the meridional parts so as to bear the same pro- portion to the diff. long, as the true diff. lat. bears to the departure. In the right-angled triangle ABC m D. long. f .^ Tan course = -^ 7 , (1) mer. D. lat. and in the triangle DEC _ *f ' = sec course D. lat. therefore dist. = D. lat. X sec course ... (2) 8 4 TEXT-BOOK ON NAVIGATION. These are the only formulae required for computing the true rhumb course and the distance on that course. Evidently this distance is longei than the distance on a great circle, but the method has the advantage of leading to the destination without change of course. Examples. 1. Find the Course and Distance from A to D on Mercator's principle. A Lat. of A, 51 36' N. ; long, of A, 8 32' W. D, 38 32' N. ; 6136'N. Mer. parts 3626 38 32 N. 2509 D, 28 38' W. 8 32' W. 28 38 W. D Lonq. Fia/81. 13 60 784 1117 20 60 1206 6 D. long. 1206 log 3-081347 Mer. D. lat. 1117 log 3'048053 Course S. 47 11| ' W log tan 10-033294 Course 47 1H' log sec 10-167780 D. lat. 784 log 2-894316 Distance 1154 miles log 3-062096 NOTE. The Course is reckoned from S because D is south of A, and towards W. because D is west of A. 2. Find the Course and Distance from P (Hawraki) to B (Hawaii). DLONG. Lat of p^ ggo 59, g . Jong of Pj 175 o 4 ; E B 5 19 50' N. ; B, 156 40' W. 35 59' S. 19 50 N. Mer. parts 2317 1215 Mer. D. lat. 3532 175 4'E. 156 40 W. 331 44 360 D. lat. 3349 FIG. 82. 28 16 60 D. long. 1696 D. long. 1696 log 3-229426 Mer. D. lat. 3532 log 3-548021 Course N. 25 39' E. Course 25 39' D. lat. 3349 log tan 9-681405 log sec 10-045056 ... log 3-524915 Distance 3715 log 3-569971 NOTE. The Course is reckoned from N. because B is north of P, and E. because it crosses the meridian of 180. THE SAILINGS. 3. Required the Course and Distance by Mercator from A to B. Lat. of A, 6 15' N. ; long, of A, 79 30' E. B, 3610'S.; B, 113 45' E. Lat. 6 15' N. Mer. parts 376 Long. 79 30' E. 36 10' S. 2330 113 45' E. A D. \ lat./ Mer. D. lat. 2706 34 15 60 2545 D. long. 2055 Mer. D. lat. 2706 .. Course S. 37 13' E. Course 37 13' ... D. lat. 2545 D. long. 2055 ... log 3-312812 ... log 3-432328 log tan 9-880484 log sec 10-098894 .. log 3-405688 Distance 3196 miles LOHG. FIG. 83. log 3-504582 4. Required the Course and Distance by Mercator's principle from A, in lat. 57 42' S. ; long. 70 3' W., to B. in lat. 46 58' S. ; long. 168 20' E. ; also the Course to steer by compass, supposing the variation to be 22 E., and the deviation of the compass 7 30' W. In Fig. 84, ABC is portion of meridian. Scale 1 inch = 2000 miles. AC = 0-53 inch is Mer. D. lat. ACD = 90. CD = 3-65 inch is D. long. AB = 0-32 inch is D. lat. BE parallel to CD is Dep. Then CAD = 82 is true Course. AE = 2-24 inches. = 4480 miles is Dist. D Lat. 57 42' S. 46 58' S. 10 44 60 FIG. 84. Mer. parts 4260 3200 Mer. D. lat. 1060 D. long. 7297 Mer. D. lat. 1060 Long. 70 3' W. 168 20' E. 238 23 360 121 37 60 D. long. 7297 ... log 3-863144 ... log 3-025306 N. 81 44' 5" W. . Course 81 44' 5" D. lat. 644 log tan 10-837838 764 74 gives 5" loj sec 10-842372 ... log 2-808886 Distance 4480 miles ... log 3-651258 86 TEXT-BOOK ON NAVIGATION. NOTE. When the Course is large, it may be advisable to work to seconds, as in this case, in order to ensure the accuracy of the distance within a mile. To find the Course by compass : True course N. 81 44' W. Variation 22 0' E. Magnetic course N. 103 44 W. Deviation ... 7 30 W. Compass course N. 96 14 W". Or .. S. 83 46 W. EXERCISES. Required the Compass Course and Distance from A to B by calculation on Mercator's principle. 1. Lat. A, 50 49' N. ; long. A, 0'. ,, B, 4817'N. ; B, 54 47' W. Variation 19 45' W. ; deviation 20 W. 2. Lat. A, 11 11' S. ; long. A, 37' W. B, 0'; B, 0'. Var. 23 30' W. ; dev. 10 17' E. 3. Lat. A, 52 57' N. ; long. A, 158 33' W. B, 3930'S.; B, 178 48' E. Var. 25 E. ; dev. 15 W. 4. Lat. A, 14 9' S. ; long. A, 180 0'. B, 00'; B, 174 57' E. Var. 9 E. ; dev. 5 30' E. 5. Lat. A, 35 15' S. ; long. A, 115 15' E. B, 51 30' S. ; ,, B, 178 52' W. Var. 7^ W. ; dev. 15J W. 6. Lat. A, 51 24' N. ; long. A, 9 36' W. B, 3836'S.; B, 170 24' E. Var. 25 W. ; dev. 14J E. 7. Lat. A, 0' ; long. A, 81 27' W. B, 6110'S.; B, 170 59' E. Var. 2 20' E. ; dev. 20 40' E. 8. Lat. A, 42 56' S. ; long. A, 100 36' W. B, 1036'N.; B, 180 0V Var. 17 10' W. ; dev. 13 50' E. 9. Lat. A, 36 19' S. ; long. A, 30 46' W. B. 3612'S.; B, 58 23' W. Var. 31 15' W. ; dev. 14 26' W. 10. Lat. A, 0' ; long. A, 31 17' W. B, 63 28' N. ; ,, B, 28 16' W. Var. 12 20' W. ; dev. 15 30' E. 11. Lat. A, 32 22' N. ; long. A, 33 15' E. B, 37 4' N.; B, 15 40' E. Var. 5 35' W. ; dev. 12 29' W. 12. Lat. A, 17' S. ; long. A, 59 47' W. B, 159'N.; B, 120 13' E. Var. 6 40' E.; dev. 13 50' W. THE SAILINGS. FIG. 85. ART. 27. Current Sailing. 1. Given the course and rate of sailing of a ship, and the "set" and "drift" of a cur- rent during the same inter- val of time, to find the course and distance made good. Let AB represent the course and the distance sailed, and AC the "set" and " drift " of the current. Draw CD parallel to AB, and BD parallel to AC, and join AD ; then AD will represent the course and distance made good during the time elapsed. Here the angle BAC is known, and therefore ABD is also known, because it is equal to 180 BAC. Also the sides AB and BD of the triangle ABD are known (because BD = AC). Hence two sides and included angle of the oblique-angled triangle ABD are known, from which the other angles and the third side AD can be computed. From the given direction of AB and the angle BAD the course of AD may be found. Example 1. A ship steers E. by N. at 10 miles per hour in a current setting S. by W. 3 miles per hour. Find the ^ course and distance made good. The angle BAC = 10 points, therefore ACD = 6 points (or 67 30'), because the two together are equal to two right angles. Applying the formula for two sides and including angle 180 0' 67 30 A + D112 30 (A + D) 56 15' "CD - AC 7 FIG. 86. tan 10-175107 log 0-845098 56 15 CD 10 AC 3 CD + AC 13 CD - AC 7 CD + AC 13 (A - D) 38 52 |(A + D)56 15 L CDA 17 23 Course of AB N. 78 45 E. 11-020205 log 1-113943 tan 9-906262 ((being the lesser angle \ opposite to lesser side) N. 96 8 E. 180 Course made good S. 83 52 E. 88 TEXT-BOOK ON NAVIGATION. To find distance made good (AD). AD sin ACD AC = therefore AD = sin CDA AC x sin ACD ACD 07 30' AC 3 CDA 17 23' AD 9-3 sin CDA ... sin 9-965615 ... log 0-477121 cosec 10-524673 log 0-967409 B 2. Given the course to a given point, the set and drift of the current, and the rate of sailing of the ship ; to find the course to steer and the distance made good towards the point in a given time. Let A be the place of the ship, and B the destination, and AC the set and drift of the current. From C set off CD equal to the distance the ship would sail in the given time, cutting AB in the point D. Then CD gives the course to steer, and AD is the distance made good towards B. In the triangle ACD, the two sides AC and CD are given, and the angle at A opposite to the latter, from which the other angles and remaining side can be found by the rule of sines FIG. 87. Example. Given the direct course from A to B, E. by S., and the set and drift of the current S.S.E. 4 miles per hour, the ship's rate of sailing being 7 miles per hour, required the course to steer in order to counteract the effect of the current, and the distance made good towards B in 2 hours ? Draw AC in a S.S.E. direction, and set off on it 8 miles (drift in 2 hours), and from C set off 14 miles (2 hours' sailing), cutting AB in the point D, and FIG. 88. join CD. The line CD will give the course to steer, and AD is the distance made good. Here AC = 8, CD = 14, and the angle CAD (opposite to CD) = 56 15'. ^ sin A CD , . T. and sin D = AC x sin A THE SAILINGS. 89 A 56 15' AC 8 CD 14 sine 9-919846 log 0-903090 10-822936 log 1-146128 D 28 22' sine 9'676808 Course from A to B S. 78 45 E. S. 107 180 7 E. Course to steer N. 72 53 E. AD. A= 56 15' P = 28 22 Q. 0*7 lf0 AD CD sn sin A sin A C= 95 23 C 95 23' CD 14 A 56 15 sin 9-998080 log 1-146128 cosec 10-080154 Dist. made good 1676 ...... log 1-224362 3. The ship's position by observation and by dead reckoning being known, the set and drift of the current may be found. (Solved by right-angled triangles.) Example. At noon on a certain day, according to dead reckoning from the previous noon, a ship was in lat. 45 30' N., long. 28 15' W. ; but by ob- servation she was found to be in lat. 46 15' N., long. 27 8' W. Required the set and drift of the current. D.R. lat. 45 30' N. Obs. 46 15 N. M. pts. 3073 Long. 28 15' W. 3137 27 8 W. B Y 45 N. 64 67 E. D. long. 67 ... log 1-826075 D /^ c. M. D. lat. 64 ... log 1-806180 Set of current N. 46 19' E. tan 10'019895 46 19' sec 10-160728 D. lat. 45 ... log 1-653213 r Drift in 24 hours 65 -1 log 1-813941 Rate per hour = - = 2'7 miles. FlG - 89 - NOTE. This is a case of finding a course and distance by Mercator, and is solved by the formulae for right-angled triangles. Examples 1 and 3 can be solved approximately by means of the Traverse Table. 90 TEXT-BOOK ON NAVIGATION. Example 1. Courses. Ptet. N. S. E. W. E. by N. S. by W. 10 3 2-0 2-9 9-8 06 2-0 2-9 2-0 9-8 06 06 0-9 92 Course and distance made good, S. 84 E. 9'3 miles. Example 3. D. lat. 64 and dep. 67 give course 46 Course 46, with D. lat. 45, give dist. 65 (1) (2) Questions in current sailing may be varied in many ways, but all of them can be solved on the principles of the three foregoing examples. ART. 28. The " Danger Angle." When sailing near the land where there are outlying dangers, a safe passage may be secured by calculating the angle sub- tended at a safe distance by the height of the top of a light- house or cliff above sea-level. Setting this angle on the sex- tant, the navigator so shapes his course that this vertical angle is not exceeded, allowance being also made for the dip of the land horizon (page 191 Norie or 35 Raper). Referring to the figure, the height BC and the safe distance AB being known (by the chart) Tan Z BAG = 5 AB Let AO be the height of observer's eye above sea-level, OE a horizontal line, and H the base of the cliff or land horizon ; then Z EOH is the dip, EB = OA, and CE = CB - EB. CE Now, tan L COE = and the dip being added to L COE, the danger angle COH is found. Example. The height of the top of a lighthouse which stands inland 500 feet is 725 feet, and, owing to the existence of a sunken rock, it is not safe to pass nearer the land than 1J mile. Required the greatest vertical angle for safe navigation which the top of the lighthouse should make with the base of the cliff on which it stands, the height of eye being 20 ft. THE SAILINGS. 9I Referring to the figure CE = 725 - 20 = 705 ft., AH = 6600 ft. and AB or OE = 6600 + 500 = 7100 ft. 705 ...... log 2 -848189 7100 ...... log 3-851258 COE = 5 40' ...... tan 8 -996931 + Dip 10 L COH 5 50 the danger angle required. To find the Width of a Kiver or Channel. Let C be a promi- nent object on the opposite bank, and A and B two stations whose distance apart is known, B being chosen so that the Z ABC is a right angle. The horizontal angle at A, subtended by B and C, can be measured by means of a sex- tant or prismatic compass. Then ?| = tan A Ar> and CB = AB x tan A Example. Given the distance AB = 600 yards, and the angle CAB = 42 30', required the width (CB) of the river. A 42 30' tan 9 -962052 AB 600 log 2-778151 BC= 549-8 log 2-740203 therefore the width of the river is 550 yards nearly. MISCELLANEOUS PROBLEMS. 1. Two ports in lats. 38 S. and 50 S. are on the same meridian ; a vessel leaves the former, and, after sailing due W. 300 miles, steers due S. until the second parallel is reached. How far is she then distant from the second port? 2. Two places, A and B, are in the same latitude, and in longs. 10 W. and 20 20' W. respectively ; a ship, sailing 10 knots, took exactly 48 hours to pass from one to the other. Find the latitude. ^ 3. A vessel sailed from lat. 35 20' K, long. 30 W., 250 miles due K, 250 miles due W., 250 miles due S., and 250 miles due E. Required her latitude and longitude in. 4. A ship which sails 8 knots per hour, within 6 points of the wind, has to reach a port 120 miles S.S.W. of her, wind S.W. How long will she take? 5. A ship, after steaming S.E. byE. 12 knots per hour for 23 hours, from a position lat. 25 36' S., long. 62 35' E., found herself in lat. 27 25' S., Ions;. 66 45' E. Find set and drift of current. 92 TEXT-BOOK ON NAVIGATION. 6. A ship has to reach a port 30 miles N.N.W. of her ; the wind is N. by E., and she can sail 7 knots within 6 points of the wind, making f- point leeway ; she starts on the starboard tack, wishing to make her port in two tacks. After what interval must she go about ? 7. An observer in a boat finds that the angle of elevation of the top of a lighthouse standing upon a cliff is 32 ; after the boat has been rowed 100 yards directly towards the lighthouse, the elevation of the same point is 71. Find the distance of the boat from the foot of the cliff at both observations ; and, if at both points of observation the angle subtended by the lighthouse is found to have the same value, find the heights of cliff and lighthouse. 8. The upper part of a mast is 30 feet long, and subtends, at a point 150 feet in a horizontal line from the base, the same angle as that subtended at the same point by a man 6 feet high standing at the base. Find height of the mast. 9. A steamer is being steered S. W. \ W., at the rate of 12 miles per hour, in a known current setting W.N.W. 2^ miles per hour. Required the course and distance made good per hour. 10. Given the course and distance to a port, N. 75 W. 196 miles ; a steady current is setting S. 15 W. 1*9 mile per hour, and the vessel steams 10 knots. What course should be steered to counteract the current, and what progress is made towards the port in 12 hours ? 11. A ship's position is, by dead reckoning, lat. 48 21' S., long. 43 28' W. ; and by observation, lat. 47 16', long. 41 3' W. Assuming the discrepancy to be due to a current, what was its set and drift during the time elapsed since the previous observations ? 12. The true course to destination is a N. 33 W., and distance 75 miles; but after steaming 5 hours on that course at 14 miles per hour, it was found to bear N. by W. (true), and distant by cross-bearings 10 miles. What was the effect of the current ? 13. How long should it take a steamer to reach her destination, bearing W.N.W. 85 miles distant, if steering the proper course to counteract the effect of a steady current setting S. 20 W. at the rate of 2 miles per hour, the ship's rate of steaming being 13 knots ? 14. A departure is taken at 6 P.M., from a position determined by cross- bearings in lat. 51 18' N., long. 9 42' W. ; her course is set W. S., and after steaming at the rate of 14 knots per hour until noon the following day, the position is found by observation to be lat. 51 5' N., long. 16 20' W. Required the set and drift of the current. 15. From a place A, two ships start at same time steaming 10 knots and 8 knots respectively due E. After two hours the two ships bear S. 46 W. and S. 66 W. from a place B on the same parallel as and 30 miles E. of A. Find set and drift of current. 16. From a ship at anchor a torpedo boat known to be steaming at 15 knots was observed, at 9 A.M., to bear N. 20 W., at 9 h 10 m she bore N. 5 E., and at 9 h 30 m N. 35 E. Determine course steamed by T. B. and her distance from ship in her last position. CHAPTER VII. TIDES. ART. 29. Observation reveals the fact that the waters around our coast are not at the same height at all hours of the day. At times great tracts are uncovered which at other times have several feet of water over them. This is a consequence of what are known as the " Tides," and is due to the attraction by the moon and sun of the waters of the ocean, combined with the rotation of the earth on its axis. The changes in the distances and relative positions of these bodies account for the general variations in the character of the tides, but local influences modify their action, and make the determination of high and low water at any place a matter of extreme difficulty. The rise and fall occurs twice in about 24f hours, which is the mean value of the lunar day, and tends to show that the moon's influence plays the most important part, being more than twice that of the sun's. Let the figure represent the earth surrounded by an envelope of water, shown by the dotted circle. FIG. 92. Under the action of the moon M, the waters assume an elliptical form, being piled up on the sides nearest to and furthest from the moon, and correspondingly reduced on the sides at right angles to this direction, thus causing the waters to be higher at A and B, in the direction of the line of centres EM ; and lower 94 TEXT-BOOK ON NAVIGATION. at C and D at right angles to this direction. As the earth rotates from west to east, this crest moves from east to west. The action of the sun is similar to the moon, and causes two solar tides in a day. When the centres of the sun and moon are in a line, that is, at new moon, the bodies being in conjunction, or at full moon, when they are said to be in opposition, the solar and lunar effects correspond, and we have the highest or Spring tides. When the bodies are 90 apart or in quadrature, the solar low water corresponds with the lunar high water, and the lowest or Neap tides occur. As the moon moves eastward faster than the sun, the com- bined effect will be to produce a crest to the westward of the moon during the first and third quarters, and thus the time of high water will be hastened ; also an opposite effect during the second and fourth quarters, when the time of high water will be retarded. This is called the priming and lagging of the tides. The middle between the levels of high water and low water at Springs is called the mean level of the sea (M.T.L.). The datum line is the level of low water of ordinary Springs, (M.L.W.S.) and the height of a tide is measured above this line unless it is stated otherwise in the Admiralty Tide Tables, where full information is given at the foot of the page for each port showing what the zero of predictions is. The interval which elapses between the moon's transit each day and the time of H.W. next following is the Lunitidal Interval; the mean of the lunitidal intervals for all days between new and full moon is the Corrected establishment of the Port; the lunitidal interval on the days of new and full moon is the Vulgar establishment, and is registered in Roman figures on Admiralty charts as H.W.F. and C. ART. 31. In the Admiralty Tide Tables are given the times of High and Low Water and Heights above the zero of predic- tions selected for the various ports, around the British Isles and at other places chosen through the world, called " standard " ports. The Tables are in two parts, at the end of Part I. is an index of all standard ports, and of ports referred to a particular standard, and full information is tabulated for each day of the month throughout the year on the pages mentioned. At the end of Part II. is a general index, and on referring to the proper pages the lat., long., H.W.F. and C., and rise at Springs and Neaps are found for each place. Part II. is used when insuffi- cient, or no information is given in Part I. for the required place. The tides in Part I. are entered in order of occurrence, P.M. times and heights in block type. TIDES. 95 The times given are STANDARD or mean time of certain meridians. Local and Greenwich times may be obtained by applying D. Long, in time. The " time differences " show how much earlier (if marked ) later (if marked +) high and low water occur at the required port than at the standard. When no time difference is entered, or place not found in Part I., turn up index Part II. and on the page or pages will be found the Greenwich times of H.W.F. and C. at the required port and at the selected standard. The difference between these is the time difference, and is added (+) when full and change at place is greater than that at standard, subtracted ( ) when less. The port to be selected as standard is the nearest one geographically to the ship. REMARK. When the difference is +, and the sum amounts to more than 12 h , reject 12 h and change from A.M. to P.M. on the same day, or from P.M. to A.M. on the day after : it may be necessary to take the P.M. time of tide for the day before, in order to obtain the A.M. time of tide on the required day. When the difference is , and the time entered in the Tables is less than the difference, borrow 12 h , and change from P.M. to A.M. on the same day, or from A.M. on the day after to P.M. on the required day. Examples. 1. Find the M.T.G. and M.T.S. of H.W. on September 20th off Pembroke Dock. M.T.G. H.W. Pemb. Dock No A.M. NOON. Long. - 20 m - 20 m /. M.T.S. H.W. Pemb. Dock ll h 40 m A.M., No P.M. 2. Find the M.T.G. and M.T.S. of H.W. on August 22nd off Llanclly. M.T.G. H.W. Pemb. Dk. O h I- A.M. O h 36 ra P.M. Diff. + 1 +1 M.T.G. H.W. Llanelly 2 A.M. 37 P.M. Long. W. - 17 - 17 M.T.S. H.W. Llanelly 11 45 P.M. 20 P.M. .*. No A.M. 3. Find the M.T.G. of H.W. on July 22nd off Limerick. Dublin T. H.W. Galway 10 h l ra A.M. 10 h 30 ra P.M. Diff. + 1 34 + 1 34 Dublin T. H.W. Limerick 11 35 A.M. 4 A.M. on 23rd. Long. W. + 25 +25 M.T.G. H.W. Limerick 00 29 A.M. on 23rd. /. Midnt. No P.M. 96 TEXT-BOOK ON NAVIGATION. 4. Find the M.T.G. of H.W. on Oct. 21st off St. Mary's. M.T.G. H.W. Devonport O h 31 m A.M. l h 9 m P.M. Diff. - 1 17 - 1 17 M.T.G. H.W. St. Mary's 11 14 P.M. 11 52 A.M. on 20th. No P.M. 5. Find the M.T.G. and M.T.S. of H.W. on Sept. 21st off Lizard Point. H.W. F. & C. Devonport 6 h O m ,, Lizard Point 5 21 Diff. -0 39 M.T.G. H.W. Devonport O h 30 m P.M. Diff. - 39 M.T.G. H.W. Lizard Point 11 51A.M. No P.M. Long. W. - 21 M.T.S. Lizard Point 11 30 A.M. No P.M. EXERCISES. 1. Find the M.T.G. of high water at Bridlington on Nov. 29th. Spurn Point on Dec 18th. Dingle on Aug. 10th. Hilbre Island on Oct. 13th. Clydebank on Sept. 10th. 2. Find the M.T.S. of high .water at Rotterdam on Jan. 19th and 20th. Ayr on Nov. 9th and 10th. Crinan on Dec. 21st and 22nd Port Carlisle on Sept. 28th and 29th. Wicklow on Sept. 30th and Oct. 1st 3. Find the M.T.G. and M.T.S. of high water at Quaco on July 8th. Strangford Quay on Oct. 29th. 4. Find the M.T.G. of high water at Ardrishaig on Nov. 9th H.W. F. and C. at Greenock 12 h 24 m . Ardrishaig 12 h 15 m . Also at Southampton on Jan. 22nd. H.W. F. and C. at Portsmouth Dockyard ll h 45 m . ( O h 39 m Southampton < -, ^ ^ Times H.W. at Portsmouth O h 44 m A.M. l h l m P.M. CHAPTER VIII. REDUCTION OF SOUNDINGS. ART. 32. Near to land in foggy weather, the greatest security for the navigator is the frequent use of the " lead/' and feeling his way by "ground navigation." A line of soundings and specimens of the bottom (brought up by the " arming " of the lead) will generally enable him to fix his position on the chart, or give him due warning of danger. In Board of Trade inquiries into cases of stranding, the first importance is properly at- tached to the taking of soundings, and woe to the captain who may be found to have neglected this precaution. The soundings marked on charts are from the Mean Low-water Level of spring tides, but the actual depth at a place varies with Springs and Neaps and the interval from high water. The difference between the sea-level at the " time of cast " and the Mean Low water of Springs is the Reduction of Soundings. The Heights of Tides in the Admiralty Tables are measured from the same zero or datum line as used for the chart soundings, but some local tables, e.g. the Liverpool or Holden's Tables, give the heights from a datum line five feet below the Mean Sea-level. In the diagram, let O (zero) be the mean level of low water spring tides, H the mean high water springs, and M the mean level of the sea or half-tide level ; also let h\, h^, h 3 , h 4 be the high- water marks of various tides, and 1, 1% 3, h the corresponding low-water marks, and B the sea-bottom. Now, if the lead were cast when the sea- level is at O, it would give the same sounding p IO 93> as the chart, but at other times there would be a difference depending on the height of the tide in the tables and the interval from high water. This difference is the Reduc- M tion of Soundings. H 98 TEXT-BOOK ON NAVIGATION The Half Range of any tide is found by subtracting the half mean spring range (H.M.S.R.) from the height, e.g. M^ = OAi - O.M. When the half-range, the time from high water and the time occupied in rising or falling are known, reference to the Tidal Diagrams No. I. to V. in the Admiralty Tables, or a diagram drawn by the student himself, gives the difference of level between M and the actual surface of the sea, which, being added to or taken from the H.M.S.R., gives the height above mean low water, or zero. ART. 33. In calculating a correction to a cast or reduction of soundings we proceed in the following order. Firstly. Obtain the interval between the time of cast and the time of H.W. : from this derive the angle of swing. Secondly. Find the Mean Tide Level (M.T.L.) and Height of Tide at ship : their difference gives the half tidal range, or half diurnal range (H.D.R.). Thirdly. With the angle and H.D.R. take either from the traverse table in the D.Lat. column using angle as Co., and H.D.R. as dist., or from the diagram drawn, distance ship is from M.T.L. Fourthly. Apply this distance to the M.T.L. for the distance the ship is from datum, which gives Reduction. To find Interval and Angle. The Interval is the elapsed time between the times of H.W. at place, and of cast: both times being reckoned from the same meridian. The time any tide takes to Rise or Fall is given at foot of column for those ports where H.W. only is predicted ; at ports for which both H.W. and L.W. are predicted it is found by subtracting preceding time of L.W. from H.W. used, for time to Rise : and by subtracting time of H.W. used from time of succeeding L.W., for time to Fall. Interval must always be less than time of Eise or Fall used. Times in Tide Table are not necessarily G.M.T. Since from H. to L.W. tide sweeps through 180, . , Interval in minutes x 180 ~ Time of Rise or Fall in minutes Examples. Find the interval and angle : 1. On August ,29th at 9 h 14 m A.M. M.T.S. off Tenby. M.T.G. Pembroke Dock 29th 6 h 46 ra A.M. Time Diif. - 35 M.T.G. H.W. Tenby 29th 6 11 A.M. Tenby Long. W. - 19 M.T.S. H.W. Tenby 29th 5 52 A.M. cast 29th 9 14 A.M. Interval 3 22 after H.W. .-. Falling. REDUCTION OF SOUNDINGS. 99 Time to fall 6" 15 m ; . 202 x 180 ^o Hence angle = ^ = 97 2. On December 16th at 9 h 40 m A.M. M.T.G. off Lamlash. M.T.G. H.W. Greenock, Dec. 16th 4 h 27 m P.M. Time Diff. - 15 Lamlash, Dec. 16th 4 12 P.M. ,, cast Dec. 16th 9 40 A.M. Interval 6 32 hefore H.W. /.Rising. H.W. Greenock 4 h 27 P.M. , preceding L.W. 9 47 A.M. Time to rise 6 40 392 x 180 Angle = ^- =175 N.B. Since only Time Diff. for H.W. is given, time of rise or fall at ship is taken to be the same as at standard port. 3. On July 23rd at 8" 15 m P.M. M.T.S. off Welchpool, N.B. Stand, time H.W. St. John, July 23rd 6 h 38 m P.M. Following L.W. l h 6 m A.M. Time Diff.+O 2 Time Diff. 10 Welchpool ... 6 40 P.M. Welchp'l L.W. 1 16 A.M. D. Long. W.-O 28 H.W. 6 40 P.M. M.T.S. Welchpool ... 6 12 P.M. To fall 636 Cast 8 15 P.M. Interval 2 3 Angle = Q ^-- = 56 after H.W. . . Falling. Oi)(j N.B. (1) Since Time Differences for both H. and L.W. are given, they must both be used to mid time of fall at ship. N.B. (2) Should no Time Difference be given, find its amount from H.W. F. and Cs. entered in Part II., and use 6 hours as time of rise or fall. EXERCISES. Find the Interval and Angle : 1. On Dec. 24th at ll h 30 m P.M. M.T.S. off Grimsby. 2. Nov. 5th at 3 h 10 m P.M. M.T.S. off Spurn Point. 3. Oct. 25th at 10 h 15 m P.M. M.T.G. off Bridlington. 4. Sept. 26th at l h 10 m P.M. M.T.S. off Donaghadee. 5. Feb. 19th at 4 h 15 m A.M. M.T.S. off Beachy Head. 6. Jan. 22nd at 8 h 15 m P.M. M.T.S. off Ranisgate. To find the Mean Tide Level and Half Daily Range. (a) The height of tides above Datum is entered for all standard ports at H.W., and at most ports for L.W. also. When H.W. heights only are predicted, the M.T.L. is given at the foot of the column. When both H.W. and L.W. heights are predicted the ioo TEXT-BOOK ON NAVIGATION. M.T.L. is half the sum of these heights, unless the L.W. height is * signifying below Datum, when half the difference must be taken. To find the H.D.R. subtract the M.T.L. from the height of tide. Thus on Dec. 2nd, A.M., off Hull. ft. in. Ht. H.W. ... 17 11 M.T.L. ... 10 5 H.D.R. ... 7 6 On Oct. 7th, A.M., off Liverpool, rise and fall. ft. in. ft. in. ft. in. ft. in. Ht. H.W. 23 8 H.W. 23 8 H.W. 23 8 H.W. 23 8 preceding L.W. 6 7 M.T.L. 15 1 following L.W. 7 3 M.T.L. 15 5J 2)30 3 H.D.R. 8 6 2)30 11 H.D.R. 8 2| M.T.L. rise 15 1| M.T.L. fall 15 5| On Sept. 13th, P.M., off Liverpool, rise and fall. Rise. Fall. ft. in. ft. in. ft. in. ft. in. Ht. H.W. 30 1 H.W. 30 1 H.W. 30 1 H.W. 30 1 preceding L.W. *0 6 M.T.L. 14 9| following L.W. 4 M.T.L. 15 2 2)29 7 H.D.R. 15 3^ 2)30 5 H.D.R. 14 10| M.T.L. 14 9 M.T.L. 15 2J (b) The height of tides for places referred to standard ports is found by applying the height differences according to sign to the heights at standard. These height differences are tabulated for Springs and Neaps, and, strictly speaking, are only correct for those days, which may be determined by reference to the Tables, when the days on which the highest H.W. occur are Springs, and those on which the lowest H.W. are entered are Neaps. These days may be on the days of full or new moon, or of quadrature, or one or two days after these phenomena. Astronomical data are entered at the beginning of the Tables. An interval of seven days may be allowed between Springs and Neaps, and so the correct height difference may be determined for any intermediate day by allowing ^th of the difference between the Spring and Neap difference for each day elapsed since last Spring or Neap. To find the M.T.L. for places referred to a standard at which only H.W. is predicted, OR to a standard at which both H.W. and L.W. are predicted, but only the height differences for H.W. are given, apply | SPRING Diff. to the M.T.L. at standard. For places at which the Difference are given for both H.W. and L.W., find the corrected Height Difference and thence the heights of H. and L.W. at the place, from which the M.T.L. for the required port may be directly obtained as for a standard. REDUCTION OF SOUNDINQ.S., ,,,,,,,,>, Examples. - Find the M.T.L. and H.D.k. off Spurn Point on Dec. 13th, P.M. On looking down the Tables, the days of Springs are seen to be on the llth and 27th, our date is two days from Springs. ft. in. Spring Diff. -22 7 Neap ,,-17 2 Diff. for 7 days - 7 7)14 2 __ o 2 Ht.Diff. for'date .'.20 2 ft. in. ft. in. M.T.L. Hull 10 5 Ht. Hull 19 1 | Spring Diff. - 1 1 ,, Diff. - 2 M.T.L. Spurn 94 Spurn 17 1 M.T.L. 9 4 H.D.R. 79 Off Dungeness on Feb. 9th, A.M., falling tide. Springs is on the 6th, hence three days from. ft. in. S. Diff. +37 13 N. ,,+48 3 Diff. for 7 days +1 1 7)39 3 +06 Ht. diff. for date +41 6 ft. in. ft. in. H.W. Ht. Dover 18 1 18 1 L.W. 4 Diff. +41 2)18 5 Ht. Dungeness 22 2 M.T.L. 11 M.T.L. Dover 9 21 | S. Diff. Dungeness + 1 9| H.D.R. 11 2 M.T.L. 11 N.B. Always use f SPRING Diff. to find M.T.L. at ship. Off Ramsgate, January 6th, A.M., rising. Our date two days to springs. ft. in. ft. in. Springs H.W. Diff. -31 L.W. Diff. +0 2 Neaps ,, -16 ,,-14 Diff. for 7 days -17 -16 2 -0 5 -05 /. Ht. Diff. H.W. -28 L.W. -03 Ht. H.W. Dover 18 '9 Ht. L.W. *6 "2 Diff. - 2 8 Diff. -0 3 Ht. H.W. Ramsgate 16 1 Ht. L.W. *0 5 M.T.L. 7 10 H.W. 16 1 H.D.R. 83. 2)15 8 M.T.L. Ramsgate 7 10 102 TEXT-&QOK ON NAVIGATION. EXERCISES. Find the M.T.L. and H.D.R. 1. On August 22nd, A.M., off Mellon. 2. Sept. 12th, P.M., off Yalentia. 3. July 20th, A.M., off Foynes Island. 4. Feb. 14th, P.M., rise and fall off Rye Bay. 5. Jan. 8th, A.M., ,, ,, Hastings. 6. August 17th, A.M., ,, Seal Cove. 7. Oct. 22nd, P.M., ,, Warren Point. 8. Sept. llth, A.M., rise off Annan Foot. To illustrate the working of the examples will now be worked in full. whole problem several Example 1. Find the Reduction of Soundings off Spurn Point on November 18th at 8" 38 m , P.M., M.T.G. M.T.G. of H.W. Hull O h 3" Time Diff. - 1 4 A.M. H.W. Spurn Cast 10 59 P.M. 8 38 A.M. Interval 2 21 before, .*. rising. Takes 6 hrs. to rise. Hence angle = 360 Ht. H.W. Hull Neap Diff. ft. in. 16 -1 7 Ht. H.W. Spurn 14 5 M.T.L. 94 H.D.R. 5 1 ft. in. M.T.L. Hull 10 5 iS. Diff. -1 1 M.T.L. Spurn 9 4 Now, see Fig. 94, draw a vertical line, and a horizontal line for the M.T.L. as OA; from O measure down 9' 4" to scale for zero, and measure up 5' 1" for H.W. ; with centre O, radius = H.D.R. draw semicircle. At O, draw OB making 7Q with vertical cutting semicircle in a point through which draw a horizontal line for " level of tide at cast," meeting vertical in V. Measure OV = 1' 10", this is distance of ship at time of cast above M.T.L. N.B. Without the aid of a protractor, OV may be found in the D. lat. column of tho Traverse Table, using angle 70^ as course, and H.D.R. 5' 1" = 61" as distance, this gives 22". M.T.L. 9' 4" OV + FIG. 94. 1 10 Reduction to zero 11 2 to be subtracted from cast. REDUCTION OF SOUNDINGS. 103 Since the zero at Hull is the level of mean low water springs, this is also the reduction to M.L.W.S. Example 2. Find the correction to cast on July 14th at 8 h 5 m , P.M., M.T.S. off Dingle. Dub. time H.W. Galway 4 h 3 m P.M. Diif. -39 Dingle 3 24 P.M. D. Long. W. -16 M.T.S. time H.W. Dingle 3 8 P.M. M.T.S. Cast 8 5 P.M. Interval 4 57 (falling) Takes 5 h 55 m to fall x 180 o /. angle = 355 ft. ins. Ht. H.W. Galway 15 2 Diff. -4 4 ft. ins. M.T.L. Galway 8 2 \ S. Diff. - 2 5 M.T.L. Dingle 5 9 3 days to Springs. ft. ins. Sp. Diff. -49 Np. -3 10 In 7 days 11 }> 3 >> ~ 5 Ht. Diff. -44 Dingle 10 10 5 9 H.D.R. 51 Now draw figure (Fig. 95). ft. ins. M.T.L. 5 9 0V -4 4 Reduction to zero 1 5 to be subtracted. FIG. 95. Since the zero is 75 foot = 9 inches below the level of M.L.W.S. the Reduction to M.L.W.S. is 8" to be subtracted from cast. Example 3. Find the Reduction on Sept. 27th at 4 h 10 m , P.M., M.T.S. off Wicklow. Stand, time H.W. Kingstown ll h 4 m P.M. Diff. - 43 Wicklow 10 21 P.M. D. Long. E. + 1 M.T.S. Wicklow M.T.S. Cast Interval To rise takes 6 h 25 m . 10 22 P.M. 4 10 P.M. 6 12 (rising) io 4 TEXT-BOOK ON NAVIGATION. H.W. M.T.L. M.T.L. Wicklow Now draw Fig. 96. ft. ins. ft in*. igstown i H.W. Kingstown 11 9 j> 11 9 Diff. - 2 9 2)12 9 ' ji Wicklow 9 M.T.L. yy 5 s.'biff. 6 -1 8 H.D.R. j > 4 5 FIG. 96. M.T.L. 5 O.V. 4 Reduction to zero 1 to be subtracted Since zero is '84 foot = 10 inches below the level of M.L.W.S. Reduction to that level is 2" to be subtracted from cast. Example 4. Find the Reduction on Sept. 12th at 5 h , A.M., M.T.S. off Donaghadee. Stand, time H.W. Kingstown 11 h l m A.M. Diff. - 2 M.T.S. To rise takes 6 h 29 ra Donaghadee D. Long. E. 10 59 A.M. + 3 "Cast 11 2A.M. 5 Interval 6 2 risim 389 ft. ins. Ht. L.W. Kingstown 6 Diff. -0 10 H.W. ft. ins. 12 4 + 6 Ht. L.W. H.W. Donaghadee 4 12 10 H.W. M.T.L. 12 10 6 3 2)12 6 H.D.R. 6 7 M.T.L. ,, 6 3 Now draw Fig. 97. ft. ins. M.T.L. 6 3 0V. 6 5 Red. to zero 2 to add to cast Since zero is *84 foot = 10 inches below the level of M.L.W.S. Reduction to M.L.W.S. is 1 foot to be added to the cast. REDUCTION OF SOUNDINGS. 105 This problem may also be utilised to find the depth of water to be found at any time at a given spot on a chart, by adding the Reduction found to the depth marked on the chart ; if from this the ship's draught is subtracted, Level areas!" i Red Fia. 97. the depth of water under the ship's bottom may be found ; or, again, the time may be found when a vessel of given draught will find sufficient water to pass an obstruction. EXERCISES. Find the correction to be applied to the cast, in order to reduce the soundings to datum and M.L.W.S. : 1. On Oct. 14th, at 5 h 56 m P.M., M.T.G. off Hull. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. July 24th, at 4 h 30 m A.M., M.T.S. off Galway. Nov. 1st, at l h 20 m A.M., M.T G. off Grimsby. Aug. 15th, at l h 10 ra A.M., M.T.S. off Limerick. Sept. 20th, at 6 h 20 m P.M., M.T.G. off Tenby. Dec. 18th, at 10 h P.M., M.T.G. off Spurn Point. Aug. 6th, at 2 h 10 A.M., M.T.S. off St. John (N.B.). Nov. 3rd, at 7 h 40 m A.M., M.T.G. off Abu Shahr. Sept. 28th, at 4 h 45 m A.M., M.T.G. off Fowey. Oct. 19th, at 5 h 20 m P.M., M.T.S. off Exmouth. Oct. 23rd, at 7 h 10 m P.M., M.T.S. off St. Mary's. Oct. 7th, at 2 h 40 m P.M., M.T.G. off Beaumaris. Sept. 13th, at 6 h 20 ra AM., M.T.S. off Workington. Oct. 12th, at O h 15 m P.M., M.T.S. off Dartmouth. Sept. 13th, at 5 h 10 m A.M., M.T.S. off Whitehaven. Nov. 20th, at 5 h 15 A.M., M.T.S. off Henjam Island. July 15th at 10 h 16 ra P.M., M.T.G. off St. Andrews (N.B.). a Spring-tide day 1^ hours from H.W. off Al Basra. a Neap-tide day 2 hours from L.W. off Kuweit. 20. The depth of water on a bar at the entrance to a harbour at mean low water springs is 10 ft. 6 in. If the time of high water is 10" 51 ra A.M., find the depth of water on the bar at 9 h A.M., the height of tide for day being 20 ft. 3 in. Spring range 22ft. io6 TEXT-BOOK ON NAVIGATION. 21. On a certain day it is high water at Falmouth at 7 h 20 m P.M. Height of tide 15 ft. 6 in.; mean tide level 8 ft. What is the earliest time in the afternoon when there will be not less than 20 feet of water over a shoal marked 2 fathoms on the chart ? 22. At St. Helier, Jersey, the M.T.L. is 15 ft. 9 in., and a certain tide rises 25 ft 6 in. Find the height of the tide l h 30 m before high water. 23. On Deal Bank the depth marked on the chart is 18 ft. What depth of water would there be over it on Feb. 6th at l h 30 m A.M. M.T.G. ? If a vessel's draught was 25 ft., what depth of water would she have under her ? 24. A vessel whose draught is 23ft. Gin. wishes to berth at inner harbour where average depth on chart is 2 fathoms. Heights of tide at high and low water on the day are 14' 10" and 1' 10" ; time of H.W. 9 h 56 m A.M., previous time L.W. 3 h 43 m A.M. What is the earliest time on the day when there will be sufficient water ? CHAPTER IX. TIME. ART. 34. Time and Arc. It is often necessary to change Time ( hras ) into Arc (' "), or the contrary. Twenty-four hours are the equivalent of 360; therefore l h is equivalent to 15 l m 15' 18 T K'/ )t *O also 4 m are 1 4 s r Two rules for converting Arc into Time are deduced from the above, viz. (1) Divide the '" by 15. (2) Multiply by 4, and divide by 60. NOTE. The latter is generally preferred, because of the easier division. Example, Convert 34 16' 30" into Time. 3)34 16' 30" 34 16' 30" 4 5)11 25 30 2 h 17 m 6 s or 6,0)13,7 6 2" 17 m 6 s EXERCISES. Find the equivalent values in Time of the following Arcs : 2 59' 30" ; 179 0' 45" ; 15 0' 15" ; 67 52' 55" ; 17' 20" ; 10 19' 40" ; 155 11' 36" ; 93 0' 45" ; 117 38' 30" ; 144 54' ; 176 39' 35" ; 3 23' ; 59' 45" ; 0' 15" ; 15' ; 48' ; 13 ; 168 9' ; 61 48' 45". By reversing the above rules, Time may be changed into Arc, viz. (1) Multiply by 15. (2) Multiply by 60 and divide by 4. io8 TEXT-BOOK ON NAVIGATION. Example. Change 7 h 32 m 15 s into arc. 7 h 32 m 15 s 7" 32 m 15 9 3 60 22 36 45 or 4)452 15 5 113 3' 45" 113 3' 45' EXERCISES. Change into Arc the following Time intervals : l h 12 m 30 f ; 10 h 19 m 55 ; 8 h O m 10' ; 6 h 59 m 18' ; 11" 59 m 58 s ; 3 h O m 19' ; 5 h 69 . 22 . . 9 h lm 37 . . O h 15 m 15 . . 2 h 25"- ; O h 10 m 35' ; O h O m 45' ; 2 h 24 ra 44" ; IP O m 19'; ll h 13 m 30'; 5 h O ra 5'; 8 h 28 52'; 10 h O ra 12 s ; O h O m 12'; O h l m 40 . ART. 35. Time and Longitude. The rules for Time and Arc may be employed to find (1) The difference between Greenwich Time and the Time at a place whose Longitude is given ; and (2) The Longitude corresponding to a given difference of Time. But it must be borne in mind that, as the Earth rotates from west to east, places in E. longitude have sunlight earlier than Greenwich, and places in W. longitude later. Therefore the Time at a place in E. longitude is always fast of or greater than the Time at Greenwich, and at a place in W. longitude it is always slow of or less than Greenwich Time. ART. 36. Difference of Time due to Longitude. Examples. 1. What is the difference between the time at a place in 72 50' E. long, and the time at Greenwich ? 3)72 50' 72 50' 4 5)24 16 40 or 6,0)29,1 20 4 h 51 m 20' 4 h 51 m 20 $ Ans. The time at place is 4 h 51 m 20" fast of Greenwich time. 2. What is the difference between Greenwich time and the time at a place in 124 10' 45" W. long. 1 3)124 10' 45" 124 10' 45" 4 5)41 23 35 or 6,0)496 43 8 h 16 m 43' 8 h 16 m 43 8 Ans. The time at place is 8 b 16 m 43" behind, or slow, of Greenwich time. TIME. 109 EXERCISES. Find the difference between time at Greenwich and corresponding time in the following longitudes, and say which is the greater : 10 20' W. ; 17 45' E. ; 110 17' E. ; 172 36' W. ; 100 16' 30" E. ; 90 4' 45" W. ; 7' 15" E. ; 179 56' 45" W. ; 1 2' 15" W. ; 115 0' 30" E. ; 2' 45" E. ; 165 0' 15" W. ART. 37. Longitude from Difference of Time. Examples. , 1. The mean time at ship is January 5 d 2 h 14 40', when mean time at Greenwich is January 5 d 9 h 45 m 20" : required the longitude. M.T.S. January 5 d 2 h 14 ra 40 1 M.T.G. 5 d 9 45 20 Difference 7 30 40 or 7 h 30 m 40 s 3 60 22 32 4)450 40 5 112 40' 0" Long. W. 112 40' 0" 2. The mean time at ship is February 14 d 17 h 16 m 25* when mean time at Greenwich is February 14 d 17 h 19 m 14" : required the longitude. M.T.S. February 14 d 17 h 16 m 25* M.T.G. ,, 14 d 17 19 14 Difference 2 49 3 or 0* 2 m 49 8 GO 8 27 5 4)2 49 049' 15' Long. W. 42' 15" 3. Given M.T.G. March 30 d 2 h 19 m 55% and M.T.S. March 29 d 21 h 16 58 s : required the longitude. M.T.G. March 30 d 2 h 19 m 55* M.T.S. 29 d 21 16 58 Difference 5 2 57 60 4)302 57 Long. W. 75 44' 15" 4. Given M.T.G. December 31 d 19 h ll m 25', and M.T.S. January l d 3 h 17'" 50' : required the longitude. no TEXT- BOOK ON NAVIGATION. M.T.G. December 31 d 19 h ll m 25" M.T.S. January l d 3 17 50 Difference 8 6 25 60 4)486 25 Long. E. 121 36' 15" 5. Given M.T.S. July 2 d 5 h 15 m 45', and M.T.G. July l d 22 h 20 m 30' : required the longitude. M.T.S. July 2 d 5M5 ra 45 i M.T.G. I d 22 20 30 Difference 6 55 15 60 4)415 15 Long.E. 103 48' 45" NOTE. The student may change Time into Longitude by any method he prefers. EXERCISES. Find the Longitudes from the following Simultaneous Times at ship and Greenwich : No. M.T.S. M.TG. h. m. 8 b. m. s. 1 1906, January 10 d 7 40 25 1906, January 10 d 1 14 42 2 H February U d 8 22 33 February 14 d 10 29 54 3 n March 3l d 19 8 11 April 1 2 28 31 4 5 April May 20 d 13 d 3 22 13 4G 48 59 April May 39 d 14 d 19 4 3 2 54 14 6 n June l d 5 3 13 May 31 d 21 11 19 7 5> July 23 d 11 12 43 July 23 1 11 10 24 8 August l d 36 18 July 31 d 23 59 48 9 5> September 30 d 23 58 45 October l d 1 5 10 ,, October 29 d 11 13 10 October 29 d 22 49 17 11 ,9 November 22 d 23 31 49 November 23 d 1 28 12 99 December l d 9 10 12 November 30 d 22 2 12 13 1905, December 31 d 18 50 40 January l d 11 50 14 M December 31 d 13 20 25 January l d 1 16 33 ART. 38. To find the Time at Greenwich when the Time at a place or at ship and the Longitude are known. Rule. Add the difference of time for West Longitude, and subtract for East. Examples. 1. When it is apparent noon at ship on May 10th in long. 96 W., what is the time at Greenwich ? A.T.S. May 10 d O h O ra 0* 9G 624 4 A.T.G. May 10 d 6 24 6,0)38,4 6 h 24 n TIME. in 2. It is apparent noon at a place in 175 40' E. on July 30th : required the time at Greenwich. A.T. at place, July 30 d O h O m 0' 175 40' 114240 4 A.T.G. July 29 d 12 17 20 6,0)70,2 40 ll h 42 m 40 3. The sun is on the Meridian of ship's place in long. 100 35' E. on May 15th : required the time at Greenwich. A.T.S. May 15 d O h O m 0' 100 35' 6 42 20 4 A.T.G. May 14 d 17 17 40 Or 15th 5 17 40 A.M. 6,0)40,2 20 6 h 42 m 20' 4. When it is 6 h 50 m P.M. apparent time on February 10th at ship in long. 75 50' W., what is the time at Greenwich ? A.T.S. February 10 d G h 50 m 0* 75 50' 5 3 20 4 A.T.G. February 10 d 11 53 20 6,0)30,3 20 5 h 3 ra 20' 5. At a place in long. 115 56' E. it is 7 h 30 m A.M. mean time on June 15th : required the Greenwich time. M.T. at place, June 14 d 19 h 30 m 0' 115 56' 7 43 44 4 M.T.G. June 14 d 11 46 16 6,0)46,3 44 7 h 43 ra 44 1 NOTE. A.M. Civil time is changed into Astronomical time by adding the interval of 12 hours between noon and midnight, which makes the Astro- nomical day one less than the Civil day ; but when it is P.M. the Astronomical day is the same as the Civil. Astronomical time is always written tlms : February 10 d 6 h 50 m s , or June 14 d ] 9 h 30 m O 1 being reckoned from noon onwards to 24 hours. Civil time is written thus : February 10th, 6 h 50 m 0' P.M., or June 15th, 7 h 30 m A.M. EXERCISES. To find the Astronomical Time at Greenwich. Civil time at ship. 1906. 1. May 16th ... 2. May 1st 3. June 13th ... 4. June 30th ... 5. June 1st 6. February 17th 7. February 28th 8. August 21st... 9. August 1st ... 10. August 10th 11. October 22nd 12. October 29th Longitude of ship. h. m. s. O 1 II .. 3 12 30 P.M. 59 16 30 W. .. 5 810A.M. 136 22 50 W. .. 82735P.M. 174 14 45 E. .. 11 249A.M. 105 31 45 W. .. 4 150P.M. 96 17 15 E. .. 74328P.M. 74 44 30 W. .. 101918A.M. 112 14 45 W. .. 61640A.M. 43 13 30 W. .. 25414P.M. 16555 OE. .. 11955P.M. 15 50 15 E. .. 9 38 12 A.M. 155 045W. .. 10 56 11 P.M. 82 17 15 W 1 12 TEXT-BOOK ON NA VIGA TION. ART. 39. To find the Time at Ship when the Time at Green- wich and the Longitude of Ship are known. Huh. Subtract the difference of time for West Longitude, and add for East, which is simply the reverse of the preceding rule. The reason has been previously explained (Art. 35). Examples. 1. Required the Time at ship in long. 65 E., when the A.T.G. is August 5th, 10" 30 ra A.M. A.T.G. August 4 d 22 h 30 m 65 4 20 4 A.T.S. August 5 d 2 50 6,0)26,0 4 h 20 m 2. What is the A.T.S. in long. 120 15' W. when the A.T.G. is 2 h 35 m P.M. on January 2nd ? A.T.G. January 2 d 2 h 35 m 120 15' 81 4 A.T.S. January l d 18 34 6,0)48,1-0 Or January 2nd 6 34 A.M. gh ] 3. Required the Mean Time at ship in long. 64 W. when the M.T.G. is August 15 d 5 h 17 m 35". M.T.G. August 15 d 5 h 17 m 35" Long. 64 W. 4 16 4 M.T.S. August 15 d 1 1 05 6,0)25,6 4 h 16 m EXERCISES. 1. What is the mean time at ship in long. 123 27' 30" E. when the M.T.G. is October 14 d 9 h 15 m 24" ? 2. What is the M.T.S. in long. 35 W. when the M.T.G. is May 6th 2 h 3 m 46 g P.M. ? 3. Find the M.T.S. in long. 143 27' 30" W. when the M.T.G. is September 1st 4 h 37 ra 23' A.M. 4. When the M.T.G. is June 2nd ll h 19 m 14' P.M., what is the mean time at ship in long. 95 46' E. ? 5. Required the M.T.S. in long. 97 16' 45" E. when the M.T.G. is October 19th 5 h 8 ra 40" P.M. 6. Find the M.T.S. when the M.T.G. is May 1st 5 h 49 m 54" A.M. in long. 164 48' E. ART. 40. Rating Chronometers. The ship's chronometer is relied on at sea to give the Mean Time at Greenwich (M.T.G.), its error on leaving port and its rate of gaining or losing per day being known. An " Error " and " Rate " are usually supplied when the chro- nometer leaves the shop, but as the Rate is liable to be affected by changes of temperature and other causes during a voyage, it TIME. 113 is advisable to ascertain the correct Error when opportunity offers at ports the ship may visit. " Eating a chronometer " means finding the rate at which it gains or loses per day by dividing the sum or difference of the errors on different dates by the number of days elapsed. Examples. 1. A chronometer was found to be 4 m 30 fast of mean noon at Greei.wich on January 15th, 1906, and on March 30th, 1906, it was 7 m 35 s fast of mean noon at Greenwich. Required the rate. m. a. Fast 4 30 on January 15. Fast 7 35 on March 30. 3 5 gained in 74 days. 60 74)185f2'-5 gain per day. 148 370 370 2. A chronometer was 14 m 25 fast of mean noon at Greenwich on September 1st, 1906, and at mean noon at Greenwich on December 4th, 1906, it was 13 m 14'-5 fast. Required the rate. Fast 14 25 on September 1. Fast 13 14-5 on December 4. 1 10-5 lost in 94 days. 60 94)70-5(75* losing daily. 658 470 470 3. A chronometer was l m 42' slow of M.T.G. at mean noon on January 1st, 1906, and on March 31st at mean noon Greenwich time it was 6 m 9* slow. Required the rate. m. s. Slow 1 42 on January 1. Slow 6 9 on March 31. 4 27 lost in 89 days. 60 89)267(3' losing daily. 267 4. On leaving Liverpool, January 19th, 1904, the ship's chronometer was 4 m 2*'5 slow of mean noon at Greenwich, and at Sydney on May 28th it was found to be 3 m 10 s - 5 slow of mean noon at Greenwich. Required the rate. r ii 4 TEXT-BOOK ON NAVIGATION. Slow 4 2 -5 on January 19. Slow 3 10-5 on May 28. 52-0 gained in 130 days. 130)52 -0(-4' gaining daily. 52-0 5. At New York on February 27th, 1904, a ship's chronometer was 18* slow at mean noon Greenwich time, and at Southampton on March 6th, at Greenwich mean noon it was 14 s fast. Required the rate. Slow 18 on February 27. Fast 14 on March 6. 32 gained in 8 days. 8)32 4" gaining daily. 6. A chronometer was fast of mean noon at Greenwich 2 m 17 s on January 14th, 1906, and on June 1st it was 3 m 55 < 5 slow of mean noon at Greenwich. Required the rate. m. s. Fast 2 17 on January 14. Slow 3 55*5 on June 1. 6 12-5 lost in 138 days. 60 138)372'5(2'-7 losing daily. 276 966 ART. 41. The time by chronometer being given, and two errors or one error and daily rate, to determine the correct G.M.T. Find the rate as explained above, multiply this by the number of days and decimals elapsed between the date of second error, and date of chronometer time. This gives the accumulated rate. Now take the time by chronometer expressed astronomi- cally, being careful to make it correspond as far as possible with the Greenwich Time deduced from the Ship Time and approximate longitude, apply the second error and accumulated rate to obtain the correct G.M.T. Examples. 1. February 24th, about 8 h 20 m A.M. at ship mlong. by account 74 15' W., the chronometer showed l h 12'" 20 8 '5, arid had been found 13 m 35" slow of mean noon at Greenwich on October 30th, and on December 24th it was 10 m 55-6 slow. Find G.M.T. TIME. 115 74 15' Ship time Feb. 23 d 20 h 20 W Slow Oct. 30th 13 m 35* 4 Long. W. + 4 57 6,0)29,7 Greenwich time 24 d 1 17 Gain in 55 days 2 39 '5 4 h 57 ra 55)159 -5(2' -9 Time by chron. Feb. 24 d l h 12 m 20 '5* 110 Second error slow 10 55 '5 495 24 d l 23 16 495 Accum. rate gained 3 M.G.T. Feb. 24 d 1 20 16 Elapsed time 62 '05 rate 2 -9 55845 12410 6,0)17,9-945 2 m 59'-9 2. October 18th, at 2 h 30 m A.M., M.T. ship nearly in long, by D.R, 50 E. ; the chronometer showed ll h 3 m 22 s , which had been found 4 m 13' 7 fast on July 24th, and losing daily 3 s "4. Find G.M.T. Ship time Oct. 17 d 14 h 30 m Elapsed time 85 "5 4 LongE. -3 20 rate 3'4 6,0)20,0 Oct. 17 d ll 10 3420 2565 3 h 20 m Time by chron. Oct. 17 d ll h 3 m 22' 6,0)29,070 Error fast - 4 137 4 m 50 3 7 Oct. 17 d 10 59 8-3 Accum. rate lost +4 507 M.T.G. Oct. 17 d 11 3 59 3. February 3rd, approximate ship time l h 8 m A.M., in long, about 31 W., when a chronometer, known to be 43 m 19 s fast on January 13th, and gaining daily 2' 5, showed 3 h 55 30 s '5. Find correct G.M.T. 31 Ship time Feb. 2 d 13 h 8 m Elapsed time 20 '6 days 4 Long. W. + 2 4 rate 2 '5 6 0)1 ; 24 Greenwich time 2 d 15 12 1030 412 2 h 4 m 51-50 Time by chron. Feb. 2 d 15 h 55 ra 30"5 Error fast - 43 19 Feb. 2 d 15 12 11-5 Accum. rate gained 51 '5 G.M.T. Feb. 2 d 15 11 20 TEXT-BOOK ON NAVIGATION. Find the chronometer Mean Noon : EXERCISES. Rate" from the following errors on Greenwich No. 1st date. Error. 2nd date. Error. h. m. B. b. m. B. 1 August 28th 072-1 fast December 9th 2 24 fast 2 September 30th 1 21 19-0 slow January 28th 1 10 43 slow 3 April 12th 25 19-0 slow July 1st 11 55-0 slow 4 April 1st 10 24-0 fast August 15th Oil 32-0 fast 5 February 27th 19-5 fast July 31st 57-5 slow 6 December 27th correct February 20th 4 57-0 fast 7 June 1st 1 8-8 slow September 30th 40-1 fast o June 16th 14 26-9 fast September 30tli 3 18-4 slow 9. November 27th, at about 10 h A.M., at ship in long. 123 15' E., when a chronometer showed l h 53 ra 34' - 6, which was slow for G.M.N. 3 m 13 s on September 10th, and fast 2 ra 30 s on October 20th. Find correct G.M.T. 10. October 31st, about 2 h 15 m P.M., at ship in long. 124 W., when a chronometer showed 10 h 7 ra 33% which was slow 10 ra 46 s '8 on September 29th, and gaining I 8 '01 daily. Find the G.M.T. 11. March 20th, about 4 h P.M., at ship in long. 63 45' W., when a chronometer showed 7 h 12 m 1', which was slow l h 2 m 4' on January 2nd, and on February 1st was slow l h 3 m 54'. Find the G.M.T. 12. December 10th, about 2 h 15 m P.M., at ship in long. 161 W., when a chronometer showed 2 h l m 23', which was fast 59 m 16' on August 12th, and on October 31st was fast l h 2 ra 16' on G.M.N. Find G.M.T. 13. July 4th, at 7 h 17 m A.M., at ship in long. 75 W., when a chronometer showed 12 o'clock, it was fast of G.M. noon 38* on April 15th, and gained 75 s daily since. Find G.M.T. 14. November 27th, at 7 h A.M., at ship in long. 4 45' W., when a chronometer showed 7 h 17 m 58' '5, which had been found 6 m 57' slow for G.M. noon on August 21st, and slow 6 m 37"'5 on September 20th. Find G.M.T. 15. December 14th, about 4 h 5 ra P.M., M.T.S. in long. 11 6' W., when a chronometer showed 5 h 20 ra 44 g , which was fast of G.M. noon 2 m 38' *1 on November 10th, and on November 27th was correct. Find G.M.T. 16. May 1st, about 6 h 30 1 " P.M., at ship in long. 91 E., when a chronometer showed 12 h . and was slow 24 m 54* on January llth for G.M. noon, and slow 24 m 34 s on March 2nd. Find G.M.T. 17. April 16th, about 4 h P.M., at ship in long. 131 25' E., when a chronometer showed 7 h 18 m 56', which was slow of G.M. noon on February 9th 9 m 21' -5, and was slow 3 19'-5 on March 17th. Find G.M.T, CHAPTER X. ELEMENTS FROM THE "NAUTICAL ALMANAC." ART. 42. A great part of the labour involved in the calcula- tion of problems in Nautical Astronomy is the Reduction of the Elements from the "Nautical Almanac;" therefore it is most important that the student should master this part of the work at the outset. The advantage of a thorough acquaintance with the " Nautical Almanac " is incalculable. The necessary extracts for working all the problems in this Book will be found in the Appendix. In the case of the stars the " mean position " only is given, and this is the one used in the calculations, as the introduction of the " apparent places " would considerably enlarge the Appendix without giving a corresponding advantage. The Sun's Elements. These are given for Noon at Greenwich, therefore it is always necessary to know the Greenwich Astronomical Time. When Apparent Time is given, the elements are taken from p. I. of the month ; but when Mean Time is given, they are taken from p. II. The variations or changes in one hour are on p. I. All minutes and seconds in the Greenwich time are reduced to decimals of an hour. (1) Declination. Examples. 1. Find the sun's declination for August 16th 8 h 36 P.M. apparent time at Greenwich. Declination at noon 13 57' 28"'0 N. H.V. 47'03 page I., app. time Cor. - 6 44 5 8'6 Dec. at 8 h 36 m 13 50 43'5 N. 28218 37624 6,0)40,4-458 6' 44";T The correction is subtracted because the declination is decreasing, ii8 TEXT-BOOK ON NAVIGATION. 2. Find sun's declination for September 29th 2 h 45 m A.M. M.T.G. ; astro. M.T.G. 28" 14 h 45 ra . Dec. for 29 d noon 2 9' 24"-l S. H.V. 58'41 page I., mean time. Cor. -9 00-3 9-25 Dec. for 28 d 14 h 45 m 2 23'8 S. 29205 11682 52569 6,0)54,0-2925 9' 0"'3 Here we work from nearest noon, and obtain multiplier for H.V. by subtracting the 14 h 45 m in G.M.T. from 24 h = 9 h 15 m . The correction is subtracted since the declination decreases from 29th to 28th. 3. Find sun's declination for September 23rd ll h 48 m P.M. M.T.G. Dec. for 23 d noon 10' 58"'l N. H.V. 58'45 Cor. - 11 297 11-8 Dec. for 23 d ll b 48 m 31 6 S. 46760 64295 6,0)68,9710 11' 29" 7 Since the declination changes in name from N. to S. , it must decrease. The declination has a maximum value twice during the year, about June 21st and December 24th, the value being about 23 27' 15" ; it has no value about March 20th and September 23rd, the sun being on the equinoctial on those dates. From March 20th to June 21st the sun's declination is N. and increasing. ,, June 21st to Sept. 23rd ,, ,, ,, ,, decreasing. Sept. 23rd to Dec. 24th ,, ,, S. ,, increasing. Dec. 24th to March 20th ,, ,, decreasing. (2) Equation of Time. Examples. 1. Find the Eq. of Time for September 24th 8 h 21 m A.M. A.T.G. Eq. of Time for Sept. 24 d noon 7 m 43 9 '36 H.V. '866 Cor. 3 '16 3*65 for nearest noon. ,, Sept. 23 d 20 h 21 m 7 40'20 4330 5196 2598 3-16090 By looking on the top of the column, we may see how to apply this amount to A.T.G. in order to obtain M.T.G. A thick line divides the heading. There is a corresponding thick line in the Eq. of Time column between l d and 2 d : this shows that all amounts above the thick line must be applied as stated above the line in the heading, and vice versa ; hence the amount found, 7 m 40 s '2, must be subtracted from the given A.T.G. Thus A.T.G. Sept. 23 d 20 h 21 m 0' Eq. of Time - 7 40'2 M.T.G. Sept. 23 d 20 13 19'8 ELEMENTS FROM THE "NAUTICAL ALMANAC." 119 2. Find Eq. of Time for September 1st 8 h 42 m P.M. M.T.G. ; and find A.T.G. Eq. of Time Sept. 1st noon O ra 9' '13 H.V. 779 Cor. 678 87 ,, I d 8 h 42 ra 2-35 5453 6232 G7773 M.T.G. Sept. l d 8 h 42" 0' Eq. - 2-35 A.T.G. l d 8 41 57-65 Correction is subtracted from Eq. because it is changing its name, and the total amount is subtracted from M.T.G. because it is above the black line in column. Had the correction been more than the original amount of Eq., the corrected amount would have turned out below the line and additive. The Eq. of Time has 4 maxima during the year, on Feb. llth, May 15th, July 27th, Nov. 4th. The Eq. of Time has 4 minima during the year, on April 15th, June 15th, Sept. 1st, Dec. 25th. On the days on which the Eq. of Time vanishes, the mean and apparent suns are on the same hour circle. (3) Eight Ascension. Examples. 1. Find the R.A. of the apparent sun on August 19th at 8 h 27 m P.M. M.T.G. R.A. August 19th noon 9 h 51 m 21"-31 H.V. 9-304 Cor. + 1 18-62 8-45 19 d 8 h 27 m 9 52 39'93 46520 37216 74432 78-61880 2. Find the R.A. of the apparent sun on August 25th at 3 h 15 m A.M. A.T.G. A.T.G. August 24 d 15 h 15 m ft. A. August 25th noon 10 h 13 m 33 s '02 H.V. 9'187 Cor. - 1 20-39 8 75 24 d 15 h 15 m 10 12 12-63 45935 64309 73496 80-38625 In correcting the above elements, the H.V. as entered in the " Nautical Almanac " has been used without alteration ; the H.V. differs, however, day by day, and therefore requires correcting itself; this is done by applying the amount that the H.V. changes in half the interval to be corrected for to the H.V. used : the element is then said to be corrected for 2nd differences. 120 TEXT-BOOK ON NAVIGATION. Example. Find the R.A., Dec., and Eq. of Time for August 1st, 8 h 30" P.M. A.T.G., correcting for 2nd Differences. R.A. for August 1st noon 8 h 42 m 55'-43 Cor. + 1 22-612 R.A. 8 h 30 m 8 44 18-042 Dec. for August 1st noon 18 11' 45"-0 N. Cor. - 5 17*8 Dec. 8 h 30 m 18 6 27'2 N. Eq. of Time August 1st noon 6 m 10 8 '22 Cor. - 1-165 8 h 6 m 6 9-055 II.V. 1st 9-724 2nd 9-698 026 J Int. 4-25 2550 850 24) -11050 005 II.V. 1st 37"'26 2nd 38-00 74 JInt. 4-25 1700 2975 24)3-1450 13 H.V. 1st '132 2nd -158 026 \ Int. 4-25 2550 850 24) -11050 005 9 s -724 Cor. -005 9-719 8-5 48595 77752 82-6115 37"'26 Cor. + '13 18695 29912 317-815 Cor. 132 + 005 137 8-5 685 1096 11645 NOTE. The Polar Distance of an object is its distance from the elevated pole, and is the complement of the declination. It is found by adding 90 to the dec. when dec. and lat. are of opposite names, and by subtracting the dec. from 90 when dec. and lat. are of same name. EXERCISES. Correct the declination, equation of time, and right ascension for the following Greenwich times. 1. Apparent time Aug. 3 d 5 h 48 m . 2. ,, Aug. 7 d 9 54. 3. Aug 14 d 14 36. 4. Mean time Sept. 14 d 10 h 24 m . 5. 6. 7. 8. 9. Sept. 23 d Sept. 23 d Sept. 23 d Sept. 23 d Sept. 29 d 6 12. 11 30. 15 9. 22 45. 18 15 ELEMENTS FROM THE ''NAUTICAL ALMANAC." 121 10. Find the A.T. ship when the Greenwich mean time is August 20 d 16" 5 m 30" in long. G4 E. 11. Find the M.T. ship when the A.T.G. is September 30 d 8 h 42 m in long. 164 30' W. 12. If the A.T.S. is September 25th 3 h 19 m 25' A.M., in long. 84 20' E., find the M.T.G. ART. 43. Sidereal Time. This is tabulated for noon of each day on p. II., and is the Sidereal Time at mean noon ; another name for it is Right Ascension of the Mean Sun. Since the Mean Sun is supposed to move uniformly in the equinoctial, the H.V. is constant. A table for correcting the Sidereal Time is found in the " Nautical Almanac ; " or in Norie, table xxxviii. The equivalents taken from the "Nautical Almanac" are given in the Appendix. Example. Find the correct Sid. Time for August 25" 13 h 15 m 25% M.T.G. S.T. for August 25 d noon 10 h ll m 21 s '50 Accel, for 13 h 2 813 15 m 2-46 25' -07 R.A.M. Sun 10 13 3216 EXERCISES. Correct the Sid. Times for following G.M.T.'s. 1. August 19 d 18" 10 m 32*. 2. 28 3 4 47. 3. September 5 14 54 14. 4. ,, 29 2 49 24. ART. 44. Moon's Elements (1) Semi- diameter. (2) Horizontal Parallax. These are entered on p. III. of " Nautical Almanac " for "noon" and "midnight" of each day. The difference between any two consecutive entries is the variation in twelve hours, from which by proportion may be found the variation corresponding to the required interval. After being corrected for "time," the S.D. requires a further amount called " Augmen- tation " to be added to it, in consequence of the apparent increase in the disc of the moon, due to her approach to the observer as she 'increases her altitude. This is found in Table D, Norie. The H.P. also requires an amount called " Reduction " to be subtracted from it, because the H.P. entered in the almanac is the Equatorial H.P., and the observer is in some latitude where the earth's radius is shorter than the equatorial radius, and so the H.P. is reduced. This correction is found in Table E, Norie. 122 TEXT-BOOK ON NAVIGATION. Example. August 16 d at 5 h 14 m 24 s M.T.G. in lat. 35 S. ; app. alt. moon, 47. Find corrected S.D. and H.P. S.D. H.P. Noon 15' l"-38 55' 2" -42 Midnight 15 5 '37 55 17 "04 Diff. in 12 hrs. 3 '99 14 "62 5-24 5-24 1596 5848 798 2924 1995 7310 12)20-9076 12)76-6088 Change in 5 h 14 m '4 1 '74 6 '38 Noon 15' 1"'38 Noon 55' 2" -42 Corr. + 1-74 Corr. + 6 '38 15 312 55 8-80 Table D +107 Table E - 3 '55 15 13-82 55 5-25 Should the time exceed 12 hours, that is, fall between midnight and noon, the element is taken out for midnight and corrected towards the next day noon, using as a multiplier the hours in the Greenwich time redwed by 12. EXERCISES. Find the S.D. and H.P. of the Moon under the following conditions : 1. August 7 d 7 h 24 m M.T.G. in lat. 40 N. Alt. 26. 2. 17 d 16 15 50 S. 36. 3. 24 d 20 36 30 N. 46. 4. 31 d 5 45 20 S. 56. (3) Meridian Passage. -This is tabulated in mean time for each day at upper and lower transit on p. IV. of the month, in Greenwich astronomical time. It is the interval between the transit of the Sun and Moon. As the Earth rotates from W. to E., and the Moon is constantly changing her angular distance eastward from the Sun (about 12 daily), it follows that this interval is less when the Moon crosses the meridian of a place in E. long., and greater for a place in W. long, than the time given in the " Nautical Almanac." The change for 24 hours is found by comparing with the following day for W. long, and the preceding day for E. long. The proportion of this change for the " Longi- tude in Time " is the correction required, and gives the time at place of the transit across the meridian of the place ; by applying " Longitude in Time " to this, we obtain the M.T.G. of the transit across the ship's meridian. NOTE. (a) It is usual to ask for the Mer. pass, on a given Civil Day ; therefore, if the corrected time of Mer. pass, exceeds 12 hours, it must be taken out for the preceding day, but for the same day if under 12 hours. (6) The "Lower" Mer. pass, is used only when the Moon is observed below the pole. ELEMENTS FROM THE ''NAUTICAL ALMANAC: 123 (c) When two asterisks (* *) are seen in the upper Mer. pass, column, it is New Moon, and therefore no observation can be made. Examples. August 25th find the M.T.G. of Moon's Mer. pass, in long. 78 W. Mer. pass. Aug. 25 d 4 h 39 m '8 Cor. for long. +'11'7 M.T. at place 25 d 4 51 "5 Long. + 5 12 M.T.G. 25 d 10 3-5 Increase to following day because of West long. 54 m 78 long, in time 5 '2 4 108 6,0)31,2 270 ( 4)280-8 24 ( 6) 70-2 117 August llth find M.T.G. of Moon's Mer. pass, in long. 156 E. Since Mer. pass, on August llth occurs at 17 h 57 m '0, we use Decrease to preceding day Mer. pass. Aug. 10 d 17 h 13 m '3 because of East long. 43 ra> 4 Cor. for long. - 18 '8 10 '4 5 h 12 156 4 M.T. atplace]0 d 16 h 54'5 Long. 10 24 M.T.G. 10 d 6 30-5 1736 6,0)62,4 434 10 h 24 ( 4)451-36 24 I 6)112-8 18-8 EXERCISES. Find the M.T.G. of the Moon's Mer. pass, under following conditions : 1. August 1st in long. 172 20' W. 2. 28th ' 96 25 W. 3. 2nd 125 25 E. 4. 27th 89 45 E. 5. 4th 175 30 W. 6. 15th 142 48 E. 7. 9th 24 30 E. 8. 7th 55 35 E. (4 and 5.) The Moon's Right Ascension and Declination. The Moon's path in the heavens is inclined to the Sun's at an approximate angle of 5 9' ; and a great circle drawn through the direction of the Moon's motion at any time cuts the ecliptic in two opposite points called the nodes. At every revolution of the moon round the earth the nodes retrograde in the ecliptic, in a manner similar to the precession of the equinoxes ; and are carried back 19 20' in one year, therefore taking 18J years to complete the circle. Her greatest latitude is therefore 5 9' ; but her greatest declination varies from 23 27' + 5 9' to 23 27' - 5 9'. This causes very large changes in her right ascension and declination from day to day, as in addition her angular motion 124 TEXT- BOOK ON NAVIGATION. in her orbit is nearly 13 per day; therefore it has been found necessary to tabulate these data for every hour of the day. This is done in the " Nautical Almanac," pp. V. to XII. of each month, and the variations are given for 10 m . In reducing these elements, it is convenient to find the var. in l m by moving the decimal point one place to the left, and then multiplying by the number of minutes and decimals of a minute in the M.T.G. Example. Correct the Moon's R.A. and Dec. for M.T.G. August 4th 12 h 24"' -8. R.A. for 4 d l h 21 h 20 m 56"87 Cor. for 24 m -8 +58-60 Corrected R.A. 21 21 55-47 Dec. for4 d !2 h 164'41"-9S. Cor. for 24 m> 8 - 3' 8'1 Corrected Dec. 16 1 33 '8 S. Using 2nd differences. R.A. for 4 d 12 h 21 h 20 m 56'-87 Cor. + 58-57 Var. in l m 2 -363 24-8 18904 9452 4726 58-6024 Var. in l m 7"'586 24-8 60688 30344 15172 6,0)18,8-1328 3' 8"-l Corrected R.A. 21 21 55'44 12" Var. in 10 m = 23 '627 13 h = 23-574 Var. in l m 2 '3616 24-8 Change in 60 m - "053 4 Int. 12-4 212 636 188928 94464 47232 58-56768 60)'6572 Ch. -Oil Req. var. = 23*616 Cor. - 3 8'6 13 Cor. Dec. 16 1 33'3 S. ^ar. in 10 m 75'86 7675 Var. in l m 7'604 24-8 60832 30416 15208 6,0)18,8-5792 3' 8-6' Ch. in60 m +'89 4 Int. 12-4 356 1068 6,0)1,1-036 Ch. -18 /. Req. var. = 76'04 ELEMENTS FROM THE "NAUTICAL ALMANAC" 125 EXERCISES. Reduce the Moon's R.A. and Dec. to the following M. Times at Greenwich : 1. August 5 d 18 h 19 m 27'. 2. 8 11 48 45. 3. 15 6 37 24. 4. 22 19 51 48. ART. 45. Planets' Elements. The planetary elements used are (1) Right Ascension. (2) Declination. These are entered for Mean noon at Greenwich on each day ; no variation is given ; and the corrections are found by using the difference between two consecutive days as diff. for 24 hours, and thence by proportion determining the change for hours and decimals in M.T.G. Example. Find the R.A. and Dec. of Jupiter for August 9 d 10 h 15" M.T.G. R.A. August 9 d noon 6 h 8 m 17 8> 24 Dec. August 9 d 23 7' 37" '9 N. 10 d ,,69 0777 10 d 237'31-3 Diff. in 24 hrs. + 50 '53 - 6 '6 Int. 10-25 10-25 25265 330 10106 132 5053 66 24)517-9325 24)67'650 Ch. 21-58 Ch. 2-8 R.A. August 9 d 6 h 8 m 17 8> 24 Dec. August 9 d 23 7' 37"'9 N. Cor. +21-58 Cor. - 2'8 Corrected R.A. 6 8 38 -82 Corrected Dec. 23 7 35-1 N. 5(3) Semi- diameter. (4) Horizontal Parallax. These elements do not require correction, and are entered for mean noon at Greenwich. The S.D. is occasionally used in " Lunar " problems. (5) Meridian Passage. This column in the "Nautical Almanac" is consulted to obtain the time for observing the Mer. Alt. of a planet for the purpose of finding the latitude. It also determines whether a planet is east or west of the meridian at a given time. The correction for longitude, when necessary, is similar to that of the Moon's Mer. Pass. 126 TEXT-BOOK ON NAVIGATION. Example. Find the G.M.T. of Mer. Pass, of Jupiter on August 9th in long. 75 E. Mer. Pass. August 8 d 21 h O m '4 Ch. to preceding day + 3'1 Cor. for long. + '6 long. 5 21 1-0 24)15-5 Long, in time 5 -6 M.T.G. 8 d 16 TO EXERCISES. 1. Correct the R.A. and Dec. of Jupiter for M.T.G. August 10 a 14 h 54 m , and find his Mer. Pass, in long. 120 W. 2. Correct the R.A. and Dec. of Saturn for August ll d 8 h 24 ra M.T.G., and find his Mer. Pass, in long. 150 E. ART. 46. Stars' Elements. The only elements used in Navi- gation are Right Ascension and Declination; these do not require correction. In the " Nautical Almanac " is a table giving the Mean Places of Stars where they are arranged in order of Right Ascension. The magnitudes are also stated by numbers, the smaller number denoting the brighter star. At the bottom of each page the proper name of the star is given ; this may also be found in Table XIV., Norie. After the Mean Places, there is a very much extended table giving the Apparent Places of Stars for every 10th day of the month. Directions. Find the star in the table of " Mean Places," and note the R.A. This R.A. is then used to find the star in the extended table of "Apparent Places." The R.A. and Dec. are then taken out for the nearest date. Examples. In our table in the Appendix we find 7 Pegasi Mag. 2'9 R.A. O h 8 m 23 1 6 Dec. 14 39' 40" N. a Arietis ,, 2'2 ,,21 52'3 ,,23 16 N. Aurigse 2'1 ,, 5 52 38 ,, 44 56 19 N. a Argus ,, -1-0 C 21 51-9 52 38 39 S. Stars like " a Argus " whose mag. is prefixed by a " - " sign are brighter than those of mag. ; and these increase in brightness as the negative number increases. Thus Sirius mag. 1'4 is brighter than a Argus. CHAPTER XI. SIDEREAL AND SOLAR TIME. ART. 47. As a consequence of the Earth's rotation from W. to E., all Celestial bodies have an apparent motion from E. to W., but not all at the same rate. A star completes its apparent revolution in exactly the same time as the Earth takes to turn once on its axis, but the Sun, Moon, and Planets all change their positions with respect to the Stars. The Sun moves to the eastward about 1 daily, and the Moon about 13 daily, whilst a planet may have a motion to the E. or W., and sometimes stands still amongst the Stars. The Apparent Motions of the Stars and the Sun are those used for measuring Time. A Sidereal Day is the interval between two successive transits of a star over the Meridian of a place, and all Sidereal Days are equal. A Solar Day is the interval between two successive transits of the Sun's centre, but Solar Days are not all equal, as they increase or decrease in length according to the time of the year. The Sun cannot therefore be a uniform time-keeper. Yet, as Day and Night depend on the Sun, it is the Natural Controller of time for ordinary purposes, and its position with respect to the Meridian marks Apparent Time, as indicated by a sun-dial. To secure a Uniform Solar Time and Equal Solar Days, a Mean Sun is adopted, which gives Equal Days of 24 Equal Hours all the year round. Its position sometimes to the E., sometimes to the W., and sometimes (four times in the year) coinciding in Right Ascension with the true Sun marks Mean Time. But on account of the progress Eastward (about 1 daily) of this Mean Sun, it is evident that it will take a longer time to come round to the Meridian again than would a fixed Star nearly 4 ra , in fact. Hence the Mean Solar Day is nearly 4 m longer than a Sidereal Day, but as both kinds of days are divided into 24 hours, it follows that Solar hours, minutes, and seconds are longer than the corresponding divisions of Sidereal Time. The exact relative values are given in two Tables in the " Nautical Almanac," on pp. 588 and 590, and in the Appendix. 128 TEXT-BOOK ON NAVIGATION. Any star would do to measure a Sidereal Day, but astrono- mers have made the Sidereal Day begin at the moment when the point where the Ecliptic cuts the Equator corresponding to the Sun's position on March 21st, and named the " First point of Aries " is on the Meridian of a place. Right Ascension is also measured from this point (see Definition). N.B. An hour of any kind, whether of Right Ascension, Sidereal Time, or Solar Time, corresponds to an arc of 15, because 24 x 15 = 360. ART. 48. To find Sidereal Time, When the Hour Angle and the Right Ascension of any Celestial Object are known, the Sidereal Time may be found as follows : (a) Hour Angle West. Add the Hour Angle to the Right Ascension (rejecting 24 if the sum exceeds 24 hours). (b) Hour Angle East Subtract the H.A. from the R.A. (adding 24 hours if necessary). Illustration. PMPi is the Celestial Meridian of a place, and the other projected semicircles from P to Pi are " Hour Circles " (15 apart). A is the " First point of Aries," and the Arc of the Equinoctial M.A. is the Sidereal Time (in this case, 4 hours). But the same Arc measured eastwards from A to M is the R.A. of the Meridian ; therefore Sidereal Time and Eight Ascension of Meridian are always equal. . 98. NOTE. The student must regard the Meridian and Hour Circles as fixed, and the " First Point of Aries," as well as all celestial objects, in motion from E. to W. in the direction of the arrow. Using the initials H.A. for Hour Angle, S.T. for Sidereal Time, R.A. for Right Ascension, and R.A.M. for Right Ascension of Meridian, the above rules may be stated concisely thus (a) H.A. West; S.T. or R.A.M. = R.A. + H.A. (b) H.A. East; S.T. or R.A.M. = R.A. - H.A. Examples. 1. Given the Sun's H.A. 3 h 10 m 30' W. of the Meridian, and Sun's R.A. 14 h 20 m 40 : required the Sidereal Time. Here S.T. = R.A. + H.A. Sun's R.A. 14 h 20 m 40 s Sun's H.A. 3 10 30 FIG. 99. S.T. 17 31 10 SIDEREAL AND SOLAR TIME. 129 2. Given the Moon's H.A. 4 h 15 m 50 s E. of Meri- dian, and Moon's R.A. 10 h 19 ra 35': required the S.T. Here S.T. = R.A. - H.A. Moon's R.A. 10 h 19 m 35* Moon's H.A. 4 15 50 S.T. 6 3 45 3. Given a star's H.A. 6 h 50 ra 25 s E., and its R.A. 2" 15 m 45': required the S.T. Star's R.A. 2 b 15 ra 45 s (add 24 h ) Star's HA. 6 50 25 FIQ. 100. S.T. 19 25 20 (always counted westward) 4. Given a Planet's H.A. 4 h 12 m 20' W., and its R.A. 22 h 44 m 14': required the S.T. Planet's R.A. 22 h 44 m 14* Planet's H.A. 4 12 20 S.T. 2 56 34 NOTE. The above calculation occurs in the "Lunar " and other problems, and should be clearly understood. ART. 49. To find Sidereal Time from Solar Time. Since Apparent Time is the Westerly Hour Angle of the Sun, and Mean Time is the Westerly Hour Angle of the Mean Sun, the formula for finding Sidereal Time from the Solar Time will be S.T. = R.A.M.S. + M.T.S. or = R.A.T.S. + A.T.S. Examples. 1. Given Apparent Time at ship in long. 78 W., August 14th, at 8 h 20 A.M. : required the Sidereal Time. A.T.S. Aug. 13 d 20 h 20 m s long. W. + 5 12 A.T.G. Aug. 14 d 1 32 Sun's R.A. at noon 9 h 32 ra 39' -09 Cor. 14-40 Reduced R.A. 9 32 53'49 A.T.S. 20 20 Sid. Time 5 52 53 -49 78 4 6,0)31,2 5 h 12 w H.V. 9-41 1-53 2823 4705 941 14-3973 i 3 o TEXT-BOOK ON NAVIGATION. 2. Given Mean Time at ship in long. 84 35' W., September 23rd, 4 h 36 m A.M. : required the Sidereal Time. M.T.S. Sept. 22 d 16 b 36 0* long. W. + 5 38 20 M.T.G. Sept. 22 d 22 14 20 84 35' 4 6,0)33,8 20 5 h 38 ra 20* R.A. Mean Sun at noon (or Sid. Time) 12 h l m 44'-97 (22 h 3 36-84 Accel, for \ 14 m 2 '30 [ see Appendix (20 s 05 Red. M.S.R.A. 12 5 24-16 M.T.S. 16 36 R.A.M. or Sid. Time 441 24-16 3. Given M.T.G. September 28 d 20 h 16 m 50', long. 75 20' E. : find S.T. M.T.G. Sept. 28 a 20 h 16 m 50 s 75 20' long. E. + 5 1 20 4 M.T.S. Sept. 29 d 1 18 10 6,0)30,1 20 5 h l m 20* S.T. at noon or R.A.M.S. 12 h 25 24*-28 (20 h 3 17-13) Accel, for < 16 m 2'63> see Appendix ( 50 s -UJ Red. M.S.R.A. 12 28 44-18 M.T.S. 1 18 10 R.A.M. or Sid. Time 13 46 54-18 NOTE. The Mean Sun's R.A. at noon is taken from the Sidereal Time column on p. II. of "Nautical Almanac," because the mean Sun being on the Meridian at Mean Noon, its R.A. at that moment is the same as the R.A.M. and the S.T. at Mean Noon. EXERCISES. Compute the S.T. from the following data : b m s 1. A.T.S. ... August 2nd 315 A.M. 2. M.T.S. ... 4th 710 P.M. 3. M.T.G , 8 d 231442 4. M.T.G. ... Sept. l d 51525 5. M.T.S. ... 14th 830 A.M. 6. A.T.S. ... 24th 540 P.M. long. 162 30' W. 105 45 E. 36 30 E. 124 37 W. 42 25 W. 155 12 E. ART. 50. To find the Hour Angle of a Celestial Object. When SIDEREAL AND SOLAR TIME. 131 the Sidereal Time and the Right Ascension of a Celestial object are known, its Hour Angle can be found by subtracting one from the other, so as to obtain the least difference between them (borrowing 24 hours when neces- sary). The object is West of the Meridian when R.A. is less than S.T., and East when greater. Illustration. A is the first point of Aries. The arc MA is the S.T. of obser- vation, ARi is the R.A. of a star west of the Meridian ; its H.A. (the arc MRi) is evidently = MA - AR b that is, S.T. - star's R.A. Again, AR 2 is the R.A. of the Moon, and its H.A. (the arc MR 2 ) = AR 2 - MA, that is, Moon's R.A. - S.T., and it is East of the Meridian. FIG. 102. Examples. 1. Given the S.T. 7 h 42 m 35', and the R.A. of a star 6 h 40 m 13 s : required its H.A. S.T. 7 h 42*" 35" Star's R.A. 6 40 13 Star's H.A. 1 2 22 W. of Mer. 2. Given S.T. 19 h 10 m 40 s , and R.A. of an object l h 2 m 50 : required its H.A. S.T. 19 h 10 ra 40 s R.A. 1 2 50 H.A. 5 52 10 E. of Mer. NOTE. In this example the Sidereal Time is subtracted from the R.A. (which is therefore the greater), because it gives a difference less than 12 hours, and no object can be farther from the Meridian than 12 hours. ART. 51. All the Celestial bodies are so related to each other in position that when the H.A. of one of them is known, the H.A. of any other can be found by means of their known Right Ascensions ; because it was shown in Art. 46 that S.T. is equal to the sum or difference of the R.A. and H.A. of a given object, and now it has been shown that the H.A. is the difference between the S.T. and the R.A. of any given object. NOTE. The H.A. can also be computed from the Altitude, as will be seen hereafter. 132 TEXT-BOOK ON NAVIGATION. FIG. 103. Examples. 1. Given R.A. of Star Aldebaran 4 h 30 m 7', and its H.A. 2 h 10 m 25' west of Meridian: what is the H.A. of the Star Regulus whose R.A. is 10 h 3 m O 9 ? R.A. Star Aldebaran 4 h 30 m 7' H.A. 2 1025W. S.T. (or R.A.M.) 6 40 32 R.A. Star Regulus 10 3 H.A. 3 22 28 E. of Mer. 2. Given R.A. of Star Achernar l h 33 ra 57', and H.A. 4 h 50 m 47' east of Meridian: required the H.A. of the Moon when its R.A. is 18 h 10 ra 25'. R.A. Star Achernar l h 33 m 57 s H.A. 4 50 47 E. S.T. (or R.A.M.) 20 43 10 R.A. of moon 18 10 25 H.A. 2 32 45 W. of Mer. 3. Given M.T.G. Aug. 16 d 8 h 14 ra 20' in long. 80 E. : find the Meridian Distance or Hour Angle of the Moon. M.T.G. Aug. 16 d 8 b 14 m 20' long. E. 5 20 M.T.S. Aug. 16 d 13 34 20 80 4 Var. in ] 2-19 14-3 6,0)32,0 657 876 219 5 h 20 m S.T. at noon 9 h 35 m 52' -51 Accel, f or { l **'** R.A.M.S. 9 37 13-66 M.T.S. 13 34 20 R.A.M. 23 11 33-66 R.A.]) 7 4 39-14 31-317 R.A. 16 d 8 h = 7 h 4 m 7' '82 Cor. 31-32 Red. R.A. 7 4 39'14 H.A. 3) 16 6 54-52 W. of Mer. or 7 53 5-48 E. of Mer. ART. 52. To find the Solar Time corresponding to given Sid. Time. Examples. 1. Find the M.T.S. in long. 65 W. on Aug. 19th, when the S.T. is 7" 55 m 20 s . SIDEREAL AND SOLAR TIME. 133 S.T. orR.A.M. 7 h 55 m 20' R.A.M.S. noon 9 47 42 Approx. M.T.S. 18 d 22 7 38 long. 4 20 Approx. M.T.G. 19 d 2 27 38 R.A.M.S. noon 9 h 47 m 42"17 f 2 h 1971 Accel, for { 27 4'44 ( 38" -10 Eed. R.A. 9 48 6'42 R.A.M. 7 55 20 M.T.S. 18 d 22 7 13-58 or 19th 10 h 7 m 13' -58A.M. 2. Find the M.T.S. in long. 82 E. on Sept. 20th, when the S.T. is 19" 14 m 35 1 . S.T. or R.A.M. 19 h 14 ra 35 R.A.M.S. Noon 11 53 52 Approx. M.T.S. 20 d 7 20 43 long. 5 28 Approx. M.T.G. 20 d 1 52 43 R.A.M.S. noon ll h 53 m 51"86 ( l h 9-86 Accel, for 52 ra 8 -54 ( 43' -12 Red R.A. 11 54 10-38 R.A.M. 19 14 35 M.T.S. 20 d 7 20 24-62 or 20th 7 h 20 m 24' '62 P.M. i EXERCISES. 1. Find the hour angle of Altair at 8 h 20 m P.M. M.T.S. in long. 126 20' E. on August 12th. 2. Find the hour angle of Capella at 7 h 15 ra A.M. M.T.S. in long. 84 35' W. on September 20th. 3. Find the hour angle of the Moon on August 22nd at 5 h 2G m P.M. M.T.S. in long. 34 W. 4. Find the hour angle of Jupiter on August 8 d 17 h 15 m M.T.G. in lono\ 56 25' E. 5. Find the M.T.S. in long. 74 E. on August 18th, when the S.T. is 6 h 14 m 24 s . 6. Find the M.T.S. in long. 95 W. on September 14th, when the S.T. is 18 h 7 m 36 s . ART. 53. To compute the Time when a given Star is on the Observer's Meridian. When a star is on the meridian, the right 134 TEXT-BOOK ON NAVIGATION. ascension of the star is the same as the right ascension of the meridian, or Sidereal Time, hence the problem is the same as finding the M.T.S. or A.T.S. corresponding to a given Sid. Time. Examples. 1. Find the M.T.S. when the star Procyon is on the observer's meridian in long. 35 W. on August 6th. R.A. of Procyon or R.A.M. = 7 h 34 m 23' 35 R.A.M.S. 8 56 27 4 Approx. M.T.S. 5 d 22 37 56 Long. W. 2 20 6,0)14,0 Approx. M.T.G. 6 d R.A.M.S. noon Accel, for 58 2 h 20 m 57 56" 56 27' 9 Red. R.A. 8 56 36 R.A.* 7 34 23 M.T.S. 5 d 22 37 47 or 6th 10 h 37 m 47 8 A.M. Care must be taken in dating the approx. M.T.S. If the hours, as in the example, exceed 12, the date is the day before the date in the question ; if the hours are less than 12, the date is the same as the question. The passages are required for the civil day, and hence the 5 d 22 h (astro, time) corresponds to the 6th 10 h A.M. (civil time). In Fig. 104, NEQW represents plane of equinoctial. NPQ is meridian, and the object X is always found between P and Q, as this represents our upper meridian. Distance to measure QX is the declination north or south. Find position of T 1st point of Aries by measur- ing to the westward the R.A. of star, which since the star is on the meridian represents the R.A.M. and Sidereal Time ; then from T measure to the eastward the R.A. of Mean Sun, this gives M. Draw hour circles through T and M. T\VQ is RA. of star and of meridian. TQM is R.A.M.S. QWEM is westerly hour angle of M is M.T.S. = 24" + TWQ - TQM. 2. Find the M.T.S. when Regulus is on the meridian in 94 E. on August 4th. R.A. Regulus or R.A.M. = 10 h 3 m 22 s 94 R.A.M.S. 8 48 34 4 Approx. M.T.S. 4 d 1 14 48 Long. E. - 6 16 6,0)37,6 6 h 16" M.T.G. 3 d 18 58 48 SIDEREAL AND SOLAR TIME. .135 R.A.M.S. noon 8 h 44 m 37 s Accel, for /18 h \59 m 2 57 10 Red. R.A. 8 47 44 R.A.* 10 3 22 M.T.S. 4 d 1 15 38 or 4th l h 15 m 38 s P.M. In Fig. 105, projected on plane of equi- noctial, measure QWT = 10 h , then TWM 8 h 48 m for M. QM is M.T.S. when X is on meridian = TWQ - YWM. FIG. 105. 3. Find the A.T.S. when Arcturus is on the meridian in 100 E. on September 2nd. Also by inspection. R.A. Arcturus or R.A.M. = 14" ll ra 22 1 App. R.A. page I. or P. A.T.S. 10 42 44 Approx. A.T.S. 2 d 3 28 38 Long. E. - 6 40 A.T.G. l d 20 48 38 App. R.A. noon 10 h 42 m 44" Cor. 29 Red. R.A. 10 42 15 R A .A.* 14 11 22 A.TS. 2 d 3 29 7 or 2nd 3 h 29 m 7" P.M. From Table XLIV., Norie, Mer. pass. Arcturus Sept. l d 6 H.V. 9-06 3-2 1812 2718 28-992 3 h 26" 3 9 Diff. in 5 days 17 1 day 3 Hence by inspection 2nd 3 23 or 2 nd 3 h 23 m P.M. In Fig. 106 NEQW plane of equinoctical. NPQ meridian. X object. P pole. A true sun. 1st point of Aries. Draw hour circles \A/ as in Fig. 106. TWQ = R. A.* and R. A.M. TWM = R.A.T.S. QM = Wly. H.A. of A'. = A T S = TWQ - TWM. 136 TEXT-BOOK ON NAVIGATION. EXERCISES. Find the ship time when the given star is on the observer's meridian from the following data : No. Name of star. Civil date. Longitude. Mean or App. Time. 1 2 o Andromeda* ft Ceti ... August 6th 8th 65 O'E. 48 W. APP. Mean 3 o Arietis ... 10th 137 E. Mean 4 5 Achernar Aldebaran 12th 14th 172 W. 37 50 E. APP- Mean 6 Capella . . . 16th 14 W. App. 7 Rigel ... 18th 25 E. Mean 8 Canopus ... 20th 100 30 E. App. 9 Betelgueuse 22nd 150 25 W. Mean 10 Regulus ... 24th 29 45 E. App. 11 Procyon ... 31st 85 W. Mean 12 Spica Sept. 3rd 39 E. App. 13 Antares ... 10th 26 45 W. Mean 14 Vega ... 17th 13 45 E. App. 15 Altair ... 24th 175 45 W. Mean ART. 54. Bright Stars and Planets. Stars which may be considered as "bright," and suitable for Navigation purposes, are those of the first and second magnitudes. In the Table of " Mean Places of Stars " in the " Nautical Almanac," there are eighteen of the former and about twice as many of the latter, which gives the Navigator a choice of fifty-four objects suitable for observations, in addition to the Sun, Moon, and Planets. It is not a work of much labour for the Student, by the help of a Star Atlas or a "Planisphere," to make himself familiar with most of the bright Stars and the Constellations to which they belong. The Planets Venus, Mars, Jupiter, and Saturn are sometimes very conspicuous objects, and well adapted for observations, but they vary very much in brightness, according to their distance from the Earth. They may easily be recognized by their wan- derings amongst the Stars, or by the times of their Meridian passage as given in the " Nautical Almanac." With such an ample choice of objects, there is no excuse for the Navigator relying upon observations of the Sun only. ART. 55. To find what Bright Stars are near the Meridian, or within a given Hour Angle at a Stated Time. Directions. Find the S.T. corresponding to the given time, then look in the "Table of Mean Places" in the "Nautical Almanac " for the names of bright Stars whose Right Ascensions do not differ from the S.T. by more than the given Hour Angle. SIDEREAL AND SOLAR TIME. 137 N.B. The star is E. of Mer. when its R.A. is greater than the S.T. (or R.A.M.), and W. when less. It is N. of the observer when its Declination reads N. of the latitude, and S. when it reads S. Remark, As the object in view is to find what stars would be suitable for finding the latitude by Meridian and Ex-Meridian Altitudes, or for finding the Time and Azimuth, in practice it will be sufficient to take the Sun's R.A. at the nearest noon at Greenwich, since the error would not amount to 2 minutes in any case; but for examination purposes the R.A.M. must be exactly determined. Examples. 1. What bright stars will cross the meridian of a place long. 48 30' E., between the hours of 6 and 9 P.M. M.T.G. on September 30th. G.M.T. Sept. 30 d 6 b O m s Sept. 30 d 9 h O m 0' Long. E. 3 14 3 14 M.T.S. 30 d 9 14 R.A.M.S. 12 33 17 Accel, for 6 h 59 30 d Accel, for 9 h 12 14 12 3317 129 R.A.M. 24 4846 RA.M. 21 48 16 Hence all stars whose RA.'s fall between 21 h 48 16' and O h 48 ra 46 s will cross the meridian during these times. These stars are o Gruis, Gruis, Fomalhaut, Alpheratz, and Ceti. 2. Sept. 20 a 6 h 19 m M.T.G. in lat. 24 S., long. 125 36' E. ; find what stars not less bright than 2 '2 magnitude are within 2 hours E. and W. of the meridian, and above the horizon ; state their meridian distances, and whether N. or S. of observer at transit. M.T.G. Sept. 20 d 6 h 19 m 0' 125 36 Long. E. 8 22 24 4 M.T.S. 20 d 14 41 24 R.A.M.S. noon 11 53 52 6 h 59 Accel, for 19 m 3 R.A.M. 2 36 18 6,0)50,2 24 8 h 22 m 24' Hence limits of R.A. are from O h 36 m 1 8 s to 4 h 36 m 18 s . Name of star. Magnitude. Meridian distance. Bearing at transit. Ceti 2-2 l h 57 m 26 1 W. N. Achernar . t 0-5 1 2 5 W. S. 7' Androinec a) 2-2 38 11 W. N. o Arietia . f 2-2 34 26 W. N. o Persei . t 1-9 41 18 E. N. Aldebaran . 1-1 1 51 14 E. N. 138 TEXT-BOOK ON NAVIGATION. In Fig. 107, NESW is Ratl. horizon. NZS is Meridian. EQW is Equinoctial. Ppole; Z zenith. 1, 2, 3, 4, 5, 6, the stars at the intersection of their respective hour circles and parallels of declination. ZP1, ZP2, etc., are the hour angles or meridian distances of the stars. EXERCISES. From following data find names of stars which are visible, and state their hour angles and bearings at transit. No. Time. Limit. Mag. Longitude. Latitude. 1 August 1st ll h 50 m P.M. M.T.S. I' 1 E. and W. 2-6 48 15' AV. 33 N. 2 3rd 3" A.M. A.T.S. ... \\ E. and W. 2-0 56 45 E. 20 8. 3 28th 3 h 40- M.T.G. ... 1 E. and W. 1-0 72 30 E. 24 8. 4 Sept. 9 d 17" 21 m 20 s A.T.G. 2 E. and W. 2-6 87 50 W. 5 N. 5 19th 6 h A.M. M.T.S. ... 1 W. of Mer. 2-0 96 45 E. 40 N. 6 29th 6 h 9 ra A.M. A.T.S. 1 E. of Mer. 1-5 162 15 W. 24 8. 7. On May 27th at a place in long 1 . 101 W., time shown by a sidereal clock was 13 h 27 m 38% error of clock on local sidereal time being 2 h 17 m 47'. Find local mean time at the instant having given sidereal time at mean noon 4 h 20 m 15'. 8. Apparent R.A. Sun on page ii for mean noon is 20 h 57 m 47 '8 s , and sidereal time on same page is 20 h 44 m 33 '2". Find equation of time and show by diagram how it is to be applied to app. time. 9. Sun's app. R.A. 011 page i for app. noon was 14 h 35 m 28*91', and on page ii for mean noon was 14 h 35 m 31 '6". Var. in R.A. for 1 hr. is 9 '883'. Find equation of time and show how it is to be applied to mean time. CHAPTER XII. CORRECTION OF ALTITUDES. ART. 56. The altitudes of all objects, whether observed at sea or on shore in an artificial horizon, must be corrected to reduce them to the true geocentric altitudes. The corrections employed are for index error, or correction; height of eye, or dip ; semi-diameter of the object ; refraction ; and parallax. The method of applying them may be understood from the following remarks. (a) Index Error. This correction is due to an imperfection in the sextant used, by which the zero mark on the arc is placed out of position. As already explained, when the " " on the vernier is in line with " " on the arc, the index and horizon glasses should be parallel ; if they (the zeros) are not in line an error is introduced ; when the "0 " on the vernier is to the left of the "0" on the arc (the glasses being parallel), it shows that the " " on the arc is to the right of its true position, hence all angles will be in excess of their true value ; the reading is " on " the arc and the error " subtractive" The " " on the arc being to the left of its true position makes all angles smaller than their true value; the reading is "off" the arc and the error "additive" The ind. err. is always applied according to the direction in the question. (6) Height of Eye. This correction allows for the depression or dip of the visible horizon below the sensible horizon ; observed angles, being measured above the visible horizon, are too large, hence the correction is always subtractive. Its amount is found in Norie, Table V., or Raper, Table 30. Should a table not be handy, an approximate correction is found by taking the square root of the height of the eye in feet ; this gives the " dip " in minutes of arc. (c) Semi-diameter. The altitude of an object's centre is the one required in the calculations, and in bodies which have a visible disc, like the sun and moon, it would be a matter of impossibility for an observer to say exactly where the centre is ; 140 TEXT-BOOK ON NAVIGATION. an observation of its limb or boundary is therefore taken. The limb nearer the horizon is called the lower limb, " L.L." \ an observation of this limb is less in altitude than that of the centre ; the S.D. is therefore added. The limb further from the horizon is called the upper limb, " U.L." ; an observation of this limb gives too great an altitude, and the S.D. is subtracted. The amount for the Sun is entered for noon of each day on page II., " Nautical Almanac." The amount for the Moon is entered for noon and midnight of each day on page III., " Nautical Almanac." The limb observed is always stated in the questions. When the corrections a, b, c are applied, the amount obtained is the Apparent altitude of the centre. (d) Refraction. Rays of light, in passing from a rarer to a denser medium, have their directions diverted towards the normal or perpendicular to the surface bounding the media. This is called the refraction or bending of the ray. Objects appear in the direction that the last portion of the rays enters the eye, and therefore, as the air is denser at the surface of the earth than higher up, the effect is to make all objects appear nearer the zenith than their true positions, hence the correction for refraction is always subtractive. Its amount is found in Norie, Table iv., or Raper, Table 31. (e) Parallax. Observations are of necessity made on the surface of the earth ; these have to be reduced to the centre or rational horizon, which is below but parallel to the visible horizon; as all objects would appear higher in altitude at the centre than at the surface, the correction for parallax is always additive. Its amount depends on the latitude of the observer, and the altitude and proximity of the object, and for the Sun is entered in Norie, Table VI., or Raper, Table 34. In the case of the Moon the amount is entered in the " Nautical Almanac," page III., and must be corrected as before directed. Refraction and parallax decrease as the altitude increases, and vanish when the object is in the zenith. Norie, Table XVIIL, gives the combined effect in the case of the Sun, and since Refraction is greater than Parallax, the amount found there is always subtractive. Norie, Table xxx., and Raper, Table 39, give the amount of the Refraction and Parallax in altitude combined in the case of the Moon, and is additive, as the Parallax is more than the Refraction. In practice the Refraction is taken out for the apparent altitude of the limb observed, that is, after the index error and dip have been applied. CORRECTION OF ALTITUDES. 141 When the Refraction and Parallax are applied to the apparent altitude, the true altitude is obtained. Examples. 1. September 23rd, theobs. alt. of the sun's L.L. was 46 25' 40" ; height of eye, 24 feet ; ind. cor. + 2' 25". Find the true alt. Obs. alt. 46 25' 40" Ind. cor. + 2 25 46 28 5 Dip -4 48 46 23 17 S.D. +15 58 App. alt. 46 39 15 lief. - 54 46 38 21 Par. + 6 True alt. 46 38 27 2. August 5th, the obs. alt. of the sun's U.L. was 14 38' 20" ; height of eye, 32 feet ; ind. cor. - 3' 37". Find the true alt. Obs. alt. 14 38' 20" Ind. cor. - 3 37 14 34 43 Dip - 5 32 14 29 11 S.D. - 15 48 App. alt. 14 13 23 Bef . - 3 38 14 9 45 Par. + 9 True alt. 14 9 54 The stars appearing merely as points of light have no S.D., and, as they are so far distant, no appreciable parallax. 3. August 6th, the obs. alt. of the star o Trianguli Australis was 27 26' 10" Find the true alt., ind. cor. - 5' 27", eye 41 feet. Obs. alt. 27 26' 10" Ind. cor. 5 27 27 20 43 Dip - 6 16 App. alt. 27 14 27 Ref . - 1 50 True alt. 27 12 37 142 TEXT-BOOK ON NAVIGATION. 4. August 19th, in lat. 46 25' N., the obs. alt. of the Moon's L.L. was 42 19' 30" ; ind. cor. + 3' 11" ; height of eye, 27 feet ; M.T.G. 19 d 4 h 23 m 50'. Find the true alt. The Par. in alt. is obtained from formula Par. in alt. = H.P. x cos app. alt. Or P.L. Par. in alt. = P.L. Hor. Par. + L. sec. app. alt. S.D. Noon 15' 28" -8 Cor. + 1-8 4" -8 4-4 H.P. 56'42"-9 Cor. +6-5 17' -7 4-4 15 30-6 Tab. D. + 10-4 192 192 56 49-4 Tab. E. - 57 708 708 15 41 Obs. alt. 1 42 19' 30" Ind. cor. + 3 11 12)21-12 56 43-7 = 3403"-7 12)77-88 1-8 6-5 42 22 41 Dip. - 5 5 42 17 36 S.D. + 15 41 3403-7 log = 3-531952 Or H P. P.L. = 5014 App. alt. 42 33 17 cos = 9-867250 sec = 1327 Ref. -12 2507 log = 3-399202 41' 48" P.L. = 6341 42 32 15 Par. +41 47 6,0)250,7 True alt. 43 14 2 41' 47" Tab. xxx. 40' 5" Cor. for 43" par. 32 3' alt. 8 Par. in alt. - ref. = cor. 40 45 5. August 9th, the obs. alt. of Jupiter's U.L. was 34 8' 50"; ind. cor. + 4' 23" ; height of eye, 25 feet. Obs. alt. 34 8' 50" Ind. cor. + 4 23 34 13 13 Dip - 4 54 34 8 19 S.D. - 16 App. alt. 34 8 3 H.P. = 1"'5 ; turn up Tab. Ref. 1 24 XLVIII., which gives par. in alt. 1". 34 6 39 Par. + 1 True alt. 34 6 40 CORRECTION OF ALTITUDES. 143 ART. 57. The Artificial Horizon is a plane horizontal mirror, or reflecting surface, used for taking altitudes when the sea horizon is obscured, or not visible. The liquid used is mercury, but treacle, oil, or tar will do as well, and they all depend on the same law in optics, viz. " The angles of incidence and reflection are equal." AB is a plane horizontal mirror, S a celestial object ; its image, I, is seen by an eye placed at E as far perpendicularly below the surface as S is above it. Join El, meeting AB in M, and join SM. The object S appears in the direction of the dotted line through E, but because of the immense distance S is off, this line and MS may be FEI is measured with the FIG. 108. considered parallel. The angle sextant, while SMA is the altitude of S. Now, observed angle FEI = TT - EMS = SMA + EMB = SMA + AMI = twice altitude of S. Hence, when an altitude is taken with an artificial horizon the index correction is first applied, and then the result halved ; moreover, as the horizon is a portion of, or parallel to, the sensible horizon, no correction for dip is necessary. Example. August 12th, the obs. alt. of the sun's L.L. in artificial horizon was 103 56' 20" ; ind. cor. - 4' 18" ; height of eye, 56 ft. Find the true alt. Obs. alt. 103 56' 20" Ind. cor. - 4 18 2)103 52 2 51 56 1 S.D. + 15 49 52 11 50 Ref. - 44 52 11 6 Par. + 5 True alt. 52 11 11 144 TEXT-BOOK ON NAVIGATION. EXEECISES. 1. August 16th, the obs. alt. of the was 50 3' 10" ; ind. cor. - 4' 19" eye, 20 ft. Find true alt. 2. August 8th, the obs. alt. of Q was 21 13' 20" ; ind. cor. + 6' 17" ; eye, 15 ft. Find true alt. 3. September 1st, the obs. alt. of G) was 19 52' 10" ; ind. cor. 3' 25" ; eye, 30 ft. Find true alt. 4. September 4th, the obs. alt. of was 32 24' 30" ; ind. cor. + 37" ; eye, 17 ft. Find true alt. 5. August 6th, the obs. alt. of Q was 47 19' 50" ; ind. cor. + 4' 29" ; eye, 24 ft. Find true alt. 6. September 27th, the obs. alt. of was 12 17' 20" ; ind. cor. - 1' 57" ; eye, 50 ft. Find true alt. 7. August 13th, at 13 d 6 h 9 m 30' M.T.G. the obs. alt. of Moon's L.L. was 55 36' 20", in lat. 24 N. ; ind. cor. 4- 2' 41" ; eye, 10 ft. Find true alt. 8. August 15th, 6 h 14 m A.M. M.T.S., in lat. 43 S., long. 45 Jl., the obs. alt. of Moon's U.L. was 15 13' 10" ; ind. cor. - 3' 17" ; eye, 32 ft. Find true alt. 9. August 17th, the obs. alt. of star Rigel was 29 17' 20" ; eye, 50 ft. ; ind. cor. + 4' 48". Find true alt. 10. August 28th, the obs. alt. of star Altair was 47 29' 30" ; eye, 35 ft. ; ind. cor. - 3' 26". Find true alt. 11. August 12th, the obs. alt. of Planet Saturn in lat. 51 N. was 23 14' ; ind. cor. + 2' 27" ; eye, 36 ft. Find true alt. 12. August 8th, the obs. alt. of Planet Jupiter in lat. 42 S. was 53 20' 10" ; ind. cor. - 37" ; eye, 25 ft. Find true alt. 13. August 27th, the obs. alt. of Sun's L.L. in artificial horizon was 78 14' 20" ; ind. cor. - 3' 36". Find true alt. 14. August 21st, the obs. alt. of Polaris in artificial horizon was 95 37' 40" ; ind. cor. + 2' 56" ; eye, 27 ft. Find true alt. 15. August 20th, when the M.T.S. was 4 h 19 32' P.M., in lat. 37 N., long. 147 30' W., the obs. alt. of Moon's L.L. in artificial horizon was 102 21' 30"; ind. cor. ; eye, 100 ft. Find true alt. CHAPTER XIII. LATITUDE BY MERIDIAN ALTITUDE. ART. 58. Latitude is the angle at the earth's centre sub- tended by the arc of the meridian intercepted between the equator and the place. The arc of the declination circle inter- cepted between the observer's zenith and the equinoctial subtends the same angle ; the latitude is therefore the declination of the observer's zenith, and may be determined easiest by meridian or ex-meridian observations. The Altitude of the Pole is equal to the Latitude of the Observer. Let ZSPiN represent the celestial meridian of the observer 0. Z is the zenith, QQi the equinoctial. E is the centre of earth. Angle OEL or latitude is subtended by arc LO or arc QZ. P is the elevated pole, SN the horizon. PEN is altitude of pole. FIG. 109. Now, PEN or altitude of pole 90 - PEZ ZEQ latitude. ART. 59. Proof of Latitude by Meridian Altitude. Let NESW re- present the plane of the horizon pro- duced to meet the celestial concave. NPZS the celestial meridian of the observer. P the elevated pole, Z his zenith, EQW the equinoctial. Let Xi, X 2 , X 3 , and X 4 be the positions of four celestial objects. Then QX 1? QX 2 , QX 3 are the decimations of the first three, and PX 4 the polar distance of the fourth. Object 1 has south declina- tion, the others north. (1) Lat. = QZ = ZXi - QXj = Her. zen. dist. dec. I 4 6 TEXT-BOOK ON NAVIGATION. (2) Lat. = QZ = ZX 2 + QX 2 = Mer. zen. dist. + dec. (3) Lat. = QZ = QX 3 - ZX 3 = Dec. Mer. zen. dist. From these the rules for finding lat. by mer. alt. may be deduced, viz. when dec. and zen. dist. are of same name the sum is the lat.; when dec. and zen. dist. are of contrary name the difference is the lat., which is of the name of the greater. The zen. dist. is always of contrary name to the bearing. (4) Lat. = NP = PX 4 + NX 4 = pol. dist. + alt. Hence when an object is observed on the meridian between the pole and the horizon, i.e. below the pole, the lat. is the sum of pol. dist. and alt. Therefore, to find lat. by mer. alt. the process is as follows : Object above the Pole. (1) Find Greenwich date, correct declination. (2) Correct altitude, and subtract from 90 for zen. dist., which mark of contrary name to the bearing of the object. (3) Dec. and zen. dist. of same name, add for lat. ; dec. and zen. dist. of contrary name, subtract for lat., which is of same name as the greater. Object below the Pole. (1) Find Greenwich date; in the case of the sun, the observation must be taken at midnight ; in the case of the moon, use the loiuer mer. pas? Correct declination, and subtract from 90 for pol. dist. (2) Correct altitude. (3) Add pol. dist. to alt. for lat., which must be of same name as dec. Examples. 1. August 17th, in long. 55 45' W. the obs. mer. alt. of the Sun's L.L. was 35 10' 30" bearing north ; ind. cor. - 4' 50" ; height of eye, 18 ft. Required the latitude. A.T.S. August 17 d O h O m 0' Dec. 13 38' 32"-5 N. H.V. 47"'59 Long. W. 3 43 Cor. 2 56 3'7 A.T.G. 17 a 3 43 13 35 36-5 N. 33313 14277 6,0)17,6-083 2' 56" LATITUDE BY MERIDIAN ALTITUDE. Obs. alt. 35 10' 30" Ind. cor. 4 50 35 5 40 Dip. - 4 9 35 1 31 S.D. + 15 50 35 17 21 Ref . - 1 20 35 16 1 Par. + 7 True alt. 35 16 8 N. 90 M.Z.D. 54 43 52 S. Dec. 13 35 37 N. Lat. 41 8 15 S. Construction. Draw a circle NESW radius f$ inch to represent rational horizon, centre point is the zenith Z. The vertical diameter is the ship's meridian on which the sun is situated (Fig. 111). Since the sun bears north, measure 35 to scale from the north point for position X of sun. His declination is 14 N., from X measure to scale 14 to the south (contrary name to declination) for the point Q where the equinoctial cuts the meridian. E and W are in the line through Z at right angles to the meridian. Through EQW draw an arc of circle for the equinoctial, the centre is some- where on the meridian produced, and also on the perpendicular bisecting the line joining EQ, and so is at their intersection. Measure 90 from Q for the position of the elevated pole P. The centres for circles of declination passing through objects whose declination is of the same name as the latitude lie between the centre for the equinoctial and P, the greater the declination, the nearer P is the centre ; when declination is of contrary name to latitude, the centre is beyond this point. Explanation. NESW represents Rational Horizon. NZS Meridian. EQW Equinoctial. Z, zenith ; P, pole ; X, object. QX is Dec. ; NX is Mer. Alt. ; ZX is zen. dist. Then Lat. = ZQ = ZX - QX. 2. August 26th, in long. 87 W., the obs. mer. alt. of the Sun's U.L. bearing south was 56 30' 30" ; ind. cor., + 3' 20" ; eye, 16 ft. Required the lat. 148 TEXT-BOOK ON NAVIGATION. 87 4 ))34,8 5 h 48 m A.T.S. August Long. A.T.G. 26 d 26 d O h 5 Qm 48 0' Dec. Cor. 10 39' 5 5" 1 N. N. H.V. 51" -92 5-8 5 48 10 34 4 41536 25960 6,0)30,1-136 Obs. alt. Q 56 30' 30" Ind. cor. + 3 20 56 33 50 Dip. - 3 55 56 29 55 S.D. - 15 51 56 14 4 Ref. - 38 56 13 26 Par. + 5 True alt. 56 13 31 S. 90 M.Z.D. 33 46 29 N. Dec. 10 34 4N. Lat. 44 20 33 N. Construction (Fig. 112). Draw NESW and NZS, as before. For X measure 56 from S, since sun bears south, and from X measure XQ to the south 10. Draw EQW ; put in P as explained. Explanation. NESW represents Rational Horizon. NZS Meridian. EQW , , Equinoctial. Z, zenith ; P, pole ; X, object. QX is Dec. ; SX is Mer. Alt. ; ZX is zen. dist. Then Lat. = ZQ = ZX + QX 3. August 1st, in long. 49 E., the obs. mer. alt. of the sun's U.L. zenith south of sun was 76 54' 40" ; eye, 25 ft. ; ind. cor. - 5' 10". Required the lat. A.T.S. August l d O h O m O' Long. 3 16 A.T.G. July 31 d 20 44 Dec. 18 11' 45" N. Cor. 2 3 18 13 48 N. H.V. 37"'26 3-3 11178 11178 6,0)12,2-958 2' 3" LATITUDE BY MERIDIAN ALTITUDE. 149 Obs. alt. Q 76 54' 40" Ind. cor. - 5 10 76 49 30 Dip. - 4 54 70 44 36 S.D. -15 47 76 28 49 Ref. - 14 76 28 35 Par. + 2 True alt. 76 28 37 90 M.Z.D. 13 31 23 S. Dec. 18 13 48 N. Lat. 4 42 25 N. Construction (Fig. 113). Draw NESW and NZS as before. Since zenith is south of sun, measure 77 from N to X, as sun must bear north. Complete figure. Explanation. As given to Fig. 111. Then Lat. = ZQ = QX - ZX. EXERCISES. 1. August 3rd, in long. 51 45' W., the obs. mer. alt. of the Sun's L.L. bearing north was 34 15' 30" ; ind. cor. - 3' 31" ; eye, 16 ft. Find the lat. 2. August 5th, in long. 33 30' W., the obs. mer. alt. of the Sun's L.L., zenith north of sun was 50 1' 20" ; ind. cor. + 4' 14" ; eye, 18 ft. Find the lat. 3. August 7th, in long. 60 E., the obs. mer. alt. of the Sun's L.L. bear- ing north was 19 32' 10" ; ind. cor. + 2' 37" ; eye, 20 ft. Find the lat. 4. August 9th, in long. 97 30' E., the obs. mer. alt. of the Sun's U.L. zenith south of sun was 69 47' 40" ; ind. cor. + 1' 10" j eye, 17 ft. Find the lat. 5. August llth, in long. 20 15' W., the obs. mer. alt. of the Sun's U.L. bearing north was 64 4' 60" ; ind. cor. - 1' 43" ; eye, 19 ft. Find the lat. 6. August 13th, in long. 70 30' E., the obs. mer. alt. of the Sun's U.L. zenith south of sun was 50 6' 10" ; ind. cor. 1' 56" ; eye, 21 ft. Find the lat. 7. August 15th, in long. 10 E., the obs. mer. alt. of the Sun's L.L. bearing north was 78 36' 30" ; ind. cor. - 2' 59" ; eye, 25 ft. Find the lat. Under the following conditions find the latitude by mer. alt. of the Sun: 150 TEXT-BOOK ON NAVIGATION. Date. Longitude. Mer. alt. Limb. Bearing. Ind. cor. lit. of eye. 8 Aug. 17th 12 0' W. 80 12' 0" u. N. -3' 42" 27 feet. 9 19th 180 W. 52 35 10 L. S. -6 25 29 10 Sept. 30th 57 19 E. 89 4 20 U S. + 5 38 30 11 23rd 48 45 W. 48 57 50 L. S. + 3 20 32 12 23rd 97 30 W. 43 22 20 L. N. -4 17 34 13 27th 13 25 E. 58 44 40 U. S. -2 34 22 14 29th 9 14 W. 36 36 30 U. N. -2 41 45 15 24th 165 E. 67 14 10 L. S. -3 58 36 16 23rd 100 40 E. 52 19 30 U. S. + 2 15 34 17 24th 24 50 W. 22 28 20 L. S. -1 52 33 18 24th 38 20 E. 30 31 U. N. +3 29 49 ART. 60. Latitude by Meridian Altitude of a Star. Example. August 5th, the obs. mer. alt. of the star Canopus (a Argus) bearing south was 54 17' 30" ; ind. cor., - 6' 10"; ye, 24 ft Required the lat. Obs. alt. 54 17' 30" Ind cor. 6 10 S FIG. 114. Given Mer. alt. = SX Dec. = QX Req. lat. = ZQ = QX - ZX = QX - (90 - SX) 54 11 20 Dip. - 4 48 Ref. 54 6 32 -41 True alt. 54 5 51 S. 90 M.Z.D. 35 54 9 N. Dec. 52 38 39 S. Lat. 16 44 30 S. EXERCISES. 1. The observed Mer. Alt. of Capella was 54 29' 15" north of the observer; index error, -3' 10"; height of eye, 20 ft. Required the latitude. 2. The observed Mer. Alt. of Rigel was 84 54' 20" north of the observer ; index error, + 1' 10" ; height of eye, 21 ft. Required the latitude. LATITUDE BY MERIDIAN ALTITUDE. 151 3. The observed Mer. Alt. of Sirius, zenith north of star, was 45 30' 30" ; index error, + 1' 30" ; height of eye, 23 ft. Required the latitude. 4. The observed Mer. Alt. of Spica bearing south of observer was 36 25' 20" ; index error, 3' 10" ; height of eye, 24 ft. Required the latitude. 5. The observed Mer. Alt. of Fomalhaut bearing south of the observer was 10 19' 30" ; height of eye, 25 ft. Required the latitude. 6. The Mer. Alt. of Aldebaran bearing south was 56 23' ; index correc- tion, + 1' 19" ; height of eye, 28 ft. Required the latitude. 7. The Mer. Alt. of Spica in an Artificial Horizon was 86 47' 20" ; index correction, 2' 18" ; zenith north of the star. Find the latitude. 8. The Mer. Alt. of Markab bearing north was 30 59' 10" ; index correction, + 4' 2" ; height of eye, 33 ft. Find the latitude. 9. The Mer. Alt. of a Arietis bearing south was 75 18' 40" ; index correction, 4' 9" ; height of eye, 19 ft. Required the latitude. 10. The Mer. Alt. of Algenib in an Artificial Horizon was 111 57' IP" ; index correction, + 8' 6" ; zenith south of the star. Find the latitude. 11. The Mer. Alt. of Regulus in an Artificial Horizon was 114 20' 30" ; index correction, 5' 16" ; zenith north of the star. Find the latitude. ART. 61. Latitude by Meridian Altitude of the Moon. It is first necessary to find the M.T.G. when the Moon is on the observer's meridian, and then reduce the " Elements " as before directed. Example. August 16th, in long. 126 5' E., the obs. mer. alt. of the Moon's L.L. bearing north was 42 26' 20" ; ind. cor. - 3' 30" ; height of eye, 16 ft. Find the latitude. The time of mer. pass, on the 16th is 21 h 55 '3, which is really about 10 o'clock on the morning of the 17th, hence we use the passage for the 15th, and since we are in E. long, work from that day towards the 14th, as the transit occurs earlier. Mer. pass. 15 d 21 h 4 m '9 49 m '2 (for 24 h ) Cor. -17-2 8-4 M.T.S. 15 d 20 47-7 1968 6,0)50,4 20 Long. 8 24-3 3936 8 h 24 m -3 M.T.G. 15 d 12 23-4 /4)413'28 16)103-32 17-2 S.D. H.P. Midngt. 14' 57"'80 3'58 54' 49"'29 131 Alt. 42 Cor. 12 -4 -44 -4 M.Z.D.48S. Dec. 20 N. 14 57-92 12)1-432 54 49*73 12)5-24 Aug. D. +9-8 Red. E. - 2'4 Approx. lat. 28 S. .jL2 '44 15 7'72 54 47-33 TEXT-BOOK ON NAVIGATION. Obs. alt. })_ 42 26' 20" Ind. cor. - 3 30 Dec. 20 11' 20" -2 N. Cor. + 26-8 42 22 50 Dip. - 3 55 42 18 55 S.D. +15 8 App. alt. 42 34 3 Cos. Ref. - 1 2 20 11 47 N. ... 9*867167 3287 log = 3-516800 Par. 42 33 1 + 40 21 True alt. 43 13 22 N. 90 M.Z.D. 46 46 38 S. Dec. 20 11 47 N. Lat. 26 34 51 S. Tar. 1-15 23-4 460 345 230 26-910 Sec ... 1328 H.P. P.L. 5166 40' 21" P.L. 6494 Given Mer. alt. = NX Deo. = QX Req. lat. = ZQ = ZX - QX = 90 - NX - QX EXERCISES. 1. August 21st, in long. 48 30' W., the obs. mer. alt. of the Moon's U.L. bearing north was 34 17' 20" ; hid. cor. + 2' 15" ; eye, 16 ft. Find the lat. 2. August 22nd, in long. 97 25' W., the obs. mer. alt. of the Moon's L.L. bearing south was 53 26' 10" ; ind. cor. - 3' 20" ; eye, 17 ft. Find the lat. 3. August 30th, in long. 62 15' E., the obs. mer. alt. of the Moon's U.L. zenith south of Moon was 61 34' 40 ' ; ind. cor. + 5' 46" ; eye, 18 ft. Find the lat. 4. August 31st, in long. 57 45' E., the obs. mer. alt. of the Moon's L.L. zenith north of Moon was 24 46' 50" j ind. cor. - 4' 27" ; eye, 20 ft. Find the lat. 5. August 31st, in long. 120 W., the obs. mer. alt. of the Moon's L.L. zenith north of Moon was 68 57' 30" ; ind. cor. + 3' 33" ; eye, 23 ft. Find the lat. Under the following conditions find the latitude by mer. alt. of the Moon : Date. Longitude. Mer. alt. Limb. Bearing. Ind. cor. lit. of eye. 6 Aug. 4th 178 30' W. 38 22' 20" L. N, -1'12" 34 feet. 7 10th 21 30 E. 42 39 20 L. S, -2 19 36 8 llth 94 15 E. 46 17 10 L. S. +3 5 43 9 12th 127 30 E. 54 29 40 U. N. -2 8 46 10 14th 139 45 W. 75 16 40 L. N. -4 57 49 11 17th 158 24 E. 73 56 10 L. N. +2 35 54 12 ., 25th 88 35 E. 84 24 30 L. S. + 1 24 56 LATITUDE BY MERIDIAN ALTITUDE. 153 A RT. 62. Latitude by Meridian Altitude of a Planet. Example. August 9th, in long. 90 W., the obs. mer. alt. of the Planet Jupiter bearing north was 42 10' 20" ; ind. cor. + 4' 15" ; eye, 27 ft. Find lat. As before explained, the G.M.T. of the Planet's mer. pass, across the observer's meridian must be first found, in a similar way to the Moon's. Mer. pass. August 8 d 21 h O m 4 Bet. - '8 M.T.S. 8 d 20 59-6 Long. + 6 Ret. in 24 hrs. 3 m l 3-1 x 6 M.T.G. 9 d 2 59-6 Dec. 23 7' 37"'9 N. Cor. - -8 23 7 37 Obs. alt. 42 10' 20" N. Ind. cor. + 4 15 42 14 35 Dip. - 5 5 App. alt. 42 9 30 Ref. -13 Par. 42 8 27 + 1 True alt. 42 90 8 28 M.Z.D. 47 51 32 S. Dec. 23 7 37 N. = '8 Diff. for 24 hours 6"-6 3 Lat. 24 43 55 S. FIG. llfi. Given Mer. alt. = NX Dec. = QX Rcq. lat. = ZQ = ZX - QX = 90 - NX - QX ART. 63. Latitude by Meridian Altitude of an Object below the Pole. An object whose Polar Distance is less than the elevation of the Pole above the horizon will never set, and will therefore be seen on the meridian below the Pole as well as above the Pole : when it has been proved that the lat. is found by adding together the Mer. Alt. and Pol. Dist. Example. August 10th, the obs. mer. alt. below the Pole in artificial horizon of the star a Ursse Majoris was 33 1' 10" ; ind. cor. + 3' 26". Find the latitude. 154 TEXT-BOOK ON NAVIGATION. Obs. alt. Ind. cor. 33 1' 10" + 3 26 2)33 4 36 16 32 18 Ref. - 3 10 True alt. 16 29 8 P.D. 27 44 29 Lat. 44 13 37 N. Dec. 62 15' 31" N. 90 Given _ Mer.alt. = NX t Pol.dist. = PX Beq. lat. = PN = NX + PX EXERCISES. 1. August llth, in long. 48 30' W., the obs. mer. alt. of the Planet Saturn bearing north was 55 16' 40" ; ind. cor. - 3' 27" ; eye, 30 ft. Find the lat. 2. September 19th, the obs. mer. alt. below the pole in artificial horizon of the star a! Crucis was 42 8' 50" ; ind. cor. - 4' 36". Find the lat. 3. August 14th, the obs. mer. alt. below the pole in artificial horizon of the star a Cygni was 18 11' 20" ; ind. cor. + 5' 42". Find the lat. 4. August 29th, in long. 135 45' E., the obs. mer. alt. of the Moon's L.L. under the south pole was 13 27' 50", taken in artificial horizon ; ind. cor. + 3' 24". Find the lat. 5. August 1st, in long. 94 25' W., the obs. mer. alt. of the Sun's U.L. under the north pole in artificial horizon was 12 36' 40" ; ind. cor. 2' 50". Find the lat. 0. September 9th, the obs. mer. alt. below the pole of )8 Centauri was 20 47' 10" ; ind. cor. + 45" ; height of eye, 37 ft. Find the lat. ART. 64. To find the Approximate Meridian Altitude of a Star, for setting the Sextant preparatory to observing the Meridian Altitude. Referring to the Rule for finding the Latitude by a Meridian Altitude (Art. 59), it was shown that the Latitude is equal to the sum of the Meridian Zenith Distance and Declination when they are of the same name, and the difference when of different names. Therefore it follows that the Meridian Zenith Distance is the difference of the Lat. and Dec. when they are of the same name, and the sum when of different names. The Meridian Zenith distance subtracted from 90 gives the Meridian Altitude. Examples. 1. Given lat. 25 30' Mer. Alt. of the Star. K, and a Star's Dec. 52 20' N., required the LATITUDE BY MERIDIAN ALTITUDE. Lat. 25 30' N. \ c Dec. 52 20 N. / me name ~ subtract. Mer. Zen. Dist. 26 50 S. 90 Mer. Alt. 63 10 North of observer. Draw NE3W and NZS as before. Lat. being north measure ZQ 25 to south for Q, draw EQW, and set off QX 52 to north for X (Fig. 118). Lat, = ZQ Dec. = QX, zen. dist. = ZX.1 Then Mer. Alt. = NX = 90 - ZX = 90 - (QX - ZQ). 2. Suppose the latitude to be 24 25' S., and the Star's Dec. 32 10' N., required its Mer. Alt. Lat. 24 25' S. \ Different names- Dec. 32 10 N. / add. 155 Mer. Zen. Dist 56 90 35 S. Mer. Alt. 33 25 North of observer. In Fig. 119, lat. being south, measure ZQ = 24 to north for Q ; make QX = 32 to north for X. Lat. = ZQ Dec. = QX, zen. dist. =ZX. Then Mer. Alt. = NX = 90 - ZX = 90 - (QX + ZQ). N 3. Givtm latitude of Ship 10 15' N., what is the Mer. Alt. of a Star whose Dec. is 20 40' S., and how does it bear from observer ? ^t. 10 15' N. I Differenfc names _ add . Dec. 20 40 S. J Mer. Zen. Dist. 30 55 N. 90 Mer. Alt. 59 5 N. South of observer. Here Fig. 120. Mer. Alt. = SX = 90 - ZX = 90 - (ZQ + QX). i 5 6 TEXT-BOOK ON NAVIGATION. N.B. A Star is north or south of observer according as its Decimation reads north or south of the latitude of observer. If the corrections for Ref., Dip., etc., are applied to the true alt. to make an obs. alt., they must all be reckoned contrary to the manner indicated in the examples on the correction of altitudes. EXERCISES. Find the Meridian Altitudes of Stars from the following data, and state whether N. or S. of observer : No. Latitude. Name of star. Declination. 1 o / 14 40 N. o Draconis O 1 64 52 N. 2 28 12 S. o Bootis 19 42 N. 3 44 10 N. a Scorpii 26 12 S. 4 14 25 S. Gruis 47 25 S. 5 52 19 S. a Piscis Aust. 30 9S. 6 Achernar 57 45 S. 7 Geminorum 28 16 N. 8 35 28 N. Sirius 16 35 S. 9 49 50 N. Polaris 88 46 N. 10 3 568. o Ursao Majoris 61 3N. CHAPTER XIV. LATITUDE BY EX- MERIDIAN ALTITUDE. ART. 65. Although the meridian altitude is the simplest method by which latitude may be determined, clouds or other causes may prevent an observation at the particular instant required; under these circumstances, the latitude is deduced from observations of the altitude taken when the object is near the meridian, that is, when its hour angle or meridian distance is small. There are two methods in use : (1) By calculating a correction which, when applied to the altitude out of the meridian, determines the object's altitude or zenith distance when on the meridian. This quantity is called a "Reduction," and the latitude by account is always used to determine its amount. The correctness of the result, therefore, depends on the accuracy with which the D.R. lat. is known. (2) By direct spherics. There are several variations of the first method, and the one we shall employ may be called the "log- rising" formula, which, according to Chauvenet, approximates more rapidly than the others. In figure, let X be object near the meridian ; then in triangle PZX PX is Polar Distance, PZ colatitude, ZX zenith distance at time of observation, P the hour angle. From the fundamental equation. Cos ZX = cos PZ . cos PX + sin PZ . sin PX . cos P. Let "a" denote the altitude, "I" the latitude, "h" the hour angle, and " d " the declination. Then above becomes Sin a = sin I . sin d + cos I . cos d . cos h = sin I . sin d + cos I . cos d - cos I . cos d (2 . sin 2 -\ = cos (l^d) cos I . cos d ( 2 . sin 2 | /. cos (If^d) = sin a + cos I . cos d (2 . sin 2 -J 158 TEXT-BOOK ON NAVIGATION. The quantity log ^2 . sin 2 ~^\ is called the " log rising," and is tabulated in Norie, Table xxix. In practice the natural number corresponding to the second term on the right-hand side is found by adding together the L. cos. lat., L. cos dec., and log-rising hour angle ; this number is added to the nat. sin of altitude ; the nat. cos of sum is the zen. dist. when on the meridian; the lat, is then found as in meridian altitude. ART. 66. Latitude by Ex-meridian Altitude. Direct Method. Given the hour angle (h), the declina- tion (8), and the true altitude (a), to find the latitude. In the figure X is the object, PX the polar distance, ZX the zenith distance, the angle ZPX(/i) the hour angle, and XA an arc of a great circle at right angles to the meridian, EQ W the equator. Then QZ is the latitude, and is equal to the sum or difference of the arcs ZA and QA. FIG. 122 In the right-angled spherical triangle PXA cos h = cot p . tan PA = tan 8 . tan PA /. tan PA = cos h . cot But QA (arc 1) is the complement of PA, because PQ = 90 /. cot arc 1 = cos h . cot 3 ..... Again, in the same triangle cos p = cos PA . cos XA i.e. sin 8 = sin arc 1 . cos XA .*. cos XA = sin 8 . cosec arc 1 ..... (1) (2) Also in the right-angled triangle XZA cos XZ = cos XA . cos ZA i.e. sin a = sin 8 . cosec arc 1 by substituting the value of cos XA in (2) r? A / o\ :. cos ZA (arc 2) = cos ZA sin o . cosec arc 1 or cos arc 2 = sin a . cosec S . sin arc 1 (3) Arc 1 is named the same as the dec., and arc 2 the same as the zenith distance, and the sum or difference, according as they LATITUDE BY EX-MERIDIAN ALTITUDE. 159 are of the same or different names, gives the latitude as in the meridian altitude problem. For computation the formulae may be arranged as follows : ... cos cot cosec arc 1 ... cot . . . sine a a ... sine arc 2 cos lat. --- N.B. When X is on the equinoctial (i.e. dec. = 0), then the perpendicular XA coincides with the equinoctial EQW and arc 1 disappears. The lat. QZ is then found from the right-angled triangle ZXQ, in which QX measures the H.A., and ZX = co-alt, or zenith distance. Then cos ZX = cos QX x cos QZ .. - cos QX = cos ZX x ses QX i.e. cos lat. = sin alt. x sec H.A. ART. 67. Latitude by Ex-meridian Altitude of the Sun, Star, etc. The hour angle of the Sun is the time from apparent noon. The hour angle of any other celestial object is the difference between the B.A.M. and R.A. of object. To allow for Run since determining error, it must be noticed that when the course has been to the eastward the ship is going to a place where the time is later ; when to the westward, where time is earlier. Hence the diff. long, in time must be added for easterly, and subtracted for westerly run. The process is as follows : (1) Find Greenwich time and hour angle. (2) Correct decimation and altitude. (S) a. Log-rising Method. Add L. cos lat., L. cos dec., and log-rising hour angle (Norie, Table XX ix.), increasing the index there found by " 1 " ; reject 10's from sum and find natural number in Table xxiv. Add this nat. no. to nat. sin true alt. in Table xxvi., and find in this same table nat. cos of sum ; this is the zen. dist., which mark of contrary name to bearing. The sum or difference of zen. dist. and dec. according as they are of same or different names gives the latitude. b. Direct Method. Add L. cos hour angle, and L. cot dec., take back L. cot of result, which call arc 1, and name it the same as the dec. 160 TEXT-BOOK ON NAVIGATION. Then add L. cosec dec., L. sin arc 1, and L. sin alt., reject 10's, and take back L. cos of result for arc 2, which name contrary to the bearing. The sum of arcs 1 and 2 when they are of the same name, or difference when they are of contrary names, gives the lat,, which is of same name as the greater. ART. 68. Latitude by Ex-meridian Altitude of Sun. Example. August 7th A.M. at ship, in lat. by account 33 30' N., long. 10 57' W., the obs. alt. of the Sim's U.L. was 72 52' 50" ; ind. cor. + 3' 4" ; eye, 16 ft. ; time by watch, 10 h 57 ra 58 1 , which was slow of A.T.S. 44 ra 10 s ; the d. long, made to the W. since the error was determined being 17| miles. Find the latitude by ex-meridian ; and supposing the course and distance to be S. by W. 4 miles since observation, find the latitude at noon. Dec. H V. Time by watch, Aug. 6 d 22 h 57 m 58' long. 10 57' 16 37' 8"-9 N. 41-50 Slow + 44 10 4 Cor. - 1C '6 6 d 23 42 8 43 m 48 s 16 36 52'3 Run. W. -1 11 Obs. alt. 72 52' 50" A.T.S. 6 d 23 40 57 Ind. cor. +3 4 Long. W. + 43 48 72 55 54 A.T.G. 7 d 24 45 Dip. - 3 55 First Method. 72 51 59 E.H.A. 19 m 3 1 log ris. = 3'538380 S-D. - 15 48 Lat. 33 30' 0" cos = 9'921107 Dec. 16 36 52 cos = 9'981479 72 36 11 R. & P. - 14 2760 log = 3-440966 T.A. 72 35' 57' ; nat. sin 954236 True alt. 72 35 57 956996 N. cos. = 16 51' 51" N. Dec. = 16 36 52 N. Obs. lat. = 33 28 43 N. Second Method. E.H.A. 19 m 3' cos = 9-998498 Dec. 16 36' 52" cot = 10'525219 cosec = 10*543740 Arc 1 16 40 8 N. cot = 10*523717 sin = 9-457640 T.A. 72 35' 57" sin = 9'979656 Arc 2 16 48 33 N cos = 9'981036 Obs. lat. 33 28 41 K (Run S. by W. 4' gives d. lat. 3'-9 Run 3 54 S. to be applied to the S , as the obs. is taken before noon.) Noon lat. 33 24 47 N. LATITUDE BY EX-MERIDIAN ALTITUDE. 161 Third Method. Directly from figure. X is position of object, AX its altitude, PX the pol. dist, P the hour angle, XD a perpendicular from X on the meridian. In right-angled spherical triangle PDX, D = 90, P the hour angle, PX the pol. dist. Find PD and DX. sin (co. P) = tan PD . tan (co. PX) /. tan PD = cos P . tan PX . . . I. Again sin DX = cos (co. P) . cos (co. PX) = sin P . sin PX . II. P = 19- 3 cos = 9-998498 PX = 73 23' 8" tan = 10-525219 sin = 8-919209 sin = 9 981479 PD = 73 19 52 tan = 10'523717 DX = 4 33' 48" sin = 8*900688 Then in right-angled triangle ZDX, given D = 90, ZX zen. dist., and DX, find ZD Sin (co. ZX) = cos ZD . cos DX /. cos ZD = cos ZX . sec DX DX = 4 33' 48" sec = 10-001379 ZX = 17 24 3 cos = 9-979656 ZD = 16 48 34 cos = 9'981035 Lat. = ZQ = 90 - PZ = 90 - (PD - DZ) = 90 - (73 19' 52 - 16 48' 34") = 90 - 56 31 18 = 33 28' 42" EXERCISES. 1. August 1st P.M. at ship, in lat. by account 37 20' S., long. 133 E., the obs. alt. of the Sun's L.L., bearing north, was 33 55' 20" ; ind. cor. + 2' 5" ; eye, 30 ft. ; time by watch, 3 h 33 m 1', which was found 8 h 50 m slow of A.T.S. the same morning; run since error was determined S. 38 E., 23 miles. Find the latitude by ex-meridian, and if the run since noon was 3 miles on the same course, find lat. at noon. 2. August 6th A.M. at ship, in lat. by account 26 20' N., long. 80 45' W., the obs. alt. of the Sun's U.L., bearing south, was 78 9' 20" ; ind. cor. - 3' 16" ; eye, 19 ft. ; time by watch, 4 h 50 ra 25', which had been found fast 5 h 22 m 18 of A.T.S. the same morning; since error was determined the diff. long, made westward was 24 '5 miles. Find the lat. by ex-meridian. 3. August 9th P.M. at ship, in lat. by account 5 35' S., long. 14 W., the obs. alt. of the Sun's L.L. near the meridian, bearing north, was 67 10' 10" ; ind. cor. - 2' 18" ; eye, 35 ft. ; time by watch, l h 23 m 7% which was 58 m fast of A.T.S. ; diff. of long, made to the eastward since determining the error was 37 miles. Find the latitude. 4. August 13th A.M. at ship, in lat. by account 53 15' N., long, 168 47' E., the obs. alt. of the Sun's L.L. near the meridian, bearing south, was 51 26' 50" ; ind. ccr. + 3' 5" ; eye, 26 ft. ; time shown by chron., 12 h 18 m 44 , which was 13 m 20 s slow of M.T.G. Find the latitude. 5. August 17th P.M. at ship, in lat. by account 44 50' S., long. 180 W., 1 62 TEXT-BOOK ON NAVIGATION. the obs. alt. of the Sun's U.L. near the meridian, bearing north, was 31 57' 50"; ind. cor. + 48"; eye, 23 ft. ; time shown by watch, l h 2 m 5', which was 50 ra 54' fast of M.T.G. Find the latitude. 6. August 22nd A.M. at ship, in lat. by account 2 35' S., long. 87 40' E., the obs. alt. of the Sun's L.L. near the meridian, bearing north, was 72 1' 20" ; ind. cor. - 3' 11"; eye, 31 ft. ; time shown by chron., 5 h 18 m 37% which was 12 ra 40 s slow of G.M.T. Find the latitude at obs., and if the run to noon was S. 46 E., 8 miles, find the lat. at noon. 7. August 26th P.M. at ship, in lat. by account 39 20' N., long, meridian of Greenwich, the obs. alt. of the Sun's L.L. near the meridian, bearing south, was 60 45' 20" ; ind. cor. + 4' 18" ; eye, 17 ft. ; time shown by chron. which had been found 14 ra 15' slow of G.M.T. was 12 h 8 m 14"; run since error was determined, S. 52 W. , 29 miles. Find lat. at sights, and if the run was 4 miles on the same course since noon, find lat. at noon. 8. September 1st A.M. at ship, in lat. by account 43 22' S., long. 93 W., the obs. alt. of the Sun's U.L. near the meridian was 38 16' 50" ; ind. cor. + 2' 13" ; eye, 40 ft. ; time by watch, 5 h 52 m 37', which was 6 h ll m 40 fast of A.T.S. ; diff. long, made to the westward since determining the error, 25 miles. Find the latitude, and if the run to noon was S. 65 W. , 5 miles, find lat. at noon. 9. September 29th P.M. at ship, in lat. by account 46 25' N., long. 40 30' W., the obs. alt. of the Sun's L.L. near the meridian was 40 45' 20" ; ind. cor. ; eye, 25 ft. ; time shown by chron. 3 h 30 m 40', which was fast 28 36 s of G.M.T.; run since determining the error, S. 42 E., 18 miles. Find the latitude, and if the run since noon was 7 miles on the same course, find the lat. at noon. 10. September 24th A.M. at ship, in lat. by account 15 38 N., long. 93 26' E., when a chron. estimated 10 m 15' fast of G.M.T. showed 5 h 25 m 11% the obs. alt. of the Sun's L.L. was 73 5' 30"; ind. cor. - 2' 5" ; eye, 18 ft. Find the latitude by ex-meridian. 11. September 23rd P.M. at ship, in lat. by D.R. 51 15' N., long. 124 30' W., when a watch showed 12 h 5 m 14', and had been found 6 m 12' slow of A.T.S. , diff. long, made to the westward since determining the error was 19 5 miles ; the obs. alt. of the Sun's U.L. was 39 5' 40" ; ind. cor. + 2' 9" ; eye, 22 ft. Find the latitude by ex-meridian. Run since noon S. by W. 4 miles. Find the lat. at noon. 12. September 23rd P.M. at ship, in D.R. lat. 40 18' S.. long. 165 54' W., when a chron. showed ll h 18 m 33 s , error on A.T.G. was 4 m 15 slow ; the obs. alt. of the Sun's L.L. near the meridian was 49 17' 10" ; ind. cor. 17" ; height of eye, 41 ft. Find the latitude at sights, and if the run since noon was N. 35 E., 6 miles, find lat. at noon. ART. 69. Latitude by Ex-meridian Altitude of a Star. Example. August 16th at ship, lat. 31 30' S., long. 175 30' 30" W. ; time by chron., 16 d O h O m 10', which was fast of M.T.G. 10' ; obs. alt. of a Cygni bearing N. and W. of the observer, 11 37' 50" ; ind. cor. + 2' 30" ; height of eye, 26 ft. Find the latitude. Time by chron. 16 d O h O m 10 175 30' 30" R.A.M.S. 9 h 35 ra 52' 5 Fast 10 4 M.T.S. 12 17 58 M.T.G. 16 d 6,0)702 2 R.A M. 21 53 50*5 Long. W. 11 42 2 R.A.* 20 38 13'6 ll h 42 m 2' M.T.S. 15 d 12 17 58 W.H.A.* 1 15 36'9 LATITUDE BY EX-MERIDIAN ALTITUDE. 163 First Method. W.H.A. l h 15 ra 37 log ris. = 4-731890 Lat. 31 30' S. cos = 9'930766 Dec. 44 56 39 N. cos = 9-849908 32551 log = 4-512564 T.A. 11 30' 48" N.S. 199596 Obs. alt. 11 37' 50" I.C. +2 30 11 40 20 Dip. - 5 Ref. 11 35 20 -4 32 232147 nat.cosZD7634'35"S. Dec. 44 56 39 N. True alt. 11 30 48 Lat. 31 37 56 S. Second Method. W.H.A. l h 15 m 37 cos = 9-975919 Dec. 44 56' 39" N. cot = 10*000846 cosec = 10150939 Arc 1 46 31 55" N. cot = 9-976765 T.A. 11 30' 48 sin= 9-860792 sin = 9-300150 Arc 2 78 10 IS cos- 9-311881 Lat. 31 38 6 S. Third Method. From figure. X is the star's position ; XD a per- pendicular on the meridian. In triangle PDX, D = 90 ; given Polar Dist. = PX ; Hour angle = P. Find PD and DX. Sin (co. P) = tan (co. PX) . tan PD /. tan PD = C , S JL = cos P . tan PX I. cot PX Again sin DX = cos (co. P) . cos (co. PX) = sin P . sin PX . II. P = l h 15 m 37* PX = 134 56' 39" 43 28' 5" 180 cos= 9-975919 tan = 10-000846 sin = 9-510526 sin = 9-849908 tan = 9-976765 13 Ib W sin = 9'360434 PD = 136 31 55 DX = 13 15' 24" Now, in triangle ZDX, D = 90; given zen. dist. = ZX and DX, find ZD. Sin (co. ZX) = cos ZD . cos DX /. cos ZD = r = cos ZX . sec DX cosDX DX = 13 15' 24" sec = 10-011730 ZX = 78 29 12 cos = 9-300153 ZD = 78 10' 1" cos = 9-311883 Lat. = 90 - PZ = 90 - (PD - ZD) = 90 - 58 21' 54" = 31 38' 6" S. 1 64 TEXT-BOOK ON NAVIGATION. EXERCISES. 1. August 1st, about 10 h P.M. at ship, in lat. by account 38 52' S., long. 10 W., when a chron. showed 10 h 45 m , and was fast l m 48 s of G.M.T., the obs. alt. of Vega near the meridian, bearing north, was 12 30' 40" ; ind. cor. + 2' 54" ; eye, 20 ft. Find lat. by ex-meridian. 2. August 4th, about ll h 15 P.M. at ship, in lat. by account 51 27' K, long. 49 30' W., when a chron. correct for A.T.G. showed 2 h 36 ra 53', the obs. alt. of Altair near the meridian, bearing south, was 46 49' 20" ; ind. cor. + 2' 14" ; eye, 45 ft. Find lat. by ex-meridian. 3. August 8th, about 10" 50 m P.M. at ship, in lat. by D.R. 2' N., long. 169 15' E., when a chron. correct for M.T.G. showed ll h 35 ra 47', the obs. alt. of a Pavonis near the meridian, bearing S. E., was 32 56' 30" ; ind. cor. 37" ; eye, 39 ft. Required the lat. 4. August 12th, at l h 5 m 32 A.M. apparent time at ship, in lat. by D.R. 42 30' N., long. 174 18' W., the obs. alt. of ;Fomalhaut near the meridian was 17 23' 40' ; ind. cor. + 1' 11" ; eye, 42 ft. Find the lat. 5. August 17th, about 2 h 45 A.M. at ship, in lat. by account 17 15' N., long. 57 12' W., when a chron. correct for M.T.G. showed 6 h 32 m 38 8 , the obs. alt. of a Andromedse near the meridian, bearing north, was 77 59' 10" ; ind. cor. 2' 7" ; eye, 50 ft. Required the lat. 6. August 26th, about 4 h 10 ra A.M. at ship, in lat. by D.R. 8 45' S., long 69 14' E., when a watch which was 4 h 38 ra 10* slow of A.T.S. showed ll h 29 ra 41' ; the diff. long, made to the westward since determining the error was 18'5 miles ; the obs. alt. of Achernar near the meridian, bearing south, was 40 11' 30" ; ind. cor. + 3' 17" ; height of eye, 38 ft. Find the lat. 7. September 4th, about 2 h 50 ra A.M. at ship, in lat. by account 31 50' S., long. 100 18' E., when a chron., whose error on M.T.G. was 10 m 57 s slow, showed 7 h 58 ra 14', the obs. alt. of a Arietis near the meridian was 34 58' 40"; ind. cor. - 1' 6"; height of eye, 41 ft. Required the lat. by ex- meridian. 8. September 9th, about 5 h 10 m A.M. at ship, in lat. by D.R. 53 22' N., long. 158 35' E. ; when a watch, which had been found 10* 32 ra 10" slow of A.T.S. , showed 6 h 36 m 15 , the diff. long, made good to the eastward since determining the error was 32'5 miles ; the obs. alt. of Aldebaran near the meridian was 52 51' 50" ; ind. cor. + 2' 11" ; eye, 33 ft. Find the lat. 9. September 13th, about 5 h 15 m A.M. at ship, in lat. by account 23 40' S., long. 50 19' W., when a chron., whose error on M.T.G. was 13 m 15* fast, showed 8 h 49 m 50% the obs. alt. of Capella near the meridian was 20 22' 50" ; ind. cor. - 2' 9" ; eye, 46 feet. Find the lat. by ex-meridian. 10. September 19th, about 5 h A.M. at ship, in lat. by D.R. 55 35' S., long. 60 15' W., when a chron. showed 9 U 10 m , error 7 m 37* fast of A.T G., the obs. alt. of Rigel near the meridian was 42 34' 30" ; ind. cor. + 2' 11" ; eye, 48 ft. Find the lat. 11. September 25th, about 8 h 30 ra P.M. at ship, in lat. by D.R. 3 N., long. 68 14' E., the obs. alt. of a Cygni (Deneb) near the meridian was 48 16' 10" ; ind. cor. - 2' 3" ; height of eye, 50 ft. ; time by chron., 3" 57 ra 19 1 correct M.T.G. Find the lat. 12. October 1st, about 9 h P.M. at ship, in lat. by account 20 10' N., long. 137 46' E., the obs. alt. of a Gruis, near the meridian, was 22 11' 30" ; eye, 55 ft. ; no ind. cor. ; time by chron., 12 h , which was 8 45* fast of A.T.G. Find the lat. by ex-meridian. ART. 70. Latitude by the Pole Star. On account of the proximity of the Pole Star to the North Pole, its azimuth is never large, and it is, therefore, a very suitable object for determining latitude by ex-meridian method at any time when LATITUDE BY EX-MERIDIAN ALTITUDE. 165 I visible. Special formula lias, however, been devised for this purpose, and tables calculated from this formula are entered in the " Nautical Almanac." Thus, if I denote the latitude, a altitude, polar distance, hour angle, it may be shown that I = a p . cos h + J sin 1" . (p sin lif . tan a p . cos h is found in Table I., calculated for different R.A.M.'s from O h to 24 h ; that is for different values of the hour angle and a constant value of the pol. dist. The third term is found in Table II. and is always added ; this is the second correction. Table ill. contains the third correction, always added, increased by 1' in order to make it so ; this 1' is a constant to be subtracted from the true alt. ; this correction in in. is an amount calculated for different dates to correct the mean values of R.A. and Dec. of Polaris used in constructing Tables I. and II. In figure, let X be position of Pole Star ; draw XD perpendicular on meridian; along NZ cut off NC = AX its altitude. PX = pol. dist., ZPX = hour angle. Latitude = NP = NO - CP = NO - PD + CD If the hour angle had exceeded 6 hours latitude = NC + PD + CD PD is first correction, and considering the triangle PDX as a plane one PD = PX . cos P or 1st cor. = p . cos h and is added or subtracted according as the hour angle is greater or less than 6 hours. 2nd cor. = CD In right-angled triangle ZDX cos ZX = cos ZD . cos DX or cos ZC = cos ZD . cos DX .-. cos (ZD + CD) = cos ZD . cos DX cos ZD . cos CD - sin ZD . sin CD == cos ZD . cos DX since CD is very small, cos CD = 1 FIG. 125. 1 66 TEXT-BOOK ON NAVIGATION. Hence cos ZD - sin ZD . sin CD = cos ZD . cos DX sin ZD . sin CD = cos ZD (1 - cos DX) or sin CD = cot ZD . (% sin 2 55} As CD and DX are very small, we may write CD" . sin 1" for DX" DX sin CD and - . sin 1" for sin - ; also ZD is very nearly 2t 2t = ZX. Hence CD" sin 1" = cot ZX . . (sin I") 2 i.e. CD = J sin 1" . (DX) 2 . cot ZX = 4 sin 1" . (PX . sin P) 2 . cot ZX. Hence approximately I = a p . cos h + 4 s i n 1" (p g i n ^) 2 tan a ART. 71. The process is as follows : (1) Find Greenwich time, and sidereal time of observation, which is Right Ascension of Meridian. (2) Correct altitude and subtract 1'. (3) Apply first correction from Table i., as there directed. (4) Add second and third corrections from respective tables for latitude. Example. August 24th, at 12 h 12 m 20" mean time at ship, the obs. alt. of Polaris out of the meridian was 58 35' 10", in longitude 28 15' W. ; ind. cor. + 2' 13" ; height of eye, 27 ft. Find the latitude. R.A.M.S. 10 h 7 m 24 -9 Accel. f 18 : M.T.S. August 24 d 12 h 12 m 20' Long. + 1 53 28 15' 4 M.T.G. August 24 d 14 5 20 6,0)11,3 Obs. alt. 58 35' 10" Ind. cor. + 2 13 58 37 2b Dip. 5 5 l h 53" 10 9 43-8 M.T.S. 12 12 20 S.T. of obs. or R.A.M. 22 22 3-8 R.A. Polaris 1 26 12'9 E.H.A. 3 4 9-1 58 32 18 Ref . - 34 T.A. 58 31 44 Const. - 1 Cor. for 22 h 20 m = 49' 48" 22 30 -0 52 1 Diff. for Red. A. 58 30 44 1st cor. - 50 15 57 40 29 2nd cor. + 37 3rd cor. + 1 24 10 = 2 13 1=00 13-3 2=00 2G-6 .'. 1st cor. = 50 14-6 Lat. 57 42 30 LATITUDE BY EX-MERIDIAN ALTITUDE. Given Alt. = AX = NO by construction Hourangle = ZPX Pol. dist. = PX 1st cor. = PD 2nd cor. = CD Req. lat. = NP = NO - PD + CD. From formula Dec. = 88 48' 9" N. Pol. dist. = 1 11 51 p = 4311" h = 3 h 4 m 9' a = 58 31' 44". p = 4311" log = 3-634578 h = 3" 4 ra 9 g cos = 9-841476 1st cor. 2993" log = 3 '47 6054 log= 3-634578 sin= 9'857208 a = 58 31' 44" i sin 1" 2nd cor. 38"'13 True alt. 58 31' 44" 1st cor. -49 53 3-491786 2 7i) 2 = 6-983572 tan = 10 '213172 - 4-384545 log = 1-581289 2nd cor. 57 41 51 + 38 Lat. 57 42 29 EXERCISES. 1. August 28th, about 5 h 50" A.M., in long. 48 30' W. ; when a chron. correct for M.T.G. showed 9 h 2 m 14 s ; the obs. alt. of Polaris out of mer. was 36 47' 20" ; ind. cor. - 2' 17" ; eye, 26 ft. Find the latitude. 2. August 5th, about l h 50 m A.M., in long. 128 50' E. ; when a chron., which was 10 m 18 s fast of A.T.G., showed 5 h 15 m 38 ; the obs. alt. of Polaris out of mer. was 46 12' 30" ; ind. cor. + 3' 46" ; eye, 32 ft. Find the latitude. 3. August 1st, about 4 h 20 m P.M., in long. 180 E. ; when a chron. correct for M.T.G. showed 4 h 19 m 50" ; the obs. alt. of the Pole star out of mer. was 29 5' 10" ; ind. cor. - 4' 15" ; eye, 37 ft. Find the latitude. 4. August 2nd, at 8 h 10 m 20 s P.M., apparent time at ship in long. 68 34' W. ; the obs. alt. of a Urs. Min. out of mer. was 53 36' 40" ; ind. cor. + 55" ; eye, 29 ft. Required the latitude. 5. September 30 d 8 h M.T.G. ; when in long. 162 36' 15" E. ; the obs. alt. of Polaris out of mer. was 23 14' 50" ; ind. cor. - 1' 8" ; eye, 42 ft. Find the latitude. 6. September 10th, about 8 h 15 m P.M., in long, of Greenwich; when a chron. correct for A.T.G. showed 8 h 15 m 38', the obs. alt. of the Pole star out of mer. was 32 26' 40" ; ind. cor + 3' 25" ; eye, 34 ft. Find the latitude. 7. September 1st A.M., at ship in long. 95 28' E. ; when a chron. correct for M.T.G. showed 7 h 3 m 18" ; the obs. alt. of Polaris in artificial horizon out of mer. was 101 50' 20" ; ind. cor. + 3' 26". Find the latitude. 8. September 30th, about 7 h 20 ra A.M., in long. 179 45' W. ; when a chron. 12 h slow of A.T.S. showed 7 h 19 m 32' ; the obs. alt. of a Urs. Min. in artificial horizon out of mer. was 96 37' 40" ; ind. cor. - 2' 48". Find the latitude. 9. August 4th, at 9 h 18 m 10 m P.M., mean time at ship in long. 127 16' W. ; the obs. alt. of Polaris in artificial horizon out of mer. was 61 15' 10" ; ind. cor. + 2' 42" ; eye, 50 ft. Find the latitude. CHAPTER XV. COMPASS ERROR BY AMPLITUDES. ART. 72. Variation is the angle between the terrestrial and magnetic meridians. It shows how much the magnetic needle is disturbed from a true north and south direction by the action of the earth. It changes with the geographical position of the observer, and also with a lapse of time; the annual change is about 6' annually. The amount of the variation for a particular year is entered on Admiralty charts. Deviation is the angle the compass needle on a ship makes with the magnetic meridian. It shows how much the compass north is deflected from magnetic north by the action of iron in the ship or cargo. The amount varies with the direction of the ship's head, with the heel of the ship, and with her geographical position. Compass Error is the angle the compass needle on a ship makes with the true meridian. It includes variation and deviation ; and is the difference between the true and compass bearing of any object. The compass bearing of an object may be observed at any time, and one method of determining true bearings is by " Amplitude," which is the bearing of an object reckoned from east when rising, or from west when setting. Thus it is possible for the sun to be used for this purpose twice each day ; and suitable stars can be selected continually at night. ART. 73. Proof of Amplitude. In the figure NESW is the horizon, X the position of an object on the horizon ; the arc WX is its ampli- N tude. In the right-angled triangle PNX, ^^ T"~~"^\ N = 90 ; PN = latitude ; PX = pol. dist. /^_ n X or co. dec. ; NX = co. amp. ^/*" \ Sin (co. PX) = cos PN . cos NX W( Z IE cos PX = cos PN . sin WX since cos NX = sin WX ^ cosPN = cos PX . sec PN FlG ^27. or sin amp. = sin dec. . sec lat. The equinoctial cuts the horizon in the east and west points, hence when an object is on the equinoctial, that is, has decimation = 0, it rises due east and sets due west. Again, when an COMPASS ERROR BY AMPLITUDES. 169 observer is on the equator the formula becomes sin amp. = sin dec., since sec = 1 ; hence in this case amp. = dec. It is also observed that when an object has north dec., it rises to the north of east, and sets to the north of west ; if south dec., it rises south of east, and sets south of west ; the name of the amplitude is therefore the same as the declination. ART. 74. To find the Time of rising or setting of a Celestial Object. In the figure the hour angle is ZPX, which is the supplement of NPX. Then from triangle NPX sin (co. P) = tan PN . tan (co. PX) cos P = tan PN . cot PX hence cos H.A. = tan lat. cot. pol. dist. ART. 75. The process for finding compass error by ampli- tude is as follows : (1) Find Greenwich time, and correct declination. (2) To L. sin declination add L. sec latitude, reject 10 from index, and take back sin true amp., which reckon from E. if A.M. at ship, and W. if P.M. at ship ; and same name as declination. Under true amp. place comp., bearing reckoned from E. or W. like the true amp. (3) When the true and comp. amps, are of the same name N. or S., their difference is the comp. err. ; when of contrary names, their sum is the comp. err. It is named E. or W. according as the true amp. reads to the right or left of the comp. amp. (4) Under the comp. err. place the variation; if of same names subtract ; if of contrary names add ; this gives the devia- tion, which is to be named like the error, unless the error is sub- tracted from the variation, when it is of contrary name to error. Examples. (1) September 28th, at 5 h 53 m A.M., mean time at ship in lat. 23 25' S., long. 12 40' W., the sun's compass bearing was E. f N. Find the comp. err. and deviation for the position of the ship's head, if the variation from chart was 28 30' W. M.T.S. Sept. 27 d 17 h 53 ra 0' 12 40' Long. W. + 50 40 4 M.T.G. Sept. 27 d 18 43 40 6,0)5,0 40 50 ra 40' Dec. 28th noon 1 46' l"-4 S. H.V. 58"'45 Cor. 58 5-27 1 40 53-4 S. 40915 11690 29225 6,0)30,8-0315 5' 8" 170 TEXT-BOOK ON NAVIGATION. Dec. 1 40' 53" Lat. 23 25 True amp. E. 1 49' 57" S. Comp. bearing E. 8 26 15 N. sin = 8-467483 sec = 10-037328 Comp. err. 10 16 12 E. Var. 28 30 W. sin = 8-504811 (Add because true and comp. amp. differ in name ; named E, because true amp. reads to the right of the comp. amp.) (Sum of comp. err. and variation Dev. 38 46 12 E. because they are of different names ; E. because comp. err. isE.) In figure X is position of sun. In triangle SPX, S = 90. Given SP = lat. PX = pol. dist. or co. dec. Keq. SX = co. amp. Formula as proved above. Sin amp. = sin dec. sec lat. (2) August 1st, at 8 h 35 m r.M. apparent time at ship, in lat. 61 45' S., long. 176 30' E., the sun's compass bearing was N. 66 30' W. If the variation from chart is 22 15' E., find compass error and deviation. A.T.S. Aug. l d 8 h 35 m 0' Long. E. - 11 46 A.T.G. July 31 d 20 49 Dec. 1st noon 18 11' 45" N. Cor. + 1 59 18 13 44 N. 6,0)70,6 ll h 46 m H.V. 37" -26 3-2 7452 11178 6,0)11,9-232 1' 59" Dec. 18 13' 44" Lat. 61 45 sin = 9-495286 sec = 10-324845 True amp. W. 41 22 5 N. sin = 9 '820131 Comp. W. 23 30 N. Err. 17 52 5 E. Var. 22 15 E. Dev. 4 22 55 W. COMPASS ERROR BY AMPLITUDES. 171 The bearing in the question N. 66 30' W. is subtracted from 90 to reckon it from W., the same point as the true amp. Since true and comp. bearings are of same name N., they are subtracted for error, which is named E., because the true amp. reads to the right of the comp. ; the difference of error and variation, because of same name E., is the dev., which is of contrary name to error, because error is subtracted from variation. In figure X is position of sun. In triangle SPX, S = 90. Given SP = lat. PX = pol. diet, or co. doo. Seq. 8X = co. amp. EXERCISES. 1. August 2nd, at 7 h 10 m P.M. mean time at ship, in lat. 42 20' S., long. 65 E., the sun's comp. bearing was N.N.W. f W., the variation being 28 30' W. Find comp. err. and dev. 2. September 28th, at 5 h 50 m A.M. mean time at ship, in lat. 50 25' S., long. 81 W., the sun rose by compass N. 64 30' E. Find true amp. and comp. err. ; also if var. is 24 15' E., find dev. 3. August 9th, at 5 h 15 m A.M. mean time at ship, in lat. 33 51' N., long. 125 W., the sun's comp. bearing was E. by N. J N. Find true amp. and comp. err.; and if var. is 11 20' E., find dev. 4. September 30th, at 6 h 23 ra P.M. A.T.S., in lat. 48 19' S., long. 76 E., the sun set by compass W. S. Find true amp. and comp. err. ; the var. being 29 45' W., find the dev. 5. August 4th, at 5 h 6 m A.M., M.T.S., in lat. 37 15' N., long. 20 30' E., the sun's comp. bearing was E. by N. N. Find true amp., comp. err., and deviation, the var. being 11 30' W. 6. September 26th, at 6 h 5 ra P.M. M.T.S., in lat. 50 15' N., long. 126 45' W., the sun's comp. bearing was S. 69 30' W. Find true amp., comp. err., and dev., the var. being 22 40' E. 7. August 7th, about 6 h A.M., in lat. 20 24' S. long. 134 W., when a chron. correct for M.T.G. showed 3 h 20 m , the sun's comp. bearing was E. Find the true amp., comp. err., and dev., the var. being 7 15' E. 8. September 7th, about 7 h P.M. , in lat. 60 14' N., long. 166 W., when a chron. correct for M.T.G. showed 6 h 4 m , the sun's comp. bearing was W. Find the true amp., comp. err., and dev., the var. being 12 20' E. 9. August 3rd, about 3 h 10 m A.M., in lat. 65 22' N., long. 170 15' W., when a chron. correct for M.T.G. showed 2 h 35 ra , the sun's comp. bearing was N. J W. Find the true amp., comp. err., and dev., the var. being 22 40' E. 10. September 23rd, at 6 h A.M. A.T.S. at ship, in lat. 0, long. 50 W., the sun's comp. bearing was E. 10 S. Find the true amp., comp. err., and dev. , the var. being 0. 11. September 25th, at 6 b 5 m P.M. A.T.S. at ship, in lat. 13 15' S., long. 95 E. , the sun's bearing by comp. was N. 76 30' W. Find the true amp., comp. err., and dev., the var. being 2 W. 1 7 2 TEXT-BOOK ON NAVIGATION. 12. September 23rd, about 6 h P.M. at ship, in lab. 39 N., long. 78 45' W., when a chron. correct for M.T.G. showed ll h 15 ra 23 s , the sun's comp. bearing was S. 83 W. Find true amp., comp. err., and dev., the var. being 6 30' E. 13. August 30th, at 5 h 59 m A.M., in lat. 0, long. 180 W., the sun's comp. bearing was N.E. by E. \ E. Find the true amp., comp. err., and dev., the var. keing 9 30' E. 14. September 24th, about 6 h A.M., in lat. 2 20' N., long. 101 11' E., the sun's comp. bearing was E., when the chron. showed ll h 15 m 14" correct M.T.G. Find the true amp., comp. err., and dev., the var. being 1 15' E. 15. Find the time of sunrise on August 16th, in lat. 53 20' N., 4 W. long. 16. If a star in the equinoctial rises at 7 h 12 m 18" by chronometer, what should the chron. indicate when star is on meridian. 17. At a certain place the sun rose E. 25 S. by compass, and set W. 15 N. by same compass. Find deviation, having given variation 22 W. 18. At a certain place a star when on the prime vertical has its altitude equal to its amplitude when setting. Find lat. 19. Find the greatest amplitude the sun can have in lat. 53 20' N". 20. In lat. 50 S. when sun's declination is 18 40' S. : find his altitude when on the prime vertical, and the approx. interval between then and setting. CHAPTER XVI. COMPASS ERROR BY AZIMUTHS. ART. 76. Another method of finding compass error and deviation is by Azimuths ; that is, by observing the bearing of any celestial object when out of the horizon. This method has the advantage of being available at all times of the day or night when the object is visible. When an altitude is observed for time, by taking the bearing very little extra trouble is involved in working the Azimuth from the altitude ; generally, however, the time is accurately taken, and the Azimuth taken from tables calculated for the purpose. Burdwood's Tables are used for latitudes 30 to 60, and Davis's Tables for Latitudes to 30 ; both for declinations from to 23. ART. 77. The Azimuth of a Celestial Object that is, its bearing reckoned from true North or South is the angle at the Zenith between the Meridian of the observer and a vertical circle passing through the object, and is measured on the Horizon. The Data required in the Calculation are the True Altitude, Latitude, and Polar Distance. Let S be the object, PS its polar distance (p\ PZ the co-lat. (90 - I), ZS the zenith distance (90 - a). The angle PZS is the azimuth from N., and its supplement SZX is the azimuth from S. PZS is a spherical triangle, of which the three sides are given to find an angle PZS. The general formula is cos A = cos a cos b . cos c sin b . sin c whence in the figure rza Q7V cos PS - cos ZS cos PZ cos PZS or - cos SZX = : ^- ; ^ sm Zb . sin PZ Putting the letters Z, a, I, p for the azimuth, altitude, latitude, and polar distance respectively i 7 4 TEXT-BOOK ON NAVIGATION. 7 __ cos p - cos (90 - a) . cos (90 - I) sin (90 - a) . sin (90 - /) __ cos p sin a . sin I cos a . cos I adding 1 to each side of the equation rj -, , cos p sin a . sin I 1 - cos Z = 1 H -- - cos a . cos I _ . o Z (cos a . cos I sin a . sin I) 4- cos /. 2 Bin* 5 = 1 - *- 2 cos a . cos I __ cos (a + + cos p cos a . cos I . 2 Z _ cos a + I + cos la + - sin , 2 cos a . cos . 2 Z _ cos 8 . cos (s ff) SJ ~ cos a . cos where s = JO + I + p) i.e. sin 2 = cos s . cos (s p) . sec a . sec rj __ _ ____ or sin = \/cos s . cos (s p) . sec a . sec I from which is derived the rule for computing the azimuth of the opposite name to the latitude, namely, add together the log secants of the altitude and latitude, and the log cosines of the half-sum and remainder; divide by two to get the square root, which gives the log sine of half the azimuth. ART. 78. To find azimuth from altitude proceed as fol- lows : (1) Find Greenwich time ; correct declination and find polar distance. (2) Correct altitude. (3) Add altitude, latitude, and polar distance, divide by two for half-sum ; difference between half-sum and polar distance is remainder. (4) Add sec alt., sec lat.,cos half-sum, and cos remainder; divide sum of logs by two, and take back sine ; this angle multiplied by two is the true azimuth which mark contrary name to lat., E. if A.M. at ship, W. if P.M. at ship. If true and comp. azimuth are different in name N. or S., subtract true from 180 to make of same name as comp. bearing. (5) The difference of the true and comp. bearings when of same name E. or W., or their sum when of different names, gives the comp. error which is marked E. or W., according as the true reads to the right or left of the comp. bearing. (6) Find deviation as in amplitude. COMPASS ERROR BY AZIMUTHS. 175 Example. September 23rd, at 3" 48 m 27* M.T.S. P.M., in lat. 56 20' S., long. 67 35' W., the obs. alt. of the Sun's U.L. was 16 54' 30"; ind. cor. + 1' 1" ; height of eye, 27 ft. The Sun's comp. bearing was W. S. Required the true azimuth, and comp. error; and, supposing the variation to be 22 E., what is the deviation ? M.T.S. Sept. 23 d 3 h 48 m 27' Long. W. + 4 30 20 M.T.G. 23 d 8 18 47 67 35' 4 6,0)27,0 20 O.A. 16 54' 30" I.C. +11 4 h 30 m 20' Dec. noon 10' 58"-l N. H.V. 58"-45 Cor. 8 5-1 8-3 16 55 31 Dip. - 5 5 16 50 26 S.D. - 15 58 2 53 N. P.D. 90 2' 53" N. 6,0)485135 8' 5"-l R. and P. 16 34 28 -2 58 T.A. 16 31 30 sec = 30-018319 Lat. 56 20 sec = 10-256208 P.D. 90 2 53 2)162 54 23 81 27 11 8 35 42 cos = 9-172076 cos =-- 9-995095 2)19-441698 sin = 9-720849 31 43' 30 /; 2 T. Az. N. 63 180 27 W. S. 116 33 W. Comp. bearing S. 87 11 15 W. Comp. error. 29 21 45 E. Var. 22 E. Dev. 7 21 45 E. The true azimuth is marked N., contrary to name of lat., and W. since it is P.M. at ship. It is subtracted from 180 to change the name to S., which reckons the angle from the same point as the comp. bearing; the difference is the error, since both are named W. ; and the error is named E., because the true reads to the right of the comp. bearing. i 7 6 TEXT-BOOK ON NAVIGATION. Draw circle with ^ ff inch radius to represent the rational horizon ; NESW : vertical diameter NZS represents meridian of observer. Mark off on scale SP = ZQ = lat. 56 20'. Since declination is so small object may be considered as on the equinoctial EQW ; with zenith distance as radius = 73 describe an arc to westward cutting EQW in X, the position of sun ; through X draw vertical circle ZXA, and hour circle PX to complete. Then Alt. is AX ; Lat. is ZQ ; Pol. dist. isPX. Hence azimuth is PZX, the formula em- ployed gives its supplement NZX. From the figure. In figure X is position of object, very nearly on the equinoctial. In triangle PZX. Given co-lat. = PZ 33 40' 0" Pol. dist. = PX 90 253 Zen. dist. = ZX 732830 Req. azimuth = Z. Z / sin s . sin (s - PX) 2 = V ~~sinl Z cos- where s = PX PZ ZX PZ . sin ZX + PZ + ZX) 90 2' 53" 33 40 cosec = 10*256208 73 28 30 cosec = 10-018319 2)197 11 23 s 98 35 41 PX 8 32 48 58 16 34 2 sin= 9-995095 sin = 9-172062 2)19-441684 cos = 9720842 True az. S. 116 33 8 W. Comp. bearing S. 87 11 15 W. Comp. error 29 21 53 E. var. 22 E. Dev. 7 21 53 E. EXERCISES. 1. August 4th, about 8 h 30 m A.M., in lat. 38 47' S., long. 92 12' E. ; when a chron. correct for M.T.G. showed 2 h 37 ra 35% the obs. alt. of Sun's L.L. was 16 42' 10" ; ind. cor., + 2' 15" ; eye, 29 ft. ; bearing by compass, N. 49 30' E. Find the true azimuth, comp. error, and deviation, the variation being 18 W. 2. August 19th, at 4 h 25 37' P.M. A.T.S., in lat. 41 24' N., long. 70 20' W., the obs. alt. of the Sun's U.L. was 26 32' 40" ; bearing by compass, W. ; ind. cor.. 2' 7" ; height of eye, 34 ft. Find the true azimuth and comp. error ; also if the variation from chart is 9 30' W., find the deviation. COMPASS ERROR BY AZIMUTHS. 177 3. August 31st, about 9 h 20 m A.M., in lat. 52 25' N., long. 3 17' 30" W. ; when a chron., which was 38 m 19 slow of M.T.G., showed 8 h 57 m 39', the obs. alt. of the Sun's U.L. was 36 26' 10" ; bearing by compass S., 28 30' E. ; ind. cor., 26" ; eye, 41 ft. Find the true azimuth, comp. error., and deviation, the variation being 15 40' W. 4. September 1st, A.M. at ship, in lat. 44 18' N., long. 10 15' W. ; when a chron. correct for M.T.G. indicated 9 h 22 m 37', the obs. alt. of Sun's U.L., 34 21' 30" ; ind. cor., -f 4' 15" ; height of eye, 23 ft. ; bearing by compass, S.E. \ E. Required the true azimuth, and comp. error ; and if variation from chart is 25 W., what is the deviation ? 5. September 22 d 18" 29 m 38' (astronomical) M.T.G., in lat. 49 40' S., long. 135 24' E. ; the Sun's comp. bearing, N. 61 50' VV.; alt. of ILL., 22 18' 50" ; height of eye, 20 ft. Find true azimuth, comp. error, and devia- tion, the variation being 0. 6. September 30th, about 3 h 15 m P.M., in lat. 49 37' S., long. 179 56' E. ; when a chron. showed 3 h 20 ra 26" correct M.T.G., the obs. alt. of Sun's L.L., 25 3' 20"; ind. cor., +3' 14"; height of eye, 37 ft.; bearing by compass, W.N.W. Find true azimuth, comp. error, and deviation, the variation from chart being 16 30' E. ART. 79. Time Azimuths of the Sun by the Tables. When the Apparent Time at Ship, the Lat., and the Sun's Dec. are known, the Sun's True Azimuth can be found by inspection in Burdwood's or Davis's Tables. The same Tables will give the true Azimuth of a Star, a Planet, or the Moon, provided the Hour Angle and Declination are within the limits of the Tables. A description of one of the books suffices, as the only differ- ence is that Burdwood is used with lats. 30 to 60, and Davis with lats. to 30. The azimuths are calculated and tabulated for each degree of lat. and dec., and at intervals of 4 mins. in time. Interpolations must be made for odd minutes of time, lat., and dec. The degrees of lat. are found at the top of the page, underneath which is a remark showing whether lat. and dec. are the same or contrary in name. The columns are headed with the degrees of dec. from to 23 consecutively; to 11 on left hand page, 12 to 23 on right; and at the sides the times are entered to every 4 mins. A.M. time on the left-hand side of the page, P.M. time on the right. The method of using the Tables will be illustrated, and the following rules must be observed : (1) Find the apparent time at ship (civil time). (2) Correct Sun's declination. (3) Open the tables at the degree of lat. in question, and observe whether dec. and lat. are of same or contrary names. Find in the A.M. or P.M. column the next less entry in point of time to the A.T.S. in question, and in this line under the degree of declination will be found the azimuth, which has to be corrected for odd minutes of time, declination, and latitude by proportion, for the true azimuth. (4) The true azimuth must be named the same name as the latitude E. if A.M. at ship, W. if P.M. at ship. N i;3 TEXT-BOOK ON NAVIGATION. Example. The apparent time at ship being 3 h 56 m 11 P.M., the correct decimation 3' N., and latitude 56 20' S. Find true azimuth from " Tables. ' J A.T.S 3 b 56 IP P.M. Dec 3' N. Lat. 5620'S. In Burdwood's Tables, with lat. 56, contrary name to dec. opposite 3 h 56 m , under dec. 0, the Sun's azimuth is found to be 116 28' ; this has to be corrected for 11" of time, 3' of dec., and 20' of lat. Az. for 3 h 56 116 28' Az. for dec. 116 28' Az. for 56 lat. 116 28' 4 115 35 1 117 1 ,, 57 Diff. in 4 m - 53 2 Diff. in 60' + 33 3 Diff. in 60' 4)10-6 , 3' 6,0)9,9 + 1-6 20' ,,11' - 2-6 Azimuth from Correction - 2'*6 + 1''6 Tables + 5'-3 = 116 28' + 4-3 -f 16 20 6,0)32,0 + 5-3 True azimuth S. 116 32'3 W. ART. 80. Time Azimuth by Calculation. Here the hour angle ZPS, the polar distance PS, and the co-latitude PZ are given to find the angle PZS, which is the azimuth of the same name as the latitude (Fig. 130). This is a case of two sides and the included angle being given to find one of the angles at the base. From the general formulae by substituting the Z and S in the figure for A and B, and the letters h, p, I' for the H.A., polar distance, and co-latitude respectively, we have therefore, tan J(Z + S) = sec %(p + I) . cos J(p - I 1 ) . cot l>h and tan J(Z - S) = cosec i(p-M') . sin~(p - I') . cot \h whence the following form for computation : + ... sec ... cosec -I') ... cos ... sine ... cot ... cot J(Z + S) ... tan ... |(Z - S) ............ tan COMPASS ERROR BY AZIMUTHS. 179 The sum of these angles is the greater of the two angles Z and S, and the difference is the lesser angle; and since the greater angle is opposite to the greater side, the azimuth Z is the greater angle when p is greater than I', and the smaller angle when p is less than V. NOTE. When \(p + I') exceeds 90, the sec is negative, and therefore tan J(Z + S) is negative, in which case the supplement must be taken for Example. The apparent time at ship being 3 h 56 II 1 P.M., correct dec. 3' N., and lat. 56 20' S., find true azimuth by calculation. This figure is drawn similarly to Fig. 131, except that as the hour angle is given instead of the altitude, we must estimate an arc of the equinoctial equal to the hour angle 3 h 56 ra to determine the point through which to draw the hour circle of the object : knowing that QW represents 6 hours : where this hour circle cuts the parallel of declination is the position of X. In triangle PZX Given P = 3" 56 ra 11" = hour angle PZ = 33 40' = co-lat. PX = 90 3' = pol. dist. Req. Z = azimuth. Y , sinJ(PX-PZ) - X= PZ = 33 40' PX- 90 3 PX + PZ = 123 43 PX-PZ= 56 23 ? = l h 58 m 5-5 cot = 10-246953 cot = 10'246953 = 61 51' 30" sec = 10-326377 cosec = 10'054638 - PZ) = 28 11 30 cos = 9-945159 sin = 9'674331 |(Z + X) = 73 8 30 tan = 10-518489 J(Z-X) = 432430 tan- 9'975922 Z = 116 33 True azimuth S. 116 33' W. EXERCISES. Under the following conditions find from the Tables or by calculation the True Azimuth, and hence the deviation : 1. A.T.S. 2 h 48'" P.M. ; dec. 10 25' S. ; lat. 40 N. 2. A.T.S. 3 h 19 ra P.M. ; dec. 14 33' N. ; lat. 49 N. 3. A.T.S. 9" 14 A.M. ; dec. 21 17' N. ; lat. 52 25' S. 4. September 1st, M.T.G. 31 d 15 h 54 m 42% in lat. 51 25' S., long. 174 25' E., the Sun's comp. bearing was W. f N. ; variation 30 E. 5. August 12, A.M. at ship, in lat. 0, long. 12 50' W. ; when a chron. showed 8 h 16 m 20 M.T.G., the Sun's comp. bearing was E. f N. ; var. 20 30' W. 6. September 3rd, at 7 h 57 A.M. M.T.S., in lat. 13 29' N., long. 60 30' E., the Sun's comp. bearing was E. by N. f N. ; var. 1 10' W. i8o TEXT-BOOK ON NAVIGATION. 7. September 23rd, about 3 h 30 m P.M. in lat. 49 40' S., long. 135 24' E. ; when a chron. showed 6 h 29 m 37 s , the Sun's comp. bearing was N. 61 50' W. ; var. 0. 8. When the M.T.G. was August 31 d 21 h 22 m 37 , in lat. 44 18' N., long. 10 15' W. ; the Sun's comp. bearing was S.E. E. ; var. 25 W. 9. About 4 h 15 m P.M. on September 23rd, in lat.O 15' S., long. 150 15' E. ; when a chron. showed 6 h 30 M.T.G. , the Sun's comp. bearing was W. ; var. 9E. 10. August 3rd, at 3 h 50 ra P.M. A.T.S., in lat. 25 16' S., long. 92 30' E., the Sun bore by comp. N. 60 J W. ; var. 8 15' W. 11. August 5th at 8 h 40 m A.M. M.T.S., in lat. 30 21' N., long. 51 15' E., the Sun's comp. bearing was S.E. by E. | E. ; var. 2 W. 12. August 9th at 4" 25 ra P.M. M.T.S., in lat. 37 27' N., long. 20 14' E., the Sun bore by comp. W. \ N. ; var. 21 W. 13. August 23rd about 9 1 20 m A.M., in lat. 52 49' S., long. 53 17' W. ; when a chron. showed l h 17 ra 12 1 G.M.T., the Sun bore by comp. N.E. by E. ; var. 15 30 ; "E. 14. August 24th about 10 h 40 ra A.M., in lat. 64 56' N., long. 170 W. ; when a chron. correct for M.T.G. showed 10 h l m 30", the Sun's comp. bearing was S. f W. ; var. 24 E. ART. 81. Star Time Azimuths. As before noticed, providing the declination of the star is less than 23, the Tables may be used for obtaining the true bearing. The Hour Angle of the star must, however, be calculated instead of the A.T.S., and always found in the P.M. column on the right-hand side of the page, otherwise the method is precisely the same. Example. September 10th at l h 36 20* A.M. mean time at ship, in lat. 38 20' N., long. 63 30' W., the star a Orionis bore by compass E. by S. | S. Required the comp. error, and deviation, the variation being 9 15' W. M.T.S. Sept. 9 d 13 h 36 m 20* C3 30' E.A.M.S. noon ll h 10 m 29'8 Long. W. + 4 14 4 Accel, for 17 h 2 47'G CAra Q.O '' / O w M.T.G. Sept. 9 d 17 50 20 6,0)25,4 11 13 25-6 4" 14 m M.T.G. 13 30 20 Dec.* 7 23' 34" N. Lat. 38 20 N. R.A.M. 49 45*6 R.A.* 5 50 4-9 E.H.A. 5 19-3 In Burd wood's Tables, with lat. 38, same name as declination, under 7 Dec., abreast 5 h on the right-hand side (always find H.A. of star on right- hand side), the star's azimuth is entered as 93 43', which requires correction for 19 time, 24' of Dec., and 20' Lat. Az. for 5 h 93 43' Az. for 7 dec. 93 43' Az. for 38 lat. 93 43' 5 h 4 m 93 5 8 ,, 92 53 Diff. ford" -38 Diff. for CO' -50 Diff. for 60' +17 3 24 20 4)11-4 6,0)120,0 6,0)34,0 19' - 2-8 24' - 20 + 57 COMPASS ERROR BY AZIMUTHS. 181 Az. from Tables 93 43' Correction - 2'8 - 20 + 57 = -17 TrueAz. N. 93 26 E. 180 S. 86 34 E. Comp. S. 73 7 E. err. 13 27 W. Var. 9 15 W. From the figure. N Dev. 4 12 W. In triangle PZX, where X is position of star ; given PZ = colat PX = Pol. dist. P = hour angle. Req. Z = azimuth. P = 5 h 19 s PX= 82 36' 30" PZ = 51 40 PX + PZ = 134 16 30 PX-PZ= 30 56 30 -7, = 2 h 30 9-5 Cot = 10-114398 KPX + PZ) = 67 8' 15" ((PX - PZ) = 15 28 15 4(Z + X) = 72 47 18 ^Z - X) - 20 38 35 Z = 93 25 53 Sec = 10-410586 Cos = 9-983972 Tan = 10-508956 Cot = 10-114398 Cosec = 10-035533 Sin = 9-426102 Tan = 9-576033 Tr. Az. N. 93 25' 53" E. EXERCISES. 1. August 18th, about l h 20 A.M., in lat. 37 50' K, long. 50 W., when a chron. showed 4 h 37 m 40 s M.T.G., the comp. bearing of a Arietis was E. Find the true azimuth from the " Tables ; " and if the variation is 14 45' W., what is the dev. ? 2. September 14th, at 2 h 19 m AM. M.T.S., in lat. 52 49' S., long. 21 17' W., the comp. bearing of Sirius was S.E. Find the true azimuth from the " Tables ; " and if var. from chart is 33 15' W., find the dev. 3. September 6th, at O h 15 m A.M. M.T.S., in lat. 47 43' N., long. 48 32' W., the comp. bearing of Altair was N. 44 30' W. Find the true azimuth from the " Tables ; " and the var. being 34 40' W., find the dev. 4. August 4th, at ll h 25 m P.M. A.T.S., in lat. 38 49' S., long. 56 W., when the comp. bearing of Vega was N.N.W., calculate the true azimuth, comp. error, and dev., the var. being 17 10' E. 5. August 2nd, at 3 h 27 m A.M. A.T.S., in lat. 40 50' K, long. 50 40' W., 182 TEXT-BOOK ON NAVIGATION. the comp. bearing of a Orionis was E. ^ N. Using the " Tables," find the true azimuth, and comp. error ; and the var. being 20 W., find the dev. 6. August 3rd, about 8 h 20 m P.M., in lat. 57 19' K, long. 3 14' E., when a chron., whose error on A.T.G. was 14 m 28 s fast, showed 8 h 44 m 32% the comp. bearing of Arcturus was S. 65 30' W. Find from " Tables " the true azimuth ; and var. being 25 W., what is the dev. ? 7. September llth, at l h 45 m A.M. M.T.S., in lat. 42 35' S., long. 85 37' E., the comp. bearing of Aldebaran was S. 50 E. Find true azimuth, comp. error, and dev., the var. being 20 W. 8. September 21st, when the M.T.G. by chron. was 20 a 20 h 38 ra 20 , the comp. bearing of Fomalhaut, in lat. 14 10' N., long. 94 50' W., was S.W. S. Find true azimuth, comp. error, and dev., the var. from chart, being 8E. 9. September 21st, at 4 h 15 m A.M. A.T.S., in lat. 21 17' S., long. 87 40' E., the comp. bearing of Pollux was N. 40 20' E., var. 2 15' W. Find true azimuth, comp. error, and dev. 10. September 21st, about ll h 30 m P.M., in lat. 48 40' S., long. 167 30' W., when a chron., whose error on M.T.G. was 19 m 24 s fast, indicated ll h 17 m 24 s , the comp. bearing of a 2 Centauri was S. E. Find true azimuth, comp. error, and dev., the var. being 19 20' E. 11. August 2nd, at 6 h 40 m P.M. A.T.S., in lat. 5 12' S., long. 35 W., the comp. bearing of Dubhe (a Urs. Maj.) was N. 8 30' E.. The var. being 11 W., find true azimuth, comp. error, and dev. 12. August 1st, at 4 h 30 m 30 s A.M. M.T.S., in lat. 21 30' S., long. 92 45' E., the comp. bearing of Antares was S.W. by W. W. Find true azimuth, and comp. error ; also, if var. is 9 W., what is the dev. 9 13. If the declination of an object be greater than the latitude of the place, both being of the same name, find the greatest azimuth of the body, the time it will have its greatest azimuth, and its altitude. 14. In lat. of Greenwich 51 28' 40" N., find greatest azimuth of a Urs. Maj. dec. 62 16' 50" N. 15. Altitude of star when on the prime vertical is 25, altitude when on the six o'clock hour circle 15. Find its declination , and latitude of observer. CHAPTER XVII. LONGITUDE BY CHRONOMETER. ART. 82. Longitude by Chronometer. The time by chro- nometer when corrected for its original Error and accumulated Rate gives the time at Greenwich, whilst the time at ship can be calculated from an observed Altitude of a Celestial Object and the known Latitude. The difference converted into Arc is the Longitude. The Longitude is E. when Greenwich time is less than ship time, and W. when greater. The following are required in the Formula : (a) The true Altitude. (6) The Polar Distance. (c) The Latitude. The Polar Distance is measured from the elevated Pole, which is the N. Pole in N. Lat., and the S. Pole in S. Lat. It is therefore 90 - Dec. when the Dec. is of the same name as the Lat., and 90 4- Dec. when of a different name. ART. 83. To find the Hour Angle of a Celestial Object, the Altitude, Latitude, and Polar Distance being known. Let O be the object, then PO is its polar distance (p\ ZO its zenith distance (=90 - a), and PZ the co-latitude (= 90 - I) ; the angle ZPO is the hour angle. Therefore, in the spherical triangle PZO, three sides are given to find an angle ZPO. The general formula for finding an angle is cos a cos b . cos c cos A = sin b . sin c which, applied to the figure, gives cos ZO cos a COS cos - : - sin PO . sin PZ i8 4 TEXT-BOOK ON NAVIGATION. Let the letters a, I, p, and h stand for altitude, latitude, polar distance, and hour angle respectively ; ,, 7 cos (90 a) cos p . cos (90 I) then cos h = - '- : -* =^ - 1 sin p . sin (90 i) _ sin a cos p . sin I sin p . cos I Subtracting each side of the equation from 1 ., -, .. sin a cos p . sin I 1 cos h = 1 -- : -- *- = - sin p . cos I o 2 /*< __ sin p . cos I + cos p . sin I sin a v@ Sill -" - ; - z - 2 Sin p . COS 6 _ sin (p + I) sin d sin . cos I CQS . sn - q) sin > . cos I Now let s = (p + Z + a), then (s a) = J(p + I - a) ; therefore . 2 7i __ cos s . sin (s a) sin - : - 2 sin p . cos 6 = cosec p . sec I . cos s . sin (s a) and sin = \/cosec ^) . sec I . cos s . sin (s a) which is the formula used for com- puting the hour angle in the longitude by chronometer and other time problems. ART. 84. To find the Time when the Sun or other Celestial Object is on the Prime Vertical, Let O be an object on the prime vertical ; then PO is its polar distance, PZ is the co-latitude, and the angle P is the hour angle. PZO is a spherical triangle, right angled at Z, of which the circular parts are PZ and ZO, and the complements of PO and the angles P and O. Taking complement of P as middle part, PZ and the com- plement of PO are adjacent parts : LONGITUDE BY CHRONOMETER. 185 then sin (comp. of P) = tan PZ x tan (comp. of PO) or cos P = tan PZ x cot PO that is, cos H.A. = cot lat. X tan dec. In the case of the Sun, the H.A. is the interval from apparent noon. For any other object, add its R.A. to its H.A. west to find the R.A. of meridian, from which subtract the sun's R.A. to find the time. ART. 85. The following rules must be observed in Sun chronometers : (1) Find the correct G.M.T. by applying the error when last found, and accumulated rate, to the time shown by chronometer. As in other problems, great care must be taken to make the time shown by the chronometer agree with the rough Greenwich time deduced from D.R. long, and ship time ; both as regards the date and hours. (2) Correct declination, equation of time, and altitude. Find polar distance ; and look at the heading of the equation of time column on page I., to see how it is to be applied. (3) Add together true alt., lat., and pol. dist. ; find the half sum, and from it subtract the alt.; this gives the chron. remainder. Add the secant lat., cosec pol. dist., cos half sum, and sin remainder, reject 10's, and look out result in Table xxxi., Norie; this is the hour angle. NOTE 1. Should the pol. dist exceed 90, use sec dec. for cosec pol. dist. NOTE 2. The lat. used must be the lat. at observation ; if not given, it must be deduced from the lat. at noon. (4) If P.M. at ship, the hour angle is the apparent time at ship of the same date as given in question. If A.M. at ship subtract the hour angle from 24 h and date the day before day in question for app. time at ship. To the A.T.S. apply equation of time for M.T.S. (5) The difference between the M.T.S. and M.T.G. is the longitude in time, which must be expressed in arc, for long, at obs. named west when G.M.T. is best, and east when G.M.T. is least. Examples. (1) August 8th, about 3 h 50 m P.M. at ship, in lat. 56 48' N., long. D.R. 7 W. ; when a chron. showed 4 b 29 m 34', the obs. alt. of the Sun's L.L. was 30 57' 10" ; ind. cor., -2' 10"; height of eye, 20 ft. Find the longitude, the chron. having been found 7 m 46" '2 fast of G.M.T. on July 18th, and losing daily 2' 7. Approx. time at ship, Aug 1 . 8 d 3 h 50 m elapsed time 21 '2 days Long. W. 28 rate 27 Greenwich 8 d 4 18 1484 424 ace. loss 57 '24 1 86 TEXT-BOOK ON NAVIGATION. Time by chron., Aug. 8 d 4" 29 m 34' Fast. -7 46-2 Dec. 16 20' 28"-7 N Cor. - 3 5-5 ace. loss 4 21 47-8 + 57-2 16 17 23-2 90 M.T.G. 8 d 4 22 45 N.P.D. 73 42 37 Eq. T. 5 m 33"2 Cor. - 1-4 II. V. 42-16 4-4 16864 16864 6,0)18,5-504 3' 6"-5 5 31-8 Obs. alt. 30 57' 10" I.C. -2 10" 307' 4-4 1228 1228 1-3508 30 55 Dip. -4 23 30 50 37 S.D. +15 48 31 6 25 R. and P. -1 27 T.A. 31 4 58 Lat. 56 48 P.D. 73 42 37 FIG. 136. 2)161 35 35 half sum remainder 80 47 47 49 42 49 sec = 10-261566 cosec = 10-017794 cos= 9-203967 sin = 9-882423 W.H.A. 3 h 50 m 26 s XXXI. = 9-365750 A.T.S., Aug. 8 d 3 h 50 m 26' Eq. T. +5 32 M.T.S. 8 d 3 55 M.T.G. 8 d 4 22 58 45 Long, in time 26 47 Long. 6 41' 45" W. Construction of Fig. 136. Draw circle NESW repr renting rational horizon, and NZS meridian. Measure NP = ZQ = 57 lat. ; set off dec. = 16 to north, and draw EQW for equinoctial, and dd parallel of declina- tion. With centre Z and radius = zen. dist. = 59 draw an arc cutting dd in X for position of sun, draw hour circle PXB, and vertical circle ZXA. ZQ = lat. PZ = co. lat. AX = alt. ZX = zen. dist. BX = dec. PX = pol. dist. ZPX = Hour angle. To find H.A. from figure. LONGITUDE BY CHRONOMETER. 187 In triangle PZX; given PZ = co-lat, PX = pol. dist., ZX = zen. dist. Req. P = Hour angle. ?- / sin s . sin (s - ZX) 2 V sin PZ . sin PX where 6 = ^(PZ + ZX + PX). ZX = 5855' 2" PZ = 33 12 cosec = 10 '261566 PX = 73 42' 37 cosec = 10'017794 2)165 49 39 s = 82 54 49 sin = 9 '996670 - ZX = 23 59 47 sin = 9 '609252 2)19-885282 _ = l h 55 ra 13 s W.H.A. 3 50 26 (2) September 1st, about 9 h 15 ra A.M., in lat. at noon 46 1' 30" S., D.R. long. 27 E. ; when a chron. showed 7 h 19 ra 43% and had been found to be fast 3 ra ir-5 on May 3rd, and slow 12 s on June 27th, the alt. of Sun's L.L. was 24 14' 40" ; ind. cor., 3' 15" ; eye, 29 ft. Find long, at sights ; and if run from obs. to noon was S. 70 E. 30 miles, find long, at noon. Approx. time at ship, Aug. 31 d 21 h 15 m May 3rd, fast 3 m 11 8 '5 Long. E. 1 48 June 27th, slow 12 ,, ,, Greenwich 31 d 19 27 Loss in 55 days 3 23'5 55)203 -5(3'7 165 Time by chron., Aug. 31 d 19 h 19 m 43' Rate 3'7 slow + 12 elapsed days 65'8 385 385 31 d 19 19 55 6,0)24,3-46 Ace. rate +4 3 4 m 3* M.T.G. Aug. 31 d 19 23 58 Dec. Cor. 8 31' 391" N. + 4 9 H.V. 54-19" 4-6 6,0)24,9-274 4' 9" Eq. T. O m 9'-l H.V. -78 3-6 4-6 8 35 48-1 N. 90 12-7 3-588 Laf, at noon 46 1' 30" S. S.P.D. 98 35 48 S. 70 E., 30' D. lat. 10 18 N. Lat. at obs. 45 51 12 S. Mid. Lat. 45 56' 21" sec = 10 '157752 S. 70 E., 30' = dep. 28'2 log = 1 '450249 D. long. 40'56 log = 1 '608001 1 88 TEXT-BOOK ON NAVIGATION. Obs. alt. 24 14' 40" I.C. -3 15 Dip. 24 11 25 -5 1C 24 6 9 S.D. +15 53 24 22 2 R. and P. -1 59 T.A. 24 20 3 Lat. 45 51 12 P.D. 98 35 48 2)168 47 3 half sum 84 23 31 '5 remainder 60 3 28 '5 sec = 10-157080 cosec = 10-004907 cos = 8-989985 sin = 9-937784 E.H.A. 2 h 44 m 13 s XXXI. = 9'089756 24 A.T.S. Aug. 31 d 21 h 15 m 47 Eq. T. + 13 M.T.S. 31 d 21 16 M.T.G. 31 d 19 23 58 1 52 2 60 4)112 2 Long, at obs. D. long. 28 0'30"E. 40 33 E. Long, at noon 28 41 3 E. To find H.A. from figure. In triangle PZX, given PZ = co-lat., PX = pol. disk, ZX = zeu. dist. Beq. P = Hour angle. Cos ? = / sin s . sin (s - ZX) 2 V sin PZ . sin PX where s = |(PZ + ZX + PX). ZX = 65 39' 57" PZ '= 44 8 48 cosec = 10*157080 PX = 98 35 48 cosec - 10*004907 2)208 24 33 s = 104 12 16-5 s - ZX = 38 32 19-5 = l h 22 m 6" -5 2 2 sin = 9-986515 sin = 9-794518 2)19-943020 cos = 9-971510 E.H.A. 2 44 13 LONGITUDE BY CHRONOMETER. 189 In Fig. 137, drawn as before explained. ZQ = lat. PZ = co. lat. AX = alt. ZX = /en. dist. BX = dec. PX = pol. dist. ZPX = Hour angle EXERCISES. 1. August 4th, about 8 k 40 ra A.M., in lat. at noon 38 15' S., D.R. long. 92 E. ; when a chron. showed l h 15 m 5 , which had been found slow l h 17 ra 9* on June 20th for G.M. noon, and slow l h 18 m 57" on July 5th, the obs. alt. of the Sun's L.L. was 16 42' 10" ; ind. cor., + 2' 15" ; eye, 29 ft. Find long. at obs. ; and if run since obs. was N. 24,E. 35 miles, find long, at noon. 2. August 19th, about 4 h 30 m P.M., in lat. at noon 42 3' N., D.R. long. 70 W.; when a chron. showed 8 h 58 m 5 , which had been found slow 15 m 8 i- 3 on June 27th for G.M. noon, and slow 14 m 22'7 on July 12th, the obs. alt. of the Sun's U.L. was 26 32' 40" ; ind. cor., 2' 7" ; eye, 34 ft., run since noon, S. 32 E. 46 miles. Find long, at obs. and noon. 3. August 31st, about 9 h 20 m A.M., in lat. at noon 52 36' N., D.R. long. 3 15' W. ; when a chron. showed 9 h 47 m 53 s , which had been found fast 12 m 18'-8 on August 1st for G.M. noon, and fast 12 m 14" on August 7th, the obs. alt. of the Sun's U.L. was 36 26' 10" ; ind. cor., - 26"; eye, 41 ft., run since obs. N. 64 W. 25 miles. Find long, at obs. and noon. 4. September 1st, about 8 h 40 ra A.?*., in lat. at noon 44 42' K, long. 11 W. ; when a chron. showed 9 h 34 3', which was slow 4 s on April 17th for G.M. noon, and on June 6th was 4 m 8' fast ; run since sights, N.W. \ N. 31 miles, the obs. alt. of Sun's U.L. was 34 21' 30" ; ind. cor., + 4' 15" ; eye, 23 ft. Find long, at sts. and noon. 5. September 23rd, about 3 h 38 m P.M., lat. at noon 49 40' S., long. 135 E. ; ship becalmed ; obs. alt. of Sun's U.L. 22 18' 50" ; eye, 20ft. ; time by chron., 6 b 27 m 28", which was slow for G.M. noon on July; 10th 2 ra 10', and has since kept mean time ; course and distance since obs.,m'Z.' Find long, at obs. and noon. 6. September 30th, about 3 h 30 ra P.M., lat. at noon 49 30' 12" S., long. 180 E., the obs. alt. of Sun's L.L. was 25 3' 20" ; ind. cor., + 3' 14" ; eye, 37 ft. ; time by chron. , 3 h 54 7 s , which had been found 8 m 16 8< 4 fast on May 6th, and fast 20 m 24 s *4 on July 15th for G.M. noon ; run since noon, S. 73 \ W. 24 miles. Find long, at obs. and noon. ART. 86. Rules for Longitude by Star Chronometers : (1) Find G.M.T. as before. (2) Take out star's R.A. and Dec., find pol. dist., and correct altitude, and R.A.M.S. (sidereal time). (3) Find H.A. as before. (4) If H.A. is west, add H.A. to R.A. of star for R.A.M. If H.A. is east, subtract H.A. from R.A. of star for R.A.M. From the R.A.M. subtract the R.A.M.S., this gives the M.T.S., which is of same date as question if P.M. at ship, and the day before date of question if A.M. at ship. (5) Finish as before for loogitude. Example. August 1st, about 2 h 39 m A.M. at ship, in lat. 30 26' N., long. 133 E. by D.R. ; the obs. alt. of Vega was 33 38' 50" ; time by chron., 4 h 41 m 58', which was slow l h 2 m 53' of G.M. noon on June 26th, and losing s '4 daily ; ind. cor., - 2' 29" ; eye, 23 ft. Find the long. Approx. ship time July 31 d 14 h 39 m Elapsed time 35 '2 Long. E. - 8 52 rate '4 Greenwich time July 31 d 5 47 14 '08 190 TEXT-BOOK ON NAVIGATION. Time by chron.., July 31 d 4 h 41 m 58* slow + 1 2 53 R.A.M.S. 8 h 36 m 44 e -2 for Aug. 1st. Cor. -2 59-9 31 d 5 44 51 Ace. loss + 14 M.T.G. 31 d 5 45 5 R A. Vega 18" 33 ra 45'-4 8 23 44-3 Dec. 38 41' 45" N. 90 N.PD. 51 18 15 In Fig. 138, constructed as before. ZQ = lat. PZ = co. lat. AX = alt. ZX = zen. dist. BX - dec. PX = pol. dist. ZPX = Hour angle. fcZX = Azimuth. Obs. alt.* 33 38' 50" I.C. - 2 29 33 36 21 Dip. - 4 42 33 31 39 Ref. - 1 26 T.A. 33 30 13 Lat. 30 26 P.D. 51 18 15 2)115 14 28 half sum remainder 57 37 14 24 7 1 sec = 10-064382 cosec = 10-107641 cos = 9728778 sin = 9-611299 W.H A.* 4 h 38 m 8' XXXI. = 9-512100 R.A.* 18 33 45 R.A.M. 23 11 53 R.A.M.S. 8 33 44 M.T.S. 31 d 14 38 9 M.T.G. 31 d 5 45 5 8 53 4 60 4)533 4 Long. 133 16' E. To find H.A. from figure. In triangle PZX, X is object, PZ = co-lat., PX = pol. dist., ZX = zen. dist. Req. P = Hour angle. r P /sin s . sin (s - ZX) 2 = V sin PZ . sin PX" where s = i(ZX + PZ + ZX). LONGITUDE BY CHRONOMETER. 191 ZX = 56 29' 47" PZ = 59 34 cosec = 10 '064382 PX = 51 18 15 cosec = 10107041 2)167 22 2 s = 83 41 1 sin = 9-997355 - ZX = 27 11 14 sin = 9-659820 2)19-829198 p = 2 h 19 4' cog = 9 . 914599 W.H.A.* 4 38 8 EXERCISKS. 1. September 10th, about 4 h 30* A.M., in lat. 35 47' 30" S., long. D.R. 111 E. ; the obs. alt. of Achernar west of meridian was 59 28' 40" ; ind. cor., + 2' 12" ; eye, 35 ft. ; when a chron. showed 9 h 7 m 12*, which had been found fast l m 3 s on May 12th of G.M. noon, and on August 20th was slow 40'. Find the long. 2. August 24th, about 4 h 40 m A.M., in lat. 10 14' 20" N., long. D.R. 75 W. ; the obs. alt. of Canopus east of the meridian was 12 15' 40"; ind. cor., - 3' 13" ; height of eye, 48 ft. ; time shown by chron. , 9 h 21 m 52 s , which had been found slow of G.M. noon 17 m 52"'7 on July 13th, and on July 31st was slow 17 m 4'-l. Find the long. 3. August 1st, P.M. at ship, in lat. 24 19' S., long, by acct. 14 30' E. ; when a chron. showed 8 h 15 m 4 , which had been found fast 28 m 19 on June 24th, and losing 1 > 2 daily; the obs. alt. of Altair east of the meridian was 42 5' 30" ; ind. cor., + 3' 54" ; height of eye, 42 ft. Required the long. 4. September llth, P.M. at ship, in lat. equator, long, of Greenwich; the obs. alt. of Vega west of the meridian was 26 18' ; ind. cor., 1' 55" ; eye, 31 ft. ; when a chron. showed ll h 10 m 54 , which had been found 12 m 29* fast of G.M. noon on July 15th, and on August 9th was 13 m 29* fast. Find the long. 5. September 21st, about l b 30 m A.M., in lat. 20 30' K, long, by acct. 31 W.; the obs. alt. of Capella east of meridian was 37 14' 30" ; ind. cor., + 3' 49"; eye, 19 ft.; when a chron. showed 3 h 21 m 26", which had been found 3* fast of G.M. noon on August 7th, and on August 31st was 3 m 21 1 slow. Find the long. 6. August 24th, about l h 15 m A.M., in lat. 18 12' 30" K, long. D.R. 91 E. ; when a chron. showed 9 h 24 m 17', which had been found fast ll m 43 > 8 on June 4th for G.M. noon, and on July 12th was fast 12 ra 18 ; the obs. alt. of Rigel east of meridian was 27 26'; ind. cor., 3' 5" ; eye, 46 ft. Find the long. 7. September 8th, about 5 h A.M., in lat. 43 24' S., long, by acct. 110 45' E. ; when a chron. showed 9 h 41 m 51", which had been found fast of G.M. noon 28"-8 on August 23rd, and on August 31st showed G.M.T. ; the obs. alt. of Procyon was 22 48' 50" ; ind. cor., + 54" ; height of eye, 30 ft. Find the long. 8. August 30th, about 10 h P.M., in lat. 35 15' K, long. D.R. 169 30' W.; when a chron. showed 9 h 12 m 34 s , which had been found slow of G.M. noon on June 14th 6 m 54 'l, and on July 25th was correct ; the obs. alt. of Fomal- haut was 15 54' 10" ; ind. cor., - 1' 26" ; eye, 25 ft. Find the long. 9. September 2nd, about 10 h 15 m P.M., in lat. 15 14' S., long. D.R. 50 E. ; when a chron. showed 6 h 50 41", which had been found correct for G.M. i 9 2 TEXT-BOOK ON NAVIGATION. noon on July 4th, and on August 18th was slow 3 m 9' ; the obs. alt. of Antares was 25 21' ; no ind. cor. ; eye, 27 ft. Find the long. 10. August 14th, about 2 h A.M., in lat. 40 27' 15" N., long, by acct. 180 \V. ; the obs. alt. of Aldebaran was 22 8' ; ind. cor., 22" ; eye, 38 ft. ; when a chron. showed 2 h 7 m 58', which had been found slow 2 m 33"'6 on June 25th, and on Aug. 4th was 4 m 10' fast. Find the long. 11. August 28th, P.M. at ship, in lat. by mer. alt. of a Ophiuchi in the art. horizon on shore, 15 55' S. , long, by cross-bearings 4 45' W. ; the obs. alt. of Arcturus west of the meridian was 35 ; ind. cor., 4' ; height of eye, 15 ft. ; time by chron. , 5 h 58 m 9", which had been found slow of G.M. noon l h 10 ra 15' on June 29th, and slow l h 3 m 33' on Aug. 8th. Find the long. 12. September 28th, about 6 P.M., in lat. 50 17' N. by mer. alt. of Vega, long. D.E. 161 E. ; the obs. alt. of a Cygni was 68 30' ; eye, 22 ft. ; time by chron., 7 h 25 m 21* '2, which was slow 2*4 of G.M. noon on June 29th, and fast 25' '6 on August 3rd. Find the long. ART. 87. Longitude and Azimuth. As the data used in the computation of Apparent Time at Ship and the Azimuth are the same, and the Formulae very similar, it is most convenient to find the Azimuth at the same time as the longitude by chronometer. Example. Using data as in star chronometer example (Fig. 138), the logs may be conveniently arranged for chron. az. as follows : True alt. 33 30' 13" sec = 10-078911 Lat. 30 26 sec = 10-064382 sec = 10-004382 P.D. 51 18 15 cosec = 10-107641 2)115 14 28 57 37 14 cos = 9-728778 cos = 9-728778 Ghron. remdr. 24 7 1 sin = 9'611299 Az. remdr. 6 18 59 cos = 9-997355 W.H.A. 4 h 38 m 8* XXXI. = 9-512100 2)19-869426 59 21' 52" sin = 9-934713 2 True az. S. 118 43 44 W. EXERCISES. 1. August 31st, about 5 h P.M., at ship in lat. at noon 21' S., long. D.R 136 W. ; the obs. alt. of the Sun's L.L. was 13 13' 40" ; ind. cor., + 7' 21" ; eye, 17 ft. ; bearing by compass W. N. ; time shown by chron., 2 h 5 ra 44', which was slow 25 m 19" on April 12th for G.M. noon, and on July 1st was ll m 55 s slow. Find the long, at sights and noon, if the run since noon was N.E., 58 miles. Also find true azimuth, comp. error, and dev., the var. being 5 40' E. 2. October 1st, about 7 h 15 m A.M. at ship, in lat. 39 30' S., long, by acct. 17 W. ; the obs. alt. of the Sun's L.L. was 16 40' 30" LONGITUDE BY CHRONOMETER. 193 eye, 20 ft.; bearing by compass S.E. by E. \ E. ; time by chron., 8 h 22 m 40 f> 4, which had been found fast l m 41 > 1 for G.M. noon on March 30th, and on June 1st was 2 m 25* '2 fast. Find the long., true bearing, and dev., the var. being 12 15' W. 3. September 1st, about 3 h 30 m P.M. at ship, in lat. at noon 51 S., long. D.R. 175 E. ; the obs. alt. of the Sun's L.L. was 14 52' 20" ; ind. cor., - 7' 10" ; eye, 15 ft. ; bearing by compass N. 50 30' W. ; time shown by chron., 3 h 44 m 40 , which had been found 13 m O f< 3 slow for G.M. noon on February 15th, and on August 1st was 10 m 30 slow. Find long, at sights and noon, the run since noon being S.W. by W. \ W. 53 miles. Also find true bearing, comp. error, and dev., the var. being 16 10' E. 4. September 23rd, about 7 h A.M. at ship, in lat. at noon 32 S., long. D.R. 55 E. ; the obs. alt. of the Sun's L.L. was 10 36' ; hid. cor., - 2' 10" ; eye, 19 ft. ; bearing by compass East ; time shown by chron., 3 h O m 35 , which had been found 5 m 45"'5 slow for G.M. noon on March 31st, and on June 14th was slow 4 m 45 < 5 ; run since sights, N. 59 W. 58 miles. Find long, at sights and noon. The var. from chart being 15 40' W., find true bearing, comp. error, and dev. 5. September 1st, about l h P.M. at ship, in lat. 11 34' 20" S., long. D.R. 144 E. ; the obs. alt. of the Sun's L.L. was 65 2' ; ind. cor., - 7' 17" ; eye, 27 ft. ; bearing by compass N. 40 W. ; time shown by chron., 2 h 32 m 48*, which had been found slow l h O m 10* for G.M. noon on April 1st, and on May 1st was 58 m 40 slow. Find the long, and the var. being 5 45' E., find comp. error and dev. 6. September 22nd, about 8 b 50 m A.M., when the chron. showed O h 20 m 32* ; the obs. alt. of the Sun's ILL. was 30 59' 30" ; ind. cor., 2' 5" off the arc ; height of eye, 30 ft. ; bearing by compass S.E. \ E. ; var., 11 30' W. ; lat. at obs., 42 16' N. ; long, by acct., 70 40' W. ; on June 16th the chron. was slow l h 14 m 30* for G.M. noon, and on August 5th it was l h ll m 8* slow. Required long., true azimuth, and dev. 7. September 23rd, about l h 45 ra P.M., lat. at noon, 40 18' S. ; the obs. alt. of the Sun's L.L. was 42 55' 40" ; eye, 7 ft. ; long by acct., 148 E. ; time by chron., 3 h 53 m 29 1 , which had been found fast 16 m 20 g for G.M. noon on May 4th, and on August 2nd was 10 m 47* fast ; run from noon to obs., W.N.W. 21 miles ; the sun's comp. bearing was N.W., and the var. 10 45' E. Find the long., true bearing, and dev. for the direction of the ship's head. 8. August 5th, about 2 h 30 P.M., in lat. at noon 30 4' 45" S., long, by acct. 94 E. ; the obs. alt. of the Sun's U.L. was 31 32' ; ind. cor., - 4' 44" ; height of eye, 19 ft. ; bearing by compass, N.W. N. ; the chron. showed 8 h 19 m 41 , and had been found to be fast of G.M.T. on May 23rd 7 m 33', and losing daily 2^*. Find the long, at sights and noon ; and if the var. from chart was 14 20' W., find true azimuth and dev. for the direction of the ship's head. Run since noon, S. 40 E. 30 miles. 9. September 1st, about 4 h P.M., in lat. 2 10' K, long. D.R. 55 45' E. ; the chron. showed 10 h 57 m 5' ; the obs. alt. of the Sun's U.L. was 29 11' 40" ; ind. cor., - 50" ; eye, 22 ft. ; bearing by compass N. 88 30' W. ; on March 24th the chron. was slow of G.M.T. 56 m 34', and on May 4th it was slow l h 3 m 27*. Find the long., true azimuth, and dev., the var. being 4 15' W. 10. September 1st, about 4 h 45 m P.M., in lat. at noon 36' N. ; long. D.R. 93 W. ; the obs. alt. of the Sun's L.L. was 18 3' ; height of eye, 32 ft. ; ind. cor., - 3' 27" ; bearing by compass S. 87 15' W. ; the chron. showed 10 h 57 m 27', and had been found correct for M.T.G. on June 30th, and slow 4 m 50 1 on July 27th ; run from noon to obs., S. 35 W. 44 miles. Find the long, at sights and noon ; also if the var. is 2 30' E., find true azimuth and dev. 11. August 20th, about 8 h 20 m A.M., in lat. at noon 12 31' S., long, by acct. 168 W., when a chron. showed 7 h 19 m 19', which had been found 18 ra 19' alow for G.M. noon on June 8th, and gaining daily *9 ; the obs. alt. of the Sun's L.L. was 29 46' 10" ; ind. cor., - 2' 22" ; eye, 45 ft. ; run since obs., 194 TEXT-BOOK ON NAVIGATION. S. 62 W. 34 miles ; Sun's comp. bearing was N. 68 30' E. ; var., 9 15' E. Find long, at obs. and noon, true azimuth, and dev. 12. September 23rd P.M., in lat. at noon 8 44' N., long. by acct. 174 W.- the obs. alt. of the Sun's U.L. was 53 29' 20" ; ind. cor., - 3' 19" ; eye, 41 ft. ; bearing by compass|W. by S. f S.; var., 8 40' E. ; time by chron., l h 47 m 27', which had been found l m 58" slow for G.M. noon on August 18th, and losing daily 1 *1 ; run from noon to obs., N. 28 E. 26 miles. Find long, at sights and noon, true azimuth, and dev. CHAPTER XVIII. SUMNER'S METHOD BY PROJECTION. ART. 88. At any given moment the Sun is vertical over some place on the Earth's surface, and if a circle be drawn with this point as the centre, the Sun would have the same altitude from all points on the circumference of this circle. Advantage is taken of this fact to find the Ship's Position by Projection on a chart. Two points on the circumference of the circle can be found from the Sun's altitude, and two assumed latitudes ; the projec- tion on the chart of the arc of the circle of equal altitudes passing through these two positions would pass through the actual position of the ship. After an interval of time another altitude is observed, and with the same assumed latitudes, two other points on a new circle of position are found, and projected on the chart in the same way. The intersection of the two projected lines gives the ship's position. In practice, the longitudes computed are marked on Mercator's Chart on the parallels of assumed latitudes, and connected by straight lines called Lines of Position. The point of intersection of these lines gives the latitude and longitude of the ship. The Sun's true bearing is at right angles to the line of position at each observation towards the east if A.M., and towards the west if P.M. If the ship changes place during the interval, the first line of position has to be moved (parallel to itself) according to the course and distance sailed. Thus in figure, NESW is plane of observer's horizon, NZS the celestial meridian, X the position of object at first observation, and Y its position at second observation ; Z the zenith, ZX and ZY the zenith distances of object. With X as centre and XZ as radius, describe a circle. This will be a circle of position for first observation, because the zenith of all places having the object at X would be situated on its circumference. Similarly, with Y as centre and YZ as radius, draw a second circle of i 9 6 TEXT-BOOK ON NAVIGATION. position cutting the first in the points Z and C. Hence either Z or C is the position of the observer. With suitably selected observations, these points are so wide apart that the true one may be chosen by comparison with the latitude by account. The lines ZXA, ZYB are portions of vertical circles, showing the bearings of the objects, and are seen to cut the circumferences of the circles of position at right angles, hence the bearing of the object is always at right angles to the line of posi- tion, which is a tangent to the circle of position at Z. ART. 89. Rules for working Sumner problem (1) Find G.M.T.'s. (2) Find correct decimation for each time, and polar distances, correct observed altitudes. (3) With the first altitude and polar distance, using the two assumed latitudes, work as in chronometer problem for two-hour angles, and hence two longitudes. These when projected on the chart and joined would give the first line of position. With the second altitude and polar distance, using the two assumed latitudes, find as before two longitudes. These positions are projected on the chart and joined for the second line of position. The direction of a line of position is taken from the chart, like a course, and may be named in either of the two ways the line trends. The true bearings are at right angles to the lines of position, and are obtained by subtracting the direction of each line from 90 ; and marked towards the east for E.H. A.'s ; and towards the west for W.H.A.'s. The run is laid off from any point of the first line, and where a parallel line to the first line, through the end of the run, cuts the second line is the position of the ship. Example. August 5th, about 9 h 15"* A.M., in long, by account 55 15' W. ; when a chron. correct forM.T.G. showed l h 2 m 21', theobs. alt of the Sun's L.L. was 44 14' 20" ; ind. cor., 2' 7" ; eye, 30 ft. ; and again P.M. at ship on the same day, when the chron. showed 7 h 5 m 23 , the obs. alt. of the Sun's L.L. was 39 21' 25" ; run in the interval, S. \ W. 63 miles. Find the line of position and Sun's true bearing at the first obs., and position of the ship at the second obs. by Sumner's method, by projection, assuming lats. 47 and 47 45' N. SUMNEFS METHOD BY PROJECTION. 197 First M.T.G., August 5 d l h 2 ra 2V. Second M.T.G., August 5 d 7 h 5 m 23'. Dec. 17 9' 52"-4 N. H.V. 40-14 Dec. 17 9' 52"-4 N. H.V. 40 14 41-3 1-03 41-3442 4 45 7-1 17 9 11 90 17 5 7 90 284-994 N.P.D. 72 50 49 N.P.D. 72 54 53 Eq. T. 5 m 52'-6 2 5 62-4 H.V. -23 1 Eq. T. 5 m 52-6 1-6 H.V. -23 7-1 23 5 51 1-633 Obs. alt. 44 14' 20" I.C. -27 Obs. alt. 39 21' 25" I.C. -27 44 12 13 Dip -5 22 39 19 18 Dip - 5 22 44 6 51 S.D. +15 48 39 13 56 S.D. +15 48 44 22 39 39 29 44 Cor. - 53 44 21' 46" 47 45 72 50 49 Cor. - 1 2 39 28' 42" 47 45 72 54 53 T.A. 44 21 46 Lat. 47 P.D. 72 50 49 39 28 42 47 72 54 53 164 12 35 164 57 35 159 23 35 160 8 35 4 Sum 82 6 18 Remdr. 37 44 30 Sees 10-166217 Cosecs 10-019760 Cos 9-137854 Sine 9-786824 82 28 48 38 7 79 41 47 40 13 6 80 4 17 40 35 35 10-172394 10-019602 9-236589 9-813369 10-172394 10-019760 9116849 9-790471 10-166217 10-019602 9-252525 9-810030 9-110655 9-099474 9-248374 9-241954 HJL's 2 h 48 m 24' 2 h 46 m 9' 3 k 19 m 8* 3 h 17 34 . A.T.S. 4 d 2111 36 Eq. T. + 5 52 4 d 21 13 51 + 552 5 d 3 19 8 + 5 51 5 d 3 17 34 + 5 51 M.T.S. 4 d 2117 28 M.T.G. 5 d 1 2 21 4 d 21 19 43 5 a 1 221 5 d 3 24 59 5 d 7 5 23 5 d 3 23 25 5 d 7 5 23 3 4453 34238 3 40 24 3 41 58 (&) 55 39' 30" W. ; (c) 55 6' 0" W. ; Long, (a) 56 13' 15" W. ; (d) 55 29' 30" W. In Fig. 140 NESW is Ratl. Horizon. NZS is Meridian. EQW is Equinoctial. P, pole. Z, zenith. X and Y positions of sun. Complete figure and I 9 8 TEXT-BOOK ON NAVIGATION. also draw circles of position with X and Y as centres and XZ and YZ as radii. AX and AY are alts. ZQ is lat. N BX and BY are decs. PX and PY are pol. dist. ZPX and ZPY are hour angles. SZX and SZY are azimuths. - Projection. Mark the longitudes (a) and (b) on their respective latitudes. The line db is the first line of position. Next set off 63 miles on a S. J W. course from any point of this line, and through the end of the run draw a parallel line to ab. This is the corrected position of the first line in consequence of the ship's change of place during the interval. At the time of taking the second observation the ship was somewhere on this line. Then mark the longitudes (c) and (d) on their latitudes ; cd is the second line of position, and at the time of taking the second observation the ship was somewhere on cd, consequently the intersection of cd with the parallel line drawn before is the position of the ship. To obtain the sun's bearing, take off the corresponding line of position, subtract from 90, and mark. 474-5 FIG. 141. Ans. Direction of first line of position, N. 28 E. or S. 28 W. ; Sun's true bearing, S. 62 E. ; ship's position, lat. 47 23' N.. long. 55 18' W. SUMNER'S METHOD BY PROJECTION. 199 EXERCISES. 1. August 4th, about 8 h 15 A.M. at ship, long, by account 64 E. ; when a chron. showed 4 h ll m 45 s , the obs. alt. of the Sun's L.L. was 36 15' 40" ; and again P.M. at ship on the same day, when a chron. showed 10 h 36 m 45', the obs. alt. of the Sun's L.L. was 43 36' 40" ; chron. correct for G.M.T. ; ind. cor. , + 2' 10" ; height of eye, 39 ft. ; the ship made a true course of W. S. for 32 miles during the interval. Find the first line of position, the Sun's true azimuth, and position of the ship at the time of taking the second observation, by Sumner's method by projection on the chart, assuming lats. 48 30' N. and 49 N. 2. August 9th, about 8 b 50 m A.M. at ship, long, by account 4 W. ; when a chron. which was correct for M.T.G. showed 9 h 8 m 39', the obs. alt. of the Sun's U.L. was 38 47' 10" ; and again P.M. of the same day, when the chron. showed 3 h 16 m 0', the obs. alt. of the Sun's U.L. was 40 40' 30" ; height of eye, 40 ft. ; ind. cor., + 2' 17" ; the ship having made 72 miles S. 60 E. by compass, var. 20 W., dev. 10 W. in the interval. Find the lines of bearing and azimuths at both observations, also ship's position at second observation by Sumner's method by projection on the chart, assuming lats. 49 20' N. and 50 N. 3. August 24th, about 7 h 40 m A.M. at ship, long. Greenwich ; when a chron. 7 m 40 slow of G.M.T. showed 7 h 30 m 4 , the obs. alt. of the Sun's U.L. was 24 33' 50" ; and again A.M. on the same day, when the same chron. showed 10 h 54 ra 1', the obs. alt. of the Sun's U.L. was 49 9' 20" ; ind. cor., 1' 15"; height of eye, 27 ft. ; the ship's true course in the interval was W. by N., 47 miles. Required the line of position and Sun's azimuth at the first observation, and position of ship at the second observation, by Sumner's method, assuming lats. 49 50' N. and 50 20' N. 4. September 1st, long, by account 180 E., with following observations find, by Sumner's method, the first line of position, the Sun's true bearing, and position of ship at second observation, assuming lats. 48 40' N. and 49 20' N. Approx. ship time. Time by chron. Obs. alts. O h 30 m P.M. 12 h 42 m 57' 48 51' 0" ind. cor. , + 1' 10" 4 40 P.M. 4 50 20 Q 19 30 20 eye, 17 ft. Run during interval W.S.W., 6 knots per hour. The chron. was fast 5 m 13* for G.M.T. on May 12th, and gaining daily 2*. 5. September 23rd, about 8 h 40 m A.M., long. D.R. 60 W. ; when a chron. showed 12 h 37 m 55 s , the obs. alt. of the Sun's U.L. was 27 33' ; and again P.M. on the same day, when a chron. showed 6 h 29 59 , the obs. alt. of the Sun's U.L. was 32 53' 40" ; ind. cor., - 3' 57" ; eye, 25 ft. ; the ship having made 23 miles on a true W. by S. f S. course in the interval. Required the line of position and true bearing of the Sun at the first observation, and ship's position at second observation, by Sumner's method by projection on the chart, assuming lats. 46 10' S. and 46 50' S. 6. September 1st, about 8 h 45 ra A.M., long, by account 110 W. ; when a chron. showed 4 h 4 m 29' G.M.T. ; the obs. alt. of the Sun's L.L. was 18 16' 40" ; and again P.M. on the same day, when the chron. showed 9 h 24 m 43", the obs. alt. of the Sun's L.L. was 25 52' 50" ; ind. cor., + 40" ; eye, 12 ft.; the ship having made 42 miles on a true S.E. by S. ^ S. course in the interval. Find the line of position and sun's true bearing when the first altitude was taken, and ship's position at the second observation, by Sumner's method by projection on the chart, assuming lats. 49 15' S. and 49 45' S. 7. August 1st, about 5 b 30 m A.M., long, by account 166 E. ; when a chron. correct for G.M.T. showed 7 h 39 m 17 s , the obs. alt. of the Sun's L.L. was 20 200 TEXT-BOOK ON NAVIGATION. 59' 40", and again A.M. on the same day, when the chron. showed 12 h 7 m 57% the obs. alt. of the Sun's L.L. was 57 6' 10" ; ind. cor., - 3' 7" ; eye, 36 ft. ; the ship having made a true N. by W. W. course 40 miles in the interval. Find the first line of position, the Sun's true bearing, and ship's position at the second observation by Sumner's method by projection on the chart, assuming lats. 49 50' K and 50 20' N. 8. September 29th, in long, by account 180 E. ; with following observa- tions, assuming lats. 47 N. and 47 50' N., find by Sumner's method first line of position and Sun's true bearing ; also position of ship at second observation. Approx. ship time. Time by chron. Obs. alt. Sun's L.L. 7 h 30 m A.M. 7 h 29 m 59 10 53' 20" eye, 16 ft. 3 30 P.M. 3 26 39 26 48 20 ind. cor., + 1' 15" Run between observations, W. by S. S. 86 miles. The chron. was 22 m 12 fast of G.M.T. on Sept. 1st, and gaining 2" < 5 daily. CHAPTER XIX. THE CHART. ART. 90. Charts used by Navigators are all constructed on Mercator's principle (except those of Harbours, Bays, etc.). Their characteristic is that the spaces between the Parallels of Latitude increase towards the Poles, whilst the Meridians of Longitude are also parallel instead of converging as they do on a globe. The reason for this construction is explained in Chap. I. Latitudes are marked on the Graduated Meridians at the sides of the chart, and longitudes on the Graduated Parallels at the top and bottom. Soundings in fathoms for Mean Low Water of Ordinary Spring Tides are given in small figures, and Roman numerals at various places give the Time of High Water, " Full and Change." The nature of the bottom is also shown by small letters and abbreviations. Since 1911 the pattern of Compasses adopted on Admiralty Charts has been altered to show both Magnetic and True. The outer or true circle being graduated from to 360. The inner or magnetic circle is divided into quadrants, each one reading from to 90, and to quarter points. The variation is marked in the circle for a given indicated year, and its decrease per annum. True north is indicated by a star, magnetic north by &fleur de Us. Bearings are given both True and Magnetic, thus 161 (S. 12 E. mag.). Variation Charts are charts of the oceans or of the globe with lines of equal variation drawn on them at intervals of one or two degrees. From these charts the variation at ship can be found, provided the latitude and longitude are known. When out of sight of land, the ship's position is fixed by the latitude and longitude found from observation of heavenly bodies. Near the coasts the position can be obtained from soundings and from bearings of lights and other known objects. ART. 91. Using the following Deviation Tables it will now be shown how to convert the course from the chart into the corresponding compass course. 202 TEXT-BOOK ON NAVIGATION. Ship's head by compass Card A. Deviation. Card B. Deviation. N. 11 E. 36 W. N. by E. 16 E. 36 W. N.N.E. 21 E. 30 W. N.E. by N. 27 E. 27 W. N.E. 31 E. 24 W. N.E. by E. 31 E. 21 W. E.N.E. 28 E. 17 W. E. by N. 26 E. 12 W. E. 24 E. 6 W. E. by S. 20 E. 2 E. E.S.E. 15 E. 6 E. S.E. by E. 9 E. 12 E. S.E. 4 E. 17 E. S.E. by S. 1 E. 22 E. S.S.E. 2 W. 27 E. S. by E. 5 W. 32 E. S. 8 W. 36 E. S. by W. 11 W. 38 E. S.S.W. 14 W. 37 E. S.W. by S. 18 W. 35 E. S.W. 21 W. 32 E. S.W. by W. 24 W. 27 E. W.S.W. 27 W. 20 E. W. by S. 30 W. 12 E. W. 26 W. 5 E. W. by N. 23 W. 3 W. W.N.W. 21 W. 10 W. N.W. by W. 19 W. 16 W. N.W. 16 W. 22 W. N.W. by N. 11 W. 27 W. N.N.W. 4 W. 31 W. N. by W. 4 E. 34 W. It is seen that the deviations are given for the compass directions of the ship's head ; when the deviation is applied to the corresponding ship's head by compass, the magnetic direction of the ship's head is obtained : thus Card A. Comp. co. N.N.E. dev. 21 E. gives mag. co. N. 44 E. Card B. N.N.E. 30 W. N. 7 W. To obtain the deviation and compass course for a given magnetic course, two consecutive compass courses must be selected from the Table, so that when converted into magnetic THE CHART. 203 courses, one will be greater and the other less than the given magnetic course, then by proportion the deviation, and hence the compass course may be obtained. Thus using Card A: Find the Compass courses for mag. courses, (a) N. 70 E. (6) S. 12 W. (c) N. 15 W. EXAMPLES. (a) N. 70 E. Comp. co. Dev. Mag. co. N.E. by N. 27 E. N. 61 E. N.E. 31 E. N. 76 E. (Diff. of dev.) 4 15 (Diff. of courses) (Diff. of N. 61 E. and N. 70 E.) 9 15)36 2 to add to 27 E. Hence deviation is 29 E. Comp. co. N. 70 E. Dev. 29 E. Mag. co. N. 41 E. (6) S. 12 W. Comp. co. Dev. Mag. co. S.S.W. 14 W. S. 9 W. S.W. by S. 18 W. S. 16 W. (Diff. of dev.) 4 7 (Diff. of courses) (Diff. of S. 9 W. and S. 12 W) 3 7)12 2 to add to 14 W. Hence deviation is 16 W. Mag. co. S. 12 W. Dev. 16 W. Comp. S. 28 W. (c) N. 15 W. Comp. co. Dev. Mag. co. N.N.W. 4 W. N. 27 W. N. byW. 4 E. N. 7 W. (Diff. of dev.) 8 20 (Diff. of courses) (Diff. of N. 27 W. and N. 15 W.) 12 20)96 5 to subtract from 4 W. Hence deviation is 1 E. Mag. co. N. 15 W. dev. 1 E. Comp. co. N. 16 W. NOTE 1. When the chart course is true, the variation must first be applied to find the magnetic course. The bearings must also be corrected for variation. 204 TEXT-BOOK ON NAVIGATION. NOTE 2. The deviation can generally be found by inspection, as the difference of deviation between two adjacent points is usually small. In the tables the deviations are purposely increased to illustrate the method. EXERCISES. 1. Using Card A. : find Comp. co. corresponding to following Mag. co. : (a) N. 79 E. (6) East, (c) S.E. (d) S. 28 E. (e) S. 85 W. 2. Using Card B. : find the courses to steer by compass to make good following chart courses : (a) North. (&) East, (c) N. 83 W. (d) West. ANSWERS. 1 (a) N. 48 E. (6) N. 60 E. (c) S. 52 E. (d) S. 27 E. (e) N. 73 W. 2. (a) N. 26 E. (ft) S. 86| E. (c) N. 83 W. (d) S. 80 W. ART. 92. When a vessel is steaming along a given course on the chart at a known rate, her estimated or expected D.R. position on that course can always be set down as the distance run is known, the observation position can be determined by station pointer, cross bearings, or two bearings and the run between, the course and distance from the D.R. to the obs. position is the set and drift of the current experienced. Thus in Fig. 142 vessel FIG. 142. proceeding from X to Y, at the end of a given time expects to be at E ; B and C are two conspicuous objects, at this time the compass bearings of B and C are taken and corrected, and then laid on the chart in the directions BO and CO, O the point of intersection of these bearings is the observed position of the ship, EO represents the set and drift of the current. Again, at a given time let the bearing of an object A be taken, corrected and laid off as AL, after a known interval another bearing of A is taken, corrected and laid off as AF, supposing the current found to be still acting, the position of the ship at the end of the time may be found thus : Measure LM along LY equal to the distance the ship steams in the interval, set off MN in the direction of the current, and equal to the drift of the current in the interval, through N draw a parallel line to the first bearing, AL : to cut the second bearing in F. F is the position of the ship allowing for the current. THE CHART. 205 FIG. 143. To set a course to counteract a given current is another problem of great importance to the navigator. In the previous paragraph the effect of a known current on the position of the ship at the end of the run has been considered ; now it is required to find what course must be steered by a ship through a known current in order that she may make a given position. In Fig. 143 the vessel is at P and wishes to get to Q 30*5 miles away : her speed is 10 knots; the current sets in the direction PR at 3 knots. Measure 3-tenths inch along PR to R, and with R as centre and 10-tenths as radius cut the line PQ in S, then RS is the direction to steer to counteract the current, that is for the ship to get to point S on the line PQ, and PS is the distance made good in one hour : this measured gives 11*8 miles. 30-5 Hence time taken to arrive at Q = =-=-5 hours 11 o = 2*5 hours. Total distance for ship to steam = 2*5 X 10 miles = 25 miles. Or by drawing a line through Q parallel to RS to cut direction of current in T, and measuring TQ, will also give total distance for ship to steam = 25*5 miles. The problem may be solved trigonometrically, as in triangle PRS are given PR, RS, and angle SPR, to find angle RSP by " rule of sines." This angle shows how much nearer south the course to steer is than the direction of PQ. N.B. Find the deviation for the new course, and apply to it to obtain the compass course. Again, in Fig. 144 let the vessel be going along the line AB, and at the end of a given time, say 1J hours, finds her position to be at C,find the course she must steer to arrive at B. As the speed of the ship is known, her DR position would be at a fixed point on AB, say E, FIG. 144. 2 o6 TEXT-BOOK ON NAVIGATION. consequently the set and drift of the current acting is the direction and length of EC in 1J hours. Join CB, this is the course required to be made good. Produce EC, make CD = CE, this represents the current for 1 hours, with D as centre, radius equal to AE the distance the ship steams in the same time, describe an arc cutting CB in F. Then the direction of DF is the required course, and CF is the distance made good in 1J hours, whence by simple proportion, the time taken to arrive at B, or the distance towards B made in any required time, may be found. N.B. When laying off a course to counteract a current, always allow the current to act, and the ship to steam for the same time, 1 hour, 2 hours, etc., as may be selected. EXERCISES. Using St. George's Channel Chart, Devia. Card B. 1. (a) Find the course to steer by compass from Bell Buoy, St. Tudwall Rds. , to New Quay Lt. Also the deviation, variation, and distance. (b) With the ship's head on the above compass course, Pengarn Mt. bore by compass N. 56 E. and Aberdovey Lt. bore S. 84 E. by the same compass. Find the ship's position, and set and drift experienced since course was set, ship steaming 6 knots and taking 2 h. 40 m. to reach position. (c) With the ship's head as above, Aberystwith Lt. bore by compass S. 65 E. and after continuing 3 h. 10 m. on the above-named course it bore N. 49 E. by the same compass. Find the ship's position and distance from New Quay Lt., making allowance for current found in (6). (d) Find the course to steer by compass from Bell buoy to New Quay in the current found in (b) vessel making 6 knots : and the distance she then makes good towards New Quay in 3 hours 50 minutes. How far would ship have to steam to reach New Quay ? (e) Horizontal angles, find position by station pointer : St. Tudwall Lt., 86| ; Aberystwith Lt., 62; New Quay Lt. (/) What do the small numbers indicate that are placed about the chart and at what time of tide ? (d) What do the Roman numerals indicate that are occasionally seen near the coast and in harbours ? (K) How would you find approximately the time of high water at any place, the Admiralty tables not being at hand, nor any other special tables available ? 2. (a) Find the courses to steer by compass and distance on each course from Dunmore Lt., Waterford Hr., to Q off Sth. Arklow Lt. V. on G. line 3} miles from Lt. V. Note. Alter course when Coningbeg Lt. bears 360, distance 3 miles, and again when Tuskar Rk. Lt. bears 288, distance 2 miles. (6) With ship's head on the 2nd compass course, Sth. Pt. S. Saltee bore by compass N. 61 W., and Carnsore Pt. bore N. 46 E. by the same com- pass. Find the ship's position, set and drift experienced, if the expected position was with Barrel's Lt. V. bearing 251, distance off 3 miles. (c) With the ship's head on the 3rd compass course, Lucifer Lt. bore by compass N. 17 E., and after continuing 56 minutes on the above-named course, it bore N. 62 W. by the same compass. Find the ship's position and distance from Black water Lt., making due allowance for a current which set N.E. mag. 3 knots, vessel steaming 15 knots. THE CHART. 207 (d) Find the courses to steer by compass from Dunmore Lt. to Q in a current which set east mag. 2 knots, the vessel making 12 knots, and the distance she then makes good towards Q in 3 hours 45 minutes. (e) Horizontal angles, find position by station pointer, Hook Pt. Lt., 58; Coningbeg Lt. V., 24 ; Tuskar Lt. 3. (a) Set the courses to steer by compass from X, lat. 51 35' N. long. 5 18' W., to Y., lat. 52 24J' N. long. 5 2' W., ship to pass 3 miles off S. Bishop Lt. ; then alter course to reach Y, with Cardigan Bay L.V. bearing 90, distance 1 miles ; maintaining your distance of 3 miles from S. Bishop Lt., and give distance run on each course. (b) With the ship's head on 1st compass course, St. Anne's Hd. Lt. bore by compass S. 17 E., and Small's Lt. bore N. 58 W. by the same com- pass. Find the ship's position, set and drift experienced, the ship having run 49 minutes on the course, steaming 11 knots. (c) With ship's head on 2nd compass course, Strumble Hd. bore by compass S. 39 E., and after continuing 1 hour 38 minutes on above course, it bore S. 31 W. by same compass. Find the ship's position and distance from Cardigan Bay L.V., allowing for a current which set S. 63 W. mag. 2\ knots, vessel steaming 11 knots. (d) After running on second course 1 hr. 22 mins., ship making 11 knots, her position was found to be lat. 52 2' N., long 5 25' W. Find course to steer by compass from this point to Y, allowing for a similar current to that experienced during- earlier part of run ; and distance made good in 2 hrs. 50 mins. towards Y. How long would she take to reach Y. ? CHAPTER XX. 8)65 USE OF NAPIER S DIAGRAM. ART. 93. In the list of questions proposed to candidates for Masters' Certificates in the Board of Trade Regulations, Nos. 11, 12, 13, and 14 refer to the use of Napier's Diagram. Example. No. 11. Having taken the following compass bearings of a distant object, find its Correct Magnetic Bearing, and thence the deviation. Correct magnetic bearing required : S. 8 E. g. 1 q gO 5 ' Ship's head by ls.T9W. standard ) compass. 12 E. P Bearing of distant object. Deviation required. Ship's head by standard compass. Bearing of distant object. Deviation required. S. 91 W. rN.89 W. N. S. 4 E. 4 W. S. S. 13 E. 5E. N.E. S. 8 W. s.w. S. 22 E. 14 E. } N. 8 W. E. S. 4 W. 12 W. W. S. 20 E. 12 E. . 12 E. S.E. W A. TT. S. 1 W. 9 W. N.W. S. 11 E. 3E. No. 12. With the deviation as above, construct a curve of deviation on a Napier's Diagram, and give the courses you would steer by the Standard Compass to make the following correct magnetic courses : Magnetic courses : S.S.W. W.N.W. N.N.E. E.N.E. S.S.E. Compass courses required : S. 14 W. N. 77 W. N. 29 E. N. 79 E. S. 20 J E. No. 13. Supposing you have steered the following courses by the Standard Compass, find the correct magnetic courses from the above curve of deviations : Compass courses : W.S.W. N.N.W. E.N.E. S.S.E. Magnetic courses required : S. 82 W. N. 23 W. N. 57 E. S. 25 E. No. 14. You have taken the following bearings of two distant objects by your Standard Compass as above, with the ship's head at W. \ S. by compass : find the correct magnetic bearings. Bearings by compass : W. by S. N. | W. Magnetic bearings required : N. 89 W. N. 3 E. Explanation. In No. 11 the Correct Magnetic Bearing is the mean of the eight observed bearings. (a) When the bearings all read one way from S. or N., the mean is found by adding the eight bearings together and dividing the sum by 8. USE OF NAPIER'S DIAGRAM 209 (6) When the bearings read different ways, viz. some towards E. and some towards W.,then the mean is found by summing up the bearings of the same Name separately (as in the above example), and dividing the difference of the totals by 8. (c) The deviations are the differences between the correct magnetic bearing and each of the eight bearings by Compass, it being E. dev. when the correct magnetic reads to the right of the compass bearing, and W. when to the left. ART. 94. Napier's Diagram and Curve of Deviations. The deviations found for the eight equidistant courses are taken with a pair of dividers from the middle line, which is graduated like a compass card, and set off from the graduated line on the dotted lines for the given courses to the left for W. dev., and to the right for E. dev. A curve is then drawn, as in the diagram (Fig. 145). The deviation for any course (compass or magnetic) may now be found by simply measuring the part of the line (dotted or plain) intercepted by the curve. For No. 12, place one foot of the dividers on the given course, and extend the other on the plain line to the curve. Keeping one foot on the middle line, turn the other up or down according as the line inclines upwards or downwards, and read off the required compass course. For No. 13, proceed exactly in the same way as in No. 12, except that the dotted lines only are to be used. NOTE. The advantage of the diagram and curve is that it affords a ready mechanical way of changing magnetic courses to compass courses, and vice versa, and so avoids the risk of error in applying deviation by adding or subtracting. For No. 14, find the Deviation for the given direction of ship's head, by measuring the part of the dotted line intercepted by the curve, and apply it in the usual way to the bearings by compass, viz. E. to right, and W. to left. Example. (a) Magnetic bearing required : S. 53 45' W. Ship's head by compass. Bearing by compass. Deviation. Ship's head by compass. Bearing by compass. Deviation. N. S. 75 W. 21 15' W. S. S. 34 W. 19 45' E. N.E. S. 64 W. 10 15' W. s.w. S. 31 W. 22 45' E. E. S. 56 W. 2 15' W. W. S. 49 W. 4 45' E. S.E. S. 50 W. 3 45' E. N.W. S. 71 W. 17 15' W. NOTE. The magnetic bearing required is the mean of all the bearings by compass, found by adding them and dividing by eight. The deviations are then marked on the dotted lines of a Napier's diagram, and a curve drawn. (6) Magnetic courses : N.N.W. S.S.E. W.N.W. E.S.E. Compass courses required : N. 1 W. S. 31 E. N. 53 W. S. 68 E. P 210 TEXT-BOOK ON NAVIGATION. DIAGRAM. "WEST?" "EAST. USE OF NAPIER'S DIAGRAM 211 Explanation. Place one leg of the compasses on the given course, and extend the other along the plain line to the curve, and turn up- wards if the line tends upwards, but downwards if it tends down- wards, and read off the required course. (c) Compass courses given : N.N.E. E.N.E. S.S.W. W.S.W. Magnetic courses required : N. 6 E. N. 62 E. S.W. S. 83 W. Explanation. Place one leg of the compasses on the given course, and extend the other along the dotted lines to the curve, and turn up or down, as the line tends. (d) Ship's head by compass : N.E. E. Compass bearings : S.E. by S. N.N.W. Magnetic bearings required : S. 44 E. N. 32 W. Explanation. Find the deviation on the given compass course from the dotted line, and apply it to both bearings (east to right, west to left). EXERCISES ON NAPIER'S DIAGRAM AND CURVE. 1. (a) Correct magnetic bearing required : Ship's head by compass. Bearing of distant object. Deviation. Ship's head by compass. Bearing of distant object. Deviation. N. N. 83 E. S. N. 50 E. N.E. S. 86 E. S.W. N. 34 E. E. E. w. N. 40 E. S.E. N. 75 E. N.W. N. 61 E. Find the correct magnetic bearing and the deviations. (6) Correct magnetic courses : N.E. J E. E.byN. S.E. by S. 8. N.JW. Required courses to steer. (c) Courses steered : N.W. f W. S. E. by S. N. by E. Required correct magnetic courses. (d) Ship's head by compass : W.S.W. Bearings of distant object : S.S.W. and N.E. Required the correct magnetic bearing. 2. (a) Correct magnetic bearing required : Ship's head by compass. Bearing of distant object. Deviation. Ship's head by compass. Bearing of distant object. Deviation. N. S. 85 W. S. N. 73 W. N.E. S. 70 W. S.W. N. 68 W. E. S. 88 W. W. N. 74 W. S.E. N. 76 W. N.W. N. 79 W. Required the correct magnetic bearing and the deviations. (6) Correct magnetic courses : E. } N. E.S.E. S. W. N.W. Required courses to steer. (c) Courses steered : N. S.E. | S. S.W. $ W. N. 35 W. Required correct magnetic courses. 212 TEXT-BOOK ON NAVIGATION. (d) Ship's head S. 65 E. Bearings by compass, W. 10 S. N. 40 E. Required the deviation and the correct magnetic bearings. 3. (a) Correct magnetic : Ship's head by compass. Bearing of distant object. Deviation. Ship's head b/ compass Baaring of distant object. Deviation. N. S. 4W. S. S. 34 W. N.E. S. 1E. s.w. S. 32 W. E. S. W. S. 11 W. S.E. S. 22 W. N.W. S. 9W. (6) Correct magnetic courses : N. \ W. W. by S. | S. S. 70 E. E. 24 N. Courses to steer by compass. (c) Courses steered by compass : W.N.W. * N. N.E. J N. S. | E. N. 40 W. Correct magnetic courses. (d) Ship's head at W. by N. by compass. Bearings by compass (of distant objects) : S. W. j W. N.W. by W. J W. Correct magnetic bearings. 4. (a) Correct magnetic. Ship's head by compass. Bearing of distant object. Deviation. Ship's head by compass. Bearing of distant object. Deviation. N. N. 32 W. S. N. 4E. N.E. N. 27 W. S.W. N. 4E. E. N. 6E. W. N. 10 W. S.E. N. 2E. - N.W. N. 15 W. E. 5 S. E. by N. N. by E. J E. S.E. } S. (6) Correct magnetic courses : S. 50 W. N. Courses to steer by compass. (c) Courses steered by compass : W. by S. S. by W. Correct magnetic courses. (d) Ship's head at N.N.W. by compass. Bearings by compass (of distant objects) : W.N.W. S.S.W. Correct magnetic bearings. 5. (a) In the following table give the correct magnetic bearing, and thence the deviation. Correct magnetic : Ship's head by standard compass. Bearing of distant object by standard compass. Deviation required. Ship's head by standard compass. Bearing of distant object by standard compass. Deviation required. N. N. 62 E. S. E. 15 S. N.E. E. 24 N. s.w. S. 81 E. E E. 7N. W. E. 5N. S.E. S. 77 E. N.W. N. 67 E. USE OF NAPIEFS DIAGRAM. 213 (6) With the deviations as above, give the courses you would steer by the Standard Compass to make the following courses correct magnetic : Magnetic courses : N.E. \ E. E. S. S. \ W. N. W. Compass (c) Supposing you have steered the following courses by the Standard Compass, find the correct magnetic courses. Compass courses : N.E. \ N. S.E. \ S. W. } N. N.W. \ N. Magnetic. (d) You have taken the following bearings of two distant objects by your Standard Compass as above : with the ship's head at N. ^ W., find the bearings correct magnetic. Compass bearings : S.S.E. \ E. E. J N. Correct magnetic. 6. (a) In the following table give the correct magnetic bearing of the distant object, and thence the deviation I- Correct magnetic : Ship's head by standard compass. Bearing of distant object by standard compass. Deviation required. Ship's bead by standard compass. Bearing of distant object by standard compass. Deviation required. N. S. 32 O'E. S. S. 6 50' W. N.E. S. 445'E. s.w. 8.26 O'E. E. S. 16 0' W. W. 8.43 O'E. S.E. S. 21 30' W. N.W. 8. 44 35' E. (6) With the deviations as above, give the courses you would steer by the Standard Compass to make the following courses correct magnetic : :: N. bv E. E. E. * N. S.S.W. 4- W. N.W. ' Magnetic : "N. by E. Courses to steer. JW. (c) Supposing you have steered the following courses by the Standard Compass, find the correct magnetic courses. Standard : N.N.E. E. E. by N. $ N. S.S.E. 1 E. N.W. N. Magnetic. (d) You have taken the following bearings of two distant objects by your Standard Compass as above : with the ship's head at S.E. J S., find the bearings correct magnetic : Compass bearings : N.E. $ E. N. by W. J W. Correct magnetic. 7. (a) Magnetic bearing required Ship's head by compass. Bearing by compass. Deviation. Ship's head by compass. Bearing by compass. Deviation. N. N. 56 E. S. N. 55 E. N.E. N. 48 E. s.w. N. 64 E. E. N. 45 E. W. N. 68 E. S.E. N. 49 E. N.W. N. 63 E. (6) Magnetic courses : N. E. S. 70 E. S. 55 W. N. 8 W. Compass courses required, (c) Compass courses : S.S.W. N.W. W. N. S. 82 E. Magnetic courses required, (c?) Ship's head by compass : S.W. Compass bearings : E. ^ N. N. by E. Magnetic bearings required. 214 TEXT-BOOK ON NAVIGATION. 8. (a) Magnetic bearing required Ship's head by compass. Bearing by compass. Deviation. Ship's head by compass. Bearing by compass. Deviation. N, N. 44 W. S. S. 79 W. N.E. N. 47 W. s.w. S. 89 W. E. N. 80 W. W. N. 69 W. S.E. S. 76 W. N.W. N. 52 W. (6) Magnetic courses : S. \ E. S.W. \ S. N. 80 W. N. 75 E. Compass course required. (c) Compass courses : N.N.W. f W. W. by S. \ S. S. N. 80 E. Magnetic courses required. (d) Ship's head by compass : W. Bearings by compass : E. 11 N. N. 6 W. Magnetic bearings required. 9. (a) Magnetic bearing required Ship's head by compass. Bearing by compass. Deviation. Ship's head by compass. Bearing by compass. Deviation. N. N. 32 W. S. N. 20 E. N.E. N. 20 W. S.W. N. 9E. E. N. 5E. W. N. 15 W. S.E. N. 17 E. N.W. N. 31 W. (6) Magnetic courses : S. 65 W. N.W. N. 32 E. E. Compass courses required. (c) Compass courses : N. N.E. by N. E. by S. S.W. J S. Magnetic courses required. (d) Ship's head by compass E. Bearings by compass : W. 1 N. S. by W. f W. Magnetic bearings required. 10. (a) Magnetic bearing required Ship's bead by compass. Bearing by compass. Deviation. Ship's head by compass. Bearing by compass. Deviation. N. 8. 43 W. S. W. N.E. S. 50 W. S.W. S. 83 W. E. S. 65 W. W. S. 65 W. S.E. S. 80 W. N.W. S. 45 W. (6) Correct magnetic courses: S.E. by E. N.N.E. N.W. i W. S. 10 W. Compass courses required. (c) Compass courses : N.W. by W. N. by W. N.E. J E. S. 60 E. Magnetic courses required. (d) Ship's head: E. $ N. Bearings by compass : S.E. f E. N.E. by N. Magnetic bearings required. USE OF NAPIER'S DIAGRAM 11. (a) Magnetic bearing required 215 Ship's head by compass. Bearing by compass. Deviation. Ship's head by compass. Bearing by compass. Deviation. N. 8. 2W. S. S. 29 K. N.E. 8. 7W. s.w. S. 26 E. E. S. 6E. w. S. 15 E. S.E. S. 16 E. N.W. S. 3E. (6) Magnetic courses : W. N. 30 E. S. 56 E. S. 3 W. Compass courses required, (c) Compass courses : E. N.E. by N. \ N. N.W. by W. S. 3 E= Magnetic courses required, (c?) Ship's head by compass : S.E. Bearings by compass : N. E. by N. N. Magnetic bearings required. 12. (a) Magnetic bearing required Ship's hrad by compass. Bearing by compass. Deviation. Sliip's head by compass. Bearing by compass. Deviation. N. N. 8. N. 21 W. N.E. N. 7E. S.W. N. 17 W. E. N. 10 W. W. N. 18 E. S.E. N. 13 W. N.W. N. 16 E. (6) Magnetic courses : E.S.E. S.S.E. W. by N. N.E. Compass courses required. (c) Compass courses : N. W. W. N. N.E. N. E. Magnetic courses required. (d) Ship's head by compass : N.E. Bearings by compass : S.W. ^ S. S.E S. Magnetic bearings required. CHAPTER XXL Fio. 146. GREAT CIRCLE SAILING. ART. 95. A circle is really a plane surface bounded by a line called the circumference, but it is common to speak of this line as the circle. A great circle of the globe is one which divides it in halves. Hence the Equator is a great circle, and as all great circles bisect one another, every other great circle must intersect the Equator in two opposite points (Fig. 146). A meridian of longitude cuts the Equator at right angles, but other great circles cross it more or less obliquely, and go to equal distances north and south. The two points on the Great Circle which are farthest from the Equator are called the Vertexes, and the meridian of longitude which passes through the Vertex is called the Meridian of Vertex. Now, the shortest distance between any two places on the globe is the arc of the Great Circle passing through them. Therefore, when this route is practicable, the navigator would save time in a sailing vessel, and coals as well as time in a steamship, by following the Great Circle track. Illustration (Fig. 147). Let A be the starting-point, and B the destination, AVB the arc of the Great Circle passing through the two places, P the north pole, PA and PB the meridians of longitude through A and B, and V the highest latitude reached on the parallel LL X and meridian of vertex PVM ; then The L APB is the Difference of Longitude of A and B. The L PAB is the Initial Course. The L PBA Final Course. AB Distance on the Great Circle. vV is the Latitude of Vertex. The L APV (or arc av) is the Longitude of A from Vertex. 147, GREAT CIRCLE SAILING. 217 The z BPV (or arc bv ) is the Longitude of B from Vertex. PA and PB, Co-latitudes of A and B. PV, Co-latitude of Vertex. 1, 2, 3, 4, 5, 6, points on the Great Circle. N.B. The Meridian of Vertex is at right angles to the Great Circle. Inspection of the figure shows that the course from A would first be northerly until the Vertex is reached, where it would be due east, and then southerly to B. From B to A these directions would be reversed. In proceeding along the Great Circle the course would change at each instant, but as this would not be possible in navigation, Great Circle sailing in practice consists in fixing the latitude and longitude of a number of points on the circle, and then shaping courses from point to point by Mercator's sailing. The greater the number of points chosen the nearer would the ship's track approximate to the Great Circle. The co-latitudes PA and PB and the distance AB are three sides of a spherical triangle, of which two sides (the co-latitudes) and the included angle (difference of longitude) are known. From these data all the other details can be found by the principles of spherical trigonometry. These details are (1) The initial and final courses. The distance on the Great Circle. The latitude and longitude of Vertex. The longitude from Vertex of A and B. The latitudes and longitudes of points on the Great (2) (3) (4) (5) Circle. (6) The courses and distances by Mercator from point to point. ART. 96. Great Circle Sailing (Fig. 148). Let AB be the arc of a great circle from the starting- point (A) to the destination (B), and CA, CB arcs of meridians of longitude ; then the angles A and B are the initial and final courses, the angle C is the difference of longitude, and CA, CB the co-latitudes of A and B. Also, let the arc CV be drawn at right angles to AB ; then V is the ver- tex, and CV the co-latitude of vertex, putting the letters a, b, c for the sides respectively opposite the angles A, B, C, and using the formulae for oblique- angled spherical triangles when two sides and included angle are known, we have FIG. 148. TEXT-BOOK ON NAVIGATION. Tan i(A + B) = cos \ ( ? " % . cot JO cos %(a 4- 6) Tan J(A - B) = s ] n & a ~~ ^ . cot J.C sin (a 4- o) 'he greater angle (A) = i(A + B) + J(A - B) lesser (B) = J(A + B) - J(A - B) ! cos A(a + 6) -ip cos Ac = ff; -^ . sin *U cos *(A 4- B) = cos \(a 4- 6) . sec J(A + B) . sin ^C The latitude of vertex, longitude from vertex, and the latitudes of points on the great circle are found by applying the rules for the solution of right-angled spherical triangles. 1. To find "Latitude of Vertex" In the right-angled triangle ACV are known the angle A and the side AC. Taking CV (co-latitude of V) as middle part, the opposite parts are comp. of A and comp. of AC ; therefore sin CV = cos (comp. of A) x cos (comp. of AC) = sin A x sin AC 2. To find " Longitude from Vertex " of A, viz. the angle ACV. In the same triangle, taking comp. of AC as middle part, comp. of ACV, and the comp. of A are adjacent parts ; .*. sin (comp. of AC) = tan (comp. of ACV) . tan (comp. of A) Hence cos AC = cot ACV . cot A or cot ACV = cos AC . tan A Similarly, the latitude of V and " longitude from V " of B can be found from the triangle BCV. The longitude of the vertex can be found by applying the " longitude from vertex " to the longitude of A or B. 3. To find the Latitude of a Point (p) whose " Longitude from Vertex " (is assumed) (viz. the angle pCV). In the right-angled triangle pVC, the side CV and the angle pCV being known sin (comp. ofpCV) = tan CV x tan (comp. of Cp) or cos >CV = tan CV x cot Cp //"\c! / y)i7 if transposing, cot Cp = ^ = cos pCV x cot CV tan OV Similarly, the latitudes of other points on the great circle track can be found. The courses and distances from point to point are calculated as in Mercator's sailing. ART. 97. Naming the Courses. When the Vertex falls on the portion of the G.C. arc joining the two places, it is said to be " in ; " when the Vertex is not situated on this arc, but on the arc produced, it is said to be " out." This can be decided GREAT CIRCLE SAILING. 219 when the initial and final cours^ have been computed; for if they are on the same side of 90 the vertex is " in," but if one is greater and the other less than 90 it is "out," and is in every case on the same side of the Equator as the greater latitude. The first course is always of the same name, N. or S., as the Pole used in the calculation ; the last course of contrary name to this ; both are named E. or W. like a Mercator course. Example. VERTEX " IN. " Required the initial and final courses and the distance on the Great Circle from (A) San Lucas to (B) Yokohama; also the position of the Vertex, and position of points on G.C. differing 10 in longitude. A lat. 22 52' N. .*. PA= 67 8' long. 109 53' W- B 35 26 N. /. PB = 5434 139 39 E. = 12142 = 6051 249 32 360 FIG. 149. PA-PB = 12 34 P = 110 28 |(PA-PB) = 6 17 = 55 14 In triangle PAB, given PA, PB, and P, required A, B, and AB. Tan 2 - cos A(PA + PB) ' cot 2 B-A sinl(PA-PB) ,P CQ AB = cos|(PA. A , ^ 2 cos (B + A) 2 BX , 60 51' sec = 10-312384 cosec = 10 "058813 cos= 9*687616 |(PA-PB)= 6 17 cos= 9-997383 sin= 9-039197 ? = 55 14 cot= 9-841457 cot= 9-841457 sin= 9-914598 KB + A) = 54 46' 45" tan = 10151224 B-A) = 4 58 30 ......... = S. 59 45 15 W. Final course = N. 49 48 15 W. Initial tan= 8-939467 sec = 10-239029 46 4' 2" cos= 9-841243 2 AB=92 8 4 = 5528-1 miles. Since A and B are of same affection, vertex is "in." To find position of vertex "V." In triangle PAV, V = 90. Given PA and A, required PV = co-lat. V, P = d long, of V from A. Sin PV = cos (co. A) .. cos (co. PA) and sin (co. PA) = tan (co. P) . tan (co. A) = sin A . sin PA or cos PA = cot P . cot A .*. cot P = cos PA . tan A 220 TEXT-BOOK ON NAVIGATION. A = 49 48' 15" sin = 9 '883004 PA = 67 8 sin = 9-964454 tan = 10-073174 cos = 9-589489 PV = 44 44 Lat. V = 45 16' N. sin = 9-847458 ... 65 18' cot = 9'662663 Long. A 109 53' W. /. Long. V 175 11 W. To find "points" on the G.C. differing 10 in long. Since no particular neridian is stated in the question as the starting-point, the simplest way is 30 calculate from the vertex, using the angles at P or differences of long, as 10, 20, 30, etc. As the total d long, between A and V is 65 18', the number of points will be f g- = 6. The latitudes of points equally distant from the vertex are the same ; hence, these lats. will do for either side of vertex. Thus, if 1, 2, 3, 4, 5, 6 represent points, respectively 10, 20, 30, 40, 50, 60 from the vertex, there are a series of triangles, PV1, PV2, PV3, etc., to solve, all right angled at V ; PV is co-lat. V, and angles at P are 10, 20, 30, etc., to find PI, P2, P3, etc. Sin (co. P) = tan PV . tan (co. PI) /. cot PI = cos P . cot PV. PV = 44 44' P = 10 cot = 10-004043 cos= 9-993351 cot = 10-004043 20 cos = 9-972986 45 10J cot = 9-997394 V = 4444' cot = 10-004043 P = 30 cos = 9-937531 46 31' cot = 9-977029 cot = 10-004043 ...40 cos= 9-884254 PV = 44 44' P = 50 48 60cot= 9-941574 cot = 10-004043 cos= 9-808067 5217i'cot= 9-888297 cot = 10-004043 cos- 9-698970 ...60 C 57 cot = 9-812110 6313J'cot = 9-703013 These are the co-lats. of the points which may now be arranged in order from A to B, thus A San Lucas lat. 22 52' N. long 109 53' W. 6 26 46A N. 115 11 W. 5 32 58* N. 125 11 W. 4 37 42| N. 135 11 W. 3 41 9l N. 145 11 W. 2 43 29~ N. 155 11 W. 1 44 49* N. 165 11 W. Vertex 45 16 N. 175 11 W. 1 44 49 N. 174 49 E. 2 43 29 N. 164 49 E. 3 41 9J N. 154 49 E. 4 37 42* N. 144 49 E. B Yokohama 35 26 N. 139 39 E. These positions set off on a chart and joined by a curve, give the G.C. track. As it is impossible to keep a ship on the track, points are taken differing very little in long., say 2, and courses steered from point to point by Mercator. Should a vessel be driven far from the track by stress of weather, the navigator should strike a new G.C. from the actual position of the ship, and not work back to the original one. GREAT CIRCLE SAILING. 221 Example. VERTEX " OUT." Required the initial and final courses, and distance on the G.C. from Start to Barbadoes ; also position of the vertex, and position of a succession of points differing 5 in long, commencing where the 5 W. meridian cuts the track. Startlat.6013'K AP= 3947'long. 3 39'W. Barb. 13 3 N. BP= 76 57 BP + AP = 116 44 = 58 22 57 37 W. P= 53 58 f = 26 59 FIG. 150. BP-AP= 37 10 - AP)= 18 35 To solve triangle APB, having given PA, PB, and P. Cos ^3. = cos flBP + AP) 2 " cos(A + B) KBP+AP) = J(BP-AP) = 35 = 26 59 7415 / 57 36 19 14 sec = 10-280270 cos= 9-976745 cosec = 10 -069855 sin= 9-503360 cot = 10 -293146 cot = 10 -293146 tan = 10 -550161 cos= 9-719730 sin= 9-656799 sec =10-566751 tan= 9-866301 A = N. 110 35 11 W. or S. 69 24' 49" W. 1st Co. B = S. 37 56 43 W. Final Course. 28 39' cos= 9-943280 2 Distance 57 18 = 3438 milea. Since A and B are of contrary affection, vertex is "out." To find position of vertex " V." In triangle APV, V = 90. Given AP and A, to find PV and P. Sin PV = cos (co. AP) . cos (co. A) and sin (co. AP) = tan (co. A) . tan (co. P) = sin AP . sin A or cos AP = cot A . cot P /. cot P = cos AP . tan A AP = 39 47' sin = 9-806103 A = 69 25 sin = 9-971351 cos= 9-885627 tan = 10-425340 PV = 36 48 sin = 9 777454 26 2V E. cot = 10 '310967 Lat. V = 53 12 N. Long. A 3 39 W. Long. V 22 23 E. To find a succession of points. First point where 5 W. meridian cuts the track. Long, from V of 1 = 22 23J' + 5 = 27 23V In triangle PV1 ; V = 90 given PV and P ; to find PI sin (co. P) = tan (co. PI) . tan PV or cos P = cot PI . tan PV .-. cot PI = cos P . cot PV 222 TEXT-BOOK ON NAVIGATION. P = 27 23' cos = 9-948355 PV = 36 48 cot = 10-126043 32 23V cos = 9-926551 " cot = 10-126043 PI = 40 7 cot = 10-074398 P2 = 41 32 cot = 10-052594 Points. 1, lat. 49 53' N. long. 5 W. 2, 48 27 N. and so on. 10 W. ART. 98. Composite Great Circle and Parallel Sailing. When the Great Circle track would take the ship into too high a latitude, then a combination of Great Circle sailing and Parallel sailing is used to obtain the shortest route. The highest practicable latitude is made the latitude of vertex for the arcs of two Great Circles, one passing through the starting-point, and the other through the destination. The route would then be along the first arc to the Vertex, next on the parallel of latitude to the Vertex of the other arc, and finally along this arc to the destination. Illustration (Fig. 151). Let A be the point of departure, and B the destination, and AVB the arc of the Great Circle whose vertex is V. The parallel LI^ is assumed to be the highest practicable latitude. AVi and BVa are arcs of two great circles touching the parallel, ViV 2 is the arc of the parallel of latitude or meridian distance, PV! and PV 2 are arcs of the meridians of vertexes and make right angles with AV! and BV 2 . The solutions, therefore, depend on right-angled spherical trigonometry and parallel sailing. Example (Fig. 152). Required the distance on the composite track from Rio to Cape Leeuwin, taking 55 as the limit of S. latitude ; also the initial and final courses. Lat. of Rio 23 4' S. Long, of Rio 43 8' W. Lat. of C. Leeuwin 34 19' S. Long, of C. Leeuwin 115 6' E. (1) To find the distances AN^ and FIG. 151. 55' BV 2 In triangle PAV 1? V l = 90 given AP and sin (co. AP) = cos AV : . cos PV l :. cos AVj = cos AP . sec PV! Similarly, cos BV 2 = cos BP . sec PV 2 GREAT CIRCLE SAILING. 223 Co-lat. of A 66 56' ... cos 9-593067 B 55 41 cos 9751099 Co-lat. of Vj and V 2 35 ... sec "086635 ... sec '086635 AV 1= 61 25i ... cos 9-679702 BV 2 = 46 30j cos 9-837734 Sum 107 56 60 6476 geographical miles. (2) To find the differences of long. APVj and BPV 2 , or longitudes from Vertex. Sin (co. P) = tan PV t . tan (co. AP) .*. cos P = tan PV l . cot AP and cos P = tan PV 2 . cot BP Co-lat. of A 66 56' ... cot 9 -629255 B 55 41 ... cot 9-834154 Co-lat. of V x and V 2 35 ... tan 9 "845227 ... tan 9 '845227 D. long of A and Y! 72 39 ... cos 9 "474482 B and V 2 61 27 cos 9'679381 Sum 134 6 Total D. long 158 14 D. long. VjV a 24 8 = 1448 geographical miles. (3) To find the meridian distance between Vj and V 2 . D. long. 1448 log 3160769 Lat. 55 cos 9-758591 Mer. dist. 830'5 log 2-919360 Arcs of Great Circles 6476 Total dist. 7306'5 NOTE. The distance on the Great Circle between A and B is 7131 miles. (4) To compute the initial and final courses. Sin PVj = cos (co. AP) . cos (co. A) = sin AP . sin A * /. sin A = sin PVj . cosec AP and sin B = sin PV 2 . cosec BP Co-lat. of A 66 56' ... cosec '036189 B 55 41 cosec 0-083054 Co-lat. of Vj and V 2 35 ... sine 9758591 ... sine 9758591 Initial co. S. 38 34 E.... sine 9794780 Final co. N. 43 59 E sine 9 '841645 (5) To find the longitudes of V x and V 2 . Long, of A 43 8' W. Long, of B 115 6' E. D. Long, to Vj 72 39 E. D. long, to V 2 61 27 W. Long, of Vj 29 31 E. Long, of V 2 53 39 E. 224 TEXT-BOOK ON NAVIGATION. Various points on the arcs AVi and BV 2 can be fixed as in G.C.S. either by assuming intervals of longitude from Vi towards A, and from V 2 towards B, or by working from a fixed meridian to destination. ART. 99. Windward Great Circle Sailing. What is called Windward Great Circle sailing applies only to sailing vessels, and consists in putting the ship on the tack on which her course would be nearest to the Great Circle track without reference to the Rhumb or Chart course, which in some cases would differ four or five points or more from the former. Illustration (Fig. 153). Let be the starting-point, and FIG. 153. suppose destination to bear E.S.E. by Chart, the Great Circle course from being S.S.E. With the wind at S.E., the course on the starboard tack would be E.N.E. and on the port tack S.S.W. It is evident that on the port tack the ship's head would only be four points off the shortest route, whilst on the starboard tack she would be steer- ing at right angles to it. In sailing 100 miles on the former course the Traverse Table shows that she would make 70 miles towards her destination, but on the latter she would be no nearer to it, having sailed on a line at right angles to the direct route on the globe. With the wind at S.E. by S., and steering S.W. by S. (ap- parently away from destination, according to chart), she would still gain 36 miles in 100 miles sailed, whilst on the starboard tack, steering E. by N., she would make good only 19 miles. Other still more striking cases might be given, but the above are sufficient to show the advantage of the method. All problems in Great Circle sailing, Composite sailing, and Windward Great Circle sailing can best be understood from a GREAT CIRCLE SAILING. 225 globe. A pair of 12-inch globes should be in every ship's equipment. EXERCISES. GREAT CIRCLE SAILING. 1. Find the first and final courses from Cape of Good Hope, lat. 34 22' S., long. 18 30' E., to Rio Janeiro, lat. 22 55' S., long. 43 9' W. ; also find the distance, latitude, and longitude of Vertex, and a succession of points for every five degrees of longitude. 2. Find the first and last courses, and distance from San Francisco, lat. 37 49' N., long. 122 29' W., to Yokohama, lat. 35 26' N., long. 139 39' E., the latitude and longitude of Vertex, and a succession of points for every twenty degrees of longitude. 3. Find the initial and final courses, distance on the Great Circle, lati- tude and longitude of Vertex, and a succession of points differing 10 in longitude. From A, lat. 33 51' S., long. 151 16' E. To B, 37 47 N., 122 30 W. 4. Find the initial and final courses, distance on the Great Circle, latitude and longitude of Vertex, and a succession of points differing 20 in longitude. From A, lat. 22 16' N., long. 114 10' E. ToB, 12 4 S., 77 16 W. 5. Find the initial and final courses, distance on the Great Circle, latitude and longitude of Vertex, and a succession of points differing 5 in longitude. From A, lat. 52 12' N., long. 6 12' W. To B, 13 7 K, 59 25 W. 6. Find the initial and final courses, distance on the Great Circle, latitude and longitude of Vertex, in sailing. From R, lat. 43 40' N., long. 150 37' E. To Q, 43 40 N., 150 37 W. Also find latitude of point where the meridian of 155 E. cuts the track, and a succession of points differing 5 in longitude, from thence to destination. 7. Find the initial and final courses, distance on the Great Circle, position of Vertex, and a succession of points starting in long. 120 W., and differing 5 in longitude to Equator. From X., lat. 38 27' N., long. 100 26' W, ToY., ,, 38 27 S., 150 47 E. 8. A ship sails from Rio de Janeiro, lat. 22 54' S., long. 43 6' W., to Madeira, lat. 32 43' N., long. 16 40' W. Find longitude of point where the track crosses the Equator. 9. The point of ''Maximum Separation in Latitude" on a Great Circle, is that point where the Great Circle course is the same as the Mercator course. Find the latitude of such a point in sailing from lat. 40 30' N., long. 120 E., to lat. 45 K, long. 172 E. 10. One Vertex of a great circle is in lat. 50 N. 70 W. : what is position of the other Vertex ? A ship sailing on a Great Circle crosses the Equator : show that the course at that time = complement lat. Vertex. 11. In sailing from lat. 34 51' S., long. 20 E., to lat. 34 19' S., long. 115 6' E., on a Great Circle arc, how far would the vessel be from Ker- gueleri island, lat. 49 54' S., long. 70 12' E., when upon same meridian as island ? 12. A ship starts from A on the course N. 64 12' E. , and sails on a Great Circle arc to B ; she increases her distance from the Equator tijl she gets into long. 19 26' 30" W. on parallel 51 50' 30" N., and then diminishes it. Find lat. and long, of A and B, their diff. long, being 43 34'. Q 226 TEXT-BOOK ON NAVIGATION. EXERCISES. COMPOSITE SAILING. 1. Find the first and last courses and distance from Bahia Blanca, lat. 38 59' S., long. 61 39' W., to Cape Lewin, lat. 34 31' S., long. 115 6' E. ; easting to be run on the parallel of 45 S. Find also the longitudes of points of arrival at and departure from the parallel. 2. On a composite track from Cape S. Mary, lat. 16 40' S., long. 50 4' E., to Mont Video, lat. 34 43' S., long. 66 16' W., the westing to be run on the parallel of 40 S. Required the first and last courses, total distance, and longitudes of points of arrival at and departure from the given parallel. 3. Find the first and last courses, distance, and longitudes of arrival at and departure from parallel of 50 N. when sailing on a composite track from Yokohama, lat. 35 26' N., long. 139 39' E., to Cape Flattery, lat. 48 23' N.. long. 124 44' W. 4. Find the first and last courses and distance when sailing on a composite track from Valparaiso, lat. 33 2' S., long. 71 28' W., to Port Lyttelton, lat. 43 36' S., long. 172 50' E., the westing being run on the latitude of Port Lyttelton ; also the position of a succession of points from long. 73 W., and every 5 to destination. CHAPTER XXII. LATITUDE BY DOUBLE ALTITUDES. ART. 100. Double Altitudes. This is a problem for determining the latitude from (a) Two different altitudes of the same object (Sun, Moon, Star, or Planet). (b) Simultaneous altitudes of two different objects. The data required for the computation are (1) The polar angle. (2) The two altitudes. (3) The declination. Definition. The polar angle is the angle at the Pole between the hour circles passing through the two positions of the object or objects. (1) For the Sun, the polar angle is the interval between the apparent times of observation ; therefore, if mean times are given, the equation of time must be applied to each, or the change of equation of time applied to the mean time interval. The rule for applying this correction is If Eq. of Time be additive to mean time and increasing add. subtractive from mean time and decreasing add. additive to mean time and decreasing subtract. tt subtractive from mean time and increasing subtract. Thus, if the mean interval on August 28th is 4 h 30 ra 18 s , what is the apparent interval ? H.V. Eq. of T. -72 Int. 4-5 360 Mean Int. 4 h 30 18" 288 Change in Int. + 3*2 3-240 App. Int. 4 30 21'2 The change being added because the Eq. of Time is subtractive from mean time and decreasing. (2) A Star's polar angle is the interval of sidereal time 228 TEXT-BOOK ON NAVIGATION. between the observations, and is found by accelerating the mean interval for the interval. Thus, if the mean interval is 4 h 30 m 18 s , what is the sidereal interval ? Mean Int. 4 h 30 m 18 9 ( 4 h 394 \ 30 m (18 s Accel, for 49 1 Sid. Int. 4 31 2-4 (3) When simultaneous altitudes of two stars are observed, the polar angle is the difference between their right ascensions. (4) For the Moon or a PJanet, the polar angle is the sidereal interval, increased or decreased by the change in right ascension during the elapsed mean time, according as the R.A. of the object is decreasing or increasing. ART. 101. Correction for " Run." Both altitudes are supposed to be observed at the same place ; therefore, if a ship changes her position during the interval, a correction depending on the course and distance run and the bearing of the object must be applied to one of the altitudes, in order to make it what it would be at the place where the other altitude is observed. B A 1 FIQ. 154. A \ B 1 FIQ. 155. B FIQ. 157. Thus, if the ship's course is the same as the bearing of the object at the first observation, that is, directly towards it, and the distance run ten miles, the first altitude would have to be increased by 10'; but if the course is opposite to the bearing, that is, directly away from the object, the firsfr altitude will have to be decreased by 10' ; whilst if the course were at right angles to the bearing, there would be no correction. When the course is obliquely towards or from the object, the angle between the bearing and the course must be found. LATITUDE BY DOUBLE ALTITUDES. 229 With this angle as a course in the Traverse Table and distance run in the Distance column, the correction will be found, in the Latitude column, and will be added to or subtracted from the first altitudes according as the course is towards or from the Sun. Illustration. Let A and B be the ship's positions at the first and second observations respectively, and AS the direction of the object. Fig. 154. Course directly towards the object; correction = AB-f 155. from ; = AB- 156. obliquely towards ; = A6-{- 157. from ; = A6- 158. at right angles to bearing ; =0 In Figs. 156 and 157 the angle at A is the 05 course, and AB the distance in the Traverse Table, and A6 the latitude, since B& is a per- pendicular upon AS. When the bearing is given at the first observation, the computation gives the lati- tude at the second observation, and vice versa. Therefore, if the bearing at the second observation is given, the "correction for run " must be applied to the second altitude, g < reversing the rule for applying it; that is, -piQ. 158. if the run is towards the object, subtract the correction from the second altitude, and if from the object, add the correction. The latitude found is the latitude at the first observation. Of the various methods for solving the problem, the best known is " Ivory's method," but the latitude found requires a correction for the change of declination in half the elapsed time. ART. 102. Double Altitudes. Ivory's Method. Let S and ^ be FIG. 159. Fin. 160. the two positions of the celestial body, and SSi the arc of a great 230 TEXT-BOOK ON NAVIGATION. circle, bisected in M ; PS or PSi, the polar distance ; ZS and ZSi, the zenith distances ; PZ, the co-lat. ; and ZA, a perpendicular on PM. The angle MPS = \ elapsed time, or \ polar angle. MS is arc 1, ZA arc 2, AM arc 3, PM arc 4, and PA arc 5. To find Arc 1 (MS). In the right-angled triangle PMS are given PS (polar distance) and the angle MPS (| polar angle), to find MS. Sin MS = cos (comp. of MPS) x cos (comp. of PS) or sin MS = sin MPS X sin PS that is, sin arc 1 = sin J polar angle x cos declination or cosec arc 1 = cosec ^ polar angle x sec declination To find Arc 2 (ZA). Applying the general formula to the triangle ZMS cos ZS = cos ZMS. sin ZM.sin MS + cos ZM.cos MS . (1) Similarly, in the triangle ZMS t cos ZSi = -cos ZMS . sin ZM . sin MS + cos ZM . cos MS . (2) (because ZMSi is the supplement of ZMS, and MSi = MS). Subtracting (2) from (1) cos ZS - cos ZS X = 2(cos ZMS . sin ZM . sin MS) i.e. sin XS - sin YS X = 2(sin AMZ . sin ZM . sin MS) . (3) (because the alts. XS and YSj and the angle AMZ are the complements of ZS, ZS b and ZMS). In the right-angled triangle AMZ sin AZ = sin AMZ . sin ZM Substituting sin AZ in equation (3) sin XS - sin YS X = 2 sin AZ . sin MS 9 *in A7 - sin XS - sin YSj .. 2smAZ-- sin AZ = co sin MS that is, sin arc 2 = cos J sum of alt-.. . sin J diff. of alts. . cosec arc 1 To find Arc 3 'AM). Adding equations (1) and (2) cos ZS + cos ZSi = 2 cos ZM . cos MS . . . . (4) But in the right-angled triangle AMZ, cos ZM = cos AM . cos AZ ; .-. cos ZS + cos ZSi = 2 cos AM .cos AZ. cos MS or sin XS + sin YSi = 2 cos AM . cos AZ . cos MS A , , sin XS + sin YSi /. cos AM = _- -- ^- l 2 cos AZ . cos MS _ sin JKXS + YSQ cos j(XS - YSQ cos AZ . cos MS that is, cos arc 3 = sin J sum of alts, x cos \ diff. alts. X sec arc 2 X sec arc 1 LATITUDE BY DOUBLE ALTITUDES. 231 To find Arc 4, (PM). In the right-angled triangle PMS cos PS = cos MS . cos PM .*. cos PM = -t^ = cos PS . sec MS cos that is, cos arc 4 = sin dec. X sec arc 1 To find Arc 5 (PA). Arc 5 is the difference or the sum of 3 and 4 according to circum stances. To find the Latitude. In the right-angled triangle PZA cos PZ = cos PA . cos ZA that is, sin lat. = cos arc 5 . cos arc 2 and therefore cosec lat. = sec arc 5 . sec arc 2 The foregoing formulae are arranged for computation as follows : o H J pol. angle fcosec Dec. sec sin Arc 1, cosec sec sec i sum alts. cos sin | diff. sin cos Arc 2, fsin *sec Arc 3, cos Arc 4, cos Arc 5, sec Arc 2, *sec Lat. cosec Logs marked thus * occur twice. Logs marked thus t are used again for finding correction to latitude. The Ambiguous Case. When the sum of arcs 3 and 4 is less than 90, it may be doubtful whether they should be added or subtracted to find arc 5. But if the lat. be first found by adding, and the lat. by account is not given, the correctness of the answer may be tested by computing the hour angle at the middle time, and seeing if it agrees with the known hour angle. If not, then subtract 3 from 4, and again compute the latitude. To compute the Hour Angle (Hi) at the Middle Time. In the right-angled triangle PZA sin AZ = sin PZ . sin ZPA . _ . sin AZ A .'. sm ZPA = -: ^7 = sin AZ . cosec PZ sin JrZi i.e. sin H.A. at mid. time = sin arc 2 . sec lat. To correct the lat. found by "Ivory" (1) Find change of decimation in half elapsed time. 232 TEXT-BOOK ON NAVIGATION. (2) Add following logs : log change of dec., secant approx. lat., cosec half int., sine arc 2, reject 10', and take back log. This is the correction. (3) When the second altitude is the greater. Allow it N, when the Sun is going N. Allow it S, when the Sun is going S. Reverse this when ihe first altitude is the greater. ART. 103. Rule for finding latitude by " Ivory" (1) Find polar angle as directed, and the declination for the Middle Greenwich time. (2) Correct the altitudes, applying correction for " Run " to alt. for which bearing of object is given, as directed. Find half sum and half difference of alts. (3) For Arc 1, add logs of cosecant half polar angle and secant dec. ; this gives cosecant arc 1. For Arc 2, add logs cosecant arc 1, cosine half sum alts., sine half diff. alts. ; this gives sine arc 2. For Arc 3, add logs secant arc 1, sine half sum alts., cosine half diff. alts., secant arc 2 ; this gives cosine arc 3. For Arc 4, add logs sine dec, and secant arc 1 ; this gives cosine arc 4. For Arc 5. This arc is always the difference of arcs 3 and 4 when lat. and dec. are of different names, it can never exceed 90, so should the sum of 3 and 4 be greater than 90, their difference must also be used for 5 ; but if the sum is less than 90, there may be some doubt, which can be decided by computing the H.A. at middle time from the formula sin H.A. = sine arc 2 x sec lat. and selecting that lat. which gives the known H.A. at middle time. For lat. add logs secant arc 5 and secant arc 2 ; this gives cosecant lat. to be corrected as directed for true lat. Remarks. (1) When dec. is 0, arc 4 is 90, because sin = 0, and there- fore cos arc 4 = 0. (2) When alts, are equal, diff. = 0, then arc 2 = 0, and arc 6 = co-lat. (3) Arc 4 is of same affection as polar distance. Therefore when lat. and dec. are of different names supplement of angle found is arc 4. Example. August 5th, lat. K, long. 55 17' 30" W. With the following observations find the latitude at second observation A.T.S. 9 h 15 m 19' A.M. Obs. alt. 44 14' 20" Ind. cor., - 2' 7" 3 18 22 P.M. 39 21 25 Eye, 30 ft. Sun's bearing at first observation, S. 62 E. ; run during the interval, S. 6 W, 63 miles. LATITUDE BY DOUBLE ALTITUDES. 233 First A.T.S. August 4 d 21 h 15 19 s Dec. 17 9' 48" '4 N. H.V. 40" '14 Second A.T.S. 5 3 18 22 Cor. - 2 39 '4 3-97 Interval 633 H.E.T. 3 1 31-5 Mid. time 5 16 50 Long. 3 41 10 17 7 9 N. 159-35 H.V. 40" -14 H.E.T. 3 Ch. of dec. in H.E.T. 120-42 Mid. A.T.G. 5 3 58 Bearing S. 62 E. KunS. 6 W. Inc. angle 68 dist. 63' gives d. lat. 23 '6 to be added to first alt., because angle is less than 90. First O.A. 44 14' 20" 1.0. - 2 7 44 12 13 Dip - 5 22 44 6 51 S.D. + 15 48 Second O.A. 39 21' 25" -2 7 39 19 18 -5 22 44 22 39 R. & P. - 51 44 21 48 Run +23 36 44 45 24 39 28 42 2)84 14 6 Half sum 42 7 3 H.E.T. 3 h l m 3r-5 cosec 10-147644 39 13 56 + 15 48 39 29 44 -12 39 28 42 44 45 24 2) 5 16 42 Half diff. 2 38 21 Dec. 17 7' 9" sec 10-019681 sin 9-468878 Arc 1. 42 51 50 cosec 10-167325 sec 10134913 sec 10-134913 sum 42 7 3 cos 9-870270 sin 9-826498 \ diff. 2 38 21 sin 8-663191 cos 9-999539 Arc 2. 2 52 41 sin 8-700786 sec 10-000548 Arc 3. 23 46 16 ... cos 9-961498 Arc 4. 66 1919 cos 9*603791 Arc 5. 42 33 3 sec 10*132723 H.E.T. cosec 10147644 Arc 2. 2 52 41 sec 10'000548 sin 8700786 Approx. lat. 47 22 13 cosec 10133271 sec 10169246 Ch. in dec. 120-4 log 2-080626 Cor. + 13 12-54 log 1-098303 True lat. 47 22 26 N. 234 TEXT-BOOK ON NAVIGATION. ART. 104. Direct Method. In the triangle PSSi are given the two sides PS and PSi and the included angle at P, from which can be found the third side SSi and the angles PSiS and PSSi. Then in the triangle ZSSi with the three sides given, the angle ZSiS can be found. Also the angle PSiZ = PSiS - ZSiS. Lastly, in the triangle PZSi the two sides PSi and ZSi and the in- cluded angle PSiZ are known, from which can be found the co-lat. PZ. The latitude found by this method requires no correction for change of declination, because the actual polar distances are used, but it is subject to the same ambiguity as Ivory's method when SSi produced passes near the zenith. In the latter case the angle PSiZ may be either the difference or the sum of PSiS and ZSiS. FIG. 161. Example. The same problem as before worked from figure. In figure X and Y are the two positions of the sun. P the polar angle = 6 h 3 m 3" PX the 1st pol. dist. = 72 50' 49" PY the 2nd = 72 54 53 ZX the 1st zen. dist. = 45 14 36 ZY the 2nd = 50 31 18 In triangle PXY, given two sides PX and PY, and included angle P, find X and XY. \s\st? 9\-*- * "l *- -A.^ TanirX y^- sin ^ PY - PX) Cot P Tan J(X - Y) _ ^gpy + px) cot 2 c XY cos|(PY C/ S "2~" = ~^' 1 ^ F sin? P = 3 h l m 31'-5 cot 9-994220 cot 9*994220 sin 9'852355 i(PY + PX) = 72 52' 51" sec 10 531122 cosec 10-019682 cos 9-468878 |(PY - PX) =022 cos 10 000000 sin 6*771935 X + Y) = 73 23 24 (X-Y)= 2 6 X = 73 25 30 tan 10 525342 tan 6785837 sec 10 543854 42 51' 51" cos 9-865087 XY = 85 43 42 LATITUDE BY DOUBLE ALTITUDES. 235 In triangle ZXY, given the three sides, to find X. X /sins, sin (s - ZY) * S "2 = V sin ZX . sin XY where a = J(ZY + ZX + XY). ZY = 50 31' 18" XY = 85 43 42 cosec 10-001208 ZX = 45 14 36 cosec 10148678 2)181 29 36 s = 90 44 48 sin 9 '999963 8 - ZY = 40 13 30 sin 9'810092 2)19-959941 17 16' 4-5" cos 9-979970 .'. ZXY = 34 32 9 PXY = 73 25 30 Hence PXZ = 38 53 21 N ow in triangle PXZ, given two sides PX and ZX, and included angle PXZ, to find PZ = co-lat. = 19 26' 40" cot 10-452191 sin 9-522305 + ZX) = 59 2 42 sec 10'288729 cos 9711271 J(PX - ZX) = 13 48 6 cos 9'987276 i(Z + P) = 73 24 32 tan 10728196 sec 10735659 21 18' 45" cos 9-969235 PZ = 42 37 30 Lat. = 47 22 30 N. Art. 105. Simultaneous Altitudes. The altitudes of two objects are observed at the same time; the problem may be worked either by Simmer's method, when the intersection of the two lines of position is the position of the ship, as there is no allowance for " Run," or by Direct Spherics. Example. September 9th, A.M. at ship in long, by account 128 W., when the G.M.T. by chron. was 8 d 20 h 52 15* ; obs. alt. of o Ceti east of meridian was 26 10' 20" ; obs. alt of o Ophiuchi west of meridian was 9 59' 30" ; ind. cor., 2' 18" ; height of eye, 30 ft. Required the ship's position, and true bearings of the stars, by Sumner's method, assuming lats. 51 and 51 30' N. , or by any other method. TEXT-BOOK ON NAVIGATION. M.T.G. Sept. 8 d 20 h 52 m 15' S.T. at noon ll h 6 m 33' -2 3 17*1 Accel. 8'6 O.A. aCeti 26 10' 20" 1.0. - 2 18 O.A. o Ophiuchi 26 0'44" 51 30 86 16 44 R.A.M.S. 11 9 59' 30" -2 18 9 58-9 9 46' 30" 51 30 77 22 20 26 8 2 Dip - 5 22 9 57 12 -5 22 26 2 40 Ref. - 1 56 9 51 50 -5 20 T.A. 26 44 Lat. 51 P.D. 86 16 44 9 46 30 61 77 22 20 163 17 28 163 47 28 138 8 50 138 38 50 sum 81 38 44 remainder 55 38 81 53 44 55 53 69 4 25 69 17 55 69 19 25 59 32 55 Sees 10-201128 Cosecs 10-000916 Cos 9-162256 Sins 9-916687 10-205850 10-000916 9-149150 9-917976 10-201128 10-010635 9-552873 9-934417 10-205850 10-010635 9-547885 9-935537 9-280987 E.H.A.'s 3 h 27 m 18 g 3 h R.A.* 2 57 22 2 9-273892 9'699053 25 30 s W.H.A/8 6 h 2 57 22 17 30 34 9-699907 6 h O m 30' 17 30 34 R.A.M. 23 30 4 23 R.A.M.S. 11 9 59 11 31 52 9 59 23 30 36 11 9 69 23 31 4 11 9 59 M.T.S. 8 d 12 20 5 8 d 12 M.T.G. 8 20 52 15 8 20 21 53 52 15 8 d 12 20 37 8 20 52 15 8 d 12 21 5 8 20 52 15 8 32 10 8 30 22 8 31 38 8 31 10 Long, (a) 128 2' 30" W. (6) 127 35' 30" W. (c) 127 54' 30" W. (d) 127 47' 30" W. By projection on a chart these longs. give- Bearing of a Ceti S. 59 E. Bearing by a Ophiuchi N. 81 W. Lat. 51 12' N. Long. 127 51 W. Instead of projecting, the position of the ship may be found thus D. long, between " a " and " c " = 8' E. ; call this A. It n " (t Tj Of?/ T> ,, ,, a o vf ,, jtJ. D. lat. lats. given = 30' The cor. to lesser lat. = A x P A + C C. D. long (a) = * A x B AC LATITUDE BY DOUBLE ALTITUDES. 237 sum of A and C is taken when they are of contrary names (as in example) diff. of A ,, C same ' ,, Cor. to lat. = 8 * 30 = 12' Lesser lat. 51 0' N. Cor. 12 Cor. to long = = 10' 48" Long, "a" 128 2' 30" W. Cor. 10 48 E. Lat. in 51 12 N. Long, in 127 51 42 W. The azimuths may be found from the Traverse Tables thus 1st obs. D. lat. = 30' D. long. = 27' This gives dep. = 17', hence co. "a" to " b " from traverse or first line of position is N. 30 E. .'. bearing of Ceti is S. 60 E. 2nd obs. D. lat. = 30' D. long. = 7' This gives dep. = 4''4, hence co. " e" to "d" from traverse or second line of position is N. 9 E. /. bearing of a Ophiuchi is N. 81. W. From the figure. In the figure X and Y are the positions of the two stars. PX is pol. dist. of 1st star = 86 16' 44" ZX is zen. = 63 59 16 PY is pol. 2nd star = 77 22 20 ZY is zen. = 80 13 30 P is diff. of R.A.'s = 9 h 26 m 48' In triangle PXY, given two sides PX and E PY, and included angle P. Find X and XY. = 4" 43 m 24' cot 9-540653 cot 9-540653 sin 9'975277 i(PX + PY) = 81 49' 32" sec 10-847138 cosec 10-004435 cos 9-152862 i(PX - PY) = 4 27 12 cos 9-998687 sin 8-890125 Y + X) = 67 40 20 tan 10-386478 sec 10'420325 Y-X)= 1 3339 tan 8-435213 X = 66 6 41 69 17' 41" cos 9-548464 XY = 138 35 22 In triangle ZXY, given three sides ZX, ZY, and XY, to find angles Z and X. 2 3 8 TEXT-BOOK ON NAVIGATION. c Z /sin s . sin ( - XY) orw1 _ X /sin a . sin (, - ZY) 2 V sin ZX . sin ZY where s = $(ZX + ZY + XY). XY = 138 35' 22" ZY= 80 13 30 ZX = 63 59 16 "2 V sin ZX . sin XY cosec 10-179503 cosec 10-006351 cosec 10-046385 cosec 10-046385 sin 9-795090 sin 9795090 sin 8-690667 sin 9-942555 2)282 48 8 s = 141 24 4 s - XY = 2 48 42 s - ZY = 61 10 34 79 17 14 Z = 158 34 28 2)18-538493 cos 9-269246 16 29' 13" ZXY = 32 58 26 PXY = 66 6 41 2)19-963533 cos 9-981766 .'. PXZ = 33 8 15 In triangle PXZ, given two sides PX and ZX, and included angle 'X ; to find PZ = co-lat., P = H.A. 1st star, Z = az. 1st star. Tan i(Z + P) = cos cos ' X - ZX >.co t f + ZX) - = 16 34' 7" 5 cot 10-526486 cot 10-526486 sin 9-455096 + ZX) = 75 8 (PX - ZX) =11 8 44 J(Z + P) = 85 33 5 KZ - P) = 33 54 35 Z = 119 27 40 P = 51 38 30 /. E.H.A. 3 h 26 m 34* R.A.* 2 57 22 BJLM.23 30 48 R.A.M.S. 11 9 59 M.T.S. 8 d 12 20 49 M.T.G. 8 d 20 52 15 sec 10-590793 cosec 10-014787 cos 9-409207 cos 9-991730 sin 9*286237 tan 11-109009 8 31 26 Long. 127 51' 30 W tan 9-827510 19 23' 45" PZ = 38 47 30 .'. lat. 51 12 30 N. sec 11-110320 cos 9-974623 LATITUDE BY DOUBLE ALTITUDES. 239 Azimuth of o Ceti = Z, or N. 119 28' E., which ia S. 60 32' E. SZX = 60 32' YZX = 158 34* Azimuth of a Ophiuchi SZY = 98 2J, or N. 81 57V W. EXERCISES. 1. August 14th, in long. 73 14' E., lat.S. ; with the following observations find the latitude at the second obs. : Approx. A.T.S., 9 h 20 ra A.M. ; time by chron., 4 h 26 ra 18'; obs. alt. Sun's L.L., 30 21' 50". Approx. A.T.S., 3 h 5 ra P.M. ; time by chron.. 10 h 12 m 12 ; obs. alt. Sun's L.L., 26 38' 10". Chron. showed correct A.T.G. ; ind. cor., - 3' 16" ; height of eye, 30 ft. ; Sun's bearing at first obs., N. 46 30' E. ; run between sights, S. 57 30' E. 8 knots per hour. 2. August 18th, lat. N., long. 124 23' W. ; with the following observations find the latitude at the second obs. : Approx. A.T.S., 10 h O m A.M. ; chron., 6 h 21 m 46' ; obs. alt. 0, 50 32' 50" ; ind. cor., + 1' 39". Approx. A.T.S., 2 h 50 ra P.M. ; chron., ll h 9 m 34 ; obs. alt. GD, 42 46' 10" ; eye, 35 ft. ; chron. correct for M.T.G. ; Sun's bearing at first obs., S. 49 30' E. ; run between obs., S. 67 30' W. 10 knots per hour. 3. August 21st, lat. N., long. 130 21' W. ; with the following observations find the latitude at the second obs. : M.T.S., 9 h 21 m 42* P.M. ; obs. alt. Yega, 75 47' 27"; bearing, S. 40 W. M.T.S., l h 52 m 50 A.M. ; obs. alt. Vega, 34 53' 50"; run, S. 50 E. 9 knots per hour; ind. cor., + 2' 58"; eye, 29 ft. 4. August 27th, lat. S., long. 84 27' E. ; with the following observations find the latitude at the second obs. ; ind. cor., + 4' 14" ; eye, 41 ft. M.T.S., 2 h 12 ra 19 s A.M. ; obs. alt. Aldebaran, 18 10' 50" ; bearing, N. 61 E. M.T.S., 5 h 12 m 30 s A.M. ; obs. alt. Aldebaran, 45 40' 10"; run, S. 6 E. 11 knots per hour. 5. August 17th, about ll h 15 m P.M. at ship; lat. by D.R. 48 15' N., long. 179 30' W. ; when a chron. 8 m 4' fast of M.T.G. showed ll h 28 m 17 , the obs. alt. of a Arietis was 27 15' 40" ; ind. cor., - V 13" ; at the same time the obs. alt. of Altair was 47 6' 30" ; ind. cor., + 2' 20" ; height of eye, 41 ft. Find the position of the ship and true bearings of the stars by Sumner's method, assuming lats. 15' on each side of D.R. lat., or by any other method. 6. September 3rd, in long. 50 25' W., lat. N.; when the A.T.S. was 7 h 14 m 20' A.M., the obs. alt. of Sun's U.L. was 18 30' 20" ; and again at ll h O m 25' A.M., A.T.S., the obs. alt. of Sun's U.L. was 47 26' 20" ; ind. cor., -2' 11"; eye, 28 ft. ; run between observations, N. 9 E. 30 miles. If the Sun bore S. 81 E. at first sight, find lat. at second sight. 7. August 16th, lat. 24 N., long. 14 W. ; with the following observations of Fomalhaut (o Pis. Aust.) find the lat. at second obs. : Approx. M.T.S., 9 b 45 m P.M. ; chron., 10 h 42 m 22' ; obs. alt., 15 8' 50" ; bear- ing, East. Approx. M.T.S., 3 h 15 m A.M. ; chron., 4 h 12 m 25' ; obs. alt., 28 14' 0" ; ind. cor., + 2' 48". Run between sights, N. 64 E. 10 knots per hour ; eye, 45ft. 8. September 15th, lat. S., long. 123 34' E., with the following observa- tions find the lat. at second obs. ; ind. cor., 3' 13" ; eye, 36 ft. : Approx. A.T.S., O h 36 m p. M. ; chron., 4 h 21 m 12' M.T.G. ; obs. alt. Sun's L.L., 38 44 A 10". Approx. A.T.S., 4 h 21 m P.M. ; chron., 8 h 6 ra 17 M.T.G. j obs. alt. Sun's L.L., 14 14' 50". 240 TEXT-BOOK ON NA VIGA TION. Bearing at first obs. N. by W. ; run between sights, N. by W. 12 knots per hour. 9. September 21st., lat. S., long. 94 26' E. ; with the following observa- tions of Achernar (o Eridani) find lat. at second obs. ; ind. cor., + 2' 34" ; eye, 33 ft. : Approx. M.T.S., 10* 6 ra P.M. ; chron., 3 h 40 0' ; obs. alt., 55 53' 50" ; bearing, S. 31 E. Approx. M.T.S., 5 h 46 m A.M. ; chron., ll h 20 m 20' ; obs. alt., 49 13' 30'" ; run, N. 31 W. 63 miles in the interval. The chron. was 8 m 23' slow of G.M.T. 10. August 27th, about 2 h 35 m A.M. at ship, in long. D.R. 156 E. ; when a chron. correct for M.T.G. showed 4 h ll m 28 ; the obs. alt. of Aldebaran was 35 11' 20" ; no ind. cor. ; at the same time the obs. alt. of Markab was 47 49' 40" ; ind. cor., nil ; eye, 45 ft. Find the position of the ship and true bearings of the stars by Sumner's method, assuming lat. 49 30' N. and 50 N., or by any other method. 11. September 10th, in long. 91 14' E., lat. D.R. 27 S. ; with following observations find lat. at second obs. ; ind. cor., nil ; eye, 25 ft. : Approx. A.T.S., 10 h 40 m A.M. ; chron., 4 h 28 m 3 ; obs. alt. Sun's L.L., 51 52' 10". Approx. A.T.S., 3 h 50 P.M. ; chron., 9 h 38 m 8' ; obs. alt. Sun's L.L., 26 25' 40". Bearing at 1st obs., N. 35 E. ; run during interval, N. 80 W. 12 knots per hour. The chron. was correct for G.M.T. 12. September 6th, lat. D.R. 13 S., long. 30 25' W. ; with following observations of Altair, find lat. at second obs. ; ind. cor., + 3' 18" ; eye, 25 ft. : Approx. M.T.S., 9 h 50 P.M. ; G.M.T., ll h 50 2' ; obs. alt., 62 13' 40" ; bearing, N. 30 W. Approx. M.T.S., 1" 20 m A.M. ; G.M.T., 3 h 20 m 15' ; obs. alt. , 18 4' 50". Run during interval N. 36 W., 14 knots per hour. 13. September 16th, about 4 h 45 A.M., in long. by acct. 127 30' E.; when a chron. which was 12 m 24 fast of G.M.T. showed 8 h 23 m 12 , the obs. alt. of Rigel was 49 28' 10" ; ind. cor., - 3' 15" ; and at the same time the obs. alt. of Fomalhaut was 27 25' 50" ; ind. cor., + 2' 11" ; height of eye, 27 ft. Find the position of the ship and true bearings of the stars by Sumner's method, assuming lats. 46 40' S. and 47 20' S., or by any other method. 14. September 28th, in lat. N., long. 6 15' W. ; with following observa- tions of the Sun's L.L. find lat. at first obs. ; ind. cor., - 2' 12" ; eye, 30 ft. : Approx.A.T.S.,ll h 10 ra A.M.; chron., ll h 10 25'; obs. alt., 34 29' 10". Approx. A.T.S., 4 h P.M. ; chron., 3 h 59 49' ; obs. alt., 16 34' 0" ; bearing, S. 50 W. ; run during interval, N. 60 W. 10 knots per hour. The chron. was 14 m 21" slow of G.M.T. 15. September 1st, in lat. N., long. 128 35' W. ; with following observa- tions of o Oygni (Deneb) find lat. at first obs. ; ind. cor., + 3' 19" ; eye, 28 ft. : Approx. M.T.S., 8 h P.M. ; chron., 4 h 40 10' ; obs. alt., 66 5' 20". Approx. M.T.S., 4 h 10 A.M. ; chron., 12 h 50 20*; obs. alt., 21 21' 30" ; bearing, N. 20 W. The chron. was 1 20* fast of M.T.G. ; run during the interval, N. 34 E. 12 knots per hour. CHAPTER XXIII. FINDING ERROR OF CHRONOMETER. ART. 106. Error of Chronometer by a Single Altitude of the Sun or a Star. For this purpose it is necessary to know the correct latitude and longitude of the place of observation. The altitude would therefore be usually observed in port, using an artificial horizon. The computation of the hour angle is exactly the same as in the problem of finding the longitude by chronometer. Thence the mean time at place and mean time at Greenwich are found, and the error of chronometer determined. Example. August 30th, at 9 h 3 m A.M. M.T.S., in lat. 40 20' N., long. 25 30' W. ; the obs. alt. of the Sun's L.L. in an artificial horizon was 79 10' 20" ; ind. cor., - 3' 18" ; time by chron., 10 h 45 ra 15*. .Required its error on M.T.G. and A.T.S. M.T.S. Aug. 29" 2l b 3 m s 25 30' Dec. 9 14' 44" -3 N. H.V. 53"-49 Long. + 1 42 4 Cor. + 1 6'9 T25 M.T.G. 29 d 22 45 6,0)10,2 9 15 51'2 26745 90 10698 l h 42 m 5349 N.P.D. 80 44 8 , 66-8625 Obs. alt. 79 10' 20" Eq. T. O m 45' -9 H.V. 75 I.C. - 3 18 Cor. 4- '9 1-2 2)79 7 2 46-8 -9,00 39 33 31 S.D. + 15 52 39 49 23 R. & P. - 1 1 T, A. 39 48 22 242 TEXT-BOOK ON NAVIGATION. T.A. 39 48 22 Lat. 40 20 sec 10 '117879 P.D. 80 44 8 cosec 10*005702 2)160 52 30 80 26 15 cos 9-220432 40 37 53 sin 9 '813708 E.H.A. 2 h 58 m 16' xxxi. =9 '157721 A.T.S. 29 d 21 1 44 Eq. T. + 46'8 M.T.S. 29 d 21 2 30-8 Long. + 1 42 M.T.G. 29 d 22 44 30'8 Chron. 29 d 22 45 15 Fast 44'2 on M.T.G. Fast 1 43 31 on A.T.S. The H.A. might also have been determined, direct from the figure, as in the case of the chronometer. In the triangle PZX, the three sides are given to find angle P. ART. 107. Error of Chronometer by Equal Altitudes. (a) Equal Altitudes of the Sun. If the times by chronometer when the Sun has equal altitudes east and west of the meridian be noted, the mean of these times, with a correction on account of the change of declination during the interval, gives the time shown by the same chronometer when the Sun's centre crosses the meridian, that is, apparent noon. This affords a means of finding the error of the chronometer. The observation can best be made on shore with an artificial horizon, the latitude and longitude of the place of observation being accurately known. If there were no change of declination, it is obvious that the hour angles E. and W. would be equal, and the mean of the times would be what the chronometer showed at apparent noon. But as the Sun's declination changes, the middle time would fall on one side or the other of apparent noon by a small interval called the " Equation of Equal Altitudes." Thus, when the Sun's polar distance increases, the east H.A. would be greater than the west H.A., because it would take less time to fall to the same altitude, and the middle time would be a few seconds before apparent noon. Again, when the polar distance decreases, the west H.A. would be the greater, because it would take a longer time for the Sun to attain the same altitude, and so the middle time would then be a few seconds after apparent noon. The object of the calculation is therefore to find the correction to be applied to the middle time to find the exact time by FINDING ERROR OF CHRONOMETER. 243 chronometer when the Sun's centre was on the meridian, and thence the error of the chronometer. FIG. 165. FIG. 366. The working out of the formula gives the correction in two parts. The first part is marked + or according as the Sun's polar distance is increasing or decreasing. The second part is marked + or according as the declination is increasing or decreasing. The algebraical sum of the two parts is the Equation of Equal Altitudes with the proper sign. The elements required for the calculation are, the Time Interval, the tfuris Declination, and the Equation of Time, with the lati- tude and longitude of the place of observation. (b) Equal Altitudes of a Star. The intervals from the time when a star crosses the meridian to the times when it has equal altitude east and west of the meridian are exactly equal, because its declination does not change. Hence the middle time by chronometer would be the time of meridian passage, and as the time of meridian passage can be found independently, the error of the chronometer can be determined. FIG. 167. Examples. (1) August 1st, at Valparaiso, lat. 33 2' S., long. 71 42' W., 4 he sun had equal alts, at 2 h 16* 12' A.M., and 7 h 20 m 42 P.M., by chron. Find its error on M.T.G. and A.T.S. 244 TEXT-BOOK ON NAVIGATION. A.T.S. Aug. l d O h O m 71 42' Long. +4 46 48 4 A.T.G. l d 4 46 48 286'48 First obs. 2 h 16 m 12' A.M. Second obs. 7 20 42 P.M. Int. 5 4 30 Int. 2 32 15 Mid. Time 4 48 27 Dec. 18 11' 45" N. H.V. 37"'26 Eq. T. 6 m 10'-22 H.V.-13 Ch.ofDec. Cor. -259 4-8 Cor. -'63 4'8 in 2 days 30' 24" 18 8 46 N. 29808 6 9'59 104 14904 52 1824 Int. 5 b 4 m 30' Lat. 33 2' 6,0)17,8-848 2' 59" log A, 7-7571 tan 9-8131 634 ... logB, 7'6533 Dec. 18 9 tan 9-6156 y ch. 1824 ... ... log 3-2610 ... log 3-2610 Pol. dist. dec. Part Dec. ,, I. II. log 0-8312 - 6-78 - 2-69 log 0-4299 Eq. of equal alts. 9'47 Mid. time by chron. 4 h 48 m 27' Watch showed at approx. noon 4 48 17'53 000 A.T.G. 4 h 46 ra 48' Eq. T. -f 6 9-59 M.T.G. 4 52 57-59 Watch 4 48 17'53 Watch fast A.T.S. 4 48 17'53 Slow M.T.G. 4 40'06 (2) September 30th, the following times by chron. were noted in lat, 7 56' S., long. 14 26' W., when Markab had equal alts. Star E. 7 h 34 m 50' Star W. 13 h 3 m 53' Find error of chron. on G.M.T. " B. A. Markab 23 h O m 4'-66 First obs. 7 h 34 50* B.A.M.S. 12 33 17-38 Second obs. 13 3 53 Approx. M.T. transit 30 d 10 h 26 m 47"28 Long. W. + 57 44 Int. 5 29 3 ilnt. 2 44 31-5 Mid. time 10 19 21-5 M.T.G. 30 d 11 24 31-28 R.A.M.S. 12 h 33 m 17'-38 1 48-42 Accel. 3-94 08 12 35 9-82 R.A.* 23 4-66 M.T.S. of transit 10 24 54'84 Long. 57 44 M.T.G. of transit 11 22 38-84 Mid. time 10 19 21*5 Chron. slow 1 3 17'34 on G.M.T. FINDING ERROR OF CHRONOMETER. 245 EXERCISES. 1. August 4th, at 8 h 40 A.M. A.T.S., in lat. 38 47' S., W 12' E. ; tie obs. alt. of the Sun's L.L. was 16 42' 10" ; ind. cor., + 2' 15"; eye, 29 ft., when a chron. showed 2 h 45 m 10*. Find its error on M.T.G. and A.T.S. 2. August 31st, at 9 h 22 m A.M. A.T.S., in lat. 52 25' N., long. 3 20' W. ; the obs. alt. of the Sun's U.L. was 36 26' 10" ; ind. cor., - 26" ; eye, 41 ft. ; if a chron. indicated 9 h 20 19 s , what was its error on M.T.G. 3. August 19th, at 4 h 26 m P.M. A.T.S., when a point in lat. 41 35' 30" N., long. 70 30' 42" W., bore N. 35 W., distant 14 miles; the obs. alt. of the Sun's U.L. in artificial horizon was 52 53' 10" ; ind. cor., - 3' 30" ; if a chron. showed 9 h 10 m , find its error on M.T.G. 4. September 30th, at 3 h 30 m P.M. A.T.S., when a point in lat. 49 42' S., long. 179 44' 30" E., bore S. 56 W., distant 9 miles ; the obs. alt. of the Sun's L.L. in artificial horizon was 49 57' 10"; ind. cor., + 3' 56"; eye, 50 ft. If a chron. showed 3 h 15 m 20 s , find its error on M.T.G. 5. August 16th, at 9 h 45 m P.M. M.T.S., when a point in lat. 24 17' 12" N. f long. 14 9' 30" W., bore S. 27 E., distant 11 miles ; the obs. alt. of Fomal- haut in artificial horizon was 31 46' 20" ; ind. cor., + 2' 12", when a chron. showed ll h . Find its error on M.T.G. 6. September 21st, at 10" 6 ra P.M. M.T.S., in lat. 43 19' S., long. 94 26' E. ; the obs. alt. of Achernar was 54 50' 50", when a chron. showed 3 h 40 m ; ind. cor., + 2' 34" ; eye, 33 ft. Find error on M.T.S. and M.T.G. 7. August 22nd, at l h 53 m A.M. M.T.S. , when a point in lat. 50 23' 24" N., long. 130 6' 30" W., bore N. 70 E., distant 10 miles ; the obs. alt. of Vega in artificial horizon was 69 44' 20" ; ind. cor., 1' 16" ; a chron. showed 10 h 35, find its error on M.T.G. 8. August 22nd, in lat. 27 10' N., long. 56 17' E., the following times were observed when the Sun had equal alts. Find the error of the chron. for A.T.S. and M.T.G. A.M. 4 h 19 27* P.M. 0" 7 m 30 4 25 32 12 Jo CHAPTER XXIV. CALCULATION OF ALTITUDES. ART. 108. Computing an Altitude. In some problems of Nautical Astronomy it is necessary to compute the altitude of a celestial object. Especially is this the case when setting ques- tions, so that the altitude shall be consistent with the other data. The data required to be known are the hour angle of the object and the latitude and longitude of the place. The following method has the advantage of requiring only one table of logarithms. In the figure, S is the object, PS the polar distance, the angle ZPS the hour angle, PZ the co-lat., and ZA an arc of a great circle perpendicular to PS. Let PA = 0; thenSA = (p~B) In the right-angled spherical triangle PAZ cos h = cot PZ. tan PA = tan I . tan .*. tan = cos h . cot I (1) In the same triangle cos PZ = cos PA . cos AZ i.e. sin I = cos . cos AZ .'. cos AZ = sin I . sec 9 (2) CALCULATION OF ALTITUDES. 247 Also in the triangle SAZ cos SZ = cos AZ . cos AS i.e. sin a = sin I . sec 9 . cos (p ~ 6) by substituting the value of cos AZ from (2). NOTE. (p ~ 6) signifies the difference between p and 6, because one or the other may be the greater according as the perpendicular ZA falls within or without the triangle PZS. For computation the following arrangement is convenient (I = lat., a = alt., p = polar distance, h = hour angle) : r h ... cos I cot sin ... tan ... sec p p ~0 ... ... ... ... cos a ... ... ... ... sm The altitude may also be calculated by the log rising method, by transposing formula proved in the ex-meridian chapter, which becomes sin alt. = cos zen. dist. cos lat. cos dec. log rising hour angle, or by direct spherics. ART. 109. The first method may be stated as follows : (1) Find the hour angle of the object. (2) Correct declination, find polar distance. (3) Add cos H. A. and cot lat. ; this gives tan arc 1. Difference between arc 1 and pol. dist. gives arc 2 : add sin lat., sec arc 1, and cos arc 2 ; the result is sin true alt. (4) Apply refraction and parallax to true alt. for app. alt. Example. September 7th, at 8 h 51 m 32 s A.M. apparent time at ship, in lat. 48 10' N., long. 149 22' 15" W. Find true and apparent altitudes of the Sun. A.T.S. Sept. 6 d 20 h 51 m 32 s 149 22' 15" Dec. 6 19' 18"'6 N. Long. W. + 9 57 29 4 Cor. - 6 22 A.T.G. Sept. 7 d 6 49 1 6,0)59,7 29 6 12 56'6 N. 90 9 h 57 m 29' N.P.D. 83 47 3 H.V. 56-01 6-82 24 h O m 0* 20 51 32 6,0)381-99 E.H.A. 3 8 28 6' 22" TEXT-BOOK ON NAVIGATION. H.A. 3 h 8 m 28' Lat. 48 10 Arc 1, 31 20 53 P.D. 83 47 3 cos = 9*832833 cot = 9-951896 tan = 9784729 Arc 2, 52 26 10 True alt. 32 7 58 Par. - 8 32 7 50 Ref. +1 31 sin = 9-872208 B ec = 10-068531 cos = 9-785078 sin = 9-725817 App. alt. 32 9 21 Rules for log rising method. (1) Find the hour angle of the object. (2) Correct declination ; find zenith distance by adding dec. and lat. if of contrary name, and subtracting them if of same name. (3) Add cos lat., cos dec., and log rising hour angle, increasing index by 1 ; reject 10's and take back log, Table xxiv. ; this gives Nat. No." Subtract " Nat. No." from Nat. cos. zen. disk, and take back Nat. sin true alt. (4) Apply ref. and par. for app. alt. Example. H.A. 3 h 8 m 28 s Lat. 48 10' N. Dec. 6 32 57 N. log rising = 5*504460 cos = 9-824104 cos = 9-997440 Z.D. 41 57' 3" T.A. 32 7 57 R. and P. + 1 23 A.A. 32 9 20 211838 N. cos 743718 log = 5-326004 5926 N. sin 531880 205)7800(38 615 1650 1640 10 From figure. In figure X is the position of the object, AX is alt. of object. In triangle PZX, Given co-lat. = PZ Pol. dist. = PX Hour angle = P Required zen. dist. = ZX CALCULATION OF ALTITUDES. 249 py _ 000 A*7' Q" p PZ= 41 50 | =lh 34m 14 cot = 10-360490 sin = 9-601715 PX+PZ = 125 37 3 J(PX + PZ) = 6248'32" sec = 10-340122 cos= 9'659878 PX-PZ= 41 57 3 J(PX-PZ) = 20 58 32 cos= 9-970222 77 57 16 tan = 10-670834 sec = 10*680503 28 56' 3" cos= 9-942096 2 ZX = 57 52 6 90 T.A. =32 7 54 R. and P. + 1 23 A.A. 32 9 17 EXERCISES. 1. September 7th, at ll h 14 35 s A.M. apparent time at ship, in lat. 48 10' N., long. 149 22' 15" W. Find true and appar. alts, of Sun. 2. September 29th, at 4 h 32 ra 46 s P.M. M.T.S., in lat. 46 50' N., long. 180 W. Find true and appar. alts, of Sun. 3. August 2nd, at 9 h 15 m 20" P.M. M.T.S., in lat. 23 12' N., long. 62 35' W. Find true and appar. alts, of Antares. 4. September 28th, at 4 h 5 m 20' A.M. M.T.S., in lat. 47 42' S., long. 84 25' E. Find true and appar. alts, of a Hydrse. 5. August 10th at 8 k 20 41' A.M. M.T.S., in lat. 37 14' K, long. 14 26' W. Find true and appar. alts, of the Moon. 6. August 12th, at 4 h 58 m 48 A.M. M.T.S., in lat. 25 19' S., long. 169 37' E. Find true and appar. alts, of Saturn. 7. August 30th, at 8 h 30 m 25' P.M. M.T.S., in lat. 10 15' S., long. 98 27' 15" E. Find true and appar. alts, of the Moon. 8. August 9th, at 5 h 4 m 2' A.M. M.T.S., in lat. 5 14' N., long. 20 27' W. Find true and appar. alts, of Jupiter. CHAPTER XXV. LONGITUDE BY LUNAR DISTANCES. ART. 110. In a lunar observation the Greenwich time is found by measuring the distance between the moon and certain bright objects which lie in or near her path. The bodies chosen are the Sun, Yenus, Mars, Jupiter, Saturn, and the bright stars a Arietis, Aldebaran, Pollux, Regulus, Spica, Antares, a Aquilae, Fomalhaut, and a Pegasi. In the NJL, pp. XIII. to XVIII., 1 are given for each day the distances east or west, between the moon and certain of these objects at intervals of three hours of M.T.G. These true distances are the angles between the centres of the Moon and object selected, as measured at the Earth's centre. The M.T.G. corre- sponding to any other distance may be found by proportion, but owing to the variation in the uniformity of the motion a correction for " second differences " found in the N.A. must be applied before the correct G.M.T. is known. Before the invention of chronometers the " Lunar Problem " was almost the only means available at sea for finding the Greenwich time, on which the determination of longitude depends. It may still be of great use in the event of chronometers " running down " or becoming unserviceable. The proportion referred to may be worked by arithmetic, but the use of " Proportional Logarithms " is simpler and more general. These are found in Norie, Table xxxiv. The proportional logarithm of a given quantity is the logarithm of the ratio of some constant greater than that quantity to the given quantity. In the N.A. the constant is 3 hours, or 10,800 seconds, and the prop, log (P.L.) for any other time, for instance, l h 42 ra 11 s , is = log lh ^ ns = log 10800 - log 6131 = 4-033424 - 3787531 = 0-245893 which is entered in tables as 2459, the nearest fourth place only being used. This table facilitates the finding of a time corre- sponding to a true distance, as by it only two references are necessary, whereas by ordinary logs four entries are made before obtaining the result. 1 Discontinued since 1908. LONGITUDE BY LUNAR DISTANCES. 251 Suppose x and y the distances in the N.A.. between which the true distance d lies, and t the number of hours elapsed since the time for which x is tabulated. Then y x is change in distance in three hours, d x the required time t. Hence y x : d x :: 3 : t or = t d x CM TTL-_ W f)l - <7* which is P.L. t = P.L. (d - x) - P.L. (y - x) The quantity P.L. (y x) is tabulated in N.A. opposite the distance x. In using these distances the instructions in the N.A. are " Take the difference between the reduced distance and the nearest distance preceding it, in order of time, in the ephemeris ; from the P.L. of this difference subtract the P.L. in the ephemeris ; the remainder will be the P.L. of a portion of time to be added to the hour answering to the nearest preceding distance, to obtain the approximate Greenwich mean time corresponding to the given distance." When selecting a suitable object for observation, it must be remembered that those objects are most favourably situated from which the moon is receding, or to which the moon is approaching most rapidly, and this is shown by the smallness of the P.L. opposite to it ; attention should also be paid to the object having as small a difference in declination from the moon as possible compared with their difference in right ascension. Examples. (1) The true distance between the Moon and o Arietis on August 13th is 37 51' 32". FindG.M.T. Tr. dist. 37 51' 32" N.A. 37 20 18 P.L. 3493 ... Aug. 13 d 12 h 0" 0* P.L. 3527 31 14 P.L. 7607 P.L. 4114 1 9 48 3463 Approx. M.T.G. 13 d 13 9 48 With diff. - 32 of P.L.'s and 2)64 Int. l h 10 ra cor. for 2nd diffs. = +9 Mean diff. 32 M.T.G. 13 d 13 9 57 (2) The true distance between the Moon and Spica on August 26th is 431'29". FindG.M.T. 25 2 TEXT-BOOK ON NAVIGATION. Tr. dist. 43 1' 29" 50 7 P.L. 2362 2352 P.L. 2357 P.L. 5553 P.L. 3196 Aug. 26 d 15 h O ra 0' 1 26 14 Approx. M.T.G. 26* 16 26 14 2)10 With diff. -5 of P.L.'s and Int. 2 h 16 ra cor. for 2nd diff. = +1*5 Mean diff. 5 M.T.G. 26 d 16 26 15'5 EXERCISES. 1. The true distance between the Moon and Aldebaran on August 6th is 80 29' 7". Find G.M.T. 2. The true distance between the Moon and Jupiter on August 7th is 89 21' 12". FindG.M.T. 3. The true distance between the Moon and a Pegasi on August 12th is 69 40' 15." Find G.M.T. 4. The true distance between the Sun and Moon on August 22nd is 42 35' 17". Find G.M.T. 5. The true distance between the Moon and Spica on August 28th is 65 32' 11." FindG.M.T. ART. 111. The Lunar. Let ZX and ZY be two vertical circles passing through the apparent posi- tions M' and S' of the Moon and another celestial object (Sun, Star, or Planet) ; the true positions M and S, as seen from the Earth's centre, will be on the same vertical circles, and differ from the appa- rent positions by the effects of re- fraction and parallax. The Moon's parallax exceeds the refraction, but the parallax of any other body is less than the refraction. Hence the Moon's apparent place is lower, and that of the other body higher, than the true positions. M'S' and MS are arcs of two great circles, and represent the apparent and true distances. XM' and XM are the apparent and true altitudes of the Moon, and YS' and YS the apparent and true altitudes of the other body. The angle Z is common to the two triangles ZM'S' and ZMS. The problem is to find the true distance MS, the apparent distance M'S', and the true and apparent altitudes being given. In the spherical triangle ZM'S' are given the three sides, Fm. 171. LONGITUDE BY LUNAR DISTANCES. 253 ZM', ZS', and M'S', to find the angle Z. Then in the triangle ZMS, the two sides ZM and ZS and the included single Z being known, the third side MS or true distance can be found. The problem can therefore be solved by simply applying the formulae of Cases I. and II. of oblique-angled spherical trigo- nometry, but as this involves two separate computations, several methods have been devised for clearing the distance by a single logarithmic computation. The method most generally used is known as Borda's, which requires only one table of logarithms. It is obtained as follows : Applying the general formula, cos A = cos - cos 6 . cos c sin b . sin c i- i 7M'' 7 cos M'S' - cos ZM' . cos ZS' .... In the triangle ZM S , cos Z = sin ZM - . sin zs - (1) In the triangle ZMS, cos Z = co* MS - cos ZM cos ZS ^ ( sin ZM . sm Zo Let m and m' represent the Moon's true and apparent altitudes ; s and s' the true and apparent altitudes of the other body; d and d' the true and apparent distances ; Then equation (1) becomes cos d' sin m' . sin s'\ cos Z = and equation (2) becomes cos m' . cos s' cos d sin m . sin s cos Z = . by putting altitudes I for zenith distances ___ cos m . cos s cos m' . cos Adding 1 to each side .. . cos d sin m . sin s n cos d' sin m . sin s' i -j- = i -f- . . COS m . COS S COS m . COS S cos d + cos 771 . cos s sin m . sin s cos m . cos s cos d' -f cos m f . cos s' - sin m' . sin ' cos m' . cos s' that is cos d + cos (m + s) cos d' + cos (m' -f s') cos m . cos 8 cos m' . cos s' .-. cos d + cos ( + *)= CQS ' { cos <*' + cos K + ')} 254 TEXT-BOOK ON NAVIGATION. But cos d = 1 - 2 sin 2 %, and cos (m + s) = 2 cos 2 m + 8 - 1 J 2 .-. 1 -2 S m 2 | + 2cos 2 ?i- s -:i = -2 sin 2 | ~M COS 771 . COS 8 . {cos d' + cos (m' + s')} cos m . cos s m + s + c - T --- . . Z COS --- - - . COS cos m . cos s 2 2 Dividing both sides by 2, and putting sin 2 6 for cos m . cos s m' + ' + cZ' w/ -f s' d' . COS ^r COS _ cos m' . cos s' we have -sin' t :. sin 2 = cos 2 - sin 2 2 2 + cos (m + s)} - J(l - cos 20) {cos (m + s) + cos 20} FORM FOR COMPUTATION. App. dist. (d') --- Moon's alt. (m') --- ... sec Sun's or Star's ,, (s) --- ... sec cos - ... cos Moon's true alt. (m) --- ... cos Sun's or Star's (s) --- ... cos (m + a) = 2) --- LONGITUDE BY LUNAR DISTANCES. 255 .(*+-'+.) (-+- ... sin cos ... cos ... sin Remark. It will be noticed that the first four logs are the same as in the alt.-azimuth problems, and that all other logs taken out are cosines, whilst the angles sought correspond to sines. ART. 112. Rule for clearing the Distance by Borda. Add together apparent distance, app. alt. of moon, and app. alt. of body observed for distance; find half sum, and the remainder, which is the difference between the app. dist. and half sum. Then add following logs : Sec app. alt. of moon, sec app. alt. of body, cos half sum, cos remainder, cos true alt. of moon, and cos true alt. of body ; reject 40 from index, and divide by 2, and look back sine of angle, which call B. Now find half sum of two true alts., which call A. Add together the cosines of the sum and difference of A and B ; divide sum by 2, and look out sine of angle, which is half true distance. Example. The apparent distance between the Sun and Moon's centre is 105 56' 46" ; Moon's app. alt., 26 7' 19", true alt., 26 55' 26" ; Sun's app. alt., 14 24' 27", true alt., 14 20' 56". Find true distance. App. dist. 105 56' 46" Moon's A.A. 26 7 19 sec 10*046792 Sun's A.A. 14 24 27 sec 10-013878 2)146 28 32 Moon's T.A. Sun's T.A. 73 14 16 32 42 30 26 55 26 14 20 56 2)41 16 22 L 20 38 11 5 29 24 14 cos 9-459997 cos 9-925019 cos 9-950174 cos 9-986236 A + B 50 2 25 B - A 8 46 3 2)19-382096 sin 9-691048 cos 9-807703 cos 9-994895 2)19-802598 52 49 2 True Dist. 105 38 sin 9-901299 2 5 6 TEXT-BOOK ON NAVIGATION. From the figure. In the figure AX and AN are the true and apparent alts, of the Sun. BY and BM are the true and app. alts, of the Moon. MN is the app. dist., and XY the true dist. In the triangle ZMN, given the three sides, ZM, Moon's a.z.d.: ZN, Sun's a.z.d. and MN, to find Z. /sin s . sin (s MN) sin ZM . sin ZN where s = J(ZM + ZN + MN). MN = 105 56' 46" ZM = 63 52 41 ZN = 75 35 33 Z _ / 2 V 2)245 25 s = 122 42 30 8 - MN = 16 45 44 cosec 10-046792 cosec 10-013877 sin 9-925019 sin 9-459995 6 45 2)19-445683 cos 9-722841 Now, in triangle ZXY, given two sides, ZY, Moon's t.z.d. ; ZX, Sun's t.z.d., and included angle Z (just found), find XY, true dist. ZX= 75 39' 4" ZY= 63 4 34 -58 6' 45" cot 9-793890 sin 9-928953 ZX+ZY = 138 43 38 (ZX + ZY) = 69 21 49 sec 10-452920 cos 9-547080 ZX-ZY= 12 34 30 (ZX-ZY) = 6 17 15 cos 9-997379 = 60 19 13-5 tan 10-244189 sec 10-305264 52 49' 1" cos 9-781297 2 True dist. 105 38 2 EXERCISES. 1. Given app. dist., 96 52' ; Sun's app. alt., 36 18' 7", true alt., 36 16' 57" ; Moon's app. alt., 24 17' 14", true alt., 25 5' 27". Find true dist. 2. Given app. dist., 107 35' 18" ; Saturn's app. alt., 9 51' 21", true alt., 9 46' 2" ; Moon's app. alt., 38 43', true alt., 39 27' 20". Find true dist. 3. Given app. dist., 101 0' 4" ; Mars' app. alt., 14 36' 7", true alt., 14 32' 35" ; Moon's app. alt,, 48 10' 5", true alt., 48 49' 29". Find true dist. 4. Given app. dist., 56 14' 32" ; Sun's app. alt., 65 15' 36", true alt., 65 15' 13" ; Moon's app. alt., 40 32' 16", true alt., 41 15' 48". Find true dist. LONGITUDE BY LUNAR DISTANCES. 257 ART. 113, Rules for Lunar Problem. (1) Find the G.M.T. (2) Correct the R.A.M.S., the Moon's R.A,, Dec., S.D., and H.P. Correct the R.A., and Dec., of object observed for distance. (3) Correct the obs. alt., finding app. and true alts., and calculate H.A. of object ; if both alts, are given, either may be chosen. (4) With H.A. just found, calculate R.A.M. and H.A. of the other object, and thence its alts, true and app. (unless given). (5) Correct observed distance for ind. cor. and S.D., adding for Near limbs, and subtracting for Far or remote limbs. In the case of a Sun Lunar, the enlightened limbs must be the near limbs. (6) Clear the distance. (7) Find corresponding G.M.T. and error of chronometer. (8) From R.A.M. in (4) find M.T.S. and longitude. Example. August 16th, P.M. at ship in lat. 20 29' S., when the time by a chron. was 16 d 2 h 14 m 50', which was estimated 2 m 56 s fast of M.T.G. ; the obs. alt. of the Sun's L.L. was 48 1' 20" ; ind. cor., + 2' 12" ; height of eye, 27 ft. ; observed distance between the enlightened limbs of Sun and Moon, 41 3' 10" ; ind. cor., 2' 4". Find error of chron. for G.M.T. and longitude. Time by chron. Aug. 16 d 2 h 14 m 50' R.A.M. S. 9 h 35 m 52'-5 Fast -2 56 - Accel. 217 M.T.G. 16 d 2 11 54 9 36 14-2 S.D. 15' l"-4 4" H.P. 55' 2"'4 14" -6 Cor. +7 2-2 Cor. +27 2 '2 15 2-1 12)8-8 55 5'1 12)32-12 Aug. +3-2 Red. -1-3 .7 . 27 15 5-3 55 3-8 Moon's Dec. 20 17' 41" -5 N. Var. '25 R.A. 6 h 51 m 0'-6 Var. 2'18 Cor. - 3 11-9 Cor. + 25*9 11-9 20 17 38 N. 2-97 6 51 26 '5 25 -942 Sun's Dec. 13 57' 31"-4 N. H.V.47'03 R.A. 9 h 40 m 9'-l H.V. 9'37 Cor. -1 43-4 2-2 Cor. 20'6 2'2 13 55 48 N. 103-46 9 40 29 '7 20 '592 O.A.0 48 1'20" I.C. + 2 12 48 3 32 Dip - 5 5 47 58 27 S.D. + 15 49 Q's A. A. 48 14 16 R. & P. - 45 Q's T.A. 48 13 31 258 TEXT-BOOK ON NAVIGATION. Q'sT.A. 48 13 31 Lat. 20 29 sec 10 "028365 P.D. 103 55 48 cosec 10-012964 2)172 38 19 86 19 10 cos 8-807492 38 5 39 sin 9 '790254 Q's W.H.A. l h 36 ra 22-5 xxxi. = 8-639075 Q'sR.A. 9 40 29-7 R.A.M. 11 16 52-2 ))'sRA. 6 51 26-5 ])'s W.H.A. 4 25 25-7 log rising 5-777405 Lat. ... 20 29' 0" S. cos 9-971635 Dec. ... 20 17 38 N. cos 9-972169 526271 log 5-721209 Z.D. 40 46' 38" Nat. cos 757255 Vs T.A. 13 21 18 N. sin 230984 Cor. - 49 33 ))'s A.A. 12 31 45 Obs. dist. 41 3' 10" I.C. - 2 4 41 1 6 Q's S.D. 15 49 D's S.D. 15 5 App. dist. 41 32 ])'sA.A. 12 31 45, sec 10-010468 Q's A.A. 48 14 16 sec 10-176498 2)102 18 1 51 9 1 cos 9-797461 9 37 1 cos 9-993854 D'aT.A. 13 21 18 cos 9*988094 Q'sT.A. 48 13 31 cos 9*823607 2)61 34 49 A. 30 47 24-5 2)19*789982 B. 51 44 27-5 sin 9-894991 A + B 82 31 52 cos 9-113903 B - A 20 57 3 cos 9-970295 2)19-084198 20 23 26 sin 9-542099 2 True dist. 40 46 52 LONGITUDE BY LUNAR DISTANCES. True dist. 40 46' 52' 259 N.A. 41 49 2 P.L. 3207 2 10 P.L. 4617 P.L. 1350 Aug. 16 d O h O m 0' 2 11 54-5 Approx. M.T.G. 16" 2 11 54'5 Cor. for, - 12, Diff. P.L.'s, and Int. 2 h 10 m from N.A. + 3 M.T.G. 16 d 2 11 57-5 Time by chron. 16 d 2 14 50 R.A.M. H b 16 m 52'-2 E.A.M.S. 9 36 14-2 Fast 2 52-5 M.T.S. 16 d 1 40 38 M.T.G. 16 d 2 11 57-5 31 19-5 Long. 7 49' 52" W. '. Prom the figure. In the figure X and A are true and app. positions of the Sun; Y and B are true and app. positions of the Moon. First, to Jin d Sun's H.A . In triangle PZX, given co-lat. = PZ, Pol. dist. = PX, and zen. dist. = ZX, to find P. r 38 ? _ /sin * - sin (s - ZX) 2 ~ V sin PX . sin PZ where s = *(ZX + PX + PZ). ZX = 41 46' 29" PX = 103 55 48 PZ = 69 31 2)215 13 17 s = 107 36 39 s - ZX = 65 50 10 n .. 5 cosec 10-012964 cosec 10-028365 sin 9-979154 sin 9-960174 2)19-980657 cos 9-990328 0'sW.H.A. 1 36 23 Q'sRA. 9 40 29-7 R.A.M. 11 16 D'slLA. 6 51 52-7 26-5 ))'sH.A. 4 25 26-2 260 TEXT- BOOK ON NAVIGATION. Second, to find Moon's True and Apparent Altitudes. In triangle PYZ, given PZ = co-lat., PY = ])'s Pol. dist. , P = Hour angle, to find ZY = )'s true Z.D. PY = 11017'38 PZ = 69 31 = 2 h 12 m 43'-l cot 10184507 sin 9738198 2 PY+PZ = 179 48 38 J(PY+PZ) = 89 54' 19" sec 12-781671 cos 7'218329 PY-PZ= 40 46 38 |(PY-PZ) = 20 23 19 cos 9*971902 89 56 2-2 tan 12-938080 sec 12 938080 38 19' 23"-5 cos 9-894607 2 J)'s true Z.D. = 76 38 47 .A. 13 21 13 Cor. - 49 33 J)'s A.A. 12 31 40 Third, to clear the Distance. In triangle ZAB, given AB = app. dist., ZA = Q's A.Z.D., ZB = })'s A. Z.D. to find angle Z. Q % _ /sin s . sin (s AB) 2 ~ V sin ZA . sin ZB where a = J(AB + ZA + ZB). AB = 41 32' 0" ZA = 41 45 44 cosec 10-176499 ZB = 77 28 20 cosec 10-010466 2)160 46 4 s = 80 23 2 sin 9-993855 8 - AB = 38 51 2 sin 9*797469 2)19-978289 T ~ = 12 45 27'5 cos 9-989144 a Finally, in triangle ZXY, given ZX = Q's true Z.D. , ZY = })'s true Z.D , and angle Z, to find XY = true dist. c XY _ cos^ZY + ZX) . Z 2 ' cosX + Y ' 2 LONGITUDE BY LUNAR DISTANCES. 261 ZY = 76 38' 47" ZX = 41 46 29 | = 12 45' 27-5" cot 10-645090 sin 9 '344053 ZY + ZX = 118 25 16 i(ZY+ZX) = 59 1238 seclO'290828 cos 9709172 ZY-ZX = 34 52 18 |(ZY-ZX) = 17 26 9 cos 9'979573 83 425 tan 10-915491 sec 10-918672 20 23 26" cos 9-971897 2 True dist. 40 46 52 Finish as before. Remark. Questions are sometimes set where it is not stated whether the object whose altitude is given is E. or W. of the meridian, but this may be determined by noticing in the r< Nautical Almanac" whether the other object is E. or W. of the Moon, and, taking the H.A. in conjunction with the distance, the wrong supposition would put one of the objects too near the horizon, or even below it. EXERCISES. 1. August 13th, about 10 h 30 m A.M. at ship, in lat. 35 46' S., long. D.R. 93 E. ; when a chron. estimated 6 ra fast of G.M.T. showed 4 h 20 m 15 s , the obs. alt. of the Sun's L.L. was 34 13' 30" ; ind. cor., - 1' 44" ; height of eye, 30 ft. ; obs. dist. between near limbs of Sun and Moon, 78 22' 50" ; no ind. cor. Find long, and error of chron. on G.M.T. 2. August 28th, about 3 h 35 m P.M. at ship, in lat. 24 14' N., long, by acct.. 163 40' W. ; when a chron. whose supposed error on G.M.T. was 9 m 18* slow showed 2 h 20 m 18 s , the obs. alt. of the Moon's ILL. was 14 27' 40" ; ind. cor., + 1' 11" ; eye, 35 ft. ; obs. dist. between enlightened limbs of Sun and Moon, 117 32' 40" ; ind. cor., + 2' 12". Find the error of chron. on G.M.T. and longitude. 3. August 10th, A.M. at ship, in lat. 37 50' K; when the M.T.G. by a chron. supposed correct was 9 d 3 h 49 m 18', the obs. alt. of the Moon's L.L. E. of meridian was 53 11' 20"; ind. cor., -3' 21"; eye, 31 ft.; obs. dist. between Aldebaran and the Moon's near limb, 49 17' 50" ; ind. cor., + 2' 13". Find long, and error of chron. on G.M.T. 4. August 22nd, about 6 h 30 m P.M. at ship, in lat. 29 45' S., long. D.R. 34 E. ; when a chron. estimated 10 m 18 s fast of G.M.T. showed 4 h 20 m 30', the obs. alt. of a Aquilae (Altair) was 28 21' 50"; ind. cor., + 2' 16"; eye, 40 ft. ; obs. dist. between Altair and the Moon's remote limb, 113 44' 10" ; ind. cor., 3' 11". Find long, and error of chron. on G.M.T. 5. August 13th, about 2 h A.M. at ship, in lat. 30 27' K, long, by acct., 41 W. ; when a chron. supposed correct for G.M.T. showed 4 h 38 m 12*, the obs. alt. of the Moon's L.L. was 23 12' 20" ; no ind. cor. ; eye, 35 ft. ; obs. dist. between Saturn's centre and the Moon's further limb., 78 43' 10" ; ind. cor., 1' 18". Find long, and error of chron. on G.M.T. 6. August 24th, at ll h 23 m A.M. at ship, in lat. 53 34' 1ST., long. 4 25' E.; when a chron. which had run down and been set going indicated ll h 20 m , the following observations were taken for determining long, and error ot chron. on G.M.T. ; height of eye, 28 ft. : 262 TEXT-BOOK ON NAVIGATION. Obs. alt. Sun's L.L. Obs. alt. Moon's ILL. Obs. dist. Sun and Moon. 46 53' 40" 10 15' 40" 56 24' 10" 1.0. + 1 18 I.C. - 2 18 I.C. -29 7. August 29th, at 8 h P.M. at ship, in lat. 48 43' S., long. 30 49' 45" W. ; when a chron. which had run down and been set going showed 10 h 15 m , the obs. alt. of Antares was 56 40' 20"; ind. cor., + 1' 57" ; eye, 35 ft. ; obs. dist. between Star and Moon's near limb was 35 57' 20" ; ind. cor., - 3' 33". Find the long, and error of chron. on G.M.T. 8. August 9th, about 5 h A.M. at ship, in lat. 5 14' N., long, by acct., 20 40' W. ; when a chron. estimated correct for G.M.T. showed 6 h 25 m 48 s , the obs. alt. of Jupiter's centre was 30 30' 20" ; ind. cor., 2' 16" ; obs. alt. of Moon's L.L. was 70 48' ; ind. cor., + 2' 9" ; eye, 32 ft. ; obs. dist. between Jupiter's centre and Moon's near limb, 78 3' 50" ; ind. cor., 2' 15". Find long, and error of chron. on M.T.G. 9. September 1st, about l h 30 m A.M. at ship, in lat. equator, long, by acct., 43 E.; when a chron. supposed to be 12 m 10* slow of G.M.T. indicated 10 h 25 m 32', the obs. alt. of a Pegasi (Marcab) was 67 52' 50" ; obs. dist. between Marcab and the Moon's remote limb, 44 55' 20" ; ind. cor. for alt., + 1' 31" ; for dist., 2' 17" ; height of eye, 31 ft. Find long, and error of chron. on M.T.G. PART II. CHAPTER XXVI. CONSTRUCTION OF CHARTS. ART. 114. Before proceeding to the actual drawing of a chart one or two matters must be considered. In the first place there are two geometrical problems which are continually in use : (1) To erect a perpendicular to a given straight line at a given point in the same. (2) To divide a given straight line into a required number of equal parts. There are several ways for performing each operation. Consider (1) when the given point is the end of the line. Let AB be line, through B draw a line perpendicular to AB. Select any point C above AB, with C as centre and CB as radius describe an arc of a circle cutting AB in D. A line through DC cuts the arc again in E, which is perpendicularly above B, since angle DBE is angle in a semicircle, and therefore a right angle. For (2) the trial method will give as good results as any to any one Fl0 ' 174 ' unaccustomed to use drawing instruments. Good practice in the above may be obtained by the construction of Diagonal Scales. Example. Construct a diagonal scale of 1 J inch to 1 long. , to measure miles. Draw a horizontal line, AB, near the bottom of the draw- ing-paper, and supposing the scale to be 1 inch to 1 long., along it measure off distances A 1, 1 2, 2 3, each = 1J inch ; at A and 3 erect perpendiculars AD, 3F, and measure along these perpendiculars any ten equal parts 2, 4, 6, 8, show the even points ; join these points by ten lines, which will be all parallel to AB. On the upper one mark off points 1, 2 to correspond with the points 1, 2 on AB, and join 1 to 1 and 2 to 2. Through D draw a line DO, making any angle with DF, and on DO measure any six equal parts at a, 6, c, d, e,f; lines 2 6 4 TEXT-BOOK ON NAVIGATION. drawn through a, b, c, d, and e parallel to /I will divide Dl into six equal parts, and since Dl represents 1, each part represents 10' of long.; with these distances divide Al into six equal parts, and join up diagonally as shown in figure, by lines similar to Ig. Considering the triangles Ihm and 1 Ig, these 2 FIG. 175. 3 B are seen to be similar, and since 1 1 is ten times Ih, therefore Ig is ten times km ; but Ig is 10' of long., hence km is 1' Of long. ; that is, the portions of the lines intercepted between 1 1 and Ig represent the single miles from 1 to 10. Angles may be set off (1) directly from a protractor ; (2) by using the scale of chords, marked Cho. On this scale the radius for describing the arc is the length to 60, and the length of the chord subtending any required angle at the centre is the length of the scale to the angle ; (3) by using the formula for finding the length of chord (I) for any convenient radius (r) to subtend yV angle (A) viz., I = 2r x sin - Thus for an angle of 37 with radius 1*5 inch I = 2 x 1'5 X sin 18i inch = 3 x -3173 inch = -952 inch. On a plan is sometimes found a fraction in the title space thus Yirhnr> called the Natural Scale, which shows the ratio that the length of a certain unit on the plan bears to the real length of that unit on the earth's surface. Hence 1 inch represents 10944 inches. 1*5 cables. Again, if 2 inches represent a nautical mile, the natural scale is 6080 x 12 36480 When the scale for a minute of longitude is given the scale for a mile of latitude may be found, and vice versa. On a Mercator Chart Mile of latitude = minute of long, x sec. lat. or min. of long. = mile of lat. x cos. lat. and so may be calculated, or found in traverse table. CONSTRUCTION OF CHARTS. 265 ART. 115. Any straight line drawn in a circle and terminated at the circumference is called a chord, and the portions into which it divides the circle are called segments. When the line passes through the centre, the chord becomes a diameter, and the segments semicircles. The angle subtended by the chord at any point in the circumference of the circle is called the angle in the segment, and all angles in the same segment of a circle are equal to one another. When the segment is equal to a semi- circle this angle is a right angle ; when the segment is greater than a semicircle this angle is less than a right angle ; when the segment is less than a semicircle this angle is obtuse. A segment of a circle may always be described on a given chord which contains an angle of a given size. When the given angle is acute, the segment is greater than a semicircle, hence the centre for describing the arc and required arc are on the same side of the chord. When the given angle is obtuse, the segment is less than a semicircle, hence the centre for describing the arc and required arc are on different sides of the chord. To find the centre set off the complement of the given angle at each end of the chord, these lines will intersect in the centre. Let A, B, and C be three conspicuous objects, A on the extreme left of the observer, C on his right, B the central object, the horizontal angle (a) between A and B may be observed and the horizontal angle (b) between B and C. If now segments of circles be described on AB and BC as chords to contain angles of a and 6 respectively, the intersection of these arcs will give the position of the observer. Care must be taken in the selection of the objects, or the cut of the two arcs is uncertain as to position. When the two arcs coincide it shows that the observer and the three points are on one circle, and there is no fix ; to avoid this the central object should be chosen as nearly in a line as possible with the other two, or should be nearer the observer than the others. The station pointer performs the operation practically and the fix is tabulated thus : A.a B.& C. ART. 116. To construct a Mercator's Chart. The advantages of such a chart are that all parallels of latitude are projected as parallel straight lines, and all meridians are projected as parallel straight lines cutting the parallels of latitude at right angles ; a rhumb line projects as a straight line, since it cuts all meridians at the same angle, and all places of small area retain their proper form on the chart, but with increased area as the latitude increases. Example 1 (Fig. 176). Construct a Mercator's chart on a scale of 1'9 inch to 1 long., to extend from lat. 53 N. to 54 30' N., and from long. 3 W. to 7 W. Insert a 266 TEXT-BOOK ON NAVIGATION. magnetic compass. Variation 17 20' W., and the following positions. Chicken Rock 54 2' N., 4 50' W., Langness 54 3' N., 4 37' W., N.W. Lt. V. 53 31' N., 3 31 W. At 2 h 20 m p.m. Chicken Rk. bore N. 32 W., and Langness N. 39 E. Shaped course S. 59 E., 15 knots, and at 4 h 40 ra p.m. altered course to S. CONSTRUCTION OF CHARTS. 267 9 W., at 5 h 50 m p.m. N.W. Lt. V. bore S. 37 E. ; and at 6 h 10 m p.m. it bore N. 64 E. ; place courses on chart, find true and D. R. positions of ship at last observation, set and drift of current. Courses and bearings magnetic. 1st. To find the size of the chart. As there are four degrees of long., 3 to 7 W., the breadth of the chart is 4 x 1*9 inches = 7'6 inches. Take out mer. pts. 53 = 3703-76 63<>J = 3813-90 54 =3864-64 " ~ 54^ =3915-99 Total ,, 152-23 60' = 1-9 inch. Hence 152'"23 = 4'82 inches. The height of the chart is 4 '82 inches. 2nd. To draw the frame, meridians, and parallels. Near the bottom of the paper draw a horizontal line 7*6 ins. long and divide into four equal parts, at each end set up perpendiculars as explained (Fig. 174). The lower line represents the parallel of 53 N. Find the distances on the chart corresponding to M.D.L.'s, viz : 50*14 = 1'59 inch, 5074 = 1-61 inch, 51*35 = 1*62 inch. Now set these distances up on each perpendicular, and measure the distance between the perpendicular at the upper divisions to see if it corresponds with the base line 7*6 inches : if so join them and the frame is complete. Measure off the whole degrees on the top line, draw the parallel of 54, and the meridian of 4, 5, and 6. Draw around the frame two parallel lines for graduation purposes, graduate the degrees, and enclose in a thicker outside line. 3rd. To draw in the compass. Select a portion of the chart clear away from the problem, and draw a circle of suitable radius. A diameter through the centre gives the true north and south line, measure the amount of the variation 17 20' W. to the left of this diameter for magnetic north, draw the diameter and top it with a half fleur-de-lis. At right angles draw another diameter for the east and west line and graduate. f. 4. To complete the problem. Insert the three given points from their lats. and longs., and from the compass set off the given bearings through Chicken Rk. and Langness, their intersection gives the point of departure. From this point set off the course S. 59 E. for 35 miles, the distance run between 2 h 20 m p.m. and 4 n 40 m p.m. ; here the course is altered to S. 9 W., which lay off for 22'5 miles, the distance run between 4 h 40 m p.m. and 6 h 10 m p.m. This gives the D.R. position which can be taken off. From the N.W. Lt. V. set off the two bearings observed, and fit in exactly the distance run by the ship between taking the observations (that is 5 miles run between 5 h 50 m p.m. and 6 h 10 m p.m.) on a line parallel to run made S. 9 W., the point of cut on the 2nd bearing is the true position of the ship which can now be taken off. Set and drift of current is the course and distance from D. R. to true position. ANSWER. True pos. lat. 53 29' N. long. 3 35' W. D.R. pos. lat. 53 26' N. long. 3 40' W. SetN. 54 E., 4j miles. Example 2 (Fig. 177). Construct a plan 7" by 5". Scale 5" = 1 nautical mile : also scale of cables and lOths. Insert compass showing 15 E , variation. /\ to be 1" from bottom and left-hand corner. From /\, A bore N. 8 W., 4 -2 cables ; B bore N. 51 E., 6'2 cables ; C bore N. 61 E., 10 cables ; all magnetic. 268 TEXT-BOOK ON NAVIGATION. At buoy " a " A 103 B 43 C. " b " A 47 B 108 C. "c" A 45 B74C. Plot buoys and give their bearings and distances from /< u - o Draw a horizontal line 7" long at bottom of paper, and at each end erect perpendiculars as in Art. 114, and make them 5" long ; measure to see if their extreme ends are 7" apart, if so join up and frame is complete. Inside the frame near the bottom draw scale of cables and lOths as in CONSTRUCTION OF CHARTS. 269 diagram, and put in a compass with magnetic north and south inclined 15 to sides of frame. Graduate compass. Plot in A I" from bottom and left-hand side ; and from it set off bearings and distances from your scale and compass to plot positions of A, B, and C. Join AB and BC. For buoy " a " it is noticed that A is on the left, as seen from the buoy, and the angle between A and B is 103, the arc is less than a semicircle, and the complement of 103, that is 13, is laid off at each end of A and B, on the side away from where the fix is to be, the centre " 1 " in Fig. is the inter- section of these lines, and the arc is drawn. Again the angle between B and C is 43, so its complement 47 is laid off at each end of BC, on the same side as the fix is to be, and intersect in the centre " 2 " ; where the former arc cuts this is buoy " a." At buoy " b " A is again on the left, since the angle between A and B is 47, set off lines making 43 with AB at each end to intersect in centre " 3 " on same side as fix, describe arc ; then set off lines making 18 with BO at each end to intersect in centre " 4," on opposite side to fix, since angle between B and C is 108 ; describe arc : these arcs cut at position of buoy "ft." For buoy " e " it will be noticed that both angles being less than 90, the centres "5" and "6," determined as before, are both on the same side as the buoy. Now take off the bearings and measure distances. From A bearing of " a " N. 53 E. "Z>" N. 58 E. "c" N. 68 E. 5'4 cables. 6-2 6 EXERCISES IN CONSTRUCTION OF MERCATOR'S CHARTS. 1. Construct a Mercator's chart on a scale 2"*4 to a degree of longitude, extending from lat. 56 N. to 57 30' N., and from long. 25 W. to 28 W. A ship from lat. 57 8' N., long. 25 10' W., sailed as follows by compass : S.W. by S. 50 miles, dev. 6 10' W. ; W.N.W. 53 miles, dev. 6 50' W. ; N.E. 40 miles, dev. 10 E. ; N. by W. 52 miles, dev. 1 10' E. The variation is two points W. Find latitude and longitude in. 2. Construct a Mercator's chart, scale 0*8" to 1 long., extending from 63 30' S. to 65 30' S., and from 120 to 123 E., putting in parallels at each j. A ship leaves A, lat. 65 7' S., long. 120 52' E., and sails the following true courses and distances : N.E. 47 miles ; N.N.W. 63 miles ; S.W. by S. 42 miles ; E. 27 miles. Find latitude and longitude in. The variation being 8 30' W., what is the magnetic bearing of A from ship 1 ? 3. Construct a Mercator's chart to extend from lat. 54 30' N. to 56 N., and long. 30 W. to 33 W., scale 1"'3 to 1 long. A point of land in lat. 54 32' N., long. 32 56' W., bears S.W. from the ship by compass, distant 7 miles; ship's head, E. f- N. ; dev. 9 55' E. ; variation, 18 W. Ship sailed as follows by compass : E. N. 90 miles, dev. 9 55' E. ; N.N.W. W. 40 miles, dev. 2 20' W. ; W.S.W. 45 miles, dev. 7 50' W. ; N.E. | E. 75 miles, dev. 10 35' E. Find latitude and longitude in, true course, and distance made good. 4. Construct a Mercator's chart to extend from lat. 54 N. to 55 30' N., and long. 3 to 6 W., scale 1"'75 = 1 long. Insert following positions : Abbey Head, lat. 54 47' N., long. 3 58' W. ; Little Ross, 54 46' N., 4 5' W. ; Burial Island, 54 29' N., 5 25' W. ; South Rock, 54 22' N., 5 2' W. At the ship, Abbey Head bore N. 46 E., and Little Ross, N. 37 W. ; she then steamed S. 63 W. 12 knots for 4 hours, when her position was fixed by cross-bearings of Burial Island W. 8 N., and South Rock S. 23 W. Find true position of ship, set and drift of current. 5. Construct a Mercator's chart, scale 2" = 1 long., extending from 2 7 o TEXT-BOOK ON NA VIGA TION. lat. 68 to 71 N., and 6 to 10 E. long. A ship leaves 69 38' N., 8 20' E., and sails as follows by compass : N.E. } E., 30', dev. 12 E. ; N.W. by W. i W., 55', dev. 8 W. ; South, 80', dev. 40 W. ; variation, 20 W. Find her latitude and longitude in, course, and distance made good. 6. Construct a Mercator's chart on a scale of 1-3 inch to 1 long., extending from lat. 48 45' S. to 52 S., and long. 20 E. to 26 E. Position of light- house, lat. 51 S. , long. 25 E. A peak lies 15 miles true North from Lt. Ho. Variation 20 E. A ship steamed from 50 S., 21 E., as follows : Comp. Co. Dev. Dist. Soundings. Bottom. S. 32 E. 2E. 120 miles 100 ftms. coral. N. 6E. 4E. 180 20 no bottom. S. 8E. 2W. 120 When the Lt. Ho. bore S. 45 E. true with an angle of 75 between it and the peak, give the lat. and long, of the position, amount of current ex- perienced, and assuming the same rate of speed to have been maintained, plot the soundings in their correct positions. 7. Draw a plan on a scale 0'85 inch = 1 mile ; and insert following : B, C, 1) are three objects, the true bearings and distances of which from A are: B, S. 81 E., 3 miles; C, S. 85 E., 6 miles; D, S.S.E. J E., 3 miles. Fix the following soundings : A 4* D 65 20' C (5 feet, mud). A 41 B 65 C (3J- fathoms, broken coral). A 116 B 123 D (7 fathoms, sand and shells. 8. Construct a plan en a scale 0'8 inch to 1 mile lat., to extend from 52 3' N. to 52 18' N., 6 to 6 25' W. Insert a compass showing 21 W. variation and graduate. A ship steering east magnetic, observed a lighthouse lat. 52 12' N., long. 6 7' W., in transit with a buoy when the horizontal angle between light vessel, lat. 52 7' K, long. 6 24' W., and Lt. Ho. was 81. Aiter steering 1'5 miles on above course, the horizontal angle was observed to be 50 between Lt. V. and Lt. Ho., and 22 between buoy and Lt. Ho. Find position of buoy with reference to its charted position, which was at first S. 31 W. of Lt. Ho. and 1-6 miles from it. 9. On a scale of 1*4 inch to 10 miles long, construct a Mercator's charb to extend from 55 15' N., to 55 45' N., and from 6 to 7 30' W., insert a magnetic compass variation 19 15' W., graduate margins. At 7 h 30 m a.m. a point (P) bore 130 from a ship proceeding on a magnetic course S. 80 E. , 10 knots, in a current setting N. 34 E. magnetic. At 9 h 45 ra a.m. it bore 206, and at 10 h 30 m a.m. it bore 225. P. lab. 55 14' N., long. 6 39J' W. Find the ship's position and distance from P when the last bearing was taken, average drift of the current, course and distance made good. CHAPTER XXVII. LAWS OF STORMS. ART. 117. The facts gathered from observations, and known as the Laws of Storms, apply more particularly to the violent Hurricanes, Cyclones, and Typhoons (which may be all classed under the name " Cyclone ") met with between the parallels of 10 and 35 of latitude ; but also in a modified degree to the strong winds, gales, and storms of higher latitudes. These facts may be summarized as follows : 1. The tropical Cyclones are never met with in the belt between 10 N. and 10 S. lat., but outside of 10 they occur in the following regions : N. Atlantic : The western part near the West Indies. N. Indian Ocean : The Bay of Bengal and Arabian Sea. N. Pacific : The China and Java Seas. S. Indian Ocean : The Western part, passing near Mauritius, and hence called " Mauritius Hurricanes." S. Pacific : Eastern part. 2. Cyclones (as the name implies) have a rotatory motion round a centre, and a progressive motion, varying from three or four miles an hour to twelve or more. Near their origin they cover only a small area, but afterwards expand to as much as a thousand or more miles in diameter. 3. The barometer stands lowest at the centre (often 28 inches or less), and gradually increases in height towards the circum- ference, whilst the wind force is generally greater the nearer the centre. 4. The bearing of the centre from the ship may be inferred approximately by counting from the wind direction 8 or 10 points to the Right in the N. Hemisphere, and the sajne number of points to the Left in the S. Hemisphere. 5. The path of the centre, or Line of Progression, divides the Storm Area into two semicircles, and by observing the change of wind it may be determined whether the ship is on the right- or left-hand side of the Line of Progression a matter of great importance in the handling of the ship. Thus, if the wind changes to the Bight, the ship is on the Right-hand side, but if to the left, she is in the Left-hand semicircle. 272 TEXT-BOOK ON NAVIGATION. If the wind is " steady," i.e. changes very slightly in direction, it indicates that the Ship is on the Line of Progression. 6. The Direction of Rotation in the N. Hemisphere is to the Left, and in the S. Hemisphere to the Right. 7. The approach of a Cyclone is often indi- Fl(J cated by well-marked signs, such as (a) An oppressive stillness, with threatening sky and much lightning. (6) A long " ground swell." (c) A dense cloud bank on the horizon. (d) A rapid motion of the upper clouds. 8. When within the Storm Area, the approach of the centre is shown by (a) A rapidly falling barometer. (6) Increase of wind, with heavy squalls, much lightning and rain, and general murkiness. (c) Heavy and confused sea. (d) Continuous " veering " or " backing " of the wind, except when on the Line of Progression. The receding of the centre would be indicated by (a) Rising barometer. (6) Wind becoming more " steady " and decreasing in force. (c) Weather clearing, but the sea still confused and dangerous. 9. Vessels, especially steamships, sometimes overtake Hurri- canes, because their speed is greater than the rate of progression of the Storm Centre. In this case, if the wind changes to the Left, the ship is on the Right-hand side of the storm, but if to the Eight, she is in the Left-hand semicircle ; contrary to par. 5. 10. In entering the Central Area, which may be up to a hundred miles in diameter, the wind suddenly ceases, with glimpses of a clear sky and a general calm of perhaps hours' duration, interrupted by puffy squalls, whilst the sea is par- ticularly confused and dangerous. After passing through the centre, the wind strikes the ship from the opposite point of the compass with renewed hurricane force. 11. The paths of Storm Centres follow the same general directions for the same regions of the Globe. The following are the usual tracks :-~ (a) N. Atlantic : Having their origin near the Windward Islands, they first travel about W.N.W., then more northerly, and about 25 N. to 30 N. they recurve to the N.E. towards Mid-Atlantic. (fc) N. Pacific: The same general direction as in the N. Atlantic. (c) N. Indian Ocean, including Bay of Bengal and Arabian Seas : A general N. and W. direction. LAWS OF STORMS. 273 (d) China Seas: Between N.W. and W.S.W. (e) Java Sea : Between N. and W. (/) S. Indian Ocean : Starting from about 12 South, they travel in a S.W. direction towards Mauritius, then more southerly, and recurve to the S.E. in 25 to 30 S. lat. 12. Seasons of Greatest Frequency of Cyclones. -The following Table of recorded Cyclones from Scott's Meteorology shows that Cyclones are most frequent in the N. Indian Ocean at the Change of Monsoons, and in the other Cyclone Regions during the hottest months : R< ;cord ;d cy clone s for each mon th. Totals Aver- Cyclone region. of re- corded age no. ,9 >j >> >? tc ^3 > obser- per 5 <5 3 >-> 5 ^ S i 1 vations. umim. West Indies (in 300 years) S. Indian Ocean (39 years) 5 9 i 13 11 10 6 8 5 4 10 42 96 80 1 69 1 17 4 7 3 355 53 1-18 136 Bombay (25 years) 1 1 1 5 9 2 4 5 8 12 9 5 62 2-5 Bay of Bengal (139 years) 2 2 9 21 10 3 4 6 31 18 9 115 0-8 China Seas (85 years) 5 1 5 5 11 10 22 40 58 35 16 6 214 2-5 The Table also shows that the most cyclone-infested regions of the globe are the W. Indies, Arabian Sea, and China Seas, and that the worst cyclone months in the various regions are (a) Bay of Bengal and Arabian Sea : May, June, October, and November. (6) N. Atlantic : July to November. (c) China and Java Seas : July to October. (d) S. Indian Ocean : January to March. The same rule would give for (e) N. Pacific : July to November. (/) S. Pacific : January to March. 13. The Rules of Action to be taken when a ship is caught in a Cyclone can be best understood by reference to the diagrams on p. 276. These Rules may be summed up as follows : (1) If on the Line of Progression of an approaching Storm Centre, run or steam before the wind until clear of the track of the centre ; then, in the N. Hemisphere, heave to on the Port Tack, but in the S. Hemisphere on the Starboard Tack. By so doing, the ship would always " come up " to the sea when hove to. (2) In the Right-hand semicircle (a) Northern Hemisphere : Head reach or heave to on the Starboard Tack. (6) Southern Hemisphere : Run with wind on Port Quarter, or heave to on Starboard Tack. T 274 TEXT-BOOK ON NAVIGATION. (3) In the Left-hand Semicircle (a) Northern Hemisphere : Run, having the wind on Star- board Quarter, or heave to on the Port Tack. (6) Southern Hemisphere: Head reach or heave to on the Port Tack. (4) When overtaking a Cyclone, it is obvious that the ship should be stopped or her course altered so that she may not approach the centre. 14. General Remarks. The most dangerous position for a ship is the track of the Centre of a Cyclone, which should be avoided if possible. This can generally be done, if there is sea room, by attention to the foregoing Rules. Observations tend to show that a Storm Area is more circular in front than in the rear; so that the Rule for finding the bearing of the centre is most reliable in the most dangerous situation. The same Rules may be applied when Heavy Gales are encountered in higher latitudes. It is common to give the name of " Dangerous Semicircle " to the Right-hand Semicircle in N. Lat., and the Left-hand Semi- circle in S. Lat., because >the direction of the air-currents is towards the Line of Progression in front of the Storm Centre. If it is suspected that the ship is overtaking a storm, the engines should be stopped, or the vessel "hove to," and the barometer carefully watched. Should the barometer begin to rise, it is a sure sign of overtaking ; but if it continues to fall (although less rapidly), it is a case of meeting a storm. As these storms have their origin in the hottest parts of the earth, their cause has been thought due to the rushing in of winds from the equatorial and polar regions to fill the partial vacuum caused by a vertical sun in rarefying the air and vaporizing the water. The path of a revolving storm is at first to the westward, but in consequence of the resistance due to meeting air carried round in an opposite direction by the rotation of the earth on its axis from W. to E., the storm trends to a place where there is less resistance, that is, in both hemispheres towards the poles. This is called the " Recurving " of the storm, and takes place in about 30 latitude. ART. 118. LAW OF STORMS. Questions to be answered by candidates for Mates' and Masters' Certificates. 1. Suppose the wind in a cyclone to blow from , state what is the probable bearing of the centre in the hemisphere. 2. If the wind in the same cyclone changes to , on what side of the line of progression is the ship, and what action would you take ? LAWS OF STORMS. yj$ 3. Under what circumstances would the change in the direction of the wind be the reverse of the above ? 4. What are the usual indications that a ship is on the line of progression of the centre of a cyclone 1 5. What are the usual indications that a ship is approaching the centre of a cyclone or receding from it ? 6. State what are the usual tracks taken by cyclones in , and at what season of the year do they most frequently occur in those regions ? LAW OF STORMS. Answers. 1. In the Northern Hemisphere the centre bears about eight to ten points to the right, and in the Southern Hemisphere eight to ten points to the left, from the given direction of the wind. 2. If the wind changes to the right the ship is on the right-hand side of the line of progression, if to the left she is on the left-hand side. These rules apply to both hemispheres. The action to take is as follows : Northern Hemisphere RIGHT-HAND SIDE. If possible, sail out on the starboard tack to a safer distance from the centre, or heave to on the starboard tack. LEFT-HAND SIDE. If possible, run with the wind on starboard quarter until in a safer position, or heave to on the port tack. Southern Hemisphere RIGHT-HAND SIDE. If possible, run with the wind on the port quarter until in a safer position, or heave to on the starboard tack. LEFT-HAND SIDE. If possible, sail out on the port tack to a safer position, or heave to on the port tack. 3. The wind would change the reverse way if the ship were on the other side of the line of progression, also if she were overtaking the storm on the same side as before. 4. In advance of the centre the barometer falls more and more rapidly as the centre approaches, the wind continues to blow from the same quarter, but increases in violence, with frequent heavy squalls, a confused angry sea, a dense cloud bank, and general threatening appearances of weather. In the central area the barometer is lowest, with a confused and dangerous sea, and calms varied by strong gusts. In the rear of the centre the barometer rises, and the wind blows at first with great violence from the opposite quarter, afterwards moderating and the squalls becoming less frequent, the sea being heavy and confused, but the weather gradually clearing. 5. (a) When approaching the centre, not on the line of progression, there are the same barometer and weather indications as stated in the preceding answer, but the wind changes in direction as it increases in force, with heavy cross- seas. (b) When receding from the centre the barometer rises, the wind becomes more steady, with less frequent squalls and less threatening appearance of weather, though the sea is still cross and dangerous. 6. North Atlantic (Tropical). The cyclones, in this region, are supposed to take their rise near the Windward or Leeward Islands, and first travel about W.N.W. towards Florida, then more northerly, and on reaching the parallel of 30 often recurve to the N.E. They occur most frequently from July to November ; worst months, August and September. South Atlantic (Tropical). They travel first about W.S.W., then 2 7 6 TEXT- BOOK ON NAVIGATION. more southerly, and sometimes recurve to the S.E. They occur most frequently in the hottest months December, January, and February. South Indian Ocean (Tropical). They are supposed to have their origin in the belt of calms, south of the Equator, often near Java, and travel first W.S.W., then more southerly towards the Mauritius, and sometimes recurve to the S.E. They occur from November to May, but most frequently in February and March. China Seas. The storms take their origin near the Philippine Islands, and travel generally between W.N.W. and W.S.W. They occur most frequently from July to November ; worst months, August and September. Bay of Bengal. The storms in this region begin in the neighbourhood of the Andaman and Nicobar Islands, and travel between W. and N.N.W. They occur most frequently in April to June, and September to November ; worst months. May, October, and November. LAWS OF STORMS. 277 Arabian Sea. Between N. and W., and most frequent in May and October. North Pacific (Tropical). The cyclones of this region begin in about 10 N., and travel first W.N.W., then more northerly, and finally recurve to the N.E. July, August, and September are the worst months. South Pacific (Tropical). The cyclones of this region first travel about W.S.W., then more southerly, and finally to the S.E. They occur most frequently from November to April. Outside of lat. 30, storms in the North Atlantic and North Pacific travel to N.E., and in the South Indian Ocean and South Pacific to the S.E. Mauritius. Storms occur from December to April, worst months being February and March ; they travel from S.S.W., recurving to S.S.E. CHAPTER XXVIII. MAGNETISM AND DEVIATION OF THE COMPASS. ART. 119. For a full and scientific discussion of the deviation of the compasses in iron ships, the student is referred to the " Admiralty Manual," or to the " Elementary Manual," edited and revised by Captain Creak, R.N. For general students Merrifield's " Magnetism and Deviation of the Compass " is recommended. The purpose of the following remarks is to enable students to clearly understand the principles involved in the Board of Trade Syllabus of Questions dealing with the practical applica- tion of the laws of magnetism to the case of iron ships and their compasses. Definition. Magnetism is a property which may be acquired by masses of iron and steel, by virtue of which they attract or repel each other under certain circumstances. Kinds of Magnets. There are three kinds of magnets, viz. natural, artificial, and electro-magnets. The natural magnets are the Earth and pieces of magnetic iron ore or " lodestone." Artificial magnets are bars or " needles " of hard-tempered steel, which have been magnetized by rubbing contact with other magnets, or by the influence of an electric current. Electro- magnets are elongated or horseshoe-shaped masses of soft iron wound round with copper wire, through which an electric current is passing. The strongest magnets are made in this way ; but the magnetic properties disappear when the current ceases. Degrees of Magnetism. Magnetism may exist in three different states or degrees in steel or iron, viz. PERMANENT, SUB-PERMANENT, and TRANSIENT INDUCED. A bar of hard-tempered steel receives magnetism slowly, but when once magnetized it retains its magnetic strength for an indefinite length of time, and is said to have permanent magnetism. Iron or steel of a less degree of hardness is more easily magnetized, but is liable to lose a considerable portion of its magnetism with lapse of time, and is therefore said to have sub-permanent m agn etism . Soft or malleable iron becomes instantly magnetized by induction from any magnetic source, but loses it when the MAGNETISM AND DEVIATION OF THE COMPASS. 279 inducing cause is removed. This temporary magnetic condition is known as TRANSIENT INDUCED magnetism. Saturation. There is a limit to the magnetic strength which FIG. 180. Variation chart of British Islands. a piece of steel or iron can acquire, and when this limit is reached it is said to be saturated. Magnetic Poles. Every magnet has two POLES, or points 2 8o TEXT-BOOK ON NAVIGATION. where the greatest energy is concentrated, distinguished as RED and BLUE poles, because it is common to paint artificial magnets in these colours. The red pole of a permanent magnet is the one which would point north when the magnet is suspended at its centre and free to turn round. This end is also called the N. or marked end. Therefore the end of a compass-needle which points N. is the red pole. ART. 120. First Law of Magnetism. It is a well-known law of magnetism that like poles repel each other, and unlike poles attract each other. This can be proved on a compass-needle, for if the red pole of a magnet be brought near to the N. or marked end, it will repel it, whilst the blue pole will attract it. Similarly, the S. end will be repelled by the blue pole, and attracted by the red. It follows from this experiment that the Earth's NORTH magnetic pole is BLUE, because the red pole of the needle is attracted towards it. The needle, if floating on water and free to move, would not, however, travel towards the N. pole, because its length is as nothing compared with the distance from the Earth's pole, its blue end being as strongly repelled as the red end is attracted. It is the same with the action of the south magnetic pole. The only result is, therefore, to give the needle a north and south direction. If a needle hundreds of miles in length could be supposed to be afloat in the North Atlantic, with its red pole towards north, no doubt it would travel bodily towards the N. pole of the Earth, because its red pole would then be attracted more strongly than the blue would be repelled. Second Law. It is another law of magnetism that the effect of a magnetic force (such as the pole of a magnet or an induced pole in soft iron) on a small magnet varies nearly in the inverse ratio of the cube of the distance ; that is, at double the distance it is one-eighth (2 x 2 x 2 = 8) ; at treble the distance it is only one twenty-seventh (3x3x3 = 27), etc. Also Third Law. A magnet placed end on to a pivoted needle causes twice the deflection that it would do if broadside on at the same distance between the centres of the magnet and needle. These facts have a very important bearing on the effect of the magnetic forces in a ship on her compasses, and show the advantage of placing the compasses as far as practicable from masses of iron having magnetic properties, especially when end on. ART. 121. Transient Induced Magnetism. When either pole of a magnet is brought near to one end of an iron rod, it develops two magnetic poles in the iron, the near end being of the opposite name to that of the pole applied. The process may be repeated through a series of rods end on, so that the N. point of a com- pass-needle would be deflected by the last one in the same manner as by the magnet itself. MAGNETISM AND DEVIATION OF THE COMPASS. 281 The same fact may be illustrated by the familiar experiment of causing a number of soft iron bars or nails to hang on to the \\3fcuanjet, S N S N S N r~ ^ i i i i i = FIG. 181. end of a strong magnet; bub whether in contact or not, the effect is the same. The above are examples of magnetic induction and the acquiring of transient induced magnetism. Lines of Force. The mysterious action of one magnet on another at a distance, and the magnetization of iron with or ithout contact, are explained the existence of lines of force, which issue from the red pole and return to the blue pole. These lines of force pass with- ou t interruption through all intervening non-magnetic sub- stances, and it is the passing of - 182< lines of force into iron which gives it magnetic properties. Their existence and direction can be shown by experiment (Fig. 182). If a sheet of white paper be laid on a magnet and iron filings sprinkled on the paper, the filings will arrange themselves in lines corresponding to the direction of the lines of force from one pole of the magnejb to the other. If a suspended magnetic needle were enclosed in a thick iron shell, the lines of force would follow round through the iron by preference to reach the blue pole, and the consequence would be that the needle would lose its directive power. This is the reason why compass-bowls are made of copper, and not of iron. Another reason is that poles would be induced in an iron bowl, which would oppose the Earth's directive action on the needle. Therefore a ship's compass which is closely surrounded by iron would have a weakened directive power. The Earth's Magnetism. The Earth itself acts as a magnet, having a blue pole towards the north and a red pole towards the south. Its action on a compass needle is merely to give a directive tendency, as before explained, but it is also a powerful inducing source. No iron can escape from its influence. Hence fireirons, knife-blades, needles, etc., are nearly always magnets with two fixed poles, as may be tested on a compass-needle, 282 TEXT-BOOK ON NA VIGA TION. whilst bars or rods of soft iron have also induced poles from the same cause, which are not, however, fixed. A fairly strong magnet can be made by holding a bar of hardened steel in a N.-and-S. direction and tapping it with a hammer ; the vibra- tion quickens the process of magnetization by the Earth's induction. A Ship's Induced Magnetism. Even a ship becomes a huge magnet from the Earth's induction in the process of building, whilst the separate parts, such as beams, stanchions, davits, rudder and stern-posts, etc., become so many temporary magnets, liable to act separately and collectively in disturbing the compasses. ART. 122. Variation. The magnetic poles of the Earth do not coincide with the true geographical poles, the N. or blue pole being in about lat. 70 N., long. 90 W., and the S. or red pole was found by Lieut. Shackle ton to be in lat. 72 S., long. 154 E. (from which it is seen they are not exactly opposite). This is the cause of the variation of the compass, which is the difference in direction between true N. and magnetic N. The amount of variation is stated on ordinary charts, and for the entire globe on special variation charts. The variation is not constant at any place, but changes at a known rate (in the United Kingdom now about 8' or 9' annually). Previous to the year 1657 the variation was E. in this country. In that year there was no variation at London, the compass pointing true N. Then it became westerly, and attained its maximum W. variation in 1816. Since then it has constantly decreased, and will again become in 1977 ; then, becoming E. again, it will reach its maximum in the year 2290. The ex- planation of this progressive change of variation is that the N. magnetic pole describes a complete circle round the true N. pole in a period of about 640 years. Twice in the course of this period the compass would point true N. at Greenwich, i.e. when the magnetic pole is on the meridian of Greenwich, and on the opposite meridian (Fig. 184). The annual change would evidently be greatest when the magnetic pole is at or near to the points and Oi in the FlG - 184 - figure, and slowest when near the points W and E, when the direction of the needle is a tangent to the small circle. At present the magnetic poles are coming towards the meridian of Greenwich. The same remarks apply to other places. A line in the direction magnetic N. and S. at any place is called the magnetic meridian; hence variation may be defined as the MAGNETISM AND DEVIATION OF THE COMPASS. 283 Fm. 185. angle between the true and magnetic meridians. An irregular line round the Earth, about midway between the poles, is the magnetic equator, which cuts the true geographical equator in two nearly opposite points about longs. 170 E. and 12 W. Magnetic Dip. A magnetized needle, poised at its centre so as to be free to move in a vertical plane, and provided with a scale of degrees, is called a dipping-needle (Fig. 185). At the magnetic equator, and placed in the plane of the magnetic meridian (i.e. pointing N. and S. magnetic), such a needle would lie horizontal ; but if carried northwards the red pole would dip more and more, until the magnetic pole is reached, where it would stand upright, red pole downwards. South of the magnetic equator the blue pole would dip, and the needle would again assume an upright position at the S. magnetic pole, blue pole downwards. At Greenwich the dip at the present time is 68, but in the West Atlantic and Gulf of St. Lawrence it is much greater. Like the variation, the dip at Greenwich is decreasing annually, and will be least in the year 1977, when the variation will be 0. These are the results of observation. The explanation is that in N. magnetic latitude the red pole of the needle points direct to the N. magnetic pole, and in S. magnetic latitude the blue pole to the S. magnetic pole of the Earth, notwithstanding the mass of the Earth intervening, whilst the horizontal plane at any place is a tangent to the Earth's surface (see Fig. 186). The dip at any place decreases when the magnetic pole, in making its circuit round the true pole, is approaching the meridian of the place, and increases when it is re- ceding. Thus in the Fig. 184, the dip at Greenwich decreases when the magnetic pole is moving from Oi to 0, and increases from to Oi, being greatest when the pole is at Oi, and least when at 0. At present the dip at Greenwich is decreasing. The position assumed by the dipping-needle may be con- FIG. 186. 284 TEXT-BOOK ON NAVIGATION. sidered the natural one for any magnetized needle which is free to move, but the horizontal position of a compass needle is a forced one. The reason why the compass card lies horizontal under the circumstances is that the point of support is above the centre of gravity of the card and needles, and the dragging down force is not sufficient to overcome their weight. It is evident from the above that an ordinary compass would be of no use for steering purposes near the magnetic poles, and is at its best on the magnetic equator. ART. 123. Parallelogram of Forces. Before considering the magnetism of iron ships and its effects on their compasses, it is necessary to understand the principle of the parallelogram of forces. It is a law in mechanics that if two adjacent sides of a parallelogram represent two given forces in direction and amount, then the diagonal from the point of meeting of these two sides will represent the resultant (or equivalent force) both C D FIG. 187. FIG. 188. in direction and amount. Thus, if AB and AC in the parallelo- gram ABCD represent two given forces, then the diagonal AD will represent the joint effect or resultant. The sides of the triangle ABD represent equally well the forces and resultant. A good illustration of this principle may be taken from the case of sailing in a current. For example, let the direction from A to B (Fig. 188) be the course steered, and the length of AB the rate of sailing per hour ; also let AC represent the " set " of the FIG 189. current and " drift " in one hour. Draw BD parallel to AC, and CD parallel to AB, and join AD ; then AD will be the course and distance made good, and repre- sents the resultant or joint effect of wind or steam and the current acting for the same length of time. It is the same with any other kind of force, and it will be seen how the principle applies in dealing with the magnetic forces in a ship. AB and AC are called components, and these can be found from the resultant, as well as the resultant from the components. The finding of the components is called "resolution," and of the resultant " composition." Thus, a force represented by AD is MAGNETISM AND DEVIATION OF THE COMPASS. 285 "resolved" into components represented by AB and AC. This expression will be frequently used hereafter. ART. 124. Magnetic Condition of Iron Ships. (a) Sub-per- manent Magnetism. It has been stated that the hull of an iron ship becomes a magnet from the Earth's induction. The effect of lying for a considerable time in one direction, and the hammering to which the iron is submitted, is to fix the acquired magnetism so that it becomes sub-permanent, i.e. less than per- manent. After launching there is some reduction of the strength of this sub-permanent magnetism, but it can often be traced throughout the whole existence of the ship, showing that a portion of it is almost of a permanent character. If the hull be supposed to be divided into halves as she lies on the stocks by a plane at right angles to the line of dip (q.v.), the half towards the N. will have red polarity, and the other half blue. Thus, if built in the United Kingdom, with head towards N., the line of dip being DP (Fig. 190) and EQ at right angles, E Q FIG. 190. the ship would be an irregularly shaped magnet, red towards the bow, and blue towards the stern. If built heading S., the red polarity would be at the stern, and blue at the bow. If built heading E., the port side would be red, and the E Q ' > P O W P JfeadW ^ JJeaclE FIG. 191. FIG. 192. starboard side blue (Fig. 192) ; but if built head W., the starboard side would be red, and the port side blue (Fig. 191). The magnetic poles or points of greatest concentration of magnetic strength might be supposed to be about the centre of the two halves. So that in all these cases the line joining the poles would be inclined, one pole being nearer the keel than the other (unless built on the magnetic equator). 286 TEXT-BOOK ON NAVIGATION. But it is only the horizontal component of the total force which causes deviation when the ship is upright. (The effect ff or Force FIG. 193. FIG. 191. of the vertical component will be considered later.) The parallelogram (or triangle) of forces will show the relation of the horizontal to the total force (Fig. 194). It is clear that the greater the dip the less will be the hori- zontal component of the ship's magnetic force, and vice versa. Hence, other things being equal, a ship built in a high latitude N FIG. 195, FIG. 196. FIG. 197. should cause less deviation from sub-permanent magnetism than one built in a low latitude, the ship being upright. If a ship were built heading magnetic N. or S., the line representing the horizontal force of sub-permanent magnetism would be parallel to the keel (Fig. 195) ; if E. or W., it would be at right angles to the keel (Fig. 196) ; but if in any other direction it would lie obliquely (Fig. 197). In the latter case, the parallelo- gram of forces would again resolve the total horizontal force into two components one represented by a line parallel to the keel, and the other at right angles to it. Induction in Horizontal and Vertical Iron. The effect of the Earth's induction on the horizontal and vertical iron of a ship may now be considered. A bar of soft iron has the greatest magnetic strength from the Earth's induction when it lies in the line of dip, but is neutral when at right angles to it. MAGNETISM AND DEVIATION OF THE COMPASS. 287 Thus, in N. magnetic latitude the end B of the horizontal bar AB and the lower end of the vertical bar CD are red poles, because towards the N. dipping-pole (Fig. 198). 'Jted Blue Red FIG. In S. magnetic latitude, the end A of the horizontal bar and the lower end of the vertical bar are blue, because towards the S. magnetic pole (Fig. 199). A bar lying at right angles to the line of dip would be neutral. This explains why the stern post, rudder post, and other vertical iron below the compass attract the N. point of the compass when north of the magnetic equator and repel it when south, whilst they have no effect when the ship is on the mag- netic equator. Also, why the ends of horizonal beams attract and repel alternately as the ship's head turns from one quadrant to another, but are neutral when her head is N. or S. magnetic. Another inference is that the magnetic strength of vertical iron in a ship decreases on going towards the magnetic equator, and increases towards the poles, whilst that of a horizontal beam lying N. and S. increases towards the equator, and decreases towards the poles. ART. 125. Coefficients. The term COEFFICIENTS is used to represent the components of magnetic forces in a ship. It may be assumed that these forces have a mechanical value, although a very ynall one, and could be measured by the same standard as other mechanical forces ; but in reference to a ship's compass they are estimated by the amount of deviation they produce when acting most effectively. This deviation itself is often called the coefficient. The letters A, B, C, D, E name the coefficients, of which B, C, and D are the most important. Coefficient B represents a magnetic force whose poles lie horizontally in the fore-and-aft midship line, and is a combina- tion of the fore-and-aft component of sub-permanent magnetism and the induced magnetism of vertical iron before or abaft the compass. It is named +B if the combined effect is to attract the N. point of the compass towards the bow, and B if towards the stern. 288 TEXT-BOOK ON NAVIGATION. Coefficient C represents a force whose poles lie horizontally at right angles to the fore-and-aft midship line, and is the athwartship component of the sub-permanent magnetism only, for vertical iron is supposed to be evenly balanced towards the sides. It is named +C when the N. point of the compass is attracted to the starboard side, and C when to the port side. The horizontal force of sub-permanent magnetism is there- fore represented by coefficients B and C, whose signs and relation to each other depend on the direction of ship's head during her construction, as illustrated in the following figures : FIG. 201. Semicircular Deviation. The deviation due to coefficient B or coefficient C is called semicircular, because it is east on courses over one half of the compass, and west in the other half. Coefficient + B (blue pole forward) causes E. deviation when the ship's head is to the eastward, and W. deviation when to the westward ; -B (blue pole aft) gives deviations of opposite name to the above. The maximum amount in both cases is found when the ship's head is E. or W. by compass. Coefficient +C gives E. deviation when the ship's head is to the northward, and W. deviation when to the southward ; - C gives the opposite, the maximum amount being found when heading N. or S. by compass. B causes no deviation when ship's MAGNETISM AND DEVIATION OF THE COMPASS. 289 head is N. or S. by compass, and G causes no deviation when her head is E. or W., because in these cases the forces act in a line with the compass needle. The kind of deviation due to coefficients B and C is seen at a glance in the following figures, where + stands for E. deviation, and for W. deviation : ax. Fio. 202. Wo -f C Fio. 203. Transient Induced Magnetism in* Ships. The transient induced magnetism of horizontal iron is represented by coefficients D and E. Beams from side to side, and elongated iron lying fore-and- aft end on to the compass, give coefficient +D; athwartship beams end on and fore-and-aft iron extending beyond the com- pass both ways, gives D (Fig. 204). Horizontal stringers from starboard quarter towards port bow give +E, and when extending from port quarter towards starboard bow they give E. When end on to the compass, they give a coefficient of the opposite sign (Fig. 205). Quadrantal Deviation : Coefficients D and E. The deviations caused by coefficients D and E are called quadrantal, because they change their signs with change of course from quadrant to quadrant. Thus, +D gives E. deviation between N. and E. and between S. and W., but W. deviation between E. and S. and between W. and N. Also, + E gives E. deviation between N.W. and N.E. and between S.E. and S.W., but W. deviation between N.E. and S.E. 290 TEXT-BOOK ON NAVIGATION. and between S.W. and N.W. D and E give deviations of opposite sign. -e- -o- Jroris(riuinrt Iron Giving + D. ' -D Fro. 204. i X X Ire \ / \ / *rL Giving Jro?vOiL vru? FIG. 205. It is an instructive exercise to prove the foregoing results by tracing the influence of one of these masses of horizontal iron on the compass as the ship's head is swung round. W.tiev. J)ev. FIG. 206. W.Deu. These figures show the magnetic polarity of a beam from side to side (giving coefficient + D) and the deviation produced in each quadrant. The zero points (i.e. points on which there is no deviation) are N., E., S., and W., and the maximum deviation is on courses halfway between these points, namely, at N.E., S.E., S.W., and N.W. Similarly, it can be shown that the zero points for coefficient MAGNETISM AND DEVIATION OF THE COMPASS. 291 E are N.E., S.E., S.W., and N.W., and the courses of maximum deviation N., E., S., and W. The student should reason out for himself the effect of the various masses of horizontal iron as the ship is swung round the compass. It is conceivable that the iron producing +D or +E might be exactly counterbalanced by iron causing D or E, but if not, then the balance on the side of the greater is the coefficient. The following figures show the quadrantal deviations due to coefficients D and E, and the zero and maximum points. The sign + stands for E. deviation, and for W. deviation : FIG. 207, f E -E FIG. 208. Coefficient A and its Causes. Coefficient A is of the nature of an index error, being of the same sign or name on all courses, and is often not due to magnetism at all. It may arise from any of the following causes : (a) The misplacing of the lubber line. (6) The magnetic axis of the needles not coinciding in direc* tion with the line through the N. and S. points of the card. (c) An error in the correct magnetic bearing used in finding the deviations. (d) Unsymmetrical arrangement of iron with respect to the compass. (e) Eetained magnetism during the process of "swinging" the ship. The first three causes, if detected, could be removed, but the 292 TEXT-BOOK ON NAVIGATION. last two can scarcely be remedied. In any case the deviation from these causes is small, and is liable to be confounded with the deviation due to coefficient E, which is also very small. The existence of a coefficient A is made evident in a table of deviations when the sum of the E. deviations is not equal to the sum of the W. deviations. When the former is greater, A is + ; when the latter, it is . ART. 126. Analysis and Compensation. The importance of making an analysis of the causes of deviation and of knowing how much is due to each, is that suitable methods of com- pensation may be employed, namely, to counterbalance the per- manent magnetism by means of magnets, and the transient induced magnetism by iron correctors which themselves have the same kind of magnetism from the same source (the Earth), but act on the compass in the opposite sense. Separation of Coefficients. It is a fortunate circumstance that the effects of the various coefficients of forces can be sorted out from the observed deviations, as explained in the answer to Question 62, and a table can be made of the deviations due to each coefficient for every point of the compass (see answer to Question 62), which, being summed up algebraically, give the total deviation for each course. The practical method of compensation is explained in answers to Question 40. Compensation of Coefficient B. The most troublesome coeffi- cient to deal with is B, because it is a combination of sub- permanent and transient induced magnetism, as before explained. When on the magnetic equator, it would be easy to compensate the former with a magnet, and then, on proceeding north or south, the amount of change in the deviation when the ship's head is E. or W. could be compensated by a " Flinders bar." But since B and G of sub-permanent magnetism stand to each other in the relationship of difference of Latitude and Departure, the^ship's head whilst building being the course, it is possible to find this B if that direction is known. For this purpose the ship's head should be steadied on N. or S. by compass, and the devia- tion from coefficient C observed. Then, with the direction of ship's head whilst being built as a course, and C in departure column (in the Traverse Table), the required B will be found in the latitude column. This is the amount to be compensated by the fore-and-aft magnet, and the balance of observed B with the Flinders bar, either before or abaft the compass as required. Effect of Magnetic Forces on Different Compasses. It may be remarked that although the same coefficients of magnetic forces in a ship disturb both steering and standard compasses, yet the 'measure of the coefficients as shown by the deviations will not be the same for both, because the forces act at different distances. MAGNETISM AND DEVIATION OF THE COMPASS. 293 Consequently, observations and corrections must be made for each compass separately. It is explained in the answer to Question 63 how this can be done for several compasses simultaneously. Compensation good for all Latitudes. If the compensations are properly made when a ship is ready for sea, they should stand good for all latitudes, provided the sub-permanent mag- netism does not change, for on this supposition the strength of the magnets and the coefficients of sub-permanent magnetism would always be in the same relation to each other, and no deviation would occur, although the directive power of the needle varies with change of latitude, whilst the horizontal force of transient induced magnetism and of the correctors would also remain in the same relation to each other. Changes in Deviation. But however carefully a compass may be compensated, changes may be expected with the lapse of time, and as these changes will be chiefly in the deviation due to sub-permanent magnetism, a readjustment of the magnets will be necessary from time to time. The effect of the vibration caused by the propeller, the buffeting of heavy seas, collisions, or stranding is to 'considerably modify the magnetic condition of the ship when she left the builder's hands. Retained Magnetism. Another troublesome cause of change of deviation is what is called retained vnagnetism ; that is, if a ship remains for a considerable time in one position in dock or on the same course at sea for several days, the induced magnetism from the earth is retained for a certain length of time, and causes a deviation when the course is altered. In sailor's language, the ship tends to follow her last course. For example, a steamer in crossing the Atlantic, bound for Liverpool, lays on an easterly course for several days together, and when on rounding Tuskar her course is changed to the northward, the retained red polarity on the port side (and blue on the starboard side) causes an E. deviation, which tends to set her to the east- ward of her proper course. To this cause is owing, no doubt, the fact that steamers from the westward, coming up Channel in thick weather, are apt to be found too near the Welsh coast, and steamers from the south, bound up the English Channel, too near the English coast. ART. 127. Uncompensated Compasses. An uncompensated compass having large deviations on some courses and very small deviations on others is very unsatisfactory for steering purposes, for two reasons : (1) The deviation will be subject to greater changes, because when a ship approaches the magnetic equator the directive force of the needle increases, and therefore the coefficients of sub- permanent magnetism, which are assumed to remain nearly 294 TEXT-BOOK ON NAVIGATION. constant, will cause less deviation. On receding from the magnetic equator, the deviation would increase for the same reasons. (2) The directive tendency of the needle would be much increased when a strong blue magnetism in the ship is towards the north from the compass, and correspondingly decreased when red is towards the north. The consequence would be that in some directions a considerable change of course would show only a small change in the compass course, and in other directions a small change of course would appear a large one by compass. Horizontal Iron and TTncompensated Compass. The deviation caused by horizontal iron would not be subject to these changes, because the horizontal component of transient induced force would vary in the same ratio as the directive force of the needle. Diminution of Directive Force. In an ordinary iron ship, the directive force is rarely more than six- sevenths of what it would be in a wooden ship or on shore, and in armoured war-ships not more than three-fourths, according to Captain Creak, R.N., who has had very large experience of such matters. The chief cause of this is the horizontal iron, for in most cases it has a red polarity to the north of the compass. Sub-permanent magnetism weakens the directive force on some courses, but strengthens it on others, according as a red pole is north or south of the compass, and so the mean effect all round is nil. These remarks nto uncompensated compasses, le consequence of a diminution of directive force is greater deviations, for the forces which cause deviation are acting on a weakened compass. The object aimed at by compensating a compass is to secure a uniform directive force on all courses, so that if any deviations remain they will be small in amount and regular in their changes. ART. 128. Heeling Error. So far it has been assumed that the ship is upright, but when she has a list from the pressure of the wind, or shifting of cargo, or unequal trimming of bunker coals, further deviations sometimes very large, and known as heeling error occur on northerly and southerly courses, but disappear on east and west courses. The Cause of Heeling Error. (a) It has been already explained that there is a vertical component of sub-permanent magnetism as well as a horizontal. The upper pole of this force, nearest to the steering compass aft, is blue if the ship is built head north (Fig. 209), and red if built head south (Fig. 210), and so long as the Bhip is upright, it only tends to attract the N. point of the needle downwards or repel it upwards, according as the upper pole is blue or red, without causing any deviation. The case is different when the ship heels, for then this upper pole is no longer in the vertical plane passing through the centre of the compass, but MAGNETISM AND DEVIATION OF THE COMPASS. 295 towards the higher side, deviating the N. point of the compass towards the higher or lower side. This effect is evidently FIG. 209. FIG. 210. greatest on N. and S. courses, and nil on E. and W. courses, because in the former case the force acts at right angles to the needle, and in the latter case in the plane of the magnetic meridian. FIG. 211. FIG. 212. (ty The upper pole of vertical iron below the compass affects the compass in exactly the same way as the sub-permanent pole when the ship heels, only it must be remembered that vertical iron has the blue pole uppermost only in N. magnetic latitude, and red pole uppermost in S. magnetic latitude, being neutral on the magnetic equator. Therefore the heeling error from this cause decreases on sailing towards the magnetic equator, and. vice versa. (c) The athwartship iron, which gives coefficient D when the ship is upright, acquires a vertical component when she heels, which is of the same nature as the magnetism of vertical iron, and is subject to the same changes the higher end being blue in N. magnetic latitude, and red in S. magnetic latitude. It is evident that in some cases these three causes of heeling error would act together, and the heeling error would be large. In other cases the first would be opposed to the other two, and they would tend to neutralize each other, and give a small heeling error. Their combined forces might be represented by a single force 296 TEXT-BOOK ON NAVIGATION. acting vertically under the centre of the compass, which would therefore be the resultant of (1) The vertical component of sub-permanent magnetism. (2) The induced magnetism of vertical soft iron. (3) The vertical component of the induced magnetism of transverse beams. This resultant force is called + if the balance on the side of the greater is blue uppermost, and if red. The total effect may be summed up by saying that a + vertical force causes a heeling error towards the higher side, and a force a heeling error towards the lower side (Figs. 213 and 214). FIG. 213. FIG. 214. Experience shows that in the majority of sailing ships the force below the compass is + in N. latitude and in S. latitude, with the result that N. of the magnetic equator the ship is drawn to windward on northerly courses, and to leeward on southerly courses. South of the magnetic equator the rule would generally be reversed. The same rules apply to steam ships, if " higher side " is substituted for " windward," etc. Navigators make a practical application of these facts by keeping " away " on northerly courses, and " closer " on southerly courses, when north of the magnetic equator, and the reverse when south of it. Since the vertical force draws the Nc point of the compass to one side or the other when the ship heels, it is called a co- efficient C, although it would be more logical to distinguish it in some way from the coefficient C already discussed, due to horizontal force, say by using Ci to represent it. 1 Like coefficient C of horizontal sub-permanent force, the new coefficient GI causes no deviation (heeling error) when the ship's head is E. or W. (magnetic), and the greatest deviation on N. and S. courses, because in the former case it acts in the plane of the magnetic meridian, and in the latter at right angles to the needle. ART. 129. Compensation. The heeling error is usually 1 For a similar reason, the part of coefficient B due to vertical iron before or abaft the compass might be distinguished as B,. MAGNETISM AND DEVIATION OF THE COMPASS. 297 compensated by means of one or more magnets placed vertically under the compass, and capable of being moved up or down. There are two modes of procedure : (1) The ship's head being at N. or S. magnetic, she is caused to heel, say about 10, and the error observed. Then the magnet is placed in position, and moved up or down, until the compass points correct magnetic N. (2) The card having been removed, and the ship upright, with her head E. or W. magnetic, a dipping needle,, weigh ted so as to lie horizontal when in the plane of the magnetic meridian on shore, is placed on the pillar in the centre of the bowl, and lying N. and S. Should the red end dip, showing a blue force below the compass, the red pole of the vertical magnet should be uppermost, and vice versa, being moved up or down until the needle is again horizontal. In either of these methods, what has been done is to counter- act the force below the compass by a force of opposite name ; by observation of the heeling error in the first case, and in the second case by observing the vertical action of the force below the compass on the dipping needle. In a compensated compass the vertical force of the athwartship iron (producing coefficient D) is counterbalanced for moderate angles of heeling by a vertical force developed in the correctors ; which leaves only the effects of vertical iron and the sub-permanent magnetism to be dealt with. The Flinders bar increases the heeling error, for its upper pole is of the same name as that of the vertical iron it is intended to counteract when the ship is upright. Therefore the compensating magnet has to neutralize the combined vertical forces of sub-permanent magnetism, vertical iron, and Flinders bar. But as the two latter change with change of latitude, it is evident that this mode of compensation is effective only so long as the ship remains in the same latitude, and constant shifting of the magnet would be necessary as the ship proceeds north or south, and probably a reversal of the poles on the other side of the magnetic equator. The only efficient compensation would be to use the magnet for the part due to sub-permanent magnetism only, and a vertical soft iron bar, having its lower end above the compass, to counteract the vertical force of soft iron below the compass. Some arrangement might possibly be made for the purpose in the construction of the binnacle. It will now be understood that the principle of compensation is to counteract the existing magnetic forces in a ship by employing counter equal forces of the same kind, to act on the compass needles. The order of procedure recommended in the "Elementary 298 TEXT- BOOK ON NA VIGA TION. Manual " for ships of the Royal Navy, where the conditions are pretty constant, is 1st. Coefficient D. 2nd. Heeling error. 3rd. Coefficients B and C. But in the Merchant Service, where cargoes change every trip, and where the compensation is often hurriedly made when ready for sea, the "Tentative Method" of adjustment is usually carried out in the following order : 1. Coefficient C (athwartship magnet). 2. Coefficient B. (1) Flinders bar, and (2) fore-and-aft magnet. 3. Coefficient D (soft-iron correctors). 4. Heeling error (vertical magnet). Coefficient E is rarely compensated, although a counter E could be introduced by moving one of the D correctors forward, and the other a little aft. For +E the starboard corrector should be moved aft, and the port one as much forward; for E, the reverse. A table of the remaining deviations is obtained by " swing- ing " the ship in dock or other convenient place. Yet, however accurately the compensation may be made, changes in the deviation, from causes already explained, will occur during a voyage, and no careful navigator will place much reliance on " deviation cards." In fact, nothing can relieve him from the obligation to find the deviation from frequent observa- tions, which is now happily made easy by Azimuth Tables. CHAPTER XXIX. SYLLABUS. QUESTIONS ON DEVIATION or THE COMPASS FOE, ORDINARY MASTERS. THE applicant is to answer correctly at least 12 of such of the following questions as are marked with a cross by the examiner. The examiner's attention is specially called to the importance of questions 11, 12, 13, 14, and 39, which must be marked in all cases. 1. State briefly the essentials of an efficient compass. 2. State briefly the chief points to be considered when selecting a position for your compass on board ship, and what should be particularly guarded against. 3. What do you mean by deviation of the compass; and how is it caused ? 4. Describe how you would determine the deviation of your compass : (1) by reciprocal bearings ; (2) by figures on the dock walls ; (3) by bearings of a distant object ; (4) by the bearings of the sun or other celestial body. 5. Having determined the deviation with the ship's head on the various points of the compass, how do you know when it is easterly and when westerly ? 6. Why is it necessary, in order to ascertain the deviations, to bring the ship's head in more than one direction ? 7. For accuracy, what is the least number of points to which the ship's head should be brought for constructing a curve or table of deviations ? 8. How would you find the deviation when sailing along a well-known coast ? 9. Name some suitable objects by which you could readily obtain the deviation of the compass when sailing along the coasts of the Channel you have been accustomed to use. 10. Supposing you have no means of ascertaining the magnetic bearing of a distant object when swinging your ship for deviations, how could you find it approximately from equidistant compass bearings ; and at what distance, as a rule, should the object be from the ship ? 11. Having taken the following compass bearings of a distant object, find the object's magnetic bearing, and thence the deviation. Magnetic bearing required : Ship's head by standard compass. Bearing of distant object by standard compass. Deviation required. Ship's head by standard compass. Bearing of distant object by standard compass. Deviation required. N. S. 4 E. S. S. 13 E. N.E. S. s.w. S. 23 E. E. S. 4 W. W. S. 21 E. S.E. S. 1 W. N.W. S. 11 E. 300 TEXT-BOOK ON NAVIGATION. 12. With the deviation as above, construct a curve of deviations on a Napier's diagram, and give the courses you would steer by the standard compass to make the following courses, correct magnetic : Magnetic courses : S.S.W. W.N.W. N.N.E. E.N.E. S.S.E. Compass courses required. 13. Supposing you have steered the following courses by the standard compass, iind the correct magnetic courses made from the above curve of deviations : Compass courses : W.S.W. N.N.W. E.N.E. S.S.E. Magnetic courses required. 14. You have taken the following bearings of two distant objects by your standard compass as above; with the ship's head at W. \ S., find the bear- ings, correct magnetic : Compass bearings : W. by S. and N. f W. Magnetic bearings required. 15. Do you expect the deviation to change ? If so, state under what circumstances. 16. How often is it desirable to test the accuracy of your table of deviations ? 17. What is meant by variation of the compass ? What is it caused by ? And where can you find the variation for any given position ? 18. The earth being regarded as a magnet, which is usually termed the blue, and which the red magnetic pole ? 19. Which end of a magnet (or compass needle) is usually termed the red or " marked " end, and which the blue 1 20. What effect has the pole of one magnet of either name on the poles of another magnet ? 21. What is meant by transient induced magnetism ? 22. Which is the red and which is the blue pole of a mass of soft vertical iron, by induction, and what effect would the upper and lower ends of it have on the compass needle (a) in the northern hemisphere, (6) in the southern hemisphere, (c) on the magnetic equator ? 23. Describe what is usually termed the sub-permanent 1 magnetism of an iron ship, and state when and how it is acquired, and which is the red and which is the blue pole, and why it is called sub-permanent magnetism. 24. Describe the meaning of the expression " coefficient A." 25. Describe the meaning of the expression " coefficient B," its signs and effects. 26. Describe the meaning of the expression "coefficient C," its signs and effects. 27. Describe the meaning of the expression "coefficient D," its signs and effects. 28. Describe the meaning of the expression ' ' coefficient E," its signs and effects. 29. Would you expect any change to be caused in the error of your compass by the ship heeling over either from the effect of the wind or the cargo, etc. ? 30. The compasses of iron ships being more or less affected by what is termed the heeling error, on what courses is this error usually at its mini- mum, and on what courses at its maximum ? 1 The term sub-permanent magnetism in these questions is used in the original sense as proposed by the late Sir G-. B. Airy, to denote the character of the perma- nent magnetism of an iron ship as distinguished from the permanent magnetism of a magnetized steel bar. The terms " sub-permanent " and " permanent " throughout these Questions may, therefore, be considered as synonymous. SYLLABUS. 301 31. Describe clearly the three principal causes of the heeling error on board ship. 32. State to which side of the ship in the majority of cases is the north point of the compass drawn when the ship heels over in the northern hemisphere. 33. Under what conditions (that is, as regards position of ship whilst building, and the arrangement of iron in the ship) is the north point of the compass needle usually drawn to windward or the high side of the ship in the northern hemisphere ; and if not allowed for, what effect has it on the assumed position of the ship when she is steering on northerly, also on southerly courses, in the northern hemisphere ? 34. Under what conditions (as in Question 33) is the north point of the compass needle usually drawn to leeward or the low side of the ship in the northern hemisphere ; and, if not allowed for, what effect would it have on the assumed position of the ship, when she is steering on northerly, also on southerly courses, in the northern hemisphere ? 35. The effects being as you state, on what courses would you keep away, and on what courses would you keep closer to the wind in the northern hemisphere in order to make good a given compass course (a) when north point of compass is drawn to windward or the high side of ship ; and (6) when drawn to leeward or the low side ? 36. Does the same rule hold good in both hemispheres with regard to the heeling error ? 37. State clearly how that part of the heeling error due to the permanent part of the magnetism of the ship varies as the ship changes her position on the globe ; and what is the reason of this. 38. State clearly how that part of the heeling error due to the induction in transverse iron (which was horizontal when ship was upright) and iron vertical to the ship's deck, varies as the ship changes her position on the globe. 39. Your steering compass having a large error, show by "Beall's Compass Deviascope " how you would correct it by compensating magnets and soft iron (as usually practised by compass adjusters in the Mercantile Marine) in order to reduce the error within manageable limits. Show also how the heeling error can be compensated. 40. As the coefficient B (capable of being corrected) usually consists of two parts, one due to the permanent magnetism of the ship, and the other to vertical induction in soft iron, how should each of the two parts, strictly speaking, be corrected when compensating the compass ? 41. If the whole of coefficient B be corrected by a permanent magnet, as is usually done, what is likely to ensue as the ship changes her magnetic latitude ? 42. Provided the needles of your compass are not so long and powerful, and so near, as to cause the soft-iron correctors to become magnetized by induction, would the coefficient D, if properly compensated, be likely to remain so in all magnetic latitudes and both hemispheres ? If so, state the reason why. 43. State at what distance, as a general rule, the magnets and soft-iron correctors should be placed from the compass needles, and what will be the consequence if they are placed too near the needles. 44. Is it necessary that the magnets used for compensating coefficients B and C should be placed on the deck ? If not, state where they may also be placed, and the rules to be observed in placing them into position. 45. Can the compensation of the heeling error be depended upon when the ship changes her latitude ? If not, state the reason. 302 TEXT-BOOK ON NAVIGATION. ANSWEBS TO THE QUESTIONS ON DEVIATION OP THE COMPASS. 1. The card should be light, accurately centred and divided, and the needles well magnetized, the magnetic axis coinciding with the N. and S. points of the card. The cap should be sound, and the pivot sharp and smooth. The point of the pivot should be level with the gimbals. The bowl should be of copper, and the mounting should not allow of any lateral play. 2. The standard compass should be in the midship line, and should afford a clear yiew all round for taking bearings, and not less than five feet from iron of any kind, especially vertical iron and any iron which is likely to change its position, such as derricks, cowls, davits, etc. Also as far as possible from a dynamo, if electric lighting is used. Steering compasses should be so placed that the card can be well seen by the steersman, and should be as far removed as possible from the influence of iron. 3. Deviation is the amount of deflection of the N. point of a compass from the correct magnetic N. caused by the magnetic influence of iron in the ship, whether in hull, equipment, or cargo. 4. (1) By reciprocal bearings : An observer is stationed at a compass on shore in a position free from local attraction, and another observer at the standard compass on board. Then, as the ship is swung round, bearings are taken of each compass by the other observer at the same time, whilst the ship's head is steadied on the required points. The difference between the bearing observed on board and the reversed shore bearing is the deviation. (2) By figures on the dock wall : These figures at Liverpool show the true bearings of the Vauxhall Chimney from the river, so that the figure in a line with the chimney as seen from the ship shows its true bearing from the observer's position. The difference between this and the bearing by the standard compass is the total error, from which the variation may be separated, and the result is the deviation. (3) By bearings of a distant object : When the correct magnetic bearing of a distant object is known, the difference between the bearing of it by compass on board, and the correct magnetic is the deviation. (4) By the bearings of a celestial object : The true azimuth of the object being found for a given moment, either by calculation or by tables, the bearing of the object at the same time is taken by compass ; the difference shows the error of compass for the actual position of ship's head, and the variation being known, the deviation may be found. 5. If the correct magnetic bearing reads to the right of the. compass bearing, the deviation is east ; if to the left, it is west. 6. Because the deviation changes with every change of course. 7. Eight equidistant points, viz. N., N.E., E., S.E., S., S.W., W., N.W. 8. By comparing the bearing of two known objects when in one (or in line) with the known correct magnetic direction of one from the other obtained from a chart. 9. (a) English Channel : Bolt Head and Start Lighthouse, Prawle Point and Start Point, etc. (6) Irish Channel: Skerries and South Stack, Smalls and Bishop Lights, Wolf ancl Longship. (c) North Sea : South Foreland Lights, Whitby Lights, and Tyne Harbour Lights. 10. The ship's head should be steadied on eight equidistant points in succession, and the bearing of a distant object observed when on each point. The mean of these may be taken as the correct magnetic. The object should be seven or eight miles distant, so that its bearing would not be sensibly affected by the swinging of the ship. 11. 12, 13, and 14 are practical. 15. Yes, the deviation may be expected to change : (a) From lapse of time j (6) By keeping the ship's head for a considerable time in an opposite SYLLABUS. 33 direction to that in which she was built ; (c) By change of latitude ; (d) By shocks or strains from heavy seas ; (e) By being kept for a long time on one course, magnetism is induced which is retained on a change of course. 16. The deviation table should be tested at every opportunity. 17. Variation is the difference in direction between the magnetic north and true north, and is due to the fact that the earth's magnetic poles do not coincide with its geographical poles. The amount, together with the annual change, may be found on a magnetic chart. 18. The earth's north magnetic pole is the blue pole, and the south the red. 19. The end of a magnetized needle, freely suspended, which points north is the red pole, and the other the blue pole. 20. Poles of the same name repel each other, and poles of different names attract each other. 21. Transient induced magnetism is the magnetic condition of soft iron caused by the influence of a magnetic force in its neighbourhood or that of the earth itself. 22. North of the magnetic equator the upper end is the blue pole and the lower the red, but south of the magnetic equator this rule is reversed, (a) In the northern hemisphere the upper end would attract the north end of the compass needle, and the lower end repel it. (6) In the southern hemi- sphere the upper end would repel and the lower attract the north end of the needle, (c) On the magnetic equator the effect would be nil. 23. The sub-permanent magnetism of an iron ship is the magnetic con- dition of a more or less enduring character which she has when she is launched. It is acquired in the course of building, from the earth's inductive influence, aided by the vibration due to hammering, riveting, etc. The red pole is that which was towards the north, and the blue towards the south when the ship was on the stocks. It is called sub-permanent magnetism to distinguish it from the magnetism of a tempered steel bar, and because it is liable to be weakened or modified by circumstances and by lapse of time. 24. Coefficient A is a deviation of the same amount and sign on all courses, usually due to a faulty or badly placed compass. 25. Coefficient B represents a horizontal magnetic force, whose poles are in the fore-and-aft midship line. It is called +B when the blue pole is towards the bow, and B when towards the stern. + B causes E. deviation on easterly courses, and W. deviation on westerly courses. B causes W. deviation on easterly courses, and E. deviation on westerly courses. The effect is nil on N. and S. courses, and greatest on E. and W. courses. 26. Coefficient C represents a horizontal magnetic force, whose poles are in a line at right angles to the fore-and-aft midship line. It is called +C when the blue pole is towards the starboard side, and C when towards the port side. +C causes E. deviation on northerly courses, and W. deviation on southerly courses. C causes W. deviation on northerly courses, and E. deviation on southerly courses. The effect is nil on E. and W. courses, and greatest on N. and S. courses. 27. Coefficient D represents the effect of transient induced magnetism in horizontal iron in fore-and-aft and athwartship directions. The deviation due to this cause changes its name in successive quadrants, and is thence called quadrantal. +D causes E. deviation from N. to E. and S. to W. courses, and W. deviation from S. to E. and N. to W. D gives W. devia- tion from N. to E. and S. to W. courses, and E. deviation from S. to E. and N. to W. The deviation due to this cause is nil on the cardinal points and greatest on N.E., S.E., S.W., and N.W. 28. Coefficient E represents the effect on the compasses of induced mag- netism in horizontal iron lying in a direction oblique to the fore-and-aft line. The deviation due to this cause is quadrantal, and is nil on N.E., S.E., S.W., and N.W., and greatest on the cardinal points. +E gives E. deviation 304 TEXT-BOOK ON NAVIGATION. from N.W. to N.E. and S.E. to S,W., and W. deviation from N.E. to S.E. and S.W. to N.W. -E gives W. deviation from N.W. to N.E. and S.E. to S.W., and E. deviation from N.E. to S.E. and S.W. to N.W. 29. Yes, because when the ship heels, horizontal athwartship iron acquires a vertical component, and besides, the upper poles of vertical iron and the poles of sub-permanent magnetism will then not be in the same vertical plane with the centre of the compass. 30. The heeling error is usually greatest on N. and S. courses, and least on E. and W. courses, because in the former case the force causing it acts at right angles to the compass needle, and in the latter case coincides with the direction of the needle. 31. The three principal causes of heeling error are : (1) Traverse iron acquires a vertical component. (2) The upper induced poles of vertical iron will not be in the same vertical plane with the centre of the compass. (3) The poles of the fore-and-aft component of the ship's sub-permanent mag- netism fall to the right or left of the vertical plane, through the centre of the compass. 32. In the N. hemisphere the N. point is usually drawn towards the higher side. 33. In ships built with head northerly, compass aft, placed over con- tinuous athwartship iron, the N. point of a steering compass will be attracted towards the higher side, and, if not allowed for, will draw the ship to windward on northerly courses, and to leeward on southerly courses. 34. In ships built with head towards south, with compass aft, placed over divided athwartship iron, the effect of the sub-permanent magnetism and soft iron would be to draw the N. point of the compass to leeward, and if these exceeded the influence of the vertical soft iron, the N. point of the compass would be drawn towards the lower side. If not allowed for, it would take the ship to leeward on northerly courses, and to windward on southerly courses. 35. (a) When the N. point is drawn to windward, I would keep away on northerly courses, and closer to the wind on southerly courses. (&) When the N. point is drawn to leeward, I would keep closer to the wind on northerly courses, and keep away on southerly courses. 36. The rule for the northern hemisphere is generally reversed in the southern, especially in ships built heading south, for the poles of vertical induction are reversed, but in ships built heading north, the same rule may still obtain, that is, when the sub-permanent vertical force exceeds the induced. 37. It decreases on approaching the magnetic equator, and increases on receding from it, because the directive force of the compass changes as the magnetic latitude changes, being greatest at the magnetic equator and nil at the magnetic poles, whilst the sub-permanent magnetism of the ship is nearly constant. 38. The induced vertical force becomes weaker on approaching the mag- netic equator, is nil at the equator and the poles are reversed on crossing it. Therefore the part of the heeling error due to this cause, drawing the N. point to windward in north mag. lat., disappears at the magnetic equator, and repels the N. point to leeward in S. mag. lat. 39. Practical. 40. The part of coefficient B due to the sub-permanent magnetism should be compensated by means of a magnet, and the part due to vertical induction in soft iron by an upright bar of soft iron before or abaft the compass in the midship line. 41. It will be found to be over-compensated or under- compensated as the ship changes her mag. lat., for the part caused by vertical induction changes with change of latitude. 42. Yes, because the horizontal inductive force of the correctors will vary in the same ratio as that of the horizontal iron which produces coefficient D. SYLLABUS. 305 in the same ratio as that of the horizontal iron which produces co- efficient D. 43. The magnets used for compensating should not be nearer the centre of compass than twice their own length, and the soft-iron correctors should not be nearer than 1 times the length of the needles. If the magnets are too near, the compensation which was made with the ship's head on the cardinal points would not be perfect on other points. If the soft-iron correctors are too near, the poles of the compass needles would have a tendency to induce opposite polarity in the correctors, rendering them useless. 44. The magnets used for compensating B and need not be on the deck, but may be placed in any convenient position near the compass, in or outside the binnacle, provided the following conditions are complied with : (1) The ship should be upright and on even keel. (2) The fore-and-aft magnets should be parallel to the deck, and bisected by a vertical plane athwartships, passing through the centre of the compass. (3) The athwartship magnets should be parallel to the deck, and placed so as to be bisected by a fore-and- aft vertical plane passing through the centre of the compass. 45. No, because a part of the heeling error is due to vertical induction in soft iron, which varies in force as the latitude changes. This part of the heeling error cannot be compensated by means of a permanent magnet fixed in position. QUESTIONS ON DEVIATION OP THE COMPASS FOR MASTERS (EXTRA). 1. Describe an artificial magnet, and how a steel bar or needle is usually magnetized. 2. Which end of the compass needle, or a magnet, is commonly termed the red, and which the blue pole ? 3. Which is the red magnetic pole of the earth, and which the blue ? and give their geographical positions. 4. What effect has the pole of one magnet of either name on the pole of the same name of another magnet ? and what would be the consequence of the pole of one magnet of either name being brought near enough to affect the pole of contrary name, if in these cases both magnets were freely suspended ? 5. By applying this law to all magnets, natural as well as artificial, describe what would be the result on a magnetic bar or needle, freely suspended, but by weight or by the nature of its mounting constrained to preserve a horizontal position ; and what would be the result, if so mounted, but free to move in every direction, the earth being regarded as a natural magnet. 6. What is the cause of the variation of the compass ? 7. What is meant by the deviation of the compass ? 8. What is meant by the term " local attraction " ? under what circum- stances have ships' compasses, from recent careful investigation, been found to be affected by it ? and name some of the localities in different parts of the world where this disturbance is to be found, and consequently where increased vigilance is necessary (see page 22, " Creak's Elementary Manual," 1908). 9. What do you understand by the term ' ' soft " iron ; and what are its properties as regards acquiring and retaining magnetism ? 10. What do you understand by the term "hard" iron; and what are its properties as regards acquiring and retaining magnetism ? 11. Describe the meaning of the term "horizontal force " of the earth ; where is it the greatest, and where the least, and what effect has it in respect to the increase or decrease of the directive force of the compass needle ? 12. Does the magnetic equator coincide with the geographical equator ? if not, state clearly how it is situated. X 3 o6 TEXT- BOOK ON NAVIGATION. 13. Where can the values of the magnetic dip, the earth's horizontal force, and the variation, be found ? 14. State in what parts of the globe lying in the usual tracks of navigation the variation changes very rapidly, and what special precautions should be observed when navigating these localities ; also why a " variation " chart is then very useful. 15. Why is a knowledge of the magnetic dip and the earth's horizontal force important in dealing with compass deviations ? 16. Describe the meaning of the term "vertical force" of the earth; where is it the greatest, and where the least ? 17. Would you expect a compass to be the more seriously affected by any given disturbing force when near the magnetic equator, or near the poles ? and state the reason. 18. State briefly (a) the essentials of an efficient compass ; and (6) what you would consider a good arrangement of the needles (that is, whether long or short, single or double, etc.) with the view to good compensation. 19. In stowing away spare compass cards or magnets, how would you place them with regard to each other, or what might be the probable consequence ? 20. State briefly the chief points to be considered when selecting a position for your compass on board ship, and what should be particularly guarded against. 21. What is meant by transient induced magnetism ? 22. Which is the red and which the blue pole of a mass of soft vertical iron (or indeed of any soft iron not in a horizontal position) by induction ; and what effect would the upper and lower ends of it have on a compass needle in the Northern hemisphere ? 23. Which is the red and which the blue pole of a mass of soft vertical iron by induction ; and what effect would the upper and lower ends of it have on the compass needle in the Southern hemisphere ? 24. What effect would a bar of soft vertical iron have on the compass needle on the magnetic equator ? 25. Describe what is usually termed the sub-permanent * magnetism of an iron ship, and state when and how it is acquired, and which is the sub- permanent red and which is the blue pole, and why it is called sub-permanent magnetism. 26. What is meant by "the composition of forces" and "the parallelo- gram of forces " ^ and show how the knowledge of these is valuable in ascertaining and compensating the sub-permanent magnetism of an iron ship. 27. Describe the nature of the coefficients B and C, plus (+) and minus ( ), and the different magnetic forces they represent; also why they are said to produce semicircular deviations. 28. Can semicircular deviations be produced by any other force than the sub-permanent magnetism of the ship ? If so, by what ? 29. On what points, by compass bearings of the ship's head, does + B give westerly deviation, and on what points does it give easterly ; also on what point does B give westerly, and on what points easterly ? 30. On what points does + give westerly deviation, and on what points easterly; also on what points does give westerly, and on what points easterly deviation ? 31. The value of either coefficient B or C being given, also the magnetic direction of the ship's head while she was being built, determine by the Traverse Tables the approximate value of the other coefficient C or B ; and, 1 See footnote, p. 300. SYLLABUS. 307 the value of both these coefficients being given, determine approximately the direction by compass of the ship's head whilst being built, assuming, of course, that these coefficients resulted altogether from sub-permanent magnetism. 32. Would you expect the greatest disturbance of the needle from the effects of sub-permanent magnetism alone to take place when ship's head is in same direction as when building, or when her head is at right angles to that direction, and in what direction of the ship's head would you expect to find the least disturbance ? 33. Describe quadrantal deviation, and state what coefficients represent it ; also on what points of the ship's head, by compass, each of these coefficients gives the greatest amount of deviation, and why it is called quadrantal deviation. 34. On what points of the compass will each of the coefficients D and E, -f and , give easterly, and on what points westerly, deviation ? 35. What conditions of the iron of a ship will produce 4-D, and what -D? 36. State clearly, then, which end of the horizontal iron running athwartship (such as beams, etc.), and of horizontal iron running fore and aft of a ship, acquires red and which blue polarity, by induction, when ship's head is at N.E., S.E., S.W., and N.W. respectively. 37. Describe the nature of the deviation represented by coefficients +A and A, and describe the errors in the construction of the compass, and other causes, that frequently produce it. 38. What is the object of compensating the compass by magnets, etc. ; and what are the general advantages of a compensated compass over an uncompensated one ? 39. Before adjusting the compass of an iron ship, what is it desirable to do with the view to eliminating, as far as possible, what may be termed the unstable part of the magnetism of the ship ? 40. Describe clearly the tentative method of compass adjustment (that is, the compensation of coefficients B, 0, and D, with ship upright) as generally practised by compass adjusters in ships of the Mercantile Marine. 41. State at what distance, as a general rule, the magnets and soft-iron correctors should be placed from the compass needles, and what will be the consequence if they are placed too near the needles. 42. Is it necessary that the magnets used for compensating co- efficients B and C should be placed on the deck? If not, state where they may also be placed, and the rules to be observed in placing them into position. 43. Does the B found on board ship usually arise altogether from sub- permanent magnetism, or does part of it usually arise from some other cause or causes ? 44. If the part of B due to induced magnetism in vertical soft iron, as well as the part due to sub-permanent magnetism, are corrected by a magnet alone, as is generally the case, what is frequently the consequence on the ship changing her magnetic latitude and hemisphere ? 45. How should each of these two parts of B then, strictly speaking, be compensated ? 46. Assuming, for the sake of clearness, that your steering compass is unavoidably placed very near to the head of the stern-post (and other vertical iron at the stern), thereby causing a very large B from induced magnetism, describe briefly any method by which the approximate position for the com- pensating vertical iron bar (Flinders' or Bundell's) could be estimated in order to reduce the error ; describe also how you would proceed, in order to improve, if not to perfect, its position after observations have been made on the magnetic equator. 47. State if standard compasses, as well as steering compasses, are 308 TEXT- BO OK ON NAVIGATION. generally subject to this disturbance from induced magnetism in vertical iron ; also whether the attraction in all cases is found to be towards the stern ; and if not, state the conditions under which it might be toward the bow, and how the compensating soft-iron bar should then be placed. 48. Generally speaking, does the magnetism induced in vertical iron usually have any effect in producing the coefficient C (ship upright), or is it generally produced by sub-permanent magnetism alone ? State also your reasons for saying so. 49. Provided the needles of your compass are not so long and powerful, and so near, as to cause the soft-iron correctors to become magnetized by induction, would the coefficient D, if properly compensated as you have described (Ans. 40), be likely to remain so in all latitudes and both hemi- spheres ? If so, state the reason why. 60. Under what circumstances does the character of A and E so change as to render it desirable that these coefficients should be disregarded or modified ? 61. Supposing your compasses were allowed to remain uncompensated, explain clearly what would be the probable changes (ship upright) in the deviations produced separately by (1) the sub-permanent magnetism of the ship alone, (2) by the induced magnetism in vertical soft iron, (a) on reaching the equator, (6) in the southern hemisphere. 52. Assuming you were able to arrive at the proper proportions to be corrected, and were then to exactly compensate the sub-permanent magnetism of the ship by means of a permanent magnet, and the induced magnetism in vertical iron by a soft-iron bar, would you expect any deviation to take place in your compass as the ship changed her latitude and hemisphere ? And state your reasons for saying so. 63. Supposing the coefficient D from horizontal soft iron were allowed to remain uncompensated, would you, or would you not, expect the D to differ in name or amount on the ship changing her magnetic latitude, and hemisphere ? And state the reason. 54. Describe how you would determine the deviation of your compass, (1) by reciprocal bearings, (2) by figures on the dock walls, (3) by bearings of a distant object. 55. Describe, in detail, how you would determine the deviation of your compass by the bearings of the Sun. Also by a star or planet. 56. Describe the uses to which the Napier's diagram can be applied, and its special advantages. 57. Describe clearly how the Napier's diagram is constructed. 58. For accuracy, what is the least number of points to which the ship's head should be brought for constructing a complete curve of deviations, or a complete table of deviations ? 59. Nearing land, and being anxious to check your deviations on a few courses you may probably require to steer, what is the least number of points it would be necessary to steady the ship's head upon, if making use of a Napier's diagram, in order to ascertain the deviation on each of the points, say, in ' a quadrant of the compass ? and describe clearly how you would do this at sea. 60. Supposing you have no means of ascertaining the magnetic bearing of the distant object when swinging your ship for deviations, how could you find it, approximately, from equidistant compass bearings ; and how far, as a rute, should the object be from the ship when swinging, or steaming round ? 61. Having taken the following equidistant compass bearings of a distant object, find the object's magnetic bearings, and thence the deviations : (a) Magnetic bearing required SYLLABUS. 3<>9 I Ship's head by standard Bearing of distant object by standard Deviation. Ship's head by standard Bearing of distant object by standard Deviation. compass. compass. compass. compass. N. S. 75 W. S. S. 34 W. N.E. S. 64 W. s.w. S. 31 W. E. S. 56 W. W. S. 49 W. S.E. S. 50 W. N.W. S. 71 W. (6) Construct a curve of deviations on a Napier's diagram, with the deviations as above, and give the courses you would steer by the standard compass to make the following courses, correct magnetic : Magnetic courses : N.N.W. S.S.E. W.N.W. E.S.E. Compass courses required. (c) Supposing you have steered the following courses by the standard compass, find the correct magnetic courses made from the above curve of deviations : Compass courses : N.N.E. E.N.E. S.S.W. W.S.W. Magnetic courses required. (d) You have taken the following bearings of two distant objects by your standard compass as above ; with the ship's head at N.E. \ E., find the bearings, correct magnetic : Compass bearings : S.E. by S. and N.N.W. Magnetic bearings required. 62. Assuming the deviations observed with the ship's head by compass to be as follows (or as in Question 61, whichever may be given), determine the value of the coefficients A, B, C, D, and E, and from them construct a com- plete table of deviations (or for as many points as the Examiner may direct). Deviation at N. (21 W.) S. (20 E. ) N.E. (10 W.) S.W. (23 E. ) E. ( 2 W.) W. ( 5 E. ) S.E. ( 4 E. ) N.W. (17 W.) 63. When swinging your ship, if it be required to construct Deviation Tables for two or more compasses situated in different parts of the vessel, describe the process, and how you would employ the Napier's diagram for this purpose. 64. State your rule for determining whether deviation is easterly or westerly. 65. Is a knowledge of the value of the various coefficients of any advantage ? If so, state why ? 66. Describe (a) what is commonly known by the term "retentive" or "retained" magnetism, and how the ship acquires it when in port and at sea ; (6) its effect on the compass needle whilst ship's head continues in the same direction; (c) the immediate consequence when the direction of the ship's head is altered ; and (d) the special precautions to be invariably observed at sea on the alteration of the ship's course. 67. Describe a " Dumb-card" or "Pelorus," and its use (a) in compen- sating a compass, (6) in determining the deviation. 68. If you determine the deviation by an azimuth or an amplitude of a heavenly body, it is then combined with variation, which together is some- times called the correction for the compass. State when the deviation is the difference between the variation and the correction, and when the sum ; and when it is of the same name as that of the correction, and when of the contrary name. 3 io TEXT- BOOK ON NAVIGATION. 69. In observing azimuths of heavenly bodies, the best method is by "time azimuths," since these can be observed without an altitude when the ship is in port, or when the horizon cannot be defined from any cause. Given the Sun's declination, the hour of the day, and the latitude, to find the true bearing of the Sun. 1 70. By night, if it be desirable to observe the correction of the compass. Given the day of the year, and time at ship, also the latitude of the place, to determine what stars will be in good position for this purpose. 71. If your correcting magnets are so mounted that their positions can be altered, describe the process by which, on open sea, you can place the ship's head correct magnetic N. (or S.), and correct magnetic E. (or W.), and can make the correction perfect. 72. Given the name of a star, the time, the place of ship, the variation of the compass, and the bearing of the star by compass, determine the deviation, and name it east or west. 73. Would you expect any change to be caused in the error of your compass by the ship heeling over, either from the effect of the wind or the cargo ? 74. Describe clearly the three principal causes of the heeling error on board an iron ship ? 75. Towards which side of the ship would that part of magnetism induced in continuous transverse iron (which was horizontal while ship was upright) help to draw the north point of the needle when ship heels over, (a) in the northern hemisphere, (b) in the southern hemisphere ? 76. Supposing the compass were placed between the two parts of a divided beam or other athwartship iron, towards which side of the ship would iron so situated help to draw the north point of the needle when ship heels over, (a) in the northern hemisphere, (6) in the southern hemisphere ? 77. Would you expect that part of the magnetism induced in iron exactly perpendicular to the ship's deck, such as stanchions, bulkheads, etc., if below the compass, to cause any part of the heeling error when ship heels over, and if so, towards which side of the ship, (a) in the northern hemisphere, (6) in the southern hemisphere ? 78. If an ordinary standard compass placed higher than the iron top sides be compensated whilst the ship is upright, what coefficient will be affected by heeling ? 79. Under what conditions (that is, as regards position whilst building and the arrangement of iron in the ship) is the north point of the compass needle usually drawn to windward, or the high side of the ship, in the northern hemisphere ? 80. Under what conditions, as a rule, is the north point of the compass needle usually drawn to leeward, or the low side of the ship, in the northern hemisphere ? 81. State to which side of the ship, in the majority of cases, is the north point of the compass drawn when ship heels over in the northern hemi- sphere ; and when this is the case, and it is not allowed for, what effect has it on the assumed position of the ship when she is steering on northerly, and also on southerly courses ? 82. On what courses would you keep away, and on what courses would you keep closer to the wind in both the northern and southern hemispheres in order to make good a given compass course, (a) when north point of compass is drawn to windward, or the high side of ship ; and (&) when drawn to leeward or the low side ? 1 The process of finding time azimuths by the ordinary formulae of spherical trigonometry is tedious, and since on board an iron ship these observations should be often repeated, the candidate will be allowed to use any table or graphic or linear method that will solve the problem within a half of a degree, the altitude of the heavenly body not being given. SYLLABUS. 311 83. If a ship is beating to windward, when she tacks, under what circumstances will the heeling error retain the same name, and under what circumstances will it take the contrary name ? 84. If a ship is placed on the opposite tack by the change of wind, the ship's course being the same by compass, will the heeling error change its name? 85. In which direction of the ship's head does the heeling error attain its maximum value, and in which direction does it generally vanish ? 86. Explain clearly how that part of the heeling error due to the per- manent part of the magnetism of the ship varies as the ship changes her geographical position, and what is the reason of this ? 87. Explain clearly how that part of the heeling error due to the induction in transverse iron (which was horizontal when ship was upright), and iron vertical to the ship's deck, varies as the ship changes her geographical position. 88. What, then, would be the probable nature of the heeling error, that is, whether to high or low side of the ship, and whether the error would be equal to the sum or difference, etc., of the forces (1) in high north lati- tudes, (2) on magnetic equator, (3) in high south latitudes ? Assuming the polarity of the sub-permanent magnetism of the ship under, and affecting, the compass to be as given below; the vertical induction in soft iron, of course, obeying the ordinary laws in the above geographical positions (1), (2), (3). (a) In cases where the effect of red vertical sub-permanent magnetism is equal to that of the vertical induction in the soft iron of the ship. (6) Where the effect of red vertical sub-permanent magnetism is greater than that of the vertical induction in the soft iron. (c) Where the effect of red vertical sub-permanent magnetism is less than that of the vertical induction in the soft iron. (d) Where the effect of blue vertical sub-permanent magnetism is equal to that of the vertical induction in the soft iron. (e) Where the effect of blue vertical sub-permanent magnetism is greater than that of the vertical induction in the soft iron. (f) Where the effect of blue vertical sub-permanent magnetism is less than that of the vertical induction in the soft iron. 89. Can the heeling error be compensated ? If so, state the means to be employed, and how the compensation may be effected. 90. Can the compensation of the heeling error be depended on in every latitude ? If not, state the reason. 91. Do the soft-iron correctors used for compensating the coefficient +D have any effect on the compass needle when the ship heels over ? and, if so, do they draw the needle towards the low or the high side of the ship, and do they counteract, or otherwise, the effect produced by the vertical induc- tion in the soft iron, (a) in the northern hemisphere, (6) in the southern hemisphere ; and what is the reason of this ? 92. Given the heel, the direction of the ship's head by compass, and the heeling error observed, to find the approximate heeling error, with a greater or less given heel, and with the ship's head on some other named point of the compass, the ship's magnetic latitude being in both cases the same. 93. Describe any instrument to show the ship's heel (generally called a clinometer), and state how and where it should be fixed. 94. Should the clinometer be observed when the ship is swung to deter- mine the deviation when the ship is upright ? If so, state the reason why. 95. Would you expect the Table of Deviations supplied by the compass adjuster from observations made in swinging the ship to remain good during the voyage, or would you expect the deviations to change ? If so, state under what circumstances. 96. Is it desirable that a record of your observations for deviations should 312 TEXT-BOOK ON NAVIGATION. be kept as a guide for any subsequent voyage, in case the ship should be in the same locality, or for further correction of the compass ? If so, describe some suitable form for keeping such record. 97. Would you, under any circumstances, consider it a safe and proper procedure to place implicit confidence in your compasses, however skilfully they may have been adjusted ? If not, what precautions is it your duty to take at all times ? ANSWERS TO THE QUESTIONS ON COMPASS DEVIATION FOB, MASTERS (EXTRA). 1. An artificial magnet is a bar or needle of hard-tempered steel which has been magnetized by artificial means, and so distinguished from a load- stone or natural magnet. Artificial magnets can be made in several ways : (a) For small needles, it is sufficient to place either pole of a strong magnet on one end of the needle and draw it in. contact to the other end, where it is taken off. If the pole used is the red pole, the end of the needle last in contact with it becomes a blue pole, and the other a red pole. (6) The method of " separate touch." The bar to be magnetized is placed with its ends resting on opposite poles of two strong magnets ; two other magnets are held, one in each hand, with opposite poles downwards, and brought together at the middle of the bar. The magnets are then inclined so as to make angles of about 30 with the bar, and are moved in contact with it towards the ends, where they are taken off. This process should be repeated several times on both sides of the bar, taking care that a red pole is moved towards the end supported by a red pole, and the blue pole towards the end resting on a blue pole. (c) By using an electro-magnet. One half of the bar is drawn from the middle to the end over one pole of a strong electro-magnet, and the other half over the other pole. , (d) By suspending the bar inside a solenoid and passing a strong electric current through the latter. 2. The end which points north when the magnet is freely suspended is named the red pole, and the other end the blue pole. 3. The south magnetic pole of the Earth is the red, and the north the blue pole. The north magnetic pole is situated about 70 3ST. latitude and 97 W. longitude, and the south magnetic pole about 72| S. latitude and 154 E. longitude. 4. Poles of the same name repel each other, whilst poles of different names attract each other. If freely suspended in the manner stated, they would assume a position of equilibrium determined by the Earth's directive force acting on the needles and the mutual attraction of their opposite poles. 5. In the first case, the needle would come to rest in the plane of the magnetic meridian, with its red pole pointing north ; in the second case, the needle would still coincide with the magnetic meridian, but its red pole would dip in N. latitude, and its blue pole would dip in S. latitude at an angle from the horizontal depending on the magnetic latitude, the dip being on the magnetic equator, and 90 at the magnetic poles. 6. The variation of the compass is due to the fact that the Earth's mag- netic poles do not coincide with the geographical poles. 7. Deviation of the compass is the angle of deflection of its no; in point from magnetic north, caused by the disturbing influence of iron. 8. " Local attraction" is jh ^ame give^ tc ^e ?^rietic influences (causing a disturbance of tlu compas, ) ^J *L^> locality in which a ship may be found ; for example, whenin dof&^ar to iron ships, iron pillars, cranes, etc., also in the neighbourhoo^^^Sj/mie Volcanic islands, It is said that the following places have a disturbing effect on ships' compasses: Ascension, SYLLABUS. 313 St. Helena, Cape St. Frances (Labrador), Cossack (North Australia), New Ireland, Solomon Islands, Tumbora, Sumbawa, coast of Madagascar, Iceland, He de Los (west coast of Africa), and Odessa Bay and Shoal. 3. Malleable iron and cast iron are termed "soft iron." Soft iron has the property of being instantaneously magnetized by induction from the magnetism of the Earth or of a natural or artificial magnet ; also by passing an electric current round it. This induced magnetism is only transient in its character, and is lost in great part or entirely when the inducing cause is removed. 10. "Hard iron " is iron combined with a certain percentage of carbon, and is usually termed steel. Tempered hard iron has the property of acquir- ing slowly and retaining permanently, or nearly so, the magnetic polarity communicated to it. 11. " Horizontal force " is the horizontal component of the Earth's total magnetic force in the line of dip, and varies as the cosine of the magnetic latitude. It is greatest at the magnetic equator, and at the magnetic poles, and therefore gives the greatest directive force to the compass needle in the former case, whilst there is no directive tendency in the latter. 12. The magnetic equator does not coincide with the geographical equator, but intersects it in two nearly opposite points situated in west longitude and east longitude. From the former point in the Pacific Ocean it runs nearly parallel to the Equator to about 100 W. long., then bends to the south on approaching South America, and in crossing Brazil it recedes from the Equator as far as 16 S. lat. ; then it bends northwards and crosses the Equator in about 6 W. long., whence it passes across the Gulf of Guinea and the African continent, receding to about 10 N. lat., and then very gradually approaches the Equator to the other point of intersection. 13. The values of the dip, etc., may be found on charts prepared for the purpose. 14. The variation changes rapidly in the English Channel, the western portion of the North Atlantic, the South Atlantic near the coast of Brazil, the South Indian Ocean, the neighbourhood of Japan, and especially in the Gulf of St. Lawrence. In navigating these localities, if the ship's course is across the magnetic meridians, the precaution should be taken of altering the courses steered by compass in accordance with the change of variation. A variation chart is useful to show what magnetic meridians are crossed and the .amount of variation. 15. Because the vertical induced magnetism which causes "heeling error " depends on the dip, and the directive force of the needle depends on the Earth's horizontal force. 16. The "vertical force" is the vertical component of the Earth's total magnetic force in the direction of the dip. It is greatest at the magnetic poles, and least at the magnetic equator. 17. The needle would be least affected near the magnetic equator, because the Earth's horizontal force, and therefore the directive power of the needle, are there at their greatest. 18. (a) The card should be accurately centred and graduated, the mag- netic axis of the needle (or the common axis, if more than one) coinciding with the north and south points of the card. There should be the least possible friction between the centre cap and the supporting pivot, the needles should have good directive power, the "lubber-point" should be exactly in the fore and aft midship line, and the deviation should be reduced by compensation to a small amount, say under 3. The gimbals should have little friction, and should not allow any lateral play. When in position in the binnacle, the card and lubber-point should be in full view of the man at the wheel. (&) It is generally agreed that two or more short needles parallel to each 314 TEXT-BOOK ON NAVIGATION. other in the same plane is the best arrangement for securing good compensation. 19. Care should be taken that the magnets are not stowed away with similar poles in contact, because in that position they would tend to weaken or demagnetize each other, or even reverse the poles. 20. A position should be chosen for the steering compass where it would not be near large elongated masses of iron, e.g. stern post, rudder post, tiller, stanchions, funnel, etc. For a standard compass the same precautions are necessary, and the position chosen should be, if possible, symmetrical with regard to the various magnetic forces in the ship. 21. Transient induced magnetism is acquired by soft iron by induction from permanent magnets, from electric currents, or from the Earth. This polarity is retained only so long as the inducing cause is present. 22. The blue pole is uppermost in north magnetic latitudes when the cause of induction is the Earth's magnetism ; therefore in the northern hemisphere the north end of the needle would be attracted by the upper end of a soft iron bar, and repelled by the lower end. 23. In south magnetic latitude the upper pole of soft iron by induction is red, and the lower end blue ; therefore the north point of a compass needle would be repelled by the upper end, and attracted by the lower. 24. No effect. 25. The sub-permanent magnetism of an iron ship is the magnetic polarity acquired in the course of construction by induction from the Earth and retained after launching. It is liable, however, to a considerable reduc- tion after the ship leaves the stocks, and is therefore termed " sub-per- manent " to distinguish it from the fixed magnetism of tempered steel. The red pole is that which was towards the north when the ship was being built, and the blue towards the south. 26. Composition of forces is the method of finding a single force which is equivalent to two or more forces of given amount and direction. In the "Parallelogram of Forces," two adjacent sides represent in direction and units of amount two given forces acting at the point of meet- ing, and a diagonal from this point represents in direction and units of amount the equivalent force, or "resultant " of the composition of the two given forces. A knowledge of this principle enables a compass adjuster to resolve the sub -permanent magnetic force in a ship into fore and aft, athwartship, and vertical components, and to compensate each independently of the others. 27. Coefficients B and C express the maximum deviations caused by magnetic forces acting horizontally in fore-and-aft and athwartship direc- tions respectively ; +B represents a blue pole at the bow, and red at the stern; B the reverse; 4-C represents a blue pole at the starboard side, and red at the port side ; C the reverse. They are said to produce semi- circular deviation, because the deviation due to each is easterly on one half of the compass, and westerly on the other half. 28. Yes; it may be caused by vertical soft iron, because the induced poles remain the same whatever the direction of ship's head. 29. +B gives E. deviation on all courses from N. through E. to S., and W. deviation from N. through W. to S. ; B gives the opposite. 30. +C gives E. deviation on all courses from W. through N. to E., and W. deviation from W. through S. to E. ; C gives the opposite. 31. See pa.ge 296. 32. The greatest disturbance from sub-permanent magnetism would be found when the ship lay at right angles to the direction in which she was when being built, and least when in the same direction. 33. Quadrantal deviation is the effect of transient induced magnetism in horizontal iron; it is represented by coefficients D and E. D gives the SYLLABUS. 315 greatest deviation on N.E., S.E., S.W., and N.W. courses, and E gives the greatest on N., E., S., and W. courses. It is termed quadrantal because it changes its name in successive quadrants. 34. + D gives E. deviation on courses from N. to E. and S. to W., and W. deviation from E. to S. and W. to N. ; - D gives the opposite. +E. gives E. deviation on courses from N.W. to N.E. and S.E. to S.W., and W. deviation from N.E. to S.E. and S.W. to N.W. ; -E gives the opposite. 35. Continuous transverse iron produces +D, and continuous fore-and-aft iron D ; but if end on to the compass, or when a beam is divided for a skylight and a compass placed between the parts, then the signs would be reversed. 36. The following table shows the results : Ship's bead. Athwartship iron. Fore-and-aft iron. Polarity acquired. N.E. End to port. End towards bow. Bed. starboard. stern. Blue. S.E. port. Bed. starboard. bow. Blue. S.W. stern. Bed. port. bow. Blue. N.W. starboard. v> Bed. port. stern. Blue. 37. Coefficient A is a constant deviation of the same amount and sign on all courses, -t-A being easterly and A westerly. It may be due to several (a) The magnetic axes of the needles not coinciding in direction with the N. and S. points of the card. (6) The lubber-point not being correctly placed. (c) Unsymmetrical arrangement of iron with respect to the compass. (d) An error in the correct magnetic bearing used for finding the deviation. (e} Retained magnetism, when swinging the ship. 38. The object of compensating compasses is to reduce the deviations to the smallest possible amounts. An uncompensated compass having large deviations would be too "sensitive" on some courses, and too "sluggish" on others, so that its indications would not correspond to the actual changes of course ; whilst a compass having its deviations reduced to a minimum by compensation would behave equally well on all courses. 39. When being equipped for sea, it is desirable that the ship's head should point the opposite way to the direction when building, the compen- sations being made just before sailing. 40. A common practice is to draw two lines on the deck where the binnacle stands one fore and aft amidships, and the other at right angles, both passing through a point vertically under the centre of the compass ; then, with the ship's head at correct magnetic N. or S., a bar magnet is laid on the deck at right angles to the fore-and-aft line, with its middle point on it and approached to the compass until the needle points correct magnetic N. Next, the ship's head is steadied on correct magnetic E. or W., and another bar is laid on the deck at right angles to the athwartship line, with its middle point on that line, and moved to or from the compass until the needle again points correct magnetic N. Both magnets are then fixed in position. Next, the ship's head is steadied on N.E., S.E., S.W., or N.W. correct magnetic for the purpose of correcting the quadrantal deviation. This is 316 TEXT- BO OK ON NAVIGATION. done by placing two soft iron correctors on the supports attached to the binnacle, and moving them towards or from the compass until the deviation is compensated. Lastly, the ship's head is brought to N. or S. by compass, and she is heeled over, say 10, and the heeling error observed. Then a small bar magnet is placed vertically under the centre of the compass, and moved up or down in its groove until the error disappears. N.B. The red pole of the magnet must be uppermost when the deviation is towards the higher side, and vice versa. In Thompson's compass the magnets are inside the binnacle. 41. The magnets should not be nearer to the centre of the card than twice their own length, and the quadrantal correctors not nearer the needles than once and a quarter the length of the needles. If the magnets are too near, they cause a new deviation called sextantal error, and if the correctors are too near they are liable to receive induced magnetic polarity from the needle, and attract either end indifferently. 42. JSTo. They may be placed either above or below the compass, on a bulkhead, or in the binnacle itself, but parallel to the deck, provided the condition is observed that they are bisected at right angles by transverse and fore-and-aft planes passing through the centre of the compass. 43. Coefficient B is partly due to sub-permanent magnetism and partly to induced magnetism in vertical iron before or abaft the compass. 44. The compass would be found to be over-compensated with respect to B on going to a lower latitude, and on crossing the magnetic equator the part due to vertical induction would be reversed, which would necessitate the shifting or removal, or perhaps reversal, of the magnet. 45. One part should be compensated by a vertical soft iron bar amidships, either before or abaft the compass, as required, and the other by a bar magnet. 46. The ship's head being at E. or W., the deviation should be observed ; then the Flinders bar should be moved into position before the compass, to correct, say, one-fourth of this deviation as a first approximation, the rest being compensated by a magnet. When on the magnetic equator the whole of coefficient B is due to sub-permanent magnetism, and should be exactly compensated by the fore-and-aft magnet. Then, as the ship proceeds N. or S., the change observed in the deviation due to coefficient B would be the amount to be compensated by the Flinders bar. A more correct method would be, when direction of ship's head whilst building is known, to enter the Traverse Table with direction of ship's head when building as a course, and coefficient in the dep. column, then B will be found in the latitude column. This would be the part of B to be com- pensated by the magnet, and the remainder by the Flinders bar. 47. Standard compasses are liable to be affected by the induced mag- netism of vertical iron in the form of stanchions, pillars, funnels, or bulk- heads near them. The attraction is not always towards the stern, because the disturbing masses of iron might be forward of the compass, in which case the Flinders bar should be abaft the compass. 48. Vertical iron could not cause any part of coefficient 0, unless it was unsymmetrically placed with regard to the compass, therefore the whole of 0, when the ship is upright, is usually due to sub-permanent magnetism. 49. Yes ; because the horizontal force of the correctors, the disturbing force, and the directive force of the needle would vary in the same ratio. 50. When the joint effect is small, and when the constant deviation is W. when the ship is swung to the right, and E. when swung to the left, and consequently the effect of temporarily retained magnetism, they may be dis- regarded when compensating the compass. 51. (1) (a) It would be found to have decreased. (6) It would increase again and tend to return to its original amount. SYLLABUS. 317 (2) (a) It would become 0. (6) It would have a different sign. 52. No deviation would occur from the sub-permanent magnetism so long as it remained constant, because it is exactly counterbalanced by the correcting magnet; nor frem the vertical induced magnetism, because it varies in the same ratio as that of the Flinders bar. 53. The deviation from this cause would not change its name and amount, because (a) the red induced pole is towards the north in all latitudes, and (6) the horizontal induced force changes in the same ratio as the directive force of the needle. 54. (1) A compass is set up on shore or in a boat at some distance from the ship, where it is not exposed to local attraction, and an observer stationed by it. The ship is then swung, and as her head comes to the desired points simultaneous bearings are taken by signal of each compass from the other. The bearings by shore compass reversed are correct magnetic, with which the bearings observed on board are compared, and the deviation found. (2) Figures on dock walls show either the correct magnetic or the true bearings of a prominent distant object from the observer's position. When, therefore, the ship swings with the tide or is specially swung, the difference is noted between the figure on the wall in line with the object and the bearing of the same object by the compass on board, as the ship's head comes to each point on which the deviation is required. These differences are either "compass errors" or deviation, according as the figures on the wall are true or magnetic bearings. (3) The correct magnetic bearings of the distant object may be found by means of a compass placed in a direct line with the ship and the object in a position free from local attraction. The deviation for the actual direction of the ship's head is the difference between the compass bearing of the object and the correct magnetic as before. 55. (a) By the sun : Take its bearing by compass and compute for the same time, or take from the Azimuth Tables its true azimuth. The difference is the compass error, from which the deviation may be found when the variation is known. (6) By a star or planet : Find the hour angle of the body and compute, or take from the Azimuth Tables (when within their limits) its true azimuth, the difference between which and the observed bearing by compass at the time for which the calculation is made will be the compass error, whence the deviation as before. 56. Uses of Napier's diagram : (a) From known deviations on a few points, either correct magnetic or by compass, a curve showing the deviation on all points may be drawn. (6) The curve so drawn may be used as a complete deviation table, from which may be found mechanically the compass course from the correct magnetic, and vice versa. Its advantages are, that a complete curve of deviations may be drawn from the deviations observed on a few points, whether by compass or correct magnetic, and whether equidistant or not. It also gives a graphic record of the deviations which appeals to the eye, and gives a much better appreciation of the results of compensation than does an ordinary deviation table. Again, if observations have been obtained of the deviations on two or three points of the quadrant in which the ship's course may lie, a curve will show the deviations on the other neighbouring points. Further, the deviations obtained by drawing a curve are free from possible errors of observation and calculation. 57. The Napier's diagram consists of a vertical line graduated to points and degrees, with two sets of parallel lines drawn across it at angles of 60, one set being continuous or plain, the other set broken or dotted, through 3 i8 TEXT-BOOK ON NAVIGATION. each of the thirty-two marks representing the "points" of the compass. The N. is at the top and bottom, the S. in the middle, and the E. and W. midway between the N. and S. marks, the other points following in order. 58. Eight equidistant points, viz. K, N.E., E., S.E., S., S.W., W., and N.W. 59. Two would be the least number, but three would be better, because a circle could be drawn through the three projected points. Assuming that deviations had been observed on three compass courses in the quadrant in which my course would lie, I would project them on the dotted lines, and then draw a fair curve through the projected points, extending it to comprise the whole quadrant. The parts intercepted by the curve of the dotted lines through the other courses would give the required deviations, measured on the vertical line. If only two observations were available, a straight line passing through the points of projection would show approximately the deviations on the intermediate courses. N.B. If the observations for deviation had been made on correct magnetic courses, the plain lines would be used instead of the dotted. 60. The mean of the equidistant bearings would be the correct magnetic bearing, nearly. The object should be so far distant that its bearing by the compass on board would not be sensibly affected by the change of position in swinging the ship. 61. See page 208. 62. See page 322. 63. An observer being stationed at each compass, bearings are taken simultaneously of a distant object as the ship's head is steadied on seven or eight points correct magnetic (by Pelorus) round the compass. Comparing these bearings with the known correct magnetic, the deviations for each compass are found, and are then marked on the plain lines of a Napier's diagram. A curve is then drawn through each set of projected marks, and from these curves a complete table of deviations on compass courses may be constructed by measuring the parts of the dotted lines intercepted by the curve. 64. If the correct magnetic bearing reads to the right of the compass bearing, the deviation is easterly ; if to the left, westerly. 65. A knowledge of the coefficients is of advantage, because (1) From the coefficients a complete table of deviations can be constructed. (2) It shows how much of the deviation is due to each of the disturbing forces in a ship. (3) It enables a compass adjuster to employ the proper means for compensating. 66. (a) " Retentive "or " retained " magnetism is the name given to the temporary magnetic polarity acquired by a ship whilst heading for a con- siderable time in one direction, and retained for a longer or shorter period after that direction is changed. It is induced by the Earth's magnetism whilst the ship is in dock or on a long-continued course at sea. (6) It has no effect on the compass needle whilst the ship's head continues on the same course. (c) A deviation will appear tending to make the ship follow the last course steered. (d) Allowance should be made for this effect, and an observation for deviation should be made as soon as possible after the course is changed. 67. The "Pelorus" has the general appearance of an azimuth compass, but it has no magnets. The "card" is usually a metal disc, which can be turned about its centre and clamped in any position. (a) Its use in compensating : First clamp the sight vanes to the known SYLLABUS. 319 correct magnetic bearing of a distant object ; then turn the card round until the desired direction of ship's head is at the lubber-point, and clamp. Next swing the ship until the object is seen through the sight vanes, and place the magnet or corrector in position to effect the compensation. (6) Its use to find the deviation : Clamp the sight vane to the known correct magnetic bearing of a distant object on land, or a celestial object, and turn the card round until the object is seen through the sight vanes. The course by "Pelorus" is the correct magnetic, the difference between which and the course by compass is the deviation for that course. 68. Deviation is the difference between the correction and variation when they 4 are of the same name, but the sum if of different names. It has the same name or sign as the correction in all cases except when the correction is less than variation of the same name. 69. See page 177. 70. See page 325. 71. When on the magnetic equator, sea smooth, and ship upright, set a watch to show apparent time at ship. Bring the ship's head to near north (or south), and clamp the sight vanes of the Pelorus to the correct magnetic bearing of the Sun found by taking out the true bearing from the tables for some minutes in advance, and applying the variation ; also bring the (N. or S.) point of the Pelorus to the lubber-point. Then steer the ship so that the centre of the Sun's image is on the sight vanes at the moment for which the bearing was computed, and steady her on that course whilst the athwartship magnet is moved into position to make the compass indicate correct magnetic. In a similar manner place and keep ship's head on correct magnetic east or west, and adjust the position of the fore-and-aft magnet. 72. See page 180. ^ 73. Yes ; a deviation appears which is known as the heeling error. 74. (1) When the ship heels, the vertical component of sub-permanent magnetism takes effect in producing deviation, the upper pole not being vertically under the centre of the compass. (2) Iron which is horizontal when the ship is upright acquires a vertical component of induced magnetism. (3) The upper ends of rudder post, stern post, or other vertical iron before or abaft the binnacle, are brought out to one side of a vertical plane passing through the centre of the compass. 75. (a) Towards the higher side. (6) Towards the lower side. 76. (a) Towards the lower side. (6) Towards the higher side. 77. Such iron would cause a heeling error, (a) Towards the higher side. (6) Towards the lower side. 78. Coefficient C. 79. In the northern hemisphere, the N. point of the compass is generally drawn to windward in ships built head north, also in ships whose compasses have + D, and in others which have large masses of vertical iron below the compass. 80. In some ships built head south, and ships having a large D. 81. In the majority of cases the N. point of the compass is drawn to windward. If not allowed for, the effect would be to draw the ship to windward on northerly courses, and to leeward on southerly. 82. (a) In N. latitude keep away by compass on either tack as the ship heels on northerly courses, and keep closer to the wind on either tack on southerly courses. (6) In N. latitude keep closer to the wind by compass on either tack on northerly courses, and keep away on southerly courses. 320 TEXT- BOOK ON NAVIGATION. This holds good in S. latitude. 83. If in tacking the ship's head is northerly on one tack, and southerly on the other, the heeling error retains the same name ; but if on both tacks her head is either northerly or southerly, the error changes its name. 84. Yes. 85. The heeling error has its maximum value when heading north or south magnetic, and nearly vanishes at east and west. 86. It diminishes when the ship goes to a lower latitude, and increases when to a higher, because the directive force of the needle increases in the first case, and decreases in the second, whilst the vertical sub-permanent force is assumed to be constant. 87. It decreases on approaching the equator, and vice versa, becomes on the magnetic equator, and changes its sign after crossing ; because the vertical force of induced magnetism decreases to at the equator, and increases with increase of latitude, the uppermost poles being blue in north magnetic latitude, and red in south. The directive force of the needle also varies with change of latitude. (1) High N. latitude. (2) Magnetic equator. (3) High S. latitude. (a) Nil. (6) Difference to low side. (c) high side, (d) Sum to (*) (/) To low side, due to sub- permanent only. 5> To high side, due to sub- permanent only. ? Sum to low side. - > ' Nil. " Difference to high side. low 89. Yes ; for a given magnetic latitude, by means of a vertical magnet which can be moved up or down in a groove or tube. There are two ways of proceeding (1) Bring the ship's head to N. or S. by compass (assumed to be correct) and give her a list of, say, 10 ; then slide the magnet up or down until the error disappears, and fix in position. (2) Bring the ship's head to E. or W. magnetic, ship upright ; remove the card and place on the supporting pillar a small dipping needle, which has been balanced by a sliding weight so as to show no dip when on shore ; then place the magnet in its groove and move up or down until the needle is again perfectly level. 90. No ; because the heeling error is partly due to vertical induced magnetism, which varies with change of latitude. 91. The soft iron correctors are found to correct the heeling error due to transverse iron for ordinary angles of heeling, (a) In the N. hemisphere they deflect the N. point of the needle to the low side ; (&) in the S. hemisphere, towards the high side. The reason is that they produce a co- efficient C due to an induced vertical force in the correctors, which has a different sign from that of the caused by transverse iron when the ship heels. 92. See page 326. 93. A clinometer consists of a graduated semicircle, which is fixed on a bulkhead with its diameter horizontal and arc downwards. The zero of the arc should coincide with a plumb-line from the middle of the diameter, so that when the ship heels the plumb-line or weighted index would show on the arc the exact degrees of inclination. SYLLABUS. 321 94. Yes ; otherwise the deviations would be complicated with the heeling errors. 95. The deviations might be expected to change from several causes (1) The place where the ship is swung may have local attraction. (2) The force of sub-permanent magnetism is liable to change from long continuance on one course, and from concussion and vibration. (3) The compensation of coefficient B can only be regarded as provisional, there being no certainty that the just proportions were assigned to the magnet and the Flinders bar. (4) Allowance must be made for a want of scientific knowledge on the part of the adjuster, or a careless performance of his work. (5) Loss of power in needles or magnets. 96. It is most necessary that such a record should be kept. The follow- ing form would serve the purpose : D^te. Lat. Long. Var. Ship's bead. Deviation. Remarks. Standard compass. Steering compass. Standard compass. Steering compass. >. 97. No, because considerable changes in the deviation may occur. There- fore it is necessary that frequent observations for deviation may be made, especially when the course is altered, or^when rapidly changing latitude. PRACTICAL QUESTIONS IN SYLLABUS OF DEVIATION FOR EXTRA MASTERS. N -B -C w + B -0 -B + + B + s Fia. 215. 31. The values of B and C being given to determine the magnetic direction of ship's head whilst building, supposing coefficients due en- tirely to sub-permanent magnetism. According to the direction of ship's head whilst building, the signs of B and C can be easily remembered from the adjoining figure. Again, let a ship be built with her head in, say, the S.E. quadrant, OB being fore-and- aft line. Then if OA, in magnetic meridian, be taken as representing the horizontal component of the total magnetic force in the ship, it can be resolved into two forces at right angles, OB in a fore-and-aft direction, and BA in an athwartship direction. OB represents coefficient B. BA C. AOB ,, magnetic direction ship's head building. 322 TEXT-BOOK ON NAVIGATION. :. cot AOB = ? jnJEj T> or cot direction = O Hence in the Traverse Table, if the direction of ship's head whilst being built be taken as a course, the latitude and departure columns will give the relative values of B and C. Therefore (1) With ship's head as a course and B in latitude column, C is found in departure column. (2) Ship's head as course and C as departure give B in latitude column. (3) B in latitude and C in departure columns give the direction of ship's head from N. or S. Examples. 1. Given the direction of ship's head whilst building N.E. by N., and coefficient B = -16'5. Find coefficient C. Here the course 3 points and lat. 16*5, give dep. in. Coefficient C is therefore + 11*1, because the ship's head was easterly. 2. Given ship's head S.S.W., and coefficient C = -3'6. Required co- efficient B. Course 2 points and dep. 3'6 give lat. 87 ; therefore coefficient B = + 87, because the ship's head was southerly. 3. Given coefficient B = +7'3, and coefficient C = -10 7. Required the direction of the ship's head when building. Lat. 7 '3 and dep. 107 give the course 56, which is S. 56 W., because B is + and C is . Exercises. 1. Given coefficient B = -11 and C = -8 0< 5. Find direction of ship's head whilst being built. j 2. Given coefficient +C = 12'5, and -B = G'8. Required the direc- tion of ship's head. 3. Given +B = 147 and +C = 87. Required ship's head. 4. Given ship's head N. 65 W., and -B = 5-4. Required coefficient 0. 6. Given ship's head S. 14 W., and -C = 3'5. Required coefficient B. 6. Given ship's head S.E. | E., and coefficient C = +7 5. Required coefficient B. 62. Example. Assuming the deviations observed with the ship's head by compass to be as follows, determine the value of the coefficients A, B, 0, D, and E, and from them construct a table of deviations. Deviation at N. (21 W.) S. (20 E. ) N.E. (10 W.) S.W. (23 E. ) E. ( 2 W.) W. ( 5 E. ) S.E. ( 4 E.) N.W. (17 W.) NOTE. In computing the coefficients, E. deviation is marked +, and W. deviation . To find coefficient A. Add (algebraically) the deviations on N., E., S., and W., and divide by 4. Deviation on N. - 21 E. - 2 S. +20 W. + 5 4)+ 2 + i = A, e. +A = SYLLABUS. To find coefficient B. Add (algebraically) the deviation on E. and W., after reversing the of the latter, and divide by 2. Deviation on E. - 2 W. 5 (reversed) 323 2) -7 -3 i.e. -B = To find coefficient 0. Add (algebraically) the deviations on N. and S., after reversing the sign of the latter. Deviation on N. -21 S. -20 (reversed) 2) -41 -20i.e. -C = 2 * area BCE = AB . BE + i BE . EC = 20 x 12 + . 12 x 10 = 240 + 60 = 300 sq. ft. The volume of a rectangular block is the product of its length, breadth, and thickness or depth, and is expressed in cubic feet or cubic inches as case may be according to the unit used. L x B x D = Vol., and since L X B = area .'. area X D = Vol., , _ Vol. Vol. and D = - , or area area The ratio of the volumes of displacement and of a block having same length, breadth, and draught as a vessel is called its block coefficient of fineness. The volume o a cylinder or pipe is its sectional area multi- plied by its length = - 4 The volume of a sphere = -^ When the volume of a solid is ascertained, its weight may be found from its specific gravity (S.G.), or in the case of a floating MISCELLANEOUS PROBLEMS. 333 body by finding the number of cubic feet of water displaced, allowing 35 cubic feet of salt water to weigh one ton. A cubic foot of fresh water weighs 1000 ounces, and contains 6J gallons approx. Hence taking the S.G. of cast iron as 7 '207, the weight of a cubic foot would be 7*207 X 1000 ounces = 7207 ounces. Examples. 1. The displacement of a ship of 15,300 tons. Her dimensions are 462 ft. long, 45 ft. beam, and 32 ft. draught : find the coefficient of fineness. p _ Displacement Coettt * = Contents of block 1700 22 taking TT = y C 2 x 7 X 6400 : 4 x 22 x 2240 t0ns ' C a , = ^4 tons, which for a 6-inch rope would work out to about 8*2 tons. Anderson, in his "Strength of Materials," says that from experiment the tenacity of this size rope is from 14 to 17 tons, MISCELLANEOUS PROBLEMS. 335 from which it seems that the ultimate strength is more than 6400 Ibs. per sq. in. The rule given estimates the safe working load at $f- tons = 5*2 tons approximately and appears to be rather excessive. The safe working strain of a wire may be considered to be two-fifths of the square of the circumference in tons ; which for 2 X 3 2 a 3 -inch wire would give ? tons = J 5 = 3*6 tons. The above rules may also be used to estimate the size rope or wire to carry a certain load. For rope, multiply the load in tons by 7 and extract the square root. For wire, multiply the load in tons by 5, divide by 2, and extract the square root. When the size rope or wire to be used with a given purchase to take out a known weight has to be determined, first find the load on the hauling part, then the size to carry that load. To find the load on the haul, add one-tenth of the weight to be lifted for each sheave in the purchase, to the weight, and divide by the number of strings at the moving block. Thus to take out a 20-ton weight with a treble purchase arid leading block : there are 7 sheaves : hence Load on haul = - - - tons = -y tons Rope to lift this load = <\/$f- % 7 = 6 inches nearly ; or, Wire to lift this load = A/_3_4 v 5. . = 3J inches nearly. Again, to find what weight can be lifted with a given pur- chase and a known rope or wire : first find the load such rope or wire will carry, multiply this load by the number of parts in the moving block, and allow one quarter of the result off for friction, etc. Thus using a 6 -inch rope and treble purchase : Safe load for 6-inch rope = -= tons. There are 7 parts in moving block, hence total weight it can lift = -f- X 7 = 36 tons ; allow one quarter or 9 tons for friction, and remainder is 27 tons ; or, using a 3 \ inch wire : 336 TEXT-BOOK ON NAVIGATION. , , , 2 /7\ 2 2 X 49 , Safe load =- X = tons = 4-9 tons. Total lift = 7x4-9 tons = 34-3 tons. For friction allow = 8'6 Remainder = 25*7 tons to be safely lifted. These rules evidently do not give similar results, but it must be remembered they are all approximations : to make them tally fy would certainly be a better multiplier than f ; that is, the allowance for friction, etc., should be rather -^ of the total lift. ART. 133. Speed, Consumption, and Displacement. When a screw turns in a solid bed, each complete turn advances the screw by an amount equal to the pitch ; hence, if " N " be the number of revolutions per minute made, and " P " be the pitch of the screw in feet, the number of nautical miles per hour is 6Q.NP t 6080 ' Since the water in which the screw of a ship turns is not solid, the actual speed over the ground, as determined by trial on measured miles, is less than this amount; the difference between the engine and actual speeds is called the " slip," and may be expressed as a percentage on the engine speed. Thus in a ship where the number of revolutions per minute is 57, the pitch 18 feet, determine the percentage of slip, if the trials give a speed of 9 '5 knots. , 60 x 57 X 18 , Engine speed = -- ~~ " = 10-1 knots. Actual speed = 9*5 Slip = -6 As 10-1 : 100 : : -6 : Hence percentage of slip = 6 % roughly. The moving power in the ship is supplied by the burning of the coal, and it may be seen that the total consumption must depend on the speed, the time occupied, and the weight moved. From various experiments made it has been found that in the same ship, under the same conditions, the consumption varies as the cube of the speed obtained, if C represents consumption in tons, s the speed in knots, and t the time in hours : C varies as s 3 x t, and since s X t = distance " d " C varies as s 2 x d ; MISCELLANEOUS PROBLEMS. 337 hence if the time is constant Consumption varies as cube of speed ; if the distance is constant Consumption varies as square of speed. From observation it is found that in modern engines, with good coal, the consumption varies from 1'2 to 1*7 Ibs. per hour per I.H.P., say 1J Ibs., and since the speed of the ship varies directly as the number of revolutions made per minute, hence the I.H.P. varies as cube of revolutions in a given time. Example. The daily consumption for a speed of 8'5 knots is 30 tons, what would it be for a speed of 10*5 knots ? Since the time is the same, consumption varies as cube of speed. As 8-5 3 : 10-5 3 : : 30 : x 10-5 3 x 30 = 34,728-75 8-5 a = 614-125 = 56 '5 tons. Again, if in above ship the consumption is reduced to 20 tons per day, what speed would be developed ? As 30 : 20 : : 8-5 3 : x 3 20 x 8-53 _ 30 V409 7 ! log = 2 ' 61 ^ 148 /. x = V4091 = 7 '42 knots. o = 0-870716 = 7-425 log. Again, if the ship has to make a voyage of 3570 miles, what would be the total consumption at the second speed ? As the distance is the same, consumption varies as square of speed. Number of days taken at 8'5 knots = ^^ ^- = 17'5 Total consumption = 17'5 x 30 = 525 tons. As 8-5 2 : 10-5 2 : : 525 : x _ 10-5 2 x 525 _ 57,881-25 8-5 2 72-25 = 801 tons. Number of days taken at 10 -5 knots = = 14 .*. Total consumption at 56 '5 tons per day = 14 x 56'5 = 800-5 tons, agreeing with amount obtained directly. From the above example it is seen that 801 tons of coal will drive the ship S570 miles at 10'5 knots, but if her speed is reduced to 8'5 knots the distance will be increased. 338 TEXT-BOOK ON NAVIGATION. 801 For 801 tons will be consumed in -^r days at 8 '5 knots ; 801 x 204 ., Hence distance = - go -- miies = 5446-8 miles. Again, if in the above ship, where the speed is 8 '5 knots for a daily consumption of 30 tons, it is found that three days' coal are left at that speed when 1000 miles from port, to what speed must she be reduced to reach port ? Consumption varies as square of speed multiplied by distance. As s 2 x 1000 : 8-5 2 x 204 : : 90 : 30 3 8'5 2 x 204 x-80". = 44-217 .*. s = 6 "65 knots. It is evident that the more deeply a ship is laden the more power is required to drive her through the water at the same speed, and it is found that the consumption or I.H.P. varies as the rd power of the displacement, that is as the cube root of the square of the displacement. Thus if in a ship of 5000 tons displacement, 4000 tons are taken on at one port, and it is found that the consumption per day is 50 tons to drive her at a speed of 9 knots, what would the consumption be per day when the 1000 tons are taken on at another port to make the same speed ? As V4000 3 : v'sooo* : : 50 : x = 50 x VSOOO 2 Viooo 2 /. log x = log 50 + f log 5000 - | log 4000 = log 50 + | (log 5000 - log 4000) 5000 log = 3-698970 4000 log = 3-602060 0-096910 2 3)0-193820 i (log 5000 - log 4000) = 0-064607 50 log = 1-698970 58-02 log = 1-763577 or x = 58 '02 tons per day. EXERCISES. 1. Find net freight on 2400 tons at 17s. Qd. per ton, after allowing 6 % loading commission and 8 % primage. 2. Find the net freight of 5600 tons at 6 dollars per ton, allowing 5 % loading commission and 7 % primage, exchange being 4'85 dollars to 1. 3. Calculate net freight on 2700 tons at 22s. Qd. per ton, if primage is 7 % and loading commission 3 %. MISCELLANEOUS PROBLEMS. 339 4. Convert 6836 10s. to dollars if 1 = 4 -66 dollars. Convert 567832'5 francs to 's. if 1 = 28 -4 francs. 5. Wine costs 3'5 francs per litre, how much is this per gallon. The price of meat is 4 francs per kilogramme, how many pence per lb. ? 6. Find displacement of a ship in tons whose dimensions are : Length 465 ft., breadth 50 ft., draught 28 ft., coefficient of fineness - 84. 7. Taking the height of a mercurial barometer to be 30 ins. when a water barometer reads 34 ft., find S.G. of mercury. 8. The sectional area at load water-level is 5600 sq. ft., how much will 320 tons of bunkers sink her ? 9. Find coefficient of displacement of a liner 525 ft. long by 51 ft. broad by 25 ft. draught, her displacement being 15,300 tons. 10. A lifeboat is 24 ft. long, by 7 ft. 3 ins. wide, and 3 ft. 6 ins. deep : how many people will it carry, and what air-tight space must be allowed ? N.B. Use '6 as multiplier for finding cubical contents, and allow 10 cub. ft. per person, of which 1 cub. ft. must be air-tight. 11. How long must a bunker be made to contain 250 tons coal, allowing 42 cub. ft. to the ton, beam of ship 56 ft., height 'tween decks 6 ft. 6 ins. 12. The length of a rectangular field is to its breadth as 3 to 2, and its area is 11,094 sq. metres. Find the cost of surrounding it with a fence at 2 '25 francs per metre length of boundary. 13. If 648 boxes are stowed close together on board a ship, find the total cost of freightage at Is. 2d. per cub. yard, when each box measures 4 ft. long, 3 ft. wide, and 2} ft. deep. 14. A rectangular tank 8 ft. long and 7 ft. wide holds when full 6 tons of water : find its depth. 15. If gold is worth 3 francs 10 centimes per gramme, find its value per ounce in English money, taking a franc as equal to 9^c?., and a gramme as 15-43 grains. 16. How many tons of coal will a bunker 10 ft. 3 ins. deep, 42 ft. 6 ins. long, and breadth at two ends 8 ft. 6 ins., and in middle 10 ft., hold ? 17. How many persons will a lifeboat hold whose length is 22 ft. 6 ins., breadth 6 ft. 6 ins., inside depth 3 ft. 6 ins., and what number of cubic feet of air-tight space should it contain ? 18. A supposed cubical iceberg is 100 feet above level of sea, S.G. of ice 921, sea water T026 : find total height of berg. 19. If the displacement of a ship is 6800 tons, her length 340 ft., breadth 35 ft., and draught 22 ft., find coefficient of fineness. 20. If 150 itons of bunker sink a ship 5 ins., find her area at that water level. 21. Taking S.G. of iron as 7*7, find weight in pounds of a plate 20' X 2' X 1|". 22. Length of ship 525 ft., beam 55 ft., coefficient of fineness '9 : how many tons per "inch " immersion 'I 23. Find approx. the daily consumption in a vessel of 5000 I.H.P. 24. Find safe working loads for chains of f in., 1* in., and in. diameter. 25. When chains are weakened by having been in use for a time, how is strength restored ? 26. Find safe working strain of manilla ropes of 3 ins., 5 ins., and 6 ; } ins. circumference. 27. Find the safe working strain of wires 1 in., 2^ ins., and 5 ins. circum- ference. 28. How many parts of a 3-in. rope will be equal in strength to a 6-in. rope? 29. What weight will a 3-in. rope hang, when rove as a tackle having 6 parts at the travelling block ? 30. What weight will a 2^-in. wire raise, using a double purchase ? 340 TEXT-BOOK ON NAVIGATION. 31. What size rope or wire would you use to get out a 15 -ton weight with a fourfold purchase and leading block ? 32. What size rope and wire would be required to take out a 14-ton weight with a treble block ? 33. If the breaking strain of a 2-in. chain is 72 tons, what is it of a 3-in. chain "? 34. Mercury is 10,574 times as heavy as air, the S.G. of mercury is 13*597 : find the weight in grains of 1 cub. ft. of air. 35. From Glazebrook's experiments the S.G. of air at 15 C. and 760 mm. pressure when J saturated was '0012. Find weight in grains of 1 cub. ft. of this air. 36. Taking the composition of air to be 77 % of nitrogen, 21 % of oxygen, and the remainder carbonic acid gas, and Ganot's estimates that 1 litre of nitrogen at C. and 760 mm. weighs 1'2561 grammes, of oxygen 1'4298 grammes, of carbonic acid 1'9774 grammes, find weight in grains of 1 cub. ft. 37. At 32 F. the velocity of sound is 1090 ft. per second, and the increase is at the rate of 10 ft. per second for each 9 F. rise in temperature. You see the flash of a gun, and 6 seconds after hear the report, tempera- ture 68 F. : how far away is the gun ? 38. If the velocity of light be reckoned at 299,860 kilometres per sec., how many miles per sec. is this ? 39. The pitch of a propeller is 15 ft., and it makes 60 revolutions per minute, allowing 5 % slip : find actual speed. 40. The pitch of a propeller is 16 ft., revolutions per minute 57, and time for measured mile is 7 minutes : find slip %. 41. The speed on the trials is 9-5 knots, propeller pitch 19 ft., revolutions per minute 56, find slip %. 42. For a speed of 10 knots, the daily consumption is 25 tons : what would the consumption be for speeds of 12 knots and 8 knots ? 43. The daily consumption for a speed of 19 knots is 150 tons : what amount would be necessary for speeds of 21 and 15 knots ? 44. For a daily consumption of 55 tons, the speed is 13 knots : what speed would be developed on consumption of 30 and 80 tons per day ? 45. On a daily consumption of 84 tons, the speed is 15 knots : what would it be for consumptions of 2 '5 and 5 tons per hour ? 46. In a ship 16 tons of fuel are consumed per hour for a speed of 24 knots : how many tons will be required for a voyage of 4 days if the speed is increased 1 knot ? 47. The total consumption for a certain distance at 9 knots is 240 tons : what would it be for same distance if speed is increased to 13 '5 knots ? 48. A ship steams 10 knots on 23 tons daily : how much coal will be required to steam 3036 miles in 11 days, and have 100 tons in reserve ? 49. At 60 revolutions per minute the I.H.P. developed was 3500, what would the I.H.P. be when revolutions were increased to 75 per minute? How much coal per day is consumed under the two conditions ? 50. For 1800 miles at 10 knots, the consumption is 500 tons : how much will be required for 2500 miles at 12 knots ? 51. The consumption per day for a speed of 12 knots is found to have been 51 tons. There are 150 tons of coal left in bunkers, and 1152 miles to port, speed is reduced to 10 knots, how much will there be at the end of the voyage ? 52. On a daily consumption of 80 tons, the vessel makes 15 knots ; with only one day's coal left at this speed and 1000 miles to go, at what speed must she proceed to reach port ? 53. There are 20 days' coal on board, when steaming at 9 knots on 35 tons per day, and 100 tons reserve. When 12 days out, owing to bad weather, it is found that 45 tons daily have been consumed : at what speed must she proceed to arrive with 50 tons in bunkers ? MISCELLANEOUS PROBLEMS. 341 54. A vessel with a displacement of 3000 tons consumes 25 tons daily : what would consumption be at displacement of 2000 and 4000 tons ? 55. A ship of 10,800 tons displacement burns 100 tons of coal daily, for a speed of 13'5 knots : what would be the effect on the consumption of taking in 300 tons extra cargo, and increasing speed knot '? 56. A vessel of 8000 tons displacement has 7000 tons on board at one port, and develops a speed of 12 knots on 50 tons per day : what consump- tion will she have to make a speed of 13 knots, when the extra 1000 tons is taken in ? APPENDIX CONTAINING EXTRACTS FROM TIDE TABLES AND NAUTICAL ALMANAC FOR THE SOLUTION OF ALL PROBLEMS IN THIS BOOK USEFUL TABLES AND EXAMINATION PAPERS 344 APPENDIX. AUGUST. AT APPARENT NOON. Date. THE SUN'S Sidereal Time of the Semi- diameter passing the Meridian.* Equation of Time, to be added to Apparent Time. Var. in 1 hour. Apparent Right Ascension Var. in 1 hour. Apparent Declination. Var. in 1 hour. h. m. s. 8. 1 II n m. s. m. s. 8. Wed. 1 8 42 55-43 9-724 N. 18 11 45-0 37-26 16-65 6 10-22 0-132 Thur. 2 8 46 48-49 9-698 17 56 41-9 38-00 1 6-56 6 6-74 0-158 Frid. 3 8 50 40-93 9-672 17 41 21-2 38-72 1 6-47 6 2-64 0-184 Sat. 4 8 54 32-76 9647 17 25 43-3 39-43 1 6-39 5 57-93 0-209 Sun. 5 8 58 23-99 9-622 17 948-4 40-14 1 6-30 5 52-61 0-234 Mon. 6 9 2 14-61 9-597 16 53 36-8 40-82 1 6-21 5 46-70 0-258 Tues. 7 9 6 4-66 9-573 1637 8-9 41-50 1 6-13 5 40-21 0-282 Wed. 8 9 954-13 9-549 16 20 24-9 42-16 1 6-Ot 5 33-14 0-307 Tlmr. 9 9 13 43-02 9-525 16 325-0 42-82 15-96 5 25-49 0-330 Frid. 10 9 17 31-34 9-502 1546 9-5 43-46 1 5-87 5 17-28 0-354 Sat. 11 9 21 19-10 9-479 15 28 38-9 44-09 1 5-79 5 8-51 0-377 Sun. 12 925 6-31 9-456 15 10 53-3 44-70 1 5-71 4 59-19 0-400 Mon. 13 9 28 52-97 9-433 14 52 53-1 45-31 1 5-63 4 49 33 0422 Tues. 14 9 32 39-09 9-411 14 34 38 6 45-90 15-55 4 38-93 0445 Wed. 15 9 36 24-68 9-389 14 16 10-1 46-47 15-47 4 27-99 0-466 Tlmr. 16 940 9-75 9-367 13 57 28-0 47-03 15-39 4 16-54 0-488 Frid. 17 9 43 54-30 9-346 13 38 32-5 47-59 1 5 31 4 4-57 0-509 Sat. 18 9 47 38-34 9-325 13 19 23-9 48-12 15-23 3 52-09 0-530 Sun. 19 9 51 21-88 9-304 13 2-6 48-65 1 5-16 3 39-11 0-551 Mon. 20 955 4-92 9-283 12 40 28-9 49-16 15-09 3 25-64 0-571 Tues. 21 9 58 47-48 9-263 12 20 43-1 49-65 15-02 3 11-68 0-591 Wed. 22 10 2 29-56 9-244 12 045-7 50-13 14-95 2 57-25 0-611 Tlmr. 23 10 6 11-17 9-224 11 40 36-8 50-60 14-89 2 42-34 0-631 Frid. 24 10 952-32 9-205 11 20 16-9 51-05 14-82 2 26-98 0-649 Sat. 25 10 13 33-02 9-187 10 59 46-2 51-49 14-76 2 11-17 0-668 Sun. 26 10 17 13-28 9-169 1039 5-2 51-92 14-70 1 54-92 0-686 Mon. 27 10 20 53-12 9-151 10 18 14-1 52-33 14-64 1 38-25 0-703 Tues. 28 10 24 32-55 9-135 9 57 13-2 52-73 14-58 1 21-17 0-720 Wed. 29 10 28 11-58 9-119 936 3-0 53-12 14-53 1 3-70 0-736 Thur. 30 10 31 50-24 9-104 9 14 43 6 53-49 1 4-48 45-K6 0-751 Frid. 31 10 33 28-55 9-089 8 53 15-5 53-85 14-43 27-66 0-765 Sat. 32 1039 6-52 9-075 N. 8 31 38'9 54-19 14-38 9-13 0-779 * Mean Time of the Seraidiameter passing may be found by subtracting 0-18 from the Sidereal Time. APPENDIX 345 II. AUGUST. AT MEAN NOON. Date. THE SUN'S Equation of Time, to be subtracted from Mean Time. Sidereal Time. Apparent. Eight Ascension. Apparent Declination. Semidiameter.* Wed. Thur. Frid. 1 2 3 h. m. s. 8 42 54-43 8 46 47-50 8 50 39-96 1 II N.18 11 48-8 17 56 45-8 17 41 25-1 15 47-14 15 47-27 15 47-40 m. s. 6 10-24 6 6-75 6 2-66 h. m. s. 8 36 44-19 8 40 40-75 8 44 37-30 Sat. Sun. Mon. 4 5 6 8 .54 31-81 8 58 23-05 9 2 13-70 17 25 47-2 17 9 52-4 16 53 40-8 15 47-53 15 47-67 15 47-81 5 57-95 5 52-64 5 46-73 8 48 33-86 8 52 30-41 8 56 26-97 Tues. Wed. Thur. 7 8 9 9 6 3-76 9 9 53-25 9 13 42-16 16 37 12-8 16 20 28-7 16 3 28-8 15 47-95 15 48-09 15 48-24 5 40-24 5 33-17 5 25-52 9 23-52 9 4 20-08 9 8 16-63 Frid. Sat. Sun. 10 11 12 9 17 30-50 9 21 18-29 9 25 5-52 15 46 13-4 15 28 42-7 15 10 57-0 15 48-39 15 48-54 15 48-69 5 17-32 5 855 4 59-23 9 12 13-19 9 16 9-74 9 20 6-29 Mon. Tues. Wed. 13 14 15 9 28 52-22 9 32 38-37 9 36 23-98 14 52 56-8 14 34 42-2 14 16 13-6 15 48-85 15 49-02 15 49-19 4 49-37 4 38-96 4 28-02 9 24 2-85 9 27 59-40 9 31 55-96 Thur. Frid. Sat. 16 17 18 9 40 9-08 9 43 53-66 9 47 37-74 13 57 31-4 13 38 35-7 13 19 27-0 15 49-36 15 49-54 15 49-72 4 16-57 4 4-60 3 52-12 9 35 52-51 9 39 49-07 9 43 45-62 Sun. Mon. Tues. 19 20 21 9 51 21-31 9 55 4-39 9 58 46-99 13 5-5 12 40 31-7 12 20 45-8 15 49-90 15 50-09 15 50-29 3 39-14 3 25-67 3 11-71 9 47 42-17 9 51 38-73 9 55 35-28 Wed. Thur. Frid. 22 23 24 10 2 2911 10 6 10-76 10 9 51-95 12 48-1 11 40 39-0 11 20 18-9 15 50-49 15 50-69 15 50-90 2 57-28 2 42-37 2 27-00 9 59 31-83 10 3 28-39 10 7 24-94 Sat. Sun. Mon. 25 26 27 10 13 32-69 10 17 12-99 10 20 52-87 10 59 48-1 10 39 6-8 10 18 15-5 15 51-11 15 51-32 15 51-54 2 11-19 1 54-94 1 38-27 10 11 21-50 10 15 18-05 10 19 14-60 Tues. Wed. Thur. Frid. 28 29 30 31 10 24 32-34 10 28 11-42 10 31 50-13 10 35 28-48 9 57 14-4 9 36 3-9 9 14 44-3 8 53 15-9 15 51-76 15 51-99 15 52-21 15 52-44 1 21-19 1 3-71 45-87 27-67 10 23 11-15 10 27 7-71 10 31 4-26 10 35 0-81 Sat. 32 10 39 6-49 N. 8 31 39-1 15 52-67 9-13 10 38 57-37 * The Semidiameter for Apparent Noon may be assumed the same as that for Mean Noon. 346 APPENDIX. AUGUST. MEAN TIME. III. Day. THE SUN'S Apparent Logarithm of the Radius Vector of the Earth. Transit of the First Point of Aries. THE MOON'S Longitude. Latitude. Semidiameter. Horizontal Parallax. Noon. Noon. Noon. Noon. Midnight. Noon. Midnight. 1 2 3 O 1 II 128 18 11-6 129 15 35-7 130 13 0-6 N.0-11 o-oo S. 0-14 0-0063890 0063306 0062709 Iu m. s. 15 20 44-55 15 16 48-64 15 12 52-73 i n 16 20-78 16 18-62 16 13-34 16 20-09 16 16-37 16 9-57 i it 59 53-36 59 45-45 59 26-08 i n 59 50 83 59 37-19 59 12-27 4 5 6 131 10 26-3 132 7 53-0 133 520-7 0-28 0-42 0-55 0-0062101 0061481 0060850 15 856-83 15 5 0-92 15 1 5-01 16 5-12 15 54-56 15 42-52 16 0-09 15 48-67 15 36-24 58 55-98 58 17-30 57 33-14 58 37-55 57 55-68 57 10-13 7 8 9 134 2 49-6 135 019-8 135 57 51-3 0-67 0-76 0-83 0-0060207 0059551 0058881 1457 9-10 14 53 13-19 14 49 17-28 15 29-96 15 17-90 15 7-18 15 23-81 15 12-33 15 2-53 56 47-15 56 2-95 55 23-65 56 24-62 55 42-51 55 6-64 10 11 12 136 55 24-1 137 52 58-3 138 50 33-9 0-88 0-91 0-90 0-0058197 0057498 0056782 14 45 21-37 14 41 25-46 14 37 29-55 14 58-46 14 52-22 14 48-72 14 55-01 14 50-12 14 48-03 54 51-72 54 28-85 54 16-05 54 39-07 54 21-16 54 13-52 13 14 15 139 48 ll'O 140 45 49-6 141 43 29-6 0-86 0-80 0-71 0-0056050 0055300 0054531 14 33 33-64 14 29 37-74 14 25 41-83 14 48-04 14 50-11 14 54-68 14 48-75 14 52-10 14 57-80 54 13-57 54 21-14 54 37-87 54 16-14 54 28-44 54 49-29 16 17 18 142 41 ll'l 143 38 54-1 144 36 38-5 0-61 0-49 0-36 0-0053744 0052936 0052108 14 21 45-92 14 17 50-01 14 13 54-11 15 1-38 15 9-70 15 19-06 15 5-37 15 14-30 15 23-93 55 2-42 55 32-91 56 7-22 55 17-04 55 49-73 56 25-05 19 20 21 145 34 24-4 146 32 11-7 14730 0-4 0-23 S. 0-09 N. 0-03 0-0051260 0050390 0049500 14 958-20 14 6 2-29 14 2 6-38 15 28-81 15 38-33 15 47-08 15 33-64 15 42-83 15 51-02 56 42-94 57 17-80 57 49-84 57 0-61 57 34-29 58 4-29 22 23 24 148 27 50-4 149 25 41-7 150 23 34-2 0-13 0-20 0-24 0-0048589 0047659 0046710 13 58 10-48 13 54 14-57 13 50 18-66 15 54-63 16 0-78 16 5-45 15 57-89 16 3-30 16 7-25 58 17-54 58 40-08 58 57-19 58 29-47 58 49-32 59 3-76 25 26 27 151 21 28-0 152 19 23-0 153 17 19-2 0-26 0-26 0-21 0-0045743 0044761 0043765 13 46 22-75 13 42 26-85 13 38 30-94 16 8-68 16 10-55 16 11-08 16 9-78 16 10-98 16 10-85 59 9-02 59 15-85 59 17-81 59 13-05 59 17-45 59 16-95 28 29 30 31 154 15 16-7 155 13 15-5 156 11 15-6 157 917-1 0-13 N. 0-03 S. 0-10 0-23 0-0042756 0041736 0040708 0039672 13 34 35-03 13 30 39-12 13 26 43-22 13 22 47-31 16 10-26 16 7-97 16 4-07 15 58-48 16 9-30 16 6-23 16 1-49 15 55-06 59 14-80 59 6-40 58 52-12 58 31-65 59 11-29 59 0-02 58 42-66 58 19-13 32 158 720-1 S. 0-36 0-0038629 13 18 51-40 15 51-25 15 47-09 58 5-16 57 49-88 APPENDIX. 347 IV. AUGUST. MEAN TIME. Day. THE MOON'S Longitude. Latitude. Age. Meridian Passage. Noon. Midnight. Noon. Midnight. Noon. Upper. Lower. 1 2 3 O 1 II 267 36 27-5 282 5 51-2 296 28 14-8 1 H 274 51 41-4 289 18 15-8 303 35 10-3 / " N. 3 40 55-7 2 36 59-9 1 23 35-1 01 II N.3 10 27-6 2 1 9-3 N. 44 57-9 d. 10-96 11-96 12-96 h. m. 9 36-2 10 36-3 11 35-3 b. m. 22 6-3 23 6-1 * * 4 5 6 310 38 27-3 324 32 7-9 338 6 161 317 37 34-9 331 21 46-3 344 45 29-9 N.O 5 58-7 S.I 10 30-5 2 21 11-7 S. 32 43-1 1 46 50-0 2 53 10-2 13-96 14-96 15-96 12 31-6 13 24-6 14 14-3 3-8 58 5 1 49 8 7 8 9 351 19 26-2 4 11 50-1 16 45 7'5 357 48 9-6 10 30 42-7 22 55 27-9 3 22 24-8 4 11 39-6 4 47 28-8 3 48 38-9 4 31 18-2 5 7-9 16-96 17-96 18-96 15 1-2 15 46-1 16 29-9 2 38-1 3 23-9 4 8-1 10 11 12 29 2 10-7 41 6 43-5 53 3 4-6 35 5 45-3 47 5 38-5 58 59 36-3 5 9 14-3 5 16 53-0 5 10 43-5 5 14 48-8 5 15 30-0 5 2 38-3 19-96 20-96 21-96 17 13-3 17 57-0 18 41-8 4 51-6 5 35-0 6 19-2 13 14 15 64 55 48-3 76 49 30-6 88 48 34-1 70 52 15-3 82 48 6-6 94 51 21-6 4 51 19-2 4 19 25-5 3 36 1-4 4 36 52-7 3 59 5-3 3 10 24-4 22-96 23-96 24-96 19 27-9 20 15-7 21 4-9 7 4-7 7 51-6 8 40-1 16 17 18 100 56 55-5 113 17 53-5 125 53 57-0 107 5 39-4 119 33 55-0 132 18 8-7 2 42 26-5 1 40 27-0 S.O 32 24-6 2 12 21-8 8.1 7 0-9 N.O 2 57-9 25-96 26-96 27-96 21 55-3 22 46-2 23 37-1 9 30-0 10 20-7 11 11-7 19 20 21 138 46 35-5 151 56 14-0 165 22 11-8 145 19 18-4 158 37 15-4 172 10 48-4 N.O 38 40-6 1 49 11-5 2 55 4-4 1 14 15-3 2 22 58-5 3 24 57-6 28-96 0-44 1-44 * * 27-7 1 17-7 12 2-5 12 52-7 13 42-5 22 23 24 179 2 48-8 192 55 41-3 206 57 57-0 185 57 53-6 199 55 50-1 214 1 40-0 3 52 7-6 4 36 25-7 5 4 45-0 4 16 5-7 4 52 45-1 5 12 11-2 2-44 3-44 4-44 2 7-4 2 57-2 3 47-8 14 32-2 15 22-4 16 13-6 25 26 27 221 6 36-7 235 18 48 9 249 31 58-6 228 12 26-4 242 25 25-6 256 38 10-7 5 14 54-0 5 5 56-9 4 38 19-2 5 12 49-1 4 54 23-5 4 17 59-5 5-44 6-44 7-44 4 39-8 5 33-8 6 29-9 17 6-6 18 1-6 18 58-6 28 29 30 31 263 43 46-1 277 52 2-9 291 54 41-0 305 49 27-6 270 48 28'8 284 54 12-5 298 53 11-9 312 43 10-9 3 53 44-3 2 55 7-5 1 46 20-8 N.O 31 55-2 3 25 57-9 2 21 44-1 N. 1 9 32-5 S.O 5 55-3 8-44 9-44 10-44 11-44 7 27-6 8 26-1 9 24-0 10 20-2 19 56-8 20 55-2 21 52-4 22 47-3 32 319 34 4-8 326 21 52-0 S.O 43 23-4 S. 1 19 55-5 12-44 11 13-7 23 39-4 The Moon's Longitude and Latitude are from HANSEN'S Tables; the Eight Ascension and Declination contain NEWCOMB'S corrections. 348 APPENDIX. AUGUST. MEAN TIME. THE MOON'S RIGHT ASCENSION AND DECLINATION. 1 Right Ascension. Var. in 10. Declination. Var. in 10 H Right Ascension. Var. in 10. Declination. Var. in iflm. Wednesday 1. Friday 3. h. m. B. a. ' " " h. m. s. s. ' ' " 17 49 48-87 25-622 S. 19 44 44-6 30-39 19 52 48-43 25-223 S. 19 30 5-1 35-89 1 17 52 22-65 25-638 19 47 42-9 29-03 1 19 55 19-67 25-190 19 26 25-9 37-18 2 17 54 56-52 25-653 19 50 33-0 27-66 2 19 57 50-71 25-156 19 22 39-0 38-46 3 17 57 30-48 25-667 19 53 14-8 26-28 3 20 021-54 25-121 19 18 44-4 39-73 4 18 4-52 25-679 19 55 48-3 24-89 4 20 2 52-16 25-086 19 14 42-3 40-98 5 18 238-63 25-692 19 58 13-5 23-51 5 20 522-57 25-050 19 10 32-6 42-24 6 18 5 12-82 25-703 20 030-4 22-13 6 20 752-76 25-013 19 615-4 43-48 7 18 747-07 25-713 20 239-0 20-73 7 20 10 22-72 24-975 19 1 50-8 44-72 8 18 10 21-38 25-723 20 4392 19-33 8 20 12 52-46 24-937 18 57 18-8 45-94 9 18 12 55-74 25-731 20 631-0 17-93 9 20 15 21-96 24-898 18 52 39-5 47-15 10 18 15 30-15 25-738 20 8 14-4 16-53 10 20 17 51-23 24-858 18 47 53-0 48-36 11 18 18 4-60 25-744 20 949-4 15-13 11 20 20 20-25 24-817 18 42 59-2 49-56 12 18 20 39-08 25-749 20 11 15-9 13-72 12 20 22 49-03 24-776 18 37 58-3 50-74 13 18 23 13-59 25-753 20 12 34-0 12-32 13 20 25 17-56 24-733 18 32 50-3 51-93 14 18 25 48-12 25-756 20 13 43-7 10-91 14 20 27 45-83 24-690 18 27 35-2 53-09 15 18 28 22-66 25-758 20 14 44-9 9-50 15 20 30 13-84 24-647 18 22 13-2 54-24 16 18 30 57-21 25-758 20 15 37-7 8-09 16 20 32 41-59 24-603 18 16 44-3 55-38 17 18 33 31-76 25-758 20 16 22-0 6-68 17 2035 9-07 24-558 1811 8-6 56-53 18 1836 6-31 25-757 20 16 57-8 5-26 18 20 37 36-29 24-513 18 5 26-0 57-65 19 18 38 40-84 25-754 20 17 25-1 3-84 19 2040 3-23 24-467 17 59 36-8 58-75 20 18 41 15-36 25-752 20 17 43-9 2-43 20 20 42 29-89 24-420 17 53 41-0 59-86 21 18 43 49-86 25-747 20 17 54-3 1-03 21 20 44 56-28 24-375 17 47 38-5 60-95 22 18 46 24-32 25-741 20 17 56-2 0-39 22 20 47 22-39 24-328 17 41 29-6 62-03 23 18 48 58-75 25-734 S. 20 17 49-6 1-81 23 20 49 48-21 24-279 S. 17 35 14-2 63-09 Thursday 2. Saturday 4. 18 51 33-13 25-726 S. 20 17 34-5 3-22 20 52 13-74 24-231 S. 17 28 52-5 64-14 1 1854 7-46 25-717 20 17 ll'O 4-62 i 20 54 38-98 24-183 17 22 24-5 65-18 2 18 56 41-73 25-707 20 16 39-1 6-03 2 2057 3-93 24-133 17 15 50-3 66-22 3 18 59 15-94 25-695 20 15 58-7 7-43 3 20 59 28'58 24-084 17 9 9-9 67-23 4 19 150-07 25-683 20 15 10-0 8-83 4 21 1 52-94 24-035 17 223-5 68-23 5 19 424-13 25-670 20 14 12-8 10-23 5 21 4 17-00 23-985 16 55 31-1 69-23 6 19 658-11 25-656 20 13 7-3 11-62 6 21 640-76 23-934 16 48 32-7 70-23 7 19 932-00 25-640 20 11 53-4 13-01 7 21 9 4-21 23-883 16 41 28-4 71-19 8 19 12 5-79 25-623 20 10 31-2 14-40 8 21 11 27-36 23-833 16 34 18-4 72-14 9 19 14 39-48 25-606 20 9 0-6 15-78 9 21 13 50-20 23-781 1627 2-7 73-09 10 19 17 13-06 25-588 20 721-8 17-16 10 21 16 12-73 23-729 16 19 41-3 74-03 11 19 19 46-53 25-568 20 534-7 18-53 11 21 18 34-95 23-678 16 12 14-4 74-95 12 19 22 19-88 25-548 20 3 39-4 19-90 12 21 20 56-87 23-627 16 441-9 75-86 13 19 24 53-10 25-526 20 1 35-9 21-27 13 21 23 18-47 23-574 15 57 4-1 76-75 14 19 27 26-19 25-503 19 59 24-2 22-63 14 21 25 39-76 23-522 15 49 20-9 77-63 15 19 29 59-13 25-478 19 57 4-3 23-99 15 21 28 0-73 23-468 15 41 32-5 78-50 16 19 32 31-93 25-454 19 54 36-3 25-33 16 21 30 21-38 23-416 15 33 38-9 79-36 17 1935 4-58 25-429 1952 0-3 26-68 17 21 32 41-72 23-364 15 25 40-2 80-20 18 19 37 37-08 25-402 19 49 16-2 28-02 18 21 35 1-75 23-311 15 17 36-5 81-03 19 1940 9-41 25-374 19 46 24-1 29-34 19 21 37 21-45 23-257 15 927-9 81-85 20 19 42 41-57 25-346 19 43 24-1 30-67 20 21 39 40-83 23-204 15 1 14-3 82-66 21 19 45 13-56 25-317 19 40 16-1 31-98 21 21 41 59-90 23-152 14 52 56-0 83-45 22 19 47 45-37 25-287 1937 0-3 33-29 22 21 44 18-65 23-098 14 44 32-9 84-23 23 19 50 17-00 25-255 19 33 36-6 34-60 23 21 46 37-07 23-044 1436 5-2 85-00 24 19 52 48-43 25-223 S. 19 30 5-1 35-89 24 21 48 55-18 22-992 S. 14 27 32-9 85-75 APPENDIX. 349 VI. AUGUST. MEAN TIME. THE MOON'S RIGHT ASCENSION AND DECLINATION. 1 Right Ascension. Var. in 10n. Declination, Var. in 10 m . jj A Right Ascension. Var. in 10n. Declination. Var. in lom. Sunday 5. Tuesday 7. h. m. s. s. ' " " h. m. s. s. ' " " 21 48 55-18 22-992 S. 14 27 32-9 85-75 23 33 26-54 20-674 S. 63236-7 107-63 1 21 51 12-97 22-938 14 18 56-2 86-49 1 23 35 30-47 20-635 6 21 50-3 107-82 2 21 53 30-43 22-884 14 10 15-0 87-22 2 23 37 34-16 20-596 611 2-9 107-99 3 21 55 47-58 22-832 14 1 29-5 87-93 3 23 39 37-62 20-558 6 14-4 108-16 4 21 58 4-41 22-778 13 52 39-8 88-63 4 23 41 40-86 20-521 5 49 25-0 108-31 5 22 020-92 22-725 13 43 45-9 89-33 5 23 43 43-87 20-483 5 38 34-7 108-46 6 22 237-11 22672 13 34 47-8 90-01 6 23 45 46-65 20-446 5 27 43-5 108-59 7 22 452-98 22-619 13 25 45-8 90-67 7 23 47 49-22 20-410 5 16 51-6 108-72 8 22 7 8-54 22-567 13 16 39-8 91-33 8 23 49 51-57 20-374 5 558-9 108-84 9 22 923-78 22-513 13 729-9 91-97 9 23 51 53-71 20-338 455 5-5 108-94 10 22 11 38-70 22-461 12 58 16-2 92-59 10 23 53 55-63 20-303 4 44 11-6 109-04 11 22 13 53-31 22-408 12 48 58-8 93-21 11 23 55 57-35 20-270 4 33 17-0 109-13 12 2216 7-60 22-356 12 39 37-7 93-81 12 23 57 58-87 20-236 4 22 22-0 109-21 13 22 18 21-58 22-304 12 30 13-1 94-40 13 0-18 20-203 4 11 26-5 109-28 14 22 20 35-25 22-253 12 20 44-9 94-98 14 2 1-30 20-170 4 030-6 109-34 15 22 22 48-61 22-200 12 11 13-4 95-53 15 4 2-22 20-138 3 49 34-4 109-40 16 2225 1-65 22-148 12 138-5 96-09 16 6 2-95 20-106 3 38 37-8 109-45 17 22 27 14-39 22-098 11 52 0-3 96-63 17 8 3-49 20-075 3 27 41-0 109-48 18 22 29 26-82 22-047 11 42 18-9 97-16 18 10 3-85 20-044 3 16 44-0 109-52 19 22 31 38-95 21-997 11 32 34-4 97-68 19 012 4-02 20-014 3 546-8 109-53 20 22 33 50-78 21-946 11 22 46-8 98-18 20 014 4-02 19-984 2 54 49-6 109-54 21 2236 2-30 21-895 11 12 56-3 98-67 21 016 3-83 19-954 2 43 52-3 109-56 22 22 38 13-52 21-845 11 3 2-8 99-15 22 18 3-47 19-927 2 32 54-9 109-55 23 22 40 24-44 21-796 S. 10 53 6-5 99-62 23 020 2-95 19-899 S. 22157-7 109-53 Monday 6. Wednesday 8. 22 42 35-07 21-747 S. 10 43 7-4 100-08 022 2-26 19-872 S. 211 0-6 109-51 1 22 44 45-40 21-698 1033 5-6 100-52 i 024 1-41 19-844 2 3-6 109-48 2 22 46 55-44 21-649 1023 1-2 100-94 2 026 0-39 19-818 149 6-8 109-44 3 2249 5-19 21-600 10 12 54-3 101-37 3 27 59-22 19-792 1 38 10-3 109-40 4 22 51 14-64 21-552 10 244-8 101-78 4 29 57-89 19-767 1 27 14-0 109-35 5 22 53 23-81 21-505 9 52 32-9 102-18 5 31 56-42 19-743 1 16 18-1 109-28 6 22 55 32-70 21-458 9 42 18-7 102-56 6 33 54-80 19-718 1 522-6 109-22 7 22 57 41-30 21-410 932 2-2 102-93 7 35 53-03 19-693 54 27-5 109-15 8 22 59 49-62 21-363 9 21 43-5 103-30 8 37 51-12 19-671 43 32-8 109-07 9 23 157-66 21-318 9 11 22-6 103-65 9 39 49-08 19-648 32 38-7 108-98 10 23 4 5-43 21-272 9 059-7 103-99 10 41 46-90 19-626 21 45-1 108-87 11 23 612-92 21-226 8 50 34-7 104-33 11 43 44-59 19-604 S. 01052-2 108-78 12 23 820-14 21-181 840 7-8 104-64 12 45 42-15 19-583 N. 0-2 108-67 13 23 10 27-09 21-137 8 29 39-0 104-95 13 47 39-59 19-563 10 51-8 108-54 14 23 12 33-78 21-093 8 19 8-4 105-24 14 49 36-90 19-543 21 42-7 108-42 15 23 14 40-20 21-048 8 836-1 105-53 15 51 34-10 19-524 32 32-8 108-28 16 23 16 46-36 21-005 758 2-1 105-81 16 053 3] -19 19-505 43 22-1 108-14 17 23 18 52-26 20-962 7 47 26-4 106-08 17 55 28-16 19-487 54 10-5 108-00 18 23 20 57-90 20-919 7 36 49-2 106-33 18 57 25-03 19-469 1 458-1 107-85 19 2323 3-29 20-878 7 26 10-5 106-57 19 59 21-79 19-451 1 15 44-7 107-68 20 2325 8-43 20-836 7 15 30-4 106-81 20 1 1 18-44 19-434 1 26 30-3 107-52 21 23 27 13-32 20-795 7 448-8 107-03 21 1 315-00 19-419 1 37 14-9 107-35 22 23 29 17-97 20-755 654 6-0 107-24 22 1 511-47 19-403 1 47 58-5 107-18 23 23 31 22-38 20-714 6 43 21-9 107-44 23 1 7 7-84 19-388 1 58 41-0 106-98 24 23 33 26-54 20-674 S. 63236-7 107-63 24 1 9 4-13 19-374 N. 2 9 22-2 106-78 APPENDIX. AUGUST. MEAN TIME. VII. THE MOON'S RIGHT ASCENSION AND DECLINATION. jj s Right Ascension. Var. In 10". Declination. Var. in 10m. 2 K Right Ascension. Var. in 10n. Declination. Var. in 10m. Thursday 9. Saturday 11. h. m. s. s. ' " " h.m. B. s. ' " i 1 9 4-13 19-374 N. 2 9 22-2 106-78 2 41 21-52 19-268 N.10 826-7 90-69 1 1 11 33 19-360 220 2-3 106-58 1 2 43 17-15 19-277 10 17 29-5 90-23 2 1 12 56-45 19-346 2 30 41-2 106-38 2 2 45 12-84 19-287 10 26 29-4 89-75 3 14 52-48 19-333 2 41 18-9 106-18 3 247 8-59 19-297 10 35 26-5 89-28 4 16 48-44 19-321 2 51 55-3 105-96 4 249 4-40 19-307 10 44 20-7 88-79 5 18 44-33 19-309 3 230-4 105-73 5 251 0-27 19-318 10 53 12-0 88-31 6 20 40-15 19-298 3'13 4-1 105-50 6 2 52 56-21 19-329 11 2 0-4 87-82 7 22 35-90 19-287 3 23 36-4 105-26 7 2 54 52-22 19-341 11 10 45-8 87-32 8 24 31-59 19-277 334 7-2 105-02 8 2 56 48-30 19-353 11 19 28-2 86-81 9 26 27-22 19-267 3 44 36-6 104-78 9 2 58 44-45 19-365 11 28 7-5 86-30 10 28 22-79 19-257 355 4-5 104-52 10 3 040-68 19-378 11 36 43-8 85-79 11 30 18-30 19-248 4 530-8 104-26 11 3 236-98 19-391 11 45 17-0 85-28 12 32 13-76 19-240 4 15 55-6 103-99 12 3 433-37 19-405 11 53 47-1 84-75 13 34 9-18 19-233 4 26 18-7 103-72 13 3 629-84 19-418 12 214-0 84-23 14 36 4-55 19-225 4 36 40-2 103-44 14 3 826-39 19-433 12 10 37-8 83-69 15 37 59-88 19-218 447 0-0 103-15 15 3 10 23-03 19-448 12 18 58-3 83-15 16 39 55-17 19-212 4 57 18-0 102-86 16 3 12 19-76 19-463 12 27 15-6 82-62 17 41 50-42 19-207 5 734-3 102-57 17 3 14 16-58 19-478 12 35 29-7 82-07 18 43 45-65 19-202 5 17 48-8 102-27 18 3 16 13-49 19-493 12 43 40-4 81-51 19 45 40-84 19-196 528 1-5 101-96 19 3 18 10-50 19-509 12 51 47-8 80-95 20 1 47 36-00 19-192 5 38 12-3 101-65 20 320 7-60 19-526 12 59 51-8 80-38 21 1 49 31-14 19-189 5 48 21-3 101-33 21 322 4-81 19-543 13 752-4 79-82 22 1 51 26-27 19-186 5 58 28-3 101-00 22 324 2-11 19-559 13 15 49-7 79-25 23 1 53 21-37 19-182 N. 6 833-3 100-68 23 3 25 59-52 19-577 N. 13 23 43-4 78-67 Friday 10. Sunday 12. 1 55 16-45 19-180 N. 6 18 36-4 100-34 3 27 57-03 19-594 N.13 31 33-7 78-09 i 1 57 11-53 19-178 6 28 37-4 99-99 i 3 29 54-65 19-613 13 39 20-5 77-50 2 1 59 6-59 19-177 6 38 36-3 99-65 2 3 31 52-38 19-631 1347 3-7 76-90 3 2 1 1-65 19-177 6 48 33-2 99-30 3 3 33 50-22 19-649 13 54 43-3 76-30 4 2 256-71 19-177 6 58 27-9 98-94 4 3 35 48-17 19-668 14 2 19-3 75-70 5 2 451-77 19-177 7 820-5 98-58 5 3 37 46-24 19-688 14 951-7 75-09 6 2 646-83 19-177 7 18 10-8 98-20 6 3 39 44-42 19-707 14 17 20-4 74-48 7 2 841-89 19-178 7 27 58-9 97-83 7 3 41 42-72 19-727 14 24 45-4 73-86 8 2 10 36-97 19-180 7 37 44-8 97-46 8 3 43 41-14 19-747 1432 6-7 73-24 9 2 12 32-05 19-182 7 47 28-4 97-08 9 3 45 39-68 19-767 14 39 24-3 72-61 10 2 14 27-15 19-184 757 9-7 96-69 10 3 47 38-34 19-787 14 46 38-0 71-98 11 2 16 22-26 19-187 8 648-7 96-29 11 3 49 37-12 19-808 14 53 48-0 71-34 12 2 18 17-40 19-192 8 16 25-2 95-88 12 3 51 36-03 19-829 15 054-1 70-69 18 2 20 12-56 19-195 8 25 59-3 95-48 13 3 53 35-07' 19-850 15 7 56-3 70-04 14 222 7-74 19-199 , 8 35 31-0 95-08 14 3 55 34-23 19-872 15 14 54-6 69-39 If) 224 2-95 19-203 845 0-3 94-67 15 3 57 33-53 19-894 15 21 49-0 68-73 1G 2 25 58-18 19-208 8 54 27-0 94-24 16 3 59 32-96 19-916 15 28 39-4 68-07 17 2 27 53-45 19-215 9 351-2 93-82 17 4 132-52 19-938 15 35 25-8 67-40 18 2 29 48-76 19-222 9 13 12-8 93-38 18 4 332-21 19-960 1542 8-2 66-73 19 2 31 44-11 19-228 9 22 31-8 92-95 19 4 532-04 19-983 15 48 46-5 66-04 20 2 33 39-50 19-235 9 31 48-2 92-51 20 4 732-01 20-007 15 55 20-7 65-36 21 2 35 34-93 19-243 941 1-9 92-06 21 4 932-12 20-029 16 1 50-8 64-67 22 2 37 30-41 19-251 9 50 12-9 91-61 22 4 11 32-36 20-052 16 816-7 63-97 23 2 39 25-94 19-259 9 59 21-2 91-15 23 4 13 32-74 20-076 16 14 38-4 63-28 24 2 41 21-52 19-268 N.10 826-7 90-69 24 4 15 33-27 20-100 N. 16 20 56-0 62-58 APPENDIX. VIII. AUGUST. MEAN TIME. THE MOON'S RIGHT ASCENSION AND DECLINATION. I Right Ascension. Var. in 10m. Declination. Var. in 10-". IM a X Right Ascension. Var. in 10-n. Declination. Var. in 10m. Monday 13. Wednesday 15. h. m. s. s. ' " " h. m. s. s. ' " " 4 15 33-27 20-100 N. 16 20 56-0 62-58 5 54 54-89 21-298 N. 19 50 36-9 22-97 1 4 17 33-94 20-123 1627 9-3 61-86 1 5 57 2-75 21-322 19 52 51-9 22-03 2 4 19 34-75 20-147 16 33 18-3 61-14 2 5 59 10-75 21-344 1955 1-3 21-09 3 4 21 35-70 20-171 16 39 23-0 60-43 3 6 1 18-88 21-367 1957 5-0 20-14 4 4 23 36-80 20-196 16 45 23-4 59-71 4 6 327-15 21-390 1959 3-0 19-19 5 4 25 38-05 20-220 16 51 19-5 58-98 5 6 535-56 21-413 20 055-3 18-24 G 4 27 39-44 20-244 16 57 ll'l 58-23 6 6 744-10 21-434 20 241-9 17-29 7 4 29 40-98 20-269 17 258-3 57-49 7 6 952-77 21-456 20 4 22-8 16-33 8 4 31 42-67 20-294 17 841-0 56-74 8 6 12 1-57 21-478 20 557-9 15-37 9 4 33 44-51 20-319 17 14 19-2 55-99 9 6 14 10-50 21-499 20 727-2 14-40 10 4 35 46-50 20-343 17 19 52-9 55-24 10 6 16 19-56 21-521 20 8 50-7 13-43 11 4 37 48-63 20-368 17 25 22-1 54-48 11 6 18 28-75 21-542 20 10 8-4 12-46 12 4 39 50-92 20-394 17 30 46-7 53-72 12 6 20 38-06 21-563 20 11 20-2 11-48 13 4 41 53 36 20-419 1736 6-7 52-94 13 6 22 47-50 21-583 20 12 26-2 10-51 14 4 43 55-95 20-445 17 41 22-0 52-17 14 6 24 57-06 21-603 20 13 26-3 9-52 15 4 45 58-70 20-470 17 46 32-7 51-39 15 627 6-73 21-623 20 14 20-4 8-53 16 448 1-59 20-495 17 51 38-7 50-61 1C 6 29 16-53 21-643 2015 8-6 7-54 17 4 50 4-64 20-521 17 56 40-0 49-82 17 6 31 26-44 21-662 20 15 50-9 6-56 18 452 7-84 20-547 18 136-5 49-02 18 6 33 36-47 21-681 20 16 27-3 5-56 19 4 54 11-20 20-573 18 6 28-2 48-22 19 6 35 46-61 21-699 20 16 57-6 4-56 20 4 56 14-71 20-598 18 11 15-1 47-41 20 6 37 56-86 21-718 20 17 22-0 3-57 21 4 58 18-37 20-623 18 15 57-1 46-60 21 640 7-22 21-736 20 17 40-4 2-56 22 5 022-19 20-649 18 20 34-3 45-78 22 6 42 17-69 21-753 20 17 52-7 1-55 23 5 226-16 20-675 N.1825 6-5 44-96 23 6 44 28-26 21-770 N. 20 17 59-0 0-54 Tuesday 14. Thursday 16. 5 430-29 20-701 N. 18 29 33-8 44-14 6 46 38-93 21-788 N. 20 17 59-2 0-47 1 5 634-57 20-727 18 33 56-2 43-32 i 6 48 49-71 21-805 20 17 53-4 1-48 2 5 839-01 20-753 18 38 13-6 42-48 2 651 0-59 21-821 20 17 41-5 2-50 3 5 10 43-60 20-778 18 42 26-0 41-64 3 6 53 11-56 21-837 20 17 23-4 3-52 4 5 12 48-34 20-803 18 46 33-3 40-79 4 6 55 22-63 21-853 20 16 59-3 4-53 5 5 14 53-24 20-829 18 50 35-5 39-95 5 6 57 33-79 21-868 20 16 29-0 5-56 6 5 16 58-29 20-855 18 54 32-7 39-10 6 6 59 45-05 21-883 20 15 52-6 6-58 7 5 19 3-50 20-880 18 58 24-7 38-23 7 7 156-39 21-898 20 15 10-1 7-60 8 521 8-85 20-905 19 2 11-5 37-38 8 7 4 7-82 21-912 20 14 21-4 8-63 9 5 23 14-36 20-932 19 553-2 36-51 9 7 619-33 21-926 20 13 26-5 9-66 10 5 25 20-03 20-957 19 929-6 35-63 10 7 830-93 21-939 20 12 25-5 10-69 11 5 27 25-84 20-981 1913 0-8 34-76 11 7 10 42-60 21-952 20 11 18-2 11-73 12 5 29 31-80 21-007 19 16 26-7 33-88 12 7 12 54-35 21-965 2010 4-8 12-75 13 5 31 37-92 21-032 19 19 47-3 32-99 13 715 6-18 21-978 20 845-2 13-78 14 5 33 44-18 21-056 1923 2-6 32-11 14 7 17 18-09 21-990 20 719-4 14-83 15 5 35 50-59 21-082 19 26 12-6 31-21 15 7 19 30-06 22-002 20 547-3 15-87 16 5 37 57-16 21-107 19 29 17-1 30-31 1G 7 21 42-11 22-013 20 4 9-0 16-90 17 540 3-87 21-130 19 32 16-3 29-41 17 7 23 54-22 22-023 20 2 24-5 17-93 18 5 42 10-72 21-154 19 35 10-0 28-50 18 726 6-39 22-034 20 033-8 18-97 19 5 44 17-72 21-179 19 37 58-3 27-59 19 7 28 18-63 22-045 19 58 36-9 20-01 20 5 46 24-87 21-203 19 40 41-1 26-68 20 7 30 30-93 22-054 19 56 33-7 21-06 21 5 48 32-16 21-228 19 43 18-4 25-75 21 7 32 43-28 22-063 19 54 24-2 22-10 22 5 50 39-60 21-251 19 45 50-1 24-83 22 7 34 55-69 22-073 ]952 8-5 23-13 23 5 52 47-17 21-274 19 48 16-3 23-90 23 737 8-15 22-081 19 49 46-6 24-18 24 5 54 54-89 21-298 N. 19 50 36-9 22-97 24 7 39 20-66 22-089 N. 19 47 18-4 25-23 352 APPENDIX. AUGUST. MEAN TIME. IX. THE MOON'S RIGHT ASCENSION AND DECLINATION. 1 Right Ascension. Var. in 10m. Declination. Var. in Iflm. fj S Right Ascension. Var. in ]0n>. Declination. Var. in lom. Friday 17. Sunday 19. h. m. s. e. ' " " h. m. s. B. ' " " 7 39 20-66 22-089 N. 19 47 18-4 25-23 9 25 37-74 22-075 N.1549 4-5 72-93 1 7 41 33-22 22-098 19 44 44-0 26-26 1 9 27 50-17 22-068 15 41 44-2 73-83 2 7 43 45-83 22-105 19 42 3-3 27-30 2 930 2-55 22-060 15 34 18-5 74-73 3 7 45 58-48 22-112 19 39 16-4 28-34 3 9 32 14-89 22-053 15 26 47-4 75-62 4 7 48 11-17 22-118 19 36 23-2 29-38 4 9 34 27-18 22-045 15 19 11-0 76-50 5 7 50 23-90 22-125 19 33 23-8 30-42 5 9 36 39-43 22-038 15 11 29-4 77-38 6 7 52 36-67 22-131 193018-2 31-45 6 9 38 51-64 22-031 15 342-5 78-25 7 7 54 49-47 22-136 19 27 6-4 32-49 7 941 3-80 22-023 14 55 50-4 79-12 8 757 2-30 22-141 19 23 48-3 33-53 8 9 43 15-91 22-014 14 47 53-1 79-98 9 7 59 15-16 22-146 19 20 24-0 34-57 9 9 45 27-97 22-007 14 39 50-7 80-82 10 8 1 28-05 22-151 19 16 53-5 35-60 10 9 47 39-99 21-998 14 31 43-3 81-66 11 8 340-97 22-154 19 13 16-8 36-63 11 9 49 51-95 21-990 14 23 30-8 82-51 12 8 5 53-90 22-158 19 933-9 37-67 12 952 3-87 21-983 14 15 13-2 83-34 13 8 8 6-86 22-162 19 544-8 38-69 13 9 54 15-74 21-973 14 6-50-7 84-16 14 8 10 19-84 22-164 19 149-6 39-73 14 9 56 27-55 21-965 13 58 23-3 84-98 15 8 12 32-83 22-167 18 57 48-1 40-76 15 9 58 39-32 21-957 13 49 51-0 85-78 16 8 14 45-84 22-168 18 53 40-5 41-78 16 10 051-03 21-948 13 41 13-9 86-58 17 8 16 58-85 22-170 18 49 26-8 42-80 17 10 3 2-70 21-940 13 32 32-0 87-38 18 8 19 11-88 22-172 1845 6-9 43-83 18 10 514-31 21-931 13 23 45-3 88-18 19 8 21 24-92 22-173 18 40 40-9 44-85 19 10 725-87 21-922 13 14 53-9 88-95 20 8 23 37-96 22-173 1836 8-7 45-87 20 10 937-37 21-913 13 557-9 89-72 21 8 25 51-00 22-174 18 31 30-5 46-88 21 10 11 48-83 21-905 12 56 57-3 90-48 22 828 4-05 22-175 18 26 46-2 47-89 22 1014 0-23 21-896 12 47 52-1 91-24 23 8 30 17-10 22-174 N. 18 21 55-8 48-90 23 10 16 11-58 21-888 N. 12 38 42-4 91-98 Saturday 18. Monday 20. 8 32 30-14 22-173 N. 18 16 59-4 49-91 10 18 22-88 21-879 N. 12 29 28-3 92-73 i 8 34 43-18 22-173 18 11 56-9 50-92 i 10 20 34-13 21-870 12 20 9-7 93-47 2 8 36 56-21 22-171 18 648-4 51-92 2 10 22 45-32 21-862 12 10 46-7 94-19 3 839 9-23 22-170 18 1 33-9 52-92 3 10 24 56-47 21-854 12 1 19-4 94-90 4 8 41 22-25 22-168 17 56 13-4 53-91 4 1027 7-57 21-845 11 51 47-9 95-61 5 8 43 35-25 22-165 17 50 47-0 54-89 5 10 29 18-61 21-837 11 42 12-1 96-32 G 8 45 48-23 22-163 17 45 14-7 55-88 6 10 31 29-61 21-828 11 32 32-1 97-01 7 848 1-20 22-161 17 39 36-4 56-87 7 10 33 40-55 21-820 11 22 48-0 97-69 8 8 50 14-16 22-158 17 33 52-2 57-85 8 10 35 51-45 21-812 11 12 59-8 98-37 9 8 52 27-09 22-153 1728 2-2 58-83 9 1038 2-29 21-803 11 3 7-6 99-03 10 8 54 40-00 22-150 1722 6-3 59-80 10 10 40 13-09 21-796 10 53 11-4 99-69 11 8 56 52-89 22-147 1716 4-6 60-77 11 10 42 23-84 21-788 10 43 11-3 100-34 12 859 5-76 22-143 17 957-1 61-73 12 10 44 34-55 21-781 1033 7-3 100-98 13 9 118-60 22-138 17 343-8 62-69 13 10 46 45-21 21-773 10 22 59-5 101-61 14 9 331-41 22-133 16 57 24-8 63-65 14 10 48 55-82 21-765 10 12 48-0 102-23 15 9 544-20 22-128 1651 0-0 64-61 15 1051 6-39 21-758 10 232-8 102-84 1G 9 756-95 22-123 16 44 29-5 65-55 16 10 53 16-92 21-751 9 52 13-9 103-45 17 910 9-67 22-118 16 37 53-4 66-48 17 10 55 27-40 21-744 9 41 51-4 104-04 18 9 12 22-36 22-113 16 31 11-7 67-43 18 10 57 37-85 21-738 9 31 25-4 104-63 19 9 14 35-02 22-107 16 24 24-3 68-36 19 10 59 48-25 21-730 9 20 55-9 105-20 20 9 16 47-64 22-100 16 17 31-4 69-28 20 11 1 58-61 21-723 9 10 23-0 105-77 21 919 0-22 22-093 16 10 32-9 70-21 21 11 4 8-93 21-718 8 59 46-7 106-33 22 9 21 12-76 22-088 16 328-9 71-13 22 11 6 19-22 21-712 849 7-1 106-88 28 9 23 25-27 22-082 15 56 19-4 72-03 23 11 8 29-47 21-705 8 38 24-2 107-41 24 9 25 37-74 22-075 N.1549 4-5 72-93 24 11 10 39-68 21-699 N. 82738-2 107-93 APPENDIX. 353 X. AUGUST. MEAN TIME. THE MOON'S RIGHT ASCENSION AND DECLINATION. 1 Right Ascension. Var. in 10m. Declination. Var. in 10-n. $ n Right Ascension. Var. in 10m. Declination. Var. iJi 10m. Tuesday 21. Thursday 23. h. m s. a. ' " " h. m. s. s. ' " " 11 10 39-68 21-699 N. 82738-2 107-93 12 54 42.34 21-786 S. 05134-4 120-89 1 11 12 49-86 21-694 8 16 49-0 108-45 1 12 56 53-09 21-798 1 339-8 120-89 2 11 15 0-01 21-689 8 556-8 108-96 2 1259 3-91 21-808 1 15 45-1 120-87 3 11 17 10-13 21-683 755 1-5 109-46 3 13 1 14-79 21-820 1 27 50-2 120-84 4 11 19 20-21 21-678 744 3-3 109-94 4 13 325-75 21-833 1 39 55-2 120-81 5 11 21 30-27 21-674 733 2-2 110-43 5 13 536-79 21-846 1 51 59-9 120-76 G 11 23 40-30 21-670 7 21 58-2 110-89 6 13 747-90 21-859 2 4 4-3 120-70 7 11 25 50-31 21-667 7 10 51-5 111-35 7 13 959-10 21-873 2 16 8-3 120-63 8 11 28 0-30 21-663 6 59 42-0 111-80 8 13 12 10-38 21-888 2 28 11-8 120-53 9 11 30 10-26 21-658 6 48 29-9 112-23 9 13 14 21-75 21-902 2 40 14-7 120-43 10 11 32 20-20 21-656 6 37 15-2 112-67 10 13 16 33-20 21-916 2 52 17-0 120-33 11 11 34 30-13 21-653 6 25 57-9 113-08 11 13 18 44-74 21-932 3 4 18-7 120-21 12 11 36 40-03 21-649 6 14 38-2 113-48 12 13 20 56-38 21-948 3 16 19-5 120-07 13 11 38 49-92 21-648 6 316-1 113-88 13 1323 8-11 21-964 3 28 19-5 119-93 11 11 40 59-80 21-645 5 51 51-6 114-27 14 13 25 19-95 21-981 3 40 18-6 119-78 15 1143 9-66 21-643 5 40 24-9 114-64 15 13 27 31-88 21-998 3 52 16-8 119-61 16 11 45 19-52 21-643 5 28 55-9 115-01 16 13 29 43-92 22-015 4 413-9 119-42 17 11 47 29-37 21-641 5 17 24-8 115-36 17 13 31 56-06 22-033 416 9-8 119-23 18 11 49 39-21 21-640 5 5 51-6 115-71 18 13 34 8-31 22-051 428 4-6 119-03 19 11 51 49-05 21-640 4 54 16-3 116-04 19 13 36 20-67 22-070 4 39 58-1 11881 20 11 53 58-89 21-640 4 42 39-1 116-37 20 13 38 33-15 22-089 4 51 50-3 118-58 21 11 56 8-73 21-640 4 30 59-9 116-68 21 13 40 45-74 22-108 5 341-1 118-34 22 11 58 18-57 21-640 4 19 18-9 116-98 22 13 42 58-45 22-129 5 15 30-4 118-08 23 12 028-41 21-641 N. 4 736-2 117-27 23 13 45 11-29 22-149 S. 527181 117-82 Wednesday 22. Friday 24. 12 238-26 21-643 N. 35551-7 117-55 13 47 24-24 22-169 S. 539 4-2 117-54 i 12 448-12 21-644 344 5-6 117-82 i 13 49 37-32 22-192 5 50 48-6 117-26 2 12 6 57-99 21-646 3 32 17-9 118-08 2 13 51 50-54 22-213 6 231-3 116-96 3 12 9 7-87 21-648 3 20 28-7 118-33 3 13 54 3-88 22-235 6 14 12-1 116-65 4 12 11 17-77 21-651 3 838-0 118-56 4 13 56 17-36 22-258 6 25 51-1 116-33 12 13 27-68 21-653 2 56 46-0 118-78 5 13 58 30-97 22-280 6 37 28-0 115-98 6 12 15 37-61 21-657 2 44 52-7 118-99 6 14 44-72 22-303 649 2-9 115-64 7 12 17 47-56 21-661 2 32 58-1 119-20 7 14 258-61 22-328 7 035-7 115-28 8 12 19 57-54 21-666 221 2-3 119-39 8 14 512-65 22-352 712 6-2 114-90 9 1222 7-55 21-670 2 9 5-4 119-57 9 14 7 26-83 22-375 7 23 34-5 114-52 10 12 24 17-58 21-674 157 7-5 119-73 10 14 941-15 22-400 735 0-5 114-13 11 12 26 27-64 21-679 145 8-6 119-88 11 14 11 55-63 22-425 7 46 24-0 113-71 12 12 28 37-73 21-685 1 33 8-9 120-03 12 14 14 10-25 22-450 7 57 45-0 113-29 13 12 30 47-86 21-692 121 8-3 120-17 13 14 16 25-03 22-476 8 9 3-5 112-86 14 12 32 58-03 21-698 1 9 6-9 120-29 14 14 18 39-96 22-503 8 20 19-3 112-41 15 1235 8-24 21-705 057 4-8 120-41 15 14 20 55-06 22-529 8 31 32-4 111-95 16 12 37 18-49 21-712 045 2-0 120-51 1C 14 23 10-31 22-555 8 42 42-7 111-48 17 12 39 28-78 21-720 32 58-7 120-59 17 14 25 25-72 22-583 8 53 50-2 111-00 18 12 41 39-13 21-728 20 54-9 120-67 18 14 27 41-30 22-610 9 454-7 110-51 19 12 43 49-52 21-737 N. 8 50-7 120-73 19 14 29 57-04 22-638 9 15 56-3 110-01 20 12 45 59-97 21-746 S. 313-9 120-79 20 14 32 12-95 22-666 9 26 54-8 109-49 21 12 48 10-47 21-756 15 18-8 120-83 21 14 34 29-03 22-694 9 37 50-2 108-96 22 12 50 21-04 21-766 27 23-9 120-86 22 14 36 45-28 22-723 9 48 42-3 108-42 23 12 52 31-66 21-775 39 29-1 120-88 28 1439 1-71 22-753 9 59 31-2 107-87 24 12 54 42-34 21-786 S. 51 34-4 120-89 24 14 41 18-31 22-782 S. 10 10 16-7 107-30 2 A 354 APPENDIX. AUGUST. MEAN TIME. XL THE MOON'S RIGHT ASCENSION AND DECLINATION. m Right Ascension. Var. in 10. Declination. Var. in 10m. | Right Ascension. Var. in 10". Declination. Var. in lom. Saturday 25. Monday 27. h. m. B. s. ' " " h. m. s. s. ' " " 14 41 18-31 22-782 S. 10 10 16-7 107-30 16 34 19-35 24-313 S. 17 18 1-9 66-68 1 14 43 35-09 22-811 10 20 58-8 106-73 1 16 36 45-31 24-342 17 24 38-7 65-58 2 14 45 52-04 22-841 10 31 37-4 106-13 2 16 39 11-45 24-371 1731 8-8 64-47 3 1448 9-18 22-871 10 42 12-4 105-53 3 16 41 37-76 24-399 17 37 32-3 63-35 4 14 50 26-49 22-901 10 52 43-7 104-92 4 1644 4-24 24-428 17 43 49-0 62-22 5 14 52 43-99 22-932 11 3 11-4 104-29 5 16 46 30-89 24-456 17 49 58-9 61-08 G 1455 1-67 22-963 11 13 35-2 103-65 6 16 48 57-71 24-483 1756 2-0 59-93 7 14 57 19-54 22-994 11 23 55-2 103-01 7 16 51 24-68 24-509 18 1 58-1 58-78 8 14 59 37-60 23-025 11 34 11-3 102-35 8 16 53 51-82 24-537 18 747-3 57-62 9 15 155-84 23-056 11 44 23-4 101-68 9 16 56 19-12 24-563 18 13 29-5 56-45 10 15 414-27 23-088 11 54 31-4 100-98 10 16 58 46-57 24-588 1819 4-7 55-27 11 15 632-89 23-119 12 4 35-2 100-29 11 17 1 14-17 24-613 18 24 32-7 54-08 12 15 851-70 23-151 12 14 34-9 99-59 12 17 341-92 24-638 18 29 53-6 52-88 13 15 11 10-70 23-183 12 24 30-3 98-87 13 17 6 9-82 24-662 1835 7-3 51-68 14 15 13 29-90 23-216 12 34 21-3 98-13 14 17 837-86 24-686 18 40 13-8 50-48 15 15 15 49-29 23-248 1244 7-9 97-38 15 17 U 6-05 24-709 18 45 13-0 49-26 1C 15 18 8-88 23-281 12 53 49-9 96-63 16 17 13 34-37 24-731 1850 4-9 48-04 17 15 20 28-66 23-313 13 327-4 95-87 17 17 16 2-82 24-753 18 54 49-5 46-81 18 15 22 48-63 23-346 13 13 0-3 95-09 18 17 18 31-40 24-774 18 59 26-6 45-57 19 1525 8-81 23-379 13 22 28-5 94-30 19 1721 0-11 24-795 19 356-3 44-33 20 15 27 29-18 23-412 13 31 51-9 93-50 20 17 23 28-94 24-816 19 8 18-6 43-08 21 15 29 49-75 23-444 13 41 10-5 92-69 21 17 25 57-90 24-836 19 12 33-3 41-83 22 15 32 10-51 23-478 13 50 24-2 91-87 22 17 28 26-97 24-853 19 16 40-5 40-57 23 15 34 31-48 23-511 S. 13 59 32-9 91-03 23 17 30 56-14 24-872 S. 19 20 40-1 39-29 Sunday 26. Tuesday 28. 15 36 52-64 23-543 S. 14 8 36-5 90-18 17 33 25-43 24-890 S. 19 24 32-0 38-02 i 15 39 14-00 23-578 14 17 35-0 89-33 i 17 35 54-82 24-907 19 28 16-3 36-75 2 15 41 35-57 23-611 14 26 28-4 88-46 2 17 38 24-31 24-923 19 31 53-0 35-47 8 15 43 57-33 23-643 14 35 16-5 87-58 3 17 40 53-90 24-939 19 35 21-9 34-18 4 15 46 19-29 23-676 14 43 59-3 86-68 4 17 43 23-58 24-953 19 38 43-1 32-88 5 15 48 41-44 23-709 14 52 36-7 85-78 5 17 45 53-34 24-967 19 41 56-5 31-58 6 15 51 3-80 23-743 15 1 8-7 84-88 6 17 48 23-18 24-981 1945 2-1 30-28 7 15 53 26-36 23-776 15 935-2 83-95 7 17 50 53-11 24-994 19 47 59-9 28-98 8 15 55 49-11 23-808 15 17 56-1 83-01 8 17 53 23-11 25-005 19 50 49-9 27-68 9 15 58 12-06 23-842 15 26 11-3 82-07 9 17 55 53-17 25-016 19 53 32-0 26-36 10 16 035-21 23-874 15 34 20-9 81-12 10 17 58 23-30 25-027 1956 6-2 25-04 11 16 258-55 23-907 15 42 24-7 80-15 11 18 053-49 25-037 19 58 32-5 23-72 12 16 522-09 23-939 15 50 22-7 79-18 12 18 323-74 25-046 20 50-9 22-40 13 16 745-82 23-972 15 58 14-8 78-18 13 18 554-04 25-053 20 3 1-3 21-08 14 1610 9-75 24-004 16 6 0-9 77-18 14 18 824-38 25-060 20 5 3-8 19-75 15 16 12 33-87 24-035 16 13 41-0 76-18 15 18 10 54-76 25-067 20 658-3 18-43 16 16 14 58-17 24-067 16 21 15-1 75-17 16 18 13 25-18 25-073 20 844-9 17-09 17 16 17 22-67 24-099 16 28 43-1 74-14 17 18 15 55-63 25-077 20 10 23-4 15-75 18 16 19 47-36 24-130 1636 4-8 73-10 18 18 18 26-10 25-081 20 11 53-9 14-42 19 16 22 12-23 24-161 16 43 20-3 72-06 19 18 20 56-60 25-084 20 13 16-4 13-08 20 16 24 37-29 24-192 16 50 29-5 71-00 20 18 23 27-11 25-087 20 14 30-9 11-74 21 16 27 2-53 24-223 16 57 32-3 69-93 21 18 25 57-64 25-088 20 15 37-3 10-40 22 16 29 27-96 24-253 17 428-6 68-85 22 18 28 28-17 25-088 20 16 35-7 9-06 23 16 31 53-56 24-283 17 11 18-5 67-78 23 18 30 58-70 25-088 20 17 26-0 7-71 24 16 34 19-35 24-313 S. 17 18 1-9 66-68 24 18 33 29-23 25-088 S. 20 18 8-2 6-37 APPENDIX. 355 XII. AUGUST. MEAN TIME. THE MOON'S RIGHT ASCENSION AND DECLINATION. 1 Right Ascension. Var. in 10. Declination. Var. in 10 it / It / // / // 1 Spica W. 72 28 41 2222 74 16 36 2223 76 4 29 2225 77 52 19 2228 Fomalhaut E. 61 27 29 2519 59 46 42 2533 58 6 15 2549 56 26 10 2568 Saturn E. 69 18 16 2170 67 29 4 2172 65 39 55 2175 63 50 49 2178 o Pegasi E. 77 8 2562 75 20 21 2570 73 40 45 2580 72 1 23 2591 2 Spica W. 86 50 19 2248 88 37 35 2254 90 24 43 2260 92 11 41 2267 Antares W. 41 19 53 2344 43 4 48 2341 44 49 48 2339 46 34 50 2340 Saturn E. 54 46 40 2199 52 58 11 2205 51 9 51 2211 49 21 40 2217 a Pegasi E. 63 48 58 2670 62 11 38 2690 60 34 45 2713 58 58 23 2740 o Arietis E. 106 25 4 2383 104 41 5 2386 102 57 10 2389 101 13 20 2394 3 Spica W. 101 3 49 2309 102 49 36 2319 104 35 8 2329 106 20 25 2340 Antares W. 55 19 19 2358 57 3 55 2364 58 48 22 2371 60 32 39 2379 Saturn E. 40 23 26 2258 38 36 25 2268 36 49 38 2277 35 3 5 2287 a Pegasi E. 51 6 8 2908 49 33 59 2952 48 2 46 3000 46 32 33 3053 o Arietia E. 92 36 15 2429 90 53 22 2439 89 10 43 2449 87 28 18 2459 4 Spica W. 115 2 42 2401 116 46 15 2416 118 29 27 2430 120 12 19 2444 Antares W. 69 11 2 2426 70 54 2437 72 36 42 2449 74 19 7 2460 Saturn E. 26 14 18 2345 24 29 24 2358 22 44 49 2371 21 33 2384 o Arietis E. 79 16 2523 77 19 35 2538 75 39 15 2553 73 59 16 2569 Aldebaran E. 110 42 15 2362 108 57 45 2374 107 13 33 2387 105 29 39 2400 5 Antares W. 82 46 49 2527 84 27 24 2541 86 7 40 2556 87 47 36 2571 o Arietis E. 65 45 8 2660 64 7 34 2681 62 30 28 2701 60 53 49 2722 Aldebaran E. 96 54 59 2470 95 13 3 2485 93 31 28 2499 91 50 14 2514 6 Antares W. 96 2 3 2649 97 39 52 2665 99 17 20 2681 100 54 26 2696 a Aquilse W. 53 13 4 3496 54 33 33 3475 55 54 25 3457 57 15 37 3442 a Arietis E. 52 58 8 2844 51 24 37 2872 49 51 42 2901 48 19 24 2931 Aldebaran E. 83 29 16 2591 81 50 8 2607 80 11 22 2622 78 32 57 2637 Jupiter E. 106 37 39 2663 105 9 2678 103 22 59 2694 101 46 11 2709 7 Antares W. 108 54 33 2779 110 29 29 2795 112 4 4 2811 113 38 17 2828 o Aquilse W. 64 4 52 3404 65 27 5 3401 66 49 20 3400 68 11 37 3400 a Arietis E. 40 48 6 3110 39 20 8 3153 37 53 1 3199 36 26 50 3249 Aldebaran E. 70 26 3 2715 68 49 43 2729 67 13 42 2744 65 38 1 2760 Jnpiter E. 93 47 20 2787 92 12 35 2802 90 38 10 2818 89 4 5 2832 8 a Aquilro W. 75 2 30 3418 76 24 27 3424 77 46 16 3430 79 7 59 3437 Saturn W. 26 59 26 2811 28 33 40 2824 30 7 37 2836 31 41 18 2849 Aldebaran E. 57 44 25 2831 56 10 37 2844 54 37 7 2857 53 3 53 2870 Jupiter E. 81 18 25 2905 79 46 12 2919 78 14 17 2931 76 42 38 2945 Pollux E. 101 50 10 2879 100 17 25 2892 98 44 56 2905 97 12 44 2917 SUN E. 124 51 22 3175 123 24 43 3190 121 58 22 3204 120 32 18 3218 9 a Aquilse W. 85 54 25 3479 87 15 13 3488 88 35 50 3498 89 56 17 3508 Saturn W. 39 25 41 2908 40 57 49 2919 42 29 44 2929 44 1 25 2939 Aldebaran E. 45 21 47 2931 43 50 7 2942 42 18 41 2952 40 47 28 2962 Jupiter E. 69 8 26 3006 67 38 21 3018 66 8 30 3028 64 38 52 3039 Pollux E. 89 35 35 2977 88 4 54 2988 86 34 26 2999 85 4 12 3009 SUN E. 113 25 56 3283 112 1 25 3295 110 37 8 3306 109 13 4 3317 358 APPENDIX. AUGUST. MEAN TIME. XV. LUNAR DISTANCES. Day. Star's Name and Position. Noon. P.L. of diff. IIP>. P.L. of diff. Vlh. P.L. of diff. JX>. P.L. of diff. O 1 II / // / ' o / // 10 a Aquilro W. 91 16 32 3519 92 36 35 3529 93 56 27 3540 95 16 7 3551 Fomalhaut W. 56 56 54 3327 58 20 34 3324 59 44 17 3323 61 8 2 3322 Saturn W. 45 32 54 2949 47 4 10 2958 48 35 15 2967 50 6 9 2976 a Pegasi W. 43 47 13 3765 45 2 52 3733 46 19 4 3705 47 35 46 3680 Aldebaran E. 39 16 28 2972 37 45 41 2981 36 15 5 2990 34 44 40 2999 Jupiter E. 63 9 27 3049 61 40 15 3058 60 11 14 3067 58 42 24 3076 Pollux E. 83 34 10 3019 82 4 21 3029 80 34 45 3039 79 5 20 3047 SUN E. 107 49 12 3328 106 25 33 3338 105 2 6 3348 103 38 50 3358 11 a Aquilge W. 101 51 17 3611 103 9 39 3624 104 27 48 3637 105 45 42 3651 Fomalhaut W. 68 6 56 3321 69 30 43 3322 70 54 29 3321 72 18 16 3321 Saturn W. 57 38 12 3010 59 8 12 3017 60 38 4 3022 62 7 50 3026 a Pegasi W. 54 5 7 3587 55 23 55 3573 56 42 59 3560 58 2 17 3549 Aldebaran E. 27 15 7 3035 25 45 38 3042 24 16 17 3047 22 47 2 3052 Jupiter E. 51 20 48 3114 49 52 55 3120 48 25 10 3126 46 57 32 3131 Pollux E. 71 40 49 3086 70 12 23 3093 68 44 5 3100 67 15 55 3105 SUN E. 96 44 58 3396 95 22 37 3403 94 24 3408 92 38 17 3413 12 Fomalhaut W. 79 17 7 3322 80 40 53 3322 82 4 39 3321 83 28 26 3321 Saturn W. 69 35 23 3043 71 4 43 3045 72 34 3046 74 3 16 3047 a Pegasi W. 64 41 40 3501 66 2 3 3494 67 22 34 3486 68 43 14 3478 Jupiter E. 39 40 48 3151 38 13 41 3154 36 46 37 3157 35 19 36 3159 Pollux E. 59 56 46 3130 58 29 14 3134 57 1 46 3138 55 34 22 3142 SUN E. 85 48 58 3432 84 27 18 3434 83 5 40 3436 81 44 4 3437 13 Fomalhaut W. 90 27 33 3316 91 51 26 3314 93 15 21 3313 94 39 18 3311 Saturn W. 81 29 31 3044 82 58 49 3042 84 28 9 3039 85 57 33 3037 a Pegasi W. 75 28 35 3444 76 50 1 3438 78 11 35 3431 79 33 16 3425 a Arietis W. 32 5 5 3652 33 22 43 3606 34 41 11 8565 36 24 3527 Jupiter E. 28 5 5 8167 26 38 16 3168 25 11 29 3168 23 44 42 3169 Pollux E. 48 18 22 3155 46 51 19 3158 45 24 20 3160 43 57 23 3163 SUN E. 74 56 10 3434 73 34 32 3432 72 12 52 3430 70 51 9 3427 U Fomalhaut W. 101 39 36 3301 103 3 47 3299 104 28 3296 105 52 16 3294 Saturn W. 93 25 33 3016 94 55 26 3011 96 25 25 3005 97 55 31 2998 a Pegasi W. 86 23 28 3394 87 45 52 3388 89 8 22 3382 90 30 59 3376 a Arietis W. 42 45 43 3382 44 8 20 3359 45 31 23 3337 46 51 52 3316 Pollux E. 36 43 31 3180 35 16 58 3185 33 50 31 3192 32 24 12 3200 SUN E. 64 1 33 3405 62 39 22 3399 61 17 4 3393 59 54 39 3386 15 Saturn W. 105 28 10 2962 106 59 10 2954 108 30 20 2946 110 1 41 2936 a Pegasi W. 97 25 47 3347 98 49 4 3342 100 12 28 3337 101 35 57 3332 a Arietis W. 53 58 3224 55 23 41 3207 56 49 41 3190 58 16 2 3174 SUN E. 53 30 3347 51 37 13 3338 50 13 45 3328 48 50 6 3319 16 a Arietis W. 65 32 28 3098 67 40 3083 68 29 10 3069 69 57 58 3055 SUN E. 41 49 2 3267 40 24 12 3255 38 59 8 3244 37 33 51 3233 17 a Arietis W. 77 26 17 2985 78 56 49 2972 80 27 37 2958 81 58 42 2945 SUN E. 30 24 3173 28 57 18 3161 27 30 22 3148 26 3 11 3137 22 SUN W. 30 47 40 2723 32 23 49 2716 34 7 2709 35 36 35 2702 APPENDIX. 359 XVI. AUGUST. MEAN TIME. LUNAR DISTANCES. Day. Star's Name and Position. Midnight. P.L. Of difif. XVb. P.L. of difif. xvmb. P.L. of diff. XXIb. P.L. of diff. O 1 H O 1 II / II 1 II 10 o Aquilse W. 96 35 35 3563 97 54 50 3575 99 13 52 3587 100 32 41 3598 Fomalhaut W. 62 31 48 3322 63 55 34 3321 65 19 21 3321 66 43 8 3320 Saturn W. 51 36 52 2984 53 7 25 2991 54 37 49 2997 56 8 5 3004 o Pegasi W. 48 52 54 3657 50 10 27 3637 51 28 21 3619 52 46 35 3602 Aldebaran E. 33 14 26 3008 31 44 23 3015 30 14 29 3022 28 44 44 3029 Jupiter E. 57 13 45 3085 55 45 17 3093 54 16 58 3101 52 48 49 3107 Pollux E. 77 36 5 3056 76 7 1 3065 74 38 8 3072 73 9 24 3079 SUN E. 102 15 45 3366 100 52 50 3374 99 30 4 3382 98 7 27 3389 11 a Aquil W. 107 3 22 3665 108 20 46 3679 109 37 55 3694 110 54 48 3709 Fomalhaut W. 73 42 3 3322 75 5 49 3322 76 29 35 3322 77 53 21 3322 Saturn W. 63 37 30 3030 65 7 4 3034 66 36 34 3037 68 6 3040 a Pegasi W. 59 21 47 3538 60 41 29 3528 62 1 22 3518 63 21 26 3509 Aldebaran E. 21 17 54 3056 19 48 51 3061 18 19 54 3065 16 51 2 3069 Jupiter E. 45 30 3136 44 2 35 3140 42 35 15 3144 41 7 59 3148 Pollux E. 65 47 52 3111 64 19 56 3117 62 52 7 3122 61 24 24 3126 SUN E. 91 16 15 3418 89 54 19 3422 88 32 28 3426 87 10 41 3429 12 Fomalhaut W. 84 52 13 3320 86 16 1 3319 87 39 50 3318 89 3 41 3317 Saturn W. 75 32 31 3047 77 1 45 3047 78 30 59 3046 80 14 3045 o Pegasi W. 70 4 3 3471 71 24 59 3464 72 46 3 3457 74 7 15 3450 Jupiter E. 33 52 38 3161 32 25 42 3163 30 58 48 3164 29 31 56 3165 Pollux E. 54 7 3 3145 52 39 48 3148 51 12 36 3150 49 45 27 3153 SUN E. 80 22 29 3438 79 55 3438 77 39 21 3437 76 17 46 3436 13 Fomalhaut W. 96 3 17 3309 97 27 18 3307 98 51 22 3305 100 15 28 3303 Saturn W. 87 27 3034 88 56 31 3030 90 26 6 3026 91 55 46 3021 o Pegasi W. 80 55 4 3419 82 16 59 3412 83 39 2 3406 85 1 12 3400 a Arietis W. 37 20 18 3493 38 40 50 3463 40 1 56 3434 41 23 34 3407 Jupiter E. 22 17 56 3171 20 51 12 3173 19 24 31 3177 17 57 54 3182 Pollux E. 42 30 30 3166 41 3 40 3169 39 36 53 3172 38 10 10 3176 SUN E. 69 29 23 3423 68 7 33 3419 66 45 38 3415 65 23 38 3410 14 Fomalhaut W. 107 16 35 3292 108 40 56 3290 110 5 19 3288 111 29 45 3286 Saturn W. 99 25 46 2992 100 56 9 2985 102 26 40 2978 103 57 20 2970 o Pegasi W. 91 53 43 3370 93 16 34 3364 94 39 32 3358 96 2 36 3352 a Arietis W. 48 18 45 3296 49 43 1 3277 51 7 39 3259 52 32 39 3241 Pollux E. 30 58 2 3209 29 32 3 3221 28 6 19 3236 26 40 52 3253 SUN E. 58 32 6 3379 57 9 25 3372 55 46 36 3364 54 23 38 3355 15 Saturn W. 111 33 14 2927 113 4 58 2918 114 36 54 2908 116 9 3 2898 a Pegasi W. 102 59 32 3327 104 23 13 3323 105 46 58 3319 107 10 48 3314 o Arietis W. 59 42 42 3158 61 9 41 3143 62 36 58 3128 64 4 34 3113 SUN E. 47 26 17 3310 46 2 17 3299 44 38 4 3288 43 13 39 3278 16 a Arietis W. 71 27 3 3041 72 56 25 3026 74 26 5 3012 75 56 2 2998 SUN E. 36 8 21 3221 34 42 37 3209 33 16 39 3198 31 50 27 3185 17 a Arietis W. 83 30 4 2933 85 1 41 2920 86 33 35 2907 88 5 45 2894 SUN E. 24 35 46 3124 23 8 6 3112 21 40 11 3101 20 12 2 3088 22 SUN W. 37 13 13 2695 38 49 59 2689 40 26 54 2683 42 3 58 2677 APPENDIX. AUGUST. MEAN TIME. XVII. LUNAR DISTANCES. Day. Star's Name and Position. Noon. P.L. of diif. IIP. P.L. of diff. I , j VI. r.L. of diff. IXh. P.L. of diff. 1 II O 1 II 1 II o I n 22 a AquilsB E. 114 52 11 3081 113 23 38 3057 111 54 36 3035 110 25 7 3015 23 SUN W. 43 41 9 2671 45 18 28 2666 46 55 53 2660 48 33 26 2655 a AquilsB E. 102 51 54 2935 101 20 19 2922 99 48 28 2912 98 16 24 2903 24 SUN W. 56 42 45 2635 58 20 53 2631 59 59 6 2627 61 37 24 2624 a Aquilse E. 90 33 31 2872 89 36 2869 87 27 37 2867 85 54 36 2867 25 SUN W. 69 49 55 2610 71 28 36 2608 73 7 20 2606 74 46 7 2604 a Aquilse E. 78 9 50 2881 76 37 7 2887 75 4 32 2895 73 32 7 2905 Fomalhaut E. 111 52 35 2588 110 13 24 2580 108 34 2 2573 106 54 30 2566 26 SUN W. 83 34 2597 84 39 33 2596 86 18 33 2596 87 57 34 2595 Spica W. 33 30 38 2393 35 14 23 2384 36 58 21 2375 38 42 31 2368 o Aquilse E. 65 53 38 2975 64 22 54 2995 62 52 35 3017 61 22 43 3041 Fomalhaut E. 98 34 46 2542 96 54 31 2539 95 14 12 2537 93 33 50 2535 Saturn E. 107 10 29 2267 105 23 41 2266 103 36 52 2265 101 50 1 2264 a Pegasi E. 113 16 54 2708 111 40 25 2697 110 3 41 2687 108 26 44 2679 27 SUN W. 96 12 53 2593 97 51 58 2594 99 31 1 2594 101 10 5 2594 Spica W. 47 25 35 2344 49 10 30 2341 50 55 30 2339 52 40 34 2337 a Aquilae E. 54 2 2 3209 52 36 3 3255 51 10 59 3305 49 46 53 3360 Fomalhaut E. 85 11 41 2536 83 31 17 2538 81 50 56 2540 80 10 38 2543 Saturn E. 92 55 33 2262 91 8 38 2263 89 21 44 2264 87 34 51 2263 a Pegasi E. 100 19 27 2649 98 41 38 2645 97 3 44 2643 95 25 47 2640 28 SUN W. 109 25 9 2599 111 4 5 2601 112 42 59 2602 114 21 51 2604 Spica W. 61 26 25 2332 63 11 38 2331 64 56 52 2332 66 42 5 2332 Fomalhaut E. 71 50 31 2569 70 10 53 2577 68 31 26 2585 66 52 11 2594 Saturn E. 78 40 32 2267 76 53 44 2268 75 6 58 2270 73 20 14 2272 a Pegasi E. 87 15 46 2643 85 37 50 2647 83 59 59 2651 82 22 13 2655 29 SUN W. 122 35 24 2618 124 13 55 2621 125 52 22 2624 127 30 45 2628 Spica W. 75 27 49 2340 77 12 51 2342 78 57 49 2344 80 42 44 2347 Antares W. 30 18 55 2525 31 59 34 2507 33 40 37 2493 35 22 2481 Fomalhaut E. 58 39 37 2660 57 2 2 2677 55 24 51 2696 53 48 6 2716 Saturn E. 64 27 17 2283 62 40 52 2285 60 54 31 2288 59 8 14 2291 o Pegasi E. 74 15 18 2692 72 38 27 2702 71 1 49 2713 69 25 26 2726 a. Arietis E. 117 25 13 2499 115 43 59 2497 114 2 42 2495 112 21 22 2494 30 Spica W. 8$ 26 6 2366 91 10 30 2371 92 54 46 2376 94 38 56 2381 Antares W. 43 52 12 2450 45 34 36 2448 47 17 3 2446 48 59 32 2445 Saturn E. 50 18 7 2311 48 32 23 2315 46 46 46 2320 45 1 16 2325 a Pegasi E. 61 28 22 2810 59 54 7 2831 58 20 20 2855 56 47 4 2882 a Arietis E. 103 54 30 2497 102 13 12 2499 100 31 58 2502 98 50 48 2506 31 Spica W. 103 17 40 2412 105 57 2420 106 44 3 2427 108 26 59 2435 Antares W. 57 31 39 2456 59 13 54 2460 60 56 3 2465 62 38 6 2470 Saturn E. 36 15 45 2355 34 31 6 2362 32 46 36 2368 31 2 16 2376 o Pegasi E. 49 10 8 3054 47 41 2 3099 46 12 51 3149 44 45 40 3204 o Arietis E. 90 26 27 2532 88 45 58 2539 87 5 39 2546 85 25 30 2554 Aldebaran E. 122 33 24 2379 120 49 19 2385 119 5 23 2392 117 21 37 2399 XVIII. APPENDIX. AUGUST. MEAN TIME. LUNAR DISTANCES. Day. Star's Name and Position. Midnight. P.L. Of diff. XVh. P.L. Of diff. XVIIIh. P.L. of diff. XXIb. P.L. of diff. O 1 II O t 'I / // / // 22 ft AquiloB E. 108 55 13 2996 107 24 55 2978 105 54 14 2962 104 23 13 2948 23 SUN W. 50 11 6 2651 51 48 52 2646 53 26 44 2642 55 4 42 2638 o AquilsQ E. 96 44 8 2894 95 11 41 2887 93 39 5 2881 92 6 21 2876 24 SUN W. 63 15 47 2621 64 54 13 2618 66 32 43 2615 68 11 17 2612 o Aquilje E. 84 21 35 2867 82 48 34 2869 81 15 35 2872 79 42 40 2876 25 SUN W. 76 24 56 2603 78 3 47 2601 79 42 41 2599 81 21 37 2597 a Aquilsa E. 71 59 54 2915 70 27 54 2927 68 56 10 2942 67 24 44 2958 Fomalhaut E. 105 14 49 2560 103 34 59 2554 101 55 1 2549 100 14 56 2545 26 SUN TV. 89 36 37 2594 91 15 40 2593 92 54 44 2593 94 33 48 2593 Spica W. 40 26 52 2362 42 11 22 2357 43 55 59 2352 45 40 44 2347 a Aquilse E. 59 53 21 3068 58 24 32 3099 56 56 21 3132 55 28 50 3168 Fomalhaut E. 91 53 25 2534 90 12 59 2534 88 32 33 2534 86 52 7 2534 Saturn E. 100 3 9 2264 98 16 16 2263 96 29 22 2263 94 42 28 2262 a Pegasi E. 106 49 36 2671 105 12 17 2663 103 34 48 2657 101 57 11 2653 27 SUN W. 102 49 8 2595 104 28 10 2596 106 7 11 2596 107 46 11 2598 Spica W. 54 25 40 2335 56 10 49 2334 57 55 59 2333 59 41 12 2332 o Aquilse E. 48 23 51 3422 47 1 59 3491 45 41 25 3568 44 22 16 3653 Fomalhaut E. 78 30 24 2547 76 50 15 2551 75 10 13 2556 73 30 18 2562 Saturn E. 85 47 57 2264 84 1 4 2264 82 14 12 2265 80 27 21 2266 o Pegasi . E. 93 47 47 2640 92 9 46 2640 90 31 45 2640 88 53 45 2641 28 SUN W. 116 40 2607 117 39 26 2609 119 18 9 2612 120 56 48 2615 Spica W. 68 27 18 2333 70 12 29 2335 71 57 38 2336 73 42 45 2338 Fomalhaut E. 65 13 8 2605 63 34 19 2617 61 55 47 2630 60 17 32 2644 Saturn E. 71 33 33 2274 69 46 54 2276 68 19 2278 66 13 46 2280 a Pegasi E. 80 44 33 2660 79 7 2667 77 29 36 2675 75 52 21 2683 29 SUN W. 129 9 2 2632 130 47 13 2636 132 25 19 2641 134 3 19 2646 Spica W. 82 27 35 2351 84 12 21 2355 85 57 1 2358 87 41 36 2362 Antares W. 37 3 40 2471 38 45 34 2464 40 27 39 2458 42 9 52 2453 Fomalhaut E. 52 11 48 2740 50 36 1 2766 49 48 2794 47 26 12 2825 Saturn E. 57 22 2 2295 55 35 55 2298 53 49 53 2302 52 3 57 2307 a Pegasi E. 67 49 21 2740 66 13 34 2755 64 38 7 2772 63 3 3 2790 a Arietis E. 110 40 2493 108 58 37 2493 107 17 14 2493 105 35 51 2495 30 Spica W. 96 22 58 2387 98 6 52 2393 99 50 37 2399 101 34 13 2405 Antares W. 50 42 2 2447 52 24 30 2448 54 6 56 2450 55 49 19 2453 Saturn E. 43 15 53 2331 41 30 38 2337 39 45 32 2342 38 34 2348 a Pegasi E. 55 14 22 2910 53 42 16 2941 52 10 49 2975 50 40 5 3013 o Arietis E. 97 9 42 2510 95 28 42 2515 93 47 50 2520 92 7 5 2525 31 Spica W. 110 9 44 2444 111 52 17 2453 113 34 37 2461 115 16 45 2470 Antares W. 64 20 3 2476 66 1 52 2481 67 43 32 2488 69 25 2 2494 Saturn E. 29 18 7 2383 27 34 8 2391 25 50 21 2399 24 6 45 2408 a Pegasi E. 43 19 35 3265 41 54 42 3333 40 31 7 3408 39 8 59 3492 a Arietis E. 83 45 32 2563 82 5 46 2572 80 26 12 2582 78 46 51 2592 Aldebaran E. 115 38 1 2407 113 54 36 2415 112 11 22 2423 110 28 19 2431 362 APPENDIX. SEPTEMBER. AT APPARENT NOON. Tta + A THE SUN'S Sidereal Time of the Semi- Equation of Time, to be added to Var. jjate. Apparent Right Ascension. Var. in 1 hour. Apparent Declination. Var. in 1 hour. passing the Meridian.* subtracted from Apparent Time. in 1 hour. Sat. 1 h. m. B 10 39 6-52 9-075 o // N. 8 31 38-9 // 54-19 m. B. 1 4-38 m. s. 9-13 0-779 Sun. 2 10 42 4417 9-063 8 9 54-2 54-53 1 4.34 9-73 0-792 Mon. 3 10 46 21-53 9-051 7 48 1-6 54-85 1 4-30 28-87 0-803 Tues. 4 10 49 58-62 9-040 7 26 1-4 55-16 4-26 48-28 0-814 Wed. 5 10 53 35-45 9-030 7 3 54-0 55-45 4-22 1 7-94 0-824 Thur. 6 10 57 12-06 9-021 6 41 39-6 55-74 4-18 1 27-83 0-833 Frid. 7 11 48-46 9-012 6 19 18-6 56-01 4-15 1 47-93 0-841 Sat. 8 11 4 24-66 9-005 5 56 51-3 56-26 4-12 2 8-22 0-849 Sun. 9 11 8 0-70 8-999 5 34 18-0 56-51 4-10 2 28-67 0-855 Mon. 10 11 11 36-59 8-993 5 11 39-0 56-74 4-08 2 49-28 0-861 Tues. 11 11 15 12-35 8-988 4 48 54-6 56-95 4-06 3 10-02 0-866 Wed. 12 11 18 48-00 8-983 4 26 5-2 57-16 4-04 3 30-87 0-871 Thar. 13 11 22 23-55 8-980 4 3 11-1 57-35 4-02 3 51-81 0-874 Frid. 14 11 25 59-03 8-977 3 40 12-6 57-52 4-01 4 12-83 0-877 Sat. 15 11 29 34-45 8-975 3 17 10-0 57-69 4-00 4 33-90 0-879 Sun. 16 11 33 9-84 8-974 2 54 3-7 57-83 1 3-99 4 55-01 0-880 Mon. 17 11 36 45-20 8-973 2 30 54-1 57-96 1 3-99 5 16-14 0-881 Tues. 18 11 40 20-55 8-973 2 7 41-4 58-08 1 3-99 5 37-28 0-881 Wed. 19 11 43 55-91 8-974 1 44 26-1 58-19 1 3-99 5 58-41 0-880 Thur. 20 11 47 31-30 8-975 1 21 8-4 58-28 1 4-00 6 19-52 0-879 Frid. 21 11 51 6-73 8-977 57 48-9 58-35 1 4-01 6 40-58 0-876 Sat. 22 11 54 42-22 8-980 34 27-7 58-41 1 4-02 7 1-59 0-874 Sun. 23 11 58 17-79 8-984 N. 11 5-3 58-45 4-04 7 22-52 0-870 Mon. 24 12 1 53-44 8-988 S. 12 17-9 58-48 4-06 7 43-36 0-866 Tues. 25 12 5 29-21 8-993 35 41-7 58-49 4-08 8 4-09 0-861 Wed. 26 12 9 5-11 8-999 59 5-6 58-49 4-10 8 24-68 0-855 Thur. 27 12 12 41-17 9-006 1 22 29-3 58-48 4-13 8 45-12 0-848 Frid. 28 12 16 17-40 9-014 1 45 52-5 58-45 1 4-16 9 5-38 0-840 Sat. 29 12 19 53-84 9-023 2 9 14-9 58-41 1 4-19 9 25-44 0-831 Sun. 30 12 23 30-50 9-033 2 32 36-0 58-35 1 4-23 9 45-28 0-822 Mon. 31 12 27 7'40 9-043 S. 2 55 55-7 58-28 1 4-27 10 4-88 0-811 * Mean Time of the Semidiameter passing may be found by subtracting 0'-18 from the Sidereal Time. APPENDIX. 363 II. SEPTEMBER. AT MEAN NOON. Date. THE SUN'Sl Equation of Time, to be subtracted from Sidereal Time. Apparent Right Ascension. Apparent Declination. Semidiameter.* added to Mean Time. Sat. Sun. Mon. 1 2 3 h. m. s. 10 39 6-49 10 42 44-19 10 46 21-60 o / // N. 8 31 39-1 8 9 54-0 7 48 1-1 / n 15 52-67 15 52-90 15 53-13 m. B. 9-13 h. m. B. 10 38 57-37 10 42 53-92 10 46 50-47 9-73 28-87 Tues. Wed. Thur. 4 5 6 10 49 58-74 10 53 35-62 10 57 12-28 7 26 0-7 7 3 52-9 6 41 38-2 15 53-36 15 53-60 15 53-83 48-29 1 7-95 1 27-85 10 50 47-03 10 54 43-58 10 58 40-13 Frid. Sat. Sun. 7 8 9 11 48-73 11 4 24-99 11 8 1-08 6 19 16-9 5 56 49-3 5 34 15-7 15 54-07 15 54-30 15 54-54 1 47-95 2 8-25 2 28-71 11 2 36-68 11 6 33-24 11 10 29-79 Mon. Tues. Wed. 10 11 12 11 11 37-02 11 15 12-83 11 18 48-53 5 11 36-3 4 48 51-6 4 26 1-9 15 54-78 15 55-03 15 55-27 2 49-32 3 10-06 3 30-92 11 14 26-34 11 18 22-89 11 22 19-44 Thur. Frid. Sat. 13 14 15 11 22 24-13 11 25 59-66 11 29 35-13 4 3 7-4 3 40 8-5 3 17 5-6 15 55-52 15 55-77 15 56-03 3 51-87 4 12-89 4 33-97 11 26 16-00 11 30 12-55 11 34 9-10 Sun. Mon. Tues. 16 17 18 11 33 10-57 11 36 45-98 11 40 21-39 2 53 59-0 2 30 49-0 2 7 36-0 15 56-28 15 56-54 15 56-80 4 55-08 5 16-22 5 37'36 11 38 5-65 11 42 2-21 11 45 58-76 Wed. Thur. Frid. 19 20 21 11 43 56-81 11 47 32-25 11 51 7-73 1 44 20-3 1 21 2-3 57 42-4 15 57-06 15 57-33 15 57-60 5 58-50 6 19-61 6 40-68 11 49 55-31 11 53 51-86 11 57 48-41 Sat. Sun. Mon. 22 23 24 11 54 43-27 11 58 18-89 12 1 54-60 34 20-9 N. 10 58-1 S. 12 25-4 15 57-87 15 58-15 15 58-42 7 1-69 7 22-63 7 43-47 12 1 44-97 12 5 41-52 12 9 38-07 Tues. Wed. Thur. 25 26 27 12 5 30-42 12 9 6-37 12 12 42-48 35 49-5 59 13-8 1 22 37-9 15 58-70 15 58-98 15 59-26 8 4-20 8 24-80 8 45-24 12 13 34-62 12 17 31-17 12 21 27-73 Frid. Sat. Sun. 28 29 30 12 16 18-77 12 19 55-26 12 23 31-97 1 46 1-4 2 9 24-1 2 32 45-6 15 59-53 15 59-81 16 0-09 9 5-51 9 25-58 9 45-42 12 25 24-28 12 29 20-83 12 33 17-38 Mon. 31 12 27 8-92 S. 2 56 5-5 16 0-37 10 5-01 12 37 13-94 * The Semidiameter for Apparent Noon may be assumed the same as that foi Mean Noon. 364 APPENDIX. SEPTEMBER. MEAN TIME. HI. Day. THE SUN'S Apparent Logarithm of the Radius Vector of the Earth. Transit of the First Point of Aries. THE MOON'S Semidiameter. Horizontal Parallax. Longitude. Latitude. Noon. Noon. Noon. Noon. Midnight. Noon. Midnight. 1 2 3 O / H 158 720-1 159 524-6 100 330-7 S. 0-36 0-50 0-62 0-0038629 0037580 0036526 h. m. s. 13 18 51-40 13 14 55-50 13 10 59-59 15 51-25 15 42-60 15 32-92 i n 15 47-09 15 37-85 15 27-88 i n 58 5-16 57 33-43 56 57-98 i n 57 49-88 57 16-04 56 39-53 4 5 6 161 138-6 161 59 48-4 16258 0-0 0-72 0-81 0-86 0-0035466 0034402 0033332 13 7 3-68 13 3 7-78 12 59 11-87 15 22-80 15 12-94 15 3-99 15 17-79 15 8-31 15 0-06 56 20-93 55 44-75 55 11-96 56 2-58 55 27-79 54 57-57 7 8 9 163 56 13-5 164 54 29-0 165 52 46-6 0-88 0-88 0-85 0-0032256 0031173 0030082 12 55 15-96 12 51 20-05 12 47 24-15 14 56-59 14 51-27 14 48-43 14 53-64 14 49-52 14 48-03 54 44-86 54 25-37 54 15-00 54 34-06 54 18-99 54 13-53 10 11 12 16651 6-2 167 49 27-9 168 47 51-7 0-81 0-74 0-63 0-0028984 0027877 0026761 12 43 28-24 12 39 32-34 12 35 36-43 14 48-34 14 51-09 14 56-63 14 49-37 14 53-52 15 0-36 54 14-66 54 24-74 54 45-01 54 18-42 54 33-64 54 58-68 13 14 15 169 46 17-6 170 44 45-5 171 43 15-6 0-51 0-38 0-25 0-0025635 0024498 0023349 12 31 40-52 12 27 44-62 12 23 48-71 15 4-68 15 14-83 15 26-42 15 9-53 15 20-49 15 32-51 55 14-51 55 51-69 56 34-16 55 32-26 56 12-45 56 56-48 16 17 18 172 41 47-8 173 40 21-9 174 38 58-1 S. 0-12 o-oo N.0-11 0-0022189 0021016 0019830 12 19 52-80 12 15 56-90 1212 0-99 15 38-64 15 50-55 16 1-18 15 44-70 15 56-08 16 5-75 57 18-95 58 2-59 58 41-54 57 41-14 58 22-85 58 58-27 19 20 21 175 37 36-2 176 36 16-2 177 34 57-9 0-19 0-24 0-28 0-0018631 0017419 0016196 12 8 5-09 12 4 9-18 12 013-27 16 9-70 16 15-50 16 18-37 16 12-96 16 17-31 16 18-73 59 12-73 59 34-01 59 44-52 59 24-70 59 40-62 59 45-84 22 23 24 178 33 41-5 179 32 26-7 180 31 13-6 0-27 0-24 017 0-0014963 0013721 0012471 11 56 17-37 11 52 21-46 11 48 25-55 16 18-43 16 16-06 16 11-82 16 17-52 16 14-15 16 9-15 59 44-73 59 36-06 59 20-51 59 41-40 59 29-03 59 10-75 25 26 27 181 30 2-2 182 28 52-6 183 27 44-6 N.0-06 S. 0-05 0-16 0-0011216 0009957 0008696 11 44 29-65 11 40 33-74 11 36 37-84 16 6-22 15 59-72 15 52-61 16 3-07 15 56-22 15 48-89 59 0-01 58 36-18 58 10-13 58 48-45 58 23-38 57 56-50 28 29 30 184 26 38-3 185 25 33-8 183 24 31-2 0-31 0-43 0-55 0-0007434 0006174 0004917 11 32 41-93 11 28 46-02 11 24 50-12 15 45-09 15 37-22 15 29-13 15 41-19 15 33-20 15 25-03 57 42-55 57 13-75 56 44-11 57 28-28 56 59-02 56 29-09 31 187 23 30-5 S. 0-66 0-0003662 11 20 54-21 15 20-93 15 16-85 56 14-06 55 59-11 APPENDIX. 365 IV. SEPTEMBER MEAN TIME. Day. THE MOON'S Longitude. Latitude. Age. Meridian Passage. Noon. Midniyht. Noon. Midnight. Noon. Upper. Lower. 1 2 3 d. 12-44 13-44 14-44 h, m. 11 13-7 12 4-4 12 52-4 h. m. 23 39-4 * * 28-7 319 34 4-8 333 6 17-3 346 24 7-0 326 21 52-0 339 47 6-2 352 57 10-5 S.O 43 23-4 1 55 0-4 2 58 57-2 S.I 19 55-5 2 28 9-3 3 27 2-8 4 5 6 359 26 10-6 12 11 54-8 24 41 49-2 5 51 4-7 18 28 46-4 30 51 17-0 3 52 8-3 4 32 28-0 4 58 48-6 4 14 0-1 4 47 25-4 5 6 36-7 15-44 16-44 17-44 13 38-4 14 23-0 15 6-9 1 15-6 2 0-8 2 45-0 7 8 9 36 57 28-0 49 1 28-5 60 57 22-2 43 43-4 55 11-2 66 53 34-6 5 10 50-7 5 8 49-7 4 53 24-9 5 11 33-6 5 2 44-6 4 40 57-5 18-44 19-44 20-44 15 50-9 16 35-4 17 21-0 3 28-9 4 13-0 4 58-0 10 11 12 72 49 23-5 84 42 15-9 96 40 56-5 78 45 24-8 90 40 33-9 102 44 0-3 4 25 30-2 3 46 10-5 2 56 41-7 4 7 11-6 3 22 36-9 2 28 36-9 21-44 22-44 23-44 18 7-8 18 56-1 19 45-6 5 44-2 6 31-8 7 20-7 13 14 15 108 50 20-4 121 15 0-1 133 58 41-9 115 30-2 127 34 16-6 140 28 33-0 1 58 36-2 S.O 53 50-5 N.O 15 5-3 1 26 54-8 S.O 19 42-8 N.O 50 9-1 24-44 25-44 26-44 20 35-9 21 26-7 22 17-5 8 10-7 9 1-3 9 52-1 16 17 18 147 4 0-7 160 31 53-3 174 21 18-1 153 45 8-8 167 24 2-8 181 23 11-4 1 25 0-4 2 32 3-7 3 31 54-3 1 59 9-3 3 3 10-0 3 57 43-6 27-44 28-44 29-44 23 8-2 23 58-8 * * 10 42-9 11 33-5 12 24-2 19 20 21 188 29 9-2 202 50 33-3 217 19 28-7 195 38 31-2 210 4 28-5 224 34 47-9 4 20 6-4 4 52 45-0 5 7 2-2 4 38 35-0 5 2 18-2 5 6 50-6 0-98 1-98 2-98 49-8 1 41-4 2 34-3 13 15-5 14 7'7 15 1-4 22 23 24 231 49 42-0 246 15 43-0 260 33 24-3 239 3 31-8 253 25 47-8 267 38 16-8 5 1 44-1 4 37 19-7 3 55 50-2 4 51 49-6 4 18 32-6 3 29 38-8 3-98 4-98 5-98 3 28-9 4 25-2 5 22-9 15 56-9 16 53-9 17 52-0 25 26 27 274 40 14-8 288 35 8-1 302 17 56-0 281 39 12-6 295 28 1-6 309 4 54-0 3 26-7 1 55 4-5 N.O 43 59-9 2 28 44-7 1 19 58-7 N.O 7 40-4 6-98 7-98 8-98 6 21-1 7 18-5 8 14-3 18 50-0 19 46-7 20 41-2 28 29 30 315 48 59-0 329 8 39-7 342 17 6-2 322 30 13-6 335 44 17-4 348 47 4-4 S.O 28 28-3 1 38 14-1 2 41 36-4 S.I 3 56-0 2 10 55-7 3 9 54-8 9-98 10-98 11-98 9 7-5 9 58-0 10 46-1 21 33-1 22 22-4 23 9-4 31 355 14 9-2 1 38 18-0 S.3 35 31-3 8.3 58 10-3 12-98 11 32-3 23 54-8 The Moon's Longitude and Latitude are from HANSEN'S Tables; the Eight Ascension and Declination contain NEWCOMB'S corrections. 366 APPENDIX. JUPITER'S ELEMENTS. Date. Apparent Eight Ascension. Apparent Declination. Polar S.D. Hor. Par. Meridian Passage. Noon. Noon. h. m. s. t H H h. m. Aug. 7 6 6 35-2 N. 23 7 48-2 15-7 1-5 21 3-5 8 6 7 26-4 23 7 43-5 15-7 1-5 21 0-4 9 6 8 17-2 23 7 37-9 15-7 1-5 20 57-3 10 6 9 7-8 23 7 31-3 15-7 1-5 20 54-2 11 6 9 58-0 23 7 23-8 15-8 1-5 20 51-1 SATURN'S ELEMENTS. b. m. a. O 1 II n h. m. Aug. 9 23 2 32-3 S. 8 21 10-0 8-7 1-0 13 51-9 10 23 2 18-4 8 22 46-9 8-8 1-0 13 47-7 11 23 2 4-3 8 24 24-9 8-8 1-0 13 43-5 12 23 1 50-0 8 26 4-0 8-8 1-0 13 39-4 13 23 1 35-5 8 27 44-1 8-8 1-0 13 35-2 MEAN PLACES OF STARS. Star's Name. Mag. Right Ascension. Declination. h m s. O 1 H a AndromedsB 2-1 3 31-6 N. 28 34 17-3 7 Pegasi 2.9 8 23-6 N. 14 39 39-6 a Phoenicia / 2-4 21 38-4 S. 42 48 59-3 /SCeti 2-2 38 52-3 S. 18 30 8-6 a Ursse Minoris 2-1 1 25 8-2 N. 88 48 18-9 a Eridani 0-5 1 34 12-8 S. 57 42 51-3 7 1 Andromedse 2-2 1 58 7-5 N. 41 52 44-3 o Arietis 2-2 2 1 52-3 N. 23 1 5-7 o Ceti 2-8 2 57 21-9 N. 3 43 16-4 aPersei 1-9 3 17 36-4 N. 49 31 37'6 a Tauri 1-1 4 30 31-5 N. 16 19 14-8 APPENDIX. 367 MEAN PLACES OF STARS (continued). Star's Name. Mag. Bight Ascension. Declination. h. m. s. 1 II a Aurigao 0-2 5 9 44-6 N. 45 54 10-8 ft Orionis 0-3 5 10 1-2 S. 8 18 35-3 7 Orionis 1-7 5 20 5'3 N. 6 15 53-8 /3 Tauri 1-8 5 20 20-9 N. 28 31 42-8 Orionis 1-7 5 31 26-6 S. 1 15 41-4 o Orionis var. 5 50 4-9 N. 7 23 24-0 ft Aurigse 2-1 5 52 38-0 N. 44 56 18-6 ft Canis Majoris 2-0 6 18 33-6 S. 17 54 31-7 o Argus -1-0 6 21 51-9 S. 52 38 39-0 7 Geminorum 1-9 6 32 16-9 N. 16 28 47-9 Canis Majoris -1-4 6 41 0-2 S. 16 35 12-2 Canis Majoris 1-6 6 54 55-9 S. 28 50 37-6 5 Canis Majoris 2-0 7 4 34-1 S. 26 14 36-9 a Geminorum 2-0 7 28 36-2 N. 32 5 43-4 a Canis Minoris 0-5 7 34 22-9 N. 5 27 57-3 ft" Geminorum ... 1-2 7 39 33-9 N. 28 15 13-3 e Argus 1-7 8 20 351 S. 59 12 24-6 5 Argus 2-0 8 42 6-3 S. 54 21 49-9 ft Argus 1-7 9 12 10-3 S. 69 19 47-8 a Hydras 2-2 9 22 58-1 S. 8 15 2-9 a Leonis 1-3 10 3 22-0 N. 12 25 36-7 a Ursse Majoris 2-0 10 57 561 N. 62 15 31-0 ai Crucis 1-0 12 21 21-8 S. 62 35 41-5 afl Crucis If Ao + 0-6 AS - 3-1 7 Crucis 1-6 12 25 56-7 S. 56 35 12-3 ft Crucis 1-5 12 42 18-4 S. 59 10 30-1 Ursse Majoris 1-8 12 49 53-8 N. 56 28 11-9 o Virginis 1-2 13 20 14-4 S. 10 40 14-8 y Urs83 Majoris 1-9 13 43 50-3 N. 49 46 56-0 ft Centauri 0-8 13 57 ll'O S. 59 55 11-0 o Bootis 0-3 14 11 22-4 N. 19 40 17-6 o 2 Centauri 1 14 33 13-2 S. 60 26 43-3 a Scorpii 1-3 16 23 38-5 S. 26 13 25-7 a Trianguli Auatralis 1-9 16 38 42-2 S. 68 51 20-9 A. Scorpii 1-8 17 27 13-5 S. 37 2 8-2 o Ophiuchi 2-1 17 30 34-2 N. 12 37 40-7 e Sagittarii 1-9 18 17 56-0 S. 34 25 45-8 o Lyraa o-i 18 33 45-4 N. 38 41 45-0 a Aquilse 0-9 19 46 11-8 N. 8 37 10-6 a Pavonis 2-0 20 18 12-9 S. 57 2 12-3 7 Cygni 2-3 20 18 51-3 N. 39 57 19-7 a Cygni 1-3 20 38 13-6 N. 44 56 38-9 a Cephei 2-0 21 16 20-2 N. 62 11 13-7 e Pegasi 2-5 21 39 34-1 N. 9 26 37-5 a Gruis 1-9 22 2 18-7 S. 47 24 59-7 ft Gruis 2-1 22 37 3-5 S. 47 22 35-0 a Piscid Australia ... 1-3 22 52 27-5 S. 30 7 14-2 o Pegasi 2-6 23 4-7 N. 14 41 57-8 368 APPENDIX. THE CORRECTION FOR SECOND DIFFERENCES, IN FINDING THE GREENWICH TIME CORRESPONDING TO A REDUCED LUNAR DISTANCE. Arguments : Interval and Mean difference of Proportional Logarithms. Interval. Mean difference of the Proportional Logarithms in the Ephemeris. 2 4 6 8 101 214 16 18 20 22 24 2C 28 30 32 34 36 38 40 42 444 648, 30 52 h. m. 10 20 30 40 50 1 1 10 1 20 1 30 b. m. 10 20 30 40 50 1 1 10 1 20 1 30 h. m. 10 20 30 40 50 1 1 10 1 20 1 30 h. m. 3 2 50 2 40 2 30 2 20 2 10 2 1 50 1 40 1 30 h. m. 3 2 50 2 40 2 30 2 20 2 10 2 1 50 1 40 1 30 8. 1 1 1 1 1 8. 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 B. 1 1 2 2 2 2 2 3 3 8. 1 1 2 2 , 3 : 3 1 3 < 3 < 3 < 8. 3 6 9 11 13 14 15 16 16 I 1 2 2 2 2 3 3 3 4 3 4 t 4 t 4 t 4 1 2 3 3 4 4 5 5 5 1 2 3 4 5 5 5 6 6 1 2 3 4 5 G 6 G 6 1 3 4 5 5 6 G 7 7 2 3 4 5 G 7 7 7 8 2 3 5 6 6 7 8 8 8 2 3 5 6 7 8 8 9 9 2 4 5 G 7 8 9 9 9 2 4 6 7 8 9 9 10 10 2 4 6 7 8 9 10 10 11 2 4 6 -8 9 10 11 11 11 2 5 7 8 9 10 11 12 12 3 5 7 9 10 11 12 12 12 3 5 7 9 10 12 12 13 13 3 5 8 101 111 121 131 141 141 3 3 6 6 3 8 010] 212] 313] 114] 115] 115] 3 G 9 Ll L3 L4 L5 L5 [6 Mean difference of the Proportional Logarithms in the Ephemeris. 51 8. 56 5860 626 466 68 70 72 74 76 7880 82 84 86 88 90 92 94 969 810( )102 8. s. B. 8. 8 . 8. . 8. 6 7 2 12 7 17 1 22 4 25 7 28 9 30 31 1 31 8. 7 13 18 22 26 28 30 31 32 5 9 12 15 17 19 21 21 22 5 9 13 16 18 20 21 22 23 5 9 13 16 19 21 22 23 23 5 5 910 1314 1617 1920 2122 2223 2424 5 10 14 17 20 22 24 25 25 5 10 14 18 21 23 24 25 25 6 10 14 18 21 23 25 26 26 6 11 15 19 22 24 25 26 27 6 11 15 19 22 24 26 27 27 G 11 16 19 22 25 27 28 28 6 11 16 20 23 25 27 28 29 6 12 16 20 23 26 28 29 29 6 121 171 212 242 272 282 293 303 7 9 12 14 5 6 7 7 7 10 12 14 16 17 17 18 7 10 13 15 16 17 18 18 7 10 13 15 17 18 19 19 8 111 131 161 171 181 192 192 8 8 112 414 616 818 919 020 021 8 12 15 17 19 20 21 21 Mean difference of the Proportional Logarithms in the Ephemeris. 04 106 108 s. 7 13 19 23 27 30 32 33 34 110 8. 7 14 19 24 27 30 32 34 34 112 114 -116 118 120] [22 12t t!26 12 130 132 134 8. 9 16 23 29 33 37 40 41 42 136 138 h. m. 3 2 50 2 40 2 30 2 20 2 10 2 1 50 1 40 1 30 B. 7 13 18 22 26 29 11 J2 8. 7 13 18 23 26 29 31 33 33 8. 7 14 19 24 28 31 33 34 35 8. 7 14 20 25 29 31 34 35 35 8. 8 14 20 25 29 32 34 35 36 B. 8 15 20 25 29 33 35 36 36 8. 8 15 21 26 30 33 35 37 37 8. 8 15 21 26 30 34 36 38 38 8. 8 15 21 27 31 34 37 38 39 8. 8 15 22 27 31 35 37 39 39 8. 8 16 22 28 32 35 38 39 40 B. 8 16 22 28 32 36 38 40 40 8. 9 16 23 28 33 37 39 41 41 B. 9 17 24 29 34 38 40 42 42 B. 9 17 24 30 34 38 41 42 43 The Correction is to be added to the approximate Greenwich Time when the Proportional Logarithms in the Ephemeris are decreasing, and subtracted when they are increasing. APPENDIX. 369 TABLES USED IN DETERMINING THE LATITUDE BY OBSERVA- TIONS OF THE POLE STAR OUT OF THE MERIDIAN. TABLE I. Containing the First Correction. Arguments : Sidereal Time of Observation. Sidereal Time. Correction. Sidereal Time. Sidereal Time. Correction. Sidereal Time. h. m. o / n h. m. h. m. / // h. m. -176 + 12 6 - 26 5 + 18 10 1 8 11 10 10 23 8 10 20 197 20 20 20 9 20 30 1 9 56 30 30 17 7 30 40 1 10 37 40 40 14 3 40 50 1 11 10 50 50 10 57 50 1 1 11 35 13 7 7 50 19 10 1 11 51 10 10 4 42 10 20 1 11 59 20 20 - 1 34 + 20 30 1 11 59 30 30 +0 1 34 - 30 40 1 11 51 40 40 4 42 40 50 1 11 35 50 50 7 50 50 2 1 11 10 14 8 10 57 20' 10 1 10 37 10 10 14 3 10 20 1 9 56 20 20 17 7 20 30 197 30 30 20 9 30 40 1 8 11 40 40 23 8 40 50 176 50 50 26 5 50 3 1 5 54 15 9 29 21 10 1 4 34 10 10 31 51 10 20 137 20 20 34 38 20 30 1 1 33 30 30 37 21 30 40 59 52 40 40 40 40 50 58 4 50 50 42 34 50 4 56 9 16 10 45 4 22 10 54 8 10 10 47 29 10 20 52 1 20 20 49 48 20 30 49 48 30 30 52 1 30 40 47 29 40 40 54 8 40 50 45 4 50 50 56 9 50 5 42 34 17 11 58 4 23 10 40 10 10 59 52 10 20 37 21 20 20 1 1 34 20 30 34 38 30 30 137 30 40 31 51 40 40 1 4 34 40 50 29 50 50 1 5 54 50 6 - 26 5 + 18 12 + 176- 24 2 B 370 APPENDIX. TABLE II. Containing the Second Correction (always to be added), Arguments : Sidereal Time and Altitude. Sidereal Time. Altitude. Sidereal Time. 5 10 15 20 25 30 35 h. m. 1 1 2 2 3 3 4 h. m. 12 30 1 1 1 1 2 30 1 13 30 30 2 1 1 14 30 1 1 1 2 2 3 30 3 1 1 2 3 3 4 5 15 30 1 2 3 4 6 7 9 30 4 2 3 5 6 8 10 12 16 30 2 4 6 9 11 14 17 30 5 3 5 8 11 14 17 21 17 30 3 6 9 13 16 20 24 30 6 3 7 11 14 18 23 28 18 30 4 7 11 16 20 25 30 30 7 4 8 12 16 21 26 31 19 30 4 8 12 16 21 26 32 30 8 4 8 12 16 21 25 31 20 30 4 7 11 15 19 24 29 30 9 3 7 10 14 18 22 27 21 30 3 6 9 12 15 19 23 30 10 2 5 7 10 13 16 19 22 30 2 4 6 8 10 12 15 30 11 1 3 4 6 7 9 11 23 30 1 2 3 4 5 6 7 30 12 1 2 2 3 3 4 24 TABLE III. Containing the Third Correction (always to le added). Arguments : Sidereal Time and Date. Sidereal Time. Jan. 1. Feb. 1. March 1. April 1. May 1. .June 1. July 1. b. / n i H / // i n / // / n / // 1 35 1 31 1 24 1 15 1 7 1 2 1 3 2 1 30 1 32 1 29 1 21 1 12 1 4 1 4 1 17 ' 1 24 1 26 1 21 1 14 1 5 57 6 1 1 10 1 15 1 16 1 12 1 4 54 8 43 53 1 1 1 7 1 7 1 2 53 10 30 38 47 56 1 1 54 12 25 29 36 45 53 58 57 14 30 28 31 39 48 56 16 43 36 34 39 46 55 3 18 1 50 45 44 48 56 6 20 1 17 1 7 59 53 53 58 7 22 1 30 1 22 1 13 1 4 1 1 6 24 1 35 1 31 1 24 1 15 1 7 1 2 3 APPENDIX. 37i TABLE II. Containing the Second Correction (always to be added). Arguments : Sidereal Time and Altitude. Sidereal Time. Altitude. Sidereal Time. 35 40 45 50 55 60 65 70 h. m. 4 5 6 7 8 10 13 16 h. m. 12 30 2 2 3 3 4 4 5 7 30 1 1 1 1 1 1 1 13 30 30 2 1 1 1 1 2 2 2 3 14 30 3 3 4 4 5 6 8 10 30 3 5 6 7 9 10 13 16 20 15 30 9 10 12 14 17 21 26 34 30 4 12 15 18 21 25 31 38 49 16 30 17 20 24 28 34 41 51 1 5 30 5 21 25 29 35 42 51 1 3 1 21 17 30 24 29 35 41 50 1 1 14 1 36 30 6 28 33 39 47 56 1 8 1 24 1 48 18 30 30 36 43 51 1 1 1 14 1 31 1 57 30 7 31 38 45 53 1 4 1 17 1 36 2 3 19 30 32 38 45 54 1 5 1 18 1 37 2 4 30 8 31 37 44 53 1 3 1 16 1 35 2 2 20 30 29 35 42 50 1 1 12 1 29 1 55 30 9 27 32 38 45 54 1 6 1 21 1 44 21 30 23 28 33 39 47 57 1 11 1 31 30 10 19 23 0-28 33 39 48 59 1 16 22 30 15 18 22 26 31 37 46 59 30 11 11 13 16 19 23 27 34 43 23 30 7 9 10 12 15 18 22 29 30 12 4 5 6 7 8 10 13 16 24 TABLE III. Containing the Third Correction (always to be added). Arguments ; Sidereal Time and Date. Sidereal Time. July l. Aug. 1. Sept. 1. Oct. 1. Nov. 1. Dec. i. Dec. 31. h. / H t n / // / n i a / // / n 1 3 1 10 1 20 1 31 1 42 1 50 1 53 2 1 1 2 1 8 1 17 1 28 1 39 1 47 4 57 53 54 58 1 7 1 17 1 28 6 54 46 41 40 44 51 1 1 8 53 43 33 27 25 27 34 10 54 45 33 23 15 12 14 12 57 50 40 29 18 10 7 14 58 52 43 32 21 13 16 3 1 7 6 2 53 43 32 18 6 1 14 19 20 1 16 1 9 59 20 7 1 17 27 33 1 35 1 33 1 26 22 6 1 15 27 37 1 45 1 48 1 46 24 3 1 10 20 31 1 42 1 50 1 53 372 APPENDIX. For converting INTERVALS of MEAN SOLAR Time into Equivalent INTERVALS of SIDEREAL Time. HOURS. MINUTES. SECONDS. Hours of Mean Time. Equivalents in Sidereal Time. 1 Minutes of Mean Time. Equivalents in Sidereal Time. 1 Minutes of Mean Time. Equivalents in Sidereal Time. (Seconds of Mean Time. Equiva- lents in Sidereal Time. 11 #! Equiva- lents in Sidereal Time. li. m. 8. m. s. m. s. 8. 8. 1 1 9-8565 1 1 0-1643 31 31 5-0925 1 1-0027 31 31-0849 2 2 19-7130 2 2 0-3286 32 32 5-2568 2 2-0055 32 32-0876 3 3 29-5694 3 3 0-4928 33 33 5-4211 3 3-0082 33 33-0904 4 4 39-4259 4 4 0-6571 34 34 5-5853* 4 4-0110 34 34-0931 5 5 49-2824 5 5 0-8214 35 35 5-7496 5 5-0137 35 35-0958 6 6 59-1388 6 6 0-9857 36 36 5-9139 6 6-0164 36 36-0986 7 7 1 8-9953 7 7 1-1499 37 37 6-0782 7 7-0192 37 37-1013 8 8 1 18-8518 8 8 1-3142 38 38 6-2424 8 8-0219 38 38-1040 9 9 1 28-7083 9 9 1-4785 39 39 6-4067 9 9-0246 39 39-1068 10 10 1 38-5647 10 10 1-6428 40 40 6-5710 10 10-0274 40 40-1095 11 11 1 48-4212 11 11 1-8070 41 41 6-7353 11 11-0301 41 41-1123 12 12 1 58-2777 12 12 1-9713 42 42 6-8995 12 12-0329 42 42-1150 13 13 2 8-1342 13 13 2-1356 43 43 7-0638 13 13-0356 43 43-1177 14 14 2 17-9906 14 14 2-2998 44 44 7-2281 14 14-0383 44 44-1205 15 15 2 27-8471 15 15 2-4641 45 45 7-3924 15 15-0411 45 45-1232 16 16 2 37-7036 16 16 2-6284 46 46 7-5566 16 16-0438 46 46-1259 17 17 2 47-5600 17 17 2-7927 47 47 7-7209 17 17-0465 47 47-1287 18 18 2 57-4165 18 18 2-9569 48 48 7-8852 18 18-0493 48 48-1314 19 19 3 7-2730 19 19 3-1212 49 49 8-0495 19 19-0520 49 49-1342 20 20 3 17-1295 20 20 3-2855 50 50 8-2137 20 20-0548 50 50-1369 21 21 3 26-9859 21 21 3-4498 51 51 8-3780 21 21-0575 51 51-1396 22 22 3 36-8424 22 22 3-6140 52 52 8-5423 22 22-0602 52 52-1424 23 23 3 46-6989 23 23 3-7783 53 53 8-7066 23 23-0630 53 53-1451 24 24 3 56-5554 24 24 3-9426 54 54 8-8708 24 24-0657 54 54-1479 25 26 25 4-1069 26 4-2711 55 56 55 9-0351 56 9-1994 25 26 25-0685 26-0712 55 56 55-1506 56-1533 27 27 4-4354 57 57 9-3637 27 27-0739 57 57-1561 28 28 4-5997 58 58 9-5279 28 28-0767 58 58-1588 29 29 4-7640 59 59 9-6922 29 29-0794 59 59-1615 30 30 4-9282 60 60 9-8565 30 30-0821 60 60-1643 APPENDIX. 373 SUMMARY OF USEFUL FORMULA. L Plane Trigonometry. 1. Sin A cosec A' cos A = Cosec A = 2. Tan A sin A' sin A sec A = cot A = sec A 1 cos A' cos A sin A tan A = cot A = cot A 1 tan A cos A' 3. Sin 2 A + cos 2 A = 1 whence, sin 2 A = 1 cos 2 A, and cos 2 A = 1 sin 2 A 4. Sec 2 A = 1 + tan 2 A, and therefore tan 2 A = sec 2 A - 1 Cosec 2 A = 1 + cot 2 A, and therefore cot 2 A = cosec 2 A 1 5. Sin A = cos (90 - A) = sin (180 - A) Cos A = sin (90 - A) = - cos (180 - A) 6. Sin (A + B) = sin A . cos B + cos A . sin B\ Sin (A B) = sin A . cos B cos A . sin B I Cos (A + B) = cos A . cos B sin A . sin B [ Cos (A B) = cos A . cos B + sin A . sin Bj 7. Tan(A + B)= tan A + tan B V Tan (A - B) = 1 tan A . tan B tan A ^- tan B 1 + tan A . tan B cot A + cot B 9. 10. cot B - cot A 8. Sin A +'sin B = 2 sin J(A + B) . cos Sin A - sin B = 2 cos $(A + B) . sin Cos A + cos B = 2 cos JiA + B) . cos Cos A - cos B = 2 sin |(A + B) . sin Sin 2A = 2 sin A . cos A ' Cos 2A = cos 2 A - sin 2 A = 1-2 sin 2 A = 2 cos 2 A - 1 Cos A = 2 cos 2 JA - 1 = 1-2 sin 2 J A Cos 2 JA = -|(1 + cos A) Sin 2 |A = J(l - cos A) NOTE. These formulae are equally true of the angles of spherical triangles. Relations of sides and angles of a plane triangle whose angles are A, B, C, and opposite sides a, 6, c. 11. A + B + = 180. -o Sin A _ sin B _ sin a b G 374 APPENDIX. 13. Cos A = 2bc Sin A = Tan A = (s-b).(s- c) s.(s-a) where = J(a + 5 4- 15. Cot JA - -- C . tan i(B - C) ft c II. Spherical Trigonometry. Relations of sides and angles of an oblique-angled spherical triangle. 1. Cos A = cos a cos 6 . cos c 2. sin b . sin c Cos a = cos 6 . cos c + sin 6 . sin c . cos A Sin A _ sin B __ sin C sin a ~ sin b ~~ sin c 3. Sin p. = x / Bin (-i.). sin (,-c) V sin 6 . sin c sin . sm c Tan i A _ A /sinQ- ~V sins. sinQ-b).sm(s-c) where s = J(a + 6 -f c) APPENDIX. 375 WEIGHTS AND MEASURES. Angular. 60 seconds (") = 1 minute. 60 minutes (') 1 degree. 90 degrees () = 1 quadrant. 4 quadrants = 1 circumference. Apothecaries' (dry). 20 grains = 1 scruple O). 3 scruples = 1 drachm (3). 8 drachms = 1 ounce (). 12 ounces = 1 pound. Apothecaries' (wet}. 60 minims or drops = 1 drachm. 8 drachms = 1 ounce (oz.). 20 ounces = 1 pint. 8 pints = 3 gallon. 4 drachms = 1 tablespoonful. 2 ounces = 1 wineglassful. 3 ounces = 1 teacupf ul. Avoirdupois. 16 drachms (dr.) = 1 ounce (oz.). 16 ounces = 1 pound (lb.). 28 pounds = 1 quarter (qr.). 4 quarters = 1 hundredweight (cwt.). 20 hundredweights = 1 ton. Troy. 24 grains (gr.) = 1 pennyweight (dwt.). 20 pennyweights = 1 ounce (oz.). 12 ounces = 1 pound. 7000 grains Troy = 1 lb. Avoirdupois. Lineal. 12 inches (in.) = 1 foot (ft.). 3 feet = 1 yard (yd.). 5 yards = 1 rod, pole, or perch (po.). 40 poles = 1 furlong (fur.). 8 furlongs = 1 mile (m.). Square. 144 square inches (sq. in.) = 1 square foot (sq. ft.). 9 square feet = 1 square yard (sq. yd.). 30 J square yards = 1 square rod, pole or perch (sq. po.). 40 poles = 1 rood (rd.). 4 roods = 1 acre (ac.). 640 acres = 1 square mile (sq. m.). 376 APPENDIX. Cubic or solid. 1728 cubic inches (cub. in.) = 1 cubic foot (cub. ft.). 27 cubic feet = 1 cubic yard (cub. yd.). Capacity or dry. 4gilla(gi.) = lpint(pt.). 2 pints = 1 quart (qt.). 4 quarts = 1 gallon (gal.). 2 gallons = 1 peck (pk.). 4 pecks = 1 bushel (bush.). 8 bushels = 1 quarter (qr.). 36 bushels = 1 chaldron (chal.). Wine or Ale. 4 noggins = 1 pint. 2 pints = 1 quart. 4 quarts = 1 gallon. 9 gallons = 1 firkin (fir.). 18 gallons = 1 kilderkin (kil.) ale. - 1 runlet (run.) spirit. 36 gallons = 1 barrel (bar.) ale. 54 gallons = 1 hogshead (hhd.) ale. 2 barrels = 1 puncheon (pun.) ale. 3 barrels = 1 butt (ale). 42 gallons = 1 Tierce (wine). 63 gallons 1 hogshead (wine). 84 gallons 1 puncheon (wine). 126 gallons = 1 pipe. 252 gallons = 1 tun. Pipes of wine vary in quantity. Metric system derives its name from the metre which is the unit of length in the system, and is equal to a ten-millionth part of the distance from the equator to the pole, or 39'3708 inches. To indicate multiples of the unit Greek prefixes are used thus Deka means 10 times unit Hekto 100 Kilo 1000 ,, Myria 10,000 To indicate sub-multiples Latin prefixes are used thus Deci means of unit Centi Milli 1 cu. centimetre Centi j&y Unit of weight is gramme ( 1 cu. of water at 4 C. \ = 15'43 grains Unit of capacity is litre = 1 cu. decimetre = 176 pints 1 litre of water weighs 1 kilogram = 2 '2 Ibs. Thus 10 millimetres make 1 centimetre 10 centimetres 1 decimetre 10 decimetres ,, 1 metre 10 metres 1 dekametre 10 dekametres ,, 1 hektometre 10 hektometres ,, 1 kilometre 10 kilometres ,, 1 myriametre. APPENDIX. 377 Specific gravity of solids at zero as compared with distilled water at 4 C. Platinum 22*069 Gold 19-362 Silver 10'474 Steel 7*816 Cast iron 7'207 Marble 2*837 Common salt 2'220 Sugar 1-600 Ice -930 Beech -852 Oak -845 Elm -800 Pine -657 Cork -240 Snow -183 Specific gravity of liquids under same conditions Mercury 13'598 Glycerine 1-260 Blood 1-060 Milk 1-029 Olive oil -915 Turpentine *870 Petroleum -836 Alcohol -793 Vapour density under same conditions Air 1-0 Hydrogen -0693 Nitrogen -8714 Oxygen 1-1057 Carbonic acid 1-5291 Water in a saturated space . . -6225 Boiling points. Freezing points. Alcohol .... 78 C. - 202 C. Turpentine . . . 157 27 Mercury . . . 358 - 38 '8 Saltwater ... 102 - 3'3 378 APPENDIX. EXAMINATION PAPERS. PAPER I. 1. Multiply 142-07 by 0'00347, by logs. 2. Divide O0004629 by 0-00000542, by logs. 3. Hour. Courses. Knts. lOtha. Winds. Lee- way. Devia- tion. Remarks. 1 2 3 4 8. 7 7 6 6 2 2 9 7 W.S.W. Pts. I 3 W. A point in lat. 50 44' N., long. 13' E. ; bear- ing by compass N.E. f E. ; ship's head, W. by S. S. ; deviation as per log, dis- tant 12 miles. Variation, 18 W. 5 6 7 8 W. by S. \ 8. 6 7 8 6 8 3 4 5 8. \ 8 W. 9 10 11 12 S.W. byW.JW. 8 7 7 7 2 7 1 N.W. \ 7W. 1 2 3 N.E. \ N. 6 6 7 2 5 3 N.N.W. If ~~ 4 5 6 7 8 W. by N. i N. 7 7 6 6 6 8 6 9 4 3 N. i T W. 9 10 11 12 N.E. i E. 7 6 6 6 4 5 3 8 N.*W. U 2E. By observation the lat. and long, of the ship was 49 30' N., 45' W. Find set and drift of the current. 4. August 3rd, in long. 68 45' E., the observed mer. alt. of the Sun's L.L. bearing S. was 58 14' 40" ; ind. err. - 3' 25", height of eye, 27 ft. Required the latitude. 5. In lat. 51 30' S., long. 37 E., the departure made good eastward was 306'5 miles. Required the long, in by parallel sailing. 6. Required the compass course and distance by Mercator's sailing From A, lat. 30 27' N. Long. 18 18' W. To B, 10 45 S. 32 54 W. Variation, 13 15 W. Deviation, 24 30 W. 7- January 19th, find the M.T.G. times of high water at Rotterdam. APPENDIX. 379 8. August 5th, at ship, in lat. 38 24' S., long. 130 45' E., when the time shown by chron. correct for G.M.T. was 10 h 13 m ; the Sun's bearing by compass at rising was N. 64 30' E. Required the true amplitude, and error of the compass ; and supposing the variation to be 6 15' W., required the deviation of the compass for the position of the ship's head. 9. August 7th, about 9 h 15 m A.M. at ship, in lat. at noon 39 31' 40" N. ; long, by account 31 W. ; the obs. alt. of Sun's L.L. was 53 53' 10" ; ind. err. + 3', height of eye, 29 ft. ; time by chron. 12 h O m 9* which was slow 10' for G.M. noon, on January 9th, and on April 29th was correct ; run since obs. W. 25 miles. Required the long, at sight, also brought up to noon. The sun's compass bearing was S. 30 E., var. 28 15' W., find the true azimuth, comp. error, and deviation. 10. September 2nd, about 3 h 30" P.M. at ship, in lat. 61 25' S., long. 174 25' E. ; when the time shown by chron. was 3 h 54 m 42 , the Sun's bearing by compass K 81 30' W., the error of the chron. on G.M.T. was 15" 12 slow. Required the true azimuth and error of the compass by the Tables ; and supposing the variation to be 30 15' E., required the deviation for the position of the ship's head. 11. August 25th, about 3 h 5 m A.M. at ship, lat. by account 26 16' S., long. 35 15' W. ; the alt. of Achernar bearing south was 57 27' 50" ; ind. err. + 2' 11" ; height of eye, 26 ft. ; time by chron. 5* 17* 32 , which had been found to^be 21 m 7' fast of G.M.T. Required the lat. by reductiom to the meridian. 12. September 23rd, A.M. at ship. If at sea and uncertain of my position when the chron. showed 23 d 9 h 51 m 13' G.M.T. ; the alt. of the Sun's L.L. 32 60' 30" ; again P.M. on the same day, when the chron. showed 23 d 13 h 51 m 9* G.M.T.; the alt. of the Sun's L.L. 32 54' 10" ; ind. err. 1' 20" ; height of eye, 17 ft. ; the ship becalmed heading east in the interval between the observations. Required the line of position when the first alt. was taken, the Sun's true azimuth when the second alt. was taken, and the position of the ship when the second alt. was observed by " Sumner's method " by projection on the chart, assuming the lats. 51 15' and 51 45' S. 13. August 14th, in long. 159 16' W., find the mean time of transit of a Arietis across the meridian. 14. On August 14th at 4 h 15" A.M. M.T.S., find what stars not less bright than the second magnitude are within two hours east of the meridian in lat. 32 14' N., long. 159 15' W., above the horizon and pole. Give their meridian distances, and state bearing N. or S. ; also find the approx. mer. alt. for setting the sextant of a Persei ; ind. cor. 4' 29" ; eye, 40 ft. 15. August 31st, about 6 h P.M. at ship, in lat. 24 19' S., long, by account 180 E., when a chron. showed 6 h 4 m 3*, which was slow on June 16th, l m 53 '5 ; and was fast l m 29"5 on Aug. 5th, for G.M.T. ; the obs. alt. of Vega was 21 29' 10" ; ind. cor. - 2' 15" ; eye, 28 ft. Find the long. 16. August 24th, when the M.T.S. was 2 b 15 A.M. in lat. equator, long. Greenwich, Rigel bore by compass S. 85 30' E. Find the true azimuth, and deviation, the variation being 20 15' W. 17. September 12th, when the M.T.S. was 4 h 40 m 15' A.M., in long. 136 24' E., the obs. alt. of Polaris out of the meridian was 39 15' 40" ; ind. cor. - 3' 48" ; eye, 27 ft. Find the lat. 18. August 7th, in long. 90 45' E., the obs. mer. alt. of the Moon's L.L. was 49 56' 20" bearing south, ind. cor. + 2' 45" ; eye, 36 ft. Find the lat. State where tables giving the approx. time of most of the principal stars passing the meridian are to be found, the numbers of the tables, and whether given in mean or apparent time. Can the approx. mer. pass, of a planet and its approx. alt. be found in the same way as in the case of a star ? State where the times of the mer. pass, of the principal planets may be found. Is the time given mean or apparent ? APPENDIX. 19. August 12th, the obs. raer. alt. of the star a Arietis was 38 47' 40" bearing north ; ind. err. - 6' 25" ; height of eye, 32 ft. Required the lat. 20. Ship's head by standard compass. Bearings of distant object by standard compass. Deviation required. Ship's head by standard compass. Bearings of distant object by standard compass. Deviation required. N. N. 7 W. S. N. 10 E. N.E. N. 27 W. S.W. N. 25 E. E. N. 21 W. W. N. 30 E. S.E. N. N.W. N. 20 E. Find correct magnetic bearing ; construct a Napier curve. With magnetic courses : E. W.S.W. S.E. by S. N.N.W. Required compass courses. With compass courses : N.W. N.E. by N. J N. S.S.W. E.S.E. Required magnetic courses. With ship's head at W.S.W. by compass, the compass bearings of distant objects were W. by S. and N. Find magnetic bearings. 21. On October llth, at 7 h 40 m A.M. M.T.S., being off Fowey by reckoning, took a cast of the lead. Required the correction to be applied to the depth obtained by the lead-line, before comparing it with the depth marked on the chart. 22. August 1st, P.M. at ship in lat. 34 28' S., when a chron. showed 7 h 20 m 35", estimated error, 6 m 15' fast of G.M.T., approx. long. 66 E. ; the following observations were taken for finding long, and error of chron., Observed alt. of a Pegasi 27 16' 20", ind. cor. + 4' 13", eye, 44 ft. ; observed distance between the star and Moon's remote limb, was 80 14' 40", ind. cor. -2' 27". 23. September 23rd, lat. N., long. 82 30' E. With the following observations of the Sun's L.L. find the lat. by double alts, at second obs. A.T.S. 9 h 30 m 10 s A.M. 2 30 20 P.M. Obs. alt. 32 13' 10". 32 13 10 Ind. cor. -2' 8". Eye, 28 ft. Sun's bearing at first obs. S. 46 E. ; course and distance during interval, N. 60 E. ; rate 5 ! 8 knots per hour. 24. August 30th, about 9 h A.M. M.T.S., a point in lat. 40 28' N., long. 25 24'-5 W., bore N. 27 E. from a ship, distant 9 miles ; the obs. alt. of Sun's L.L. in an artificial horizon was 79 10' 20" ; ind. cor. - 3' 18". Time by chron. 10 h 30 m , estimated slow 15 m 15 s of M.T.G. Find error on M.T.G. 25. Find the first course and distance on a G.C. from Wolf Rock, lat. 49 57' N., long. 5 48' W., to Charleston lat. 32 45' N., long. 79 52' W. ; also position of Vertex, lat. of points 5, 10, 15, 20 from Wolf ; and lat. at which the 61st W. mer. cuts the G.C. arc. 26. In a triangle PQR, Q = 90, PR = 364, P = 28 45'. Find PQ. 27. In triangle PQR, PQ = 417, PR = 243, P = 58 26'. Find R. 28. In spherical triangle ABC, A = 36 38', B = 101 42', C = 90. Find AB. 29. In spherical triangle, A = 90, B = 104 25', AC = 107 39'. Find BC. 30. Coefficient B = -15 36', C = +9 24'. Find direction of ship's head whilst building. Direction of ship's head whilst building S. 49 W., coefficient B = +20 15'. FindC. 31. September 20th, ll h 48 m P.M. M.T.S. ; in lat. 49 S., long. 168 15' W. Name some stars which would be suitable for finding the correction of the compass. APPENDIX. 381 32. Given deviations with ship's head by compass at N. N.E. E. S.E. S. S.W. to be 17 W. 28 W. 24 W. 9 W. 16 E. 32 E. W. 26 E. N.W. 5E. find coefficients A, B, C, D, and E, and construct a table of deviations for all points from N to S, through W. 33. With a ship steering N.E. by N. on the port tack, heeling 12; the heeling error was 15. Required the error when steering E.S.E. on the port tack, and heeling 10. 34. Construct a Mercator's chart, scale 1"'5 = 1 long., extending from lat. 58 to 61 K, and 6 to 10 E. long. Graduate a degree of lat. and long, and place on it a magnetic compass showing points, variation 20 W. PAPER IL 1. Multiply 47290 by 0-0045, by logs. 2. Divide 4765*07 by 32406 '5, by logs. 3. lira. Courses. Knts. lOths. Winds. Lee- way. Devia- tion. Remarks. Pts. 1 S.E. by E. \ E. 10 7 S.S.W. f 25 W A point in lat. 2 11 3 25 15' S., long. 3 9 9 4720'W.; bear- 4 5 11 11 1 ing by compass N.N.E. } E. ; 6 E. by S. | S. 10 4 S. u 24 W. by E. \ E. ; devi- 7 9 7 ation as per log, 8 10 5 distant 15 miles. 9 9 8 Variation, 20 W. 10 9 6 11 S.E. by S. 9 7 S.W. 1 18 W. 12 10 2 1 10 4 2 11 7 3 S.W. by S. 9 6 S.S.E. 1* 25 E. 4 10 2 5 11 2 6 & 9 5 E.S.E. u 5E. 7 10 7 A current set N". 8 9 8 14 W. true, 54 miles during the day. 9 S. by E. | E. 10 4 E. 1 9 W. 10 11 3 11 10 9 12 9 4 Find the course and distance made good ; lat. and long, of the ship. 4. September 23rd, in long. 28 45' W., the obs. mer. alt. of the Sun's L.L. bearing north, was 64 53' 10"; ind. err. -4' 26"; height of eye, 37 ft. Required the latitude. 5. From lat. 40 30' N., long. 39 43' W., the departure made good was 89 miles to the west. Required long, in by parallel sailing. 382 APPENDIX. 6. Required the compass course and distance by Mercator's sailing From A, lat. 52 50' S. Long. 81 10' W. To B, 41 12 N. 127 55 W. Variation, 18 25 E. Deviation, 22 15 E. 7. January 30th. Find the M.T.G. of high water at Dungeness. 8. September 29th, about 6 h 5 m P.M. at ship, in lat. 42 10' N., long. 146 58' E., when a chron. correct for M.T.G. showed 8* 2 m 33 , the Sun's bearing by compass was N. 84 30' W. Required the true amplitude, and error of the compass; and supposing the variation to be 0, required the deviation of the compass for the position of the ship's head. 9. August 4th, about 7 h 15 m A.M. at ship, in lat. at noon 15 22' N., long, by account 37 30' E. ; the obs. alt. of sun's L.L. was 22 31' 10" ; ind. err. -1' 30"; height of eye, 25 ft. ; time shown by chron. 4 h 42 m 11*, which was 9" 44 -2 slow for G.M. noon on April 20th, and on June 3rd was 10 m 15 s slow ; run since sights, S. 67 E. 46 miles. Required the long, at sight, also brought up to noon ; and if the compass bearing be E". 87 30' E., find true azimuth and deviation, the variation being 5 W. 10. September 1st, about 4 h P.M. at ship, in lat. 44 23' S., long. 145 17' fi. ; when a chron. correct for G.M.T. showed 6 h 24 m 35 s , the Sun's bearing by compass W. \ N. Required the true azimuth and error of the compass by the tables; and supposing the variation to be 10 15' E., required the deviation for the position of the ship's head. 11. August 7th, A.M. at ship, lat. by account 33 15' N., long. 10 57' W.; the alt. of the Sun's U.L. 72 53' ; ind. err. +3'; height of eye, 16 ft.; time by watch 10 h 57 m 58', which had been found to be 44 m 10 1 slow of A.T.S. ; the diif. long, made westward 17 f miles after the error was determined. Required the lat. by reduction to the meridian, and if the run since observation was S. 11 W., 4 miles, find the lat. at noon. 12. September 29th about l h 20 m P.M. at ship. If in long, by account 125 30' W., when the chron. showed 9 h 26 12' G.M.T. ; the alt. of the Sun's L.L. 36 33' 40" ; again P.M. on the same day, when the chron. showed 12 h 2 m 10' G.M.T. ; the alt. of the Sun's L.L. 18 39' 20" ; ind. err. -2' 5" ; height of eye, 34 ft. ; the ship having made 43 miles on a true E. N. course in the interval between the observations. Required the line of 'position when the first alt. was taken, the Sun's true azimuth and the position of the ship when the second alt. was observed by " Sumner's method " by projection on the chart, assuming the lats. 48 and 48 30' N. 13. On Sept. 28th. At what apparent ship time will Capella cross the meridian of a place in long. 73 48' E. ? Also by inspection? 14. What stars of brightness not less than 1'5 will be within 2 hours of the meridian of a place above the horizon and pole in lat. 30 55' S., long. 73 48' E., on September 28th at 5 b 27 m A.M. M.T.S. ? Give meridian distances, and if ind. cor. be 3' 47" to add, eye 35 ft. , find approx. mer. alt. of Rigel for placing on sextant. 15. September 7th, about 7 h P.M. at ship, in lat. 35 20' N., long, by account 165 W., when a chron. showed 7 h 54 m 50 , which was fast l h 58 m 5 '4 on June 29th, and was fast 2 h 3 m 13' '4 on August 3rd for G.M. noon, the obs. alt. 6f Altair was 53 1' 40" ; ind. cor. +1' 11" ; eye, 29 ft. Find the longitude. Are you acquainted with any maps, diagrams, or any other means for facilitating the recognition of the fixed stars and planets? If so, give a short description of the one you prefer, and how you use it. 16. August 27th at l h 15 m A.M. A.T.S. in lat. 47 24' K, long. 68 19' W., the compass bearing of Altair was S. 81 W. Find the true azimuth from the "tables," compass error, and deviation, supposing the variation to be 14 30' W. 17. September 7th, when the G.M.T. was noon in long. 125 40' E., the APPENDIX. 383 obs. alt. of Polaris out of the meridian in an artificial horizon was 58 37' 20", ind. cor. 5' 40". Required the latitude. 18. August 30th, in long. 49 36' W., the obs. mer. alt. of the Moon's L.L. was 58 57' 50" ; zenith south of moon; ind. cor. +6' 24" ; eye, 29 ft. Required the latitude. 19. September 19th, the obs. mer. alt. of the star ft Argus was 50 26' 40" bearing south ; ind. err. 7' 46" ; height of eye, 66 ft. Required the latitude. 20. Ship's head by standard compass. Bearings of distant object by standard compass. Deviation required. Ship's head by standard compass. Bearings of distant object by standard compass. Deviation required. N. S. 43 W. S. W. N.E. S. 50 W. s.w. S. 83 W. E. S. 65 W. W. S. 65 W. S.E. S. 80 W. N.W. S. 45 W. Find correct magnetic bearing, and construct a curve. Given magnetic courses : S.E. by E. N.N.E. N.W. J W. S. 10 W. Find compass courses. Given compass courses : N.W. by W. N. by W. N.E. E. S. 50 E. Find magnetic courses. With ship's head at E. |- N. by compass, the comp. bearings of distant objects were S.E. f E. and N.E. by N. Find magnetic bearings. 21. On December 15th, at 11 h 36 m P.M. M.T.G., being off Ardrossan by reckoning, took a cast of the lead. Required the correction to be applied to the depth obtained by the lead-line, before comparing it with the depth marked on the chart. 22. August 2nd, at 9 h 20 m P.M. M.T.S., in lat. 25 14' N., long, by account, 100 37' 30" W. ; the following observations were taken for determining the error of a chron. which had run down, and set going on G.M.T. and showed 4 h ; obs. alt. of the Moon's L.L., 39 8' 30"; ind. cor. -4' 12"; height of eye, 17 feet ; obs. distance between Spica and the Moon's near limb, 88 59' 40" ; ind. cor. +7' 43". 23. September 29th, in long. 125 35' W., lat. N. With following observations, find lat. at time of taking second obs. : A.T.S. l h 13 m 40 f P.M. 3 49 40 , Alt. Sun's L.L. 36 33' 40", bearing S. 23 W. 18 39 20 The ship's true course, N. 87 E., 16'5 knots per hour during the interval ; ind. cor. -2' 3" ; height of eye, 34 feet. 24. September 23rd, 2 h 49 m P.M. M.T.S., at a place in lat. 23 28' S., long, 158 35' E. ; the obs. alt. of Sun's ILL., 41 32' 10"; ind. cor. -2' 25" ; eye, 34 feet ; the chron. shows 4 h 17 m 7*. Find its error on M.T.G. 25. Find initial and final courses and distance, easting to be run down on parallel of 55 S., from lat. 42 53' S., long. 147 21' E., to 33 2' S., long. 71 38' W. Find latitudes of points where the meridians 160 E., 170 E., 90 W., and 80 W. cut the track. 26. In triangle ABC, A = 90, C = 68 40', AC = 37'2. Find AB. 27. In triangle OST, OS = 379, OT = 325, ST = 85. Find T. 28. In spherical triangle ABC, C = 90, B = 103 20', BC = 75. Find AB. 29. PQR, P = 90, RQ = 98 56', R = 7245'. Find Q. 30. If coefficient B = +16 54', coefficient C = -18 36', find direction of ship's head whilst building. Coefficient C = - 20 30' ; direction ship's head building, S. 39 W. Find B. APPENDIX. 31. In lat. 45 N., long. 58 W., find what stars are suitable for finding compass error by azimuth on September 7th at 2 h 30 m A.M. 32. The deviations with ship's head by compass at N., N.E., E., S.E., S., S.W., W., and N.W. being 6 E., 5 E., 2 W., 13 W., 15 W., 3 W., 13 E., 8 E., find the coefficients A, B, C, D, E, and construct a table of deviations for every alternate point of the compass, starting from North. 33. With the wind at S.E., the ship heads N.E. by E. \ E., having a heel of 14, and heeling error 21. Find what the error will be when heeling 15 with her head S.S.W., wind in same direction. PAPER III. 1. Multiply 0-0765 by '000006201, by logs. 2. Divide 0*00652 by 0'9876, by logs. 3. Hour Courses. Knts. 10th. Winds. Lee- way. Devia- tion. Remarks. 1 2 3 4 W. 7 7 7 7 N.N.W. Pts. 10 W. Departure taken from a point in lat. 60 45' N., long. 4 43' E. ; bearing by com- pass,S.W.; ship's head, W. ; devi- ation as per log, distant 15 miles. Variation, 17 W. 5 6 7 8 9 10 11 12 N.W. 7 7 7 7 5 5 5 5 N.N.E. J 20 W. W. by N. J N. 9 9 9 9 S.W. by S. I 13 W. 1 2 3 4 5 6 7 8 w.s.w. 10 10 10 10 S. 2 7W. A current set N.N.E. correct magnetic all day at \ mile per hour. s.w. 10 10 10 10 5 5 5 5 W.N.W. \ 9 10 11 12 W. by S. 8 8 8 8 S. by W. u 8W. Find the course and distance made good ; lat. and long, of the ship. 4. August 1st, in long. 124 27' E., the obs. mer. alt. of the Sun's U.L. bearing north was 81 27' 10"; ind. err. -1' 25"; height of eye, 33 ft. Required the latitude. 5. In lat. 54 37' K, the diff. long, made good was 5 47'. Required the distance sailed by parallel sailing. 6. Required the compass course and distance by Mercator's sailing. From A, lat. 50 10' N. Long. 13 25' W. To B, 47 50 N. 62 30 W. 'Deviation, 20 30 E. Variation, 25 W. APPENDIX. 385 7- November 5th. Find the M.T.G. of high water at Bridlington. 8. August 5th, about 5 h 40 ra A.M., in lat. 31 14' S., long. 157 20' E., when the chron. showed correct G.M.T., 7 h ll m 40', the Sun's bearing by compass was N. 75 30' E. Required the true amplitude and error of the compass ; and supposing the variation to be 6 20' E., required the deviation of the compass for the position of the ship's head. 9. August 17th, about 8 h A.M. at ship, in lat. at noon 35 48' N"., long, by account 36 W., the obs. alt. of Sun's L.L. was 30 25' 50"; ind. err. + 2' 14" ; height of eye, 30 ft. ; time by chron. 10 h 39 m 14% which was fast 20 m 12 1 for G.M. noon, on June 9th and on July 19th was fast 19 m 40 1 ; run since sights, S. 25 E. 43 miles. Required the long, at sight, also brought up to noon ; and if the compass bearing was S. 80 30' E., and variation 23 30' W., find the true azimuth and deviation. 10. August 4th, about 10 k 14 m A.M., in lat. 38 27' N., long. 84 20' W., when a chron. supposed 15 m 12 fast of M.T.G. showed 4 h 7 m 47 s , the Sun's bearing by compass S. 52 30' E. Required the true azimuth and error of the compass by the tables ; and supposing the variation to be 7 30' E., required the deviation for the position of the ship's head. 11. August 19th, P.M. at ship, lat. by account, 17 10' S., long. 177 30' E., the alt. of the Sun's L.L. 57 15' 20 /y ; ind. err. + 1' 18" ; height of eye, 18 ft. ; time by chron. 12 h 51 m 39', which had been found to be slow ll h 55 m of A.T.S. ; the course and dist. were S. 54 E., 23' since the error was determined. Required the lat. by reduction to the meridian; and if the run from noon to obs. on the same course was 3 miles, find the lat. at noon. 12. September 7th, about 9 b A.M. at ship. If in lat. by account 48 5' N., long. 149 30' W., chron. showed 6 h 47 m 8 1 G.M.T. ; the alt. of the Sun's ILL. 32 26' 50"; again A.M. on the same day, when the chron. showed 9* 10 ra 9 1 G.M.T. ; the alt. of the Sun's ILL. 47 20' 30" ; ind. err. -3' 13" ; height of eye, 35 ft. ; the ship having made 23 miles on a true N. 49 E. course in the interval between the observations ; required the line of position, the Sun's true azimuth, and the position of the ship when the second alt. was observed by " Sumner's method" by projection on the chart, assuming the lats. 47 30' K and 48 N. 13. Find the A.T.S. of the meridian passage of a Cygni in long. 96 45' E. on September 7th ; also by inspection. 14. What stars not less than magnitude 2 are within 1 hour east of the meridian of a place in lat. 20 30' S., long. 80 40' E., above the horizon and pole at 7 h P.M. M.T.S. on August 7th ? 15. Calculate the mer. alt. of a Pavonis for placing on the sextant ind. cor. -2' 13" ; eye, 27 ft. in lat. 12 40' N. 16. September 26th, in lat. 36 32' S., long. 125 40' W., at l h 35 m A.M. M.T.S. ; the compass bearing of Aldebaran was N. 40 E. Find from the tables the true azimuth ; and if the variation be 9 E. , find the deviation. 17. September 28th, about 4 h A.M. at ship, in lat. 47 42' S.,long. by account, 84 15' E., the obs. alt. of a Hydrse was 17 15' 20"; no ind. cor. ; eye, 29 ft. ; time by chron., 9 h 13 m 25 s , which was slow l h 16 m 12* of G.M. noon on September 2nd, and gaining 4* '6 daily. Find the longitude. 18. August 12th, when the G.M.T. was 12 d ll h 36 ra 31 s , in long. 23 49' W. ; the obs. alt. of Polaris out of the meridian was 42 17' 50" ; ind. cor. 3' 11" ; eye, 28 ft. Find the latitude. 19. August 17th, in long. 33 45' E. ; the obs. meridian alt. of the Moon's L.L. bearing north was 80 37' 40" ; ind. cor. -1' 20" j eye, 35 ft. Find the latitude. 20. September 19th, the obs. meridian alt. of the star Tauri was 53 25' 10" bearing north; ind. err. -3' 17"; height of eye, 24 ft. Required the latitude. 2 C 386 APPENDIX. 21. Ship's head by standard compass. Bearings of distant object by standard compass. Deviation required. Ship's head by standard compass. Bearings of distant object by standard compass. Deviation required. N. N. 41 E. s. N. 33 W. N.E. N. 30 E. s.w. N. 27 W. E. N. 11 E. W. N. S.E. N. 12 W. N.W. N. 28 E. Find correct magnetic bearing ; construct a Napier's curve. With mag. courses : N. by E. \ E. E. | S. S.S.W. N.W. by N. J N. Find comp. courses. Given comp. courses : N. \ E. E.N.E. S.E. by E. J E. S.W. | W. Find mag. courses. With ship's head at E. f S. by compass, the comp. bearings of distant objects were E.S.E. and W. by N. f N. Find magnetic bearings. 22. On October 13th, at 2 h 47 m A.M. M.T.S., being off Fleetwood by reckoning, took a cast of the lead. Required the correction to be applied to the depth obtained by the lead-line, before comparing it with the depth marked on the chart. 23. August 10th, about 8 b 15 m A.M, in lat. 37 14' N., long, by account 14 W., when a chron. estimated 5 m 25 s slow of G.M.T. showed 9 h 13 m ; the following observations were taken for determining long, and chron. error : obs. alt. of Moon's L.L. 29 38' 20" ; ind. cor. + 3' 14" ; height of eye, 30 ft. ; obs. distance between enlightened limbs of Sun and Moon. 109 17' 50" ; ind. cor. 2' 35". 24. August 2nd, about 9 h P.M. in lat. 23 15' N., approx. long. 62 20' W., when a chron. showed l h 25 m 40 s correct M.T.G. ; the obs. alt. of Antares was 35 42' 40" ; and at the same time the obs. alt. of Spica was 14 18' 20" ; ind. cor. 3' 18" ; eye, 35 ft. Find the true bearings of the stars and ship's position, either by "Sumner's method," assuming lats. 15' on each side of lat. by account, or by direct spherics. 25. August 5th, at 2 h 30 m P.M. M.T.S. in lat. 30 27' 45" S., long. 93 55' E. ; the obs. alt. of Sun's U.L. in artificial horizon was 62 50' 50" ; ind. cor. 4' 44" ; eye, 19 ft. If the chron. showed 8 h 19 m 41", required its error on M.T.G. 26. Find the first and last courses and distance on a G.C. arc from lat. 11 51' N., long. 51 16' E., to lat. 52 33' S., long. 169 9' E., position of vertex, and latitudes of intersection of meridians of 60 E., 70 E., 80 E., 90 E., and 160E.with the track. 27. In triangle ABC, A = 90, AC = 1950, C = 1 3'. Find BC. 28. In triangle XYZ, XY = 210, X = 47 48', Y = 67 26'. Find XZ. 29. In spherical triangle, R = 90, QR = 119 8', PR = 168 20'. Find P and PQ. 30. The deviations with ship's head by compass, on N., N.E., E., S.E., S., S.W.,W.,andN.W. are 20 W., 16 W., 5E.,12 E., 12 E., 9E., 2 E., 4 W. Find coefficients A, B, C, D, and E, and compute deviations for points from E. to W. through north. 31. With ship steering N. 35 E., and heeling 14 to port, the heeling error is 15. Find the amount when steering N. 70 W., and heeling 18 to starboard. APPENDIX. 387 PAPER IV. 1. Multiply 300-07 by 0'024G, by logs. 2. Divide 5623'87 by 48960-02, by logs. 3. Hour Courses. luits lUths. Winds. Lee- way. Devia- tion. Remarks. 1 w.s.w. 12 8 N.W. Pts. M 26 E. Departure taken 2 13 1 from a point in 3 12 5 lat. 37 3' N., 4 11 G long. 4 25' E. ; 5 11 bearing by com- G w. 12 9 N.N.W. 2 19.}E. ship's head S.W.; 7 8 13 12 8 deviation 29 E., distant 20 miles. 9 12 3 Variation 21 W. 10 W. N. 12 2 N. 11 17 E. 11 12 7 12 12 1 1 W. by S. 11 9 N.N.W. 3 23 E. 2 11 8 3 11 9 4 11 8 5 11 6 G W. | S. 11 5 N.N.W. 14 20 E. A current set 7 12 9 S. 21 E. true, 42 8 12 8 miles during the 9 12 8 day. 10 W.N.VV. 13 1 N. 1 11 E. 11 13 6 12 13 3 Find the course and distance made good ; lat. and long, of the ship. 4. August 4th, in long. 82 35' E., the observed meridian altitude of the Sun's L.L. zenith south of Sun was 15 32' 30" ; ind. err. -3' 4"; height of eye, 35 feet. Required the latitude. 5. How far due east must a vessel sail on the parallel of 39 27' S. to change her longitude 1 29' 1 6. Required the compass course and distance by Mercator's sailing From A, lat. 42 35' N. To B, 53 22 N. Variation, 4 30 W. Long. 71 18' W. 12 30 W. Deviation, 19 20 E. 7. July 24th. Find the M.T.S. of high water at Limerick. 8. September 8th P.M. at ship, in lat. 23 40' S., long. 65 50' E., when the A.T.S. was 5 h 50 m 20*, the Sun's bearing by compass was W. S. Required the true amplitude, and error of the compass ; and supposing the variation to be 10 30' W., required the deviation of the compass for the position of the ship's head. 9. August 10th, about 8 h 15 m A.M. at ship, in lat. at noon 37 29' K, long, by account 14 30' W., the obs. alt. of the Sun's L.L. was 36 6' 40" ; ind. 3 88 APPENDIX. err. ; height of eye, 30 feet ; time by chron. 9 h 20 m 53', which was slow l m 12' for Gr.M. noon on June 25th, and on July 10th was correct; run between obs. and noon, N.E. by E. 27 miles. Required the long, at sight, also brought up to noon ; also, if the compass bearing was East and variation 21 15' W., find true azimuth and deviation. 10. September 2nd, about 4 h 20 ra P.M. at ship, in lat. 43 16' S., long. 14 5' W. ; when the correct G.M.T. by chron. was 5 h 25 m 10% the Sun's bearing by compass N. 65 15' W. Required the true azimuth and error of the compass by the tables ; and supposing the variation to be 14 W., required the deviation for the position of the ship's head. 11. September 25th P.M. at ship, lat. by account 29 10' S., long. 7 29' E. ; the alt. of Algenib bearing N. was 46 5' 50" ; ind. err. -2' 8" ; height of eye, 19 ft. ; time by watch ll h 25 ra 4 s , which had been found to be fast of M.T.S. 2 m 22'; the diff. long, made eastward was 12'7' after the error was determined. Required the lat. by reduction to the meridian. 12. September 1st, about O h 33 ra P.M. at ship. In lat. by account 49 K, long. 180 E., when the chron. showed 12 h 42 m 57 s , and was fast 8 m 55* of G.M.T. ; the alt. of the Sun's L.L. 48 51' ; again P.M. on the same day, when the chron. showed 4 h 50 m 20 s ; the alt. of the Sun's U.L. 19 30' 20" ; ind. err. +1' 10" ; height of eye, 17 ft. ; the ship having made 6^ knots per hour on a true E.N.E. course in the interval between the observations. Required the line of position, the Sun's true azimuth, and the position of the ship when the second alt. was observed by "Sumner's method" by projection on the chart, assuming the lats. 20' on each side of lat. by account. 13. In long. 127 45' E., find the mean time of transit of Fomalhaufc on Sept. 14th. 14. What stars not less bright than mag. 2 5 are within 1 hour E. and W. of the meridian, above the horizon of a place in lat. 45 N., long. 95 W. on Sept. 8th at 9 h P.M. M.T.S. Give their meridian distances and bearings. 15. Calculate for placing on the sextant the mer. alt. of a Gruis in lat. 30 24' S., ind. cor. + 1' 17" ; eye, 30 ft. 16. August 16th, midnight mean time at ship in lat. 30 20' N., long. 64 20' W., the compass bearing of a Arietis was E. % S. Find true bearing from the "tables ;" and if the variation was 6 W., required the deviation. 17. August 2nd, about 9 h 30 m P.M. in lat. 25 14' N., long. 97 15' W., the obs. alt. of Spica was 12 19' 20"; ind. cor. -4' 12" ; eye, 17 ft. ; when a chron. showed 3 h 55 m 58* '5, and had been found slow 3 m 31" '5 on April 29th of G.M. noon, and fast 2 m 15' on June 23rd. Find the longitude. 18. September 8th, A.M. at ship, about 3 h in long. 157 E., when the A.T.G. was 4 h 36 ra 26, the obs. alt. of Polaris out of the meridian was 29 26' 40"; ind. cor. +6' 4" ; height of eye, 41 ft. Find the latitude. 19. August 15th, in long. 136 15' E., the obs. alt. of the Moon's L.L. bearing north was 46 19' 10" ; ind. cor. +2' 28" ; height of eye, 34 ft. Required the latitude. 20. August 22nd, the observed meridian alt. of the star j8 Canis Majoris was 51 19' 20" bearing north; ind. err. -4' 37"; height of eye, 35 ft. Required the latitude. 21. Ship's head by standard compass. Bearings of distant object by standard compass. Deviation required. Ship's head by standard compass. Bearings of distant object by standard compass. Deviation required. N. N. 32 W. S. N. 20 E. N.E. N. 20 W. s.w. N. 9 E. E. N. 5E. W. N. 15 W. S.E. N. 17 E. N.W. N. 31 W. APPENDIX. 389 Find correct magnetic bearing ; construct a curve. Given mag. courses : S.E. \ E. N.E. by N. W. by K i N. S. 72 W. Find comp. courses. Given comp. courses : E. by N. \ N. S.E. f S. S.S.W. W. 15 N. Find mag. courses. With ship's head at E S.E. by compass, comp. bearings of distant objects were S. 50 W. and S. 35 E. Find magnetic bearings. 22. On September 7th, at 5 h A.M. M.T.S., being off Wicklow by reckoning, took a cast of the lead. Required the correction to be applied to the depth obtained by the lead-line before comparing it with the depth marked on the chart. 23. August llth, at 4 h 6 m A.M. M.T.S., in lat."40 17' N., long, by account 21 39' 45" W., when a chron. which had run down and been set going showed 5 h 30 m , the following observations were taken for finding the long, and error of chron. for G.M.T. : obs. alb. of Moon's L.L., 54 33' 50" ; hid. cor. 2' 2" ; height of eye, 27 ft. ; obs. dist. between Jupiter's centre and the Moon's near limb, 54 16' 10" ; ind. cor. -3' 3". 24. September 28th, about 4 h A.M. at ship, in lat. 47 40' S., long. 84J E. by account ; when a chron. correct for M.T.G. showed 10 h 27 m 40 s ; the obs. alt. of e Orionis was 41 42' 20" ; and at the same time the obs. alt. of Alphard was 17 15' 20" ; eye, 29 ft. Find the true bearings of the stars, and position of ship, either by " Sumner's method " assuming lats. 47 and 47 40' S., or by direct spherics. 25. September 22nd, at 2 h 17 m 27' P.M. at ship, from which a place in lat. 9 12' K, long. 173 44' W., bore N. 33 W., distant 6 miles ; the obs. alt. of Sun's L.L. was 53 29' 20"; ind. cor. -3' 19"; height of eye, 21 ft. If a chron. showed I b 47 m 27*, find its error on G.M.T. 26. Find the initial course and distance on a Great Circle, and position of vertex in sailing from lat. 34 36' S., long. 58 22V W., to lat. 33 56' S., long. 18 29' E. ; and lat. of place where meridian of Greenwich cuts the track. 27. AB is a horizontal line 1300 ft. long. ; a vertical line is drawn from B upwards, and in it two points P and Q are taken such that BQ = 3 BP, BAP = 10 30' 20". Find BP and BAQ. 28. XY is 2000 ft. long ; Y is due east of X ; at Y a point Z bears N. 46 20' W., at X it bears N. 8 45' E. Find XZ. 29. Two places, A and B, are on the same parallel 35 N., and their difference of longitude is 85. Find the distance between them on the G.C. , also lat. of vertex. 30. From a rock 7 ft. above high water, the angle of elevation to the summit of a cliff distant 5320 ft. in a horizontal direction is 1 30' ; height of eye, 5 ft. Find height of cliff above high water. 31. September 12th, in lat. 25 17' S., long. 100 14' E. at 8 h 30 m P.M. M.T.S. , the compass bearing of 7 Crucis was S. 28 30' W. Find the deviation, the variation being 8 15' W. 32. Coefficient B = -18 30', coefficient C = +27 48'. Find direction ship's head. Ship's head S. 74 W., coefficient B = +5'7. Find coefficient C. 33. Ship's head by compass K, N.E., E., S.E., S., S.W., W., N.W. Deviations 10 W., 15 W., 14 W., 8 E., 20 E., 18 E., 3 W., 5 W. Find coefficients A, B, C, D and E, and thence compute a table of deviations from E. to W. through S. 34. When steering S. 28 E., and heeling to port 15, the error was -10. Find its amount when steering west and heeling 12 to starboard. 390 APPENDIX. PAPER V. 1. Multiply 1-0009 by 627014, by logs. 2. Divide 7 '5 by 1000, by logs. 3. lire. Courses. Knts. lOths. Winds. Lee- way. Devia- tion. Remarks. 1 N.E. by E. 10 N. by W. J'ts. \ 1GE. Departure taken 2 10 from a point in 3 10 lat. 62 30' N., 4 10 long. 178 54' E.; 5 E. by N. 10 5 S.S.E. n 18 E. bearing by com- 6 10 5 pass W.N.W. ; 7 10 5 ship's head 8 10 5 N.N.E.; devia- 9 N.E. 11 E.S.E. I 10 E. tion 5 E., dis- 10 11 tant 20 miles. 11 11 Variation, 15 E. 12 11 1 E.N.E. 11 5 S.E. i 17 E. 2 11 5 3 11 5 4 11 5 5 E. 12 S. by E. 2i 20 E. A current set 6 12 E.N.E. correct 7 12 magnetic, 2^ 8 12 knots per hour 9 N.E. by E. \ E. 11 5 S.E. by E. i 16 E. for all the day. 10 11 5 11 11 5 12 11 5 Find the course and distance made good ; lat. and long, of the ship. 4. August 6th, in long. 19 IT W. ; the obs. mer. alt. of the Sun's U.L., 78 16' 50", zenith south of object ; ind. err., -5' 11" ; height of eye, 37 ft. Required the latitude. 5. In lat. 45 50' N. What is the departure made good to cause a change of 800' of long, by sailing due east ? 6. Required the compass course and distance by Mercator's sailing ; var. 30 W. j dev. 10 E. From A, lat. 33 24' S. To B, 53 40 S. long. 20 15' E. 63 37 W 7. August 15th. Find the M.T.S. of high water at Seal Cove. N.B. 8. August 15fch, about 5 h P.M. at ship, in lat. 44 29' S., long. 129 35' E., when a chron. known to be 15 m 50 slow of M.T.G. showed 7 h 56 ra 50% the Sun's bearing by compass was S. 88 30' W. Required the true amplitude, and error of the compass ; and supposing the variation to be 2 20' W., required the deviation of the compass for the position of the ship's head. 9. September 29th, about 4 h 45 m P.M. at ship, in lat. at noon 46 23' N., long, by account 179 50' W., the obs. alt. of Sun's L.L. was 11 12'; ind. err., nil ; height of eye, 25 ft. ; time by chron., 4 h 43 m 46 s , which was fast 13 m 47* for G.M. noon on August 10th, and losing 3' '3 per day. The run since noon, N. 50 W. 42 miles. Required the long, at sight, also brought up to noon ; and if the compass bearing was W. J N., variation 33 15' E., find the deviation. APPENDIX. 10. August 14th, about 9 h A.M. in lat. 58 49' K, long. 32 56' W., when a chron. correct for G.M.T. showed ll h 22 m 38* ; the sun's bearing by compass, S. 5 30' E. Required the true azimuth and error of the compass by the "Tables ; " and supposing the variation to be 50 W., required the deviation for the position of the ship's head. 11. August 19th P.M. at ship, lat. by account 30 22' S., long. 170 48' E. ; the alt. of the Sun's L.L., 45 47' 50" ; ind. err., + 3' 55" ; height of eye, 19 ft. ; time by chron., l h 16 58', which had been found to be fast 10 m 6 s of M.T.G. ; the run was S. 48 W. 35 miles after the error was determined. Require the lat. by reduction to the meridian ; and if the run since noon was 4 miles on the same course, find the lat. at noon. 12. September 29th, about O h 30 m P.M. at ship, lat. by account 46 55' IS"., long. 179 50' W., when the chron. showed 12 h 20 m 39 s G.M.T. ; the alt. of the Sun's L.L., 40 30' 40" ; again P.M. on the same day, when the chron. showed 4 h 32 m 46* G.M.T. ; the alt. of the Sun's L L., 11 7' ; ind. err., + 5' ; height of eye, 25 ft. ; the ship having made 42 miles on a true N. 45 W. course in the interval between the observations. Required the line of position when the first alt. was taken, the sun's true azimuth and position of the ship when the second alt. was observed by " Sumner's method" by projection on the chart, assuming the lats. 46 and 46 40' N. 13. Find the apparent time of transit of Markab across the meridian of a place long. 121 45' E. on September 21st ; also by inspection. 14. Find what bright stars of at least 2nd mag. are within 1 hour E. and W. of mer., above horizon and pole, on August 14th at l h 30 m A.M. A.T.S. in lat. 29 N., long. 130 E., and state their hour angles. 15. Calculate mer. alt. of a Cephei for placing on sextant in lat. 12 15' N. ; ind. cor., - 4' 26" ; height of eye, 35 ft. 16. August 2nd, about 9 h P.M. at ship, in lat. 23 12' N., long. 62 30' W. ; the obs. alt. of Antares was 35 37' 10" ; ind. cor., + 2' 14" ; eye, 35 ft. ; when a chron. showed l h 13 m 21 s , which was slow 15 m 47* on May 2nd, and slow 14 m 17 s on June llth for G.M. noon. Required the longitude. 17. August 26th, at 3 h 42 m A.M. A.T.S., in lat. 42 18' S., long. 91 14' W. ; the compass bearing of Markab was N.W. J N., the variation being 18 45' E. Find true azimuth from "Tables," and deviation. 18. September 18th, about 5 h 15 m A.M. at ship, in long. 148 15' W., when a chron. which showed G.M.T. indicated 3 h 14 m 12% the obs. alt. of Polaris out of the meridian was 51 7' 20"; ind. cor., 5' 37"; eye, 19 ft. Required the latitude. 19. August 10th, in long. 29 30' W. ; the obs. mer. alt. of the Moon's ILL. was 48 13' 20"; zenith north of Moon; ind. cor., -3' 16"; height of eye, 32 ft. Find latitude. 20. The obs. mer. alt. of the star Argus under the South Pole was 30 47' 50" ; ind. err., + 5' 26" ; height of eye, 12 ft. Required the latitude. 21. Ship's head by standard compass. Bearings of distant object by standard compass. Deviation required. Ship's head by standard compass. Bearings of distant object by standard compass. Deviation required. N. N. 56 E. S. N. 56 E. N.E. N. 48 E. s.w. N. 63 E. E. N. 45 E. W. N. 68 E. S.E. N. 49 E. N.W. N. 64 E. Find correct magnetic bearing ; construct a curve. Given magnetic courses : E. by N. S.E. by E. W.S.W. N.N.W. Find compass courses. 392 APPENDIX. Given compass courses : S. 79 W. S. 23 E. N. 34 E. N. W. Find magnetic courses. With ship's head at E.N.E. by compass, the compass bearings of distant objects were S. by W. \ W., and S. 59 E. Find magnetic bearings. 22. On August 6th, at 6 h P.M. M.T.S., being off Tenby by reckoning, took a cast of the lead. Required the correction to be applied to the depth obtained by the lead-line, before comparing it with the depth marked on the chart. 23. August 17th, about 8 h A.M. at ship, in lat. 36 27' N., long, by account, 36 W., when a chron. estimated 12 m 10 fast on G.M.T. showed 10 h 31 ra 40 s ; the obs. alt. of the Sun's L.L. was 30 25' 50"; ind. cor., +2' 14" ; height of eye, 30 ft. ; obs. distance between the enlightened limbs of Sun and Moon, 30 12' 40"; ind. cor., +3' 16". Find the true and apparent altitudes of the Moon, long, of ship, and error of chron. on G.M.T. 24. September 7th, in lat. N. , long. 149 30' W., the following observations were made for finding latitude by " Ivory " or " Direct," at second obs. : .A pp. times ship nearly. G.M.T. by chron. Obs. alts. Sun's L.L. 8 h 50 m A.M. 6 h 47 m 7' 32 26' 50" 11 15 A.M. 9 10 8 47 20 30 Bearing at first obs., S. 60 \ E. ; run between obs., N. 48| E., 9 '5 knots per hour ; ind. cor., -3' 13" ; eye, 35 ft. 25. September 4th, at 4 b 30 ra P.M., when Cape Farewell in lat. 40 29' 50" S., long. 172 40' 6" E. , bore from ship west 13 miles distant, the obs. alt. of Sun's ILL. in artificial horizon was 24 0' 30" ; ind. cor., 2' 8" ; time shown by chron. , 4 h 37 m . Find its error on M.T.G. 26. Find the initial and final courses from lat. 34 21' S., long. 18 30' E., to lat. 38 18' S., long. 144 39' E., by composite sailing; easting to be run on the parallel of 46 S. Also find total distance, and lats. where meridians of 50 E. and 140 E. cut the track. 27. From a boat the angle of elevation of the truck, known to be 165 ft. above the water-level, was found to be 9 17'. Find distance of observer from ship. 28. A vessel sails S.S.E. \ E., 137 miles by log, but by observation she is known to be 101 miles S. \ W. of her former position. Find the set and drift of the current. 29. In a right-angled spherical triangle, hyp. = 105 20' ; one angle is 35 4'. Find side opposite given angle, and remaining angle. 30. Coefficient B = -3 12 ; coefiicient C = +8 48'. Find direction of ship's head on stocks. With ship's head at building S. 14 W., and coefiicient C= -1 -47, find B. 31. Using deviations found in question 21, find coefficients A, B, C, D, and E ; and construct a table of deviations for quadrants N. to E and S. to W. 32. Steering S. W. by W., heeling 14, wind N.N.W., obs. heeling error, 15. Required the error when course is changed to E. by N., heeling 10. APPENDIX. 393 PAPEK VI. 1. Multiply -00483506 by 100'0706 by logs. 2. Divide 6789 by 78900 by logs. 3. Hrs. Courses. Knts. lOths. Winds. Lee- way. Devia- tion. Remarks. 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 G 7 8 9 10 11 12 N.E. J E. E.N.E. E. by N. \ N. E. by S. J S. E.S.E. E.N.E. 14 14 13 14 14 6 8 9 3 4 N.W. & W. W.N.W. W. S.W. s.s.w. B.S.W. Pts. 1 i 1} 1 I 20 E. 21 E. 22 E. 19 E. 18 E. 14 E. Departure taken from a point in lat. 37 45' S., long. 178 32' E.; bearing by com- pass, E. \ N.; ship's head, N.E. \ E. ; devia- tion as per log, distant 17 miles. Variation, 14 E. A current set N. \ B., correct magnetic, 2^ knots per hour for 24 hours. 14 14 13 9 5 6 13 13 14 14 14 13 14 14 9 8 3 7 6 3 4 14 13 14 13 2 8 1 9 14 14 13 13 2 1 9 8 Find the course and distance made good ; lat. and long, of the ship. 4. September 23rd, in long. 8 45' W., the obs. mer. alt. of the Sun's L.L. bearing north was 47 4' 20" ; ind. err., + 4' 32" ; height of eye, 39 ft. Required the latitude. 5. Along what parallel is a vessel sailing when for a distance sailed due west of 76 miles she alters her long. 1 30' ? 6. Required the compass course and distance by Mercator's sailing From A, lat. 55 13' S. Long. 43 30' E. To B, 10 19 S. 120 12 E. Variation, 36 40 W. Deviation, 14 10 W. 7. July 13th. Find the M.T.G. of high water at Quaco. 394 APPENDIX. 8. September 18th, in lat. 35 18' K, long. 156 17' W., when the A.T.S. (civil time) was 5 h 54 m A.M., the Sun's bearing by compass was E.N.E. Required the true amplitude, and error of the compass ; and supposing the variation to be 14 20' E., required the deviation of the compass for the position of the ship's head. 9. September 7th, about 8 h 50 m A.M. at ship, in lat. at noon 48 37' K, approx. long. 149 30' W., the obs. alt. of Sun's U.L. was 32 35' 20" ; ind. err., - 4' 19" ; height of eye, 35 ft. ; time by chron., 6 h 41 m 30 s , which was fast 3 m 10 s '6 for G.M. noon on July 19th, and on August 8th was slow 19 s '4 ; run since obs. N. 50 E. 42 miles. Required the long, at sight, also brought up to noon; and if the compass bearing was N. 84 30' E., and variation 22 15' E., find true azimuth and deviation. 10. September 12th, about 2 h 15 m P.M., in lat. 35 33' S., long. 43 27' E., when a chron. correct for G.M.T. showed ll h 21 m 21 s , the Sun's bearing by compass, N. 8 30' W. Required the true azimuth and error of the compass by the " Tables ; " and supposing the variation to be 23 30'_W., required the deviation for the position of the ship's head. 11. September 17th, about 5 h 15 ra A.M. at ship, lat. by account 40 35' S., long. 88 16' W., the alt. of Aldebaran bearing north was 32 58' 40"; ind. err., -4' 18" ; height of eye, 37 ft. ; time by watch, 10 h 42 m 24 s , which had been found to be 24 m 7 s slow of M.T.G. ; the diff. long, to the westward 17 miles after the error was determined. Required the lat. by reduction to the meridian. 12. September 23rd, about 9 h 30 m A.M. at ship, in lat. 47 30' K, long. 82 20' E. by D.R., when the chron. showed 3 h 52 m 39 s G.M.T. ; the alt. of the Sun's L.L., 32 13' 10"; again P.M. on the same day, when the chron. showed 8 h 52 m 46 s G.M.T. ; the alt. of the Sun's L.L., 32 13' 10" ; ind. err., 2' 8" ; height of eye, 28 ft. ; the ship having made 21 miles on a true N. 60 E. course in the interval between the observations. Required the line of position when the first alt. was taken, and the Sun's true azimuth ; also the position of the ship when the second alt. was observed by " Sumner's method" by projection on the chart, assuming the lats. 20 miles on each side of D.R. lat. 13. Find the mean time of passage of Arcturus across the meridian of 38 15' W. on September 28th. 14. September 28th. What bright stars not less than 2nd mag. are within 2 hours of the meridian, above the horizon and pole in lat. 30 S., long. 140 W., at 2 h 15 A.M. M.T.S. ? State their mer. distances and bearings. 15. Calculate for placing on the sextant the mer. alt. of 7 Cygni in lat. 24 12' S. ; ind. cor., +3' 27" ; eye, 32 ft. 16. September 16th, at O h 15 m A.M. A.T.S. , in lat. 38 47' N., long. 14 29' W., the compass bearing of Altair was West. Find from the "Tables" the true azimuth; and if the variation be 23 30' W., find the deviation. 17. August 5th, about 9 h 30 m P.M. at ship, in lat. 48 30' N., long, by account 50 W., the obs. alt. of a Cygni, 67 14' 30" ; ind. cor., -3' 13"; eye, 39 ft. ; when a chron. showed ll h 57 m 25 s , which had been found to be 45 m 32 s slow of G.M. noon on June 6th, and 48 m 13 s slow on June 26th. Find the longitude. 18. August 26th, at 9 h 14 m 23 s P.M. M.T.S., in long. 64 45' W., the obs. alt. of the Pole star out of the meridian was 38 29' 20" ; ind. cor., + 2' 25" ; eye, 37 ft. Find the latitude. 19. August 7th, in long. 48 45' W., the obs. mer. alt. of the Moon's L.L. was 32 17' 50" bearing north ; ind. cor., - 5' 5", height of eye, 48 ft. Required the latitude. 20. August 13th, the obs. mer. alt. of the star a Ursss Majoris was 62 15' 30" zenith south of star; ind. err,, - 4' 45" ; height of eye, 36 ft. Required the latitude. APPENDIX. 395 21. Ship's head by standard compass. Bearings of distant object by standard compass. Deviation required. Ship's head by standard compass. Bearings of distant object by standard compass. Deviation required. N. S. 5W. S. S. 10 E. N.E. S. 6E. s.w. S. 7W. E. S. 18 E. w. S. 18 W. S.E. S. 17 E. N.W. S. 21 W. Find correct magnetic bearing, and construct a curve. Given magnetic courses : N. by W. W. | S. S.E. \ E. N. 32 E. Find compass courses. Given compass courses : E. by S. J S. S.W. } S. W. N. N. 8 W. Find magnetic courses. With ship's head at E. by S. by compass, the compass bearings of distant objects were N.W. J W. and S.W. f W. Find magnetic bearings. 22. On January 10th, at 8 h 15 m A.M. M.T.G., being off Beachy Head by reckoning, took a cast of the lead. Required the correction to be applied to the depth obtained by the lead-line, before comparing it with the depth marked on the chart. 23. August 30th, at 8 h 29 30' P.M. at ship, in lat. 10 15' S., long by D.R. 98 14' E., when a chron. which had stopped and been set going showed l b 45 m , the obs. alt. of the Moon's L.L. was 76 38' 40" ; obs. alt. of Antares was 49 33' 50" ; ind. cor. to alts., - 2' 15" ; height of eye, 40 ft. ; obs. distance between the Star and Moon's near limb, 44 44' 30" ; ind. cor., 4- 2' 47". Find longitude and error of chron. on M.T.G. 24. August 19th, in lat. 48 30' S., long. 157 E. by D.R., the following obs. were made for finding lat. at second obs. by "Ivory" or "Direct" method : Mean ship times. 8 h 13 m P.M. 12 33 G.M.T. by chron. Obs. alt. Crucis. 9 h 42 m 40' 44 19' 30" bearing S. 40* W. 2 2 40 23 38 50 Run between obs. N. 20| W., 12 knots per hour; ind. cor., - 2' 12" ; eye, 36 ft. 25. September 4th, at 2 h A.M. A.T.S., Cape Lobos in lat. 18 45' S., long. 70 24' W., bore S. 40 E., distant 17 miles; the obs. alt. of a Gruis in artificial horizon was 79 34' 40" ; ind. cor., + 4' 16". If a chron. showed 6 h 40 m , find its error on M.T.G. 26. Find initial course and distance on a composite track, highest lat. 50 S., and lats. of points where meridians of 120 W. and 180 W. cut the track From lat. 33 3' S., long. 71 38' W. To 41 16 S., 174 53 E. 27. In the triangle ABC, C = 90, AC = 15,866 ft., BC = 13,000 ft. Find the angles, and length of line drawn from B to AC, bisecting angle 28. The elevation of a tower at A is 28 17' ; at B, in the same horizontal plane and 155 ft. nearer the tower, the elevation is 62 37'. Find height of tower. 29. A vessel sails for 18 hours S.E. by S. 95 miles by the log, in a current setting S. by W. W. 2 knots per hour. What is the course and distance made good ? 30. Using deviations from question 21, find coefficients A, B ; C, D, and E, 396 APPENDIX. and construct a table showing deviations for each point from N.N.E. to S.E. through east. 31. The Sun's decimation being 18 30' ST., and his true bearing at rising N.E. by E. |E., find the lat. and hour angle at rising. 32. Construct a Mercator's chart, scale 1'8 in. to 1 long, extending from 59 to 61 S., and 28 30' W. to 31 30' W. A ship sails by compass from lat. 59 40' S., long. 30 30' W., as follows : S.E. by S., 57 miles (deviation 3 W.) ; W., 25 miles (dev. 8 30' E.) ; variation, 10 E. Find position of ship, course, and dist. made good. 33. Steering W. \ N"., on the port tack, heeling 12 ; the error obs. was 1. Find the amount when steering south on the starboard tack, and heeling 15. PAPER VII. 1. Multiply 13-5426 by 51400, by logs. 2. Divide 7560009 by 387'003, by logs. 3. Hrs. Courses. Knts. lOths. Winds. Lee- way. Devia- tion. Remarks. 1 N.W. 10 W.S.W. 1 10 E. Departure taken 2 10 from a point in 3 N.N.W. 11 W. i 15 E. lat. 50 10' S., 4 11 Jong. 110 15' W, 5 w.s.w. 12 N.W. \ 8 W. Variation, 20 E. 6 12 7 N.N.W. \ W. 11 5 W. 15 E. 8 11 5 9 11 5 10 11 5 11 12 12 1 12 12 Position by obser- vation at end of 2 12 run, lat.4610'S., 3 4 12 12 long. 110 30' W. 5 12 6 12 7 12 8 12 9 12 10 12 11 12 12 12 Find the set and drift of the current. 4. September 30th, in long. 124 26' E., the obs. mer. alt. of the Sun's ILL. bearing south was 88 56' 40" ; ind. err., + 1' 44" ; height of eye, 41 ft. Required the latitude. 5. From lat. 51 23' K, long. 9 36' W., the distance sailed due west was 250 miles. Required present position. 6. Required the compass course and distance by Mercator's sailing From A, lat. 38 13' S. Long. 178 43' W. To B, 52 10 N. 165 15 E. Variation, 13 E. Deviation, 5 30 W. APPENDIX. 397 7. September 28th. Find the M.T.S. of H.W. at Port Carlisle. 8. August 25, in lat. 48 30' S., long. 49 35' E., when the M.T.S. was 6 h 50 m A.M., the Sun's bearing by compass was E. by S. \ S. Required the true amplitude and error of the compass ; and supposing the variation to be 33 20' W., required the deviation of the compass for the position of the ship's head. 9. September 24th, about 2 h 45 m P.M. at ship, in lat. 23 28' S., long. 158 30' E., the obs. alt. of Sun's U.L. was 41 32' 20" ; ind. err., - 2' 35" ; height of eye, 34 ft. ; time shown by chron. , 3 h 58 m 53 , which was slow 21 m 41 for G.M. noon on April llth, and on July 20th was slow 18 m 14 ; the compass bearing was W.N.W. Required the long., compass error, and deviation, the variation being 6 30' E. 10. August 24th, about 8" 30 m A.M. in lat. 44 17' S., long. 15 45' W., when a chron. which was 8 ra 43' slow of G.M.T. showed 9 h 29 m 32*; the Sun's bearing by compass N.E. f E. Required the true azimuth and error of the compass by the "Tables;" and supposing the variation to be 12 40' W., required the deviation for the position of the ship's head. 11. September 29th, A.M. at ship, lat. by account 44 15' N., long. 47 42' W. ; the alt. of the Sun's L.L., 43 0' 30" ; ind. err., - 3' 3" ; height of eye, 32 ft. ; time by chron., 2 h 33 m 56% which had been found to be 3 h 2 m 38' fast of A.T.S. ; the run was N. 34 E. 25 miles, after the error was determined. Required the lat. by reduction to the meridian ; and if the run on the same course was 4 miles, find the lat. at noon. 12. August 4th, about 8 h 15 m A.M. at ship, in lat. 47 20' S., long. Green- wich when the chron. showed 8 h 20 m 51* G.M.T. ; the alt. of the Sun's U.L., 9 9' 20" ; again P.M. on the same day, when the chron. showed 2 h 22 m 42* G.M.T. : the alt. of the Sun's U.L., 18 45'; ind. err., - 2' 23"; height of eye, 23 ft. ; the ship having made 48 miles on a true S.W. f W. course in the interval between the observations. Required the line of position, the Sun's true azimuth, and the position of the ship when the second alt. was observed by "Sumner's method" by projection on the chart, assuming the lats. 46 50' and 47 20' S. 13. Find the apparent time of the meridian passage of Castor on August 23rd in long. 49 30' E. ; also by inspection. 14. Calculate the meridian alt. of e Sagittarii for placing on the sextant in lat. 24 30' N. ; ind. cor., - 5' 14"; eye, 40 ft. 15. Find what stars of not less than second mag. are within 2 hours east of the meridian of a place lat. 40 N., long. 150 E., above the horizon and pole on August 21st at 5 h 50 m A.M. M.T.S., and state their meridian distances. 16. August 12th, about ll h P.M. at ship, lat. 54 26' S., long. 160 25' E., when a chron. correct for M.T.G. showed 12 h 48 20% the compass bearing of a Ophiuchi was N.W. \ W. The variation being 17 E., find from the " Tables " the true bearing, and thence the deviation. 17. August 30th, about 8 h 30 m P.M. at ship, in lat. 10 15' S., long. 98i E. by D.R., the obs. alt. of Antares was 49 33' 50" ; ind. cor., - 2' 15".; eye, 40 ft. ; when a chron. showed ll h 34 m 24% which was 2 h 17 m 38* slow on June 21st, and slow 2 h 19 m 44 on July 21st for G.M.T. Find the longitude. 18. September 25th, at ll h 30 m 50' P.M. A.T.S. , in long. 129 19' E., the obs. alt. of Polaris out of the meridian was 21 24' 50" ; ind. cor., + 1' 14" ; eye, 38 ft. Find the latitude. 19. August 30th, in long. 14 55' E., the obs. mer. alt. of the Moon's upper limb zenith north of Moon was 29 46' 40"; ind. cor., + 2' 29"; eye, 61 ft. Find the latitude. 20. August 10th, the obs. mer. alt. of the star Arcturus was 37 50' 20" bearing south ; ind. err., 4' 37" j height of eye, 39 ft. Required the latitude. 39 8< APPENDIX. 21. Ship's head by standard compass. Bearings of distant object by standard compass. Deviation required. Ship's head by standard compass. Bearings of distant object by standard compass. Deviation required. N. N.E. E. S.E. N. 44 W. N. 47 W. N. 80 W. S. 76 W. S. s.w. W. N.W. S. 79 W. S. 89 W. N. 69 W. N. 52 W. Find correct magnetic bearing, and construct a curve. Given magnetic courses : N. 30 W. S. 55 E. S. 66 W. N. 35 E. Find compass courses. Given compass courses : E. } S. S.S.W. J W. W. \ N. N.W. by W. Find magnetic courses. With ship's head at N. \ E. by compass, the compass bearings of distant objects were S. 55 W. and N. 32 W. Find magnetic bearings. 22. On September 21st, at l h P.M. M.T.S., being off Torbay by reckoning, took a cast of the lead. Required the correction to be applied to the depth obtained by the lead-line, before comparing it with the depth marked on the chart. 23. August 12th, A.M. at ship, in lat. 25 19' S., when the time by a chron. estimated 10 m 30 fast of M.T.G. was ll d 5 h 50 m 50 s ; the obs. alt. of Saturn was 41 1' ; ind. cor., + 2' 14" ; height of eye, 29 ft. ; observed distance between Saturn's centre and the Moon's remote limb was 61 2' 10" ; ind. cor., 3' 15". Find the true and apparent alts, of the moon, longitude, and error of chron. for G.M.T. 24. September 29th, P.M. at ship, in lat. N., long. 180 W., with the following observations, find the lat. at second obs. by " Ivory" or " Direct " method : Approx. A.T.S. O h 30 m P.M. 4 42 P.M. M.T.G. by chron. O h 20 ra 39' 4 32 46 Obs. alt. Sun's L.L. 40 35' 40" No ind. cor. 11 12 Height of eye, 25ft. Sun's bearing at first obs., S. 10 W. ; run between obs., N. 45 W. 42 miles. 25. September 28th, at l h 30 m A.M., M.T.S. when Cape San Roque, lat. 5 29' S., long. 35 15' W., bore N. 64 W. distant 16 miles ; the obs. alt. of Capella in artificial horizon was 45 20' 30"; ind. cor., +1' 16". If a chron. showed 4 h 10 ra 5% find its error on M.T.G. 26. Find the first and last courses on a great circle, from Tarifa, lat. 36 0' N., long. 5 37' W., to Monte Video, lat. 34 57' S., long. 55 55' W., position of vertex, distance, the course to steer crossing the equator, and the long, to cross the equator. 27. The summit of a spire is vertically over the middle point of a horizontal square enclosure whose side is 1000 ft. long ; the shadow of the spire reaches a corner of the square when the sun has an altitude of 27 29' 48". Required the height of the summit of the spire above the level of the square. 28. An island bears from a cape S.E. \ E. 35 miles ; a ship bound to the island meets a tide running to the eastward, and finds herself after some time 28 miles from the island, and 40 miles from the cape. Required the bearing of the island and set of the tide. 29. Calculate angle A and side AC of a spherical triangle, having given C = 90, AB = 74 49', and BC = 30. 30. Taking the deviations as found in question 21, find coefficients A, B, C, D, and E ; and thence calculate a table of deviations for each point from W. to E. through south. APPENDIX. 399 31. Steering N.W. \ W., heeling 10 to port, the heeling error was found to be 15. What will be the amount when steering E. \ N., and heeling 20 to starboard ? 32. August 26th, about 5 h A.M., in lat. 5 30' S., long. 40 30' W., the obs. compass bearing of a Persei was North, the variation being 6 15' W. Find the deviation, when a chron. correct for G.M.T. showed 7 h 45 m . PAPER VIII. 1. Multiply -0043 by "00025, by logs. 2. Divide 9999 '99 by 100,000, by logs. 3. Hrs. Courses. Knts. 10th. Winds. Lee- way. Devia- tion. Remarks. pt9. 1 W.N.W. 12 N. }E. \ 25 W. Departure taken 2 12 from a point in 3 12 lat. 59 49' N., 4 12 long. 43 54' W.; 5 13 bearing by com- 6 13 pass, E. by N. ; 7 W. 13 N.N.W. i 30 W. ship's head, 8 13 N.W. J N. ; de- 9 14 viation as per 10 14 log, distant 30 11 14 miles. 12 14 1 N.W. by W. } W. 14 s.w. \ s. 1 21 W. Variation: first 8 2 14 hours, 51 W. ; 3 12 second 8 hours, 4 12 49 W. ; third 8 5 15 hours, 47 W. 6 15 7 15 8 15 A current set true 9 15 S. 65 W., 10 10 N.W. \ N. 15 N.E. by N. \ 20 W. miles for the 24 11 15 hours. 12 15 Find the course and distance made good ; lat. and long, of the ship. 4. September 24th, in long. 172 15' E., the obs. mer. alt. of the Sun's L.L. zenith south of Sun was 56 42' 10" ; ind. err., - 2' 23".; height of eye, 43 ft. Required the latitude. 5. At noon the lat. is 51 20' S. ; by morning sights the long, was found to be 85 40' W. ; run since sights, East 76 miles. Find noon longitude. 6. Required the compass course and distance by Mercator's sailing From A, lat. 35 7' N. To B, 54 53 S. Variation, 1 20 E. Long. 141 6' E. 75 35 W. Deviation, 18 40 W. 400 APPENDIX. 7. September 13th. Find the M.T.S. of high water at Dartmouth. 8. September 28th, about 6 h P.M., in lat. 50 5' N., long. 175 30' W., when a chron. correct for M.T.G. showed 5 h 43 m 24', the Sun's bearing by compass was N. 84 30' W. Required the true amplitude, and error of the compass ; and supposing the variation to be 18 20' E., required the deviation of the compass for the position of the ship's head. 9. August 4th, about 8 h 20 m A.M. at ship, in lat. 47 24' S., long. Greenwich, the obs. alt. of Sun's U.L. was 8 16' 30' ; ind. err., + 3' 12" ; height of eye, 28 ft. ; time by chron., 8 h 18 ra 59 s , which was 3 m fast for G.M. noon on May 31st, and on July 10th was correct. Find the long. ; also, if the compass bearing was N. 50 30' E. and variation 10 30' W., find the deviation. 10. September 22nd, about 5 h P.M., in lat. 57 11' N., long. 169 28' E., when a chron. correct for M.T.G. showed 5 h 42 m 23*, the Sun's bearing by compass, S. 60 W. Required the true azimuth and error of the compass by the "Tables ;" and supposing the variation to be 10 40' E., required the deviation for the position of the ship's head. 11. September 21st, about 10 h 20 m P.M. at ship ; lat. by account 43 30' N., long. 146 27' E. ; the alt. of Fomalhaut, 16 5' 40" ; ind. err., - 4' 11" ; height of eye, 27 ft.; time shown by chron., 12 h 12 m 54 , which had been found to be 16 m 44* slow of M.T.G. Required the latitude by reduction to the meridian. 12. October 1st, about ll b 30 m A.M. at ship, in lat. 49 20' N., long. 179 50' E. by D.R., when the chron. showed ll h 21 m 24* G.M.T., the alt. of the Sun's L.L., 37 30' 50"; again P.M. on the same day, when the chron. showed 3 h 4 m 26' G.M.T. ; the alt. of the Sun's L.L. 23 6' 30" ; ind. err., 2' 9" ; height of eye, 32 ft. ; the ship having made 50 miles on a true S. 81 W. course in the interval between the observations. Required the line of position when the first alt. was taken, the Sun's true azimuth, and the position of the ship when the second alt. was observed by " Sumner's method" by projection on the chart, assuming the lats. 48 40' N. and 49 20' N. 13. Find the mean time of transit of Sirius across the meridian of 6215'W. on August 30th. 14. Find what bright stars up to mag. 2 '5 are within 1 hour E. and W. of the meridian, above the horizon and pole, in lat. 35 S., long. 160 W., at 4 h 30* A.M. A.T.S. on September 23rd. State their hour angles, and bearings when on meridian. 15. Calculate the mer. alt. for placing on sextant of a Ophiuchi in lat. 55 30' N. ; ind. cor., 5' 14" on the arc ; eye, 40 ft. 16. September 14th, about 9 h P.M., in lat. 35 37' 1ST., long. 74 20' W., when a chron. showed 2 h 12 m 10* G.M.T. , the compass bearing of y Pegasi was E. by ET. Find from the " Tables " the true azimuth ; and if the variation was 5 15' W. , find the deviation. 17. August 20th, about O h 30 m A.M., in lat. 48 20' S., long. 157 E. by D.R., the obs. alt. of a 1 Crucis W. of mer. was 23 38' 50" ; ind. cor., 2' 8"; eye, 36 ft. ; time shown by chron., 2 h ll m 9', which had been found fast 7 m 29' of G.M. noon on June 5th, and fast 8 m 13* on July 30th. Find the longitude. 18. August 30th at 2 h 47 m 23' A.M. M.T.S., in long. 165 15' W., the obs. alt. of Polaris out of the meridian was 30 2' 10" ; ind. cor., - 2' 16" ; eye, 25 ft. Required the latitude. 19. September 1st, in long. 174 30' E., the obs. mer. alt. of the Moon's L.L. was 81 11' 24", zenith north of Moon; ind. cor., - 4' 15" ; eye, 27 ft. Required the latitude. 20. September 20th. The obs. mer. alt. of the star \ Scorpii was 41 13' 10" bearing south; ind. err., -1' 49"; height of eye, 25 ft Required the latitude. APPENDIX. 401 21. Ship's head by standard compass. Benrings of distant object by standard compass. Deviation required. Ship's head by standard compass. Beat ings of distant object by standard compass. Deviation required. N. N. s. N. 21 W. N.E. N. 7 E. s.w. N. 17 W. E. N. 10 \V. W. N. 18 E. 8.E. N. 13 W. N.W. N. 16 E. Find correct magnetic bearing ; construct a curve. Given magnetic courses : N. J E. S.E. by E. E. Find compass courses. Given compass courses : E. Find magnetic courses. With ship's head at E. W.S.W. W.12N. N.W. } W. S. 65 W. S. 32 E. objects were S. by W. Obs. alts, of a Cygni. 67 21' 40" Ind. cor., + 2' 12" 68 26 40 Eye, 39 ft. S. by compass, the compass bearings of distant W., and S.E. by E. Find magnetic bearings. 22. On September 5th, at O h 30 ra A.M. M.T.G., being off Fowey by reckoning, took a cast of the lead. Required the correction to be applied to the depth obtained by the lead-line before comparing it with the depth marked on the chart. 23. August 15th, about l h 25 m P.M. at ship, in lat. 23 27' 1ST., long, by account, 170 30' W., when a chron. estimated 7 m slow of G.M.T. showed 12 h 40 m , the alt. of the Sun's L.L. was 68 37' 10"; ind. cor., +4' 9"; height of eye, 28 ft. ; obs. distance between the enlightened limbs of the Sun and Moon, 47 22' 40" ; ind. cor., 3' 17". Find the true and apparent alts, of the Moon, longitude, and error of chron. on G.M.T. 24. August 5th, in lat. 48 N., long. 50 W., with following observations, find lat. at second obs. by " Ivory " or " Direct " method : M.T.S. nearly. Time shown by chron. 5th 9 h 30 m P.M. 12 h 46 m 4' 6th 1 48 A.M. 5 3 59 The chron. was 5 ro slow of M.T.G. ; bearing at first obs., 1ST. 86 E. ; run between obs., S. 13 W., 10 knots per hour. 25. September 9th, at 9 h 15 m A.M. A.T.S., when Cape Comino lat 40 31' 30" N., long. 9 50' 30" E., bore N. 35 E. dist. 14 miles, the obs. alt. of Sun's L.L. in artificial horizon was 77 51' 10" ; ind. cor., + 3' 18" ; time shown by chron., 8 h 43 m . Find its error on M.T.G. 26. Find the initial course and distance on a composite track, easting run on parallel of 42 S., from lat. 33 56' S., long. 18 29' E., to lat. 35 2' S., long. 117 54' E. ; and lats. of points in which meridians of 50 E., 90 E and 100 E. cut the track. 27. If the diff. lat. is 224'6 N., and dep. 157*9 miles E., find course and distance by calculation. 28. In triangle PQR, PQ = 137, PR = 257, R = 30 40'. Find QR. 29. The angular elevation of a tower at A is 30, at B in the same horizontal plane, and 100 ft. nearer the tower, the elevation is 60. Find height of the tower. 30. In spherical triangle PQR, P = 90, Q = 68 30', PQ = 101 20' Find R and RQ. 31. In spheric! triangle ABC. C = 90, AC = 43 45', BC = 81 27' Find A and AB. 32. Construct a Mercator chart, scale 1"'2 = 1 long., extending from 67 to 70 N., and 2 W. to 2 E. A ship sails from 69 30' N., 30' E. the following true courses : S.W. by S. 30' ; E. by S. 40'; S.S.W. 56' ; W. by N. 25'. Find her latitude and longitude in. 2 D 402 APPENDIX. 33. Given coefficient B = 15 7, coefficient C = 14 '5. Find direction of ship's head at building. 34. Using deviations found in question 21, find coefficients A, B, C, D, and E ; and construct a table of deviations for every alternate point of the compass, starting at N. 35. Steering N.W. \ N., heeling 16 on the port tack, the error was observed to be 14 '5. Find its amount when steering east on the starboard tack and heeling 20. 36. August 17th, at ll h 35 m 20* P.M. M.T.S., in lat. 5 24' tf., long. 68 30' E., the compass bearing of ft Gruis was S.S.E. Find the true bearing; and if the variation is 1 15' W., find the deviation. ANSWERS. CHAPTER III. LOGARITHMS. 1. 0-477121 1-672098 1-698970 2-000000 2-439333 3-922206 1-758079 2-563173 0-000389 4-876391 1-797278 2-243058 4-361779 2-215643 4-333447 4-370328 1598 139-9 50-15 5-199 1-010 48026-9 490000 1408961 10969874 NUMBERS. MULTIPLICATION BY LOGS. 1. 19639 2. 23360 3. 279088 4. 358900 5. 709-3934 6. 47-808 7. 2110-41 8. 288-651 9. 3540707-3 10. 5-9585 11. -0000081005 12. 6-4831 0-532627 0-875061 2-213385 2-847579 4-513697 9-698970 8-690196 7-556302 7-080626 7-827886 6-447794 5-876795 8-539087 8-878545 9-564667 149899-7 3400015-6 10010023-1 230100 11000 1121000 1-001 4679-01 18880043-5 239901-1 13. -0000129 14. -000259 15. -5 16. 24951-6 17. 100000 18. 20-42 19. -2303 20. -001795 21. 1267-618 22. 229500 23. 18-16 24. 143-21 404 ANSWERS. DIVISION BY LOGS. 1. 46-3572 2. 293-6 3. 9040 4. 176020-7 5. 4-613 6. 24-7069 7. 1328856-7 8. 1026-903 9. 1-424 10. 142199-67 11. 1-02217 12. -00001029 1490996-5 44826082-5 000001 20 1-2187 5-8916 57-454 20967-1 21. 20-399 22. -032725 23. 7-9876 24. 957-94 13. 14. 15. 16. 17. 18. 19. 20. INVOLUTION BY LOGS. 1. 547-56 2. 6-602 3. -000005797 4. -000000001411 5. -000002191 EVOLUTION BY LOGS. 1. 17-5 2. 3-6 3. -30012 4. 230 5. -08 ADDITIONAL EXERCISES. 1. 143-8, 2774, 360-8 2. 891-9 and 9754-5 3. 172974-6, -03528 4. 1-439 CHAPTER IV. RIGHT-ANGLED PLANE TRIANGLES. 1 A = 44 47' 10" ; B = 45 12' 50" ; BC = 354-3. 2. P = 28 48' 30" ; R = 61 11' 30" ; PR = 423'4. 3. B = 10 25' 30" ; P = 79 34' 30" ; OB = 683 -5. 4. AC = 443*4 ; BC = 937 '4. 5. AB = 1095; AC = 617. 6. A = 52 6' 11" ; B = 37 53' 49" ; AC = 90 '3. 7. PR = 5227 ; QR = 2864. 8. A = 54 15' 30" ; B = 35 44' 30" ; AB = 1208-5. 9. Height, 91*55 ft. 10. Course, S. 42 22' W. ; dist. 320 '1 miles. 11. Height, 15 ft. 7 in. 12. First dist. 9 '08 miles; second dist. 5 '75 miles. 13. 6 15.V. 14. 113'8'ft. ANSWERS. 405 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. OBLIQUE-ANGLED PLANE TRIANGLES. A = 47 7' 8" ; B = 14 25' 32" ; C = 118 27' 20". A = 42 16' 41" ; B = 23 38' 38" ; C = 114 4' 41". A = 83 25' 54" ; B = 55 52' 48" ; C - 40 41' 18". = 67 41' 0" ; 6 = 164 ; c = 157'5. A = 64 27' 0" ; a = 10787 ; c = 1172-8. B = 32 10' 0" ; a = 20'62 ; 6 = 11 '08. A = 96 16' 25" ; B = 44 3' 35" ; c = 4466. A = 45 31' 53" ; C = 65 42' 7" ; 6 = 3487. B = 46 50' 2" ; C = 35 51' 58" ; a = 82-95. B = 42 50' 18" ; C = 84 43' 42" ; c = 1086. A = 19 4' 29" ; B = 121 51' 31" ; a = 180'5 ; or 82 47' 31" ; or 58 8' 29" ; or 547 "8. B = 90 30' 11" ; C = 39 53' 49" ; b = 1398 '4. B = 35 4' 5" ; = 123 32' 55" ; c = 1342 ; or 144 55' 55" ; or 13 41' 5" ; or 380'9. A = 48 11' 22" ; B = 58 24' 42" ; C = 73 23' 56". A = 27 7' 36" ; B = 22 19' 54" ; C = 130 32' 30". A = 44 24' 54"; B = 57 7' 20" ; C = 78 27' 46". a = 6291 ; c = 4044. a = 206-3 ; & = 1847. = 33 30' 59"; a = 201 -2. C = 731'61"; 6 = 2015. ; B = 16 13' 31" ; c = 1591-8. B = 62 31' 25" ; C = 102 17' 35" ; c = 5919; or 117 28' 35" ; or 47 20' 25" ; or 4455. B = 59 57' 53" ; C = 84 47' 7" ; c = 3451 ; or 120 2' 7" ; or 24 42' 53" ; or 144-9. 212 ft. 134-7 ft. MISCELLANEOUS PROBLEMS; G = 11 40' 0" B = 61 25' 0" B = 88 30' 1" A = 48 41' 9" A = 46 35' 31' 1. 56 18' 36" and 33 41' 24". 2. 610-5 yards. 3. N. 36 12' 15" W. 4. 71-36 ft. 5- Height 74 ft. Width 477 ft. 6. 367-9 ft. 7- 4 67V. 8. 16287 ft. 9. S. 53 11' 24" W. 10. S. 80 W. 11. 115 ft. 12. S. 52", 1'36"E. 13. 578-5 ft., 818-5 ft., 54 39' 28' 14. 64 26' 30". 29-66 yards. CHAPTER V. RIGHT-ANGLED SPHERICAL TRIANGLES. 1. A = 54 15' 52" ; B = 70 5' 36" ; AB = 74 53' 52". 2. BO = 71 20' 56" ; AC = 38 18' 16" ; AB = 75 27' 57". 3. A = 62 17' ; B = 79 8' 34" ; AC = 77 42' 53". 4. B = 77 2' 15" ; BC = 50 17' 53" ; AC = 73 20' 39". 5. A = 45 40' 14" ; AC = 39 6' 20" ; AB = 49 18' 54". 6. B = 139 26' 36", or 40 33' 24" ; AC = 145 5' 45", or 34 54' 15" ; AB = 118 21' 6", or 61 38' 54". 7. C = 70 55' 17" ; A = 101 55' 44" ; AC = 94 11' 25". 8. C = 110 3' 55" ; A = 98 13' 51" ; AC = 86 58' 17". 9. AB = 117 2' 58" ; BC = 65 5' 37" ; AC = 101 2' 29". 10. AB = 92 52' 13" ; BC = 118 45' 49" ; AC = 88 37' 9" 11. C = 78 10' 28" ; A = 144 52' 34" : BC = 146 41'. 406 ANSWERS. 12. C = 96 57' 4" ; A = 101 19' 54" ; AB = 97 5' 24". 13. B = 55 2' 52" ; C = 96 0' 20" ; AC = 54 49' 36". 14. B = 144 52' 47" ; AC = 148 4' 7" ; AB = 62 23' 2". 15. B = 129 14' 9" ; AC = 129 43' 36" ; AB = 100 40' 35". .16. C = 95 41' 50"; AC = 76 8' 49" ; BC = 91 24' 9". 17. B = 106 24' 22" ; AB = 94 18' 33" ; BC = 88 47'. 18. B = 49 37', or 130 23' ; AC = 46 57' 45", or 123 2' 15" BC = 106 21' 15", or 73 38' 45". OBLIQUE-ANGLED SPHERICAL TRIANGLES. 1. A = 82 42' 30" ; B = 90 9' 16" ; C = 104 19' 36". 2. A = 153 7' 14" ; B = 41 22' 24" ; C = 43 50' 32". 3. A = 116 19' 7"; B = 104 58' 43" ; AB = 137 29' 6". 4. BC = 91 29' 30". 5. A = 119 15' ; C = 73 59'. 6. A = 91 30' 10". 7. 24 20' 38". 8. 249 51' 15". 29 37' 12" N. 10. 41 0' 20" N. 42 36' 50" W. CHAPTER VI. PARALLEL SAILING 1. 174' -8. 2. 116' -7. 3. 373'-!. 4. 70 31' 44". 5. 12'-01. 6. 57 52' K 41 52' S. Diff. lat. 5984 miles. 7. 34 44' W. 8. 45 20' N. Dep. 210'9. 9. 13-09. 10. 38 8' S. 11. 1266. 12. 60. 13. 136 49' 24" E. 14. 44 21' 18" E. 15. 96-24. 16. 50 44'. 17. 92-99. 18. 41 241'. DAYS' WORKS. 1. N. 31 W., 49'; N. 34 W., 61': N. 17 E., 63'; N. 60 W., 55'; N. 15 W., 57' ; S. 27 W. , 62' ; S. 25 E. , 57' ; S. 30 E. , 4' ; D. lat. 125' -0, dep. 105' -3 ; lat. in 1 31' N. ; long, in 2 2' W. Course, N. 40 W., dist. 163'. 2. N. 37 W., 7' ; S. 81 W., 99' -6 ; S. 79 W., 49' -4 ; S. 80 W., 47''2 ; S. 78W., 96'-8; S. 56 E., 18'; S. 58E.,18'; S. 60E.,18'; D. lat. 76'-4, dep. 246'-5. Course, S. 73 W. ; dist. 258; lat. in 53 57''6 N. ; long, in 15 25' -5 W. ANSWERS. 407 3. N. 41 W., 19' ; N. 38 E., 37' ; S. 87 E., 25' ; S. 7S E., 23' ; S. 88 W. , 21' ; S. 9 W., 34' ; N. 62 E. , 40' ; S. 22 E. , 15' ; D. lat. 8'-0, dep. 72' -4. Course, N. 83.V 3 E., 73' ; lat. in 18 51' S. ; long, in 178 36' W. 4. S. 14 E., 60' ; S. 31 W., 53' ; N. 82 W., 51' ; S. 64 W., 58' ; N. 44 E. , 49' ; N. 26 E. , 65' ; S. 84 W. , 21' ; N. 75 W. , 39' ; D. lat. 20' '4, dep. Ill' '5. Course, S. 80 W. , 113' ; lat. in 58 50' S. ; long, in 68 28' W. 5. S. 23 W., 15'; N. 49 W., 12'; S. 37 E., 13'; S. 16 W., 25'; N. 34 E. , 20' ; N. 78 W. , 15' ; N. 15 E., 27' ; D.R. lat. 30 4' "5 N. ; long. 132 41| E. Set N. 46 E. ; drift 30'. 6. S. 33 E.,8'; S. 6 W., 49' ; N. 24 E., 53' ; S. 65 E., 49'; S. 30 E., 44' ; S. 25 W. , 54' ; S. 66 E., 52' ; D.R. lat. 52 6'1 N. ; long. 178 26'-9 W. Set S. 62 E. ; drift 15'. MERCATOR SAILING. No. True course. Compass course. Distance. 1 K o > n 1 S. 855541 W N. 54 19 19 W. 2141 miles. 2 N. 3 8 15 E. N. 16 21 15 E. 672 3 S. 12 5 34 W. S. 2 534W. 5673 4 N. 19 27 W. N.3357 OW. 900-4 5 - S. 71 5 33 E. S. 48 5 33 E. 3009 6 S. 60 27 26 E. or W. /S. 49 27 36 E. or \ \S. 71 27 26 W. j 10951-7 7 S. 54 6 30 W. S. 31 6 30 W. 6261 8 N.5343 OW. N.5023 OW. 5428 ;; 9 N. 89 41 19 W. N.44 019W. 1292 10 N. 2 5 OE. N. 1 5 OW. 3810 11 N.7156 OW. N.5352 OW. 909-3 12 N. 89 16 43 E. or W. /S. 833317 E.or\ \N.82 643W. / 10801 MISCELLANEOUS EXERCISES. 1. 244-7 miles. 2 39 16" 3! Lat. 35 20' N., long. 30 17''5 W. 4. 36-2 hours. 5. N. 7 54' W., 44-8 miles. 6. 6 h 33. 7. Cliff, 131-6 ft. ; lighthouse, 107'2 ft. 8. 315-3 ft. 9. S. 60 9' W., 13-36 miles. 10. 116-82 miles. 11. N. 56 13' E., 116-9 miles. 12. S. 7 50' W., 5 "6 miles. 13. 6 h 45 m . 14. 12-34 miles. 15. S. 63 20' E. ; 7'38 miles. 16. N. 71 15' E. ; 7-27 miles. 408 ANSWERS. CHAPTER VII. TIDES. No. A.M. P.M. h. m. h. m. 1 6 48 7 15 M.T.G. 10 31 10 58 13 49 11 53 11 35 11 57 ., 2 4 42 5 18 11 8 11 47 1 8 2 20 Midnight 32 11 45 4 5 5 11 1 2 M.T.S. 36 28 5> 46 50 23 54 20 3 9 4 1 1 1 3!> 28 6 9 5 1 1 32 M.T.G. 10 M.T.S. 49 M.T.G. 27 M T.S. 4 11 1? 31 1 38 11 {J 54 M.T.G. 18 55 CHAPTER VIII. Interval. Angle. h. m. 1 5 48 R. 174 2 1 43 F. 48 3 5 24 R. 162 4 2 48 F. 86 5 4 53 F. 117 6 7 11 F. 175 M.T.L. H.D.R. ft. ius. ft. ins. 1 10 2 11 2 5 11 6 6 '3 8 1 6 10 4 11 2R. 5 8 11 OF. 5 10 5 12 5 R. 12 6 12 2 F. 12 9 6 10 11 R. 9 6 11 2F. 9 3 7 8 7 R. 3 9 8 5F. 3 11 8 11 R. 11 ANSWERS. 409 No. Angle. H.D.R. Red. to M.L.W.S. ft. ins. ft. ins. 1 61 10 4 15 6 2 154 3 3 - 5 3 3 144 9 10 - 1 6 4 168 10 4 + 1 2 5 166 3 11 - 8 2 6 29 5 7 14 3 7 47 9 3 -17 4 8 105 11 - 1 10 9 31 7 6 -13 10 10 176 1 5 4 1 11 158 5 8 - 2 8 12 145 5 9 - 7 4 13 180 14 3 + 1 11 14 170 7 7 + 05 15 151 13 8 - 2 11 16 11 2 9 - 7 10 17 176 12 5 - 2 3 18 45 4 6 7 8 19 120 2 3 - 4 7 20. 26 ft. 10 ins. 21. 5 h 8 m p.m. 22. 22 ft. 8 ins. 23. depth 34 ft. 9 ins. ; under 9 ft. 9 ins. 24. 7 h 33 m a.m. CHAPTER IX. TIME EQUIVALENTS. 11 58'; ll h 56 3'; l h O m 1'; 4 h 31 m 31-6; 1 9"3; 4T- 18' 6; 10" 20 m 46"4; 6 h 12 m 3 s ; 7 h 50 m 34'; 9 h 39 m 36'; ll h 46 ra 38"3; 13' '53; 3 m 59' ; !; l m ; 3 12' ; ll h 12 m 36' ; 4 h 17 m 15'. ARC EQUIVALENTS. 18 7' 30" ; 154 58' 45" ; 120 2' 30" ; 104 49' 30" ; 179 59' 30" ; 45 4' 45"; 89 50' 30"; 135 24' 15"; 3 48' 45"; 36 15'; 2 38' 45"; 11' 15"; 36 11'; 165 4' 45"; 168 22' 30"; 75 1' 15"; 127 13'; 150 3'; 3'; 25'. DIFFERENCE OF TIMES. 41 m 20' less ; 1" ll m more ; 7 h 21 ra 8 more ; ll h SO 24' less ; 6 h 41 6 s more ; 6 h O m 19' less ; 29' more ; ll h 59 m 47' less ; 4 m 9' less ; 7 h 40 m 2' more ; 11 more ; ll h O m 1' less. ANSWERS. LONGITUDES. 1. 96 25' 45" E. 2. 106 50 15 W. 3. 110 5 W. 4. 122 27 30 E. 5. 78 48 45 W. 6. 117 58 30 E. 7. 34 45 E. 8. 9 7' 30" E. 9. 35 W. 10. 174 1 45 W. 11. 22 9 45 W. 12. 167 E. 13. 80 17 30 W. 14. 179 2 W. GREENWICH TIMES. 1. 16 d 7 h 9 m 36'. 2. 1 2 13 41-3. 3. 12 20 50 36. 4. 30 6 4 56. 5. 31 21 36 41. 6. 17 12 42 26. 7. 28 d 5 h 48 m 17 g . 8. 20 21 9 34. 9. 31 15 50 34. 10. 10 16 34. 11. 22 7 58 15. 12. 29 16 25 20. SHIP TIMES. 1. October 15th, 5" 29 14 1 A.M. 2. May 6th, 11 43 46 A.M. 3. August 31st, 7 3 33 P.M. 4. June 3rd, 5" 42 m 18' A.M. 5. October 19th, 11 37 47 P.M. 6. May 1st, 4 49 6 P.M. CHRONOMETER RATES AND G.M.T. 1. Rate 2' -7 losing. 2. 5 '3 gaining. 3. 10 '05 gaining. 4. '5 gaming. 5. *5 losing. 6. 5 '4 gaining. 7. '9 gaming. 8. 10-05 losing. 9. G.M.T. Nov. 26 d 13 h 45 m 45*. 10. Oct. 31 10 17 47. 11. Mar. 20 8 18 49. 12. Dec. 10 12 57 36. 13. July 3 23 58 22. 14. Nov. 26 19 23 52. 15. ,, Dec. 14 5 23 24. 16. May 1 24 10. 17. April 15 19 17 16. CHAPTER X SUN'S ELEMENTS. 1. Dec. 17 37' 36" -2 N. Eq. of Time 6 m r-73 2. 16 30 16-8 N. 5 37-39 3. 9 14 23 25-7 N. n 4 32-33 4. > 3 30 9-9 N. }} 4 22 5. 9 4 55-7 N. }J 7 28-01 6. | 14-1 S. n 7 32-62 7. 9 3 47-9 S. ) > 7 35-80 8. } 11 12-3 S. 7 42-39 9. I 2 27 10 S. 9 40-69 10. A.T. S. Aug . 20 a 20" 18 ra 13"6. 11. M.T.S. Sept. 29 21 34 7-4. 12. M.T .G. Sept. 24 9 34 13. R.A. 8 b 51 m 9 7 9 34 11 27 11 59 12 12 12 12 1 22 37-01. 39-28. 56-38. 33-02. 14-6. 2-2. 35-1. 43-4. 40-04. ANSWERS. SIDEREAL TIME OR R.A M.S. 10 50 m 41 .. 32 23 41-51. 3. 10 h 57 ra 10 s -48. 4. 12 29 48-66. MOON'S S.D. AND H.P. 1. S.D. 15' 33" 2. 15 24-9 3. 16 20-5 4. 16 10-5 H.P. 56' 28"-6. 55 49-4. 59 47. 58 24-4. MOON'S MERIDIAN PASSAGE. 1. M.T.G. Aug. l d 21 h 34 m -3. 2. 28 14 8-92. 3. ,,21 537. 4. 27 16-9. 5. M.T.G. Aug. 3 d 23 h 44 m -8. 6. ,, ,, 14 10 25-5. 7. ,, ., 8 14 5-0. 8. 6 10 24-3. MOON'S R.A. AND DEC. 1. R.A. 22 h 30 m 9' 7 2. ,, 45 20 3. 69 4-27 4. 12 45 42-13 Dec. 11 39' 9"-8 S. ,,02 2-1 S. 20 3 45-9 N. 1 34-8 S. 1. R.A.Jupiter 6 h 9 ra 38-94 2. Saturn 23 1 59 '32 PLANET'S ELEMENTS. Dec. 23 7'26"-6N. 8 24 59-6 S. Mer. Pass.lO a 4 h 56 n -27. 11 3 45-3. CHAPTER XL SIDEREAL TIME OR R.A.M. 1. O h 2 m 8* -66. 2. 15 58 34-45. 3. 10 48 51-2. 4. 7 h 36 m 46"18. 5. 8 5-93. 6. 17 41 11-29. 1. E.H.A. 2 h 2. W.H.A. 1 3. W.H.A. 3 6 6-41. 59 16-02. 7 29-19. HOUR ANGLES. 4. E.H.A. 0" O m 12-84. 5. M.T.S. Aug. 17* 20 h 32 n 1-49. 6. M.T.S. Sept. 14 6 35 1575. STAR'S MERIDIAN PASSAGE. 1. A.T. 2. M.T 3. M.T 4. A.T. 5. M.T 6. A.T. 7. M.T 8. A.T S. .S. .S. S. .S. S. .S. S. 3 m 25' 35 23 52 20 A.M. A.M. A.M. 8 3 30 27 28 33 46 9 17 22 A.M. A.M. A.M. A.M. A.M. 9. M.T.S. 10. A.T.S. 11 11. M.T.S. 8 12. A.T.S. 2 13. M.T.S. 5 14. A.T.S. 6 15. M.T.S. 7 7 h 49 35' 53 58 33 8 56 33 A.M. A.M. A.M. P.M. P.M. P.M. 23 P.M. 49 56 53 5 6 412 ANSWERS. 1. 9' 35" 50 2. 20 57 44 3. 19 56 46 4. 32 35 36 5. 47 2 57 6. 12 20 15 7. 56 20 43 8. 15 9. 29 10. 47 11. 23 12. 53 13. 39 38 28 13 31 19 24 8 22 13 51 20 11 BRIGHT STARS NEAR MERIDIAN. 1. Altair 43 ra 1' W. and S. 7 Cygni 10 11 W. N. Deneb 9 IE. N. oCephei47 7 E. N. 2. Fomalhaut 56 ra 8' W. and S. 3. Vega 20 2' W. and K Altair 52 25 E. N. 4. a Cephei l h 24 IF W. and N. e Pegasi 1 1 4 W. , N. o Gruis 38 18 W. S. Gruis 3 34 W. S. Fomalhaut 11 50 E. S. Markab 19 28 E. N. Alpheratz 1 22 55 E. N. j8 Cassiopeise 1 23 32 E. N. a Phoenicia 1 41 1 E. S. jBCeti 1 58 15 E. S. 5. Capella O h 38 m 7 ' W. andfl Rigel Bellatrix 37 27 51 48 W. W. ,, k- Tauri 27 31 W, ) > c e Orionis 16 26 W. s. 6. Sirius O h 11' 21 ! E . and N. 7. ll h 22 ra 11 s . 8. + 13 m 14 6 s . 9. + 16 m 25-9*. CHAPTER XII. ALTITUDES. S.D. 15' 0" ,, 14 57 H.P. 54' 13" 54 26 Cor. 29' 47" , 49 4 14. 47 49 27 15. 52 1 47 S.D. 15' 56" H.P. 57' 33" Cor. 35' 6" ANSWERS. 413 CHAPTER XIII. LATITUDE BY SUN'S MERIDIAN ALTITUDE. No. Declination. True altitude. Latitude. 1 II o ' n O 1 II 1 17 39 10 N. 34 22 44 37 58 6 S. 2 17 8 20 N. 50 16 30 56 51 50 N. 3 16 39 55 N. 19 43 40 53 36 25 S. 4 16 8 3N. 69 28 41 4 23 16 S. 5 15 27 39 N. 63 42 38 10 49 43 8. 6 14 53 16 N. 49 43 13 25 23 31 S. 7 14 6 41 N. 78 50 4 3 6 45 N. 8 13 37 54 N. 79 47 14 3 25 8 N. 9 12 50 19 N. 52 38 40 50 11 39 N. 10 2 28 54 S. 88 48 35 1 17 29 S. 11 7 55 N. 49 10 53 40 57 2 N. 12 4 45 N. 43 27 24 46 27 51 S. 13 1 21 36 S. 58 21 2 30 17 22 N. 14 2 9 50 S. 36 10 5 55 59 45 S. 15 1 35 S. 67 19 57 22 38 28 N. 16 17 37 N. 51 59 25 38 18 12 N. 17 13 55 S. 22 34 39 67 11 26 N. 18 9 48 S. 30 10 10 59 59 38 S. LATITUDE BY STAR'S MERIDIAN ALTITUDE. 1. 10 15' 12" N. 2. 13 27 39 S. 3. 27 58 26 N. 4. 43 3 40 N. 5. 49 43 18 N. 6. 50 45 N. 7. 35 58' 14" N. 8. 44 22 3 S. 9. 37 51 6 N. 10. 19 18 20 S. 11. 45 18 37 N. LATITUDE BY MOON'S MERIDIAN ALTITUDE. No. M.T.G. S.D. H.P. Declination. True alt. Latitude. d. b. m. / // i n O 1 II o t H / // 1 21 4 38 4 15 58 57 49 7 37 IN. 34 46 16 47 36 43 S. 2 22 8 50-6 16 10 58 22 2 10 58 N. 54 8 55 38 2 3N. 3 30 5 5-0 16 17 58 42 19 40 28 S. 61 47 40 47 52 48 S. 4 31 6 20-2 16 4 58 19 17 42 1 S. 25 44 55 46 33 4 N. 5 31 18 38-0 16 8 58 11 16 19 18 S. 69 32 47 4 7 55 N. 6 3 23 57-2 16 16 58 46 17 29 13 S. 39 16 31 68 12 42 S. 7 9 15 1-3 15 11 54 56 4 47 13 N. 43 25 36 51 21 37 N. 8 10 10 44-9 15 3 54 34 8 4 24 N. 47 5 35 50 58 50 N. 9 11 9 11-5 15 4 54 21 11 29 47 N. 54 37 2 23 53 11 S. 10 14 5 5-5 15 5 54 24 18 50 58 N. 75 33 28 4 24 26 N. 11 16 10 59-1 15 19 55 16 20 11 19 N. 74 21 44 4 33 3 N. 12 24 22 32-7 1C 25 59 8 9 54 37 S. 84 40 30 4 35 7 S. 414 ANSWERS. LATITUDE BY MERIDIAN ALTITUDE BELOW POLE. Dec. 8 23' 56" S. Lat. 43 16' 44" S. 1. M.T.G. 10" I7 h 1-1. 2. 48 23' 69" S. 3. 54 6' 9" N. 4. M.T.G. 28 d 10 h 31 n '9. S.D. 16' 12 True Alt. 7 53' 4". Lat. 77 55' 38" S. 5. Dec. 18 0' 19" N. True Alt. 5 53' 11". 6. 50 44' 17" S. H.P. 59' 1". Dec. 19 57' 2G" S. Lat. 77 52' 52" N. STAR'S CALCULATED ALTITUDES. 1. 39 48' 2. 42 6 3. 19 38 4. 57 5. 67 50 6. 32 15' 7. 61 44 8. 37 57 9. 51 4 10. 25 1 CHAPTER XIV. LATITUDE BY EX-MERIDIAN ALTITUDE OP THE SUN. No. Hour angle. Nat. No. Arc I. Arc II. Latitude sights. Latitude at noon. 1 m. B. 24 13 W. 4213 18 22 46 55 41 42 37 19 S. 37 16 36 S. 2 33 31 E. 9161 17 56 9 16 21 26 17 N. 3 27 36 W. 6927 16 9 4 21 46 7 5 37 S. 4 17 42 E. 1725 15 4 6 38 7 50 53 12 N. 5 7 13 W. 342 44 53 S. 6 41 4E. 15638 12 17 42 14 51 4 2 33 S. 2 39 4 S. 7 18 36 W. 2503 10 40 49 28 39 8 39 20 N. 39 22 30 N. 8 20 43 E. 2932 8 28 25 51 53 20 43 25 S. 43 27 6 S. 9 30 42 W. 6173 2 13 34 48 37 35 46 24 N. 46 29 12 N. 10 23 42 E. 5134 5 53 15 41 42 15 36 N. 11 10 8W. 611 2 41 51 11 6 51 14 N. 51 17 54 N. 12 19 12 W. 2675 ~~ ~~ 40 20 S. 40 25 S. LATITUDE BY EX-MERIDIAN ALTITUDE or A STAR. No. Hour angle. Nat. No. Arc I. Arc II. Latitude. m. B. O t II o / ;/ / // 1 7 56-5 365 38 42 45 77 34 45 38 52 S. 2 29 35-0 5126 8 41 27 42 45 50 51 27 17 N. 3 21 9-7 2318 57 8 54 57 8 55 IN. 4 21 42-6 2861 30 13 56 72 40 57 42 27 1 N. 5 19 13-7 2950 28 39 23 11 25 19 17 14 4 N. 6 47 43-0 11403 58 16 35 49 34 31 8 42 4 S. 7 23 18-4 4037 23 7 31 54 59 32 31 52 1 S. 8 14 32-4 1152 16 21 7 37 3 44 53 24 51 N. 9 28 43-0 5000 46 7 42 69 43 43 23 36 1 S. 10 25 8-7 3365 8 21 34 47 12 24 55 33 58 S. 11 6 15-0 264 44 57 15 41 53 3 4 16 N. 12 32 53-7 6532 47 42 46 67 51 30 20 8 44 N. ANSWERS. LATITUDE BY POLE STAR. 415 No. R.A.M. First cor. Second cor. Third cor. Latitude. h m. s. O 1 II ; n / // / II 1 4 10 55-9 -0 53 56 15 53 35 44 59 2 22 36 2 -0 53 17 22 1 16 45 17 10 3 12 55 19 + 1 11 23 54 30 4 31 4 16 59 12 +0 42 49 40 1 10 54 15 16 5 7 25 1 000 19 31 23 4 59 6 19 28 29 -016 28 1 27 32 22 41 7 1 21 -1 7 15 7 1 20 49 49 19 8 7 44 9 4-0 5 59 51 28 48 22 55 9 18 9 40 +0 23 13 24 1 15 31 1 12 CHAPTER XV. COMPASS ERROR BY AMPLITUDE. No. Declination. True amplitude. i Compass error. Deviation. 1 H O 1 II 1 II o / // 1 17 55 N. W. 24 35 30 N. 34 28 30 W. 5 58 30 W. 2 1 45 17 S. E. 2 45 16 S.' 28 15 16 E. 4 16 E. 3 16 2 20 N. E. 19 25 53 N. 5 22 8 W. 16 42 8 W. 4 2 33 52 S. W. 3 51 28 S. 1 46 2 E. 31 31 2 E. 5 17 31 14 N. E. 22 13 15 N. 5 20 45 W. 6 9 15 E. 6 1 13 23 S. W. 1 54 46 S. 18 35 14 E. 4 4 46 W. 7 16 34 56 N. E. 17 43 39 N. 17 43 39 W. 24 58 39 W. 8 6 2 21 N. W. 12 14 4 N. 12 14 4 E. 5 56 W. 9 17 39 44 N. E. 46 42 45 N. 46 6 OE. 23 26 E. 10 13 43 N. E. 13 43 N. 10 13 43 W. 10 13 43 W. 11 35 35 S. W. 36 33 S. 14 6 33 W. 12 6 33 W. 12 000 West 2 OE. 4 30 W. 13 9 9 23 N. E. 9 9 23 N. 18 58 7 E. 9 28 7 E. 14 000 East 000 1 15 W. 15. 4 h 41 m 30'A.M. 16. l h ll m 17 . 17. 2 E. 18. 45. 19. 41 49'. 20. 24 42'. CHAPTER XVI. COMPASS ERROR BY ALT. AZIMUTH. 2 h 40 m 50'. No. Polar distance. True altitude. True azimuth. Compass error. Deviation. 1 H 1 II o / // O 1 II O / H 1 107 31 58 16 51 58 N. 49 39 6 E. 9 6E. 18 9 6E. 2 77 7 20 26 7 13 S. 84 10 50 W. 5 49 10 W. 3 40 50 E. 8 81 4 35 36 2 26 S. 50 53 8 E. 22 23 8 W. 6 43 8 W. 4 81 26 34 3 54 S. 65 53 E. 15 15 30 W. 9 44 30 E. 5 90 16 20 21 56 18 N. 61 9 54 W. 40 6 E. 40 6E 6 87 35 42 25 14 42 N. 61 8 10 W. 6 21 50 E. 10 8 10 W. 4 i6 ANSWERS. COMPASS ERROR BY TIME AZIMUTH OF SUN. No. A.T.S. Corrections. True azimuth. Deviation. h. ra. s. / 1 o / 1 2 48 P.M. + 16 N. 132 45 W. 2 3 19 P.M. +14-25 N. 112 35 W. 3 9 14 A.M. +25+7+2 S. 141 25 E. 4 3 32 P.M. +21+5 N. 54 43 W. 3 9W. 5 7 20 A.M. -14 N. 73 51 E. 12 47 E. 6 7 57 22 A.M. +4+5+17 N. 88 36 E. 19 27 E. 7 3 38 31 P.M. -33+10+14 N. 61 24 W. 26 E. 8 8 41 26 A.M. + 19-27+11 S. 65 47 E. 9 51 E. 9 4 38 18 P.M. -17+6 S. 89 49 W. 9 11 W. 10 3 50 P.M. -18+37+5 S. 121 34 W. 10 19 E. 11 8 34 6 A.M. +17-18+18 N. 93 24 E. 19 55 W. 12 4 19 35 P.M. +4-1+15 N. 91 54 W. 16 18 E. 13 9 41 21 A.M. +19+16+11 S. 143 55 E. 35 40 W. 14 10 39 9 A.M. 82 59'+72 45' N. 155 44 E. 56 43 W. COMPASS ERROR BY TIME AZIMUTH OP STARS. No. Hour angle. Corrections. True azimuth. Deviation. b. in. s. / / 1 3 1 39 E. -17' -2' +59' N. 97 37 E. 22 22 E. 2 4 53 9 E. , -12-25+20 S. 92 51 E. 14 42 W. 3 3 26 4 W. -28 -27+24 N. 114 36 W. 35 26 W. 4 1 48 14 W. 89 59'+68 58' S. 158 57 W. 15 43 W. 5 5 37 7 E. -11-19+8 N. 88 5E. 23 43 E. 6 3 23 41 W. +5-27+13 N. 113 30 W. 26 OE. 7 3 29 46 E. -21+12+8 S. 128 24 E. 21 36 E. 8 3 23 47 W. 94 5'+38 37 N. 132 42 W. 5W. 9 3 35 29 E. 91 57+39 34 S. 131 31 E. 10 25 E. 10 9 16 22 W. 24 30-3 47 S. 20 43 W. 7 IE. 11 4 30 20 W. 111 1+43 22 S. 154 23 W. 23 10 W. 12 8 41 20 W. 48 49-1 12 S. 47 37 W. 2 26 W. 14. 15. N. 48 20' E. Dec. 19 19' N. ; Lat. 51 30' N. CHAPTER XVII. LONGITUDE BY SUN CHRONOMETER. 1 M.T.G. Pol. dlst. Eq. Time. True altitude. Hour angle. Longitude at Bights. Longitude at noon. d. h. m. s. / // m. s. o / // h. m. s. o t n / // 1 3 14 37 35 107 31 58 + 6 16 51 58 3 19 35 92 12 30 E. 92 30 40 E. ( 2 19 91031 77 7 22 +3 34 26 7 13 4 25 34 70 20 45 W. 70 53 26 W. 3 30 21 35 58 81 4 35 + 29 36 2 26 2 37 42 3 17 30 W. 3 54 28 W. 4 31 21 22 37 81 26 + 11 34 3 54 3 19 30 1029 OW. 10 56 37 W. 5 22 18 29 38 90 16 20 -7 18 21 56 18 3 37 24 135 7 OE. G 29 15 20 26 87 35 42 -9 38 25 14 42 3 29 48 179 56 OE. 179 28 30 W. ANSWERS. LONGITUDE BY STAR CHRONOMETER. 417 No. M.T.G. E.A.M.S. True altitude. Hour angle. Longitude. d. b. m. s. h. m. a. / // b. m. s. o / // 1 9 9 8 13 11 12 59 24 30 2 9 2W. 110 45 30 E. 2 23 21 37 52 10 7 2 12 1 18 3 36 40 E. 74 55 15 W. 3 1 7 47 31 8 38 1 42 2 2 22 59 E. 14 25 15 E. 4 11 10 56 5 11 20 11 26 8 42 3 42 30 W. 000 5 20 15 27 42 11 56 24 37 12 48 3 48 43 E. 30 46 15 W. 6 23 9 11 21 10 4 59 27 14 27 3 50 37 E. 90 46 E. 7 7 9 42 18 11 4 12 22 42 6 3 25 33 E. 110 35 E. 8 30 21 6 21 10 34 32 15 44 30 2 29 43 E. 169 32 15 W. 9 2 6 54 54 10 44 2 25 13 55 4 36 W. 50 10 45 E. 10 14 2 2 6 9 28 19 21 59 16 4 59 E. 179 43 30 W. 11 28 6 58 18 10 24 20 34 50 51 2 51 28 W. 4 56 45 W. 12 27 19 24 11 12 24 39 68 25 1 2 5 9E. 161 3 45 E. 2 E 4i8 ANSWERS. * CM O O ryi fe ryj F1 M M M CM O OS O -^ CO O O CO O CM (M 08 4 P4 r4 H H H CM CO 8 g I _H_ d a ^ rH rH * CO T*< O ^ "f( CO rH CM CO OS CO s S8 CO l> rH go o os o o o 3 ^ os o So OS CM 2 S S3 CO o os O 00 O rH rH rH CO OS O rH ANSWERS. 419 CHAPTER XVIII. SUMNER'S METHOD. 1. First observation : T.A. 36 26' 20" ; P,D. 72 29' 5" ; eq. time, + 6 m 0* ; H.A.'s, 3 h 37 ra 17' and 3 h 36 ra 25' ; longs., (a), 64 14' 30" E. ; (6), 64 27' 30" E. Second observation : T.A. 43 47' 38" ; P.D. 72 33' 18" ; eq. time, + 5 m 58' ; H.A.'s, 2 h 49 m 12' and 2 h 47 m 37' ; longs., (c), 64 36' 15" E. ; (d), 64 12' 30" E. First line: {s'l^w}'' second azimuth, S. 63 W. ; lat. 49 27' N. ; long. 63 52' E. 2. First observation: T.A. 38 26' 23" ; P.D. 73 54' 28"; eq. time, + 5 m 26 ; H.A.'s, 3 h 15 m 40 and 3 h 13 m 56 s ; longs., (a), 4 43' 15" W. ; (6) 4 17' 15" W. Second observation ; T.A. 40 19' 48" ; P.D. 73 58' 52" ; eq. time, + 5 m 24'; H.A.'s, 3 h 2 m 35' and 3 h O m 34 ; longs., (c), 2 0' 15" W. ; (d), 2 30' 30" W. First line : {$'30 ^ } ; azimuth, S. 67 E. ; second line : ||' 250 ^j ; lat. 50 0' N. ; long. 2 30' 30" W. 3. First observation: T.A. 24 9' 42"; P.D. 78 35' 56"; eq. time, + 2 m 30'; H.A.'s, 4 h 23 m 33' and 4 h 22 m 57'; longs., (a), 18' 15" E. ; (6), 27' 15" E. Second observation : T.A. 48 46' 26" ; P.D. 78 38' 51" ; eq. time, + 2 m 28'; H.A.'s, l h ll m 42 g and l h 5 m 19 s ; longs., (c), 2 43' 45" W. ; (d}. 1 8' 0" W. First line: ' w' 5 azimuth, S. 79 E. ; lat. 50 28' N. ; long. 41' 30" W. 4. First observation: T.A. 48 31' 31"; P.D. 81 18' 3"; eq. time, + O m 18*; H.A.'s, O h 52 m 40* and O h 39 m 41* ; longs., (a), 175 15' 45" W. ; (b), 178 30' 30" W. Second observation: T.A. 19 9' 3"; P.D. 81 21' 45" ; eq. time, + O m 15' ; H. A.'s, 4 h 43 m 0' and 4 h 42 ra 21' ; longs., (c), 179 32' 15" W. ; (d), 179 42' 0" W. First line: |^' Jp iT}; second azimuth s - 81 w - J lat - ^ W N. ; long. 179 40' W. ' ' 5. First observation: T.A. 27 6' 30"; P.D. 90 10' 21"; eq. time, -7 m 23'; H.A.'s, 3 h 14 ra 28' and 3 h ll m 57 ; longs., (a), 59 56' 30" W.; (6), 59 18' 45" W. Second observation : T.A. 32 27' 30" ; P.D. 90 4' 38" ; eq. time, - 7 m 28 ; H.A.'s, 2 h 36" 18' and 2" 32 m 46'; longs., (c), 60 17' 15" W. ; (d), 61 10' 15" W. First line : ' ' 5 azimuth, N. 57 E. ; lat. 46 18' S. ; long. 60 27' W. 6. First observation : T.A. 18 27' 6" ; P.D. 98 27' 58" ; eq. time, + O m 6' ; H.A.'s, 3 h 13 m 54 1 and 3 h ll m 24'; longs., (a), 109 34' 15" W.; (6), 108 56' 45" W. Second observation: T.A. 26 4' 10" ; P.D. 98 23' 10" ; eq. time, + O m 2 ; H.A.'s, 2 h 6 m 27' and 2 h 2 m 0' ; longs., (a), 109 33' 30" W. ; (d), 110 40' 15" W. First line : |^' * jy ^'\ ; azimuth, N. 49 E. ; lat. 49 21' S. ; long. 109 45' W. 7. First observation /T.A. 21 4' 7" ; P.D. 71 38' 7" ; eq. time, + 6 12 ; H.A.'s, 5 h 15 m 16' and 5 h 15 ra 29' ; longs. , (a), 167 54' 45" E. ; (6), 167 51' 30" E. Second observation : T.A. 57 12' 25" ; P.D. 71 40' 48" ; eq. time, + 6 m 12' ; H.A.'s, O h 45 m 14' and O h 35 m 36' ; longs., (c), 168 15' 15" E. ; (d), 170 39' 45" E. First line : ' ' ; second azimuth, S. 73 E. ; lat. 49 44' N. long. 166 13' E. 420 ANSWERS. 8. First observation : T. A. 11 1' 58" ; P.D. 91 52' 56" ; eq. time, - 9 m 12' ; H.A.'s, 4 h 46 m 19' and 4 h 45 m 0' ; longs., (a), 179 27' 30" E. ; (6), 179 47' 15" E. Second observation: T.A. 26 59' 56"; P.D. 92 0' 41"; eq. time, - 9 m 18' ; H.A.'s, 3 h I 6' and 2 h 57 m 16' ; Iong3., (a), 177 7' 30" E. ; (d), 176 10' 0" E. First line : (f '^o^' } ; azimuth, S. 75 W. ; lat. 46 40' K ; long. 177 31' E. . CHAPTER XIX. CHART. 1. (a) S. 17 E., 30 E. } 18 W. ; 35| miles. (b) Lat. 52 35J' K ; long. 4 17' W. ; set N. 70 E. ; dist. 5J- miles. (c) Lat. 52 20f' N. ; long. 4 17$' W. ; dist. 7| miles. (d) S. 5 E. ; dist. 20 miles ; she steams 40 miles. (e) Lat. 52 24J' N. . long. 4 31' W. 2. (a) Comp. Co. S. 49 E. ; dev. 15 E. ; var. 18 W. ; dist. 15| miles. N. 85 E. ; dev. 7 W. ; var. 18 W. ; dist. 23 miles. N. 58 E. ; dev. 21 W. ; var. 18 W. ; dist. 31* miles. (6) Lat. 52 6J' N. ; long. 6 24J' W. ; set S. 80 W. ; dist. 3} miles. (c) Lat. 52 23J N. ; long. 5 51^' W. ; dist. 10 miles. (d) Comp. Co. S. 42 E. ; N. 87 E. ; N. 49 E. ; dist. 52 miles. (e) Lat. 51 56J' N. ; long. 6 44J' W. 3. (a) Comp. Co. N. 23 E. ; dev. 30 W. ; var. 18 W. ; dist. 17| miles. N. 63 E. ; dev. 18 W. ; dist. 37 miles. (b) Lat. 51 45f ' N. ; long. 5 26 W. ; set N. 14 W. ; dist. 3 miles. (c) Lat. 52 17' N. ; long. 5 6' W. ; dist 8 miles. (d) Comp. Co. N. 71 E. ; dist. made good 22 miles. Total dist. steamed 32 miles. CHAPTER XX. NAPIER'S CURVE. 1. Correct magnetic bearing : N. 66 E. Deviations: 17 W. 28 W. 24 W. 9W. 16 E. 32 E. 26 E. 5E. , Compass courses : N. 76 E. S. 80 E. S. 28 E. N. 16 E. Magnetic courses : N. 44 W. S. 16 W. N . 79^ E. N. 9 W. Deviation: 31 E. Bearings : S. 53 W. N. 76 E. 2. Correct magnetic bearing : N. 82* W. Deviations: 121 E. 18* E. 9i E. 6* W. 9i W. 14 W. 8i W. 3| W. Compass courses : N. 71 E. S. 69 E. S. 20 W. N. 42 W. Correct magnetic bearings : N. 12 E. S. 44 E. S.37W, N. 35 W. Deviation: 0. Correct bearings : S. 80 W. N. 40 E. ANSWERS. 421 3. Correct magnetic bearing : S. 14 W. Deviations : 10 E. 15 E. 14 E. 8 W. 20 W. 18 W. 3 E. 6E. Compass courses : K 16 W. S. 75 W. S. 75 E. N. 51 E. Correct magnetic courses : N. 56 W. N. 56 E. S. 27 E. N. 33 W. Deviation : 4 E. Correct bearings : S. 67 W. N. 58 W. 4. Correct magnetic bearing : N. 8i W. Deviations: 23iE. 18* E. 14i W. 101 W. 12* W. 12i W. 1JB. 6|E. Compass courses : S. 57 W. N. 18^ W. S. 73 E. S, 88 E. Correct magnetic courses : S. 80 W. N. 40 E. S. 45 E. S. 1 E. Deviation: 16 E. Bearings : N. 52 W. S. 39 W. 6. Correct magnetic bearing : N. 83| E. Deviations : 21 45' E. 17 45' E. 45' E. 19 15' W. 21 15' W. 15 15' W. 1 15' W. 16 45' E. Compass courses : N. 29 E. S. 86 E. S. 25 W. N. 26 W. Correct magnetic courses : N. 58E. S. 61E. N.87iW. N.21W. Deviation : 22 E. Bearings : S. 3 E. S. 73^ E. 6. Correct magnetic bearing : S. 13J E. Deviations: 18|E. 8W. 29iW. 34|W. 20 5' W. 12|E. 291 E. 31 20' E. Compass courses : N. 2 E. S. 60i E. S. 26 W. N. 8U W. Correct magnetic courses : N. 32 E." N. 50 E. S. 46J E. S. 11 W. Deviation : 35 W. Correct bearings : N. 15 38' E. N. 49 4' W. 7. (a) Magnetic bearing, N. 56 E. (6) N. | E., S. 80 E., S. 66 W., N. 7 W. (c) S. 19 W., N. 59 W., N., S. 71 E. (d) E. 11 N., N. 3 E. 8. (a) N. 73 W. ft) S. 37 E., S. 18 W., N. 65 W., N. 78 E. (c) N.W. by W., W. by S., S. 28 W., N. 80 E. (d) N. 74 E., N. 10 W. 9. (a) N. 6 W. (6) W.S.W., N. 65 W., N. 7 E., S. 75 E. (c) N. 26 E., N. 51 E., N. 87 E., S. 21 W. (d) S. 78 W., S. 5 W. 10. (a) S. 65 W. (6) S. 40 E., N., N. 63 W., S. 21 W. (c) N. 40 W., N. 10 E., N. 64 E., S. 63 E. (d) S. 51 E., N. 36 E. 11. (a) S. 11 E. (6) S. 84 W., N. 48 E., S. 58 E., S. 11 E. (c) N. 85 E. , N. 13 E., N. 61 W.. S. 14 W. (d) N. 5 E., N. 75 E. 12. (a) N. 2 W. (6) S. 76 E., S. 35 E., N. 58 W., N. 50 E. (c) N. by W., S. 76 W., N. 27 E., S. 82 E. (d) S. 32iW., S. 46 E. 422 ANSWERS. CHAPTER XXI. GREAT CIRCLE SAILING. 1. First course, S. 84 44' 50" W. ; last course, K 63 10' 40" W. ; lat. of vertex, 34 43' S., long. 9 15' E. ; distance, 3269 miles. Points. Longitude. Latitude. Long, from V. o / 1 O t 9 15 18 30 E. 34 22 S. 5 14 15 E. 34 37 S. 9 15 E. 34 43 S. 5 4 15 E. 34 37 S. 10 45 W. 34 19 S. 15 5 45 W. 33 47 S. 20 10 45 W. 33 4S. 25 15 45 W. 32 7J S. 30 20 45 W. 30 57 S. 35 25 45 W. 29 34 S. 40 30 45 W. 27 57J S. 45 35 45 W. 26 6| S. 50 40 45 W. 24 S. 52 44 43 9W. 22 55 S. 2. Initial course, N. 56 54' W. ; final course, S. 54 18' W. ; distance, 4469 miles ; lat. and long, of vertex, 48 34' K, 169 14' W. Points 1. 40 57V N., 129 14' W. 2. 46 47| K, 149 14 W. 3. 48 34 K, 169 14 W. 4. 46 47 * N., 170 46 E. 5. 40 57| K, 150 46 E. 3. Courses, N. 55 43' E. , and N. 60 15' E. ; dist. C441 miles : vertex, lat. 46 40' N. , long. 79 29' W. Points 1. 27 55|' S., 160 31' E. 2. 19 56 S., 170 31 E. 3. 10 26 S., 179 29 W. 4. 169 29 W. 5. 10 26 N.,159 29 W. 6. 19 56 N. , 149 29 W. 7. 27 551 N.,139 29 W. 8. 34 16j N.,129 29 W. 4. Courses, K 48 47*' E. , and S. 45 23' E. ; dist. 9904 miles ; vertex , lat. 45 52' K, long. 179 14' W. Points 1. 27 16' K, 120 46' E. 2. 38 18 N., 140 46 E. 3. 44 51 N. 160 46 E. 4. 45 52| N. 179 14 W. 5. 44 6} N. 159 14 W. 6. 38 18" N. 139 14 W. 7. 27 16 N. 119 14 W. 8. 10 9 N. 99 14 W. 9. 10 9 S.,' 79 14 W. ANSWERS. 423 5. Courses, S. 67 35' W., and S. 35 34 W. ; dist. 3452 miles ; vertex, lat. 55 29' N., long. 21 21 J 1 E. Points 1. 51 33' K, 8 38' W. 2. 49 59 N., 13 38 W. 3. 48 5 N., 18 38 W. 4. 45 48 N., 23 38 W. 5. 43 4 K, 28 38 W. 6. 39 50 1ST., 33 38 W. 7. 36 1| N., 38 38 W. 8. 31 341 N., 43 38 W. 9. 26 26 N., 48 38 W. 10. 20 37i N., 53 38 W. 11. 14 10 N., 58 38 W. 6. Courses, N. 68 45' E., and S. 68 45' E. ; dist. 2494*6 miles; Vertex, lat. 47 36*' N., long. 180 E. Points 1. 44 47 J' K, 155 E. 2. 45 49| K, 160 E. 3. 46 37 N., 165 E. ,, 4. 47 10 N., 170 E. ,, 5. 47 30 K, 175 E. 6. Yertex. ,, 7. 47 30 N., 175 W., etc. 7. Courses, S. 66 W., and S. 66 W. ; dist. 7544*6 miles; vertex, lat. 44 19' N., long. 64 49^' W. Points 1. 29 9' N., 120 W. 2. 25 54i N., 125 W. ,, 3. 22 18 K, 130 W. 4. 18 19J N., 135 W. 5. 14 2 N., 140 W. 6. 9 28 N., 145 W. 7. 4 42 K, 150 W. 8. 154 49' W. 8. Long. 32 39J' W. 9. Lat. 45 52^' N. 10. 50 S., 110 E. 11. 119 miles north. 12. A. Lat. 46 40' N. long. 53 3' W. B. Lat. 51 25' N. ; long. 9 29' W. COMPOSITE TRACKS. No. First course. Last course. Distance. Longitude arrival. Longitude departure. o / / / 1 1 2 3 S. 65 27J E. S. 53 5J W. N.52 5 E. N. 59 6E. N. 68 44J W. S. 75 26 E. 7832-4 5600-6 4096-1 25 40 W. 19 1W. 167 1 W. 68 32$ E. 21 55 W. 143 54 W. 4. Courses, S. 59 45' W., and West ; longs. 118 24' W., and 172 50' E. ; dist. 2988 + 2266'2 = 5254'2 miles. Points 1. 33 46' 2. 35 57 3. 37 49 4. 39 24 73 W. 78 W. 83 W. S., 88 W. S., 424 ANSWERS. Points 5. 40 42 S., 93 W. 6. 41 45 S., 98 W. 7. 42 33 S., 103 W. 8. 43 n S., 108 W. ,, 9. 43 28} S., 113 W. 10. 43 36 S., 118 W., etc. CHAPTER XXII. LATITUDE BY DOUBLE ALTITUDES. 1. H.E.T. 2 h 52 57 s ; mid. dec. 14 38' 13" N. ; cor. for ran, + 11' 6" ; first alt. 30 38' 37" ; second alt. 26 43' 34" ; XY = 83 1' 34" ; X, = 103 19' 32" ; X 2 = 63 0' 32". Cor. for lat. + 12". Lat. 30 21' 30" S. 2. H.E.T. 2 h 23 m 55"5 ; mid. dec. 13 12' 25" N. ; cor. for run, - 21' 48" ; first alt. 50 22' 0"; second alt. 42 57' 0"; XY = 69 46' 30"; X t = 80 37' 38" ; X 2 = 46 5' 30". Cor. for lat. = 21". Lat. 43 27' 45" N. 3. H.E.T. 2 h 15 m 56' -7; dec. 38 41' 45" X.; cor. for run, 0; first alt. 75 44' 55" ; second alt. 34 50' 11" ; XY = 51 43' 52" ; Y x = 67 8' 50" ; Y 2 = 17 14' 56". Lat. 50 20' N. 4. H.E.T. l h 30 m 20 s "3 ; dec. 16 19' 15" N. ; cor. for run, - 12' 54" ; first alt. 17 53' 0" ; second alt. 45 37' 12" ; XY = 43 15' 18" : Y, = 96 40' 1" : Y 2 = 41 9'. Lat. 25 31' 8" S. 5. R.A.M.S.9 h 43 BI 39 t ; T.A. o Arietis, 276'20"; longs., (a) 179 57' W. ; (6) 179 59' E. Bearing, N. 85 6' E. T.A. Altair, 47 1' 41" ; longs., (c) 179 4' 30" W. ; (d) 179 35' 15" E. Bearing, S. 28 50' W. XY - 90 12' 41" ; X t = 80 32' 23" ; X,, = 34 32' 3". Lat. 48 21' 0" N ; long. 180 W. 6. H.E.T. I h 53 m 2-5; mid. dec. 7 47' 35" N. ; run,0; first alt. 18 4' 22" ; second alt. 47 2' 15"; XY = 55 57' 8" ; X a = 85 53' 6" ; X 2 = 44 53'. Cor. for lat. - 2' 30". Lat. 48 51' N. 7. H.E.T. 2" 45 m 28' '6 ; run, + 49' 24" ; first alt. 15 50' 58" ; second alt. 28 8' 24" ; XY = 69 44" ; Xj = 113 50' 32" ; X 2 = 65 18' 15". Lat. 24 27' N. 8. H.E.T. l b 52 m 34'; mid. dec. 3 22' 34" N. ; run, +45'; first alt. 39 34' 56" ; second alt. 14 18' 3" ; XY = 56 10' 39" ; Y, = 91 51' 40" ; Y 2 = 51 37' 54". Lat. 46 23' 48" S. 9. H.E.T. 3 b 50 m 47-8 ; run, - 1 3' 0" ; first alt. 54 47' 7" ; second alt. 49 9' 37" ; XY = 53 40' 42" ; Y : = 36 47' 22" ; Y 2 = 45 33' 48". Lat. 43 19' 6" S. 10. R.A.M.S. 10 h 15 m 59 ; T.A. Aldebaran, 35 3' 22"; longs., (a) 156 7' 30" E. ; (6) 156 22' E. Bearing, S. 72 9' E. T.A. Markab, 47 42' 15 ; longs., (c) 157 39' 45" E. ; (d) 156 52' E. Bearing, S. 43 W. XY = 79 0' 30" ; ^ = 77 43' 59" ; X 2 = 38 21' 14". Lat. 50 14' N. ; long. 156 29' E. 11. H.E.T. 2 h 35 m 4-5; mid. dec. 5 16' 18" N. ; run, - 26' 12"; first alt. 51 36' 17" ; second alt. 26 34' 54". XY = 77 8' 54" ; Y t = 94 16' 13" ; Y 2 = 38 18' 8". Cor. for lat. - 1' 7". Lat. 27 14' S. 12. H.E.T. l h 45 m 23-8 ; run, + 49' 0" ; first alt. 63 0' 37" ; second alt. 18 37' 27". XY = 52 3' 28" ; Y, = 94 14' 43" ; Y 2 = 21 37' 35". Lat. 1325'S. ANSWERS. 425 13. R. A.M.S. 11* 35 m 30" ; T. A. Rigel, 49 19' 1" ; longs., (a) 125 4' 30" E. ; (6) 127 25' 15" E. Bearing, K 23 E. T.A. Fomalhaut, 27 21' 6" ; longs., (c) 127 23' 15" E. ; (d) 127 39' 45" E. Bearing, S. 74 30' W. XY = 89 35' 57" ; Y, ; 80 36' 55" ; Y 2 = 31 45' 47". Lat. 47 24' 30" S. ; long. 127 41' 30" E. 14. H.E.T. 2 h 24 ra 44 s ; mid. dec. 1 47' 48" S. ; first alt. 34 36' 15"; run, - 16' 24"; second alt. 16 22' 59"; XY = 72 19' 30"; Y t =. 91 15' 47"; Y 2 = 58 9' 30". Cor. for lat. - 1' 25". Lat. 52 35' HT. 15. H.E.T. 4 h 5 m 45"2; first alt. 66 3' 3" ; run, - 57' 36"; second alt. 20 19' 39". XY = 76 53' 45"; Y, = 37' 37" ; Y 2 = 23 52' 18". Lat. 34 12' 32" N. CHAPTER XXIII. ERROR OP CHRONOMETER BY SINGLE ALTITUDE. 1. Fast M.T.G. 7 m 35 s ; slow A.T.S. 5 h 55 m 13'. 2. Slow 15 m 39 s . 3. Slow 31'. 4. Slow5 m 6'. 5. Fast 17 m 38*. 6. Slow 8 m 23'. 7. Fast 46'. 8. Slow A.T.S. 3 h 43 m 43' ; fast M.G.T. 18 s . CHAPTER XXIV. CALCULATION OP ALTITUDES. 1. E.H.A. O h 45 m 25'; dec. 6 10' 42" N. ; T.A. 46 54' 43"; A.A 46 55' 30". 2. W.H.A. 4 h 42 m 25 s ; dec. 2 25' 30" S. : T.A. 11 18' 30"; A.A 11 23' 6". 3. W.H.A. l h 34 m 34-6 ; T.A. 35 32' 16" ; A.A. 35 33' 36". 4. E.H.A. 4 h 54 ra 27 8< 3 ; T.A. 17 6' 1" ; A.A. 17 9' 4". 5. W.H.A. 3 h 42 m 21-3; H.P. 54' 51"; dec. 5 51' 28" K ; T.A. 30 37' 13"; A. A. 29 51' 19". 6. W.H.A. 3" 13 ra 52-6; dec. 8 24' 48"; T.A. 40 56' 54"; A.A. 40 57' 58". 7. E.H.A. O h 36 m 28' ; H.P. 58' 50" ; dec. 19 49' 51" S. ; T.A. 76 59' 48" : A.A. 76 46' 33". Q 8. E.H.A. 3 h 56 m 40-9; dec. 23 7' 40" K ; T.A. 30 20' 56"; A.A. 30 22 <32 CHAPTER XXV. LONGITUDE BY LUNAR DISTANCE. O.M.T. 1. 6 d 17 h 27 m 36-5. 2. 7 d 20 b 27 m 13 s . 3. 12 d ll h 7 m 1-5. 4. 22 d 21 h 58 m 2'. 5. 28 d 7 h O m 25'. 426 ANSWERS. TIIUE DISTANCE. 1. Z = 119 37' 16" ; B = 25 26' 45" ; true dist. = 96 18' 32". 2. Z = 122 9' 50" ; B = 24 56' 56" ; true dist. 107 17' 39". 3. Z = 125 55' 28" ; B = 21 16' 40" ; true dist. 100 39' 17". 4. Z = 96 14' 22" ; B = 21 59' 24"; true dist. 55 37' 4". LONGITUDES AND ERRORS. 1. Sun's elements: Dec. 14 58' 51" N. ; R.A. 9 h 27 m 39 s ; R.A.M.S. 9 h 22 m 46 s -3 ; A.A. 34 22' 13" ; T.A. 34 20' 57" ; E.H.A. l h 36 m 32 s . Moon's elements : R.A. 4 h O m 1-3; dec. 15 30' 16" N. ; H.P. 54' 9"'l; S.D. 14' 51"-6; W.H.A. 3 h 51 m 6' ; T.A. 15 6' 45" ; A. A. 14 17' 56" ; app. dist. 78 53' 30" ; B = 40 41' 23" ; Z = 86 11' 1" ; true dist. 78 27' 16 ' ; error, 6 ra fast; long. 93 31' 45" E. 2. Sun's elements: R.A. 10 h 26 m 44-8; dec. 9 44' 29" N. ; R.A.M.S. 10 h 25 m 34'; W.H.A. 3 h 34 m 17"2 ; T.A. 37 5' 42" ; A. A. 37 6' 51". Mom's elements: R.A. 18 h 9 m 38"7; dec. 20 6' 2" S. ; S.D. 16' 13"; H.P. 59' 8" -3; E.H.A. 4 h 8 m 36"7 ; A.A. 14 6' 50" ; T.A. 15 0' 27" ; app. dist. 118 6' 57" ; B = 16 7' 49" ; Z = 143 5' 37" ; true dist. 117 22' 44 s ; error, 9 m 18 s slow ; long. 163 32' 0" W. 3. Star's elements: R.A. 4 h 30 m 31"5; dec. 16 19' 15" N. ; R.A.M.S. 9 h 8 m 54-3 ; E.H.A. 3 h 53 m 49 s ; T.A. 34 40' 39" ; A.A. 34 42' 0". Moon's elements: R.A. I h 16 m 27"6; dec. 2 50' 3" N. ; S.D. 15' 17" '7 ; H.P. 55' 14" -5 ; A.A. 53 17' 50" ; T.A. 53 50' 9" ; E.H.A. O h 39 m 45'; app. dist. 49 35' 21" ; B = 35 30' 39" ; Z = 67 1' 7" ; true dist. 49 32' 54" ; error, 3' slow ; long. 174 36' 53" E. 4. Star's elements: R.A. 19 h 46 m 11-8; dec. 8 37' 11" N. ; R.A.M.S. 10 h O m 12 ,. 9 . A A 28 o ft, 54 // . T A 2g o 16 , 8 // . E>H 16'; error 14 m 31' slow. 25 First course, 1ST. 78 3' W. ; dist. 3346 miles ; vertex lat., 50 59' N. ; long. 21 16' W. ; 5 W. of Wolf, lat. 50 31' N. ; 10 W. of Wolf, 50 51*' N. ; 15 W, 50 591' N. ; 20, 50 53i N.; 61 W., 43 30' N. 26. 319-1. 27. 86 1' 21". 28. 106 10' 16. 29. 100 17' 28", or 79 42' 32". 30. N. 31 4' E.; 0, - 23 18'. 31. R.A.M. 23 h 45 m 39"; E., Rigel ; W. a 1 Crucis, a 2 Centauri, Markab. 32. A + '25; B - 25; - 16 -5; D + 2; E - '75. Deviations, North, 17 W. ; 12 29' W. ; 7 22' W. ; 1 43' W. ; 4 15' E. ; 10 19' E. ; 16 10' E. ; 21 28' E.; West, 26 E. ; 29 26' E. ; 31 36' E.; 32 21' E.; 31 36' E. ; 29 25' E. ; 25 58' E. ; 21 24' E. ; South, 16 E. 33. 5 45' E. PAPER II. 1. 212-8 log = 2-327982. 2. -147 log = 9-167437. 3. S. 17 E., 15' ; N. 67| E., 54' ; N. 52 E., 50'; S. 83 E., 42' ; S. 56 W., 31' ; S. 1 E., 30' ; S. 37 E., 42' ; N. 14 W.. 54' ; D. lat. 3' '2 N., dep. 122' -6 E. ; course, N. 88 E. ; dist. 123' ; lat. in 25 12' IS. ; long. 45 4V W. 4. Dec. 9' 13" N. ; T.A. 64 58' 19" ; lat. 24 52' 28" S. 5. 41 40' W. 6. D. lat. 5642 ; M.D.L. 6465 ; D. long. 2805 ; true course, N. 23 27J' W. ; comp. course, N. 64 7' W. ; dist. 6150 miles. 7. 5 h 40 m A.M. j ,6 h 23 m p.M. 8. Dec. 2 5' 33" S. ; true amp. W. 2 49' 25" S. ; dev. 8 19' 30" W. 9. M.T.G. 3 d 16 h 53 m 9' ; dec. 17 30' 27" N. ; eq. time, + 5 m 59' ; T.A. 22 38' 24" ; E.H.A. 4 h 42 m 44' ; st. long. 37 31' 30" E. ; noon long. 38 15' 30" E. ; true azi. S. 102 48' E. ; error 10 18' W. ; dev. 5 18' W. 10. W.H.A. 4 h 5 m 30' ; dec. 8 36' 43" N. ; cor. - 17 + 26 + 6 ; true azi. S. 116 50' W. ; , 150 10' W. ; (6), 149 44' 15" W. ; second T.A. 46 54' 48" ; P.D. 83 49' 18" ; long, (c), 152 16' 46" W. ; (d\ 150 10' 30" W. Second line : ' 5 second azi. S. 20 E. ; lat. 48 12' N. ; long. 149 21' W. 13. 9 U 36 m 57' P.M. ; by inspection, 9 h 31 m P.M. 14. RAM. 16 h O ra 39' ; Antares, S. O h 23 m E. ; a Triang. Aust. S. O h 38 ra E. 15. Alt. 20 27' 40". 16. E.H. A. 2 h 38 ra 20' ; cor. - 26 + 12 + 10 ; true azi. S. 137 32' E. ; dev. 6 32* W. 17. M.T.G. 27 d 10 h 27 m 40 s ; R.A.M.S. 12 b 23 m IV ; E.H.A. 4 h 54 m 21' ; long. 84 26i' E. 18. Cor. + 33", + 41", + 1' 17" ; lat. 42 9' 56" N. 19. M.T.G. 16 a 19 h 35 ra -6 ; S.D. 15' 22" ; H.P. 56' 27'' ; dec. 19 57' 25" N. ; T.A. 80 54' 39" ; lat. 10 52' 4" N. 20. Lat. 8 11' 54" S. 21. Mag. bearing, N. 4 45' E. ; comp. course, N. 43 E., S. 81 E., S. 11 E., N. 7 E. ; mag. course, N. 30 W., N. 50 E., S. 49 E., S. 81 W. ; mag. bearings, S. 68^ E., and N. 71 W. 22. Red, - 14 ft. 8 ins. ' 23. Sun's elements: R.A. 9 h 17 m 4"9 ; dec. 15 48' 11" N. ; R.A.M.S. 9 u ir n 46 .. 7 . E.H.A. 3 h 44 ra 37' ; T.A. 36 15' 56" ; A.A. 36 17' 6". Moon's elements: R.A. l h 50 m 6"4; dec. 5 51' 28" N. ; S.D. 15' 7"; H.P. 54' 51" ; A.A. 29 51' 19" ; T.A. 30 37' 13" ; W.H.A. 3 h 42 M 21* ; app. dist., 109 46' 10" ; B = 10 25' 56" ; Z = 154 49' 24" ; true dist. 109 4' 27" ; error, 5 m 29' slow ; long. 14 27' W. 24. Antares H.A.'s l h 36 m 20% and l h 31 m 53' ; long, (a), 62 8' 45" W., (6), 63 15' 30" W. ; Spica H.A.'s 4 h 38 m 15', and 4 h 37 m 32 s ; long, (c), 62 31' W., (d), 62 41' 45" W. ; bearings S. 26 W., and S. 72 W. ; lat. 23 12' N. ; long. 62 35' 15" W. ; Y t = 116 19' 47" ; Y 2 = 53 45' 30" ; XY = 45 54' 16" ; bearings S. 26 14' W., and S. 71 37' W. ; lat. 23 12' 6" N. : long. 62 35' 7" W. 25. W.H.A. 2 h 23 m 50'-5 ; error, 5 m 37' fast. 26. First course, S. 36 48' E. ; final, N. 74 35' E. ; dist. 6971 miles ; vertex lat. 54 6i S. ; long. 150 E. : points : lats. 0, 13 30' S.. 25 18' S., 34 38' S., 63 42' S. 27. 1950-33. 28. 214-37. 29. P = 96 25' 50" ; P.Q. = 61 31' 28". 30. A = - -25 ; B = + l-5 ; C = - 16 ; D = - 3'7o ; E = - 3'75. Deviations, East, + 5 ; + 8' E. ; - 4 58' ; - 9 55' ; - 14 14 ; - 17 37' ; - 19 47' ; - 20 34' ; North, - 20 ; - 18 16' ; - 15 37' J - 12 22' ; - 8 52' ; - 5 30' j - 2 26' ; + 6' ; West, + 2. 31. +8 3'. PAPER IY. 1. 7-382 log = 0-868158. 2. -1149 log = 9-060194. 3. S. 87 W., 20'; S. 53 W., 61'; S. 66 W., 51'; S. 75 W., 37', S. 72 W., 59' ; S. 69 W., 50' j N. 89 W., 40' j S. 21 E., 42' ; D. lat. ANSWERS. 433 142-6 S. ; dep. 2787 W. ; course, S. 63 W. j dist. 313 miles; lat. 34 40' N. ; long. 1 19' W. 4. Dec. 17 29' 20" N. ; lat. 56 54' 31" S. 5. 68-73 miles. 6. True course, N. 74 36V E. ; compass course, N. 59 46' E. ; dist. 2433 miles. 7. 10 h 49 m A.M. ; ll h 26 ra P.M. 8. True amp. W. 6 28' N. ; error, 9 17' E. ; dev. 19 47' E. 9. M.T.G. 9 d 21 h 18 m 25' ; at. lat. 37 14' S. ; E.H. A. 3 h 44 m 36' ; st. long. 14 25' 45" W. ; noon long. 13 57' 34" W. ; true azi. S. 82 20' 20" E. ; dev. 28 54' 40" E. 10. Cor. - 11 + 4 + 2 ; true azi. S. 111 47' W. ; dev. 11 2' E. 11. E.H.A. 29" 29 ; nat. no. 6980. Arc. I. 14 46' 41" ; Arc. II. 43 33' 38" ; lat. 28 46' 57" S. 12. First T.A. 49 3' 18" ; P.D. 81 18' 3" ; H.A.'s O h 42 m 16', and O h 24 m 3' : long, (a), 177 52' W. ; (6), 177 34' 45" E. ; second T.A. 19 9' 3" ; P.D. 81 21' 45" ; H.A.'s 4 h 43 m 0', and 4 h 42" 21' ; long, (c), 179 32' 30" W. ; (d), 179 42' 15" W. Second line: { N g. 9 K '} ' azi. S. 81 W; lat. 49 11' N. ; long. 179 40' W. 13. ll h 21 m 47 P.M. 14. Altair, S. O h 22 52* W. ; 7 Cygni, S. O b 9 m 47* E. ; Deneb, S. O h 29' 10" E. 15. Alt. 73 3' 20". 16. E.H.A. 4 h 23 m 19 s ; cor. -19-1+13; true azi. N. 80 5' E. ; dev. 9 33' W. 17. M.T.G. 2" 15 h 49 m 27* ; W.H.A. 4 h 44 m 10'-5 ; E.A.M.S. 8 h 43 m 17 ; long. 97 5' 15" W. 18. K A.M . 2 h 5 m 56' ; cor. - 1 10' 51" +!'+!' 10" ; lat. 28 14' 10" N. 19. M.T.G. 14 d 10 h 52 ra -6 ; S.D. 15' 2' ; H.P. 54' 26" ; dec. 19 12' 35" N. ; T.A. 47 7' 32" ; lat. 23 39' 55" S. 20. Lat. 56 46' 23" S. 21. Mag. bearing, N. 6 W. ; comp. course, S. 27i E., N. 10 E., N. 84 W., S. 75 W. ; mag. course, N. 74 E., S. 61 E., S., N. 60 W. ; mag. bearings, S. 30 W., and S. 54 E. 22. Red, - 5 ft. 1 in. 23. Jupiter' segments: R.A. 6 h 9 m 44'-5; dec. 23 7' 26" N. ; R.A.M.S. 9 h 15 m 6-2 ; E.H.A. 4 h 50 m 39"5 ; T.A. 27 34' 42" ; A.A. 27 36' 29". Moon's elements: R.A. 2 h 28 m 56' '1 ; dec. 9 8' 57" N. ; S.D. 15' 6"; H.P. 54' 30" ; A.A. 54 41' 48" ; T.A. 55 12' 37" ; E.H.A. l h 9 m 51 ; app. dist. 54 28' 13" ; B = 36 27' 21" ; Z = 66 39' 32" ; true dist. 54 30' 24" ; error, 2 ra 36' '5 slow ; long. 22 9' 15" W. 24. e Orionis : E.H. A.'s l h 12 ra 41 , and l h 3" 26* ; long, (a), 81 58' 30" E., (6), 84 17' 30" E. Alphard: E.H.A. 's 4 b 54 ra 50', and 4 h 54 m 23' ; long, (c), 84 19' 30" E., (d), 84 26' 15" E. ; bearings, N. 23 E., and N. 84 E. ; lat. 47 43' S. ; long. 84 27' E. ; XY = 58 2' 52" ; X a = 93 40' 57" ; X 2 = 51 12' 10" ; bearings N. 21 13' E., and N. 83 22' E. ; lat. 47 42' 48" S. ; long. 84 26' 30" E. 25. Eq. time - 7 m 13'-8; P.D. 89 39' 10"; H.A. 2 h 21 ra 54'; error, l m 55-8 slow. 26. First course, S. 66 26|' E. ; dist. 3708 miles ; vertex, 41 1' S. ; long. 20 51' W. Point, lat. 39 6f N. ; long. 0. 27. BP = 241-1 ft.; BAQ = 29 5' 30". 28. XZ = 1684 ft. 29. Dist. 4032 miles ; lat. 43 31' 30" N. 30. 151-3 ft. 2 434 ANSWERS. 31. W.H.A. 7 h 26 m 40'-5 ; true azi. 44 56' - 13 38' = S. 31 18' W. ; dev. 11 3' E. 32. K 56 22' E. ; = - 19 53'. 33. A = - 1'75 ; B = - 5'5 ; C = - 15 ; D = 0; E = + 6-75 ; Deviations, East, - 14 ; - 10 29' ; - 5 54' ; - 36' ; + 4 56' ; + 10 16' ; + 14 49' ; + 18 10' ; South, + 20 ; + 20 16' ; + 19 2' ; + 16 23' ; + 12 44' ; + 8 35' ; + 4 16' ; + 18' ; West, - 3. 34. 0. PAPER V. 1. 62-7577 log = 1-797667. 2. 0-0075 log = 7-875091. 3. S. 48 E., 20' ; E. 40' ; S. 82 E., 42' ; N. 64 E., 44' ; N. 88 E., 46' ; S. 80 E.. 48' ; E. 46' ; N. 83 E., 54' ; D. lat. 0, dep. 328'9 E. ; course E., dist. 329 miles ; lat. 62 30' N. ; long. 169 13' '7 W. 4. Dec. 16 52' 44" N. ; T.A. 77 49' 43" ; lat. 4 42' 27" N. 5. 557 '4 miles. 6. True course, S. 71 18' W. ; compass course, N. 89 12' W. ; dist. 3793 miles. 7. No. A.M. ; O h 4 m P.M. 8. Dec. 14 19' 10" N.; true amp. W. 20 17' N. ; error, 21 47' E. ; dev. 24 7' E. 9. M.T.G. 29 d 16 h 32 m 46 8 ; P.D. 92 25' 31" ; eq. time, - 9 m 39 s ; st. lat. 46 50' : W.H.A. 4 h 42 m 25 1 ; sight long. 180 W. ; noon 179 13' 40" W. ; true azi. S. 73 57' 42" W. ; dev. 32 6' W. 10. Cor. + 36 - 19 + 28 ; true azi. N. 125 17' E. ; dev. 47' E. 11. W.H.A. 24 m 19 s ; nat. no. 4724. Arc I. 13 13' 14" ; Arc. II. 43 38' 22" ; sight lat. 30 25' 5" S. ; noon 30 22' 23" S. (c), 179 55' 30" W. ; (d), 179 39' 15" W. /"XT ftfi W "k First line : | g g^ j/j J second azi. S. 74 W ; lat. 46 45|' N. ; long. 179 58' W. 13. A.T.S. ll h 8 m 31' P.M. ; inspection, ll b l m P.M. 14. a Gruis S. 57 m 20* W. ; Fomalhaut, S. 7 m 10 s W. 15. Alt. 40 15' 10". 16. M.T.G. 2 d 13 h 25 ra 40 ; W.H.A. l h 34 m 35" ; R.A.M.S. 8 h 42 m 53' ; long. 62 35' W. 17. W.H.A. 2 h 58 48 s ; cor. -35 + 24 + 5; true azi. S. 133 43' W. ; dev. 22 51' W. 18. Cor. - 40' 35", + 37", + 50" ; lat. 50 16' 33" N. 19. M.T.G. 9 d 18 h 31 m -5; dec. 5 23' 11" N. ; S.D. 15' 11" ; H.P.54'51"; T.A. 48 25' 19" ; lat. 46 57' 52". 20. Lat. 51 28' 29" S. 21. Mag. bearing, N. 56 E. ; compass course, N. 68 E., S. 67 E., S. 79 W., N. 19 W. ; mag. course, S. 68 W., S. 18 E., N. 40 E., N. 53 W. ; mag. bearings, S. 21 W., and S. 49 E. 22. - 8 ft. 5 ins. 23. Sun's elements: RA. 9 h 43 m 37" '8 ; dec. 13 39' 57" N. : RA.M.S. 9 h 39 m 32-5 ; A.A. 30 38' 31" ; T.A. 30 37' 1" ; E.H.A. 4 h 7 m 20'. Moon's elements: R.A. 7 h 35 m 38 '5 ; dec. 19 51' 24" N. ; S.D. 15' 22" : H.P. 55' 27" ; E.H.A. 1" 59 m 21' ; T.A. 59 6' 1" ; A.A. 58 37' 44" ; app. dist. 30 47' 7" ; B = 40 59' 0" ; Z = 18 49' 30" ; true dist. 31 12' 20" ; error, 12 m 15' fast ; long. 35 40' W. 24. H.E.T. l h ll m 31 s -5 ; mid. dec. 6 11' 49" ; cor. run + 7' 25" ; first ANSWERS. 435 alt. 32 39' 46" ; second alt. 47 26' 36" ; XY = 35 32' 54" ; X, = 88 3' 46" ; X 2 = 52 34' 2" ; lat. correction, - 1' 47" ; lat. 47 42' 53" N. 25. P.D. 97 32' 29" ; eq. time, - O m 42"7 ; W.H.A. 4 h 30 ra 26 s ; error, 20 m 54 s -4 slow. 26. First course, S. 57 17' 30" E. ; final, N. 62 17' 30" E. : first long. 67 12' E. ; second long. 104 21' E. ; dist. = 2300 + 1549 + 1830 = 5679 miles ; points : lat. 44 41|' S., long. 50 E., lat. 40 4f S., 140 E. 27. 1009. 28. N. 70 7' 25" W. ; drift 67 '55 miles. 29. Side 33 38' 50" ; angle 100 30' 56". 30. N. 70 E. ; B= + 5-86. 31. A = - '25 ; B = + ll-5 ; C = ; D= + '5 ; E= +25 ; Deviations, North, + 2 25' ; + 4 41' ; + 6 42' ; + 8 23' ; + 9 40' ; + 10 33' ; + 10 59' ; East, + 11 ; South, ; - 2 4' ; - 4 7' ; - 6 5' ; - 7 53' ; - 9 27' ; - 10 43' ; - 11 34' ; West, - 12. 32. -3 45'. PAPER VI. 1. -4839 log = 9-684706. 2. -08604 log = 8-934729. 3. N. 62 W.> 17'; S. 84 E., 72'; S. 75 E., 43'; S. 68 E., 56'; S. 54 E., 57' ; S. 47 E., 56' ; E. 56' ; N. 22 E., 60' ; D. lat. 47'7 ; dep. 315-6 ; course, S. 81 E. ; dist. 318 miles ; lat. 38 33' S. ; long. 174 47' W. 4. Dec. 10' 30" N. ; T.A. 47 17' 56" ; lat. 42 31' 34" S. 5. 32 23'. 6. True course, N. 53 48' E. ; compass course, S. 75 22' E. ; dist. 4561 miles. 7. l h 8 ra A.M. ; l h 54 m P.M. 8. Dec. 2 3' 32" N. ; true amp. E. 2 31* N. ; dev. 5 38' 37" E. 9. M.T.G. 7 d 6 b 47 m 7" ; P.D. 83 47' 4" ; st. lat. 48 10' ; eq. time, - l m 54' ; T.A. 32 7' 56" ; E.H.A. 3" 8 28 s ; st. long, 149 22' 15" W. ; noon, 148 33' 45" W. ; true azi. S. 59 20' 20" E. ; dev. 13 54' 40" E. 10. W.H.A. 2 h 18 m 39 s ; cor. -41 + 21 + 20 ; true azi. S. 133 36' W. ; dev. 14 24' W. 11. W.H.A. 23 m 4P ; nat. no. 3884. Arc 1. 16 24' 13" ; Arc II. 57 2' 9" ; lat. 40 37' 56" S. 12. First P.D. 89 41' 9" ; eq. time, - 7 m 16' ; T.A. 32 20' 25" ; H.A.'s 2 h 34 m 37 and 2 h 30 m 56 s ; long, (a), 81 22' E. ; (6), 82 17' 15" E. ; second P.D. 89 46' 1" ; eq. time, - 7 m 20 s ; T.A. 32 20' 25" ; H.A.'s 2 h 34 m 3 s and 2 h 30 21' ; long, (c), 83 29' 15" E. ; (d), 82 33' 45" E. First line: {^' ^jf^'} J first azi. S. 47 E; lat. 47 51' N. ; long. 82 32' E. 13. M.T.S. l h 45 ra 15 s P.M. 14. Achernar, S. l h 6 ra 7 1 W. ; a Persei, N. 37 m 16 3 E. ; Aldebaran, N. 1" 50 m 12* E. 15. Alt. 25 54' 40". 16. W.H.A. 4 h O m 20 s ; cor. - 5 - 33 + 25 ; true azi. N. 102 36' W. ; dev. 10 54' E. 17. M.T.G. 5 d 12 h 51 ra 4 s ; E.H.A. 2 h 13 ra 13' ; R.A.M.S. 8 h 54 m 37 s ; long. 50 10' W. 18. Cor. - 2' 12", + 35", + 1' 25" ; lat. 38 23' 24" N. 19. M.T.G. 6 d 17 h 35 m '6 ; S.D. 15' 42" ; H.P. 56' 50" ; dec. 7 41' 9" S.; cor. + 46' 30" ; T.A. 33 8' 9" ; lat. 64 33' 0" S. 20. Lat. 34 19' 53" N. 21. Mag. bearings, South ; comp. course, N. 3 W., N. 80 W., S. 66 E., 436 ANSWERS. N. 31 . ; mag. course, S. 55 E., S. 30 W., S. 77 W., N. 19 W. ; mag. bearings, N. 30 W., S. 72 W. 22. Red, 2 ft. 9 ins. 23. Star's elements: R.A. 16 h 23 m 38"5 ; dec. 26 13' 26" S. ; R.A.M.S. 10 h 31 23"5 ; A.A. 49 25' 23" ; T.A. 49 24' 34" ; W.H.A. 2 h 38 m 10 s . Moon's elements : R.A. 19 h 38 m 16-3 ; dec. 19 49' 51" S. ; S.D. 16' 20" ; H.P. 58' 50" ; A.A. 76 46' 33" ; T.A. 76 59' 48" ; E.H.A. 36 m 28' ; app. dist. 45 3' 37" ; B = 13 48' 1" ; Z = 102 49' 30" ; true dist. 44 58' 32" ; error, ll m 38 s slow ; long. 98 26' 45" E. 24. H.E.T. 2" 10 21-5; run cor. - 25' 12"; first T.A. 43 45' 14"; second 23 28' 34" ; XY = 28 43' 33" ; X, = 60 25' 42" ; X 2 = 126 49' 36" ; lat. 48 20' S. 25. Ship 18 32' S.; 7035'-5W.; W.H.A. 3 h 22 m 64'; error, 37 m 40' slow. 26. First course, S. 50 4.V W. ; final, N. 58 46 J' W. ; longs. 128 32J W., 142 32' W. ; dist. 5050*1 mile's ; lat. 49 41' S. ; long. 120 W. ; lat. 43 244 S. ; long. 180 W. 27. B = 50i40' 12" ; A = 39 19' 48" ; BD = 14383 ft. 28. 115-6 ft. 29. Course, S. 78 30' E. ; dist. 92 miles. 30. A = + l-25 ; B = + 18 ; C = - 7'5 ; D = + '75 ; E. = + l-25 ; Deviations, + 2 37' ; + 6 11' ; + 9 26' ; + 12 16' ; + 14 40' ; +16 34' ; East, + 18 ; + 18 55' ; + 19 20' ; + 19 13' ; + 18 32'. 31. Lat. 42 5' ; H.A. 7 h 10 m 21-5. 32. Lat. 60 22' S. ; long. 30 26' W. ; course S. 3 E. j dist. 43 miles. 33. - 25 30'. PAPER VII. 1. 696090-3 log = 5-842665. 2. 19534-82 log = 4-290809. 3. N. 12 W., 20'; N.15E., 22'; S.71W.,24'; N. 7 E., 214' ; D. lat. 245' -5 ; dep. 4-9 ; D.R. lat. 46 4-'5 S. ; long. 110 7' 40" W. ; set S. 70 W. ; drift 16^ miles. 4. Dec. 2 24' 32" S. ; T.A. 88 36' 7". Lat. 1 0' 39" S. 5. 51 23' N. ; 16 16' 30" W. 6. Compass course, N. 16 22' W. ; dist. 5489 miles. 7. Midnt. ; O h 23 P.M. 8. True amp., E. 16 55' N. ; dev. 2 21' E. 9. M.T.G. 23 d 16 h 14 m 51' ; P.D. 89 55' 8" ; H.A. 2 h 57 m 0" ; long. 158 38' E. ; true azi. N. 67 51' 28" W. ; dev. 6 51' 28" W. 10. A.T S. 8 h 33 ra 13 s A.M. ; cor. + 21 + 14 + 5 ; true azi. S. 126 15' E. ; dev. 12 59' E. 11. E.H. A. 27 m 24 s ; nat.no. 5109. Arc. I. 2 12' 50" ; Arc. II. 46 29' 45". Lat. st. 44 16' 55" N. ; noon, 44 20' 13" N". 12. FirstP.D. 107 28' 11"; T.A. 8 40' 49" ; eq. time, + 5 m 59" ; H.A.'s 3 h 41 m 54 s and 3 h 39 m 39 s ; long, (a), 49' E. ; (6), 1 22' 15" E. ; second P.D. 107 24' 12" ; T.A. 18 19' 28" ; eq. time, + 5 m 57" ; H.A.'s 2 h 21 m 22 s and 2 h 17 m 9 s ; long, (c), 1 9' 30" E. ; (d), 6' E. Second line: {^ 55 El'}; second azL K 35 W< ; lat ' 47 25/ S ' ; long. 3' W. 13. A.T.S. 9 h 23 ra 19" A.M. 14. Alt. 31 17' 20". 15. Aldebaran, S. 47 m 36" E. ; Capella, N. l h 26 M 49' E. ; Bigel, S. l h 27 m 5 s E. ; Bellatrix, S. l h 37 m 9' E. ; Tauri, S. l h 37 m 25 8 E. ; Orionis, S. l h 48 m 31" E. 16. W.H.A. 3 h 19 m 40' ; cor. - 49 + 18 4- 4 : true azi. S. 130 32' W. ; dev. 18 39' W. ANSWERS. - 437 17. M.T.G. 30 d l h 56 m 56' ; W.H.A. 2 h 38 m 10 s ; B.A.M.S. 10 h 31 m 24' ; long. 98 22' 15" E. 18. Cor. - 1 4' 5", + 2", + 1' 25". ; lat. 20 14' 2" N. 19. M.T.G. 30 d 8 h 21 m -9 ; S.D. 16' 11" ; H.P. 58' 41" ; dec. 19 29' 21" S. ; T.A. 30 15' 24" ; lat. 40 15' 15" N. 20. Lat. 72 1' 56" K 21. Mag. bearing, N. 73 30' W. ; comp. course, N. 2 W., S. 74 E., S. 82 W., N. 55 E. ; mag. course, S. 67 E., S. 20 W., S. 86| W., K 74 W. ; mag. bearings, S. 25 W., and N. 61 30' W. 22. - 9 ft. 5 ins. 23. Planet's elements: R.A. 23 h 2 m 1'; dec. 8 24' 48" S. ; R.A.M.S. 9 h 17 m 5 .. 6 . A A 4QO 57 / 58 /i . T A 40 o 56 / 54 // . W.H.A. 3 h 13 m 54'. Moon's elements: KA. 2 h 52 m 18"! ; dec. 10 59' 8" N. ; S.D. 15' 3"; H.P. 54' 23" ; E.H.A. 36 m 23 s ; T.A. 52 37' 46" ; A.A. 52 5' 6" ; app. dist. 60 43' 53" ; B = 27 38' 28" ; Z = 93 30' 2" ; true dist. 60 28' 13" ; error, 10 m 34' '5 ; long. 169 38' 30" E. 24. H.E.T. 2 h 6 m 5' ; mid. dec. 2 23' 28" S. ; run cor., - 24' 6" ; first alt. 40 21' 10" ; second alt. 11 18' 30" ; XY = 62 58' 56" ; ^ = 91 31' 18" ; X 2 = 98 18' 12" ; lat. cor., + 2' 28" ; lat. 46 50' 30" N. 25. Ship lat. 5 36' S. ; long. 35 0' 32" W. j E.H.A. 3 h 15 m 41' ; error, 20 m 3 fast. 26. First course, S. 39 16J' W. ; final S. 38 40J' W. ; vertex lat. 59 11J', N. ; long. 58 43' E. ; dist. 5101 miles ; course at equator, S. 31 W. : in long. 31 16|' W. 27. 368 ft. 28. Bearing S. 27 19' 49" W. ; set E. 3 49' 42" N. 29. A = 31 12' 14" ; AC = 72 23' 48". 30. A =0 ; B = + 5 5 ; C = - 28-5 ; D = - 4'5 ; E = - 1. De- viations, West, - 4 30' ; - 38' ; + 3 22' ; + 7 29' ; + 11 46' ; + 16 6' ; + 20 20' ; + 24 14' ; South, + 27 30' ; + 29 49' ; + 30 55' ; + 30 32' j + 28 32' ; + 24 56' ; + 19 53' ; + 13 35' ; East. + 6 30'. 31. +4 38'. 32. H.A. 0; dev. 6 15' E. PAPER VIII. 1. -000001075 log = 4-031408. 2. -1 log = 9-000000. 3. S. 3 W., 30'; S. 34 W., 74'; S. 1 W., 26'; S. 3 W., 56'; S. 54 W., 52' ; S. 56 W., 75' ; S. 71 W., 45' ; S. 65 W., 10' ; D. lat. 264-3, dep. 204-9 ; course, S. 37 47' W. ; dist. 334-5 miles ; lat. 55 24' 42" N. : long. 50 16' 30" W. 4. Dec. 1' 5" S. ; T.A. 56 48' 49" ; lat. 33 12' 16" S. 5. Noon long, 83 38' -4 W. 6. True course, S. 54 10' E. ; compass course, S. 36 50' E. ; dist. 9224 miles. 7. 6 h 54 m A.M. ; 7 h 15 m A.M. 8. Dec. 2 3' 16" S. ; true amp. W. 3 12' 10" S. ; dev. 27 2' 10" W. 9. M.T.G. 3 d 20 h 20 m 51' ; P.D. 107 28' 12" ; eq. time, + 5 m 59 s ; T.A. 7 52' 33" ; E.H.A. 3 h 45 m 18' ; long. 2' 30" W. j true azi. N. 53 16' E. ; dev. 13 16' E. 10. A.T.S. 5 h 7 m 10 P.M. ; cor. - 41 - 22 + 1 ; true azi. N. 100 47' W. ; dev. 8 33' E. 11. E.H.A. 39 m 8 ; nat. no. 9125. Arc I. 30 29' 10" ; Arc. II. 73 56' 9". Lat. 43 26' 59" N. 12. First T.A. 37 38' 2" ; P.D. 92 43' 47" ; eq. time, - 9 ra 55* ; H.A.'s O h 46 ra 6" and O h 25 m 57 s ; long, (a), 175 38' 45" E. ; (6), 179 19' W. ; second 438 ANSWERS. T.A. 23 12' 42" ; P.D. 92 47' 27" ; eq. time, - 9 m 58' ; H.A.'s 3 h 16 m 57* long, (c), 17 IN. 79 E.I ; eq. t and 3 h 14 m 6' ; long, (c), 179 21' 45" W. ; (rf), 179 55' 30" E. First line: {g 790^'}; second azi. S. 55 W ; lat. 49 16J' N. ; long. 179 59V E. 13. M.T.S. 8 h 9 m 53 s A.M. 14. Aldebaran, N. l m 44' E. ; Capella, N. 40 ra 58 s E. ; Rigel, N. 41 m 14 s E. ; Bellatrix, N. 51 m 18 s E. ; ft Tauri, N. 51 m 34 s E. 15. Alt. 47 20'. 16. E.H.A. 3 h 21 m 1 : cor.. -11-43 + 31 ; true azi. N. 101 43' E. ; dev. 28 13' E. 17. M.T.G. 19 d 2 h 2 ra 40 s ; W.H.A. 9 h 59 m 20 s ; R.A.M.S. 9 h 48 m 2 g , long. 157 30' E. 18. Cor., - 1 11' 58", + 0, + 1' 12" ; lat. 28 41' 36" K 19. M.T.G. 31 d 23 h 9 m -9 ; S.D. 16' 8" ; H.P. 58' 6" j dec. 15 45' 11" S. ; cor. + 8' 37" ; T.A. 81 26' 50" ; lat. 7 12' 2" S. 20. Lat. 11 52' 30" N. 21. Mag. bearing, N. 2 W. ; comp. course, N. 8 E., S. 71 E., S. 79 W., N. 57 W.; rnag. course, N. 85 E., N. 72 W., S. 65 W., S. 20 E. ; mag. bearings, S. 20 W., and S. 50 E. 22. - 2 ft. 9 ins. 23. Sun's elements: R.A. 9 h 38 m 24' -2 ; dec. 14 6' 19" N. ; R.A.M.S. 9 b 34 m 2' ; A.A. 68 51' 57" ; T.A. 68 51' 38" ; W.H.A. l h 20 m 17'. Moon's elements: R.A. 6 h 22 m 19"5; dec. 20 12' 13" N. ; S.D. 15' 4" H.P. 54' 48" ; W.H.A. 4 h 36 ra 21' ; T.A. 26 24' 26" ; A.A. 25 36' 59" app. dist. 47 50' 16" ; B 32 52' 29" ; Z = 34 29' 14" ; true dist. 47 4' 21" error, 7 m 3' slow ; long. 170 36' W. 24. H.E.T. 2 h 9 ra 18 a '7; run cor., -12' 36"; first alt. 67 4' 46" second alt. 68 22' 20" ; XY = 44 28' 55" ; X t = 65 54' 45" ; X 2 = 3 12' 0" lat. 48 29' 42" N. 25. P.D. 84 22' 30" ; eq. time, - 2 m 25" '8 ; E.H.A. 2 h 45 m 48 s ; error 9 m 53-7 fast. 26. First course, S. 63 36' E. ; longs. 60 8' E., and 79 2' E. ; dist 4705-6 miles ; lat. 41 33' ; long. 50 E. ; lat. 41 28' 30" S. ; long. 90 E. lat. 40 3' 30" S. ; long. 100 E. 27. N. 35 6J' E. ; dist. 274'5 miles. 28. QR = 260-9 or 181-2. 29. 86-6 ft. 30. R = 100 32' 7" ; RQ = 94 12' 4". 31. A = 84 4' 12" ; AB = 83 50' 5". 32. Lat. 68 10' 30" N. ; long. 34' W. 33. N. 42 43' 30" W. 34. A = + -75 ; B = + 14 ; C = - 10'5 ; D = + 3'25 ; E = + 7'25. Deviations, North : - 2 30' ; + 3 50' ; + 6 29' ; + 6 50' ; East, + 7 30' ; + 10 16' ; + 14 49' ; + 18 38' ; South. + 18 30' ; + 12 31' ; + 1 31' ; - 10 59' ; West, - 20 30' ; - 23 38' ; - 19 49' ; - 11 29'. 35. 0. 36. E.H.A. l h 20 m 45': true azi. 94 4' + 69 30' = N. 163 34' E. ; dev. 7 19' E. THE END. PRINTED IN GREAT BRITAIN BY WILLIAM CLOWES AND SONS, LIMITED, BECCLES. INDEX FOR TIDAL EXEECISES. INDEX FOR TIDAL EXERCISES. SHOWING STANDARD POETS IN SMALL CAPITALS. Name. Standard Port Name. Standard Port ABU SHAHR page (12) KINGSTOWN page (9) Al Basra Abu Shahr Kuweit Abu Shahr Annan Foot Liverpool Ardriahaig Lamlash Greenock Ardrossan Greenock Limerick Galway Ayr Greenock LIVERPOOL P*8 e (7) Llanelly Pembroke Dock Beachy Head Dover Beaumaris Liverpool Mellon Galway Bridlington Hull PEMBROKE DOCK page (8) Carlisle Port Liverpool Clydebank Greenock Quaco St. John Crinan Greenock Ramsgate Dover Dartmouth Devonport Rotterdam Dover DEAL page (4) Rye Bay Dover DEVONPORT Dingle page (3) Galway St. Andrews St. John Donaghadee Kingstown ST. JOHN, N.B. . page (11) DOVER Dungeness page (4) Dover St. Mary's Seal Cove Devonport St. John SOUTH ROCK page (9) Exmouth Devonport Spurn Point Hull FLEETWOOD Strangford Quay, page (7) Kingstown Fowey Foynes Island Devonport Galway Tenby Tor Bay Pembroke Dock Devonport GALWAY GREENOCK page (10) page (6) Valentia Galway Grimsby Hull Warren Point Kingstown Hastings Dover Welchpool St. John Henjam Island Abu Shahr Whitehaven Liverpool Hilbre Island Liverpool Wicklow Kingstown HULL page (5) Workington Liverpool -' < f ENGLAND, SOUTH COAST DEVON PORT. time). (Standard time). . Long. 4 10* W. | \VKKK DAY. j 1 MONTH DAY. SEPTEMBER. 1 I >. 5 OCTOBER. Time. Ht. H. M. F. J. Time. Ht. H. XJ 9. L Time. H X. Ht. F. I. Time.1 B. M. Ht. F- I. S Time. Ht H. M.' F. I. Time.) Ht. H. M. F. I. Time.! H. M. Ht. F. I. Time. 1 H. M. Ht. F I. F. 1 1 47 2 7 45 15 2 1 8 8 015 S. 1 1 52 6 7 53 15 1 O Q 9 8 11 14 7 S. 2 2 16 9 8 1614 8 2 32 1 1 83414 6 L 1 2 26 1 2 8 30 14 8 2 44] 1 3 8 4913 11 1 Tu. 3 3 2 2 9 8 4 2 3 22 2 1 9 31 13 3 S. 3 2 48 1 4 8 49 14 3 3 4 1 8 9 713 11 V. 4 3 42 3 3 9 58 13 6 4 7 3 10 3212 3 M. 1 3 21 2 3 9 25113 9 339 2 5 9 47 13 2 Th. 4 43 4 1 10 11 9 5 27 3 10 11 5511 9 Tu. o 3 57 3 2 10 813 2 4 17 3 4 10 36 12 5 j> 6 6 13 6 4612 11 7 li 3 9 W. 6 4 47 4 3 11 11 12 7 5 26 4 2 11 5211 9 ;. . 1 32 12 3 7 46 4 3 2 15 13 9 8 30 2 7 Th. " 6 10 5 1 4312 6 6 58 4 3 F. S 1 3512 7 48 4 7 2 1913 4 8 34 3 S. 8 2 54 13 4 9 12 2 11 3 29 14 8 9 46 1 5 S. 9 3 5,13 9 23 3 3 3 44 14 8 10 1 1 5 M. 9 3 54114 4 10 11 1 8 4 2015 5 10 35 5 Tu. 10 4 45:15 2 10 59 7 5 816 11 22 *0 4 S. 10 4 16 14 2 10 32 1 10 4 4215 8 10 57 2 W. 11 9 28 15 9 11 43 *0 2 5 4716 2 - M. 11 5 8*15 11 22 7 5 3316 3 11 48 *0 7 Th. 2 3*0 8 6 8 15 11 23 *0 6 6 2715 11 Tu. i 5 55 15 9 11*0 3 6 18 16 7 7 f j 41 *0 6 6 46 15 9 1 0*0 31 7 415 6 W. i 33 *1 2 6 38 16 2 53*0 8 6 59 16 6 >. [4 1 19 1 7 18 15 4 1 34 3 7 3514 10 Th. 14 1 12*1 ( 7 16 16 1 31 *0 5 7 31 16 15 1 47 *0 6 7 49 15 7 2 5 1 8 615 4 ST. 15 1 61 10 7 51 14 5 2 7 1 5 8 81311 B.' 16 2 22 5 8 22 14 11 2 37 Oil 83814 5 M. 16 2 23 2 2 8 24 14 1 2 38 1 11 8 3813 Cu. 2 62 3 2 8 52 13 4 3 6 2 11 9 11 12 1 17 2 53 1 6 8 51 14 1 3 5 2 1 9 713 4 W. IS 3 22 4 4 9 33 12 6 3 42 4 1 9 5911 2 M. 18 3 21 3 9 24 13 3 3 36 3 2 9 4512 2 Th. lf 4 9 5 6 10 31 11 10 4 45 4 11 11 610 7 iTu. 19 3 57 4 3 10 2 12 3 412 4 4 10 31 11 1 7, 20 5 24 6 2 11 47 11 7 6 4 5 2 20 4 44 5 6 11 3 11 6 5 21 5 3 11 4410 6 S. 31 10 10 6 46 6 10 1 9 12 2 7~23 4 6 Th. 21 6 2 6 4 3011 6 6 45 5 i 22 1 1310 8 7 27 5 10 1 5612 2 8 10 4 81 S. 22 1 48 11 7 8 2 4 10 2 24 12 10 8 40 3 5 2( 2 35 11 6 8 51 4 9 3 9 13 1 927 3 1 M. ru. 23 24 2 53 3 42 12 6 13 7 9 11 9 59 3 8 2 6 3 20 13 8 4 514 4 9 37, 2 5 10 21 1 7 k. 24 3 4012 7 9 57 3 6 4 513 11 10 21 2 1 w. 25 4 26 14 6 10 42 1 6 4 46 15 11 11 8 M. 25 4 26 13 5 10 42 2 6 44714 7 11 2 1 3 Th 26 5 6 15 1 11 20 7 5 2415 4 11 39 3 Iru 26 5 614 3 11 20 1 6 5 2415 2 11 38 5 F. 27 5 43 15 6 11 59 6 115 6 2" 5 40 15 11 55 7 55715 7 S. 17 *0 1 6 20 15 7 351*0 2 6~41 15~~6 th 28 13 *0 1 6 1515 4 29 1 6 30 15 6 2! 45,*0 1 6 4815 6 1 1 7 515 6 ST. 2 54 6 59 15 8 1 14 7 1915 2 i 30 1 19 1 7 1915 4 1 35 3 7 3615 2 M. 31 1 33 6 7 37 15 6 1 53 3 7 5914 8 1 TU. 3 2 15 1 1 8 21 15 236 010 8 4314 1 * Signifies below Datum. The tides are placed in order of occurrence, p.m. times and heights being printed in large block A comparison of consecutive heights will indicate whether it is high or low water. The times given are STANDARD, or mean time of the meridian of Greenwich. To obtain local time subtract 17 minutes. TIDAL DATUM, ETC. The zero of the predictions, which is also the datum of the Admiralty charts of Devonport, is 8*42 feet low Ordnance datum, and is the level of mean low water spring tides. TIDAL DIFFERENCES ON DEVONPORT. Place. Lat. N. Long. W. Time of Height of H. W. Height of L. W, H. W. L. W. Springs. Neaps. Springs. Neaps. St. Mary's . . Fowey . . . Dartmouth . . Tor Bay . . . Exmouth . 49 55 50 20 50 21 50 25 50 37 619 4 38 3 34 3 32 3 25 H. M. -1 17 27 + 30 + 14 + 41 H. X. F. I. + 4 -0 5 -1 3 21 -4 5 F. i. -0 5 -0 6 -15 -2 -3 8 F. I. F. I. -0 5 -0 5 -0 4 -1 ( 3 ) ENGLAND, SOUTH COAST DOVER. H. W. F. & C. ( XI H - 12M. 1 XL 7. (Local mean time). (Standard time). Lat. 51 7' N. Long. 1 19' E. i 14 *i P v H a JANUAKY. ^ M 1 H <] P a Q FEBKUARY. Time. H. M. Ht. F. 1. Time. H. M. Ht. F. I. Time. H. M. Ht. F. I. Time. H. M. Ht. F. I. Time. H M. Ht. P. I. Time. H M Ht. F, I Time. H. M. Ht. Time. Ht. s. 1 1 19 3 9 6 63 15 8 2 3 7 28 15 4 Tu. 1 3 21 2 10 8 51 16 3 4 4 9 Q a O 9 20 16 : W. 2 4 37 1 7 9 49 17 3 5 9 1 2 10 12 .7 ' 8. o 2 31 3 1 7 58 6 3 3 9 2 4 8 30 16 2 ?h. 5 36 6 40 18 2 6 2 4 10 59 18 M. j 3 39 2 3 8 57 17 1 413 1 5 9 24 17 3 4 6 28 *0 4 1 29 18 9 6 52 *0 3 11 45 19 : Tu. ( 4 42 1 2 9 61 17 11 5 12 8 17 17 11 u 5 7 17 *0 11 16 19 1 7 37 *o W. 5 5 40 6 10 44 18 6 6 8 3 11 9 18 7 Th. 6 6 36 *0 2 11 38 18 9 7 2 *0 2 S. 6 29 19 6 8 1 *1 2 1 19 1 8 19 *0 F. 1 19 7 30 *0 6 31 18 11 7~53 *0 3 M. P 1 11 19 6 8 40 *1 1 1 40 18 9 857 *0 S. y 61 19 2 8 19 *0 8 1 20 18 9 8 40 *0 2 Tu. 8 1 51 19 9 17 *0 7 2 18 18 9 32 W. 9 2 30 18 1 9 53 4 2 55 16 10 10 6 1 8. 9 1 38 19 9 5 *0 6 2 7 18 4 9 23 4 Ih. ^0 3 10 16 11 10 27 1 6 3 33 15 8 10 38 2 M. Q 2 23 18 3 9 46 1 2 53 17 4 10 3 1 1 <\ 3 62 15 8 11 6 2 10 4 20 14 5 11 23 3 Tu. 3 8 17 6 10 28 11 3 39 16 3 10 42 2 1 S. 12 4 50 14 5 11 65 4 2 5 28 13 5 W. 3 64 16 6 11 9 2 1 4 29 15 3 11 27 3 2 Th 4 48 16 6 11 59 3 2 5 30 14 4 S. Li 27 4 10 6 6 13 8 1 10 4 11 6 50 13 F. 24 4 2 5 51 14 9 1 1 3 11 6~36 13 11 M. 14 1 52 5 7 33 13 9 2 39 4 9 8 16 13 S. 1 33 4 6 7 6 14 7 2 16 4 749 14 1 Tu. 15 3 23 4 6 8 54 14 5 4 5 4 9 19 14 W. 16 4 35 3 4 9 48 15 6 5 5 3 10 5 15 8. [ 3 61 4 3 8 17 14 11 330 3 8 855 14 8 Th 17 5 26 2 3 10 28 16 3 5 46 2 3 10 38 16 M. I 4 2 3 9 9 16 15 6 4 35 3 1 9 44 15 5 F. 18 6 5 1 6 11 1 16 10 622 1 8 11 9 17 Tu 4 69 2 11 10 3 16 5 24 2 6 10 24 16 S.' 19 6 40 10 11 33 17 6 654 1 11 40 18 W I 6 45 2 4 10 42 16 6 6 6 2 2 11 16 6 Th 2 6 26 1 10 11 18 16 9 6 44 1 10 11 34 17 S. 2 7 12 2 4 17 10 726 F. 2 7 2 1 < 11 63 17 7 17 1 6 M. 2 12 18 6 7 43 *0 3 35 18 1 7 57 S. 2 7 17 6 7 36 1 27 17 4 7~50 1~3 Tu 2 45 18 9 8 15 *0 5 1 8 18 3 8 28 W 2 1 20 18 9 8 48 *0 5 1 42 18 1 9 S. 2! 40 17 10 8 9 8 1 2 17 6 8 23 1 1 Th 2 1 67 18 6 9 21 2 17 17 6 9 34 M. 2< 1 13 18 1 8 41 6 1 37 17 6 8 53 11 F. 2 2 36 17 8 9 56 9 2 57 16 8 10 10 Tu 2 1 46 18 2 9 13 5 2 13 17 4 9 26 1 1 S. 2 3 19 16 7 10 34 1 9 3 46 15 7 10 54 2 W 2 2 22 17 10 9 47 9 2 50 16 9 9 59 1 5 Th 2 3 1 17 2 10 22 1 4 3 29 16 1 10 33 2 1 . 2 4 20 15 4 11 25 3 1 4 57 14 7 __ F. 2 3 48 16 6 11 1 2 3 4 18 15 3 11 20 2 11 M. 2 Midt 3 6 6 45 14 6 46 4 6 31 14 S. 4 49 15 6 11 57 3 3 5 26 14 7 Tu 2 1 36 3 9 7 26 14 9 2 29 3 8 8 7 15 S. 3 26 3 9 6 f 15 1 11 3 9 6 49 14 6 M. 3 1 61 3 9 7 33 15 3 2 41 3 3 8 14 15 5 * Signifies below Datum. The tides are placed in order of occurrence, p.m. times and heights being printed in large bloc!: type. A comparison of consecutive heights will indicate whether it is high or low water. The times given are STANDARD, or mean time of the meridian of Greenwich. To obtain Iocs, mean time add 5 minutes. TIDAL DATUM, ETC. The zero of the predictions, which is also the datum of the Admiralty chart of Dover, is 8'42 fee below Ordnance datum, and is 0'15 foot above the level of mean low water spring tides. TIDAL DIFFERENCES ON DOVER. Place. Lat. N. Long. Time of Height of H. W. Height of L. W. H. W. L. W. Springs. Neaps. Springs. Neaps. Dungeness . . . Deal 50 54 51 13 51 20 51 56 50 44 50 51 50 56 58 1 25 1 25 4 30 14 35 45 H. H. -0 26 + 2 + 31 + 6 + 12 -0 16 + 10 H. H. F. I. + 3 7 -22 -31 + 4 8 -1 10 -16 + 6" 2 - i" 4 Ramsgate Rotterdam . Beachy head Hastings .... Rye bay .... ... + 1 10 + 5 9 + 3 10 + 8 + 3 2 + 2 11 + 6" 2 ... + b" 7 ( 4 ) ENGLAND, EAST COAST HULL. H. W. F. & C / VI- 29M ' ( Local mean me). \VI. 30 (Standard time). Lat. 53 44' N. Long. 19' W. > q 1 OCTOBER. S H NOVEMBER. ! S DECEMBER. 1 1 HIGH WATER 1 1 HIGH WATER. g a H HIGH WATER. X Time. H M Ht. F I Time. Ht. p 3 Time. Ht. Time. Ht. ^ g Time. Ht. Time. Ht. . M. Tu 3 8 14 8 52 936 20 11 19 11 18 7 832 912 10 1 20 6 19 2 18 W. Th. F. 1 2 3 H. H. 9 32 10 30 11 36 V. I 18 9 17 9 17 1 H. M 959 11 18 3 17 4 F. S. 1 2 H. M 10 27 11 23 F. I 18 8 17 11 H. M 1055 11 51 F. I. 18 2 17 8 W. Th 4 fi 1031 11 36 17 6 16 7 11 1610 S. 4 12 17 058 17 2 S. 3 022 17 7 F. 8. 7 020 2 15 16 6 17 116 258 16 8 17 9 S. M. 5 6 1 48 3 4 17 6 18 4 231 337 1710 1810 M. Tu. W. 4 5 6 1 225 3 33 17 7 17 9 18 3 144 3 1 4 2 17 7 18 1ft R S. M. Tu. 8 9 10 3 35 427 5 6 18 5 19 10 20 9 4 4 446 528 19 2 20 4 21 1 Tu. W. Th. F. 7 8 9 10 4 3 4 45 5 26 6 7 19 4 20 1 20 5 20 7 424 5 5 547 627 19 9 20 3 20 6 20 6 Th. F. S. 7 8 9 425 5 10 5 62 18 10 19 2 19 5 447 531 611 19 19 3 19 5 W. Th. F. S. 11 12 13 14 5 48 6 30 7 8 742 21 4 21 8 21 6 21 6 9 649 725 758 21 7 21 7 21 4 20 9 S. S. M. 11 12 13 6 47 7 25 7 58 20 6 20 2 19 9 7 8 741 813 20 4 1911 19 6 S. M. Tu. W. 10 11 12 13 6 31 7 14 7 47 8 18 19 6 19 6 19 5 19 4 653 730 8 4 836 19 6 19 6 19 5 19 1 S. M. Tu. 15 16 17 8 15 8 52 930 20 4 18 10 17 6 832 910 951 19 7 18 2 1611 Tu. W. Th. F. 14 15 16 17 8 31 9 10 9 52 10 37 18 10 17 10 17 16 4 851 930 1016 11 18 3 17 4 16 8 16 1 Th. F. S. 14 15 16 853 929 10 9 18 8 18 17 6 911 950 1032 18 4 17 9 17 3 W. Th. 18 19 10 15 11 9 16 4 15 6 1039 1144 1510 15 4 S. 18 11 31 16 S. M 17 18 1051 11 35 17 16 7 1111 16 9 F. S. 20 21 1 11 15 5 019 2 15 3 15 9 S. M. 19 20 3 1 18 16 16 3 039 2 2 16 1 16 7 Tu. W. 19 20 2 1 7 16 6 16 7 033 1 49 16 6 16 9 S. M. Tu. 22 23 24 2 40 3 43 423 16 2 17 4 18 6 314 4 5 441 16 9 1711 19 1 Tu. W. Th. F. 21 22 23 24 2 38 337 419 4 57 17 17 11 18 11 19 8 3 8 358 437 519 17 6 18 5 19 4 20 Th. F. 8. 21 22 23 2 31 3 40 433 17 17 10 18 11 3 5 410 459 17 5 18 4 19 5 W. ?' 25 26 27 28 4 57 5 29 6 6 6 45 19 6 20 3 20 9 21 1 512 547 624 7 6 1910 20 6 21 21 1 S. S. \C 25 26 27 541 623 7 12 20 3 20 9 21 6 2 647 733 20 6 2011 21 1 S. M. Tu. W. 24 25 26 27 5 19 6 12 7 2 7 50 19 9 20 6 21 2 21 7 546 637 728 813 20 2 2011 21 5 21 8 vt. Tu. 29 30 31 723 8 1 8 42 21 1 20 11 19 11 742 821 9 6 21 20 7 19 4 Tu. W. Th. 28 29 30 8 55 8 38 9 30 21 20 7 19 6 816 9 5 958 21 20 19 Th. F. S. 28 29 30 8 34 921 10 11 21 8 20 9 19 9 857 946 1033 21 3 20 3 19 3 S. 31 1058 18 9 1121 18 3 Mean tide level 10*- P.M. times and heights are printed in large block type. The times given are STANDARD, or mean time of the meridian of Greenwich. To obtain local mean ime subtract I minute. The zero of the predictions, which is also the datum of the Admiralty chart of Hull, is 10-07 feefc )elow Ordnance datum, and is the level of mean low water spring tides. TIDAL DIFFERENCES ON HULL. Place. LAI N. "f Time of H. W. Height of H. W. Springs. Neaps. 53 35 53 35 54 5 / 4 E. 7 W. 11 H. If. -0 54 -1 4 -1 50 F. I. -1 9 -2 2 -4 11 F. I. -1 6 -1 7 -4 7 Spurn point .... Bridlington . ... ( 5 ) SCOTLAND, WEST COAST GREENOCK. ~ ( XII H - 5 M - (Local mean time), T O A *KO 97' TSJ T.nna 4. 4-fi' W TT ^7V T^ fir C^ \ ^_. - . . . * ' J_j and is ' 24 fo0 ' below the level of mean L ENGLAND, WEST COAST LIVERPOOL. H W F H. W. F. I XI H - 23 M - (Local mean time). {XL 35. (Standard time). Lat. 53 24' N. Long. 3 0' W. t* 2 B SEPTEMBER. ! 1 MONTH DAY. | OCTOBER. Time. H. M Ht. P. I. Time. H. M. Ht. P. I. Time. Ht. Time. H. M. Ht. p i. Time. H. M. Ht. P. I. Time. H. M. Ht. P. I. Time. H. M. Ht. Time. Ht. F. 1 1 1 28 11 7 56 1 9 1 22 27 9 8 5 2 7 S. 1 1 8 28 10 7 59 2 4 1 27 28 1 8 13 2 9 S. 2 1 33 28 5 8 27 2 6 1 53 27 2 8 36 3 3 M. 2 1 46 27 8 8 33 3 7 2 5 27 8 51 4 Tu. 3 2 30 26 1 9 10 5 4 2 50 25 6 9 36 5 7 S. 3 2 8 27 6 8 58 3 7 2 27 26 3 9 10 4 4 W. 4 3 24 24 3 9 69 7 1 3 50 24 1 10 38 7 M. 4 2 48 26 1 9 31 6 1 3 9 25 9 51 5 7 Th. 5 4 40 22 10 11 14 8 4 5 15 23 4 Tu. 6 3 35 24 6 10 16 6 9 4 4 23 8 1047 7 1 F. 6 7 7 6 6 13 22 7 49 8 5 6~51 23 9 W. 6 4 43 22 11 11 22 8 5 20 22 10 S. 7 1 46 6 7 7 47 23 8 2 20 7 3 8 12 25 5 Th. 7 6 7 7 6 10 22 6 51 8 3 6 5723 2 F. 8 1 45 7 7 49 23 2 2 27 7 4 8 2424 9 S. 8 3 5 4 10 8 55 25 7 3 23 5 4 9 9 27 5 S. 9 3 11 6 3 9 5 25 1 3 41 5 5 9 2827 M. 9 4 2 11 9 44 27 6 4 15 3 5 9 54 29 2 Tu. 10 4 47 1 5 10 24 28 11 5 1 2 10 35 30 3 S. 10 4 15 3 1 10 27 3 4 36 3 5 10 15^9 2 W. 11 5 30 6 11 2 29 9 5 43 1 1 11 1530 7 M. 11 6 8 1 2 10 44 28 10 5 24 1 9 10 5830 6 Th. 12 6 10 3 11 39 29 10 6 24 11 11 5430 2 Tu. 12 5 54 11 26 29 9 6 8 9 11 4031 2 F. 13 - 6 48 9 15 29 6 7 4 1 3 W. 13 6 37 *0 6 6 30 1 6 50 4 S. 14 32 29 6 7 24 1 8 50 28 9 7 41 2 2 Th. 14 20 31 1 7 16 *0 4 44 29 10 7 28 7 F. 15 58 30 5 7 52 6 1 21 29 8 4 1 6 S. 15 1 9 28 1 7 58 3 1 26 27 6 816 3 7 S. 16 1 36 29 8 26 2 1 1 57 27 6 8 39 3 1 M. 16 1 47 26 5 8 30 4 9 2 3 26 1 8 52 5 3 Tu. 17 2 28 24 6 9 4 6 9 2 43 24 4 9 31 7 1 6?. 17 2 15 27 2 8 59 4 1 2 33 25 11 9 15 5 W. 18 3 12 22 8 9 47 8 7 3 34 22 9 10 23 8 6 M. 18 2 66 24 11 9 35 6 3 3 14 23 11 9 58 7 1 Th. 19 4 15 21 3 10 48 9 10 4 42 21 10 11 36 9 2 Tu. 19 3 43 22 9 10 22 8 4 4 10 22 3 10 56 8 8 F. 20 5 32 20 10 6 10 3 6 4 21 9 W. 20 4 49 21 2 11 28 9 8 5 25 21 4 S. 21 67 8 10 6 66 21 4 127 9 9 7 23 22 7 Th. OJ 12 9 5 6 14 20 8 51 10 1 6 54 21 5 F. 22 1 43 9 7 44 21 3 2 20 9 5 8 1522 6 S. 22 2 18 7 10 8 9 22 8 240 8 5 8 26 24 S. 23 3 5 7 9 8 54 22 9 3 30 8 9 1324 2 M. 23 3 20 6 3 9 2 24 4 3 32 6 9 9 10 25 9 Tu. 24 4 4 4 8 9 43 26 4 12 5 1 9 4627 3 S. 24 4 1 6 9 44 24 6 4 16 6 4 9 5325 11 W. 25 4 40 3 3 10 15 27 5 450 3 8 10 19 28 6 M. 25 4 44 4 6 10 21 26 1 455 4 9 10 2727 5 Th. 26 5 14 2 3 10 46 28 5 5 27 2 7 10 53 29 3 Tu. 26 5 21 3 10 61 27 3 5 30 3 7 10 5728 6 F. 27 5 48 1 8 11 18 29 6 4 1 11 11 31 29 5 W. 27 5 53 2 11 19 28 1 6 2 2 8 11 2529 2 S. 28 6 24 1 6 11 63 29 3 6 42 1 8 Th. 28 6 23 1 6 11 48 28 7 6 33 2 1 11 5729 6 F. 29 6 54 1 4 19 28 9 7 5i 1 10 S. 29 11 29 3 7 1 1 11 31 29 1 7 21 1 10 S. 30 32 29 5 7 26 1 7 52 28 9 7 38 2 M. 30 52 28 9 7 38 2 6 1 10 28 8 8 1 2 5 Tu. 31 1 35 27 7 8 16 3 9 1 51 27 8 8 45 3 8 1 * Signifies below Datum. The tides are placed in order of occurrence, p.m. times and heights being printed in large block type. A comparison of consecutive heights will indicate whether it is high or low water. The times given are STANDARD or mean time of the meridian of Greenwich. To obtain local mean time subtract 12 minutes. TIDAL DIFFERENCES ON LIVERPOOL. Time of Height of H. W. Height of L. W. Place. Lat. N. Long. W. H. W. L.W. Springs. Neaps. Springs. Neaps. H. M. H. M. P. i. r. i. F. I. F. I. Port Carlisle . 54 57 3 10 + 48 Annan foot . . 54 58 3 16 + 43 -6 5 -3 2 ... ... Workington . . 54 39 3 34 -0 1 -3 2 -3 2 Whitehaven . . 54 33 3 36 -0 7 -3 2 -4 11 + 1 1 + 10 Fleetwood, \ Wyre lighthouse/ 53 57 3 2 -0 12 ... -1 5 -1 8 + 3 -0 1 Hilbre island . . 53 23 3 13 -0 22 -1 2 -0 2 -0 1 Beaumaris . . . 53 16 4 5 -0 49 ... -5 8 -6 8 ... ... The zero of the predictions, which is also the datum of the Admiralty chart of Liverpool, is 14-66 feet below Ordnance datum, and is 1-40 feet below the level of mean low water spring tides. ( 7 ) WALES PEMBROKE DOCK. T trr -ci ,,n JVI H -12 M - (Local mean time). I. W. F. & C. | VI 32 (Standard time). Lat. 51 42' N. Long. 4 57' W. I S JULY. 3 1 AUGUST. 3 & SEPTEMBER. M g HIGH WATER. H i HIGH WATER. 1 o HIGH WATER. i Time. Ht. Time. Ht. Time. Ht. Time. Ht. ^ Time. Ht. Time. Ht. H. M. F I. H. M. F. I. H. M. F. I. H. M. F. I. S. T H. H. 7 1 F. I. 20 2 H. M. 720 F. I. 20 3 Tu. W. 2 7 45 8 15 21 4 21 5 8 831 21 5 21 5 F. S.' 1 2 8 17 8 52 22 2 21 6 834 9 9 2110 21 2 s. M. Tu. W. Th. F. S. 2 3 4 5 6 7 8 7 35 8 7 841 9 18 9 53 1029 11 9 20 4 20 4 20 4 20 2 19 9 19 2 18 7 751 824 859 935 1011 1048 1128 20 4 20 4 20 3 1911 19 6 1811 18 2 Th. F. S. S. M. Tu. W. 3 4 6 6 7 8 9 8 48 923 9 57 1034 11 19 53 21 4 20 10 20 1 19 3 18 1 17 1 9 3 940 1015 1055 1144 016 141 21 3 20 6 19 8 18 8 17 6 17 3 17 4 S. M. Tu. W. Th. F. S.' 3 4 5 6 7 8 9 927 10 6 1056 11 59 044 2 20 346 20 7 19 4 18 16 11 16 10 17 8 19 8 946 1029 1124 136 311 421 20 18 8 17 3 18 7 20 7 S. 9 11 50 17 10 Th. 10 2 31 17 9 314 18 6 M. Tu. W. Th. F. S. 10 11 12 13 14 15 19 1 23 249 4 7 6 14 6 15 17 8 17 8 18 4 19 9 21 2 22 6 047 2 7 327 441 545 642 17~8 1711 19 20 6 2111 23 1 F. S. S. M. Tu. W. 11 12 13 14 15 16 3 65 5 6 6 3 6 49 7 33 8 14 19 6 21 4 22 10 23 10 24 2 23 9 432 534 628 713 753 834 20 5 22 1 23 6 24 1 24 23 5 S. M. Tu. W. Th. S." 10 11 12 13 14 15 16 4 49 5 42 626 7 5 7 43 8 19 8 55 21 6 23 1 23 11 24 2 23 8 22 8 21 5 515 6 5 644 725 8 2 838 915 22 4 23 6 24 1 24 23 3 22 1 20 6 S. M. Tu. 16 17 18 7 6 762 8 40 23 6 23 8 23 2 730 816 9 1 23 7 23 5 22 11 Th. F. S. 17 18 19 8 52 9 30 10 7 22 11 21 7 19 11 911 947 1025 22 4 20 9 19 S. M. 17 18 933 10 10 19 8 18 950 1032 1810 17 1 W. 19 924 22 6 945 21 11 Tu. 19 1054 16 4 1124 15 8 Th. 20 10 3 21 3 1024 20 7 S. 20 10 42 18 3 11 3 17 4 W. 20 Noon 15 4 F. 21 10 43 19 11 11 3 19 3 M. 21 11 28 16 6 Th. 21 41 15 3 130 15 4 S. 22 11 24 18 5 1144 17 7 Tu. 22 1 15 10 036 15 7 F. 22 2 13 16 8 254 16 4 W. 23 1 20 15 6 2 8 15 7 S." 23 330 17 356 1711 S. 23 9 17 Th. 24 2 53 16 331 16 7 M. 24 041 16 7 1 16 16 3 F. 26 4 3 17 3 431 18 S. 24 422 18 9 442 19 5 Tu. 25 1 59 16 2 244 16 5 S.' 26 457 18 8 517 19 4 M. 25 5 3 20 2 521 20 9 W. 26 3 23 16 10 357 17 3 Tu. 26 5 39 21 4 556 21 9 Th. 27 4 28 17 9 455 18 4 S. 27 537 20 556 20 6 W. 27 6 13 22 2 629 22 6 F. 28 621 18 10 543 19 3 M. 28 6 14 21 629 21 5 Th. 28 6 44 22 9 7 3 2210 S. 29 6 5 19 8 625 20 Tu. 29 6 46 21 10 7 2 22 1 4* 29 7 17 22 10 734 22 8 W. 30 7 18 22 3 731 22 5 S.' 30 762 22 6 8 9 22 2 S 30 6 43 20 5 658 20 9 Th. 31 7 46 22 4 8 2 22 4 M. 31 715 21 730 21 2 Mean tide level 12*- P.M. times and heights are printed in large block type. The times given are STANDARD, or mean time of the meridian of Greenwich. To obtain local mean time subtract 20 minutes. TIDAL DIFFERENCES ON PEMBROKE DOCK. The Tidal observation station was at H.M. dockyard. The zero of the predictions, which is also the datum of the Admiralty charts of Milford Haven, is 11-82 feet below Ordnance datum, and is T50 feet below the level of mean low water spring tides. Place. Lat. N. Long. Time of H.W. Height of H. W. Springs. Neaps. Tenby . . . 51 41 51 40 14 4 10 H. M. -0 35 + 1 f. i. + 3 + 2 6 T. I. + 2 2 + 1 8 Llanelly .... IRELAND, EAST COAST KINGSTOWN. TT TXT -ci a n \ XIH- 12M ' (Local mean time). H. W. F. & C. | XI u [ Stan d a rd time). Lat. 53 18' N. Long. 6 7' W. i g H ! & SEPTEMBER. 3 W , ! MONTH DAY. I OCTOBEK. Time. H M. Ht. F I Time. H. M. Ht. F. I. Time. Ht. F I. Time. H. M. Ht. F. I. Time. H. M. Ht. F. I. Time. Ht. F. I. Time.l Ht. H. M. F. I. Time. H. M. Ht. F. I. F. i 37 11 8 6 24 1 053 11 7 6 40 1 1 8. 1 46 11 7 6 33 1 1 1 811 5 6 55 1 3 S. 2 1 12 11 6 6 69 1 3 1 32 11 4 718 1 5 M. 2 1 32 11 3 7 18 1 5 1 5310 11 7 41 1 9 Tu. 3 2 24 10 8 8 7 2 1 2 55 10 5 8 36 2 4 S. 3 1 51 11 2 7 37 1 7 2 14 10 11 758 1 10 W. 4 3 31 10 2 9 9 2 8 4 9 9 11 9 46 2 11 M. 4 2 39 10 8 8 22 2 1 3 7 10 5 848 2 4 Th. 5 4 53 9 10 10 30 3 2 541 9 9 11 18 3 4 Tu. 5 3 40 10 2 9 18 2 8 4 17 9 11 9 53 2 11 F. 6 6 26 9 11 5 3 4 7 10 10 W. 6 4 58 9 9 10 35 3 3 5 40 9 9 11 17 3 5 S. 7 49 3 2 7 44 10 4 1 26 2 10 8 12 10 7 Th. r 6 29 9 9 7 3 5 7 15 9 11 F. 8 54 3 4 7 63 10 1 1 35 3 8 26 10 5 S. 8 1 59 2 6 8 37 10 11 2 27 2 2 9 1 11 3 S. 9 2 13 2 8 8 53 10 10 2 43 2 3 9 20 11 2 M. 9 2 49 1 9 9 20 11 6 3 6 1 6 9 40 11 8 Tu. 10 3 27 1 2 10 11 10 347 11 10 1812 S. 10 3 7 1 10 9 41 11 6 3 28 1 5 10 2 11 9 W. 11 4 7 9 10 37 12 1 4 26 7 105712 1 M. 11 3 50 1 1 10 23 12 4 12 9 10 44 12 2 Th. 12 4 47 6 11 13 12 1 5 3 6 11 3112 Tu. 12 4 32 6 11 1 12 4 451 4 11 19 12 4 F. 13 6 20 7 11 49 11 11 5 37 9 W. 13 5 9 3 11 38 12 4 5 26 3 11 58 12 4 S. 14 8 11 9 6 55 11 28 11 7 6~15 1 1 Th. 14 5 46 4 18 12 2 6 5 5 F. 15 37 12 6 24 8 057 11 10 6 44 11 S. 15 49 11 4 6 36 1 6 1 9 11 1 6 55 1 8 S. 16 1 17 11 6 7 4 1 2 1 37 11 3 7 23 1 6 M. 16 1 32 10 9 7 18 2 1 55 10 6 7 39 2 3 Tu. 17 2 lg 10 3 8 1 2 6 2 45 9 11 8 26 2 10 S. 17 2 10 11 7 45 1 10 2 24 10 7 8 7 2 2 W. 18 3 16 9 8 8 53 3 1 346 9 6 9 22 3 5 M. 18 2 49 10 3 8 30 2 6 3 17 9 11 8 56 2 11 Th. 19 4 22 9 4 9 59 3 8 5 4 9 3 10 40 3 10 Tu. 19 3 48 9 7 9 25 3 3 4 20 9 4 9 56 3 7 F. 20 5 40 9 3 11 17 3 11 6 21 9 4 W. 20 5 3 9 3 10 39 3 10 544 9 2 11 20 4 S. 21 Midt. 3 11 6 59 9 5 038 3 9 7~30 9~7 Th. 21 6 26 9 2 4 4 7 10 9 3 F. 22 49 4 7 44 9 5 1 25 3 9 8 15 9 7 S. 22 1 11 3 7 7 67 9 10 1 41 3 4 820 10 1 S. 23 2 3 6 8 42 9 10 2 30 3 3 9 1 10 2 M. 23 2 7 3 8 38 10 4 2 28 2 9 8 56 10 8 Tu. 24 2 45 2 6 9 14 10 11 3 1 2 1 9 31 11 1 S. 24 2 51 2 11 9 21 10 6 3 9 2 6 9 38 10 9 W. 25 3 17 1 10 9 45 11 4 332 1 7 10 1 11 6 M. 25 3 24 2 3 9 53 11 3 40 1 11 10 7 11 2 rh. 26 3 48 I 4 10 17 11 7 4 6 1 2 10 35 11 8 Tu. 26 3 55 1 8 10 21 11 4 4 9 1 5 10 35 11 6 F. 27 4 23 1 10 51 11 9 4 41 10 11 10 11 10 W. 27 4 24 1 2 10 60 11 8 4 39 1 11 4 11 9 S. 28 5 9 11 29 11 10 5 17 9 11 47 11 10 Th. 28 4 65 10 11 19 11 10 5 9 9 11 36 11 10 F. 29 5 24 9 11 51 11 10 5 39 9 S. 29 5 35 10 8 11 9 5 55 11 S. 30 9 11 10 5 55 10 27 11 9 6^4 0~11 M. 30 31 11 7 6 17 1 05511 6 6 42 1 3 Tu. 31 1 20 11 3 7 7 1 6 1 50 11 C 7 35 1 9 * Signifies below Datum. The tides are placed in order of occurrence, p.m. times and heights being printed in large block type. A comparison of consecutive heights will indicate whether it is high or low water. The times given are STAND AED, or mean time of the meridian of Dublin, or 6 20' W. from Green- wich. To obtain local mean time add 1 minute. The zero of the predictions, which is also the datum of the Admiralty chart of Kingstown, is T76 feet above Ordnance datum, and is 0'84 foot below the level of mean low water spring tides. TIDAL DIFFERENCES ON KINGSTOWN. Time of Height of H. W. Height of L. W. Place. Lat. N. Long. w: 1 H. W. L. W. Springs. Neaps. Springs. Neaps. H. H. H H. r. i. F. I. F. I. r. i. Wicklow . . . 52 59 6 2 -0 43 -2 9 -3 4 Warren point . 54 6 6 15 -0 2 + 4 + 1 7 + 2 + 2 1 Strangford quay . 54 22 5 33 + 1 16 -1 5 -1 3 -0 10 -18 South rock . . 54 24 5 25 -0 17 et + 1 3 + 1 -0 10 -11 Donaghadee . 54 38 5 32 -0 2 + 6 + 5 -0 10 -0 11 ( 9 ) IRELAND, WEST COAST GALWAY. TT w w *r JIVH-35M. (Local mean time), H. W. F. & C. ^ iy 46 (Standard t i me ). Lat. 53 16' N. Long. 9 3' W. i 1 1 MONTH DAY. | JULY. >' S I t MONTH DAY. 1 AUGUST. g I 1 MONTH DAY. SEPTEMBER. HIGH WATER. HIGH WATER. HIGH WATER. Time. Ht. Time. Ht. Time. H. M. Ht. F. I. Time. H. M. Ht. F. I. Tune. H. M. Ht. F. I. Time. H. M. Ht. F I. S. 1 5 15 13 11 534 14 1 Tu. 1 6 1 14 10 616 1411 ? t 1 6 34 15 5 651 15 2 W. 2 632 14 11 648 1411 3. 2 7 10 14 11 727 14 7 /s. 6 50 14 1 6 7 14 1 rh. 3 7 6 14 9 721 14 8 M. 6 24 14 1 641 14 1 4 741 14 5 8 14 2 S. 3 7 47 14 3 8 9 13 9 Tu. 4 6 58 14 1 717 14 s.' 5 8 19 13 10 841 13 5 M. 4 8 32 13 2 9 12 7 W. 7 36 13 10 755 13 9 Tu. 6 9 31 12 1 10 8 11 9 Th. 6 8 16 13 7 837 13 4 S. 6 9 3 13 927 12 7 W. 6 10 50 11 7 1142 11 8 F. 8 57 13 920 12 9 M. 7 9 56 12 3 1029 1111 Th. 7 030 1111 S. 8 9 44 12 6 10 8 12 4 Tu. 8 11 8 11 10 1152 1110 r. 8 1 12 12 5 150 13 W. 9 036 12 1 5. 9 220 13 9 249 14 4 S. 9 1037 12 2 1111 12 2 Th. 10 Tl7 12 6 153 1211 M. 10 11 45 12 3 F. 11 227 13 7 259 14 2 &'. 10 3 14 15 336 15 7 Tu. 11 19 12 4 056 12 7 S. 12 3 29 14 10 353 15 4 M. 11 359 16 1 421 16 5 W. 1 32 12 11 2 3 13 4 Tu. 12 4 40 16 9 459 1611 Th. g 2 37 13 9 3 8 14 3 s. 13 419 15 11 443 16 5 W. 13 5 19 17 540 1610 F. 14 3 36 14 8 4 3 15 2 M. 14 6 4 16 9 527 1611 Th. 14 5 59 16 7 619 16 3 S. 15 431 15 8 456 16 2 Tu. 15 5 48 17 6 9 1611 F. 15 6 36 16 10 655 15 4 W. 16 631 16 8 651 16 4 5. 16 7 13 14 10 734 14 2 S. 16 520 16 6 545 16 7 Th. 17 7 10 16 729 15 6 M. 7 6 8 16 7 633 16 5 F. 18 7 49 14 11 8 9 14 4 S. 17 7 55 13 6 815 1210 Tu. 18 667 16 3 719 1511 S. 19 8 32 13 8 853 1211 M. 18 8 39 12 1 9 5 11 5 W. 19 742 16 8 8 5 15 3 Tu. 19 9 32 10 11 1012 10 7 Th. -0 825 14 9 850 14 2 S. 20 9 14 12 3 939 11 8 W. 20 1063 10 5 1140 10 5 F. L] 9 12 13 7 936 13 M. 21 10 12 11 1 1053 10 9 Th. 21 025 10 7 S. L2 10 1 12 6 1030 12 Tu. 22 11 33 10 8 F. 2i 1 2 10 11 137 11 4 W. 2S 17 10 8 ^59 1010 8. 23 2 7 11 10 228 12 5 S. 23 11 1 11 8 1137 11 5 Th. 24 1 37 11 1 2 9 11 6 M. 2<: 014 11 4 F. 25 2 36 11 11 3 1 12 5 S. 24 2 50 12 11 3 9 13 5 Tu. 25 51 11 3 129 11 5 S. 26 3 24 12 10 340 13 4 M. 25 3 27 13 11 342 14 4 W. 26 2 3 11 8 231 1111 Tu 26 3 67 14 9 412 15 1 Th. 27 2 58 12 3 322 12 7 S. 27 3 67 13 9 413 14 2 W. 27 429 15 5 443 15 8 F. 2S 3 45 12 11 4 3 13 2 M. 28 4 30 14 6 444 1411 Th 28 459 15 11 517 16 S. 29 422 13 6 441 1310 Tu. 29 5 15 2 \ 516 15 5 F. 29 5 32 16 550 1511 W. 30 533 15 7 546 15 8 S. 30 6 9 15 8 626 15 5 S. 30 457 14 2 513 14 5 rh. 31 6 2 15 8 619 15 7 M. 31 529 14 8 545 14 9 Mean tide level 8 F - 2 1 - P.M. times and heights are printed in large block type. The times given are STANDARD, or mean time of the meridian of Dublin, or 6 20' W. from Greenwich. To obtain local mean time subtract 11 minutes. The zero of the predictions, which is also the datum of the Admiralty chart of Galway, is 1'30 feet above Ordnance datum, and is 0'75 foot below the level of mean low- water spring tides. TIDAL DIFFERENCES ON GALWAY. Place. Lat. N. Long. W. Time of H. W. Height of H. W. Springs. Neaps. Foynes island 52 37 52 40 52 38 52 7 51 54 9 '7 8 50 8 38 10 15 10 20 H. M. + 1 + 1 25 + 1 34 -0 39 -0 48 P. i. -0 2 + 3 8 + 3 -4 9 -4 6 F. I. + 5 + 2 3 + 2 10 - 3 10 -3 3 Mellon ....... Limerick .... Dingle .... Valentia. . . . CANADA, ATLANTIC COAST ST. JOHN, N.B. w w - P SXI*21*. (Local mean time). H. W. F. & G. J XI 45 (Standard time). Lat. 45 16' N. Long. 66 3' W. tH* & 1 * S K O JULY. tH S w ! i AUGUST. Time. H. M. Ht. F I, Time. H M. Ht. F. I. Time. H. M. Ht. F. I. Time. Ht F. I Time. Ht. Time. Ht. F 1 Time. Ht. Time. Ht s. 1 6 25 4 5 23 22 5 645 5 10 Tu. i 43 24 7 10 3 11 1 10 23 1 7 24 W. 2 1 2023 11 7 45 3 10 1 46 23 5 8 4 A 2 3324 1 7 2 4 5 1 1 22 6 720 5 8 Th. 3 1 5823 10 8 22 3 10 2 23 23 7 8 39 4 M. I 1 1224 7 39 4 5 1 39 22 7 7 56 5 6 F. 4 2 3823 6 9 2 3 11 3 2 23 8 9 22 4 Tu. 4 1 6223 10 8 17 4 5 2 18 22 8 8 33 5 4 S. 5 3 21,23 1 9 44 4 1 345 23 8 10 8 4 W. B 2 33 23 6 8 56 4 4 2 59 22 10 9 12 5 1 Th. 6 3 15 23 2 9 36 4 5 3 4323 9 55 5 S. 6 4 8 22 6 10 29 4 7 4 36 23 10 58 4 F. 7 4 22 10 10 18 4 7 4 30 23 10 42 4 11 M. 7 5 021 10 11 20 5 5 34 23 11 55 4 S. 8 4 49 22 5 11 4 4 10 5 20 23 1 11 33 4 11 Tu. 8 6 3 21 4 23 5 6 3823 W. 9 1 2 4 7 7 14 21 5 1 33 5 7 4323 1 S. 9 5 42 22 11 55 6 1 6 14 23 4 rh. 10 2 14 4 8 22 22 2 2 40 4 8 462* 1 M. 10 29 4 10 6 39 21 10 53 5 2 7 11 23 8 F. 11 3 20 3 9 23 23 4 3 41 3 9 4525 1 Tu. 11 1 29 4 5 7 40 22 1 55 5 8 9 24 6 s. 12 4 20 1 10 10 19 24 7 438 2 10 41126 W. 12 2 33 3 8 8 42 22 8 258 4 6 9 6 25 6 Th. 13 3 35 2 8 9 42 23 7 3 59 3 7 10 1 26 4 s. 13 5 13 1 11 12 25 7 533 1 11 34 27 F. 14 4 34 1 8 10 38 24 7 456 210 10 55 27 1 M. 14 6 3 6 3 26 6 26 1 S. 15 5 29 11 11 30 25 6 5 51 2 2 11 48 27 6 Tu. 15 25 27 4 6 52 5 053 26 7 16 1 W. 16 1 1526 11 7 40 10 1 42 26 8 5 1 S. 16 6 22 7 21 25 11 6 44 1 10 rh. 17 2 626 2 8 28 1 7 2 30 26 8 53 1 1 M. 17 40 27 5 7 14 6 1 11 26 7 36 1 7 F. 18 2 57:25 2 9 17 2 8 3 19 25 9 42 2 1 Tu. 18 1 32 27 8 5 8 2 3 26 8 29 1 8 S. 19 3 4923 11 10 7 4 4 9 24 1033 4 W. 19 2 25 26 4 8 55 1 4 2 57 25 8 9 21 2 2 Th. 20 3 20 25 6 9 45 2 4 352 25 4 10 14 3 g. 20 4 42 22 7 10 59 5 4 5 3 23 11 28 5 F. 21 4 16 24 4 10 37 3 6 446 24 7 11 8 4 M. 21 5 3821 5 11 55 6 5 6 1 22 S. 22 6 13 23 11 33 4 10 5 41 23 8 Tu. 22 29 6 6 39 20 7 56 7 7 2 22 W. 23 I 35 6 4 7 44 20 4 2 3 7 8 222 flf. 23 5 4 11 6 12 21 11 34 6 6 38 23 1 Th. 24 2 39 6 1 8 46 20 7 3 5 6 8 5922 M. 24 1 6 5 6 7 13 21 2 1 39 6 7 7 3822 8 F. 25 3 35 6 6 9 37 21 2 3 56 6 9 4822 1 Tu. 25 2 12 5 10 8 15 21 2 41 6 8 8 36|22 8 S. 26 4 20 4 10 10 19 21 10 4 36 5 10 2923 W. 26 3 14 5 6 9 14 21 1 3 37 6 6 9 2723 Th. 27 4 8 5 1 10 4 21 5 4 26 6 1 10 1123 5 S. 27 4 58 4 4 10 56 22 6 5 12 4 8 11 8 23 1 F. 28 4 52 4 8 10 46 21 10 5 7 5 8 10 50j23 8 M. 28 5 34 3 11 11 31 23 1 5 47 4 4 11 45 24 S. 29 6 29 4 5 11 23 22 2 5 43 5 5 11 2823 11 Tu. 29 6 9 3 8 5 23 6 6 21 4 W. 30 21 24 2 6 43 3 7 40 23 11 6 55 3 S. 30 6 3 4 2 11 59 22 6 6 17 5 1 Th. 31 56 24 2 7 17 3 6 1 16 24 2 7 31 3 M. 31 6 24 6 36 4 034 22 10 6 50 4 10 * Signifies below Datum. The tides are placed in order of occurrence, p.m. times and heights being printed in large block typ I A comparison of consecutive heights will indicate whether it is high or low water. The times given are STANDARD, or mean time of the meridian of 60 W. from Greenwich. 1 | obtain local mean time subtract 24 minutes. TIDAL DIFFERENCES ON ST. JOHN. The zero of the predictions, which is also the datum of the Admiralty chart of St. John, is 55-60 fei ' DW a bench mark near the south-east corner of the Custom House, and is approximately 2| fe< low the level of mean low water spring tides. Time of Height of H. W. Height of L.W. Place. Lat. N. Long. W. H.W. L. W. Springs. Neaps. Springs. Neaps. H. M. H. H. F. I. F. I. F. I. F. I. Seal Cove . . 4437 66 5l -0 22 -5 6 -6 6 Welchpool . . 44 53 66 58 + 2 + 10 -2 -16 St. Andrews. . 45 4 67 3 + 8 + 18 -0 6 ... Quaco . . . 45 20 65 32 + 12 ... + 4 6 + 3 6 ... ... II PERSIAN GULF ABU SHAHR (BUSHIRE}. H. W. F. &C. VII H - 43?- Lat. 28 54' N. Long. 50 43' E. 1 1 MONTH DAY. I NOVEMBER. 1 1 * fi 1 DECEMBER. Time. Ht. Time. Ht. Time. Ht. Time. Ht. F. I. Time. H. M. Ht. F. I. Time. H. M. Ht. F. I. Time. H. M. Ht. F. I. Time. H. M. Ht F. W. 1 5 43 9 46 2 4 3 22 2 3 10 28 5 9 K 1 6 14 7 1 16 2 11 5 9 2 5 11 11 5 Th. 2 6 44 11 2 35 2 6 4 15 2 6 11 21 5 4 S. 2 7 10 2 9 3 4 6 44 2 r. 3 7 51 1 l 3 37 2 10 6 18 2 8 s. 4 28 4 9 8 53 1 3 4 10 3 3 8 9 2 8 S. 3 14 4 3 7 47 1 2 256 3 9 8 26 2 M. 4 1 37 3 7 8 34 1 5 3 38 4 2 10 10 2 s. 5 2 2 4 2 9 47 1 5 4 40 3 9 10 2 3 r u . 5 3 31 3 9 21 1 8 4 17 4 8 11 34 M. 6 3 47 3 10 10 34 1 6 5 10 4 3 11 23 1 9 W. 6 5 14 2 9 10 7 1 10 4 54 5 1 Tu. 7 5 15 3 8 11 14 1 7 5 39 4 9 Th. 7 41 1 6 38 2 8 10 48 2 5~30 5 W. 8 25 1 3 6 21 3 5 11 48 1 8 6~~6 5~2 8 1 33 7 7 38 2 7 11 26 2 1 6 5 5 Th. 9 1 16 10 7 16 3 3 18 1 10 634 5 7 S.' 9 2 18 5 8 31 2 6 Noon 2 1 6 40 5 F. 10 2 4 7 8 4 3 1 47 1 11 7 4 5 9 S. 2 60 5 8 50 2 11 1 15 2 736 5 11 S. 10 2 59 3 9 17 2 6 33 2 2 7 15 5 M. 11 3 37 3 9 59 2 6 1 7 2 3 750 5 S. 12 3 34 5 9 35 2 9 1 40 2 1 810 5 11 Tu. 12 4 12 4 10 36 2 7 1 42 2 3 8 25 5 M. 13 4 17 7 10 22 2 7 2 1 2 2 8 46 5 10 W. 13 4 43 6 11 7 2 8 2 19 2 4 9 1 5 Tu. 14 4 59 9 11 14 2 6 2 23 2 4 9 23 5 7 Th. 14 5 9 7 11 36 2 9 2 59 2 5 9 38 5 W. 15 5 39 11 13 2 7 2 49 2 5 10 1 5 4 F. 16 5 30 9 7 2 11 347 2 6 10 16 4 Th. 16 6 16 1 1 1 16 2 8 3 23 2 7 10 40 411 S. 16 6 51 11 042 3 2 445 2 7 10 55 4 F. 17 6 51 1 3 2 20 2 10 4 22 2 9 11 23 4 6 S. 18 7 28 1 4 3 6 3 1 6 19 211 8. 17 6 15 1 1 22 3 5 555 2 8 11 37 3 M. 18 6 43 1 2 2 4 3 8 7 24 2 I S. 19 23 4 1 8 9 1 6 3 43 3 5 811 2 11 Tu. 19 28 3 4 7 16 1 4 2 47 4 1 9 20 2 M. 20 1 46 3 8 8 52 1 6 4 14 3 10 10 5 2 7 W. 20 1 63 2 10 7 64 1 6 3 30 4 7 11 25 2 r Tu. 21 3 15 3 4 9 35 1 7 4 42 4 3 11 25 2 1 Th. 21 4 2 2 5 8 38 1 8 4 12 5 W. 22 4 41 3 2 10 15 1 8 5 10 4 9 F. 22 31 1 6 5 45 2 3 9 28 1 9 4~53 5 Th. 23 21 1 7 6 51 3 10 51 1 9 5~39 5~3 S. 23 1 18 10 6 59 2 3 10 22 1 10 5 35 5 F. 24 1 7 1 1 6 51 2 11 11 25 1 9 6 10 5 8 S. 25 1 51 8 7 48 2 9 Noon 1 10 6 44 6 1 S. 24 1 59 5 8 2 3 11 18 1 11 6 18 6 M. 25 2 37 1 8 52 2 4 15 1 11 7 3 6 S. 26 2 35 5 8 43 2 7 37 1 10 7 21 6 4 Tu. 26 3 14 *0 1 9 36 2 6 1 13 1 11 7 50 61 M. 27 3 19 2 9 37 2 6 1 17 1 11 8 1 6 5 W. 27 3 50 *0 2 10 15 2 8 2 12 1 11 8 37 61 Tu. 28 4 2 1 10 31 2 6 2 2 8 43 6 4 Th. 28 4 26 *0 1 10 53 2 11 3 13 1 11 9 25 5 W. 29 4 45 2 11 26 2 6 247 2 2 9 28 6 1 F. 29 5 2 2 11 31 3 3 4 16 1 11 10 14 5 Th. 30 5 29 4 021 2 8 3 50 2 3 1017 5 7 S. 30 5 38 5 10 3 7 5 23 2 11 4 4P S. 31 6 14 9 051 3 11 6 39 2 1 11 57 3 * Signifies feeZow Datum. The tides are placed in order of occurrence, p.m. times and heights being printed in large bio type. A comparison of consecutive heights will indicate whether it is high or low water. The times given are LOCAL MEAN TIME. PERSIAN GULF. Position. H. W . F. & C. Ris e. Place. Lati- tude. Longi- tude. Long, in Time. M. T. Place. G. M. T. Springs. Neaps. N. E. H. M. H. M. H. M. Feet. Feet Kuweit 29 22 48 3 12 13 9 1 ill 8 Shatt al Arab : Al Basra . 30 30 47 53 3 12 6 2 48 9 ABU SHAHR or BUSHIBE . 29 50 43 3 23 7 43 4 20 6 5 Hen jam island anchorage . 26 41 55 54 3 44 1130 7 46 10J 7i 12 RETURN CIRCULATION DEPARTMENT TO ^ 202 Main Library LOAN PERIOD 1 HOME USE 2 3 4 5 6 ALL BOOKS MAY BE RECALLED AFTER 7 DAYS Renewals and Recharges may be made 4 days prior to the due date. Books may be Renewed by calling 642-3405. DUE AS STAMPED BELOW APR 04 1988 .-.> UNIVERSITY OF CALIFORNIA, BERKELEY FORM NO. DD6, BERKELEY, CA 94720 $ LD 21-10m-5, '43 (6061s) 03276 UNIVERSITY OF CALIFORNIA LIBRARY j