VIBRARK OF THE UNIVERSITY ^y^ONOMY UBRARt '^ Alexander Montgomery Library Astronomical Society o[ tlie Pacific, SAN KRA.NCISCO. LIBR ARY OF THE ASTRONOMICAL SOCIETY OF THE PACIFIC ^?y Digitized by the Internet Archive in 2007 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementarytreatiOOrichrich ELEMENTARY TREATISE OW NAVIGATION AND NAUTICAL ASTRONOMY BY EUGENE L. RICHARDS, M.A. PROFKSSOB OF MATHEMATICS IN YALB UNIVKRSITY •ismimiiictf sm' NEW YORK.:. CINCINNATI-:. CHICAGO AMERICAN BOOK COMPANY AStRONOMYUBRARy CoPTRiGnr, 1901, by EUGENE L. RICHARDS. Entkeed at Stationers' Hall, London. NAVIGATION. W. p. 2 PREFACE The following pages are the outcome of the author's own teaching. To understand the prin- ciples set forth in them a knowledge of elementary Plane and Spherical Geometry and Trigonometry is all that is needed. The author wishes to acknowledge special obli- gations to Martin's " Navigation " and Bowditch's " Navigator." To either of these works the present book might serve as an introduction. Most of the examples have been worked by means of Bowditch's "Useful Tables," pubhshed by the United States Government. The corrections to Mid- dle Latitude have been taken from the table (pages 172, 173) prepared by the author. References to Elements of Plane and Spherical Trigonometry by the author and to Elements of Geometry by Phillips and Fisher are indicated by (Trig.) and (P. and F.) respectively. WGTVl^ CONTENTS OHAPTIB PAOB I. Plane Sailing. Middle Latitude Sailing. Mercator*s Sailing 7 II. Great Circle Sailing 42 III. Courses 50 IV. Astronomical Terms 63 V. Time 74 VI. The Nautical Almanac 88 VII. The Hour Angle 94 VIII. Corrections of Altitude 110 IX. Latitude 122 X. Longitude 138 Definitions of Terms used in Nautical Astronomy . . . 149 Examples 153 Parts of the Ephemeris for the Year 1898 158 Table of Corrections to Middle Latitude 172 NAVIGATION AND NAUTICAL ASTRONOMY CHAPTER I PLANE SAILING — MIDDLE LATITUDE SAILING — MERCATOR's SAILING In navigation the earth is regarded as a sphere. Small parts of its surface (as in surveying) are con- sidered 2i^ planes. Art. 1. The axis of the earth is the diameter about which it revolves. The extremities of this axis are called poles, one being named the North Pole and the other the South Pole. 2. The meridian of any point, or place, on the earth is the great circle arc passing through the point, or place, and through the poles of the earth. The meridian of a point, or place, may be said to be the intersection of a plane with the surface of the earth, the plane being determined by the axis and the point (Phillips and Fisher, Elements of Geometry, 526, 807). (a) Meridians are, therefore, north and south lines. 3, The earth's equator is the circumference of the great circle, whose plane is perpendicular to the axis. (a) The equator is perpendicular, therefore, to the meridians (P. and F., 837). 7 8 NAVIGATION AND 4. Parallels of latitude on the earth are circum- ferences of small circles, whose planes are perpen- dicular to the axis. The planes of these parallels are parallel to each other and to the plane of the equator (P. and F., 559). (a) Parallels of latitude are east and loest lines. 5. The longitude of a point, or place, is the angle between the plane of the meridian of the point, or place, and the plane of some fixed meridian. This angle is measured by the arc of the equator inter- cepted between these planes, since this arc measures the plane angle of the dihedral angle of the planes (P. and F., 836). This arc, intercepted between the two meridians, is spoken of as the longitude, as its degree measure is the same as that of the dihedral angle. One assumed meridian from which longitude is reckoned is the meridian of the Observatory of Greenwich, England ; another is the meridian of the Observatory of Washington. The French, also, have a fixed meridian from which longitude is reckoned. (a) Longitude is reckoned, on the arc of the equa- tor, east and west of the assumed meridian, from 0° to 180°. (6) The difference of longitude of two places is the angle between the planes of their meridians, and is measured by the arc of the equator intercepted between these meridians. This arc is evidently the difference of the two arcS; which measure the longitudes of the two places, NAUTICAL ASTRONOMY 9 if the places are either both E. or both W. of the assumed meridian. (c) If we give to E. longitudes the sign + , and to W. longitudes the sign — , the arc which measures the difference of longitude of two places will always be the algebraic difference of the longitudes of the places. {d) To find, then, the difference of longitude of two places whose longitude is given, we subtract the less from the greater if both are E. or both are W., but add the tivo if one is E. and the other W. 6. The latitude of a point, or place, is the angle made with the plane of the equator by a line drawn from this point, or place, to the center of the earth. The latitude is measured by the arc of the meridian (of the point) which subtends the angle. This sub- tending arc is spoken of as the latitude, as its degree measure is the same as that of the inclination of the line to the plane of the equator. Latitude is reckoned from 0° to 90°, north and south of the equator. 7. The difference of latitude of two places is the difference between the latitudes of the two places, difference being understood as algebraic, and north latitudes having the sign + and south latitudes the sign -. (a) To find, then, the difference of latitude of two places whose latitudec^is- given, take the less from the greater if both are N, or both are S.^ but add the two if one is iV, and the other S» 10 NAVIGATION AND (b) The difference of latitude of two places is measured on a7iy arc of a meridian intercepted be- tween the parallels of latitude of the places. Let the figure represent a hemisphere of the earth. Let N and S be the poles; C the center; and WDE the equator. Suppose A and B to be two points on the surface; VAH to be the parallel of latitude of A, and NAS to be its meridian ; ELB to be the par- allel of latitude, and NBS the meridian of B. Then it is to be proved that the dif- ference of latitude of A and B is measured by AL, HB, VB, or any other meridian arc intercepted be- tween VAH and RLB. Let the meridian NAS intersect the parallel ELB in the point L, and the equator in the point D. Also, let the meridian NBS intersect the parallel VAH in the point H, and the equator in the point G. Draw the straight lines CA, CD, CB, and CG. The plane of the meridian NAS is perpendicular to the plane of the equator (Arts. 2 and 3), and is, therefore, the plane which projects the line CA upon that plane ; CD is the inter- section of these two planes (P. and F., 528), and contains (as a part of it) the projection of the line CA. The angle ACD is, therefore, the angle made by the line AC with the plane of the equator (P. and F., 586), and is, consequently, the latitude of the point A (Art. 6). Also, the plane of the meridian, NGS, is perpendicular to the plane of the equator, NAUTICAL ASTRONOMY 11 and by its intersection with that plane determines the pro- jection of the line CB upon the plane. Therefore, BCO is the latitude of the point B. Now, ACD, or the latitude of A, is measured by AD^ and BCO is measured by BG] therefore, the difference of latitude of A and B is measured by the difference between AD and BG ; that is, by AD - BG. AD = ND-NA = NG- NH= NW- NV (P. and R, 817). Also, BG = ND-NL = NG-NB=NW-NR. .'. AD- BG = NL - NA = NB - NH= NR- NV\ = AL = HB=VR, etc. If the point P be taken on a parallel of latitude south of the equator, the difference of latitude would be measured by HP or by Aa, an arc of a meridian intercepted between the parallels. 8. It is evident that the position of any point or place on the earth's surface is determined if the lati- tude and longitude of the point, or place, are known. Thus, suppose NWSE to represent a hemisphere of the earth; NWS to be the meridian from which longitude is reckoned ; WDE to be the arc of the equa- tor ; RLB and VAH to be parallels of lati- tude ; and NDS, NBS to be meridians. Suppose the lati- tude of the point to be 30° N., and the longitude to be 40° E. If, now, RLB be a parallel of latitude, of which the polar distance NR, NL, 12 NAVIGATION AND or NB is 60°, since NW, ND, or NG is 90^ TFK, DL, or GB is 30° ; therefore, RLB is a parallel of latitude, every point of which is 30° N. of the equator. Consequently, the point whose latitude is given must be found somewhere on this arc RLB. Again, if NDS be a meridian, whose plane NDS makes with the plane WNS an angle of 40° (measured by the arc WD of the equator), the longitude of every point on NDS is 40° E. Therefore, the point whose longitude is given must be some- where on the meridian NDS. Since the point is on the arc RLB^ and at the same time on the arc NDS, it must be at their intersection, L. Therefore, the point is determined when its latitude and longitude are given. It might be said that two circles intersect twice, and there- fore that the point of the circle RLB diametrically opposite to L would be indicated by lat. 30° N., long. 40° E. This is evidently false, since the other half of NDS and the other half of RLB, which, by their intersection, determine this second point, are on the other hemisphere. The latitude of this second point is 30° N., but its longitude is 140° W. of the assumed meridian (Art. 5, (a)). 9. As charts of the earth's surface are constructed for the use of navigators with meridians and parallels of latitude either drawn on them, or indicated, if a ship's latitude and longitude are known, the position of the ship is determined. It is im- portant that this position should be determined from day to day, and therefore it is important that the ship's latitude and longitude should be known. Latitude and longitude are best obtained by observations of the heavenly bodies. This is a depm-tment of navigation which belongs to astronomy. It is necessary to have other methods of determining a ship's position when it is impossible to resort to the methods of astronomy. These other methods are now to be considered. 10. (a) The distance sailed by a ship, in going from one point to another, is the length of the line traversed by the ship between the two points. NAUTICAL ASTRONOMY 13 (h) The hearing or course of a ship, at any point, is the angle which the line traversed by the ship (that is, the distance) makes with the meridian passing through that point. If a ship cuts every meridian at the same angle, she is said to continue on the same course. If a ship is said to sail a given distance on a given course, it is assumed that in that distance she continues on the same course. The path made by a ship continuing on the same course is called a rhumb line, or simply a rhumb. (c) The departure of a ship, in sailing from one point to another, is the whole east or west distance she makes measured from the meridian from which she sails, and is an easting or westing according as she sails in an easterly or westerly direction. If the distance sailed is small, it may be considered a straight line, and the departure might also be re- garded as a straight line measuring the perpendicular distance between the meridians of the two points. In this case the meridians may be considered parallel straight lines (as in surveying), since they are lines on a small portion of the earth's surface, and are perpendicular to the same line. If the distance is not small, it may be divided up into such a number of small parts that each of them may be considered as a straight line. The departure of each of these small distances will then also be a straight line, and the departure of the lohole distance will be the sum of the departures of the parts. 14 NAVIGATION AND (d) Difference of latitude of two points has already been defined (Art. 7). If a given distance sailed by a ship is small, it may be regarded as a straight line, and then the difference of latitude of the two extremities of the line, repre- senting this distance, is measured by a line, which may be also regarded as a straight line. The differ- ence of latitude is then a northing or southing (as in surveying). If the distance is not small, it may be divided into such a number of parts that each part may be small enough to be considered a straight line. The differ- ence of latitude of each part will then be a straight line, and the difference of latitude of the whole dis- tance will be the sum of the differences of latitude of the parts. Thus, suppose AC to be a small distance on the earth's sur- face. Let AK and BC be meridians of the points A and C, and let these lines be consid- ered parallel. If CK be a perpendicular to AK drawn from C, it will be the departure of AC, and AK will be the dif- ference of latitude of A and C, or the difference of latitude for the dis- tance AC. If the distance be a long distance, as from A to D, then it can be divided into such a number of short distances — as, for instance, AC, CE, EG, and OD — that each one of them can be considered as a straight line. If AK, CB, EF, and OH be the NAUTICAL ASTRONOMY 15 meridians of the points A, C, E, and G, and if CK be the per- pendicular from C to AK, EB be perpendicular to CB, OF to EF, and DH to OH, then the departure for AD will be KC + BE-\-FO + HD, and the difference of latitude will be AK-\- CB -^-EF+OH. 11. Plane sailing is the art of determining the position of a ship at sea by means of a right-angled plane triangle. Of this triangle the hypotenuse is the distance, the base is the difference of latitude, the perpendicular is the departure, and the angle between the base and the hypotenuse is the course. When the distance sailed is short, it is evident from the figure that the four quantities mentioned are the parts of a right-angled triangle ; for then AC is the dis- tance, AK is the differ- ence of latitude, KC at right angles to AK is the departure, and CAK is the course. If the distance sailed is not short, — as, for instance, the distance AD, — then divide it into such a number of small distances, AC, CE, EO, and OD, that each may be considered a straight line. Complete the figure as in the preceding article. Suppose the ship's course to be the same in sailing from A to D, then the angles CAK, ECB, OEF, and DGH are equal (Art. 10, (6)). Now, take any straight line A'N, and on it lay oE A'C\ CE', E'O', and O'D', equal respectively to AC, CE, EO, and OD, and on these lines A'C, CE', E'O', and O'D' construct right- angled triangles A'K'C, C'B'E', E'F'G\ JR S T H' N P O K 17 w^ 2^ 16 NAVIGATION AND and G'H'D' equal respectively to AKC, CBE, EFG, and GHD, then AD' = AD, the distance, and A'K' + C'B' + E'F' + G'H' = AK-{- CB^EF+ GH =dif> ference of latitude ; K'C + B'E' + F'G' + H'D' = KCi- BE + FG -i- HD = de- parture. Since the ship sails on the same course, the angles K'A'Cy B'C'E', F'E'G', and H'G'D' are all equal, and, therefore, the lines A'K\ C'B\ E'F, G'H' sue parallel; also, the lines K'C, B'E', FG', and H'D' are parallel (P. and F., 44). Produce A'K' and D'H' to meet at R-, produce C'B' and E'F to meet D'R at S and T; and produce E'B' and G'F' to meet A'R at O and P. i? is a right angle, since it is equal to K'. A'RD' is consequently a right-angled triangle. A'D' represents dis- tance sailed. D'A'R represents the course. A'R represents the distance of latitude, for A'R=A'K'-{-K'0 4-0P-{-PR = A'K'+C'B'-^ E'F'-h G'H'. RD' represents the departure, for RD'=RS + ST+TH'+H'D'=K'C' + B'E' ■\-FG'+H'U. 12. Any two parts of a right-angled triangle being given, in addition to the right angle, the other parts may be found ; therefore, of the four quantities, the distance, the course, the departure, and the difference of latitude, any two being given, the other two may be found, since these quantities may be represented by the parts of a right-angled triangle, as has been shown in the preceding article, and will, therefore, have the same relation to one another as the corre- sponding parts of the right-angled triangle. When the distance is small, this is evident. If the distance is great, it may be divided, as before, into such a number of NAUTICAL ASTRONOMY 17 small distances, AC, CE, EG, and GD, that each may be con- sidered a straight line. Let the differences of latitude for these small distances be AK, CB, EF, and GH, and let the departures be KG, BE, EG, and HD. As the course is sup- posed to be the same for the whole distance AD, the angles CAK, ECB, GEE, and DGH are all equal. In the right-angled triangle AKC, course ; In the right-angled triangle QBE, course ; In the right-angled triangle EFG, course : and AK AC CB CE EF EG GH cos CAK= cos cos BCE = cos = cos FEG = cos In the right-angled triangle GHD, —— = cos HGD= cos GD course. Therefore (P. and F., 265), AK+CB^EF+GH (1) cos course. R P O K A. AC -^CE^ EG -f GD -^ Now, if we construct the right-angled 5 T g* triangle A'RD', as in the preceding article, having A'D' = AD, then, as in Art. 11, it may be shown that A'E = AK+ CB-{-EF+ GH= dif. of latitude, and CE -h EG -\-GD = AD = dist. A'D' = AC Substituting these values in (1), we have : A'R difference of latitude A'D' distance NAV. AND NAUT. ASTR. — 2 cos course ; or, 18 NAVIGATION AND (2) difference of latitude = distance x cos course. In the same manner it can be shown (3) departure = distance x sine of course ; (4) departure = difference of latitude x tan course ; and that the other relations shown to hold between the parts of a right-angled plane triangle hold between the quantities in navigation represented by these parts. 13. If at a given time it is required to find the position of a ship by plane sailing^ the rate of speed per hour at which she is sailing is first ascertained. This rate, multiplied by the number of hours elapsed since the last ascertained position, will give the dis- tance from that position. The angle made by the direction in which the ship is headed, and the N. and S. line of the mariner's compass (with correction, if necessary), will furnish the course. From these data the difference of latitude and the departure are found (Art. 12, (2) and (3)), and thus the position of the ship is known. For example, suppose the average rate of sailing is ascer- tained to be 9 miles an hour, and that 12 hours have elapsed since the last ascertained position, then the distance is 108. If the course is observed to be N. 30° E., the ship's position N. of her last position will be, in miles, 108 x cos 30°, or 93.5 miles, and her position E. will be 108 x sin 30°, or 54 miles. 14. The rate of sailing is ascertained by means of the log. The log, in one of its simplest forms, is a triangular piece of wood, so weighted as to assume, when attached to its line and placed in water, a position calculated to oppose the most resist- NAUTICAL ASTRONOMY 19 ance to force applied to the line. The line is a rope knotted at regular intervals. When the log is thrown overboard and the line is reeled" out by the forward motion of the ship, the number of knots passing over a given point in a given period of time will give the rate of sailing for that period of time. Moreover, if the interval between the knots be the same part of a mile that the period of time is of an hour, the number of knots passed out during the period of time will give the number of miles per hour sailed by the ship. For instance, let the period of time be J minute or j^-^ hour, then the interval between the knots must be y^^ mile. Sup- pose, then, 4 such knots (counting the intervals by the knots) should be reeled out by the forward motion of the ship during i minute, we should find the distance sailed per hour (that is, the rate per hour) by the proportion ^ min. : 60 min. ; : j^-^ mile : x (the distance per hour). .-. X = 2 X 60 X yf ^ = 4 miles, the same number of miles per hour as knots per half minute. 15. The mariner s compass consists of a circular card attached to a magnetic needle, which generally points N. and S.* Each quadrant of this card is divided into eight equal parts, called points, to which names are given as represented in the accompanying figure, t * The magnetic needle does not at all places on the earth's surface point N. and S. Charts for the use of navigators, however, give the amount of variation for places where the needle is subject to variation, so that for such places a correction can be applied to the direction indi- cated by the needle, so as to obtain a true N. and S. line. \ The naming of the points in each quadrant will be seen to be not without method. Thus, in the quadrant between N. and E., the point midway between N. and E, takes its name from both these points , then the point midway between N. and N.E., and the point midway between 20 NAVIGATION AND Also, to express courses between the points, the points are subdivided into half points and quarter points. The points are read (taking the quadrant between the N. and E. points), North by East, North North East, North East by North, North. East, etc., etc. The angle between two adjacent points is %^°, or 11° 15' 16. Distance, departure, and difference of latitude are all expressed in nautical miles. E. and N.E., take their names respectively from the two points between which each is situated, as one of these is north and the other is east ofN.E. The remaining points are named from the nearest main point (calling N., E., and N.E. main points), with the addition of N. or E. as the point to be named is north or east of this nearest point, with the word hy placed between the two. Thus, the point between N. and N.N.E. is N. hy E. ; the point between N.N.E. and N.E. is N.E. hy N. ; the point between N.E. and E.N.E. is N.E. hy E. ; and the point between E.N.E. and E. is E. hy N. The points of the other quadrants may be shown to be named on the same method. NAUTICAL ASTRONOMY 21 A nautical mile is equal to a minute of an arc of the circumference of a great circle of the earth. As there are 69.115 common miles in a degree of such an arc (Trig., Art. 173, Ex. 5), a nautical mile is longer than the common statute mile. Differences of latitude expressed in degrees and minutes is, therefore, easily converted into miles, or, when expressed in nautical miles, is easily changed into degrees and minutes. Thus, 5"^ 33' difference of latitude = 333 miles ; and 656 miles difference of latitude = 10° 56'. Ex. 1. A ship sails N.E. b. N. a distance of 70 miles. Ee- quired her departure and difference of latitude at the end of that distance. The course is 3 points from N. toward E., and is, therefore, 3 x (11° 15') or 33° 45'. Ans. Dep. = 38.89 miles ; dif. lat. = 58.2 miles. Ex. 2. A ship from lat. 33° 5' N. sails S.S.W. 362 miles. Required her departure and the latitude arrived at. Ans. Dep. = 138.5 miles ; lat. 27° 30.6' N. Ex. 3. A ship, leaving port in lat. 42° N., sails S. 37° W. till her departure is 62 miles. Required the distance sailed and the latitude arrived at. Ans. Dist. = 103 miles ; lat. 40° 38' N. Ex. 4. A ship sails S. 50° E. from lat. 7° N. to lat. 4° S. Required her distance and departure. Ans. Dist. = 1026.78 miles; dep. = 786-56 miles. Ex. 5. A ship sails from the equator on a course between S. and W. to lat. 5° 52' S , when her departure is found to be 260 miles. Required her course and the distance sailed. Ans. Course = S. 36° 27' W. ; dist. = 437.6 miles. Ex. 6. A ship sails from lat. 3° 2' N. on a course between N. and W. a distance of 382 miles, when her departure is found to be 150 miles. Required her course and the latitude arrived at. Ans. Course = N. 23° 7^' W. ; lat. 8° 53' N. 22 NAVIGATION AND 17. A traverse is the path described by a ship which changes its course from time to time. The object of traverse sailing is to find the posi- tion of a ship at the end of a traverse ; the distance sailed from the position left to the position reached ; and the course for this distance. The method of accomplishing this object will best be seen by means of an example. Suppose a ship to start from A and sail to B, then from B to (7, and then from C to D. It is required to find her position at 2); that is, to find the difference of latitude and the departure made in going from A to D. These quantities being found, and tlie course DAk, can be the distance calculated. AD, (Remark. — The distances AB, BC, and CD are all sup- posed, in traverse sailing, to be short distances, and therefore are to be treated, like similar distances in plane sailing, as straight lines.) Through JB, A, and C suppose meridians pn, Ik, and Ch, and through B, A, C, and D parallels of latitude Bf, me, Cp, and Dkn to be drawn. Ak is the difference of latitude of AD, and kD is the departure of AD. (1) Ak = mil = Ch -Cm = Ch- {Cf-^fm)= Ch - (Bp + eB). Now, Ch is a north latitude, and Bp and eB are south lati- tudes, therefore, the difference of latitude of ^D is equal to the difference between the N. latitude of one distance of the traverse and the sum of the S. latitudes of the other distances. NAUTICAL ASTRONOMY 23 (2) kD = nD-kn = nh + JiD - Ae = pC -\- hD — Ae. But pC and JiD are west departures, and Ae is an east departure; therefore, the departure for AD is equal to the difference between the sum of the west departures of two distances of the traverse and the east departure of the third distance. In the case given above, the number of the parts of the traverse is only three, but if a fourth distance on a course between N. and E. were given, a second north latitude and a second east departure would enter our figure, so that the difference of latitude between the first and last position of the ship would, in that case, be equal to the difference hetween the sum of the north latitudes and the sum. of the south lati- tudes ; and the departure, in passing directly from the first to the last position, would be equal to the difference hetween the sum of the east departures and the sum of the ivest departures. The same principle would hold true for a traverse of any number of distances greater than four. The proof would be similar to that given above for a traverse of three distances. The principle stated may, therefore, be taken as a general one. Ex. 1. Suppose a ship sailing on a traverse makes courses and distances as follows: from A to B, E. b. S. 16 miles; from B to C, W. b. S. 30 miles ; and from C to Z), N. b. W. 14 miles. Required the distance from ^ to Z> and the course for that distance. Before solving these examples the student is advised to plot the figures for them by means of a protractor and a plane scale. 24 NAVIGATION AND Course Distance N. s. E. W. 1 2 3 S. 78° 45' E. S. 78° 45' W. N. 11° 15' W. 16 30 14 13.73 3.12 5.85 15.69 29.42 2.73 Sum 13.73 8.97 15.69 32.15 8.97 15.69 dif. of lat. = 4.76 N. dep. = 16.46 W. /. (In the figure, page 22) Ak = 4.76, and kD = 16.46. Course = kAD, ^ = ^^^ = tan. 73° 52' 15" = tan. kAD Ak ■ 4.76 .-. Course = N. 73° 52' 15" W., or N. 73° 52' W., as the result is generally given only to the nearest minute. kD 16.46 Dist. = AD. sin DAk sin 73° 52' = 17.13 miles. Ex. 2. A ship sails on a traverse, making the following courses and distances: S.E., 25 miles; E.S.E., 32 miles; E.. 17 miles ; N. b. W., 63 miles. Required the distance from her first to her last position, and the course. Ans. Dist. = 60.94 miles ; course = N. 58° 29' E. Ex. 3. A ship sails on a traverse, making the following courses and distances : N„E., 25 miles ; E.S.E., 40 miles ; E. b. N., 35 miles ; N. b. W., 33 miles. Required the course and distance from her first position to her last position. Ans. Course = N. 63° 16' E. ; dist. = 92.41 miles. 18. Parallel sailing is sailing on a parallel of lati- tude. In parallel sailing, therefore, a ship sails east or west (Art. 4, {a)). The distance is the same as her departure, and the difference of latitude disap- pears. The problem in parallel sailing is to convert di$' NAUTICAL ASTRONOMY 25 tance on a parallel into difference of longitude ; that is, given a distance between two meridians measured on a parallel of latitude, to find from it the distance between the same meridians measured on the equator (Art.5,(&)). The method of solving this problem will be under- stood by means of the accompanying figure, which represents a part of the earth. In this figure let C represent the center of the earth ; P be one of the poles ; EF a part of the equator, CE and CF its radii; AB a part of a paral- lel of latitude inter- cepted between two meridians, PAE and PBF', and let DA and DB be the radii of this parallel. Draw the radius AC. AB^AD_ AD EF EC AC = cos D AC = cos ACE. But ACE is the latitude of A (Art. 6), or of the parallel AB. distance on parallel between two meridians distance on equator between same meridians = cos lat. of parallel. 26 NAVIGATION AND Or, expressing this in other terms, f-, X dist. on a parallel i . i? n i (^^ dif. of lon gitude = '°' ^"*- "^ P^^"""^^- (2) .-. dif. of longitude = dist. on parallel COS lat. of parallel = dist. on parallel x sec lat. 19. Since for a short distance departure is meas- ured on a parallel of latitude (Art. 10, (c)), in (1) of last article substituting departure for distance on a parallel, we have (1) departure = dif. of long, x cos lat. ; and (2) dif. of long. = — 1 — -^ — = departure x sec lat. cos lat. 20. In plane sailing, when the distance sailed is short, the departure can be converted into difference of longitude by formula (2) of the preceding article, or, when the difference of longitude is given, it can be changed into departure by formula (1); in both cases the parallel of latitude being supposed to be known. But if the distance is not short, there is danger of error, since the latitude varies from point to point of the distance, and the departure is neither the distance on a parallel through the point from which the ship sails, nor on a parallel through the point arrived at. This will be understood from the figure. NAUTICAL ASTRONOMY 27 Suppose the figure to represent a part of the earth's surface, and that AC represents the distance sailed by a ship. The departure for that distance would be KE + LF -{- MG, etc. (Art. 10, (c)), which is, evidently, equal to neither AD nor RC, since, as the meridians PE, PF, etc., meet at the pole P, the distance between them measured on a parallel diminishes as we proceed ^\ from the equator. The total departure is conse- quently less than AD and greater than EC. It would, also, be incorrect to convert this departure into difference of longitude by using the latitude of PC or the latitude of AD, as we really ought to use the lati- tude of the part departure, KE for KE, the latitude of LF for LF, etc., and then take the sum of the differences of longi- tude corresponding to these departures for the whole difference of longitude. If this method were practicable, and we could make the distances AE, EF, etc., small enough, we should find the difference of longitude without appreciable error. As this method is not practicable, two other methods are used for changing departure into difference of longitude. One is the method of middle latitude sailing, the other is the method of Mercator^s sailing. 21. In middle latitude sailing, departure is eon- verted into difference of longitude by using, in Art. 19, (2), the latitude, whose parallel is midway between the parallel of the point sailed from and the parallel of the point arrived at. This latitude is equal to the half sum of thfe lati- tude sailed from and the latitude arrived at, if both 28 NAVIGATION AND latitudes are on the same side of the equator, but to the half differe^ice, if one is north and the other south of the equator. Thus, suppose aST is the parallel midway between RC and AD ; that is, suppose AS = SE, A and C being both north of the equator. WS is the measure of the latitude of the parallel ST (Art. 6). 7|r^ ^ ^(^-^ + ^S) ^ WA-\-WR In a similar manner it may be shown that, in case one place is north and the other south of the equator, the middle latitude is half the difference of the latitudes of the two places. The method of middle latitude sailing is not perfectly exact, but is made nearly so by applying corrections taken from a table prepared for that purpose.* For short distances or for sailing near the equator it is practically correct. 22. By Art. 19, (1) dep. = dif . of long, x cos lat., and (2) dif. of long. = i-- = dep. x sec lat. cos lat. In middle latitude sailing, for latitude we substitute mid. lat., and (1) becomes (a) dep. = dif. of long, x cos mid. lat., and (2) becomes (6) dif. of lonff. = r^^ — = dep. x sec mid. lat. ^ ^ ^ cos mid. lat. ^ Equations (a) and (&) can be represented in terms of base and hypotenuse of a right-angled triangle. * Table of Corrections to Middle Latitude, pages 172, 173. NAUTICAL ASTRONOMY 29 This triangle can be combined in one figure with the triangle for plane sailing, as will be seen by the accompanying diagram. ^73 Ex. 1. From lat. 40° N. and long. 50° W. a vessel sails on a course N.W. b. N. to lat. 50° 12' N. Re- quired distance sailed, and the r* longitude of point of arrival. AB = 10° 12' = 612. Angle A = 33° 45'. Angle DCB = mid. lat. =40° + 50°12' = 45°6'-f cor*of 2' = 45°8'. ^e-dist.--^^- ^12 cos^ COS 33° 45' L. L. = 2.78675 = 9.91985 log 736.3 = 2.86690 dist. = 736.3 miles. dep. = AB tan A = 612 tan 33° 45'. BC 612 tan 33° 45' BC CD = dif . of long. COS DCB COS 45° 8' log612 = 2.78675 log tan 33° 45' = 9.82489 colog. cos 45° 8' = 0.15153 log 579.7 2.76317 dif. of long. = 579'.7 W. = 9°39'.7W. long, of pt. of departure = 50° W. long, of pt. of arrival =59°39'.7 W. Table of Corrections to Middle Latitude, pages 172, 173. 80 NAVIGATION AND Ex. 2. From lat. 32° 22' N., long. 64° 38' W., a ship sails S.W. by W. a distance of 375 miles. Kequired the latitude and longitude of point of arrival. Ans. 28°53'.7 N.; 70°38'.6 W. Ex. 3. From lat. 40° 28' N., long. 74° 1' W., a ship sails S.E. b. S. a distance of 450 miles. Kequired the latitude and longi- tude of point of arrival. Ans. 34° 13'.8 N. ; 68° 47 '.5 W. Ex. 4. From lat. 40° 28' N., long. 74° 1' W., a ship sails S.E. b. E. to lat. 31° 10' N. Required the distance sailed and longi- tude of point of arrival. Ans. 1004 miles; 56° 53'.5 W. Ex. 5. From lat. 32° 28' N., long. 64° 48' W., a vessel sails on a course between S. and W. to lat. 28° 54' N., making a dis- tance of 475 miles. Required the course and the longitude of the point of arrival. J^ns. S. 63° 13' 22" W. ; 72° 59' W. Ex. 6. If from lat. 46° 40' N., long. 53° 7' AV., a ship sails to lat. 32° 38' N., long. 16° 40' W., required the course and distance sailed. Ans. S. 63° 23' E. ; 1879 miles. 23. In Mercator's sailing departure is converted into difference of longitude by means of the principles of Mercator's chart. As the meridians all pass through the poles, a chart, in order to represent correctly the earth's surface, should make the meridian lines curved and approach- ing one another toward either pole. The parallels of latitude being: circles smaller and smaller the nearer they are to the poles, should, on a correct chart, be shorter and shorter curves the farther they are from the equator. On Mercator's chart the equator, the meridians, and parallels of latitude are all represented as straight lines. Meridian lines are all drawn at right angles to NAUTICAL ASTRONOMY 81 the equator, and are, therefore, parallel to each other. Parallels of latitude are made parallel to the equator, and therefore, like parts of any parallel, are equal to like parts of the equator. On Mercator's chart, there- fore, the east and west dimensions of any part of the earth's surface are made too large, except near the equator. To preserve the true proportion existing between the dimensions of any particular part of the earth, the north and south dimensions are lengthened in proportion to the lengthening of the east and west dimensions. The method of accomplishing this will be understood by means of the accompany- ing figures. On a globe representing the earth, the meridians PE, FA, etc., make with the equator and with parallels of latitude a number of quadrilaterals, all of whose sides are curved lines. Thus, in the figure, if the equator, represented by WE, be supposed to be divided into a number of parts of 10°, each equal to AE, and on the meridian FE, we lay ofe EG, GK, KM, etc., each also equal to 10'', drawing parallels of latitude through tlie points of division G, K, M, etc., w^e should divide the surface of the globe into several tiers of quadri- laterals ; one tier composed ot* quadrilaterals each equal to FGAE, a second tier of quadrilaterals each equal to HFGK, a third tier of figures each equal to 32 NAVIGATION AND LHKM, etc. Supposing the earth to be a sphere, on the globe representing it, AE, EG, GK, and KM would all be equal, as they are equal parts of equal great circles. Also, the ratio of EG to GF = sec 10° (Art. 18, (2)), and GK KH = sec 20°; KM ML sec 30^ If we desire to represent these various tiers of quadrilaterals on Mercator's chart, with the features of the earth which they inclose, we draw a straight line of the same length as the curved line represent- ing the equator on the globe ; that is, we make we equal to WE, and divide it into parts wc, cb, ha, and ae, each equal to AE, and at the points iv, c, h, a, and e erect perpendiculars wp, ed, hn, etc., to represent the meridians FW, FC, FB, etc. The lines wp, cd, hn, etc., being at right angles to we, are parallel. If we draw a series of lines par- allel to ive to represent parallels of latitude, as dm, hk, and fg, we form tiers of rectangles ; one tier of rectangles each equal to fgea, a second tier of rectan- gles each equal to hkgf, and so on. By this construc- tion fg, hk, and hn are all made equal to ae. To make the quadrilaterals, like fgea, represent the corresponding quadrilaterals, like FGEA, we must f> i n 1 vn. ^ l f % u J i I , c X. NAUTICAL ASTRONOMY 33 lengthen eg as much as we have lengthened fg. We have made fg^ae = AE = FG sec 10° (Art. 18, (2)), therefore we must make eg = EG sec 10° (or, ae sec 10°), Consequently, if on the line em we take a point g so that eg = ae sec 10°, and through g draw a straight line parallel to ivae, we shall form a tier of quadri- laterals each equal to afge, whose sides af and eg have the same ratio to fg which AF and FG bear to FG. In like manner, if we make gk = ae sec 20°, and through k draw another straight line parallel to wae, we shall form a second tier of quadrilaterals each equal to fhkg, whose sides fh and gk have the same ratio to hk which FH and GK bear to HK. Through m, if Z:m = ae sec30°, we draw another straight line parallel to wae, making a third tier of quadrilaterals, and so on for the rest of the chart. If, instead of taking the parts, like ae, equal to 10° of the equator we make them 1° or V, then the parallels of latitude will be drawn at smaller intervals on the meridian me. If ae = r, then em = 1' (sec 1' + sec 2' + sec 3'). NAV. AND NAUT. ASTR. — 3 34 NAVIGATION AND In the same way the length of Mercator's meridian up to 30° would equal the sum of sec 1' + sec 2' + sec 3' -. + sec 29° 59' + sec 30% or the sum of the series of se(?ants, from sec 1', in- creasing by intervals of V up to sec 30°. Mercator's chart is, therefore, a chart of the earth's surface on which the unit of the scale of representa- tion is continually changing. Near the equator the parts of the earth's surface are correctly represented. As we go north or south to any distance from that line, the parts of the earth are enlarged, as compared with the parts near the equator. As the earth is not a perfect sphere, but a spheroid with its shorter diameter connecting the poles, the meridians are all smaller curves than the equator, so that in the later Mercator's charts, and in the tables of the lengths of Mercator's meridians for different latitudes (called Tables of Meridional Parts), this fact is taken into account. However, with this modifica- tion, the method of construction of a Mercator's chart just given is substantially correct. In Mercator's sailing the unit of measure, or the nautical mile, is 1' of the equator. Tables of Meridi- onal Parts accordingly give in minutes, or nautical miles, the length of Mercator's meridian from the equator to any point of latitude denoted by the table. 24r, The path of a ship continuing on the same course is, on Mercator's chart, a straight line, since to continue on the sam^e course the ship must cut each NAUTICAL ASTRONOMY 35 of the meridians at the same angle, and the meridians are parallel straight lines. As Mercator's meridian is longer than the tnie meridian on a chart representing a curved surface, and is continually lengthening, the number of parts in a certain number of degrees and minutes of the table will generally be greater than the number of minutes in the corresponding number of degrees and minutes of true meridian. Thus, for example, the number of parts of 16° of the table of meridional parts is 966.4, while the number of minutes of 16° of true meridian is 960. Near the equator the number of minutes of true meridian is greater than the number of meridional parts of the same degree measure. Thus, 4° of true meridian = 240°, while meridional parts of 4° = 238.6. (a) Meridional difference of latitude is the distance on Mercator's meridian between two parallels of latitude. Where the latitudes of two places are given, the meridional difference of latitude is found by taking the meridio7ial parts of the less latitude //'om the meridi- onal parts of the greater, if both are north, or both are south latitudes ; but, by adding the meridional parts of the two latitudes, if one is north and the other south latitude. The rule is the same as for finding the true difference of latitude, except that meridional parts of latitude are used instead of latitude. 36 NAVIGATION AND Ex. 1. lat. of Newport, E.I., is 41° 29' N. lat. of Savannah, Ga., is 32° 5' N. dif. of lat. 9° 24' Ex. 2. lat. of Pato Island is 10° 38' N. lat. of Cape St. Eoque is 5° 29' S. dif. of lat. 16° V merid. parts 2725.0 merid. pprts 2022.1 merid. dif. lat. 702.9 merid. parts 637.5 merid. parts 327.3 merid. dif. lat. 964.8 If one latitude is given, and the meridional differ- ence of latitude is found, the latitude required is found by adding the meridional parts of the given latitude to the meridional difference of latitude, if the place whose latitude is required is farther from the equator than the place whose latitude is given, and if both places are on the same side of the equator ; but, by subtracting the meridional difference of latitude from the meridional parts of the given latitude if the place whose latitude required is nearer the equator than the place whose latitude is given ; the degrees and minutes, answering to the result as found in the table of meridional parts, will be the latitude re- quired. This will be evident from the figure, in which WE repre- sents the equator on Mercator's chart. If the latitude of A is given, the meridional parts, or the dis- tance AW, can be found from the table. AB being the merid- NAUTICAL ASTRONOMY 37 ional difference of latitude, the meridional parts of Cor the dis- tance, EC = WB =WA + AB. If the latitude G is given, then from the table CE{=WB) is found ; then, A W= WB-AB=CE- AB. Ex. 1. lat. of place left is 25° 6' N. merid. parts = 1546.9 ship sails northerly till she makes merid. dif. of lat. 750.0 meridional parts of place arrived at = 2296.9 Therefore, from table, latitude arrived at is (nearly) 35° 54' N. Ex. 2. lat. of place left is 46° 10' S. merid. parts = 3113.4 ; ship going northerly, her merid. dif. of lat. is found to be = 825.6 merid. parts of lat. arrived at = 2287.8 Therefore, latitude arrived at is 35° 46'.4 S. If a ship starting from one side of the equator sails to a point on the other side, the latitude of the point arrived at is found by subtracting the meridional parts of the given latitude from the meridional difference of latitude ; the result will be the meridional parts of the required latitude. Ex. The meridional difference of latitude is . . . 1805.8 which is made by a ship going no)^thj starting from lat. 8°41'S merid. parts = 519.5 Therefore, latitude arrived at is 21° 5' N. Merid. parts 1286.3 It is evident, therefore, if the meridional difference of latitude made by a ship sailing from a point on either side of the equator toward a point on the oppo- site side, exceeds the meridional parts of the latitude left, that the ship has crossed the line and has arrived at a N. latitude, if the latitude left was S., but has arrived at a S. latitude if the latitude left was N. 88 NAVIGATION AND Wf B Thus, on the figure, WE representing the equator, a ship sails from B toward A. BW repre- sents the meridional parts of the latitude left, BD is the meridional difference of latitude, AC represents the meridional parts of the lati- tude arrived at. A / y / E / c AC^WD^BD-BW. 25. Combining Mercator^s sailing with plane sailing. Let the figure represent a part of Mercator's chart, on which WE represents the equator, and AC \s> the lengthened distance be- tween two points A and C CAB is the course for that distance. If, from ^ C, CB be drawn perpen- dicular to tho meridian WB, AB will be the me- ridional difference of lati- tude, and BC will be the lengthened departure (Art. 10, (c)). Now, BC^WE', that is, departure, on Merca- tor's c\i2irt, equals difference of longitude (Art. 5, (??)). If, in plane sailing, the same course and distance were represented by the hypotenuse and acute angle of a right-angled triangle, A'RD, AR would be true difference of latitude, RD would be depar- ture. n c U ^ y / X y / J\ 'V^ c X p NAUTICAL ASTRONOMY 39 Now, ABC and A'RD are similar triangles, since the angles A and A are equal, as they both repre- sent the same course, and the angles i^ and ^ are right angles. Placing the angle A upon A\ the two triangles may be combined, as in the figure ; then AR RB (1) AB BC dif. of lat. that is, dep. merid. dif. of lat. dif. of long. Also, BC = AB X tan A' ; that is, (2) dif. of long. = merid. dif. of lat. x tan course. By means of these two triangles all cases of Merca- tor's sailing may be solved, and the position of a ship at sea may be determined from the usual data. The latitude of one position of the ship, either of the point left or the point arrived at, must always be known in order to use Mercator's sailing. Tlie line A'C is 7iot required in calculations. A'D represents the tme distance. Ex. 1. A ship starting from lat. 37° N., long. 10° W., sails on a course between N. and E. to lat. 41° N., making a distance of 300 miles. Eequired the course and the longitude arrived at. lat. 41° N. merid. parts = 2686.5 lat. 37° N. merid. parts = 2378.8 dif. of lat. = 4° = 240' merid. dif. of lat. = 307.7 40 NAVIGATION AND Taking the figure of the preceding article, A'D = 300, A'R = 240, and ^'^ = 307.7. -— — = — — = cos A' = cos course. BC = A'B x tan A'. A'D 300 dif. long. = 307.7 x tan 36° 52' 12" log 240 = 2.38021 log 307.7 = 2.48813 log 300 = 2.47712 log tan 36° 52' 12" = 9.87506 log cos 36° 52' 12" 9.90309 log 230.8 2.36319 dif. long. = 3° 50'.8 E. 9.90309 course = N. 36° 52' E. long, left, 10° W. dif. of long. 3° 50'.8 E. long, arrived at = 6° 9'.2 W. Ex. 2. A ship leaving lat. 50° 10' N., long. 60° E., sails E. S. E. till her departure is 957 miles. Required latitude and longitude arrived at, and the distance sailed. 957 = dist.* log 957 = 2.98091 log sin 67° 30' = 9.96562 log 1035.8 3.01529 dist. = 1035.8 miles. 957 dif. of lat. tan 67° 30' log 957= 2.98091 log tan 67° 30' = 10.38278 log 396.4= 2.59813 dif. lat. = 6° 36'.4 S. lat. left = 50° 10' N. lat. reached = 43° 33'.6 N. merid. parts of 50° 10' = 3472.4 merid. parts of 43° 33'.6 = 2893.4 merid. dif. of lat. = 579 dif. long. = 579 x tan 67° 30'. log 579 = 2.76268 dif. long. = 23° 17'.8 E. long, tan 67° 30' = 10.38278 long, left = 60^ E. log 1397.8 = 3.14546 long, reached = 83° 17'.8 E. * No figure is given for this example, but the student is advised to plot the figure for it, and the figure for each of the examples which follow. NAUTICAL ASTRONOMY 41 Ex. 3. From a point in lat. 49" 57' N., long. 5° 14' W., a vessel sails on a course S. 39° W. to a point in lat. 45° 31' N. Required the distance sailed and the longitude reached. Ans. Dist. = 342.28 miles ; long. = 10° 33'.5 W. Ex. 4. From a point in lat. 49° 57' W., long. 5° 14' W., a vessel goes to lat. 39° 20' N., making a W. departure of 789 miles. Required the course sailed, the distance made, and the longitude reached. Ans. Course = S. 51° i'W.; dist. = 1014 miles; long. = 23°43'.8 W. Ex. 5. From a point in lat. 14° 45' N., long. 17° 33 W., a vessel sails S. 28° 7i' W. to a point in long. 29° 26' W. Required the latitude reached and the distance sailed. Ans. Lat. reached = 7° 26'.5 S ; dist. = 1509.8 miles. Ex. 6. From a point in lat. 20° 22' N., long. 45° 24' W. to a point in lat. 40° 30' N., long. 20° 10' W., it is required to find the course and distance. A71S. Course = N. 47° 6^' E. ; dist. = 1774.9 miles. 42 NAVIGATION AND CHAPTER II GREAT CIRCLE SAILING 26. To find the distance on the arc of a great circle between two points on the earth, the latitude and longi- tude of each point being given. Suppose A and C to represent the two points. If F represents the pole of the earth, WE a part of the equator, FU the me- ridian from which longitude is reckoned, and FW and F£ me- ridians through A and (7, then , WB will be the difference of longi- tude between A and C ; WA will measure the latitude of A, and BC will measure the latitude of C. In the spherical triangle AFC, AF=FW-AW= 90° - lat. of A ; FC=FB-BC= 90° - lat. of C; angle AFC is measured by arc WB, or, degrees of ^PC= degrees of difference of longitude; therefore, we have given two sides and included angle of a spherical triangle to find the thii'd side. NAUTICAL ASTRONOMY 43 27. If it is required to find the distance only, we may proceed in the following manner: Denote the sides opposite A, F, and C by a, p, and c, respectively. From C draw an arc, CD, perpen- Denote dicular to PA at D. the segment PD by x. Then the segment AD will be c-cc, if D falls within the triangle ; if D falls on PA produced, AD will be x — c. (1) Take the case where the perpendicular falls within the triangle. Applying Napier's Rule of the Circular Parts to triangle CDP, we find cosP cota ' cos a tana; = alsoy cos CD cosx (a) (&) In the triangle CD A, from Napier's Rule, p cos/) = cos J.Z>cos (7Z) _ cos (c - x) cos a " cos a: (c) (2) If the perpendicular falls without the triangle, then PD = x, and equations for tan x and cos CD remain the same; but for cos^D we have cos (x — c), so that the C equation for P becomes 44 NAVIGATION AND COS (x — c) COS a cos p = — ^ ^ (d) To find V, therefore, it is necessary only to compute X from equation (a), and to substitute its value in (c) or {d). Ex. 1. It is required to find the distance, on the arc of a great circle, between a point in lat. 40° 28' N., long. 74° 8' W., and a point in lat. 55° 18' N., long. 6° 24' W. Let the first point be represented by A and the second point by C in a figure similar to the first figure of the preceding article. Then, c = P^ = 90° - 40° 28' = 49° 32', a = PC = 90° - 55° 18' = 34° 42', angle APC =P=WB = 74° 8' - 6° 24' = 67° 44'. cos 67° 44' \ofr= 9.57855 tanic cot 34° 42' log = 10.15962 log tan 14° 42' 7" = 9.41893 c = 49° 32' a: =14° 42' 7" cos JO = c- a; = 34° 49' 53" cos 34° 49' 53" cos 34° 42' cos 14° 42' 7" = cos 34° 49' 53" cos 34° 42' sec 14° 42' 7". log cos 34° 49' 53" = 9.91425 log cos 34° 42' = 9.91495 log sec 14° 42' 7" ^ 10.01445 log cos 45° 45' 37" 9.84365 p = 45° 45'f^ = 2745.6 nautical miles. NAUTICAL ASTRONOMY 45 Ex. 2. It is required to find the distance, on the arc of a great circle, between a point in lat. 32° 44' N., long.- 73° 26' W., and a point in lat. 8° 14' S., long. 14° W. c = 90° - 32° 44' = 57° 16', a = 90°+ 8° 14' = 98° 14', P=73°26' -14° = = 59° 26'. tan PD = tan X = cos 59° 26' cot 98° 14' log = = 9.16046 log = = 9.70633 log tan 105° 52' 57^" = 10.54587 tan X = minus quantity. .-. X or PD is > 90°. a; =105° 52' 57V' c= 57° 16' x-c= 48° 36' 57V' cos AC = cos p = cos 48° 36' 57 Jr" cos 98° 14' cos 105° 52' 574" log cos 48° 36' 57V' = 9.82027 log cos 98° 14' = 9.15596 log sec 105° 52' 57V' = 10-o6278 log cos 69° 45' 37" = 9.53901 p = 69° 45fJ' = 4185.6 nautical miles. Ex. 3. Required to find the distance, on the arc of a great circle, between a point in lat. 41° 4' N., long. 69° 55' W., and a point in lat. 51° 26' N., long. 9° 29' W. Ans. 2507.5 miles. Ex. 4, Required to find the distance, on the arc of a great circle, between a point in lat. 37° 48' N., long. 122° 2S' W., and a point in lat. 6° 9' S., 8° 11' E. Ans. 7516.3 miles. 46 NAVIGATION AND 28. When a ship sails between two points, making the shortest distance between these points, it sails on the arc of a great circle. To do this, it cannot continue on the same course, as an arc of a great circle between two points, of different latitudes and longitudes, does not cut the meridians at the same angle. Thus taking Ex. 1 of the previous article and solving by Napier's Analogies, we have : 2 ~ 9 ~ ' ' *^ ' p - = 29° 43'. ^-a^ 40^58^ ^20° 29'; ^ ^ z tan \{C ^-A) = cos 20° 29' x cot 29° 43' sec 77° 45', and tan \{C - A) = sin 20° 29' cot 29° 43' cosec 77° 45'. log cos 20° 29' = 9.97163 log sin 20° 29' = 9.54399 log cot 29° 43' = 10.24353 log cot 29° 43' = 10.24353 log sec 77° 45' = 10.67330 log cosec 77° 45' = 10.01000 log tan 82° 38' 10.88846 log tan 32° 6' 10"= 9.79752 i{C + A)= 82° 38' ilC-A)= 32° 6' 10" A= 50° 31' 50" C= 114° 44' 10" We see, therefore, that the distance AC makes an angle with the meridian PA of 50° 31' 50", and with the meridian PC, of 114° 44' 10". Consequently, the vessel starts on a course N. 50° 31' 50" E., and ends with a course N. 65° 15' 50" E. (the supplement of 114° 44' 10"). Between the points A and C the course would be continually changing. In practice, the course is altered at certain intervals, as, for instance, at points NAUTICAL ASTRONOMY 47 10° in longitude apart, for which the new course is calculated, and the distance between the points is run by Mercator's Sailing. 29. In great circle sailing, the arc of the circle might lead to too high a latitude, or to some obstacle like land or ice, which it would be necessary to avoid. In such cases composite sailing is adopted, or a combi- nation of sailing on the arcs of great circles and on a parallel of latitude. Thus, suppose it were de- sired to sail from A to C hy composite sailing, and that BD were the parallel of high- est latitude to be reached. The great circle starting from A and tangent to the paral- lel is first found ; then the great circle through C and tangent to BD at D is found, tangent to BD, AB is perpendicular to the meridian PB,*^ and CD is perpendicular to the meridian PD. We have, therefore, two right-angled spherical tri- angles, APB and CDP, in each of which an hypote- nuse and a side are given ; PA from the latitude of A and PC from the latitude of C are known ; PB and PD, since each is the complement of the latitude of the highest parallel to be reached, are also known. Conse- Since these circles are * PB is the least line which can he drawn from P to arc AB, and therefore passes through the pole of AB. Consequently, by geometry, PB cuts AB at right angles. 48 NAVIGATION AND quently, the other parts of these triangles can be com- puted by Napier's Rule of the Circular Parts. We can thus ascertain the courses at A and C and the angles AFB and DFC. The angles ^P^and DFC will give us the difference of longitude between A and B, and between C and JD. Since the longitudes of A and C are known, the longitudes of J3 and D are also known. By this method of sailing the vessel goes on the arc of a great circle from A to B, on a parallel of latitude from B to D (in the figure due E.), and then on a great circle from D to C. Ex. 1. A ship sails on a composite track from lat. 37° 15' N., long. 75° 10' W. to lat. 48° 23' N., long. 4° 30' W., not going north of lat. 49° N. Required, the longitude of the point of arrival on the parallel of 49° N., the longitude of the point of departure from the parallel, the initial and final courses, and the total distance sailed.. In the triangle ABP right- In the triangle PDC right- angled at B, P^ = 52° 45', angled at i>,. PC = 41° 37', PB = 41°. PD = 41°. cos APB = cot 52° 45' tan 41° cos DPC = cot 41° 37' tan 41° log cot 52° 45' = 9.88105 log cot 41° 37' = 10.05141 log tan 41° = 9.9391G log tan 41° = 9.93916 log cos 48° 37' 21" = 9.82021 log cos 11° 53' 40" = 9.99057 sin^i^ sin 41° log = 9.81694 sin 52° 45' log = 9.90091 log sin 55° 30' 27" = 9.91603 . ^^ sin 41° log = 9.81694 ®^^ sin 41° 37' log = 9.82226 log sinSr 3' = 9.99468 NAUTICAL ASTRONOMY 49 cog ^^^ cos 52° 45^ log = 9.78197 cos 41° log = 9.87778 log cos 36° 40' 33" = 9.90419 cosOZ> = ^^^ill5^ log = 9.87367 cos 41° log = 9.87778 log cos 7° 52' = 9.99589 long, of ^ = 75° 10' W. long, of C = 4° 30' W. dif. of long. = 48° 37'.4 E. * dif. of long. = 11° 53'f W. long, oi B = 26° 32'.6 W. long, oi D = 16° 23'.7 W. Course at ^ = N. 55° 30' 27" E. Course at C = S. 81° 3' E. long, oi B = 26° 32'.6 W. dist. AB = 2200.55 long, of i> = 16° 23'.7 W. dist. BD = 399.5 dif. of long. = 10° 8 '.9 W. dist. CD = 472.0 = 608.9 log = 2.78455 total dist. = 3072.05 miles log cos 49° =9.81694 log 399.5 = 2.60149 Ex. 2. A vessel sails on a composite track from a point in lat. 46° 10' S., long. 45° E. to a point in lat. 43° 40' S., long. 71° 15' W., not going S. of parallel of 50° S. Required the longitude of the point of arrival on the parallel of 50° S., the longitude of the point of departure from that parallel, the initial and final courses, and the total distance sailed. Ans. 15°55'.4E.,34°27'.9 W.; S. 68°8'48" W.;N. 62°42' W.; 4663.2 miles. NAV. AND NAUT. A8TK. — 4 60 NAVIGATION AND CHAPTER III COURSES The magnetic needle of the compass is supposed to give a north and south line, but in point of fact it rarely points north and south. It is subject to influ- ences which deflect it from a north and south line ; so that the north point of the magnet is sometimes east and sometimes west of a true north and south line. The most important deflecting influences cause two errors, as they are called ; namely, an error of Variation, and an error of Deviation. The error of Variation is due to the magnetic action of the earth. The error is greater or less, or even nothing, according to the position of the compass at various points on the earth's surface. Variation may therefore be called a geographical error. It is known and calculable, and allowance can be made for it at any point on the earth. Tlie error of Deviation is due to the magnetic action of the ship and its cargo, and changes according to the direction in which the ship is headed. Each ship has its own error of Deviation. This error can be known, and, to a certain extent, can be counteracted by proper arrangements, but must always be taken into account. NAUTICAL ASTRONOMY 51 30. The True Course of a ship is the angle between the distance, or the line traversed by the ship, and the meridian or true north and south line. 31. The Magnetic Course of a ship is the angle between the distance and a north and south line, as indicated by the magnet of a compass lohich is not affected by the error of Deviation. 32. The Compass Course of a ship is the angle between the distance and a north and south line, as indicated by the compass of a ship, 33. For a steamship, in calm weather, or in a sail- ing vessel with a wind directly astern, the Compass Course, when corrected for variation and deviation, will give the True Course ; but when the wind blows from any direction, except from right ahead or astern, it pushes the vessel aside from the course on which she is headed, so that her track is not in the direction in which she is headed, but makes an angle with that direction. This angle is called leeivay, because the push of the wind on the vessel is to leeward. Thus, in the figure NS is a true meridian ; NCB is the apparent course ; NCE is the true course ; the wind, shown by the direction of the arrow, diverting the vessel from the track AB, in which she is headed, to the true track DE. The angle between these tracks, ECB, leeioay ; is given in points ; and is estimated by the eye. 52 NAVIGATION AND . Although leeway is not an error of the compass, the effect of it is the same as if it were, and allow- ance must be made for it in order to determine the true course of a ship. In navigation it is important to be able to con- vert a true course into a compass course and also a compass course into a true course, by applying corrections for the various errors, which have been mentioned. The method of doing this will be best ascertained by applying the errors one by one. In expressing, or converting courses, the observer is supposed to be at the center of the compass card. 34. To convert a true course into a magnetic course, the variation being given. "^ Both variation and deviation are given in terms which are applied to the north point of the compass needle. For instance, if the variation is given as 8° E., the north point of the needle points 8° east of a true N. and S. line, or, looking from the center of the compass, 8° to the right of that line. Looking south from the center, the variation would still be 8° to the right. In works on Navigation it is customary to give rules for converting courses, but it is best to draw a diagram, which will illustrate the example given, and after a little practice, rules can be derived by the learner himself. * Variation charts are published by the Government Coast Survey. NAUTICAL ASTRONOMY 53 Ex. 1. Let the true course be N.E. b. N. and the variation be 8° E. Required the magnetic course. Suppose the observer to be at ; the line NS = true N. and S. line ; N^S^ = magnetic N. and S. line. true course= iVO^ =33° 45' to right of N. variation = iVOiV^= 8° to right of K mag. course = iV^OJ. = 25° 45' to right of N. or N.N.E. \ E., nearly. Ex. 2. Let the true course be N.W., and the variation be 12° W. ^ NS = true N. and S. line ; N^S„ = magnetic N. and S. line. true course = 5 OiV =45°, or 4 pts. left of N. variation = ^^0^=12°, or 1 pt., nearly, left of N. mag. course =^^0J3= 33°, or 3 pts. left of N. =N.W. b. N. iViV„ In converting courses sometimes the work is expressed in degrees, and sometimes to the nearest points, half points, or quarter points. Ex. 3. If the course is N.W., and the variation b, is 12° E., to obtain the magnetic course we add the 12° to the true course. true course = ^05 =45°, or 4 pts. left of N. variation = iVOiV^= 12°, or 1 pt. right of N. mag. course =-^„0B = 57°, or 5 pts. left of N. '"•>S 54 NAVIGATION AND jf Ex. 4. Let true course be S.E. b. S. and vaxia- tion be 22° E. NS = true N. and S. line. ^m>Sm = magnetic N. and S. line. Suppose observer to be at 0. true course SOA — 3 pts. left of S. = 33° 45' left of S. variation SOS^ = 22° right of S. magnetic course = Sr^OA = 55° 45' left of S. = nearly 5 pts. left of S. Ex. 5. Let true course be S.E. b. S., but vari- ation be 22° W. true course SOA = 33° 45' left of S. variation SOS^ = 22° left of S. magnetic course = S^OA = 11° 45' left of S. From these examples and by an inspec- tion of the figures, supposing the observer to be at center of compass, it will be seen that ivhen the true course and the variation are both to the right or both to the left of either the N. or S. points, the magnetic course is the difference of the two; but when one is to the right and the other to the left of the N. or S. points, the magnetic course is the sum of the two. 35. To change a magnetic course into a true course : if the given course and variation are both to the right or both to the left of either iV. or S. points, add the tioo ; if one is to the right and the other to the left, take the difference. NAUTICAL ASTRONOMY 55 This rule for changing magnetic to true courses would naturally follow, from what has been said of converting true courses into magnetic courses, as the processes are reversed, and we should, therefore, reverse the former rule. We will illustrate by examples. Ex. 1. Magnetic course is N.N.E. Vari- ation is 22° E. Find true course. mag. course = N^AB = 22° 30' right of N. variation = N^AN= 22° right of K true course = NAB = 44° 30' right of N. = 4 pts., nearly. = N.E., nearly. Ex. 2. Let the magnetic course be S.E. b. S., and the variation be 11° W. Find true course. mag. course = S,,AB = 33° 45' left of S. variation = S^AS = 11° left of S. true course = SAB = 44° 45' left of S. = S.E., nearly. Ex. 3. Let the magnetic course be S.W. b. W., and the variation be 11° 15' W. Find true course. mag. course = S^AB = 5 pts. right of S. variation = S^AS = 1 pt. left of S. true course SAB S.W. 4 pts. right of S. 66 NAVIGATION AND 36. To convert magnetic courses into compass courses, or compass courses into magnetic courses, it is necessary to have a list of deviations correspond- ing to the different directions in which the ship heads. This is determined before the ship leaves port. De- viation acting on the compass needle to deflect it from a magnetic N. and S. line, tables of deviation give the amounts of deviation E. and W. of the mag- netic north point. Though each ship has its own Deviation Table, the table here given will serve to illustrate the subject. Deviation Table I. Direction in Degrees and Minutes. I. Course by Ship's Compass. II. Deviation of the Compass. I. Course by Ship's Compass. n. Deviation of the Compass. North 3° 10' W. South 3°10'E. 11° 15' N. b. E. 2 35 E. S. b. W. 5 E. 22 30 N.N.E. 8 10 E. s.s.w. 3 W. 33 45 N.E. b. N. 13 10 E. S.W. b. s. 6 30 W. 45 N.E. 16 50 E. s.w. 9 40 W. 56 15 N.E. b. E. 19 30 E. S.W. b. w. 13 W. 67 30 E.N.E. 20 30 E. w.s.w. 16 10 W. 78 45 E. b. N. 21 5 E. W. b. s. 19 15 W. 90 East 20 20 E. West 21 10 W. 78 45 E. b. S. 19 15 E. W. b. N. 23 20 W. 67 30 E.S.E. 18 5 E. W.N.W. 24 W. m 15 S.E. b. E. 16 30 E. N.W. b. W. 23 35 W. 45 S.E. 14 40 E. N.W. 22 W. 33 45 S.E. b. S. 12 5 E. N.W. b. N. 19 W. 22 30 S.S.E. 9 40 E. N.N.W. 14 50 W. 11 15 S. b. E. 6 E. N. b. W. 9 15 W. South 3 10 E. North 3 10 W. NAUTICAL ASTRONOMY 57 37. To find the magnetic course, having given the compass course and the deviation. Nm^, Ex. 1. Let the compass course be N.N.E. By the table the deviation is 8° 10' E. Let N^S^ be magnetic N. and S. line ; N^S^ be compass N. and S. line ; and AB be ship's track. com. course = N,AB = 22° 30' right of N. deviation = N,AN„ = 8° 10' right of N. mag. course = N^AB = 30° 40' right of N. = N.N.E. I E. Ex. 2. Let the compass course be N. 80° W. This is li° W. of W. b. N. The deviation for W. b. N. is 23° 20' W. The deviation for N. 80° W. will be a little less. As in steering a vessel it is impossible to hold her head to a minute of correction, if we call the deviation 23° W. we shall not be much out of the way. com. course = N,AB = 80° left of N. deviation = N,AN^ = 23° left of N. mag. course = N^AB = 103° left of N. = 77° right of S. mag. course = S. 77° W. 38. To find the compass course, the magnetic course and deviation being given. Ex. 1. Let the magnetic course be E.KE. magnetic course =6 pts. right of N. or N. 67° 30' E. deviation from page 56 = If pts. right of N. or N. 20° 30' E. B<- approximate compass course =4:| pts. right of N. or N. 47° E. 58 NAVIGATION AND This is only an approximate answer, as will be evident ; for if we steer by coinpa^ss N. 47° E., the deviation for that course is nearly 17° 30'. Thus : compass course = 47° right of N. deviation = 17° 30' right of N. magnetic course would then = 64J° right of N. or 3 J° less than the given course. But, if we apply to the given magnetic course the correc- tion due to deviation for the approximate compass course, the example will prove. Thus : magnetic course = 6 pts. or 67° 30' right of N. . deviation for N. ^ E. = 1^ pts. or 17° 30' right of N. compass course N.E. | E. = 4^ pts. or 50° right of N. Proof: compass course = 4 J pts. or 50° right of N. deviation = 1 J pts. or 17° 30' right of N. magnetic course = 6 pts. or 67° 30' right of N. Courses and deviations, when given in points, are given to nearest points, half points, or quarter points. Since the Deviation tables are made for angles indicated by the compass courses, we get only an approximate result by applying the deviation corre- sponding to the magnetic course. Hence, to be accu- rate, we first find this approximate compass course, and then apply the correction, which corresponds to this approximate course in the table, to the original magnetic course. We have considered the applications of variation and deviation separately, for the sake of clearness; but in practice, their action on the magnet of the NAUTICAL ASTRONOMY 59 JV compass is combined. We have to convert compass courses into true courses, and also true courses into compass courses. In changing a compass course into a true course tlie result is the same, whether we apply corrections for variation and deviation separately, or together; but in converting a true course into a compass course we must apply correction for variation first, and then correction /or deviation, Ex. 1. Find true course; variation being 25° E.; compass course being N.N.E. ; and deviation being taken from table on page 56. In figure let notation of lines be the same as in pre- ceding figures. compass course = N,AB = 22° 30' right of N. variation = NAN^ = 25° right of N. deviation = N^AN, = 8° 10' right of N. sum = NAN, = 33° 10' right of N. true course = NAB = 55° 40' right of N. = nearly N.E. b. E. Ex. 2. Find true course, variation being 25° W. ; compass course being S.E. b. E. ; and deviation being taken from table. variation = NAN„ = SAS^ = 2-1- pts. left of S. deviation = S^AS^ = 1^- pts. right of S. difference = SAS, = f pt. left, of S. compass course = S^AB = 5 pts. left of S. true course = SAB = 5| pts. left of S. = E.S.E. i S. >ss. 60 NAVIGATION AND Ex. 3. Let the true course be N. 35° W. and the variation be 10° E. Find the compass course. true course = NAB = 35° left of N. variation = NAN^ = 10° right of N. magnetic course = N^AB = 45° left of N. deviation (approximate) = 22° left of N. N^N approx. compass course = 23° left of N. deviation = 14° 50' left of N. compass course N,AB = 30° 10' left of N. = N. 30° 10' W. Ex. 4. Let the true course be N.E. b. E. and the variation be 20° W. Find the compass course. true course = NAB = 5 pts. right of N. variation = NAN„ = If P^s. left of N. magnetic course = N^AB = 6 j pts. right of N. By table, page 56 : approximate deviation = IJ pts. right of N. approx. compass course = 5 pts. right of N. deviation = If pts. right of N. compass course = 5 pts. right of N. Same examples by degrees : true course = NAB = 56° 15' right of N. variation = NAN^ = 20' left of N. magnetic course = N„AB = 76° 15' right of N. approximate deviation = 21° right of N. approximate compass course = 55° 15' right of N. deviation (to be taken from 76° 15') = 19° 30' right of N. compass course = N^AB = 56° 45' right of N. NAUTICAL ASTRONOMY 61 Ex. 5. Find the true course; the compass course being S.E., the variation being 28° W., leeway being 2 pts., and the wind blowing E.N.E. Take deviation from table on page 56. compass course = -iS^^jB' =45° left of S. deviation=AS,^^^=14° 40' right of S. variation = aS^^>S =28° left of S. dif. = ^,^>S' =13° 20' left of S. appar. true course=>S'^^' =58° 20' left of S. But the influence of the wind, whose direc- s Sc tion is shown by arrow in figure, changes this apparent true course to the leeward by two points, represented by the angle BAB'. Thus : apparent true course = SAB' = 5S° 20' left of S. leeway = BAB' = 22° 30' toward S., or right of S. true course = SAB = 35° 50' left of S., or S.E. } S. Ex. 6. Compass course is S.W. i S. Variation is 6° E. ; wind is S.S.E., and leeway IJ pts. Deviation being taken from table on page 56. Find true course. Example can be worked without figure thus : course by compass = 3| pts. right of S. variation = 6° right of S. 1 deviation = 9° left of S. J dif. = 3° left of S. = J pt. left of S. apparent true course = 3^ pts. right of S. leeway = If pts. right of S. true course = 5^ pts. right of S. W.S.W. f s. 62 NAVIGATION AND The preceding examples could all have been worked without figures, but, until the learner has become familiar with the methods of applying the different corrections, it is best to check the numerical work by means of a diagram. OF THE ^ACIFIU NAUTICAL ASTRONOMY 63 CHAPTER IV ASTRONOMICAL TERMS 39. Before giving definitions of tke terms used in Nautical Astronomy, we must first "consider the effects of the earth's revolution around the sun, as they appear to an observer on the earth. In the figure, let ABCD represent the orbit in which the earth revolves about the sun, S, and S^r^^ A, B, (7, and D represent the positions of the earth at the beginning of the seasons of spring, summer, fall, and winter, respectively. If the figure repre- sents the plane of the earth's orbit, the axis of the earth is not at ricrht ano-les to that orbit, but mnkes an angle with it of about 66° 33'. The plane of the equator therefore makes an angle with it of 23° 27'. 64 NAVIGATION AND To the observer on the earth the heavenly bodies, the sun included, appear to be on the interior sur- face of a very large sphere, of which the center is his own point of observation, or his own eye. This imaginary interior surface of a sphere is called the celestial concave. The poles of the heavens are the points of the celestial concave, toward which the axis of the earth is directed. The celestial pole ahove the horizon is called the elevated pole. Considering the earth as motionless, to the observer on it, the sun appears to travel daily in the celestial concave from east to west. If from a standpoint on the earth we could watch the sun in the heavens during the whole year, it would appear to describe a circle on the celestial concave. This circle is called the ecliptic. The plane of the earth's equator, being supposed produced, would cut the celestial concave in the celestial equator or equinoctial. The ecliptic and the equinoctial intersect in two ptc^^ points, known as the first point of Aries and the first point of Libra. About March 21 the center of the sun is at the first point of Aries, where the equinoc- B NAUTICAL ASTRONOMY 65 tial crosses the ecliptic : and about September 23 it is at the first point of Libra, where the equinoctial intersects the ecliptic a second time. These points of intersection are called equinoctial points, because, at the seasons of the year when the sun reaches them, the days and nights are of nearly equal length. Thus in the figure, ABCD is the ecliptic. AECF is the equinoctial. A is the first point of Aries, where the sun changes its declination from S. to N. ; C is the first point of Libra, where the sun changes its declination from N. to S. The equinoctial is a fixed circle on the celestial concave, and the first point of Aries is considered a fixed point,''^ as it is the point of intersection of the ecliptic and the equinoctial. The positions of heavenly bodies may therefore be expressed with reference to them, just as the positions of places on the earth's surface are expressed in latitude and lon- gitude by reference to the equator and the meridian of Greenwich. 40. Let the accompanying figure represent the earth, PWP'E, surrounded by the celestial sphere, pivp'e. If the axis of the earth, PP\ be produced to meet the celestial concave in the points p and p\ these points are called the celestial poles, and the line pp is called the axis of the celestial sphere. ♦First point of Aries moves yearly 50" (nearly) to westward. NAV. AND NAUT. ASTR. 5 66 NAVIGATION AND The plane of the equator, WDE, produced, inter- sects the celestial sphere in the celestial equator, wSe. The planes of the meridians PEP', PDP\ intersect the celestial sphere in great circles, pep\ pTp, which are called hour circles and also circles of declination. Since the earth revolves upon its axis once in 24 hours, every point on a ce- lest.'al meridian would appear, to an observer on the earth's surface, to move through a complete circumference, or 360°, during that time. If, now, the celestial meridians are drawn at intervals of 15° (on the equator), there will be 24 such meridians. Since the time in which all these meridians pass by an observer is 24 hours, the interval of time of passage between two successive meridians will be one hour, since 24 of them pass by him in 24 hours. If meridians are drawn at intervals of 1°, the inter- val of time of passage of two such meridians will be — , or 4 minutes. Thus, the passage of 15 these meridians or of points on them being measured by time or degrees, we can convert one measure into the other. NAUTICAL ASTRONOMY 67 The angles made by these meridians at p and p' are called hour angles, and these angles are measured by the arcs which they intercept on the arc of the celestial equator wSe. The celestial horizon of any place, on the earth's surface, is the circle made by a plane passing through the center of the earth parallel to the plane of the horizon at that place, and intersecting the celestial sphere. The celestial horizon of the point L is HSK. If a straight line be drawn from the center, (7, to Z, and this line be produced through L to meet the celestial sphere at Z, Z will be the zenith of Z ; Zp will measure the zenith distance of L {i.e. the dis- tance of the zenith of the point L from the pole), and Ze will measure the celestial latitude of L. The zenith distance is the complement of the celestial latitude. The degree measure of the celestial latitude is the same as that of the terrestrial latitude, since they both subtend the same angle at the center of the earth. Thus, Ze and LE both subtend the angle LCE. Since Z is the extremity of the diameter perpen- dicular to the plane of the horizon HSK, Z is the pole of HSK, and therefore every point on HSK is 90^ from Z. If the line CZ be produced to meet the surface of the celestial sphere again at n, n will be the nadir of the observer at X. The declination of a heavenly body is the arc of 68 NAVIGATION AND the circle of declination, intercepted between the equinoctial and the position of the body. Declination is measured in degrees, minutes, etc., N. or S. from the equinoctial, toward the pole. Thus in the preceding figure, TR is the declination of R and is S. declination. The polar distance of a heavenly body is the dis- tance of that body from the elevated pole, and is 90° T the declination : the minus sign being taken if the declination of the body is of the same name with the pole, that is, both being N. or both S. ; but the plus sign being used if the declination and the pole are not of the same name, that is, one being N. and the other S. In the preceding figure, calling p the N. pole, and consider- ing it the elevated pole, the polar distance of R is 90° + TR. If 2>' were taken as the elevated pole, p'R would be the polar distance and would be 90° — TR. The altitude of a heavenly body is the angle of elevation of the body above the plane of the horizon. A distinction is made between an observed altitude of a body and its true altitude. By an observed altitude^ in Navigation, is generally understood the angle of elevation of a body above the visible horizon, as represented by the horizon line of the sea. A true altitude is an observed altitude corrected, so as to represent the angle of elevation of the body above the celestial horizon. NAUTICAL ASTRONOMY 69 Circles of altitude are great circles of the celestial sphere which pass through the zenith of the observer. Circles of altitude are also called vertical circles because their planes are perpendicular or vertical to the plane of the horizon. The altitude of a body is measured on the arc of a circle of altitude between the horizon circle and the position of the body. This measure is generally used in calculations as the altitude. In the preceding figure, Ze/fand ZTWare circles of altitude. MT is the altitude of T. The zenith distance of a body is its distance from the zenith measured on a circle of altitude. ZT is zenith distance of T and equals 90° - MT or 90°- altitude of T. The celestial meridian of any place is the circle on the celestial concave in which the plane of the terrestrial meridian of that place produced cuts the concave. It is the circle of altitude which passes through the celestial poles. In the preceding figure, if X be a place on the earth's surface, and the plane of the meridian PLEP be produced to cut the celestial concave in HpZeK, HpZeK is the celestial meridian of L. It coincides with the circle of altitude through Z. The points in which the celestial meridian cuts the horizon are the N. and S. points of the horizon. H and K are the N. and S. points of the celestial horizon of the place L, supposing P and P to be N. and S. poles. 70 NAVIGATION AND The prime vertical is the circle of altitude whose plane is at right angles to the plane of the celestial meridian. It intersects the horizon in the E. and W. points. If, in the preceding figure, a plane be passed through Cz at right angles to the plane of HpZeK, the circle in which it cuts the celestial concave will be the prime vertical. The right ascension of a heavenly body is the arc of the equinoctial intercepted between the first point of Aries and the circle of declination which passes through the center of the body. Right ascension is measured eastward from the first point of Aries from 0° to 360°; or, in hours, from h. to 24 h. Let the figure represent the celestial sphere projected on the plane of the horizon NWE\ P will represent the N. pole; WDE will represent the equinoc- tial ; ^C will represent the eclip- tic ; and A^ the intersection of the ecliptic with the equinoctial, will represent the first point of Aries. If B represent the position of a heavenly body, draw the arc of a circle of declination, PB, and produce the arc to meet the equi- noctial at D. AD will represent the right ascension of B. 41. The earth being inside the celestial concave, the observer sees the heavenly bodies from the inside. Astronomical diagrams are drawn on the supposition NAUTICAL ASTRONOMY 71 that the observer is on the outside of the celestial concave, as the relations and positions of celestial bodies can best be represented on this supposi- tion. The representations are made on different planes, according to the supposed different points of view. Thus, if the point of view is directly above the zenith^ the representation of the heavenly bodies is made on the plane of the horizon. This is a very useful mode of representation. If the point of view is at either the E. or W, points, the representation is made on the plane of the celestial meridian. If the poi7it of view is directly above the celestial pole, the representation is made on the plane of the equinoctial or celestial equator. If NWSE represent the horizon, and if the point of view is directly above the zenith, the zenith will be projected on the center of the circle, and the circles of alti- tude^ passing through the zenith, will be projected as straight lines. If N, S, E, W be the N., S., E., and W. points of the horizon, NS will be the celestial meridian of the observer whose zenith is Z. The prime vertical, or circle of altitude at right angles to the celestial meridian, in the figure will be WE. 72 NAVIGATION AND 42. To represent, on the plane of the horizon, the celestial pole and the celestial equator for a given latitude. Suppose the latitude to be 42° N. Let ZL represent 42% and LP represent 90°. If an arc of a great circle WLE be drawn with P as a pole, it will pass through TT, L, and E, and represent the celestial equator, or equi- noctial. For, since by defi- nition, the planes of the celestial meridian and prime vertical are at right angles to each other, the diameter joining E and W lies in the plane perpendicular to the plane of NS. Therefore, E and W are poles of NS. Consequently, ^ and H^ are each at a quadrant's distance from P, for the polar distance of a great circle is a quadrant. But PL is a quadrant by construction. Therefore, P repre- senting the celestial pole, WLE will represent the equinoctial or celestial equator. 43. The azimuth of a heavenly body is the angle, at the zenith of the observer, between the celestial meridian and the circle of altitude passing through the body. It is measured by an arc of the horizon between the N. and S. points and the point in w^hich the circle of altitude intersects the horizon. Azimuth is meas- ured from the N. and S. points E. and W. from 0° to 90^ NAUTICAL ASTRONOMY 73 Azimuth is sometimes called the true bearing of a heavenly body. To represent on the plane of the horizon the altitude, zenith distance, and azimuth of a heavenly body. Let NWSE represent the plane of the horizon. Let the azimuth be S. 50° W., and the altitude be 30°. Measure SA = 50° ; through A draw the circle of altitude, ZA. On ZA take ^^=30° to represent the altitude. This will give B as the place of the heavenly body. ZB is the zenith distance. If P be supposed to be the celestial pole, PB will represent the polar distance of the body. SZB is the azimuth, measured by arc SA, 74 NAVIGATION AND CHAPTER V TIME 44. Time is measured by the intervals between the appearances of certain celestial bodies on the meridian of the observer. Thus, sidereal time is measured by the successive appearcinces of the first point of Aries on the meridian. The period elapsing between two successive appear- ances of the first point of Aries on the same part of the meridian is called a sidereal day. The transit of any heavenly body is its passage across the celestial meridian. The instant when the first point of Aries, or when any heavenly body, is on the meridian is called the time of its transit. As the celestial meridian passes through the zenith and nadir, the first point of Aries is really on the celestial meridian twice ; but a sidereal day is meas- ured by the interval between two successive transits on that part of the meridian ivhicJi contains the zenith. Transits on this part of the meridian are called upper transits^ while transits on the part of the meridian which contains the nadir are called lower transits. The terms meridian passage and culmination are sometimes used in place of the term transit NAUTICAL ASTRONOMY 75 Besides sidereal time, we have solar time. Apparent solar time is measured in terms of an apparent solar day. An apparent solar day is the interval between two successive upper transits of the center of the sun over the meridian of the observer. These successive returns of the real sun have not always equal intervals between them : first, because the sun does not move in the plane of the equinoctial, but in the ecliptic, which is inclined at an angle of 23° 27' to the equinoctial ; and, second, because the sun's movement in the ecliptic is not uniform. Thus, when the earth is nearest to the sun it moves in its orbit a little over 61' daily, or, considering the earth as still, the sun moves in the ecliptic the same amount ; but when the earth and sun are farthest from one another, the sun moves in the ecliptic about 57' daily, and, at all other times, at rates varying between these two amounts. To secure an invariable unit of time, mean solar time is used, measured in terms of the mean solar day, which is equal in length to the average of all the apparent solar days of the year. Mean solar time is supposed to be regulated by the movements of a fictitious sun, moving in the equinoc- tial or celestial equator, at a rate which is the average or mean rate of movement of the true sun in the ecliptic. If the imaginary or mean sun and the true sun are supposed to start from the same circle of declination, and return to the same circle at the end 76 NAVIGATION AND of the year, in the interval they are sometimes on the same circle of declination, but generally on different circles, the mean sun being sometimes ahead of the true sun and sometimes behind it. The equation of time is the difference between time measured by the mean sun and time measured by the real sun. This equation of time for every day is always to be found in the Nautical Almanac on pages I and II of each month. To illustrate, by a figure, the meanings of sidereal time, apparent solar time, mean solar time, and the equation of time. Let NWSE represent the horizon; P the pole; WRE the celestial equator or equinoctial ; A the first point of Aries; and ABQ the ecliptic. Let B represent the place of the true sun on the eclip- tic, and m the place of the mean sun on the equinoctial. Draw circles of declination, PBT and Pm. Sidereal time is represented by the angle RPA, or by its measuring arc RA, Apparent solar time is the angle RPB, or its meas- uring arc RT. Mean solar time is RPm, or the arc Rm. The equation of time is mPT, or arc mT. Thus we may define time by angles measured from the celestial meridian westward. Sidereal time is the angle at the pole of the equi- NAUTICAL ASTRONOMY 77 noctial between the meridian and a circle of declina- tion passing through the first point of Aries. Apparent solar time is the angle at the pole be- tween the meridian and a circle of declination passing through the center of the trit^e sun. Mean solar time is the angle at the pole between the meridian and a circle of declination passing through the position of the mean sun. A sidereal clock is adjusted so as to mark 24 hours between two successive transits of the first point of Aries. A mean solar clock is adjusted to mark 24 hours between two successive transits of the mean sun. Clocks and watches in ordinary use are adjusted to mean solar time. 45. The daily motion of the mean sun, in the equi- noctial, is found to be 59' 8''.33. This is easily deter- mined from the time it takes the true and the mean suns, starting from the meridian of any point, to return to the same meridian. This time is found to be 365.2422 mean solar days, during which the mean sun travels through a complete circle, or 360°. In one day, therefore, it would travel through gj^", or 59' 8".33. 46. In order to find the arc described by a merid- ian of the earth in a mean solar day, let P and P^ represent two positions of the center of the earth in its orbit, separated by an interval of time equal to a mean solar day. 78 NAVIGATION AND Suppose a plane to be passed through the celestial equator; and that the small circles represent the terrestrial equator of the earth in its two positions ; and S to be the position of the mean sun. FA and F^A^ will be the two projections of the same meridian. As the fixed stars are at such immense distances from the earth, rays of light from such a star, represented by TA and TiAi would fall in parallel lines on the earth, in its two positions. Thus, the meridian FA, having the light from the star on it, in its first position, would receive the same light in its. second position F^A^, having in the interval made a complete rotation, or having gone through an arc of 360°. Now if S, on the line TA, be supposed to be the position of the mean sun, we join SF^ Since by Art. 45 FF, is 59' 8".33, the angle FSF, is also 59' 8".33. Therefore the alternate angle SF^T^ is an angle of 59' 8".33, and the arc AB is an arc of 59' 8".33 ; that is, the earth in passing from F to F^ in its rotation on its axis, carries the meridian FA past its position F^A^ to the position F^B, and, therefore, the meridian moves through an arc of 360° 59' 8".33 in a mean solar day, or 59' 8".33 more than in a sidereal day. NAUTICAL ASTRONOMY 79 47. In a sidereal day of 24 hours the meridian of any place on the earth revolves through 360°. In one hour it passes through -^2^^° = 15° ; in one minute it passes through \^° = \° = 15'; in one second it passes through l^'= 15"; consequently, in passing through an arc of 59' 8".33, it takes an amount of time equal to (|^) m. + (~) s*., or equal to 3 m. 56.555 s. In a mean solar day of 24 hours, the meridian of any place revolves through 360° 59' 8". 33. A day of 24 hours of mean solar time is therefore longer than a day of 24 hours of sidereal time by the amount of time (sidereal) which it takes the meridian to pass through an arc of 59' 8".33 ; that is, 3 m. 56.555 s. Therefore, 24 h. mean solar time = 24 h. 3 m. 56.555 s. sidereal time. Thus the sidereal day is shorter than a mean solar day. 48. To convert sidereal time into mean solar time, and mean sola time into sidereal time. Let St = any interval of sidereal time, and M^ = the same interval expressed in mean solar time. As the sidereal day is shorter than the mean solar day, a given interval of time will have more sidereal hours in it than solar hours, and the ratio of the hours sidereal to the hours mean solar will be the ratio between the number of hours, minutes, and seconds in a sidereal day, and the 24 hours in a mean solar day. 80 NAVIGATION AND Thus S, 24 h. 3 m. 56.555 s. M. and --^ = M, S, 24 h. 24 h. = 1.0027379, 0.9972697. and 24 h. 3 m. 56.555 s. S, = M, X 1.0027379 = Jf,+ .0027379 M„ M,= S, X 0.9972697 = aS, - .0027303 ;S,. By means of these formulae the tables of the Nauti- cal Almanac, and those in Bowditch's Tables, for converting sidereal into mean solar time or mean solar into sidereal time, can be computed. 49. To convert a given mean solar time into appar- ent solar time ; and, conversely, to convert given apparent time into mean time ; given also the equa- tion of time. Ex. 1. Let mean time be 3 h. 14 m. ; and the equation of time be 3 m. 4 s., to be subtracted. Required apparent time. mean time = 3 h. 14 m. equation of time = 3 m. 4 s. apparent time =>3 h. 10 m. 56 s. To illustrate this example by a figure, suppose in addi- tion to the given terms, the declination of the sun is 15° N. Let NWSE be the plane of the horizon ; Z the zenith ; P the pole; and WBE the celestial equator ; AS^ the NAUTICAL ASTRONOMY 81 ecliptic ; aS^ the center of the true sun ; M the posi- tion of the mean sun on the equinoctial. Through S^ draw the circle of declination FS^C; and draw FM to M. S,C= 15°. Then MFB = mean time = 3 h. 14 m. SJPM= equation of time = 3 m. 4 s. S^FB = apparent time = 3 h. 10 m. 56 s. Ex. 2. Let apparent time be 4 h. ; and equation of time be 2 m. 56 s., to be added ; and declination of sun be 20° N. Re- quired Mf In figure above, SiC = 20°. apparent time = SiPB = 4 h. equation of time = S^PM = 2 m. bQ s. Mt = MPB = 4 h. 2 m. m s. Sometimes the equation of time is additive, and at other times subtractive. It is given for every day of the year, on pages I and II (for the month), in the Nautical Almanac, and whether additive or subtractive. 50. Given mean time, and the right ascension of the m.ean sun, to find sidereal time at any place ; that is, the right ascension of the meridian of the observer. Let NWSE represent the plane of the equinoctial ; NFS the projection on it of the celestial meridian ; A the position of the first point of Aries ; and M the position of the mean sun. (Defs. pages 76 and 77.) NAV, AND NALT. ASTR. 6 82 NAVIGATION AND (1) S^ = SPA = MPA + SPM = right ascension of mean sun + mean time. If Ml be position of mean sun, S, = SPA = M^PA - M,PS, But JfiP;S=360° (or 24 h.)- angle measured by SANMi = 24 h. - mean time. .-. /S< = R.A. mean sun -(24 h. — mean time), i.e. (2) St = R. A. of mean sun + mean time — 24 h. From equations (1) and (2) we see that sidereal time = R. A. mean sun + mean time, but that when the sum of R.A. mean sun and mean time is greater than 24 h., we subtract 24 h. from that sum. Ex. 1. Given 3f^ = 7 h. 10 m. and R.A. mean sun = 2h. 38 m. 42 s. Find sidereal time. ^, = 2 h. 38 m. 42 s. + 7 h. 10 m. = 9 h. 48 m. 42 s. Ex. 2. Given mean time 10 h. 32 m. 40 s. and R.A. mean sun = 18 h. 45 m. 35 s. Find sidereal time. M, = 10 h. 32 m. 40 s. R.A. mean sun = 18 h. 45 m. 35 s. Sid. time = 29 h. 18 m. 15 s. - 24 h. = 5 h. 18 m. 15 s. 51. To convert sidereal time into meaii time ; given the right ascension of the mean sun. Since by the preceding article sidereal time = ^.A. mean sun + mean time, or = R.A. mean sun + mean time - 24 h. NAUTICAL ASTRONOMY 83 Mean ^me = sidereal time — R. A. mean sun, or = sidereal time — R. A. mean sun -1-24 h. sidereal time = 15 h. 30 m. 12 s. E/.A. mean sun = 6 h. 24 m. 13 s. mean time = 9 h. 5 m. 59 s. sidereal time = 4 h. 20 m. 18 s. R.A. mean sun = 7 h. 50 m. 10 s. mean time = 20 h. 30 m. 8 s. In this example we add 24 h. to 4 h. 20 m. 18 s. before sub- tracting R.A. mean sun. Thus, sidereal time = 4 h. 20 m. 18 s. 24 h. Ex. 1. Let and Then Ex. 2. Let and Then 28 h. 20 m. 18 s. R.A. mean sun = 7 h. 50 m. 10 s. mean time = 20 h. 30 m. 8 s. 52. Civil time and astronomical time. The civil day begins at midnight and ends at mid- nighty after the lapse of 24 hours in two periods of 12 hours each, one period beginning at midnight, and the other at noon. The astronomical day begins at noon^ or 12 hours later than the civil day of the same date, and ends at the next noon, after a lapse of 24 hours. The two periods of the civil day are distinguished from each other by placing, after the figures denot- ing time between midnight and noon, the letters a.m. (Ante Meridian) ; and, after the figures denoting the time between noon and midnight, the letters p.m. (Post Meridian). 84 NAVIGATION AND Thus it will be seen that to convert civil time into astronomical time, the p.m. is dropped if the given civil time is after noori ; but if the time is a.m., 12 hours is added to the given civil time and the date is changed to the preceding day. Ex. 1. Given civil time = 3 h. 10 m. p.m., August 10. Astronomical time = 3 h. 10 m., August 10. Ex. 2. Given civil time = January 8, 10 h. 15 m. a.m. Add 12 h., drop the a.m., and astronomical time = January 7, 22 h. 15 m. Conversely, to convert astronomical time into civil time.. If the given time is under 12 hours, put on p.m. If the given time is over 12 hours, subtract from it 12 hours, add a.m. to the remainder, and add one day to the date. Thus, January 10, 4 h. 15 m. astronomical time = January 10, 4 h. 15. m. p.m. civil time. February 11, 17 h. 16 m. astronomical time = Feb- ruary 12, 5 h. 15 m. a.m. civil time. 53. In every problem of Nautical Astronomy it is necessary to find either the apparent or mean time, at Greenwich, of the instant of taking an observa- tion ; since the calculated positions of the heavenly bodies are made for definite times at the meridian of Greenwich. These positions, with the definite times corresponding, are published in the Nautical Almanac. 54. The hour angle of the sun, at the celestial meridian of any place, is the local time of the place. NAUTICAL ASTRONOMY 85 The hour angle of the sun, at the same instant, at the meridian of Greenwich is the Greenwich time. 55. As the earth makes one complete rotation on its axis in 24 hours, so that the same meridian, on its surface, is opposite the first point of Aries, or oppo- site the same fixed star at the beginning and end of this period of time, and as a complete rotation is measured by 360°, 24 hours in time corresponds to 360°, or we can say: 24 h. = 360° and 360° = 1 h. = 15° 15° = 1 m. = 15' 1° = 1 s. = 15'' Y = r = We can use the first table to convert time into angu- lar measure, and the second table to convert angular measure into time measure. Thus3h. 10m. 30 s. = 3 x 15° = 45 + 10 X 15' = 2° 30' + 30 X 15"= 7'30'' = 47° 37' 30" Again, 48° 15' 38" = 3 h. 13 m. 2^^ s. For 48° = 3 h. 12 m. 15 X 4 s. = 1 m. 38 X ^3^ s. = 2j\ s. 24 h. Ih. 4 m. 4 s. tV s.- = 3 h. 13 m. 2^\ s. 86 NAVIGATION AND 56. In the case of a mean solar day, it was shown in Art. 46, that the meridian of any place moved tlirough an arc of 360° 59' 8/'33 during 24 mean solar hours. If we suppose 24 meridians drawn on the earth's surface, these meridians will be each 15° apart, and, in the rotation of the earth on its axis, will follow each other at an hours interval; so that we can use the tables in the preceding article to convert mean solar time into angular measure, or angular measure into time measure.* The same tables will give the relation of apparent solar time to angular measure. 57. These facts have an important bearing in the determination of longitude by means of time. This will be understood by means of a figure. Let GWE be the plane of the earth's equator, P the projection of the pole on that plane, PG the pro- jection of the meridian of * As each meridian between two transits of the sun passes through an arc of 360° 59' 8 "33, on first thought it might seem that in order to make intervals correspond to hours, the space to equal one hour should be 15° 2' +. The difficulty will be cleared by remembering that though it is true each meridian moves in space 15° 2' + for an hour, before it comes to the position occupied by the meridian immediately preceding it, all the meridians here spoken of are 15° apart on the earthy corresponding to the division of a great circle of 360° by 24. NAUTICAL ASTRONOMY 87 Greenwich, FA and FB the projections of the merid- ians of two places, each 15° from the meridian FG. If the sun is on the line FG produced at 12 noon, as the direction of the arrows shows the direction of the earth's rotation to be from W. to E., FA will be 15° west, and FB 15° east of FG. Consequently, when it is 12 noon at any place on the meridian FG, it will be 11 a.m. at any place on the meridian FA, and 1 p.m. at any place on the meridian FB ; for there is an hour's interval of time required to bring FA to the place oi FG and FG to the place of FB, Now the longitude of any place on FA is 15° W., and the longitude of any place on FB is 15° E. of Greenwich : consequently, 1 h. = 15° dif, of longitude; 1 m. = 15' dif. of longitude; 1 s. = 15" dif. of longitude; or, 15° dif. of longitude = 1 h. dif. in time; 1° dif. of longitude = 4 m. dif. in time ; 1' dif. of longitude = 4 s. dif. in time ; 1" dif. of longitude = ^V s. dif. in time. 88 NAVIGATION AND CHAPTER VI THE NAUTICAL ALMANAC 58. As the calculated positions of the heavenly bodies, recorded in the Nautical Almanac, are given in Greenwich time, the relations established in the preceding chapter between time and angular measure, and between time and difference of longitude, become important in determining the Greenwich date of any observation. The Greenioich date is the apparent or mean time at Greenioich, corresponding to the time at which an observation of a heavenly body is taken at any other place on the earth. Ex. 1. Given ship time June 8, 8 h. 16 m. p.m. (mean time), and longitude 40° 18' W. Required the Greenwich date. ship time June 8 8 h. 16 m. long. 40° 18' W. reduced to time = 2 h. 41 m. 12 s. Ans. Greenwich, June 8 10 h. 57 m. 12 s. The time of an observation is always expressed as astronomical time (Art. 52). Ex. 2. Given ship time Jan. 18, 3 h. 20 m. a.m., and longi- tude 43° 25' E. Required Greenwich date. ship time = Jan. 17 15 h. 20 m. long, in time = 2 h. 53 m. 40 s. Ans. Greenwich, June 17 12 h. 26 m. 20 s. NAUTICAL ASTRONOMY 89 59. From the Nautical Almanac, to take the declina- tion of the sun for any place and date, the longitude of the place being given. Ex. 1. Required sun's declination for Jan. 3, 1893, 8 h. 15 m. A.M., mean time, at a place in longitude 42° 18' W. ship, Jan. 2 20 h. 15 m. long, in time 2 h. 49 m. 12 s. Greenwich, Jan. 2 23 h. 4 m. 12 s. = 23.07 h. = Jan. 3 - 0.93 h. Jan. 3, dif. for 1 h. = 15".3 Jan. 3, sun's .93 dec. at M.N. = 22° 46' 46" S. 459 14.2 1377 22° 47' 00".2 S. 14".229 to be added. In this example, the correction for 0.93 h. we add to 22° 46' 46", because, as the declination is S. and decreasing S., that is, tend- ing N., it must be further S. 0.93 h. before noon than it is at noon. Ex. 2. In longitude 72° 54' W., on June 15, 1897, at 4.30 p.m., mean time, it is required to find the sun's declination. ship, June 15 4 h. 30 m. long. 5 h. 51 m. 36 s. Greenwich, June 15 10 h. 21 m. 36 s. = 10.36 h. sun's declination mean noon, June 15 = 23° 20' 33".7 N". correction = 10.36 x 5".66 = 58".6+ sun's declination at time of observation = 23° 21' 32".3 N. difference for 1 h. 15th = 5".87 difference for 1 h. 16th = 4".84 decrease 24 h. = 1".03 90 NAVIGATION AND decrease 5 h. = -^-^ x 1".03 change for 5 h. = — 0.21 hourly difference for 5 h. after noon = ^".^^ 10.36 3396 1698 566 58".6376 As the difference per hour is changing, where great accuracy is required it is customary to find the change of difference for the hour midioay between noon and the time of observation, and apply this change to the hourly difference, as in this example. For ordinary observations at sea, the hourly difference opposite the noon nearest the time of observation is used. Thus, O's dec. June 15 noon = 23° 20' 33".7 N. correction 5".87x 10.36= Y 0".8 O's dec. at time of obs. = 23° 21' 3r.5 N. From these examples it is seen that, in order to obtain from the Nautical Almanac the sun's declina- tion for any time and place, the longitude of the place being given, we first : Find the Greenwich date; and, second, apply the correction for time elapsed since noon to the declination given opposite the nearest noon. 60. From the Nautical Almanac, to find the equation of time for a given date, the longitude of the place being given. NAUTICAL ASTRONOMY 91 Ex. 1. In longitude 56° 10' W., March 3, 1897, 6 h. 15 m. P.M., mean time, it is required to find the equation of time. ship, March 3 6 h. 15 m. longitude 3 h. 44 m. 40 s. Greenwich, March 3 9 h. 59 m. 40 s. = 9.994 h. dif. 1 h. = 0.541 s. eq. of time = 12 m. 0.75 s. 9.99 correction 5.40 4869 11 m. 55.35 s. = eq. of time. 4869 4869 5.40458 to be subtracted. If it were required in this example to obtain ap- parent time, we subtract the 11 m. 55 s. from mean time. Thus : March 3, 1897, 6 h. 15 m. p.m. mean time equation of time 11m. 55 s. March 3, 1897, 6 h. 3 m. 5 s. p.m. apparent time Ex. 2. Given longitude 75° 18' W., Sept. 13, 1897, 6 h. 30 m. A.M., apparent time. Required equation of time and corre- sponding mean time. ship, Sept. 12 18 h. 30 m. longitude 5 h. 1 m. 12 s. Greenwich, Sept. 12 23 h. 31 m. 12 s. Sept. 12 23.52 h. = Sept. 13 - 0.48 h. eq. of time, Sept. 13, apparent noon = 4 m. 16.73 s. 0.882 X 0.48 = correction -.42 eq. of time to be subtracted = 4 m. 16.31 s. apparent time 6 h. 30 m. a.m. Ans. Sept. 13, 1897 6 h. 25 m. 43.69 s. a.m. 92 NAVIGATION AND In all the foregoing examples the general method of arriving at the required result is : 1. Express the ship time in astronomical time. 2. Find the corresponding Greenwich date. 3. Take the required quantity opposite the nearest Greenioich noon, and apply corrections corresponding to the number of hours by which the given time exceeds or falls short of this nearest noon. 61. Given mean solar time and the longitude; by means of the Nautical Almanac, to find the corre- sponding sidereal time (Art. 50). Thus, Jan. 20, 1895, 3 h. 19 m. p.m., mean time, in longitude 48° 40' W., it is required to find the sidereal time. ship, Jan. 20 3 h. 19 m. longitude 3 h. 14 m. 40 s. Greenwich, Jan. 20 6 h. 33 m. 40 s. Jan. 20, 1895, Greenwich mean noon : R.A. mean sun = 19 h. 58 m. 27 s. Table 9, Bowditch : correction for 6 h. 33 m. = Im. 4.56 s. correction for 40 s. = 0.11 R.A.M.O = 19 h. 59 m. 31.67 s. M.T. 3h. 19 m. sidereal time = 23 h. 18 m. 31 .67 s. NAUTICAL ASTRONOMY 93 62. Given apparent solar time and the longitude; from the Nautical Almanac, to obtain the correspond- ing sidereal time. 1. Convert apparent into mean time. 2. Proceed as in previous article to convert mean time into sidereal time. Ex. July 15, 1895, 6 h. 14 m. a.m., apparent time, iii longi- tude 20° 12' E., required corresponding sidereal time. ship apparent time, July 14 18 h. 14 m. longitude 1 h. 20 m. 48 s. Greenwich apparent time, July 14 16 h. 53 m. 12 s. July 14, 16.887 h. = July 15 - 7.113 h. July 15, noon, equation of time = 5 m. 41.34 s. correction = 0.26 s. x 7.113 = 1.85 s. eq. of time to be added to apparent time = 5 m. 39.49 s. apparent time = 18 h. 14 m. ship mean time = 18 h. 19 m. 39.49 s. longitude = 1 h. 20 m. 48 s. Greenwich mean time, July 14 = 16 h. 58 m. 51.49 s. R.A.M. sun, July 14, noon = 7 h. 28 m. 24.34 s. correction for 16 h. 58 m. = 2 m. 47.23 s. correction for 51.5 s. = 0.14 s. R.A.M. 0= 7h. 31m. 11.71s. ship mean time = 18 h. 19 m. 39.49 s. 25 h. 50 m. 51.2 s. sidereal time = 1 h. 50 m. 51.02 s. 94 NAVIGATION AND CHAPTER VII THE HOUR ANGLE The hour angle of any celestial body is the angle, at the nearer celestial pole, made hy the celestial meridian of the place with the circle of decliriation which passes through the body. Hour angles are measured westward from the me- ridian from Oh. to 24 h. Let the figure represent the plane of the equinoc- tial, P the projection of the celestial pole, and PA and PB the projections of circles of declination, PA being to the W. and PB to the E. of the meridian NPS, If C and D represent the positions of two heavenly bodies, SPA, measured by the arc SA, is the hour angle of (7, and the salient angle SPB, measured by the arc SWNEB, is the hour angle of D. If C and D represent two positions of the sun, then SPA and SPB would be apparent solar time. SPA and SPB would be mean solar time if A and B represented the positions of the mean sun. NAUTICAL ASTRONOMY 95 Also if A and B represented two positions of the first point of Aries, the angles &FA and &PB would be sidereal time (defs. pages 76, 77). 63. Gwen the altitude ^ the declination of a heavenly body, and the latitude of the place of observation ; to find the hour angle of the body. Let the figure represent the plane of the horizon ; iV^S' the projection on it of the meridian; and Z the projec- tion of the zenith of the observer. Let P be the ele- vated or nearer celestial pole ; A the position of a heavenly body ; and let WDE be the equinoctial. Draw the circle of declination FAB, and the circle of altitude ZAC. Then AC = the altitude of A ; AB = the declination of A ; ZD = latitude of the observer. Consequently, in the triangle APZ, in order to find the hour angle DPB, we have given : ZA=^W-AC= 90° - altitude, PA = 90° -AB= 90° - declination, and PZ = W- ZB= 90° - latitude ; that is, to find P, in the triangle APZ, we have the three sides given. 96 NAVIGATION AND Ex. 1. Given, in lat. 41° 24' N., the declination of Venus = 24° 19' N., and the altitude = 24° 14'. Find the hour angle. In the figure ZA = 90° - 24° 14' = ^h"" 46'. PA = 90° - 24° 19' = ^^'^ 41', PZ = 90° - 41° 24' = 48° 36'. Denoting the sides of the triangle by a, p, and z,a = 48° 36', p = 65° 46', 2=65° 41 ' ; we can solve for P by the formula, sin ip=J si^(^-«)sm(g-2r) » ain n sin v Sin a Sin z = Vsin (s — a) sin (s — z) cosec a cosec z a= 48° 36' z= 65° 41' p= 65° 46' s=180° 3' log cosec = 10.12487 log cosec = 10.04035 = 90° 1'30" s_ a = 41° 25' 30" log sin s - 2 = 24° 20' 30" log sin 9.82062 9.61508 2)19.60092 log sin 39° lOf = 9.80046.= log sin ^ P 78° 20f 5 h. 13 m. 21f s. Ex. 2. In lat. 41° 23' N., the altitude of the sun was found to be 26° 38' 44", and its declination to be 19° 20' 26" S. Re- quired the hour angle, supposing the sun to be east of the meridian; that is, that the observation was taken in the morning. a = PZ= 48° 37' z = PA = 109° 20' 26" p = ZA= 63° 21' 16" s = 221° 18' 42" 110° 39' 21" NAUTICAL ASTRONOMY 97 s_a = 62° 2'21" s-z= 1° 18' 65" s-p = 47°18' 5" 'IP- /sin62°2'21"xsmri8'55" sin 2 ^ -\gijj 4^0 37, ^ sijj -L090 20' 26" log sin 62° 2' 21"= 9.94609 log sin ri8'55"= 8.36084 logcosec 48° 37' = 0.12476 log cosec 109° 20' 26" = 0.02523 2 )18.45692 log sin ^ 1 h. 17 m. 56 s. = 9.22846 = log sin ^ acute angle ZPAj but astronomical time = salient angle ZPA. .-. hour angle = 24 h. - 1 h. 17 m. 56 s. = 22 h. 42 m. 4 s., or civil apparent time = 10 h. 42 m. 4 s. a.m. Ex. 3. Suppose in addition to the data of the preceding example, the longitude of the place of observation was given as 72° 56' W., and it was required to find the mean time at the instant of the observation on Nov. 19, 1894, at 10 h. 42 m. 4 s. apparent time. By definition on page 77 apparent solar time is the angle, at the pole, between the meridian and a circle of declination passing through the center of the true sun. Consequently, the answer in the preceding example is apparent time, and we have to apply the equation of time for the given date. ship, Nov. 18, 22 h. 42 m. 4 s. long, in time = 4 h. 51 m. 44 s. Greenwich, Nov. 19 = 3 h. 33 m. 48 s. = 3.56 h. eq. of time, Nov. 19, Green., noon = 14 m. 26.88 s. sub. (dif. 1 h.) 0.574 s. x 3.56 = 2.04 s. equation of time = 14 m. 24.84 s. sub. apparent time = 10 h. 42 m. 4 s. a.m. mean time = 10 h. 27 m. 39.16 s. a.m. NAV. AND NAUT. ASTR. — 7 98 NAVIGATION AND 64. To find the time of suririse or sunset for a given day, at any place on the earth, the latitude and longitude of the place, and the suns tleclination for the day being given. Let the figure represent, as in Art. 63, the projec- tion of the celestial sphere on the plane of the horizon. Suppose A to represent the position of the sun on the eastern horizon when it is first visible to an observer whose zenith is Z ; and suppose A' to represent the position of the sun on the western horizon when it is last visible to the same observer. NPZS being the celestial meridian of the observer, when the sun is on that meridian, the time is apparent noon. The angle ZPA, expressed in time, would give the hours, minutes, and seconds which the sun, in its passage across the heavens, would take to go from its position at A to its position on the meridian. In other words, the angle ZPA gives the hours, minutes, and seconds of apparent time between sunrise and noon. In the same way, the angle ZPA' gives the apparent time between noon and sunset, or in common language, the apparent time of sunset. 24 h. - angle ZPA (expressed in time) would give the astronomical 'Su^^duX- ent time of sunrise. 12 h. - angle ZPA (expressed in time) would give the civil apparent time. NAUTICAL ASTRONOMY 99 In the preceding figure, the declination BA is given as S. declination, while the elevated pole F is supposed to be N. The zenith distance to ^, a point on the horizon, is 90°. But as the time of sunrise is calculated from the instant when the ujpjper rim of the sun is first visible, and as measurements are made to the center of the sun, 16' is added to 90°, as the center of the sun is about that distance below the horizon. More- over, as by refraction the sun, though helow the horizon, is made to appear above it, 34' is added also to 90° for refraction. Consequently, for problems in sunrise and sunset the distances ZA and ZA' are generally taken to be each 90° 50'. Though the declination of the sun is continually changing, so that the declination is not exactly the same at sunrise and sunset, yet the change is so small that it is assumed to be the same both at those times and at noon. For convenience of calculation, therefore, the declination of the sun for noon is used in the solution of problems in sunrise and sunset. Ex. 1. January 28, 1898, in lat. 42° 18' N., long. 72° 55|' W., it is required to find the apparent time of sunrise and sunset. local time at noon = h. Cm. s. long, in time = 4 h. 51 m. 43 s. Greenwich, Jan. 28 = 4 h. 51 m. 43 s. = 4.86 h. declination of sun, Greenwich noon January 28 = 18°6'25".8 S. cor. = 39 ".85 x 4.86 = 3' 13^7 K declination of sun at local apparent noon = 18° 3' 12".l S. 100 NAVIGATION AND hourly difference of declination of sun = 39".85 N. 4.86 23910 31880 15940 193".6710 = 3' 13".7 In preceding figure, PZ = a = 90°-41°18' = 48° 42' PA = z = 90° -j- 18° 3' 12" = 108° 3' 12" Z^=i> = 90° + 50' = 90° 50' 247° 35' 12" '= 2 = 123° 47' 36" 5-a= 75° 5' 36" s- z= 15° 44' 24" s-p= 32° 57' 36" sin iP= Vsin (s — a) sin {s — z) cosec a cosec z. log sin 75° 5' 36"= 9.98513 log sin 15° 44' 24" = 9.43341 log cosec 48° 42' = 10.12421 log cosec 108° 3' 12" = 10.02191 2 )19.56466 log sin 37° 17"^ = 9.78233 = log sin | P P = 74° 34 '3^ = 4 h. 58 m. 17^ s. = apparent time of sunset. 12 h. — 4 h. 58 m. 17| s. = 7 h. 1 m. 42J s. = apparent time of sunrise. Ex. 2. In preceding example, required the mean times of sunrise and sunset ; also eastern standard time of sunrise and sunset. January 28, equation of time Greenwich noon = 13 m. 13.97 s. difference for 1 h. = 0.457 s. x 4.86 = 2.22 + local equation of time at noon = 13 m. 16.19 s. NAUTICAL ASTRONOMY 101 0.457 4.86 2742 3656 1828 2.22102 local mean time of apparent noon = 12 h. 13 m. 16.19 s. subtract hour angle = 4 h. 58 m. 17.5 s. local mean time of sunrise = 7 h. 14 m. 58.69 s. a.m. local mean time of sunset = 5 h. 11 m. 33.69 s. p.m. eastern standard time = time of meridian of 75° W. local meridian = 7 2°55'|W. difference = 2° 4'i = 8 m. 17 s. taking 8 m. 17 s. from the mean times calculated above eastern standard time of sunrise = 7 h. 6 m. 41.69 s. a.m. eastern standard time of sunset = 5 h. 3 m. 16.69 s. p.m. In this example we have used the noon equation of time to be applied to time of sunrise and sunset. A more exact calcu- lation would apply the equation of time as derived for the instant of apparent time of sunrise or of sunset. For sunrise. Greenwich, 27th 19 h. 1 m. 42| s. longitude in time 4 h. 51 m. 43 s. Greenwich, Jan. 27 23 h. 53 m. 251 g. or Jan. 28 - Oh. 6 m. 34.5 s. = - .011 eq. of time, Greenwich, noon 13 m. 13.97 s. correction 0.457 s x .011 h. = 0.01 equation of time for sunrise = 13 m. 13.96 s.+ apparent time of sunrise = 7 h. 1 m. 42.5 s. exact mean time = 7 h. 14 m. 56.46 s. a.m. 102 NAVIGATION AND For sunset. Jan. 28 4 h. 58 ra. 17.5 s. longitude 4 h. 51 m. 43 s. Greenwich, Jan. 28 9 h. 50 m. 0.5 s. = 9.83 h. 0.457 6881 4915 3932 correction = 4.49231 s. eq. of time, Greenwich, noon = correction 13 m. 13.97 4.49 s. equation of time = apparent time of sunset = 4h. 13 m. , 58 m. . 18.46 17.5 s s. exact mean time = 5 h. 11 m. 35.96 s. p.m. Since the time of sunrise and the time of sunset are generally calculated to the nearest minute only, the first method of apply- ing the local noon equation of time is generally used. By com- paring the results by the two methods it will be seen that the difference in the answers does not much exceed two seconds. Ex. 3. June 1, 1898, in latitude 41° 18' N., longitude 72° 55'} W., required the eastern standard times of sunrise and sunset. local noon Oh. m. s. longitude 4 h. 51 m. 43 s. . Greenwich, June 1 4 h. 51 m. 43 s. = 4.86 h. Declination of sun. Greenwich, noon =22° 6' 0".7 N. correction 20".12 x 4.86= 137.8+ declination of sun=22° V 38".5 N. polar distance=67° 52' 22'^ NAUTICAL ASTRONOMY 103 equation of time, Greenwich noon= 2ni. 24.55 8.— correction 0.375 s. x 4.86 h. = 1.82- equation of time, local noon= 2 m. 22.73 s.— apparent noon = 12 h. mean time of apparent noon =11 h. 57 m. 37.17 s. a.m. deduct for eastern standard time 8 m. 17 s. eastern standard time of apparent noon =11 h. 49 m. 20.17 s. a.m. Projecting the celestial concave on the celestial meridian. PZ=a= 48° 42' log cosec = 10.12421 P4 = 2!= 67° 52' 22" log cosec = 10.03322 ZA=i>= 90° 50' . = ?5Z121^=103°42'1^ 2 «_a= m"" O'll" log sin = 9.91338 8-z= 35° 49' 49" log sin = 9.76744 2 )19.83825 log sin 56° 6' 33"= 9.91912^ 08_ P=112°13' 6" 4J = 7h. 28 m. 52.4 s. eastern standard time of apparent noon =11 h. 49 m. 20.2 s. eastern standard time of sunrise = 4 h. 20 m. 27.8 s. a.m. eastern standard time of sunset= 7h. 18 m. 12.6 s. p.m. Ex. 4. Jan. 10, 1898, in latitude 39° 57' N., longitude 75° 9' W., required mean time of sunrise and of sunset. An8. 7 h. 21 m. 38 s. a.m. ; 4 h. 54 m. 16 s. p.m. Ex. 5. May 16, 1898, in latitude 42° 36' N., longitude 70° 40' W., required eastern standard time of sunrise and of sunset. Ans. 4 h. 18 m. 44 s. a.m. ; 6 h. 58 m. 16 s. p.m. 104 NAVIGATION AND 65. Given a stars hour angle, to find mean time. Let the figure represent the plane of the equi- noctial ; P the projection of the pole on the plane ; C the position of the star ; A the position of the first point of Aries ; and M the position of the mean sun. If NFS be the projection Im of the celestial meridian, and FCB be the projection of the circle of declination passing through C, SFC will be the hour angle of the star, and SB will measure that angle. Now SM= SB + AB - AM; that is, mean time = star's hour angle + R.A. of star - R. A. of mean sun. In the case just given the star is W. of the meridian. Suppose the star is at C% and east of the meridian ; that Ai is first point of Aries, and Mi is position of mean sun; then SMi = SB^i- A^M^- A^B^ or (24 h. — mean time) = (24 h. - star's hour angle) 4- R.A. mean sun — star's R.A. .*. mean time = star's hour angle + R.A. of star -R.A. mean sun. 66. To find the mean time at any place, having given the hour angle of a star ; the longitude of the place; the date; and the approximate local mean time. By the previous article we have to add to the hour angle the stars B.A., and from the sum subtract the NAUTICAL ASTRONOMY 105 R.A. of the mean sun for the given date and approxi- mate time. Ex. 1. Nov. 22, 1891, 7 h. 15 in. p.m., approximate mean time in long. 87° 56' W., the hour angle of Aldebaran (a Tauri), was 18 h. 55 m. 15 s. (E. of meridian). Star's R.A.= 4 h. 29 m. 41.5 s. Required mean time at the place. ship, Nov. 22 = 7 h. 15 m. longitude = 5 h. 51 m. 44 s. Greenwich, Nov. 22 = 13 h. 6 m. 44 s. Green., Nov. 22, noon, R.A. mean sun = 16 h. 4 m. 44.5 s. correction for 13 h. 6 m. = 2 m. 9.1 s. correction for 44 s, = .1 s. R.A. mean sun at time of observation = 16 h. 6 m. 53.7 s. star's H.A. = 18 h. 55 m. 15 s. star's R.A. = 4 h. 29 m. 41.5 s. 23 h. 24 m. 56.5 s. R.A. mean sun = 16 h. 6 m. 53.7 s. Ans. 7 h. 18 m. 2.8 s. p.m. Ex. 2. June 23, 1891, at 4 h. 12 m. a.m. mean time, nearly, in long. 50° 15' W., the hour angle of a Lyrse was 3 h. 41 m. W. of meridian. Required mean time. Star's R.A = 18 h. 33 m. 15.8 s. ship, June 22 = 16 h. 12 m. longitude = 3 h. 21 m. Greenwich, June 22 = 19 h. 33 m. Green., June 22, noon, sid. time = 6 h. 1 m. 31.55 s. correction for 19 h. 33 m. = 3 m. 12.69 s. R.A. mean sun = 6 h. 4 m. 44.24 s. star's H.A. = 3 h. 41 m. star's R.A. = 18 h. 33 m. 15.8 s. 22 h. 14 m. 15.8 s. 6 h. 4 m. 44.2 s. June 22 16 h. 9 m. 31.6 s. ast. time June 23 4 h. 9 m. 31.6 s. a.m. m. t. 106 NAVIGATION AND 67. Given mean, or apparent time at place of given longitude; to find what star of \st or 2d magnitude will imss the meridian next after that time. The solution of this problem is simply to find the sidereal time corresponding to the given time, and then, from list of fixed stars in Nautical Almanac, to choose the star of required magnitude whose right ascension is the next greater than the sidereal time found. Ex. In long. 72° 56' W., Dec. 7, 1897, at 11 h. 30 m. p.m. mean time, what star of 1st or 2d magnitude passed the merid- ian shortly after that time ? ship, Dec. 7, 1897 = 11 h. 30 m. longitude = 4 h. 51 m. 44 s. Greenwich, Dec. 7 = 16 h. 21 m. 44 s. Dec. 7, mean noon R.A.M. O = 17 h. 6 m. 3.95 s. correction for 16 h. 21 m. = 2 m. 41.15 s. correction for 44 s. = .12 s. R.A.M. sun = 17 h. 8 m. 45.22 s. ship, Dec. 7 = 11 h. 30 m. = 28 h. 38 m. 45.2 s. = 24h. sidereal time or R.A. of meridian = 4 h. 38 m. 5.8 s. In catalogue of fixed stars (Capella), a Aurigae has R.A. 5 h. 9 m. 4.8 s., and is, therefore, star required. 68. 7b find at what mean time any star will pass a given meridian. Let the figure represent the plane of the equinoc- tial ; P the pole ; NFS the celestial meridian ; A the NAUTICAL ASTRONOMY 107 first point of Aries ; m the position of the mean sun ; and B the position of the star at instant of crossing the meridian. jsr Then mS = mean imiQ = AS— Am^ or mean time = sidereal time of star — R.A. of mean sun = R.A. of star -R.A. of mean sun. Ex. To find at what time Sirius passed the meridian in longitude 72° b& W., Dec. 8, 1897. R.A. of Sirius = 6 h. 40 m. 39 s. add 24 h. 30 h. 40 m. 39 s. R.A. of sun (noon) = 18 h. 10 m. 0.5 s. ship approximate mean time = 13 h. 30 m. 38.5 s. longitude = 4 h. 51 m. 44 s. Greenwich, Dec. 8 = 17 h. 22 m. 22.5 s. R.A. M.S. noon = 17 h. 10 m. 0.5 s. correction for 18 h. 22 m. = 3 m. 1.03 s. correction for 22.5 s. = .06 s. R.A. M. sun = 17 h. 13 m. 1.59 s. subtract from R.A. Sirius = 30 h. 40 m 39 s. 13 h. 27 m. 37 s. ast. time 12 h. Ans. Dec. 8. 1 h. 27 m. 37 s. 3 a.m. 69. To find the meridian altitude of a heavenly body for a given place, and whether it will pass N. or S. of the zenith, the declination of the body and the latitude of the place being given. 108 NAVIGATION AND Ex. 1. At a place in latitude 42° N., it is required to find the meridian altitude of a star whose declination is 25° N. ; also whether it passes N. or S. of the zenith. Let NZS represent the plane of the celestial meridian; P the upper or N. pole; Z the zenith; N and S the north and south points of the horizon; and E the point where the equinoctial inter- sects the meridian. Let ^^=25°, then A is the position of the star at transit. Let ZE = latitude 42° N. ZA = ZE-AE = 42° - 25° = 17°. .-. star's transit is south of zenith. Again, altitude of star = AS = ZS-ZA = 90° - 17° = 73°. Ex. 2. Dec. 9, 1897, at what time did a Orionis pass the meridian of longitude 72° 56' W. in latitude 42° 18' N. ; and did it pass N. or S. of zenith ? Required its altitude also. given the declination of star = 7° 23' 16" N. R. A. of star =5 h. 49 m. 40 s. ; R. A. M.S. = 17 h. 13 m. 57 s. R.A. of star + 24 h. = 29 h. 49 m. 40 s. R.A. of sun (Greenwich noon) = 17 h. 13 m. 57 s . mean time (approximately) = 12 h. 35 m. 43 s. longitude = 4 h. 51 m. 44 s. Greenwich mean time (approximately) = 17 h. 27 m. 27 s. R.A. M. sun (Greenwich noon) = 17 h. 13 m. 57 s. correction for 17 h. 27 m. = 2 m. 51.9 s. correction for 27 s. = As. R.A. M. sun = 17 h. 16 m. 49 s. R.A. of star = 29 h. 49 m. 40 s. star on meridian = 12 h. 32 m. 51 s. = 32 m. 51 s. after midnight Dec. 10 NAUTICAL ASTRONOMY 109 latitude = 42° 18' N. declination of star = 7° 23' 16" N. 34° 54' 44" S. of zenith 90^ 55° 5' 16" = altitude Ex. 3. At what time, Dec. 10, 1897, in latitude 42° 18' N., longitude 72° 56' W., did rj Ursae Majoris pass the meridian? Was the transit N. or S. of the zenith ? R.A. of star = 13 h. 43 m. 31 s. declination of star = 49° 49' 2" N. Let NPZES be the meridian; P the pole ; Z the zenith ; A be the position of star at transit. ^^ = 49° 49' 2" ZE = 42° 18' ZA= 7° 31' 2" star N. of zenith ZiV^=90^ altitude = ^iV= 82° 28' 58" To find at what time the star passed the meridian Dec. 10, we must begin one day hack, and take out the R.A. of M. O for Dec. 9. thus, K.A. of star + 24 h. = 37 h. 43 m. 31 s. R.A. of M. sun, Dec. 9, noon = 17 h. 13 m. 57 s. approximate mean time = 20 h. 29 m. 34 s. longitude = 4 h. 51 m. 44 s. Dec. 10, Greenwich mean time = 1 h. 21 m. 18 s. " " R.A. M. O noon = 17 h. 17 m. 54 s. correction for 1 h. 21 m. = 13.3 s. correction for 18 s. = .05 s . R.A. M. O = 17 h. 18 m. 7.4 s. R.A. star = 37 h. 43 m. 31 s. Dec. 9 20 h. 25 m. 23.6 s. ast. time Dec. 10 8h. 25 m. 23.6 s. a.m. no NAVIGATION AND CHAPTER VIII CORRECTIONS OF ALTITUDE 70. In order to obtain the true altitude of a heav- enly body, a number of corrections must be applied to the observed altitude, namely : Index correction, due to some error in the instru- ment used ; and corrections for dip, refraction, semi- diameter, and parallax, corrections required by the fact that, to combine observations made at any place on the earth's surface with the elements from the Nautical Almanac, those observations must all be reduced to a common point of observation. This common point of observation is considered to be the center of the earth. The sectant is an instrument for measuring angles in any plane. At sea it is used chiefly to measure the altitudes of heavenly bodies. The accompanying figure will serve to explain the prin- ciples of the construction of the sextant. AB is a circular arc a little longer than a sixth of the whole circumference. ^iVand CM are two glasses whose NAUTICAL ASTRONOMY HI planes are perpendicular to the plane of the arc AB, EN is fixed in position, and its glass is silvered on the half next to the frame of the instrument. EN is called the horizon glass, because through it the horizon is viewed in taking observations. CM is called the index glass. It is entirely silvered (on one face). By means of the index bar, CB, it is mov- able about the point (7, which is the center of the arc AB. When the index bar is at the zero point on the arc AB, the planes of the two glasses, EN and CM, are parallel. If it is required to find the altitude of any body, S, above the horizon, the observer looks at the horizon line through the plain part of the glass EN, and moves the instrument and the index bar till an image of S reflected from CM upon EN appears to coincide with a point upon the horizon. Let K be the point of the horizon with which S appears to coincide. Let CM' be the position of the index glass and CD be the position of the index bar when K and S appear in coincidence. Join SC, CN, and KN. Produce SC and KN to meet at J. JK wiU represent the plane of the horizon, and the angle SJK will be the altitude of S. Produce EN to meet CD (in this case) at D, The arc DB measures the angle DCB. But DCB = NDC, since ^iVand (7if are parallel. When a ray of light is reflected from a plane sur- face, the angle of incidence is equal to the angle of reflection : 112 NAVIGATION AND therefore SCff=NCD, but SCH=M'CJ, these being vertical angles ; therefore, NCJ=2iNCD). Also, since angle of incidence is equal to angle of reflection., ENC=DNJ, but DNJ^ENK', therefore (1), KNC^ 2 {ENC) = 2 {{NCD) + D], because ENC is exterior angle of triangle NCD. Also (2), KNC=NCJ+J=2[NCD) + J', consequently, 2 {NCD) + 2 Z) = 2 [NCD) -f J; that is, - D = ^J; but as D= DCB, and DB measures DCB, DB meas- ures half of t7, or half the altitude of S. The whole arc AB, however, is so graduated that each half degree counts as a degree, and the reading of the arc DB gives the measure of the whole angle J. Index error. The planes of the index glass and horizon glass should be parallel when the index bar is at the zero point on the graduated arc AB. The distance, either on the arc (that is, to the left of the zero point), or off the arc (that is, to the right of the zero point), to which the index bar must be moved to make these planes parallel, is called the index error. This error demands a correction for every angle measured. NAUTICAL ASTRONOMY 113 To determine the index error for any instrument, the simplest method is to measure at successive instants the angle subtended by the sun near the zero point. As the diameter of the sun is the same, these measurements should agree if there is no error, but if they do not agree, there is an error in the instrument. This will be understood by means of the figure. Let A OB be a part of the arc of the sextant having the zero point at 0. Suppose that in measuring the diameter of the sun on the arc the index bar is moved to A, and that in meas- uring the same diameter off the arc the index is moved to D. Then, denothig the measure of the diameter by d, AD =2 d ; consequently B, the middle point of AD, should be the real zero point of the graduated arc. OB would represent the error, which is off the arc, in this case, and the correction for the error, called index correction, must be added. Denote OB by e ; the reading OA by r ; the read- ing OD by / ; then AB=BD, or AO+OB=OD-OB', that is, r+e = / — e; therefore, €= — - — NAV. AND NAUT. ASTR. 8 114 NAVIGATION AND If the reading AG, on the arc, is greater than the reading OD, off the arc, since AB = BD, r — r' In this case the index correction must be suhtracted. 71. The dip of the horizon is the angle of depres- sion of the visible horizon below the horizontal plane of the observer. This depression of the visible horizon is due to the elevation of the eye of the observer above the level of the sea. Let the figure represent a section of. tlie earth by a plane passed through A, the point of observation, and C, the center of the earth. The small circle, of which BD is the diameter, would represent the plane of the observer's visible horizon. If AE be the line in which the plane ABC intersects the horizontal plane through A^ then EAB would be the dij), or angle of depression of the visible hori- zon, BD, below the horizontal plane of the observer NAUTICAL ASTRONOMY 115 at A, If aS be a celestial body, the angle SAE would be its true altitude, SAB its measured or observed altitude. Dip must always be subtracted from the observed altitude to obtain the true altitude, for SAB-EAB = SAE. AB is tangent at B. Join C and B by straight line, CB. EA is parallel to tangent at G, and there- fore is perpendicular to CA. Angles EAB and ACB are complements of BAO and therefore equal ; that is, ACB = dip. Let AG = hsiudCG=^ E. Then AB = VAC' - CB' = ^{R + hf - R' = V2 Rh + h\ .', tan dip = t'dn ACB = = EC R =v 2 Rh + h' R But since h is small compared with i?, h^ may be neglected, and tan dip = \—fr nearly. But as the dip is usually a very small angle, and since for a very small angle the circular measure of the angle is approximately equal to the tangent of the angle, we can say circular measure of dip = \^—. R 116 NAVIGATION AND Now circular measure of dip = — — ^ 180 wliere n = number of degrees in angle, n being integral or fractional ; therefore reducing to minutes. 60 7277 J2h .4 180 X 60 ^ R' or, since 60 n = dip in minutes, dip in minutes = 10800 J 2h TT '3960 X 5280' reducing li to feet, R being 3960 miles. n- . . , 10800 V2 /y- Dip m minutes = — v/i. 7rV3960 X 5280 log 10800 = 4.03342 log V2 = 0.15051 colog 77 = 9.50285 - 10 colog V3960 = 8.20115 - 10 colog V5280 = 8.13868 - 10 log 1.063 = 0.02661 ' /. dip in minutes = 1.063 V^. This value of dip is diminished by refraction. The amount by which it is diminished is variously estimated. If we take that amount as ^, we shall ob- tain the true value of dip ; allowing for refraction, dip = 1.063 VA - ^(1.063 VA) = .984 Va, approximately. NAUTICAL ASTRONOxMY 117 72. Refraction, To understand the effect of refrac- tion, we represent, by the Bgure, a great circle section of the earth AMN, made by a plane passing through A, the point of observa- , tion, and through the at- mosphere surrounding the earth. A ray of light from a distant object, as a star, *S', entering the atmosphere obliquely at d, and passing through strata of varying density, is bent out of its course into a curve, defcjA, concave to the earth's surface. The object itself appears at A on AS\ which is curve defgA at A. If we join the center C with A and produce the line to z, z will represent the zenith of the observer. Produce the line Sd (supposed to be a straight line before it enters the atmosphere at d) to meet CZ at G. If we draw AD parallel to GS, DAB would represent the true altitude of S\ DA and GS, repre- senting rays of light from an object so remote as one of the celestial bodies, may be regarded as paral- lel straight lines. If there were no refraction, the light would come on the line AD. S'AD is the angle of refraction. The correction for refraction, therefore, is to be sub- a line tangent to the 118 NAVIGATION AND tracted from the observed altitude to give the true altitude, for S'AB - S'AD = DAB. Rays of light from an object in the zenith, falling on the strata of the atmosphere, are not refracted. The more obliquely the light enters the atmos- phere, the greater the refraction. Consequently, refraction increases, the nearer the body is to the horizon. 73. Correction for semidiameter. The positions of heavenly bodies indicated in the Nautical Almanac are given for their centers. Observations of heavenly bodies of perceptible size are generally made to the upper or lower edge of the body, called respectively the upper or lower limb. If an observed altitude is one of the lower Ihnh, the semi-diameter expressed in minutes or seconds of the body must be added to give the altitude of the center. If an observation is taken of the upp^er limb, the semidiameter must be subtracted to give the true altitude. 74. Parallax. Altitudes of celestial objects are observed at the surface of the earth, or slightly above it. They are taken with reference to the sensible horizon, that is, with a plane tangent to the earth's surface vertically beloiv the point of observation. But to these observed altitudes we have to apply correc- tions in order to obtain the altitudes of the same bodies if the observations were made at the center of the earth, and with reference to the rational horizon, NAUTICAL ASTRONOMY 119 that is, a plane passed through the center of the earth parallel to the sensible horizon. Let the figure represent a section of the earth made by a plane passed through its center C, and through the point of observation at A. Produce line CA to zenith Z. Let S be position of heavenly body. Its altitude with reference to the sensible horizon, repre- sented by line AB drawn perpendicular to AC, is the angle SAB. Its alti- tude with reference to the ratio7ial horizon, represented by line CD, drawn parallel to AB, is the angle SCD. Let E be the point where AB and SC intersect. Since AB and CD are parallel, (1) SCD = SjEB=SAB + ASC. The angle ASC is called the parallax in altitude of S, or simply 'parallax of S. To obtain the true alti- tude of a heavenly body (in addition to the other cor- rections to be applied to the observed altitude), from equation (1) it is evident that parallax must be added to the observed altitude. Let R denote AC, the radius of the earth; let d denote CS, the distance of the heavenly body from 120 NAVIGATION AND the center of the earth. Denote observed altitude SAB by L sin ASC_B^ ., . sin parallax _ E^ sin SAC~d' ^^ ^^' sin (90° + /?)"^' 7? 7? or (2) sin (parallax) = — sin (90° -\-h)= — cos h. (X 66 Suppose the celestial body to be in the horizon at B ; then 7-> sin parallax = sin ABC = — • a In this case the parallax is called the horizontal parallax ; that is, (3) sin horizontal parallax = — • Hi Substituting in (2) this equivalent of — , we have d (4) sin parallax = sin (horizontal parallax) cos h. Since parallax and horizontal parallax are always small angles (except in the case of the moon), we may substitute for the sines the measures of these angles, at any altitude, and (4) becomes parallax = horizontal parallax x cos h. Both from the equation and from the figure it is evident that parallax is greatest when the heavenly body is in the horizon ; decreases as the altitude of the body increases ; and vanishes at the zenith. NAUTICAL ASTRONOMY 121 Also, from the figure, if aS be at a very great dis- tance from the earth, d may be so large that the ratio — approaches ; in that case, sin parallax in (2) will vanish. For the^a:;^^? stars, which are supposed to be at such immense distances from the earth that rays of light from them fall on any two points of the earth in nearly parallel lines, no correction for paral- lax is applied. Again, the nearer S is to the earth, the greater the 7? value of — , and consequently the greater the parallax. Of the heavenly bodies, the moon is the nearer to the earth and has the greatest parallax. 122 NAVIGATION AND CHAPTER IX LATITUDE 75. Latitude. Let lopAe represent a great circle section of the earth through the meridian of the observer at A ; and let NFS be the celestial meridian of the same ob- server. wCe will then be the projection of the ter- restrial equator, and WCE will be the projection of the celestial equator, or equinoctial on the same plane, viz. the plane of the terrestrial and celestial meridians. Let p be the pole of the earth, and P the corresponding elevated pole of the celestial con- cave. Join CA, and produce the line to meet the celestial concave at Z, the zenith of the observer. Through C at right angles to CA draw NCS, which will represent the projection of the rational horizon of the observer. If at ^ a line be drawn tangent to the circle pAe, cutting the celestial meridian at H and 0, this line would represent the sensible horizon of the observer (Art. 74). In consequence of the immense distances of the NAUTICAL ASTRONOMY 123 heavenly bodies on the celestial concave, smd S and // and N are supposed to coincide, and altitudes of objects are observed with reference to HAO. Where accuracy is required, such observations have to be corrected so as to equal the true altitude with refer- ence to NCS (Art. 74). Ae measures the latitude of A, viz. the angle ACe. This angle is also measured by ZE. NZ = OC^ = PE. If from these equals we take away the common part PZ, we have PN= ZE ; or, the elevation of the nearer celestial pole above the horizon of the observer is equal to his latitude. 76. To find the latitude. Latitude is found hy observing the altitude of any heavenly body while on the meridian, the declination of the body being given. The altitude of the body may be observed either at its upper transit or at its loiver transit, in case it moves in a small circle on the celestial concave, and always above the horizon. Let WPZS represent the celestial con- cave projected on the merid- ian of the observer; P will be the nearer (in this case N.) pole; Z the zenith; WEC the projection of the equinoctial ; NES the projection of the horizon. Z (7 or PiV" will measure the latitude (Art. 75). 124 NAVIGATION AND Suppose A to be the position of the heavenly body on the meridian at its upper transit. If the angle AES is observed, the arc AS, which measures this angle, is known. CA is the declina tion, and in this figure is a S. declination. (a) lat. = ZC= ZS - (AS+ AC) = 90° - (alt. + dec). If the object observed is at B, and having a N. declination, BS is the measure of its altitude, and (h) lat. = ZC= 90° - {BS- BC) = 90° - (alt. - dec). The observer is supposed to be in the N. hemisphere, and the latitude required is a N. latitude. In this case, therefore, it is easily seen that when the altitude of a body is taken at its upper transit, if the latitude required is N., and the declination is S., (a) lat. = the complement of the sum of the altitude and declination ; but if the latitude required and declination are both N., (h) lat. = complement of the altitude diminished by the declination. If the observer were in the S. hemisphere, since CA would then be a N. declination and CB a S. declination, (c) lat. = 90° - (alt. + dec), if lat. is S. and dec N. (d) lat. = 90° -(alt. -dec), if lat. is S. and dec S. We can bring these four cases under one rule, viz. : NAUTICAL ASTRONOMY 125 If latitude and declination are of the same naine (either N. or S.), (e) the lat. = 90° - (alt. - dec.) ; but, if of different names, (/) lat. = 90° -(alt. 4- dec). Since the zenith distance of a heavenly body is the complement of its altitude, {g) {e) becomes lat. = (90° - alt. + dec.) = zenith dist. -h dec. (A) (/) becomes lat. = zenith dist. — dec. 2. Considering now the case of the lower transit of a celestial body, Let the figure represent, as before, the celestial merid- ian. Let A be the position of a heavenly body at its lower transit, and NA the measure of its altitude, and WA the measure of its decli- nation. Then lat. = ZC ^ NP = NA + PA = alt. + (90 - dec.) or lat. = alt. 4- polar dist. Ex. 1. June 10, 1895, in long. 87° 10' W., the observed meridian altitude of the sun's lower limb was 69° 24' (zenith N.); the index correction was + 2' 20"; height of the eye above the sea was 20 ft. Kequired the latitude. 126 NAVIGATION AND Local apparent time June 10 h. m. obs. alt. = 69° 24' longitude in time 5 h. 48 m. 40 s. in. cor. 2' 20" + Gr. app. time 5 h. 48 m. 40 s. = 5.81 h. sun's dec. at app. noon 23° 1' 27" N. cor.= ll".5x 5.81= 1' 6".8 + dec. at time of obs. = 23° 2' 33".8 N. 5.81 11.5 2905 6391 66.815 or 1' 6".8 In figure, p. 123, BS = 69° 37' 25" ^C = 23° 2' 34" ^C= 46° 34' 51" dip 22"- 3" + . diaii alt.= dist. 69° 26' 20" 4'23"- ref.; par. 69° 21' 57" 19"- sem 69° 1. 21'38" 15' 47"+ true = 69= 90° *37'25" zen. dec. 20° 23° 22' 35" 2' 34" latitude 43° 25' 9"N. 90°-^O= latitude =ZC = 43° 25' 9"N. Ex. 2. In long. 85° 14' W., Feb. 10, 1897, the observed meridian altitude of the sun's upper limb was 36° 42' (zenith N.); index correction was — 1'40"; height of eye above sea was 16 ft. Required latitude. local time Oh. m. obs. alt. 36° 42' longitude in time 5 h. 40 m. 56 s. in. cor. 1'40"- Gr. app. time 5 h. 40 m. 56 s. = 5.68 h. sun's dec. at app. noon, 14° 9' 32 ".6 S. cor. = 49".ll X 5.68 = 4' 38".9- dec. at time of obs. = 14°4'53".7 S. 36° 40' 20" dip 3'55"- 36° 36' 25" ref.l par. 'T;l '•''"- 36° 35' 14" sem. diam. 16' 14"- true alt. 36° 19' NAUTICAL ASTRONOMY 127 In figure, p. 123, 49.11 SA = 36° 19' 5.68 CA = 14° 4' 54" ' 39288 ^0=50° 23' 54" 90** 29466 true alt. 36° 19' 24555 90°- ■SC: = ZC = 39° 36' 6" zen. dist. 53° 41 ' 278.9448 = =4'38".9. latitude = 39° 36' N. dec. 14° 4' 54" 39° 36' 6" Ex. 3. March 22, 1898, the observed meridian altitude of Arcturus was 66° 42' (zenith N.); index correction was 2' 20" + ; height of eye 16 ft. Declination of star was 19° 42' 44" N. Required latitude. obs altitude = 66° 42' index cor. = 2' 20"+ dip ref. 25"- *true alt. zen. dist. declination 66° 44' 20" 3' 55"- 66° 40' 25" 25"- = 66° 40' 90° = 23° 20' 19° '42' '44" 43° In figure, p. 123, SB = 66° 40' CB=19°42'44" C5 = 46°57'16" >SO=43° 2' 44" 2' 44" latitude =90° latitude = 43° 3' N. 77. To find the latitude by an observation of a heavenly body near the meridian, the declination and the time of the observation being known. Let NWSE represent the projection of the celestial concave on the plane of the horizon ; Z will be the * For fixed star, parallax is 0. 128 NAVIGATION AND zenith ; P will be the pole ; and WDE will be the equinoctial. Suppose A to be the posi- tion of the object observed. Draw the circle of altitude ZAC, and the circle of declination PAD. From A draw the arc, AF, per- pendicular to PH. NPH will represent the meridian. Denote the altitude of A, AC, by a, and the declination, AD, by d. In the figure, A is represented with N. declination. In this case, PA is 90° — c?. But if the object had a S. declination, A would be below Z), and PA would be 90°4-6Z. Z^ = 90°-a. ZPA represents the time elapsed since noon. Denote this by t. If the object observed were at A\ the time would be before noon, and the angle ZPA would be 12 — t, if the time given were civil time, or 24i — t, if the given time were astronomical. Let PF= X, and ZF=ij; then PZ = x-y. Lat. = ZH= PH- PZ = 90° - (x - y). In right-angle triangle PAF, by Napier's rule, cos ZPA (1) or. tan PF = tan X cot PA cos t cot (90° - d) = cos t cot d. NAUTICAL ASTRONOMY 129 /ox A T.1 COS FA cos (90° — d) sin d (2) cos AJ^ = =r-p, = ^ — ■ ^ = , cos Fi^ cos ic cos X ryjp COS ZA cos (90° — a) cos cc cos ZF= -— = 5^ — : — -^ ; cos Ar sin a that is, (3) cos y = sm a cos x cosec d. By means of (1) we obtain the value of x, and by means of (3) we obtain y. Then latitude = 90° - (a; -?/) . As this method of obtaining latitude depends upon the time (before or after noon), an error in time introduces an error into the result, which is almost unavoidable, so that the method is not very reliable, when the object observed is far from the celestial meridian.* Ex. 1. July 15, 1896, in long. 73° 45' W. at 12 h. 45 m. p.m., mean time, the observed altitude of the sun's lower limb was 58° 42' (zenith N. of sun); index correction was +2' 20"; height of eye was 15 ft. Required the latitude. ship time, July 15 = h. 45 m. longitude = 4 h. 5^ m. Greenwicli, July 15, Mt = 5 h. 40 m. = 5.67 h. equation of time = 5 m. 46.16 s. correction (.245) x 5.67 = 1.39 5.67 5 m. 47.55 s. = equation of time 1715 45 m. 1470 39 m. 12.45 s. = time = apparent time 1225 39 m. = 9° 45' 1.38915 12.45 s. = 3' 7" apparent time = 38m. 12.45 s. = 9° 48' 7" ♦Bowditch. NAV. AND NAUT. ASTB.--9 130 NAVIGATION AND observed altitude = 58* 42' index correction = 2' 20"+ 58° ' 44' 20" ref. par. dip 35"- ) 4"+) 58° 58° 3'48"- 40' 32" 31"- 40' 01" sem L. diam. = = 15' 47"+ true altitude : = 58^ 55' 48" 90° zenith distance = 31° 4' 12" declination of sun at noon, Gr. mean time = 21° 25' 17".l N. correction = 24 ".33 x 5.67 = 2' 18"- declination of sun at time of observation = 21° 22' 59" N. 90° polar distance = 68° 37' 1" In the preceding figure, Z^PZ= 9° 48' 7"; PA = 6S° 37' 1"; Z^ = 31°4'12". tan X = cos 9° 48' 7" cot 21° 22' 59" log cos 9° 48' 7" = 9.99362 log cot 21° 22' 59" = 10.407 21 log tan 68° 19' 47" = 10.40083 x = 68° 19' 47" . cos 2/ = sin 58° 55' 48" cos 68° 19' 47" cosec 21° 22" 59' log sin 58° 55' 48"= 9.93275 log cos 68° 19' 47"= 9.56734 log cosec 21° 22' 59" = 10.43818 log cos 29° 49' 51"= 9.93827 y = 29° 49' 51" NAUTICAL ASTRONOMY 131 x=PF=6S°19'A7" y = ZF = 29° 49' 51" x-y = FZ = SS°29'56" PH= 90° ZH = lat. = 51° 30' 4" N. Ex. 2. Jan. 16, 1895, at 12 h. 42 m. 30 s. p.m., mean time, in long. 64° 20' W., the observed altitude of the sun's lower limb was 17° 50' 20" (zenith N.) ; index correction was —2' 10"; height of eye 12 ft. Required the latitude. ship time=0 h. 42 m. 30 s. longitude =4 h. 17 m. 20 s. Greenwich, Jan. 16=4 h. 59 m. 50 s. =4.997 h. =5 h. nearly declination of sun Jan. 16, noon = 20° 55' 58" S. correction = 28 ".68 x 5 = 2' 23"- declination of sun at time of observation = 20° 53' 35" S. .-. FA = 110° 53' 35". equation of time at noon = 9 m. 58.25 s. correction = 0.849 x 5 = 4.25 s. equation of time for observation = 10 m. 2.5 s. mean time = 42 m. 30 s. apparent time = 32 m. 27.5 s. = 8°6'52i"=Z^P2^ 132 NAVIGATION AND observed altitude = 17° 50' 20" index correction = 2' 10" 17° 48' 10" dip = 3' 24" ref. par. 3'-U 8"+ i 17° 17° 44' 46" 2' 52" 41' 64" sem. , diam. = 16' 18" rue altitude = 17° 58' 12" % ZA = :72° ' 1'48" In triangle PAF, ^ or, . cos8°6'52"i tan PF = tan x = ^ — cot 110° 53' 35" log cos 8° 6' 52"^ = 9.99563 log cot 110° 53' 35" = 9.58175 log tan 111° 5' 10" = 10.41388 • eos^ = cosll0°53'35>^ cos a; In triangle ZAF, cos ZF= cos y = ^^^-^y ^ cos AF cos 72° 1' 48" _ iiiof;Mnfr ^^^^ = cosll0°53'35" ^^^'''''^ log cos 72° 1'48"= 9.48927 log cos 111° 5' 10"= 9.55602 log sec 110° 53' 35" = 10.44778 log cos 71° 52' 1"= 9.49307 NAUTICAL ASTRONOiMY 133 x=nV 6' 10" =PF y= 71° 52' 1" =ZF jr x-y= 39° 13' 9" 90° 0\ r \ H'^ ^^ \ TT^^'^ J lat.= 50°46'51"N. = ZJ/ In case the perpendicular meets the meridian at F^ a point between F and Z, as in the figure, then FZ = cc + ?/ and Z//= lat. = 90° - (cc + y). In this case FA = (90 - ^). Ex. 3. If in long. 60° 10' W., on Jan. 3, 1895, at 5 h. 42 m. 13 s. P.M., mean time, the declination of a star was found to be 72° 12' N., and its true altitude to be 58° 42' 40" (zenith N.), required the latitude. ship time, Jan. 3 = 5 h. 42 ra. 13 s. longitude = 4h. Om. 40 s. Greenwich, Jan. 3. mean time = 9 h. 42 m. 53 s. R.A. of mean sun 3d noon = 18 h. 51 m. 25.5 s. Correction for 9 h. 42 m. 53 s. = 1 m. 35.7 s. R.A. mean sun = 18 h. 53 m. 01 s. mean time = 5 h. 42 m. 13 s. 24 h. 35 m. 14 s. 24 h. sidereal time = h. 35 m. 14 s. = APZ. ^0=68° 42' 40" 90^^ AZ = 31° 17' 20" cos 35 m. 14 s. tan x = cot 17° 48' log tan 17° 36' 11" ^Z) = 72°12' P^ = 17°48' log= 9.99485 log = 10.49341 = 9.50144 134 NAVIGATION AND cos y = cos 31° 17' 20" cos 17° 36' 11" sec 17° 48' log cos 31° 17' 20"= 9.93174 log cos 17° 36' 11"= 9.97917 log sec 17° 48' = 10.02130 log cos 31° 11' 15" = 9.93221 PF=17°36'll" = a; ZF= 31°ll'15" = y PZ=48°47'26" = x + y 90^ ZH= lat. = 41° 12' 34" N. = 90° - (a: + y). 78. To find the latitude by observing the altitude of the Pole Star (Polaris). This method is confined to northern latitudes. Let the figure represent the projection of the celes- tial concave on the celestial meridian ; P tlie N. pole ; Z the zenith ; QEC the pro- jection of the equinoctial ; NUS the projection of the horizon. Since PC =90° and ZN = 90°, FC=ZN. If from these equals we take PZ, FN=ZC, but ZC= the latitude of the observer ; that is, FN, the altitude of the nearer pole above the horizon, is equal to the latitude (a principle already shown in Art. 75). The star called Polaris is very near the N. pole of celestial sphere. It moves in a small circle about that pole. The polar distance of this circle is very nearly NAUTICAL ASTRONOMY 135 1° 14' (1898). By observing its altitude, at its upper and lower culminations, and subtracting or adding its exact polar distance, the latitude may be obtained. As this method is not always practicable, its altitude is observed at any moment, and to this altitude cor- rections are applied which are arranged in tables for the purpose of obtaining the true latitude. Let the figure represent the projection of the celes- tial concave on the plane of the horizon. In order to understand the correc- tions required, draw ASBS' to represent the circle in which Polaris moves each 24 hours (sidereal). If, with Z as a pole and a distance ZP we describe a circle, cut- ting ASBS' in the points A and B, these points will be the points where the altitude of Polaris will be the same as the altitude of P. Since ZL, ZN, and ZR each equals 90°, and ZA = ZP=^ZB, therefore AL = P]Sf= BR. If we take any other position of the star, as S, on the arc A SB, its alti- tude w411 evidently be greater than that of the pole P, or if we take S' on the arc AS'B, its altitude will be less than that of the pole P. If, with Z as a pole and polar distance ZS we de- scribe a circle cutting the meridian ZN in D, the 136 NAVIGATION AND altitude of S will be the same as that of D ; and if with polar distance ZS' we draw a circle cutting me- ridian at D', the altitude of aS" will be the same as that of I>\ Join FS and FS% and from S and S' draw SC and S'C, perpendiculars to the meridian. If we denote the hour angle of the star in any posi- tion by t, then at position S the angle SFC will be t, and at S' the salient ande S'FC will be t. The tri- angles SFC and SFC may be considered as plane triangles, since their sides are such small arcs. Con- sequently, (1) FC =FS cos SFC =FScost, and (2) FC = FS' cos SFC = FS cost. Now, FS and FS" are the polm^ distances of the star, and therefore are the complements of its declina- tion. As the declination is given in the Nautical Almanac, FS and jPaS" are known. Denote Pas' and FS' by,^; then PC and FC from equations (1) and (2) can both be" expressed by one equation, viz. : FC, or FC'=pcost. In this expression attention must be paid to the sign of cos t. From h. to 6 h. and from 18 h. to 24 h. the sign is + ; between 6 h. and 18 h. the sign is — . From the figure it is evident that for an observed altitude of the star in any position on the arc A TB, except at the points A, T, and B, the latitude, FN=ND-DF = ND-{FC-CD) ^ altitude —p cos t + CD. NAUTICAL ASTRONOMY IS7 At A and B the latitude = altitude, since by construc- tion ZA, ZF, and ZB are equal. At 2" the latitude = NF = NT- FT= altitude -p. For star observed in any position on arc AKB, except A, K, and B, latitude, FN= NU + UF = NU +{FC'+ C'U), or latitude = altitude +/> cos t + C'U. At K the latitude = FN= NK+ FK= altitude -{-p. The values of p cos t and of CD, for all positions of Polaris, are calculated and arranged in tables. When the latitude is desired within 2' of the true latitude, the table for p cos t is used.* If, however, the correct latitude is required, the cor- rections for CD must also be applied. The method of using the table for pcost, only, "is suffi- ciently precise for nautical purposes." f Ex. April 1, 1898, 10 p.m. (mean time) nearly, in longitude 72° 56' W., the altitude of Polaris was observed, and, corrected, was found to be 40° 22'. Required the latitude. local time = 10 h. m. s. longitude = 4 h. 51 m. 44 s. Greenwich, April 1, mean time = 14 h. 51 m. 44 s. Greenwich, April 1, R.A. mean sun = h. 39 m. 27.9 s. correction for 14 h. 51 m. 44 s. = 2 m. 26 s. R.A. M. sun at time of observation = h. 41 m. 54 s. local mean time = 10 h. local sidereal time = 10 h. 41 m. 54 s. R.A. Polaris = 1 h. 21m. 48 s. - hour angle = 9 h. 20 m. 06 s. for hour angle of 9h. 20 m. correction from page 170 is -f 56'.9 approximate latitude = 41° 19' N. * Martin. t Bowditch. 138 NAVIGATION AND CHAPTER X LONGITUDE 79. By Art. 54 the local time was defined as the hour angle of the sun at the celestial meridian of the place ; and the Greemvich time at the same instant was defined as the hour angle of the sun at the me- ridian of Greenwich, both angles being made at the pole by the hour circle passing through the sun with the respective meridians of the place and of Greenwich. The difference of these angles can be expressed either in degree measure or in time measure. Expressed in degree measure, it is called the longitude of the place. The longitude of a place can always be determined, therefore, by comparing the local time with the Greemvich time at the same instant. All sea-going vessels are furnished with a fixed chronometer set to Greenwich time. Its rate, or the average amount of time which it loses or gains in a day, is ascertained, and applied to the time indi- cated. The error of the clock is the amount of time by which it is fast or sloio, as compared with true Green- wich time. Both the rate and error of the clock are kept on record, and taken into account in calculating longitude. NAUTICAL ASTRONOMY 139 The local, or ship time, is determined by observing the altitude of some celestial body. When the object observed is not on the meridian of the observer, the latitude of the place of observation, and the declination of the object being known, the hour angle is calculated (Art. 63). Observations for latitude are generally made when the object observed is on the meridian, or near it. Observations for longitude are preferred to be taken at the time the object is near the prime vertical. The latitude used in determining the hour angle for longitude is the latitude last observed, corrected for change due to the run of the ship in the interval between the two observations. This change of lati- tude is found by dead reckoning. 80. When the Greemvich time is greater than the ship time, the longitude of the ship is West ; when the Greenwich time is less than the ship time, the longitude of the ship is East. Let the figure represent the earth, piope, and the celestial concave, P WP'E, projected on the plane at right angles to the meridian of Greenwich. pgp will represent the terrestrial me- ridian, and PGP' the celestial meridian of Greenwich. If wge represent the terrestrial equator, its plane when 140 NAVIGATION AND produced will intersect the celestial concave in the celestial equator, WGE. If Z) be a place on the earth's surface west of Green- wich, the plane of its meridian phap produced will intersect the celestial concave in the meridian PAP'. If U be a place east of Greenwich, ph'ap will be its terrestrial meridian, and PAP' its celestial meridian. Now, if the meridian PMP' be the meridian pass- ing through the mean sun at M, at the time of an observation, GPM = Greenwich mean time, at that instant. APM = mean time at &, at that instant. A'PM^ mean time at ¥, at that instant. GPM-APM=GPA = gph; because GPA and gph are two arc angles, which are each equal to the diedral angle of the same two planes. But gph is measured by ga, and is the longitude of h west. Therefore, Greenwich mean time — local mean time = longitude west. In the same wsiy, A'PM- GPM=gph' ; hut gph' is measured by ga, and is the longitude of h' east. Therefore, local mean time - Greenwich mean time = longitude east. Ex. 1. At 9.13 P.M. (mean time) nearly, June 24, 1898, in longitude 16° 18' W. (by account), a ship's chronometer in- dicated lOh. 11m. 3 s. (Greenwich time). On June 14, at Greenwich mean noon, the chronometer was slow 1 m. 15.8 s., and its mean daily rate was 6.4 s., losing. Required the correct Greenwich mean time, corresponding to ship time. NAUTICAL ASTROlJiOMi: 141 ship time June 24 longitude Gr. June 24, M. time 9h. 13 m. Ih. 5 m. 12 s. 24 10 h. 18 m. 12 s. approximately. Interval of time between June 14 noon, and 10 h. 18 m June = 10 d. 10 h. 18 m. = 10 d. 8 h. + 2 h. + 15 m. daily rate . 10 d. . . . 8 h. = 1 d. 2h. =,Vd. 18m. = ji^d. 6.4 s. 64.00 2.13 0.53 0.00 66.7 accum. rate = 1 m. 6.7 s. slow. .-. to be added, chronometer showed 10 h. 11m. 3 s. original error cor. Green, mean time 10 h. 12 m. 9.7 s. 1 m. 15.8 s. 10 h. 13 m. 25.5 s. Ex. 2. April 19, 1898, 4 p.m. (mean time) nearly, in latitude 41° 19' N., longitude (by account) 41° 18' W., the altitude of the sun's lower limb was 29° 48' 20", when a chronometer showed 6 h. 49 m. 49 s. The index correction was — 2' 30" ; heisfht of eye above sea level, 25 feet. On April 10 at noon, Greenwich mean time, the chronometer was fast 5 m. 10 s., and its daily rate was 2.5 s., gaining. Required the longitude. ship, April 19 4 h. m. s. longitude 2 h. 45 m. 12 s. Green. April 19 6 h. 45 m. 12 s. 6.75 h. dec. of sun noon m. t. 11° 16' 44 ".2 N. correction for 6.75 h. 5'49".3+ eq. of time m. 57 76 s. correction 3.br) s eq. of time 1 m. 01.45 3. 0.546 to he sub. 6.75 from ap. time. 2730 ^ 3822 3276 declination of sun 11° 22' 33".5 N. 3.6855 142 NAVIGATION AND 51".75 6.75 25875 36225 31050 349.3125 5'49".3 observed altitude of sun 29° 48' 20" I. C. 2' 30"- 29° 45 dip ref. 1'42"- ) par. 8"+ I S. I). 29° 45' 50" 4'54"- 29° 40' 56" 1'34"- 29° 39' 22" 15' 57"+ true altitude 29° 55' 19" Interval from April 10, noon, to date of observation, 9 d. 6.75 h. daily rate .... 2.5 s. 9 9d 22.5 id 6 Ad _1 accum. gain . . . 23.2 to be subtracted. chronometer 6 h. 49 m. 49 s. 6 h. 49 m. 26 s. original error 5 m. 10 s. correct Greenwich time 6 h. 44 m. 16 s. PZ = 90° - 41° 19' = 48° 41' P^ = 90°- 11°22'33".5 = 78° 37' 27" ^Z= 90° -29° 55' 19" = 60° 04 '41" a= 48° 41' 2= 78° 37' 27" p= 60° 04' 41" 5 =187° 23' 8" = 93° 41 '34" NAUTICAL ASTRONOMY 143 s-a= 45° 0'34" s-z= 15° 04' 7" s-p= 33° 36' 53'' sin ^P = Vsin (s — a) sin (s — z) cosec a cosec z log sin 45° 0' 34"= 9.84956 log sin 15° 04' 7" = 9.41493 log cosec 48° 41' = 10.12432 log cosec 78° 37' 27" = 10.00862 2 )19.39743 log sin i (3 h. 59 m. 44 s. + 7 s.) = 9.69871J 53_ correction 7 s. = 18^ ship apparent time = 3 h. 59 m. 51 s. equation of time = 1 m. 01 s.— ship mean time = 3 h. 58 m. 50 s. Greenwich, mean time = 6 h. 44 m. 16 s. longitude = 2 h. 45 m. 26 s. = 41° 21' 30" W. Ex. 3. Feb. 13, 1898, 6.30 a.m. (mean time) nearly, in lat. 45° 16' S., and long. 28° 42' E. (by account), a chronometer showed 4 h. 41 m. 48 s., when an observed altitude of the sun's upper limb was 14° 18' 20". Index correction was — 1'13", height of eye, 12 ft. Feb. 7, at noon (G.M.T.), the chronome- ter was slow 3 m. 6 s., and its daily rate was 1.4 s., losing. ship time, Feb. 12 18 h. 30 m. s. longitude 1 h. 54 m. 48 s. Greenwich, Feb. 12 16 h. 35 m. 12 s. 16.59 h. or Greenwich, Feb. 13 —7.41 h. 144 NAVIGATION AND hourly difference of declination 60 ".65 7.41 5065 20260 35455 375".3165, or 6' 15".3 declination of sun, Feb. 13, noon 13° 14' 45" S. correction for — 7.41 h. = 6' 15" declination of sun = 13° 21' si equation of time, Feb. 13, noon = 14 m. 24.2 s. correction for — 7.41 h. = 0.6 s. equation of time = 14 m. 24.8 s. to he added to ap. t. hourly dif. of eq. of time 0.078 7.41 78 312 526 .55798 Interval from Feb. 7 noon to obs. alt. of sun 14° 18' 20" time of observation 5 d. index cor. 1' 13"- 16.59 h. 14° 17' 07" dip 3' 24" daily rate 1.4 s. ^ 14° 13' 43" _5_ ref. 3''46"-l o, o.,, 5d 7. par. 9"+ j ^ '^^ |d 0.7 14° 10' 06" |d 0.2 sem. diam. 16' 14" accumulated loss . . 7.9 s. true alt. of sun 13° 53' 52" chron. showed 4 h. 41 m. 48 s. 4h. 41m. 56 s. orig. error 3 m. 6 s. cor. G.M.T. 41i.45m. 2 s. NAUTICAL ASTRUJSOMr 145 a = 44° 44' Z=z 76° 39' P = 76° 06' 8" s = 197° 29' 8" 2 = 98° 44' 34" s — a = 54° 0'34" s — z = 22° 05' 34" s-p = 22° 38' 26" tan \ P = Vsin (s — a) sin (s — z) cosec s cosec (s — p) log cosec s = 10.00507 log sin (s- a) = 9.90801 log sin (s — z)= 9.57531 log cosec (s —p) = 10.41460 2 )19.90299 log tan ^ (6 h. 25 m. 34 s.) 9.95149^ equation of time 14 m. 25 s. 64 mean time of ship 6 h. 39 m. 59 s. 14j- Greenwich mean time 4 h. 45 m. 2 s. longitude 1 h. 54 m. 57 s. = 28° 44' 15" E. Ex. 4. Jan. 20, 1898, 8.30 a.m., (mean time) nearly, latitude 39° 58' N., longitude, by account, 30° 15' W., a chronometer showed 10 h. 53 m. 9 s,, when an observed altitude of the sun's upper limb was 13° 2' 30". Index correction was — 3' 50", height of eye was 18 ft. Jan. 12, noon, Greenwich mean time, the chronometer was 10 m. 36 s. fast and its daily rate was 1.2 s., gaining. Required the longitude. ship, Jan. 19 20 h. 30 m. dec. of sun Jan. 20 noon 20° 3' 9".8 S. longitude 2 h. 1 m. correction for — 1.48 48".6 Gr. Jan. 19 22 h. 31m. declination of sun 20°3'58".4S. = 22.52 h. or Jan. 20 - 1.48 h. NAV. AND NAIT. ASTR. — 10 146 NAVIGATION AND 32".84 1.48 interval from Jan. 12 noon 26272 eq. of time 11 m. 19.98 s. to time of obs. 7 d. 22^ h. 13136 correction 1.07 s. daily rate 1.2 s. 3284 eq. of time 11 m. 18.91 s. 7 48.6032 0.724 to be added to 7d. = 8.4 1.48 apparent time. id.= .6 5792 4d.= .4 2896 Ad.= .1 724 • accum. gain 9.5 s. 1.07152 accum. gain = 9.5 s. cliron. showed 10 h. 53 m. 9. s. obs. alt. of sun 13° 2' 30" I.e. 3' 50" 10 h. 52 m. 59.5 s. original error 10 m. 36 s. Gr. M. time 10 h. 42 m. 23.5 s. a= 50° 2' « = 110° 3' 58" p= 77° 25' 46" s = 237°3r44" 2 = 118° 45' 52" s_a= 68° 43' 52" s-z= 8° 41' 54" s-i)= 41° 20' 06" log tan i (8 h. 29 m. 52 s ship apparent time 8 h. 29 m. 5^ s. equation of time 11 m. 19 s. 8 h. 41 m. 14 s. 12° 58' 40" dip 4' 09" 12° 54' 31" ref. 4' 9"- 1 par. 9"+ J 4'00"- 12° 50' 31" S.D. 16' 17"- true alt. of sun 12° 34' 14"- log cosec = 10.05720 log sin = 9.96936 log sin = 9.17966 log cosec = 10.18016 2)19.38638 . + 3 s.) 9.69319 3 s. cor. 29 for 10 NAUTICAL ASTRONOMY 147 Greenwich mean time = 10 h. 42 m. 23.5 s. ship mean time 8 h. 41 m. 14 s. longitude 2 h. 01 m. 9.5 s. longitude 30° 17' 221" W. Ex. 5. April 9, 1898, 4 p.m. (mean time) nearly, in latitude 46°o2'N., longitude (by account), 50° 35' W., a chronometer shewed 7 h. 28 m. 4 s., when the altitude of the sun's lower limb was 23° 58' 40". Index correction was + 2' 48" ; height of eye above sea level, 14 ft. April 1, noon, Greenwich mean time, the chronometer was slow 6 m. 35 s., and its daily rate was 1.2 s., losing. Required the longitude. Ans. 50° 39' W. Ex. 6. June 13, 1898, 6 p.m. (mean time) nearly, in latitude 42° 4' N., longitude (by account), 36° 22' W., the observed alti- tude of sun's lower limb was 15° 7' 30", when a chronometer showed 8 h. 16 m. 28 s. Index correction was — 3' 14" ; height of eye, 20 ft. June 1, noon, Greenwich mean time, chronom- eter was slow 8 m. 13 s., and its daily rate was 1.3 s., gaining. Required the longitude. Ans. 35° 57' W. Ex. 7. May 2, 1898, 5 p.m. (mean time) nearly, in lat. 50° 16' N., longitude (by account) 40° 18' W., the observed altitude of the sun's lower limb was 21° 16' 50", when a chronometer showed 7 h. 44 m. 2 s. Index correction was + 1' 12" ; height of eye above sea level was 15 ft. April 25, noon, G.M.T., chro- nometer was fast 6 m. 18 s., and daily rate was 0.6 s., losing. Required the longitude. Ans. 40° 16' W. Ex. 8. May 14, 1898, 6 a.m. (mean time) nearly, in lat. 44° 48' N., longitude (by account) 33° 22' W., the observed altitude of the sun's lower limb was 13° 5' 40", when a chronometer showed 8 h. 23 m. 28 s. Index correction was — 2' 25" ; height of eye above sea level was 18 ft. May 6, at noon, G.M.T., the chronometer was fast 12 m. 36 s., and its daily rate was 1.6 s., gaining. Required the longitude. Ans, 33° 24 J' W. 148 NAVIGATION AND Ex. 9. Feb. 28, 1898, 8 a.m. (mean time) nearly, in lat. 46° 22' N., longitude (by account) 50° 42' W., a chronometer showed 11 h. 30 m. 54 s., when the observed altitude of the sun's upper limb was 14° 25' 30". Index correction was +2' 20" ; height of eye above sea level was 20 ft. Feb. 20, noon, G.M.T., chronometer was slow 4 m. 30 s., and its daily rate was 0.8 s., gaining. Required the longitude. Given dec. of sun, Feb. 28, Green., noon, 7° 50' 24" S. Hourly dif. 56".79 N. Equation of time at Green., noon, 12 m. 40.7 s. to be added to mean time. Hourly dif. 0.479 s., decreasing from Feb. 28 to March 1. Ayis. 50° 39' 15" W. NAUTICAL ASTRONOMY 149 DEFINITIONS OF TERMS USED IN NAUTICAL ASTRONOMY Altitude. The altitude of a heavenly body is the angle of ele- vation of the body above the horizon, and is measured on the circle of altitude passing through the body. This measured distance is generally used for the altitude. Observed Altitude. The observed altitude of a heavenly body is the altitude of the body above the sea horizon taken with a sextant or other instrument. True Altitude. The true altitude of a heavenly body is its observed altitude corrected for index error, dip, refraction, parallax, and semi-diameter. First Point of Aries. The first point of Aries is the point on the celestial concave in which the ecliptic cuts the equi- noctial, where the sun passes from the south to the north of the equinoctial. Axis. The axis of the celestial sphere is the diameter about which the celestial concave appears to revolve from east to west. It is coincident with the earth's axis produced. Azimuth. The azimuth or true bearing of a heavenly body is the angle at the zenith made by the celestial meridian and the circle of altitude passing through the body. Celestial Concave. The celestial concave is the surface of a very large sphere of which the center is the center of the earth. Apparent Solar Day. An apparent solar day is the interval of time between two successive transits of the sun over the same celestial meridian. 150 NAVIGATION AND Mean Solar Day. A mean solar day is the interval of time between two successive transits of the mean sun over the same celestial meridian. Sidereal Day. A sidereal day is the interval of time between two successive transits of the first point of Aries over the same celestial meridian. Declination. The declination of a heavenly body is the arc of a circle of declination between the body and the equi- noctial, or celestial equator. Circles of Declination. Circles of declination are great circles of the celestial concave which pass through its poles. Circles of declination are also called hour circles. Angle of Depression. The angle of depression of any body below the observer is the angle between a line drawn to it from the observer's eye, and the horizontal plane through the observer's eye. Ecliptic. The ecliptic is the great circle in which the plane of the earth's orbit cuts the celestial concave. Angle of Elevation. The angle of elevation of any body above the observer is the angle at the observer's eye, between a line dravs^n from it to the body and a horizontal plane through the eye. Celestial Equator and Equinoctial. The equinoctial is the celes- tial equator and is the great circle of the celestial con- cave made by producing the plane of the terrestrial equator to cut the concave. Greenwich Date. The Greenwich date is the astronomical time at Greenwich, when an observation is taken at any place on the earth. Horizon. The celestial horizon or simply the horizon at any place is the great circle of the celestial concave, in which a plane tangent to the earth at that place meets the con- cave. This plane is known as the plane of the horizon. NAUTICAL ASTRONOMY 151 Rational Horizon. The rational horizon is a plane passed through the center of the earth parallel to the sensible horizon. Sensible Horizon. The sensible horizon is a plane tangent to the earth at a point vertically below the point of observa- tion. Visible Horizon. The visible horizon is the small circle which bounds the vision of the observer. Hour Angle. The hour angle of any heavenly body is the angle at the pole between the celestial meridian of the observer and the hour circle passing through the body. Hour Circles. Hour circles are circles of declination. Celestial Meridian. The celestial meridian of any place is the great circle in which the plane of the terrestrial meridian cuts the celestial concave. Apparent Noon. Apparent noon is the instant when the center of the real sun is on the celestial meridian. Mean Noon. Mean noon is the instant when the mean sun is on the celestial meridian. Poles of the Heavens. The poles of the heavens are the extremi- ties of the axis of the celestial concave. Prime Vertical. The prime vertical is the circle of altitude, whose plane is at right angles to the plane of the celestial meridian. Right Ascension. The right ascension of a heavenly body is the arc of the equinoctial, or celestial equator, between the first point of Aries and the circle of declination pass- ing through the body. Right ascension is measured in time eastward from h. to 24 h. Apparent Time. Apparent time is the hour angle of the real sun. Mean Time. Mean time is the hour angle of the mean sun. 152 NAVIGATION AND Equation of Time. The equation of time is the difference between apparent time and mean time. Astronomical Time. Astronomical time is reckoned in periods of twenty-four hours, each period beginning at noon. Civil Time. Civil time is reckoned in two periods of twelve hours, named a.m. and p.m. according as they come before or after noon of the day, which, in this method of reckon- ing time, begins at midnight. Zenith. The zenith is the pole of the celestial horizon directly above the observer. NAUTICAL ASTRONOMY 163 EXAMPLES CHAPTER III In the following examples, deviation is to be taken from table on page 56. Find the true courses: Ex. 1. Compass course = N. 47° E.; variation = 9° W.; lee- way = 0°. Ans. N. m° E Ex. 2. Compass course = E. b. N. J N. ; variation = 21° E. leeway = 1\ pt. and wind N. Ans. S.E Ex. 3. Compass course = S. 51° E.; variation = 18° E.; lee way = 0. , Ans. S. 17° E Ex. 4. Compass course = S. f W. ; variation = 21° W. ; lee way = 1 pt; wind E.S.E. Ans. S. i E Ex. 5. Compass course = W. b. S. f S. ; variation = 11° W. leeway = 1 pt. ; wind S. Ans. S. W. | W Ex. 6. Compass course = N.N. W. \ W. ; variation = 30° W. leeway = J pt. ; wind W. Ans. W.N.W Find the compass course : Ex. 7. True course = N.N.E. J E.; variation being 21° E. Ans. N. 5° E. Ex. 8. True course = N. 62° E. ; variation being 11° W. Ans. N. 54° E. Ex. 9. True course = E. | S. ; variation being 12° W. Ans. East. Ex. 10. True course = S. b. W. J W. ; variation being 19° E. Ans. S. 10° E. Ex. 11. True course = N.W. J W.; variation being 34° W. Ans. N. 9° W. 154 NAVIGATION AND CHAPTERS V AND VI Ex. 1. May 28, 1898, in long. 72° 55|' W., required mean time of apparent noon, and declination of sun at that time. Ans. Mean time, 11 h. 57 m. 4.15 s. a.m.; dec. of sun, 21° 32' 42" N. Ex. 2. May 28, 1898, in long. 72° 55|' W., given mean times 10.15 A.M. and 1.45 p.m., required corresponding sidereal times. Ans. 14 h. 39 m. 42 s. ; 6 h. 10 m. 17 s. Ex. 3. May 27, 1898, in long. 72° 55|' W., given mean times 9.45 A.M. and 1.30 p.m., required corresponding sidereal times. Ans. 2 h. 5 m. 41 s.; 5 h. 51 m. 18 s. Ex. 4. May 27, 1898, in long. 72° 55|' W., required the mean time of apparent noon ; also declination of sun at that time. Ans. 11 h. 56 m. 57 s. a.m.; 21° 23' 2" N. Ex. 5. March 15, 1898, in long. 72° 55|' W., given apparent times, 6.30 a.m. and 5 p.m., to find corresponding mean times. Ans. 6.39 a.m. ; 5 h. 8 m. 52 s. p.m. Ex. 6. In long. 72° 55|' W., March 19, 1898, 10.45 a.m. mean time, required apparent time, sidereal time, and declination of sun. Ans. Apparent time, 10 h. 37 m. 13 s. ; sidereal time, 22 h. 33 m. 48 s. ; declination of sun, 0° 22' 10" S. CHAPTER VII Ex. 1. In lat. 41° 18' N., long. 72° 55|' W., May 2, 1898, 3.19 P.M. apparent time, nearly, the true altitude of the sun was 40° 14'; required its hour angle. Ans. 3 h. 18 m. 31 s. Ex. 2. In lat. 41° 18' N., long. 72° 55|' W., Jan. 10, 1898, 10 A.M. mean time approximately, the true altitude of sun was 20° 40'; required mean time. Ans. 10 h. 4 m. 53 s. a.m. Ex. 3. In lat. 41° 18' N., long. 72° 5oi> W., Jan. 10, 1898, 11 a.m. mean time approximately, the true altitude of the sun "was 24° 40' ; required mean time. Ans. 10 h. 56 m. 23 s. a.jvj. NAUTICAL ASTRONOMY 155 Ex. 4. April 1, 1898, at 7 p.m. mean time nearly, in long. 72° 55|' W., the hour angle of a Orionis was 1 h. 50 m. h^ s., W. of meridian. Required mean time. Ans. 6 h. 59 m. 11 s. p.m. Ex. 5. Nov. 22, 1898, 7.15 p.m. mean time nearly, in long. 87° 56' W., the hour angle of Aldebaran (a Tauri) was 18 h. ^o m. 15 s. (E. of meridian). Nov. 22, noon Greenwich R.A. mean sun was 16 h. 5 m. 58.42 s. Ans. 7 h. 17 m. 14 s. p.m. Ex. 6. Find at what time Procyon (« Canis Minoris) passed the meridian of 72° m' W., April 5. 1898. Ans. 6 h. 36 m. 51 s. p.m. Ex. 7. Find at what time Sirius passed the meridian 72° 55|' W., April 6, 1898. If the place is in lat. 41° 18" K, required also its meridian altitude at transit. Ans. 5 h. 39 m. 45 s. p.m.; 32° 7 27". Ex. 8. In lat. 41° 18' N., long. 72° h^^ W., April 6, 1898, find at what time Regulus passed the meridian, and at what altitude. Ans. 9 h. 1 m. 29 s. p.m.; 61° 9' 52". Ex. 9. In lat. 41° 18' N., long. 72° 55|' W., April 5, 1898, 10 P.M. mean time nearly, the altitude of y8 Geminorum was 48° 17', and its declination was 28° 16' 19" N. Required mean time. Ans. 9 h. 57 m. 33 s. p.m. CHAPTER IX Ex. 1. In long. 72° 55|' W., April 20, 1898, the observed meridian altitude of the sun's lower limb was 33° 22' 30" (zenith N.); index correction was —2' 10"; height of eye above sea level was 25 ft. Required the latitude. Ans. 68° 11' 27" N. Ex. 2. April 21, 1898, in long. 72° 5oJ' W., the observed meridian altitude of the sun's lower limb was 56° 10' 20" (zenith N.); index correction was +2' 25"; height of eye was 18 ft. Required the latitude. Ans. 45° 37' 52" N. 156 NAVIGATION AND NAUTICAL ASTRONOMY Ex. 3. Jan. 2, 1898, the observed altitude of Vega (a Lyras) (zenith N.) was 70° 2' 30"; index correction was +2' 16"; height of eye above sea level was 14 ft. Required the latitude. Ans. 58° 40' 34" N. Ex. 4. April 20, 1898, the observed meridian altitude of Arcturus was 62° 40' 30"; index correction was -|-3' 16"; height of eye above sea level was 20 ft. Required the latitude. Ans. 47° 3' 50" N. Ex. 5. March 14, 1898, at 2 a.m. (nearly), in long. 45° 40' W., the observed altitude of Polaris was 43° 16'; index correction was — 2' 22"; height of eye was 18 ft. Required the latitude. Ans. 44° 22' N. Ex. 6. April 22, 1898, at 3 a.m. (nearly), in long. 50° 10' W., the observed altitude of Polaris was 46° 38'; index correction was -h 1' 40"; height of eye was 13 ft. Required the latitude. Ans. 47° 18' N. Ex. 7. In long. 16° 16' W., June 16, 1898, 12 h. 12 ra. 26 s. P.M. mean time, the observed altitude of the sun's upper limb (zenith K) was 61° 40' 10"; index correction was +2' 25"; height of eye above sea level was 17 ft. Required the latitude. Ans. 51° 54' 34" N. ASTROiNOMICAL EPHEMERIS FOB THB MERIDIAN OF GREENWICH 158 NAVIGATION AND JANUARY, 1898 At Greenwich Apparent Noon I 1 5 § THE SUN'S Equation of 1 1 o 1 Time, to be Added to Apparent Time Diflf. for 1 iiour 1 Apparent Declination Diflf. for 1 hour Semi- diameter Sat. 1 » '/ S. 22 59 1.6 II + 12.81 16 18.37 m. 8. 3 55.32 1.177 SUN. 2 22 53 40.7 13.94 16 18.37 4 23.36 1.160 Mon. 3 22 47 52.4 15.07 16 18.37 4 51.02 1.143 Tues. 4 22 41 37.1 + 16.20 16 18.36 5 18.26 1.126 Wed. 5 22 34 54.8 17.82 16 18.35 6 45.08 1.108 Thur. 6 22 27 45.7 18.43 16 18.33 6 11.42 1.088 Frid. 7 22 20 10.2 + 19.53 16 18.30 6 37.29 1.067 Sat. 8 22 12 8.3 20.02 16 18.20 7 2.64 1.045 SUN. 9 22 3 40.3 21.71 16 18.22 7 27.47 1.023 Mon. 10 21 54 46.4 + 22.78 16 18.18 7 51.74 1.000 Tues. 11 21 45 26.9 23.84 16 1».13 8 15.46 0.975 Wed. 12 21 35 42.0 24.89 16 18.07 8 38.57 0.950 Thur. 13 21 25 32.0 + 25.93 16 18.00 9 1.08 0.925 Frid. 14 21 14 57.1 26.96 16 17.93 9 22.96 0.899 Sat. 15 21 3 57.7 27.98 16 17.86 9 44.21 0.871 SUN. 16 20 52 34.1 + 28.98 16 17.78 10 4.79 0.842 Mon. 17 20 40 46.5 29.97 16 17.70 10 24.69 0.814 Tues. 18 20 28 35.3 30.95 16 17.61 10 43.89 0.785 Wed. IP 20 16 0.9 +31.91 16 17.52 11 2.38 0.755 Thur. 20 20 3 3.6 32.86 16 17.42 1120.12 0.724 Frid. 21 19 49 43.7 33.79 16 17.32 1137.11 0.693 Sat. 22 19 36 1.6 + 34.71 16 17.22 11 53.36 0.661 SUN 23 19 21 57.8 35.61 16 17.11 12 8.82 0.(528 Mon. 24 19 7 32.6 36.49 16 17.00 12 23.47 0.595 Tues. 25 18 52 46.4 +37.35 16 16.89 12 37.34 0.561 Wed. 26 18 .37 39.6 38.20 16 16.77 12 50.37 0.526 Thur. 27 18 22 12.6 39.04 16 16.65 13 2.59 0.492 Frid. 28 18 6 25.8 +39.85 16 16.53 13 13.97 0.457 Sat. 29 17 50 19.7 40.(55 16 16.40 13 24.53 0.422 SUN 30 17 S3 54.6 41.43 16 16.27 13 34.23 0.387 Mon. 31 17 17 10.9 42.20 16 16.13 13 43.10 0.352 Tues. 32 S. 17 9.0 + 42.95 16 15.99 13 61.12 0.318 NAUTICAL ASTRONOMY 159 II, JANUARY, 1898 At Greenwich Mean Noon 1 5 a o ® THE SUN'S Equation of Time, to be Subtracted from Mean Time Diff. for 1 hour Sidereal Time, or 1 Apparent Declination Diff. for 1 hour Right Ascension of Mean Sun Sat. SUN. Mon. 1 2 3 • " S. 22 59 2.4 22 53 41.7 22 47 53.7 + 12.79 13.03 15.07 m. s. 3 55.24 4 23.27 4 50.92 i.no 1.100 1.143 h. m. 8. 18 44 37.92 18 48 34.48 18 52 31.04 Tues. Wed. Thur. 4 5 G 22 41 38.5 22 34 50.4 22 27 47.7 + 10.20 17.31 18.42 5 18.10 5 44.97 11.31 1.120 1.108 1.088 18 50 27.00 19 24.15 19 4 20.71 Frid. Sat. SUN 7 8 9 22 20 12.4 22 12 10.7 22 3 43.0 + 19.52 20.01 21.09 37.17 7 2.52 7 27.34 1.007 1.045 1.023 19 8 17.27 19 12 13.83 19 10 10.39 Mon. Tues. Wed. 10 11 12 21 54 49.4 21 45 30.2 2135 45.0 + 22.76 23.83 24.88 7 51.01 8 15.32 8 38.43 1.000 0.975 0.950 19 20 0.95 19 24 3.50 19 28 0.06 Thur. Frid. Sat. 13 14 15 21 25 35.9 21 15 1.4 21 4 2.3 + 25.92 20.95 27.97 9 0.94 9 22.82 9 44.07 0.925 0.809 0.871 19 31 £6.62 19 35 53.18 19 39 49.73 SUN. Mon. Tues. 10 17 18 20 52 39.0 20 40 51.8 20 28 40.9 + 28.97 29.90 30.94 10 4.05 10 24.55 10 43.75 0.843 0.814 0.785 19 43 46.29 19 47 42.85 19 51 39.40 Wed. Timr. Frid. 19 20 21 20 16 6.8 20 3 9.8 19 49 50.3 + 31.90 32.84 33.77 11 2.24 11 19.98 11 30.98 0.755 0.724 0.093 19 55 35.96 19 59 32.52 20 3 29.08 Sat. SUN. Mon. 22 28 24 19 36 8.6 19 22 5.1 19 7 40.2 +34.09 35.59 30.47 1153.23 12 8.09 12 23.35 0.001 0.028 0.595 20 7 25.63 20 11 22.19 20 15 18.75 Tues. Wed. Thur. 25 20 27 18 52 54.3 18 37 47.8 18 22 21.1 + 37.34 38.19 39.02 12 37.22 12 50.20 13 2.48 0.501 0.520 0.492 20 19 15.30 20 23 11.80 20 27 8.42 Frid. 8at. SUN Mon. 28 29 30 31 18 6 34.7 17 50 28.8 17 34 4.0 17 17 20.0 + 39.84 40.04 41.42 42.19 13 13.87 13 24.43 13 34.14 13 43.02 0.457 0.422 0.387 0.352 20 31 4.97 20 35 1.53 20 38 .^8.09 20 42 54.64 Tues. 32 S. 17 19.0 + 42.94 13 51.05 0.318 20 46 51.20 160 NAVIGATION AND MARCH, 1898 At Greenwich Apparent Noon § 1 THE SUN'S Equation of Time, to be Added to Apparent Time Diflf. for Ihour 1 Apparent Declination Diff. for 1 liour Semi- diameter Tues. Wed. Thur. 1 2 3 S. 7 27 25.8 7 4 33.4 6 41 35.2 + 57.05 57.30 57.54 16 10.86 16 10.12 16 9.88 ni. s. 12 28.85 12 16.55 12 3.78 0.501 0.522 0.542 Frid. Sat. SUN. 4 5 6 6 18 31.4 5 65 22.6 5 32 8.9 + 57.76 57.97 58.16 16 16 16 9.64 9.39 9.14 11 50.52 11 36.80 11 22.65 0.561 0.580 0.598 Mon. Tues. Wed. 7 8 9 5 8 50.8 4 45 28.7 4 22 2.9 + 58.34 58.50 58.65 16 16 16 8.88 8.62 8.36 11 8.09 10 53.11 10 37.79 0.615 0.631 0.645 Thur. Frid. Sat. 10 11 12 3 58 33.7 3 35 1.5 3 11 26.7 + 58.78 58.90 59.00 16 16 16 8.10 7.84 7.57 10 22.14 10 6.14 9 49.86 0.659 0.672 0.684 SUN. Mon. Tues. 13 14 15 2 47 49.6 2 24 10.5 2 29.9 + 59.09 59.16 59.22 16 16 16 7.30 7.03 6.75 9 33.30 9 16.48 8 59.44 0.695 0.705 0.714 Wed. Thur. Frid. 16 17 18 1 36 48.2 1 13 5.6 49 22.7 + 59.26 59.28 59.29 16 16 16 6.48 6.20 5.92 8 42.19 8 24.76 8 7.13 0.722 0.730 0.737 Sat. SUN. Mon. 19 20 21 25 39.7 S. 1 57.0 N. 21 44.8 + 59.28 59.26 59.22 16 16 16 5.64 5.36 5.09 7 49.37 7 31.46 7 13.44 0.743 0.749 0.753 Tues. Wed. Thur. 22 23 24 45 25.6 1 9 4.8 1 32 42.2 + 59.17 59.10 59.01 16 16 16 4.81 4.54 4.26 6 55.32 6 37.13 6 18.85 0.756 0.759 0.762 Frid. Sat. SUN. 25 26 27 1 56 17.2 2 19 49.6 2 43 18.9 + 58.91 58.79 58.65 16 16 16 3.99 3.72 3.45 6 0.53 5 42.19 5 23.81 0.764 0.765 0.766 Mon. Tues. Wed. Thur. 28 29 30 31 3 6 44.8 3 30 7.0 3 53 25.1 4 16 38.8 + 58.50 58.34 58.16 57.97 16 16 16 16 3.18 2.91 2.64 2.37 5 5.44 4 47.10 4 28.78 4 10.54 0.765 0.763 0.762 0.760 Frid. 32 N. 4 39 47.7 + 57.77 16 2.10 3 52.37 0.756 II. NAUTICAL ASTRONOMY MARCH, 1898 At Greenwich Mean Noon 161 +-> B O 1 THE SUN'S Equation of Time, to be Subtracted from Mean Time Diflf. for 1 hour Sidereal Time, or Eight Ascension of Mean Sun 1 Apparent Declination Diff. for 1 hour Tues. Wed. Thur. 1 2 3 O ' " S. 7 27 37.7 7 4 45.2 6 41 40.8 + 57.06 5731 57.55 m. s. 12 28.95 12 16.66 12 3.89 0.501 0.522 0.542 h. m. 8. 22 37 14.73 22 41 11.29 22 45 7.84 Frid. Sat. SUN, 4 5 6 6 18 42.9 5 55 33.8 5 32 20.0 + 57.77 57.98 58.17 11 50.63 11 36.91 11 22.76 0.501 0.580 0.598 22 49 4.39 22 53 0.95 22 56 57.50 Men. Tues. Wed. 7 8 9 5 9 1.7 4 45 39.4 4 22 13.3 + 58.35 58.51 68.66 11 8.20 10 53.23 10 37.91 0.615 0.631 0.645 23 54.05 23 4 50.61 23 8 47.16 Thur. Frid. Sat. 10 11 12 3 58 43.9 3 35 11.5 3 1136.4 + 58.79 58.91 59.01 10 22.25 10 6.25 9 49.97 0.659 0.672 0.684 23 12 43.71 23 16 40.27 23 20 36.82 SUK Mnn. Tues. 13 14 15 2 47 59.0 2 24 19.7 2 38.9 + 59.10 59.17 59.23 9 33.41 9 10.59 8 59.55 0.695 0.705 0.714 23 24 33.37 23 28 29.93 23 32 26.48 Wed. Thur. Frid. IP) 17 18 1 36 56.8 1 13 14.0 49 30.8 + 59.27 59.29 59.30 8 42.30 8 24.86 8 7.23 0.722 0.730 0.737 23 36 23.03 23 40 19.58 23 44 16.14 Sat. SUN. Mon. 19 20 21 25 47.4 S. 2 4.5 N. 21 37.7 + 59.30 59.28 59.24 7 49.47 7 31.56 7 13.53 0.743 0.749 0.753 23 48 12.69 23 52 9.24 23 56 5.80 Tues. Wed. Thur. 22 23 24 45 18.7 1 8 58.3 1 32 35.9 + 59.18 59. 1 1 59.02 6 55.41 6 37.21 6 18.93 0.756 0.759 0.762 2.35 3 58.90 7 55.46 Frid. Sat. SUN. 25 26 27 1 56 11.3 2 19 44.0 2 43 13.6 + 58.92 58.80 58.66 6 0.61 5 42.26 5 23.88 0.764 0.765 0.766 Oil 52.01 15 48.56 19 45.12 Mon. Tues. Wed. Thur. 28 29 30 31 3 6 39.9 3 30 2.4 3 53 20 8 4 16 34.8 + 58.51 58.35 58.17 57.98 6 5.51 4 47.16 4 28.84 4 10.59 765 0.763 0.762 0.760 23 41.67 27 38.22 31 34.78 35 31.33 Frid. 32 K 4 39 44.0 + 57.78 3 52.42 0.756 39 27.88 NAV. AND NALT. ASTR, U 162 NAVIGATIOX AND APRIL, 1898 At Greenwich Apparent Noon 1 S3 B O JS "o >> THE SUN'S Equation of Time, to be Added to Diff. J3 1 Apparent Declination DlflF. for 1 hour Semi- diameter Subtracted from Apparent Time for 1 hour Frid. Sat. SUiV. 1 2 3 N. 4 39 47.7 5 2 51.5 5 25 49.9 + 57.77 57.55 57.31 16 2.10 16 1.82 16 1.55 m. 8. 3 52.37 3 34.29 3 16.33 0.756 0.751 0.745 Mon. Tues. Wed. 4 5 C 5 48 42.5 6 1129.1 6 34 9.3 + 57.06 56.81 56.54 16 1.28 10 1.01 16 0.73 2 58.52 2 40.86 2 23.39 0.739 0.731 0.723 Thur. Frid. Sat. 7 8 9 6 56 42.8 7 19 9.3 7 41 28.4 + 56.25 55.95 55.64 16 0.46 16 0.18 15 59.90 2 6.14 1 49.09 1 32.30 0.714 0.705 0.694 SUN. Mon. Tues. 10 11 12 8 3 30.9 8 25 43.4 8 47 38.7 + 55.31 54.97 54.62 15 59.62 15 59.. 34 15 59.07 1 15.79 59.56 43.64 0.682 0.069 0.656 Wed. Thur. 13 14 15 9 9 25.2 9 31 2.8 9 52 31.0 + 54.25 53.87 53.47 15 58.79 15 58.52 15 58.25 28.06 12.81 0.642 0.628 Frid. 2.08 0.012 Sat. SUiY. Mon. 16 17 18 10 13 49.5 10 34 53.0 10 55 56.0 + 53.06 52.64 52.20 15 57.98 15 57.71 15 57.44 16.60 30.71 44.44 0.596 0.580 0.563 Tues. Wed. Thur. in 20 21 11 16 4.3.4 11 37 19.6 11 57 44.4 + 51.74 51.27 50.79 15 57.18 15 56.92 15 56.66 57.75 1 10.64 123.11 0.646 0.528 0.510 Frid. Sat. SUN-. 22 23 24 12 17 57.3 12 37 58.1 12 57 46.4 + 50.29 49.77 49.24 15 56.40 15 56.15 15 55.90 135.12 1 46.70 1 57.80 0.491 0.472 0.453 Mon. Tues. Wed. 25 26 27 13 17 21.8 13 36 44.1 13 55 52.9 +48.70 48.15 47.58 15 55.65 15 55.41 15 55.17 2 8.45 2 18.63 2 28.33 0.434 0.414 0.394 Thur. Frid. Sat. 28 29 30 14 14 47.8 14 33 28.7 14 51 55.1 +47.00 46.40 45.79 15 54.93 15 54.70 15 54.47 2 37.53 2 46.24 2 54.44 0.373 0.352 0.331 SUK 31 N.15 10 6.8 + 45.18 15 54.24 3 2.13 0.310 II. NAUTICAL ASTRONOMY APRIL, 1898 At Greenwich Mean Noon 163 1 1 c c 1 1 THE SUN'S Equation of Time to be Subtracted from Diff. for 1 hour Sidereal Time, or 5 Apparent Declination Diff. for 1 hour Right Ascension of Mean San 1 Added to Mean Time Frid. Sat. SUiY. 2 3 O f " N. 4 39 44.0 5 2 48.1 5 25 40.8 + 57.78 57.56 57.33 m. s. 3 52.42 3 34.33 3 16.37 0.756 0.751 0.745 h. m. P. 39 27.88 43 24.44 47 20.99 Mon. Tues. Wed. 4 5 6 5 48 39.7 6 11 26.6 6 34 7.0 + 57.08 66.82 56.55 2 58.56 2 40.89 2 23.42 0.739 0.731 0.723 51 17.54 55 14.10 59 10.65 Thur. Frid. Sat. 7 8 9 6 56 40.8 7 19. 7.6 7 41 27.0 + 56.26 55.96 55.65 2 6.16 1 49.11 1 32.32 0.714 0.7U5 0.694 1 3 7.20 1 7 3.76 1 11 0.31 SUN. Mon. Tues. 10 11 12 8 3 38.8 8 25 42.5 8 47 38.0 + 55.32 54.98 54.63 1 15.81 59.57 43.65 0.682 0.669 0.656 1 14 56.86 1 18 53.42 1 22 49.97 Wed. Thur. 13 14 15 9 9 24.8 9 31 2.6 9 52 31.0 + 54.26 53.88 53.48 28.07 12.81 0.642 0.628 0.612 1 26 46.52 1 30 43.08 Frid. 2.U8 1 34 39.63 Sat. SUN Mon. 16 17 18 10 13 49.7 10 31 58.4 10 55 56.7 + 53.07 52.61 52.20 16.60 30.72 44.45 0.596 0.580 0.563 138 36.19 142 32.74 1 46 29.30 Tues. Wed. Thur. 19 20 21 11 16 44.2 1137 20.6 11 57 45.5 + 51.75 51.28 50.79 57.76 1 10.65 1 23.12 0.546 0.528 0.510 1 50 25.85 1 54 22.40 1 58 18.96 Frid. Sat. SUN 22 23 24 12 17 58.6 12 37 59.6 12 57 48.0 + 50.29 49.78 49.25 1 35.13 1 46.71 1 57.82 0.491 0.472 0.453 2 2 15.51 2 6 12.07 2 10 8.62 Mon. Tues. Wed. 25 2« 27 13 17 23.6 13 36 46.0 13 55 54.9 +48.71 48.15 47.58 2 8.47 2 18.65 2 28.35 0.434 0.414 0.394 2 14 5.18 2 18 1.73 2 21 58.29 Thur. Frid. Sat. 28 29 30 14 14 49.9 14 33 30.9 14 51 57.4 + 47.00 46.41 45.80 2 37.55 2 46.26 2 54.46 0.373 0.352 0.331 2 25 54.84 2 29 51.40 2 33 47.95 SUN 31 N. 15 10 9.1 + 45.18 3 2.15 0.310 2 37 44.51 164 NAVIGATION AND MAY, 1898 At Greenwich Apparent Noon I. Xi a o i i 1 THE SUN'S Equation of Time, to be Subtracted from Apparent Time Diff. for 1 hour 1 1 Apparent Declination DiflF. for 1 hour Serai- diameter suy. 1 • '/ N. 15 10 6.8 + 45.18 15 54.24 m. 8. 3 2.13 0.310 Mon. 2 15 28 3.4 44.55 15 54.01 3 9.29 0.288 Tues. 3 15 45 44.7 43.90 15 53.78 3 15.93 0.265 Wed. 4 16 3 10.4 +43.24 15 53.55 3 22.03 0.242 Thur. 5 16 20 20.2 42.57 15 53.32 3 27.56 0.219 Frid. 6 16 37 13.7 41.89 15 53.10 3 32.54 0.195 Sat. 7 16 53 50.8 + 41.19 15 52.87 3 36.94 0.172 SUN-. 8 17 10 11.0 40.48 15 52.65 3 40.78 0.148 Mon. 9 17 26 14.2 39.77 15 52.43 3 44.03 0.124 Tues. 10 17 42 0.0 + 39.04 15 52.21 3 46.70 0.099 Wed. 11 17 57 28.2 38.30 15 51.99 3 48.78 0.075 Thur. 12 18 12 38.3 37.55 15 51.78 3 50.26 0.050 Frid. 13 18 27 30.2 +36.78 15 51.57 3 61.16 0.025 Sat. 14 18 42 3.6 3(5.00 15 51.37 3 51.46 0.000 SUJ>^. 15 18 56 18.1 35.21 15 51.16 3 51.15 0.025 Mon. 16 • 19 10 13.4 + 34.41 15 50.96 3 50.28 0.049 'lues. 17 19 23 49.4 33.59 15 50.76 3 48.82 0.073 Wed. 18 19 37 5.6 32.76 15 50.57 3 46.79 0.096 Thur. 19 19 50 1.8 + 31.92 15 50.38 3 44.19 0.120 Frid. 20 20 2 37.8 31.07 15 50.20 3 41.05 0.143 Sat. 21 20 14 53.2 30.21 15 50.02 3 37.34 0.165 SUN-. 22 20 26 47.9 + 29.34 15 49.85 3 33.12 0.187 Mon. 23 20 38 21 5 28.46 15 49.68 3 28.38 0.208 Tues. 24 20 49 33.8 27.57 15 49.52 3 23.12 0.229 Wed. 25 21 24.7 + 26.67 15 49.36 3 17.37 0.249 Thur. 2(> 21 10 53.8 25.76 15 49.20 3 11.16 0.269 Frid. 27 2121 1.0 24.84 15 49.05 3 4.48 0.288 Sat. 28 21 30 46.0 + 23.91 15 48.91 2 57.34 0.306 SUN-. 20 21 40 8.6 22.98 15 48.77 2 49.77 0.324 Mon. 30 2149 8.8 22.04 15 48.63 2 41.77 0.342 Tues. 31 21 57 46.2 21.08 15 48.49 2 33.36 0.359 Wed. 32 N. 22 6 0.7 + 20.12 15 48.36 2 24.55 0.376 NAUTICAL ASTRONOMY 165 II MAY, 1898 At Greenwich Mean Noon ^ ^ o 1 c 1 THE SUN'S Equation of Time, to be Added to Mean Time Diff. for 1 hour Sidereal Time, or 1 Apparent Declination Diff. for 1 hour Eight Ascension of Mean Sun SUN. Mon. Tues. 1 2 3 N. 15 10 9.1 15 28 5.8 15 45 47.1 +45.18 44.54 43.89 m. s. 3 2.15 3 9.31 3 15.94 0.310 0.288 0.265 h. m. 8. 2 37 44.51 2 41 41.06 2 45 37.62 Wed. Thur. Frid. 4 6 6 16 3 12.8 16 20 22.6 16 37 16.2 +43.24 42.57 41.89 3 22.04 3 27.57 3 32.55 0.242 0.219 0.196 2 49.34.18 2 63 30.73 2 57 27.29 Sat. SUN Mon. 7 8 9 16 53 53.3 17 10 13.5 17 26 16.7 +41.20 40.49 39.77 3 36.95 3 40.79 3 44.04 0.172 0.148 0.124 3 1 23.84 3 6 20.40 3 9 16.95 Tues. Wed. Thur. 10 11 12 17 42 2.5 17 57 30.6 18 12 40.8 +39.04 38.30 37.54 3 46.71 3 48.79 3 50.26 0.099 0.075 0.050 3 13 13.51 3 17 10.07 3 21 6.62 Frid. Sat. SUN. 13 14 15 18 27 32.6 18 42 5.9 18 56 20.4 + 36.77 35.99 36.20 3 51.16 3 51.46 3 51.15 0.025 0.000 0.025 3 25 3.18 3 S8 69.74 3 32 66.29 Mon. Tues. Wed. 16 17 18 19 10 15.7 19 23 51.6 19 37 7.7 +34.40 33.59 32.76 3 50.28 3 48.81 3 46.78 0.049 0.073 0.096 3 36 f 2.85 3 40 49.40 3 44 45.96 Thur. Frid. Sat. 19 20 21 19 50 3.8 20 2 39.7 20 14 55.1 +31.92 31.07 30.21 3 44.18 3 41.04 3 37.33 0.120 0.143 0.165 3 48 42.62 3 52 39.08 3 56 36.63 SUN Mon. Tues. 22 23 24 20 26 49.6 20 38 23.2 20 49 35.4 +29.34 28.46 27.56 3 33.11 3 28.37 3 23.11 0.187 0.208 0.229 4 32.19 4 4 28.75 4 8 25.30 Wed. Thur. Frid. 25 26 27 21 26.2 21 10 55.2 21 21 2.3 + 26.66 25.75 24.83 3 17.36 3 11.14 3 4.46 0.249 0.269 0.288 4 12 21.86 4 16 18.42 4 20 14.98 Sat. SUN Mon. Tues. 28 29 30 31 21 30 47.2 2140 9.8 2149 9.8 2157 47.1 + 23.91 22.98 22.03 21.08 2 57.32 -2 49.75 2 41.75 2 33.34 0.306 0.324 0.342 0.359 4 24 11.53 4 28 8.09 4 32 4.65 4 36 1.21 Wed. 32 N. 22 6 1.6 + 20.12 2 24.63 0.375 4 39 57.76 166 NAVIGATION AND JUNE, 1898 At Greenwich Apparent Noon I, c o o THE SUN'S Equation of Time, to be Subtracted from Diff. for Apparent Declination Diff. for 1 iiour Semi- diameter o 1 Added to Apparent Time Wed. Thur. Frid. 1 2 3 N. 22 6 0.7 22 13 52.2 22 21 20.4 + 20.12 19.16 18.19 15 48.36 15 48.23 15 48.11 m. 8. 2 24.55 2 15.36 2 5.80 0.375 0.390 0.405 Sat. SUN Mou. 4 5 6 22 28 25.3 22 35 Q.Q 22 41 2i.3 + 17.21 16.23 15.24 15 47.98 15 47.86 15 47.74 1 55.87 1 45.59 1 35.00 0.420 0.435 0.448 Tues. Wed. Thur. 7 8 9 22 47 18.1 22 52 48.0 22 57 53.8 + 14.24 13.24 12.24 15.47.62 15 47.51 15 47.40 1 24.00 1 12.88 1 1.39 0.461 0.473 0.485 Frid. Sat. SUN. 10 11 12 23 2 35.3 23 6 52.6 23 10 45.5 + 11.23 10.21 9.19 15 47.29 15 47.19 15 47.09 49.62 37.63 25.41 0.405 0.504 0.513 Mon. Tues. 13 14 15 23 14 13.8 23 17 17.6 23 19 50.7 + 8.17 7.14 6.11 15 47.00 15 46.91 15 46.82 12.98 0.38 0.521 0.528 Wed. 12.37 0.534 Tliur. Frid. Sat. 16 17 18 23 22 11.0 23 24 0.6 23 25 25.4 + 5.08 4.05 3.02 15 46.74 15 46.67 15 46.60 25.26 38.25 51.29 0.538 0.542 0.545 SUN Mon. Tues. 19 20 21 23 26 25.4 23 27 0.6 23 27 10.9 + 1.99 + 0.95 - 0.09 15 46.54 15 46.48 15 46.43 1 4.39 1 17.50 1 30.61 0.546 0.546 0.545 Wed. Thur. Frid. 22 23 24 23 26 56.4 23 26 17.1 23 25 13.0 - 1.12 2.16 3.19 15 46.39 15 46.35 15 46.31 1 43.68 1 56.68 2 9.59 0.643 0.540 0.535 Sat. SUN Mon. 25 26 27 23 23 44.2 23 21 50.8 23 19 32.7 - 4.22 5.24 6.26 15 46.28 15 46.26 15 46.24 2 22.37 2 35.02 2 47.49 0.530 0.524 0.516 Tues. Wed. Thur. 28 29 30 23 16 50.1 23 13 43.0 23 10 11.6 - 7.28 8.30 9.32 15 46.22 15 46.21 15 46.20 2 59.80 3 11.87 3 23.72 0.507 0.498 0.488 Frid. 31 N. 23 6 15.9 -10.32 15 46.19 3 35.32 0.478 NAUTICAL ASTRONOMY 167 II. JUNE, 1898 At Greenwich Mean Noon 1 § « ■5 THE SUN'S Equation of Time, to be Added to Diff. for 1 hour Sidereal Time, or « 5 Apparent Declination Diff. for 1 hour Kight Ascension of Mean Sun o 1 Subtracted from Mean Time Wed. Thur. Frid. 1 2 3 N. 22 6 1.6 22 13 53.0 22 21 21.1 + 20.12 19.16 18.19 m. s. 2 24.53 2 15.34 2 5.78 0.375 0.390 0.405 b. m. s. 4 39 57.76 4 43 54.32 4 47 50.88 Sat. SUN. Men. 4 5 6 22 28 25.9 22 35 7.1 22 41 24.7 + 17.21 16.22 15.23 1 55.86 1 45.58 1 34.99 0.420 0.435 0.448 4 51 47.44 4 65 43.99 4 59 40.55 Tues. Wed. Thur. 7 8 9 22 47 18.4 22 52 48.2 22 57 54.0 + 14.24 13.24 12.23 1 24.08 1 12.87 1 1.38 0.461 0.473 0.485 5 3 37.11 5 7 33.07 5 11 30.23 Frid. Sat. SUN. 10 11 12 23 2 35.5 23 6 52.7 23 10 45.6 + 11.22 10.21 9.19 49.61 37.62 25.40 0.495 0.504 0.513 5 15 26.78 5 19 23. .34 5 23 19.90 Men. Tues. 18 14 15 23 14 13.9 23 17 17.6 23 19 56.7 + 8.17 7.14 6.11 12.98 0.38 0.521 0.528 0.534 5 27 16.46 5 31 13.02 Wed. U 12.37 6 35 9.58 Thur. Frid. Sat. 16 17 18 23 22 11.0 23 24 0.6 23 25 25.4 + 5.08 4.05 3.02 25.26 38.24 51.28 0.538 0.542 0.545 5 39 6.13 5 43 2.69 5 46 59.25 SUN. Mod. Tues. 19 20 21 23 26 25.4 23 27 0.6 23 27 10.9 + 1.98 + 0.94 - 0.09 1 4.38 1 17.49 1 30.60 0.546 0.546 0.545 5 50 55.81 5 54 52.37 5 58 48.92 Wed. Thur. Frid. 22 23 24 23 26 56.4 23 26 17.2 23 25 13.1 - 1.12 2.15 3.18 1 43.66 1 56.66 2 9.57 0.543 0.540 0.535 6 2 45.48 6 6 42.04 6 10 38.60 Sat. Men. 25 26 27 23 23 44.4 23 21 51.0 23 19 33.0 - 4.21 5.24 6.26 2 22.35 2 35.00 2 47.47 0.530 0.524 0.516 6 14 35.16 6 18 31.72 6 22 28.28 Tues. Wed. Thur. 28 29 30 23 16 50.5 23 13 43.5 23 10 12.1 - 7.28 8.30 9.31 2 59.77 3 11.84 3 23.69 0.507 0.498 0.488 6 26 24.83 6 30 21.39 6 34 17.95 Frid. 31 N. 23 6 16.5 -10.32 3 35.29 0.478 6 38 14.51 168 NAVIGATION AND FIXED STARS, 1898 Mean Places for the Beginning of Name of Star Mag- nitude Right Ascension Annual Variation Declination Annual Variation a Ursae Min. {Polaris) 2 h. m. s. 1 21 43.70 + 24!832 • " + 88 45 49.1 + 18.79 a Tauri (^Aldebaran) 1 4 30 4.02 3.438 + 16 18 15.0 + 7.48 a Aurigae (Capella) 5 9 9.19 4.426 +45 53 38.7 + 3.98 /S Orionis (Bigel) 6 9 38.13 2.882 - 8 19 10.5 + 4.36 a Orionis (var.) 5 49 38.96 3.247 + 7 23 16.6 + 0.91 a Canis Maj. {Sirius) 6 40 39.21 2.644 -16 34 34.6 - 4.74 a Canis Min. {Procyon) 7 33 67.77 3.143 + 6 29 10.7 - 9.03 /5 Gerainorum {Pollux) 7 39 4.63 3.678 +28 16 20.9 - 8.46 a Leonis {Reguliis) 10 2 66.43 3.200 + 12 27 56.5 -17.49 a Virginis {Spica) 13 19 49.10 3.166 -10 37 44.5 -18.89 a Bootis {Arcturus) 14 11 0.53 2.735 + 19 42 48.1 -18.86 a Scorpii {Antares) 16 23 9.13 3.672 -26 12 20.5 - 8.26 aLyrsB (Vega) 18 33 29.12 2.031 + 38 41 18.8 + 3.19 a AquilsB {Altair) 19 46 48.41 2.927 + 8 36 65.7 + 9.30 NAUTICAL ASTRONOMY 169 Table for Finding the Latitude by an Observed Altitude of Polaris Reduce the observed altitude of Polaris to the true altitude. Reduce the recorded time of observation to the local sidereal time. less than 1 h. 21.8 m., subtract it from 1 h. 2L8 m. ; between 1 h. 21.8 m. and 13 h. 21.8 m., sub- tract 1 h. 21.8 m. from it; greater than 13 h. 21.8 m., subtract it from 25h. 21.8 m. ; and the remainder is the hour angle of Polaris. With this hour angle, take out the correction from Table (next page), and add it to or subtract it from the true altitude,»according to its sign. The result is the approximate latitude of the place. Example. —1898, Oct. 1, at 10 h. 40 m. 30 s. p.m., mean solar time, in longitude 29° east of Greenwich, suppose the true altitude of Polaris to be 43° 20' ; required the latitude of the place. If the sidereal time is h. in. 8. 10 40 30 + 145 12 40 58 - 19 23 22 54 Local astronomical mean time Reduction for 10 h. 40. m. 30 s Greenwich sidereal time for mean noon, Oct. 1 . Reduction for longitude ( = 11). 56 m. east, or minus), Sum (having regard to signs) is equal to local sidereal time b. m. 8. 25 2148 Subtract sidereal time 23 22 54 Remainder is equal to hour angle of Polaris . . 1 58 54 True altitude -I- 43 20 Correction from table (next page), -14 Approximate latitude . . . -f 42 16 170 NAVIGATION AND NAUTICAL ASTRONOMY JB ^ »0 O ® <© CO « O «0 CO « «0 t-; n^ 05 W 1-; r^ lO C. 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I- I- o o o o 1 1 1 1 1 1 CO-^O^OO'tH — OQOO ->1<'*rtiCOC0o:)CO(M(>4(M 1 1 1 1 1 1 1 1 1 1 vO(N — OSXCOOrtlCOO? 1 1 1 1 1 1 i 1 1 1 '« 1 1 1 1 1 1 Ot^COOt- hCOOCSOO uOTtiTti-i*COCOCOCO(N**CO 1 1 1 1 1 1 1 1 1 1 °^ Ci -t^ 00 CO oo CO . l^ t- O CO O uO i 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 i 1 %> 5D r^ -^ r^ 00 CO . t- t- t^ •£ o o 1 1 1 1 1 1 c-^^^ao — oo>'^-*'-^i— »Ci lOO-^Tt^COCOCOCOCO-M 1 1 1 1 1 1 1 1 1 1 r-xcooi'-'--x?oot^ OJi<0?0> iO-rJirt<'^rJ0 W05eoeoeoeoeoeoc9'«« NAUTICAL ASTRONOMY 173 iHeteo^iACOt^aooo ^ ,«< ^ ^ Tl4 * Tfi -^ uo o o o o X) CO <:o to I- I- t- CO CO CO c: cr. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + OOOCiO'MCO-tiOt--GO O^COTjH::;cOCi— 'CO'^'tJ t-O^MOCOi-i'-tCO'Mt^ 'N (M Ol CO CO CO CO CO CO CO rf "* -* ^r rr -^r '^ O O i-O i-O O O O CC t- t^ t^ CO QC' + + + + + + + -f + + + + + + + + + + + + + + + + + + + + + + %0 1-i 'Mco-^->cr^coc:c:ot^Ocooc:coi^ S* -t^ 'rti -^ O O »-0 O CC <© CC ++++++++++++++++++++++++++++++ ^ r^ -^t.OXil^OiCJ^'NC0-+i OOOOOO'-tCO-fOt^ OO-M-^COOOCCOCCCi ^^^^^r-i(^'7^7^(^ C^ C<« C^ M CO CO CO CO CO CO CO -^ -+ -^ ^ rf tC O UO O + + + + + + + + + + + + + + + + ++ + + +++ + + + + + + + s ©(Mo^-^oiot-cor-o 1-H Ol CO -t* '0 t^ CO C^ O 'N CO >0) O CO C 0 1 1 1 1 1 1 ++ + + + + + + + + + + % ^ ^ ^ o cr. c: CO I- -j:! o 7777 1 1 1 1 1 1 >-0-!tiOC03-0 77777 1 1 1 1 1 <©!0 0-<:t.OCO'^-^COCOiOtOiO>O<0 ssssss^ssg 13 Text-Books in Trigonometry CROCKETT'S ELEMENTS OF PLANE AND SPHERICAL TRIGONOMETRY AND TABLES. 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