《航海学》课程参考文献（地文资料）CHAPTER 24 THE SAILINGS

CHAPTER 24THE SAILINGSINTRODUCTION2400.Introductionof these vertices the direction changes progressively untilthe intersection with the equator isreached,90°in longitudeDead reckoning involves the determination of one'saway,where the great circle crosses the equator at an anglepresent orfutureposition byprojectingthe ship's courseequal tothe latitudeofthevertex.and distancerun from aknownposition.Aclosely relatedOn a Mercator chart a great circle appears as a sineproblem is that of finding the course and distancefrom onecurve extending equal distances each side of the equator.Therhumb line connecting any two points of thegreat cir-knownpointtoanotherknownpoint.Forshortdistancestheseproblems areeasilysolved directlyon charts, but forcle on the same side of the equator is a chord of the curve.longdistances,apurelymathematical solution is oftenaAlong any intersecting meridian thegreat circle crosses atbetter method.Collectively,these methods are called Thea higher latitude than the rhumb line.If the two points areSailings.on oppositesides of the equator,thedirection ofcurvatureNavigational computerprogramsandcalculatorscomof thegreat circle relative to the rhumb line changes at themonlycontainalgorithmsforcomputingalloftheproblemsequator.The rhumb line and greatcircle may intersecteachof thesailings.For those situations whenacalculator is notother,and if the points are equal distances on each side ofavailable,this chapter also discusses sailing solutions bythe equator, the intersection takes place at the equator.Table 4, the Traverse Tables.Greatcirclesailingtakesadvantageoftheshorterdistancealong thegreatcirclebetweentwo points,ratherthan2401.RhumbLinesAndGreatCirclesthelongerrhumbline.Thearcofthegreatcirclebetweenthe points is called the great circle track.If it could befol-The principal advantage of a rhumb line is that it main-lowed exactly,the destination would be dead aheadtains constant true direction.A shipfollowing therhumbthroughout the voyage (assuming course and heading werelinebetween two places does not changetruecourse.Athe same).The rhumb line appears the more direct route onrhumb line makes the same angle with all meridians ita Mercator chart because of chart distortion.The great cir-crosses and appears as a straight lineon a Mercator chart.clecrossesmeridiansathigherlatitudes.wherethedistanceFor any other case, the difference between the rhumb linebetween them is less.This is why the great circle route isand thegreat circle connecting two points increases()asshorter than the rhumb line.the latitude increases, (2)as the difference of latitude be-The decision as to whether or not to use great-circletween the two points decreases,and (3)as the difference ofsailing depends upon the conditions.The saving in distancelongitudeincreases.should be worth the additional effort, and of course the greatA great circle is the intersection of the surface of acircleroutecannot cross land, nor should it carry the vesselsphereand aplanepassingthrough thecenterofthesphereintodangerous waters.Composite sailing(see section2402It is thelargest circlethat can bedrawn on the surface oftheand section 2411)may save time and distance over thesphere, and is the shortest distance along the surface be-rhumb linetrack withoutleading thevessel intodangerSince great circles other than a meridian or the equatortweenanytwopoints.Anytwopointsareconnectedbyonly one great circle unless the points are antipodal (1800are curved lines whose true direction changes continuallyapart on the earth), and then an infinite number ofgreat cir-the navigator does not attempt to follow it exactly.Rather,he selects a number of points along the great circle, con-clespasses throughthem.Everygreatcirclebisects everyother greatcircle.Thus,exceptfor the equator,everygreatstructs rhumb lines between the points, and follows thesecirclelies exactly half in the Northern Hemisphere and halfrhumb lines frompointtopoint.inthe SouthernHemisphere.Anytwopoints180°apart on2402. Kinds Of Sailingsa great circle have the same latitude numerically,but con-trary names, and are180°apart in longitude.The point ofgreatest latitude is called the vertex. For each great circle,There are seven types of sailings:there is a vertex in each hemisphere,180° apart in longi-1.Plane sailing solves problems involving a singletude.At these points thegreat circle is tangent to a paralleloflatitude.anditsdirectionisdueeast-west.Oneachsidecourse and distance, difference of latitude,and de-355

355 CHAPTER 24 THE SAILINGS INTRODUCTION 2400. Introduction Dead reckoning involves the determination of one’s present or future position by projecting the ship’s course and distance run from a known position. A closely related problem is that of finding the course and distance from one known point to another known point. For short distances, these problems are easily solved directly on charts, but for long distances, a purely mathematical solution is often a better method. Collectively, these methods are called The Sailings. Navigational computer programs and calculators com￾monly contain algorithms for computing all of the problems of the sailings. For those situations when a calculator is not available, this chapter also discusses sailing solutions by Table 4, the Traverse Tables. 2401. Rhumb Lines And Great Circles The principal advantage of a rhumb line is that it main￾tains constant true direction. A ship following the rhumb line between two places does not change true course. A rhumb line makes the same angle with all meridians it crosses and appears as a straight line on a Mercator chart. For any other case, the difference between the rhumb line and the great circle connecting two points increases (1) as the latitude increases, (2) as the difference of latitude be￾tween the two points decreases, and (3) as the difference of longitude increases. A great circle is the intersection of the surface of a sphere and a plane passing through the center of the sphere. It is the largest circle that can be drawn on the surface of the sphere, and is the shortest distance along the surface be￾tween any two points. Any two points are connected by only one great circle unless the points are antipodal (180° apart on the earth), and then an infinite number of great cir￾cles passes through them. Every great circle bisects every other great circle. Thus, except for the equator, every great circle lies exactly half in the Northern Hemisphere and half in the Southern Hemisphere. Any two points 180° apart on a great circle have the same latitude numerically, but con￾trary names, and are 180° apart in longitude. The point of greatest latitude is called the vertex. For each great circle, there is a vertex in each hemisphere, 180° apart in longi￾tude. At these points the great circle is tangent to a parallel of latitude, and its direction is due east-west. On each side of these vertices the direction changes progressively until the intersection with the equator is reached, 90° in longitude away, where the great circle crosses the equator at an angle equal to the latitude of the vertex. On a Mercator chart a great circle appears as a sine curve extending equal distances each side of the equator. The rhumb line connecting any two points of the great cir￾cle on the same side of the equator is a chord of the curve. Along any intersecting meridian the great circle crosses at a higher latitude than the rhumb line. If the two points are on opposite sides of the equator, the direction of curvature of the great circle relative to the rhumb line changes at the equator. The rhumb line and great circle may intersect each other, and if the points are equal distances on each side of the equator, the intersection takes place at the equator. Great circle sailing takes advantage of the shorter dis￾tance along the great circle between two points, rather than the longer rhumb line. The arc of the great circle between the points is called the great circle track. If it could be fol￾lowed exactly, the destination would be dead ahead throughout the voyage (assuming course and heading were the same). The rhumb line appears the more direct route on a Mercator chart because of chart distortion. The great cir￾cle crosses meridians at higher latitudes, where the distance between them is less. This is why the great circle route is shorter than the rhumb line. The decision as to whether or not to use great-circle sailing depends upon the conditions. The saving in distance should be worth the additional effort, and of course the great circle route cannot cross land, nor should it carry the vessel into dangerous waters. Composite sailing (see section 2402 and section 2411) may save time and distance over the rhumb line track without leading the vessel into danger. Since great circles other than a meridian or the equator are curved lines whose true direction changes continually, the navigator does not attempt to follow it exactly. Rather, he selects a number of points along the great circle, con￾structs rhumb lines between the points, and follows these rhumb lines from point to point. 2402. Kinds Of Sailings There are seven types of sailings: 1. Plane sailing solves problems involving a single course and distance, difference of latitude, and de-

356THE SAILINGSparture, in which the earth is regarded as a planeused:surface.This method,therefore,provides solutionforlatitudeofthepointofarrival, but notforlongi-1.Latitude (L).The latitude of the point of departuretude. To calculate the longitude, the sphericalisdesignated L,that of the destination,L,middlesailings are necessary.Do not use this method for(mid)ormeanlatitude,Lm,latitudeofthevertexofdistances ofmore than a few hundred miles.a great circle, Ly, and latitude of any point on a2.Traverse sailing combines the plane sailing solu-great circle, Lxtions when there are two or more courses and2. Mean latitude (Lm).Half the arithmetical sum of thedetermines the equivalent course and distancelatitudesoftwoplacesonthesamesideoftheequator.madegoodbyavesselsteamingalongaseriesof3. Middle or mid latitude (Lm).The latitude atrhumblines3Parallel sailing is the interconversion ofdeparturewhich the arc length of the parallel separating theand difference of longitude when a vessel ispro-meridians passing through two specific points isceeding due east or due west.exactlyequal tothedeparture inproceedingfrom4Middle- (or mid-) latitude sailing uses the mean lat-one point to the other. The mean latitude is useditude for converting departureto difference ofwhen there is no practicable means of determininglongitudewhen the course is not due east orduewestthemiddle latitude.5.Mercatorsailingprovidesamathematical solution4. Difference of latitude (I or DLat.).oftheplotasmadeonaMercator chart.It issimilar5. Meridional parts (M).The meridional parts ofthetoplane sailing,butuses meridional difference andpoint of departure are designated M, and of thedifferenceof longitude inplaceof differenceof lat-pointofarrival or the destination, M2itude and departure.Great circle sailing involves the solution of cours-6. Meridional difference (m).6es, distances, and points along a great circle7. Longitude (2). The longitude of the point ofdepar-betweentwopoints.ture is designated that of the point of arrival or7.Composite sailing is a modification of great-circlethedestination,2,ofthevertex ofagreatcircle,lvsailingto limitthe maximumlatitude,generallytoand of any point on a great circle, xavoid ice or severe weather near the poles.8.Difference of longitude (DLo)9. Departure (p or Dep.).2403.TermsAnd Definitions10. Course or course angle (Cn or C).In solutions ofthe sailings, thefollowing quantities are11. Distance (D or Dist.)GREAT CIRCLE SAILING2404.GreatCircle SailingByChartparture replaces the assumed position of the observer, thedestination replaces thegeographical position of the body,difNavigators can most easily solve great-circle sailingferenceoflongitude replacesmeridian angleor local hourangleproblems graphically.DMAHTC publishes several gno-initial course angle replaces azimuth angle,and great circle dis-monicprojections coveringtheprincipal navigablewaterstance replaces zenith distance (90-altitude).SeeFigure2405oftheworld.Onthesegreatcirclecharts,anystraightlineTherefore,anytable ofazimuths (ifthe entering values areme-is a great circle.The chart, however, is not conformal;ridian angle, declination, and latitude) can be used fortherefore, the navigator cannot directlymeasure directionsdetermining initial great-circle course. Tables which solve foranddistancesas on a Mercator chart.altitude,such asPub.No.229,canbeusedfor determininggreatTheusual method of using agnomonic chart is to plotcircle distance.The required distance is 90- altitudethe route and pick points along thetrack every5of longi-In inspection tables such as Pub. No. 229, the giventude using the latitude and longitude scales in the immediatecombination of L,L2,and DLo may not be tabulated. Invicinityofeachpoint.Thesepointsarethentransferredtoathis casereverse thename ofLanduse180-DLofor en-Mercator chart and connected by rhumb lines.The coursetering the table.The required course angle is then 1800and distanceforeach leg is measuredontheMercator chartminus the tabulated azimuth,anddistance is 90°plus the al-SeeChapter25fora discussionof thisprocess.titude.If neither combinationcanbefound, solutioncannotbe made by that method.By interchanging L, and L2, one2405.Great Circle Sailing By Sight ReductionTablescan find the supplement of the final course angleAny method of solving a celestial spherical triangle can beSolution by table often provides a rapid approximateused for solving great circle sailing problems.The point of de-check, but accurate results usually require triple interpola-

356 THE SAILINGS parture, in which the earth is regarded as a plane surface. This method, therefore, provides solution for latitude of the point of arrival, but not for longi￾tude. To calculate the longitude, the spherical sailings are necessary. Do not use this method for distances of more than a few hundred miles. 2. Traverse sailing combines the plane sailing solu￾tions when there are two or more courses and determines the equivalent course and distance made good by a vessel steaming along a series of rhumb lines. 3. Parallel sailing is the interconversion of departure and difference of longitude when a vessel is pro￾ceeding due east or due west. 4. Middle- (or mid-) latitude sailing uses the mean lat￾itude for converting departure to difference of longitude when the course is not due east or due west. 5. Mercator sailing provides a mathematical solution of the plot as made on a Mercator chart. It is similar to plane sailing, but uses meridional difference and difference of longitude in place of difference of lat￾itude and departure. 6. Great circle sailing involves the solution of cours￾es, distances, and points along a great circle between two points. 7. Composite sailing is a modification of great-circle sailing to limit the maximum latitude, generally to avoid ice or severe weather near the poles. 2403. Terms And Definitions In solutions of the sailings, the following quantities are used: 1. Latitude (L). The latitude of the point of departure is designated Ll ; that of the destination, L2; middle (mid) or mean latitude, Lm; latitude of the vertex of a great circle, Lv; and latitude of any point on a great circle, Lx. 2. Mean latitude (Lm). Half the arithmetical sum of the latitudes of two places on the same side of the equator. 3. Middle or mid latitude (Lm). The latitude at which the arc length of the parallel separating the meridians passing through two specific points is exactly equal to the departure in proceeding from one point to the other. The mean latitude is used when there is no practicable means of determining the middle latitude. 4. Difference of latitude (l or DLat.). 5. Meridional parts (M). The meridional parts of the point of departure are designated Ml , and of the point of arrival or the destination, M2. 6. Meridional difference (m). 7. Longitude (λ). The longitude of the point of depar￾ture is designated λ1; that of the point of arrival or the destination, λ2; of the vertex of a great circle, lv; and of any point on a great circle, λx 8. Difference of longitude (DLo). 9. Departure (p or Dep.). 10. Course or course angle (Cn or C). 11. Distance (D or Dist.). GREAT CIRCLE SAILING 2404. Great Circle Sailing By Chart Navigators can most easily solve great-circle sailing problems graphically. DMAHTC publishes several gno￾monic projections covering the principal navigable waters of the world. On these great circle charts, any straight line is a great circle. The chart, however, is not conformal; therefore, the navigator cannot directly measure directions and distances as on a Mercator chart. The usual method of using a gnomonic chart is to plot the route and pick points along the track every 5° of longi￾tude using the latitude and longitude scales in the immediate vicinity of each point. These points are then transferred to a Mercator chart and connected by rhumb lines. The course and distance for each leg is measured on the Mercator chart. See Chapter 25 for a discussion of this process. 2405. Great Circle Sailing By Sight Reduction Tables Any method of solving a celestial spherical triangle can be used for solving great circle sailing problems. The point of de￾parture replaces the assumed position of the observer, the destination replaces the geographical position of the body, dif￾ference of longitude replaces meridian angle or local hour angle, initial course angle replaces azimuth angle, and great circle dis￾tance replaces zenith distance (90° - altitude). See Figure 2405. Therefore, any table of azimuths (if the entering values are me￾ridian angle, declination, and latitude) can be used for determining initial great-circle course. Tables which solve for altitude, such as Pub. No. 229, can be used for determining great circle distance. The required distance is 90° - altitude. In inspection tables such as Pub. No. 229, the given combination of L1, L2, and DLo may not be tabulated. In this case reverse the name of L2 and use 180° - DLo for en￾tering the table. The required course angle is then 180° minus the tabulated azimuth, and distance is 90° plus the al￾titude. If neither combination can be found, solution cannot be made by that method. By interchanging L1 and L2, one can find the supplement of the final course angle. Solution by table often provides a rapid approximate check, but accurate results usually require triple interpola-

357THESAILINGSPnPnLHADistaPnPnGP(LgA)PLANEOFCELESTIALEQUATORFigure 2405.Adapting the astronomical triangleto the navigational triangle of great circle sailingtion.ExceptforPub.No.229,inspectiontablesdonotcomes thedistance;the supplementof thetabularazimuthprovide a solution for points alongthegreat circle.Pub.No.angle becomes the initial great-circle course angle229provides solutionsforthesepointsonlyif interpolationWhentheContrary/Same(CS)Lineiscrossed ineitheris not required.direction, the altitude becomes negative; the body lies be-low the celestial horizon.For example:If thetables are2406.Great Circle Sailing ByPub.No.229entered with the LHA (DLo)at the bottom of a right-handpage and declination (L2) such that the respondents lieByenteringPub.No.229with the latitudeof the pointabove the CS Line, the CS Line has been crossed. Then thedistance is 90° plus the tabular altitude; the initial courseof departureas latitude,latitude of destination as declina-tion, and difference of longitude as LHA, the tabularangle isthe supplement of thetabular azimuth angle.Simi-altitudeand azimuth anglemaybeextracted and convertedlarly, ifthe tables are entered with the LHA (DLo)at the topto great-circle distance and course. As in sight reduction,of a right-hand page and the respondents arefound belowthe tables are entered according to whether the name of thethe CS Line,the distance is 90°plus the tabular altitude; thelatitude of thepointof departure is the same as or contraryinitial course angleisthe supplement ofthe tabularazimuthto the name of the latitude of the destination (declination)angle. If the tables are entered with the LHA (DLo) at theIf thevalues correspond to thoseofa celestial bodyabovebottom of aright-hand page and the name of L, is contrarythe celestial horizon, 90°minus the arc of thetabular alti-to Li, the respondents are found in the column for L on thetude becomes the distance,the tabular azimuth anglefacing page. In this case, the Cs Line has been crossed, thebecomes the initial great-circle course angle.If the respon-distance is 9oplus thetabular altitude;the initial coursedents correspond to those of a celestial body below theangleisthe supplementof thetabular azimuth angle.celestial horizon, the arc ofthe tabular altitude plus 90°be-The tabular azimuth angle, or its supplement, is pre-

THE SAILINGS 357 tion. Except for Pub. No. 229, inspection tables do not provide a solution for points along the great circle. Pub. No. 229 provides solutions for these points only if interpolation is not required. 2406. Great Circle Sailing By Pub. No. 229 By entering Pub. No. 229 with the latitude of the point of departure as latitude, latitude of destination as declina￾tion, and difference of longitude as LHA, the tabular altitude and azimuth angle may be extracted and converted to great-circle distance and course. As in sight reduction, the tables are entered according to whether the name of the latitude of the point of departure is the same as or contrary to the name of the latitude of the destination (declination). If the values correspond to those of a celestial body above the celestial horizon, 90° minus the arc of the tabular alti￾tude becomes the distance; the tabular azimuth angle becomes the initial great-circle course angle. If the respon￾dents correspond to those of a celestial body below the celestial horizon, the arc of the tabular altitude plus 90° be￾comes the distance; the supplement of the tabular azimuth angle becomes the initial great-circle course angle. When the Contrary/Same (CS) Line is crossed in either direction, the altitude becomes negative; the body lies be￾low the celestial horizon. For example: If the tables are entered with the LHA (DLo) at the bottom of a right-hand page and declination (L2) such that the respondents lie above the CS Line, the CS Line has been crossed. Then the distance is 90° plus the tabular altitude; the initial course angle is the supplement of the tabular azimuth angle. Simi￾larly, if the tables are entered with the LHA (DLo) at the top of a right-hand page and the respondents are found below the CS Line, the distance is 90° plus the tabular altitude; the initial course angle is the supplement of the tabular azimuth angle. If the tables are entered with the LHA (DLo) at the bottom of a right-hand page and the name of L2 is contrary to L1, the respondents are found in the column for L1 on the facing page. In this case, the CS Line has been crossed; the distance is 90° plus the tabular altitude; the initial course angle is the supplement of the tabular azimuth angle. The tabular azimuth angle, or its supplement, is pre￾Figure 2405. Adapting the astronomical triangle to the navigational triangle of great circle sailing

358THE SAILINGSfixed N or S for the latitudeof thepoint of departureandbecomes the distance:the supplement of tabularazimuthangle (180°-69.00=111.0°)becomessuffixedEorWdependinguponthedestinationbeingeastorwestofthepointofdeparturethe initial great circle course angle, prefixed NIf all entering arguments are integral degrees, the dis-for the latitude of the point of departure and suffixed W since the destination is west of the pointtanceand courseangle areobtained directlyfromthetablesofdeparture.without interpolation.Ifthelatitudeofthedestination isNote that the data is extracted from across the CS Linenonintegral,interpolationfortheadditionalminutes of lati-tude is done as in correcting altitude for any declinationfrom the entering argument (LHA 87°), indicatingincrement, if thelatitude of departureordifference of lon-thatthecorrespondingcelestial bodywouldbebe-gitude is nonintegral, theadditional interpolation is donelowthecelestial horizongraphicallySincethelatitudeofdestinationbecomesthedeclina-Answer:tion entry, and all declinations appear on every page, theD=6137nautical milesgreat circle solution can always be extracted from the vol-C= NII1.0°W = 249°ume which covers the latitude of thepoint ofdeparture.2407.Great Circle Sailing By ComputationExampleI:UsingPub.No.229findthedistanceandinitial great circlecoursefrom lat.32°S, long.InFigure2407.1 is thepoint of departure.2thedesti-I16°E to lat. 30°S, long. 31°E.nation, Pthe pole nearer 1,I-X-V-2 the great circle through1 and 2, V the vertex, and X any point on the great circle.Solution:RefertoFigure2405.ThepointofdepartureThe arcsP1,PX,PV,andP2are the colatitudes ofpoints1,(lat.32°S, long.116°E)replaces theAPof theob-X,V,and2,respectively.If1and2areon oppositesidesofserver; the destination (lat. 30°S, long. 31°E)theequator,P2 is 90°+L2.The length ofarc 1-2 is thegreat-replacestheGPof the celestial body;thediffercircle distance between 1 and 2.Arcs 1-2,P1,and P2formence of longitude(DLo 85°)replaces local houra spherical triangle.The angle at 1 is the initial great-circleangle (LHA) of the body.coursefrom1to2,thatat2thesupplement of thefinalEnterPub.229,Volume3with lat.32°(SameName),great-circle course (or the initial course from 2 to 1), andLHA85°,anddeclination 30°.TherespondentsthatatPtheDLobetween1and2correspond to a celestial bodyabove the celestialGreat circle sailing by computation usually involveshorizon.Therefore,90°minus the tabular altitudesolving for the initial great circle course; the distance; lati-(900-19°12.4'=70°47.6)becomesthedistancetude and longitude,and sometimes the distance, of thethe tabular azimuth angle (S66.0°W) becomes thevertex,and thelatitudeand longitudeofvarious points (X)oninitialgreatcirclecourseangle,prefixedSforthethegreat circle.The computation for initial courseand thelatitudeof thepoint of departureand suffixedWdistance involves solution of an oblique spherical triangle,due to the destination being west of the point ofdeparture.Answer:D=4248nauticalmilesC=S66.0°W=246.0°Example2:UsingPub.No.229findthedistanceandinitial great circle course from lat. 38°N, long.122°Wto lat.24°s, long.151°E.VSolution:RefertoFigure 2405.The point of departure(lat. 38°N, long.122°W) replaces the AP of the ob-server:thedestination(lat.24°S, long.151°E)replaces theGPof thecelestial body:thedifferenceof longitude(DLo 87)replaces local houEQUATORangle (LHA)ofthebodyEnter Pub.No.229Volume3with lat.38°(ContraryName), LHA 87°, and declination 24°. The re-spondentscorrespondtothoseofacelestialbodyFigure 2407.The navigational triangle and great circlebelow the celestial horizon. Therefore,the tabu-laraltitudeplus90°(12°17.0+90°=102°17.0)sailing

358 THE SAILINGS fixed N or S for the latitude of the point of departure and suffixed E or W depending upon the destination being east or west of the point of departure. If all entering arguments are integral degrees, the dis￾tance and course angle are obtained directly from the tables without interpolation. If the latitude of the destination is nonintegral, interpolation for the additional minutes of lati￾tude is done as in correcting altitude for any declination increment; if the latitude of departure or difference of lon￾gitude is nonintegral, the additional interpolation is done graphically. Since the latitude of destination becomes the declina￾tion entry, and all declinations appear on every page, the great circle solution can always be extracted from the vol￾ume which covers the latitude of the point of departure. Example 1: Using Pub. No. 229 find the distance and initial great circle course from lat. 32°S, long. 116°E to lat. 30°S, long. 31°E. Solution: Refer to Figure 2405. The point of departure (lat. 32°S, long. 116°E) replaces the AP of the ob￾server; the destination (lat. 30°S, long. 31°E) replaces the GP of the celestial body; the differ￾ence of longitude (DLo 85°) replaces local hour angle (LHA) of the body. Enter Pub. 229, Volume 3 with lat. 32° (Same Name), LHA 85°, and declination 30°. The respondents correspond to a celestial body above the celestial horizon. Therefore, 90° minus the tabular altitude (90° - 19°12.4’ = 70°47.6’) becomes the distance; the tabular azimuth angle (S66.0°W) becomes the initial great circle course angle, prefixed S for the latitude of the point of departure and suffixed W due to the destination being west of the point of departure. Answer: D = 4248 nautical miles C = S66.0°W = 246.0°. Example 2: Using Pub. No. 229 find the distance and initial great circle course from lat. 38°N, long. 122°W to lat. 24°S, long. 151°E. Solution: Refer to Figure 2405. The point of departure (lat. 38°N, long. 122°W) replaces the AP of the ob￾server; the destination (lat. 24°S, long. 151°E) replaces the GP of the celestial body; the differ￾ence of longitude (DLo 87°) replaces local hour angle (LHA) of the body Enter Pub. No. 229 Volume 3 with lat. 38° (Contrary Name), LHA 87°, and declination 24°. The re￾spondents correspond to those of a celestial body below the celestial horizon. Therefore, the tabu￾lar altitude plus 90° (12°17.0’ + 90° = 102°17.0’) becomes the distance; the supplement of tabular azimuth angle (180° - 69.0° = 111.0°) becomes the initial great circle course angle, prefixed N for the latitude of the point of departure and suf￾fixed W since the destination is west of the point of departure. Note that the data is extracted from across the CS Line from the entering argument (LHA 87°), indicating that the corresponding celestial body would be be￾low the celestial horizon. Answer: D = 6137 nautical miles C = N111.0°W = 249°. 2407. Great Circle Sailing By Computation In Figure 2407, 1 is the point of departure, 2 the desti￾nation, P the pole nearer 1, l-X-V-2 the great circle through 1 and 2, V the vertex, and X any point on the great circle. The arcs P1, PX, PV, and P2 are the colatitudes of points 1, X, V, and 2, respectively. If 1 and 2 are on opposite sides of the equator, P2 is 90°+ L2. The length of arc 1-2 is the great￾circle distance between 1 and 2. Arcs 1-2, P1, and P2 form a spherical triangle. The angle at 1 is the initial great-circle course from 1 to 2, that at 2 the supplement of the final great-circle course (or the initial course from 2 to 1), and that at P the DLo between 1 and 2. Great circle sailing by computation usually involves solving for the initial great circle course; the distance; lati￾tude and longitude, and sometimes the distance, of the vertex; and the latitude and longitude of various points (X) on the great circle. The computation for initial course and the distance involves solution of an oblique spherical triangle, Figure 2407. The navigational triangle and great circle sailing

359THE SAILINGSAnswer:andanymethodof solvingsuchatrianglecanbeused.If2isthegeographical position (GP)ofacelestial body (thepointD (NM)3006009006600at which the body is in the zenith), this triangle is solved in501001501100D (arc)celestial navigation, except that 90o- D (the altitude) is de-85°80°75°20decsired instead of D.The solution for thevertexand anypoint39.6°N40.9°N41.9°N3.1°NLat.X usually involves the solution ofright spherical trianglesDep.125°W125°W125°W125°W6.1018.90118.50DLo12.402408.Points Along The Great Circle131.1°W137.4°W143.9°W116.5°ELong2409.FindingTheVertexIf the latitude of thepoint of departure and the initialgreat-circle course angleare integral degrees,points alongUsing Pub. No.229 to find the approximate position ofthe greatcircle are found by entering the tables with the lat-the vertex of a great circle track provides a rapid check onitude of departure as the latitude argument (always SameName), the initial great circle course angle as the LHA ar-thesolutionbycomputation.Thisapproximatesolutionisalso useful for voyage planning purposes.gument, and 90o minus distance to a point on the greatUsingtheproceduresforfindingpoints alongthegreatcircle as the declination argument. The latitude of the pointcircle,inspectthecolumnofdataforthelatitudeoftheon thegreat circle and the differenceof longitudebetweenpoint of departure and find the maximum value of tabularthat pointand thepointofdeparture arethetabularaltitudealtitude.This maximum tabular altitude and thetabular az-and azimuth angle, respectively.If, however, the respon-imuthanglecorrespondtothelatitudeofthevertexandthedents are extracted from across the CS Line, the tabulardifference of longitude of thevertexandthepoint ofaltitudecorrespondstoalatitudeonthesideoftheequatordeparture.opposite from that ofthe point ofdeparture; the tabular az-imuth angle is the supplement of the differenceofExamplel:Find thevertex of the great circle tracklongitude.from lat.38°N, long.125°Wwhen the initial greatcirclecourseangleisN69ow.Exanmple I: Find a number of points along the greatcircle from latitude38°N.longitude125°WwhenSolution:Enter Pub.No.229with lat. 38°(Samethe initial great circle course angle is NllloW.Name),LHA69o,andinspectthecolumnforlat.38°tofindthemaximumtabularaltitude.Themax-Solution:Entering the tables with latitude38°(Sameimum altitude is 42038.1'at a distance of 1500Name),LHA 111°, and with successive declina-nautical miles (900-650=25°)from the point oftionsof 85,800,75°,etc,thelatitudes anddeparture.The corresponding tabular azimuth an-differencesin longitudefrom125°Warefound asgle is 32.40.Therefore, the difference of longitudetabularaltitudesandazimuthanglesrespectively:of vertex and point of departure is 32.40Answer:Answer:Latitudeofvertex=42°38.1N.300600900D (NM)3600Longitudeofvertex=1250+32.40=157.4°W5°D (arc)10°15°60°85°750800300dec2410.Altering A Great Circle Track To Avoid36.1° N33.9° N31.4°N3.6°NLat.Obstructions125°WDep.125°W125°W125°W5.8011.3054.10DLo16.5°Land,ice,orsevereweathermayprevent theuse ofLong130.8°W136.3°W141.5°W179.1°Wgreat circle sailing for some or all ofone's route. One oftheprincipal advantages of solution by great circlechart is thatExample 2: Find a number of points along the greatthe presence of any hazards is immediately apparent.Thecircle trackfrom latitude 38°N, long.125°Wwhenpilot charts are particularly useful in this regard. Often arel-the initial great circle course angle is N 69w.atively short runbyrhumbline is sufficient toreach a pointfrom which thegreat circle track can be followed. Where aSolution: Enter the tables with latitude 380 (Samechoice ispossible,therhumb line selected should conformName),LHA69andwithsuccessivedeclinationsas nearly as practicableto the direct great circle.as shown.Find thelatitudes and differences ofIf thegreat circleroute crosses anavigation hazardlongitudefrom125°Was tabularaltitudesandaz-changethetrack.It maybe satisfactorytofollowagreatcir-imuthangles,respectively:cleto thevicinityof the hazard, one or more rhumb lines

THE SAILINGS 359 and any method of solving such a triangle can be used. If 2 is the geographical position (GP) of a celestial body (the point at which the body is in the zenith), this triangle is solved in celestial navigation, except that 90° - D (the altitude) is de￾sired instead of D. The solution for the vertex and any point X usually involves the solution of right spherical triangles. 2408. Points Along The Great Circle If the latitude of the point of departure and the initial great-circle course angle are integral degrees, points along the great circle are found by entering the tables with the lat￾itude of departure as the latitude argument (always Same Name), the initial great circle course angle as the LHA ar￾gument, and 90° minus distance to a point on the great circle as the declination argument. The latitude of the point on the great circle and the difference of longitude between that point and the point of departure are the tabular altitude and azimuth angle, respectively. If, however, the respon￾dents are extracted from across the CS Line, the tabular altitude corresponds to a latitude on the side of the equator opposite from that of the point of departure; the tabular az￾imuth angle is the supplement of the difference of longitude. Example 1: Find a number of points along the great circle from latitude 38°N, longitude 125°W when the initial great circle course angle is N111°W. Solution: Entering the tables with latitude 38° (Same Name), LHA 111°, and with successive declina￾tions of 85°, 80°, 75°, etc., the latitudes and differences in longitude from 125°W are found as tabular altitudes and azimuth angles respectively: Answer: Example 2: Find a number of points along the great circle track from latitude 38°N, long. 125°W when the initial great circle course angle is N 69° W. Solution: Enter the tables with latitude 38° (Same Name), LHA 69°, and with successive declinations as shown. Find the latitudes and differences of longitude from 125°W as tabular altitudes and az￾imuth angles, respectively: Answer: 2409. Finding The Vertex Using Pub. No. 229 to find the approximate position of the vertex of a great circle track provides a rapid check on the solution by computation. This approximate solution is also useful for voyage planning purposes. Using the procedures for finding points along the great circle, inspect the column of data for the latitude of the point of departure and find the maximum value of tabular altitude. This maximum tabular altitude and the tabular az￾imuth angle correspond to the latitude of the vertex and the difference of longitude of the vertex and the point of departure. Example 1: Find the vertex of the great circle track from lat. 38°N, long. 125°W when the initial great circle course angle is N69°W. Solution: Enter Pub. No. 229 with lat. 38° (Same Name), LHA 69°, and inspect the column for lat. 38° to find the maximum tabular altitude. The max￾imum altitude is 42°38.1’ at a distance of 1500 nautical miles (90° - 65° = 25°) from the point of departure. The corresponding tabular azimuth an￾gle is 32.4°. Therefore, the difference of longitude of vertex and point of departure is 32.4°. Answer: Latitude of vertex = 42°38.1’N. Longitude of vertex = 125° + 32.4° = 157.4°W. 2410. Altering A Great Circle Track To Avoid Obstructions Land, ice, or severe weather may prevent the use of great circle sailing for some or all of one’s route. One of the principal advantages of solution by great circle chart is that the presence of any hazards is immediately apparent. The pilot charts are particularly useful in this regard. Often a rel￾atively short run by rhumb line is sufficient to reach a point from which the great circle track can be followed. Where a choice is possible, the rhumb line selected should conform as nearly as practicable to the direct great circle. If the great circle route crosses a navigation hazard, change the track. It may be satisfactory to follow a great cir￾cle to the vicinity of the hazard, one or more rhumb lines D (NM) 300 600 900 3600 D (arc) 5° 10° 15° 60° dec 85° 80° 75° 30° Lat. 36.1° N 33.9° N 31.4° N 3.6° N Dep. 125° W 125° W 125° W 125° W DLo 5.8° 11.3° 16.5° 54.1° Long 130.8°W 136.3°W 141.5°W 179.1°W D (NM.) 300 600 900 6600 D (arc) 5° 10° 15° 110° dec 85° 80° 75° 20° Lat. 39.6° N 40.9° N 41.9° N 3.1° N Dep. 125° W 125° W 125° W 125° W DLo 6.1° 12.4° 18.9° 118.5° Long 131.1°W 137.4°W 143.9°W 116.5°E

360THE SAILINGSalong the edge of the hazard, and anothergreat circle to thefrom thepoint of departure and the destination,tangent todestination.Anotherpossiblesolution is theuseofcompos-the limitingparallel.Then measurethe coordinates of vari-ite sailing, still another is the use of two great circles, oneous selected points alongthe compositetrack andtransferfrom thepointofdeparturetoa pointnear the maximum lat-them to a Mercator chart,as in greatcircle sailing.Compos-itudeof unobstructed water and the second fromthis pointite sailingproblems can also besolvedby computation,to the destination.using the equation:cos DLow= tanL,cot L2411.CompositeSailingThe point ofdeparture and the destination are used suc-When the great circle would carry a vessel to a highercessively as pointX,Solve thetwogreat circles at each endlatitude than desired, a modification of great circle sailingof the limiting parallel, and use parallel sailing along thecalled composite sailing may be used to good advantage.limiting parallel. Since both great circles have vertices atThecompositetrackconsistsofagreatcirclefromthepointthesameparallel,computationforC,D,andDLoyxcan beof departure and tangent to the limiting parallel, a coursemadeby consideringthem parts of the same great circlelinealongtheparallel,andagreatcircletangenttothelim-itingparallel and through thedestinationwith Li, L2, and Lv as given and DLo =DLov1+DLov2Solution of composite sailing problems is most easilyThetotal distance is the sum of thegreat circle and paralleldistances.made with a great circle chart.For this solution,draw linesTRAVERSETABLES2412.UsingTraverseTablespoints ofdeparture and arrival,respectively.The course an-gle and the three sides are as labeled. From this triangle:Traverse tables can be used in the solution of any of1ptan C= PsinC=cos C=the sailings exceptgreat circle and composite.They consistDDof thetabulationofthe solutions of planerighttriangles.Because the solutions are for integral values of the courseangleandthedistance,interpolationforintermediatevaluesmay be required.Through appropriate interchanges of theheadingsofthecolumns,solutionsforotherthanplanesail-ing can be made.For the solution ofthe planeright triangleDep. (p)QVany value N in the distance (Dist.) column is the hypote-nuse; the value opposite in the difference of latitude (D(L2, 入2)Lat.)column is the product of N and the cosineofthe acuteangle,and the other number opposite in the departure(Dep.)column is theproductof Nandthesineoftheacuteangle.Or,the number in theD.Lat.column is thevalue ofthe side adjacent, and the number in the Dep.column is the(2) 187value of the side opposite the acute angle. Hence, if theacute angle is the course angle, the side adjacent in the DDLat.column is meridional differencem,the side oppositein0the Dep.column is DLo.If the acute angle is the midlati-Stude of theformulap=DLocos Lm, thenDLo is anyvalueN in the Dist.column, and the departure is the value Nx coscLmintheD.Lat.columnTheexamples below clarifytheuse of thetraverseta-bles for plane,traverse,parallel,mid latitude,and Mercatorsailings.2413.Plane SailingPI (L入,)In plane sailing the figure formed by the meridianthrough the point ofdeparture, the parallel through thepointof arrival, and the course line is considered aplane righttri-angle.This is illustrated inFigure2413a.Pand P2are theFigure2413a.Theplane sailingtriangle

360 THE SAILINGS along the edge of the hazard, and another great circle to the destination. Another possible solution is the use of compos￾ite sailing; still another is the use of two great circles, one from the point of departure to a point near the maximum lat￾itude of unobstructed water and the second from this point to the destination. 2411. Composite Sailing When the great circle would carry a vessel to a higher latitude than desired, a modification of great circle sailing called composite sailing may be used to good advantage. The composite track consists of a great circle from the point of departure and tangent to the limiting parallel, a course line along the parallel, and a great circle tangent to the lim￾iting parallel and through the destination. Solution of composite sailing problems is most easily made with a great circle chart. For this solution, draw lines from the point of departure and the destination, tangent to the limiting parallel. Then measure the coordinates of vari￾ous selected points along the composite track and transfer them to a Mercator chart, as in great circle sailing. Compos￾ite sailing problems can also be solved by computation, using the equation: The point of departure and the destination are used suc￾cessively as point X. Solve the two great circles at each end of the limiting parallel, and use parallel sailing along the limiting parallel. Since both great circles have vertices at the same parallel, computation for C, D, and DLovx can be made by considering them parts of the same great circle with L1, L2, and Lv as given and DLo = DLov1 + DLov2. The total distance is the sum of the great circle and parallel distances. TRAVERSE TABLES 2412. Using Traverse Tables Traverse tables can be used in the solution of any of the sailings except great circle and composite. They consist of the tabulation of the solutions of plane right triangles. Because the solutions are for integral values of the course angle and the distance, interpolation for intermediate values may be required. Through appropriate interchanges of the headings of the columns, solutions for other than plane sail￾ing can be made. For the solution of the plane right triangle, any value N in the distance (Dist.) column is the hypote￾nuse; the value opposite in the difference of latitude (D. Lat.) column is the product of N and the cosine of the acute angle; and the other number opposite in the departure (Dep.) column is the product of N and the sine of the acute angle. Or, the number in the D. Lat. column is the value of the side adjacent, and the number in the Dep. column is the value of the side opposite the acute angle. Hence, if the acute angle is the course angle, the side adjacent in the D. Lat. column is meridional difference m; the side opposite in the Dep. column is DLo. If the acute angle is the midlati￾tude of the formula p = DLo cos Lm, then DLo is any value N in the Dist. column, and the departure is the value N × cos Lm in the D. Lat. column. The examples below clarify the use of the traverse ta￾bles for plane, traverse, parallel, mid latitude, and Mercator sailings. 2413. Plane Sailing In plane sailing the figure formed by the meridian through the point of departure, the parallel through the point of arrival, and the course line is considered a plane right tri￾angle. This is illustrated in Figure 2413a. P1 and P2 are the points of departure and arrival, respectively. The course an￾gle and the three sides are as labeled. From this triangle: cos DLovx Lx cot L v = tan Figure 2413a. The plane sailing triangle. cos C l D = sin C - p D = tan C - p l = -

361THE SAILINGSFrom the firsttwo oftheseformulasthefollowingre-Answer:lationships can be derived:Dif. Lat. = 3° 07.3'NI= D cos CD =I sec Cp= Dsinc.departure=16.4miles(2) Difference of latitude and departure by traverseLabel I as Nor S,and pas Eor W,toaid in identificatable:tion of the quadrant of the course.Solutions by calculationsandtraversetables areillustratedinthefollowingexamplesRefer to Figure 2413b. Enter the traverse table andfind course 005°at the top of the page. Using theExamplel:Avesselsteams188.0milesoncourse0050column headings at the top of the table, opposite188 in the Dist. column extract D. Lat. 187.3 andRequired:(l) (a)Differenceof latitudeand (b)deparDep. 16.4.turebycomputation.(2)(a)differenceoflatitudeand (b) departure by traverse table.(a) D. Lat. = 187.3'N.(b) Dep. = 16.4 mi. E.Solution:Example2:Ashiphassteamed136.0milesnorthand203.0 miles west.()(a)Difference of latitudebycomputation:=DXCOS Cdiff latitudeRequired: (I) (a) Course and (b) distance by computa-=188.0miles×cos (005)tion. (2) (a) course and (b) distance by traverse=187.3arcmintable.= 3° 07.3'NSolution:(l) (b) Departure by computation:(l)(a)Coursebycomputation:=DxsinCdeparturedeparature=188.0miles×sin(005)C= arctandiff. lat.=16.4miles355*005TABLE 4355*005"5°185*175*185°175°TraverseTableotDist,Depist.Dep.n.LatDpDist.D, Lat.DepDist.D.La.Dep.D.Lat.D,Lat1389888018135.3121120.510.5#设部防的180.3祖设信化公信8司240.12101284667899201838588998121.510.624188387888181.318888888383888882#装装区盛送医#的#的a00N-31187.164L28-18816.53128.51010.0ILL3248.189o21810WNAMA8059.8239.120.9130026.16.220119.5179.3010.52D.Lat.Dist.Dep.DepD.Lat.DistDistDepD.LaDesDep.1LatDistDep, LatpDist.B, Lat.27508585°NNx Cos.NS265°095Siafe Adj, Side OpeHypoteneFigure2413b.ExtractfromTable4

THE SAILINGS 361 From the first two of these formulas the following re￾lationships can be derived: Label l as N or S, and p as E or W, to aid in identifica￾tion of the quadrant of the course. Solutions by calculations and traverse tables are illustrated in the following examples: Example 1: A vessel steams 188.0 miles on course 005°. Required: (1) (a) Difference of latitude and (b) depar￾ture by computation. (2) (a) difference of latitude and (b) departure by traverse table. Solution: (1) (a) Difference of latitude by computation: (1) (b) Departure by computation: Answer: Diff. Lat. = 3° 07.3’ N departure = 16.4 miles (2) Difference of latitude and departure by traverse table: Refer to Figure 2413b. Enter the traverse table and find course 005° at the top of the page. Using the column headings at the top of the table, opposite 188 in the Dist. column extract D. Lat. 187.3 and Dep. 16.4. (a) D. Lat. = 187.3’ N. (b) Dep. = 16.4 mi. E. Example 2: A ship has steamed 136.0 miles north and 203.0 miles west. Required: (1) (a) Course and (b) distance by computa￾tion. (2) (a) course and (b) distance by traverse table. Solution: (1) (a) Course by computation: diff latitude = D × cos C = 188.0 miles × cos (005°) = 187.3 arc min = 3° 07.3’ N departure = D × sin C = 188.0 miles × sin (005°) = 16.4 miles l= D D cos C = p D sin C. l sec C = C arctandeparature diff. lat. = - Figure 2413b. Extract from Table 4

362THE SAILINGS203 beside each other in the columns labeled D.203.0Lat. and Dep., respectively.This occurs mostC = arctan 36.0nearly on the page for course angle 56°There-fore,thecourse is 304°Interpolating forC=N56°10.8'Wintermediatevalues,the correspondingnumber intheDist.column is244.3miles.C=304°(to nearest degree)Answer:Draw the course vectors to determine the correct(a) C = 3040course.In this case the vessel has gone north 136(b) D=244.3 mi.miles and west 203 miles. The course, therefore,must have been between 270°and 3600.No solu-2414.Traverse Sailingtionotherthan304°isreasonable.A traverse is a series of courses or a track consisting(l)(b)Distancebycomputation:ofa number ofcourse lines, suchasmightresultfroma sail-ing vessel beating into the wind. Traverse sailing is theD=diff latitude × sec Cfinding ofa single equivalent course and distance.=136miles×sec (304°)Though the problem can be solved graphically on the=136 miles ×1.8chart, traverse tables provide a mathematical solution.The=244.8milesdistanceto thenorth or south andto the east or weston eachcourse is tabulated, the algebraic sum of difference of lati-Answer:tude and departure isfound, and converted to course anddistance.C = 304°D=244.8milesExample:Ashipsteamsas follows:course158°,dis-tance15.5miles;course 135°distance 33.7(2) Solution by traverse table:miles; course 2590 distance 16.1 miles; course293°,distance 39.0miles; course 1690,distanceRefertoFigure2413c.Enterthetableandfind136and40.4miles.TABLE4326"034°32603434°214-146°146*214°TableTraverseD.Lat.D. Lat.D.Lat.Dep-Dist.DepDist.DepDist.Dep.Dist.D.Lat.Dep.Dist.D.Lat188#883100.867.7181101-224110846150.1199.81348006福135136.4163.4[103.5]2001187.0MVWMNA?WMW806049.733.62099,567.1149.2100.740199,0[1342300248.7167.8DDep.n.Lat.DepD.LaL.Dep.DisDistDist.DepD.LatD,Lat3DstDiLDep.D,Lat.30405656°N.Ns Cos.NxSin.236124°spotenuseSideOppSideAdjFigure2413c.ExtractfromTable4

362 THE SAILINGS Draw the course vectors to determine the correct course. In this case the vessel has gone north 136 miles and west 203 miles. The course, therefore, must have been between 270° and 360°. No solu￾tion other than 304° is reasonable. (1) (b) Distance by computation: Answer: C = 304° D = 244.8 miles (2) Solution by traverse table: Refer to Figure 2413c. Enter the table and find 136 and 203 beside each other in the columns labeled D. Lat. and Dep., respectively. This occurs most nearly on the page for course angle 56°. There￾fore, the course is 304°. Interpolating for intermediate values, the corresponding number in the Dist. column is 244.3 miles. Answer: (a) C = 304° (b) D = 244.3 mi. 2414. Traverse Sailing A traverse is a series of courses or a track consisting of a number of course lines, such as might result from a sail￾ing vessel beating into the wind. Traverse sailing is the finding of a single equivalent course and distance. Though the problem can be solved graphically on the chart, traverse tables provide a mathematical solution. The distance to the north or south and to the east or west on each course is tabulated, the algebraic sum of difference of lati￾tude and departure is found, and converted to course and distance. Example: A ship steams as follows: course 158°, dis￾tance 15.5 miles; course 135°, distance 33.7 miles; course 259°, distance 16.1 miles; course 293°, distance 39.0 miles; course 169°, distance 40.4 miles. D = diff. latitude × sec C = 136 miles × sec (304°) = 136 miles × 1.8 = 244.8 miles C arc 203.0 136.0 = tan - C = N 56° 10.8’ W C = 304°( ) to nearest degree Figure 2413c. Extract from Table 4

363THE SAILINGSDLo=210arcminRequired:Equivalent single (l) course (2) distance.P= DLo X cos LP=210 arcminutes×cos (49.5%)Solution: Solve each leg as a plane sailing and tabu-late each solution as follows.For course 1580p=136.4milesextract the values for D. Lat. and Dep. opposite155 in the Dist. column. Then, divide the values byAnswer:l0androundthemofftothenearesttenth.Repeatthe procedure for each leg of the vessel's journey.p= 136.4 milesEWDist.NsCourse(2) Solution by traverse table:mi.mi.mi.mi.mi.degreesRefer to Figure 2415a.Enter the traverse table withlatitude as course angle and substituteDLo as the15815.55.814.4heading of theDist.column and Dep.as thehead-13533.723.823.8ing of the D. Lat. column. Since the table is2593.1 15.816.1computed for integral degrees of course angle (or29339.015.235.9latitude),thetabulations in the pagesfor 49°and16939.77.740.450°mustbe interpolated for the intermediateval-ue (49°30').Thedeparture for latitude 49°and15.281.037.351.7SubtotalsDLo210'is137.8miles.The departureforlatitude-15.2-37.350°andDLo210is135.0miles.InterpolatingforN/S Total65.8S14.4 Wthe intermediate latitude,the departureis136.4miles.Thus, thelatitudedifference is S65.8miles and thedeparture is W 14.4 miles. Convert this to a courseAnswer:and distance using the formulas discussed in section2413.p=136.4milesAnswer:Example2:TheDRlatitudeofa shipon course2700is 38°15'S. The ship steams on this course for a(I) C = 192.30distanceof215.5miles.(2) D = 67.3 miles.Required:The change in longitude by (l)computation2415.Parallel Sailingand (2) traverse table.Parallel sailing consistsof the interconversion of deSolution:parture and difference of longitude.It is the simplest formof spherical sailing.Theformulasforthesetransformations(l)Solution by computationare:DLo=215.5arcmin×sec(38.25°)p=DLocosLDLo =p sec LDLo=215.5arcmin×1.27DLo=274.4minutesofarc (west)Example 1:The DR latitude of a ship on course 0900DLo=4°34.4°Wis 49°30'N. The ship steams on this course untilthe longitude changes 3°30.Answer:Required: The departure by (l) computation and (2)DLo = 4°34.4 Wtraverse table.(2)SolutionbytraversetableSolution:Refer to Figure 2415b.Enter the traverse tables with(l)Solutionbycomputation.latitude as course angle and substitute DLo as the headingof the Dist. column and Dep.as the heading of the D.Lat.column.As the tableis computed for integral degrees ofDLo= 3° 30'courseangle(orlatitude),thetabulationsinthepages for

THE SAILINGS 363 Required: Equivalent single (1) course (2) distance. Solution: Solve each leg as a plane sailing and tabu￾late each solution as follows. For course 158°, extract the values for D. Lat. and Dep. opposite 155 in the Dist. column. Then, divide the values by 10 and round them off to the nearest tenth. Repeat the procedure for each leg of the vessel’s journey. Thus, the latitude difference is S 65.8 miles and the de￾parture is W 14.4 miles. Convert this to a course and distance using the formulas discussed in sec￾tion 2413. Answer: (1) C = 192.3° (2) D = 67.3 miles. 2415. Parallel Sailing Parallel sailing consists of the interconversion of de￾parture and difference of longitude. It is the simplest form of spherical sailing. The formulas for these transformations are: Example 1: The DR latitude of a ship on course 090° is 49°30' N. The ship steams on this course until the longitude changes 3°30'. Required: The departure by (1) computation and (2) traverse table. Solution: (1) Solution by computation: Answer: p = 136.4 miles (2) Solution by traverse table: Refer to Figure 2415a. Enter the traverse table with latitude as course angle and substitute DLo as the heading of the Dist. column and Dep. as the head￾ing of the D. Lat. column. Since the table is computed for integral degrees of course angle (or latitude), the tabulations in the pages for 49° and 50° must be interpolated for the intermediate val￾ue (49°30'). The departure for latitude 49° and DLo 210' is 137.8 miles. The departure for latitude 50° and DLo 210' is 135.0 miles. Interpolating for the intermediate latitude, the departure is 136.4 miles. Answer: p = 136.4 miles Example 2: The DR latitude of a ship on course 270° is 38°15'S. The ship steams on this course for a distance of 215.5 miles. Required: The change in longitude by (1) computation and (2) traverse table. Solution: (1) Solution by computation Answer: DLo = 4° 34.4' W (2) Solution by traverse table Refer to Figure 2415b. Enter the traverse tables with latitude as course angle and substitute DLo as the heading of the Dist. column and Dep. as the heading of the D. Lat. column. As the table is computed for integral degrees of course angle (or latitude), the tabulations in the pages for Course Dist. N S E W degrees mi. mi. mi. mi. mi. 158 15.5 14.4 5.8 135 33.7 23.8 23.8 259 16.1 3.1 15.8 293 39.0 15.2 35.9 169 40.4 39.7 7.7 Subtotals 15.2 81.0 37.3 51.7 -15.2 -37.3 N/S Total 65.8 S 14.4 W DLo = 3° 30' DLo = p sec L p = DLo cos L DLo = 210 arc min p = DLo × cos L p = 210 arc minutes × cos (49.5°) p = 136.4 miles DLo = 215.5 arc min × sec (38.25°) DLo = 215.5 arc min × 1.27 DLo = 274.4 minutes of arc (west) DLo = 4° 34.4' W

364THE SAILINGSTABLE4319°041°319°041°41°221°221-139°139°TraverseTableist.n.Lat.Dist.Ip.Lat.Bepnist.DistD.LaDkpDep.n,LatepDisLnLat.pAWMnWM1888887888款国141106.92.5201261197.06151131.171.2-61.9107.918855858862197.753.84293.2152-5132.171.9-就好好62.68#5招摄#8153.2133,2188888882198.58154.0133199.2173.1055.8109.4154.7200.0173.9L8174..5.?文175.2国175.o019.710177.1WWNVAbist.Dep-D.LatDep.D. Lat.Ip,1DepDist1aIist.B.1a.Dep.311°04949°DLaPap229°131mDLaFigure2415a.ExtractfromTable4.TABLE4322°038322"038°38°218*218°142°142°TableTraversemattintRLaDin.p.LatDepDiatDLatepDpDrpDistDLatDepOLa可082#设B冰1证B8话口新国国41311215.1.131.8215.9168.2.27.668.565122.196.415169.75216.7169.3413DuL.LBrPEDIaLapDFigure2415b.ExtractfromTable438°and39°must be interpolatedfor theminutes of latitude.normally usedforwant ofapracticablemeans ofdetermin-ing the middle latitude, the latitude at which the arc lengthCorrespondingtoDep.215.5milesintheformerisDLooftheparallel separatingthemeridians passingthroughtwo273.5'and in the latterDLo277.3:Interpolatingfor min-specific points is exactlyequal to thedeparture inproceed-utesof latitude,theDLois274.4W.ingfrom onepointtotheother.Theformulasforthesetransformationsare:Answer:DLo = 4°34.42416.Middle-Latitude SailingDLo = psec Lmp= DLo cos LmMiddle-latitude sailing combines plane sailingand par-allel sailing.Plane sailing is used to find difference oflatitudeanddeparture when course anddistanceareknown,or vice versa. Parallel sailing is used to interconvert depar-Themean latitude (Lm)is half the arithmetical sumofture and difference of longitude.The mean latitude (Lm)isthe latitudesof twoplaces on the same sideof the equator

364 THE SAILINGS 38° and 39° must be interpolated for the minutes of latitude. Corresponding to Dep. 215.5 miles in the former is DLo 273.5’, and in the latter DLo 277.3’. Interpolating for min￾utes of latitude, the DLo is 274.4’W. Answer: DLo = 4° 34.4’ 2416. Middle-Latitude Sailing Middle-latitude sailing combines plane sailing and par￾allel sailing. Plane sailing is used to find difference of latitude and departure when course and distance are known, or vice versa. Parallel sailing is used to interconvert depar￾ture and difference of longitude. The mean latitude (Lm) is normally used for want of a practicable means of determin￾ing the middle latitude, the latitude at which the arc length of the parallel separating the meridians passing through two specific points is exactly equal to the departure in proceed￾ing from one point to the other. The formulas for these transformations are: The mean latitude (Lm) is half the arithmetical sum of the latitudes of two places on the same side of the equator. Figure 2415a. Extract fromTable 4. Figure 2415b. Extract from Table 4. DLo p sec Lm = p DLo cos Lm · =