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 composite sailing; still another is the use of two great circles, one from the point of departure to a point near the maximum latitude 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 limiting 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 various selected points along the composite track and transfer them to a Mercator chart, as in great circle sailing. Composite sailing problems can also be solved by computation, using the equation: The point of departure and the destination are used successively 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 sailing can be made. For the solution of the plane right triangle, any value N in the distance (Dist.) column is the hypotenuse; 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 midlatitude 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 tables 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 triangle. This is illustrated in Figure 2413a. P1 and P2 are the points of departure and arrival, respectively. The course angle 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 relationships can be derived: Label l as N or S, and p as E or W, to aid in identification 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) departure 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 computation. (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