Great Circle and Sailings Problems

TheCutterman’sGuidetoNavigationProblems PartFive:GreatCircleandSailingsProblems

Whenplanningavoyageofsignificantdistance,thecurvatureoftheearthmustbe accountedfor.Additionally,greatcircletechniquesallowanavigatortoaccurately dead-reckontoalocation/timemanyhoursinthefuture. Theabilitytodead-reckonviamid-latitudeorparallelsailingtoalaterlocationis criticaltosolvingmorecomplexproblemssuchastimeofphenomenonorcelestial sightproblemsinlaterParts. Definitions

Lator L–Latitude. L1–t –th hela latitudeof of the poi pointof ofde departure. L2 – the latitude of the destination. Lm–themiddlelatitude((L1+L2)/2) Lm–themiddlelatitude((L1+L2)/2) Lv –thelatitudeofthevertexpoint(thepointatthe“peak”ofthetrackline). –thelatitudeofthevertexpoint(thepointatthe“peak”ofthetrackline). Lx –thelatitudeofanypointalongthetrackline. –thelatitudeofanypointalongthetrackline. –thedifferencebetweentwogivenlatitudes. DLat or DLat  orl –thedifferencebetweentwogivenlatitudes. m–Meridionalparts.Theactuallengthofanarcfromtheequatortoagiven latitude,givenalongalongitude/meridianline.Table6inBowditchprovides meridionalparts. –themeridionalpartsofthepointofdeparture. m1–themeridionalpartsofthepointofdeparture. m1 –themeridionalpartsofthedestination. m2–themeridionalpartsofthedestination. m2

orLong–Longitude.  λorLong–Longitude. Long Lo ng1 1–t –th helo elong ngiitude tudeo of fth the epo poiinto ntof fde depa parrtur ture. Long2 ng2–t –th helo longitudeof ofth the destination. λV– λV–th the elo long ngit itud ude eof oft the hev ver erte tex xpo poin int( t(th thep epoi oint ntat atth the“ e“pe peak” ak”of ofth the etr trac ack) k).. λX λX–t –the hel lon ongi gitu tude deo of fan any ypo poin int tal alon ong ga atr trac ackl klin ine. e. DLo–thedifferenceintwolongitudes.

Dep–Departure.Thedistancebetweentwomeridiansatanygivenparallelof  por Dep–Departure.Thedistancebetweentwomeridiansatanygivenparallelof latitude. orCn–Courseangle.Acoursemeasuredfrom000°clockwiseoranti-clockwise C or through180°labeledwithprefixandsuffixdirectionstoclarifytheintendedcourse. Forexample,atraditionalcourseof120°Twouldbeexpressedasacourseangleof E120°N,whileatraditionalcourseof330°Twouldbeexpressedasacourseangleof

W30°N.FurtherdiscussioncanbefoundinBowditch(intheglossaryorin“The Sailings.) Distor D–Distance. TrigonometricIdentitiesandI Trigonometri cIdentitiesandInverseFunction nverseFunctions s

TheseformulaearealsolocatedinBowditch,whichisfoundonthebridgeofmost ocean-goingvessels.Additionally,theseformulaeareprovidedvia“Bowditch VolumeII,”whichislocatedinthetestingcenterformerchantmarinerexams. TrigonometricIdentities: sin 

1 =

csc

cos



tan 



1 =

sec

csc





1 =

sec

cot  

1 =

cos

1 =



cot  

sin 



1 =

tan 

InverseTrigonometricFunctions.Often,equivalentidentitiesareexpressed differentlyindifferenttexts.Also,sometimesitisnecessarytoperformalgebraic tasksongivenformulaetosolveforyourdesiredvalue,necessitatinginversetrig functions.Asarefresher,allofthefollowingareequivalent: sin 



=



=

=



arcsin 

!!

sin



TheSailingsFormulae

Notethatthebelowformulaearere-createdfromBowditch(availableinmerchant marinertestcentersorvessel’sbridges).Theygenerallydonotneedtobe memorized.Althoughlistedundercertaincategories(e.g.“planesailing”or “Mercatorsailing”),theyareofteninterchangeable,dependingonthetypeof problemtobesolved. ThePlaneSailingFormulae: cos



sin 

tan 

1 =

=











cos





sec



=



 =

=





=

 sin 

CourseAngleNotation

Problem5-1(CG-452).ThefollowingquestionistakendirectlyfromtheUSCGtest bankandillustratestheconceptofcourseangle. Fromlatitude7°12’N,longitude80°00’Wtoapositionatlatitude47°12’S,longitude 169°18’E,theinitialgreatcirclecourseangleis137.25°.Howwouldyounamethis course? Answer:N137.25°W.FromBowditch(glossary):“Courseangleisthecourse measuredfrom0°atthereferencedirectionclockwiseorcounter-clockwise through90°or180°Itislabeledwiththereferencedirectionasaprefixandthe directionofmeasurementfromthereferencedirectionasasuffix.” Inthisexample,thevesselisproceedingfromapointinthenorthernandwestern hemisphere(nearthePanamaCanal)toapointinthesouthernandeastern hemisphere.Thecourseangleisgivenas137.25°. Sincethereferencehemisphereisnorth,thereferencedirectionisnorth(000°). Sincethevesselisheadingsouthwestfromthedeparturepointtothearrivalpoint, thecourseismeasured“westofnorth”bythegiven137.25°.Thecorrectnotation isthereforeN137.25°W.Instandardnotation(notaskedforinthisproblem),the coursewouldbe360°-137.25°=222.75°True. Mid-LatitudeSailingProblems

Problem5-2(CG-7).ThefollowingquestionistakendirectlyfromtheUSCGtest bankandillustrateshowtosolvemid-latitudesailingsproblems.  Avesselinlatitude20°00’N,longitude107°30’Wistoproceedtolatitude24°40’N, longitude112°30’W.Whatisthecourseanddistancebymid-latitudesailing? Answer:Course315.27°T,394.4nm. Step1:

Convertstandardlatitudeandlongitudestodecimalform: DepartureLocation 20°00’N=20.0°N 107°30’W=107.5°W ArrivalLocation 24°40’N=24.67°N 112°30’W=112.5°W

Step2:

Determinethemid-latitude(Lm). Lm=(20.0°N+24.67°N)÷2=22.33°N

Step3:

DeterminetheDifferenceofLatitude(DLat )andconverttoarc minutestodeterminel.

DLat =24.67°-20.0°=4.67° l =4.67°×60=280.2’ Step4:

DeterminetheDifferenceofLongitude(DLo)andconverttoarc minutes. 112.5°W–107.5°W=5.0°=300’

Step5:

Applymid-latitudeformulaetodeterminetheDeparture( p).

Step6:



=





=

300′



=



=

cos



(22.33) (300) (0.9250) cos

!

277.49

DeterminetheCourseAngle(C )usingtheMercatorFormula. 

tan 

=

tan 

=

tan 

=

 277.49 280.2 0.9903

C =tan!! 0.9903 C =44.722° Step7:

Step8:

DeterminethedistancegiventhePlaneSailingFormula. 





=



=

(280.2)



=

280.2

(



=

280.2

(



=

sec

(44.722)

sec

1

cos

1

44.722

)

)

0.7105

.4nm

DeterminetheactualcoursegiventhesolvedCourseAngle(C ).Note thatthevesselclearlyheadedinanorthwesterlydirectionbasedon theoriginanddestinationpoints.Thereforethecourseisgivenby: Course=360°-C  Course=360°-44.722° Course=315.278°

Problem5-3(CG-30).ThefollowingquestionistakendirectlyfromtheUSCGtest bankandillustrateshowtosolvemid-latitudesailingproblems.  Avesselsteams640milesoncourse047°Tfromlatitude34°45’N,longitude140°00’ E.Whatarethelatitudeandlongitudeofarrivalbymid-latitudesailing? Answer:42°01.5’N,149°57.2’E

Step1:

Convertstandardlatitudeandlongitudestodecimalform: DepartureLocation 34°45’N=34.750°N 140°00’E=140.000°E

Step2:

DeterminetheDifferenceinLatitude(l )usingtheplanesailing formulaandconverttodecimalnotation. 

=





cos

(640) cos(047°) (640) (0.6820)   436.5′ 436.5’ ÷60=7.275°(tothenorthbasedoninitialcourse). 

= = =

Step3:

DeterminethearrivallatitudeusingtheDifferenceinLatitude(l ). Converttostandardnotationforpartialproblemanswer. Latitude1=34.750°N DifferenceinLatitude=7.275°tothenorth. Latitude2=34.750°+7.275°=42.025°N=42°01.5’N

Step4:

DeterminethemidlatitudeusingtheDifferenceinLatitude(l )and arrivallatitude. Latitude1=34.750°N Latitude2=42.025°N MidLat=

Step5:

Step6:

!".!"#°!!".!"#° !

=

38.388°

DeterminetheDeparture( p)usingtheplanesailingformula. 

=



=



=



=

 sin 

(640) sin 47° (640) (0.7313) 468.1′

DeterminetheDifferenceinLongitude(DLo)giventhedepartureand mid-latitude. 

=



sec





=



sec



 

=

=



(

!

!"#

!"

=



(

! !"#

!"

)

(468.1 ) ( !

!"#

!

)

!".!""° !



=

(468.1′) (



=

(468.1′) (1.2758)



=

597.2′

(!.!"#")

)

)

Step7:

ConverttheDifferenceinLongitude(DLo)todecimalnotationand determinethearrivallongitude.Converttostandardnotationfor partialproblemanswer. 597.2′ ÷60=9.953°(totheeastbasedoninitialcourse).  Longitude 1 = 140.000° E DifferenceinLongitude=9.953°totheeast. Longitude2=140.000°+9.953°=149.953°E 149.953°E=149°57.2’E =

MercatorSailingProblems

Problem5-4(CG-777).ThefollowingquestionistakendirectlyfromtheUSCGtest bankandillustrateshowtosolveMercatorSailingproblemsforarrivallocation givenaninitialcourseanddistance. Youdepartlatitude40°42.0’N,longitude074°01.0’Wandsteam3365.6mileson course118°T.WhatisthelongitudeofyourarrivalbyMercatorSailing? Answer:17°40.62’W.Usetable6inBowditchtoobtainthenecessarymeridional parts. Step1:

Convertstandardlatitudeandlongitudestodecimalform: DepartureLocation 40°42.0’N=40.70°N 74°01.0’W=74.02°W

Step2:

DeterminetheDifferenceofLatitude(l )giventheknowncourseand distanceusingthePlaneSailingformula.Thenconvertthemileageto arc.

Step3:



=



=



=



=

1580.05!



=

1580.05!



cos



(3365.6) cos(118°) (3365.6) (0.4695) ÷

60

=

26.3342°

Giventhattheinitialcoursewastothesoutheast,subtractthe DifferenceofLatitude(l )fromtheinitiallatitudetodeterminethe arrivallatitude. DepartureLatitude:40.70°N DifferenceofLatitude:-26.3342° ArrivalLatitude=40.70°-26.3342°=14.366°=14°22.0’N

Step4:

UseTable6inBowditchtoobtainthemeridionalparts(m1andm2) forthegivenlatitudes.Besuretousethetablecorrectly.

Latitude1:40°42.0’ MeridionalParts1:m1=2662.8 Latitude2:14°22.0’ MeridionalParts2:m2=865.4 Step5:

Determinethedifferenceinmeridionalparts(m). m1–m2=m 2662.8–865.4=m=1797.4

Step6:

DeterminetheDifferenceinLongitude(DLo)usingtheMercator Sailingformula.Convertthemileagetoarc.

Step7:

tan 



=



=



=



=

3380.42′



=

3380.42



(1797.4) tan 118° (1797.4) (1.8807) !

÷

60

=

56.34°

DeterminethearrivallongitudeusingtheDifferenceofLongitude (DLo). DepartureLongitude:74.017°=74°01.0’W DifferenceofLongitude:56.34°=56°20.4’ ArrivalLongitude=74.017°-56.34°=17.677°=17°40.62’N

Problem5-5(CG-22).ThefollowingquestionistakendirectlyfromtheUSCGtest bankandillustrateshowtosolveMercatorSailingproblemsforcourseanddistance givenastartingandendinglocation.  Avesselatlatitude45°36.0’N,longitude011°36.0Wheadsforadestinationat latitude24°16.0’N,longitude073°52.0’W.Determinethetruecourseanddistance byMercatorsailing. Answer:247.2°T,3299.2nm.Usetable6inBowditchtoobtainthenecessary meridionalparts. Step1:

Convertstandardlatitudeandlongitudestodecimalform: DepartureLocation 45°36.0’N=45.600°N 11°36.0’W=11.600°W Arrival Location 24° 16.0’ N = 24.267° N 73°52.0’W=73.867°W

Step2:

DeterminetheDifferenceofLatitude(l ).Converttominutesofarc. l=45.600°N-24.267°N=21.333° l =21.333°×60=1279.98’

Step3:

DeterminetheDifferenceinLongitude(DLo).Converttominutesof arc. DLo=11.600°W-73.867°W=62.267° DLo=62.267°×60=3736.02’

Step4:

UseTable6inBowditchtoobtainthemeridionalparts(m1andm2) forthegivenlatitudes. Latitude1:45°36.0’ MeridionalParts1:m1=3064.7 Latitude2:24°16.0’ MeridionalParts2:m2=1492.1

Step5:

Determinethedifferenceinmeridionalparts(m). m1–m2=m 3064.7–1492.1=m=1572.6

Step6:

DeterminetheCourseAngle(C )usingthemodifiedMercatorSailing formula.Determinetheactualcourse. 

tan 



=

 tan 

=



tan 

=

3736.02′

1572.6 2.3756 tan  !! tan 2.3756  =

=



=

67.172

Step7:

Determinetheactualcoursesteeredgiventhesolvedcourseangle. NotethatthevesselclearlyheadedinaSWdirectiongiventheinitial andfinalpositions.Thereforethecourseangle67.172°isexpressed asS67.172°W.Youcanroundfinalanswerstonearesttenth. C =S67.172°W 180°+67.172°W=247.172°T=247.2°T 

Step8:

DeterminethedistancetravelledusingthePlaneSailingformula. 

=



=



=



=



=



=

 

sec

(



!

!"#

!

)

(1279.98) ( (1279.98) (

! !"#

!"#.! !

(!.!"#$)

)

)

(1279.98) (2.5806) . nm

VertexProblems

Problem5-6.ThefollowingquestionismodifiedfromaquestionintheUSCGtest bankandillustrateshowtofindthelatitudeofthevertex.Findingthelongitudeof thevertexisgiveninthenextproblem. Thegreatcircledistancefromlatitude25°50.0’N,longitude077°00.0’Wtolatitude 35°56.0’N,longitude006°15.0’Wis3616nauticalmiles.Theinitialcourseis061.7° T.Determinethelatitudeofthevertex. Answer:37°34.92’N Step1:

Convertstandardlatitudeandlongitudestodecimalform: DepartureLocation 25°50.0’N=25.834°N 77°00.0’W=77.000°W Arrival Location 35° 56.0’ N = 35.934° N 06°15.0’W=06.250°W

Step2:

DeterminetheLatitudeoftheVertexusingtheVertexFormulae. 

Step3:

=



=



=



=



=

!!

((cos 1)(sin  )) cos cos 25.834°)(sin 61.7°)) !! cos (0.900060)(0.880477) ! ! cos (0.792482) cos

!! ((

37.582°

Convertthedecimallongitudetostandardnotation. 37.582°N=37°34.92’N

Problem5-7(CG-551).ThefollowingquestionistakendirectlyfromtheUSCGtest bankandillustrateshowtofindthelongitudeofthevertexofaGreatCircletrack. NotethattheUSCGtestquestionprovidesthelatitudeofthevertex.Solvingforthe latitudeofthevertexisshowninthepreviousproblem. Thegreatcircledistancefromlatitude25°50.0’N,longitude077°00.0’Wtolatitude 35°56.0’N,longitude006°15.0’Wis3616nauticalmiles.Theinitialcourseis061.7° T.Thelatitudeofthevertexis37°34.9’N.Determinethelongitudeofthevertex. Answer:025°59.4’W Step1:

Convertstandardlatitudeandlongitudestodecimalform: DepartureLocation

25°50.0’N=25.834°N 77°00.0’W=77.000°W ArrivalLocation 35°56.0’N=35.934°N 06°15.0’W=06.250°W LatitudeoftheVertex 37°34.9’N=37.582°N Step2:

DeterminetheDifferenceinLongitudeoftheVertex(DLo(v))given theVertexFormulae. !!





=

sin





=

sin





=

sin





=





=

!!

!!

cos



sin  cos 61.7° sin(37.582)

(0.4741)

(0.6099) ! ! sin 0.7773 51.01°

Step3:

DeterminetheLongitudeoftheVertexgiventheinitiallongitudeand thedirectionoftravel(easterlyinthiscasebasedongivenpositions). Longitude1=77.000°W DifferenceofLong(vx)=51.01° LongitudeoftheVertex=77.00°-51.01°=25.99°W

Step4:

Convertthedecimallongitudetostandardnotation. 25.99°W=25°59.4’W

Problem5-8(CG-552).ThefollowingquestionistakendirectlyfromtheUSCGtest bankandillustrateshowtofindthelatitudeintersectingthegreatcircletrack,away fromthevertex.Notethisexampleisacontinuationoftheprevioustwoexamples (havingpreviouslyfoundthelatitudeandlongitudeofthevertex).Alsonotethatthe examplesareslightly“off”duetoroundingdifferencesbetweenthistextandthe officialCGanswers. Thegreatcircledistancefromlatitude25°50.0’N,longitude077°00.0’Wtolatitude 35°56.0’N,longitude006°15.0’Wis3616nauticalmiles.Theinitialcourseis061.7° T.Thepositionofthevertexis37°34.9’N,25°59.0’W.Determinethelatitude intersectingthegreatcircletrack600mileswestofthevertex,alongthegreatcircle track. Answer:36°54.9’N

Step1:

Convertstandardlatitudeandlongitudestodecimalform: DepartureLocation 25°50.0’N=25.834°N 77°00.0’W=77.000°W ArrivalLocation 35°56.0’N=35.934°N 06°15.0’W=06.250°W LatitudeoftheVertex 37°34.9’N=37.582°N Longitude of the Vertex 25° 59.0’ W = 25.983° W

Step2:

Convert600milestoarc. 600÷60=10°

Step3:

Determinethelatitudeataposition600miles(10°ofarc)westofthe vertexalongthegreatcircletrackusingtheVertexFormulae.Note that“D(vx)” standsforthe“distancefromthevertex”and“Lx ”stands for“latitudeofanypointalongthegreatcircletrack.” !! sin ((sin  ( ))(cos  ( )))  !! sin ((sin(37.582°)(cos(10°))  !! sin ((0.6099)(0.9848))  !! sin (0.6006)   36.915° = = = = =

Step4:

Convertthedecimallatitudetostandardnotation.  36.915°=36°+0.915° 0.915° ×60=54.9’  36°54.9’N 36.915 =

=

=

PlaneandParallelSailingsProblems

Typically,theseproblemsarepartoflargerproblems.Forexample,when calculatingatimeofphenomenonproblem(coveredinPart9)itisnecessaryto dead-reckontheship’sfutureposition. Planeandparallelsailingformulascanbeusedtoaccuratelycalculateafutureship position,andthesecalculationsareoftenincorporatedintomorecomplextimeof phenomenaproblemslater.Thebasicformulaeforplaneandparallelsailingsare shownhere,andthissectionshouldbereviewedbeforebeginningtheadvanced problemsinPart9:TimeofPhenomenonProblems. Thereareseveralplanesailingformulaegiveninthe“formulae”sectionabove,as wellasinBowditch,butthemainformulaenecessaryformostproblemsare:



=







cos





sec



=

=

 sin 

Themainparallelsailingformulaisidenticaltothemid-latitudeformula,giventhat whendeterminingDifferenceinLongitude(DLo),itisassumedavesselis proceedingalongaparallel,e.g.parallelsailing.Themainparallelsailingformula neededis: 

 =

cos



Problem5-9.ThefollowingquestionismodifiedfromaquestionintheUSCGtest bankandillustrateshowtosolveforasecondpositiongiveninitialcourse,speed, andposition. Youareoncourse082°T,speed19knots.Your0830DRpositionislatitude24°14.8’N andlongitude133°35.5’W.Youwishtomakeacelestialobservationat1146local time.Whatwilltheship’spositionbeatthattime? Answer:Latitude2=24°23.5’N,Longitude2=132°28.0’W Step1:

DeterminethetransittimeandtheDistance(D)steamedoverthat time. 0830to1146=3hoursand16minutes=3.27hours. 3.27hoursat19knots=D=62.13nmcovered

Step2:

DeterminetheDifferenceinLatitude(l )usingtheplanesailing formula. 

=





=

62.13



=

(62.13) (0.1392)



=

8.65′

cos



(82°)

cos

Step3:

Determinelatitude2giventheinitialpositionandtheDifferencein Latitude(l ). Lat1=24°14.8’N l=0°08.65’ Lat2=24°14.8’+0°08.65’=24°23.5’N

Step4:

DeterminetheDeparture( p)usingtheplanesailingformula. 

=

 sin 



=

(62.13)



=

62.13



=

61.53

sin(82°)

0.9903

Step5:

Determinethemid-latitudeandconverttodecimalnotation. Lat1 = 24° 14.8’ N Lat2=24°23.5’N Lm=

Step6:

(!"° !".!’ ! !"° !".!) !

=24°19.15’N=24.319°

DeterminetheDifferenceinLongitude(DLo)usingtheparallelsailing formula.Thenconverttoarc. 

! =

!"#



=

! !".!"

(!".!"#°)

!"#

 

Step7:

=

=

!".!" !.!""#

67.52’=1°07.5’

Determinelongitude2giventheinitialpositionandtheDifferencein Longitude(DLo). Long1=133°35.5’W DLo=1°07.5’(eastwardgiventheinitialcourse). Long2=133°35.5’–1°07.5’=132°28.0’W

Example5-10.ThefollowingquestionismodifiedfromaquestionintheUSCGtest bankandillustrateshowtosolveforasecondpositiongiveninitialcourse,speed, andposition. Yourship’s0400zonetimeDRpositionis22°31.’0Nand031°45.0’W.Youareon course240°Tat16.5knots.Sunriseisatapproximately0505.Whatpositionwillthe shipbeatthattime? Answer:Latitude2=22°22.1’N,Longitude2=134°01.7’W Step1:

DeterminethetransittimeandtheDistance(D)steamedoverthat time. 0400to0505=1hourand05minutes=1.083hours. 1.083hoursat16.5knots=D=17.87nmcovered

Step2:

DeterminetheDifferenceinLatitude(l )usingtheplanesailing formula.

Step3:



=





=

17.87



=

(17.87) (−0.5000)



=

−8.935′

cos



(240°)

cos

Determinelatitude2giventheinitialpositionandtheDifferencein Latitude(l ). Lat1=22°31.0’N

l =-0°08.935’ Lat2=22°31.0+(-0°08.935’)=22°22.1’N Step4:

Step5:

DeterminetheDeparture( p)usingtheplanesailingformula. 

=

 sin 



=

(17.87)

sin(240°)



=

17.87

−0.8660



=

−15.475

Determinethemid-latitudeandconverttodecimalnotation. Lat1 = 22° 31.0’ N Lat2=22°22.1’N Lm=

Step6:

(!!° !".!’ ! !!° !!.!) !

=22°26.55’N=24.443°

DeterminetheDifferenceinLongitude(DLo)usingtheparallelsailing formula.Thenconverttoarc. 

! =

!"#



=

!

!!".!"# (!!.!!"°) !!".!"# !"#

 

Step7:

=

=

!.!"#$

16.742’=0°16.742’

Determinelongitude2giventheinitialpositionandtheDifferencein Longitude(DLo). Long1=031°45.0’W DLo=0°16.742’(westwardgiventheinitialcourse). Long2=133°35.5’+0°16.742’=133°61.742’W=134°01.7’W

 Problem5-11(CG-768).ThefollowingquestionistakendirectlyfromtheUSCGtest bankandillustrateshowtosolveparallelsailingproblems. Youdepartlatitude25°54’N,longitude009°38’Eandsteam592milesoncourse  270°T.Whatisthelongitudeofarrival? Answer001°20.1’W.Sincethecourseis270°,thisisstrictlyaparallelsailing problem. Step1:

Convertstandardlatitudeandlongitudestodecimalform: DepartureLocation 25°54’N=25.900°N 009°38’E=9.633°E

Step2:

DeterminetheDeparture( p). 

=

 sin 



=

(592)

sin(270°)

Step3:



=

(592) (−1.0)



=

−592′

DeterminetheDifferenceofLongitude(DLo)usingtheparallelsailing formula.  

=

=

 sec 

=

(−592′)(



!

(

!"#

!

)

! !"#

)

!".!""°

!

=

(−592′)(



=

(−592′)(1.1116)



=

−658.1′



!.!""#

)

Step4:

ConverttheDifferenceofLongitude(DLo)intodecimalnotation. -658.1’÷60=-10.968°=10.968°tothewest.

Step5:

Sumthelongitudestodeterminethefinallongitude. Departurelongitude=9.633°E Differenceinlongitude=10.968°tothewest. Arrivallongitude=9.633°E–10.968°tothewest=-1.335° -1.335°=1.335°W

Step6:

Convertthearrivallongitudetostandardnotation. 1.335°W=001°20.1’W

Problem5-12(CG-775).ThefollowingquestionistakendirectlyfromtheUSCGtest bankandillustrateshowtosolveparallelsailingproblems. Youdepartlatitude38°12’S,longitude012°06’Wandsteam1543milesoncourse  270°T.Whatisthelongitudeofarrival? Answer:044°49.3’W.Sincethecourseis270°,thisisstrictlyaparallelsailing problem. Step1:

Convertstandardlatitudeandlongitudestodecimalform: DepartureLocation 38°12’S=38.200°S 012°06’W=12.100°W

Step2:

DeterminetheDeparture( p). 

=



=



=



=

 sin 

(1543) sin(270°) (1543) (−1.0)

−1543′

Step3:

DeterminetheDifferenceofLongitude(DLo)usingtheparallelsailing formula. 

=

 sec 

=



(

!

!"#

!

)

!

=

(−1543′)(

=

(−1543′)(



=

(−1543′)(1.2724)



=

−1963.3′

 

)

!"# !".!"" ° !

)

!.!"#$

Step4:

ConverttheDifferenceofLongitude(DLo)intodecimalnotation. -1963.3’÷60=-32.722°=32.722°tothewest.

Step5:

Sumthelongitudestodeterminethefinallongitude. Departurelongitude=12.100°W Differenceinlongitude=32.722°tothewest. Arrivallongitude=12.100°W+32.722°tothewest=44.822°W

Step6:

Convertthearrivallongitudetostandardnotation. 44.822°W=044°49.3’W

AdditionalProblemsandAnswers

Allofthefollowingquestionsweretakendirectlyfromthe2013USCGtestbankand illustratetheconceptsinthisPart.Note–notallproblemshavebeenworkedand aresubjecttooccasionalerrorsinthedatabase.Formoreproblemsandanswers, seetheUSCGdatabaseofquestions(databaseinformationlocatedinthepreface). ProblemCG-5.Avesselatlatitude14°10’N,longitude61°00’Wistoproceedto latitude10°00’N,longitude53°23’W.Whatisthecourseanddistancebymidlatitudesailing? a) b) c) d)

117.3°Tfor503miles 117.9°Tfor504miles 118.6°Tfor508miles 119.2°Tfor512miles-correct

ProblemCG-7.Avesselatlatitude20°10’N,longitude107°30’Wistoproceedto latitude24°40’N,longitude112°30’W.Whatisthecourseanddistancebymidlatitudesailing? a) b) c) d)

314.0°Tfor389miles 315.3°Tfor394miles-correct 317.2°Tfor397miles 318.3°Tfor399miles

ProblemCG-8.Avesselatlatitude20°10’N,longitude122°00’Eistoproceedto latitude26°18’N,longitude128°20’E.Whatarethecourseanddistancebymidlatitudesailing? a) b) c) d)

041.2°Tfor501miles 041.9°Tfor503.6miles 043.5°Tfor507.3miles-correct 044.7°Tfor509.7miles

ProblemCG-28.Avesselsteams576milesoncourse260°Tfromlatitude40°36’N, longitude50°24’W.Whatarethelatitudeandlongitudeofthepointofarrivalbymidlatitudesailing? a) b) c) d)

39°12’N,62°28’W 39°06’N,62°34’W 39°02’N,62°37’W 38°56’N,62°42’W-correct

ProblemCG-29.Avesselsteams580milesoncourse083°Tfromlatitude13°12’N,71° 12’W.Whatarethelatitudeandlongitudeofthepointofarrivalbymid-latitude sailing?

a) b) c) d)

14°17’N,61°23’W 14°20’N,61°21’W 14°23’N,61°19’W-correct 14°25’N,61°17’W

ProblemCG-30.Avesselsteams640milesoncourse047°Tfromlatitude34°45’N, longitude140°00’E.Whatarethelatitudeandlongitudeofthepointofarrivalby mid-latitudesailing? a) b) c) d)

41°57’N,150°02’E 42°01’N,149°57’E-correct 42°06’N,149°53’E 42°09’N,149°50’E

ProblemCG-9.Avesselatlatitude21°18.5’N,longitude157°52.2’Wheadsfora destinationatlatitude8°53.0’N,longitude79°31.0’W.Determinethetruecourseand distancebyMercatorsailing. a) b) c) d)

081°T,4617.5miles 081°T,4915.8miles 099°T,4617.5miles-correct 099°T,4915.8miles

ProblemCG-13.Avesselatlatitude29°38.0’N,longitude93°49.0’Wheadsfora destinationatlatitude24°38.0’N,longitude82°55.2’W.Determinethetruecourse anddistancebyMercatorsailing. a) b) c) d)

115°T,637miles 117°T,658miles-correct 122°T,648miles 126°T,665miles

ProblemCG-16.Avesselatlatitude32°14.7’N,longitude66°28.9’Wheadsfora destinationatlatitude36°58.7’N,longitude75°42.2’W.Determinethetruecourseby Mercatorsailing. a) b) c) d)

058.2°T  235.2°T 301.8°T-correct 348.3°T

ProblemCG-6.Avesselatlatitude18°54’N,longitude73°00’Eheadsforadestination atlatitude13°12’N,longitude54°00’E.Determinethetruecourseanddistanceby Mercatorsailing.

a) b) c) d)

 247°T,1161miles  250°T,1172miles  253°T,1154miles-correct  256°T,1136miles

ProblemCG-3.Avesselatlatitude10°22’S,longitude7°18’Eheadsforadestination atlatitude6°54’N,longitude57°23’W.Determinethecourseanddistanceby Mercatorsailing. a) b) c) d)

 285°T,3825.3miles  285°T,4025.7miles-correct  296°T,3825.3miles  296°T,4025.7miles

ProblemCG-24.Avesselsteams1082milesoncourse047°Tfromlatitude37°18.0’N, longitude24°40.0’W.Whatisthelatitudeandlongitudeofthepointofarrivalby Mercatorsailing? a) b) c) d)

49°30’N,6°22’W 49°33’N,6°25’W 49°36’N,6°28’W-correct 49°39’N,6°31’W

ProblemCG-26.Avesselsteams1650milesoncourse077°Tfromlatitude12°47’N, longitude45°10’E.Whatisthelatitudeandlongitudeofthepointofarrivalby Mercatorsailing? a) b) c) d)

18°54’N,72°58’E 18°58’N,72°52’E-correct 19°02’N,72°44’E 19°06’N,72°36’E

ProblemCG-779.Youdepartlatitude49°45.0’N,longitude6°35.0’Wandsteam3599 milesoncourse246.5°T.WhatisthelongitudeofyourarrivalbyMercatorsailing? a) b) c) d)

76°36.2’W 77°02.8’W-correct 78°14.0’W 78°22.6’W

ProblemCG-914.Yourvesseldepartslatitude32°45’N,longitude79°50’W,andis boundforlatitude34°21’S,longitude18°29’E.DeterminethedistancebyMercator sailing. a) 5,021miles b) 6,884miles-correct

c) 6,954miles d) 7,002miles ProblemCG-988.Yourvesselreceivesadistresscallfromavesselreportingher  positionaslatitude35°01’S,longitude18°51’W.Yourpositionislatitude30°18’S, longitude21°42’W.Determinethetruecoursefromyourvesseltothevesselin distressbyMercatorsailing. a) b) c) d)

135°T 149°T 153°T-correct 160°T

ProblemCG-100.Determinethedistancefromlatitude59°12’N,longitude14°00’W tolatitude59°12’N,longitude03°20’Wbyparallelsailing. a) b) c) d)

324.2miles 325.4miles 327.7miles-correct 328.9miles

ProblemCG-99.Determinethedistancefromlatitude34°18’S,longitude172°40’Eto latitude34°18’S,longitude152°38’Ebyparallelsailing. a) b) c) d)

993.0miles-correct 995.2miles 996.4miles 998.6miles

ProblemCG-765.Youdepartlatitude15°48’N,longitude174°06’Eandsteam905 milesoncourse090°T.Whatisthelongitudeofarrival? a) b) c) d)

165°41’W 170°13’W-correct 172°47’W 179°06’E

ProblemCG-987.Yourvesselreceivesadistresscallfromavesselreportingher  positionaslatitude35°01.0’S,longitude18°51.0’W.Yourpositionislatitude35° 01.0’,longitude21°42.0’W.Determinethetruecourseanddistancefromyourvessel tothevesselindistressbyparallelsailing. a) b) c) d)

090°T,140.0miles-correct 090°T,189.2miles  270°T,140.0miles  270°T,189.2miles

ProblemCG-775.Youdepartlatitude38°12’S,longitude12°06’Wandsteam1543 milesoncourse270°T.Whatisthelongitudeofarrival? a) b) c) d)

44°49’W-correct 45°12’W 45°37’W 45°42’W

ProblemCG-1006.Theinitialgreatcirclecourseanglebetweenlatitude23°S, longitude42°W,andlatitude34°S,longitude18°Eis063.8°.Whatisthetruecourse? a) b) c) d)

063.8°T 116.2°T-correct  243.8°T  296.2°T

ProblemCG-549.Thegreatcircledistancefromlatitude24°N25.3’W,longitude83° 02.6’Wtolatitude35°57.2’N,longitude5°45.7’Wis3966.5miles.Determinethe latitudeofthevertex. a) b) c) d)

38°46.2’N 38°16.4’N 38°09.4’N-correct 37°57.3’N

ProblemCG-548.Thegreatcircledistancefromlatitude8°50.0’N,longitude80°21.0’ Wtolatitude22°36.0’N,longitude128°16.0’Eis7801milesandtheinitialcourseis 318°45’T.Thelatitudeofthevertexis49°20.6’N.Whatisthelongitudeofthevertex? a) b) c) d)

156°43’W 162°41’W-correct 159°32’W 161°18’W

ProblemCG-642.Youareonagreatcircletrackdepartinglatitude25°50’N, longitude77°00’W,andyourinitialcourseis061.7°T.Thepositionofthevertexis latitude37°35.6’N,longitude25°57.8’W.Whatisthedistancealongthegreatcircle trackbetweenthepointofdepartureandthevertex? a) b) c) d)

 2735.1miles  2664.9miles-correct  2583.2miles  2420.0miles

ProblemCG-555.Thegreatcircledistancefromlatitude35°08’S,longitude19°26’E tolatitude33°16’S,longitude115°36’Eis4559milesandtheinitialcourseis121°T. Determinethelatitudeofthevertex. a) b) c) d)

44°29.1’S 45°30.9’S-correct 46°18.2’S 43°41.8’S

ProblemCG-556.Thegreatcircledistancefromlatitude35°08’S,longitude19°26’E tolatitude33°16’S,longitude115°36’Eis4559milesandtheinitialcourseis121°T. Determinethelongitudeofthevertex. a) b) c) d)

 26°50.9’E 65°45.9’E-correct 69°19.1’E 72°18.3’E

ProblemCG-643.Youareonagreatcircletrackdepartingfrompositionlatitude25° 50’N,longitude77°00’W.Thepositionofthevertexislatitude37°35.6’N,longitude  25°57.8’W.Thedistancealongthegreatcircletrackfromthevertextoapoint(x)is 600mileswestward.Determinethepositionofpoint(x)onthegreatcircletrack. a) b) c) d)

36°47.5’N,38°21.8’W 36°50.4’N,38°25.6’W 36°55.6’N,38°30.0’W-correct 37°02.3’N,38°34.4’W

ProblemCG-105.Determinethegreatcircledistanceandinitialcoursefromlatitude  24°52.0’N,longitude78°27.0’Wtolatitude47°19.0’N,longitude6°42.0’W. a) b) c) d)

3593miles,048.1°T-correct 3457miles,053.3°T 3389miles,042.4°T 3367miles,045.0°T

ProblemCG-108.Determinethegreatcircledistanceandinitialcoursefromlatitude  26°00’S,longitude56°00’Wtolatitude34°00’S,longitude18°15.0’E. a) b) c) d) 

3705miles,153°T 3841miles,068°T 3849miles,248°T 3805miles,117°T-correct