13

1 CLOSE‐QUARTERSITUATIONDISTANCEAND

IMMEDIATEDANGERDISTANCE

Theunderstandingofthetermssuchas“closequarter

situation”and“immediatedanger”isveryimportant

toship’sofficers.Howtousethemcorrectlyisstilla

questionnotquitewellanswered.

Thedistance ofclose quartersituation(Dclose) is

definedasanact

iontakingdistancetoatargetship,

whenownshiptakesalargealteringcourse(suchas

90°)toavoidcollisionwiththetargetship,whichcan

makeownshippassthetargetshipattheminimum

safetydistance.

The distance of immediate danger (Dcollid) is

definedasanact

iontakingdistancetoatargetship,at

which own ship canʹt avoid collide with the target

shipbasedonownshipalonealteringcourseaction.

Thispaperwilldiscussownship’srelative motion

to a target shipand define DCPA as plus or minus.

When the target ship is on the left side of relative

motion tra

ck, define DCPA asʺpositiveʺ. When the

targetshipisontherightsideoftherelativemotion

track,defineDCPAasʺnegativeʺ.

1.1 Theshipmotionmathematicalmodels

Asshowninfigure1,OandTrepresentrespectively

theownshipandtheta

rgetship.Letthevelocityand

courseofownshipastheV

0andtheC0respectively,

thevelocityandcourseofthetargetshipareV

1andC1

respectively,azimuthforB,distanceforDist,

VCA 

.

So,therelativevelocityV

01andrelativecourseC01of

own ship to target ship, will show respectively as

follows

[1,2,and3]

:

0011

sin( ) sin( )







VV C V C (1)

0011

cos( ) cos( )







VV C V C

 (2)

(/) ( 0, 0)

90 ( 0, 0)

180 ( / ) ( 0)

270 ( 0, 0)

360 ( / ) ( 0, 0)

(0, 0)

































xy x y

xy y

xy x y

tg V V V V

tg V V V

tg V V V V

nil V V

(3)

221/2

()

VVV

(4)

Research on Double Collision Avoidance Mechanism

of Ships at Sea

X.Y.Bi&X.J.Liu

GuangzhouMaritimeInstitute,Guangzhou,China

ABSTRACT:Whentwopowerdrivenvesselsencounteratseasoastoinvolveriskofcollision, theyneedto

avoidcollisioneffectively.TheconceptofRightShipmaymisleadthisship’sofficersthinkinghisorherdirect

navigatinghasabsolutepowerwiththisspecialship.Thispa

perwilldefineDCPAsymbols;givethecauseand

themethodofdoublecollisionavoidancemechanismofshipsatsea.

http://www.transnav.eu

the International Journal

on Marine Navigation

and Safety of Sea Transportation

Volume 9

Number 1

March 2015

DOI:10.12716/1001.09.01.01

14



Figure1.Ownship’srelativemoti on

DCPA (distance of the closest point of approach)

and TCPA (time to closest point of approach) will

showrespectivelyasfollows

[1]

:

sin( ) DCPA Dist C B  (5)

cos( )



Dist C B

TCPA

 (6)

1.2 Thedistanceofclose‐quartersituation

Asshowninfigure2,letsafetyencounterdistancebe

DSPA,namelytheradiusofthecircleTinthisfigure,

theturningradiusofownshipbeR.Whenownship

arriveatpointAinrelativemotiontothe

targetship,

it begin to give way altering course tostarboard.

TheownshipwillgettopointEinthiscourseafter

doubleship’slengthwhichiscalledthelagdistance.

The own ship begin turning from point E, reached

pointE

1intruemotionontheturningcircleH,while

target ship along her course C

1 to get point T1. The

own ship turning circle H tangents the target ship

safetyencounterdistanceT

1atpointE1.So, segment

ATisthedistanceofclose‐quartersituationshowas

D. This close‐quarter situation distance can be

calculated by the relative motion equation of own

ship.

Letthecycleofownship’sturningcircleunderfull

rudderbet

0minutes,turningangularvelocitybe

,

the own ship takes t

0 minutes completing relative

movementtothetargetfrompointOtopointE.Then

theownshipbyturningmovementfromthepointE

to the point E

1, at this time the target ship carrying

safetydistance circlemovein linearfrom point Tto

point T

1, it takes time interval of t1‐t0, the turning

circleradius(R)isgenerallydoubletheshipʹslength,

then









 (7)



Figure2. The distance to target ship in close‐quarter

situation

In Cartesian coordinate system XOY, the relative

motiontrajectoryequationofownshipfrompointO

(t=0),thenpassingE(t=t

0),tothepointE2(t= t1)can

beexpressedasfollows.

000

10 010 1

10 010 01

1 cos( ( )) sin( ( )) cos( )

sin( ( )) cos( ( )) 1 sin( )

() sin()



 



 







 









 













tt tt C

Vt t Vt C

(8)

222

( sin()) (Y cos())



 

tt tt

DSPA X Dist B Dist B  (9)

Decoding formula (7)‐(9), making the ship’s

distancetothetargetatthetangentpointE

1equalto

thesafetyencounterdistance,wecangett

0andt1,so

we can solve the close‐quarter situation distance

(Dclose=D).

01 0

((( )))





DDCPA VTCPAt

 (10)

The own ship’s maximum altering course to the

rightwillbeasfollow.

2( )



 



 (11)

In the same way, considering the own ship as a

onepoint,theextentofthetargetshipwillexpandto

thesumoftwo shipʹslength, L

0 + L1,replace itwith

the DSPA in formula (9), we can get the collision

distance(Dcollid=D)andtheownship’smaximum

alteringcoursetostarboard.Ifthelengthofthetarget

shipisunknown,wecantakeadesirableL

1=330mto

getaconservativeestimatecollisiondistance.

2 EXAMPLES

2.1 Theactionofgive‐wayship

LetownshiplengthbeL

0=190m,velocityandcourse

as V

0=16 kns and C0=000° respectively, velocity and

15

courseofthetargetshiparerespectivelyV

1=18kns，

1=240°,the azimuthof target ship is B=30°,

distanceD=8nauticalmiles.

So, according to the formula above, we can get

owe ship’s relative speed to the target ship V

01=29.5

knots, the relative course C

01=32°.0. Target ship is

locatedontheleftsideoftherelativemovementline, 

theDCPA=+0.27nmiles,theTCPA=16.3minutes.



Figure3.Thecollisionriskandrelativemotiontotargetof

give‐wayship

TakingDSPA=1.0nauticalmiles,letthesumofthe

two ship length be the maximum, L

0+L1= 190+330=

520m= 0.28 nautical miles, the turning circle radius

(R) be double the shipʹs length, 380m=0.21 nautical

miles, the period of own ship turning circle be 5

minutes, wecan calculate the close‐quarter situation

distanceDclose=2.13nauticalmiles,atthatpointown

shipwillaltercourse

2°tostarboardside,wecanalso

calculate the collision distance Dcollid=0.59 nautical

miles,atthatpointownshipshouldaltercourse7°.2

tostarboardsidetoavoidcollision.

Anothercurveinthefigureiscollisionriskcurve.

This curve can be used to determine collision

avoidanceopportunity and

thevalueof thecollision

avoidanceaction.Forthisexampletheopportunityof

actiontakingis3.8nauticalmilesfromthetargetship,

theactionvalueisalteringcourse24°tostarboard.

2.2 Theactionofstand‐onship

LetownshiplengthbeL

0=190m,velocityandcourse

as V

0=18 kns and C0=240° respectively, velocity and

courseofthetargetshiparerespectivelyV

1=16kns，

1=000°,the azimuthof target ship is B=210°,

distanceD=8nauticalmiles.

So, according to the formula above, we can get

owe ship’s relative speed to the target ship V

01=29.5

knots, the relative course C

01=212°.0. Target ship is

locatedontheleftsideoftherelativemovementline, 

theDCPA=+0.27nmiles,theTCPA=16.3minutes.



Figure4.Thecollisionriskandrelativemotiontotargetof

stand‐onship

TakingDSPA=1.0nauticalmiles,letthesumofthe

two ship length be the maximum, L

0+L1= 190+330=

520m= 0.28 nautical miles, the turning circle radius

(R) be double the shipʹs length, 380m=0.21 nautical

miles, the period of own ship turning circle be 5

minutes, wecan calculate the close‐quarter situation

distanceDclose=1.94nauticalmiles,atthatpointown

shipwill alter course

180°tostarboard side, we can

also calculate the collision distance Dcollid=0.55

nautical miles,at that point own ship should alter

course14°.4tostarboardside toavoidcollision.The

opportunityofactiontakingis3.3nauticalmilesfrom

thetargetship;theactionvalueisalteringcourse24°

tostarboard.

3 THERELATIONSHIPBETWEENTHESETWO

DISTANCES

In case of crossing situation, the distance of close‐

quartersituationofgive‐wayshipisgreaterthanthat

ofthestand‐onship.Inthecaseaboveitisabout0.19

nauticalmiles,whichisequivalentto23seconds.The

minimum

 limit of the give‐way ship’s collision

avoidance action must not drag to the close‐quarter

situation distance in order to give stand‐on ship

enoughtimetotakenecessaryactionalone.Basedon

theaboveexample,takingdifferentbearingsoftarget

ship,wegetasetofnumericalsimulation;

theresults

of own ship’s action as give‐way ship are shown in

table1.

Table1. Crossing simulation calculation give‐way ship’s

action, (safety passing distance DSPA=1’.0, V

1=18, C1 =240,

D=8.0,V

01=29.5,C01=32,TCPA=16.3)

_______________________________________________

No.TargetshipOwnship (give‐wayship) action

___________________________________________

B DCPA Dclose C0 Dcollid C0 DactC0

_______________________________________________

1 28 0.55 1.89 81.4‐‐2.920

2 29 0.41 2.02 95.0‐‐3.223

3 30 0.27 2.13 108.7 0.59 7.2 3.824

4 31 0.13 2.21 113.8 0.96 50.4 4.127

5 32 ‐0.01 2.29 126.7 1.15 72 4.230

6 33 ‐0.15 2.35 137.5 1.19 93.6 4.532

 34 ‐0.29 2.40 153.4‐‐4.734

_______________________________________________



Also, considering the stand‐on ship taking action

alone,takingdifferentbearingsoftargetship,weget

16

a set of numerical simulation; the results of own

ship’sactionasstand‐onshipareshownintable2.

Table2. Crossing simulation calculation stand‐on ship’s

action, (safety passing distance DSPA=1’.0, V

1=16, C1 =0,

D=8.0,V

01=29.5,C01=212,TCPA=16.3)

_______________________________________________

No. Targetship Ownship (stand‐onship) action

___________________________________________

B DCPA Dclose C0 Dcollid C0DactC0

_______________________________________________

1 208 0.55 1.83 177.8‐‐2.8 18

2 209 0.41 1.90 178.6‐‐3.0 21

3 210 0.27 1.94 180.0 0.55 14.4 3.3 24

4 211 0.13 1.97 175.0 0.96 172.8 3.6 26

5 212‐0.01 1.99 112.3 1.01 93.6 3.8 28

6 213‐0.15 1.99 108.0 1.01 86.4 4.2 29

 214‐0.29 1.98 108.7‐‐4.4 31

_______________________________________________



Comparing the collision avoidance action results

intable1andtable2,thegive‐wayship’sdistanceof

close‐quartersituationislonger0.06to0.42milesthan

thatofthestand‐onship,whichisequivalentto7 to

51 seconds. The duty officer on the give‐way ship

should

 fully consider the stand‐on ship officer’s

psychology bearing ability and take early collision

avoidanceaction.

Also, we can get the result that the distance of

give‐way ship’s close‐quarter situation is larger 1.16

to1.46milesthanitscollisiondistance.Such ashort

time interval will not allow

the officer take any

hesitationandrequirehimorherinatimelymanner

to make correct collision avoidance decision‐making

andtakedirectactions.

4 CONCEPTOFDOUBLECOLLISION

AVOIDANCEANDCLOSE‐QUARTER

SITUATION

Ship’s collision segments will include the free

navigationatadistance,riskofcollision,close‐quarter



situation, imminent danger and collision

[5, 6]

. From

late segment of the risk of collision to the early

segmentclose‐quartersituation is theforming  phase

of close‐quarter situation and is the most important

momentofthemanipulatingactionaloneforthetwo

ships. After the two ships coordinated action, by

avoiding the close‐quarter situation,

 these two ships

can pass in a safety distance and can sufficiently

avoidcollision.

Inrule8ActiontoavoidCollision paragraph(a),

theminimumlimitofʺampletimeʺshouldbenotto

form the close‐quarter situation. The stand‐on ship

ʺmay however take action to avoid collision by

 her

manoeuvrealoneʺisthekeyofnotletthevesselsfall

inclose‐quartersituation.Sincethen,thereisnomore

absoluteRightrouteforthestand‐onship.Thestand‐

on ship must bear the obligation of action alone to

avoidclose‐quartersituation.Theconceptofʺdouble

collision avoidanceʺ in the modified collision

regulationshasbeenclearly revealedwhich requires

ship officers to have clear quantitative distance

figures.

Thekeystepand thefirst priorityof establishing

practice for preventing collision at sea are to avoid

close‐quarter situation. The outlook, judgment,

decision‐making, action and validation segments



shouldbearoundthisfirstprioritysoastograspthe

crucial point of collision avoidance. If this concept

becomescommoncognitiveandwidelybeused,more

andmoreshipmanipulations couldleadtosafepass

inadesirabledistance.

5 CONCLUSION

Through calculating the distance of close‐quarter

situation,

wefindthattheactionofgive‐wayvesselis

easier to achieve the desired effect of collision

avoidancethanthatofthestand‐onship.Wesuggest

that the give‐way vessel’s  collision avoidance action

should be strictly observed to make it not lose the

good anti‐collision opportunity. In

 fact, all ships are

responsibletoensurenavigationsafetyandprotection

ofthe marineenvironment and arethe main partof

the obligations. The division of stand‐on ship and

give‐wayshipbythecollisionavoidancerulesisonly

thedivisionofobligationforcollisionavoidanceand

not to

exempt the stand‐on ship from liability of

complyingwiththeobligations.Thegive‐wayvessel,

however,shouldfullyrecognizeherownadvantages,

give the stand‐on ship more behavior space and

relieve the psychological  pressure of stand‐on ship

officer.Becausethepurposeofpracticeforpreventing

collisionatsea

istoavoidclose‐quartersituation, we

should advocates the concept ofʺdouble collision

avoidanceʺ. Estimating the close‐quarter situation

distanceandtheimmediatedangerdistancecorrectly

will helpthe navigator fullyunderstand the process

of collision avoidance and take correct collision

avoidance action in a timely manner. This

massage,

nodoubt,will increasethe navigator’sresponsibility

andself‐confidenceofanti‐collisionmanipulationand

will be the basis of analyzing specific encounter

situationandcollisionavoidancedecision‐making.

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