Nav Solving Problem 2 (1-20)

8/13/2019 Nav Solving Problem 2 (1-20)

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1. You are steaming on a course of 133Tat 16 knots . A t 2216Hrs. you observe

lighthouse bearing 086T. At 2223Hrs.the lighthouse bears 054T. What is yourdistance of f the second bear ing?

Ans. 2.6 miles


3/48

1. Solu t ion :

2216 Hrs.

2223 Hrs.

A

B

C

Ltho

Find : Dist.

A t 2ndBrg.


4/48

Co. = 133

1stBrg. = 086 (-)

< a = 47 Time of 1stBrg.= 2216

Time of 2ndBrg =2223

Steam ing Time = 7 m ins.

AB = Speed x Time / 60

= 16 x 7 60

AB = 1.87 nm

Co. = 133

2ndBrg. = 054( - )

< b = 79< a = 47 ( - )

( b a ) = 32

BC = AB x Sin < a

Sin ( ba )

= 1.87 x Sin 47Sin 32

BC = 2.58 nm


5/48

2. You are steam ing on a cou rse of

071T at 19 knots. At 1907Hrs. you

observe a lighthouse bearing 122T. At1915Hrs. the lighthouse bears 154T.What is you r distance of f the second

bear ing?

Ans. 3.7 miles


6/48

2. Solu t ion :

1907 Hrs.

1915 Hrs.

Lighthouse

Find: Dist at 2ndBrg.

A

B

C


7/48

Co. = 071

1stBrg. = 122

< a = 51 Time of 1stBrg.= 1915

Time of 2ndBrg =1923

Steam ing Time = 8 m ins.

AB = Speed x Time/ 60

= 19 x 8 60

AB = 2.53 nm

Co. = 071

2ndBrg = 154 ( - )

< b = 83< a = 51 ( - )

( b - a ) = 32

BC = AB x Sin < a

Sin ( ba )

= 2.53 x Sin 51Sin 32

BC = 3.71 nm


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3.You are steaming on a course 168Tat a speed o f 18 kno ts. A t 1426Hrs. you

sight a buoy bearing 144T. At 1435Hrs.you sight the same buoy bearing 116T.What is your d istance of f at the second

bear ing and predicted distance when

abeam?

Ans. 2.3 miles 2nd bearing, 1.8 milesabeam


9/48

3. Solu t ion :

1426 Hrs

1435 Hrs.

A

CB

DFind: Dist. At 2nd

Bearing and Predicted

Dist. when Abeam

Buoy


10/48

Co. = 168

1stBrg. = 144 ( - )

< a = 24

Time of 1stBrg. = 1426

Time of 2ndBrg.= 1435

Steam ing Time = 9 m ins .

AB = Spd x Time / 60

AB = 18 x 9

60

AB = 2.7 nm

Co. = 168

2ndBrg. = 116 ( - )

< b = 52< a = 24

( b a ) = 28

BC = AB x Sin aSin (ba )= 2.7 x Sin 24

Sin 28

BC = 2.33 nm

CD = BC x Sin < b

= 2.33 x Sin 52

CD = 1.83 (Dis t Abeam )


11/48

4. You are go ing from a po int in ZD( -2) to

a po int in (ZD +4) over a dis tance o f

6,589 m iles at a speed o f 25.0 kno tsZone time of departu re is 0536Hrs. on

November 21, 2001. What is the ETA and

date of arr ival?

Ans. 2310 Hrs., December 1


12/48

4. Solu t ion :

Dep ZT = 0536 21 Nov.

S.T. = 2334 10 days

E.T.A. = 0510 02 Dec. T/ Diff. = 06 ( - )

E.T.A . = 2310 01 Dec .

Steaming Time = Distance / Speed

= 6,589 nm / 25 knots

= 263.56 / 24

Steaming Time = 10d 23h 34m


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5.You are on a voyage from Limon , Cos taRica, to Los Angeles , U.S.A . The dis tance

from p i lot to pi lo t is 3,150 m iles. Thespeed o f advance is 14.0 kno ts. You

est imate 24.0 hours for bunker ing at

Colon , and 12.0 hours fo r the PanamaCanal trans it . If you take departure at 1836

hours (ZD+6), on 28 January. What is your

ETA (ZD+8) at Los Angeles?

Ans. 1336 Hrs., 8 February


14/48

5. Solu t ion :

Dep Time = 1836 28 Jan.

S.T. = 0900 09 days E.T. A. = 0336 07 Feb.

T/Diff. = 02 ( - )

E.T.A. = 0136 07 Feb.

Bunkering = 2400 ( + )

E.T.A. = 0136 08 Feb. Transit = 1200 ( + )

E.T.A L .A = 1336 08 Feb.

ST = Dist / Speed

= 3150 / 14= 225 / 24

ST = 09d 09h 00m


15/48

6. A ship is at DR longitude 6725 W at1815 ZT steam ing on a wester ly course

when the navigator inform s the Captainthat the sh ip is about to enter a new t ime

zone. The Captain orders that the shipsclocks be set to the new zone t ime when

the next hour is s truck. What w i l l be the

new zone descr ip t ion , and what wi l l be

the zone time by ships clock

immediately on being set to new zone

t ime?

Ans. New ZD (+5); New ZT 1800


16/48

67 25 W

1815 ZT

67 30W

ZD + 4ZD + 5


17/48

7. A vessel at Lat. 32 14.7 N, Long. 06628.9 W heads for a destination at Lat. 36

58.7 N, Long. 075 42.2 W. Determine thet rue cou rse and distance by the mercator

sai l ing?

Ans. 302 T, 538.2 nm.


18/48

7. Solu t ion :

Lat1= 32 14.7N

Lat2= 36 58.7N Dlat = 04 44N

x 60

Dlat = 284

MP1 = 2033.31

MP2 = 2376.99 ( - )

DMP = 343.68

Long1 = 066 28.9 W

Long2 = 075 42.2 W

Dlo = 9 13.3 Wx 60Dlo = 553.3

Tan Co. = Dlo / DMP

= 553.3 / 343.68

= 1.60992 inv Tan

Co. = N 58.15 W- 360

T/Co. = 301.8

Dist. = Dlat / Cos. Co.= 284 / Cos 58.15

Dist = 538.2 nm


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How To Acquire Meridional Parts

Meridional Parts= Lat. / 2 + 45 = Tanlog x 7915.7Lat Sin x 23.3


20/48

8. You departed Lat. 1854 N, Long. 07300 E and steamed 1,150 miles on course

253T. What are the latitude andlongitude of arrival by mercatorssai l ing?

Ans. Lat. 1318 N, Long. 053 03 E


21/48

8. Solu t ion :

Dlat =Dist x Cos. Co.

=1150 x Cos 73 Dlat = 336.22 / 60

Dlat = 5 36.2 S

Lat1 = 18 54 N

Lat2 = 13 17.8 N

MP1 = 1147.59

MP2 = 781.52 ( - )

DMP = 366.07

Dlo = DMP x Tan Co.= 366.07 x Tan 73

= 1197.36 / 60

Dlo = 19 57.3 WLong1= 73 00 E ( - )Long2 = 53 02.7 E


22/48

9. You r vessel receives a dis tress cal lf rom a vessel repo rt ing her pos i tion as

LAT 3501.0 S, LONG 1851.0 W. Yourposi t ion is LAT 3501.0 S,LONG 2142.0 W. Determine the true course and

distance from your vessel to the vesselin d istress b y paral lel sai l ing .

Ans. 090 T, 140.0 miles


23/48

9. Solu t ion :

Long1 = 18 51 W

Long2 = 21 42 W

Dlo = 2 51 W

x 60

Dlo = 171

Dist = Dlo x Cos Lat.

= 171 x Cos 35 01Dist = 140 nm

Cou rse = 090


24/48

10. A vessel at Lat. 2000 N, Long.10730 W is proceeding to Lat. 2440 N,

Long. 11230 W. What is the courseand distance by m id-lat i tude sai l ing?

Ans. 315.4 T, 394.2 nm


25/48

10. Solu t ion :

Lat1 = 20 00 N

Lat2 = 24 40 N( - ) Dlat = 4 40 N

x 60

Dlat = 280 (N)

Long1 = 107 30 W

Long2 =112 30 W( - )

Dlo = 5 00 W

x 60

Dlo = 300 ( W )

Mlat = Lat1 + Lat2 / 2

= (20 + 24 40 ) / 2

= 44 40 /2Mlat = 22 20

Dep. = Dlo x Cos Mlat= 300 x Cos 22 20

Dep . = 277.50 ( W )


26/48

DLAT

280 Inv. R-P 277.5=

394.2 Inv. X-Y 44.74

DEP DIST COURSE

360

Course = 44.74 ( - )T/ Co. = 315. 26 or 315.4


27/48

11. A vessel steams 666 miles on the

course 295T from Lat. 24 24 N, Long.

08300 W. What are the latitude andlong i tude of the point o f arr ival by m id-lat i tude sai l ing?

Ans. Lat. 29 05.5 N, Long. 094 15 W


28/48

11. Solu t ion :

Dist. Course Dlat. Dep.

666 Inv P - R 295 = 281.46Inv x - y 603.6 Dlat = 281.46 / 60 = 4 41.45 N

Lat1 = 24 24 N

Dlat = 4 41.45 N

Lat2 = 29 05.45 N

Mlat = Lat1 + Lat2 / 2

= 24 24 + 29 05.45 / 2

Mlat = 53 29.45 / 2

Mlat = 26 44.7 N


29/48

Dlo = Dep / Cos Mlat

= 603.6 / Cos 26 44.7 N

Dlo = 675.9 / 60 Dlo = 11 15.1 W

Long1 = 83 00 W

Long2 = 94 15.1 W


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12. The great circ le d istance from LAT

0850.0 N, LONG 08021.0 W to LAT

1236 N, LONG 12816 E is 8,664m iles and the in it ial course is 306.6T.Determ ine the lat i tude of the vertex.

Ans. 37 30.3 N


31/48

12. Solu t ion:

Cos Lat V = Cos Lat1 x Sin I.C.

= Cos 8 50 N x Sin 53.4 = 0.793295 inv. Cos

Lat V = 37 30.3N

NOTE:

The name of the lat i tude of vertexistaken from the name of lat i tude of

departure.


32/48


33/48

13. Solu t ion :

Long. = 50 20 E

CM = 45 00 E ( Central Meridian) Diff = 5 20

x 4

T/ Diff. = 21m 20s

LMT M.P. =12h 10m 00s ( - )

ZT M. P. = 11h 48m 40s


34/48

RULES FOR ZONE TIME

CM > Long and named East ( + ) CM < Long and named East ( - )

CM > Long and named West ( - )

CM


35/48

14. The local mean time o f sunrise at

Lat. 2020 S, Long. 3540 E is 0857H.

Find the co rrespond ing zone time ofsunr ise?

Ans. 08h 34m 20s


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15. The GMT of LAN was found to be 22h

36m 24s. From the dai ly pages o f the

Naut ical Almanac, the GHA of the sun at22h ind icates 15348.5. Find theco rresponding increments for 36m 24s?

Ans. 9 06


38/48

15. Solu t ion :

36m 24s 4

9 06


39/48

16. The GHA of the sun at 1200 GMT was

found in the Nautical almanac to be 358

28.4. What is the Equation of time?

Ans. 6m 06s


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41/48

17. The GHA of the sun at 1200 GMT was

found in the Naut ical Almanac to be 356

28.4. What will be the GMT of MeridianPassage?

Ans. 12h 14m 06s GMT


42/48

17. Solu t ion :

GHA = 356 28.4

CM = 360 00 ( - )

Diff = 3 31.6

x 4

T/Diff = 14m 06s

GMT = 12h 00m 00s ( + )

GMT MP = 12h 14m 06s


43/48

18. Whi le proceeding up a channel on

course 076 PGC, you notice a range in

l ine dead ahead . Check ing w ith the chartyou f ind the true direct ion of the range is

075T. Variation for the locality is 11 W.If the vessels course is 089 PSC. Whatis the deviat ion fo r the present heading?

Ans. 3 West


44/48

18. Solu t ion :

T / Brg. = 075

Var. = 11 W ( + ) M/ Co = 086

Dev. = 3 W ( + )

C/Co = 089


45/48

19. Enroute from the Valpariso to Cal lao ,

the true course is 005. Variation is 13

East, deviation is 4 West. A NW windproduce 5leeway. Which of the followingcourses wou ld you s teer PSC to make

good the true course?

Ans. 351


46/48

19. Solu t ion :

T/Co. = 005

Var. = 13 E M/Co. = 352

Dev. = 4 W

C/Co. = 356

L/W = - 5 ( NW )

CTS = 351

NW Wind

5 LW000

C/Error

20 A l i h di 110 b


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20. A vessel is heading 110 bycompass. Var iat ion for the local ity is

9East, deviation is 5 West. Alighthouse bears 225PSC. The deviationon a heading of 225 is 2 West.Determ ine the compass error.

Ans. 4 East


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20. Solu t ion :

Variation = 9 E

Deviation = 5 W ( - ) C/Er ro r = 4 E

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Nav Solving Problem 2 (1-20)