1. You are steaming on a course of 133˚T 1. You are steaming on a course of 133˚T at 16 knots. At 2216Hrs. you observe at 16 knots. At 2216Hrs. you observe lighthouse bearing 086˚T. At 2223Hrs. lighthouse bearing 086˚T. At 2223Hrs. the lighthouse bears 054˚T. What is your the lighthouse bears 054˚T. What is your distance off the second bearing?distance off the second bearing?

Ans. 2.6 miles

1. Solution:

Co. 133º T

1st Brg. 086º2216 Hrs.

2nd B

rg. 0

54º

2223 Hrs.

A

B

C

Ltho

Find: Dist. Find: Dist. At 2At 2ndnd Brg Brg.

• Co. = 133º• 1st Brg. = 086º (-)• < a = 47º• Time of 1st Brg.= 2216• Time of 2nd Brg =2223• Steaming Time = 7 mins.

• AB = Speed x Time / 60• = 16 x 7• 60• AB = 1.87 nm

Co. = 133º2nd Brg. = 054º( - ) < b = 79º < a = 47º ( - ) ( b – a ) = 32º

BC = AB x Sin < a Sin ( b – a ) = 1.87 x Sin 47º Sin 32ºBC = 2.58 nm

2. You are steaming on a course of 2. You are steaming on a course of 071˚T at 19 knots. At 1907Hrs. you 071˚T at 19 knots. At 1907Hrs. you observe a lighthouse bearing 122˚T. At observe a lighthouse bearing 122˚T. At 1915Hrs. the lighthouse bears 154˚T. 1915Hrs. the lighthouse bears 154˚T. What is your distance off the second What is your distance off the second bearing? bearing?

Ans. 3.7 miles

• 2. Solution:

Co. 071º T1907 Hrs.

1 st Brg. 122º T

2 nd Brg. 154º T

1915 Hrs.

Lighthouse

Find: Dist at 2nd Brg.

A

B

C

• Co. = 071º• 1st Brg. = 122º• < a = 51º• Time of 1st Brg.= 1915• Time of 2nd Brg =1923• Steaming Time = 8 mins.

• AB = Speed x Time/ 60• = 19 x 8• 60• AB = 2.53 nm

Co. = 071º2nd Brg = 154º ( - ) < b = 83º < a = 51º ( - )( b - a ) = 32º

BC = AB x Sin < a Sin ( b – a ) = 2.53 x Sin 51º Sin 32º BC = 3.71 nm

• 3.You are steaming on a course 168˚T 3.You are steaming on a course 168˚T at a speed of 18 knots. At 1426Hrs. you at a speed of 18 knots. At 1426Hrs. you sight a buoy bearing 144˚T. At 1435Hrs. sight a buoy bearing 144˚T. At 1435Hrs. you sight the same buoy bearing you sight the same buoy bearing 116˚T. What is your distance off at the 116˚T. What is your distance off at the second bearing and predicted distance second bearing and predicted distance when abeam?when abeam?

• Ans. 2.3 miles 2nd bearing, 1.8 miles

abeam

• 3. Solution:

C0 168º T

1426 Hrs

Ist Brg. 144º T2nd Brg. 116º

1435 Hrs.

A

CB

DFind: Dist. At 2nd Bearing and Predicted Dist. when Abeam

Buoy

• Co. = 168º

• 1st Brg. = 144º ( - )

• < a = 24º

• Time of 1st Brg. = 1426

• Time of 2nd Brg.= 1435

• Steaming Time = 9 mins.

• AB = Spd x Time / 60

• AB = 18 x 9

• 60

• AB = 2.7 nm

Co. = 168º2nd Brg. = 116º ( - ) < b = 52º < a = 24º( b – a ) = 28º

BC = AB x Sin a Sin (b – a ) = 2.7 x Sin 24º Sin 28º

BC = 2.33 nm

CD = BC x Sin < b = 2.33 x Sin 52º

CD = 1.83 (Dist Abeam)

4. You are going from a point in ZD( -2) to 4. You are going from a point in ZD( -2) to a point in (ZD +4) over a distance of a point in (ZD +4) over a distance of 6,589 miles at a speed of 25.0 knots 6,589 miles at a speed of 25.0 knots Zone time of departure is 0536Hrs. on Zone time of departure is 0536Hrs. on November 21, 2001. What is the ETA and November 21, 2001. What is the ETA and date of arrival?date of arrival?

Ans. 2310 Hrs., December 1

• 4. Solution:

• 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 / 24Steaming Time = 10d 23h 34m

• 5.You are on a voyage from Limon, Costa 5.You are on a voyage from Limon, Costa Rica, to Los Angeles, U.S.A. The distance Rica, to Los Angeles, U.S.A. The distance from pilot to pilot is 3,150 miles. The from pilot to pilot is 3,150 miles. The speed of advance is 14.0 knots. You speed of advance is 14.0 knots. You estimate 24.0 hours for bunkering at estimate 24.0 hours for bunkering at Colon, and 12.0 hours for the Panama Colon, and 12.0 hours for the Panama Canal transit. If you take departure at 1836 Canal transit. If you take departure at 1836 hours (ZD+6), on 28 January. What is your hours (ZD+6), on 28 January. What is your ETA (ZD+8) at Los Angeles?ETA (ZD+8) at Los Angeles?

Ans. 1336 Hrs., 8 February

• 5. Solution:

• 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 / 24ST = 09d 09h 00m

6. A ship is at DR longitude 67˚25’ W at 6. A ship is at DR longitude 67˚25’ W at 1815 ZT steaming on a westerly course 1815 ZT steaming on a westerly course when the navigator informs the Captain when the navigator informs the Captain that the ship is about to enter a new time that the ship is about to enter a new time zone. The Captain orders that the ship’s zone. The Captain orders that the ship’s clocks be set to the new zone time whenclocks be set to the new zone time when the next hour is struck.the next hour is struck. What will be the What will be the new zone description, and what will be new zone description, and what will be the zone time by ship’s clock the zone time by ship’s clock immediately on being set to new zone immediately on being set to new zone time?time?

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

67º 25’ W

1815 ZT1815 ZT

67º 30’W

ZD + 4ZD + 4ZD + 5ZD + 5

• 7. A vessel at Lat. 32˚ 14.7’ N, Long. 066˚ 7. A vessel at Lat. 32˚ 14.7’ N, Long. 066˚ 28.9’ W heads for a destination at Lat. 36˚ 28.9’ W heads for a destination at Lat. 36˚ 58.7’ N, Long. 075˚ 42.2’ W. Determine the 58.7’ N, Long. 075˚ 42.2’ W. Determine the true course and distance by the mercator true course and distance by the mercator sailing?sailing?

Ans. 302˚ T, 538.2 nm.

• 7. Solution:• Lat1= 32º 14.7’N

• Lat2= 36º 58.7’N• Dlat = 04º 44’N• x 60• Dlat = 284’

MP1 = 2033.31MP2 = 2376.99 ( - )DMP = 343.68

Long1 = 066º 28.9 WLong2 = 075º 42.2’ WDlo = 9º 13.3’ W x 60Dlo = 553.3’

Tan Co. = Dlo / DMP = 553.3 / 343.68 = 1.60992 inv Tan Co. = N 58.15º W - 360º T/Co. = 301.8º

Dist. = Dlat / Cos. Co. = 284 / Cos 58.15ºDist = 538.2 nm

• How To Acquire Meridional Parts

• Meridional PartsMeridional Parts = Lat. / 2 + 45 = Tan = Lat. / 2 + 45 = Tan log x 7915.7 – Lat Sin x 23.3log x 7915.7 – Lat Sin x 23.3

8. You departed Lat. 18˚54’ N, Long. 073˚ 8. You departed Lat. 18˚54’ N, Long. 073˚ 00’ E and steamed 1,150 miles on course 00’ E and steamed 1,150 miles on course 253˚T. What are the latitude and 253˚T. What are the latitude and longitude of arrival by mercator’s longitude of arrival by mercator’s sailing?sailing?

Ans. Lat. 13˚18’ N, Long. 053˚ 03’ E

• 8. Solution:

• Dlat =Dist x Cos. Co.

• =1150 x Cos 73º

• Dlat = 336.22 / 60

• Dlat = 5º 36.2’ S

• Lat1 = 18º 54’ N

• Lat2 = 13º 17.8’ N

• MP1 = 1147.59

• MP2 = 781.52 ( - )

• DMP = 366.07

Dlo = DMP x Tan Co. = 366.07 x Tan 73º = 1197.36 / 60 Dlo = 19º 57.3’ WLong1= 73º 00’ E ( - )Long2 = 53º 02.7’ E

• 9. Your vessel receives a distress call 9. Your vessel receives a distress call from a vessel reporting her position as from a vessel reporting her position as LAT 35° 01.0’ SLAT 35° 01.0’ S, LONG 18° 51.0’ W. Your , LONG 18° 51.0’ W. Your position is position is LAT 35° 01.0’ S,LAT 35° 01.0’ S, LONG 21° LONG 21° 42.0’ W. Determine the true course and 42.0’ W. Determine the true course and distance from your vessel to the vessel distance from your vessel to the vessel in distress by parallel sailing.in distress by parallel sailing.

Ans. 090° T, 140.0 miles

• 9. Solution:

• Long1 = 18º 51’ W

• Long2 = 21º 42’ W

• Dlo = 2º 51’ W

• x 60’

• Dlo = 171’

Dist = Dlo x Cos Lat.

= 171 x Cos 35º 01’

Dist = 140 nm

Course = 090º

10. A vessel at Lat. 20˚00’ N, Long. 10. A vessel at Lat. 20˚00’ N, Long. 107˚30’ W is proceeding to Lat. 24˚40’ N, 107˚30’ W is proceeding to Lat. 24˚40’ N, Long. 112˚30’ W. What is the course Long. 112˚30’ W. What is the course and distance by mid-latitude sailing?and distance by mid-latitude sailing?

Ans. 315.4˚ T, 394.2 nm

• 10. Solution:

• Lat1 = 20º 00’ N

• Lat2 = 24º 40’ N( - )

• Dlat = 4º 40’ N

• x 60

• Dlat = 280’ (N)

• Long1 = 107º 30’ W

• Long2 =112º 30’ W( - )

• Dlo = 5º 00’ W

• x 60

• Dlo = 300’ ( W )

Mlat = Lat1 + Lat2 / 2 = (20º + 24º 40’ ) / 2 = 44º 40’ /2Mlat = 22º 20’

Dep. = Dlo x Cos Mlat = 300 x Cos 22º 20’Dep. = 277.50 ( W )

• DLAT

• 280 Inv. R-P 277.5

=

394.2 Inv. X-Y 44.74

DEP DIST COURSE

360ºCourse = 44.74º ( - )T/ Co. = 315. 26º or 315.4º

11. A vessel steams 666 miles on the 11. A vessel steams 666 miles on the course 295˚T from Lat. 24˚ 24’ N, Long. course 295˚T from Lat. 24˚ 24’ N, Long. 083˚00’ W. What are the latitude and 083˚00’ W. What are the latitude and longitude of the point of arrival by mid-longitude of the point of arrival by mid-latitude sailing?latitude sailing?

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

• 11. Solution:

• Dist. Course Dlat. Dep.

• 666 Inv P - R 295º = 281.46 Inv x - y 603.6

• Dlat = 281.46 / 60 = 4º 41.45’ N

• Lat1 = 24º 24’ N

• Dlat = 4º 41.45’ N

• Lat2 = 29º 05.45’ N

• Mlat = Lat1 + Lat2 / 2

• = 24º 24’ + 29º 05.45’ / 2

• Mlat = 53º 29.45’ / 2

• Mlat = 26º 44.7’ N

• Dlo = Dep / Cos Mlat

• = 603.6 / Cos 26º 44.7’ N

• Dlo = 675.9’ / 60

• Dlo = 11º 15.1’ W

• Long1 = 83º 00’ W

• Long2 = 94º 15.1’ W

12. The great circle distance from LAT 12. The great circle distance from LAT 08° 50.0’ N, LONG 080° 21.0’ W to LAT 08° 50.0’ N, LONG 080° 21.0’ W to LAT 12° 36’ N, LONG 128° 16’ E is 8,664 12° 36’ N, LONG 128° 16’ E is 8,664 miles and the initial course is 306.6° T. miles and the initial course is 306.6° T. Determine the latitude of the vertex.Determine the latitude of the vertex.

Ans. 37˚ 30.3’ N

• 12. Solution:

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

• = Cos 8º 50’ N x Sin 53.4º

• = 0.793295 inv. Cos

• Lat V = 37º 30.3’N

• NOTE:

• The name of the latitude of vertex is taken from the name of latitude of departure.

13 .On March 12, the LMT of meridian 13 .On March 12, the LMT of meridian transit is 1210H. What will be your zone transit is 1210H. What will be your zone time of meridian transit if you are on time of meridian transit if you are on Longitude 50˚ 20’ E?Longitude 50˚ 20’ E?

Ans. 11h 48m 40s ZT

• 13. Solution:

• Long. = 50º 20’ E

• CM = 45º 00’ E ( Central Meridian )

• Diff = 5º 20’

• x 4’

• T/ Diff. = 21m 20s

• LMT M.P. =12h 10m 00s ( - )

• ZT M. P. = 11h 48m 40s

• RULES FOR ZONE TIME

• CM > Long and named East ( + )

• CM < Long and named East ( - )

• CM > Long and named West ( - )

• CM < Long and named West ( + )

14. The local mean time of sunrise at 14. The local mean time of sunrise at Lat. 20° 20’ S, Long. 35° 40’ E is 0857H. Lat. 20° 20’ S, Long. 35° 40’ E is 0857H. Find the corresponding zone time of Find the corresponding zone time of sunrise?sunrise?

Ans. 08h 34m 20s

• 14. Solution:14. Solution:

• Long = 35º 40’ E

• CM = 30º 00’ E ( - )

• Diff = 5º 40’

• x 4

• T/Diff = 22m 40s

• LMT ☼↑ = 08h 57m 00s ( - )

• Z.T. ☼↑ = 08h 34m 20s

•

15. The GMT of LAN was found to be 22h 15. The GMT of LAN was found to be 22h 36m 24s. From the daily pages of the 36m 24s. From the daily pages of the Nautical Almanac, the GHA of the sun at Nautical Almanac, the GHA of the sun at 22h indicates 153° 48.5’. Find the 22h indicates 153° 48.5’. Find the corresponding increments for 36m 24s?corresponding increments for 36m 24s?

Ans. 9° 06’

• 15. Solution:15. Solution:

• 36m 24s

• 4

• 9º 06’

16. The GHA of the sun at 1200 GMT was 16. The GHA of the sun at 1200 GMT was found in the Nautical almanac to be 358° found in the Nautical almanac to be 358° 28.4’. What is the Equation of time?28.4’. What is the Equation of time?

Ans. 6m 06s

• 16. Solution16. Solution::

• GHA ☼ = 358º 28.4’

• CM = 360º 00.0’ ( - )

• Diff = 1º 31.6’

• x 4

• Eq. of Time = 06m 06.2s

17.17. The GHA of the sun at 1200 GMT The GHA of the sun at 1200 GMT was found in the Nautical Almanac to be was found in the Nautical Almanac to be 356° 28.4’. What will be the GMT of 356° 28.4’. What will be the GMT of Meridian Passage?Meridian Passage?

Ans. 12h 14m 06s GMT

• 17. Solution:

• GHA ☼ = 356º 28.4’

• CM = 360º 00’ ( - )

• Diff = 3º 31.6’

• x 4

• T/Diff = 14m 06s

• GMT = 12h 00m 00s ( + )

• GMT MP = 12h 14m 06s

18. While proceeding up a channel on 18. While proceeding up a channel on course 076˚ PGC, you notice a range in course 076˚ PGC, you notice a range in line dead ahead. Checking with the chart line dead ahead. Checking with the chart you find the true direction of the range you find the true direction of the range is 075˚T. Variation for the locality is 11˚ is 075˚T. Variation for the locality is 11˚ W. If the vessel’s course is 089˚ PSC. W. If the vessel’s course is 089˚ PSC. What is the deviation for the present What is the deviation for the present heading?heading?

Ans. 3˚ West

• 18. Solution:

• T / Brg. = 075º

• Var. = 11º W ( + )

• M/ Co = 086º

• Dev. = 3º W ( + )

• C/Co = 089º

19. Enroute from the Valpariso to Callao, 19. Enroute from the Valpariso to Callao, the true course is 005˚. Variation is 13˚ the true course is 005˚. Variation is 13˚ East, deviation is 4˚ West. A NW wind East, deviation is 4˚ West. A NW wind produce 5˚leeway. Which of the following produce 5˚leeway. Which of the following courses would you steer PSC to make courses would you steer PSC to make good the true course?good the true course?

Ans. 351˚

• 19. Solution:

• T/Co. = 005º

• Var. = 13º E

• M/Co. = 352º

• Dev. = 4º W

• C/Co. = 356º

• L/W = - 5º ( NW )

• CTS = 351º

C/C

o 356º

CTS 351º

NW Wind5º LW

000º T/C

o 0

05º

C/Error

20. A vessel is heading 110˚ by 20. A vessel is heading 110˚ by compass. Variation for the locality is compass. Variation for the locality is 9˚East, deviation is 5˚ West. A 9˚East, deviation is 5˚ West. A lighthouse bears 225˚PSC. The deviation lighthouse bears 225˚PSC. The deviation on a heading of 225˚ is 2˚ West. on a heading of 225˚ is 2˚ West. Determine the compass error.Determine the compass error.

Ans. 4˚ East

• 20. Solution:

• Variation = 9º E

• Deviation = 5º W ( - )

• C/Error = 4º E