Q1.

Calculate the course and distance by Mercator Sailing from: F (45° 00' N, 140°
00'E) to T (65° 00' N 110 ° W).
Lat T 65° 00' N. M.P Lat T 5157.72 Long. T 110 ° W.
Lat F 45° 00' N. MP Lat.F 3013.4 Long. F + 140 ° E.
D.lat 20 ° 00' N. D.M.P 2144.24 D.Iong. 250 ° W
.- 360 ° E.

110° E.

x 60 = 6600 Miles

Tan course = D.Long = 6600 =3.078

D.M.P 2144.24

# COURSE =N 72 ° E # T.COURSE= 072 °

Distance = D.Lat = d.lat. sec course

Cos. Co

= 1200 sec 72 ° = 3883.658 = 3883.7 miles.

Q02. A ship in fix (31° 30'N, 029° 30'E) nearest to Alexandria Iwbor, she wishes to
be in position (41° 20'N, 002° 20'E) nearest to Barcelona harbor. Calculate the
course and distance.
Lat B 41° 20' N. M.P Lat. B 2712.70 Long. B 29° 30' E
Lat A 31° 30' N. MP lat. A 1980.88 Long. A 02° 20' E.
D. lat. 09 ° 50' N. D.M.P 731.82 D. long 27° 10' W ( 1630 Miles).
(590 Miles).
Tan course = d.long = 1630 = 1.872637
D.M.P 731.82

#course =N 65.8° W = 360-65.8 =294.2°

Distance = d.lat = 590 =1440.5 miles

Cos co cos 298.1°

 Q
0
Q03.A ship in fix (39° 37'S,47° 28'W). - She .sailed on True Comse 040° for a 1
.,
;
distance of 2300 miles. Find the arrival position.

Intial lat. 39° 37' S. MP = 2577.821

D.Lat. 29° 21.9' N. . M.P= 614.25

Arrival lat. 10° 15.1 S D.M.P= 1963.57

.: Tan course = D.long / D.M.P

:. D.long = Tan course x D.M.P = Tan 40 °x 1 963.57 = 1647.6
1647.60 / 60 = 27 ° 27.6"

Initial long. 47° 28' W.

D.Long. 27° 27.6' E. ·

Final long. 20° 00.4' W.

Final position (10 ° 15.1' S, 020° 00.4' W).

Q04 .A ship in fix A- (12° 40' N , 35° 20' E ) she sailed due north till she reached
position

B-(18° 50' N, 35° 20' E) find the distance run.

Lat.A 112° 40' N

LatB I 18° 50' N

dlat 16° 10' N

(6° x 60) + 10' = 370 Miles.

distance = d.lat

distance = 370 Miles

Q05.A ship in fix A- (60° 00' N , 29° 18' E ) she sailed due east till she reached
position

B- (60° 00' N , 38° 30' E) find the distance run.

Long. A 29° 18' E

Long. B38° 30' E

d.long. 00 9° 12' E

(9° x 60) + 12' = 552 Miles.

Distance = departure= D.Long cos (lat)

Distance = 552 cos (60°) = 276 Miles.

15- A Mercator chart has a natural scale of 1:4840000 at Lat. 15⁰00’ N,

What will be its dimensions to enable the area between the parallels 07 ⁰00’
S and 34⁰21’ N and between the meridians 103⁰10’ E and 131⁰35’ E (Using
the geographical mile 1855.356 meters).

a) Calculate the D. Long:

Long A. 103⁰10’ E
Long B. 131⁰35’ E
D. Long 28⁰25 E = 1705 ‘

b) Finding Horizontal Border:

Length of 1705’ of Long. = 1705’ x 1855.356 x Cos 15⁰00’ = 3055592.353

On Chart On Earth
1 4840000
X 3055592.353

X=195622.2323 / 308250 = 0.63132 m = 631.32 mm

Horizontal Border = 631.32 mm

c) Length of 1’ Long.
 631.32 / 1705 = 0.37 mm for each 1’ Long.

d) Find The DMP:

Lat. A 07⁰00’ S M.P. 418.20

Lat. B 34⁰21’ N M.P. 2183.66
D.M.P. 2601.86 (Different Name Add)
e) Vertical Border:
Vertical Border = Length of 1’ of Long. X DMP
= 0.37 X 2601.86 = 962.7 mm

Chart Dimensions (962.7 mm X 631.32 mm)

16- Two ships at lat. 56˚ N with 75 n. miles apart sailed due north at 12 knots. When shall
the distance between them close to 63 n. miles?

Departure= D. long * Cos. Lat1

D.long = Departure1 = 75 = 134.12
Cos. Lat. Cos. 56

Departure2= D. long * Cos. Lat2

63 = 134.12 * Cos. Lat2
Cos. Lat2 = 63 = 0.46972
134.12
Cos. ^-1 0.46972 = 61.983============== Lat2=61˚ 59' N
Lat1 56˚ 00 N
Lat2 61˚ 59' N
D.Lat 5 ˚ 59’ N=========x60 =359 n. miles
Dist.=D.Lat=359 n. mile
Time = 359/12= 30 hrs.

17- A vessel steams 040° T for 2300 miles from a position 39° 37' S 47° 28' W. Find the
arrival position.

Answer:

d'lat = distance x cosine course

= 2300 x cos 40°
= 1761.9
 = 29° 21.9'
initial latitude 39° 37.0' S MP 2577.82
d'lat 29° 21.9' N MP 1963.67
final la ti tude 10° 15.1' S DMP 614.15
d'long = DMP x tan course
= 1963.57 x tan 40°
= 1647.7
= 27° 27.7' E
Initial longitude . 47° 28.0'W
d.long 27° 27.6' E
Final longitude 20° 00.4' W

Final position (10° 15.1' S 20° 00.4' W)

18- A ship Steamed from A (38⁰00’ N, 062⁰58’ E) to B (42⁰29’ N, 120 ⁰ 03’ E) On Great
Circle Track. Find the Following:
(1) Distance
(2) Initial and Final Course
(3) Position of Vertex

1) Distance
Long. A 62⁰58’ E
Long. B 120⁰03’ E
D. Long 57⁰05’ E
Cos Dist. = (Cos Lat. A * Cos Lat. B * Cos D. Long) + (Sin Lat. A* Sin Lat. B)
=(0.315800856)+(0.415802807)
=0.731603663
Dist. = 42.98⁰ = 2578.74 Miles
2) Initial and final Course

Sin I. Co = (Sin D. Long. * Cos Lat. B) /Sin Dist.

=(0.839461826 * 0.737473826) / 0.681743027
=0.619081124 / 0.681743027
t =0.908085745
I.CO = N 65.24⁰ E = 065.24⁰ (T)

Sin F. Co = (Sin D. Long. * Cos Lat A) / Sin Dist.

= =(0.839461826 * 0.788010753) / 0.681743027= 0.970314209
F. Co = S 76.00⁰ E = 104⁰ (T)
3) Position of Vertex
fgCos Lat. Vertex = Cos Lat. A * Sin I. Codcd
= 0.788010753 * 0.90807009
=0.715568995
Lat. Vertex = 44⁰18.6’ N

Tan D. Long Vertex = 1 / (Sin Lat. A * Tan I.Co)

= 1 / (0.615661475 * 2.168172302)
 =0.749142143
D. Long Vertex = 36⁰50.3’ E
Long. A 062⁰58’ E
D. Long. 036⁰50.3’ E
Long. B 099⁰48.3’ E
Vertex Position (44⁰18.6’ N, 099⁰48.3’ E)

Q6. Use chart 5052 to solve the following task:-

a) At 1200 the H.S.A between Le Treport Lt.Ho. and Cayeux-sur-Mer Lt.Ho. was 47˚ and at
the same time the relative Brg. of Le Treport Lt.Ho. Was 104˚ green, ship steers 034˚g
and Gyro error (2˚H), speed 12 knots?
Fix the ship position at 1200?

Answers:-
-The HAS is less than 90˚ then :
90˚ - 47˚= 43˚
-Draw a line joining the two light houses.
- Draw a line from each L H. with angle 43 ˚.
- Draw a circle from the intersection of the two lines passing through the two L.H. 's.
- course tosteer = 034 ˚g
- Gyro error = 2˚H
T.Course = 032 ˚ T
Relative bearing = 104 Green(+)
T. bearing = 136 ˚ T
- Draw the bearing line to cut the circle .

b) At 1400 hrs a ship in fix position (50° 45.5'N , 000° 51' E) , log speed 8 kts. , the ship
affected by both N.Westerly wind causing 6° leeway angle and Current
Setting 340°/2.5 kts., height of eye 15 meters, meteorological visibility 3 miles
If a ship require to pass 7 miles distance off Royal sovereign Lt.Ho. FIND:-

1) Course made good

2) Speed made good
3) Leeway Track
4) True course to steer
5) Gyro course to steer if gyro error 2° H
6) Compass Course to steer if Deviation is 4.9°W ( use year 2006)
7) The time when Royal Sovereign Lt.Ho. will be at the shortest distance .
8) The time when Royal Sovereign Lt.Ho. will be abeam and the beam distance
9) The time when Royal Sovereign Lt.Ho. will just appear & on what angle on the bow.

Answer

1) Course made good

- plot position at 1400
- Measure 7 miles from Royal Sovereign Lt.Ho. and draw a circle
- from position at 1400 draw a tangent to the previous circle
- measure this line on the rose it will be the C.M.G = 236 T

2) Speed made good

- From position 1400 draw the direction & speed of current within one hr.
- from the end of current direction measure 8 miles (ship speed) to intersect the
C.M.G at 1500
- Measure the distance from position at 1400 to the intersect point on C.M.G it will
be the speed made good within one hr.
- Speed made good = 7 kt.

3) Leeway Track
- draw a line from the end of current to the intersect point at 1500 hrs. and
measure this line on the rose it will be the Leeway track
- Leeway Track = 218 T

4) True course to steer

- wind direction is N.W that mean we will go against the wind direction( 6 deg.
leeway angle)
L.W.T = 218 T
L.W.A = + 6
T.CO.TO.STEER = 224 T

5) Gyro course to steer if gyro error 2 H

T.Co. = 224T
G.Error = +2 H
Gyro.Co.= 226G
6) Compass Course to steer if Deviation is 4.9W ( use year 2006)
Var. on chart 5 55 W on 1979 and decreasing 4' annually
2006 – 1979 = 27 x 4 = 108' (01 48' W)
05 55 – 01 48 = 04 07' W
Var. = 4.1 W
C.E = dev + var. = 4.9 W + 4.1 W= 9 W
To.co.to.steer = 224 T
C.E = +9
Compass .co. = 233 C
7) The time when Royal Sovereign lt.ho. will be at shortest distance
- from the lt.ho. make a perpendicular with C.M.G., it will be the shortest distance
point .(236+90=326)
- measure the distance from 1500hrs to this point = 7.3miles
- s.m.g(7) ----------- 60
7.3----------- x
X = 7.3 *60/ 7 = 63 = 1 h 03 minutes
- Shortest distance Time = 1500 +0103 = 1603
8) The time when Royal Sovereign lt.ho. will be abeam and the beam distance
- T.Co. To. Steer = 224 T
90
T.brg.= 314 T
- from the lt.ho. draw T.brg. 314 to intersect with C.M.G ( it will be the beam
point)
- measure the distance from 1500 to this point = 5.8miles
- s.m.g(07) -------------- 60m
5.8 -------------- x
 X = 5.8*60/7 = 50 minutes
- Time of abeam = 1500 + 0050 = 1550
- Beam distance = 7.2miles
9) The time when Royal Sovereign lt.ho. will just appear & on what angle on the
bow
- to find the Luminous range by using the diagram( Nominal range = 28M)
And visibility=3 miles
The Luminous range = 9.5 miles
- to find the geographical range
G.R = 2.03√H + 2.03√h
G.R = 2.03√28 + 2.03√15=18.6
G.R > Luminous range , therefore we will use range 9.5miles
- from Royal lt.ho. measure 9.5' and make a circle to intersect with C.M.G( this
point is the first appear point)
- measure the distance from 1400hrs to this point =8 miles
- s.m.g(07) ------------- 60m
8 ------------ x
X= 8 * 60 / 7 = 68= 1 h 8 minutes
- Time of first appear = 1400 + 0108 = 1508hrs
- to find the angle on the bow .measure the T.brg. of Royal .lt.ho from the point
of first appear = 286T
- Angle on the bow = T.brg. - T.Co.to.steer
Angle on the bow = 285 – 224 = 61 green

Q7.Use chart 5052 to solve the following task:-

A) Courses & Bearings are by magnetic Compass (C) use deviation table No 1, Course to
steer 080 ° C, ships speed 8 kts. (Use year 2002)

Assuming a continuous effect (during the total sailing period) of the tidal stream of

area M of the one hour before H.W spring at Dover

At 1900 pt du haut Banc Lt.Ho. brg. 116 ° C & Le Touquet lt.ho brg. 053 ° C

FIND: The ship's position at 1900 hrs

1) Compass Course to steer to pass 6 miles off Cap d, Alpreach lt.ho.

2) When will Cap d, Alpreach Lt.Ho. be abeam.
1)
- to calculate the Compass Error
var. !979 = 05 55 W decreasing 4' annually
2002 – 1979 =23 x 4 = 92' (01 32' W)
Var. on 2002 = 05 55' – 01 32' =04 23' =4.4W
C.E = 5.4W + 4.4W = 9.8 W
Compass.Co.= 080 Comp.brg1=116 C Comp.brg.2= 053 C
C.E = 9.8 W C.E = 9.8 W C.E = 9.8 W
T.Co.To.Steer = 070.2T T.brg.1= 106.2T T.brg.2 = 043.2T

Position at 1900 hrs. ( 50 25' N & 001 26.4E)

2) Compass Course to steer to pass 6 miles off Cap d,Alpreach lt.ho.

- from the tidal stream table which is on the chart area M (Dir.022/ Rate 1.6KT)
 - from the Cap. d. Alprech draw a circle (6 miles )
- from position at 1900 hrs draw a tangent with a circle and measure this line it will
be (Co.made good)
- from position at 1900 hrs draw the dir. Of current within one hr. (022/1.6') at point
(a)
- from point (a) measure 11 miles(ships speed) and intersect C.M.G at point (b)
and draw a line between a&b it will be T.Co.To.Steer = 351 T
- Compass Co. To. Steer = 351˚T +0˚.4W = 351˚.4 C

DEVIATION TABLE NO. (1)

SH. C. Deviation Magnetic Course Variation TRUE COURSE
Co.
350˚ 4˚.0 E 354˚ 4˚.4W 349˚.6

351.4 4˚.0 E 355.4 4˚.4W 351

000˚C 4˚.0 E 004˚ 4˚.4W 359.6

3) When will Cap d,Alpreach lt.ho.be abeam.

- add 90 deg. To T.co.to.Steer ( 90+ 351=441 -360=081 T)
- from Cap d. Alprech draw brg. 081T to intersect with C.M.G at point ( c )
- measure the distance from position 1900 to point (b) = 9.5' this is speed made good
- measure the distance between point b at 2000 and point c = 6.5'
- s.m.g (9.5) ---------- 60min
- 6.5 ---------- x

X= 6.5*60 /9.5 = 41 minutes

- Time of abeam = 2000 + 0041 = 2041hrs

B) A ship steers Gyro Course 232°g (gyro error 2°H) by speed 8 kts.

At 1220 Dungeness Lt.Ho. bore 347°g and distance off 6 miles

At 1340 Hasting Lt.Ho. Bore 332°g and distance off 10 miles.

Find the actual set and rate of Drift.

Answers:-

- G.brg.1= 347 g G.brg.2=332 g

G.error= -2 H G.error= -2 H
T.brg.1= 345 T T.brg.2=330 T

- plot the position at 1220 & 1340 hrs.

-
- draw a line between two position and measure it . this is C.M.G=238T
-
- from position at 1220 draw T.co.to.steer (T.co.= 232 – 2 = 230 T)
 -
- 1340 – 1220 = 0120 (80 minutes)
-
- Distance run on T.co. = 8*80/60 = 10.7 miles
-
- From position at 1220 measure 10.7 miles and plot D.R position at 1340hrs
-
- From D.R position at 1340 draw a line to reach to the fixed position at 1340 and measure
this line on the chart rose this line is the current direction within 80 minutes = 276 deg. T
-
-Measure the distance between D.R & fix position at 1340 = 3.1miles
E
- 3.1 --------------80 min.
X ---------------60 min.

X= 60*2.3 / 80 = 2.3'

Actual Current set 276 T / 2.3 kt.

Q8.
- Courses and Bearings by Magnetic Compass.
- Deviation table (1).
- Compass Course 045° (C).
- Ship's Speed 8 knots.
- Use the Tidal Stream of area
3 hours before H.W (Spring tide) at Dover.
- Variation = 03° 20 W (1993) decreasing 7 annually. ( Use 2005)
- AT 0900
Cap Gris Nez Lt. Ho. Brg. 161° (C).
CALAIS Lt. Ho. Brg. 103° (C).
Find:
a) Position at 0900.
b) Compass Course to steer to pass one mile distance off Ruytingen S.W. buoy.
c) When Will Ruytingen S.W buoy be abeam?

Answers:-
1. Calculate Variation in 2005.
2005-1993 = 12 year × 7 = 84 = 01° 24° W)
Variation = 03° 20°W decrease = - 01° 24’
Variation = 01° 56° W = 1.9° W (2005).
2. Deviation table (1).
C.CO Dev. M.CO
040° 1.2°W
050° 2.2°W
060° 3.0°W
070° 4.2°W

040° ------1.2°
045° ----- X
050° ----- 2.2°
10° ---1.0°
05° --- X
X= 5*1.0/10= 0.5°
3. Dev. Of C.CO 045° = 1.2°+0.5°=1.7°W
C.error= Var. + Dev.
1.9°W + 1.7°W = 3.6° W
4. Cap Gris Nez Lt. Ho. Brg. 161° (C) -3.6° W= 157.4°(T)
CALAIS Lt. Ho. Brg. 103° (C) -3.6° W=099.4°(T)
5. from the lt.ho. draw T.brg.
Cap Gris Nez Lt. Ho. Brg. 157.4°(T)
CALAIS Lt. Ho. Brg. 099.4°(T)
A) PSN. At 0900 (50° 59.7°N, 001° 30°E).
6. draw a circle (1 miles ) From Ruytingen S.W. buoy.
Course made good .
- from position at 0900 Tangential with a circle Ruytingen S.W. buoy
- get it CMG.

7. from PSN 0900 draw current 220/2.8.

End of current measure log SPD. 8 knots. Will get ( b ) and T.co = 054°T

8. T.CO 050°+1.9° W (Var. Above) = 051.9° M.CO

9. after doing interpolation Dev. of C.co. is 2.6° W +051.9°= 054.5° ©

10. Abeam with Ruytingen S.W. buoy.

T.Brg = T. Co + 90° ( STBD) = 050°+90°= 140° (T)

 Draw A line to CMG.

11. SMG was 5.4 from the 1000 PSN. To abeam the Dist. Is 11.4 mile.
5.2 ----60
11.4 --- x 131.5 min
0900 + 02: 11: 30 = 11: 11:30. Abeam time.

- Measure 7 miles from Royal Sovereign Lt.Ho. and draw a circle

- from position at 1400 draw a tangent to the previous circle
- measure this line on the rose it will be the C.M.G = 236 T

Q9.At 1200 hrs. Bassurelle RC Racon Bore 070°g. and the same time
Vergoyer SW Bouy Bore 090 °g (gyro error Nill).
log speed 12 kts, the ship affected by both NNW wind causing 6° leeway angle and
Current Setting 230°/3 kts., height of eye 15 meters, meteorological visibility 3 miles
If a ship require to pass 2 miles distance Vergoyer SW Buoy.

FIND: -
1. Ship position at 1200.
2. Course made good.
3. Speed made good.
4. Leeway Track.
5. True course to steer
6. Compass Course to steer if Deviation is 3.9°W ( use year 2006)
7. The time when. Bassurelle RC Racon will be at the shortest distance.
8. The time when Bassurelle RC Racon. will be abeam and the beam distance.
9. Find the Alteration course AC ( WP PSN) for the ship if you know its planning with
abeam point with Bassurelle RC Racon.

1. draw tow Brg. And find the position at 1200.

2. Course made good

- Measure 2 miles from Vergoyer SW Buoy and draw a circle
- from position at 1200 draw a tangent to the previous circle
- measure this line on the rose it will be the C.M.G 084.

3. Speed made good

- From position 1200 draw the direction & speed of current within one hr.
- from the end of current direction measure 12 miles (ship speed) to intersect the
C.M.G at 1300
- Measure the distance from position at 1300 to the intersect point on C.M.G it will
be the speed made good within one hr.
- Speed made good = 9.4 kt.

4.Leeway Track
- draw a line from the end of current to the intersect point at 1300 hrs. and
measure this line on the rose it will be the Leeway track
- Leeway Track = 075
 5.True course to steer
- wind direction is N.W that mean we will go against the wind direction( 6 deg.
leeway angle)
L.W.T = 075 T
L.W.A = - 6
T.CO.TO.STEER = 069 T

6.Compass Course to steer if Deviation is 4.5W ( use year 2006)

Var. on chart 5 55 W on 1979 and decreasing 4' annually
2006 – 1979 = 27 x 4 = 108' (01 48' W)
05 55 – 01 48 = 04 07' W
Var. = 4.1 W
C.E = dev + var. = 3.9 W + 4.1 W= 8 W
To.co.to.steer = 069 T
C.E = +8
Compass .co. = 077 C
7. The time when. Bassurelle RC Racon will be at shortest distance
- from the lt.ho. make a perpendicular with C.M.G., it will be the shortest distance
point .(084+90=176)
- measure the distance from 1300hrs to this point = 7.0miles
- s.m.g(9.4) ----------- 60
7.0----------- x
X = 7.0 *60/ 9.4 = 63 = 00h 45 minutes
- Shortest distance Time = 1300 +0045 = 1345
8.The time when Bassurelle RC Racon will be abeam and the beam distance
- T.Co. To. Steer = 069 T+360
-90 red
T.brg.= 339T
- from the lt.ho. draw T.brg. 339 to intersect with C.M.G ( it will be the beam
point)
- measure the distance from 1300 to this point = 7.6miles
- s.m.g(9.4) -------------- 60m
7.6 -------------- x
X = 7.6*60/9.4 = 49 minutes
- Time of abeam = 1300 + 0049 = 1349
- Beam distance = 4.0miles

9. Find the Alteration course AC ( WP PSN) for the ship if you know its planning
with abeam point with Bassurelle RC Racon.
AC will be Same Abeam point in position 050 25.8 N,001 00 E.

DEVIATION TABLE NO. (1)

.SH. C. Co Deviation .SH. C. Co Deviation

C˚000 E 4.0 C ˚180 E ˚3.1

010 E 3.0 190 E 2.2

020 E 2.0 200 E 0.6

030 E 0.4 210 W 2.0

040 W 1.2 220 W 3.2

050 W 2.2 230 W 4.4

060 W 3.0 240 W 4.2

070 W 4.2 250 W 4.0

080 W 5.4 260 W 5.6

090 W 4.3 270 W 2.1

100 W 3.1 280 W 0.8

110 E 2.2 290 E 0.2

120 W 0.6 300 E 1.5

130 E 2.2 310 E 2.9

140 E 3.3 320 E 4.1

150 E 4.5 330 E 5.6

160 E 5.4 340 E 4.2

170 E 4.2 350 E 4.0

Q10- In what latitude will one minute (1') of the latitude scale on a Mercator Chart equal two minutes
(2') of the longitude scale?
- Scale of 1' of lat. = Scale of 1' long x sec. lat.
1/2 = sec lat.
.. lat = 60°
Q11. On a Mercator Chart 1 ° of longitude measures 152.5mm. What is the length along the meridian
between lat. 48° 13'N and lat. 49° 02'N?
1' of long = 1 meridianal part
m.p. for lat. 49° 02. = 3367.45.
m.p. for lat. 48° 13. = 3239.54.
-----------------------------------------
D.m.p. 73.91
length along the meridian = 73.91 x 152.5
-------------- --- = 187.85 mm.
60
Q12.- what is the limitations of the following :

1- Mercator charts.
2- Gnomonic charts.
.
1) The limitation of Mercator charts:
1- Irrepresentative for polar regions.
2-Great circles other than meridians & equator appear as curved lines.
3- Unequal area representation.
4- Straight lines are not the shortest distances except equator and meridians.
5- Distance scale changes with the change of latitude.

2) Limitations of Gnomonic charts:

1- Distance scale changes with the change of latitude.
:
2- In correct angular relationship (not orthomorphic).

3- Unequal area representation.

4- Rhumb lines are represented as curved lines.

Q013.Calculate the dimensions of a Mercator chart for long 100°
to 105°E., lat. 30° to 33°N.
If the long scale is 1/500000 (G.M..= 1855.356 meters).
(G.M.=Geographical Mile)

D. long = 5° = 300' = 300 x 1855356 m.m. on equator Length of transverse border of

the chart (length of the part of equator or parallel of latitude)

300 x 1855356
= -------------------- = 1113.2136 m.m. = 111.32 cm.
500000

1113.2136
1 M.P. = -------------- = 3.710712 mm.
300

M.Ps of lat. 33 ° N 2086.78

M.Ps of lat. 30 ° N 1876.67
----------------------------------------------
D.M.P. 210.11

Length of longitudinal border (length of the meridians) = 210.11 x 3.710712 = 77.97 cm.

Q14.State the desirable properties of preferable projection.

- The desirable properties of preferable projection:

1- True shape of physical features.
2- Correct angular relationship. A projection with this characteristics is said to be conformal or
orthomorphic.
3- Equal area, or the representation of areas in their correct relative proportions.
4- Constant scale values for measuring distances.
5- Great circles represented as straight lines.
6- Rhumb lines represented as straight lines.
Q15.Vessel sailed from A (38⁰00’ N, 062⁰58’ E) to B (42⁰29’ N, 120⁰ 03’ E) On Great Circle
Track. Find the Following:
(4) Distance
(5) Initial and Final Course
(6) Position of Vertex

4) Distance

Long. A 62⁰58’ E
Long. B 120⁰03’ E
D. Long 57⁰05’ E
Cos Dist. = (Cos Lat. A * Cos Lat. B * Cos D. Long) + (Sin Lat. A* Sin Lat. B)

Dist. = 42.9⁰ = 2574 Miles

5) Initial and final Course

Sin I. Co = (Sin D. Long. * Cos Lat. B) /Sin Dist.

I.CO = N 65.4⁰ E = 065.4⁰ (T)

Sin F. Co = (Sin D. Long. * Cos Lat A) / Sin Dist.

F. Co = S 76.4⁰ E = 103.6⁰ (T)

6) Position of Vertex
Cos Lat. Vertex = Cos Lat. A * Sin I. Co

Lat. Vertex = 44⁰14.1’ N

Tan D. Long Vertex = 1 / (Sin Lat. A * Tan I.Co)

D. Long Vertex = 36⁰38.1’ E

Long. A 062⁰58’ E
D. Long. 036⁰38.1’ E
Long. B 099⁰36.1’ E

Vertex Position (44⁰14.1’ N, 099⁰36.1’ E).

 Q16.A ship Steamed from A (22⁰00’ S, 060⁰00’ E) to B (38⁰33’ S, 127⁰ 43’ E)
On Great Circle Track. Find the Following:
• 1)Distance
• 2)Initial and Final Course
• 3)Position of Vertex

Distance
A 22⁰00’ S 060°00’ E
B 38⁰33’ S 127°43’ E
D. Long 67°43’ E
Cos Dist. = (Cos D. Long *Cos Lat. A * Cos Lat. B )+ (Sin Lat. A* Sin Lat. B)
= (cos 67°43’*cos 22˚*cos 38˚ 33′*)+(sin 22˚* sin 38˚ 33′)
= (0.379187*0.927184*0.782065)+(0.374606*0.623197)
=0.274955+0.233453
=0.508408
Dist. = 59.44°*60 = 3566.5 nm
Initial and final Course
A 22⁰00’ S 060°00’ E
B 38⁰33’ S 127°43’ E
D. Long 67°43’ E
Dist= 59.44˚
Sin I. Co = (Sin D. Long. * Cos Lat. B) /Sin Dist.
=sin 67°43’ *cos 38⁰33’ /sin 59.44˚
=(0.92532*0.782065)/0.861097=0.840411
I.CO = S 57.2° E = 122.8⁰ (T)
Sin F. Co = (Sin D. Long. * Cos Lat A) / Sin Dist.
=sin 67°43’ * 22⁰00’ /sin 59.44˚
=(0.92532*0.927184)/0.861097=0.996336
F. Co = N85.1° E =085.1⁰ (T)
Position of Vertex
A 22⁰00’ S 060°00’ E
 B 38⁰33’ S 127°43’ E
D. Long 67°43’ E
Dist= 59.44˚ I.CO = S 57.2° E
Cos Lat. V = Cos Lat. A * Sin I. Co
=cos 22⁰00’ *sin 57.2°
= 0.927184*0.840567=0.779359
Lat. Vertex = 38⁰47.8’ N
coTan D.Long V = Sin Lat. A * Tan I.Co
= sin 22⁰00’ * tan 57.2°
= 0.374606*1.551696=0.581276
D. Long V 59⁰49.9’ E
Long. A 60⁰00’ E
Long. V 119⁰49.9’ E
Vertex Position (38⁰47.8’ N, 119⁰49.9’ E)

Q17.Find the great circle distance & initial course and the Vertex ?
From a position (45° 00' s, 140° 00' E) to a position (65° 00' N, 110° 00' W).

Distance
To Find The Distance: D. Lat=110° 00' N ,D. Long. 110° 00' E
:Distance By Cosine Formula
Cos, Dístense = Cos. D Long* Cos. lat. A* Cos. Lat. B + Sin Lat. A Sin Lat. B
° Cos 110 ° Cos45 COS. 65 °- Sin 45 ° Sin 65=
0.90631 * 0.7071 - 0.42262 * 0.7071 * 0.34202 -=
0.74306 - =
Distance = 137.993 ° *60 = 8279.58 Miles

Initial and final Course

:To Find The Initial Course

Sin In. Co. = Sin D. long Cos Lat B Cosec Distance. A= 0.36397 S
Sin 110 ° Cos 65 ° Cosec 137.993 ° = 0.593423 B= 2.28213 =
N
Initial Course == 36.4 ° == N 36.4 ° E C= 1.918
Tan. Co.= 0.7372============== Co.=N 36.4 E
:To Find The Final Course
Sin Final Course = Sin d. Long Cos Lat. A Cosec Dístense
° Sin 110 ° Cos 45 ° Cosec 137.993 =
0.99289 =
Final Course
N 83.163 ° E = 083.16 = ° 83.163 =

Position of Vertex
:To Find The Vertex
.Cos. Lat. Vertex =Cos. Lat. A x Sin. Co
° Cos. 45 ° x Sin 36.4 =
41961. =
65.1866 =
N 4.'11 ° 65=
.Tan D. Long Vertex =Cosec. Lat. A x Cotan. Co
x1.35636 1.414213 =
1.91819 =
D. Long Vertex = 62.46588 °E====62 ° 27.95 'E
E 117.534 = 62.46588 -180 =
Long Vertex = 140 00 E
D. Long Vertex = 117.534 E
E 534. 257
360
w 102.466°

Vertex position = (65 ° 11'.4 N 102.466° W)

Q18.Find the course and distance between the following positions:

A- (37° 01' N,009° 00' W. B-( 36° 11' N. 006° 02' W).

,/

A 37° 01' N. 9° 00' W.

B 36° 11' N. 6° 02' W. 36° 11'N.

D.lat. 00° 50' S. d.long. 2° 58' E. 1/2 d.lat. 25'

= 178' Mean lat. 36° 36' N.

Dep. = d. long. X cos. lat.

= 178 x 0.80282
 = 142.9

Tan co. = dep

d.Lat

Tan co. = 142.9 = 2.858

50

Course = 70° 43'

dist. = d.lat. X sec. co.

= 50 x sec. 70° 43'

= 50 x 3.02789 = 151.4 Miles

Answer: course = S. 70° 43' E., distance = 151.4 miles.

Q19.The course and distance from A to B is 055° T. 720 nautical miles. Find the d. lat. and departure

made good.

d. lat. = dist. x cos. (course)

= 720 x cos. 55°

=720x0.57358
= 412.96

=6° 53' N

Dep. = dist. x sin. (Course) = 720 x sin. 55°

= 720 x 0.81915

= 589.8 M.======589.9/60= 6° 53'

D. Lat. = 6° 53' N

Dep. = 589.8 nautical miles.

Q20.From lat. 50° 28' N, a vessel steamed 156° T for 1550 nautical miles. Find the Latitude in which

she arrived.

d. lat. = dist. x cos. (course)

= 1550 x 0.91355 = 1416' =============1416/60= 23° 36' S.

= 23° 36' S.

Initial lat. 50° 28.0' N.

D. Lat. 23° 36.0' S.

Final lat. 26° 52.0' N.

Q21.A vessel steers 327° T. and makes a departure of 396.7 nautical miles. How far did she steam?

Dist. = dep. X cosec. (Course) = 396.7 x cosec. 33°

= 396.7 x 1.83608

= 728.4 M

Dist. steamed = 728.4 nautical miles.

Q 22.A ship in fix position (40° 30' N, 035° 15' W) - She steams the fol1owing courses and distances:

056 ° T Distance 45 miles

020 ° T Distance 20 miles

335 ° T Distance 35 miles

300 ° T Distance 50 miles

Remember:

We calculate d.lat and departure from the formula: d.lat = distance cos. co.

dep. = distance sin co.

Course Distance D.lat Departure

N. S. E. W.

56° 45 25.2 37.7

20° 20 18.8 6.8

335° 35 31.7 14.8

300° 50 25 43.3

100.7 00.00 44.1 58.1

100.7 N 14.0 W

d.lat =100.7=====/60= 01° 40.7

Initial lat. 40° 30.0' N.

D. Lat. 01° 40.7' N.

Final lat. 42° 10.7' N.

Mean lat. = 40° 30.0' N + 42° 10.7' N.

= 41° 20.3' N
Dep. = d. long. X Cos.Mean lat.

d. long. = Dep. = 14 = 18.6 W

Cos.Mean lat. Cos 41 20

Initial Long 35 ° 15.0' W

D.Long 00 ° 18.6' W

Final Long 35 ° 33.6 W

* FINAL POSITION (42 ° 25' N, 015 ° 33.6' W)

Q23.A ship in position (45 ° 25' N, 15 ° 05' W) she sailed the following courses and distances. What is

her final position?

Course Distance

1. 045° 11.5'

2. 312° 8'

3. 217° 16.5'

4. 103° 24.1'

5. 190° 7'

D.1at Departure

No. Course Distance

+N -S +E -W

1. 045° 11.5' 8.132 8.132

2. 312° 8' 5.336 -5.927

3. 217° 16.5' -13.154 -9.912

4. 103° 24.1 ' -5.417 23.466

5. 190° 7' -6.894 -1.216

13.5 -25.5 31.6 -17.1

 -12 (S) 14.5 (E)

D lat= 12' S dep. = 14.5 E

initial latitude 45° 25.0' N

D. lat. 00 ° 12.0 S

Arrival lat. 45 ° 13.0 N

Mean lat. = Initial lat. + Arrival lat. = 45 ° 25' + 45 ° 13.0'

2 2

= 90 ° 38.0' = 45 ° 19.0 N

D. Long = Dep. = 14.5 = 20.6' E

Cos M. Lat. Cos. 45 ° 19.0'

Initial Long 15 ° 05.0' W

D.Long 00 ° 20.6' E

Final Long 14 ° 44.4 W

* FINAL POSITION (45 ° 13.0 N, 14 ° 44.4 W)

Q24.Between what morning hours on 23rd November 1998 can a ship whose draft is 9
meters pass over shoal at DEVONPORT (charted depth 7 m) with Clearance 1 m?
The height of tide of DEVONPORT as following:-

On 23rd.Nov ------- 0133 1.4 on 24th.Nov. --------- 0204 1.5

0751 5.2 0834 5.1
1354 1.4 1437 1.5
2008 4.9 2051
Answers:-

The height of tide of DEVONPORT as following:-

On 23rd.Nov ------- 0133 1.4 On 24th.Nov. --------- 0204 1.5

0751 5.2 0834 5.1
 1354 1.4 1437 1.5
2008 4.9 2051 5.0

Ht.of tide required = draft + clearance – charted depth

Ht. of tide required = 9 + 1 – 6 = 3.0m

Ht.o.t is 3.0m that’s mean it will be between the following times (afternoon):-

23rd.Nov. 0133 1.4

0751 5.2

1354 1.4

FROM DEVONPORT CURVE:-

-The first time of passing over shoal will be at 0352

-The last time of passing over shoal will be at 1133
Q25: Between what daylight on 14th.April 1998 at BOMBAY can a ship whose Mast head
height is 11 meters. Pass under bridge (charted height 9.5m with clearance 1.5m) if
M.H.W.S is 4.4m.?
On 14th.April at BOMBAY the time of tide as the following:-
0104 4.1
0709 0.6
1335 4.2
1939 1.4

Answers:-
On 14th.April at BOMBAY the time of tide as the following:-
0104 4.1
0709 0.6
1335 4.2
1939 1.4

Max.Ht.of.tide = charted ht. of bridge + M.H.W.S – Masthead ht. – Clearance

Max.Ht.of tide = 9.5 + 4.4 – 10 – 1.5 = 2.4m

To find the first time of passing under bridge

Ht.o.t 2.4m will be between the following times at daylight

0104 4.1

0709 0.6
Duration = time of L.W – time of H.W = 0709 – 0104 = 0605

By using the curve of 6 hrs. ..the first time of passing under bridge will be at 0400

To find the last time of passing under bridge

Ht.o.t 1.4m will be between the following times at daylight
0709 0.6
1335 4.2
Duration = time of L.W – time of H.W = 1335 – 0709 = 0626
By using the curve between 6 hrs. and 7hrs the last time of passing under bridge will be
at 1035
Q26.Between what daylight on 14th.April 1998 at BOMBAY can a ship whose Mast head height
is 11 meters. Pass under bridge (charted height 9.5m with clearance 1.5m) if M.H.W.S is
4.4m.?
On 14th.April at BOMBAY the time of tide as the following:-
0104 4.1
0709 0.6
1335 4.2
1939 1.4

Answers:-

On 14th.April at BOMBAY the time of tide as the following:-

0104 4.1
0709 0.6
1335 4.2
1939 1.4

Range1 = 4.1 -0.6 =3.5

Range1 = 4.2 -0.6 =3.6

Max.Ht.of.tide = charted ht. of bridge + M.H.W.S – Masthead ht. – Clearance

Max.Ht.of tide = 9.5 + 4.4 – 11 – 1.5 = 1.4m

To find the first time of passing under bridge

Ht.o.t 1.4m will be between the following times at daylight
0104 4.1
0709 0.6
 Duration = time of L.W – time of H.W = 0709 – 0104 = 0605

By using the curve of 6 hrs. ..the first time of passing under bridge will be at 0512
To find the last time of passing under bridge

Ht.o.t 1.4m will be between the following times at daylight

0709 0.6
1335 4.2
Duration = time of L.W – time of H.W = 1335 – 0709 = 0626
By using the curve between 6 hrs. and 7hrs the last time of passing under bridge will be at
0910
Q27.A ship whose draft is 8m .grounded off LONDON BRIDGE At 0520 on 20th.Sep.1998

1) When would she Re-float? Tacking in consideration that the draft will decrease by 0.7m
2) Find the least sounding during the grounding.
On 20th.Sep. the time of tide of LONDON BRIDGE as following:-
0122 7.2
0750 0.4
1342 7.1
2021 0.1
Answers:-
1) * To find the Ht.o.t at grounding moment
The grounding time (0520) is happened between the following times
0122 7.2
0750 0.4
By using the curve of LONDON BRIDGE the Ht.o.t at 0520 will be 2.55m
Range = 7.2 – 0.4 = 6.8m
* To find the Ht.o.t at re-floating time
The floating time will be between the following times
0750 0.4
1341 7.1
Ht.o.t at re-floating time = 2.55 – 0.7 = 1.85m
Range = 7.1 – 0.4 = 6.7m
By using the curve of LONDON BRIDGE the ship will re-float at 0926
2) The least depth = draft – Ht.of.tide + Ht.of L.W

The least depth = 8 – 2.55 + 0.4 = 5.85m

Q28.Between what morning hours on 23rd November 1998 can a ship whose draft is 9
meters pass over shoal at DEVONPORT (charted depth 7 m) with Clearance 1 m?
The height of tide of DEVONPORT as following:-

On 23rd.Nov ------- 0133 1.4 on 24th.Nov. --------- 0204 1.5

0751 5.2 0834 5.1
1354 1.4 1437 1.5
2008 4.9 2051
Answers:-

The height of tide of DEVONPORT as following:-

On 23rd.Nov ------- 0133 1.4 On 24th.Nov. --------- 0204 1.5

0751 5.2 0834 5.1
1354 1.4 1437 1.5
2008 4.9 2051 5.0

Ht.of tide required = draft + clearance – charted depth

Ht. of tide required = 9 + 1 – 6 = 3.0m

Ht.o.t is 3.0m that’s mean it will be between the following times (afternoon):-

23rd.Nov. 0133 1.4

0751 5.2

1354 1.4

FROM DEVONPORT CURVE:-

-The first time of passing over shoal will be at 0352

-The last time of passing over shoal will be at 1133
Q29: Between what daylight on 14th.April 1998 at BOMBAY can a ship whose Mast head
height is 11 meters. Pass under bridge (charted height 9.5m with clearance 1.5m) if
M.H.W.S is 4.4m.?
On 14th.April at BOMBAY the time of tide as the following:-
0104 4.1
0709 0.6
1335 4.2
1939 1.4

Answers:-
On 14th.April at BOMBAY the time of tide as the following:-
0104 4.1
0709 0.6
1335 4.2
1939 1.4

Max.Ht.of.tide = charted ht. of bridge + M.H.W.S – Masthead ht. – Clearance

Max.Ht.of tide = 9.5 + 4.4 – 10 – 1.5 = 2.4m

To find the first time of passing under bridge

Ht.o.t 2.4m will be between the following times at daylight

0104 4.1

0709 0.6
Duration = time of L.W – time of H.W = 0709 – 0104 = 0605

By using the curve of 6 hrs. ..the first time of passing under bridge will be at 0400

To find the last time of passing under bridge

Ht.o.t 1.4m will be between the following times at daylight
0709 0.6
1335 4.2
Duration = time of L.W – time of H.W = 1335 – 0709 = 0626
By using the curve between 6 hrs. and 7hrs the last time of passing under bridge will be
at 1035
Q30Between what daylight on 14th.April 1998 at BOMBAY can a ship whose Mast head height
is 11 meters. Pass under bridge (charted height 9.5m with clearance 1.5m) if M.H.W.S is
4.4m.?
On 14th.April at BOMBAY the time of tide as the following:-
0104 4.1
0709 0.6
1335 4.2
1939 1.4

Answers:-

On 14th.April at BOMBAY the time of tide as the following:-

0104 4.1
0709 0.6
1335 4.2
1939 1.4

Range1 = 4.1 -0.6 =3.5

Range1 = 4.2 -0.6 =3.6

Max.Ht.of.tide = charted ht. of bridge + M.H.W.S – Masthead ht. – Clearance

Max.Ht.of tide = 9.5 + 4.4 – 11 – 1.5 = 1.4m

To find the first time of passing under bridge

Ht.o.t 1.4m will be between the following times at daylight
0104 4.1
0709 0.6
 Duration = time of L.W – time of H.W = 0709 – 0104 = 0605

By using the curve of 6 hrs. ..the first time of passing under bridge will be at 0512
To find the last time of passing under bridge

Ht.o.t 1.4m will be between the following times at daylight

0709 0.6
1335 4.2
Duration = time of L.W – time of H.W = 1335 – 0709 = 0626
By using the curve between 6 hrs. and 7hrs the last time of passing under bridge will be at
0910
Q31. A ship whose draft is 8m .grounded off LONDON BRIDGE At 0520 on 20th.Sep.1998

1) When would she Re-float? Tacking in consideration that the draft will decrease by 0.7m
2) Find the least sounding during the grounding.

On 20th.Sep. the time of tide of LONDON BRIDGE as following:-

0122 7.2
0750 0.4
1342 7.1
2021 0.1

Answers:-

1) * To find the Ht.o.t at grounding moment

The grounding time (0520) is happened between the following times

0122 7.2
0750 0.4

By using the curve of LONDON BRIDGE the Ht.o.t at 0520 will be 2.55m

Range = 7.2 – 0.4 = 6.8m

* To find the Ht.o.t at re-floating time

The floating time will be between the following times

0750 0.4
1341 7.1

Ht.o.t at re-floating time = 2.55 – 0.7 = 1.85m

Range = 7.1 – 0.4 = 6.7m

By using the curve of LONDON BRIDGE the ship will re-float at 0926

2) The least depth = draft – Ht.of.tide + Ht.of L.W

The least depth = 8 – 2.55 + 0.4 = 5.85m