MERCATOR SAILING

Problem: A vessel from Manila Latitude 14°34’N Longitude 120°57’E sailed on course
254° a distance of 1,500 miles. Find the Latitude in and Longitude of arrival.

Solution:

254° Cos c
= DLat
- 180°
Dist
74° (S74°W) DLat
= Cos C Dist

74° Log Cos =

9.44034
1,500 Log =
3.17609

DLat Log = 2.61643

DLat = 413.45 miles
Or DLat = 6°53’30” S

Lat fr 14°34’00”N DMP 877.7 (Table 5)

DLat 6°53’30”S DMP 459.3 Tan C = DLo
Lat in 7°40’30”N m 418.4 DLo = Tan C m

74° Log Tan = 0.54250 Longitude fr 120°57’E

418.4 Log = 2.62159 DLo 24°19’W
DLo Log = 3.16409 Longitude in 96°38’E
DLo = 1,459 mi
Or DLo = 24°19’W
 MERCATOR SAILING BY TABLE 3 & 5

Example: A vessel from Latitude 2°30’N Longitude 65°22’W sailed on course 111° a
total distance of 305° miles. Require the Latitude and Longitude of arrival.

Solution:

180° Table 3: D I
- 111° 300 107.5
69° (S69°E) 5 1.792
305 109.292

109.292 = 1°49.3S Lat fr 2°30.0’ DMP 149.0

60 DLat 1°49.3’S
Lat in 0°40.7’N DMP 40.4
m 108.6

m DLo
100.0 260.5 Long fr 65°22.0’W
8.0 20.841 DLo 4°42.9’E
0.6 1.5631
_____ ________ __________
108.6 282.9041 (60) Long in 60°39.1’W

DLo = 4° 45.9’E
 CURRENT SAILING

CURRENT is a body of water moving steadily in a horizontal direction. It tends to carry any
floating object within at the same speed and direction it is flowing.

CURRENT SET – is the direction toward which it is flowing.

CURRENT DRIFT – is the speed of current in knots; or the number of miles it covers within
time interval.

DISTANCE MADE GOOD (DMG) any distance measured along made good.

COURSE MADE GOOD (CMD) actual course followed by a ship when affected by current.

SPEED MADE GOOD (SMG) is the actual speed with the effect of current.

Example: A vessel is steering a course of 310° Psc, speed 12 knots is affected by current set
of 015° and a drift of 4 knots. Variation 5°E. Deviation of 1°W. Required (1) Plot
the CMG of the vessel after 1 hour (2) Find the speed made good (SMG). Use scale
½ cm = 1 knot or 1 mile.

Procedure:

1. Draw the true course steered and plot the DR every hour or per minute interval.

2. At the end of measure distance, plot current set and drift for the same time interval.

3. Connect point of departure and end of current drift with a line. This is the course
made good.

Solution:

Co 310° psc

Te 4° psc

TC 314°

Ans: Cmg = 328°

Smg = 14.5 Kts.
Example 2: Captain desires to make good course of 095° through a current setting 170°
and drift 2.5 knots. Required the course to steer and speed made good.

Procedure: 1. Draw AB 095° to a convenient scale

2. Draw AC.
3. With C as center, swing arc
CD (12 kts) to intersect AB
at D.
4. CD is the course to steer.
5. AD is the speed made good.
Ans: Co = 083° pd = 12.4 kts

Example 3: Captain desires to make good a course of 265° and speed of 15 knots through
current having set of 185° and drift 3 knots. Required course to steer and
speed to use.

Procedure: 1. Draw AB 265°

2. Draw AC set and drift
3. Draw C to B

Answer : Course to Steer = 276°

Speed = 14.8 Knots

Answer: At 1300H a ship is 100 miles due west of her desired destination. If the ship is to
arrive at her destination 1800 H. Find the course and speed to order if a 2 knot
current setting southeast is predicted.

Answer: Course = 086° ; Speed = 18.6 Knots

 Hav A = Hav PB – Hav (AB AP) Csc AB Csc AP

PB 56°11.2’ nat hav 0.22175 To find the Latitude and

(AB AP) 25°51.1’ nat hav 0.05004 Longitude of the Vertex:

nat hav 0.17171 (Solution by Napier’s

Analogy)
log hav 9.23480
AB 103°05.9’ log csc 0.01145
AP 77°14.8’ log csc 0.01085

A log hav 9.25710

A = 50°19.3’
Or Initial Course = N50°19.
‘E

Sin PV =Cos Co-AP Cos Co-A ;

Sin Co-A Tan Co-A Tan Co-P
Sin PV = Sin AP Sin A
Cos AP = Cot A Cot P
Cot P = Cos AP;Cos AP Tan A
A 50°19.3’ log sin 9.98915 ______
AP 77°14.8 log sin 9.88629 Cot A
_______ AP 77°14.8 log cos 9.34392
Pv log sin 9.87544 A 50°19.3 log tan 0.08114
Pv = 48°38.8’ _______
90° P log cot 9.42505
______ P (DLo) = 75°05.9’E
Latitude of Vertex= 41°21.2’N Longitude of A = 12°20.1’E
_________
199°26.0’
360°
_________

Longitude of Vertex 160°34.0’W

Exercise: From Latitude 37°42’N Longitude 123°00’W to Latitude 12°45’N Longitude

124°17’E.
Required:
(1) Great circle distance
(2) Initial course
(3) Final course
(4) Latitude and Longitude of the vertex
(5) Possibly computing points along the circle track for every 10° of
Longitude.

Answers: (1) 5,963 miles (2) N65°46’W or 294°.2 (3) S47°43’W or 227.7°
(4) 43°49’N 159°21’W (5) To be solve if time permits.
 COMPOSITE SAILING

COMPOSITE SAILING is a combination of a great circle sailing and parallel sailing which
is used when it is desired to limit the highest latitude.

Example: A vessel sails from Latitudes 34°49’S Longitude 20°10’E to a point of Latitude
35°52’S
Longitude 174°25’E. Using the Latitude of 54° S as limiting parallel, find the
initial and final course and the composite distance.

Solution: 90° 90°

- -

A 34° 49’S D35°52’S

AP 55° 11’ DP 54° 08’

90°
- 54°
36° = CP . or BP.

In spherical triangle ABP:

Sin Co-AP = Cos AB Cos BP

= Cos AB Cos BP
Cos Ab = Cos AP
______
Cos BP
Cos AB = Cos Ap Sec BP 55° 11’

AP 55° 11’ Log Cos 9.75660

BP 36° Log Sec 0.09294
______
AB Log Cos 9.84864
AB 45°06.7’
AB 2,706.7 miles
Sin Co-P = Tan Co-AP Tan BP
Cos P = Cot AP Tan BP

AP 55°11’ Log Cot 9.84227

BP Log Tan 9.86126
______
P Log Cos 9.70353
P 59°39’
In Sheprical triangle DCP² :

Given: DP = 54°08’CP = 36°

Sin Co-DP = Cos CD Cos CP

Cos DP = Cos CD Sec CP

DP 54°08’ Log Cos 9.76782

CP 36° Log Sec 0.09204 Sin Co-P = Tan Co-DP Tan CP
______ Cos P = Cot DP Tan CP
CD Log Cos 9.85986
CD 43°35.85 DP 54°08’ Log Cot 9.85913
or CD 2,615.85 miles CP 36° Log Tan 9.86126

P² Log Cos 9.72039

Longitude fr 20°10’E P² 58°18.8’
Longitude in 174°25’E P 59° 39.0’

P + P² 17° 57.8’
DLo 154° 15’E
P + P² 117°57.8’

(DLo) Ps 36°17.2’ or 2,177.2 miles

In Spherical triangle BPsC: Departure

Cos Latitude = ________

DLo

Departure = DLo Cos Lat

Lat 54° Log Cos 9.76922 To Find the Composite Distance:
DLo 2,177.2 Log 3.33787
______
Dep Log 3.10709 AB 2,706.7
BC or Dep 1,279.7 miles CD 2,615.8
BC 1,279.7
______
6,602.2 miles

Initial Course: FINAL COURSE:

Sin BP = Cos co-A Cos co-AP Sin CP = Cos co-D Cos Co-DP
Sin A = Sin BP Csc AP Sin D = Sin CP Csc DP
A = S 45° 43.3’E D = N46° 29.7’E
 GREAT CIRCLE SAILING

Example 1: A ship proceeding from Manila to Los Angeles. The Captain wishes to use
Great circle sailing from Latitude 12° 45.2’N Longitude 124° 20.1’E to
Latitude 33° 48.8’N120° 07.1’W. Find (1) Initial and great circle course
(2) Great circle distance (3) Latitude and Longitude of the vertex.

Solution:
Long fr 124°20.1’E
Long in 120°07.1’W
__________
244°27.2’
360°
___________
P (DLo) 115°32.8’E
90°

L1 12°45.2’N
___________
PA 77°14.8’

90°

L2 33°48.8’N
___________
PB 56°11.2’
(PA PB) 21°03.6’

Hav AB-hav (PA PB)

Hav P = ___________________
Sin PA Sin PB
Hav AB - Hav (PA PB) = Hav P Sin PA Sin PB
Hav AB = Hav P Sin PA Sin PB + Hav (PA PB)

P 115°32.8’ log hav 9.85468 To get the initial Course:

PA 77°14.8’ log sin 9.98915
PB 56°11.2’ log sin 9.91953 AB 103°05.9
______ AP 77°14.8’
log hav 9.76336 _______
_______ (AB-AP) 25°51.1’
nat hav 0.57991
(PA~PB) 21° 03.6 Nhav 0.03340
_______ Hav PB Hav (AB AP)
AB nat hav 0.61331 Hav A = _________________
AB 103°03.9’ Sin AB Sin AP
or AB 6,185.9 miles

THE CELESTIAL SPHERE

CELESTIAL SPHERE - The sphere of infinite radius with the earth at its center. It is an
Imaginary sphere from the framework of celestial system.

The sun,moon,stars and planets are outlined against the sky and
Appears to be moving slowly across it. Its appearance is that of
A vast hallow sphere with the earth as center and with various
Celestial bodies located on inner surface of the sphere.

THE EQUINOCTIAL SYSTEM OF COORDINATES – The equinoctial system of coor-

dinates is similar to the Latitude and Longitude system used on
the earth. Before the significance of the coordinates themselves
can be understood clearly, it is necessary to be familiar with the
terms used in the definitions. The various reference circle of the
equinoctial system are illustrated in the figure below together
with other circles of points that are important in determining
the coordinates.

Illustration:

Fig.1 Terrestrial and celestial circles and points

Definition of Circles and Points:

CELESTIAL POLES - are those points where the axis of the earth extended at both
direction for an infinite distance intersects the celestial sphere.

NORTH CELESTIAL POLE - the north end of the extended axis.

SOUTH CELESTIAL POLE- the south extended end of the axis.

EQUINOCTIAL OR CELESTIAL EQUATOR – the great circle of the celestial sphere

where the plane of the earth’s equator intersects when
extended to that sphere.

CELESTIAL MERIDIAN - is a great circle of the celestial sphere which passes through
celestial pole, the plane of which passes through a particular
point on the surface of the earth. This is divided into two equal
parts or branches ,the UPPER and LOWER branch .The upper
branch is that which passes through a point directly overhead or
that which contains the zenith. The other half which contains
the Nadir is the lower branch.

HOUR CIRCLE - is the similar to a great circle of the celestial sphere which passes
through the poles and a heavenly body which is considered to
rotate with the body.

DIURNAL CIRCLE OR CIRCLE OF DECLINATION – is any small circle of the celestial

sphere which is parallel to the equinoctial. Any celestial body
appears to revolve along a diurnal circle as a result of the daily
rotation of the earth on its axis. The diurnal circle is actually a
parallel of latitude projected into the celestial sphere.

ECLIPTIC - the earth makes a complete revolution around the sun once each
Year. But the sun as observed on the earth appears to revolve
around the earth instead. The apparent path of the sun on the
celestial sphere is called the ecliptic.

VERNAL EQUINOX - this is sometimes called the FIRST OF ARIES (Y), the point where
the sun in its apparent annual journey crosses the equinoctial when
moving from south to north declination on or about March 21st.
( See also Equinoxes…)

AUTUMINAL EQUINOX - that point of intersection of the ecliptic and the celestial equator
occupied by the sun as it changes from north to south declination,
on or about September 23, or the instant this occurs.

DECLINATION (decl.) - is the angular distance of the body north or south of the equinoctial
 This correspond to the latitude of a point on the earth’s surface.

POLAR DISTANCE - of a body is its angular distance from the elevated pole measured
along the hour circle of the body ( in degrees, minutes, and seconds).
If declination of the body is same as the elevated pole, Polar
Distance is equal to 90° minus declination (PX=09° - decl.)Contrary
name, Polar Distance equals 90° plus declination or expressed as
PX=90° + decl.

Coordinates of the Equinoctial System :

GREENWICH HOUR ANGLE (GHA) – the angle between the plane of the meridian of
Greenwich and the plane of the circle that is passing through a body.
It is measured from 0° - 360° or from 00h-24h westward. It may also
be defined as local hour angle measured from the celestial meridian
of Greenwich.

Greenwich Meridian – the meridian through which Greenwich.

England serving as the Prime Meridian and the reference meridian
for Greenwich Time.

Fig. 2. Showing Polar Distance, Declination and Greenwich Hour Angle

LOCAL HOUR ANGLE (LHA) – for a given place is the angle at the celestial pole between
the meridian of the place and the hour circle of the body, measured
westward from the meridian 0-360 degrees or 00h-24h to the hour
circle passing through the body.
MERIDIAN ANGLE (t) – the angle at the celestial pole between the meridian of the place
and the hour circle of the body measured EASTWARD or
WESTWARD from the meridian 0-180 degrees or 00h-12h if
the body is East or West of the meridian.

RIGHT ASCENSION (R.A.) – is the angle at the celestial pole between the hour circle of the
vernal equinox and the hour circle of the body measured
EASTWARD from 0-360 degrees or 00h-24h.

SIDEREDIAL HOUR ANGLE (SHA) – is an angular measurement from the hour circle of
the vernal equinox westward to the hour circle of the body 0-360
degrees or 00h-24h. SHA= 360° 0 RA

SHA Method of obtaining GHA:

GHA = GHA Aries +SHA Star (If the sum exceeds 360 degrees,
GHA is found by subtracting 360 degrees).

Illustration

Fig.3 Equinoctial coordinates cont’d.

TIME
TIME is a definite period of duration, or and elapsed interval. It is measured by the apparent
movement of celestial bodies around the earth.

GREENWICH TIME is the time based upon the Greenwich meridian.

GREENWICH MEAN TIME (GMT) is the local mean time at Greenwich meridian and is of
particular interest to a navigator because it is the principal time used in entering arguments
for the almanacs.

ZONE TIME the local mean time reference or zone meridian whose time is kept throughout
a designed zone.

ZONE DESCRIPTION (ZD) is found by mentally dividing the ship’s longitude by 15 (15° of
longitude equals 1 hour) to the nearest whole number. If the remainder is 7°30’ or more add
1 to the ZD then indicate it with the plus or minus sign as appropriate.

Examples Problems: Find the zone description of the time used by the ship in each of the
following longitudes:

(a) 99° 27’E (b) 52°19’W (c) 172°47’E

(d) 6°57’W (e) 141°33’E (e) 128°02’W

Answers: (a) -7 (b) +3 (c) -12 (d) 0 (e) -9 (f) +9

Exercises:

1. The ship is in Longitude 60°W. The zone time is 2200hr. August 7. Find the GMT
and date. Similarly what will be the ZT at Longitude 60°E?
Ans. GMT is 0200H Aug. 8; ZT 0600h Aug at Long 60°E

2. A ship is Longitude 120°E, Zone time 0300 May 5. find GMT.

Ans. 1900H May 4.

3. At longitude 74°30W the ZT is 21H 15m 40s April 14. Required GMT and date, also
ZT and date Longitude 108°20’W.
Ans. GMT = 02h 15m 40s Apr 15; ZT= 19h 15m 40s Apr,14.

4. On Sept. 30 a ship is on DR Longitude 50°30’W zone time 1500H. 10 hours later the
DR Longitude is 54°29.5’W. Find the ZT and date at the second Longitude
Ans. ZT = 0000H October 1.

Note the crossing the L 180th Meridian (INTERNATIONAL DATE LINE) sailing
EASTWARD, subtract one day. Sailing WESTWARD, add one day to the date.

INTERCONVERSION OF TIME AND ARC

 Since the sun performs an apparent revolution of 360˚ of longitude in 24 hrs, 15 degrees
of arc= 1 hour of time, 1 degree of arc= 4 minutes of time, 15 minutes of arc= 1 minute of
time and 1 minute of arc= 4 seconds of time.

To Convert Time to Arc

1. Multiply the hours by 15 to get degrees

2. Divide minutes of time by 4 obtain degrees, multiply remainder by 15 to obtain
minutes of arc.
3. Divide seconds of time by four to obtain minutes and tenths of minutes of arc or
multiply remainder by 15 to obtain seconds of arc.
4. Add degrees, and tenths ( seconds )

Example: Convert 7h 50m 18s to arc

Solution: (1) 7h x 15 = 105º

(2) 50m ÷ 4 = 12 remainder 2 x 15 = 12 º30’
(3) 18s ÷ 4 = 4 remainder 2 x 15 = 4.5’ or 4’30”
(4) 7h 50m 18s =117 º 34.5’

To Convert Arc to Time

1. Divide degrees of arc by 15 to obtain hours, multiply remainder by 4 to obtain

minutes of time.
2. Divide minutes of arc by 15 to obtain minutes of time, multiply remainder by 4 to
obtain seconds of time.
3. Divide seconds of arc by 15 to obtain seconds of time.
4. Add hours, minutes and seconds.

Example: Convert 84º17’33”

Solution: (1) 84 º ÷ 15 = 5h remainder 9 x 4 = 5h 36m

(2) 17’ ÷ 15 = 1m remainder 2 x 4 = 1m 08s
(3) 33” ÷ 15 = 2.2s = 0.02s
(4) 84 º17’33” = 5h 37m 10.2s