INFINITY MARINER: Great circle sailing

Great circle sailing

The distance between any two points is said to be shortest if the distance is measured along a straight line. Suppose a straight line is not possible between them, the distance would be the arc of a circle; it will be passing through the two given points and having the maximum possible radius. Radius will get increased as the distance along the arc between the two point’s increases.

On the earth’s surface, the straight line is not possible between two points (which can be considered spherical). Therefore, on the Earth’s surface the distance between two points is a shorter arc of a circle passing between them and having the greatest radius. By definition a great circle has a radius, that equals the radius of the sphere itself; therefore the great circle has the radius as the largest possible radius for that sphere. Hence the shortest distance between any two points on the surface of the earth is the shorter arc of the great circle; and it is passing through those two points.

By navigating along a great circle path, we can reduce the travelling distance when compared to navigating between the same two positions on a rhumb line path.

The following media explains about the great circle sailing

Spherical Triangles

A great circle track is the shortest distance, measured along the earth's surface, between two places. A great circle track cuts successive meridians at different angles because the meridians are not parallel to one another. The course, therefore, would change slightly whilst crossing each meridian. Practically at sea, the departure and arrival positions are marked on a Gnomonic chart and joined by a straight line. This gives an overview of the entire passage where islands and other features can be seen at a glance. The position of the vertex and suitable points are taken from there and transferred to a Mercator chart. The ship can then sail from point to point thereby following the Great Circle track effectively and conveniently. Most GPS receivers have the facility to do great circle calculations.

A spherical triangle is a triangle, on the surface of a sphere, formed by the intersection of three great circles.

Properties of a spherical triangle
The magnitude of the side of a spherical triangle is the angle subtended by it at the centre of the sphere and is expressed in degrees and minutes of arc.
The maximum value of a side of a spherical triangle is 180°.
The maximum value of an angle of a spherical triangle is 180°.
The sum of the three sides of a spherical triangle is less than 360°.
The sum of the three angles of a spherical triangle is any value between two right angles and six right angles (i.e., between 180° and 540°).
The sum of any two sides of a spherical triangle is greater than the third.
The greater side has the greater angle opposite to it.
If two sides of a spherical triangle are equal, the angles opposite to them are also equal to each other.
Types of spherical triangles

Right angled spherical triangle:

A right angled spherical triangle is one in which one angle equals to 90°. In a spherical triangle, it is possible for more than one angle to be equal to 90°.

Quadrantal spherical triangle:

A quadrantal spherical triangle is one in which one side equals to 90°. In a spherical triangle, it is possible for more than one side to be equal to 90°.

Oblique spherical triangle:

An oblique spherical triangle is one which is not a right angled or a quadrantal spherical triangle.

Symmetrical spherical triangle:

Two spherical triangles are said to be symmetrically equal when each of the six elements (i.e. three sides and three angles) of one are equal in value to each of the six elements of the other.

Because spherical triangles lie on the surface of a sphere, and are hence three dimensional, `symmetrically equal' does not necessarily mean congruent.
Two spherical triangles are symmetrically equal if any of the following conditions are met:
Three sides of one are correspondingly equal to the three sides of the other.
Two sides and the included angle of one are respectively equal to the two sides and included angle of the other.
Three angles of one are respectively equal to the three angles of the other.
Two angles and the included side of one are respectively equal to the two angles and included side of the other.
Congruent spherical triangles:

Two spherical triangles are said to be congruent only if it is possible to superimpose one on the other.

In the foregoing figure, ABC and DEF are two spherical triangles as seen from OUTSIDE the sphere. All six elements of triangle ABC are correspondingly equal to the six elements of triangle DEF. However, the triangles are latterly inverted (are mirror images) and, both being convex, it is not possible to superimpose one on the other. The two triangles are, therefore, symmetrically equal but not congruent.

Initial and final courses

Solution of Great Circle Sailing Problems

To solve the great circle sailing problems, we need to solve the spherical triangle. One side of the spherical triangle is the great circle path between the points, whereas, the other two sides are along the meridians through the two points. It can be seen from given diagram.

The great circle path between two positions in the same hemisphere will curve towards the pole of that hemisphere. It can be seen from circle 1. The great circle path from a point on one hemisphere to a point on the other hemisphere will curve towards the pole of the hemisphere at which the point with the higher latitudes lies. It can be seen from the 2^nd and 3^rd circle.

Construct spherical D PXY
Data given - Lat. and Long. of initial position X and final position Y of the ship.
Mark X at a convenient place on the answer sheet.
Mark Y approximately by comparing the lat. and long. of both the positions.
If Lat. of X and Y are North, then mark P as N-pole above the positions of X and Y.
If Lat. of X and Y are South, then mark P as S-pole below the positions of X and Y.
If X and Y are in different hemispheres, then either the N-pole or S-pole can be marked either above or below the positions of X and Y, but preferably mark the N-pole.
Draw arcs PX and PY, which represent the meridians of X and Y respectively.
Draw a GC in the form of an arc from X to Y such that the curvature points towards the pole. If X and Y are in different hemispheres then draw the GC as straight line from X to Y.

Calculate great circle distance from X to Y
Calculate $\angle \text{P},$ which is D’Long from X to Y.
Cos XY = (Cos P $\times$ Cos Lat X $\times$ Cos Lat Y) $\pm$ (Sin Lat X $\times$ Sin Lat Y)
If both Lat. are in same hemisphere, sign is +ve. If both Lat. are in opposite hemispheres, sign is –ve.
XY is the great circle distance, which should be stated in min. of arc or nautical miles.

Calculate initial course by A, B, C Tables theory
A = Tan Lat X $\div$ Tan P
If $\angle \text{P} < 90^{\circ},$ then name of A will be opposite to the name of Lat X.
If $\angle \text{P} > 90^{\circ},$ then name of A will be same as name of Lat X.
B = Tan Lat Y $\div$ Sin P
Name of B will be same as name of Lat Y.
C = A $\pm$ B
If A and B have same names, add. If A and B have different names, subtract.
Name of C will be same as name of A or B, whichever is higher in value.
Tan Course = 1 $\div$ (C $\times$ Cos Lat X)
Quadrant of the initial course will be named according to the name of C and the direction of D’Long from X to Y.

Calculate final course by A, B, C Tables theory
A = Tan Lat Y $\div$ Tan P
If $\angle \text{P} < 90^{\circ},$ then name of A will be opposite to the name of Lat Y.
If $\angle \text{P} > 90^{\circ},$ then name of A will be same as name of Lat Y.
B = Tan Lat X $\div$ Sin P
Name of B will be same as name of Lat X.
C = A $\pm$ B
If A and B have same names, add. If A and B have different names, subtract.
Name of C will be same as name of A or B, whichever is higher in value.
Tan Course = 1 $\div$ (C $\times$ Cos Lat Y)
This course is from Y to X, and the quadrant of the course will be named according to the name of C and the direction of D’Long from Y to X.
Quadrant of the final course will be opposite to the quadrant calculated above.

When the values of angles A and B are obtained, one can feel difficulty in naming the course and representing it in the three figure notation. This can be done easily, as we know that, in three figure notations, the course is calculated clockwise from the north meridian to the ship’s head. In the examples the internal angles A and B are given, the reader should observe, how the ship course is expressed in the three figure notation.

1. If proceeding from A to B, initial course =045° and final course 110°. If proceeding from B to A, initial course 290° and final course 225°.

2. If proceeding from A to B, initial course = 120°, final course = 030°. If proceeding from B to A, initial course = 210° and final course = 300°.

3. If proceeding from A to B, initial course = 070°, final course = 040°. If proceeding from B to A, initial course = 210° and final course = 250°.

By using ABC Tables, the initial and final courses can be calculated without the necessity of first calculating the great circle distance. Even those who have not already studied calculation of Azimuth and zenith distance, will find the calculations very easy to follow the working shown here.

Worked example:

Find the initial & final courses and the distance by a great circle track:
From Q 24°00'N 74°15'W to R 46°00'N 53° 45'W.

Step 1: Obtain d’lat & d’long and draw a simple sketch

Latitude

Longitude

From Q

24° 00'N

074° 15'W

To R

46° 00'N

053° 45'W

Difference

22° 00'N

020° 30'E

Q to R

d'lat

d'long

Notes:
D’long must be less than 180°; in case it is, you are going in the wrong direction!
Since R lies to the East of Q, all great circle courses from R to Q must be Easterly (i.e., North Easterly or South Easterly) and not Westerly.
Although d'lat is N and d'long is E, the courses may be either NE'ly or SE'ly. This is described later under ‘Vertex’.

Step 2: Calculate initial course

In order to solve this by ABC Tables, imagine that:
Latitude Q = latitude, Latitude R = declination, and d’long = Angle P Eastwards.
Angle PQR, expressed in the three figure notation, is the Azimuth = initial course.

By scientific calculator or ABC tables:

$\text{A} = \frac{\text{Tan\ lat}}{\text{Tan\ P}}$

1.191S

Named opposite to latitude

$\text{B} = \frac{\text{Tan\ dec}}{\text{Sin\ P}}$

2.957N

Named same as declination

C

1.766N

Same name add, different names subtract & retain name of larger one

Az (see note below)

N31.791°E

$\text{Tan\ Az} = \frac{1}{\text{C} \cdot \text{Cos\ lat}} = 0.619839$

Initial course

031.791°

Note: Prefix of Azimuth is the name of ‘C’ and suffix is the same as d’long.

Step 3: Calculate final course

Imagine latitude R = latitude; Latitude Q = declination; d’long = Angle P Westwards

Angle PRQ, in the three figure notation, is the Azimuth or bearing of Q from R. Since the ship at R is going towards C, and NOT towards Q, final course will be Az±180°.

By scientific calculator or ABC tables:

$\text{A} = \frac{\text{Tan\ lat}}{\text{Tan\ P}}$

2.770S

Named opposite to latitude

$\text{B} = \frac{\text{Tan\ dec}}{\text{Sin\ P}}$

1.27N

Named same as declination

C

1.498S

Same name add, different names subtract & retain name of larger one

Az of Q from R

S43.860°W

$\text{Tan\ Az} = \frac{1}{\text{C} \cdot \text{Cos\ lat}} = 0.960986$

Az (see note below)

223.860°

Initial course

043.860°

Note: Prefix of Azimuth is the name of ‘C’ and suffix is the same as d’long.

Step 4: Calculate the distance

Imagine latitude Q = latitude; Latitude B = declination; d’long = Angle P.

Imagine that QR is the Zenith Distance (ZD). Use the intercept formula and calculate the ZD which is the great circle distance AB.

Cos ZD = Cos P $\cdot$ Cos lat $\cdot$ Cos dec $\pm$ Sin lat $\cdot$ Sin dec

Note: If Lat and dec are same name (+), contrary names (-). Here, both are same name so in the formula, the (+) has been put above the (-).
P = 20° 30.0’ lat A = 24° 00.0’N lat B = dec = 46° 00.0’N

Cos ZD = Cos 20° 30’ $\cdot$ Cos 24° 00’ $\cdot$ Cos 46° 00’ + Sin 24° 00’ $\cdot$ Sin 46°00’ = 0.886996
ZD = 27.502° = 27° 30.1’ = 1650.1’
Great Circle distance AB = 1650.1 Miles

Step 5: Answer
Initial course = 031.791°; final course = 043.860°; Great Circle distance = 1650.1 M

GC Distances

Plot the position of V in the spherical $\Delta \text{PXY}$
Construct the $\Delta \text{PXY}$ on the answer sheet as explained before.
Calculate initial and final courses as explained before.
If the two courses lie in any two adjacent quadrants, then V will lie on the GC between the initial and final positions X and Y.
If the two courses lie in the same quadrant, then V will lie on the GC outside the $\Delta \text{PXY}$ as follows
If initial course > final course, then V will lie on the side of position X.
If final course > initial course, then V will lie on the side of position Y.
Arc PV represents the meridian of V.

Calculate position of V

If V lies inside the $\Delta \text{PXY}$ or outside it on the side of X
Consider the spherical $\Delta \text{PXV}$ , wherein $\angle \text{V}$ is $90^{\circ}$ .
Calculate initial course as explained above, and from it calculate $\angle \text{X}$ in the $\Delta$ .
Length PX is Co-Lat. of position X.
By the theory of Napier’s Rule, $\text{Sin} \text{PV} = \text{Cos} (90 - \text{X}) \times \text{Cos} (90 - \text{PX})$
Distance PV is Co-Lat of V. From PV calculate Lat. of V.
Again by Napier’s Rule, $\text{Sin} (90 - \text{PX}) = \text{Tan} (90 - \text{X}) \times \text{Tan} (90 - \text{P})$
$\angle \text{P}$ is the D’Long between X and V. Knowing Long. of X and $\angle \text{P}$ , calculate Long. of V.

If V lies outside the $\Delta \text{PXY}$ on the side of Y
Consider the spherical $\Delta \text{PXV}$ , wherein $\angle \text{V}$ is $90^{\circ}$ .
Calculate final course as explained above, and from it calculate $\angle \text{Y}$ in the $\Delta$ .
Length PY is Co-Lat. of position Y.
By the theory of Napier’s Rule, $\text{Sin} \text{PV} = \text{Cos} (90 - \text{Y}) \times \text{Cos} (90 - \text{PY})$
Distance PV is Co-Lat. of V. From PV calculate Lat. of V.
Again by Napier’s Rule, $\text{Sin} (90 - \text{PY}) = \text{Tan} (90 - \text{Y}) \times \text{Tan} (90 - \text{P})$
$\angle \text{P}$ is the D’Long between Y and V. Knowing Long. of Y and $\angle \text{P}$ , calculate Long. of V.

Composite great circle
If a ship sails on a GC between two positions X and Y,it is likely to reach latitudes higher than the latitudes of both X and Y. This may not be considered safe due to adverse weather conditions or presence of icebergs in that region.
If the ship sails along a rhumb line between the two positions, it will not go into high latitude but then the distance run will increase considerably.
Hence a composite track is adopted wherein part of the distance is covered along GC and balance along a rhumb line.
The total distance covered on this track will be more than the distance on a GC from X to Y, but less than the distance along a rhumb line.

Construct composite great circle track
Master will decide the maximum limiting latitude upto which it is safe to proceed. This will obviously be less than the Lat. of V of the GC between positions X and Y.
Plot the approximate positions of X and Y, relative to each other, on the answer sheet. Join X and Y by a GC. From pole P draw the meridians of X and Y.
Draw the limiting latitude as a rhumb line across the GC.
From X draw a second GC such that its vertex $(\text{V}_1)$ lies on the limiting latitude.
From Y draw a third GC in the reverse direction such that its vertex $(\text{V}_2)$ also lies on the limiting latitude.
Draw $\text{PV}_1$ and $\text{PV}_2$ as meridians of $\text{V}_1$ and $\text{V}_2$ .
The composite track consists of the distance from X to $\text{V}_1$ along the second GC, $\text{V}_1$ to $\text{V}_2$ along the rhumb line on the limiting latitude, and finally from $\text{V}_2$ to Y along the third GC.

Calculate initial course, final course, total distance from X to Y along composite track and Long. of both vertices.
In spherical $\Delta \text{PXV}_1, \angle \text{V}_1$ is $90^{\circ}$ , PX is Co-Lat of X and $\text{PV}_1$ is Co-Lat of $\text{V}_1$ .
By Napier’s Rule, $\text{Sin\ PV}_1 = \text{Cos} (90 - X) \times \text{Cos} (90 - PX)$
From $\angle \text{X}$ and figure of $\Delta \text{PXV}_1,$ calculate the quadrant of initial course.
By Napier’s Rule, $\text{Sin} (90 - \text{P}) = \text{Tan\ PV}_1 \times \text{Tan} (90 - \text{PX})$
From $\angle \text{P}$ and Long. of X, calculate Long $\text{V}_1$ .
By Napier’s Rule, $\text{Sin} (90 - \text{PX}) = \text{Cos\ XV}_1 \times \text{Cos\ PV}_1$
$\text{XV}_1$ is the GC distance from X to $\text{V}_1$ .

In spherical $\Delta \text{PXV}_2, \angle \text{V}_2$ is $90^{\circ}$ , PX is Co-Lat of X and $\text{PV}_2$ is Co-Lat of $\text{V}_2$ .
By Napier’s Rule, $\text{Sin\ PV}_2 = \text{Cos} (90 - \text{X}) \times \text{Cos} (90 - \text{PX})$
From $\angle \text{Y}$ and figure of $\Delta \text{PXV}_2,$ calculate the quadrant of initial course.
By Napier’s Rule, $\text{Sin} (90 - \text{P}) = \text{Tan\ PV}_2 \times \text{Tan} (90 - \text{PX})$
From $\angle \text{P}$ and Long. of Y, calculate Long $\text{V}_2$ .
By Napier’s Rule, $\text{Sin} (90 - \text{PX}) = \text{Cos\ XV}_2 \times \text{Cos\ PV}_2$
$\text{YV}_2$ is the GC distance from Y to $\text{V}_2$ .

D’Long (between $\text{V}_1$ and $\text{V}_2$ ) = Long $\text{V}_1$ ~ Long $\text{V}_2$
Calculate Dep. $(\text{V}_1 \text{V}_2)$ = D’Long $\times$ Cos (Limiting Lat.)
Total distance $= \text{XV}_1 + \text{YV}_2 + \text{V}_1 \text{V}_2$

Examples for calculating initial course, final course, GC distance between the two positions X and Y, and position of vertex

$1. \ \ \ \text{X} - 30^{\circ} 20^{\circ}'\ \text{S}, 142^{\circ} 45'\ \text{W} ;\ \text{Y} - 50^{\circ} 40^{\circ}'\ \text{S}, 170^{\circ} 30'\ \text{E}$

$\text{D'Long} =\angle \text{P} = 360^{\circ} - (142^{\circ} 45' + 170^{\circ} 30') = 46^{\circ} 45'\ \text{W}$
$\text{Cos\ XY} = (\text{Cos}\ 46^{\circ} 45' \times \text{Cos}\ 30^{\circ} 20' \times \text{Cos}\ 50^{\circ} 40') + (\text{Sin}\ 30^{\circ} 20' \times \text{Sin}\ 50^{\circ} 40')$
$\text{GC\ distance\ XY} = 40^{\circ} 03.1' = 2403.1 \ \text{miles}$

By A, B, C Tables theory

Ship is sailing from X to Y in SWly direction.
$\text{A} = \text{Tan}\ 30^{\circ} 20' \div \text{Tan}\ 46^{\circ} 45' = 0.550\ \text{N} ;\ \text{B} = \text{Tan}\ 50^{\circ} 40' \div \text{Sin}\ 46^{\circ} 45' = 1.675\ \text{S}$
$\text{C} = 1.125\ \text{S} ; \text{Tan\ Course} = 1 \div (1.125 \times \text{Cos}\ 30^{\circ} 20') ; \text{I. Co.} = \text{S}\ 45^{\circ}\ 50.6'\ \text{W}$

Assume that the ship is sailing from Y to X in NEly direction.

$\text{A} = \text{Tan}\ 50^{\circ} 40' \div \text{Tan}\ 46^{\circ} 45' = 1.148\ \text{N} ;\ \text{B} = \text{Tan}\ 30^{\circ} 20' \div \text{Sin}\ 46^{\circ} 45' = 0.803\ \text{S}$
$\text{C} = 0.345\ \text{N} ;$ Tan Course $= 1 \div (0.345 \times \text{Cos}\ 50^{\circ} 40') = 1 \div (0.345 \times \text{Cos}\ 50^{\circ} 40') ;$
$\text{Reverse\ Co.} = \text{N}\ 77^{\circ} 39.9'\ \text{E}$
$\text{F. Co.} = \text{S}\ 77^{\circ}\ 39.9'\ \text{W}$

I.Co. and F. Co. are in the same quadrant and F. Co. > I. Co., hence vertex lies outside the GC on the side of Y i.e. West of Y.

By Napier’s Rule, in $\Delta \text{PYV}$

$\text{PY} = 90^{\circ} - 50^{\circ} 40' = 39^{\circ} 20'$
$\text{Sin\ PV} = \text{Cos} (90 - 77^{\circ} 39.9') \times \text{Cos} (90 - 39^{\circ} 20') ;$ $\text{PV} = 38^{\circ} 15.5' ; \text{Lat\ V} = 51^{\circ} 44.5'\ \text{S}$

$\text{Sin} (90 - 39^{\circ} 20') = \text{Tan} (90 - 77^{\circ} 39.9') \times \text{Tan} (90 - \text{P}) ; \angle \text{P} = 15^{\circ} 47.2'\ \text{West\ of\ Y}$
$\text{Long\ V} = 154^{\circ} 42.8'\ \text{E}$

$2.\ \ \ \text{X} - 69^{\circ} 30'\ \text{N}, 060^{\circ} 50'\ \text{E} ; \text{Y} - 44^{\circ} 40'\ \text{N}, 120^{\circ} 20' \ \text{E}$

$\text{D'Long} = \angle \text{P} = 59^{\circ} 30'\ \text{E}$
$\text{Cos\ XY} = (\text{Cos}\ 59^{\circ} 30' \times \text{Cos}\ 69^{\circ} 30' \times \text{Cos}\ 44^{\circ} 40') + (\text{Sin}\ 69^{\circ} 30' \times \text{Sin}\ 44^{\circ} 40')$
$\text{GC\ distance\ XY} = 38^{\circ} 17.4' = 2297.4 \ \text{miles}$

By A, B, C Tables theory

Ship is sailing from X to Y in SEly direction.
$\text{A} = \text{Tan}\ 69^{\circ} 30' \div \text{Tan}\ 59^{\circ} 30' = 1.575\ \text{S} ;\ \text{B} = \text{Tan}\ 44^{\circ} 40' \div \text{Sin}\ 59^{\circ} 30' = 1.147\ \text{N}$
$\text{C} = 0.428\ \text{S} ; \text{Tan\ Course} = 1 \div (0.428 \times \text{Cos}\ 69^{\circ} 30') ; \text{I. Co.} = \text{S}\ 81^{\circ}\ 28.5'\ \text{E}$

Assume that the ship is sailing from Y to X in NWly direction.
$\text{A} = \text{Tan}\ 44^{\circ} 40' \div \text{Tan}\ 59^{\circ} 30' = 0.582\ \text{S} ;\ \text{B} = \text{Tan}\ 69^{\circ} 30' \div \text{Sin}\ 59^{\circ} 30' = 3.104\ \text{N}$
$\text{C} = 2.522\ \text{N} ; \text{Tan\ Course} = 1 \div (2.522 \times \text{Cos}\ 44^{\circ} 40') ; \text{Reverse\ Co.} = \text{N}\ 29^{\circ} 08.4'\ \text{W}$
$\text{F. Co.} = \text{S}\ 29^{\circ}\ 08.4'\ \text{E}$

I. Co. and F. Co. are in the same quadrant and F. Co. < I. Co., hence vertex lies outside the GC on the side of X. i.e. West of X

By Napier’s Rule, in $\Delta \text{PXV}$

$\text{PX} = 90^{\circ} - 69^{\circ} 30' = 20^{\circ} 30'$
$\text{Sin\ PV} = \text{Cos} (90 - 81^{\circ} 28.5') \times \text{Cos} (90 - 20^{\circ} 30') ; \text{PV} = 20^{\circ} 15.8' ;$ $\text{Lat\ V} = 69^{\circ} 44.2' \text{N}$
$\text{Sin} (90 - 20^{\circ} 30') = \text{Tan} (90 - 81^{\circ} 28.5') \times \text{Tan} (90 - \text{P}) ; \angle \text{P} = 09^{\circ} 05.5' \ \text{East\ of\ X}$
$\text{Long\ V} = 051^{\circ} 44.5' \text{E}$

$3.\ \ \ \text{X} - 49^{\circ} 50' \text{N}, 005^{\circ} 15' \text{W} ; \text{Y} - 32^{\circ} 29' \text{N}, 064^{\circ} 00' \text {W}$

$\text{D'Long} = \angle \text{P} = 58^{\circ} 45' \text{W}$
$\text{Cos\ XY} = (\text{Cos}\ 58^{\circ} 45' \times \text{Cos}\ 49^{\circ} 50' \times \text{Cos}\ 32^{\circ} 29') + (\text{Sin}\ 49^{\circ} 50' \times \text{Sin}\ 32^{\circ} 29')$
$\text{GC\ distance\ XY} = 46^{\circ} 09.5' = 2769.5\ \text{miles}$

By A, B, C Tables theory

Ship is sailing from X to Y in Wly direction.
$\text{A} = \text{Tan}\ 49^{\circ} 50' \div \text{Tan}\ 58^{\circ} 45' = 0.719\ \text{S} ;\ \text{B} = \text{Tan}\ 32^{\circ} 29' \div \text{Sin}\ 58^{\circ} 45' = 0.745 \ \text{N}$
$\text{C} = 0.026\ \text{N} ; \text{Tan\ Course} = 1 \div (0.026 \times \text{Cos}\ 49^{\circ} 50') ; \text{I. Co.} = \text{N}\ 89^{\circ}\ 02.4'\ \text{W}$

Assume that the ship is sailing from Y to X in NWly direction.
$\text{A} = \text{Tan}\ 32^{\circ} 29' \div \text{Tan}\ 58^{\circ} 45' = 0.386\ \text{S} ;\ \text{B} = \text{Tan}\ 49^{\circ} 50' \div \text{Sin}\ 58^{\circ} 45' = 1.386\ \text{N}$
$\text{C} = 1.0\ \text{N} ; \text{Tan\ Course} = 1 \div (1.0 \times \text{Cos}\ 32^{\circ} 29') ;$ $\text{Reverse\ Co.} = \text{N}\ 49^{\circ} 51.0'\ \text{E}; \text{F. Co.} = \text{S}\ 49^{\circ}\ 51.0'\ \text{W}$

I. Co. and F. Co. are in different quadrants hence vertex lies inside the GC. i.e. West of X.

By Napier’s Rule, in $\Delta \text{PXV}$
$\text{PX} = 90^{\circ} - 49^{\circ} 50' = 40^{\circ} 10'$
$\text{Sin\ PV} = \text{Cos} (90 - 89^{\circ} 02.4') \times \text{Cos} (90 - 40^{\circ} 10') ; \text{PV} = 40^{\circ} 09.6' ;$ $\text {Lat\ V} = 49^{\circ} 50.4'\ \text{N}$
$\text{Sin} (90 - 40^{\circ} 30') = \text{Tan} (90 - 89^{\circ} 02.4') \times \text{Tan} (90 - \text{P}) ; \angle \text{P} = 01^{\circ} 15.4'\ \text{West\ of\ X}$
$\text{Long\ V} = 006^{\circ} 30.4'\ \text{E}$

$4. \ \ \text{X} - 06^{\circ} 00'\ \text{N}, 079^{\circ} 00'\ \text{W} ; \ \ \ \text{Y} - 38^{\circ} 00'\ \text{S}, 179^{\circ} 00'\ \text{E}$

Select the S-pole as P for constructing the $\Delta \text{PXY}$ .
$\text{D' Long} = \angle \text{P} = 360^{\circ} - (079^{\circ} 00' + 179^{\circ} 00') = 102^{\circ} 00' \text{W}$
$\text{Cos XY} = (\text{Cos}\ 102^{\circ} 00' \times \text{Cos}\ 06^{\circ} 00' \times \text{Cos}\ 38^{\circ} 00') - (\text{Sin}\ 06^{\circ} 00' \times \text{Sin}\ 38^{\circ} 00')$
$\text{GC\ distance\ XY} = 103^{\circ} 08.3' = 6188.3\ \text{miles}$

By A, B, C Tables theory

Ship is sailing from X to Y in SWly direction.
$\text{A} = \text{Tan}\ 06^{\circ} 00' \div \text{Tan}\ 102^{\circ} 00' = 0.022\ \text{N} ; \text{B} = \text{Tan}\ 38^{\circ} 00' \div \text{Sin}\ 102^{\circ} 00' = 0.799\ \text{S}$
$\text{C} = 0.777\ \text{S} ; \text{Tan\ Course} = 1 \div (0.777 \times \text{Cos}\ 06^{\circ} 00') ; \text{I. Co.} = \text{S}\ 52^{\circ} 18.3'\ \text{W}$

Assume that the ship is sailing from Y to X in Ely direction.
$\text{A} = \text{Tan}\ 38^{\circ} 00' \div \text{Tan}\ 102^{\circ} 00' = 0.166\ \text{S} ; \text{B} = \text{Tan}\ 06^{\circ} 00' \div \text{Sin}\ 102^{\circ} 00' = 0.107\ \text{N}$
$\text{C} = 0.059\ \text{S} ; \text{Tan\ Course} = 1 \div (0.059 \times \text{Cos}\ 38^{\circ} 00') ; \text{Reverse\ Co.} = \text{S}\ 87^{\circ} 20.3'\ \text{E}$
$\text{F. Co.} = \text{N}\ 87^{\circ} 20.3'\ \text{W}$

The two courses lie in different quadrants, hence V lies within the $\Delta$ and close to Y.

By Napier’s Rule, in $\Delta \text{PYV},$ using P as the S-pole

$\text{PY} = 90^{\circ} - 38^{\circ} 00' = 52^{\circ} 00' ; \angle \text{Y} = 87^{\circ} 20.3'$
$\text{Sin\ PV} = \text{Cos} (90^{\circ} - 87^{\circ} 20.3') \times \text{Cos} (90^{\circ} - 52^{\circ} 00')$
$\text{PV} = 51^{\circ} 55.3' ; \text{Lat. V} = 38^{\circ} 04.7'\ \text{S}$

$\text{Sin} (90 - 52^{\circ} 00') = \text{Tan} (90 - 87^{\circ} 20.3') \times \text{Tan} (90 - \text{P}) ; \angle \text{P} = 04^{\circ} 19.9' \text{East\ of\ Y}$
$\text{Long V} = 176^{\circ} 40.9' \ \text{W}$

$5. \ \ \ \text{X} - 45^{\circ} 54'\ \text{S}, 170^{\circ} 45'\ \text{E} ; \text{Y} - 49^{\circ} 06' \ \text{S}, 075^{\circ} 50'\ \text{W} ; \text{Limiting\ Lat.} 55^{\circ} 00' \ \text{S} ;$ Calculate initial course, final course, total distance on composite GC track and Long. of both vertices.
Construct the $\Delta\ \text{PXY}$ with P as S-pole, as explained before.
Draw the limiting latitude. The GC from X to Y will not be used in this calculation.
Draw the second GC from X such that its vertex $\text{V}_1$ is on the limiting latitude.
Similarly draw the third GC from Y in reverse direction such that its vertex $\text{V}_2$ is also on the limiting latitude.
Draw the meridians of $\text{V}_1$ and $\text{V}_2$ .

By Napier’s Rule

$\text{In} \Delta \text{PXV}_1, \angle \text{V}_1 = 90^{\circ} ; \text{PX} = 90^{\circ} - 45^{\circ} 54' = 44^{\circ} 06' ; \text{PV}_1 = 90^{\circ} - 55^{\circ} = 35^{\circ} 00'$

$\text{Sin}\ 35^{\circ} 00' = \text{Cos} (90 - \text{X}) \times \text{Cos} (90 - 44^{\circ} 06') ; \angle \text{X} = \text{I. Co.} = \text{S}\ 55^{\circ} 30.5' \text{E}$
$\text{Sin} (90 - 44^{\circ} 06') = \text{Cos\ XV}_1 \times \text{Cos}\ 35^{\circ} 00' ; \text{XV}_1 = 28^{\circ}45.4' = 1725.4\ \text{miles}$
$\text{Sin} (90 - \angle \text{XPV}_1) = \text{Tan}\ 35^{\circ} 00' \times \text{Tan} (90 - 44^{\circ} 06') ; \angle \text{XPV}_1 = 43^{\circ} 44'\ \text{East\ of}\ \text{X}$
$\text{Long.\ V}_1 = 170^{\circ} 45' + 43^{\circ} 44' - 360^{\circ} = 145^{\circ} 31' \text{W}$

$\text{In}\ \Delta \text{PYV}_2, \angle \text{V}_2 = 90^{\circ} ; \text{PY} = 90^{\circ} - 49^{\circ} 06' = 40^{\circ} 54' ; \text{PV}_2 = 90^{\circ} - 55^{\circ} = 35^{\circ} 00'$

$\text{Sin}\ 35^{\circ} 00' = \text{Cos} (90 - \text{Y}) \times \text{Cos} (90 - 40^{\circ} 54')$
$\angle \text{Y} = \text{Reverse\ Co.} = \text{S}\ 61^{\circ} 10.1'\ \text{W} ; \text{F. Co.} = \text{N}\ 61^{\circ} 10.1'\ \text{E}$
$\text{Sin} (90 - 40^{\circ} 54') = \text{Cos\ YV}_2 \times \text{Cos}\ 35^{\circ} 00' ; \text{YV}_2 = 22^{\circ} 40.3' = 1360.3 \text{miles}$
$\text{Sin} (90 - \angle \text{YPV}_2) = \text{Tan}\ 35^{\circ} 00' \times \text{Tan} (90 - 40^{\circ} 54') ; \angle \text{YPV}_2 = 36^{\circ} 03.9'\ \text{West\ of\ Y}$
$\text{Long.\ V}_2 = 75^{\circ} 50' + 36^{\circ} 03.9' = 111^{\circ} 53.9' \text{W}$

$\text{D'Long\ (between} \text{V}_1 \text{and}\ \text{V}_2) = 145^{\circ} 31' - 111^{\circ} 53.9' = 33^{\circ} 37.1' = 2017.1'$
$\text{Dep.} = 2017.1' \times \text{Cos}\ 55^{\circ} = 1157.0'$

Total distance = 1725.4 + 1360.3 + 1157.0 = 4242.7 miles

Napier's Rules

Napier’s Rules for right angled spherical triangles.

Right angled spherical triangles can be solved by the Sine, Cosine and Haversine formulae using the value of Sin 90° as unity and of Cos 90° as zero, wherever they occur. However, it is much simpler by the use of Napier's Rules.

Formulae:

Draw a circle, divide it into five parts and number them as shown below:

In segments (1) and (2) insert the sides containing the right angle. In (3), insert complement of angle opposite the side in (1). In (5), insert complement of angle opposite the side in (2). In (4), insert the complement of the hypotenuse.

Formulae

Sin mid part = Tan adjacent part x Tan adjacent part
Sin mid part = Cos opposite part x Cos opposite part
(Remember Tangent and adjacent go together)
If segment number (2) is taken to be the mid part, (1) & (3) are its adjacent parts whereas (4) & (5) are its opposite parts.
If segment number (5) is taken to be the mid part, (1) & (4) are its adjacent parts whereas (2) & (3) are its opposite parts, and so on.

Worked example:

In spherical triangle PZX, right angled at Z, p = 110° 20' and z = 84° 12'. Find the values of P, X and x.

Sin mid part = Cos opp part $\cdot$ Cos opp part
Sin (90 - z) = Cos p $\cdot$ Cos x
Cos z = Cos p $\cdot$ Cos x

$\text{Cos}\ x=\frac{\text{Cos \ z}}{\text{Cos\ p}}=\frac {\text{Cos}\ 84^\circ 12'}{\text{Cos} \ 110^\circ20'}= \frac {\text{+ Cos}\ 84^\circ 12'}{\text{- Cos}\ 69^\circ 40'}$

By scientific calculator, x = 106° 54.4’   Answer (i)

Sin mid part = Cos opp part $\cdot$ Cos opp part
Sin p = Cos (90 - P) $\cdot$ Cos (90 - z)
Sin p = Sin P $\cdot$ Sin z

$\text{Sin\ P} =\frac{\text{Sin\ p}}{\text{Sin\ z}}=\frac{\text{Sin}\ 110^\circ 20'}{\text{Sin}\ 84^\circ 12'}=\frac{\text{Sin}\ 69^\circ 40'}{\text{Sin}\ 84^\circ 12'}$

By scientific calculator, P = 70° 28.7’ OR 109° 31.3’

Solution of this ambiguity:

The greater side must have the greater angle opposite to it.

Opposite side opposite angle
z = 84° 12' Z = 90° 00'
p = 110° 20' P > 90° 00'.

P = 109° 31.3' NOT 70° 28.7'   Answer (ii).

Sin mid part = Tan adj part $\cdot$ Tan adj part
Sin (90 - X) = Tan p $\cdot$ Tan (90 - z)
Cos X = Tan 110° 20' $\cdot$ Tan 5° 48'

By scientific calculator, X = 105° 54.5'   Answer (iii).

Napier’s Rules for quadrantal spherical triangles

Quadrantal spherical triangles can be solved by the Sine, Cosine and Haversine formulae using the value of Sin 90° as unity and of Cos 90° as zero, wherever they occur. However, it is much simpler by the use of Napier's Rules.

Formulae

Draw a circle, divide it into five parts and number them as shown below:

In segments (1) and (2) insert the angles containing the quadrant. In (3), insert complement of side opposite the angle in (1). In (5), insert complement of side opposite the angle in (2).

important: In segment (4) insert:
(angle opposite the quadrant minus 90).

Formulae

Sin mid part = Tan adjacent part $\cdot$ Tan adjacent part
Sin mid part = Cos opposite part $\cdot$ Cos opposite part
(Remember Tangent and adjacent go together).

If segment number (2) is taken to be the mid part, (1) & (3) are its adjacent parts and (4) & (5) are its opposite parts.
If segment number (5) is taken to be the mid part, (1) & (4) are its adjacent parts whereas (2) & (3) are its opposite parts, and so on.

Worked example

In spherical triangle PQR, PQ = 52° 11', Q = 69° 47' and QR = 90°. Calculate P, R and PR.

Sin mid part = Tan adj part $\cdot$ Tan adj part
Sin (90 - r) = Tan Q $\cdot$ Tan (P - 90)
Tan (P - 90) = Cos r ÷ Tan Q
Tan (P - 90) = Cos 52° 11' ÷ Tan 69° 47'
By scientific calculator, (P – 90) = 12° 43.4’
P = 12° 43.4' + 90° = 102° 43.4'

P = 102° 43.4' Answer (i).

Sin mid part = Tan adj part $\cdot$ Tan adj part
Sin Q = Tan R $\cdot$ Tan (90 - r)
Sin Q = Tan R $\cdot$ Cot r
Tan R = Sin Q $\cdot$ Tan r
Tan R = Sin 69°47' $\cdot$ Tan 52° 11'

By scientific calculator, R = 50° 24.3' Answer (ii).

Sin mid part = Cos opp part $\cdot$ Cos opp part
Sin (90 - q) = Cos Q $\cdot$ Cos (90 - r)
Cos q = Cos Q $\cdot$ Sin r
Cos q = Cos 69° 47' $\cdot$ Sin 52° 11'

By scientific calculator, q = 74° 09.5' Answer (iii).

You must take the quiz only after you study the eBook contents in this chapter, including watching the videos. The self-assessment quizzes are drawn from the complete chapter. Please take this as often as you wish. Please make sure that you return here after you have finished reviewing this topic.

Composite Great Circle sailing

As explained in earlier topics, the curvature of a GC course drawn on a Mercator’s chart will point towards the pole. Hence the vertex of the GC in the Northern hemisphere is likely to lie in high North Latitude. A ship proceeding along such a GC course between two positions is likely to experience rough weather, icebergs and severe cold conditions, which may not be acceptable from the point of view of safety of the ship. At the same time, sailing along a Rhumb line course between the same two positions will considerably increase the distance, which may not be acceptable from commercial point of view. Hence a compromise is adopted between the GC and Rhumb line courses, in the form of Composite Great Circle Track, as follows ---

Refer to the sketch drawn for the example below.
First the Master shall decide the highest Latitude called the Limiting Latitude, upto which it will be reasonably safe to proceed, which will be less than the Latitude (V) of the Vertex.
From the Initial position X a new GC is drawn such that the vertex ( $\text V_1$ ) of this GC lies on the Limiting Latitude.
From the final position Y a second GC is drawn in the reverse direction such that the vertex ( $\text V_2$ ) of this GC also lies on the same Limiting Latitude.
A Rhumb line course is drawn between $\text V_1$ and $\text V_2$ , which will necessarily lie along the East-West direction on the Limiting Latitude.
The ship will now sail along the first GC from X to $\text V_1$ , then along the Rhumb line course from $\text V_1$ to $\text V_2$ , and finally along the second GC from $\text V_2$ to Y.
Student should note that the two GCs drawn from X and Y, as stated above, have no relationship with each other, and also no relationship with the original GC from X to Y.
Composite GC track is drawn on a Gnomonic chart as follows ---
On Gnomonic chart all Parallels of Latitude are concentric arcs. Draw astraight line from X tangential to the Limiting Latitude, meeting it at $\text V_1$ . This is the first GC course from X to $\text V_1$ .
Similarly draw the second GC course in the reverse direction from Y, tangential to the Limiting Latitude, meeting it at $\text V_2$ .
The course from $\text V_1$ to $\text V_2$ is a Rhumb line along the Limiting Latitude.
This composite track is then transferred to the Mercator chart as explained earlier.
Distance between two positions along a GC will be less than the total distance along the composite GC, which will be less than the distance along the Rhumb line course.
Example

Calculate the initial course, final course and distance along the composite GC track from A $51^\circ$ 20' N $010^\circ$ 00' E to B $52^\circ$ 00' N $055^\circ$ 00' E having a limiting latitude of $53^\circ$ 00' N.

Notes :
GC drawn directly from A to B, as shown in the sketch, will have its vertex (V) at higher latitude than the Limiting Latitude given in the example.
The spherical triangles $\text {PAV}_1$ and $\text {PAV}_2$ are right-angled at $\text V_1$ and $\text V_2$ respectively. Hence Napier’s Rules can be applied to these triangles.
In $\Delta \text {PAV}_1$ ---
PA = $90^\circ$ - $51^\circ$ 20' = $38^\circ$ 40' ; $\text {PV}_1$ = $90^\circ$ - $53^\circ$ = $37^\circ$
Sin $\text {PV}_1$ = Cos (90 – A) $\times$ Cos (90 – PA) ; Sin A = Sin $\text {PV}_1\div$ Sin PA = Sin $37^\circ\div$ Sin $38^\circ$ 40'
$\angle A$ = Initial course = $\text N74^\circ$ 24.9'E

Sin (90 – P) = Tan $\text {PV}_1\times$ Tan (90 – PA) = Cos P = Tan $\text {PV}_1\div$ Tan PA
= Tan $37^\circ\div$ Tan $38^\circ$ 40'
$\angle \text P$ = D’Long between A and $\text V_1$ = $19^\circ$ 39.5' E

Sin (90 – PA) = Cos $\text {AV}_1\times$ Cos $\text {PV}_1$ ; Cos $\text {AV}_1$ = Cos PA $\div$ Cos $\text {PV}_1$ = Cos $38^\circ$ 40' $\div$ Cos $37^\circ$
$\text {AV}_1$ = Distance between A and $\text V_1$ = $12^\circ$ 08' = 728 miles

In $\Delta \text {PBV}_2$ ---

PB = $90^\circ$ - $52^\circ$ = $38^\circ$ ; $\text {PV}_2$ = $90^\circ$ - $53^\circ$ = $37^\circ$
Sin $\text {PV}_2$ = Cos (90 – B) $\times$ Cos (90 – PB) ; Sin B = Sin $\text {PV}_2\div$ Sin PB = Sin $37^\circ\div$ Sin $38^\circ$
$\angle \text B$ = Final course = $\text S77^\circ$ 49.5'E

Sin (90 – P) = Tan $\text {PV}_2\times$ Tan (90 – PB) = Cos P = Tan $\text {PV}_2\div$ Tan PB = Tan $37^\circ$ Tan $38^\circ$
$\angle \text P$ = D’Long between B and $\text V_2$ = $15^\circ$ 18.7' E

Sin (90 – PB) = Cos $\text {BV}_2\times$ Cos $\text {PV}_2$ ; Cos $\text {BV}_2$ = Cos PB $\div$ Cos $\text {PV}_2$ = Cos $38^\circ$ Cos $37^\circ$
$\text {BV}_2$ = Distance between B and $\text V_1$ = $09^\circ$ 21.4' = 561.4 miles

D’Long between A and B = $55^\circ$ - $10^\circ$ = $45^\circ$ E
D’Long between $\text V_1$ and $\text V_2$ = $45^\circ$ – $19^\circ$ 39.5' - $15^\circ$ 18.7' = $10^\circ$ 01.8' = 601.8'
Dep. = Distance between $\text V_1$ and $\text V_2$ = D’Long $\times$ Cos Limiting Lat. = 601.8' $\times$ Cos $53^\circ$
= 362.2 miles

Total distance = 728 + 561.4 + 362.2 = 1651.6 miles.

Transferring GC Course to Mercator Chart

Transfer position along a Great Circle Track to Mercator Chart

On a gnomonic chart, the parallels of latitude are curved and the meridians of longitude are straight lines converging towards the pole. So a parallel ruler cannot be used to measure off position.

Gnomonic chats are ocean charts and the positions on them in terms of latitude and longitude is somewhat lacking as they are small scale charts.
Draw the GC course as a straight line on the Gnomonic chart.
Select meridians cutting the GC at suitable intervals of $5^\circ$ or $10^\circ$ , and mark the points of intersection between the meridians and the GC.
For latitude: Use a divider and read off the d’lat from nearest parallel marked on the chart. Take that to the eastern or western boundary of the chart and read off the latitude.
For longitude: The straight lines are converging towards the poles. Use a divider and measure off the distance in millimetres between the nearest meridian on each side next to the position. Measure the distance in millimetres from the position to the nearest meridian and interpolate the value of longitude.
Plot the latitudes and longitudes of each point of intersection on the Mercator chart and draw Rhumb lines by joining the points.
The ship will sail along these Rhumb line courses, which is the closest representation of a GC course.

Chart projections

The following media explains Chart Projections:

Mercator Chart
The angles between any three real objects in the chart should be same as the angles between the real objects on land.

Although a straight course on a chart is an arc on the globe it should appear as a straight line in the chart.

To understand the projection in two dimensions, imagine the earth is marked with the parallels of latitude and meridians of longitude. Wrap a paper sheet in cylinder form around the planet earth so that it touches the equator closely. On this cylinder, the surface of earth is projected with a lamp placed at the center; the projected light will cause images of all lines on the surface along with land masses on the enclosed cylindrical paper. If the enclosing cylinder paper is cut open and laid flat, we get the two dimensional mercator chart. The mercator chart of the world in reality is done by mathematical calculation.

This is the most frequently used chart as it shows the direct course between two points as a straight line, know as a rhumb line. On a mercator chart the direction is correctly represented. This type of projection is said to be orthomorphic and the mercator chart is a cylindrical orthomorphic projection.

The mercator chart is named after Gerard Mercator. A nautical chart requires a two dimensional representation of a three dimensional earth sphere of the world, therefore such charts were developed with mathematical precision.

The cylindrical projection

To understand the mercator projection, imagine the earth wrapped by a cylinder of paper, touching along the equator. Imagine that the sphere is marked with parallels of latitude and meridians of longitude and that your eye is at the centre of the sphere, looking at the projection of these lines onto the cylinder.

Projection of parallels of latitude

The parallels of latitude – small circles on the surface of the sphere, i.e. the planes of the circles that do not pass through the centre of the sphere – are projected so that their distance apart increases as the latitude increases. The projected distance between the parallels of $60^\circ$ and $70^\circ$ is much greater than the distance between the parallels of $30^\circ$ and $40^\circ$ . In high latitudes the projected distance becomes very large, and infinitely large at the poles. So the mercator projection is only of practical use in latitudes of less than about $70^\circ$ north or south of the equator.

Projection of meridians of longitude

The meridians of longitude – great circles on the surface of the sphere, i.e. the planes of the circles that pass through the centre of the sphere, are projected as equidistantly spaced straight lines. If there are six meridians, the angle at the pole or the arc of the equator between each of them will be $60^\circ$ $(360^\circ/6)$ . If one of the meridians is at Greenwich i.e the Prime Meridian, (the meridian passing through the Greenwich Observatory in London) and the cylinder is removed and laid flat, the meridians will appear as straight lines representing longitudes $0^\circ$ (Greenwich), $60^\circ \text{E} \ \& \ \text{W}, 120^\circ \text{E} \ \& \ \text{W}$ and $180^\circ$ .

The mercator graticule

The complete mercator graticule thus appears as shown in the chart projections graphic, with the parallels of latitude becoming progressively further apart as the latitude increases, but with the meridians of longitude equidistant. Because of the increasing distance between the parallels, distance measurements are always made in the latitude under consideration.

The measurement of direction on a mercator chart

The rhumb line course from A to B cuts every meridian at the same angle. If this angle is $62^\circ$ , the course is at an angle of $62^\circ$ from true north as the meridians run true north and south. The direction could also be expressed as the direction of B from A is 062º. A protractor printed on the chart, known as a compass rose, is used to read the angle between a rhumb line or bearing and the meridians. Parallel rulers are normally used to transfer the direction of the rhumb line or bearing to the nearest compass rose, as shown in the graphic. A plastic protractor, oriented to the meridian, can also be used. Often a protractor is printed on the parallel rulers.

The measurement of distance on a mercator chart

The nautical mile is defined as the arc of a meridian subtended by one minute at the centre of the sphere and the latitude scale also represents distance along the meridians. Since the nautical mile is also represented by the length of one minute of latitude, the latitude scale is used for measuring distance. In the graphic, the distance between A and B is placed upon the dividers and then measured on the latitude scale adjacent to the same latitude. The points lie on $36^\circ 39.3 ' \text{S}$ and $36^\circ 43.0 ' \text{S}$ . A difference of 3.7’ of latitude, i.e. a distance of 3.7 nautical miles.

Only the latitude scale is used for measuring distance, except where there is a special distance scale, common on large scale charts such as harbour charts. The longitude scale is not used as it does not indicate distance. It is used for reading longitudes.

Summary

Mercator charts are used because:
Direction is correctly represented; the projection is orthomorphic
A rhumb line or bearing is represented by a straight line cutting all meridians at the same angle
The scale of a mercator chart expands as the latitude increases, causing areas to be exaggerated in higher latitudes. The expansion of the latitude scale is in the same ratio as the separation of the meridians with increasing latitude and is equal to the secant of the latitude. The secant of $90^\circ$ is infinity, hence the poles cannot be shown. Distance must therefore be measured at the mean latitude of the latitudes of the two positions between which the distance is being measured.

Features

The main properties of the mercator chart are courses appearing as straight lines on the chart, which allows drawing a straight line from the departure point to the destination.

The angles between the course lines and meridians allow us to measure the steady course to sail to the destination.

The parallels of latitude appear as straight lines parallel to the equator; spacing between the parallels increases as the latitude increases.

Distances are measured on the latitude scale, which are always on the left and right hand margins of the chart.

The meridians appear as parallel straight lines at right angles to the equator. Spacing between meridians is equal. Longitudes are measured on the upper and lower margins of the chart.

The longitude scales should not be used for measuring distance.

A straight line drawn on the chart joining two points does not represent the shortest distance between them.

The chart is orthomorphic. It means the shape of the land and the directional angles are preserved.

Limitations of the mercator chart are that this projection is only used for charts covering areas between 70 degrees north and 70 degrees south.

Thus the north and south polar regions cannot be projected in this method. They are projected on polar gnomonic charts.

Gnomonic Projection Charts
A gnomonic projection is the projection of the surface of the earth from the centre to the surface for a given point. It can be any point.

This is useful for projecting less than one hemisphere at a time.

A chart drawn on the gnomonic projection will have all great circles appearing as straight lines.

Hence, these charts are useful for great circle sailing.

The gnomonic projection is useful for polar charts as the points of the tangency may be located at one of the earth poles.

The projections are also used for charts having a natural scale greater than 1 to 50,000.

The limitation of this projection is that it is not orthomorphic; and hence it is hard to measure the direction and distance.

Meridians of longitude appear as straight lines converging at the nearest pole.

Plan Charts

These charts show small areas such as harbours, ports docks etc. They display details of jetties, berths, docks, canal, rivers etc. They facilitate navigation in narrow and confined waters. The natural scale may be from 1: 50,000 down to about 1: 12,500

Ocean Charts

The ocean charts cover large areas of earth’s surface and are very useful in planning of ocean voyages. They cannot be used for small period position plotting like an hour or fraction of it. As the areas covered are very large they do not show details of undersea dangers such as shoals or wrecks. They are available in natural scale from 1:600 000 to 1 : 14 000 000.

Mercator Chart

Mercator projection
This has been covered earlier. However, a short recapitulation is given here.

Chart is orthomorphic – Though areas appear larger at higher latitudes, directions are correct.
The equator appears as a straight line.
All meridians appear as straight lines parallel to one another.
The distance between consecutive meridians is constant.
All meridians cross the equator at right angles.
All meridians & and all parallels of latitude cross at right angles.
Parallels of latitude would appear as straight lines parallel to one another.
The distance between consecutive parallels increases as latitude increases.
One minute of d'long is the same size in all parts of the chart (see (4) above).
One minute of d'lat increases in size steadily as we go away from the equator (see (8) above).
Polar regions do not appear.
Rhumb lines appear as straight lines.
The great circle path (GCP) always lies on the pole ward side of the rhumb line, on the chart as well as on a globe.

Tangential Mercator Chart

Tangential Mercator projection
This is very similar in principle of the normal Mercator projection except that the globe’s axis is rotated by 90º making the plane of the equator coincide with the axis of the cylinder.
The latitude and longitude of a position, called the chosen position, that is required to be the centre of the chart are noted.
The globe is then rotated along its N-S axis until the chosen meridian of longitude appears at the centre of the chart.
The globe is then rotated along its equatorial axis until the chosen latitude touches the cylinder.
The tangential point is thus the latitude and longitude of the chosen position.
The parts of the meridians visible on the chart would appear straight and parallel.
The parallels of latitude visible on the chart would also appear as straight lines but the distance between them would gradually increase as we go towards the equator but this distortion over a few hundred miles would be negligible.

Both, Normal Mercator projection and Tangential Mercator projection, share the same properties.:
Both projections are orthomorphic – directions and shapes are well preserved though sizes get distorted with latitude,
Both projections have constant scale on the line of tangency (the equator for the normal Mercator and the central meridian for the transverse).
Since the latitude and central meridian of the transverse Mercator can be chosen at will, it may be used to construct highly accurate maps (of narrow width) anywhere on the globe. This projection is used for accurate large scale maps of small areas.

Universal Transverse Mercator Chart

Universal Transverse Mercator (UTM) Projection

Gnomonic Chart

Gnomonic Projection
Like Mercator projection, Gnomonic projection is another method of depicting on a flat surface, parts of the curved surface of the earth.
Imagine the earth to be a transparent globe with a source of light at its centre.
Imagine that a flat, semi-transparent surface is placed on its top, tangential to the north geographic pole. If you now look at the flat surface from top, the picture you will see is shown in the accompanying picture.

Gnomonic map projection displays all great circles, including Meridians, as straight lines. Thus the shortest route between two locations corresponds to that on the chart. The least distortion occurs at the point of contact (the tangential point), the North Pole in this case. Less than half of the sphere can be projected on a gnomonic chart. In the above case, since the tangential plane is the pole, equatorial regions cannot be shown.
The accompanying pictures show a trans-Pacific passage being planned on a Gnomonic Chart & then on a Mercator Chart.

The main use of gnomonic charts is for large ocean passages. The starting & finishing points are plotted & a straight line is drawn joining them. This represents the great circle track between the two points. The highest latitude during the passage, suitable way points, the different islands situated nearby the track, etc. can be seen at a glance.
Though GPS can do all the above mathematical calculations, on a gnomonic chart you can see the entire ocean passage at a glance, especially islands & land masses near the route.
Gnomonic charts are also used in high latitudes where distortions on a Mercator chart would be unacceptable.

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Sunday, June 9, 2024

Great circle sailing

Spherical Triangles

Initial and final courses

GC Distances

Napier's Rules

Composite Great Circle sailing

Transferring GC Course to Mercator Chart

Chart projections

Mercator Chart

Tangential Mercator Chart

Universal Transverse Mercator Chart

Universal Transverse Mercator (UTM) Projection

Gnomonic Chart

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