# Plane Parallel Mercator Sailing Formula

## Parallel Sailing

• Departure and Arrival positions are on the same latitude. This is called parallel sailing.
• In parallel Sailing, dep is the distance.
• Departure = dlong * cos latitude
• Assume that the earth is a perfect sphere.

Find the distance travelled by a ship on a course of 090° T at latitude 45° N. If its longitude changed by 20°. If latitude was 60° N, find the distance.

20° = 20 * 60 = 1200′ of arc

For 45° N:

dep = dlong * cos lat = 1200 * cos 45° = 848.5′

For 60° N:

dep = dlong * cos lat = 1200 * cos 60° = 600′

A ship in position 41° 10′ S 032° 45′ W is steaming a course of 090° T at a speed of 16 knots. Find the longitude reached after 22 hours of steaming.

Distance covered in 22 hours = 22 * 16 = 352′

In parallel sailing, dep = distance = 352′

dep = dlong * cos lat

dlong = dep / cos lat = 352 / cos 41° 10′ = 7° 47.6′ E (Direction E because course is 090° T)

Initial longitude – 032° 45′ W , dlong – 7° 47.6′ E

So, Final longitude = 024° 57.4′ W

## Plane Sailing

In plane sailing, dlat – dep – dist – course are considered to form the plane of a right angled triangle.

Plane sailing formula should not be calculated for distance more than 600′.

In plane sailing formula:

• sinco = dep / dist
• cosco = dlat / dist
• tanco = dep / dlat
• dep = dlong * cos m’lat
• mean lat = (N + N)/2 or, (N – S)/2
• dlat same name (-) and opposite name (+); direction towards to destination
• Final lat = Initial lat – dlat (if name opposite)
• Final lat = Initial lat + dlat (if name same)

Find the distance travelled and course steered by a ship that has moved 45′ to the south and 30′ to the west of its initial position.

Here, dlat = 45′ and dep = 30′

tanco = dep / dlat = 30 / 45 = 33.7° = S 33.5° W

cosco = dlat / dist

So, dist = dlat / cosco = 45 / cos 33.7° = 54.1′

If a ship covers a distance of 35′ in a general north easterly direction and changes its latitude by 20°, find the course that it has steered.

Here, dlat = 20° and dist = 35′

We know, cosco = dlat / dist = 20 / 35 = 55.1° = N 55° E or 55° T

If a ship departs from position 24° 30′ N 038° 20′ W for 22° 40′ N 039° 35’W, find the course and distance travelled by the ship.

Direction – SW

dlat = 24° 30′ N – 22° 40′ N = 1° 50 ‘ S = 110’

dlong = 038° 20′ W – 039° 35′ W = 001° 15′ W = 75′

mean latitude = (24° 30′ + 22° 40′) / 2 = 23° 35′ N

dep = dlong * cos m’lat = 75 * cos 23° 35′ = 68.74′ W (direction as dlong)

tanco = dep / dlat = 68.74 / 110 = 32° = S 32° W = 212° T

cosco = dlat / dist;

So, dist = dlat / cosco = 110 / cos 32 = 129.7′

If a ship departs from position 34° 20′ S 040° 30′ W on a course of 033° T for 350′; find the position reached.

Here, Course = 033° T = N 33° E

dist = 350′

cosco = dlat / dist

So, dlat = cosco * dist = cos 33° * 350′ = 04° 53.5′ N

Final lat = Initial lat – dlat = 34° 20′ S – 04° 53.5′ N = 29° 26.5′ S

mean lat = (34° 20′ S + 29° 26.5′ S) / 2 = 31° 53.25′ S

sinco = dep / dist

So, dep = sinco * dist = sin 33° * 350 = 190.6′

dep = dlong * cos m’lat

So, dlong = dep / cos m’lat = 190.6 / cos 31° 53.25′ = 3° 44.5′ E

Final longitude = 040° 30′ W- 3° 44.5′ E = 036° 45.5′ W

Position reached = 29° 26.5′ S 036° 45.5′ W

## Mercator Sailing

• Mercator sailing uses DMP and dlong instead of dlat and dep.
• DMP = Difference of meridional parts
• MP value can be found in nautical tables.
• IF same latitude, DMP = MP1 – MP2 (same names subtract AND different names add)
• dlat = DMP
• dlong = departure
• Mercator sailing provides greater accuracy
• In Mercator projection, meridians appear @ right-angles to the earth’s equator, and lines of latitude are horizontal lines whose actual distance increases from the equator towards the pole.

Determine the Mercator course and distance between 20° 24′ S 057° 26′ E and 34° 10′ S 112° 28′ E.

Direction – SE

dlat = 20° 24′ S – 34° 10′ S = 13° 46′ = 826′ S

dlong = 057° 26′ – 112° 28′ = 055° 02′ = 3302′ E

We have got MP, 1242.56 and 2170.41

So, DMP = 1242.56 – 2170.41 = 927.85

tanco = dlong / DMP = 3302 / 927.85 = 74.3° = S 74.5° E = 105.5° T

cosco = dlat / dist

So, dist = dlat / cosco = 826 / cos 74.3° = 3053′

If a ship departs from position 46° 14′ N 125° 36′ W on a course of 237° T and covers 7076′, find the position reached.

Course = 237° T = S 57° W

dlat = distance x cos course = 7076′ x cos 57° = 64° 13′.9 S

Arrived lat = 46° 14′ N – 64° 13′.9 S = 17° 59.9′ S
46° 14′ N – MP 3118.83
17° 59.9′ S – MP 1090.885
So, DMP 4209.715

d.long = tan course x DMP = tan 57° x 4209.715 = 108° 02.4′ W

Arrived long = departure long + d.long = 125° 36.0′ W + 108° 02.4′ W = 126° 21.6′ E (Subtraction of 360° has been carried out)

Position reached = 17° 59.9′ S 126° 21.6′ E