Parallel sailing consists of the interconversion of departure and difference of longitude. It is the simplest form of spherical sailing. The formulas for these transformations are:
 $DL_o = p \sec L$ $p = DL_o \cos L$

Example 1: The DR latitude of a ship on course 090° is 49°30' N. The ship steams on this course until the longitude changes 3°30'.

Required: The departure by (1) computation and (2) traverse table.

Solution:

(1) Solution by computation:

$DL_o &=3^{\circ}{30}'$

$DL_o = 210 \;arc\;min$

$p = DL_o \times \cos L$

$p = 210\; arc\;minutes \times \cos 49.5^{\circ}$

$p = 136.4\;miles$

p = 136.4 miles

(2) Solution by traverse table:

Refer to Figure A.

Enter the traverse table with latitude as course angle and substitute $DL_o$ as the heading of the Dist. column and Dep. as the heading of the D. Lat. column. Since the table is computed for integral degrees of course angle (or latitude), the tabulations in the pages for 49° and 50° must be interpolated for the intermediate value (49°30'). The departure for latitude 49° and $DL_o \;{210}'$is 137.8 miles. The departure for latitude 50° and $DL_o\; {210}'$ is 135.0 miles. Interpolating for the intermediate latitude, the departure is 136.4 miles.

p = 136.4 miles

Example 2: The DR latitude of a ship on course 270° is 38°15'S. The ship steams on this course for a distance of 215.5 miles.

Required: The change in longitude by (1) computation and (2) traverse table.

Solution:

(1) Solution by computation

$DL_o &=3^{\circ}{30}'$

$DL_o = 210\; arc\;min$

$p= DL_o \times \cos L$

$p= 210\; arc\;minutes \times \cos 49.5^{\circ}$

$p = 136.4\;miles$

Answer: $DL_o = 4^{\circ} {34.4}' W$

(2) Solution by traverse table Refer to Figure B.

Enter the traverse tables with latitude as course angle and substitute $DL_o$ as the heading of the Dist. column and Dep. as the heading of the D. Lat. column. As the table is computed for integral degrees of course angle (or latitude), the tabulations in the pages for 38° and 39° must be interpolated for the minutes of latitude. Corresponding to Dep. 215.5 miles in the former is $DL_o \;273^{\circ}{4}'W$, and in the latter $DL_o\; {277.3}'$. Interpolating for minutes of latitude, the $DL_o = 274^{\circ}.{4}'W$

$DL_o = 4^{\circ} {34.4}'W$