NAVIGATION. 
2.57 
will be less than the departure ; and lastly, if 
the course is just 4 points, the difference of la- 
titude will be equal to the departure. 
5. Since the distance, difference of latitude, 
and departure, form a right-angled triangle, in 
which the oblique angle opposite to the depar- 
ture is the course, and the other its complement; 
therefore, having any two of these given, we 
can (by plane trigonometry) find the rest ; and 
hence arise the cases of plane-sailing, which are 
as follow : 
Case I. Course and distance given, to find 
the difference of latitude and departure. 
Example. Suppose a ship sails from the lati- 
tude of 30° 25' north, N. NE. 32 miles (fig. 3). 
Required the difference of latitude and depar- 
ture, and the latitude come to. Then (by right- 
angled trigonometry) we have the following ana- 
logy for finding the departure, viz. 
As radius ----- 10.00000 
to the distance AC - 32. 1.50515 
so is the sine of the course A 22° 30' 9.58284 
to the departure BC - 12.25 1.08799 
so the ship has made 12.25 miles of departure 
easterly, or has got so far to the eastward of her 
meridian. Then for the difference of latitude or 
northing the ship has made, we have (by rect- 
angular trigonometry) the following analogy, 
viz. 
As radius ----- 10.00000 
is the distance AC - 32 1.50515 
so is the co-sine of course A 22° 30' 9 58284 
to the difference of lat. AB 29.57 1.47077 
so the ship has differed her latitude, or made of 
northing, 29.57 minutes. 
And since her former latitude was north, and 
her difference of latitude also north ; therefore, 
To the latitude sailed from - 30°, 25' N 
add the difference of latitude 00°, 29.57 
and the sum is the latitude come to 30°, 54.57'N. 
By this case are calculated the tables of dif- 
ference of latitude, and departure, to every de- 
gree, point, and quarter-point, of the compass. 
Case II. Course and difference of latitude 
given, to find distance and departure. 
Example. Suppose a ship in the latitude of 
45° 25' north, sails NEiN£ easterly (Plate Na- 
vigation, fig. 4), till she comes to the latitude of 
46° 55' north: required the distance and depar- 
tnre made good upon that course. 
Since both latitudes are northerly, and the 
course also northerly ; therefore, 
From the latitude come to - 46°, 55' 
subtract the latitude sailed from 45°, 25' 
and there remains - - - 01°, 3<y 
the difference of latitude, equal to 90 miles. 
And (by rectangular trigonometry) we have 
the following analogy for finding the departure 
BD, viz. 
As radius ----- 10.00000 
is to the diff. of latitude AB 90 1.95424 
so is the tangent of course A 39°, 22' 9.91104 
to the departure BD - 73.84 1.86328 
si* the ship has got 73.84 miles to the eastward 
of her former meridian. 
Again, for the distance AD, we have (by rect- 
angular trigonometry) the following proportion, 
viz. 
As radius ----- 10.00000 
is to the secant of the course 39°, 22' 10.1 1 176 
so is the diff of latitude AB 90 1.95424 
to the distance AD - 116.4 2.06G00 
Case III. Difference of latitude and distance 
given, to find course and departure. 
Example. Suppose a ship sails from the lati- 
tude of 56° 50' north, on a rhumb between south 
and west, 126 miles, and she is then found by 
observation to be in the latitude of 55°, 40' 
north : required the course she sailed on, and 
her departure from the meridian. (Fig. 5.) 
Since the latitudes are both north, and the 
\ OL. II. 
ship sailing towards the equator; therefore, 
From the latitude sailed from - 56°, 50' 
subtract the observed latitude - 55°, 10' 
and the remainder - 01°, 40' 
equal to 70 miles, is the difference of latitude. 
By rectangular trigonometry we have the fol- 
lowing proportion for finding the angle of the 
course F, viz. 
As the distance sailed DF 126 2.10037 
is to radius - - - - 10.00000 
so is the diff'. of latitude FD 70 1.84510 
to the co-sine of the course F 56°, 15' 9.74473 
which, because she sails between south and west, 
will be south 56° 15' west, or SW/AV, Then, 
for the departure, we have (by rectangular tri- 
gonometry) the following proportion, viz. 
As radius' - 10.00000 
is to the distance sailed DF 126 2.10037 
so is the sine of the course F 56°, 15' 9.91985 
to the departure DE - 104.8 2.02022 
consequently she has made 104.8 miles of de- 
parture westerly. 
Case IV. Difference of latitude and departure 
given, to find course and distance. 
Example. Suppose a ship sails from the latitude 
of 44° 50' north, between south and east, till 
she has made 64 miles of easting, and is then 
found by observation to be in- the latitude of 
42° 56' north: required the course and distance 
made good. 
Since the latitudes are both north, anti the 
ship sailing towards the equator; therefore. 
From the latitude sailed from - 44°, 50' N 
take the latitude come to - 42°, 56' 
and there remains - - - 01°, 54' 
equal to 114 miles, the difference of latitude or 
southing. 
In this case (by rectangular trigonometry) we 
have the following proportion to find the course 
KGL (fig. 6), viz. 
As the diff. of latitude GK 114 2.05690 
is to radius - - - - 10.00000 
so is the departure KL 64 1.80618 
to the tangent of course G 29°, 19' 9.74928 
which, because the ship is sailing between south 
and east, will be south 29° Iff east, or SSE^ 
east nearly. 
Then for the distance, we shall have (by rect- 
angular trigonometry) the following analogy, 
viz. 
As radius ----- 10.00000 
is to the diff. of latitude GK 114 2.05690 
so is the secant of the course 29°, 19' 10.05952 
to the distance GL - 130.8 2.11642 
consequently the ship has sailed on a SSE^ 
east course 130.8 miles. 
Case V. Distance and departure given, to 
find course and difference of latitude. 
Example. Suppose a ship at sea sails from the 
latitude of 34° 24' north, between north and 
west, 124 miles, and is found to have made of 
westing 86 miles : required the course steered, 
and the difference of latitude or northing made 
good. 
In this case (by rectangular trigonometry) we 
have the following proportion for finding the 
course ADB, (fig. 7), viz. 
As the distance AD - 124 2.09342 
is to radius - - 10.00000 
so is the departure AB 86 1.93450 
to the sine of the course D 43° 54' 9.84108 
so the ship’s course is north 33° 45' west, or 
NW7;N - -west nearly. 
Then for the difference of latitude, we have 
(by rectangular trigonometry) the following 
j analogy, viz. 
• As radius ----- 10.00000 
j is to the distance AD 124 2.09312 
! so is the co-sine of the course 43°, 54' 9.81766 
to the diff of latitude BD 89.35 T.9J103 
which is equal to 1 degree and 29 min. nearly. 
K. k 
Hence, to find tlm latitude the ship is in, since 
both latitudes are north and the ship sailing from 
the equator ; therefore 
To the latitude sailed from - 34°, 24' 
add the difference of latitude - 1°> 29' 
the sum is - 35°, 53 
the latitude the ship is in north. 
Case VI. Course and departure given, to find 
distance and difference of latitude. 
Example. Suppose a ship at sea, in the latitude 
of 24° 30' south, sails SE^S, til! she has made cf 
easting 96 miles : required the distance and dit- 
ference of latitude made good on that course. 
In this case (by rectangular trigonometry and 
by case 2,) we have the following proportion for 
finding the distance (fig. 8), viz. 
As the sine of the course G 33°, 45' 9.74474 
is to the departure HM 96 1.98227 
so is radius - IO.OOOCO 
to the distance GM - 172.8 2.23753 
Then, for the difference of latitude, we have 
(by rectangular trigonomety) the following ana- 
logy, viz. 
As the tangent of course 33°, 45' 9.82489 
is to the departure HM 96 - 1.9S227 
so is radius ... - IO.OOOCO 
to the difference of lat. GH 143.7 2.15738 
equal to 2°, 24' nearly. Consequently, since the 
latitude the ship sailed from was south, and she 
sailing still towards the south, 
To the latitude sailed from - 24°, 30' 
add the difference of latitude - 2°, 25' 
and the sum - 26°, 5T 
is the latitude she is come to south. 
6. When a ship sails on several courses in 24 
hours, the reducing all these into one, and 
thereby finding the course and distance made 
good upon the whole, is commonly called the 
resolving of a traverse. 
7. At sea they commonlv begin each day’s 
reckoning from the noon of that day, and from 
that time they set down ail the different courses 
and distances sailed by the snip till noon next 
day upon the log-board ; then from these several 
courses and' distances, they compute the differ- 
ence of latitude and departure- for each course 
(by Case I. qf Plane Sailing) ; and these, toge- 
ther with the courses and distances, are set down 
in a table, called the Traverse Table, which 
consists of five columns : in the first of which 
are placed the courses and distances ; in the two 
next, the differences of latitude belonging to 
these courses, according as they are north or 
south ; and in the two last are placed the de- 
partures belonging to these courses, according 
as they are east or west. Then they sum up all 
the northings, and all the southings ; and taking 
the difference v-f these, they know the difference 
of latitude made good by the ship in the last 24 
hours, which will be north or south, according 
as the sum' of the northings or southings is 
greatest : the same way, by taking the sum of 
all the eastings, and likewise of all the westings, 
and subtracting the lesser of these from the 
greater, the difference will he the departure 
made good by the ship last 24 hours, which will 
he east or west according as the sum of the east- 
ings is greater or less than the sum of the west- 
ings ; then from the difference of latitude and 
departure made good by the sh : p last 24 hours, 
found as above, they find the true course and 
distance made good upon the whole (by Case 4 
of Plane Sailing), as also the. course and distance 
to tire intended port 
E ■ ample. Suppose a ship at sea, in the latitude 
of 48°, 24' north, ;>.t noon anv day, is bound to a 
port in the latitude of 43° 40' north, whose de- 
parture from the hip is l i t miles east ; conse- 
quently the direct course and distance of the 
ship is SSE. \ east 81 1 miles ; but by reason of 
the shifting of ti c winds she is obliged to steer 
