Wnmjtcs ; alas the distance between the parallel 
of 5(5°, and the proposed parallel of 56° 44', is 
4 1 minutes: then, by the preceding proportion, 
it will be, As 60 is to 87, so is 44 to 638, the dif- 
ference between a degree on the parallel of 56° 
and a degree on the parallel of 56° 44' ; which, 
therefore, taken from 33.55, the miles answering 
to a degree on the parallel of 56°, leaves 32 912, 
the miles answering to a degree on the parallel 
'of 56° 44'’, as was required. 
Case II. The distance sailed in any parallel of 
latitude, or the distance between any two places 
on that parallel, being given ; to find the differ- 
ence of longitude. 
Example. Suppose a ship in the latitude of 
55° 36' north, sails directly east 685,6 miles : 
required how much she has differed her lo»- 
gitude. 
By Cor. 4. Art. 1. of this section, it will be 
As the co-sine of the lat. 55° 36' 9.75202 
is to radius - - - 10.00000 
*o is the distance sailed 68.5.6 2.83607 
to minute of dilferenceof Ion. 1213 3.08405 
which reduced into degrees, by dividing by 60, 
makes 20° 13', the difference of longitude the 
ship has made. 
This also may be solved by help of the pre- 
ceding table, viz. by finding from it the miles 
answering to a degree on the proposed parallel, 
and dividing with this the given number of 
miles, the quotient will be the degrees and mi- 
nutes of difference of longitude required. 
Thus in the last example, we find, from the 
foregoing table, that a degree on the parallel of 
55° 33' is equal to 33.89 miles ; by this we divide 
the proposed number of miles 685.6, and the 
quotient is 20.13 degrees, i. e. 20° 13', the dif- 
ference of longitude required. 
Case III. The difference of longitude between 
two places on the same parallel, and the distance 
between them, being given ; to find the latitude 
of that parallel. 
Example. Suppose a ship sails on a certain 
parallel directly west 624 miles, and then has 
■differed her longitude 18° 46', or 1126 miles: 
required the latitude of the parallel she sailed 
upon ; it will be by Cor. 3. before 
As the min. of diff. long. 1. 126 3.05154 
is to the distance sailed 624 2.7954 8 
so is radius - 10.00000 
to the co-sine of the lat. 56°, 21' 9.74364 j 
consequently the latitude of the ship, or parallel ] 
she sailed upon, was 56° 21'. 
From what has been said, may be solved the 
following problems : 
Prob. I. Suppose two ships in the latitude of 
46° 30' north, distant asunder 654 miles, sail 
both directly north 256 miles, and consequently 
are come to the latitude of 50° 46' north: re- 
quired their distance on that parallel. 
By Cor. 5. Art. 1. of this section, it will be, 
As the co-sine of 46 Q , 30' 9 83781 
is to the co-sine of 50°, 46' 9.80105 
so is 654 - 2.81558 
to - - 601 - 2.77882 
the distance between the ships when on the pa- 
rallel of 50° 46'. 
Prob. II. Suppose two ships in the latitude of 
45° 48' north, distant 846 miles, sail directly 
north till the distance between them is 624 
miles : required the latitude come to, and the 
distance sailed. 
By Cor. 5. Art. 1. of this section, it will be, 
As their first distance 846 2.92737 
is to their second distance 624 2.79518 
so is the co-sine of - 45°, 48' 9.84334 
to the co-sine - - 59°, Off 9.71115 
the latitude of the parallel the ships are come to. 
Consequently, to find their distance sailed, 
From the latitude come to - 59°, 04' 
subtract the latitude sailed from 45 , 48 
and there remains! - .- 13 , 10 
NAVIGATION 
equal to 796 miles, the difference of latitude or 
distance sailed. 
Of Middle-latitude Sailing. 1 . When two places 
lie both on the same parallel, we have shewn 
how, from the difference of longitude given, to 
find the miles of easting or westing between 
them, at e contra . But when two places lie not 
on the same parallel, then their difference of 
longitude cannot be reduced to miles of easting 
or westing on the parallel of either place : for 
if counted on the parallel of that place tliat has 
the greatest latitude, it would be too small ; 
and if on the parallel of that place having the 
least latitude, it would be too great. Hence the 
Qommon way of reducing the difference of lon- 
gitude between two places, lying on different 
parallels, to miles of easting or westing, ct e con- 
tra, is by counting it on the middle parallel be- 
tween the places, which is found by adding the 
latitudes of the two places together, and taking- 
half the sum, which will be the latitude of the 
middle parallel required. And hence arises the 
solution of the following cases : 
Case I. The latitudes of two places, and their 
difference of longitude, given ; to find the direct 
course and distance. 
Example. Required the direct course and dis- 
tance between the Lizard in the latitude of 
50° 0' north, and longitude of 5° 14' west, and 
St. Vincent in the latitude of 1 7° ' 1C/ N. and lon- 
gitude of 24° 20' W. 
First, To the latitude of the Lizard 50° 00' N 
add the latitude of St. Vincent 17 10 
Proiff the latitude she 5va$ fit 
take the difference of latitude 
50 °, 00 * 
1 , 36 
The sum is - - 67 10 
Half the sum or latitude of the 
middle parallel is - 33 35 N 
Also the difference of latitude is 33 50 
equal to 1970 miles of southing. Again, 
From the longitude of St. Vincent 24 20 W 
take the longitude of the Lizard 05 14 
33°, 35 
9.92069 
there remains - - - 16 06 
equal tol 146 min. of diff'. of Ion. west. 
Then for the miles of westing, or departure, 
it will be (by Case 1. of Parallel Sailing), 
As radius - - - "10.00000 
is to the co-sine of the? 
middle parallel $ 
so is min. cliff, of Ion. - 1146 3.05918 
to the miles of westing 954.7 2.97987 
And for the course it will be (by Case 4. of 
Plane Sailing), 
As the diff of lat. - 1970 3.29447 
is to radius - - - - 10.00000 
so is the departure - 954.7 2.97987 
to the tang, of the course 25°, 51' 9.68540 
which, because it is between south and west, 
will be SSW ~ west nearly. 
For the distance, it will be, by the same case, 
As radius - 10.00000 
is to the diff. of lat. - 1970 3.29447 
so is the secant of the course 25°, 5 1' 10.04579 
to the distance - - 2189 3.34026 
whence the direct course and distance front the 
Lizard to St. Vincent are SSW i 2189 W. miles. 
Case IT. One latitude, course, and distance 
sailed, being given; to find the other latitude, 
and difference of iongitude. 
Example. Suppose a ship in the latitude of 
50° 00' north, sails south 50°0(>' west, 150 miles: 
required the latitude the ship has come to, and 
how much she has differed lur longitude. 
First, For the difference of latitude, it will be, 
(by Case 1. of Plane Sailing,) 
As radius - - 10.00000 
is to the distance - 150 2 17609 
so is the co-sine of the course 50°, 06' 9.80716 
to the diff of latitude - 96.22 ] .98825 | 
equal to 1°, 36', And since the ship is sailing j 
toward.* the equator : therefore, 
is. k y 
and there remains - - 48 , 24 
the latitude she has come to north. Conse- 
quently the latitude of the middle parallel will 
be 49° 12'. 
Then for departure or westing it will be, by 
the same Case, 
As radius - - 10.00000 
is to the distance - 150 2.17609 
so is the sine of the course 50°, 06' 9.88489 
to the departure - 115.1 2.06098 
As for the difference of longitude, it will be, 
(by Case 2. of Plane Sailing,) 
As the co-s. of the middle par. 49° 12' 9.81519 
is to radius - 10.00000 
so is the departure - 115.1 2.06098 
to the min. diff. of longitude 176.1 2 24579 
equal to 2° 56', which is the difference of longi- 
tude the ship has made westerly. 
Case III. Course and difference of latitude 
given ; to find the distance sailed, and differ- 
ence of longitude. 
Example. Suppose a ship in the latitude of 
53° 34' north, sails SE4S, till by observation she 
is found to be in the latitude of 51° 12', and 
consequently has differed her latitude 2° 22', or 
142 miles: required the distance sailed, and 
the difference of longitude. 
First, for the departure, it will be, (by Case 2. 
of Plane Sailing,) 
As radius - - - 10 00000 
is to the diff. of latitude 142 2.15229 
so is the- tang, of course 33°, 45' 9.82489 
to the departure - 94.88 1.9771S 
And for the distance it will be, (by the same 
Case,) 
As radius __ - - 1 0.00000 
is to the diff. of latitude 142 2.15229 
so is the secant of the course 33°, 45' 10.080l£ 
to the distance - 170,8 2.23244 
Then, since the latitude sailed from was 53° 
34' north, and the latitude come to 51° 12’ 
north ; therefore the middle parallel will be 
52° 23' ; and consequently, for the difference of 
longitude, it will be, (by Case 2. of Parallel 
Sailing,) 
As the co-sine of the mid. par. 52°, 23' 9.78560 
is to the departure - 94.88 1.97718 
so is radius - - 10.00000 
t» min. of diff. of longitude 155.5 2.1917S 
equal to 2° 35', the difference of longitude 
easterly. 
Case IV. Difference of latitude and distance 
sailed, given ; to find the course and difference 
of longitude, 
Example, Suppose a ship in the latitude of 
43° 26' north, sails between south and east, 246 
miles, and then is found by observation to be 
in the latitude of 41° 06' north : required the. 
direct course and difference of longitude. 
First, for the course, it will be, (by Case 3. of 
Plane Sailing,) 
As the distance - 246 2.30094 
is to radius - - 10.00000 
so is the diff. of latitude 140 2.1461.3 
to the co-sine of the course 55 Q , 19' 9.75519 
which, because the ship sails between south and 
east, will be south 55° 19' east, or SE4E nearly. 
Then, for departure, it will be, by the same 
Case, 
As radius _ ... lO.OOOQO 
is to the distance - 213 2.39094 
so is the sine of the course 55°, 19' 9.91504 
to the departure - 202,3 2,3050? 
Lastly, For the difference of longitude, it wil^ 
will be, (by Case 2. of Parallel Sailing.) 
As the co-sine of the mid. par, 42°, Iff 9.86924 
is to the departure - 803.8 2.3059a 
so is radius ^ - = IOGGjoQ 
te min. of diff. of longitude 273,3 Vvff-ff,-* 
