2 JO 
equal to 4° 33', the difference of longitude 
easterly. 
C.\ se V. Course and departure given ; to find 
difference of latitude, difference of longitude, 
and distance sailed. 
Example. Suppose a ship in rite latitude of 
48° 23' north, sails SW/?3, till she has made of 
westing 123 miles: required the latitude come 
to, the difference of longitude, and the distance 
sailed. 
First, For the distance, it will be, (by Case C. 
of Plane Sailing,) 
As the sine of the course 33°, 45' 6.74474 
is to the departure - 123 2 08991 
so is radius - 10.00000 
to the distance - 221.4 2.34,517 
And for the difference of latitude, it will be, 
by the same Case, 
As the tang, of course 33°, 45' 9.82489 
is to the departure - 123 2.08991 
so is radius - 10.00000 
to the diff. of latitude 184 2.26502 
equal to 3° 04'; and since the ship is sailing to- 
wards the equator, the latitude come to will be 
45° 19' north; and consequently the middle 
parallel will be 46° 51'. 
Then, to find the difference of longitude, it 
will be, (Case 2. of Parallel Sailing,) 
As the co-sine of middle par. 46°, 51' 9.83500 
is to the departure - 123 2.08991 
so is radius - 10.00000 
to min. of diff. of longitude 180 2.25491 
which is equal to 3° 00', the difference of longi- 
tude westerly. 
Case VI. Difference of latitude and departure 
given ; to find course, distance, and difference of 
longitude. 
Example. Suppose a ship in the latitude of 
46° 37' north, sails between south and east, till 
she has made of easting 146 miles, and is 
then found by observation to be in the latitude 
of 43° 24' north : required the course, distance, 
and difference of longitude. 
First, by Case 4. of Plane Sailing, it will be 
for the course, 
As the diff. of latitude 193 2.28556 
is to the departure - 146 2.16137 
so is.-radius - - 10.00000 
9.87581 
to the tang, of the course 36°, 55' 
which, because the ship is sailing between south 
and east, will be south 36° 55' east, or SE£S ^ 
east nearly. 
For the distance, it will be, by the same Case, 
As radius - - J 10 00000 
is to the diff. of latitude 193 2.28556 
so is the secant of the course 3 6°, 55' 10.09718 
to the distance - 241.4 2.38274 
Then, for the difference of longitude, it will 
be, by Case 2. of Parallel Sailing, 
As the co-sine of the mid. par. 45°, 00' 9.84949 
is to the departure - 146 2.16137 
so is radius - - 10.00000 
to min. of diff. of longitude 205 2.31188 
equal to 3° 25', the difference of longitude 
easterly. 
Case VII. Distance and departure given ; to 
find difference of latitude, course, and difference 
of longitude. 
Example. Suppose a ship in the latitude of 
S3 0 40' north, sails between south and east 165 
miles, and has then made of easting 1 12.5 miles: 
required the difference of latitude, course, and 
difference of longitude. 
First, for the course, it will be, by Case 5. of 
Plane Sailing, 
As the distance - 165 2.21748 
is to radius - - 10.00000 
so is the departure - 102.5 2.05115 
to the sine of the course 42°, 59' 9.83367 
which, because the ship sails between south and 
cast, will be south 42° 59' east, or SFtfE j east 
nearly. 
NAVIGATION*. 
And for the difference of latitude, it will be, 
by the same Case, 
As radius - . 10.00000 
is to the distance - 165 2.21748 
so is the co-sine of the course 42° 59' 9.86436 
to the difference of latitude 120.7 2.08184 
equal to 2° 00'; consequently the latitude come 
to will be 3 !° 40' north, and the latitude of the 
middle parallel will be 32° 40'. Hence, to find 
the difference ot longitude, it will be, by Case 2. 
ot Parallel Sailing, 
h.s the co-sine of the mid. par. 32°, 40' 9.92522 
is to the departure - 112.5 2.05115 
so is radius ... 10.00000 
to min. of diff. of long. 133.6 2.12593 
equal to 2° 13' nearly, the difference of longi- 
tude easterly. 
Case VIII. Difference of longitude and de- 
parture g; veil : to find difference of latitude, 
course, and distance sailed. 
Example. Suppose a ship in the latitude of 
50° 46' north, sails between south and west, till 
her difference of longitude is 3° 12', and is then 
found to have departed from her former meri- 
dian L26 miles: required the difference of la- 
titude, course, and distance sailed. 
First, for the latitude she has come to, it will 
be, by Case 3. of Parallel Sailing, 
As min. of diff. of long. 192 2.28330 
is to the departure - 126 2.10037 
so is radius - - 10.00000 
to the co-sine of mid. par. 48°, 59' 9.81707 
Now, since the middle latitude is equal to 
half the sum of the two latitudes (by art. 1. of 
this sect.) and so the sum of the two latitudes 
equal to double the middle latitude ; it follows, 
that if from double the middle latitude we sub- 
tract any one of the latitudes, the remainder 
will be the other. Plence from twice 48° 59', 
viz. 97° 58', taking 50° 46' the latitude sailed 
from, there remains 47° 12' the latitude come 
to ; consequently the difference of latitude is 
3° 34', or 214 minutes. 
Then, for the course, it will be, by Case 4. of 
Plane Sailing, 
As difference of latitude 214 2.33041 
is to radius - 10.00000 
so is the departure - 126 2.10037 
to the tang, of the course 30° 29' 9.7699S 
which, because it is between south and west, 
will be south 30° 29' west, or SSW f west 
nearly. 
And for the distance, it will be, by the same 
Case, 
As radius - - _ 10.00000 
is to the difference of lat. 214 2.33041 
so is the secant of the course 30°, 29' 10:06461 
to the distance - - 248.4 2.39502 
2. From what has been said, it will be easy to 
solve a traverse by the rules of Middle-latitude 
Sailing. 
Example. Suppose a ship in the latitude of 
43° 25' north, sails upon the following courses, 
viz. SW//S 63 miles, SSW \ west 45 miles, S5E 
54 miles, and SW4W 74 miles: required the 
latitude the ship has come to, and how far she 
has differed her longitude. 
First, By Case 2. of this sect, find the differ- 
ence ot latitude and difference of longitude be- 
longing to each course and distance, and they 
will stand as in the following table : 
Courses. Distances. 
Diff. of Lat. 
Diff.oJ 
Long. 
SWbS - 63 
SSWfW - 45 
S//F. - 54 
SWtSW - 74 
Diff. 
North. 
South. 
East. 
West. 
>f Lat. 
52.4 
39.7 
53.0 
41.1 
14.75 
47.85 
28.62 
81.08 
157.55 
13.75 
186.2 
Diff. of Long. 
143.80 
Hence it is plain the ship has differed her 
latitude 186.2 minutes, or 3* 6', and so has 
come to the latitude ol 40° 19' north, and lias- 
made of difference of longitude 143.8 minutes, 
or 2° 23' 48", westerly. 
3, I his method ot sailing, though it is not. 
strictly true, yet comes very near the truth, 
as will be evident by comparing an example 
wrought by ties method with the same- 
wrought by the method delivered in the next 
section, which is strictly true; and it serves, 
without any considerable error, in runnings- 
ol 450 mile's between the equator and parallel 
of 30 degrees, oi 300 miles between that and 
the parallel of 60 degrees, and of 150 miles 
as far as there is any occasion, and conse- 
quently must be sufficiently exact for 24 
hours run. 
Of Mercator’.'; sailing. Though the me- 
ridians do all meet at the pole, and the pa- 
rallels to the equator do continually decrease, 
and that in proportion to the co-sines of their 
latitudes ; vet m old sea-charts the meridians 
weie drawn parallel to one another, and con- 
sequently the parallels of latitude made equal 
to the equator, and so a degree of longitude, 
on any parallel as large as a degree on the 
equator ; also in these charts the degrees of 
latitude were still r epresented (as they are in 
themselves) equal to each other, and to those 
of the equator. Rv tnese means-the degrees 
of longitude being increased beyond their 
just proportion, and the more so the nearer 
they approach the pole, the degrees of lati- 
tude at the same time remaining the same, it 
is evident places must be very erroneously 
marked down upon these charts with respect 
to their latitude and longitude, and conse- 
quently their bearing from one another very 
false. J 
1 o remedy this inconvenience, so as still to 
keep the meridians parallel, it is plain we 
must protract, or lengthen, the degrees of 
latitude in the same proportion as those of 
longitude are, that so the proportion in east- 
ing and westing may be the same with that 
of southing and northing, and consequently 
the bearings of places from one another are the 
same upon the chart as upon the globe itself. 
Let ABD (tig. 10.) be a quadrant of a me- 
ridian, A the pole, D a point on the equator,. 
AC half the axis, B any point upon the me- 
ridian, from which draw BE perpendicular 
to AC, and BG perpendicular to CD: then 
BG will he the sine, and BF or CG the co- 
sine, of BD the latitude of the point B ; draw 
I) E the tangent and C E the secant of the arch 
CD. It has been demonstrated,, that any 
arch of a parallel is to the like arch of the 
equator, as the co-sine of the latitude of that 
parallel is to radius. Thus any arch, as a 
minute on the parallel described bv the point 
B, will be to a minute on the equator, as BF 
°r GGis to C D ; but since the triangles CGI?,. 
CDE, are similar, therefore CG will be to 
C D as CB is to CE, i. e. the co-sine of any 
parallel is to radius as radius is to the secant 
of the latitude of that parallel. But it has 
been just now shown, th t the co-sine of any 
parallel is to radius, as the length of any arch 
(as a minute) on that parallel is to the length 
of the like arch on the equator ; therefore the 
length of any arc!) (as a minute) on any pa- 
rallel, is to the length of the like arch on. the 
equator, as radius is to the secant of the 
latitude of that parallel ; and so the length 
