263 
V'h'c'n, because the ship sails between south and 
west, will be south 37° 27' west, or SW/5S 
6° 3& westerly. 
Lastly, For the distance AB, it will be, (by 
rectangular trigonpmetry,) 
s, a : bc :: r : ab. 
i. e. As the sine of the course 37° 27' 9.78395 
js to the departure - 126.4 2.10175 
so is radius - - 10.00000 
to the direct distance - 207.9 2.31780 
Case IX. One latitude, distance sailed, and 
1 departure, given ; to find the other latitude, dif- 
ference of longitude, and course. 
Example. Suppose a ship in the latitude of 
4S° 33' north, sails between south and east 138 
miles, and has then made of departure 112.6: 
required the latitude come to, the direct course, 
and difference of longitude. 
Geometrically. 1. Draw BD (fig. 19) for 
the meridian of the ship at B ; and parallel to it 
draw FE, at the distance of 112.6, the departure. 
Take 138, the distance, in your compasses, and 
fixing one point of them in B. with the other 
cross the line FE in the point E ; then join B 
and E, and from E let fall upon BD the perpen- 
dicular ED ; so BD will be the proper difference 
of latitude, and the angle B will be the course ; 
to find which by calculation, 
First, For the course it will be, (by rectaiigular 
| trigonometry,) 
BE : R :: DE ; S. B. 
I i. e. As the distance - 138 2.13988 
| is to radius - - - 10.00000 j 
I so is the departure - 112.6. 2.05154 I 
[ to the sine of the course 54° 4V 9.91166 
I which, because- the ship sails between south and 
f east, will be south 54° 41' east, or SE 0° 41' 
* easterly. 
Then, for the difference of latitude, it will be, 
j (by rectangular trigonometry,) 
r ; be : : Co. s. b : bd. 
i. e. As radius - 10.00000 
is to the distance - 138 2.13988 
so is the co-sine of the course 54° 41' 9.76200 
i to the difference of latitude 79.8 1.90188 
equal to 1° 19'. Consequently the ship has come 
I to the latitude of 47° 13'. Hence the meridional 
difference of latitude will be 117.7. 
2dly. Produce B to A, till BA is equal ' to 
117.7 ; and through A draw AC parallel to DE, 
meeting BE produced in C ; then AC will be the 
difference of longitude ; to find which by cal- 
culation, it will be, 
BD : DE BA : AC. 
i. e. As the proper diif. of lat. 79.8 1.90180 
is to the departure - 112.6 2.05154 
so is the enlarged diff. of lat. 117.7 2.07078 
to the diff. of longitude - 166.1 2.22014 
equal to 2° 46' 03", the difference of longitude 
easterly. 
Having shewn under the article Maps how 
to construct a Mercator’s chart, we shall now 
proceed to point out its several uses. 
Prob. I. Let it be required to lay down a 
place upon the chart, its latitude, and the differ- 
ence of longitude between it and some known 
place upon the chart being given. 
Example. Let the known place be the Lizard, 
lying on the parallel of 50° 00' north, and the 
place to be laid down St. Katherine’s on the east 
coast of America, differing in longitude from 
the Lizard 42° 36', lying so much to the west- 
ward of it. 
Let L represent the Lizard on the chart, (fig. 
20,) lying on the parallel of 50 J 00' north,- its 
meridian. Set off AE from E upon the equator 
EQ 42 r> 36', towards O, which will reach from 
E to F. Through F draw the meridian EG, and 
this will be the meridian of St. Katharine’s then 
set off from Q to H upon the graduated meri- 
dian QB, 28 degrees ; and through FI draw the 
parallel of latitude FIM, which will meet the 
NAVIGATION. 
former meridian in K, the place upon the chart 
required. 
P;;ob. II. Given two places upon the chart, to 
find their difference of latitude and difference of 
longitude. 
Through the two places draw parallels of la- 
titude ; then the distance between these parallels, 
numbered in degrees and minutes upon the gra- 
duated meridian, will be the difference of lati- 
tude required ; and through the two places 
drawing meridians, the distance between these, 
counted in degrees and minutes on the equator, 
or any graduated parallel, will be the difference 
of longitude required. 
Prob. PI. To find the bearing of one place 
from another, upon the chart. 
Example. Required the bearing of St. Kathe- 
rine’s at K, from the Lizard at L. 
Draw the meridian of the Lizard AE, and 
join K and L with the right line KI. ; then, by 
the line of chords, measuring the angle KLE, 
and with that entering the tables, we shall have 
the thing required. 
This may also be done, by having compasses 
drawn on the chart (suppose at two of its cor- 
ners) ; then lay the edge of a ruler over the two 
places, and let fall a perpendicular, or take the 
nearest distance from the centre of the compass 
next the first place, to the ruler’s edge ; then, 
with this distance in your compasses, slide them 
along by the ruler’s edge, keeping one foot of 
them close to the ruler, and the other as near as 
you can judge perpemfcular to it, which will 
describe the rhumb required. 
Prob. IV. To find the distance between two 
given places upon the chart. 
This problem admits of four cases, according 
to the situation of the two places with respect 
to one another. 
Case I. When the given places lie both upon 
the equator. 
In this case their distance is found by convert- 
ing the degrees of difference of longitude inter- 
cepted between them into minutes. 
Case II. When the two places lie both on the 
same meridian. 
Draw the parallels of those places ; and the 
degrees upon the graduated meridian, inter- 
cepted between those parallels, reduced to mi- 
nutes, give the distance required. 
Case III. When the two places lie on the same 
parallel. 
Example. Required to find the distance be- 
tween the points K and N, both lying on the 
parallel of 28° 00' north. 1’ake from your scale 
the chord of 60 or radius, in your compasses, 
and with that extent on KN as a base make the 
isosceles triangle KPN : then take from the line 
of sines the co-sine of the latitude, or sine of 72°, 
and set that off from P to S and T. Join S and 
T with the right line ST, and that applied to 
the graduated equator will give the degrees and 
minutes upon it equal to the distance ; which, 
converted into minutes, will be the distance re- 
quired. 
The reason of this is evident from the me- 
thod of Parallel Sailing ; for it has been there 
demonstrated, that radius is to the co-sine of 
any parallel, as the length of any arch on the 
equator, to the length of the same arch on that 
parallel. Now, in tins chart KN is the distance 
of the meridians of the two places K and N 
upon the equator; and since, in the triangle 
PNK, ST is the parallel .to KN, therefore PN 
• pq' • • NK ; TS. Consequently TS will be 
the distance of the two places K and N upon 
the parallel of 2S°. , . 
If the parallel the two places lie on is not far 
from the equator, and they not far asunder : 
then their distance may be found thus : Take 
the distance between them in your comp isses 
and apply that to the graduated meridian, so 
that one foot may he as many minutes above as 
the other is below the given parallel ; and the 
degrees and minutes intercepted, reduced to 
minutes, will give the distance. 
Or it may also be found thus : Take the 
length of a degree on the meridian at the given 
parallel, and turn that over on the parallel from 
the one place to the other, as oft as you can ; 
then, as often as that extent is contained between 
the places, so many times 60 miles will be con- 
tained in the distance between them. 
Case IV. When the places differ both in lon- 
gitude and latitude. 
Example. Suppose it was required to find the 
distance between the two places a ar.d e upon the 
chart. By Prob. If. find the difference of latitude 
between them ; and take that in your compasses 
from the graduated equator, which set off on 
the meridian of a, from a to b ; then through b 
draw A parallel to de ; and taking ac in your 
compasses, apply it to the graduated equator, 
and it will shew the degrees and minutes con- 
tained in the distance required, which multiplied 
by 60 will give the miles of distance. 
The reason of this is evident ; for it is plain 
ad is the enlarged difference of latitude, and ab 
the proper ; consequently ae is the enlarged dis- 
tance, and ac the proper. 
Prob. V. To lay down a place upon the chart, 
its latitude and bearing from some known place 
upon the chart being known ; or (which i> the 
same) having the course and difference of lati- 
tude that a ship has made, to lay down the run- 
ning of the ship, and find her place upon the 
chart. 
Example. A ship from the Lizard in the lati- 
tude of 50 D 00' north, sails SSW till she has dif- 
fered her latitude 36° 40' : required her place 
upon the chart. 
Count from the Lizard at L, on the gradu- 
ated meridian downwards (because the course is 
southerly) 36° 40' to g ; through which draw a 
parallel of latitude, which will be the parallel 
the ship is in ; then from I. draw a SSW line 
Lf, cutting the former parallel in f, and this will 
be the ship’s place upon the chart. 
Prob. VI. One latitude, course, and distance 
sailed, given ; to lay down the running of the 
ship, and find her place upon the chart. 
Example. Suppose a ship at a in the latitude 
of 20° 00' north, sails north 37° 20', east 191 
miles: required the ship’s place upon the chart- 
Having drawn the meridian and parallel of 
the place a, set off the rhumb-line ae,. making 
with ab an angle of 37° 20' ; and upon it set oiF 
191 from a to c through c draw the parallel ch ; 
and taking ab in your compasses, apply it to the 
graduated equator, and observe the number of 
degrees it contains; then count the same num- 
ber of degrees on the graduated meridian from. 
C to h, and through h draw the parallel he, which 
will cut ac produced in the point e, the ship’s 
place required. 
Prob. VII. Both latitudes and distance sailed 
given ; to find the ship’s place upon the chart. 
Example. Suppose a ship sails from a, in the 
latitude of 20° 00' north, between north and east 
191 miles, and is then in the latitude of 45° 00' 
north: required the ship’s place upon the chart. 
Draw de the parallel of 45°, and set off upon 
the meridian of a upwards, ab equal to the pro- 
per difference of latitude taken from the equator 
or graduated parallel. Through b draw be pa- 
rallel to de then, with 191 in your compasses, 
fixing one foot of them in a, with the other cross- 
be in c. Join a in c with the right line ac ; which- 
produced will meet de in e, the ship’s place re- 
quired. 
Pkob. VIII. One latitude, course, and differ- 
ence of longitude, given ; to find the ship’s pla.ee- 
upon the chart . 
i IQ 
