258 
cT^;c'°-’ 0W '. n ^ C0l ' rse3 til! noon next day, viz. 
t 6 ru,Jcs ’ 64 miles, NWiW 48 miles, 
* . e v ,cst 54 miles, and SEiS 4 east 74 miles : 
re.juired t he course and distance made good the 
', lSL "1 Il0urs > an< 4 the hearing and distance of 
tne sh'p from the intended port. 
i h j solution of this traverse depends entirely 
on the 1st and 4th Cases of Plane Sailing: ; ana 
fmst we must (by Case 1.) find the difference of 
latitude and departure for each course. Thus, 
1. Course SE/.S distance 53 miles. 
For departure. 
As radius _ 
is to the distance - 5g 
so is the sine of the course S3 0 , 45' 
to the departure' - 31.11 
For difference of latitude. 
As radius _ 
is to the distance - ,%• 
so is the co-sine of the course 33°, 45' 
to the diif of latitude 46.57 
2. Course SSE and distance G-l miles, 
For departure. 
As radms _ 
is to the distance - 
so is the sine of the course 22°, 30' 
to the departure - 24.5 
For difference of latitude. 
As radius - _ _ 
is to the distance - 64 
so is the co-sine of the course 22°, 30' 
to the difference of latitude 59.13 .... 
3. Course NW7>W and distance 48 miles. 
For departure. 
As radius _ . _ 
is to the distance - 48 
so is the sine of the course 56°, 15 ' 
to the departure - 39.91 
For difference of latitude. 
As radius - 
is to the distance - 48 
so is the co-sine of the course 56°, 15' 
to the difference of latitude 26.67 
4. Course S6W -§ west and distance 54 miles. 
For departure. 
As radius - 
is to the distance - 54 
so is the sine of the course 16°, 52' 
to the departure - 15.67 
For difference of latitude. 
As radius - 
is to the distance - 54 
so is the co-sine of the course 16°, 52' 
to the difference of latitude 51.67 
5. Course SE^S ^ east and distance 74 miles. 
For departure. 
As radius - 
is to the distance - 74 
so is the sine of the course 39°, 22' 
to the departure - 46.94 
For difference of latitude. 
A’sTadius - 
is to the distance - 74 
to is the eo-sine of the course 39°, 22' 
to the difference of latitude 57.21 
Now these several courses and distances, n 
g-ether with the differences of latitude and d 
partures deduced from them, being set down 
the proper columns in the traverse table, wi 
Stand as follow : 
The Traverse Table. 
10 00000 
1.74819 
9.74474 
1.49293 
10.00000 
1.74819 
9.91985 
1.66804 
10.00060 
1.80618 
2.58284 
1.38902 
10 00000 
1.80618 
9.96562 
1.77280 
10.0000C 
1.68124 
9.91985 
1.60109 
10.00000 
1.68124 
9.74474 
1.42598 
10.0000( 
1.73231 
9.46261 
8.19501 
1 0.00CXX 
1.73231 
9.9809C 
1.71321 
lO.OOOQi 
1.8692: 
9.80221 
1.6715 
10.0000! 
1.8092: 
9.8882- 
1.75*4' 
Coicrses. 
Distances. 
Diff. of Lat. 
1 
| Depa 
rtufe. 
N. 
S. 
E. 
W. 
SE/, S 
— 56 

46.57 
31.11 
S SE 
— §4 
— 
59.13 
24.5 

NW b W 
— 48 
26.67 

39.91 
fi/W-fW 
— - 54 
— 
51,67 

15.67 
SEAS 
— • 94 
— 
57.21 
46.94 
— 
26.67 
214. 2 
108.55 
55 58 
1 
26.67 
55.5 8 
Diff. of Lat. '187.91 | 46.97 | Cep. 
NAVIGATION'. 
From the above table it is plain, since the 
sum of the northings is 26.67, and of the south- 
ings 214.58, the difference between these, viz. 
187.91, will be the southing made good by the 
shi p the last 24 hours ; also the sum of the east- 
ings being 102.55, and of the westings 55-58, the 
difference 46.97 will be the easting or departure 
made good by the ship’s last 24 hours ; conse- 
quently, to find the true course and distance 
made good by the ship in that time, it will be 
(by Case 4. of Plane Sailing), 
As the difference of latitude 187.91 2.27393 
is to the radius - - 10.00000 
so is the departure - 46.97 1.67182 
to the tangent of the course 14°, 03' 9.39789 
which is S/>E ^ east nearly. Then for the dis- 
tance, it will be, 
As radius 10.00000 
is to the difference of latitude 187.91 2.27393 
so is the secant of the course 14°, 03' 10.01319 
to the distance - 193.7 2.28712 
consequently the ship has made good the last 
24 hours, on a S/>E -J east course, 193.7 miles: 
and s:nce the ship is sailing towards, the equator; 
therefore, 
From the latitude sailed from - 48°, 24' N 
take the diff. of latitude made good 3 , 08 S 
there remains - - 45 , 16 N 
the latitude the ship is in north. And because 
the port the ship is bound for lies in the latitude 
of 43° 40' N. and consequently south ef the 
ship ; .therefore, 
From the latitude the ship is in 44°, 16' N 
take the latitude she is bound for 43 , 40 N 
and there remains - - - 1 , 36 
or 96 miles, the difference of latitude or south- 
ing the ship has to make. Again, the whole 
easting the ship had to make being 144 miles, 
and she having already made 46.97, or 47 miles 
of easting ; therefore the departure or easting 
she still has to make will be 97 miles : consA 
quently, to find the direct course and distance 
between the ship and the intended port, it will 
be (by Case 4. of Plane Sailing), 
As the difference of latitude 96 1.98227 
is to radius - 10.00000 
so is the departure - 97 1.98677 
to the tangent of the course 45°, 19' 10.00450 
And 
As radius - - 10.00000 
is to the difference of latitude 96 1.98227 
so is the secant of the course 45°, 19' 10.15293 
to the distance - - 136.5 2.13620 
whence the true bearing and distance of the in- 
tended port is, SE 136.5 miles. 
Of Parallel Sailing. Since the parallels of lati- 
tude do always decrease the nearer they ap- 
proach the pole, it is plain ?. degree on any of 
them must be less than a degree upon the equa- 
tor. Now in order to know the length of a de- 
gree on any of them, let PB (fig. 9) represent 
half the earth’s axis, PA a quadrant of a meri- 
dian, and consequently A a point on the equa- 
tor, C a point on the meridian, and CD a per- 
pendicular from that point upon the axis, which 
plainly will be the sine of CP the distance of 
that point from the pole, or the co-sine of CA 
its distance from the equator ; and CD will be 
to AB, as the sine of CP, or co-sine of CA, is to 
the radius. Again, if the quadrant PAB is 
turned round upon the axis PB, it is plain the 
point A will describe the circumference of the 
equator whose radius is AB,and any other point 
C Upon the meridian will describe the circum- 
ference of a parallel whose radius is CD. 
Cor. 1. Hence (because the circumference of 
circles are as their radii) it follows, that the cir- 
cumference of any parallel is to the circumfer- 
ence of the equator, as the co-sine of its latitude 
is to radius. 
Cor. 2. And since the wholes a e as teir 
similar parts, it will be. As the leng h of a de- 
I gree on any parallel, is to the length of a degrer 
I upon the equator, so is the co-sine of the latb» 
tude of that parallel, to radius. 
Cor. 3. Hence, As radius, is to the co-sine ef 
any latitude, so are the minutes of difference of 
longitude between two meridians, or their dis- 
tance in miles upon the equator, to tire distance 
of these two meridians on the parallel in miles. 
Cor. 4. And, As the co-sine of any parallel, is 
to radius, so is the length of any arch on that 
parallel (Intercepted between two meridians) in 
miles, to the length of a similar arch on the 
equator, or minutes of difference of longitude. 
Cor. 5. Also, As the co-sine of anyone pa- 
rallel, is to the co-sine of any other parallel, so 
is the length of any arch on the first in miles, 
to the length of the same arch on the other in 
miles. 
From what has been said, arises the solution 
of the several cases of parallel sailing, which are 
as follow : 
Case I. Given the difference of longitude be- 
tween two places, both lying on the same pa- 
rallel ; to find the distance between those places. 
Example 1. Suppose a ship in the latitude of 
54° 20' north, sails directly west on that parallel 
till she lias differed her longitude 12° 45'; re- 
quired the distance sailed on that parallel. 
First, The difference of longitude reduced into 
minutes, or nautical miles, is 765', which is the 
distance between the meridian sailed from, and. 
the meridian come to, upon the equator ; then 
to find the distance between these meridians on 
the parallel of 54? 20/, or the distance sailed, 
it will be, by Cor. 3. of the last article, 
As radius - - - 10.00000- 
is to the co-sine of the lat. 5 4° 20' 9.76572 
so are the minutes of diff. Ion. 7 65 2.18366 
to the distance on the parallel 446.1 2.64938; 
Example 2. A degree en the equator being GO 
minutes or nautical miles ; required the length 
of a degree on the parallel of 51° 32'. 
By Cor. S. of the last article, it will be 
As radius ... 10.00000 
is to the co-sine of the latitude51°, 32' 9.79383 
so are the min. in l°on the equa. 60 1.77815 
to - - 37.32 1.57198 
the miles answering to a degree on the parallel 
of 51° 32'. 
By this problem a table is constructed, shew- 
ing the geographic miles answering to a degree 
on any parallel of latitude; in which you may 
observe, that the columns marked at the top 
with D. L. contain the degrees of latitude be- 
longing to each parallel : and the adjacent co- 
lumns marked at the top Miles, contain the 
geographic miles answering to a degree upon, 
these parallels. See the table in the article Map- 
Though the table does only shew the miles 
answering to a degree of any parallel, whose la- 
titude consists of a whole number of degrees; 
yet it may be made to serve for any parallel 
whose latitude is some number of degrees and 
minutes, by making the following proportion, 
viz. 
As 1 degree, or 60 minutes, is to the difference 
between the miles answering to a degree in the 
next greater and next less tabular latitude than 
that proposed ; so is the excess of the proposed 
latitude above the next tabular latitude, to a 
proportional part ; which, subtracted from the 
miles answering to a degree of longitude in the 
next less tabular latitude, will give the miles 
answering to a degree in the proposed latitude. 
Example. Required to find the miles answering 
to a degree on the parallel of 56° 44'. 
First, The next less parallel of latitude in the 
table than that proposed, is that of 56°, a de- 
gree of which (by the table) is equal to 35.55 
miles ; and the next greater parallel of latitude 
in the table, than that proposed, is that of 5.7°, 
a degree of which is (by the table) equal to 32.68 
miles ; the difference of these is 87, and the dis- 
tance between these parallels is 1 degree, or CO 
