\ 
262 
{• (' As radius - . 10.03000 
is to the distance - 156 2.19812 
so is the co-sine of the course 35° 40' 9.90978 
to the proper difF. of lat. 127 2.10290 
equal to 2° 07'; and since the ship is sailing 
from a north latitude towards the south, there- 
fore the latitude come to will be 47° 53' north. 
Hence the meridional difference of latitude will 
be 198.4, 
2. Produce BK to D, till BD is equal to 
193.4 ; through D draw DL parallel to MK, 
meeting DM produced in L ; then DL will be 
the difference of longitude : to find which by 
calculation, it will be, (by rectangular trigono- 
metry,) 
R : BD ;; T. LBD : DL, 
i. e, As radius . - 10.00003 
is to the meridional diff. of lat. 193.4 2.28646' 
so is the tang, of the course 35°40 / 9.85594 
to minutes of diff. of long. 188.8 2,14240 
equal to 2° 18' 48", the difference of longitude 
the ship has made westerly. 
• C’ a se IV. Given course and both latitudes, 
viz. the latitude sailed from, and the latitude 
come to; to find the distance sailed, and the dif- 
ference of longitude. 
Example. Suppose a ship in the latitude of 
5' - 20 north, sails south 33° 45' east, until by 
observation she is found to be in the latitude of 
■5’E 4;> north: required the distance sailed, apd 
tire difference of longitude. 
Geometrically, Draw AB (fig. 14), to re- 
present the meridian of the ship in the first la- 
titude j and set off from A to B 155 the minutes 
of the proper difference of latitude, also AG 
equal to 257.9 the minutes of the enlarged dif- 
ference of latitude. Through B and G, draw 
the lines BC and GK perpendicular to AG; also 
draw AK, making with AG an angle of 83° 45', 
which will meet the two former lines in the 
points C and K ; so the case is constructed, and 
AC and GK may be found from the line of 
equal parts ; to find which, 
Ey Calculation ; 
First, For the difference of longitude, it will 
be, (by rectangular trigonometry,) 
r : ag t. gak • gx, 
i. *. As radius - - 10.00000 
is to the enlarged diff. of lat, 257.2 2,41 145 
so is the tang, of the course 38° 45 / 9.84289 
to min. of diff. of longitude 172.3 2.23634 
equal to 2° 52' IS", the difference of longitude 
the ship has made easterly. 
This might also have been found, by first 
finding the departure BC (by Case 2. of Plane 
Sailing), and then it would be AB ) BC ; ; AG 
* GK, the difference of longitude required. 
Then, for the direct distance AC, it will be, 
(by rectangular trigonometry,) 
r : ab" : : sec. a : ac, 
i. e. As radius - 10.00000 
is to the proper diff. of lat. 155 2,19038 
ao is the secant of the course 33° 45' 10.08015 
to the direct distance 186.4 2.27048 
consequently the ship has sailed south 83° 45' 
east 186.4 miles, and has differed her lon°-itude 
2° 52' 13" easterly. 
Case V. Both latitudes and distances sailed, 
given ; to find the direct course, and difference 
of longitude. 
Example. Suppose a ship from the latitude of 
45° 26' north, sails between north and east 195 
miles, and then by observation she is found to 
bo in the latitude of 48° 6' north : required the 
direct course, and difference of longitude. 
Geometrically. Draw AB (fig, 15), equal 
to 160, the proper difference of latitude, and 
from the point B raise the perpendicular BD; 
then take 195 in your compasses, and setting 
one foot of them in A, with the other cross the 
Hi no BD in D. Produce AB, till AC is equal 
lo 23S.6 the enlarged difference of latitude. 
NAVIGATION’. 
Through C d raw CK parallel to BD, meeting 
AD produced in IC: so the case is constructed ; 
and the angle A may be measured by the line 
of chords, and C1C by the line of equal parts : 
to find which, 
By Calculation ? 
Tii st. For the angle of the course BAD, it will 
be, (by rectangular trigonometry,) 
ab ; r :: ad ; s cc . a. 
{• e. As the proper diff. of lat. 160 2.20412 
is to radius - - 10.00000 
so is the distance - 195 2.29003 
to the secant of the course 34° 52' 10.08591 
which, because the ship is sailing between north 
and east, will be north 34° 52' east, or NEiN 
1° 7' easterly. 
Then, for the difference of longitude, it will 
he, (by rectangular trigonometry.) 
r : ac ; ; t. a : ck, 
'• As radius _ - - 10.00000 
is to the merid. diff. of lat. 233.6 2.36847 
so sthe tang, of the course 34° 52' 9,34307 
to min. of diff. of longitude 162.8 2.21 154 
equal to *2° 42' 48", the difference of longitude 
easterly. 
Case VI. One latitude, course, and difference 
of longitude, given ; to find the other latitude 
and distance sailed. 
Example. Suppose a ship from the latitude of 
48° 50' north, sails south 34° 40' west, till her 
difference of longitude is 2° 42' : required the 
latitude come to, and the distance sailed. 
Geometrically. 1 . Draw AE (fig. 16), to 
represent the meridian of the ship in the first 
latitude, and make the angle EAC equal to 
34° 40', the angle of the course ; then draw EC 
parallel to AE, at the distance of 164 the mi- 
nutes of difference of longitude, which will 
meet AC in the point C. From C let fall upon 
AE the perpendicular CE ; then AE will bethe 
enlarged difference of latitude. To find which, 
by calculation, it will be, (by rectangular tri- 
gonometry,) 
'T. A : R ;; CE ) AE, 
{■ As 'the tang, of the course 34° 40' 9.83984 
is to the radius _ - - 10.00000 
so is min. of diff longitude 164 2.21484 
to the enlarged diff of latitude 2.37.2 2.37500 
arid because the ship is sailing from a north la- 
titude southerly, therefore 
From the merid. parts of 1 
the latitude sailed from 5 4 3866.9 
take the merid, difference of latitude 237.2 
and there remains ... 3f29.7 
the meridional parts of the latitude come to, viz 
46° 09'. 
Hence, for the proper difference of latitude, 
From the latitude sailed from - 48© 50' N 
take the latitude come to - - 46 09 N 
and there remains - - 2 41 
equal to 161, the minutes of difference of lati- 
tude. 
2. Set off upon AE the length AD equal to 
161 the proper difference of latitude, and 
through D draw DB parallel to CE ; then AB 
will be the direct distance. To find which, by 
calculation, it will be, (by rectangular trigono- 
metry,) 
R : AD : : Sec. A : AB. 
'• <?• As radius - - * 10.00000 
t3 to the proper diff of lat. 161 2.20683 
so is the secant of the course 34° 40' 10.08488 
to the direct distance - 195.8 2.29171 
i Case VII One latitude, course, and departure, 
given ; to find the other latitude, distance sailed, 
and difference of longitude. 
Example. Suppose a ship sails from the latitude 
of 54° 86' north, south 42 ' 33' east, until she has 
made of departure 116 miles: required the la- 
titude she is in, her direct distance sailed, and 
how mueh she ha« altered her longitude. 
9.33010 
2.06446 
10.00000 
2.23486 
, GroMETMCAMY. 1. Havingdrawn the merG 
dian AB (fig. 17), make the Tingle BAD equal 
to 42 33'. Draw FD parallel to AB at the dis- 
tance of 116, which will meet AD in D. Let 
fall upon AB the perpendicular DB. Then AB 
will be the proper difference of latitude, and 
AD the direct distance : to find which by cal- 
culation, first, for the distance AD it will be, 
(by rectangular trigonometry,) 
s. a ; bd : : r : ad. 
I. c. As the sine of the course 42 3 33' 
is to the departure - 116 
so is radius 
to the direct distance - 171.5 
Then, for the proper difference of latitude, it 
will be, (by rectangular trigonometry,) 
T. A : ed ;; ‘r ; ab/ 
[■ <?• As the tang, of the course 4*2° 33' 9.9628 1 
is to the departure - 116 2.06446 
so is radius - . 10.00000 
to the proper diff of latitude 126.4 2.10165 
equal to 2" O': consequently the ship has come 
to the latitude of 5 2° SO' north ; and so the me- 
ridional difference of latitude will be 212.2. 
2. Produce AB to E, till AE be equal to 
212.2; and through E draw EC parallel to BD, 
meeting AD produced in C ; then EC will be 
the difference of longitude; to find which by 
calculation, it will be, ^hy rectangular trigono- 
metry,) 
R : AE T, A ; EC. 
i. e. As radius 
is to the merid. diff. of lat. 212.2 
so is the tang, of the course 42° 33' 
to the min. of diff. of long, 194,8 
equal to 3 J 14' 48", the difference of longitude 
easterly. 0 
This might have been found otherwise, thus: 
Because the triangles ACE, ADB, are similar; 
therefore it will be, 
AB ; BD AE * EC. 
h c. As the proper diff, of lat, 1 26.4 
is to the departure - 116 
so is the enlarged diff of lat. 212.2 
to min. of diff'. of longitude 194.8 
Case Vllf. Both latitudes and 
given ; to find course, distance, and difference 
of longitude. 
Example. Suppose a ship from the latitude of 
46° 20' N. sails between south and -west, till she 
has made of departure 126.4 miles; and is then 
found by observation to be in the latitude of 
"4 3o north : required the course and distance 
sailed, and difference of longitude. 
Geometrically. Draw AK (fig. IS), to re- 
present the meridian ox the ship m her first lati- 
tude ; set off - upon it AC, equal to 165, the pro- 
per difference of latitude. Draw BC perpendi- 
culai to AC, equal to 126.4 the departure, and 
join AB. Set off from A, AK equal to 233.3, 
the enlarged difference of latitude ; and thrbugh 
K draw KD parallel to BC, meeting AB pro- 
duced in D ; so the case is constructed, and DiC 
will be the difference of longitude, AB the dis- 
tance, and the angle A the course ; to find 
which, 
By Calculation ; 
First, For DC the difference of longitude, it 
will be, 
ac : cb :: ak ; kd. 
/. e. As the proper diff. of lat. 165 
is to the departure - 126.4 
so is the enlarged diff of lat. 233.3 
to min. of diff. of longitude 178,7 
equal to 2° 58' 42", the difference of longitude 
westerly. 
Then, for the course it will be, (by rectan- 
gular trigonometry,) 
ac : bc :: r : t. a. 
e. As the proper diff. of lat. 165 2.21748 
is to the departure - 446,1 2,10175 
so is radius - - 1Q.0Q00C 
to the tang, of the course 87° 27' 9.88 127 
10.00000 
2.32675 
9.96281 
2.2S956 
2.10165 
2.06446 
2.32675 
2.28956 
departure 
2.21748 
2.10175 
2.36791 
2.25218 
