NAVIGATION.

II

Excamp. Suppose a ship sails from the latitude of 30? 23' north,N. N. £.32 miles, tig. 4. Required the difference of latitudeand departure, and the latitude tome tor Then (by right-angledtrigonometry,) we have the following analogy, tor. finding the de-parture, viz.

As radius..

to the distance AC.32 1-50515

So is the sine of the course A.22° 30'—.. 9.58284

to the departure BC.12'25 . 1-08799

s ° the ship lias made 12-25. miles of departure easterly, or lias got?° far to the eastward of her meridian. Then for the difference oflatitude or northing, the ship has made, we have (by rectangulartr'gonoinetrv) the following analogy, viz.

A s radius...'.... 10 00000

’ s to th e distance AC.;32 . 1-50515

0 is the to sine of course A.22° 30'. 9'58284

t° the difference of lat. AB.,...29'57 . 1-47077

s° tlie ship h as differed her latitude, or made of northing, 29.57tfiinutes, And since her former latitude was north, and her diff-erence of latitude also north ; therefore,

o the latitude sailed from.30° 25'N.

a «d the difference of latitude. 0 29-57'

a J'd the sum is the latitude come to...30« 3.4.07' N.

this case a>-e calculated the tables of difference of latitude, anddeparture, to every degree, point, and quarter-point, of the com-pass.

Case II. Course and difference of latitude given, to find dis-tance and departure ?

Examp. Suppose a ship in lat. 45° 25' N. sails NE6N easterly,hg* 5, till she come to lat. 46° 55' J4. Required the distance anddeparture'made good upon that course? Since both latitudes arenortherly, and the course also northerly ; therefore,

from (be latitude come to..49° 35

subtract the latitude sailed from... .45 25

and there remains... 1° 30'

the difference of latitude, equal to 90 miles.

And,.by rectangular trigonometry, rad. : diff. of lat. AB : : tang,of the course A : dep. BID, 73.84 miles eastward. Also, rad. :sec. of the course: : diff. of lat. AB : distance AD 116-4.

Case III. Difference of latitude and- distance given, to findcourse and departure ?

Examp. Suppose a ship sails from the latitude of 56° 50' N.ona-rhumb between south and west, 126 miles, and she is then foundny observation to he in the latitude of 55° 40' north : Required thecourse she sailed on, and her departure from the meridian, fig. 6.nince the latitudes are both north, and the ship sailing towards theoejuator ; therefore,

. orn the latitude sailed from..36° 30'

subtract the observed latitude.33° 10

1' 40'

and the remainder.;••••••*••...

equal to 70 miles, is the difference of latitude.

By rectangular trigonometry w-e have the distance • •••

diff. of lat. : co-sme of the course I’ = 56° 15' or feVUVV • Andrad. : distance DF::sine of the course: dep. 104-8 miles wes-terly. . , r I

Case IV. Difference of latitude and departure given, to tmcicourse and distance ? , . , ,, a yr ,i „a,-i 1 >

Examp. Suppose a, ship sails from the latitude of 44 j >

between south and east, till she has made 04-miles of easting, .fs then found by observation to be in the latitude o - A *’Required the course and distance made good i ng- 7- U1C ;.

latitudes are botli north, and the ship sailing towards the equatoi ,therefore,

Irom the latitude sailed from.

take the latitude come to.

44° 50' N..42' 56'

1° 54'

and there remains. a ..."Wuivcr In this

equal to 114 miles, the difference of latitude or soul tv q

case, diff. of lat. GK. : rad. :: dep. KL: tang, of ‘ * C ^st29° 19', which, because the ship is sailing between soul I ’

will be south 29° 1 9' east, SSE-£ east nearly. 1 * ie ’ ^ _ etance, we shall have as rad. -. diff. of lat. OK:: sec.

-: distance GL 130-8 ; consequently the ship has sailed on a SSE-Jeast course 130-8 miles.

Case V. Distance and departure given, to find course and dif-ference of latitude ?

Examp. Suppose a ship at sea sails from lat. 34° 24' N. betweenN, and VV. 124 miles, and is found to have made of westing 86miles: required the course steered, and the difference of latitudeior northing made good ? Fig. 8. In (his case, the distanc e AD :rad. : : departure AB : sine of the course D 43° 54' so that theship’s course is north 43° 54' west, or NIV^N? west nearly. Andrad. ; distance AD : : co-sine of the course : diff'. of latitude BD89"35 which is equal to 1 degree and 29 minutes nearly, Henceto find the latitude the ship is in, since both latitudes are north,and the ship sailing from the equator, therefore;

To the latitude sailed from.34° 24'

add the difference of the latitude. 1° 29’

the sum is......35° 53'

the latitude the ship is in north.

r-, t , i *. 1 ii , • j _ r.. -i i ?_j_ _ _ t

Case VI. Course and departure given, to find distance anddifference of latitude ?

Examp. Suppose a ship at sea, in the latitude of 24? 30' south,sails SF5S, till she lias made of easting 96 miles : required the dis-tance and difference of latitude made good on that course. Fig. 9:In this case by rectangular trigonometry and by Case 2, we havethe following proportion for finding the distance. Sine of thecourse G : tiie dep. IlM : : rad. : distance GM 172'8. Then,for the difference of latitude, we have, tangent of course : dep.11M : : rad. : diff'. of lat. Gil. 143 7 equal to 2° 24' nearly. Con-sequently, since the latitude the ship sailed from was south, and-she sailiug still towards the south,

To the latitude sailed from...24° 30'

add the difference of latitude. 2° 24!

and the sum.26° 54*

is the latitude she is come to south. When a ship sails on severalcourses in 24 hours, the reducing of all these into one, and therebyfinding die course and distance made good upon the whole, iscommonly called the resolving of a traverse. At sea they com-monly begin each day’s reckoning from the noon of that day, andfrom that time they set down all the different courses and dis-tances sailed by the ship till noon next day upon the log-board;then from these several courses and distances, they compute thedifference of latitude and departure tor each course (by Case 1 ofPlane Sailing) ; and these, together with the courses and distances,are set down in a table, called the traverse table, which'consists oflive columns : in the first of which are placed the courses and dis-tances ; in the two next, the difference of latitude belonging tothese courses, according.as they are nortli or south ; and in thetwo last are placed the departures belonging to these courses, ac-cording as they are castor west. Then they sum up all thenorthings, and all the southings, and taking the diflerence of these, 1they know the difference of latitude ivade good by the ship in thelast 24 hours, which will be north or south, according as the sumof the northings or southings is greatest: the same way, bv takingthe sum of all the eastings, and likewise of all the westings, andsubtraeting the less of these from the greater, the difference will bethe departure made good by the ship in the last 24 hnurs, -which willbe east or west according as the sum of the eastings is greater orless than thesuin of the westings ; then from the difference of lati-tude and departure made good by the ship in the last 24 hours, foundas above, they find the true course and distance made good uponthe whole (by Case 4-of Plane Sailing), as-also, the course and,distance to the intended port.

Examp. Suppose a ship at sea, in the" latitude of 48° 24'north--at noon any dfiv, is bound tp a port in the latitude of 43° 40’'north, whose departure from.the ship is 144-miles east’; conse-quently; the direct course and distance of the ship is-SSE. -E east:315 miles; but by reason oftlie shifting of the winds she is obiim-dlto steer the following courses till noon next dav, viz, S Eh l? 56miles, SSE 64 miles, NWiW ,48 miles, S6W 4 w'-est 34 miles, andiSEAS -J.east 74 miles: required the course and distance made:good the last 24 hours, and the bearing and distance of the ship:from the intended port ? The solution of this traverse depends),entirely, on the-1st and-4th Cases of Plane Sailing; and first wet

must