NAVIGATION.

n

Courses. Distances.

SSW.

Diff. of Lat.jDiff. of Lon.

N.

S.

E.

W.

40-6534-4646-56| 27-46

25.6

16.16

49.27

s i> w \ w .36

SW b S... 56

S b E.

8-59

Diff. of Lat. 149-13

8-59

88.03

8.59

Diff. of Lon

. 79.44

tude resulting from the several departures on different parallels ;and therefore we have chosen, in the last example of a traverse, tofind the difference of longitude answering to each particular courseand distance, the sum of which nr list be the true difference of lon-gitude, made good by the ship on these several courses and dis-tances. We shewed above howto construct a Mercator ’s chart;

- contained in the follow-

Hence it is plain that the ship has made of' «> ° anl j so

suites, and consequently Has come to tat. 47 . ,. ' . w ;n be

meridional difference of lat. between that and nei ■ rep 44' W.£26-1: and since she has made of difference o lizard

therefore, for die direct course and distance be w For the di-and the ship, it will be, (by Case 2 of tins section). 1rect course, 2-35430

As the mevid. diff. of lat.. ..50-00000

is to radius..... l-QOOO't

so is the diff. of Ion. 79’44...... 9-54593

to the tana, oftiie course. 19° 22'....-..VlVrpnce of

which, because the ditference of lat. is S. am 1 r r

Ion. W. will be S. 19° 22' W. or SfcW 8° f **. H«i>, for

direct distance, ’ 10-00000

As .•••••■■■•.;;; 2-17349

is to the proper dirt, of lat.... uj -02530

so is the secant of the course.19 2/ .2-19879

to the direct distance..^470 31' N.

From the latitude the ship is in...'32 20 N.

subtract the lat. of the Madeira...

now vve shall proceed to its several uses,ins; problems:

Prob. f. Let it be required (o lay down a place upon the chart,its latitude, and the difference of longitude between it and someknown place upon the chart being given.

Ex-amp. Let the known place be the Lizard K ing on the parallelof 50° 0' N. and the place to be laid down St. Katharine’s on theAmerica , differing in longitude, from the Lizard

15 11

equal to 911 minutes, the proper difference of lat. between theship and the Madeira.

Again, from the rnerid. parts answering to the lat. the )ship is in S

Take the meridional parts answering to the latitude of)the Madeira 5

3248-4

2052-0

_ And there remains 1196-4

ddra* 1 a n? e lf difference of latitude between the ship and the Ma-

Also, from the difference of longitude between ) j ^the Lizard and the Madeira )

Take the difference of longitude between the Li-)zard and the ship 5

40

W.

1 19 W.

10 20 W.

equal to 620' 36’ of difference of longitude between the ship andthe Madeira W.- Then for the direct course and distance be-tween the ship and the Madeira, it will be for the direct cou ^5’

As the merid. diff. of lat.1196-4. 3'0/7Jj9

is to radius... 10 '2?222

so is the diff. of Ion. 620-56... 2dL'-

to the tang, of the course...27° 25'... J-714J0

For the direct distance, „„„„ ,

As radius.!.™°?000

is to the proper diff. of lat.>.,..9l 1... 2 9a9-’-

so is the secant of the course.27° 20'.1® '^ot-to the direct distance.102-7 .. 3.01UO

It is very common in working a day’s reckoning at sea, to hudthe difference of latitude and departure to caeh course and dis-tance ; and adding all the departures together, and all the ditler-ences of latitudes, for the whole departure, and difference 01 lati-tude made good that day, from thence (by Case 8 of tins section)to find the difference of longitude, Ac. made good that day.Now that this method is false, will evidently appear,_ if we consi-der that the same departure, reckoned on the two different paral-lels, will give unequal differences of longitude: and consequently•when several departures are compounded together and reckonedon the same parallel, the difference of longitude resulting fromthat cannot be the same with the sum' of the differences of longi--vot,. iv.— so. 156.

east coast of America , differing in longitml42° 36', lying so much to the westward of it. Let L, elateCXXVI.fig. 1, represent the Lizard on the chart, lying on theparallel of 50° 0' N. on its meridian. Set off AE from E upon the-equator EQ, 42° 36', towards Q, which will reach from E to F.Through F draw the meridian FG, and this will be the meridian ofSt. Katharine’s; then set off from Q to II upon the graduated me-ridian QB 28 degrees; and through H draw' the parallel of latitudeItM, which will meet the former meridian in K, the place uponthe chart required.

Prob. II. Given two places upon the chart, to find their dif-ference of latitude and difference of longitude ? Through the twoplaces draw parallels of latitude ; then the distance between theseparallels, numbered in degrees and minutes upon the graduatedmeridian, will be the difference of latitude required ; and throughthe two places drawing meridians, the distance between these,counted in degrees and minutes on the equator, or any graduatedparallel, will be the difference of longitude required.

Prob. III. To find the bearing of one place from another upon.

the chart ?

Examp. Required the bearing of St. Katharine’s at K, the samefigure, from the Lizard at L ? Draw the meridian of the LizardAE, and join K and L with the right line KE: then by the line ofchords measuring the angle KLE, and with that entering the ta-bles, vve shall have the. thing required. This may also be done byhaving 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 falla perpendicular, or take the nearest distance from the centre of thecompass next the first place to the ruler’s edge ; then with this dis-tance in your compasses, slide them along by the ruler’s edge,keeping one foot of them close to the ruler, and the other as nearas you can judge perpendicular to it, which will describe therhumb required.

Prob. IV. To find the distance between two given placeson the chart ? This problem adinitsof four cases, according to the si-tuation of the two places with respect to one another.

Case I. When the given places lie both upon the equator. In-dus case their distance is found by converting the degrees of dif-ference of longitude intercepted between them into minutes.

Case II. When the two places both lie on the. same meridian.Draw the parallels of those places; and the degrees upon the gra-duated meridian, intercepted between those parallels, reduced to-minutes, give the distance required.

Case III. When the two places lie on the same parallel.

Examp. Required to find the distance between the points K andN (same figure), both lying on the parallel of 28° 0' N. Takefrom your scale the chord of 60° or radius in your compasses, andwith that extent on KN as a base make the isosceles triangleKPN : then take from the line of sines the co-sine of the latitude,or sine of 62°, and set that off from P to S and T. Join S and Twith the right liufe S P, and that applied to the graduated equatorwill give the degrees and minutes upon it equal to the distance re-quired. The reason of this is evident from the section of ParallelSailing: where it is demonstrated, that radius is to the co-sine ofany parallel, as the length of any arch on the equator to- thelength of the same, arch on that parallel. Now in this- chartKN is the distance of the meridians of the two places K andN upon the equator; and since, in the triangle PNK, STis the parallel to KN, therefore PN : PT : : NK : TS. Conse-

.1 __.:n 1. _

quently TS will he the distance of the two places K and N uponthe parallel ot 28°. If the parallel the two places lie on he not bu-ffo m the equator, and they not far asunder-; then their distanceG may