)9
rse
9
to
9
ilu; point A, ami set the several courses of the ship in older on tcircumference as at 1, 2, 3, 4, 5, on the hret course seUhetanee AP, 56miles; trom P draw PQ parallel to A > lj e j
to the second distance 64 miles ; in like manner diaw Ito A3, 49 miles, and so of the other courses ; and t A vyirwill determine the point D, for the place of the ship. D-" ' ,;perpendicular, and DP paraual to AB, and i„ nartu ™
then Alt is the difference of the latitude, and ED lmade good; also the angle EDC is the direct course, ai ■
distance to the intended port. In like manner uom , ^ •
day’s traverse may be constructed. f l he angle loimei > , ,
rktian and rhumb that a ship sails upon, is exhibitedlowing table for every point and \ point of the compass.
North.
South.
Points.
D. M.
North.
South.
I
4
t
2.49
5.37
N. by E.
2
8.26
S. by E.
1
11.15
N. by W.
S. by W.
1
t
14. 4
1
i
■f,
16.52
N. N. E.
1
2
19.41
S. S.E.
2
22.30
N. N. W.
S. S. W.
2
i
4
25.19
2
i
4
28. 7
N.EbyN.
2
i
30 56
S.E.byS.
3
33.45
N.W.byN.
S. W. by S.
3
J.
36.34
3
i
39.22
N. E.
3
i
42.11
S. E.
4
45. 0
N. W.
s. w.
4
X
47.49
4
I
76
50.37
4
i
53.20
S. E. by E.
5
56.15
N.W.byW.
S.W.byW.
5
i_
59. 4
5
61.52
E. N. E.
5
2
64.42
E. S. E.
6
67.30
VV. N. W.
w. s. w.
6
t
4
70.19
6
i
4
73. 7
E. by N.
6
2
75.56
E. by S.
7
78.45
W. by N.
W. by S.
7
l
4
81.34
7
i
4
84.22
East.
7
2
87.11
8
90. 0
West.
:
Or Parallel Sailing.
Since the parallels of latitude do always decrease the nearer theyapproach the pole, it is plain a degree on any of them must beless lhan a degree upon the eqnator. Now in order to know thelength of a degree on any of them, let PB, fig. 11, represent half theearth’s axis, PA a quadrant of a meridian, and consequently A apoint on the equator, C a point on the meridian, and CD a per-pendicular from that point upon the axis, which plainly will be thesine of CP the distance of that point from the pole, or the co-sineof CA its distance from the equator; and CD will be to AB, asthe sine of CP, or co-sine of CA, is to the radius. Again, if thequadrant P AB is turned round upon the axis PB, it is plain thepoint A will describe the circumference of the equator whose ra-dius is AB, and any other point C upon the meridian will describethe circumference of a parallel whose radius is CD.
Cor. 1. Hence (because the circumferences of circles are as theirVOL. iv.— no. 156.
radii) it follows, that the circumference of any parallel is to thecircumference of the equator, as the co-sine of its latitude is toradius.
Cor. 2. And since the whole are as their similar parts, it will be,As the length of a degree on any parallel is to the iengtli of a de-gree upon the equator, so is the co-sine of the latitude ot that pa-rallel to radius.
Cor. 3. Ifence, as radius is to the co-sine of any latitude, soare the minutes of diffircnce'of longitude between two meridians,or their distance in miles upon the equator, to the distance of thesetwo meridians on the parallel in miles.
Cor. 4. And as the co-sine of any parallel is to radius so is thelength of any arch on that parallel (intercepted between two me-ridians) 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 any one parallel is to the co-sineof any other parallel, so is the length of any arch on the first, inmiles, to the length of the same arch on the other in miles. Fromwhat has been said, arises the solution of the several cases of pa-rallel sailing, which are as follow :
Case I. Given the difference of longitude between two places,both lying on the same parallel; to find the distance between thoseplaces ?
Examp. 1. Suppose a ship in the latitude of 54° 20' N. sails di-rectly W. on that parallel till she has differed her longitude 12*45'; required the distance sailed on that parallel ? Tire differenceof longitude reduced into minutes, or nautical miles, is 765', whichis the distance between the meridian sailed from, and the meridiancome to, upon the equator ; then to find the distance betweenthese meridians on the parallel of 54° 20' or the distance sailed, itwill be, by Cor. 3,
As radius...10-00000
is to the co-sine of the lat.54° 20'. 9 - 76572
so are the minutes of diff. Ion....765 . 2-88366
to the distance on the parallel.446- L ..... 2-64938
Examp. 2. A degree on the equator being 60 minutes (or nau-tical miles) required the length of a degree on the parallel of 51®32' —By Cor. 3. of the last article, it will be
As radius.........10.00000
is to the co sine of the lat.51° 32'. 9-79583
so are the minutes in 1 degree on
to .37°.32’. 1-57198
the miles answering (o a degree on the parallel of 51° 32'.—Bythis problem the table, under the article Geography , is construct-ed, for shewing the geographic miles answering to a degree oflongitude on the different parallels.
Case N. The distance sailed in any parallel of latitude, or thedistance between any two places on that parallel, being given ; tofind the difference of longitude?
F.xamp. Suppose a ship in the lat. of 55° 36' N. sails directlyE. 685 6 miles: Required how much she has differed her longi-tude? By Cor. 4, it will he
As the co-sine of the lat.55° 36'. 9-75202
is to radius....,.10 00000
so is the distance sailed.683-6 . 2.83607
to min. of diff of Ion.1213. 3-08405
which reduced into degrees, by dividing by 60, makes 20° 13',the difference of longitude the ship has made. This also may besolved by help of the forementioned table, viz. by finding from itthe miles answering to adegree on the proposed parallel, and divid-ing with this the given number of miles, the quotient will he thedegrees and minutes of difference of longitude required. Thus inthe last example, we find, from the said table, that a degreeon the parallel of 55° 36' is equal to 33-89 miles; by this we di-vide the proposed number of miles 685 6, and the quotient is20-13 degrees, i. c. 20° 13', the difference of longitude required.
Case ill. The difference of longitude between two places onthe same parallel, and the distance between them, being given ; tofind the latitude of that parallel ?
Examp. Suppose a ship sails on a certain parallel directly west624 miles, and then has differed her longitude 18° 46', or 1126miles: Required the latitude of the parallel she sailed upon ?—ByC or. 3, it will be,
As the min. of dill'. Ion. 1.1126. 3-04154
E iw