15
Case V. Course and departure given, to hud the difference oflatitude, difference of longitude, and distance sailed ?
^ Examp. Suppose a ship in the latitude of 48 5 23' N. sailsSW6S, till she has made of westing 123 miles: Required the lati-tude come to, the difference of longitude, and the distance sailed;hirst, f or the distance, it will be, (by Case 6 of Plane Sailing,)As the sine of the course : depaiture : : radius: the distance 221-4.And for the difference of \at. it will be, by the same Case, As thetang, of course : departure: : radius: cliff, of lat. 1S4 equal to 34' : And since the ship is sailing towards the equator, the lat. cometo will be 45° 19' N. and consequently the middle parallel will be46° 51'. Then to find the difference of longitude, it will be, (byCase 2 of Parallel Sailing,) As the co-sine of mid. par. : the de-parture :; radius : min. of diff. of longit. 180, which is equal to3° O', the difier-nce of longitude westerly.
Case Vi. Difference of latitude and departure given, to findcourse, distance, and difference of longitude?
Examp. Suppose a ship in lat. 46° 37' N. sails between S. and E.till she has made of easting 146 miles, and is then found by obser-vation to be in lat. 43° 24' N. Required the course, distance, anddifference of longitude ?—-First, By Case 4 of Plane Sailing, itwill be for the course, As the diff. of lat. : departure : : radius :tang, of the course 36° 55' which, because the ship is sailing be-tween S. and E. will he S. 36° 55' E. or SEiS -J east nearly. Forthe distance, it will be, by the same Case, As radius : diff. of lat. : :secant of the course : the distance 241-4. Then for the differenceof Ion . it will be, by Case 2 of Parallel Sailing, As the co-sine ofthe mid. par. : the departure: : radius : min. of diff. of Ion. 205,equal to 3° 25'the difference of Ion . E.
Case \T1. Distance and departure given, to find difference oflatitude, course, and difference of longitude?
Examp. Supposeaship in the latitude of33°40'N. sails betweenS-and E. 165 miles, and has then made of easting 112-5 miles;Required the difference of latitude, course, and difference of lon-gitude? First, for the course, it will be, by Case 5 of Plane Sail-ing, As the distance: radius : : trie departure : sine of the course42° 59', which, because the ship sails between S. and E. will be S.42° 59' E. or SE6E | east nearly. And for the difference of lati-tude, it will he, bv the same case, As radius : distance : : co-sine ofthe course: the difference of lat. 120‘7 equal to 2° 0'; consequentlythe latitude come to will be 31° 40' N. and the latitude of themiddle panTh ', will be 32° 40'. Hence, to find the difference ollongitude, it wi. ; be, by Case 2 of Parallel Sailing, As the co-sine of the mid. par. : departure : : radius : min. of diff. of long,133-6, equal to 2° 15' nearly, the difference of Ion . E.
Case VIIf. Difference of longitude and departure given; tcfind difference of latitude, course, and distance sailed? __
Examp. Suppose a ship in lat. 50° 46' N. sails between S antW. (ill her difference of Ion. is 3° 12', and is then found to havedeparted from her former meridian 126 miles: Required the dif-ference of latitude, course, and distance sailed : — first, for t hilatitude she has come to, it will be, by Case 3 of Parallel SailingAs min. of diff. of long. : depaiture : : radius: co-sine of mid. par4S° 59'. Now since the middle latitude is equal to half the sunof the two latitudes (by art. 1 of tliissect.) am! so the sum of the twilatitudes equal to double the middle latitude: it follows, that ifrom double the middle latitude we subtract any one of the latitildes, the remainder will be the other. Hence from twice 4859', viz. 97» 53', taking 50“ 46' the latitude sailed from, there remains 47“ 12' the latitude come to ; consequently the difference olatitude is 3° 34', or 214 minutes. Then for the course, it will beby Case 4 of Plane Sailing, As diff. of lat. : radius : : departuretang, of the course 30° 29', which because it is between S. and Wwill be S. 30° 29' W. or SSVV $ west nearly. And for the distanceit will be, by the same Case, As radius : diff'. of lat. : : secant othe course : the distance 248-4.
II From what has been said, it will be easy to solve a traverseby the rules of Middle-latitude Sailing.
Examp. Suppose a ship in lat. 43° 25' N. sails upon the following courses, viz. SW6S 63 miles, SSW west 45 miles, S&E 5miles, andSW&W 74 miles;: Required the latitude the ship hacome to, and how far she has differed her longitude ?— First, By Cas2 of this Sect, find the difference of latitude and difference of lorgitude belonging to each course and distance, and they will stan-as in the following table : i
SWiS.63
SSW'-W.45
,S6E.5
SW6W.74,
52- 439-7
53- 04T1
Diff. of Lat. 186-2
47-85
28-62
14-75 !
-I 81-08
157*55
13.75
Diff. of Lon.
143-80
Hence it is plain the ship has differed her latitude 186° 2'or3° 6', and so has come to the lat. of 40° 19' N. and has made ofdifference of Ion. 143' 8" or 2° 23' 48" W.
III. This method of sailing, though it be not strictly true, yetit comes very near the truth, as will be evident, by comparingau ex-ample wrought by this method and that delivered in the next section,which is strictly true ; and it serves, without any considerable error,in runnings of 450 miles between the equator and parallel of 30 de-grees ; of 300 miles, between that and the parallel of 60 degrees, andof 150 miles as far as there is any occasion, and consequently mustbe sufficiently exact lor 24 hours run. The construction of thecases of middle-latitude sailing is performed by means of plane andparallel sailing, as will be manifest from a consideration of fig. 14and 13 ; in which AC is the depaiture, AD the distance, DC thedifference of latitude, and AE the difference of longitude ; also theangle ADC is the course, and CAE the middle-latitude. Thetriangle ACD is constructed as in plane sailing, and ACE as inparallel sailing.
Though the meridians all meet at the pole, and the parallels tothe equator continually decrease, in proportion to the co-sines oftheir latitudes; yet in old sea-charts the meridians were drawnparallel to one another, and consequently the parallels of latitudemade equal to the equator, and so a degree of longitude on anyparallel as large as a degree on the equator : also in these chartsthe degrees of latitude were still represented (as they are in them-selves) equal to each other, and to those of the equator. Bythese means the degrees of longitude being increased beyond theirjust proportion, and the more so the nearer they approach the pole,the degrees of latitude at the same time remaining the same, it isevident places must be very erroneously marked down upon thesecharts with respect to their relative situations, and conse-quently their hearings from one another very false. To remedythis inconvenience, so as still to keep the meridians parallel, it isplain we must protract, or lengthen, the degrees of latitude in thesame proportion as those of longitude are, that so the pioportion ineasting and westing may be the same with that ot southing andnorthing, and consequently the hearings of places from one anotherbe the same upon the chart as upon the globe itself. Let ABDfig. 16, be a quadrant, of a meridian, A the pole, D a point on theequal or, AC half the axis, B any point upon the meridian, fromwhich draw BF perpendicular to AC , and BG perpendicular toCD; then BG will be the sine, and BF or CG the co-sine, of BDthe latitude of the point B; draw DE the tangent and CEthe se-cant of the arch DC, Then, as shewn in parallel sailing, any arch,as a minute on the parallel described by the point B, will lie to aminute on the equator as BF or CG is to CD ; but since the tri-angles CGB, CDE, are similar, therefore CG will be to CD as CBis to CE, i. e. the co-sine of any parallel is to radius as radius is tothe secant of the latitude of that parallel. But it has been justshewn, that the co-sine of any parallel is to radius as the lengthof any arch, as a minute on that parallel, is to the length ofthe like arch on the equator; therefore the length of any arch rsa minute on any parallel, is to the length of the like arch on theequator, as radius is to the secant of the latitude of that parallel:and so the length of any arch, as a minute on the equator, is longerthan the like arch of any parallel in the same proportion as the se-cant of the latitude of that parallel is to radius. But since in thisprojection the meridians are parallel, and consequently each pa-rallel of latitude equal to the equator, it is plain ihelength of anyarch, as a minute or a degree on any parallel, is increased beyond
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