the Sphere, and therefore truly Circular in every Part; in them the Meridians being all straight Lines, are also parallel to each other; but here being Great Circles they incline to each other as they approach the Poles. The Solutions given by Mercator's Chart, in the Cafe of Departure or Meridian Distance, are Ambiguous, or falfe rather; the fame Analogy being ufed for the Departure, either North or South of the Place departed from; but by this Method the Departure being determined by the Difference of Longitude, and the Latitude come to, is most certain and infallible. In Great Circle Sailing, we ufe neither Rumbs nor Parallels; the first being a kind of Spiral, or Heli-fpherical Lines, and not Circles; and the latter only fmall Circles, and not great ones, cannot be thus ufed. Wherefore Arches of Meridians, of the Equinoctial, and other Great Circles, are the only Means here used for Solutions of this Kind of Navigation. I fhall now exemplify the moft eminent Cafes of this fort of Sailing in the following Series of Problems. Note, In the Chapter of Geograghy, you will find Tables for converting Degrees of the Equinoctial, and Meridian, as alfo of any Parallel, into Miles and Leagues; and the Contrary. The PROBLEM I. Suppofe two Places lie under the Equinoctial; or be in the fame Meridian. Then in the firft Cafe is their Pofition from each Eaft and Weft; and in the other Cafe, North and South; and in each Cafe, the Distance between them is the Difference of Longitude or Latitude, turned into Miles or Leagues, and herein does Circular Sailing agree with the Rectilineal. PROBLEM II. Suppofe one Place under the Equinoctial, and another in any given Latitude, with the Difference of Longitude given; to find the Distance between them, the mutual Pofition of one from the other, and the Departure. Practice. Suppose the Place A in Latitude 51° 32', and D the Place under the Equinoctial, and their Difference of Longitude DC = 40° 00'. 10.0000000 1. To find the Distance in a Great Circle AD, fay, As Radius Is to the Co-fine of Lat. 51° 32' AC 9.7938317 So is the Co-fine of Long. 40° 00' To the Co-fine of the Dift. 61° 33' But 61° 33' 3693 Miles = 3693 Miles mon Measure. CD 9.8842540 AD 9.6780857 1231 Leagues, Com 2. To find the Pofition of D to A, or Angle at A. 10.0000000 As the Sine of Latit. C A 510 321 9.8937452 Is to Radius So is the Tang. of Long. CD = 40° co' 9.9238135 To the Tang. of the Ang. at A=46° 58' 10.0300683 That is S. W. and a little more by W. 3. To find the Pofition of A to D, or the Angle at D. 10.0000000 As Sine of Diff. of Long. CD= 40 co 9.8080675 Is to Radius So is the Tangent of Lat. AC=51 32 10.0999135 To the Tang. of the Ang. at D=62 56 10.2918460 So A bears from D, N. E. by N. and fomewhat more. 4. To find the Departure, fuppofing the Ship failed from D to A; ufe this Analogy. As Radius 10.0000000 9.7938317 Is to Co-fine of the Latitude 51° 32′ So is the Sine of Diff. of Long. 20° col 9.5340517 To the Sine of half the Departure 12° 17' 9.3278834 Wherefore the Departure is 24° 34' in the Arch of a Great Circle; which reduced to common Italian Meafure, make 1474 Miles 491 Leagues; which is much lefs than by the Table of Parallel Miles. 5. To find the Ship's Course. This is already determined by the fecond and third Articles; for if a Ship fail from A to D, the Course is the the Angle at A; if from D to A, the Courfe is the Complement of the Angle D; both which are known as aforefaid. Note, That if of the Latitude, Difference of Longitude, Course, or Pofition, Diftance, and Departure, any two being given, the reft may be found by the Rightangled Triangle AC D. PROBLEM III. Suppofe a Ship in the Latitude of 30° 06′ at G, fail 'till the arrives at the fame Degree of Latitude in H, and then obferves her Difference of Longitude to 'be 10° 00'; I demand her Distance failed, and greatest Difference of Latitude made in her Course ? Practice. In the Quadrantal Triangle G P H, there are givert the Sides GP, and P H, each equal to 60° 00'; and the Angle P = 10° 00', to find G H, and the Perpendicular let fall from P on G H, as the dotted Line P O. I. To find G H the Distance failed; fay, As the Radius 10.0000000 Is to Co-fine of Latitude 30° 00' 9.9375306 So is Sine of half Diff. of Long. 5° 00' 8.9402960 To the Sine of half Distance 39 26 8.8778266 Wherefore the Distance failed is 6o 52' = GH = 412 Miles 137 Leagues common Measure. 2. To find the Arch O P, the Complement of the greatest Latitude of the Courfe; fay thus, As |
