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1. By the Globe. XXVI.
Apply the Quadrant of Altitude, or Brass Plate, to the Two Places, and the Number of Degrees thereof contained between the Two Places, is their Distance, which will be found to be 50 Deg: 32 Min.
II. By Trigonometrical Calculation.
The Canon for Calculation. By CASE I. of R. A. S. T.
30 Deg. 07 Min..
PRO B. V.
1. If one of the Places lye under the Aquinoctial, and so have no Latitude ; and the other under fome Parallel of Latitude be: tween the Aquino&tial, and one of the Poles: As London, in 51 Deg. 30 Min. of North Latitude at L; and St. Thomas lisand under the Æquino&tial at T, but differ in Longitude 18. Deg. For finding the Distance of these. Two Places.
In , .
1. Upon the Terrestrial Globe. Bring London to the Brass Meridian, and over it, fix the Qua. XXVI. drant of Altitude : The Globe being in this position, bring the Quadrant of Altitude to lye just over St. Thomas Island, and you will find it cut the Quadrant of Altitude, in 54 Deg. 45 Min. for the Distance of the Two Places.
II. By Trigonometrical Calculation. The Globe resting in the former Position, you will find constituted upon it a Right-angled Spherical Triangle L ÆT, composed of, (1.) L Æ, an Arch of the Brass, Meridian. (2.) Æ T, an Arch of the Aquino&tial. And, (3.) LT, an Arch of a Great Circle (made by the Quadrant of Altitude) paffing through both the Places: And in this Triangle, you have given, (besides the Right Angle at Æ) (1.) The Perpendicular LÆ, the Latitude of London, 51 Deg. 30 Min. (2.) The Angle at L, the Difference of Longitude 18 Deg. 10 Min. To find the Hypotenuse LT, the Distance.
The Canon for Calculation. By CAS E XIV. of R. A. S. T.
Is to the Radius;
To the Co-tangent of LT, 52 Deg. 55 Min.
tween London, and St. Thomas Island.
Upon the Terrestrial Globe. Bring one of the Places, as London, to the Brass Meridian, and over it screw the Quadrant of Altitude, and keep the Globe there fixed, then move the Quadrant of Altitude, till it lye over HieruSalem, and you shall find it to lye under 39 Deg. 32 Min. of the Quadrant : And that is the Distance of the Two Places.
By Trigonometrical Calculation. XXVI.
The Globe being in the former Position, you will find upon it an Oblique-angled Spherical Triangle, composed of P L, an Arch of the Brass Meridian: OfP H, an Arch of the Meridian that paf. seth over Hierufalem ; and of L H, an Arch of the Quadrant of Altitude. In which Triangle you have given, (1.) PL, the Complement of the Latitude of London, 38 Deg. 30 Min.' (2.) The Side P H (the Complement of the Latitude of Hierusalém 59 Deg. 20 Min. (3.) The Angle L PH, the Difference of Longi. tude of the Two Places, 46 Deg. To find the Two Angles of Poo sition at L and H: And, (2.) The Side L H, the Distance of the Two Places.
The Canons for Calculation.
50 The Half Sum is
48 25 The Half Difference is
Deg. 25 Min.
half the Angle L PH, 23 Deg.
Which is the half Difference of the Angles P L H and PH L. (2.) As the Co-sine of half the Sum of the Sides P L and P H,
41 Deg. 35 Min.
half the Angle L PH 23 Deg.
31 equal to C PLH.
45 32 equal to <PHL.
So is the Sine of L PH, the Differ. of Longitude, 46 Dug.
for the Distance of the Two Places.
Upon the Terrestrial Globe.
By Trigonometrical Calculation.
Fig. The Canons for Calculations, as in the last, by CASE III and XXYI. IX. of A. O.S. T.
(1.) As the Sine of the Half Sum of the Sides PV and PC, 84 D.
Ís to the Sine of half their Difference, 41 Deg.
(half the Angle V PC) 29 Deg. 30 Min.
Which is the Half Difference of the Angles PVC and PCV. . (2.) As the Co-fine of half the sum of the Sides, P. V and P.C,
49 23 Their Sum
134 55 equal to <P CV. Their Difference
36 09 equal to PVC.
As the Sine of PVC, 36 Deg. 9 Min.
So is the Sine of V'PĆ, 59 Deg.
Side V C, which is the Distance of the Two Places. Which
in Miles is 5862. And these are all the Varieties of Positions that any Two
Places upon the Globe can be fituate.