6. Set 24°20', the Plane's Declination, from the Scale of Chords, on the Primitive Circle from H to c, from D to e, and from 0 to f. 7. Then a Ruler laid from Z, to the Point c, e, f, will give the Points W, B, E for the Weft, South, and Eaft Points of the Horizon of London. 8. Lay a Ruler from W to Z, and it will cut the Primitive in b; then lay off 38° 28′ (the Co-latitude of London) from the Chords from b to d. 9. Take the Tangent of 54° 00' and fet from 2 to K, and draw thro' Ka Line parallel to HO, infinitely extended, and thereon fet off the Half-Tangent of 24° 20', from K to M; then with the Tangent of 65° 40′ ( fet the other way from K, on the fame Line extended) you find the Center whereon to draw the Circle Z BM, which fhall be the Meridian of London. 10. A Ruler laid from W to d will cut the Meridian in P for the North Pole of the World. II. Thro' P and draw the Right Line NQT, which fhall be the Meridian of the Place fought after, and the Axis of the World; and, being produced, will meet the Meridian of London in S, the South Pole of the World. 12. Cross NT at Right Angles with V X, and a Ruler laid from 0 to V will give the Point Æ, where the Equinoctial interfects the Meridian of the enquired Place. 13. Thro' V, Æ, and X, draw the Equinoctial Circle, as hath been taught; which alfo will pafs thro' W and E, the Eaft and West Points of the Horizon of London. 14. Divide P S into two equal Parts in E, and draw E G at Right Angles thro' the fame; extending it infinitely make P E Radius, and lay off the Tangents of 15°, 30°, 45°, &c. both ways from E on the the Line E G; and thus you will have the Centers of the Hour Circles to be projected; which when done your Projection is finished. 15. Now to find the Requifites, take QP in your Compaffes, and measure it on a Scale of Half-Tangents, and you will find it equal to 72° 34' whose Complement PN, or 2, is 17° 26'; and that is the Latitude in which this Reclining Plane will be an Horizontal Dial. 16. Lastly, lay a Ruler from P to a, and it will cut the Primitive in b; then Th measured on the Chords will be found 14° 41', equal E a, the Dif ference of Longitude required. Therefore a Plane declining 24° 20', and Reclining North 36° oo' in the Latitude of London, will be an Horizontal Plane in the Latitude 17° 26', and Difference of Longitude from London 14° 41', which is about Caffena in Negroland. The CHAP. VII. Theorems for the Explication of the Doctrine of Spherical Triangles, and the Manner of their Solution. in common with the other Great Circle therefore the common Section ABC, D, AFC; A C, fhall be a Diameter of each; and fo will cut them into two equal Parts. Q, E. D. THEOREM II. If from the Pole B, of any Great Circle AFC, be drawn a Right Line B D, to the Center thereof, the faid Line will be perpendicular to the Plane of that Circle. Demonftration. Let be drawn the Diameters n the Circle EF, GH, Then BDF, BDE, then because in the Triangles the Sides are equal to the Sides and the Base therefore fhall the Angle and fo each is a right one; thus it is demonftrated that Angles BDG, BDH, ЛЕСЕ, Q, E. D. are right ones; and fo is the Line perpendicular to the Plane of the Circle by Euclid 11. Prop. 4. THEOREM III. If a great Circle EB F, be defcribed about the Pole A then the Arch BF, intercepted between AB, AF, is the Measure of the Angle BAF, or BDF. Demonftration. By Theorem 2, the Angles are right ones; because the Arches AD B, ADE are Quadrants; Confequently the Angle (whofe Measure is the Arch BDF, is equal to the Inclination of the Planes ABC, AFC; and fo equal to the Spherical Angle BAF, or BCF, COROLLARY I. ་ 2. E, D. Because AB, A F, are Quadrants, A fhall be the Pole of the Circle paffing thro' the Points B, F; for A is at Right Angles to the Plane FD B, by Euclid 11. Prob. 14. COROLLARY II. The Vertical Angles are equal; being each equal to the Inclination of the Circles; alfo the adjoining VOL. II. |