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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 O to f.
7. Then a Ruler laid from Z, to the Point c, e, f, will give the Points W, B, E for the West, South, and East 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 set from 2 to K, and draw thro' Ka Line parallel to HO, infinitely extended, and thereon set off the Half-Tangent of 24° 20', from K to M; then with the Tangent of 65° 40' ( set the other way from K, on the fame Line extended) you find the Center whereon to draw the Circle Z B M, which shall 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.
11. Thro' P and 2.draw the Right Line N 2 T, which shall 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 N T 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 also will pass thro' W and E, the East and West Points of the Hori. zon 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 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 Requisites, take QP in your Compasses, and measure it on a Scale of Half-Tangents,
you will find it equal to 72° 34' whose Complement PN, or QÆ, is 179 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 h; then Ih measured on the Chords will be found 14° 41', equal Æ a, the Difference of Longitude required.
Therefore a Plane declining 24° 20', and Reclining North 360 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.
CH A P. VII.
Theorems for the Explication of the Doc
trine of Spherical Triangles, and the Manner of their Solution.
NY two Great Circles of the Sphere mutually bisect each other.
Demonstration. For since one Great Circle
ABC, hath the Center
D in common with the other Great Circle AFC; therefore the common Section
AC, shall be a Diameter of each ; and so will cut them into two equal Parts. 2. E, D.
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.
Demonstration. Let be drawn the Diameters
EF, GH, n the Circle
then because in the Triangles BD F, BDE, the Sides
BD, BF, are equal to the Sides
BD, DE, and the Base
BF= BE; therefore shall the Angle
BDF=BDE, and so each is a right one ; thus it is demonstrated that Angles
BDG, BDH, are right ones, and so is the Line
- BD, perpendicular to the Plane of the Circle
AECF, by Euclid 11. Prop. 4.
2. E. D.
If a great Circle E B F, be described about the Pole A ; then the Arch B F, intercepted between AB, A F, is the Measure of the Angle B AF, or BD F.
Demonstration. By Theorem 2, the Angles
ADB, ADE are right ones ; because the Arches: A B, AF, are Quadrants ; Consequently the Angle B DF, ( whole Measure is the Arch
BF, ) is equal to the Inclination of the Planes A B C, AFC; and so equal to the Spherical Angle B AF, or BCF,
2. E, D.
COROLLARY I. Because AB, AF, are Quadrants, A shall be the Pole of the Circle passing 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; also the adjoining VOL. II.