erect the Perpendicular D I, G D B E cutting the first Diameter produced at pleasure in 1; then thro' the three Points P, O, I, defcribe the Circle FO GI; and it is the Great Circle required. See Euclid, Lib. Prop. 25. 3. Note, If the Diameter FG, paffeth thro' the Center C, the Circle is drawn true; otherwife not. PROBLEM V. To Measure any Spherical Angle in the Projection. Practice. In this Problem are four Cafes. Cafe 1. When the Angle is at the Center of the Pri mitive C, As the Angle BCH ; Take the Arch B H, and Ad, of the Oblique Circle EDF; then draw the Lines E d, Eg; fo fhall the Arch Fg measured on the Line of Chords, give the Angle AED = 44° 30'; and thus the Arch d F, will give the Angle DEC = 45° 30'. Cafe 3. When the Angle is neither in the Center, nor Periphery of the Primitive; but at P, Thus; let the Angle to be measured be BP2; from the Angular Point P, lay a Ruler from the Pole C, of the Oblique Circle BO H, and it will cut the Primitive in b; But the Point D, is the Pole of the Right Circle QR; wherefore the Arch Db, is is the Measure of the Angle BP 2, and will be found on the Chords to contain 659 15' as required. Cafe 4. When the Angle, as A, is contained under two Circles oblique to the Primitive, as BOH, DFE, Thus; to measure the Angle BAD, find the Poles of the Oblique Circles a, C; lay a Ruler from rhe Oblique Point A, to each Pole, and it will cut the Primitive in, b, e; wherefore the Distance of the Arch be, applied to the Chords, gives the Angle BAD = 35o; as was required. PROBLEM VI. To lay off any Number of Degrees on any Great Circle. Practice. In this Problem are three Cafes. Cafe 1. To divide the Pri mitive A E B C this is done by a Line of Chords thus ; fit the Sector to Radius AF; E xh e n fet off the Chord of 46° 30′ A from A to e, g FDB This is performed by a Line of Half-Tangents thus ; from the Center F, fet the Half-Tangent of 46° 30' to f, in the Right Circle FE, and and it is done; for the Arch: Ff46° 30°. Cafe 3. To lay any Number of Degrees on an Ob lique Circle, as Firft, find the Pole of that Circle then from the Line of Chords, lay off on the Primitive, from then Tay a Ruler on the Points ED C. g; 50° 30% B to bi Dn = 50° 30′; and it will cut the Oblique Circle in fo fhall the Arch as was required. Note, That to measure any Part of a great Circle, is only the Reverse of this Problem in each Cafe; and cannot be difficult to any one, who hath orderly come thus far. Cafe 2. as To draw a Parallel to any Right Circle, Set the Half-Tangent of its Distance, (fuppofe from the Right Circle) from Then fet the Tangent of (the Complement of its Distance) from in the Diameter produced to On that Point, as a Center, draw AD. CN, 30% C to N; 60°, N to e, e; LNM, E g af B and it fhall be the Parallel Circle required. Cafe 3. To draw a Paral lel to any Oblique Cir EDG; it will be 32° 30'; G which fubftract from (the Complement of the given Distance and there will remain fet the Half-Tangent of this from from the Point and fet the Half-Tangent of then draw the Line bifect this Line in on which, as a Center, defcribe and that will be the Parallel Circle required. By this Problem, the Path of the Vertex of any Place is defcribed in Aftronomical Schemes, PRO |