PROBL. II. The Suns Distance from the next Equinoctial-point; and his greatest De∣clination being given, to find the Declination of any Point required.
VVIth your Compasses take the Chord of 60 degr. upon the Centre C, describe the Circle HZON, and draw the Diameter HCO, which represents the Horizon, and at Right-Angles, or perpendicular thereunto, draw ZCN, the Vertical Azimuth of East and West, and take the Latitude of the Place, as in this Example, is 51 d. 28 m. and prick it from O to N, and from H to S, and draw the Axis or Meridian of the Hour of Six NCS; then prick from Z to AE, and from N to Q 51 degr. 28 min, and draw the Equinoctial-Line AECQ, then the Suns place being given, take 23 deg. 31 m. and prick from AE to ♋, and from Q to P, and draw the prickt Line ♋ CP; then take the Suns Distance from the next Equin.-Point, which in this Example shall be 61 deg. 18 m. out of the Line of Signs, and prick it from C to ☉, and through ☉ ♁ draw a Parallel-Line to the Equinoctial, as TD, and it shall be a Parallel of Decli∣nation, and where it cuts the outward Meridian, as at T; apply the Distance AET to the Line of Chords, and you have the Declination 20 degr. 30 min. which was required; Or you may take the nearest Distance from ☉ to the Equator, and apply it to the Line of Signs, and that will give you the Declination 20 degr. 30 min. as before; and if through ☉ ♁ you draw a Line Parallel to the Horizon HO, as ef, it is a Parallel of the Suns Altitude, and so have you the Sphere Orthographically in Right-Lines in the Convex-Sphere; and if you follow the directions of the use of Tangents, and half Tan∣gents in the 12 Chap. of the fourth Book of the Description of the Globe in Plano, you have the Sphere projected in Plain and Circular Lines, and fitted for the use of divers Questions; the Direction in both Spheres by the Letters signify the same thing; but observe what you are directed by Signs in the Convex-Sphere, is likewise to be done by ½ Tangents in the Concave-Sphere.
By the Tables in the Right-Angled Triangle CK O; we have given, first the Hy∣pothenase C ☉ 61 degrees 18′; secondly, the Angle KC ☉ 23 degr. 31′, hence to find K ☉, the Rule is, as the Radius is in proportion 10
to the Sign of the Suns greatest Decl. 23 d. 31′ KC ☉ | 960099 |
So is the Sign of the Suns distance from the next Equinoctial-Point 61 degrees 18 min. C ☉ | 994307 |
to the Sign of his Declination required 20 degrees, 30′ K ♁ | 954406 |
Or extend the Compasses in the Line of Artificial Signs from 90 degr. to 23 degr. 30 min. the same extent will give the distance from the Suns Place, to his Declination.