PROBL. III. Having the Suns greatest Declination, and his Distance from the next Equi∣noctial-Point; to find his Right-Ascention.
IN the Foregoing Scheme, having drawn the Parallel of the Suns Declination TD, passing through the Place at ✶ the extent S ♁, is the Sign of the Suns Right-Ascen∣tion from the next nearest Equinoctial-Point, to the Radius of the Parallel TD; and therefore place the extent ST from C to X, and upon X as a Centre with the extent S ♁, describe the Arch at k, a Ruler laid from the Centre just touching the extremity of that Arch, finds the Point N in the Limb of the Meridian or Quadrant, and the Arch ON, applyed to the Line of Chords, is 59 degr. 09 min. and so much is the Suns Right-Ascension in the first quarter of the Ecliptick.
In the Triangle CK ♁ we have given as before, (1.) the Angle of the Suns grea∣test Declination KC ♁ 23 degr. 31 min. (2.) the Longitude of the Sun from the next Equinoctial-Point Aries C ♁ 61 degr. 18 min. hence to find the Suns Rght-As∣scention, the Rule is,
As the Radius | 10 |
to the Tangent of the distance 61 degr. 18 min. C ♁ | 1026162 |
So is the Co-Sign of the Suns greatest Decl. 23 deg. 31′ KC ♁ | 996234 |
to the Tangent of the Right-Ascention CK 59 deg. 9 m. | 1022396 |
Or in the Concave-Sphere; if you draw the Meridian from N through ♁ to S, whose Centre will be found upon the Equator, it will cut the Equinoctial in K; mea∣sure the distance GK on the Line of half-Tangents, and you have 59 d. 09′, as before.
Or extend the Compasses from 90 d. to 66 d. 29, the same distance will reach from 61 deg. 18 m. to 59 deg. 9 min. which is the Suns Right-Ascention in 61 deg. 18 ♊.