I 2 II Fifthly, Add all the Longitudes together, and they make 14 Sig. 04 Deg. 29 Min. (from which abate 12 Signs, and there remains only Two Signs.) --Also Add the Anomalies together, and they make 10 Sig. 25 Deg. oo Min. Sixtbly, (in Table IV.) Look for 10 Signs at the Bottom of the Table, and 25 Deg. in the last Column towards the Right Hand, so against it, over io Signs, you shall find i Deg 6 Min. to be Added. Set them under the Mean Longitude, and add them to it, so will the Sum be 2 Sig. 05 Deg. 35 Min. “And that is the true Place of the Sun in the Ecliptick, which is 35 Deg. distant from the Aquino&tial Point r Aries. Note, that Signs 01 3 4 5 6 7 8 9 * In this example, it 9 20 07 6 12 21 the Day of the Month out of the Column that Hours 61 15 15 hath Bisex. at the Mean Motion Il 18 31 Head of it. 1 52 S. D. M.S.D. M. | the Sum of the Sun's Year 1713 9 20 48 Anomalies be less than 6 13 10 April 23 3 21 23 3 21 22 Six Signs, they will At Noon 0 00 00 0 00 00 ; be found at the Head Mean Morion 10 04 32 of the Æquation TaÆqui. Add. ble, and the Degrees OL 35 in the First Column toI 13 46 | 8 Taurus. wards the Left Hand, Longit. O Anom. O and the Æquation must S. D. M. S. D. M. (always) be Substract ed from the Mean Lon. Year 1761 9 20 10 6 12 42 Jan. Day 12 11 50 11 50 gitude ; whereas in all Hours these Examples it hath 39 39 been Added. Mean Motion 3 39 6 25 II 51 O true Place, 10 4 30 8 jo 45 o true Place. o true Place 16 Æquat. Add. In all the Problems in this Se&tion, it is to be understood, That Pole of the World. Aquinoctial Circle, or Æquator. Parallels (or smaller Circles) of Declination of the Sun, or of a Star. Pole of the Ecliptick: Ecliprick. Parallels (or lesser Circles) of Altitude of the a or Country. Υ Vernal 2 Interfe&tion of the Ecliptick and Æ: Autumnal) quator. Ascension Repre- Descension of any point of the Ecliptick. the Nadir. Horizon. Prime Vertical Circle, or Azimuth of East and Pole of the Meridian : Or the Place 2 East in the Horizon, where the Sun, or 2 West Star, Rijos or Sets. Right Angle. Star. Side, Sun (and sometimes) Star. North? } Plain Triangles. Or Aftrono Astronomical Problems. 1 PROBLEM I. 1. The Sun's Right Afcenfion. Meridian and the Ecliptick. I. By the Celestial Globe. rus 8, (that is, 59 Deg. from the beginning of Aries r, which is the nearest Æquino&tial Point.) The Globe being in any Position, (for in this Problem there is no regard to be had to the Latitude) count the Sun's Place in the Ecliptick upon the Ecliptick Circle, from r; and bring that Point to the Graduated Side of the Brass Meridian. Then, (1.) Will the Brass Meridian cut the Æquino&tial Circle in 56 d. 46 m. counted also from r; and that is the Right Ascension.' And, (2.), The Number of Degrees of the Brass Meridian, comprehended between the Æqui noctial and Ecliptick Circles, will be 20 d. which is the Sun's Declination. —And, (3.) The Angle made by the Interfe&tion of the Brass Meridian, and the Ecliptick Circle in the Point of the Sun's Place, will be 77 d. 23 m. which is the Angle of Position, in respect of the Meridian and Ecliptick. II. By Trigonometrical Calculation. The Globe being in this position, there is represented upon the Superficies thereof, Two Right-angled Spherical Triangles, such as are expressed in the Diagrams; and are there noted Fig. with OR r and OR; in which, the Sides Or and O XXVII. are Arches of the Ecliprick Circle, and is the Sun's Longitude : XXVIII, The H h 2 Fig. The Sides r R and R are Arches of the Æquino&ial, and XXVII. is the Sun's Right Afcenfion; and the Sides OR are Arches XXVIII. of the Brass Meridian, and is the Sun's Declination. Also, the Angle at r. or is the Angle of the greatest Obliquiry of the Ecliprick, and is equal to the Sun's greatest Declinarion 23 d. 31 m.) The Angle R is a Right Angle, and the Angle o is the Angle at the Sun's Position, in respect of the Ecliptick and Meridian Circles. To resolve this Problem Trigonometrically, you are to consider the Quadrant of the Ecliptick, in which the Sun is, which in the Figures are signified by one of these Numbers, 1. 2. 3. 4. Of which, the first is of the Spring, from the beginning of Aries or, to the beginning of Cancer $, &c. Then in the Triangle R r o or R = 0. (1. The Angle at por , 23 d. 31 m. There is given, 2. R r or RA: The Sun's Diftance from the besides the Right next Æquino&ial Point; to be numbred Angle at R, from ror, unto the Degrees of the Sun's Longitude or Place in the Ecliptick given. And, 1. RO, The Sun's Declination for that Sign which he is in; whether North or South. C1. The Degrees found by the Canon, are the Degrees of 2. The Degrees found, must be 2. The Arch substracted from 180, and the r R, or remainded are the Degrees of AR,which 3. The Degrees found must be Right if it be in added to 180, and the Sum | Ascension. Quadrant, are the Degrees of 4. The Degrees found must be fubftraéted from 360 d. and the Remainder are the Degrees of The Canons for Calculation. There is required, 1. Fos ୨୦ w mo 7 1. For the Right Ascension r or R. By Cafe III. of R. A. S. T. Fig. XXVII. As the Sine of 90 d. XXVIII. 66 d. 29 m. d. Afcenfion; because his Place given was in the Firft Qua- 46 m. D. M. 2. For the Sun's Declination R O. By Cafe II. of R. A. S. T. Is to the Sine of the greatest Obliquity of the Ecliptick 23 d.31m. Which is North, becaufe he is in a Nortbern Sign. A.S.T. Is to the Sine of greatest Declination 23 d. 31 m. PROB. II. . The Right Ascension, or Declination of the Sun given; To HIS is the Converse of the foregoing Problem ; for in this, the Sun's Place, in respect of the Æquinoctial Circle, is given : And his Place, in respect of the Ecliptick Circle, is re. quired : And may be resolved as followeth, |