PROB. XXI. At what Hour (any time of the Year) Aldebaran comes to the L ET the time be the First of January, at which time the Sun is in 22 d. of Capricorn. Bring 22 d. of Capricorn to the Meridian, and fet the Hour Index to 12. Then turn the Globe about till Aldebaran be under the Meridian, and then you shall find the Index to point at 42 m. after 8 of the Clock, at which time Aldebaran will be upon the Meridian that Night. In like manner you may find that H. M. Upon Fanuary 21 Syrius the Meridi 9 33 Fanuary 1 Algol an at The fame Trigonometrical Calculation as for the Sun. PROB. XXII. At what Hour (at any time of the Year) Aldebaran (or any other L ET the time be January 1. By the laft before-going, you found that Aldebaran came to the Meridian at 8 h. 42 m. And by the last but one you found his Semidiurnal Arch to be 7 h. 27 m. This being taken from 8 h. 42 m. the time of his being South, leaveth 1 h. 15 m. the time of its rifing; fo that upon the First of January Aldebaran did rife at 15 m. after 1 in the Afternoon. Again, if you add his Semidiurnal 7 h. 27 m. to the time of its being South 8 h. 42 m. the Sum will be 16 h. 9 m. from which take 12 h. and the Remainder will be 4 h. So that Aldebaran did fet at 9 m. after 4 of the Clock the next Morning. 9 m. And in like manner you may find that October 282 Arcturus Upon January 21 Syrius H. M. H. M. 3 82 Rifes at 5 3 3Sets at 2 3 January Algol Algol never Rifes nor Sets. The fame. Trigonometrical Calculation as for the Sun. Fig. XXXIII. PROB Fig. XXXIII. PROB. XXIII. At what Horary Distance from the Meridian, Aldebaran will be dat BR Ring Aldebaran to the Meridian, and the Hour-Index to 12, and the Quadrant of Altitude to the Weft Point of the Horizon: Then turn the Globe Eaftward, till the Centre of the Star be juft under the Edge of the Quadrant; then shall the Index point at 5 h. and 40 m. So that when Aldebaran is due Eaft or Weft, he will be 5 h. 40 m. of time fhort of, or gone beyond, the Meridian. And when the Centre of Aldebaran is juft under the Edge of In like manner may you find that Arcturus2 will be upon the Eaft or Weft 4 492 and his Syrius 22 D. M. 27 13 20 56 54 37 The fame Trigonometrical Calculation as for the Sun. PROB. XXIV. What Altitude and Azimuth Aldebaran (or any other Star) fhall have, when Six Hours diftant from the Meridian. Bring Aldebaran to the Meridian, and the Index to 12. Then turn the Globe about till the Index point at 6; then lay the Quadrant of Altitude over the Centre of the Star, and you fhall find it to lye under 12 d. 18 m. of the Quadrant: And fuch is the Altitude of Aldebaran. At the fame time look what Degrees of the Horizon are cut by the Quadrant of Altitude, and you fhall find 8 d. between it and the Eaft or Weft Points Northwards. And fuch is the Azimuth of Aldebaran. And according to this Rule you fhall find that when Altit. Azim. D. M. D. M. 15276 36 from the 38579 435 North. Arcturus is 6 h. distant from S16 Syrius is never 6 h. diftant from the Meridian, nor any other Fig. Star that hath South Declination. The Trigonometrical Calculation as for the Sun. PROB. XXV. To find what Altitude and Azimuth any Star bath when he is at any horary Distance from the Meridian. THI HIS is no other than the laft. For having brought the Star to the Meridian, and the Index to 12, move the Globe till it come to the defigned Hour. Then the Quadrant of Altitude being laid over the Star, fhall at the fame time fhew you both the Altitude and Azimuth thereof as before. This needeth no Example. The fame Trigonometrical Calculation as for the Sun. PROB. XXVI. Having the Azimuth of a Star, to find at what Horary Difiance that Star is from the Meridian, and what Altitude that Star then bath. Bring the Star to the Meridian, the Index to 12, and the Qua drant of Altitude to the Given Azimuth; then turn the Globe about till the Centre of the Star lye juft under the Quadrant of Altitude, the Index at that time fhall give the Horary Diftance, and Quadrant the Altitude of the Star. Example. Aldebaran being feen upon 80 d. of Azimuth from the North Westward, that is, near upon the W. by N. Point of the Compass, the Star brought to the Meridian, and the Quadrant of Altitude to 80 d. and the Hour-Index to 12. If you bring the Star to the Quadrant of Altitude, you fhall find the Index to point at 6 h. which is the Star's Horary Distance from the Meridian. And the Quadrant of Altitude will fhew 12 d. 18 m. the Altitude of Aldebaran at that time. Thefe Twelve laft Problems may be refolved by Trigonometrical Calculation in all Respects as thofe of the Sun. if (in all of them) instead of the Word [Sun] there, read [Star] here. And the Canons for Calculation will be the fame in both. XXXIII. Fig. XXXIII Fig. PROB. XXVII. The Longitude and Latitude of a Star, which is fituate out of the ERE is given the Situation of the Star, in refpect to the I. By the Cœleftial Globe. Screw the Quadrant of Altitude over the Pole of the Ecliptick, and lay it to the Degree of the Star's Longitude in the Ecliptick. Then count the Star's Latitude given upon the Quadrant of Altitude: And obferve what Meridian paffeth through that Point: For the Degrees of that Meridian, comprehended between that Point, and the Pole of the World, are the Degrees of the Star's Declination: And the fame Meridian continued, will cross the Equinoctial Circle in the Point of the Star's Right Afcenfion. II. By Trigonometrical Calculation. Sun falls to be in any Sign make choice of the Triangle given Star if it be North, the which the Letters b and m written within them will determine. In which Triangle there is Given, 1. The Side PF, the Distance of the Pole of the Aqua- 3. The Side FS, the Complement of the given Star's Required, Required, 1. The Side P S, the Complement of the Star's Fig. Example. Let there be given the Head of Andromeda, which is a Star of the Second Magnitude; whofe Longitude is in 9 d. 38 m., and Latitude 25 d. 42 m. b. The Longitude of the Star falls out to be in an Afcending Sign Aries; therefore, make choice of the Triangle P FS, Fig. XXXV. the Latitude being North, in which there is given, (1) P F, the Distance of the Pole of the Equator from the Pole of the Ecliptick, 23 d. 30 m. (2.) The Angle F, which the Ark of the Ecliptick intercepted between the Sides F P and FS, being produced, meafureth, and gives the Longitude of the Star to be 9 d. 38 m. , fo that the Solftitial Point is neareft; fubftract therefore 9 d. 38 m. Y from 90 d. the beginning of, the Remainder will be 80 d. 22 m. which is the Angle at F. (3.) FS, the Complement of the Latitude of the given Star, 64 d. 18 m. -And let there be fought, (1.) PS, the Complement of the Declination 62 d. 45 m. therefore the Declination of the Star is 27 d. 15 m. (2.) The Angle P, 87 d. 47 m. Which added to 270 d. makes the Right Afcenfion of the Star to be 357 d. (1.) As the Sine of half the Sum of the Sides SF and F P, 43 d. 54 m. Is to the Sine of half the Difference of thofe Sides, 20 d. 24 m. So is the Co-tangent of half the Angle at F, 40 d. 11 m. To the Tangent of half the Difference of the Angles at S and (2) As the Co-fine of half the Sum of the Sides F S and FP, Is to the Co-fine of half the Difference of thofe Two Sides, 39 d. 36 m. M m 2 So |