226 Fig. XXXIII. Ancilla Mathematica. PROB. ΧΧΙΙΙ. At what Horary Distance from the Meridian, Aldebaran will be due the and and the Quadrant of Altitude to the West Point of the Horizon: Then turn the Globe Eastward, till the Centre of the Star be just 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 West, he will be 5 h. 40 m. of time short of, or gone beyond, the Meridian. And when the Centre of Aldebaran is just under the Edge of the Quadrant of Altitude, you shall find it to touch 20 d. 21 m. And fuch is the Altitude of Aldebaran when he is upon the Eaft or West Azimuth. In like manner may you find that H. M. D. M. Ariturus 2 will be upon the East or West 54 492 and his 527 13 Syrius Algol Azimuth, when he is di 04 { 20 56 stant from the Meridian. (3 15 will be 54 37 The fame Trigonometrical Calculation as for the Sun. PROB. XXIV. What Altitude and Azimuth Aldebaran (or any other Star) Shall have, when Six Hours distant 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 shall find it to lye under 12 d. 18 m. of the Quadrant: And such is the Altitude of Aldebaran. At the same time look what Degrees of the Horizon are cut by the Quadrant of Altitude, and you shall find 8 d. between it and the East or West Points Northwards. And such is the Azimuth of Aldebaran. And according to this Rule you shall find that when Altit. Azim. D. M. D. M. 6 h. distant from 516 15276 367 from the 5 the Meridian his 212 38579 435 North. Arcturus Zis Algol Syrius Syrius is never 6 h. distant 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. T For having brought the Star HIS is no other than the last. to the Meridian, and the Index to 12, move the Globe till it come to the designed Hour. Then the Quadrant of Altitude being laid over the Star, shall at the same time shew you both the Altitude and Azimuth thereof as before. This needeth no Example. The same Trigonometrical Calculation as for the Sun. PROB. XXVI. Having the Azimuth of a Star, to find at what Horary Distance that Star is from the Meridian, and what Altitude that Star then hath. BRing the Star to the Meridian, the Index to 12, and the Quadrant of Altitude to the Given Azimuth; then turn the Globe about till the Centre of the Star lye just under the Quadrant of Altitude; the Index at that time shall give the Horary Distance, and Quadrant the Altitude of the Star. Example. Aldebaran being seen 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 shall find the Index to point at 6 h. which is the Star's Horary Distance from the Meridian. And the Quadrant of Altitude will shew 12 d. 18 m. the Altitude of Aldebaran at that time. These Twelve last Problems may be resolved by Trigonometrical Calculation in all Respects as those 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. 1 Fig. XXXIIL PROB. XXVII. The Longitude and Latitude of a Star, which is fituate out of the Ecliptick, being given; To find the Right Ascension and Declination of that Star. ERE is given the Situation of the Star, in respect to the H Ecliptick, altho its Siruation be out of the Ecliptick his Place, in respect of the Equinoctial, is required I. By the Cœleftial Globe. and 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 observe what Meridian passeth 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 same Meridian continued, will cross the Equinoctial Circle in the Point of the Star's Right Afcenfion. II. By Trigonometrical Calculation. Fig. When the Longitude of the Sun falls to be in any Sign XXXIV. SAscending, as in Fig. XXXIV. 2 as in Fig. XXXV. S XXXV. make choice of the Triangle: } Defcending, North, 2 South, will deter 1. The Side PF, the Distance of the Pole of the Æquator from the Pole of the Ecliptick, 23 d. 30. m. 2. The Angle F, whose Measure is the Arch of the Ecliptick, intercepted between the Sides F Pand Given, FS, from whom the Longitude of the Star is pro duced, and the Solstitial Point found out. 3. The Side FS, the Complement of the given Star's Latitude. Required, Required, 1. The Side PS, the Complement of the Star's Fig. { Declination. XXXIV. 2. The Angle P, which the Ark of the Æquator XXXV. Example. Let there be given the Head of Andromeda, which is a Star of the Second Magnitude; whose Longitude is in 9 d. 38 m. r, and Latitude 25 d. 42 m. b. The Longitude of the Star falls out to be in an Ascending Sign Aries; therefore, make choice of the Triangle PFS, Fig. XXXV. the Latitude being North, in which there is given, (1.) PF, the Distance of the Pole of the Æquator from the Pole of the Ecliptick, 23 d. 30 m. (2.) The Angle F, which the Ark of the Ecliptick intercepted between the Sides FP and FS, being produced, meafureth, and gives the Longitude of the Star to be 9 d. 38m. r, so that the Solstitial Point is nearest, substract therefore 9 d. 38 m. r 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. 47 m. The Canons for Calculation. (1.) As the Sine of half the Sum of the Sides SF and FP, 43 d. 54 m. Is to the Sine of half the Difference of those 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 FS and FP, 46 d. 6 m. Is to the Co-fine of half the Difference of those Two Sides, 39 d. 36 m. Mm2 So Fig. XXXIV. XXXV. So is the Co-tangent of half the Angle at F, 40 d. iim. I m. Which 57 d. 1 m. added to the aforefound Difference, gives 87 d. 47 m. for the greater Angle at P; and 30 d. 46 m. the Difference aforefound substracted from 57 d. leaves 26 d. 15 m. for the leffer Angle at S. Then, I m. 15 m. is the Star's Declination. PRO B. XXVII. The Declination and Right Afcenfion of a Star being given : To find the Longitude and Latitude of that Star. T HIS is the Converte of the former Problem, and may be refolved as followeth. I. By the Cœleftial Globe. Screw the Quadrant of Altitude over the Pole of the Eclip tick: Then count upon the Equinoctial Circle the Degrees of Rigbt Ascension; and upon that Meridian which passes through that Point, count the Star's Declination from the Equinoctial, towards the Pole, and to that Point bring the Quadrant of Altitude; then will the Quadrant cut the Ecliptick in the Degrees of the Star's Longitude; and the Degrees of the Quadrant, comprehended between the Ecliptick Circle and its Pole, will give the Degrees of the Star's Latitude. II. By Trigonometrical Calculation. When the Right Afcenfion of the Star falls out to be in the First or Fourth Quadrant of the Equinoctial; as in Fig. XXXIV. or in the Second or Third: As in Fig. XXXV. make use of the Triangle PFS for the Star's Declination North or South. In which there is |