Altitude of a star in the prime vertical. so that when the declination and latitude are of the same name, the hour angle is less than 6 hours, and the star is above the horizon; but when the declination and latitude are of different names, the hour angle is greater than 6 hours, and the star is below the horizon. 36. Scholium. The problem is, by Sph. Trig. § 27, impossible, when the declination is greater than the latitude; so that, in this case, the star is never exactly east or west of the observer. 37. Scholium The problem is, by Sph. Trig. § 28, indeterminate, when the latitude and declination are both equal to zero; so that, in this case, the star is always upon the prime vertical. 38. EXAMPLES. 1. Find the hour angle and altitude of Aldebaran, when it is exactly east or west of an observer at Boston, in the year 1840. Ans. The hour angle = 4" 45m 44". The altitude = 24° 26'. 2. Find the hour angle and altitude of Fomalhaut, when it is exactly east or west of an observer at Boston, in the year 1840. Ans. The hour angle =8h 40m 51'. The depression below the horizon 48° 49'. 3. Find the hour angle and altitude of Dubhe, when it is Time of a star's rising. exactly east or west of an observer at Boston, in the year 1840. Ans. Dubhe is never upon the prime vertical of Boston. 4. Find the hour angle and altitude of Canopus, when it is exactly east or west of an observer at Boston, in the year 1840. Ans. Canopus is never upon the prime vertical of Boston. 39. Problem. To find the hour angle and amplitude of a star, when it is in the horizon. Solution. In this case the side ZB (fig. 35) of the triangle ZPB is 90°. The corresponding angle of the polar triangle is, therefore, a right angle, and the polar triangle is a right triangle, of which the other two angles are The hypothenuse of the polar triangle is 180° h, and the leg, opposite the angle, 90° D, is 180°-a. Hence, by Sph. Trig. § 40, and Pl. Trig. § 60 and 62, in which the upper sign is used when the latitude and declination have the same name, and the lower sign when they have different names; so that in the former case the hour angle is greater than 6 hours, and the azimuth is counted Time of a star's rising. from the direction of the elevated pole; but in the latter case, the hour angle is less than 6 hours, and the azimuth is counted from the direction of the depressed pole. The amplitude is the difference between the azimuth a and 90°. Hence cos. 's azim. = sin. ✯'s amp. = sin. D sec. L. (389) 40. Scholium. The problem is, by Sph. Trig. § 41, impossible, when the sum of the declination and latitude is greater than 90°; so that, in this case, the star does not rise or set. 41. EXAMPLES. 1. Find the hour angle and amplitude of Aldebaran, when it rises or sets, to an observer at Boston, in the year 1840. Ans. The hour angle 7h 1m 21. 2. Find the hour angle and amplitude of Fomalhaut, when it rises or sets, to an observer at Boston, in the year 1840. 3. Find the hour angle and amplitude of Dubhe, when it rises or sets, to an observer at Boston, in the year 1840. Ans. Dubhe neither rises nor sets at Boston. 4. Find the hour angle and amplitude of Canopus, when it rises or sets, to an observer at Boston, in the year 1840. Ans. Canopus neither rises nor sets at Boston. Determination of the meridian line. CHAPTER III. THE MERIDIAN. 42. The intersection of the plane of the meridian with that of the horizon, is called the meridian line. 43. Problem. To determine the meridian line. Solution. First Method. Stars obviously rise to their greatest altitude in the plane of the meridian; so that if their progress could be traced with perfect accuracy, and the instant of their rising to their greatest height be observed, the direction of the meridian line could be exactly determined. But stars, when they are at their greatest height, change their altitude so slowly, that this method is of but little practical value. Second Method. A star is evidently at equal altitudes when it is at equal distances from the meridian on opposite sides of it. If, therefore, the direction and altitude of a star are observed before it comes to the meridian; and if its direction is also observed, when it has descended again to the same altitude, after passing the meridian; the horizontal line, which bisects the angle of the two horizontal lines drawn in the direction thus determined, is the meridian line. Third Method. [B. p. 147.] The time which elapses between the superior and inferior passage of a star over the meridian is just half of a sideral day. If, then, a telescope Meridian determined by circumpolar stars. were placed so as to revolve on a horizontal axis in the plane of the meridian, the two intervals of time between three successive passages of a star over the central wire, must be exactly equal. But if the vertical plane of the telescope is not that of the meridian, these two intervals will not be equal, and the position of the telescope must be changed until they become equal. Thus, if Z Mm N (fig. 38) is the plane of the meridian, ZSS T that of the vertical circle described by the telescope, MSWsm E the circle of declination described by the star about the pole P; this star will be observed at the points S and s instead of at the points M and m. Now the star describes the circle of declination with an uniform motion, and therefore the arc SP moves uniformly with the star around the pole, so that the angle SPM is proportional to the time of its description; that is, the angle SP M, reduced to time, denotes the sideral time of its description. Let then T = the sideral time of describing the arc S M, I interval from the observation at S to that at s, si the difference of these two intervals; we have then, in sideral time, = I = 12 - T-t 12 - (T+t) i = 12h+T+ t = 12 + (T+ t) dii I 2 (T+t); (390) so that if T and t were equal to each other, and they are nearly so in the case of the pole-star, we should have |
