PZB (fig. 35), the three sides, to find the hour angle ZPB, which is thus found by (336), The hour angle may also be found by (341); thus, if we put s = } (180° — A — L − p) — 90° — s'+p=90° — A — L + s′ = - == 8 — p = 90° — s', s — (90° — L) = s' — A, Third Method. By the distance from a fixed terrestrial object. If the position of the terrestrial object has been before determined, its hour angle and polar distance may be considered as known. Hence, if T (fig. 40) is the position of the terrestrial object projected upon the celestial sphere, P the pole, and S the star. distance TS be observed, and let Let the (565) sin. (s S- p) sin. (s sin. P sin. p (566) If the polar distance and hour angle of the terrestrial object is not known, but only its altitude and azimuth, the polar distance and hour angle can be easily found by solving the triangle PZT. Fourth Method. By a meridian transit. [B., p. 221.] If the passage of a star is observed over the different wires of a transit instrument, the mean of the observed times is the time of the meridian transit, which should agree with the known time of this transit. This method surpasses all others in accuracy and brevity. Fifth Method. By a disappearance behind a terrestrial object. If the instant of a star's disappearance behind a vertical tower has been observed repeatedly with great care, the observed time of this disappearance may afterwards be used for correcting the chronometer. For this purpose, the position of the observer must always be precisely the same. Any change in the right ascension of the star does not affect the star's hour angle, that is, the elapsed time from the meridian transit; this change, consequently, affects the observed time exactly as if the observation were that of a meridian transit. A small change in the declination of the star affects the hour angle, and therefore the time of observation. Thus, if P (fig. 44) is the pole, Z the zenith, ZSS' the vertical plane of the terrestrial object; then if the polar distance PS is diminished by RS & D, the hour angle ZPS is diminished by the angle SPS' 8 h, But SR is nearly perpendicular to SP, and the sides of SSR are so small, that their curvature may be neglected, whence RS&D tang. S⇒ 15 cos. D . 8 h, 1. On May 20, 1823, in latitude 54° 20′ N., the sun was at equal altitudes, the observed interval was 6' 1" 36; find the correction for 2. On September 1, 1824, in latitude 46° 50' N., the interval between the observations, when the sun was at equal altitudes, was 7h 46m 35o; the sun's declination was 8° 14′ N., and his daily increase of declination-21' 49"; what is the correction for the mean of the observations? Ans. 16$.4. 3. On March 5, 1825, in latitude 38° 34' N., the interval between the observations, when the sun was at equal altitudes, was 8h 29m 28s ; the sun's declination was 6° 2′ S., and his daily increase of declination was 23' 9"; what is the correction for the mean of the observations? Ans. 15$.4. 4. On March 27, 1794, in latitude 51° 32′ N., the interval between the observations, when the sun was at equal altitudes, was 7h 29m 55s; the sun's declination was 2° 47′ N., and his daily increase of declination 23′ 26′′; what is the correction for the mean of the observations? 5. In latitude 20° 26' N., the altitude of Aldebaran, before arriving at the meridian, was found to be 45° 20', and, after passing the meridian, to be 45° 10′; the interval between the observations was 7h 16m 35o, and the declination of Aldebaran was 16° 10' N.; what is the correction for the mean of the observations? Ans. 195. 6. In latitude 36° 39' S., the sun's correct central altitude was found to be 10° 40′, when his declination was 9° 27' N.; what was the hour angle? Ans. 4h 36m 9s. 7. In latitude 13° 17' N., the sun's correct central altitude was found to be 36° 37', when his declination was 22° 10' S.; what was the hour angle? Ans. 2h 42m 52s. 8. In latitude 50° 56′ 17′′ N., the zenith distance of a terrestrial object was found to be 90° 24′ 28′′, and its azimuth 35° 47′ 4′′ from the south; what were its polar distance and hour angle è 9. From the preceding terrestrial object, three distances of the sun were found to be 78° 9′ 26′′, 77° 39′ 26′′, and 77° 29′ 26′′, when his declination was 14° 7′ 18′′ S.; what were the sun's hour angles, if he was on the opposite side of the meridian from the terrestrial object? Ans. 2h 45m 49, 2h 43m 26, and 2h 42m 40o. CHAPTER VIII. LONGITUDE. 102. Problem. To find the longitude of a place. First Method. By terrestrial measurement. 1 If the longitude of a place is known, that of another place, which is near it, can be found by measuring the bearing and distance; whence the difference of longitude may be calculated by the rules already given in Navigation. Second Method. By signals. . The stars, by their diurnal motion, pass round the earth once in 24 sideral hours; hence they arrive at each meridian by a difference of sideral time equal to the difference of longitude. In the same way, the sun passes round the earth once in 24 solar hours; so that it arrives at each meridian by a difference of solar time equal to the difference of longitude. The difference of longitude of two places is, consequently, equal to their difference of time. Now if any signal, as the bursting of a rocket, is observed at two places; the instant of this event, as noticed by the clocks of the two places, gives their difference of time. Third Method. By a chronometer. The difference of time of two places can, obviously, be determined by carrying a chronometer, whose rate is well ascertained, from one place to the other; and if the chronometer did not change its rate during the passage, this method would be perfectly accurate, Fourth Method. By an eclipse of one of Jupiter's satellites. The signal of the second method cannot be used, when the places |