to the combined influence of several elements, are not altogether uniform; some stars which had once north declination have now south declination, and vice versa. The latitudes also, have a small annual variation. (167.) The time which has been used in the preceding problems is apparent time, and not that which is shewn by a clock or chronometer. It may, perhaps, be proper in this place, to explain the difference. A solar day is the interval of time between two successive noons; it is greater than a revolution of the heavens, which constitutes a sidereal day; for if we imagine the sun and a star to pass the meridian at the same instant, the next day the sun will pass later, in consequence of his proper motion, from west to east. The solar days are not equal, owing to the obliquity of the ecliptic, and the unequal motion of the sun in the ecliptic. Hence we obtain the idea of a mean solar day:—the year being supposed to be divided into as many mean solar days as there are real days. The time measured by a clock so adjusted, is denominated mean or true time. The time shown by the sun is called apparent time; this is that which is used in the Nautical Almanac, and the difference between these is called the equation of time. The equation of time is given in the Almanac for every day in the year. (168.) We have also supposed the heavenly bodies to rise and set when they are in the rational horizon; but they are really visible when they are 33′ below it; this effect, which is called refraction, diminishes from the horizon to the zenith, where it vanishes: it is the principal correction to be applied to the apparent places of the fixed stars. But for the heavenly bodies in our system, another correction called parallax, must be taken into account, to make their calculated places correspond with observation. A popular illustration of this subject may be obtained by considering the difference of situation in a body supposed to be viewed from the surface and from the centre of the earth. This last position alone is the real place of the body with respect to that of our planet (e). (169.) Problem XVI. Given the right ascensions and declinations of two heavenly bodies, or their latitudes and longitudes, to find their distance asunder. Example I. At apparent noon at Greenwich, Sept. 1st, 1831, the sun's longitude was 158° 15′ 16′′, that of the moon 95° 25′ 43′′, and the moon's latitude 3o 48′ 42′′ south; required their true distance. In the spherical triangle ABC, (fig. 34.) let A M represent the place of the sun, B that of the moon, and C the pole of the ecliptic; then we have given the side a=93° 48′ 42′′, the side b=90o, and the included angle C, the difference of longitude,= 62° 49' 33", to determine the side c, the required distance. For this purpose we have the equation, sin 93° 48′ 42′′, log...... 9 9990382 Example II. Required the distance between the fixed stars Aldebaran, and a Lyræ, the right ascension of the former being 4h 26m 21s, and its declination α 16° 10′ N., the right ascension of the latter being 18h 31m 17s, and its declination 38° 37′ 59′′ N. In the spherical triangle ABC (fig. 35.), in which C represents the pole of the equator, A the position of Aldebaran, and B that of a Lyræ; we have given the sides a and b, the north polar distances of the two stars, and the included angle C=148° 46′; to find the opposite side c, the distance between the two stars. To obtain this directly, either of the formulæ in (91.) or (92.) may be used, but as the solution by the latter will give rise to some observations which may be useful to the student, we shall employ it. We have, tan cos C. tan b, cos c= which are thus computed, cos b. cos (a—◊), cos cos 148° 46', log......... 99319980 tan ... 10.0293237 The number corresponding to this logarithmic tangent, is 46° 55′ 58′′; but since cos C is negative, we must consider tan negative, and therefore we may take for the value of , either-46° 55′ 58′′, or 133° 4′ 2′′ (9.) or (13.), the solution on either hypothesis being the same. Adopting the latter value we have (a−q) = -59° 14' 2"; and we finish the computation as follows: cos 59° 14′ 2′′, log...... 9 7088751 cos 51° 22′ 1′′............ 9.7954145 Now since cos is negative, cos c is also negative; therefore the number answering to this logarithmic cosine must be the supplement of that in the tables; hence we have for the value of c the arc 117° 53′ 5′′. By this problem the lunar distances in the Nautical Almanac are calculated. The right ascension and declination of the moon being obtained from the lunar tables, and those of the nine fixed stars, viz. a Arietis, Aldebaran, Pollux, Regulus, Spica m, Antares, a Aquila, Fomalhaut, and a Pegasi, whose situations are convenient to show the rapid motion of the moon, are determined by observation with more than ordinary care, the annual variations |