The length of an apparent solar day is variable from two causes: 1st. From the variable motion of the sun in the ecliptic. 2nd. From the motion of the sun being in a circle inclined to the equator. 19. To explain briefly these causes of variation, let us suppose the two circles A Q, A I, to represent the celestial equator and ecliptic, and s s' the arc described by the sun in one day. The angle at P, between the two circles of declination, is measured, not by the arc s s' described by the sun, but by the arc R R' of the equator. Now, Ꭱ R 1st. The velocity or motion of the sun in the ecliptic is variable, on account of the earth moving in an elliptic orbit; it sometimes describes an arc of 57' in a day; at other times the arc described is about 61': this is one cause of the inequality in the length of the solar days. P Fig. 2. Ꭱ Ꭱ 2nd. But even supposing the arcs of the ecliptic described by the sun to be equal, yet the angles at P between the meridians as R P R' (in the three figures) will not be so, since these angles are measured by the arc R R of the equator to which s s' will be differently inclined according to the place of the sun in the ecliptic. At the equinoxes, or when the sun is at A, the arcs s s' and R R will be inclined to each other at an angle of about 23° 27' (see fig. 2). At the solstices they are parallel (see fig. 3). This is the second cause of the inequality. 20. To obtain a proper measure of time, we must proceed therefore as follows: an imaginary, or as it is called a mean sun, is supposed to move uniformly in the equator with the mean velocity of the true sun. A mean solar day may therefore be defined to be the interval between two successive transits of the mean sun over the same meridian. It begins when the mean sun is on the meridian. To find the daily motion of the mean sun in the equator. 21. The mean solar year, or the time the sun takes to return again to the first point of Aries, has been found to be equal to 365d.2422. Let us suppose the mean sun to describe the equator in this time, then we shall find its daily motion in the equator as follows:-Let x = daily motion, 365d.2422: 1a :: 360°: x = 0°·9856472 = 59′ 8′′·33 or, the mean sun's daily motion in the equator from west to east is 59' 8"-33. To find the arc described by a meridian of the earth, in a mean solar day. E 22. Let P A M P' represent the meridian of a spectator a, drawn in some plane; as, for instance, that of the paper; E Q the celestial equator which must therefore be supposed at right angles to the paper. Sup m m M pose the mean sun to be at m on the meridian of ▲, and therefore in the plane of the paper; and let m m' be the arc of the equator described by the mean sun in one day, namely, 59′ 8′′-33. Now, let the earth be supposed to revolve about P P', from west to east, until the meridian again passes through the mean sun, which has arrived at м'. Then the whole number of degrees described by the meridian of the spectator will evidently be one complete revolution, or 360° (by which it is again brought into the plane of the paper), together with the arc м M' m m', or 59' 8"-33. Hence in a mean solar day a meridian describes 360° 59′ 8′′-33. Sidereal time, apparent solar time, and mean solar time. 23. Sidereal time is the angle at the pole of the heavens m 4 Q between the celestial meridian and a circle of declination passing through the first point of Aries, measuring from the meridian westward; thus, if PQ be the celestial meridian, A the first point of Aries, a Q the equator, then the angle Q.PA is sidereal time. 24. Mean solar time is the angle at the pole between the celestial meridian, and a circle of declination passing through the mean sun, measuring from the meridian westward: thus let m be the mean sun considered as a point, then QPm is mean solar time. Similarly, if s be the place of the true sun in the ecliptic A 1, the angle Q P S measured from P Q westward is apparent solar time. 25. The equation of time is the difference in time between the places of the true and mean sun: thus, the angle m PS is the equation of time. Sidereal clock, and mean solar clock. 26. A sidereal clock is a clock adjusted so as to go 24 hours during one complete revolution of the earth; that is during the interval of two successive transits of a fixed star: or supposing the first point of Aries to be invariable between two successive transits of the first point of Aries. A mean solar clock is a clock adjusted to go 24 hours during one complete revolution of the mean sun; or while a sidereal clock is going 24h 3m 56.555. 65 CHAPTER III. INTRODUCTORY OR PRELIMINARY RULES IN NAUTICAL ASTRONOMY. Nautical day and Astronomical day. 27. The nautical or civil day begins at midnight and ends the next midnight. The astronomical day begins at noon (Art. 20) and ends at noon, and is later than the civil day by 12 hours. Again, in the astronomical day the hours are reckoned throughout from Oh to 24h; in the nautical day there are twice 12 hours, the first 12 hours being before noon, or before the commencement of the astronomical day (denoted by A.M., ante meridiem); the latter are afternoon, and distinguished by the letters P.M. (post meridiem.) Rule I. Given civil or nautical time at ship, to reduce it to astronomical time. 1. If the nautical time at ship be P.M., it will be also astronomical time, P.M. being omitted. 2. If the nautical time be A.M., add 12h thereto, and put the day one back; thus (1.) April 27, at 4h 10m P.M. (civil) is April 27, at 4h 10m (astro.) (2.) April 27, at 4 10 A.M. (civil) is April 26 at 16 10 (astro.) EXAMPLES. Reduce the following civil or nautical times to astronomical times. Rule II. Given astronomical time at the ship, to reduce it to civil or nautical time. 1. If the astronomical time is less than 12 hours, it will also be nautical time P.M. 2. If the astronomical time be greater than 12 hours, reject 12 and put the day one forward; the result will be civil time A.M.; thus (1.) April 27, at 4h 10m (astro.) is April 27, at 4h 10m P.M. (civil). (2.) April 27, at 16 10m (astro.) is April 28, at 4 10 A.M. (civil). EXAMPLES. Reduce the following astronomical times to nautical or civil times. Given the time at Greenwich, to find the time at the same instant at any other place, and the converse. 28. To find the time at any place, as Greenwich, corresponding to a given time at any other place, or the converse, we must remember that since the earth revolves through 360° in 24 hours, from west to east, or 15° in 1 hour, and therefore through 1o in 4 minutes, or l' of arc in 4 seconds of time, at a place 15° to the eastward of a spectator the sun will be on the meridian 1 hour before, and at a place 15° to the westward, the sun will be on the meridian 1 hour later than at the place of the spectator: hence, when it is 10 o'clock at a given place, it will at the same instant be 11 o'clock at a place 15° to the eastward, and 9 o'clock at a place 15° to the westward. If, therefore, the longitude of a place is known, that is, the number of degrees it is to the east or west of Greenwich, we can readily tell what time it |