Remark. If the equated or mean times of the sun's rising and setting be required, then, to the apparent times, found as above, apply the reduced equation of time, as directed in Problem I., page 415; and the result will be the mean times of that object's rising and setting. PROBLEM II. Given the Latitude of a Place, and the Height of the Eye above the Level of the Horizon; to find the apparent Times of the Rising and the Setting of a fixed Star. RULE. Compute the apparent time of the star's transit, or passage over the meridian of the given place, by Problem XII., page 317; then, To the aggregate of 90 degrees, the horizontal refraction,† and the dip of the horizon, add the star's polar distance and the co-latitude of the place of observation: take half the sum; the difference between which and the first term, call the remainder. Now, to the logarithmic co-secants, less radius, of the polar distance, and the co-latitude, add the logarithmic sines of the half sum and of the remainder half the sum of these four logarithms will be the logarithmic co-sine of an arch; which, being doubled and converted into time, will be the star's semi-diurnal arc, or half the time of its continuance above the horizon; which is to be reduced to apparent solar time, by subtracting therefrom the proportional part corresponding to it and the variation of the sun's right ascension for the given day: this is done by Problem V., page 298. Now, the apparent semi-diurnal arc, thus found, being applied by subtraction and addition to the apparent time of the star's transit over the given meridian, will give the respective apparent times of its rising and setting at that meridian: these may be reduced to the mean times of rising and setting, by Problem I., page 415. Example 1. Required the apparent times of the rising and setting of the star a Arietis, January 1st, 1824, in latitude 50:48: N., and longitude 30:0. E.; the height of the eye above the level of the sea being 16 feet? This is the star's distance from the zenith when its centre is in the horizon. + The horizontal refraction of a celestial object is 33 minutes of a degree. The fixed stars have no sensible parallax. *'s dec., red. to given day, 22:37:33 "N., & its R.A.=1:57:16: Sun's right ascension at noon of the given day = 18.43.58 Approx. time of star's transit over the meridian = 7:13 18.. 7:13 18: Longitude of the given place 30:0: E., in time=2. 0. 0 Reduc. of trans. aus. to 5:13" 18: &4" 24: the var. of sun's R.A.= Apparent time of star's transit over merid. of the given place = 712" 21: 90 degrees + 33 +3:50 = 90:38:50% Star's north polar distance = 67.22.27 Co-latitude of the given place=39.12. 0 Log. co-secant=0.034781 61: 0:58 Log. co-sine = 9.685350 Star's semi-diurnal arc = 122: 1:56%, in time = 8*6#38: Star's apparent semi-diurnal arc = 7.12.21 Apparent time of the star's rising past noon, Dec. 31st, 1823, 23 543: Apparent time of the star's setting past noon of the given day 15:18:59: Example 2. Required the apparent times of the rising and setting of the star Sirius, January 1st, 1824, in latitude 40:30: N., and longitude 120:0: W.; the height of the eye above the level of the horizon being 46 feet? *'s dec., red. to given day, 16:28:53"S., & its R.A.=6'37" 23! Approximate time of star's transit over the merid. Corresponding time at Greenwich = 18.43.58 11:5325: 11:53:25: + 8. 0. 0 19:5325: Reduc.of trans. ans. to 19:53" 25:& 4" 24! the var. of sun's R.A.= Apparent time of the star's transit over the given meridian 11'49" 46: Prop. part of var. of sun's R. A. 424 answering to 5:511 = 54 15: Star's apparent semi-diurnal arc = = Apparent time of the star's rising past noon of the given day 6:45:31: Apparent time of the star's setting past noon of the given day=16:54" 1: See Example 2, page 129; and, also, the Example, page 130. PROBLEM III. Given the Latitude of a Place, and the Height of the Eye above the Level of the Horizon; to find the apparent Times of a Planet's Rising and Setting. RULE. Compute the apparent time of the planet's transit over the meridian of the given place, by Problems X. and XI., pages 313 and 315; reduce this time to the meridian of Greenwich, by Problem III., page 297; to which let the planet's declination be reduced, by Problem VII., page 307; then, To the logarithmic tangent of the latitude add the logarithmic tangent of the planet's reduced declination, and the sum (abating 10 in the index,) will be the logarithmic sine of an arch; which, being converted into time, and added to 6 hours when the latitude and the declination are of the same name, but subtracted from 6 hours when of contrary names, the sum or difference will be the planet's approximate semi-diurnal arc, or half the time of its continuance above the horizon. Let this time be applied, by subtraction and addition, to the apparent time of transit; and the approximate times of the planet's rising and setting will be obtained. Reduce the approximate times of rising and setting, thus found, to the correspondent times at Greenwich, by Problem III., page 297; to which times, respectively, let the planet's declination be reduced, by Problem VII., page 307; then, To the aggregate of 90 degrees,* the horizontal refraction,† and the dip of the horizon, diminished by the planet's horizontal parallax,‡ add the planet's polar distance at the approximate time of rising or setting, and the co-latitude of the given place: take half the sum; the difference between which and the first term, call the remainder. Now, to the logarithmic co-secants, less radius, of the polar distance and the co-latitude, add the logarithmic sines of the half sum and of the remainder half the sum of these four logarithms will be the logarithmic co-sine of an arch; which, being doubled and converted into time, will be half the time of the planet's continuance above the horizon, or its semidiurnal arc. Find the proportional part of the variation of the planet's transit over the meridian, answering to half its continuance above the horizon, by Problem XI., page 315, in the same manner as if it were the reduction of transit to a different meridian that was under consideration. Now, this proportional part being added to half the time of the planet's continuance above the horizon when the planet's transit is increasing, but subtracted therefrom when decreasing, the sum or difference will be the planet's apparent semi-diurnal arc; which being applied by subtraction to the apparent time of transit, the remainder will be the apparent time of the planet's rising. In the same manner let the apparent semi-diurnal arc for the time of setting be computed; which, being added to the apparent time of transit, will give the apparent time of the planet's setting. The apparent times of rising and setting, thus found, may be reduced to the mean times of rising and setting, if necessary, by Problem I., page 415. Example 1. Required the apparent times of Jupiter's rising and setting, January 4th, 1824, in latitude 36: N., and longitude 135: W.; the height of the eye above the level of the horizon being 23 feet? The apparent time of Jupiter's transit over the meridian of the given This is the zenith distance of a planet when its centre is in the horizon. + The horizontal refraction of a celestial object is 33 minutes. For the parallaxes of the planets, see page 326. place is 112030; his declination, being reduced to this time and the given longitude, is 23:18:55" north. Latitude of the given place = 36° 0′ 0′′N. Log. tangent 9. 861261 Jupiter's reduced declination=23. 18.55 N. Log. tangent = 9.634461 Arch, converted into time, = 11259; this, being added to 7:1259: 6 hours, gives the approximate semi-diurnal arc = Apparent time of Jupiter's transit over the given meridian. 11. 20.30 The longitude, in time, being applied to those times by addition, because it is west, shows the approximate time of the planet's rising at Greenwich to be 13731 past noon of the given day, and that of its setting 3:33:29: past noon, January 5th. The declination reduced to these times respectively, is 23:18:46 N. at the time of rising, and 23:19:4" N. at the time of setting. To find the apparent Time of Rising : which, answering to 7:16:32, is = Apparent semi-diurnal arc = Apparent time of Jupiter's transit = 54:34 3 Log. co-sine = 9.763235 109: 8 6", in time = 7:16:32: Variation of transit=30" decreasing; the proportional part of If the transit had been progressive, or increasing, the proportional part would be additive. |