Difference = 20:10:40" . +11.32. . 20:22:12" Apparent altitude of the moon's centre = True altitude of the moon's centre = Example 2. 21:15: 7: In a certain latitude, March 26th, 1836, at 3:30:47: mean time, the observed altitude of the moon's upper limb was 30:17:30, and the height of the eye above the level of the sea 30 feet; required the true altitude of the moon's centre; the longitude of the place of observation being 94:15:30 east of the meridian of Greenwich? Mn. time of observation 3:30:47: || D's reduced semidiameter=14:53" Longitude 94:15:30: Augmentation, Table IV.= + 7 Correction answering to ditto and hor. par. in Table XVIII. = +45. 40 True altitude of the moon's centre PROBLEM XXV. .30:42:55% Given the observed Altitude of a Planet's Centre, to find its true Altitude. RULE. From the planet's observed central altitude* (corrected for index * See Article 64, page 328. error, if any), subtract the dip of the horizon, and the remainder will be the apparent central altitude. Find the difference between the planet's parallax in altitude* (Table · VI.) and its refraction in altitude (Table VIII.); now, this being subtracted from the apparent altitude, the remainder will be the true central altitude of the planet. Note.-The difference between the refraction and parallax constitutes the correction of a planet's apparent altitude in the lunar observations. Example. Let the observed central altitude of Mars be 17:29:40%, the index error 3'45" additive, and the height of the eye above the surface of the water 26 feet; required the true central altitude of that planet, allowing his horizontal parallax to be 14 seconds? Observed central altitude of Mars = Dip of the horizon for 26 feet - 17:29:40% + 3.45 Remark. In taking the altitude of a planet at sea, its centre is to be brought down to the horizon; in this case its semidiameter need not be taken into account. But, if the altitude be taken on shore, by means of an artificial horizon, it is the lower limb of the object, particularly of Venus and Jupiter, whose semidiameters are considerable, that should be brought in contact with the apparent upper limb, seen by reflection in the artificial horizon in this instance, the semidiameter of the planet is to be added to half the altitude observed in the artificial horizon;-this shall be shown presently.-See Explanatory Article 64, page 328, and 67, page 329. : PROBLEM XXVI. Given the observed Altitude of a fixed Star, to find the true Altitude. Rule. To the observed altitude of the start apply the index error, if any; * The parallaxes and semidiameters of the planets are now given in the Nautical Almanac, between pages 359 and 361. + See Article 63, page 327. from which subtract the dip of the horizon, and the remainder will be the star's apparent altitude. From the apparent altitude, thus found, let the refraction corresponding thereto be subtracted, and the remainder will be the true altitude of the star, Note. The refraction constitutes the correction of a star's apparent altitude in the lunar observations. Example. Let the observed altitude of Spica Virginis be 18:30, the index error 3:20 subtractive, and the height of the eye above the level of the water 18 feet; required the true altitude of that star? Observed altitude of Spica Virginis = Dip of the horizon for 18 feet = 18:30: 0" 3.20 4. 4 Note. The fixed stars do not exhibit any apparent semidiameter, nor any sensible parallax; because the immense and inconceivable distance at which they are placed from the earth's surface causes them to appear, at all times, as so many mere luminous indivisible points in the heavens. PROBLEM XXVII. To deduce the true Altitude of a Celestial Object from its DOUBLE Altitude, observed by means of an Artificial Horizon. GENERAL RULE. First. To find the sun's true central altitude, Correct the observed angle of altitude for the index error of the sextant, if any; to the half of which apply the sun's semidiameter by addition, if the lower limb be observed, but by subtraction if it be the upper limb; the result will be the apparent central altitude; from which let the difference between the refraction and parallax corresponding thereto be subtracted (Table VII. and VIII.), and the remainder will be the correct altitude of the sun's centre, Second. To find the Moon's true Central Altitude. Find the moon's apparent central altitude in the same manner as if it were the sun that was under consideration; observing to correct her semidiameter by the equation contained in Table IV. ;-then, to the apparent altitude, thus found, let the correction in Table XVIII. be added, and the sun will be the true altitude of the moon's centre. Third.-To find the true Altitude of a Star. Correct the observed angle for the index error of the sextant, if any; from the half of which subtract the refraction corresponding thereto (Table VIII.), and the remainder will be the star's true altitude. Fourth. To find a Planet's true Central Altitude. Correct the observed angle for the index error of the sextant, if any; to the half of which let the planet's semidiameter be added, and the sum will be its apparent central altitude; from which let the difference between the refraction and parallax (Tables VI. and VIII.) be subtracted, and the remainder will be the correct altitude of the planet's centre. Example. July 24th, 1836, the measure of the observed angle between the lower limb of the planet Venus, and the apparent upper limb thereof, reflected from an artificial horizon, was 33:20:10%, the index error of the sextant was 2:30′′ subtractive; required the true central altitude of the planet; her semidiameter, on the given day, being 28", and her horizontal parallax 30"? Note. It is presumed that the examples to the three preceding problems render any further illustration unnecessary respecting the altitudes of the sun, moon, and stars. PROBLEM XXVIII. To find the Obliquity of the Ecliptic. The obliquity, or inclination of the ecliptic, signifies the angle which is made by the intersection of the planes of the ecliptic and the equinoctial: the two points of intersection are diametrically opposite to each other, or 180: asunder, one being at the first point of Aries, and the other at the first point of Libra. The solstices are those two points of the ecliptic which are equidistant, or 90% from the above-mentioned points of intersection (Definitions 10 and 11, pages 298 and 299). When the sun enters a solstitial point which is on the same side of the equator with any given place, his meridional altitude will be the greatest possible at that place; and when he enters the opposite solstitial point, his meridional altitude will be the least at the same place. From these premises the obliquity of the ecliptic may be readily determined by the following RULE. Let the sun's meridional altitude be carefully observed on the 21st of June, and again on the 22nd of December, or on the days on which that great luminary will be the nearest to the solstices at noon.-Reduce the observed meridional altitudes to the true central altitudes by Problem XXIII., page 374.-Now, half the difference of the true meridional altitudes, thus found, will be the obliquity of the ecliptic. Example. June 21st, 1834, in latitude 50:47:27" north, the meridional altitude of the sun, duly corrected, was 62:40:9"; and on the 22nd December following, it was 15:44:54", required the obliquity of the ecliptic? which is the obliquity of the ecliptic, as required. |