tude (Table IV.), and the remainder will be the moon's correction, which, added to the apparent altitude, will give the true altitude. EXAMPLE I. On the 18th of July 1840, in longitude 113° 10′ W. at 20h 45m 30s, if the observed altitude of the moon's lower limb be 45° 22′ 3′′, error of the quadrant 58′′ subtractive, and the eye 21 feet above the level of the sea; what is the true central altitude? Time at the given place 20h 45m 30° = 7 32 40 W. 28 18 10 24 Time at Greenwich on the 19th of July 4 18 10 = D's semidiameter at noon, 19th 15 30", Horizontal parallax 12: 18" :: 4h 18m 10s: Horizontal parallax at noon Horizontal par. at 4h 18m 10=56_57 60 In seconds=3417 Cos's app. alt. 45°32′ 25′′ = 9.84535 Parallax in altitude 2393′′log=3.37899 45 36 47 Then 2393" = -4 22 D's correction App. altitude)'s centre = 45 32 25 38 57 2. On the 22d of April, 1840, in longitude 105° E. of Greenwich, at 13h 30m, if the observed altitude of the moon's upper limb be 23° 48′, and the eye 18 feet above the level of the sea; required the moon's true central altitude. The moon's semidiameter and horizontal parallax at noon [Naut. Alm.), being 14′ 49′′ and 54′ 21′′; and at midnight, 14′ 51′′ and 54'′ 30′′. Answer. The time at Greenwich is 6h 30m; D's aug. semidiameter 14′ 57′′, horizontal parallax 54′ 26′′; apparent altitude of the D's centre 23° 29′ 0′′; the D's correction 47′ 45′′, and the true altitude of her centre 24° 16′ 45′′. PROBLEM XIII. (566) Given the sun's meridian altitude, to find the latitude of the place of observation. RULE. Reduce the sun's declination to the meridian of the place of observation * (558). Subtract the sun's corrected altitude (564) from 90°, the remainder is the zenith distance. If the sun be south of the observer when the altitude is taken, call the zenith distance north; but, if north, call it south. Then, if the zenith distance and declination have the same name, their sum is the latitude; but, if they have contrary names, their difference is the latitude, and it is always of the same name with the greater of the two quantities. EXAMPLE I. On the 17th of October, 1840, in longitude 52° W., suppose the meridian altitude of the sun's lower limb to be 28° 40', the observer at sea, the sun to the south of him, and the eye 14 feet above the surface of the water. Required the latitude of the place of observation. * The same rule, with a little variation, will answer for a planet; and for the moon, correcting for the declination as at (560).-See Dr. Mackay's Navigation, 1st edition, page 151-155. 2. On the 18th of March, 1840, in longitude 106° E., suppose the meridian altitude of the sun's lower limb, when north of the observer, to be 74° 56', and the height of the eye 16 feet above the level of the sea; required the latitude. The sun's semidiameter (Naut. Alm.) being 16' 5"; the declination at noon on the 17th 1° 11′ 37′′ S., and on the 18th 0° 47′ 56′′ S. Answer. The sun's true altitude = 75° 8' 3"; time at Greenwich 16h 56m on the 17th, sun's reduced declination 0° 54′ 54′′ S., latitude 15° 46′ 51′′ S. 3. On the 28th of September, 1840, suppose the meridian altitude of Arcturus to be 36° 18′ north of the observer, and the eye 20 feet above the level of the sea; required the latitude. Answer. 33° 46′ 41′′ S. CHAP. XI. THE APPLICATION OF RIGHT-ANGLED SPHERICAL TRIANGLES TO ASTRONOMICAL PROBLEMS. (567) The celestial sphere is represented by Plate III. Fig. 1. 1. Let the circular space, South, Zenith, North, Nadir, represent the brazen meridian* of a celestial or terrestrial globe, having its north pole elevated above the horizon. 2. Imagine the globe to be cut in halves by the brass meridian, and the semi-globe to be of transparent glass with the circles of the sphere drawn on it. Now if a sheet of paper be put upon the section, and a light be placed in the point aries (the eye being in libra), the shadows of all the most useful circles of the sphere will form a plane figure similar to Figure 1. Plate III. 3. NAS will present the axis of the globe, N the north pole, s the south pole. 4. EAQ the equator. 5. Ho the horizon. 6. Zenith A the quadrant of altitude screwed on the zenith, and passing through aries. Or, Zenith ▲ Nadir, the prime vertical passing through aries A. 7. WOAO the ecliptic, n its north pole, m its south pole. It is here presumed that the learner has some knowledge of the globes, and of the circles described thereon. It would be a very good exercise to solve the succeeding problems both by the globes and by calculation. 8. the tropic of cancer, or any parallel of the sun's north declination. 9. vvs the tropic of capricorn, or any parallel of the sun's south declination. 10. On the elevation of the north pole above the horizon, equal to Æ zenith, the latitude of the place. 11. HE the elevation of the equator above the horizon, equal to zenith N, the complement of the latitude. 12. TOW a parallel circle 18 degrees above the horizon, or the boundary between twilight and dark night. 13. Zenith SEC, Zenith ad, &c. azimuth, or vertical circles. 14. NSDS, NGO cws, &c. meridians or circles of terrestrial longitude. 15. nvxm, ncvm, &c. circles of celestial longitude. NOTE. The several triangles in this general figure, and the lines which form them, are explained at the head of each problem, to which the triangle is applied. PROBLEM I. (Plate III. Fig. 1.) (568) Given the obliquity of the ecliptic and the sun's longitude, to find his right ascension and declination. In the right-angled spherical triangle AGO, or ABO. 1. A sun's longitude, an arc of the ecliptic from aries. 2. AG, or AB sun's right ascension, an arc of the equinoctial; the point в always comes to the meridian with the sun. 3. GO, or Bosun's declination, an arc of a meridian passing through the sun's place. This arc likewise comes to the meridian with the sun. 4. GAO, or BA=the obliquity of the ecliptic, or the angle formed between the equinoctial and the ecliptic. 5. AG, or AB the angle formed by the ecliptic, and the meridian passing through the sun's place. Any two of these five quantities being given, the rest may be found. EXAMPLE. On the 17th of May, 1840, the sun's longitude at mean noon is 56° 35′ 20′′; required his declination, right ascension, and the angle formed between the ecliptic and the meridian passing through the sun's place, the obliquity of the ecliptic being 23° 27′ 44′′. 1. To find the sun's declination, in the triangle ABO. * rad x sin BO=sin BAO X sin AO. .. log sin BA+log sin A-10=log sin B. or, rad sin A= 56° 35′ 20′′:: sin BA 23° 27′ 44′′ : sin BO 19° 24' 40". 2. To find the sun's right ascension, in the triangle AB○• * rad x cos BA=cot AOX tan AB. .. 10+log cos BAO-log cot A=log tan AB. or, cot A56° 35′ 20′′: rad :: cos BA=23° 27′ 44′′ : tan AB=54° 16′ 46′′=3h 37m 7o•1. 3. To find the angle AOB between the ecliptic and meridian+. * rad x cos Acot BAG X cot AOB. .. 10+log cos AO-log cot BAO=log cot AOB. or, cot BA=23° 27′ 44′′: rad: : cos A = 56° 35′ 20′′ : cot AB=76° 33′ 31′′. (569) Astronomers reckon the sun's longitude and right ascension from aries quite round the globe. Hence, though at equal distances from the equinoctial points, aries and libra, the sun may have the same quantity of declination and of the same name, viz. from aries to libra, north; and from libra to aries, south; yet the longitudes and right ascensions differ materially at these points where the declinations agree. Hence, If the sun be in the second quadrant of the ecliptic, subtract the longitude from 180°; use the remainder to find the right ascension; and take the right ascension, when found, from 180°. If the sun be in the third quadrant, subtract 180° from his longitude; use the remainder to find his right ascension; and add 180° to the right ascension when found. If the sun be in the fourth quadrant of the ecliptic, subtract his longitude from 360°; use the remainder to find the sun's right ascension; and take the right ascension, when found, from 360°. In the first quadrant of the ecliptic between the 20th March and 21st June, the sun's declination is north, and increasing; and in the third quadrant, between the 22nd September and 21st December, the sun's declination is south and increasing. In the second quadrant of the ecliptic, from 21st June to 22nd September, the sun's declination is north, and decreasing; and in the fourth quadrant, from 21st December to the 20th March, the sun's declination is south and decreasing. PRACTICAL EXAMPLES. 1. On the noon of the 17th of May, 1840, the obliquity of the ecliptic=23° 27' 44", the sun's declination=19° 24' 40" N. and increasing; required his longitude, right ascension, and + The finding of this angle is of no other use than as an exercise of right-angled spherical triangles. |