30. Angle of position is the angle at a star, formed by two great circles, one passing through the pole of the equinoctial and the other through the pole of the ecliptic. 31. Hour of the day, or hour angle, is the arc of the equinoctial, contained between the meridian of the place and the star's meridian. 32. Nodes are the points, where the orbits of the primary planets cut the ecliptic, and where the orbits of the secondaries cut the orbits of their primaries. The ascending node is the point, where the planet crosses, moving northward; and the descending the point, where it crosses, moving southward. 33. Two heavenly bodies are in conjunction, when they have the same longitude; in opposition, when the difference of their longitudes is 180°; in quadrature, when the difference is 90°. And syzygy denotes either conjunction or opposition. NOTE. The character is used to denote the Sun, D the Moon, Mercury, Venus the Earth, 8 Mars, 2 Jupiter, Saturn, Herschel, a fixed star ;-6 Conjunction, & Opposition, □ Quadrature ;- the ascending Node, the decending Node. PROBLEMS. PROBLEM I. Given the sun's longitude, and the obliquity of the ecliptic; to find the sun's right ascension, and declination. EXAMPLE. Given the sun's longitude 1° 11° 1' 44", and the obliquity of the ecliptic=23° 28'; required the sun's right ascension and declination. points P, O, S, draw a circle, which will be a circle of right ascension. 10 CALCULATION. In the right-angled spheric triangle YOB, YO=41° 1′ 44", ZOYB=23° 28'. 2. To find the right ascension YB, Rx cos. BYO = cot. VO x tang. B. Hence the right ascension = 38° 35′ 49′′, which, reduced to time, 2h. 34′ 23′′ 16"".* PROBLEM II. Given the obliquity of the ecliptic, and the sun's declination; to find the sun's longitude and right ascension. EXAMPLE. The obliquity of the ecliptic being 23° 28′, and the sun's declination 15° 9′ 12′′ N; required the sun's longitude and right ascension. * Degrees may be converted into time by the following rule: 1. Divide the given degrees by 15 for hours; and multiply the remainder by 4 for minutes. 2. Divide the minutes, seconds, &c. in the same manner by 15 for minutes, seconds, &c. and multiply each remainder by 4 for the next lower denomination. The reason of the rule is, that 15° 1h. and consequently 1° =4', &c. CALCULATION. In the rectangular spheric triangle YOB, BYO =23a 28', and BO=15° 9′ 12′′. NOTE 1. Other Problems may be formed by varying the data in this triangle. Thus, the obliquity of the ecliptic and the sun's right ascension, or the sun's longitude and right ascension, may be given to find the rest. By a reverse operation, time may be reduced to degrees; that is, multiply the hours by 15, to the product add of the minutes, seconds, &c. each being removed one place higher, or toward the left. 16"", Thus, 2h. 34' 33" 16" NOTE 2. While the sun is ascending from Aries to Cancer, its longitude is YO, or the hypotenuse of the triangle YOB. But when the sun has passed the solstitial point of, and is descending toward, its longitude being subtracted from 180°, the remainder is the hypotenuse ; and B is then the supplement of the right ascension. When the sun's longitude is less than 180°, the declination is north. While the sun is decending from the beginning of toward, the excess of its longitude above 180° will be the hypotenuse; and A, the corresponding arc of right ascension, must be added to 180° for the whole right ascension. When the sun has passed the solstitial point of , and is ascending toward P, the longitude being subtracted from 660°, the remainder is the hypotenuse ; and the corresponding arc of right ascension A must be subtracted from 360°, to give the right ascension. While the longitude is between 180° and 360° the declinazion is south. PROBLEM III. Given the latitude of the place, and the sun's declination; to find the sun's altitude and azimuth at 6 o'clock. EXAMPLE. Suppose the latitude of the place to be 42° 23′ 28′′ N, and the sun's declination 23° 27′ 57′′ N; required the sun's altitude and azimuth at 6 o'clock. |