2. For the longitude. June 7, 1795. Lun. dist. from the sun at midnight 119° 27' converted into degrees, is equal to 75° 5′ 45′′ NOTE. The time of the observation, if not shewn by a time keeper, may be calculated from the latitude of the place, the sun's altitude and declination. 5 o 23, which, the lon PROBLEM PROBLEM XXVI. Given the latitude of the place, the obliquity of the ecliptic, the sun's longitude, and the hour; to find the angle, formed by the ecliptic and horizon, or the height of the nonagesimal degree, the ascending point of the ecliptic, the point of the nonagesimal degree, the azimuth of the ascending point, the culminating point, its altitude, &c. tic, or the nonagesimal degree, and ZNG a vertical circle, In the right-angled spherical triangle SB, the data are the sun's longitude S, and the obliquity of the ecliptic SYB, to find the sun's right ascension B. Hence, the hour being given, the distance of the meridians, or the arc of the equinoctial BE, is known, and consequently its complement Bo. And hence YoYB -Bo is determined. Therefore in the oblique spherical triangle Avo, the data are Avo, the obliquity of the ecliptic, Aov, the complement of the latitude, and Yo, to find the rest. 1. The angle Ao, and consequently its supplement SAo, which is the angle, formed by the ecliptic and horizon; or, which the same, the arc NG, the height of the nonagesimal degree. 2. The 2. The side Av, and consequently the point A, which is the descending point of the ecliptic, and the opposite or ascending point. 3. The side Ao, which is the azimuth of the descending point in the west, or of the ascending point in the east. By the addition of AN, or 90°, to Av the arc YN is found, or N the highest point of the ecliptic, or point of the nonagesimal degree. Since AG and oO are both quadrants, GOAo; therefore GO is known, that is, the azimuth of the nonagesimal degree from the south. Ao+o0AO. Then, in the right-angled spherical triangle AOM, the data are the side AO and the angle MAO, to find the hypotenuse AM. YA+AM❤M; whence the culminating point M of the ecliptic is known. And MO the altitude of the culminating point of the ecliptic, together with the AMO, formed by the ecliptic and meridian, may also be found from the same data. Or, in the right-angled spherical triangle YEM, VE and EM are given, to find M, ME, and ME. Whence MO is known. In the right-angled spherical triangle ZSN, the sides ZN, NS, are given; for ZN is the complement of NG, and NS is the complement of AS, or of S—YA. Whence the angle ZSN may be found, that is, the angle, formed by the ecliptic and vertical circle passing through the centre of the sun; and ZS, the sun's zenith distance. These may also be found in the oblique spherical triangle 2MS. For ZM 90-MO, ZMS=180°-AMO, and MSMS, are given; to find ZSM, and ZS. COR. Hence may be found the angle ZIF, formed by the vertical circle ZS and the small circle FIK, parallel to the ecliptic. From From the angular point I draw IL perpendicular to MSA. Then in the right-angled spherical triangle ISL, the data are the angle ISL, found before, and the side IL, the distance of the parallel from its great circle; to find the angle SIL, the complement of KIS or ZIF. NOTE. The right ascension of the meridian, medium Cœli, or mid-heaven, at any given time, is equal to the apparent time, reckoned astronomically in degrees, added to the sun's right ascension. Or, the mean longitude of the sun, added to the mean time in degrees, is equal to the right ascension of the midheaven. FINIS. ERRATA. N. B. N. signifies notes at the bottom, 1. line, and b. from the bottom of the text, or notes, according as the erratum is in the text or notes. Page 1. VOLUME FIRST. 13 1 N. For easily read easy. 15 The braces in the table fhould be inverted. N. 3d paragraph b. Dele the commas after n, n—1, 7-2, 1-3; and the apostrophes between o and s, 11, 10 and 9 b. Dele —1 after n; and for 2, 3, after y 185 After shillings insert sterling. 40 5 and 3 b. For make one read one maker. 41 11 b. After 18s. insert sterling. 42 2 For 9grs. read 3dwt. 18grs. 19 For 8" read 48". 58 5 b. For 6f. z′ read 3f. 2′. 64 11 For the numerator 30 read 36. 66 4 and 5 For o in the numerators read 9. 715 For s. and d. read d. and q. 19 For 10oz. read 1 oz. 73 8 b. Dele 9 at the end. 90 5 b. For read 4. 98 2 b. N. 90 read og. 101 13 For 16 read 16. 202 8 For 6 read 4. 1 VOL. II. 210 Exce |