12 Hours, is 2 Hours, 57' 44", the true Time of the Break of Day. And thus it begins and ends on the 17th of August. In this Manner you calculate the fame Things, when the Sun has South Declination. To find when the shortest Twilight happens in all the Year, is a Problem of a distinct Nature from this above ; and is very abstruse and perplexed in its Demonstration, as may be seen by the Inquisitive in the Works of Gregory, Keil, &c. However I shall give the Analogy for finding it in any Latitude, as suppose this of Chichester 50° 56', and it is as follows. As Radius 10.0000000 Is to the Sine of the Latitude 50° 56' 9.8900929 So the Tang. of half the Depression 9° 00' 9.1997125 To the Sine of the Sun's Declination 7e 3' 9.0898054 Now the Times of the Year, when the Sun is upon that Parallel of Declination are March the 28th, and Odober the ist, and on those two Days the Twilight is shortest of all, PROBLEM XX, Given the Latitude and Longitude of a Star or Planet, to find its Declination. Practice. Pratice. Admit the Longitude of the Goat P Star, called Capella, 'A be I 18° 2' 30", and its Latitude 220 51' 47" North ; I o demand its Declination? Having constructed the Scheme from what is given, there will be formed the S Oblique Triangle CPE, in which there is known the Side P E = 23° 29', constantly ; the Side C E= 67° 08' 13", the Co-Latitude ; and the included Angle CEP = 11° 57' 30", the Complement of the Star's Longitude from Aries r ; to find the Side PC = the Complement of its Declination. Let fall the Perpendicular P R ; then by Case 4 of Oblique Triangles, Variety 1, thus fay; As Radius 10.0000000 Is to the Co-fine of PEC= 11° 57' 9.9904699 So is the Tangent of P E = 23° 29' 9.6379563 To the Tangent of ER= 23° 02/ 9.6284.262 VI ; Then CE-ER=CR= 44° 06'; Whence fay; As the Co-line of R E=23° 02' Co. Ar. 0.0360313 Is to the Co-fine of RC=449 061 9.8562088 So is the Co-sine of P E= 23° 29' 9.9624527 To the Co-fine of PC = 44° 16' 9.8547428 VOL. II. Kk Whose I P Whose Complement is 45° 42! = A C the Declination North, as was required. Example 2. Let it be required to find the Declination of Fomebant; whose Longitude is more 29° 59' 501, and Latitude 21° 4' 51" South. According to these Z Data, let the Scheme be prepared, and there will be forınid d A the Oblique Triangle CEP, in which there is given the Side P E=230 29" ; the Side EC=INIO 454" ; and the Angle included PEC= 1 200 00'; to find the Side PC, which is done as before, by N letting fall the Perpendicular SR, in the Complemental Triangle C Se. For therein is given the Side Se = P E = 23° 29'; the Side Ce= Complement of Ec to 1800 = 68° 55' 6" ; and the included Angle SeC = 60° 00'; to find the Side S C, or its Complement CA = the Declination South. Thus ; As Radius 10.0000000 Is to the Co-line of Se C. = 60° oo! 9.6989700 So is the Tangent of Se = 23° 29' 9.6379563 To the Tangent of e R = 12° 15' 9.3369263 Then Then CeeR=CR= 56° 401; then say ; As the Co-line of ė R=12° 15' Co. Ar. 0.0100023 Is to the Co-sine of RC= 5640' 9.7399748 So is the Co-line of Se = 23° 29' 9.9624527 To the Sine of C A = 31° 02' 9.7124298 the Declination sought, and is South. PPOBLEM XXI. Given the Longitude and Latitu le of a Star er Planet, to find its Right Ascension ? PraEtice. The Right Ascension is required of the Goat Star Capella, whose Longitude is I 18° 02'30", and its Latitude 220 57' 4711 North. In the first Schenie to the last Problem, in the Oblique Triangle C P E, there is given ( as is there fpecified ) the Sides P E and C E, and the included Angie at E ; to find the Angle C P E, or its Complement Æ P A = the Complement of the Right Afcension to 90°. This I shall do by the Method without a Perpendicular, by Prob. 1, and 2. Thus; СЕ+ЕР 45° 18'C. A. 0.1482529 2 CE-EP Is to the Sine of = 21° 50' 9.5704355 2 So is the Co-Tangent of PEC = 5° 58' 10.9809169 Р- С To the Tangent of 7842' 10.6996053 2 VOL. II. Kk 2 Then Then say again by Prob. 2. As Co-fine of Sum of Sides 45° 18' C. A. 0.1528c09 Is to the Co-line of their Differ. 21° 50' 9.9676741 So is Co-tang. of 1 inclu. Ang. E=5° 58' 10.9809169 P+C To the Tangent of = 85° 28' - 11.1013919 2 Now to the half Sum 85 28 Add the half Difference 78 42 The Sum is 164 10 = the Angle CP E. From which deduct 90 The Remainder is Av = 74 10 = the Right Alcension from Aries, as was required. In this Manner also the Right Ascension of Fomabant (in the second Scheme to the foregoing Problem ) will be found 3400 35' from Aries. PROBLEM XXII. Given the Right Ascension and Declination of a Star or Planet, to find its Longitude and Latitude. Practice. This Problem is the Converse of the two last exactly ; For admit the Right Ascension of the Goat Star Capella be 740 10', and its Declination 45° 42' North ; then in the Triangle C P E, there is given the Side P E = 23° 29'; the Sides C P = 44° 16', the Co-Declination; and the included Angle at P = 164° 10' = Right Ascension from Capricorn, to find the Angle at E = the Longitude ; and C E = the Co-Latitude of the Star. From |