250 Fig. XLI. EXAMPLE. But, The Head of Andromeda (at London) fets Achronically with XLII. 26 d. 46 m. of r: - Cofmically, with 26 d. 46 m. of. XLIII. the Angle r D Oc. is found to be 121 d. 32 m. and its CompleXLIV. ment to 180 d. the Angle R D O, 58 d. 28 m. Wherefore, the Head of Andromeda fets Heliacally, when the Sun is in 11 d. The Latitude of a Place, the Oblique. Afcenfion or Defcenfion of a Star, and the Longitude of the Sun being given: To know whether a Star may be feen Heliacally. 1. By the Cœleftial Globe. O perform this by the Globe, fee before Prob. XXXI. II. By Trigonometrical Calculation. This is but the Converse of the Two foregoing Problems: And the Spherical Triangles upon the Globe will be the fame; and therefore, in the Triangle R AO or R DO There is given, The Oriental Angle at the Point A, or the Occidental at the Point D: The one being found in the Triangle Or. A, or Or. A, as in the Problem: The other in the Triangle r Oc. D, or Oc. D, as in the Problem. 2. The Side A O, or DO; the Arch which is intercepted between the Longitude of the Sun, and the Afcending or Defcending Point of the Ecliptick. To find out the Arch R O. Which Arch, if it be greater, or equal, to the vifible Arch of the Given Star, that Star doth appear: But, if it be lefs, it lyes hid under the Sun's Beams. [Note.] Sometimes the Oriental Angle is lefs than the Arch of Vifion; and then (even at Midnight) fuch a Star cannot be seen. But, if you fubftract the Arch of Vifion from the greateft_Depreffion of the Equator under the Horizon (which is the Complement of the Latitude (or Elevation of the Pole at London 38 d. 30 m.) there remains the Northern Declination of a Degree of the Ecliptick, unto which the Sun muft first come, before that Star will begin to appear at Midnight. As Fig. As for Example. All Stars of the Fifth Magnitude, have an Arch of Vifion of 16 d. therefore fubftract 16 from 38 d. 30 m. XLI. the Depreffion of the Equator, and there will remain 22 d. XLII. 30 m. the Northern Declination of the Sun, viz. as much as is XLIII. required that it be depreffed 16 d. below the Horizon; that the XLIV. Stars of the Fifth Magnitude may appear. But the Sun hath as great Declination in 13 d. 26 min. of Gemini II, and in 16 d. 33 m. of Cancer : Therefore, whilft the Sun goes through that Arch of the Ecliptick (which will be between the 24th of May, and the 28th of June) the Stars of the Fifth Magnitude (are not vifible at London (even at Midnight) because the Twilight hinders. Here needs no Canons for Calculation. PROB. XXXIV. The Elevation of the Pole, and the Longitude of the Sun: To find the Beginning of the Morning, and End of the Evening, Twilight. I. By the Coeleftial Globe. N this and the following Problem, the Situation of the Sun is given, in respect of the Equator, and its Situation is fought for in refpect of the Meridian. But, because, at every Ceffation of the Twilight, there is commonly required a Depreffion of the Sun, of 18 d. below the Ho rizon. Now, in the Latitude of 48 d. 30 m. the Sun being in the beginning of Cancer, at Midnight, is deprefs'd below the Horizon, precifely 18 d. Hence it follows, that in Places, having a greater Elevation of the Pole than 43 m. the Sun being in the Tropick, muft (at Midnight) be depreffed lefs than 18 d. from whence it follows, that Twilight will neither Begin nor End, but will laft all Night. Therefore, to know when this Problem is in Ufe, fubftra&t 18 d. from the greatest Depreffion of the Equator, (that is, at London, 38 d. 30 m.) and there remains the Northern Declination 20 d. 30 m. (as by the Problem converfed): So the Sun, when he hath 20 d. 30 m. of North Deciination, which he will have when he is in 1 d. 20 m. of Gemini II, and in 28 d. 40 m. of Cancer: Therefore, whilft the Sun goes through that Arch of the Ecliptick (which is between the 12th Day of May, and P p the Fig. XLV. the 11th Day of July) the Twilights lafts all Night at London, all which Time there is no Ufe of this Problem there. At other Times of the Year it will be in Ufe, and may be refolved as followeth.. II. By Trigonometrical Calculation. Upon the Globe, the Triangle refolving this Problem, is the Oblique-angled Spherical Triangle PNO in Fig. XLV. for the Declination of the Sun, Northern or Southern. In which there is Given, 1. The Side P N, the Distance of the Pole of the Aqua- Leffer Degrees, When in SouthernS Signs. Required, The Angle at P. Which must be turned into Time, the Shews the Beginning of the Morning Counted from Midnight And fubftracted from 12 Hours, fhews the End of the Evening Twilight. The Canon for Calculation. The Sum of the Three Sides of the Triangle is Then fay, The Half Sum is 99 d. 24 m. Or. The Difference between the Half Sum 99 d. 24 m. And the Side oppofite to P is (1.) As the Radius, To the Sine of P N, (one of the Sides containing the enquired Angle P) 38 d. 30 m. So is to the Sine of P (the other Side containing the enquired Angle P) 88 d. 48 m... (2.) As |