344 Fig. LXXIII. F CHAP. VII. To find the Place of a Planet in the Ecliptick, &c. OR that a Planets Place in Longitude in the Ecliptick, differeth In order whereunto, in the Scheme; Let S represent the Sun, The Trigonometrical Calculation. 1. In the Spherical Triangle BHV, Right-angled at V, there is given. (1) The Side B H 26.65 de. (the Complement of the Argument of Latitude.)-(2) The Angle HBV, the Greatest Inclination of Mars, 1.85 de. By which you may find the Side BV: (By CASE I. of R. A. S. T. Thus As Radius, 90 de. To Co-fine HBV, 88. 15 de. So is the Tangent of BH, 26.63 de. To the Tangent of BV 26.64 de. 10. 9-999773 9.700577 9.700350 The Difference between BH and BV, is 0.01 de. is to be fub stracted from 201.43 d. the place of Mars, because the Arch BV is lesser then the Arch B H. Heliocen. Place of Mars, in his Ellipfis. 201.43 deg. Reduction Difference Subst. 10 201.42 215.85 14.43 Which is the Anomaly of Commutation. 2. In the fame Triangle, there is given as before, whereby the Inclination of the Orbit from the Plain of the Ecliptick H V, may be found (by CASE II. of R. A. S. T. Thus As : Unto which, the Angle HSV is equal. 10. Fig. 3. For the Curtated Distance S.V. In the Right-angled Plain Triangle HSV, there is given, (1) The Angle V SH, 0.83 d. (2) The Side SH 15204, the distance of Mars in his Orbit, from the Sun: By which you may find the Side S V, (By CASE IV.. of R. A. S. T..) Thus As Radius, Sine 90.00 de HVS, Is to the Side HS, 15204. So is the Sine of V HS, 89.17 d. To the Side SV, 15202. 4:181958 14.181912 The Ecliptic Place of the Planet Mars, with his Inclination and Curtation thus attained: The next thing to be enquired after is, the Parallax of the Earths Orb; and his Geocentrick Place, in Longitude and Latitude. 4. In order whereunto, in Figure LXXIII. the Circle OXEZ, Fig. representing the Earths Orbit; Number the quantity of the Anoma- LXXIII. ly of Commitation 14.43 de. from X (the opposite Place of Mars from the Sun) to E, drawing the Line EV; which will conftitute an Oblique-angled Plain Triangle SVE: in which there is Given, (1) The Side SV 15202. (2) The Side SE, 10000, (3) The included Angle VSE 165.57 de. (the Complement of the Angle of Commitation to 180 deg.) By which you may find (1.) the Angles SVE and VES. And (2) the Side V E. As the Sum of VS and SE 25202. Is to their Difference 5202. 3.716170 So is the Tang. of half the Angles at E and V, 8.22 d. 9.159743 To the Tangent of 1.71 deg. their difference. 12.875913 8.474478 Which added to the half Sum of the Angles at V and E (8.22 d.) gives 9.03 de.. for the Angle VES: And substracted, therefrom, leaves 6.51 de. for the Angle SVE: Which is the Parallax of the Orb.. Then Fig. Then for the Side V E, LXXIII. As the Sine of VES 9.93 de. Fig. LXXIV. Is to the Sine of VSE 163.56 (16.44) So is the Side VS, 15202 CHAP VIII. Of the Proportions of the Semidiameters of the Sun, Earth, Moon, and the other Planets. I. Of the Sun, Earth, &c. Moon; B Y the best Telescope-Observations. The Semidameter of the Sun is of those Parts 46300 The Semidiameter of the Earth 727 To the mean diftance of the Moon from the Earth 1.00000 1650 446 1 The Semidiameter of the Earth is The Semidiameter of the Moon And from hence, at all times, the Distance of the Luminaries being first found, their apparent Semidiameters may - be obtained: And for the 'Semidiameter of the Earths Shadow, in Lunar Eclipses; I have here inserted the Diagram of Hyparcus. ! In which Diagram. A denotes the Centre of the Sun. Lee. BD B E her Semidiameter. AED, or A BD, the apparent Semidiameter of the Sun. AEH, or BDE, the Horizontal Parallax. CGF equal to HED, the Semi-Angle of the Cone of the Earths B Cand BF, being equal to the Distance of the Moon from BTE, her Horizontal Parallax. CBF, the apparent Semidiameter of the Earths Shadow. From hence, 1. The Semidiameter of the Sun, the Horizontal Parallax being - substracted, is equal to the Semi-angle of the Cone of the Earths Shadow. So A EDAEHHED 2. The Horizontal Parallax of the Moon, the Semi-angle of the ン Cone of the Earths Shadow being substracted, is equal to the apparent Semidiameter of the Shadow. So BF ECGE=GB F. 3. The Sum of the Horizontal Parallaxes of the Sun and Moon, is equal to the Sum of the apparent Semidiameter of the Sun and Shadow of the Earth. So, BDF BFD=ABD+CBF. Therefore, From the Sum of the Horizontal Parallax of the So, BFD+BDF-ABDCBF. I. For the Proportional Magnitudes of these Three Bodies. Fig. LXXV The Logar. of 258309 5.412141 So that the Body of the Sun exceeds the Body of the Earth 258309 times. To the Logar. of 50.63 1.704447 So that the Earth is greater than the Moon 50 times, and parts. III. Of the Semidiameters, and Proportions of the other Primary From the accurate Observations of Hugenius, Gassendus, and Horrox are determined. From hence may be gathered, That the Body of Saturn is Greater than the Earth 298 times. The Body of Jupiter 577 times. But the Earth is Greater than the other Three: For it. |