Nine Geometricall Exercises

344

Fig. LXXIII.

CHAP. VII.

To find the Place of a Planet in the Ecliptick, &c.

OR that a Planets Place in Longitude in the Ecliptick, differeth
fomething from the Longitude in its own Orb, I shall there-
fore shew (by Trigonometry,) How to find the Reduction, and
the Ecliptick Place, with the Curtation and Parallax of the Orb,
and confequently, the Geocentrick Place and Latitude.

In order whereunto, in the Scheme; Let S represent the Sun,
H the Place of Mars in his Orbit; V his place in the Ecliptick;
E the Earth: A the North Node of Mars; B the South Node; (the
two Interfections of the Planets Orb with the Ecliptick; ADB, a
Semicircle of the Ecliptick: DR, the limit of the Planets Greatest
North Latitude. These things premised, I proceed to

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.

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

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
9.999954

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;

Y the best Telescope-Observations.
To the mean diftance of the Earth from the Sun 10.00000

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

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.
AD his Semidiameter.
B the Centre of the Earth.

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
Shadow.

B Cand BF, being equal to the Distance of the Moon from
the Earth.

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
Sun and Moon, fubstract the apparent Semidiameter of the Sun;
and there will remain the apparent Semidiameter of the Earths
Shadow.