I

2

II

Fifthly, Add all the Longitudes together, and they make 14 Sig. 04 Deg. 29 Min. (from which abate 12 Signs, and there remains only Two Signs.) --Also Add the Anomalies together, and they make 10 Sig. 25 Deg. oo Min.

Sixtbly, (in Table IV.) Look for 10 Signs at the Bottom of the Table, and 25 Deg. in the last Column towards the Right Hand, so against it, over io Signs, you shall find i Deg 6 Min. to be Added. Set them under the Mean Longitude, and add them to it, so will the Sum be 2 Sig. 05 Deg. 35 Min. “And that is the true Place of the Sun in the Ecliptick, which is 35 Deg. distant from the Aquino&tial Point r Aries. Note, that Signs 01 3 4 5 6 7 8

9
Is r 8 I to of me M Þ w

*
Other Examples.
| Longit. O Anom. O

In this example, it
S. D. M. S. D. M. being Leap-Year, I take
Year
1720.

9 20 07 6 12 21 the Day of the Month
Febr. Day 29
I 28 09 I 28 09

out of the Column that Hours 61

15

15 hath Bisex. at the Mean Motion Il 18 31

Head of it.
Equat. Add.

1 52
11 20 23 1 * Pisces. |
Longit. O Anom. O Note also, That if

S. D. M.S.D. M. | the Sum of the Sun's Year 1713 9 20 48

Anomalies be less than

6 13 10 April 23 3 21 23 3 21 22 Six Signs, they will At Noon

0 00 00

0 00 00 ; be found at the Head Mean Morion

10 04 32

of the Æquation TaÆqui. Add.

ble, and the Degrees OL 35

in the First Column toI 13 46 | 8 Taurus.

wards the Left Hand, Longit. O Anom. O

and the Æquation must S. D. M. S. D. M. (always) be Substract

ed from the Mean Lon. Year 1761

9 20 10

6 12 42 Jan. Day 12 11 50

11 50 gitude ; whereas in all Hours

these Examples it hath 39

39

been Added. Mean Motion

3 39 6 25 II

8 jo 45

o true Place.

o true Place

16

Æquat. Add.

In all the Problems in this Se&tion, it is to be understood, That
(in all the Spherical Schemes, or Figures following.
P

Pole of the World.
ÆQ

Aquinoctial Circle, or Æquator.
PP

Parallels (or smaller Circles) of Declination of the

Sun, or of a Star.
F

Pole of the Ecliptick:
EC

Ecliprick.
FF

Parallels (or lesser Circles) of Altitude of the
Sun, or of a Star, or the Latitude of a Place

a

or Country. Υ

Vernal 2 Interfe&tion of the Ecliptick and Æ:

Autumnal) quator.
A

Ascension
D

Repre- Descension of any point of the Ecliptick.
Z sents Zenith.

the Nadir.
но

Horizon.
ZN

Prime Vertical Circle, or Azimuth of East and
Weft.

Pole of the Meridian : Or the Place

2 East

in the Horizon, where the Sun, or 2

West Star, Rijos or Sets.
R

Right Angle.
Sun.

Star.
S

Side, Sun (and sometimes) Star.
b

North?
Sourb Latitude, Declination, Amplitude.

}
R. A. P. T. Right-angled ca > Plain Triangles.
0. A. P.T. Oblique-angled
R. A. S. T. Right-angled ?
0. A. S. T. Oblique-angled

Plain Triangles.

Or
OC

Aftrono

Astronomical Problems.

1

PROBLEM I.
The Longitude, or Place of the Sun, in the Ecliptick, being
given; To find,

1. The Sun's Right Afcenfion.
2. The Sun's Declination.
3. The Angle of Position made by the Interfe&tion of the

Meridian and the Ecliptick.
N this Problem there is given the Sun's or a Star's Place,
I in
the Æquinoctial, is required,

I. By the Celestial Globe.
Example. Let the Place of the Sun be in 29 Deg. of Tau-

rus 8, (that is, 59 Deg. from the beginning of Aries r,

which is the nearest Æquino&tial Point.) The Globe being in any Position, (for in this Problem there is no regard to be had to the Latitude) count the Sun's Place in the Ecliptick upon the Ecliptick Circle, from r; and bring that Point to the Graduated Side of the Brass Meridian. Then, (1.) Will the Brass Meridian cut the Æquino&tial Circle in 56 d. 46 m. counted also from r; and that is the Right Ascension.' And, (2.), The Number of Degrees of the Brass Meridian, comprehended between the Æqui noctial and Ecliptick Circles, will be 20 d. which is the Sun's Declination. —And, (3.) The Angle made by the Interfe&tion of the Brass Meridian, and the Ecliptick Circle in the Point of the Sun's Place, will be 77 d. 23 m. which is the Angle of Position, in respect of the Meridian and Ecliptick.

II. By Trigonometrical Calculation. The Globe being in this position, there is represented upon the Superficies thereof, Two Right-angled Spherical Triangles, such as are expressed in the Diagrams; and are there noted Fig. with OR r and OR; in which, the Sides Or and O XXVII. are Arches of the Ecliprick Circle, and is the Sun's Longitude : XXVIII,

The

H h 2

Fig. The Sides r R and R are Arches of the Æquino&ial, and XXVII. is the Sun's Right Afcenfion; and the Sides OR are Arches XXVIII. of the Brass Meridian, and is the Sun's Declination. Also,

the Angle at r. or is the Angle of the greatest Obliquiry of the Ecliprick, and is equal to the Sun's greatest Declinarion 23 d. 31 m.) The Angle R is a Right Angle, and the Angle o is the Angle at the Sun's Position, in respect of the Ecliptick and Meridian Circles.

To resolve this Problem Trigonometrically, you are to consider the Quadrant of the Ecliptick, in which the Sun is, which in the Figures are signified by one of these Numbers, 1. 2. 3. 4. Of which, the first is of the Spring, from the beginning of Aries or, to the beginning of Cancer $, &c. Then in the Triangle R r o or R = 0.

(1. The Angle at por , 23 d. 31 m. There is given, 2. R r or RA: The Sun's Diftance from the besides the Right next Æquino&ial Point; to be numbred Angle at R,

from ror, unto the Degrees of the Sun's Longitude or Place in the Ecliptick given.

And, 1. RO, The Sun's Declination for that Sign which he is in; whether North or South.

C1. The Degrees found by the

Canon, are the Degrees of

2. The Degrees found, must be 2. The Arch

substracted from 180, and the r R, or remainded are the Degrees of AR,which 3. The Degrees found must be Right if it be in added to 180, and the Sum | Ascension. Quadrant, are the Degrees of

4. The Degrees found must be

fubftraéted from 360 d. and the

Remainder are the Degrees of
3. The Angle o, or the Angle of Position, in respect of
the Meridian and Ecliprick Circles.

The Canons for Calculation.
The Sun being in 29 d. of Taurus 8, which is 59 d. from
Aries r.

There is required,

୨୦

w mo

7

1. For the Right Ascension r or R. By Cafe III. of R. A. S. T.

Fig.

XXVII. As the Sine of 90 d.

XXVIII.
Is to the Co-line of the greareft Obliquity of the Ecliptick

66 d. 29 m.
So is the Tangent of the Sun's Distance from r or a 59

d.
To the Tangent of 56 d. 46 m. Which is the Sun's Right

Afcenfion; because his Place given was in the Firft Qua-
drant :-But if the Sun had been in i d. of Leon, or
29 d. of Scorpio me, or 1 d. of Aquarius de to All which
Points are 59 d. diftant from ror; the Right Afcen-
fion would be found the same, as before, viz. 56 d.

46 m.
But by the Rule before given in this Problem.

D. M.
If the Sun's S1 d. of Leo, in Quad. 2. The R. Afcen-S123 14
Place given 29 d. of Scorpio, in Qu.3. fion would 236 46
had been in Zid. of Aquarius, in Q0.4.S have been 1303 is

2. For the Sun's Declination R O. By Cafe II. of R. A. S. T.
As the Sine of 90 d.

Is to the Sine of the greatest Obliquity of the Ecliptick 23 d.31m.
So is the Sine of the Sun's Diftance from r or – 59 d.
To the Sine of 20 d. 'the Sun's prefent Declination.

Which is North, becaufe he is in a Nortbern Sign.
3. For the Angle of the Sun's Position o. By Cafe JII. of R.

A.S.T.
As the Sine of the Sun's present Declination 20 di

Is to the Sine of greatest Declination 23 d. 31 m.
So is the Sine of the Sun's R. Ascension 56 d. 46 m.
To the Sine of 77 d. 23 m. The Sun's Angle of Position, mide
by the Meridian and Ecliptick Circles.

PROB. II.

.

The Right Ascension, or Declination of the Sun given; To
find his Longitude ( or Place ) in the Écliptick.

HIS is the Converse of the foregoing Problem ; for in

this, the Sun's Place, in respect of the Æquinoctial Circle, is given : And his Place, in respect of the Ecliptick Circle, is re. quired : And may be resolved as followeth,