_To find the Suns true Place_ May _the 28th at 4 hours 3 min. 42 sec. in
the afternoon, the year before Christ 585, which was a Leap year_[82].
+---------------------------------------------+-------------+-------------+
| | Sun’s Long. | Sun’s Anom. |
| +-------------+-------------+
| | s ° ʹ ʺ | s ° ʹ |
| From the Sun’s mean Longitude and Anomaly +-------------+-------------+
| at the beginning of the year Christ 1 | 9 7 53 10 | 6 29 54 |
| Subtract his mean Motion and Anomaly for | | |
| 600 years | 0 4 32 0 | 11 24 2 |
| + ----------- | ---------- |
| And the remainder, or radix, is | 9 3 21 10 | 7 5 52 |
| To which add what 585 wants of 600, | | |
| _viz._ 15 years | 11 29 22 27 | 11 29 7 |
| { _May_ | 3 28 16 40 | 3 28 17 |
| { days 28 Bissextile | 0 28 35 2 | 0 28 35 |
| And also those of { hours 4 | 0 9 51 | 0 10 |
| { min. 3 | 0 7 | ---------- |
| { sec. 42 | 2 | 0 2 1 |
| | ----------- | Sun’s Anom. |
| Sun’s mean Long. _May_ 28th, at 4 hour | +-------------+
| 3 min. 24 sec. afternoon | 1 29 45 19 | |
| Equation of the Sun’s mean Place subtract | 2 2 | 2ʹ 22ʺ |
| | ----------- | Sun’s Equat.|
| Rem. his true Place for the same time, | | subtract. |
| _viz._ ♉ Taurus 29° 43ʹ 17ʺ | 1 29 43 17 | |
+---------------------------------------------+-------------+-------------+
_N. B._ As the Longitudes or Places of all the visible Stars in the
Heavens are well known, we have an easy method of finding the Sun’s true
Place in the Ecliptic: for the Sun is directly opposite to that Point of
the Ecliptic which comes to the Meridian at mid-night.
_To find the Sun’s Declination._
[Sidenote: Fourth Element.]
362. PRECEPT. Enter Table XVII with the Signs and Degrees of the Sun’s
Place; and making proportions, take out his Declination answering
thereto. If the Signs are at the head of the Table, the Degrees are at
the left hand; but if the Signs are at the foot of the Table, the
Degrees are at the right hand. So, the Sun’s Declination answering to
his true Place (found by § 359 to be 0^s 12° 9ʹ) is 4 degrees 48 minutes
54 seconds, making allowance for the 9ʹ that his Place exceeds 12°.
_To find the Angle of the Moon’s visible Path with the Ecliptic._
[Sidenote: Fifth Element.]
PRECEPT. This we may state at 5 degrees 38 minutes, as near enough for
the purpose; since it is never above 8 minutes of a degree more or less.
_To find the Moon’s Latitude._
[Sidenote: Sixth Element.]
363. PRECEPT. Having found the Sun’s distance from the Ascending Node by
§ 357, at the mean time of New Moon, and his Anomaly for that time by §
359, find the Equation of the Node in Table XIII, by the Sun’s Anomaly,
and the Equation of the Sun’s mean Place in Table XII by his Anomaly:
these two Equations applied (as the titles direct) to the Sun’s mean
distance from the Ascending Node, give his true distance from it, and
also the Moon’s true distance at the time of Change: but when the Moon
is Full, this distance must be increased by the addition of 6 Signs,
which will then be the Moon’s true distance from the same Node.
The Moon’s true distance from the Ascending Node is called the _Argument
of the Moon’s Latitude_; with the Signs of which, at the head of Table
XIV, and Degrees at the left hand, or with the Signs at the foot of the
Table and Degrees at the right hand, take out the Moon’s Latitude: which
is _North Ascending_, _North Descending_, _South Ascending_, or _South
Descending_, according to the letters _NA_, _ND_, _SA_ or _SD_, annexed
to the Signs of the said Argument.
[Illustration: Plate XII.
_The Geometrical Construction of Solar and Lunar Eclipses._
_J. Ferguson delin._ _J. Mynde Sculp._]