383. Having shewn how to compute the times and project the phases of a Solar Eclipse, we now proceed to those of the Lunar. And it has been already mentioned § [317], that when the Full Moon is within 12 degrees of either of her Nodes, she must be eclipsed. We shall now enquire whether or no the Moon will be eclipsed May 18, 1761, N. S. at 32 minutes past 10 at Night. See page [193].
| s | ° | ʹ | |
|---|---|---|---|
| Sun from Node at Full Moon in March 1761 | 9 | 25 | 27 |
| Add his distance for two Lunations, to bring it into May | 2 | 1 | 20 |
| And his distance at Full Moon in that month is | 11 | 26 | 47 |
Subtract this from a Circle, or 12 Signs, and there will remain 3° 13ʹ; which is all that the Sun wants of coming round to the Ascending Node; and the Moon being then opposite to the Sun, must be just as near the Descending Node: consequently, far within the limit of an Eclipse.
384. Knowing then that the Moon will be eclipsed in May 1761, we must find her true distance from the Node at that time, by applying the proper Equations as taught § [363], and then find her true Latitude as taught in that article.
[Table IV.]
[Table XIII.]
[Table XII.]
| s | ° | ʹ | |
|---|---|---|---|
| Sun’s mean distance from the Node at F. Moon in May 1761 | 11 | 26 | 47 |
| Add the Equation of the Node, for the Sun’s Anomaly 10s 18° 15ʹ[[85]] | + | 6 | |
| Sun’s mean distance from the Node corrected | 11 | 26 | 53 |
| Add the Equation of the Sun’s mean Place | + | 1 | 15 |
| Sun’s true distance from the Ascending Node | 11 | 28 | 8 |
| To which add 6 Signs, See § [363] | 6 | ||
| The sum is the Moon’s true distance from the same Node | 5 | 28 | 8 |
Or the Argument of her Latitude; which in [Table XIV], gives the Moon’s true Latitude, viz. 9ʹ 56ʺ North Descending.
385. Having by the foregoing precepts § [355] found the true time of Opposition of the Sun and Moon in a lunar Eclipse, with the Moon’s Anomaly enter [Table XV] and take out her horizontal Parallax, also her true horary Motion and Semi-diameter: and likewise those of the Sun by his Anomaly, as already taught § [364] & seq. Then add the Sun’s horizontal Parallax, which is always 10 Seconds, to the Moon’s horizontal Parallax, and from their sum subtract the Sun’s Semi-diameter; the remainder will be the Semi-diameter of that part of the Earth’s shadow which the Moon goes through.