_To find the time of New Moon in_ April 1764, _N. S._
+------------------------------------+----------+-------------+-------------+
| | | Sun’s Anom. | Moon’s Ano. |
| | D. H. M. +-------------+-------------+
| | | s ° ʹ | s ° ʹ |
| +----------+-------------+-------------+
| Tab. II. Mean time of New Moon | | | |
| in _March_ | 2 8 57 | 8 2 23 | 10 13 32 |
| Add, for Lunation, from Tab. VI. | 29 12 44 | 0 29 6 | 0 25 49 |
| | -------- | ---------- | ---------- |
| Mean New Moon and Anomaly | 31 21 41 | 0 1 29 | 11 9 21 |
| To which Time add the Moon’s | +-------------+ |
| Ann. Equ. Tab. VIII. | + 0 22 | Equ. Moon’s Anom. - 20 |
| And it gives the Mean time | -------- | ---------- |
| corrected | 31 22 3 | Anom. cor. 11 9 1 |
| From which subtract the Moon’s | | Sun’s Equat. + 1 56 |
| elliptic Equ. Tab. X. | - 3 10 | ---------- |
| | -------- | Moon’s Ano. 11 10 57 |
| And it gives the Mean time equated | 31 18 53 +---------------------------+
| To which add the Sun’s Equation, | | h. m. |
| Tab. XI. | + 3 32 | Moon’s ann. Equ. 0 22 add |
| And it gives the true time | -------- | Her ellipt. Equ. 3 10 sub.|
| of Conjunction | 31 22 25 | Sun’s Equation 3 32 add |
| +----------+---------------------------+
| |
| Which true time answers to the first of _April_, at 25 minutes past 10 in |
| the forenoon: for, as the Astronomical Day begins at Noon, then 22 |
| hours 25 min. after the Noon of _March 31_, is _April 1_, at 10 hours |
| 25 min. in the Forenoon. |
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