**

# EPACTS

#### Analysis of Sacred Chronology

- Contents**

# Search Results

- Results
- Related
- Featured

- Weighted Relevancy
- Content Sequence
- Relevancy
- Earliest First
- Latest First

- Exact Match First, Root Words Second
- Exact word match
- Root word match

- EGW Collections
- All collections
- Lifetime Works (1845-1917)
- Compilations (1918-present)
- Adventist Pioneer Library
- My Bible
- Dictionary

- Reference
- Short
- Long
- Paragraph

No results.

##
EGW Extras

##
Directory

#### EPACTS

These are additional days, given to find the Moon’s age. A Lunar year of twelve moons contains two hundred and fifty-four days-eleven days less than the solar year-a deficiency extending through each year of the Lunar Cycle. This would require eleven days to be added for the *Epact* of the first year, 22 for the second, 33 - 30 = 3 for the third, (because the moon’s age cannot exceed 30 days,) 14 for the fourth, and so on through each year of the cycle; the Epact of the last year of which will be 29; and 11, again, that of the first year of the next cycle, as before.ASC 15.3

As the vulgar era commenced in the second year of the Cycle of the Moon, the year of the Cycle corresponding with any year of the era is found, by adding 1 to the given year, and dividing the sum by 19-the remainder being the Golden Number for the year. If there is no remainder, 19 is the required number.ASC 16.1

The Epact for any year previous to 1752 is found *by multiplying the Golden Number by* 11. If the product is less than 30, it will be the Epact for the year. If it is greater, *divide it by* 30, *and the remainder will be the Epact required.*ASC 16.2

As 11 days were struck from the calendar in 1752, the Epact of any year, since then, is found by subtracting 11 from the Epact as before found, if it is greater than that number; and if it is less, by adding 30 - 11 = 19. The number thus resulting is the Epact for any given year.ASC 16.3

With these numbers, the Moon’s age, for any day in any year, is found *by adding the Epact for the year*, *the number of the month from March inclusive*, *and the day of the month.* If the sum is less than 30, it is the moon’s age for that day; if it is greater, its remainder, when divided by 30, is its age.ASC 16.4

Thus, to find the moon’s age on the 19th of May, 1780-the Dark Day-we find, first, the Golden Number: (1780+1) ÷ 19 = 93, with a remainder of 14 for the number sought. Then find the Epact: 14 × 11 ÷ 30 = 5, with a remainder of 4. Then 4 + 30 - 11 = 23, the Epact sought. Then 23 + 3 + 19 = 45. And 45 + 30 = 1, with a remainder of 15, for the age of the moon at that date. As this is the day of the full moon, and as an eclipse of the sun can only occur at the new moon, the darkness on that occasion could not be the result of an eclipse of that luminary.ASC 16.5