18. More on the course of the Moon: for those who are ignorant of the signs
THE RECKONING OF TIME
circle, the longest night [of the year] sends [the Moon] forth to climb up the [summer] solstitial circle. And when [the Sun] keeps to one equinox, the Moon, when she is full, keeps to the other. And the distance by which the Sun has passed the equinox or solstice which it has most recently illumined obviously corresponds to the distance by which the Moon has passed the opposite solstice or equinox.187
18. MORE ON THE COURSE OF THE MOON: FOR THOSE
WHO ARE IGNORANT OF THE SIGNS
Anyone who is ignorant of the [zodiac] signs, but who wishes nonetheless to discover the course of the Moon, should know that every day the Sun completes one part of the zodiac ^ for we should understand that the ''parts'' of the zodiac are nothing but the daily advances of the Sun in the heavens ^ and that every day the Moon completes thirteen parts of the same zodiac, that is, four puncti and one part. Because the Sun completes one part while [the Moon] completes thirteen, the result (as we taught earlier) is that [the Moon] in its daily progress distances itself from the Sun exactly four puncti, that is, twelve parts.188
Choose, then, where you wish to calculate the [position of] Moon ^ say, on the kalends of January [1 January]. The ¢rst [Moon], when a night and a day has passed, /340/ occupies that position in the sky which the Sun does on the 13th day of the same month. When it is the second [day of the] Moon, multiply 2 by 4, which is 8. Then, so that you may proceed from puncti to parts, multiply 8 by 3,189 which makes 24.
The Moon will occupy that part of the sky on the 4th nones of January [2 January] in which the Sun will on the twenty-fourth day thereafter.
When it is the third [day of the Moon], multiply 3 by 4, which makes 12.
Divide by 10: 10 times 1 is 10, with 2 puncti (that is, 6 parts) left over.
Therefore the third Moon occupies that part of the sky which the Sun 187 ''For to be sure . . . solstice or equinox.'' Cf. Anatolius 2, as reported in Ru¢nus' trans. of Eusebius, HE 7.32.18 (725.8^14); and Victor of Capua, De pascha. On the relationship of Bede's text to thatof Victor of Capua and Anatolius, see Jones, BOT 353.
188 It will be recalled from ch.17 that every zodiac sign contains 10 puncti. Since it is also divided into 30 parts (what we would call ''degrees''), this means that each punctus contains 3 parts. Therefore 4 puncti plus 1 part equals 13 parts. Therefore the Moon advances each day 13 parts along the zodiac. Since the Sun advances by one part, this means that the Moon's net gain on the Sun is 4 puncti.
189 There are 3 parts in 1 punctus.
TRANSLATION
will [occupy] when a whole month and six days have passed from the third nones of January [3 January],190 that is, at the end of the sixth day following the 3rd nones of February [3 February]. When it is the fourth
[day of the Moon], multiply by 4, which makes 16. Divide by 10: 10 times 1 is 10, with 6 puncti remaining, that is, 18 parts. Therefore the fourth Moon will occupy that part of the heavens which the Sun will when a month and eighteen days have passed after the 2nd nones of January [4 January]. When it is the ¢fth [day of the Moon], multiply 5 by 4, which makes 20. Divide by 10: 2 times 10 is 20. Therefore when two months are over, that is, on the nones of March [5 March] the Sun will arrive in the part of the sky which the Moon occupies when ¢ve days old. When the eighth [day of the Moon] is on the 6th ides of January [8 January], multiply 8 by 4, which is 32. Divide by 10: 3 times
10 is 30 with two puncti, that is six parts, remaining. On the 6th ides of January, the Moon eight days old will occupy that part of the sky which the Sun will arrive at three months and six days thence, that is, on the sixth day after the 6th ides of April [8 April]. When the nineteenth
Moon falls on the 14th kalends of February [19 January], /341/ multiply by 4, which makes 76. Divide by 10: 7 times 10 is 70, with 6 [puncti] remaining. Multiply by 3 and they make 18 [parts]. On the 14th kalends of February, the nineteenth Moon circles around that part of the heavens where the Sun will be seven months and eighteen days thence, that is on the eighteenth day after the 14th kalends of September
[19 August], or the 8th ides of September [6 September].
In case you are worried about a fallacious argument, apply [this formula] to the diameter of the year: rare is the person who does not know that the Moon is in this position when it is ¢fteen days old.191
Multiply 15 by 4, which makes 60. Divide by 10: 6 times 10 is 60. When six months are over, that is, when half his journey through the circuit of the year is complete, the Sun will enter that part of the heavens in which the Moon is borne about when ¢fteen days old. When that takes place on the ¢fteenth day of January, without doubt the Sun will arrive at that 190 10 puncti equal 30 parts, or a whole zodiac sign. Therefore when the product exceeds 10, one must divide by 10 to ¢nd the number of signs (i.e. months) distant the corresponding position of the Sun will be. The remainder, converted into parts, is the number of days in addition to these months.
191 The full Moon is in opposition to the Sun,and is therefore at the exactopposite point of the zodiacal circle, i.e.where the Sun will be six months thence (''the diameter of the year'').
THE RECKONING OF TIME
part on the ¢fteenth day of July.192 And so that we may come to a conclusion, take the thirtieth [day of the] Moon, which will then be on the 3rd kalends of February [30 January]. Multiply by 4, which makes 120.
Divide by 10: 12 times 10 is 120. The Moon when thirty days old runs in that part of the heaven where the Sun will be visible when twelve months are over, that is, when he has passed through the whole circle of the year. This is the very same part where he abides now /342/ and receives the Moon in conjunction.193
So that these matters might be more evident to everyone, even those of slower understanding, we have taken the trouble of carefully noting down, for each age of the Moon, how far distant from the Sun it is. So the Moon one day old, when its age is complete, is distant from the Sun by a space of twelve days;194 when two days old by a space of twentyfour days; when three days old by a space of one month and six days; when four days old by a space of one month and eighteen days; when
¢ve days old by a space of two months; when six days old by a space of two months and twelve days; when seven days old by a space of two months and twenty-four days; when eight days old by a space of three months and six days; when nine days old by a space of three months and eighteen days; when ten days old by a space of four months; when eleven days old by a space of four months and twelve days; when twelve days old by a space of four months and twenty-four days; when thirteen days old by a space of ¢ve months and six days; when fourteen days old by a space of ¢ve months and eighteen days; when ¢fteen days old by a space of six months; when sixteen days old by a space of six months and twelve days; when seventeen days old by a space of six months and twenty-four days; when eighteen days old by a space of seven months and six days; when nineteen days old by a space of seven months and eighteen days; when twenty days old by a space of eight months; when twenty-one days old by a space of eight months and twelve days; when twenty-two days old by a space of eight months and twenty-four days; when twenty-three days old by a space of nine months and six days;
192 As Jones observes, Bede's formula is very approximate, and works on the assumption that a Julian month is always 30 days long: see BOT 354.
193 The Moon when 30 days old is in conjunction with the Sun, that is, in exactly the same part of the zodiac. The Sun will be in that same part of the zodiac again after one solar year.
194 I.e.the 13 days of the Moon's progress, minus the 1 dayof the Sun'sprogress in the same direction.