An Introduction to the Study of the Maya Hieroglyphs

by Sylvanus Griswold Morley

Applying this rule to 31,741, we have seen above that its division by 13 gives as the fractional part of the quotient 8⁄13. Assuming that the count is forward from the starting point, 4 Ahau 8 Cumhu, if 8 (the numerator of the fractional part of the quotient) be counted forward from 4, the day coefficient of the starting point (4 Ahau 8 Cumhu), the day coefficient of the resulting date will be 12 (4 + 8). Since this number is below 13, the last sentence of the above rule has no application in this case. In counting forward 31,741 from the date 4 Ahau 8 Cumhu, therefore, the day coefficient of the resulting date will be 12; thus we have determined our first unknown. Let us next find the second unknown, the day sign to which this 12 is prefixed.

It was explained on page [37] that the twenty day signs given in Table [I] succeed one another in endless rotation, the first following immediately the twentieth no matter which one of the twenty was chosen as the first. Consequently, it is clear that the highest multiple of 20 which the given number contains may be deducted from it without affecting in any way the name of the day sign of the date which the number will reach when counted from the starting point. This is true because, no matter what the day sign of the starting point may be, any multiple of 20 will always bring the count back to the same day sign.

Returning to the number 31,741, let us deduct from it the highest multiple of 20 which it contains, found by dividing the number by 20 and multiplying the whole number part of the resulting quotient by 20; 31,741 ÷ 20 = 1,5871⁄20. Multiplying 1,587 by 20, we have 31,740, which is the highest multiple of 20 that 31,741 contains, and which may be deducted from 31,741 without affecting the resulting day sign; 31,741 - 31,740 = 1. Therefore in the present example 1 is the number which, if counted forward from the day sign of the starting point in the sequence of the 20 day signs given in Table [I], will reach the day sign of the resulting date. In other words, after dividing by 20 the only part of the resulting quotient which is used in determining the new day sign is the numerator of the fractional part. Thus we may formulate the rule for determining the second unknown on page [138] (the day sign):

Rule 2. To find the new day sign, divide the given number by 20, and count forward the numerator of the fractional part of the resulting quotient from the starting point in the sequence of the twenty day signs given in Table [I], if the count is forward, and backward if the count is backward, and the sign reached will be the new day sign.

Applying this rule to 31,741, we have seen above that its division by 20 gives us as the fractional part of the quotient, 1⁄20. Since the count was forward from the starting point, if 1 (the numerator of the fractional part of the quotient) be counted forward in the sequence of the 20 day signs in Table [I] from the day sign of the starting point, Ahau (4 Ahau 8 Cumhu), the day sign reached will be the day sign of the resulting date. Counting forward 1 from Ahau in Table [I], the day sign Imix is reached, and Imix, therefore, will be the new day sign. Thus our second unknown is determined.

By combining the above two values, the 12 for the first unknown and Imix for the second, we can now say that in counting forward 31,741 from the date 4 Ahau 8 Cumhu, the day reached will be 12 Imix. It remains to find what position this particular day occupied in the 365-day year, or haab, and thus to determine the third and fourth unknowns on page [138]. Both of these may be found at one time by the same operation.

It was explained on pages [44]-[51] that the Maya year, at least in so far as the calendar was concerned, contained only 365 days, divided into 18 uinals of 20 days each, and the xma kaba kin of 5 days; and further, that when the last position in the last division of the year (4 Uayeb) was reached, it was followed without interruption by the first position of the first division of the next year (0 Pop); and, finally, that this sequence was continued indefinitely. Consequently it is clear that the highest multiple of 365 which the given number contains may be subtracted from it without affecting in any way the position in the year of the day which the number will reach when

counted from the starting point. This is true, because no matter what position in the year the day of the starting point may occupy, any multiple of 365 will bring the count back again to the same position in the year.

Returning again to the number 31,741, let us deduct from it the highest multiple of 365 which it contains. This will be found by dividing the number by 365 and multiplying the whole number part of the resulting quotient by 365: 31,741 ÷ 365 = 86351⁄365. Multiplying 86 by 365, we have 31,390, which is the highest multiple that 31,741 contains. Hence it may be deducted from 31,741 without affecting the position in the year of the resulting day; 31,741 - 31,390 = 351. Therefore, in the present example, 351 is the number which, if counted forward from the year position of the starting date in the sequence of the 365 positions in the year, given in Table [XV], will reach the position in the year of the day of the resulting date. This enables us to formulate the rule for determining the third and fourth unknowns on page [138] (the position in the year of the day of the resulting date):