An Introduction to the Study of the Maya Hieroglyphs

by Sylvanus Griswold Morley

It now remains to join the tonalamatl, which gives the complete names of the 260 Maya days, to the haab, which gives the positions of the days in the divisions of the year, in such a way that any one of the days whose name-part is Ik, Manik, Eb, or Caban shall occupy the first position of the first division of the year; that is, 0 Pop, or, as we should write it, the first day of Pop. It matters little which one of these four name parts we choose first, since in four years each one of them in succession will have appeared in the position 0 Pop.

Perhaps the easiest way to visualize the combination of the tonalamatl and the haab is to conceive these two periods as two cogwheels revolving in contact with each other. Let us imagine that the first of these, A (fig. [21]), has 260 teeth, or cogs, each one of which is named after one of the 260 days of the tonalamatl and follows the sequence shown in plate [5]. The second wheel, B (fig. [21]), is somewhat larger, having 365 cogs. Each of the spaces or sockets between these represents one of the 365 positions of the days in the divisions of the year, beginning with 0 Pop and ending with 4 Uayeb. See Table [IV] for the positions of the days at the end of one year and the commencement of the next. Finally, let us imagine that these two wheels are brought into contact with each other in such a way that the tooth or cog named 2 Ik in A shall fit into the socket named

0 Pop in B, after which both wheels start to revolve in the directions indicated by the arrows.

Fig. 21. Diagram showing engagement of tonalamatl wheel of 260 days (A), and haab wheel of 365 positions (B); the combination of the two giving the Calendar Round, or 52-year period.

The first day of the year whose beginning is shown at the point of contact of the two wheels in figure [21] is 2 Ik 0 Pop, that is, the day 2 Ik which occupies the first position in the month Pop. The next day in succession will be 3 Akbal 1 Pop, the next 4 Kan 2 Pop, the next 5 Chicchan 3 Pop, the next 6 Cimi 4 Pop, and so on. As the wheels revolve in the directions indicated, the days of the tonalamatl successively fall into their appropriate positions in the divisions of the year. Since the number of cogs in A is smaller than the number in B, it is clear that the former will have returned to its starting point, 2 Ik (that is, made one complete revolution), before the latter will have made one complete revolution; and, further, that when the latter (B) has returned to its starting point, 0 Pop, the corresponding cog in B will not be 2 Ik, but another day (3 Manik), since by that time the smaller wheel will have progressed 105 cogs, or days, farther, to the cog 3 Manik.

The question now arises, how many revolutions will each wheel have to make before the day 2 Ik will return to the position 0 Pop. The solution of this problem depends on the application of one sequence to another, and the possibilities concerning the numbers or names which stand at the head of the resulting sequence, a subject already discussed on page [52]. In the present case the numbers in question, 260 and 365, contain a common factor, therefore our problem falls under the third contingency there presented. Consequently, only certain of the 260 days can occupy the position 0 Pop, or, in other words, cog 2 Ik in A will return to the position 0 Pop in B in fewer than 260 revolutions of A. The actual solution of the problem

is a simple question of arithmetic. Since the day 2 Ik can not return to its original position in A until after 260 days shall have passed, and since the day 0 Pop can not return to its original position in B until after 365 days shall have passed, it is clear that the day 2 Ik 0 Pop can not recur until after a number of days shall have passed equal to the least common multiple of these numbers, which is (260/5)×(365/5)×5, or 52×73×5 = 18,980 days. But 18,980 days = 52×365 = 73×260; in other words the day 2 Ik 0 Pop can not recur until after 52 revolutions of B, or 52 years of 365 days each, and 73 revolutions of A, or 73 tonalamatls of 260 days each. The Maya name for this 52-year period is unknown; it has been called the Calendar Round by modern students because it was only after this interval of time had elapsed that any given day could return to the same position in the year. The Aztec name for this period was xiuhmolpilli or toxiuhmolpia.[^[34]]

The Calendar Round was the real basis of Maya chronology, since its 18,980 dates included all the possible combinations of the 260 days with the 365 positions of the year. Although the Maya developed a much more elaborate system of counting time, wherein any date of the Calendar Round could be fixed with absolute certainty within a period of 374,400 years, this truly remarkable feat was accomplished only by using a sequence of Calendar Rounds, or 52-year periods, in endless repetition from a fixed point of departure.