146 BUREAU OF AMERICAN ETHNOLOGY [bull. 57 



Rounds possible. By turning to Table XVI we see that 17 Calendar 

 Rounds, or 322,660, may be deducted from our number: 322,920 — 

 322,660 = 260. In other words, we can use 260 exactly as though it 

 were 322,920. Dividing by 13, we have 260 -^ 13 = 20. Since there 

 is no fraction in the quotient, the numerator of the fraction will be 

 0, and counting forward from the day coefficient of the starting 

 point, 13, we have 13 as the day coefficient of the terminal date 

 (rule 1, p. 139). Dividing by 20 w^e have 260 ^20 = 13. Since there 

 is no fraction in the quotient, the numerator of the fraction ^vi]\ be 

 0, and counting forward from the day sign of the starting point, Ik" 

 in Table I, the day sign Ik will remain the day sign of the terminal 

 date (rule 2, p. 140). Combining the two values just determined, 

 we see that the day of the terminal date will be 13 Ik, or a day of the 

 same name as the day of the starting point. This follows also from 

 the fact that there are only 260 differently named days (see pp. 41-44) 

 and any given day will have to recur, therefore, after the lapse 

 of 260 days.^ Dividing by 365 we have: 260-^365 = f|f. Counting 

 forward the numerator of the fraction, 260, from the year position of 

 the starting point, Zip, in Table XV, the position m the year of the 

 day of the terminal date will be found to be Pax. Since 260 days 

 equal just 13 uinals, we have only to count foi-ward from Zip 13 

 uinals in order to reach the year position; that is, Zotz is 1 uinal; 

 to Tzec 2 uinals, to Xul 3 uinals, and so on in Table XV to Pax, 

 which will complete the last of the 13 uinals (rule 3, p. 141). 



Combining the above values, we find that in counting forward 

 322,920 (or 260) from the starting point 13 Ik Zip, the terminal 

 date reached is 13 Ik Pax. 



In order to illustrate the method of procedure when the coimt is 

 backward, let us assume an example of this kmd. Suppose we count 

 backward the number 9,663 from the starting point 3 Imix 4 Uayeb. 

 Since this number is below 18,980, no Calendar Round can be deducted 

 from it. Dividing the given niunber by 13, we have: 9,663-7-13 = 

 743^^. Counting the numerator of the fractional part of tliis quo- 

 tient, 4, backward from the day coefficient of the starting point, 3, 

 we reach 12 as the day coefiicient of the terminal date, that is, 2, 1, 

 13, 12 (rule 1, p. 139). Dividing the given number by 20, w^e have: 

 9,663^20=483^. Counting the numerator of the fractional part 

 of this quotient, 3, backward from the day sign of the starting point, 

 Imix, in Table I, we reach Eznab as the day sign of the terminal 

 date (Ahau, Cauac, Eznab) ; consequently the day reached in the 

 count wdll be 12 Eznab. Dividing the given number by 365, we have 



1 The student can prove this point for himself by turning to the tonalamatl wheel in pi. 5; after selecting 

 any particular day, as 1 Ik for example, proceed to count 260 days from this day as a starting point, in 

 either direction around the wheel. No matter in which direction he has counted, whether beginning 

 with 13 Imix or 2 Akbal, the 260th day will be Ilk again. 



