THOMAS] I^^UMBER OF CYCLES IN GKEAT CYCLE 241 



cycles, 5 cycles, 19 katuiis, L'! ahaus, 12 cliueus, and 8 days, we should, 

 if riy theoi-y be true, reach the same date (-1 Eb 5 Chen, year Lamat) 

 as by counting- the whole series, thus obtaining a check on our 

 calculation. 



Days 



Multiply 1 cycle 144.000 



by 20." - 20 



1 great cycle of 30 cycles 2,880.000 



Subtract 151 calendar rouuds 2,865,980 



Remainder 14,020 



Countinu; forward this number of days from 9 Kan 12 Kayab, year 

 3 Ben, we reach 2 Kan 17 Xul, year 3 Lamat. This should be the 

 initial day of the 3rd great cycle, as numbered above. Counting 

 forward li,020 days from 2 Kan 17 Xul, year 3 Lamat, brings us to 

 8 Kan 7 Kankin, year 2 Ezanali. This should be the first day of the 

 2nd great cycle, as numbered above. Counting forward 11:,020 days 

 from the latter date (8 Kan 7 Kankin, year 2 Ezanab), we reach 1 Kan 

 2 Zip, year 2 Ben. This should be the first day of the 1st great 

 cj'cle, as numbered above, and with the subordinate periods gives the 

 series 1-5-19-13-12-8, or 1 great cycle, 5 cycles, 19 katuns, 13 ahaus, 

 12 chuens, 8 days. Counting forward from 1 Kan 12 Zip, year 2 Ben, 

 should bring us to 4 Eb 5 Chen, j'ear 9 Lamat, the date obtained bj' 

 counting the entire series from 9 Kan 12 Kayab, year 3 Ben. 



In order to test it we make the calculation; reduced to days, the 

 result is as follows: 



Days 



1 great cycle (of 20 cycles) 3,880.000 



5 cycles 720,000 



lOkatuns 136,800 



13 ahaus ... 4,680 



12 chuens : 240 



8 days 8 



Total 3.T41.T28 



Subtract 197 calendar rounds 3, 739. 060 



Remainder 2. 668 



Counting forwartl this number of days from 1 Kan 12 Zip, year 2 

 Ben, we reach 4 Eb 5 Chen, year 9 Lamat, the date at the bottom of 

 the series, and the same as that obtained by using the entire series 

 and counting from 9 Kan 12 Kayab. 



As a further test, we count forward 14,020 days from 1 Kan 12 Zip, 

 year 2 Ben, and reach 7 Kan 2 Zac, j^ear 1 Akbal. This should be the 

 first day of the incomplete great cj^cle in Avhich the minor periods 

 fall. Therefore, by taking the sum of the.se periods and counting 

 forward from this date, we should reach 4 Eb 5 Chen, yeai 9 Lamat. 

 22 ETH— 04 Ki 



