. An introduction to the study of the Maya hieroglyphs . completerevolution; and,further, that whenthe latter (B) hasreturned to itsstarting point, 0Pop, the corre-sponding cog in Bwill not be 2 Ik,but another day (3 Manik), since by that time the smaller wheel willhave progressed 105 cogs, or days, farther, to the cog 3 Manik. The question now arises, how many revolutions will each wheelhave to make before the day 2 Ik will return to the position 0 solution of this problem depends on the application of onesequence to another, and the possibilities concerning the numbers ornames which
. An introduction to the study of the Maya hieroglyphs . completerevolution; and,further, that whenthe latter (B) hasreturned to itsstarting point, 0Pop, the corre-sponding cog in Bwill not be 2 Ik,but another day (3 Manik), since by that time the smaller wheel willhave progressed 105 cogs, or days, farther, to the cog 3 Manik. The question now arises, how many revolutions will each wheelhave to make before the day 2 Ik will return to the position 0 solution of this problem depends on the application of onesequence to another, and the possibilities concerning the numbers ornames which stand at the head of the resulting sequence, a subjectalready discussed on page 52. In the present case the numbers inquestion, 260 and 365, contain a common factor, therefore our prob-lem falls under the third contingency there presented. Consequently,only certain of the 260 days can occupy the position 0 Pop, or, inother words, cog 2 Ik in A will return to the position 0 Pop in B infewer than 260 revolutions t)f A. The actual solution of the problem. Fig. Diagram showing engagemeatof tonalamatl wheel of260days(A),and haab wheel of 365 positions(B); the combination of the two givingthe Calendar Round, or 52-year period. 58 BTJEEAtr OF AMERICAN ETHNOLOGY [bcll. 57 is a simple question of arithmetic. Since the day 2 Ik can not returnto 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 Buntil after 365 days shall have passed, it is clear that the day 2 Ik0 Pop can not recur until after a number of days shall havepassed equal to the least common multiple of these numbers, which is ^X^X5, or 52X73X5=18,980 days. But 18,980 days = 52X 365 = 73X260; in other words the day 2 Ik 0 Pop can not recuruntil after 52 revolutions of B, or 52 years of 365 days each, and 73revolutions of A, or 73 tonalamatls of 260 days each. The Mayaname for this 52-year period is unknown; it has been called theCalendar Eo
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